Theory of Aerospace Propulsion, 2nd Edition Pasquale M. Sforza Solution Manual complete update

SOLUTION MANUAL

SOLUTION MANUAL

Theory of Aerospace Propulsion, Second edition: Solutions Manual E1. PROPULSION PRINCIPLES AND ENGINE CLASSIFICATION

E1.1 A jet airplane is producing 10kN of net thrust flying at a constant speed of 500 m/s at 18km altitude (ρ0=0.1kg/m3, T0=216K). Determine the drag of the airplane.

Solution At constant speed the net thrust and drag are equal: D = Fn =10kN

E1.2 A large turbofan engine has a bypass ratio of 10 and produces a static thrust of 500kN with a jet exit velocity of 800m/s and a fan exit velocity of 120m/s. Assuming ideal conditions, what is the total mass flow m passing through the engine?

Solution ⎛ m fan ⎞ m = m jet + m fan = m jet ⎜1 + ⎟ = 11m jet ⎜ m jet ⎟⎠ ⎝ F = F jet + F fan = m jetV jet + m fanV fan = m jet ⎡⎣( 800m / s ) + 10 (120m / s ) ⎤⎦ = 500kN

m jet =

500, 000 N = 250kg / s 2000m / s

m = 11m jet = 11( 250kg / s ) = 2750kg / s

E1.3 A jet engine on a test stand is operating at its maximum thrust of 100kN with an exit velocity V7 = 1000m/s. What is the mass flow of the exhaust?

Solution On a static test stand the thrust developed is the gross thrust and

m7 =

Fg V7

=

100, 000 N = 100kg / s 1000m / s

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Theory of Aerospace Propulsion, Second edition: Solutions Manual E1.4 A rocket using a 4 to 1 mixture of LOX and LH2 as a propellant has a specific impulse of Isp= 400s. The LH2 mass flow rate is 10kg/s. What is the thrust?

Solution ⎛ m ⎞ m = mo + m f = m f ⎜ 1 + o ⎟ = (10kg / s )(1 + 4 ) = 50kg / s ⎜ m ⎟ f ⎠ ⎝

Fn = mgI sp = ( 50kg / s ) (10m / s 2 ) ( 400s ) = 200, 000 N = 200kN

E1.5 A ramjet provides net thrust Fn=10kN to drive a missile at V0=1000m/s at an altitude of 33.5km (ρ0=0.01kg/m3 and T0=232K). If the missile inlet captures a streamtube of air A0=1m2 in area, the fuel to air ratio is 0.04, and the fuel heating value is 50MJ/kg, what is the overall efficiency?

Solution

ηo = mf =

mf mo

(10kN )(1000 m s ) = 1 kg s FnV0 FV = n 0 = mΔQ m f Q f m f ( 50, 000 kN ⋅ m kg ) 5m f

mo = 0.04 ( ρ AV )0 = 0.04 ( 0.01kg / s ) (1m 2 ) (1000m / s ) = 0.4kg / s

ηo =

1 1 = = 50% ( 5 s kg )( 0.4 kg s ) 2

E1.6 A 3.57m diameter propeller on an airplane requires 581.6kW at a flight speed of 100m/s at an altitude of 6.7km (ρ= 0.5kg/m3) while operating at a propulsive efficiency of 95%. What is the net thrust produced?

Solution Fn (100 m s ) = 581.6kW = 5.816 × 105 N ⋅ m s 0.95

P=

FnV0

Fn =

( 0.95) 5.816 ×105 N ⋅ m s = 5,525 N

ηp

=

100 m s

2

Theory of Aerospace Propulsion, Second edition: Solutions Manual Another approach involves solving the propulsive efficiency equation for Ve to obtain

Ve =

2V0

ηp

− V0 =

200 m s − 100 m s = 110.5 m s 0.95

Then, knowing that at the propeller V=Vavg and assuming the density is constant solve the thrust equation: Fn = m (Ve − V0 ) = ρ AVavg (10.5 m s ) = ( 0.5 kg m3 )(10m 2 ) (105.25 m s )(10.5 m s ) = 5,525 N

E1.7 How does the propulsive efficiency depend on the effective jet velocity Ve and the flight velocity V0?

Solution The propulsive efficiency ηp is directly proportional to the flight velocity and inversely proportional to the average velocity (Ve+V0)/2 and is given in Eq. (1.9) as

ηp =

V0 V0 = Vavg 1 V + V ( e 0) 2

E1.8 How does the net thrust of a rocket engine differ from that of an airbreathing jet engine?

Solution

The rocket does not take any fluid on board in flight therefore does not suffer the penalty of ram drag. The net thrust of the rocket is equal to the gross thrust.

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Theory of Aerospace Propulsion, Second edition: Solutions Manual E1.9 Use expressions for thrust power required and total heat addition to form an overall efficiency measure.

Solution

The thrust power required is FV0 and the total heat input is mQ . The ratio of the two is the overall efficiency which is given in Eq. (1.15) as

ηo =

FV0 mQ

E1.10 (a) The addition of heat to the flow passing through an airbreathing turbojet engine raises the stagnation temperature of that flow. How does this serve to produce thrust? (b) Use the equations for stagnation temperature and gross thrust to illustrate the answer given to part (a).

Solution

(a) Raising the stagnation temperature by adding heat increases the internal energy of the exhaust gas. The exhaust nozzle transforms this increased internal energy per unit mass into increased kinetic energy per unit mass Ve2/2. The gross thrust is proportional to Ve and therefore it also increases. (b) The equation for gross thrust, using subscript 7 for the exit plane and subscript 0 for the surroundings, is

Fg = m7V7 + A7 ( p7 − p0 ) The one-dimensional energy equation for adiabatic flow with constant specific heat is 1 c pT7 + V72 = c pTt 7 2

The gross thrust may then be written as

Fg = m7 2c p (Tt 7 − T7 ) + A7 ( p7 − p0 ) For a fixed mass flow the equation above suggests that the gross thrust increases with stagnation temperature. It is perhaps more convincing to use the equation for gross thrust in terms of Mach number:

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Theory of Aerospace Propulsion, Second edition: Solutions Manual Fg = p7 A7 (1 + γ 7 M 72 ) + p0 A7

For a given exit area A7 and a given nozzle area distribution the exit Mach number M7 is fixed and therefore Fg~p7. If the mass flow is fixed we will see in later chapters that m7 =

pt 7 A7 p f (γ 7 , M 7 ) ∼ t 7 Tt 7 Tt 7

Then to keep the mass flow fixed when the stagnation temperature is increased it is necessary for the stagnation pressure to also increase. Because the ratio p7/pt7 is fixed for a given nozzle, the pressure p7 must rise as pt7 and Tt7 rise. Because Fg~p7 and p7~(Tt7)1/2 for fixed mass flow, the gross thrust must increase with stagnation temperature. Note that Fg~(T7)1/2~V7 as was suggested in the earlier, simpler explanation.

E1.11 Define each of the following terms and indicate, wherever possible, what each term is a function of: (a) effective exhaust velocity, (b) ram drag, (c) specific fuel consumption, (d) gross thrust, (e) heating value of fuel, (f) additive drag, (g) specific impulse Solution

(a) The effective exhaust velocity is the velocity a jet exiting at station 7 would reach once its pressure is equilibrated with that of the surroundings, station 0. It is a function of V7, A7, p7, p0, and the mass flow according to A7 ( p7 − p0 ) m (b) Ram drag is the momentum penalty incurred by an airbreathing jet by taking onboard Ve = V7 +

air from the surroundings. It is a function of the capture mass flow and the flight velocity according to Fr=(ρ0A0V0)V0.

(c) The specific fuel consumption is the ratio of the mass flow of fuel consumed to the thrust produced. It is a function of the average velocity (Ve+V0)/2, the burner efficiency

ηb, the thermal efficiency ηth, and the heating value of the fuel Qf according to

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Theory of Aerospace Propulsion, Second edition: Solutions Manual

cj =

Vavg

ηthηb Q f

(d) The gross thrust is the thrust produced by the exiting jet alone; it does not include ram drag. The equation for gross thrust is

Fg = mVe = mV7 + A7 ( p7 − p0 )

(e) The heating value of a fuel Qf is the heat energy released per unit mass in a complete combustion process. It depends upon the chemical composition of the fuel.

(f) Additive drag is the force arising from pressure on the streamtube entering the engine.

It is of particular concern in the design of inlets for supersonic flight.

(g) Specific impulse is the ratio of thrust force produced to the mass flow of propellants.

It has the dimensions of velocity and is generally used to describe the fuel efficiency of rockets. It is essentially the reciprocal of the specific fuel consumption with one difference: it is the mass flow of propellants rather than of fuel alone. Using this definition for rockets properly accounts for the fact that rockets carry both fuel and oxidizer whereas airbreathing engines carry only fuel, taking air on board in flight to serve as oxidizer.

E1.12 A sea-level static test of an engine with a 0.5m exit diameter showed a thrust level of about 31kN with an air flow of 59kg/s while consuming 0.8kg/s of fuel. The static pressure at the inlet was measured to be p1=69.8kPa. For this engine calculate (a) the effective exhaust velocity, (b) the exhaust gas mass flow, (c) the specific fuel consumption cj, (d) the ram drag, and (e) the inflow velocity at the 0.6m inlet face diameter.

Solution (a) The effective exhaust velocity is given by Ve =

Fg m7

=

Fg m0 + m f

=

31, 000 N = 518m / s 59 kg s + 0.8 kg s

(b) The exhaust gas mass flow is 6

Theory of Aerospace Propulsion, Second edition: Solutions Manual m7 = m0 + m f = 59 kg s + 0.8 kg s = 59.8 kg s

(c) The specific fuel consumption is cj =

mf Fn

=

mf Fg

=

( 0.8 kg s )( 3600 s hr ) = 0.0929 kg hr N 31, 000 N

Note that because this is a static test where V0=0, the net thrust is equal to the gross thrust. The specific fuel consumption may be converted to English units by multiplying through by 9.807m/s2 to obtain 0.911lbm/hr/lb. (d) The ram drag is Fr = m0V0 = 0 (e) Assuming incompressible flow into the engine, the inlet velocity is V1 =

m0 59 kg s = = 170 m s ρ1 A1 (1.225 kg m3 )( 0.283m 2 )

The inlet velocity V1=170m/s corresponds to M1= (170m/s)/(340m/s)=0.5. This Mach number is high enough to make us concerned about compressibility effects. Therefore we assume that the flow into the inlet is isentropic so that 1

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⎛ p ⎞γ ⎛ 69.8kPa ⎞ 7 3 ρ1 = ρ 0 ⎜ 1 ⎟ = (1.225 kg m3 ) ⎜ ⎟ = 0.938 kg m p 101.3 kPa ⎝ ⎠ ⎝ 0⎠ Using this density we find V1=222m/s which corresponds to about M1=0.76. The maximum mass flow that can enter the inlet occurs when M1=1 and may be calculated by equations in Chapter 5 and it turns out to be about 68.4kg/s.

E1.13 McDonnell Aircraft’s F2H Banshee fighter saw wide service in the Korean War. It first flew in 1947 and some served into the 1960s. The aircraft used 2 Westinghouse J34WE-30 turbojets delivering 3,150 lbs (14,000N) of thrust each, giving a maximum speed of 587 mph (945km/hr) and a range of almost 1,300 miles (2,090km). The F2H could climb at about 7,380 feet per minute (2,250m/min) and reach a service ceiling of 48,500 feet (14,800m). If the airplane is operating at an altitude of 20,000 feet (6.1km) with a choked nozzle exit (M7=1, γ7=1.33) and the nozzle exit area A7=180 in2 (0.116m2) and the exit pressure is p7=9psia (62kPa) find the following: (a) gross thrust produced, (b) percentage of gross thrust caused by the pressure imbalance at the nozzle exit, (c)exit mass flow if T7=1000F (537.8C), (d) ram drag at a flight velocity of 550 ft/s (167.6m/s) if f/a =0.02, (e) net thrust, (f) thrust power produced, (g) exit velocity V7, (h) effective exhaust velocity Veff, (i)propulsive efficiency ηp, (j) ηth and ηo for Qf=18,900 Btu/lb (43.96MJ/kg), (k) the specific fuel consumption, cj.

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Theory of Aerospace Propulsion, Second edition: Solutions Manual Solution

(a) The gross thrust is required and only the exit pressure (p7) and area (A7) are known; the far field area (Ae) where pe=p0 is not known. The momentum Eq. (1.2) may be written between station 0 and station 7 rather than between stations 0 and e by accounting for the pressure acting on the exit area and is given in Eq. (1.65) which shows the gross thrust to be Fg = m7V7 + A7 ( p7 − p0 ) . The ambient pressure at an altitude of 20kft (6.1km) is 6.75psia (46.5kPa) for the 1976 U.S. standard atmosphere, the exit Mach number M7=1, the exit ratio of specific heats is γ7=1.33, the exit pressure p7=9psia (62.1kPa), and the exit area A7=180in2=1.25ft2 (0.116m2). Then the momentum term is

⎛ p ⎞ ⎛ lb ⎞ 2 m7V7 = ( ρ Aa 2 M 2 ) = ⎜ ⎟ A7 ( γ RT )7 (1) = γ 7 p7 A7 = 1.33 ⎜ 9 2 ⎟ (180in ) = 2,155lb 7 ⎝ RT ⎠7 ⎝ in ⎠ The pressure term is A7(p7-p0)=(180in2)(2.25lb/in2)=405 lbs and the sum of the two is the gross thrust Fg=2560 lbs (11.39kN). (b) The percentage of the gross thrust due to the pressure imbalance at the exit plane is %Fg=405 lbs/2560 lbs =15.8% (c) The gas constant for the exhaust gas is taken to be that for air, R=53.34 ft-lb/lb-R (0.287kJ/kg) = 1716ft2/s2R (287m2/s2K) because the amount of fuel added is small and the change in temperature does not make a very large difference in the molecular weight of the exhaust gases. Then the exit weight flow rate for the case where the exit temperature is T7=1000F (537.8C) is

⎛ p ⎞ m7 = ( ρ AV )7 = ⎜ ⎟ A7 ( γ RT )7 M 7 ⎝ RT ⎠7 1296 lb ft 2 ) ( ⎛ 180in 2 ⎞ m7 = 1.33 ( 53.34 ft 2 s 2 ⋅ R ) (1460 R ) ⎜ 2 2 ⎟ ⋅ ⋅ 53.34 ft lb lbm R 1460 R 144 in ft ( )( )⎝ ⎠ The mass flow is therefore 38lbm/s (17.3kg/s)

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Theory of Aerospace Propulsion, Second edition: Solutions Manual (d) The ram drag at a speed V0 = 550ft/s when using a fuel to air ratio f/a =0.02 may be calculated by correcting the exit weight flow using the fact that

m7 = m0 + m f = (1 + 0.02 ) m0 and then finding Fr = m0V0 =

m7 V0 = 1.02

38 lbm s ft ⎞ ⎛ 550 ⎟ ⎜ lbm ⋅ ft ⎞ ⎝ s⎠ ⎛ 1.02 ⎜ 32.17 2 ⎟ lb ⋅ s ⎠ ⎝

Thus Fr=637 lbs (2833N) (e) The net thrust Fn = Fg - Fr so that Fn=2560lb - 636lb =1923 lbs (8553N) (f) The thrust power produced is Pthrust=FnV0=(1923lbs)(550ft/s)/(550ft-lb/s-hp) so Pthrust=1923hp (1.434MW) (g) The exhaust velocity V7=a7M7=[(γRT)7]1/2M7=[(1.33)(1716ft2/s2-R)(1460R]1/2(1) which yields V7=1825fps (556m/s) (h) The effective exhaust velocity is the exhaust velocity that would occur if the exit pressure was matched to the ambient pressure, that is, p7=p0. In that case we find Veff =

Fg m7

=

2560lb ( 32.17 lbm ⋅ ft lb ⋅ s 2 ) 38 lbm s

Therefore the effective exhaust velocity is Veff = 2167fps (661m/s) (i) The propulsive efficiency is η p =

550

ft s

V0 = = 40.5% ft ft ⎞ Vavg 1 ⎛ ⎜ 550 + 2167 ⎟ s s⎠ 2⎝

(j) The thermal efficiency and overall efficiency may be found from the net thrust and the heat release. Note that

η0 =

FnV0 FnV0 = m f ηbQ f ⎛ 0.02m7 ⎞ ⎜ ⎟ηb HQ f ⎝ 1.02 ⎠

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Theory of Aerospace Propulsion, Second edition: Solutions Manual Assuming the burner efficiency to be 100% (it is usually quite high, around 95%) and Qf=18,900Btu/lbm (44MJ/kg) we find

(1923lb ) ⎛⎜ 550

ft ⎞ ⎟ s ⎠ ⎝ = 9.65% ηo = ( 0.02 1.02 )( 38 lbm lb )(1)(18,900 Btu lbm ⋅ 778 ft ⋅ lb Btu )

The thermal efficiency is related to the overall efficiency by ηo=ηpηth so ηth=9.65/40.5 or

ηth=23.8% (k) The specific fuel consumption

cj =

( 3600 s hr )( 0.02 1.02 )( m7 ) = ( 70.58 s hr )( 38 lbm s ) = 1.39 lbm hr lb Fn

1923lb

This is equivalent to (1.39/9.807)=0.141kg/hr/N

E1.14 A General Electric J79-GE-15 turbojet engine is one of two propelling a McDonnell F4C airplane like that shown in Figure E1.2 cruising at a constant Mach number M0 = 0.82 at an altitude of 35,000ft (10,670m). The wing area is S = 530ft2 (49.24m2) and the inlet area for each engine is A1= 6.5 ft2 (0.604m2). The drag coefficient of the aircraft under these conditions is 0.045. Determine (a) the net thrust of the engine, (b) the gross thrust of the engine, and (c) the mass flow through the engine. Then, if the exhaust pressure p7 = p0 (d) estimate the exhaust velocity V7 and, assuming that the turbine exit stagnation temperature, which is the nozzle entry stagnation temperature, Tt,5=Tt,7=1350F (732C), (e) estimate the exit static temperature T7 and (f) the exit Mach number M7 Solution

(a) Constant Mach number at constant altitude implies constant velocity cruise.

Equilibrium requires that the thrust equals the drag and the lift equals the weight. The drag D=cDqS and the dynamic pressure q=0.5ρV2=0.5γpM2. The standard atmosphere yields p0=23.9kPa so that q0=11.25kPa and D=24.9kN. Since the thrust equals the drag and there are two engines, each engine must provide a net thrust Fn=12.45kN

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Theory of Aerospace Propulsion, Second edition: Solutions Manual (b) The gross thrust is Fg = Fn + Fram = 12.45kN + m0 V0=12.45kN+2A0q0. At this high subsonic speed A0 ~ A1 so that Fg =12.45kN +13.59kN or Fg=26.04kN

(c) m0 =

2 A0 q0 2 A1 q0 ≈ and a0 for the standard atmosphere at 10.67km altitude is V0 a0 M 0

296.6m/s so m0 = 55.9kg/s (d) Fg = m7V7 ≈ m0V7 , assuming that the fuel to air ratio is very small (typically 0.02), and therefore V7 ~ 465.8m/s (e) For adiabatic flow in the nozzle the stagnation temperature is constant throughout the nozzle and the exit temperature T7 = Tt,5 – V72 / 2cp = 1006K – (465.8)2/2cp. We may write cp = γR / (γ-1) and for air in the range of 1000K γ ~ 1.33 and therefore cp ~ 4R = 1.148kJ/kg-K. Therefore T7 ~ 911K (f) M7=V7(γ7RT7)-1/2 and using the previous answers we find M7 ~ 0.79. Note that although M7~M0 the exit velocity V7~2V0.

E1.15 The Redstone missile, shown in Figure E1.3 launching the Freedom 7 spacecraft on May 5, 1961, was the launch vehicle for America’s first man in (sub-orbital) space, Alan Shepard. This rocket can boost a 2,850 lb (1,290 kg) payload to an altitude of 115 miles (185 km). It is a one stage rocket vehicle using a single A-6 engine with a burn time of approximately 155 seconds. The propellant, which is comprised of alcohol and LOX, is consumed at the rate of about 300 lbm/s (136kg/s). Experiments have shown that the exhaust velocity of the combustion products is V7=8,400 ft/s (2,560m/s). Assuming that the exit pressure of the nozzle is matched (p7=p0) determine: (a) the thrust and thrust power under static conditions (V0=0), (b) the thrust and thrust power at a flight speed V0=5700 mph (9,172km/hr), (c) the specific impulse Isp’, (d) the overall efficiency η0 at the given flight speed if the heating value of the alcohol is Qf=12,800Btu/lbm (29.77MJ/kg). The optimal oxidizer to fuel ratio is O/F = 1.47 for the alcohol-LOX propellant combination.

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Theory of Aerospace Propulsion, Second edition: Solutions Manual Solution

(a) The thrust of the A-6 rocket engine may be found from the equation F = Fn = m7V7 because there is no ram drag with a rocket engine. Given that the propellant consumption rate is 300 lbm/s and V7=8400ft/s then the thrust is Fn = (300lbm/s)(8400ft/s)/(32.17lbmft/s2lb) and therefore Fn = 78,330 lbs (348kN). The thrust power is Pthrust = FnV0 so that at the point of lift-off, where V0=0, the thrust power is Pthrust=0 (b) The thrust and thrust power at a flight speed of V0 = 5,700mph = 8,362ft/s may be calculated directly from the previous equations. The thrust doesn’t depend on the flight speed so Fn =78,330 lb (348kN). However the thrust power is now Pthrust=(78,330lb)(8,362ft/s) / (550ft-lb/s-hp) or Pthrust=1.191x106hp (888MW) (c) The specific impulse is Isp= V7/g = (8400ft/s)/(32.17ft/s2) and Isp=261s (d) The overall efficiency is

ηo =

FnV0 m VV ⎛ O⎞ VV = e e 0 = ⎜1 + ⎟ e 0 m f ηb Q f m f η b Q f ⎝ F ⎠ η b Q f

For a burner efficiency ηb = 100% and Qf =(12,800Btu/lbm)(778ft-lb/Btu)(32.17lbm-ft/lb-s2)=3.204x108ft2/s2

ηo = (1 + O F )

( 8400 ft s )( 8362 ft s ) = 0.219 1 + O F ( ) (1) ( 3.204 ×108 ft 2 s 2 )

For alcohol the optimum mixture ratio is O/F ~ 1.47 so ηo = 54%

E1.16 The Breguet equation for the range of a jet airplane is

⎛ L ⎞V ⎛W ⎞ R = ⎜ ⎟ ln ⎜ 1 ⎟ ⎝ D ⎠ c j ⎝ W2 ⎠

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Theory of Aerospace Propulsion, Second edition: Solutions Manual The total weight of the airplane at the start of cruise is W1=W2+Wf, where W2 is the weight of the airplane at the end of the cruise and Wf is the weight of fuel used for the cruise segment of the mission. Assume flight in the stratosphere at constant velocity V0 =500mph (805km/hr) and lift to drag ratio L/D=15. What is the effect of specific fuel consumption cj on the range R when the fuel fraction Wf/W1 is varied between 0 and 1?

What has a more important effect on the cruise range, specific fuel consumption (consider values of cj between 0.5 and 1.5lbm/hr/lb) or the fuel fraction carried, Wf/W1?

Substantiate your conclusions with graphs and/or sensitivity analyses.

Solution

The simplest form of the Breguet range equation is

R=

( L / D )V ln ⎛ W1 ⎞

⎜ ⎟ ⎝ W2 ⎠

cj

(E1.1)

The subscripts 1 and 2 denote conditions at the start and end of the cruise, respectively, and W2=1-Wf. Then we may rewrite Equation (E1.1) as

R=

( L D )V ln ⎛ cj

( L D )V ln 1 − M W1 ⎞ ⎜⎜ ⎟⎟ = − ( f) cj ⎝ W1 − W f ⎠

(E1.2)

The change in the maximum range with respect to changes in cj and in the weight fraction Mf =Wf/W1 is

dR =

∂R ∂R dc j + dM f ∂c j ∂M f

(E1.3)

This may be written as

⎤ dM f dc dM f dc ⎡ −M f dR 1 =− j − = [ −1] j + ⎢ ⎥ R c j ln (1 − M f ) 1 − M f c j ⎢ (1 − M f ) ln (1 − M f ) ⎥ M f ⎣ ⎦

(E1.4)

Equation (E1.3) is in the form of percent change of range as a function of percent change in cj and in weight fraction Mf. The coefficients of the two terms determine their relative importance. The coefficient in square brackets for the cj term, -1, indicates a one-for one percent increase in range for a percent decrease in cj while the coefficient in square

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Theory of Aerospace Propulsion, Second edition: Solutions Manual brackets of the second term, f = -Mf[(1-Mf)ln(1-Mf)]-1, indicates a relatively larger percent increase in range for a given percent increase in Mf as Mf gets closer to unity. The behavior of the coefficient is shown in Figure E1.1 below.

Because the coefficient of the cj term is equal to unity we see from Figure E.1 that increasing Mf has a greater effect in increasing range than does decreasing cj. It is interesting that in the region of Mf typical of practical aircraft both cj and Mf are reasonably close in importance. This is why efforts are made to reduce specific fuel consumption of engines as well as to reduce structural weight of airframes in order to improve range performance.

4.0 3.5 3.0 2.5 (1/R)dR/dMf 2.0 1.5 1.0 0.5 0.0

practical range

0

0.2

0.4

0.6

0.8

1

fuel mass fraction Mf

Figure E1.1 Variation of the influence coefficient f of the Mf term is shown as a

function of Mf. The horizontal ellipse shows the range of Mf for typical aircraft.

Specific fuel consumption for ramjets, turbojets, and turbofan range from cj=2 or 3 lbm/hr/lb down to cj=0.5 lbm/hr/lb while fuel weight fractions may be between about 0.25 and 0.45. One common result is shown in Figure E1.2, illustrating the variation of range with Mf using cj as a parameter using the cruise conditions specified in the problem statement. The conclusions drawn from this figure will of course be qualitative.

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Theory of Aerospace Propulsion, Second edition: Solutions Manual

10000 cj=0.5lbm/hr/lb 8000 1

1.5

6000 R (nm) 4000 2000 0 0

0.2

0.4

0.6

0.8

1

Mf=Wf/W1

Figure E1.2 The variation of range with Mf using cj (in units of lbm/hr/lb) as a

parameter. The dotted ellipse shows the range of Mf for typical aircraft.

E1.17 The 46 ft long BOMARC interceptor missile shown in Figure E1.4 was powered by an Aerojet General LR59-AG-13 liquid fuel booster rocket of 35,875 lb (159.6kN) thrust and two Marquardt RJ43-MA-3 sustainer ramjets of 11,500 lb (51.15kN) thrust each.

The maximum speed was 1,975mph (3,178km/hr) and the ceiling was 65,000ft (19.81km), with a range of 260mi (418.4km). The nominal inlet diameter of the engine is 24in (0.6096m), its length is 145in (3.683m), and its mass is 485lbm (219.7kg). For flight at M0=1.5 and an altitude of 50,000ft (15.24km) with an effective exhaust velocity of the engine Ve = 2,500ft/s (762m/s) and assuming the flow conditions at the inlet are equal to those of flight calculate the following: (a) air flow m0 , (b) fuel flow rate m f for a fuel to

air ratio of 0.03, (c) exhaust gas flow m7 , (d) gross thrust, (e) ram drag, (f) net thrust, (g) thrust power, (h) propulsive efficiency ηp, (i) thermal efficiency ηth, (j) specific fuel consumption, and (k) overall efficiency Solution

(a) The air mass flow into the inlet of the Marquardt ramjet is

m0 = ρ0 A0 ( a0 ) M 0 = 53.4 lbm s = 24.2 kg s (b) The fuel mass flow

m f = m0 ( f a ) = ( 53.4 lbm s )(.03) = 1.60 lbm s = 0.726 kg s

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Theory of Aerospace Propulsion, Second edition: Solutions Manual (c) The exit mass flow m7 = m0 + m f = 55.0 lbm s = 24.93 kg s

(d) Fg = m7Veff =

(e) Fr = m0V0 =

55 lbm s ( 2500 ft s ) = 4274lb = 19.01kN 32.17 ft ⋅ lbm s 2 ⋅ lb

53.4 lbm s (1.5 )( 968.07 ft s ) = 2410lb = 10.72kN 32.17 ft ⋅ lbm s 2 ⋅ lb

(f) Fn = Fg – Fr = 1864 lbs =8.29kN (g) P = FnV0 =(1864lb)(1452ft/s)/550ft-lb/s-hp= 49210hp (3.67MW) (h) ηp = V0/Vavg = 2V0/(V0+Veff) = 73.5%

(i) ηth = (j) c j =

(1864lb )(1452 ft s ) FnV0 = = 11.5% m f ηbQ f (1.6 lbm s )(1)(18,900 Btu lbm )( 778 ft ⋅ lb Btu ) 3600m f Fn

=

( 3600 s hr )(1.6 lbm s ) = 3.09 lbm hr lb (1864lb )

(k) ηo = ηpηth = 8.45%

E1.18 The Space Shuttle Transportation System (SSTS) shown in Figure E1.5 incorporates 3 Space Shuttle Main Engines (SSME, shown in Figure E1.6). Each uses liquid hydrogen (LH2) as the fuel and liquid oxygen (LOX) as the oxidizer in the ratio of 1:6. The heating value of LH2 is Qf=51,600 Btu/lb (120MJ/kg). The thrust produced by each engine in the near vacuum conditions of the upper atmosphere is 512,000 lbs (2.28MN) at a specific impulse of 454 seconds. The typical burn time for the engines is 8.5 minutes. Determine the following: (a) the effective exhaust velocity of the engine, (b) the mass flow of fuel and of oxidizer for the engine, (c) the total mass of the fuel and oxidizer required for the cluster of 3 engines, (d) the volume for the fuel and oxidizer tanks, (e) the length of the fuel and oxidizer tanks assuming they are cylinders equal in diameter, 27.6ft (8.4m), to that of the 153.8ft (46.9m) long Space Shuttle External Tank, (f) the overall efficiency ηo at a flight speed of 6000fps (1.83km/s).

Solution

16

Theory of Aerospace Propulsion, Second edition: Solutions Manual

(a) The exhaust velocity Ve=Ispg = 454s (32.17ft/s2) = 14,605ft/s= 4452m/s (b) For each engine m =

F 512, 000lb = 32.17 ft ⋅ lbm s 2 ⋅ lb ) = 1128 lbm s ( Ve 14, 605 ft s

(c) Total mass of propellant for each engine mp = mf+mo which may be calculated from

( m + m ) t = (1128 lbm s )(8.5 × 60s ) = 575, 280lbm = 260, 600kg f

o

Then the total mass for 3 engines mp,total=1.726x106lbm =0.7818x106kg. The total mass of propellant may be written as mp,total = mf (1+O/F) = 7mf . Solving for mf yields mf = 246,550lbm =111,640kg Because mf= 6mo the mass of oxidizer is mo = 6mp,total / 7 = 1.479x106 lbm =670,000kg (d) The volume of the fuel tank is υf = mf /ρf = 246,550lbm/4.34lbm/ft3 = 56,800ft3 (1608m3), and the volume of the oxidizer tank υo = mo /ρo =1.479x106/71.2lbm/ft3 = 20,770ft3 (588.2m3) (e) The length of the fuel tank is Lf =υf /0.785(27.6ft)2 = 95ft =29m and the length of the oxidizer tank is Lo=υo/0.785(27.6ft)2=34.7ft =10.6m. The total length LT = 129.7ft = 39.5m. (f) Assuming a burner efficiency of 100% the overall efficiency ηo= IspV0(1+O/F)/ηbQf = 47.5%

17

Theory of Aerospace Propulsion, Second edition: Solutions Manual

ηo =

FV0 mVeV0 = m f ηb Q f m f ηb Q f ⎛ ⎝

ηo = ⎜1 +

(14, 605 ft s )( 6000 ft s ) O⎞ = 47.4% ⎟ F ⎠ (1)( 51, 600 Btu lbm )( 778 ft ⋅ lb Btu ) ( 32.17 ft ⋅ lb s 2 ⋅ lbm )

E1.19 Five F-1 engines were used in the first stage of the Saturn launch vehicle (shown in Figure E1.7 of the text) which lifted astronauts on a flight to the moon in the Apollo program. The F-1, shown in Figure E1.8 of the text, the most powerful rocket engine ever built, uses RP-1 (kerosene) as the fuel and LOX as the oxidizer in the ratio of 1:2.27. The heating value of RP-1 is Qf=18,900 Btu/lb (43.96MJ/kg). The thrust developed is 1.52x106 lbs (6.761MN) at a specific impulse of 265 seconds and with a typical burn time of about 2.5 minutes. Determine the following: (a) the effective exhaust velocity of each engine, (b) the mass flow of fuel and of oxidizer for the engine, (c) the total mass of the fuel and oxidizer required for the cluster of 5 engines (d) the volume for the fuel and oxidizer tanks, (e) the length of the fuel and oxidizer tanks assuming they are cylinders equal in diameter to that of the first stage, 33ft (10.06m), (f) the overall efficiency at a flight speed of 6000ft/s (1,829m/s) if the burner efficiency is 95%.

Solution

(a) For each engine Ve= gIsp= (32.17ft/s2)(265s) = 8,525ft/s (2,598m/s) (b) For each engine mp =

F 1.52 ×106 lb = 32.17 ft ⋅ lbm s 2 ⋅ lb ) = 5, 736 lbm s = 2,598 kg s ( Ve 8,525 ft s

(c) Total mass of propellant for each engine is mp = mf + mo m p = ( m f + mo ) t = ( 5, 735 lbm s )(161s ) = 9.233 × 105 lbm = 4.183 × 105 kg

The total mass for 5 engines is mp,total = 4.617x106lbs (2.09x106kg). The total mass of propellant is mp,total = mf(1+O/F) = 3.27mf and therefore the mass of fuel is mf = 1.412x106 lbs = 6.40x105kg)

18

Theory of Aerospace Propulsion, Second edition: Solutions Manual The mass of oxidizer is mo =2.27mf and therefore mo = 3.205x106 lbs =1.452x106kg) (d) The volume of the fuel tank is

υf = mf / ρf = (1.412x106lbm)/(45.47lbm/ft3) = 31,050ft3 (879m3), The volume of the oxidizer tank is

υo=mo / ρo = (3.205x106lbm)/(71.2lbm/ft3) = 45,010ft3 (1275m3) (e) The volume of the hemisphere-capped cylinders may be written as υ=πr3[(l/r)+(4/3)] and their lengths as L=l+2r. Knowing r=d/2=16.5ft (5.03m)and the volume of the fuel and oxidizer tanks from part (d) we may solve for the length of each tank, in feet, from the equation 4⎤ 4⎤ ⎡ υ ⎡ υ L = r ⎢ 3 − ⎥ + 2r = 16.5 ⎢ − ⎥ + 33 ⎣π r 3 ⎦ ⎣14,112 3 ⎦ We find Lf=47.31ft (14.42m) and Lo=63.63ft (19.39m) so that the total length L=110.9ft (33.8m). This answer is consistent with the actual dimensions of the Saturn S-IC first stage shown in Figure E1.7 of the textbook. (f) Assuming a burner efficiency ηb=95% the overall efficiency ⎛ O⎞ VV ηo = ⎜ 1 + ⎟ e 0 ⎝ F ⎠ ηb Q f

ηo = (1 + 2.27 )

( 8,525 ft s )( 6000 ft s ) = 37.2% 0.95 (18,900 Btu lbm )( 778 ft ⋅ lb Btu ) ( 32.17 ft ⋅ lbm s 2 ⋅ lb )

19

Theory of Aerospace Propulsion, Second edition: Solutions Manual E.20 A ramjet engine operates at a flight Mach number M0=0.8 at 9,000m altitude where p0=30.14kPa and T0=229K. For an effective exhaust velocity Ve=760m/s and inlet crosssectional area A0=0.187m2 calculate (a) the air mass flow m0 , (b) the fuel mass flow m f

for a fuel to air ratio of 0.03, (c) the exhaust gas mass flow m7 , (d) the gross thrust, (e) the ram drag, (f) the net thrust, (g) the thrust power, (h) the propulsive efficiency ηp, (i) the thermal efficiency ηth, (j) the specific fuel consumption cj, and (k) the overall efficiency ηo.

Solution

(a) The mass flow at any station is ρAV=(p/RT)AMa=(p/RT)AM(γRT)1/2. The free stream mass flow may then be written as m0 = p0 M 0 A0

γ RT0

( 30,140 N m ) 0.8 ( 0.187m ) 1.4 = 20.8 kg s ( 287 m s ⋅ K ) ( 229 K ) 2

=

2

2

2

(b) The fuel mass flow is ⎛ f ⎞ m f = m0 ⎜ ⎟ = 20.8 kg s ( 0.03) = 0.624 kg s ⎝a⎠

(c) The mass flow at the exit station is m7 = m0 + m f = 20.8 kg s + 0.624 kg s = 21.42 kg s

(d) The gross thrust

Fg = m7Ve = ( 21.42 kg s )( 760 m s ) = 16.28kN (e) The ram drag is based on the flight speed V0=M0(γRT0)1/2 and is given by

Fr = m0V0 = ( 20.8 kg s )( 0.8 ) 1.4 ( 287 J kg ⋅ K ) 229 K = 5.05kN

(f) The net thrust Fn = Fg - Fr = 11.23kN (g) From part (e) the flight velocity is V0=243m/s so the thrust power

20

Theory of Aerospace Propulsion, Second edition: Solutions Manual FnV0 = (11,230N)(243m/s) = 2,729kW (h) Propulsive efficiency ηp=2V0/(V0+Ve)=48.5%

(i) Because no net power is added to the ramjet flow the thermal efficiency is 1 2 2 he − h0 2 m7 (V7 − V0 ) = ηth = 1 − Q mf Qf

( 21.42 kg s ) ⎡⎣( 760 m s ) − ( 243 m s ) ⎤⎦ ηth = = 20.5% 2 ( 0.624 kg s )( 43, 400 kJ kg ) (103 m2 s 2 kJ ) 2

2

Here we have assumed a typical hydrocarbon fuel with Qf=43,400kJ/kg. (j) The specific fuel consumption is

cj =

( 3600 s hr ) m f g ( 3600 s hr )( 0.624 kg s ) = = 0.2 kg hr N Fn

11, 230 N

In English units this would be cj=9.807(0.2kg/hr/N)=1.96lbm/hr/lb (g) The overall efficiency ηo = ηpηth = (0.485)(0.205) = 9.9%

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

E2. QUASI-ONE-DIMENSIONAL FLOW EQUATIONS E2.1 Show that the change in the impulse function I =A ( p + ρ V 2 ) between two stations in an idealized flow machine yields the force acting on the fluid in that interval (it is suggested to put the impulse function in terms of Mach number). Use this result to determine the net thrust acting between stations 1 and 7 Solution The impulse function may also be written in terms of Mach number and the difference in the impulse function between the exit station and the undisturbed free stream station becomes I = Ap (1 + γ M 2 ) I 7 − I 0 = ( Apγ M 2 ) − ( Apγ M 2 ) + ( p7 − p0 ) A7 + p0 ( A7 − A0 ) 7

0

Assuming that the mass flow of fuel entering the control volume is negligible with respect to the air mass flow through the control volume permits one to write

m 7 = ( ρ AaM )7 = m 0 = ( ρ AaM )0 In general, we have = ρ Aa 2 M 2 = ρ A mV

γp 2 M = γ pM 2 A ρ

Thus the difference in impulse function is I 7 − I 0 = m7V7 − m0V0 + p7 A7 − p0 A0 This is exactly the result shown in Equation (1.52) in the text where the force on the fluid in the control volume F1+F2 = I7 – I0.

Note that when V=0 and p=patm the difference in impulse function I7-I0=patm(A7-A0) is not zero. Therefore one should be using gauge pressure pg=p-patm when carrying out force calculations. The difference in impulse function becomes I 7 − I 0 = pg ,7 A7 (1 + γ M 72 ) − pg ,0 A0 (1 + γ M 02 )

Then, under static conditions pg=0 and I7-I0=0. This is also seen in the net thrust Eq. (1.65) in the text.

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E2.2 Consider frictionless flow in a constant area duct shown in Figure E2.1, in which work may be added between stations 2 and 3. (a) using Table 2.1 of the text develop equations for the pressure, temperature, density, velocity, and work (per unit mass) done on the fluid as a function of Mach number in the duct, (b) for air entering station 1 at p1 = p2 =101kPa, T1 = T2 = 288K, and M1 = M2 =0.4 find the velocity, stagnation pressure, and stagnation temperature at station 2, (c) If 169.3kJ/kg of work is done on the flow find the Mach number, pressure, temperature, density, velocity, stagnation pressure and stagnation temperature at station 3.

M3

M1

2

1

3 Wk

Figure E2.1 Flow in a constant area duct with work input Solution (a) The equations for pressure, temperature, density, velocity and work per unit mass done on the fluid may be written using the influence coefficients of Table 2.1 as follows: dp −γ M 2 ⎛ −dWk dW ⎞ = − ⎜ ⎟ p 1 − M 2 ⎜⎝ c pT W ⎟⎠ 2 dT 1 − γ M 2 ⎛ − dWk ( γ − 1) M dW ⎞ = + ⎜ ⎟ 1 − M 2 ⎜⎝ c pT 1 − γ M 2 W ⎟⎠ T

dρ

ρ

=

−1 ⎛ −dWk dW ⎞ − ⎜ ⎟ 1 − M 2 ⎜⎝ c pT W ⎟⎠

1 ⎛ −dWk dW ⎞ dV = − ⎜ ⎟ 1 − M 2 ⎜⎝ c pT V W ⎟⎠ dM 2 1 + γ M 2 ⎛ − dWk dW ⎞ d γ = − ⎜ ⎟− 1 − M 2 ⎜⎝ c pT M2 W ⎟⎠ γ

The equation for M can be rearranged to read as follows: 2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

d (γ M 2 )

γM2

1 + γ M 2 ⎛ −dWk dW ⎞ = − ⎜ ⎟ 1 − M 2 ⎜⎝ c pT W ⎠⎟

(E2.1)

Using this form of the equation we may find 2 dp − d ( γ M ) = p 1+ γ M 2

dρ

ρ

=

−d (γ M 2 )

γ M 2 (1 + γ M 2 )

Using these two equations and the equation of state to find the temperature yields dT 1 − kM 2 d ( kM ) dW = + T 1 + kM 2 kM 2 W 2

Integrating these equations leads to the following results:

p3 1 + γ 2 M 22 = p2 1 + γ 3 M 32

2 2 ρ3 γ 2 M 2 (1 + γ 3 M 3 ) = ρ 2 γ 3 M 32 (1 + γ 2 M 22 )

(E2.2) 2

T3 γ 3 M 32 ⎛ 1 + γ 2 M 22 ⎞ W3 = ⎜ ⎟ T2 γ 2 M 22 ⎝ 1 + γ 3 M 32 ⎠ W2

Note that these are the same equations derived in Chapter 4 for the constant area combustor, Equations (4.15), (4.16), and (4.17) in the text. The equation for the velocity follows from the conservation of mass equation so that V3 ρ 2 = V2 ρ3

We assume that W and γ remain constant through the compression process, which should be quite accurate because the change in pressure will not cause high enough temperatures

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual to change the chemical composition of the gas, but this can be checked a posteriori. Then Equation (E2.1) may be rewritten as follows:

⎛ 1 − M 2 dM 2 ⎞ −dWk = c p T ⎜ 2 2 ⎟ ⎝1+ γ M M ⎠

(E2.3)

The equation for T in Equation (E2.2) is applicable to any point in the flow between 2 and 3 so that T =C

γM2

(1 + γ M 2 )

(E2.4)

2

Here the constant

(1 + γ M ) T C= 2 2 2

γ M 22

2

Thus Equation (E2.3) becomes

(1 − M ) dM − dW = c C (1 + kM ) 2

k

p

2

2 3

(E2.5)

This equation may be integrated, noting that dWk,2=0, to yield

⎡1 + γ ( 2M 2 − 1) 1 + γ ( 2 M 2 − 1) ⎤ 3 2 ⎥ −dWk ,3 + dWk ,2 = − dWk ,3 = c p C ⎢ − 2 2 2 ⎥ ⎢ 2γ (1 + γ M 2 ) M + 2 γ 1 γ ( 3 2 ) ⎦ ⎣

(E2.6)

(b) Conditions at station 2 and station 1 are identical since the area is constant and neither heat nor work is transferred across the boundaries between those two locations. Using

γ=7/5 the stagnation pressure, stagnation temperature and velocity are given by

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

⎛ ( 0.4 )2 ⎞ ⎛ γ −1 2 ⎞ Tt ,2 = T2 ⎜1 + M 2 ⎟ = ( 288 K ) ⎜1 + ⎟ = 297.2 K ⎜ 2 5 ⎟⎠ ⎝ ⎠ ⎝ γ

⎛ ( 0.4 )2 ⎞ ⎛ γ − 1 2 ⎞ γ −1 pt ,2 = p2 ⎜ 1 + M 2 ⎟ = (101kPa ) ⎜1 + ⎟ ⎜ ⎟ 2 5 ⎝ ⎠ ⎝ ⎠

3.5

= 112.8kPa

V2 = M 2 a2 = γ RT2 M 2 = 1.4 ( 287 J kg ⋅ K ) 288 K ( 0.4 ) = 136.1 m s

(c) If the work added is –dWk,3=169.3kJ/kg and cp=1kJ/kg-K, and, from Equation (E2.4) evaluated at station 2 we find C=1926.2K. Using Equation (E2.6) we find M3=0.6107, from Equation (2.4) we find T3= 433K, and from Equation (E2.2) we calculate p3 = 81.2kPa. The velocity is V3 = a3M3 = [(1.4)(287J/kg-K)(433K)]1/2(0.6107)=255m/s Using the equation of state one finds the density is

ρ3 = p3/RT3 = (81,200N/m2) / [(287J/kg-K)(433K)]=0.653kg/m3 The stagnation pressure and temperature at station 3 may be found using the equations in part (b) above, leading to pt,3=104.5kPa and Tt,3=465K. Note: The temperature changes 146oK which corresponds to an increase in cp of 1.4% so that the assumption regarding constant cp and γ is reasonable.

E2.3 Assume that a simple converging duct is attached to station 3 of problem E2.2 and that the exit of that duct is called station 7, with the area ratio A7/A3=0.85, as shown in Figure E2.2. Using the results of problem 2 for the conditions of station 3 find the Mach number, pressure, temperature, density, velocity, stagnation pressure and stagnation temperature at station 7. The flow in this duct may be considered frictionless, but no transfer of heat or work occurs, so that the flow is isentropic in the converging duct.

Determine the force acting on the entire structure if the radius r1=25cm.

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

M7

M1

2

1

3

7

Wk

Figure E2.2 Schematic diagram of duct with a nozzle-like reducer

Solution

(a) Because the flow in the reducer is assumed to be isentropic the stagnation conditions at station 7 are equivalent to those of station 3 so pt,7=104.4 kPa and Tt,7=465K. The equation for the Mach number variation as a function of area alone may be obtained from the table of influence coefficients (Table 2.1 in the text). Here we note that the flow conditions are such that it is accurate to set γ = constant, which leads to: 2

dM = −2 M2

1+

γ −1

M2

2 1− M 2

dA A

This equation may be integrated directly, and, if the constant of integration is determined at the critical condition, that is, at M=1, the Mach number variation with area is given by γ +1

⎛ γ − 1 2 ⎞ 2(γ −1) M ⎟ 1+ A 1 ⎜ 2 = ⎜ γ + 1 ⎟⎟ A* M ⎜ ⎝ 2 ⎠

(E2.7)

The quantity A* is the critical area, that is, the area at which M=1. The stagnation conditions are fixed and therefore the area A* is fixed for this flow and we may use Equation (E2.7) to calculate A3/A* = (1/0.6107)[(5+.61072)/6]3 = 1.1759. Then A7/A* =

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (A7 / A3)(A3 / A*)=(0.85)(1.1759)=0.999, which indicates that M7=1. Then the pressure and temperature at station 7 are as follows: −3.5 ⎛ p7 ⎞ ⎛ γ −1 ⎞ p7 = pt ,7 ⎜ = 104.4kPa ) ⎜1 + ⎟ = 55.15kPa ⎜ p ⎟⎟ ( 2 ⎠ ⎝ ⎝ t ,7 ⎠ −1 ⎛ T7 ⎞ ⎛ γ −1 ⎞ = 465 K ) ⎜ 1 + T7 = Tt ,7 ⎜ ⎟ = 387.5 K ⎜ T ⎟⎟ ( 2 ⎠ ⎝ ⎝ t ,7 ⎠

Because M7=1 we have V7 = a7 = (γRT7)1/2 = [(1.4)(287J/kg-K)(387.5K)]1/2 = 394.5m/s.

The density follows from the state equation:

ρ = p/RT = (55,150N/m2) / [(287J/kg-K)(387.5K)]= 0.495kg/m3. (b)The force on the duct unit may be found by the momentum theorem applied to the control volume occupied by the fluid. We already determined in exercise E2.1 that the force on the fluid may also be found by calculating the difference in the impulse function between two stations: I7 – I1 = p7A7(1+γM72) – p1A1(1+γM12) =(9.204kN)[1+1.4(1)2] – (19.83kN)[1+1.4(0.4)2]=-2.18kN

Then the force on the fluid in the duct is F = I7 – I1 = -2.18kN. The force on the interior of the duct is equal and opposite in direction to the force on the fluid. Therefore the force on the interior of the duct due to the fluid motion is F=2.18kN which is in the flow direction and thus constitutes a negative thrust force, or a drag force. Note that the ambient exerts a force on the exterior of the duct equal to patm(A7-A0) which balances the force of the fluid on the interior of the duct when the motion ceases. This is why one should use gauge pressures in determining the forces due to fluid motion.

If the converging duct had not been added to the constant area duct the net force on the duct would turn out to be zero. Note that the impulse function in differential form is given by the following:

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 2 dI dp dA d ( γ M ) = + + I p A 1+ γ M 2

This may be shown to reduce to dI 1 dA = 2 I 1+ γ M A Introducing the relationship between Mach number and area for flows with simple area change permits the above equation to be integrated from the nozzle entrance (here station 3) to the nozzle exit (here station 7), assuming the flow is isentropic, that is, with no change in stagnation pressure, yielding I7 M 3 1 + γ 7 M = I3 M 7 1 + γ 3 M

2 7 2 3

1+

γ 3 −1

M 32 2 γ −1 2 1+ 7 M7 2

For a constant value of γ=7/5 and a choked exit this becomes

I7 12M 3 = I 3 5 + 7 M 32

5 + M 32 6

The ratio I7/I5 <1 so that isentropic flow in a choked converging nozzle, considered as an isolated component, always produces a net drag force across it, as illustrated, for example, in Figure 9.30 in the text. Note that across an entire engine, of which the nozzle is just one component, there is a substantial change in stagnation pressure such that the change in impulse is positive.

E2.4 A constant area duct has heat added to the flow between two stations 3 and 4, while work is removed from the flow between the two downstream stations 4 and 5. Develop an equation for the pressure ratio p5/p3 in terms of the Mach numbers M3 and M5 and γ. It may be assumed that γ = constant and that the molecular weight of the gas remains constant.

Solution

From the Table of Influence Coefficients (Table 2.1) we find

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual dp −γ M 2 ⎛ dQ − dWk − dH ⎞ = ⎜ ⎟⎟ p 1 − M 2 ⎜⎝ c pT ⎠ dM 2 1 + γ M 2 ⎛ dQ − dWk − dH ⎞ = ⎜ ⎟⎟ 1 − M 2 ⎜⎝ M2 c pT ⎠

Combining these equations yields

dp −γ dM 2 = p 1+ γ M 2 This may be integrated to yield p=

C 1+ γ M 2

Here C=constant of integration and the ratio p5 1 + γ M 32 = p3 1 + γ M 52

E2.5 An airplane flying at M0=0.8 under standard sea level conditions has instrumentation embedded at the stagnation point on its nose. Assuming that the flow brought to rest at this point has undergone an isentropic process calculate the temperature and pressure there. Determine the difference between these values and the corresponding static values in the undisturbed atmosphere ahead of the airplane.

Solution

For isentropic flow we have the following relations: Tt γ −1 2 M = 1+ 2 T γ

pt ⎛ γ − 1 2 ⎞ γ −1 M ⎟ = ⎜1 + 2 p ⎝ ⎠ The standard sea level conditions are p=101.3kPa and T=288K so that for γ=1.4 we find pt=154.4kPa and Tt=325K. The temperature difference Tt – T = 37K (a 12.8% increase) and the pressure difference is pt – p = 53.1kPa (a 52.4% increase).

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E2.6 A diffuser in a jet airplane intake system is instrumented to give readings of stagnation temperature and stagnation pressure. If these readings are 333K and 138kPa, respectively, and the static pressure is 101.3kPa calculate the Mach number, static temperature, and air velocity at the instrumented station.

Solution

We may use the same equations as used in problem E2.5 so that γ

pt ⎛ γ − 1 2 ⎞ γ −1 138kPa = ⎜1 + = 1.362 M ⎟ = 2 101.3kPa p ⎝ ⎠ This equation may be solved for Mach number and yields M=0.679. Using this Mach number in the temperature equation leads to T=305K. The velocity may be found from the definition of Mach number M=V/a where a is the local speed of sound given by a=(γRT)1/2. We find that a=[(1.4)(287m2/s2-K)(305K)]1/2 leading to a=350m/s. Therefore the velocity is V=238m/s.

E2.7 Consider the flow of air through a constant area duct as shown in Figure E2.3. The gas constant is R=0.287 kJ/kg-K and γ=1.4. (a) provide an analysis that determines the type of flow occurring in the duct, i.e., Fanno, Rayleigh, normal shock, isentropic, or some combination thereof, (b) give the simplest set of equations that govern the model flow determined in the previous part, (c) find the Mach number at station 2, M2, (d) determine as many pertinent quantitative physical factors of importance for this flow as possible, such as friction factor, duct length, amount of heat addition or removal, shock location, etc., (e) If the pressure at station 2 was now reported to be p2=712 kPa while the temperature T2 was the same as found in the previous parts, what new answers would you give for parts (a) and (c)?

M1=3 p1= 69 kPa T = 297 K

p2 = 226 kPa V2 = 850 m/s

Figure E2.3 Supersonic flow entering a constant area duct Solution

(a) The given information shows p increasing from 69 kPa to 226 kPa. We may calculate a1=(γRT1)1/2= 345.4 m/s which yields V1=345(3)=1036 m/s. Thus V is decreasing down

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual the duct. From conservation of mass, ρ1V1=ρ2V2, find T2=(p2/p1)(V2/V1)T1=798K and a2= (γRT2)1/2 =566.3 m/s so M2=V2/a2=1.5. We now have p and T increasing along the duct while V decreases; M is also decreasing from one supersonic Mach number to a lower, but still supersonic, one. Therefore the flow cannot be a normal shock flow but can be either a Rayleigh or Fanno flow. It cannot be an isentropic flow because there is no area change to alter flow variables. In a Fanno flow the process is adiabatic so the stagnation temperature Tt = constant = T [1 + (γ-1)M2/2]. Here Tt,1 = 832 K while Tt,2 = 1157 K so the flow cannot be Fanno flow and must be a Rayleigh flow with a heat addition of Q=cp2Tt,2cp,1Tt,1)=(1.17kJ/kg-K)(1157K)-(1.106kJ/kg-K)(832K)=433.5kJ/kg assuming appropriate cp values for air at the given temperatures as presented in Appendix B of the text. (b) Equations of motion: ρV=constant; dp +ρVdV =0; cpdT + d(u2/2) = Q

(c) M2=1.5, already calculated in part (a) (d) Because this is a flow with simple heat addition the pertinent flow property is Q=433.5 kJ/kg, as calculated in part (a)

(e) If p2 is now 712kPa, it appears likely that the flow would have to be a normal shock flow, i.e., an adiabatic flow with Q=0. The conservation of mass leads to pM(γ/RT)1/2=constant and this yields M2=0.476, which is subsonic. To determine if the stated pressure p2=712 kPa is the pressure behind a normal shock at M1=3 and p1=69 kPa one might check the normal shock tables in Appendix A and find that p2/p1=10.31 indeed corresponds to a normal shock at M1=3. The temperature ratio in the same table is T2/T1= 2.679. Thus the new temperature at station 2 would be T2= 2.679(297) = 796K.

E2.8 Consider a long smooth circular pipe 1.46 cm in diameter connected to a large high pressure air (γ=1.4) reservoir with pt=13MPa and Tt=420K. The contoured contraction section enters the pipe section smoothly with Mach number M1=0.14. A schematic diagram of the facility is shown in Figure E2.4. (a) Find the static and stagnation values of pressure and temperature at stations 1 and 2, (b) Find the length L of the smooth pipe section required to have M2=0.5, (c) If the initial Mach number was increased to M1=0.4 what would be the new value of L?

11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

reservoir d=1.46 cm

L 1

2

contraction

smooth pipe

Figure E2.4 Schematic diagram of reservoir and pipe facility Solution

(a) At station 1, the inlet of the smooth pipe, the Mach number is given as M1=0.14.

Assume that the flow from the stagnation conditions in the reservoir (pt=13 MPa and Tt=420K) through the contoured contraction is isentropic up to location 1. Then we know from the isentropic flow equations or tables that at station 1, for M1=0.14, the static to stagnation ratios of pressure and temperature are p1/pt,1=0.9864 and T1/Tt,1=0.9961, respectively. We also assume that in the smooth pipe we have adiabatic flow with friction (Fanno flow) and from the appropriate equations or tables we find pt,1/pt,1*=4.182, p1/p1*=7.809, T1/T1*=1.195, and 4cFL1,max/d=32.511. From these relations we find p1=12.833 MPa

p1*=1.642 MPa

pt,1=13 MPa

T1=418.36 K

T1*=350.10 K

Tt,1=420 K

The mass flow through the pipe is

m = (ρAV)1=p1A1M1[γ /RT1]1/2=1.026 kg/s

At station 2 we expect the Mach number to be M2=0.5 and the adiabatic flow with friction (Fanno flow) equations or tables we find pt,2/pt,2*=1.340, p2/p2*=2.138, T2/T2*=1.143, and 4cFL2,max/d=1.0691. Using isentropic relations or tables we calculate that at station 2, for M2=0.50, p2/pt,2=0.8430 and T2/Tt,2=0.9524. The critical conditions at

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual station 1 are the same throughout the smooth pipe section right up to station 2 and the total temperature is a constant in the pipe. From these relations and the critical conditions determined at station 1 we find

p2=3.511 MPa

p1*= p2*= 1.642 MPa

pt,2=4.165 MPa

T2=400.20 K

T1*= T2*= 350.10 K

Tt,2=Tt,1= 420 K

(b) The length of the smooth pipe section is determined from Equation (2.69) as follows:

4c L ⎛ 0.0146m ⎞ 0.1148 ⎡ 4c L ⎤⎛ d ⎞ (E2.8) L = ⎢ F 1,max − F 2,max ⎥ ⎜ ⎟ = ( 32.51 − 1.0691) ⎜ ⎟= d d cF ⎣ ⎦ ⎝ 4cF ⎠ ⎝ 4cF ⎠ The friction coefficient cF is assumed to be constant over the smooth pipe section. To find cF we may use Equation (2.70) in the text for which we need to enter the Reynolds number based on the mass-averaged flow velocity Vavg which is defined by m = ρVavg A .

The viscosity varies with temperature but because the temperature changes little between stations 1 and 2 we use the value for μ=2.38x10-5 kg/m-s which is the viscosity for air at 150C (423K). The Reynolds number based on diameter and mass averaged velocity is given by Equation (2.71) in the text and is repeated below:

Re d =

m d 4m = = 3.76 ×106 A μ πμ d

Using Equation (2.70) for a smooth pipe yields cF=0.00234 which leads to L=48.94m, a long pipe indeed. The effect of friction on pipe flows at very low Mach numbers, say, M<0.25, is very small.

(c) If the initial Mach number is M1=0.4 the quantity 4cFL1,max/d decreases to 2.309 and Equation (E2.8) becomes

⎛ 0.0146m ⎞ 0.004524 L = ( 2.309 − 1.0691) ⎜ ⎟= cF ⎝ 4cF ⎠ 13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual For fixed mass flow Red and cF remain the same (assuming m remains at its assumed value) resulting in L=1.93m, thereby illustrating the substantial effect of friction as the initial flow Mach number increases.

E2.9 A gaseous fuel-air mixture enters a constant area duct like that in Figure E2.3 with the following conditions: M1=2, p1=69 kPa, and T1=166K. (a) Assuming that the duct is frictionless and that the gas mixture has the properties of air with cp=1.007kJ/kg-K, find (i) the maximum heat addition possible, Qmax and (ii) the stagnation pressure for that case of maximum heat addition, pt,max, (b) The flow in part (a) now encounters a normal shock wave just after entering the duct with the conditions stated. With the same assumptions, find (i) the maximum heat addition possible, Qmax and (ii) the stagnation pressure for that case of maximum heat addition, pt,max.(c) Using a temperature-entropy (T-s) diagram, show the processes described in parts (a) and (b) on a Rayleigh line, assuming that the entropy at the entrance is s1=0. Make a careful graph showing at least three points on the T-s plane in addition to the entry point (166,0). Fair in the Rayleigh curve passing through these points and indicate, by arrows, the directions the processes are taking.

Solution

(a) At the entrance M1=2, p1=69 kPa, and T1=166 K and from the isentropic flow equations or tables we find pt,1=7.825(69)=539.9 kPa

Tt,1=1.800(166)=298.8K.

Then, from the flow equations (or Rayleigh flow tables) we find

Tt,1/Tt*=0.7934 T1/T*=0.5289

pt,1/pt* = 1.5031

p1/p* = 0.3636.

Tt* = 298.8/0.7934 =376.6 K

pt* = 539.9/1.5031=359.1 kPa

(i) The condition of maximum heat addition requires the flow at the exit to be choked and all the thermodynamic variables will be at their critical values, e.g., T=T*, and so on. Then the heat addition is Qmax=cp (Tt* – Tt,1)= (1.007kJ/kg/K)(376.6K 298.8K)=78.34 kJ/kg. (ii) The stagnation pressure under these conditions is pt*=359.1 kPa.

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b)The same conditions occur at the entrance as in part (a), after which the flow encounters a normal shock wave. Note that a normal shock is considered adiabatic so no heat is added by it. Using the flow equations (or normal shock tables) yields the following conditions behind the normal shock (denoted by the subscript y):

My=0.5774 pt,y=0.7209(539.9kPa)=389.2 kPa py=4.5(69kPa)=310.5kPa

Tt,y=Tt,1=298.8K Ty=1.6875(166K)=280K

From the flow equations (or Rayleigh flow tables) we find the ratios of the flow variables downstream of the shock to the critical values as follows:

Tt,y/Tt*=0.7934 Ty/T* = 0.8926

pt,y/pt* = 1.0835

py/p* = 1.6363.

Tt* = 376.6 K

pt*=359.2 kPa.

p*=189.8kPa

T*=314K

(i) For maximum heat addition the flow at the exit will be choked with M2=1 and all the thermodynamic variables will be critical so the heat addition is Qmax=cp (Tt* – Tt,1) = (1.007kJ/kg-K)(376.6K -298.8K) = 78.34 kJ/kg. (ii) The stagnation pressure under these conditions is pt*=359.2 kPa (c) The entropy change between two states 1 and 2 may be written as follows:

⎛ T ⎞ γ − 1 ⎛ p2 ⎞ Δs s2 − s1 ln ⎜ ⎟ = = ln ⎜ 2 ⎟ − cp cp ⎝ T1 ⎠ γ ⎝ p1 ⎠ The plot for parts (a) and (b) on a T,s plane is shown in Figure E2.5. Note that both processes considered end up at the same point and the same amount of heat is added in both cases. The difference is in the pressure and temperature levels in the duct in each case. The normal shock adds no heat but boosts the temperature to as much as 100C higher than in the supersonic deceleration process.

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

350

subsonic acceleration M=1

300 shock

250 T (K) 200

supersonic deceleration 150 100 0

0.1

0.2

0.3

0.4

(s2-s1)/cp

Figure E2.5 T-s diagram for the two flow processes in Exercise E2.9 E2.10 Consider the circular pipe extension connected to the exit plane of a M=2.5 nozzle which in turn is connected to a large air reservoir operating at a pressure of 1725 kPa and a temperature of 677K as shown in Figure E2.6. The relatively short length of the nozzle section permits the assumption that the flow within it is frictionless. The purpose of this pipe addition is to act as a simple diffuser to reduce the Mach number exhausting into the atmosphere at station 3 to a value of M3=1.5. (a) If the friction factor is cf = 0.0025, what is the length, L, of pipe needed and what is the corresponding exit stagnation pressure pt,3? (b) Assuming that the inlet Mach number to the circular pipe, M2, is permitted to become arbitrarily large while keeping the friction factor constant at cf =0.0025 what is the maximum length to diameter ratio L/d possible and what is the associated maximum stagnation pressure ratio required, pt*/pt,2?

L

y, j 2.67 cm

4.34cm

1

minimum

2

x, i

3

Figure E2.6 A circular pipe diffuser for a supersonic nozzle

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Solution

(a) We may check the entrance Mach number M2=2.5 by noting that the area ratio at the nozzle exit A2/Amin=A2/A*=(d2/dmin)2=2.642. For air with γ=7/5 the isentropic compressible flow relations or the tables in Appendix A show that this corresponds to M2=2.5. Using Equation (2.69) in the text, with D=d2 because the pipe is circular, the friction between two stations can be written as

⎛z 1 ⎞ z ⎞ 6 ⎛ 5 + M 32 M 22 ⎞ 5 ⎛ 1 4c f ⎜ 3 − 2 ⎟ = ln ⎜ + ⎜ 2− 2⎟ 2 2 ⎟ ⎝ d2 d2 ⎠ 7 ⎝ 5 + M 2 M 3 ⎠ 7 ⎝ M 2 M 3 ⎠

(E2.9)

At station 2 we have M2=2.5 and at station 3 we desire the exit Mach number to be M3=1.5 and substituting those values into Equation (E2.9) we find z3 − z 2 L 1 5 ⎡6 ⎤ = = ln1.79 + ( 0.1600 − 0.4444 ) ⎥ = 29.54 ⎢ d2 d 2 4 ( 0.0025 ) ⎣ 7 7 ⎦ The required length of pipe is therefore L=1.284m. At station 2 the stagnation pressure ratio pt,2/pt*=2.6367 and at station 3 the stagnation pressure ratio is pt,3/pt*=1.1762. Then the stagnation pressure at station 3, pt,3= (pt,3/ pt,2) pt,2=(1.1762/2.6367)1725=769.5 kPa (b) Equation (E2.9) gives a relationship for the length between any two stations and the Mach numbers corresponding to each for γ=7/5. Letting the inlet Mach number M2 become arbitrarily large, i.e., M2>>1, while keeping the flow at station 3 choked, i.e., M3=1, leads to the following simplification of that equation: z3 − z 2 1 ⎛6 5⎞ ln 6 − ⎟ = 82.15 = ⎜ M 2 →∞ d 0.01 ⎝ 7 7⎠ lim

Table 2.2 gives the following relation for stagnation pressure in terms of Mach number pt 1 ⎛ 5+ M 2 ⎞ = ⎜ ⎟ pt* M ⎝ 6 ⎠

3

Then for M=M2>>1 we find pt M 5 = 3 M 2 →∞ p* 6 t lim

This ratio demonstrates the great loss in stagnation pressure due to friction at high Mach numbers. Note that a duct length of about 82 diameters would accommodate a very high entrance Mach number but the required stagnation pressure grows like M5.

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E.2.11 The axisymmetric pipe shown in Figure E2.7 is connected to a large volume air supply (γ=1.4, R=0.287 kJ/kg-K) held at a pressure of 3 atmospheres and a temperature of 27C. The contoured nozzle section (stations 0-1-2) has A2/A1=1.687 and may be considered to operate without friction. However, the long constant diameter (d = 5cm) section (stations 2 to 3) is characterized by a friction coefficient cf = 0.0025. The duct exits into a large receiving tank which is at a pressure of 0.5atm and a temperature of 27C. A pitot probe placed at the centerline of the exit plane reads a pressure of 1.78 atmospheres. (a) What is the maximum mass flow that can pass through the duct under the given conditions? (b) What is the length, L, of the constant area duct for maximum mass flow, under the given condition? (c) What are the pressures p2 and p3? (d) Sketch on the T-s plane the process you assume to have occurred between stations 2 and 3. On the same sketch show another valid process between the same points.

Pitot probe

L 0

1

2

3

4

Figure E2.7 Schematic diagram of contour nozzle with a long exhaust pipe Solution

(a) The mass flow ρAV = pAM(γ /RT)1/2 is a maximum at A1 when M1=1 and can be evaluated there as

m =

pt ,0 A1M 1 ⎛ p1 ⎞ Tt ,0 = 0.825kg / s ⎜⎜ ⎟ pt ,0 ⎟⎠ T1 R ⎝ Tt ,0

γ

The area can be found from the given information as A1=0.0116m2, the stagnation pressure and temperature are given and are constant through the contoured nozzle, R and

γ are given, and the static to stagnation ratios at M=1 can be read from the isentropic flow tables.

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) At station 2 we know, from the given information, that A2/A1=A2/A*=1.687. Using isentropic flow relations or tables we find that this area ratio corresponds to either M2=2 and p2/pt,2= 0.1278 or M2=0.38 and p2/pt,2=0.9052. In either case the stagnation pressure is given as pt,2 = pt,0 =3(101.3kPa)=303.9kPa and we find that for M2=2

p2=33.84kPa

pt,2 = 303.9kPa

M2=0.38

p2= 275.1kPa.

pt,2 = =303.9kPa

From equations or tables for Fanno line flow we obtain M2=2

pt,2/pt*=1.6875

p2/p* = 0.40825

M2=0.38

pt,2/pt*=1.6587

p2/p* = 1.665

For maximum mass flow through the pipe the flow must also be choked at the pipe exit, station 3. Therefore pt,3*/pt* = 1 and because the Pitot probe measurement yields pt,3=1.78atm, we know that pt,3* = 1.78atm. The stagnation pressure has therefore dropped from 3atm to 1.78atm, that is, pt,0/pt,3= pt,2/pt,3*= 1.685. Note that this value for Fanno flow once again corresponds to either M2 =2 or M2=0.38. Then we also can compile the following results: M2=2

T2=167K

(s-s*)/cp=-0.1495

4cfLmax/d=0.305

M3=1

T2=250K

(s-s*)/cp=0

4cfLmax/d=0

M2=0.38

T2=292K

(s-s*)/cp=-0.1446

4cfLmax/d=2.705

M2=2

Lmax=1.53m

M2=0.38

Lmax=13.5m

(c) From part (b) we know that M2=2

p2=33.84kPa

pt,2 = 303.9kPa

M2=0.38

p2= 275.1kPa.

pt,2 = =303.9kPa

In addition we calculate p3 as follows: M3= 1

p3 = (p3/pt,3)pt,3 = (0.5283)1.78=0.94atm=95.26kPa

(d) The process on the T-s plane is shown in Figure E2.8. The possible paths are as follows: (A) supersonic flow (M2=2) at the entrance of the duct with friction slowing the

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual flow down to M=1 at station 3, (B) subsonic flow (M2=0.38) at the entrance of the duct, with acceleration to M3=1 for maximum mass flow, (C) normal shock at end of nozzle reducing Mach number to 0.58 from 2.0, and then friction accelerating the flow to M=1 at station 3. The starting points for the two processes are both shown on the figure as originating at (s-s*)/cp=0.15 for clarity. The actual values, which are quite close, are found in part (b),

275kPa

T (K) 291 281

152kPa

M2 =0.38 B Mshock =0.58

95kPa

M=1

33kPa

250 C

A

167

-0.3

M2=2

-0.15

-.093

0

(s-s*)/cp

Figure E2.8 Sketch of a T-s plot for three possible processes for the flow through the long exhaust pipe including notional isobars. Not to scale.

E2.12 An experimental rig for determining friction coefficients in supersonic flow of air in constant area tubes consists of a short converging-diverging nozzle attached to a long smooth round tube as shown in Figure E2.9. Pressure taps are located at x1/d=1.75 and x2/d=29.60, and the diameter of the tube d=12.72 mm. The throat diameter of the nozzle dt=6.137 mm and the stagnation conditions are pt=687.9 kPa and Tt=89.5 C. During a test the pressures measured at the two locations are p1=24.33 kPa and p2=49.46 kPa. Assuming that the flow is isentropic only up to the nozzle throat and that beyond the throat the flow is adiabatic, find: (a) M1 (b) M2, (c) the average friction factor cF between the two measurement stations, (d) the entropy change (s2-s1)/cp.

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 12.72mm

6.137mm

Stagnation chamber

throat

2

1

1.75d 29.6d

Figure E2.9 Schematic diagram of a converging-diverging nozzle attached to a long smooth round tube. Pressure taps are located at z1/d=1.75 and z2/d=29.6.

Solution

(a) We may use the given information to calculate A1 /A*=4.2960; (p/pt)1=0.03537. The flow is not assumed to be isentropic between the throat and station 1, but for adiabatic flow from the throat to station 1, the stagnation temperature is constant and equal to that in the chamber, Tt=89.5 C=362.5K. Then for the mass flow at the throat to be equal to the mass flow at station 1 we have γ +1

p p * Tt p ⎛ 2 ⎞ 2(γ −1) p1 A1M 1 = t A* = t A * ⎜ ⎟ T1 Tt pt T * Tt ⎝ γ +1⎠ −

γ +1

1

1 ⎛ γ − 1 2 ⎞ 2 ⎛ 2 ⎞ 2(γ −1) p1 A1 = M1 ⎟ ⎜ = 0.1520 ⎜1 + ⎟ 2 pt A * M 1 ⎝ ⎠ ⎝ γ +1⎠ From this equation we find M1=2.524. (b) Then from friction flow tables or 1

1 ⎛ ⎞2 + 1) γ ( ⎜ ⎟ 1 p 2 = ⎜ 1 ⎟ p * M ⎜ 1+ γ −1 M 2 ⎟ ( ) ⎝ ⎠ 2

( γ + 1) M 4cF Lmax 1 − M 2 γ + 1 = + ln 2 2γ γM d ⎡ 1 ⎤ 2 ⎢1 + ( γ − 1) M 2 ⎥ ⎣ 2 ⎦ 2

From these equations we find p1 /p*=0.2878 and (4cFLmax/d)1=0.4371. We may determine the pressure ratio p2/p2* from

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

( p / p *)2 p2 49.46kPa = = 2.032 = p1 24.33kPa ( p / p *)1 Solving this equation results in p2/p2*=0.5850. Again, using tables or the equations given above we find M2=1.542 and 4cFLmax,2/d=0.1512. (c) Then (4cFLmax,1/d) - (4cFLmax,2/d) =0.2859, and since ΔL/d = 29.6-1.75 =27.85, the average friction coefficient is cF = 0.00257. (d) The entropy change (s2 – s1)/cp = ln (T2/T1) – [(γ-1)/γ]ln (p2/p1) = ln [(T2/T*)/(T1/T*)] – (2/7)ln[(p2/p*)/(p1/p*)] =ln(0.814/0.528) – (2/7)ln(0.5850/0.288) = 0.433 - 0.202 = 0.231

E2.13 Consider a smooth circular pipe 14.6 cm in diameter connected to a large high pressure air (γ=1.4) reservoir (pressure pt=13 MPa and temperature Tt=420 K) having a contoured contraction section with an exit Mach number M1=0.5. Downstream of this inlet is a heat exchanger used to raise the temperature of the air flow passing through the 14.60 cm diameter pipe. The heated flow passes through a contoured nozzle to supply supersonic flow to a circular test section 30cm in diameter. A schematic diagram of the facility is shown in Figure E2.10. (a) Find the static and stagnation values of pressure and temperature at stations 1 and 2 (b) Find the thermal power required from the heater.

reservoir

heat exchanger

d=30cm

d=14.60 cm L 1

contraction

2

heater

3

4

nozzle

Figure E2.10 Schematic diagram of facility with heater-driven flow

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Solution

The Mach number at station 1, the inlet to the heater section of the pipe, is given as M1=0.5. Assuming the flow is subjected to heating alone we can use the equations or tables for flow with stagnation temperature change (Rayleigh flow) to find pt,1/pt,1*=1.114, p1/p1*=1.7778, T1/T1*=0.79012 and Tt,1/Tt*=0.69136. Assuming that the flow from the stagnation conditions in the reservoir (pt=13 MPa and Tt=420K) through the contoured contraction is isentropic up to location 1 we may use the isentropic flow equations or tables to find p1/pt,1=0.843 and T1/Tt,1=0.9524. From these relations we find p1=10.96 MPa

p1*=6.164 Mpa

pt,1=13 Mpa

pt,1*=11.66MPa

T1=400 K

T1*=506.3 K

Tt,1=420 K

Tt,1*= 607.4K

Although the Mach number at station 2 is not specified, we can reasonably infer that it must be M2=1 in order to provide supersonic flow through the nozzle and into the larger pipe section attached to the nozzle. Thus we have

M2=1.0; pt,2/pt,2*=1.0, p2/p2*=1.0, T2/T2*=1.0, and Tt,2/Tt,2*=1.0 We also know from the isentropic flow tables that at station 2, for M2=1.0, we have p2/pt,2=0.5283 and T2/Tt,2=0.8334. From these relations we find p2= p1*= 6.164 Mpa

pt,2=pt,1*= 11.66 MPa

T2= T1*= 506.3 K

Tt,2=Tt,1* = 607.4K

Assuming that the flow through the contoured nozzle is isentropic we may deduce from the given pipe diameters that the Mach number at the nozzle exit is M3=3, since A4/A3=(30/14.60)2=4.222. We then know, from the isentropic flow tables, that at station 3, for M3=3.0, we have p3/pt,3=0.02722 and T3/Tt,3=0.3571. Since the stagnation properties are constant from station 2 to station 3 the flow variables at 3 are found to be

23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual p3= 317.4 kPa;

pt,3 = pt,2= 11.66MPa

T3= 216.9 K;

Tt,3 = Tt,2 = 607.4 K

(b) The power required from the heater may be found by writing the energy conservation equation across the heated section to find the heat transferred to be

Q/cp= Tt,2 -Tt,1 = 607.4K - 420K=187.4K The value for cp is taken to be the value at T=453K=180C, the average temperature between T2 and T1 and is given by cp = 1.02kJ/kg-K. Using this value we find Q=191.1 kJ/kg. The mass flow may be determined at any station since it is constant throughout.

Then, using station 3, we obtain

m = ρ3 A3V3 = p3 A3 M 3

γ RT3

= 320 kg s

Then the thermal power is given by

= ( 320 kg s )(191.1 kJ kg ) = 61.15MW P = mQ

E2.14 Flow at M1=2 enters a constant area duct at station 1 with Tt,1=520K and pt,1=1MPa and is cooled by some method not involving friction so as to accelerate the flow to M2=3. Assuming that the gas is air (γ = 7/5 and cp=1.03kJ/kg-K) and that no shocks are encountered, find (a) the heat removal per unit mass required, and (b) the values of T, Tt, p, and pt at station 2.

Solution (a) Cooling a supersonic flow speeds it up. The most heat that can be removed from the flow corresponds to the static temperature T2=0 and M2 becoming infinite. For the present case where M2=3 the stagnation temperature ratio may be deduced from Table 2.3 and becomes

⎛T ⎞ ⎛T ⎞ M 2 5 + M 22 ( 5 + 7 M 1 ) = ⎜ t* ⎟ ⎜ t* ⎟ = 22 = 0.824 Tt ,1 ⎝ Tt ⎠2 ⎝ Tt ⎠1 M 1 5 + M 12 ( 5 + 7 M 2 )2 2

Tt ,2

−1

2 2

Then the heat removed from the flow is given by 24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

⎛T ⎞ Q = c pTt ,1 ⎜ t ,2 − 1⎟ = (1.03 kJ kg ⋅ K )( 520 K )( 0.824 − 1) = −94.3 kJ kg ⎜T ⎟ ⎝ t ,1 ⎠

Note that the heat of vaporization of water is about 2257kJ/kg so that of an amount of liquid water equal to about 4.6% of the gas mass flow would provide the amount of cooling required. Indeed, water is often injected into the subsonic inlet flow of jet engines to reduce the temperature of the air before it enters the compressor. (b) The new stagnation temperature is then given by Tt ,2 = 0.824Tt ,1 = 428 K

The static temperature and pressure at station 1 may be found from the isentropic relations −1

⎛ M2 ⎞ T1 = Tt ,1 ⎜1 + 1 ⎟ = 289 K 5 ⎠ ⎝ ⎛ M2 ⎞ p1 = pt ,1 ⎜ 1 + 1 ⎟ 5 ⎠ ⎝

−3.5

= 128kPa

The equations for T2/T1, p2/p1, and pt,2/pt,1 may be found using Table 2.3 for γ =7/5: 2 −1 M 22 ( 5 + 7 M 1 ) ⎛T ⎞ ⎛T ⎞ = 153K T2 = T1 ⎜ * ⎟ ⎜ * ⎟ = ( 289 K ) 2 M 1 ( 5 + 7 M 2 )2 ⎝ T ⎠2 ⎝ T ⎠1 2 2

−1

⎛ p⎞ ⎛ p⎞ 5 + 7 M 12 p2 = p1 ⎜ * ⎟ ⎜ * ⎟ = (128kPa ) = 62.1kPa 5 + 7 M 22 ⎝ p ⎠ 2 ⎝ p ⎠1 −1

⎛p ⎞ ⎛p ⎞ 5 + 7 M 12 ⎛ 5 + M 22 ⎞ pt ,2 = pt ,1 ⎜ *t ⎟ ⎜ *t ⎟ = (1.0 MPa ) ⎜ ⎟ 5 + 7 M 22 ⎝ 5 + M 12 ⎠ ⎝ pt ⎠2 ⎝ pt ⎠1

25

3.5

= 2.28MPa

Theory of Aerospace Propulsion Second Edition: Solutions Manual E3. IDEALIZED CYCLE ANALYSIS OF JET PROPULSION ENGINES E3.1 A turbojet engine with a convergent nozzle is operating at 100% power at M0=0.9 and an altitude of 11km. The engine characteristics are such that the compressor, burner, and turbine efficiencies may be taken as 0.85, 0.96, and 0.90, respectively. The inlet pressure recovery may be taken as the MIL-E-5008B standard. The mass flow captured by the inlet is 29.17kg/s and the heating value of the fuel is 44MJ/kg.The compressor pressure ratio is pt,3/pt,2=18, the combustor pressure loss is (pt,3-pt,4)=0.02pt,3, and the turbine inlet temperature Tt,4=1330K. Calculate the properties of the flow at each station in the engine, the gross thrust, and the net thrust.

Solution (i) Free stream, Station 0: For the 1976 U.S. standard atmosphere as given in Appendix C p0= p0,s.l.σ = (101.3kPa) (0.224) = 22.69kPa, T0=216.8K, a0=295.2m/s, and therefore V0=a0M0= (295.2m/s)(0.9) = 265.7m/s.

For air with γ =1.4 the stagnation to static pressure ratio is pt/p = (1+M2/5)3.5 and the stagnation to static temperature ratio is Tt/T = (1+M2/5). Therefore pt.0=1.691p0=38.38kPa Tt,0=1.162T0=251.9K (ii) Inlet, Stations 1-2: The MIL-E-5008B standard pressure recovery for M0=0.9 is pt,2/pt,0 =1.0 and the inlet flow is assumed to be adiabatic so that Tt,2/Tt,0=1. Therefore pt,2=38.38kPa Tt,2=251.9K (iii) Compressor, Stations 2-3: The pressure ratio is given as pt,3/pt,2=18 and assuming a constant effective specific heat cp2 = cp3 =cp,e and γ =1.4 the work done is given by either of the following: WC=cpTt,2[(pt,3/pt,2)0.286-1]/ηc

(E3.1)

WC = cp(Tt,3-Tt,2).

(E3.2)

The approximate equation for the equivalent specific heat of air is given in Chapter 3 as c p ,e = 0.94 + 0.2 (Tavg / 1000 )

Similarly, an approximate equation for the ratio of specific heats of air is

1

(E3.3)

Theory of Aerospace Propulsion Second Edition: Solutions Manual

γ = 1.4 − 0.006 (Tavg 1000 )

(E3.4)

The quantity Tt,avg =(Tt,3+Tt,2)/2 but we must estimate Tt,3 because it is as yet unknown.

Here we do know the compressor pressure ratio so we may find the ideal value Tt,3’ from (Tt,3’/Tt,2)=(pt,3/pt,2)(γ-1)/γ . As a first approximation we may use γ=1.4. Noting that pt,3= 18pt,2 = 690.8kPa we find Tt,3’=575.8 and therefore Tt,avg = 413.9K which, in turn, yields cp,e=1.023kJ/kg-K and γ=1.382. Then the work done by the compressor is given by Equation (E3.1) WC=(1.023kJ/kg-K)(251.9K)[(18)0.276-1]/0.85=370.0kJ/kg Using the improved value for cp,e in Equation (E3.2) permits us to calculate a better approximation to the actual compressor exit temperature as Tt,3=(370.0 kJ/kg)/(1.023kJ/kg-K)+251.9=613.6K The power absorbed by the compressor is C = (29.17kg/s)(370.0kJ/kg)=10.79MW PC = mW

(iv) Combustor, Stations 3-4: Here we know both Tt,3 and Tt,4 so that we find Tt,avg = (Tt,4+Tt,3)/2 = 971.8K Once again we estimate an effective specific heat using the slightly different coefficient of Tt,avg (0.215 as opposed to 0.2) in Equation (E3.3), as discussed in the text, to account for the presence of a small amount of combustion products present in the heated air: cpe=0.94+0.215(Tt,avg) =1.149kJ/kg/K Similarly, the effective ratio of specific heats is found to be

γ=1.4-0.0667(Tavg/1000)=1.335 Assuming that the liquid fuel enthalpy is negligible with respect to the gas enthalpy the energy balance across the combustor is [1+(f/a)]cp,e Tt,4 =cp,eTt,3 +(f/a)ηbQf (1.149kJ/kg-K)(1330K-613.6K) = (f/a)[(0.96)(44x103kJ/kg)-(1.149kJ/kg-K)(1330K)] (f/a) = 0.02022 Therefore the fuel flow rate is m f = 0.5898 kg s

We are given that the stagnation pressure loss in the combustor is (pt,3-pt,4) = 0.02pt,3 so pt,4 = 0.98pt,3 = 0.98(690.8kPa) and pt,4= 677.0kPa 2

Theory of Aerospace Propulsion Second Edition: Solutions Manual (v) Turbine, Stations 4-5: The matching condition for the turbine is that the work it extracts is equivalent to the work required by the compressor so WC=[1+(f/a)]WT. This expression accounts for the additional mass flow arising from the combustion of the fuel.

However, it is common to bleed air from the compressor for various vehicle requirements like air conditioning, etc., and engines are designed to provide such airflow with the bleed rate typically kept to less than 5% of the total air flow. As a result it is necessary to keep this in mind for detailed calculations. Keeping the increased mass flow due to the addition of fuel we obtain WT = cp,e (Tt,4 –Tt,5) = (1.149kJ/kg)(1330K– Tt,5) WT=WC/(1+f/a)=(370.0kJ/kg)/1.0202=362.7kJ/kg Using the value of cp,e=1.149 from part (iv) leads to Tt,5=1014K. If we now calculate Tt,avg= (1330+1014)/2 = 1172K we find cp,e = 0.94+0.215(Tt,avg/1000) = 1.192kJ/kg-K and

γ=1.4-0.0667(Tt,avg/1000)=1.322. Using the second approximation for cp,e yields Tt,5=1026K. The stagnation pressure at the turbine exit may be found from the turbine work equation, using the new value of γ =1.322, as follows: WT = ηtcp,eTt,4[1 – (pt,4/pt,5)-0.244]=362.7kJ/kg WT=(0.90)(1.192kJ/kg-K)(1330K) [1 – (pt,4/pt,5)-0.244] Therefore (pt,4/pt,5)-0.244=0.7458 and (pt,4/pt,5)=3.327. Then pt,5 = pt,,4/3.327 = (677kPa)/3.327 and pt,5=203.5kPa (vi) Nozzle, Stations 5-6: The nozzle pressure ratio pt,5/p0 =203.5kPa/22.69kPa = 4.672 so the nozzle is clearly operating in a supercritical state and therefore will be choked with M6=1.0. The exit pressure will be the critical pressure p* and the ratio of specific heats is chosen to be γ =1.33, consistent with an expected average temperature in the nozzle about 900K. The pressure and temperature at any point in the nozzle are given by the isentropic relations p/pt,5=[1+0.165M2] -4.03 and T/Tt,5=[1+0.165M2]-1 Therefore the exit pressure is p6 = p* = (0.5404)pt,5 =110.0kPa The exit temperature

3

Theory of Aerospace Propulsion Second Edition: Solutions Manual T6 = T* = Tt,5/1.165 = 880.7K. The exit velocity V6 = a* = (γRT*)1/2 = [(1.33)(287m2/s2-K)(880.7K)]1/2 = 579.8m/s.

The required nozzle exit area is found from the conservation of mass equation

m 6 = ρ 6 A6V6 = p* A6 γ RT * = (1.02022)(29.17kg/s)=29.76kg/s.

Then the exit area required is A6=0.1179m2. The density is found from the state equation

ρ*=p*/RT* = 0.4352kg/m3.

6V6 + A6 ( p6 − p0 ) so that (vii) Performance: The gross thrust Fg = m Fg= (29.76kg/s)(579.8m/s) + (0.1179m2 ) (110,000 N/m2 – 22,690N/m2) Fg =17,250N + 10,290N=27.54kN The ram drag

Fr = m 0V0 = (29.17kg/s)(265.7m/s) = 7.750kN The net thrust Fn = 27.54kN-7.75kN = 19.79kN. The specific fuel consumption is cj =

m f Fn

=

( 0.5892 kg s )( 3600 s hr ) = 0.1072 kg hr N = 1.051lbm hr lb (19, 790 N )

The overall efficiency of the engine is

ηo =

(19, 790 N )( 265.7 m s ) = 20.28% FnV0 = m f Q f ( 0.5892 kg s ) ( 44 × 106 J kg )

E3.2 A turbofan engine with the same component efficiencies as the turbojet engine in problem 3.1 is operating at the same flight conditions. The total mass flow, core flow plus fan flow, is 50kg/s and the bypass ratio β=1. The fan operates in a bypass duct without mixing with the core flow. The fan pressure ratio pt,3F/pt,2=2.2 and its efficiency is 0.85.

The nozzle of the fan bypass duct may be assumed to operate critically, that is, with an exit M3F=1. Calculate the properties of the flow at each station in the engine, the gross thrust, and the net thrust.

Solution (i) Free stream, Station 0: For the 1976 U.S. standard atmosphere as given in Appendix C

4

Theory of Aerospace Propulsion Second Edition: Solutions Manual p0= p0,s.l.σ = (101.3kPa) (0.224) = 22.69kPa, T0=216.8K, a0=295.2m/s, and therefore V0=a0M0= (295.2m/s)(0.9) = 265.7m/s.

For air with γ =1.4 the stagnation to static pressure ratio is pt/p = (1+M2/5)3.5 and the stagnation to static temperature ratio is Tt/T = (1+M2/5). Therefore pt.0=1.691p0=38.38kPa Tt,0=1.162T0=251.9K The total mass flow, core flow plus fan flow, is 50kg/s and the bypass ratio β=1, therefore the mass flow is 25kg/s through the core engine and 25kg/s through the fan. (ii) Inlet, Stations 1-2: The MIL-E-5008B standard pressure recovery for M0=0.9 is pt,2/pt,0=1.0 and the inlet flow is assumed to be adiabatic so that Tt,2/Tt,0=1. Therefore pt,2=38.38kPa Tt,2=251.9K (iii) Compressor, Stations 2-3: The pressure ratio is given as pt,3/pt,2=18 and assuming constant specific heat as calculated in Problem E3.1 we have cp2 = cp3 =cp,e= 1.023kJ/kgK. With γ =1.382 the work done is given by either WC=cpTt,2[(pt,3/pt,2)0.286-1]/ηc or WC = cp(Tt,3-Tt,2). Therefore pt,3=18pt,2=690.8kPa WC=(1.023kJ/kg-K)(251.9K)[(18)0.276-1]/0.85=370.0kJ/kg Tt,3=(370.0kJ/kg)/(1.023kJ/kg-K)+251.9=613.6K The power absorbed by the compressor is PC = mW C = (25kg/s)(370.0kJ/kg)=9.25MW (iv) Fan, Stations 2-2.5: The fan pressure ratio is given as pt,2.5/pt,2=2.2 and, because of the low pressure ratio we assume constant specific heat cp,2 = cp,2.5=1.004kJ/kg-K and γ =1.4. The work done is given by either WC=cpTt,2[(pt,2.5/pt,2)0.286-1]/ηF or WC = cp(Tt,2.5Tt,2). Therefore pt,2.5=2.2pt,2=84.44kPa WF=(1.004kJ/kg-K)(251.9K)[(2.2)0.286-1]/0.85=75.26kJ/kg Tt,2.5=(75.26kJ/kg)/(1.004kJ/kg-K)+251.9=326.9K The power absorbed by the fan is PF = m FWF = (25kg/s)(75.26kJ/kg)=1.882MW

5

Theory of Aerospace Propulsion Second Edition: Solutions Manual (v) Combustor, Stations 3-4: Here we know both Tt,3 and Tt,4 so that we find Tt,avg = (Tt,4+Tt,3)/2 = 971.8K Once again we need to estimate an effective specific heat using the slightly different coefficient of Tt,avg (0.215 as opposed to 0.2) which accounts for the presence of a small amount of combustion products present in the heated air: cpe=0.94+0.215(Tt,avg) =1.149kJ/kg/K Assuming that the liquid fuel enthalpy is negligible with respect to the gas enthalpy the energy balance across the combustor is [1+(f/a)]cp,e Tt,4 =cp,eTt,3 +(f/a)ηbQf (1.149kJ/kg-K)(1330K-613.6K) = (f/a)[(0.96)(44x103kJ/kg)-(1.149kJ/kg-K)(1330K)] (f/a) = 0.02022 Therefore the fuel flow rate is 0.02022(25kg/s) and m f = 0.5050 kg s

We are given that the stagnation pressure loss in the combustor is (pt,3-pt,4) = 0.02pt,3 so pt,4 = 0.98pt,3 = 0.98(690.8kPa) and therefore pt,4 = 677.0kPa (v) Turbine, Stations 4-5: The matching condition for the turbine is that the work it extracts is equivalent to the work required by the compressor plus the fan so WC + βWF = [1+(f/a)]Wt. This expression accounts for the additional mass flow arising from the combustion of the fuel. However, it is common to bleed air from the compressor for various vehicle requirements like air conditioning and engines are designed to provide such airflow with the bleed rate typically kept to less than 5% of the total air flow. As a result it is necessary to keep this in mind for detailed calculations. Keeping the increased mass flow due to the addition of fuel we obtain WT = cp,e (Tt,4 –Tt,5) = (1.149kJ/kg-K)(1330K– Tt,5) WT = (WC +βWF) /(1+f/a)= (370.0+75.26)kJ/kg/(1.02022)=436.4kJ/kg Using the value of cp,e=1.149kJ/kg-K from part (iv) leads to Tt,5=950.2K. If we now calculate Tt,avg= (1330+950.2)/2 = 1140K we find cp,e = 0.94+0.215(Tt,avg/1000) = 1.185kJ/kg-K. Then a second approximation yields Tt,5=961.7K.

6

Theory of Aerospace Propulsion Second Edition: Solutions Manual The stagnation pressure at the turbine exit may be found from the turbine work equation, with an average value of γ =1.4-0.0667(Tt,avg/1000)=1.336, as follows: WT = ηtcp,eTt,4[1 – (pt,4/pt,5)-0.251]=436.4kJ/kg WT=(0.90)(1.185kJ/kg-K)(1330K) [1 – (pt,4/pt,5)-0.251] Therefore (pt,4/pt,5)-0.251=0.6923 and (pt,4/pt,5)=4.328. Then pt,5 = 677kPa/4.328 and pt,5=156.4kPa (vi) Core nozzle, Stations 5-6: The nozzle pressure ratio pt,5/p0 = 156.4kPa/22.69kPa = 6.812 so the nozzle is clearly operating in a supercritical state and therefore will be choked with M6=1.0. The exit pressure will be the critical pressure and the ratio of specific heats is chosen to be γ =1.33, consistent with an average temperature in the nozzle of about 900K. The pressure and temperature at any point in the nozzle are given by the isentropic relations p/pt,5=[1+0.165M2] -4.03 and T/Tt,5=[1+0.165M2]-1.

Therefore the exit pressure is p6=p*=(0.5404)pt,5 or p6=84.52kPa The exit temperature T6 =T* = Tt,5/1.165 or T6= 825.5K. The exit velocity V6=a*=(γRT*)1/2=[(1.33)(287m2/s2-K)(825.5K)]1/2 =561.3m/s.

The required nozzle exit area is found from the conservation of mass equation m 6 = ρ 6 A6V6 = ρ ∗ A6 a ∗ = (1.02022)(25kg/s)=25.51kg/s.

The density is found from the state equation

ρ*=p*/RT* = 0.3567kg/m3.

Then the exit area required is A6=0.1274m2. (vi) Fan nozzle, Stations 2.5-3F: The fan nozzle pressure ratio pt,2.5/p0 = 84.44kPa/22.69kPa = 3.721 so the nozzle is clearly operating in a supercritical state and therefore will be choked with exit Mach number M3F =1.0. Because the fan operates at relatively cool temperatures we take cp=constant=1.004kJ/kg-K and γ =1.4. The pressure and temperature at any point in the nozzle are given by the isentropic relations p/pt,2.5 = [1+0.2M2]-3.5 and T/Tt,2.5=[1+0.2M2]-1. The fan exit pressure is then p3F = p* = (0.5283)pt,2.5 = 44.61kPa and the fan exit temperature T3F = T* = Tt,2.5/1.2 = 272.4K. The 7

Theory of Aerospace Propulsion Second Edition: Solutions Manual fan exit velocity V3F = a* = (γRT*)1/2 = [(1.4)(287m2/s2-K)(272.4K)]1/2 = 330.8m/s. The required nozzle exit area is found from the conservation of mass equation

m 3 F = ρ 3 F A3 FV3 F = ρ ∗ A3 F a ∗ = 25kg/s. The density is found from the state equation ρ* = p*/RT* = 0.5706kg/m3. Then the exit area required is A6=0.1324m2. (vii) Performance: The gross thrust

Fg = m 6V6 + A6 ( p6 − p0 ) + m FV3 F + A3 F ( p3 F − p0 ) Fg= (25.51kg/s)(561.3m/s) + (0.1274m2 ) (84,520 N/m2 – 22,690N/m2) +(25kg/s)(330.8m/s)+(0.1324m2)(44,610 N/m2 – 22,690N/m2) =14.32kN+7.877kN+8.270kN+2.902kN =33.37kN

0V0 =(50kg/s)(265.7m/s)=13.29kN and net thrust The ram drag Fr = m Fn=33.37kN-13.29kN = 20.08kN. The specific fuel consumption is

cj =

m f Fn

=

( 0.5050 kg s )( 3600 s hr ) = 0.09054 kg hr N = 0.8879 lbm hr lb ( 20, 080 N )

The overall efficiency of the engine is

ηo =

( 20, 080 N )( 265.7 m s ) = 24.01% FnV0 = m f Q f ( 0.5050 kg s ) ( 44 × 106 J kg )

E3.3 Use the results of problems 3.1 and 3.2 to compare the performance of the two engines in terms of net thrust, specific fuel consumption, overall efficiency, and total mass flow. Discuss the significance of the comparison.

Solution The desired results are shown in Table E3.1 below. The total net thrust of the turbofan is about equal to that of the turbojet, but the specific fuel consumption of the turbofan is 15.5% less than that of the turbojet. The overall efficiency of the turbofan is 18.4% greater than that of the turbojet as a result of moving almost 72% more airflow. The efficiency arising from accelerating a large mass flow of air a relatively small amount in

8

Theory of Aerospace Propulsion Second Edition: Solutions Manual the fan is the main factor in the improvement of the specific fuel consumption in the turbofan engine. This performance gain in the turbofan is the reason for its dominance in aircraft applications.

Table E3.1 Comparison of turbojet and turbofan engines Property

Turbojet Turbofan

Net thrust, Fn (kN)

19.79

20.08

Specific fuel consumption 1.051

0.8879

cj (lbm/hr/lb) Overall efficiency, ηo

20.28%

24.01%

Airflow (kg/s)

29.17

50.0

E3.4 Consider adding an afterburner to the turbojet engine of problem 3.1 with the additional information that the afterburner combustion efficiency is 0.85 and frictional losses in the afterburner yield (pt,5-pt,5b)=0.05pt,5. In the afterburner the fuel flow is m f =1.17 kg/s. Calculate the properties of the flow at each station in the engine, the gross thrust, and the net thrust.

Solution The results of problem 3.1(i)-(v) apply to this problem and we may start with the afterburner section. (vi) Afterburner, Stations 5-5b: The total pressure at the exit of the afterburner, station 5b, may be found from the specified loss equation (pt,5 - pt,5b)=0.05pt,5 so that pt,5b= 0.95pt,5 =0.95(203.5kPa) and pt,5b=193.3kPa. The total temperature at the afterburner exit Tt,5b is given by the energy balance

( m + m ) c T + m 0

f

p , e t ,5

η AB Qeff = ( m 0 + m f + m f , AB ) c p ,eTt ,5b

f , AB

9

Theory of Aerospace Propulsion Second Edition: Solutions Manual The heating value in the afterburner Qeff is lower than Qf because the flow into the afterburner has had some oxygen depletion due to fuel burn in the main combustor. This is given approximately by Equation (3.26) in the text:

Qeff = ⎡⎣1 − 14.7 ( f a ) ⎤⎦ Q f The afterburner fuel flow is given as 1.17kg/s so that (f/a)AB=1.17/29.17=0.4. Then, using the value for cp,e=1.192 determined previously for the turbine as a first approximation here, the energy balance becomes

⎡⎣1 + ( f a ) ⎤⎦ c p ,eTt ,5 + ( f a ) AB η AB ⎡⎣1 − 14.7 ( f a ) ⎤⎦ Q f = [1 + ( f a ) + ( f a ) AB ] c p ,eTt ,5b Substituting in the appropriate values yields (1.02022)(1.192kJ/kg-K)(1026K)+(0.04)(0.85)[1-14.7(0.02022)](44,000kJ/kg) = [1.02022+0.04](1.192kJ/kg)Tt,5b [1.06022](1.192)Tt,5b= 1248kJ/kg+1051kJ/kg= 2299kJ/kg Then as a first approximation Tt,5b = 1819K and we use this in calculating a new average stagnation temperature: Tt,avg=(1026+1819)/2=1423K. Then cp,e= 0.94+(0.215Tt,avg)/1000 =1.246 and γ=1.4-0.0667(Tt,avg)=1.305. The second approximation to the stagnation temperature in the afterburner is then Tt,5b = (2299kJ/kg)/ [(1.06022)(1.246kJ/kg-K)] = 1740K. (vi) Nozzle, Stations 5-6: The nozzle pressure ratio pt,5b/p0 = 193.3kPa/22.69kPa =8.519 so the nozzle is clearly operating in a supercritical state and therefore will be choked with M6=1.0. The exit pressure will be the critical pressure p* and the ratio of specific heats is chosen to be γ =1.4-0.0667(Tt,avg)=1.29, consistent with an average temperature in the nozzle of about 1600K. The pressure and temperature at any point in the nozzle are given by the isentropic relations p/pt,5 = [1+0.145M2] -4.45and T/Tt,5 = [1+0.145M2]-1. Therefore the exit pressure is p6=p*=(0.5474)pt,5=105.8kPa The exit temperature T6 =T* = Tt,5b/1.145 = 1494K. The exit velocity V6 = a* = (γRT*)1/2 = [(1.29)(287m2/s2-K)(1494K)]1/2 = 743.7m/s.

The required nozzle exit area is found from the conservation of mass equation

10

Theory of Aerospace Propulsion Second Edition: Solutions Manual m 6 = ρ 6 A6V6 = ρ ∗ A6 a ∗ = (1.06022)(29.17kg/s)=30.93kg/s.

The density is found from the state equation ρ*=p*/RT* = 0.2467kg/m3. Then the exit area required is therefore A6=0.1686m2. (vii) Performance: The gross thrust Fg = m 6V6 + A6 ( p6 − p0 ) Fg= (30.93kg/s)(743.7m/s) + (0.1686m2 ) (105,800 N/m2 – 22,690N/m2) Fg=23.00kN + 14.01kN=37.01kN The ram drag Fr = m 0V0 =(29.17kg/s)(265.7m/s)=7.750kN and net thrust Fn=37.01kN-7.75kN = 29.26kN. The total mass flow of fuel is 0.06022(29.17kg/s)=1.757kg/s and the specific fuel consumption is cj =

m f Fn

=

(1.757 kg s )( 3600 s hr ) = 0.2162 kg hr N = 2.120 lbm hr lb 29, 260 N

The overall efficiency of the engine is

ηo =

( 29, 260 N )( 265.7 m s ) = 10.06% FnV0 = m f Q f (1.757 kg s ) ( 44 × 10 6 J kg )

E3.5 Consider adding an afterburner to the turbofan engine of problem 3.2 with the additional information that the afterburner combustion efficiency is 0.85 and frictional losses in the afterburner yield (pt,5-pt,5b)=0.05pt,5. In the afterburner the fuel flow is m f , AB =1.0 kg/s. The bypass duct flow is assumed to mix internally with the turbine exit flow such that the resulting stagnation pressure is the average of the two. Calculate the properties of the flow at each station in the engine, the gross thrust, and the net thrust.

Solution The results of problem 3.2(i)-(vi) apply to this problem and we may start with the afterburner section. Recall that the bypass ratio β=1, and that the results of problem 3.2 include (f/a)=0.02022, pt,5=156.4kPa, Tt,5=961.7K, and Tt,F=326.9K. (vii) Afterburner, Stations 5-5b: The fan flow is assumed to mix internally with the core flow and the mass-averaged stagnation pressure entering the afterburner is now pt5 = 0.5(84.44kPa + 156.4kPa) =120.4kPa. The energy balance across the afterburner requires the sum of the total enthalpies of the fan flow (subscript F) and the core flow (subscript

11

Theory of Aerospace Propulsion Second Edition: Solutions Manual C) plus the energy released by combustion of the afterburner fuel to be equal to the total enthalpy passing through the nozzle and is given by

m 0, F ht ,5 F + ⎡⎣1 + ( f a ) ⎤⎦ m 0,C ht ,5 + m f , ABηbQeff = m 5,b ht ,5b Note that the ratio of the fuel flow in the afterburner to the core air flow may be written:

m f , AB m f , AB ⎛ 1 kg s ⎞ = (1 + β ) = 2 ( f a ) AB = 2 ⎜ ⎟ = 0.04 m 0,C m 0,C + m 0, F ⎝ 50 kg s ⎠ Similarly, the ratio of the mass flow entering the nozzle to the core air flow is

m 5,b m 0, F + ⎡⎣1 + ( f a ) ⎤⎦ m 0,C + m f , AB = = β + 1 + ( f a ) + 2 ( f a ) AB = 2.060 m 0,C m 0,C The effective heating value may be estimated on the basis that the core mass flow has combustion products while the bypass flow that mixes with it is pure air. As discussed in Chapter 3 of the textbook, the effective heating value for such flows is ⎡ ( f a) ⎤ Q Qeff = ⎢1 − 14.7 ⎥ f 1+ β ⎦ ⎣

Assuming a constant effective specific heat for the process the energy balance equation becomes ⎡ ⎣

β c p ,eTt , F + 1.02022c p ,eTt ,5 + 0.04 ( 0.85 ) ⎢1 − 14.7

0.02022 ⎤ ( 44, 000 ) = 2.06c p ,eTt ,5b 2 ⎥⎦

(1)(326.9K) + (1.02022)(961.7K) +(1274kJ/kg)/cp,e = (2.06)Tt,5AB (326.9K) + (981.1K) +(1274kJ/kg)/cp,e = (2.06)Tt,5b Tt,5b=635.0+(618.4kJ/kg)/cp,e Assume cp,e~1.16 to estimate Tt,5b~1168K. Then Tt,avg= (326.9K+944.5K+1168K)/3 = 813.1K and the effective specific heat becomes cp,e= 0.94+0.215(Tt,avg/1000) =1.115kJ/kg-K. With this value we find Tt,5b=1190K. Stagnation pressure losses in the afterburner are such that pt,5-pt,5b=0.05pt,5 or pt,5b=0.95pt,5=0.95(120.4kPa)=114.4kPa. (viii) Nozzle, Stations 5-6: The exit pressure will be the critical pressure p* and the ratio of specific heats is chosen from Table B.2 to be γ =1.33, consistent with an average temperature in the nozzle of about 1100K. The pressure and temperature at any point in

12

Theory of Aerospace Propulsion Second Edition: Solutions Manual the nozzle are given by the isentropic relations p/pt,5b = [1+0.165M2]-4.03and T/Tt,5b = [1+0.165M2]-1. Therefore the exit pressure is p6=p*=(0.5404)pt,5b=61.81kPa The exit temperature T6 =T* = Tt,5b/1.165 = 1021K. The exit velocity V6 = a* = (γRT*)1/2 = [(1.33)(287m2/s2-K)(1021K)]1/2 = 624.3m/s.

The required nozzle exit area is found from the conservation of mass equation

m 6 = ρ6 A6V6 = ⎡⎣1 + ( f a ) ⎤⎦ m 0,C + m 0, F + m f , AB m 6 = ⎡⎣1 + ( f a ) + β ⎤⎦ m 0,C + ( f a ) AB (1 + β ) m 0,C m 6 = ⎡⎣1 + β + ( f a ) + (1 + β )( f a ) AB ⎤⎦ m 0,C = ⎡⎣ 2.02022 + 2 ( 0.02 ) ⎤⎦ ( 25 kg s ) = 51.51 kg s The density is found from the state equation

ρ∗ = p*/RT* = 0.2109kg/m3.

Then the exit area required is A6=0.3912m2. (ix) Performance: The fan is internally mixed so the gross thrust

Fg = m 6V6 + A6 ( p6 − p0 ) Fg= (51.51kg/s)(624.3m/s) + (0.3912m2 ) (61,810 N/m2 – 22,690N/m2) Fg= 32.16kN + 15.30kN =47.46 kN

0V0 = (50kg/s)(265.7m/s) =13.29kN and net thrust The ram drag Fr = m Fn = 47.46kN-13.29kN = 34.17kN. The fuel mass flow is 0.5055kg/s for the core engine plus 1.0kg/s for the afterburner. The specific fuel consumption is

cj =

m f Fn

=

(1.5055 kg s )( 3600 s hr ) = 0.1586 kg hr N = 1.555 lbm hr lb 34,170 N

The overall efficiency of the engine is

ηo =

( 34,170 N )( 265.7 m s ) = 13.71% FnV0 = m f Q f (1.5055 kg s ) ( 44 ×106 J kg )

13

Theory of Aerospace Propulsion Second Edition: Solutions Manual E3.6 Use the results of problems 3.4 and 3.5 to compare the performance of the two engines in terms of net thrust, specific fuel consumption, overall efficiency, and thrust augmentation ratio. Discuss the significance of the comparison.

Solution The desired results are shown in Table E3.2 below. The net thrust of the afterburning turbofan is 16.8% greater than that of the afterburning turbojet while its specific fuel consumption is 26.7% less than that of the afterburning turbojet. The overall efficiency of the turbofan is 36.3% better than that of the turbojet in spite of moving almost 72% more airflow. Indeed, the efficiency arises from accelerating a large mass flow of air a relatively small amount in the fan bypass. The thrust augmentation of the afterburning turbofan is about 15% greater than that achieved by the afterburning turbojet, with both providing about a 50% to 70% improvement over the non-afterburning case. The specific fuel consumption gain in the afterburning turbofan is the major reason for its dominance in aircraft applications.

Table E3.2 Comparison of afterburning turbojet and turbofan engines Property

Turbojet

Turbofan

Net thrust, Fn (kN)

29.26

34.17

Specific fuel consumption, cj (lbm/hr/lb)

2.120

1.555

Overall efficiency, ηo

10.06%

13.71%

Airflow (kg/s)

29.17

50.0

Thrust augmentation ratio due to afterburning

1.479

1.702

It may be noted that using the conventional values of cp and γ for the “cold” and “hot” sections of the engine yields results that are within about 5% of the results presented here.

14

Theory of Aerospace Propulsion Second Edition: Solutions Manual E3.7 The Garrett TFE 1042 low-bypass turbofan can operate with or without afterburning. The internal fan has a bypass ratio of 0.75 and a pressure ratio of 2.5; the overall pressure ratio for the engine is 20 and the inlet airflow rate is 75lbm/s. Assuming all efficiencies are 100%, Qf=18,900Btu/lbm, the turbine inlet temperature Tt,4 = 1830R, and the afterburning temperature Tt,5 =3500R, find the net thrust and specific fuel consumption cj with and without afterburning for a static test, i.e., M0=0, under standard day atmospheric conditions.

fuel lines fan cowling inlet

fan flow

afterburner core flow

combustor

nozzle

combustor

fan flow core flow fuel injection

fan flow

1

2

3F

3

4

5

5b

6

7

Figure E3.1 Notional schematic diagram of the Garrett TFE 1042 low-bypass turbofan Solution Consider the schematic diagram of the engine shown in Figure E3.1. Because the engine is assumed to be ideal we will simply use the standard values of cp=0.24Btu/lbm and

γ =1.4 for the “cold” sections of the engine and cp=0.276Btu/lbm and γ =1.33 for the “hot” sections of the engine. (i) Compressor-Fan-Turbine match: The work done by the turbine is assumed to be equal to the work required by the fan-compressor combination. The engine is on a test stand under standard sea level day conditions so M0=0, pt,0=p0=2116lb/ft2, Tt,0=T0=520R. For 100% efficiency the work absorbed by the fan (stations 2 to 3F) is given by WF = cpTt,2[(pt,2.5/pt,2)0.286-1]=(0.24Btu/lbm-R)(520R)[(2.50.286-1]=37.39Btu/lbm We assume here that cp and γ are constant at 0.24Btu/lbm-R and 1.4, respectively. For the compressor, in which the temperature will rise to a level that can be estimated by an

15

Theory of Aerospace Propulsion Second Edition: Solutions Manual isentropic compression using standard day conditions (noting that Tt,2=Tt,0), e.g., Tt,3~Tt,2(pt,2.5/pt,2)(γ-1)/γ= 520R(20)0.286=1225R. Then the compressor work, for 100% efficiency, is WC = cpTt,2[(pt,3/pt,2)0.286-1]=(0.24Btu/lbm-R)(520R)[(200.286-1]=169.2Btu/lbm The power that must be delivered by the turbine is

PT = m FWF + m CWC The total mass flow is 75lbm/s and for β=0.75 the mass flow split is 42.86lbm/s through the compressor and 32.14lbm/s through the fan so that the power is PT = (32.14lbm/s)(37.39Btu/lbm) + (42.86lbm/s)(169.2Btu/lbm)= 8454Btu/s PT= 11,959hp (8.922MW) (ii) Fan, Stations 2 – 3F: The stagnation pressure at the fan exit is pt,3F = 2.5pt,2 = 2.5pt,0 = 5,290lb/ft2 and the stagnation temperature at the fan exit, for isentropic flow, is Tt,3F = Tt,2(pt,3F/pt,2)0.286 = (520R)(1.3) = 675.7R. (iii) Compressor, Stations 2 – 3: The stagnation pressure at the compressor exit is pt,3 = 20pt,2 = 20pt,0 = 43,320 lb/ft2 and the stagnation temperature at the compressor exit, for isentropic flow, has already been found to be Tt,3=1225R (iv) Combustor, Stations 3 – 4: The turbine inlet temperature is given as Tt,4=1830R and for an ideal combustor there is no stagnation pressure loss so pt,4 = pt,3=43,320psf. The energy balance requires that the fuel to air ratio be consistent with the following: cp,3Tt,3+(f/a)Qf = [1+(f/a)]cp,4Tt,4 (0.24 Btu/lbm-R)(1225R)+(f/a)(18,900Btu/lbm)=[1+(f/a)](0.276Btu/lbm-R)(1830R) Therefore (f/a)= 0.0115. With this fuel to air ratio, recognizing that only the compressor air is involved in the combustion process, the fuel flow rate is (0.0115)(42.86lbm/s) = 0.4913lbm/s (v) Turbine, Stations 4 – 5: The turbine work is given by WT =

Pt

⎡⎣1 + ( f a ) ⎤⎦ m c

=

11,959hp × 0.707 Btu lbm hp = 195 Btu lbm (1.0115 ) 42.86 lbm s

WT= cp,4Tt,4 – cp,5Tt,5 = (0.276Btu/lb-R)(1830R) – cp,5Tt,5 =195Btu/lbm

16

Theory of Aerospace Propulsion Second Edition: Solutions Manual Then cp,5Tt,5 = 310.1Btu/lbm=(0.276Btu/lbm-R)Tt,5 and Tt,5 = 1123R. For 100% efficiency the turbine exit pressure may be obtained from the isentropic expansion equation (pt,5/pt,4)=(Tt,5/Tt,4)γ/(γ-1). Using γ =1.33 in this hot section of the engine yields pt,5 = 43,320lb/ft2(1123R/1830R)4.03 = 6,053lb/ft2. (v) Nozzle, no afterburning, Stations 5 – 7: The stagnation pressure is assumed constant through the nozzle and the nozzle is assumed to be matched so as to obtain maximum thrust. Therefore the nozzle pressure ratio is pt,7 / p7 = pt,5 / p0. The stagnation pressure at the fan and turbine exits are almost equal so that, after mixing the fan and turbine streams, the stagnation pressure is assumed to be the mass-averaged value: pt,5avg = [42.86pt,5 + 32.14pt,2.5]/75 =[0.2539x106+0.1700x106]/75 = 5,734b/ft2 The stagnation temperature at the exit is found from the following energy balance across the nozzle:

p ,7Tt ,7 m F c p ,3 F Tt ,3 F + m 5 (1 + f a ) c p ,5Tt ,5 = mc

( 32.14 )( 0.24 )( 675.7 ) + ( 42.86 )(1 + 0.0115)( 0.276 )(1123) = ( 75)( 0.276 ) Tt ,7 The resulting stagnation temperature is Tt,7=901R. For a matched nozzle the pressure ratio is therefore (pt,7/p7)-1 = (5734/2116)-1 = 0.3692. Assuming γ =1.33 the exit Mach number M7=1.3. For this value the static to stagnation temperature ratio is T7 / Tt,7 = 0.7802 and therefore T7 = 702.9R. The exit sound speed is a7 = (γRT7)1/2 = [(1.33)(1716ft2/s2-R)(702.9R)]1/2 or a7 =1267ft/s. The exit velocity is V7 = (1267ft/s)(1.3) = 1647ft/s. (vi) Performance without afterburning: The gross thrust Fg = [(75lbm/s) / (32.17ft/s2)] (1647ft/s) = 3,839lb.

Since the flight Mach number M0=0, the net thrust equals the gross thrust, Fn=3,839lb.The specific fuel consumption cj = (0.0115)(42.86lbm/s)(3600s/hr)/(3,839lb) = 0.462lbm/hr/lb. (vi) Afterburner, Stations 5 – 5b: The mixing of the fan and turbine flow may be treated by first considering the energy balance prior to fuel injection assuming that the two streams mix completely:

17

Theory of Aerospace Propulsion Second Edition: Solutions Manual

βcp,FTt,3F +cp,5Tt,,5+(f /a)ABQeff = [1+β+(f /a)+(f /a)AB)](cp,5bTt,5b) βcp,FTt,3F +cp,5Tt,,5-[1+β+(f/a)]cp,5bTt,5b = (f /a)AB)[(cp,5bTt,5b)-Qeff] Substitution of known properties yields

β(0.24)(675.7)+0.276(1123)-[1+β+ (f/a)](0.276)(3500) = (f /a)AB{(0.276)(3500)-[1-14.7(f /a) /(1+β)](18,900)} With the known values β=0.75 and f/a=0.0115 we find (f/a)AB=0.0793. Since this fuel-air ratio is based on the compressor air flow, the fuel weight flow for the afterburner is 3.986lb/s. If there is no further pressure loss the nozzle pressure ratio remains the same as does the Mach number. However, the static temperature T7 = (0.7819)3500R = 2736R. The exit sound speed is a7 = (γRT7)1/2 = [(1.33)(1716ft2/s2-R)(2736R)]1/2 =2499ft/s The exit velocity is V7=(2499ft/s)(1.3)=3249ft/s. (vii) Performance: The gross thrust is Fg=[(75lbm/s) / (32.17ft/s2)] (3249ft/s) = 7,575lb.

Since the flight Mach number M0=0, the net thrust equals the gross thrust, Fn=7,575lb.

The specific fuel consumption cj = (0.0115+0.0793)(42.86lb/s)(3600s/hr)/(7575lb) = 1.85lbm/hr/lb. Thus afterburning doubles the thrust while quadrupling the specific fuel consumption.

E3.8 Consider an ideal turbofan engine with a bypass ratio β and fixed turbine entry stagnation temperature Tt,4. Assuming that the working gas has constant thermal properties with γ=1.4 determine the optimum value of β, that is, the β for which the net thrust Fn is maximum, under the following conditions: T0=216.7K, M0=0.9, fan pressure ratio pt,3F/pt,0=2.2, compressor pressure ratio pt,3/pt,2=8, and Tt,4=1333K. The compressor mass flow m C is to be held constant and may be assumed to be equal to the turbine mass flow (fuel addition is considered negligible). Determine the bypass ratio β for which the normalized net thrust Fn* = Fn m CV0 is a maximum.

Solution Consider the turbofan engine as shown below in Figure E3.2. Assuming an ideal engine with a matched nozzle and neglecting the mass flow of fuel the net thrust is given by Fn = m C ⎡⎣(V7 − V0 ) + β (V3 F − V0 ) ⎤⎦

18

(E3.5)

Theory of Aerospace Propulsion Second Edition: Solutions Manual For fixed compressor mass flow we may normalize Equation (E3.5) as follows:

⎛V ⎞ ⎛V ⎞ Fn = ⎜ 7 ⎟ + β ⎜ 3 F ⎟ − ( β + 1) m CV0 ⎝ V0 ⎠ ⎝ V0 ⎠

(E3.6)

fan cowling fan flow fan flow

cold flow

combustor core flow

nozzle

hot flow

combustor

cold flow fan flow

fuel line

fan

1

2

2.5

3

3F

4

5

6,7

Figure E3.2 Notional schematic diagram of a turbofan engine The thermodynamic properties cp and γ are assumed to be constant and the flight velocity V0=M0(γRT0)1/2 and the quantity Γ=(γ−1)/γ. Then the exit velocities for the core nozzle and the fan nozzle are given, respectively, by V7 = V0 V3 F = V0

Γ 2c pTt ,7 ⎡ ⎛ p7 ⎞ ⎤ ⎢1 − ⎜ ⎟ ⎥ M 02γ RT0 ⎢ ⎜⎝ pt ,7 ⎟⎠ ⎥ ⎣ ⎦ Γ 2c pTt ,3 F ⎡ ⎛ p3, F ⎞ ⎤ ⎢1 − ⎜ ⎟ ⎥ M 02γ RT0 ⎢ ⎜⎝ pt ,3 F ⎟⎠ ⎥ ⎣ ⎦

Both nozzles are assumed to be perfect so that the stagnation pressures within them are constant and therefore pt,7=pt,5 and pt,3F=pt,2.5. Furthermore in keeping with the ideal behavior of the engine we assume both nozzles are matched to the ambient atmospheric pressure so that p3F=p7=p0. Using the definition for the quantity 2cp/γR = 2/(γ-1) the normalized net thrust of Equation (E3.6) is given by

19

Theory of Aerospace Propulsion Second Edition: Solutions Manual

Fn = m CV0

2 Γ Γ ⎫ + M 02 ⎧ Tt ,3 F ⎡ ⎛ p0 ⎞ ⎤ ⎪ ⎪ Tt ,7 ⎢⎡ ⎛ p0 ⎞ ⎥⎤ γ −1 ⎢1 − ⎜ 1− ⎜ ⎟⎟ + β ⎟ ⎥ ⎬ − (1 + β ) ⎨ ⎜ M 02 Tt ,2 ⎢ ⎜⎝ pt ,3 F ⎟⎠ ⎥ ⎪ ⎪ Tt ,2 ⎢⎣ ⎝ pt ,7 ⎠ ⎥⎦ ⎣ ⎦⎭ ⎩

Here we have also assumed the inlet to be perfect so that

Tt ,2 T0

=

Tt ,0 T0

= 1+

γ −1 2

M 02

Because the nozzle is perfect Tt,7=Tt,5. Furthermore, if we let γ=7/5 so that Γ=2/7 and (γ 1)/2=1/5, then the normalized net thrust is 2/7 ⎧ ⎡ ⎛ p ⎞ 2/7 ⎤ ⎪⎫ 5 + M 02 ⎪ Tt ,5 ⎡ ⎛ p0 ⎞ ⎤ Fn 2/7 0 ⎢1 − ⎜ = ⎟ ⎥ + β RF ⎢1 − ⎜⎜ ⎟ ⎥ ⎬ − (1 + β ) (E3.7) ⎨ m CV0 M 02 ⎪ Tt ,2 ⎢ ⎜⎝ pt ,7 ⎟⎠ ⎥ ⎢ ⎝ pt ,3 F ⎟⎠ ⎥ ⎪ ⎣ ⎦ ⎣ ⎦⎭ ⎩

Here we will use RF = (pt,2.5/pt,2), RC = (pt,3/pt,2), and RT = (pt,4/pt,5) to denote the fan, compressor, and turbine stagnation pressure ratios, respectively. The connection between the fan and the core flow parameters is the turbine work equation, written below (neglecting the fuel addition): c p (Tt ,3 − Tt ,2 ) + β c p (Tt ,2.5 − Tt ,2 ) = c p (Tt ,4 − Tt ,5 )

Since cp is constant and the flow through the fan nozzle and the core nozzle is assumed to be isentropic we may write this as ⎡⎛ p ⎞ Γ ⎤ ⎡⎛ p ⎞ Γ ⎤ T Tt ,5 t ,3 t ,3 F t ,4 ⎢⎜ − 1⎥ + β ⎢⎜ − 1⎥ = − ⎟ ⎟ ⎢⎜⎝ pt ,2 ⎟⎠ ⎥ ⎢⎜⎝ pt ,2 ⎟⎠ ⎥ Tt ,2 Tt ,2 ⎣ ⎦ ⎣ ⎦

This matching equation thus becomes Tt ,4 Tt ,2

− RCΓ − β RFΓ + (1 + β ) =

Tt ,5 Tt ,2

Using the chain rule we may write the temperature ratio as Tt,7/Tt,2 = (Tt,7/Tt,5)(Tt,5/Tt,4)(Tt,4/Tt,2) = (1)(RT)-Γ(Tt,4/Tt,2)

Then Equation (E3.8) may be solved for the turbine pressure ratio as follows:

20

(E3.8)

Theory of Aerospace Propulsion Second Edition: Solutions Manual ⎧⎪ T ⎫⎪ RT = ⎨1 − t ,2 ⎡⎣( RcΓ − 1) + β ( RFΓ − 1) ⎤⎦ ⎬ ⎩⎪ Tt ,4 ⎭⎪

−1/ Γ

(E3.9)

Note that for fixed flight conditions, turbine inlet temperature, and fan and compressor pressure ratios, the turbine pressure ratio is controlled by the bypass ratio β. As β increases the required turbine pressure ratio RT = (pt,4/pt,5) also increases. However, for fixed compressor pressure ratio the value of pt,4 is likewise fixed, which means that the turbine exit stagnation pressure pt,5, and therefore the nozzle exit stagnation pressure pt,7, must also decrease. This in turn ensures that the nozzle thrust will suffer, as can be seen from Equation (E3.7).The core nozzle equation may be developed using the chain rule as follows: p0/pt7 = (p0/pt0)(pt0/pt2)(pt2/pt3)(pt3/pt4)(pt4/pt5)(pt5/pt7) p0/pt7 = (1+M02/5)-1/Γ(1)(RC)-1(1)(RT)(1)

Then the nozzle pressure function becomes Γ

⎛ p ⎞ ⎛ RT ⎞ 1 1− ⎜ 0 ⎟ = 1− 2 ⎜p ⎟ (1 + M 0 / 5) ⎜⎝ RC ⎟⎠ ⎝ t ,7 ⎠

Γ

(E3.10)

Dividing Equation (E3.9) by RCΓ yields Γ

⎧ ⎫⎪ ⎛ RT ⎞ −Γ Tt ,4 ⎪ Tt ,4 − ⎡⎣( RCΓ − 1) + β ( RFΓ − 1) ⎤⎦ ⎬ ⎨ ⎜ ⎟ = RC Tt ,2 ⎪⎩ Tt ,2 ⎪⎭ ⎝ RT ⎠

−1

(E3.11)

Then, using Equations (E3.10) and (E3.11) in Equation (E3.3) permits the term under the first radical sign in braces in Equation (E3.7) to be written as Γ Γ ⎫⎪ ⎪⎧ Tt ,7 ⎡ ⎛ p0 ⎞ ⎤ ⎧⎪ Tt ,4 ⎛ RT ⎞ ⎪⎫ 1 Γ Γ ⎢1 − ⎜ − ⎡( RC − 1) + β ( RF − 1) ⎤⎦ ⎬ ⎨1 − ⎟ ⎥=⎨ ⎜ ⎟ ⎬ (E3.12) 1 + M 02 / 5 ⎝ RC ⎠ ⎪ Tt ,2 ⎢ ⎜⎝ pt ,7 ⎟⎠ ⎥ ⎩⎪ Tt ,2 ⎣ ⎪ ⎪ ⎭ ⎩ ⎭ ⎣ ⎦

21

Theory of Aerospace Propulsion Second Edition: Solutions Manual The first term in braces on the right hand side of Equation (E3.12) may be denoted by (A - βB), where A=[(Tt,4/Tt,2)-(RCΓ-1)] and B=(RFΓ-1). Then Equation (E3.11) may be rewritten as Γ

⎛ RT ⎞ 1 Tt ,4 ⎡ 1 ⎤ ⎜ ⎟ = Γ RC Tt ,2 ⎢⎣ A − β B ⎥⎦ ⎝ RC ⎠

The second term in braces in Equation (E3.12) may be rewritten as Γ

⎡ Tt ,4 1 ⎤ ⎛ RT ⎞ 1 1 1 β = − − 1− A B ( ) ⎢ ⎥ ⎜ ⎟ 1 + M 02 / 5 ⎝ RC ⎠ 1 + M 02 / 5 Tt ,2 RCΓ ⎦⎥ ( A − β B ) ⎣⎢

Then Equation (E3.12) becomes Γ Tt ,7 ⎡ ⎛ p0 ⎞ ⎤ ⎡ 5 Tt ,4 1 ⎤ ⎢1 − ⎜ ⎥ = ⎢( A − β B ) − ⎟ ⎥ Tt ,2 ⎢ ⎜⎝ pt ,7 ⎟⎠ ⎥ ⎣⎢ 5 + M 02 Tt ,2 RCΓ ⎦⎥ ⎣ ⎦

The fan nozzle pressure term is developed in the same fashion and is given by Γ

⎛ p ⎞ 1 1 1− ⎜ 0 ⎟ = 1− ⎜p ⎟ (1 + M 02 / 5) RFΓ ⎝ t ,3 F ⎠

Then the term under the second radical sign in braces in Equation (E3.7) may be written as ⎡ ⎛ p ⎞Γ ⎤ ⎛ ⎞ 1 RFΓ ⎢1 − ⎜ 0 ⎟ ⎥ = ⎜ RFΓ − ⎟=C 2 ⎜ ⎟ 1+ M0 / 5 ⎠ ⎢ ⎝ pt ,3 F ⎠ ⎥ ⎝ ⎣ ⎦

Here the term C = RFΓ – (1+M02/5)-1. Then the normalized net thrust becomes

22

Theory of Aerospace Propulsion Second Edition: Solutions Manual Fn 5 + M 02 ⎧⎪ Tt ,4 ⎡ 5 5 ⎫⎪ −Γ ⎤ Γ Γ Γ = − − − − − + − R R R R 1 1 1 β β ( ) ( ) ⎨ ⎬ − (1 + β ) ⎢ C ⎥ C F F m CV0 M 02 ⎪ Tt ,2 ⎣ 5 + M 02 5 + M 02 ⎪ ⎦ ⎩ ⎭

There are several important points to consider in this normalized net thrust equation. It shows the core engine nozzle contribution as the first term within the braces and the fan contribution as the second term within the braces. First, the core engine nozzle contribution decreases as β increases. This is because the turbine must provide increasingly large power to drive the fan as the fan flow rate grows and the resulting reduction in turbine exit stagnation pressure and temperature diminish the nozzle thrust.

Second, the nozzle contribution will equal zero when

β=

⎤ 1 ⎡ Tt ,4 ⎛ 5 −Γ ⎞ Γ 1 1 R R − − − ( ) ⎢ ⎥ ⎜ ⎟ C C RFΓ − 1 ⎣⎢ Tt ,2 ⎝ 5 + M 02 ⎠ ⎦⎥

Thus there is a maximum allowable bypass ratio for given flight conditions, pressure ratios, and turbine inlet temperatures. In essence the engine becomes a ducted fan rather than a turbofan. The normalized thrust equation may be put into more compact form, as follows: Fn 5 + M 02 = m CV0 M 02

{ [ D + Eβ ] + β C } − (1 + β )

Here A = [(Tt,4/Tt,2)-(RCΓ-1)] B = (RFΓ-1) C = RFΓ – (1+M02/5)-1 D = A- [5/(5+M02)](Tt,4/Tt,2)(RC)-Γ

E=–B

23

(E3.13)

Theory of Aerospace Propulsion Second Edition: Solutions Manual Then the limiting case of β=0 yields

Tt ,4 ⎡⎛ 5 + M 02 ⎞ 5 1 ⎤ ⎛ 5 + M 02 ⎞ Γ Fn = ⎢⎜ ⎥ −⎜ ⎟− ⎟ ( RC − 1) − 1 lim CV0 Tt ,2 ⎣⎝ M 02 ⎠ M 02 RCΓ ⎦ ⎝ M 02 ⎠ β →0 m The limiting case of maximum bypass ratio is given by

Fn 5 + M 02 ⎛ D ⎞ ⎛ D⎞ = − ⎟ C − ⎜1 − ⎟ ⎜ lim 2 CV0 M0 ⎝ E ⎠ ⎝ E⎠ β → β max m

The conditions to be considered here are as follows: (a) Flight conditions: M0=0.9, T0=216.7K, V0=266.1m/s. Therefore1+M02/5=1.162 so that Tt,2=251.7K. Furthermore, [(1+M02/5)/ (M02/5)]=7.1728 (b) Fan pressure ratio RF=2.2, RF2/7=1.25267, (c) Compressor pressure ratio RC=8, RC2/7=1.81145, (d) Turbine inlet stagnation temperature Tt,4=1333K Then the coefficients are as follows: A = [(Tt,4/Tt,2)-(RCΓ-1)] =5.2937 – 0.81145=4.4823 B = (RFΓ-1) =0.25267 C = RFΓ – (1+M02/5)-1 = 0.39208 D = A- [5/(5+M02)](Tt,4/Tt,2)(RC)-Γ =4.4823 – 0.86*5.2937/1.81145=1.97 E = - B = -0.25267

The normalized thrust equation thus becomes Fn = (7.1728)1/2{(1.97 – 0.25267β)1/2 +β(0.39208)1/2} – (1+β) m CV0

The limiting cases are: (a) β=0:

Fn = 2.759 m CV0

(b) β=7.8:

Fn = 4.28 m CV0

24

Theory of Aerospace Propulsion Second Edition: Solutions Manual The variation of the normalized net thrust with bypass ratio is shown in Figure E3.3. The maximum thrust is found to occur at a bypass ratio of about β=6.8, somewhat prior to the loss of all jet thrust. The exact value may be found by differentiating Equation (E3.13) with respect to β and setting the result to zero. This operation yields

⎤ M 02 E⎡ C β Fn ,max = ⎢ − ⎥ 4 ⎢⎣ 5 + M 02 ⎥⎦

−2

−

D E

6 5 4 Fn/mcV0 3 constant core mass flow

2 1 0 0

1

2

3

4

5

6

7

8

Bypass ratio β

Figure E3.3Variation of the normalized net thrust with bypass ratio for fixed core

mass flow for RF=2.2, RC=8, Tt,4=1333K and flight at M0=0.9 at 11km altitude Though it wasn’t requested in the problem statement, the specific fuel consumption may be found. The energy conservation equation across the ideal burner, neglecting the fuel flow with respect to the core air flow, requires that

c pTt ,3 + ( f a ) Q f = c pTt ,4

We may solve for the fuel to air ratio and find that

( f a) =

cp

⎤ c pTt ,2 ⎡ Tt ,4 − RCΓ ⎥ ⎢ ⎢⎣ Tt ,2 ⎥⎦ f

(T − T ) = Q Q t ,4

t ,3

f

25

Theory of Aerospace Propulsion Second Edition: Solutions Manual The specific fuel consumption is then given by

cj =

m f Fn

=

( f a ) m C = f a ⎛ m CV0 ⎞⎛ 1 ⎞ ( )⎜ ⎟⎜ ⎟ ⎝ Fn ⎠⎝ V0 ⎠

Fn

For the conditions of this problem

( f a) =

(1.004 kJ kg ⋅ K )( 251.7 K ) ⎛ 1333K − 80.286 ⎞ = 0.02 ⎜ ⎝ 251.7 K

44, 000 kJ kg

⎟ ⎠

For the specific fuel consumptions in the conventional units of lbm/hr/lb we merely multiply cj given above by 9.807m/s2 and obtain

( 9.807 m s ) ( 3600 s hr ) ⎛ F ⎞ = 2.653 ⎛ F ⎞ lbm hr lb c = 0.02 −1

2

−1

⎜ ⎟ ⎝ m CV0 ⎠

⎜ ⎟ ⎝ m CV0 ⎠

n

j

266.1 m s

n

The results for specific fuel consumption as a function of by pass ratio are shown in Figure E3.4.

1 0.8 0.6 cj (lbm/hr/lb) 0.4 0.2 0 0

1

2

3

4

5

6

7

8

Bypass ratio β

Figure E3.4 Variation of the specific fuel consumption with bypass ratio for a

fixed core mass flow for RF=2.2, RC=8, Tt,4=1333K and flight at M0=0.9 at 11km altitude

26

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4. COMBUSTION CHAMBERS FOR AIRBREATHING ENGINES E4.1. Consider a combustion chamber designed for constant static temperature operation. (a) Develop the quasi-one-dimensional equations to describe the thermodynamic and flow properties at the exit station under the assumptions of heat addition by stagnation temperature change, variable cross-sectional area, negligible friction, and constant chemical composition. (b) Show a plot of the variation of Mach number M4 with stagnation temperature ratio Tt,4/Tt,3 for M3=0.4, 0.8, and 2; (c) Determine if this combustion chamber is limited by thermal choking.

Solution (a) From the array of influence coefficients given in Table 2.1 of the text we may form the equation for the change in static temperature, which we require to be zero, as follows: 2 γ − 1) M 2 dA (1 − γ M ) ⎛ −dH ⎞ ( dT =0= + ⎜ ⎟ T 1− M 2 A 1 − M 2 ⎜⎝ c pT ⎟⎠

From this relation we find that

⎛ −dH ⎞ − ( γ − 1) M 2 dA ⎜⎜ ⎟⎟ = 2 ⎝ c pT ⎠ (1 − γ M ) A From Table 2.1 we find the influence coefficients for Mach number change which lead to 2

dM = −2 M2

1+

γ −1

M2

2 1− M 2

dA 1 + γ M 2 ⎛ −dH ⎞ + ⎜ ⎟ A 1 − M 2 ⎜⎝ c pT ⎟⎠

Eliminating (-dH/cpT) and carrying out some manipulation of the resulting equation leads to the following expression for the area change: dA 1 dM 2 = − (1 − γ M 2 ) 2 A 2 M From this equation we find the relation between the areas at stations 3 and 4 to be

γ A4 M 3 = exp ( M 42 − M 32 ) A3 M 3 2 The equation for the pressure change is formed in a similar manner yielding 2 dp γ M 2 dA γ M 2 ⎡ 1 2 dM ⎤ M 1 γ = = − − ( ) p 1 − γ M 2 A 1 − M 2 ⎢⎣ 2 M 2 ⎥⎦

This leads to

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

p4 ⎡ γ ⎤ = exp ⎢ − ( M 42 − M 32 ) ⎥ p3 ⎣ 2 ⎦ Then the change in stagnation temperature and pressure are simply

γ −1 2 γ −1 2 M 4 1+ M4 T 2 2 = 4 = Tt ,3 T3 1 + γ − 1 M 2 1 + γ − 1 M 2 3 3 2 2 1+

Tt ,4

⎡ γ −1 2 ⎤ pt ,4 p4 ⎢1 + 2 M 4 ⎥ = ⎢ ⎥ pt ,3 p3 ⎢1 + γ − 1 M 2 ⎥ 3 2 ⎣ ⎦

γ ( γ −1)

(b) A plot of M4 as a function of Tt,4/Tt,3 is shown below

5 4.5 4 3.5 3 M4 2.5 2 1.5 1 0.5 0 1

1.5

2

2.5

3

Tt,4/Tt,3

Figure E4.1 The variation of Mach number M4 with stagnation temperature ratio

Tt,4/Tt,3 is shown for M3=0.4, 0.8, and 2 (c) There is no possibility of choking because, as can be seen in Figure E4.1, flow passes smoothly through M=1 from any starting Mach number and the entropy change may be shown to be a monotonically increasing function of M4 given by

s4 − s3 γ − 1 = M 42 − M 32 ) ( 2 cp

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.2 The SCORPION (Statoréacteur Cruciforme comme ORgane Portant Intégré à Orifices Nasaux) was a prototype of a long range ground to air missile (M0=6, range=600 km); 220mm diameter liquid fueled ramjet (demonstrator) designed by ONERA of France. Successful ground tests of the fully integrated vehicle in their S4 MA wind tunnel (1973-1974). Assume that the ramjet combustion has a constant crosssectional area and that the air from the intakes enters it at a speed of 100fps (30.5m/s) and a stagnation temperature Tt,3=780F (416C). The combustor is designed to burn a mixture of fuel and primary air in the ratio of 0.06:1 with a burner efficiency ηb=90%.

The kerosene fuel has a heating value of 18,900 Btu/lb (44MJ/kg) and the ratio of secondary air flow to total airflow m 3, s / m 3 can be varied by using different combustor

liners. For values of m 3, s / m 3 =0, 0.2, 0.4, and 0.6 determine the stagnation temperature (Tt,4) leaving the combustor, the percentage loss in stagnation pressure across the combustor, and the velocity leaving the combustor, V4. Show your results on graphs using m 3, s / m 3 as the abscissa and briefly explain the behavior of the flow. Assume that friction and fuel injector strut drag are negligible and that the average cp=0.276Btu/lb-R (1.16kJ/kg-K) and the average γ=1.33.

Solution

We will assume that the thermodynamic properties of the gases involved have the following characteristics: k = γ = constant = 1.33, cp = constant = 0.276Btu/lb-R, and the molecular weight of the gases W3=W4=constant. The relationship between the rate of heat addition and the heating value of the fuel is given by m 3c p (Tt ,4 − Tt ,3 ) = m f ηb Q f

(E4.1)

The mass flow of air entering the combustor is related to the fuel mass flow as follows: m 3 = m 3, p + m 3,s

(E4.2)

The combustor burns the fuel with the primary air flow in the ratio mf/m3,p=0.06 so that m 3 =

m f 0.06

+ m 3,s

(E4.3)

Then the stagnation temperature at station 4 may be determined from Equations (E4.1) and (E4.3) to be (results for both the Kelvin and Rankine scales are shown)

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Tt ,4 = Tt ,3 +

m f ηb Q f ⎛ m ⎞ η Q = Tt ,3 + 0.06 ⎜1 − 3, s ⎟ b f m 3 c p m 3 ⎠ c p ⎝

⎛ m ⎞ 0.9 ( 44 MJ / kg ) ⎛ m ⎞ = 689 K + ⎜1 − 3, s ⎟ 2048 K Tt ,4 = 689 K + 0.06 ⎜1 − 3, s ⎟ m 3 ⎠ 1.16kJ / kg − K m 3 ⎠ ⎝ ⎝ ⎛ m ⎞ Tt ,4 = 1240 R + ⎜1 − 3, s ⎟ 3686 R m 3 ⎠ ⎝

(E4.4)

The variation of Tt,4 with percent cooling air is given in Equation (E4.4) for both the Rankine and Kelvin temperature scales and is shown in Figure E4.2.

4924R (2736K) and 394 ft/s (137m/s)

5000R (2737K)

400ft/s

335 ft/s (102m/s)

Tt,4 3750R (2083K)

V4

4187R (2326K)

300 ft/s

276 ft/s (84 m/s) 3450R (1917K) 217 ft/s (75.5m/s)

2500R (1389K)

200 ft/s

2714R (1508K)

100 ft/s

1250R (694K)

0

0 0

0.2

0.4

0.6

0.8

1.0

m3,s/m3

Figure E4.2 Variation of combustor exit total temperature and velocity as a

function of percent cooling air flow

The temperature at station 3 may be found from the energy equation for adiabatic flow at station 3, that is, 1 c pT3 + V32 = c pTt ,3 2

( 30.5m / s ) V2 T3 = Tt ,3 − 3 = 689 K − = 689 K − 0.4 K ≈ 689 K 2c p 2 (1.16kJ / kg − K ) 2

4

(E4.5)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Note that the velocity entering the combustor is so low that T3~Tt,3 and therefore the speed of sound is

a3 = γ RT3 = 1.33 ( 287m2 / s 2 − K ) 689 K = 513m / s

(E4.6)

Then the Mach number at station 3 is M3= 30.5/513=0.059. The total temperature ratio across the combustor is then given by (assuming ⎡ γ −1 2 ⎤ 2 1+ 4 M4 ⎥ 1 + 0.167 M 42 ) ( M 42 ⎡1 + γ 3 M 32 ⎤ ⎢ 2 2 = ⎥ ≈ 287 M 4 ⎢ ⎥ ⎢ 2 Tt ,3 M 32 ⎣1 + γ 4 M 42 ⎦ ⎢ γ 3 − 1 2 ⎥ 1 + 1.33M 42 ) ( M3 1+ ⎢⎣ ⎥⎦ 2

Tt ,4

(E4.7)

Since the maximum value of Tt,4/Tt,3 ~ 4 the Mach number at station 4 must be very small, as it is at station 3. Then we may write Equation (E4.7) as 1 + 0.167 M 42 ≈ 287 M ≈ 287 M 42 (1 − 0.44 M 42 + ...) ≈ 287 M 42 2 Tt ,3 1 + 2.66 M 4 + ...

Tt ,4

2 4

(E4.8)

Because T4=Tt,4[1+(γ-1)M42/2]-1 ~ Tt,4 we find the exit velocity V4=M4(γRT4)1/2 ~ M4(γRTt,4)1/2 = (γR/287Tt,3)1/2Tt,4 = 0.0439Tt,4 (in m/s and K).

The stagnation pressure loss is (pt,4-pt,3)/pt,3= (pt,4/pt,3) -1 and the stagnation pressure ratio is given by γ4

⎛ γ 4 − 1 2 ⎞ γ 4 −1 4 M4 ⎟ 2 ⎜1 + 1 + 0.165M 42 ) pt ,4 1 + γ 3 M 3 ⎝ ( 2 ⎠ = ≈ 1.0023 γ3 1 + 1.33M 42 pt ,3 1 + γ 4 M 42 γ − 1 − γ 1 3 ⎛ 2⎞ 3 M3 ⎟ ⎜1 + 2 ⎝ ⎠

(E4.9)

We have shown in Equation (E4.8) that M4 ~ (Tt,4/287Tt,3) = 0.00168(Tt,4)1/2 where temperature is measured on the Rankine scale. The drop in stagnation pressure is shown as a function of the amount of cooling airflow in Figure E4.3.

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Cooling airflow (%)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 -0.8

-0.7

-0.6

-0.5

-0.4

-0.3

-0.2

Drop in stagnation pressure (%)

Figure E4.3 Percent drop in stagnation pressure as a function of percent cooling

airflow

If we hadn’t neglected the weight flow of fuel then our stagnation temperature would be given by (results for both the Kelvin and Rankine scales are shown) ⎛ m ⎞ η HV 1 1 = Tt ,3 + 0.06 ⎜1 − 3, s ⎟ b m f

m c ⎡ ⎛ m 3, s ⎞ ⎤ p 3 ⎠ ⎝ 1+ ⎢1 + 0.06 ⎜1 − ⎟⎥ m 3 m 3 ⎠ ⎦ ⎝ ⎣ ⎛ m ⎞ 0.9 ( 44 MJ / kg ) ⎛ m ⎞ 1 2048K Tt ,4 = 689 K + 0.06 ⎜1 − 3, s ⎟ = 689 K + ⎜1 − 3, s ⎟ m 3 ⎠ 1.16kJ / kg − K ⎡ m 3 ⎠ ⎡ ⎛ m 3, s ⎞ ⎤ ⎛ m 3,s ⎞ ⎤ ⎝ ⎝ ⎢1 + 0.06 ⎜1 − ⎢1 + 0.06 ⎜1 − ⎥ ⎟⎥ ⎟ m 3 ⎠ ⎦ m 3 ⎠ ⎦ ⎝ ⎝ ⎣ ⎣ Tt ,4 = Tt ,3 +

m f ηb HV m 3 c p

⎛ m ⎞ 3684 R Tt ,4 = 1240 R + ⎜1 − 3, s ⎟ m 3 ⎠ ⎡ ⎛ m 3, s ⎞ ⎤ ⎝ ⎢1 + 0.06 ⎜1 − ⎟⎥ m 3 ⎠ ⎦ ⎝ ⎣

Table E4.1 shows that neglecting the mass flow of fuel results in an over-estimate of the stagnation temperature leaving the combustor. However, by including the weight flow of fuel while neglecting cp and γ variations one isn’t necessarily improving the accuracy of the solutions thus obtained.

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Table E4.1 Effects on stagnation temperature of including fuel flow m 3, s m 3

Tt,4 (R)

Tt,4 (R)

% difference

(fuel neglected) (fuel included) 0

4924

4645

6

0.2

4187

3995

4.8

0.4

3450

3330

3.6

0.6

2714

2671

1.6

E4.3 An S-shaped exponential curve may be used to model the spatial distribution of the heat released in the combustion chamber. One such model is as follows: ΔH = H eff exp − B1ς B2

(

ζ =

)

z −1 z1

Here z is the axial location in the combustor and z1 is the axial location where combustion is complete. The quantities B1 and B2 provide the shape that is considered to best characterize the heat release as a function of the axial coordinate. The quantity ΔH represents the heat released up to a given axial station in the combustor while Heff represents the total heat release per unit mass during combustion and is proportional to the heating value of the fuel. Using this information: (a) Describe the significance of the term z1 and how it influences the design of a combustor for a practical aerospace vehicle, (b) Relate the term z1 to the Damkohler similarity parameter, (c) Consider a combustor typical of the Pratt & Whitney TF30 engine used on the Grumman F-14A Tomcat in which the combustor length L=50cm, the combustor height d=20cm, and Heff can be taken as 40MJ/kg. Assuming that z1=50cm and B1=4.605 plot the heat release as a function of axial distance for B2=2, 3, and 4. (d) Define a combustion zone length Lc based on your results and show a plot of Lc as a function of B2 (you may wish to carry out calculations for additional values of B2 to clarify the functional relationship sought. (e) Discuss how your results for Lc are related to the general concept of the chemical production rate of the combustion reactions taking place in the combustor.

Solution

(a) The term z1 represents the axial distance over which the entire chemical energy release is achieved. It influences the design of a combustor for a practical aerospace vehicle by describing the minimum length of the combustor as described in part (b) below.

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) We may see how the term z1 is related to the Damkohler similarity parameter Da=τf /τc , which is the ratio of the typical residence time of a fluid particle to the characteristic time for the chemical reaction to go to completion, by considering a characteristic flow speed V in a combustor of length L. The typical residence time of a fluid particle in the combustor is then τf =L/V. However, based on the answer to part (a) above, the characteristic time for the chemical reaction to go to completion is z1/V. Therefore the Damkohler parameter may be given by Da=(z1/L)-1 so that when z1/L=0 there is equilibrium flow in the combustor and when (z1/L)>>1 there is frozen flow in the combustor. (c) Using the information given we may develop the chart shown in Figure E4.4

40 35 30

B2=4

3

2

25 ΔH (MJ/kg) 20 15 10 5 0 0

10

20

30

40

50

z (cm)

Figure E4.4 Energy release profile

(d) By choosing the start of the combustion zone to be that point where ΔH/H=0.01 and the end to be at the location where ΔH/H=0.99 we may find an expression for the combustion chamber length as follows:

Lc = 1 − ( 2.162 ×10

8

1 −3 B 2

)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The combustion zone length Lc based on the previous results is shown as a function of B2 B

(calculations for additional values of B2 are used to clarify the functional relationship) in B

Figure E4.5. (e) The results for Lc are related to the general concept of the chemical production rate of the combustion reactions taking place in the combustor by the general speed of the reaction. Therefore it is important that the three “T’s” of combustion are appropriately considered: sufficiently high temperature of the reactants, sufficiently high level of turbulence to ensure good mixing, and, most important, sufficient time (length of combustor) for reactions to go to completion.

5 4.5 4 3.5 3 B2 2.5 2 1.5 1 0.5 0 20

25

30

35

40

45

50

Combustion zone length Lc (cm)

Figure E4.5 Variation of the energy release exponent as a function of axial distance

E4.4 A combustor is to accept air from a compressor at a pressure p3 = 1MPa, a speed V3 = 150m/s and a temperature T3 = 500C. It is to operate with a stoichiometric primary zone fuel to air ratio m ( f / a ) primary = f = 0.066 m a , p

and a burner efficiency ηb = 96%. The hydrocarbon fuel has a heating value Qf = 44MJ/kg and secondary air mass flow m a , s can be supplied to the combustor through m appropriate combustor liner design over the range 0.8 ≥ a , s ≥ 0 .Two designs are being m a considered: a constant area combustor design and a constant pressure combustor design.

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Show the variation of (a) the temperature leaving the combustor and entering the turbine T4, (b) the stagnation pressure loss across the combustor, and (c) the velocity leaving the combustor and entering the turbine V4 as functions of the cooling secondary air flow ratio m a , s m a . Compare the performance of the two combustor designs for this range of cooling flow and discuss the results. For the purposes of this problem the thermodynamic properties of the combustion products may be assumed to be constant with average values cp=1.37 kJ/kg-K, γ =1.27 and W=28.96. Furthermore you may assume that the mass flow of fuel is much less than that of the air so that m = m f + m a , p + m a , s m a , p + m a , s

Solution

(a) The energy conservation for one-dimensional adiabatic flow is given by 1 ⎞ 1 ⎞ ⎛ ⎛ m 3 ⎜ h3 + V32 ⎟ + ΔH = m 4 ⎜ h4 + V42 ⎟ 2 ⎠ 2 ⎠ ⎝ ⎝

The quantity ΔH is the heat addition per unit mass due to combustion. If we let h=cpT the equation becomes m 3c p ,3Tt ,3 + ΔH = m 4 c p ,4Tt ,4

We may develop an equation for the stagnation temperature rise in a combustor Tt,4-Tt,3 in terms of the heating value of the fuel by noting that ΔH = m f ηb Q f m 3 = m a = m a , p + m a ,s m 4 = m a + m f

Assuming that cp,3=cp,4= cp,avg = constant we may write the rise in stagnation temperature as follows:

m f Q ⎛ m f ⎞ + − = 1 ηb f T T ⎜ ⎟ t ,4 t ,3 m a c p ,avg ⎝ m a ⎠ If we further assume that m f m a = f / a 1 then the stagnation temperature rise is Tt ,4 − Tt ,3 = ( f a )ηb

The fuel to air ratio may be written as

10

Qf c p ,avg

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ⎛ m f ⎞ ⎛ m a , p ⎞ m a , p f a=⎜ = 0.066 ⎟ ⎜ ⎟ ⎜ m ⎟ m m a ⎝ a, p ⎠ ⎝ a ⎠

The primary air to secondary air ratio may be written as 1=

m a , p m a , s + m a m a

Then the fuel to air ratio is then given by

⎛ m ⎞ f a = 0.066 ⎜1 − a , s ⎟ m a ⎠ ⎝ The stagnation temperature rise may be rewritten as

⎛ m ⎞ Q Tt ,4 − Tt ,3 = 0.066 ⎜1 − a , s ⎟ηb f m a ⎠ c p ⎝ Using the values given in the problem statement this becomes

⎛ m ⎞ Tt ,4 − Tt ,3 = ( 2035K ) ⎜1 − a , s ⎟ m a ⎠ ⎝ We may write this as a stagnation temperature ratio

⎛ m ⎞ = 1 + 2.604 ⎜1 − a , s ⎟ Tt ,3 m a ⎠ ⎝

Tt ,4

For the constant area or constant pressure combustors we have equations for all the ratios of thermodynamic properties across them in terms of γ, M3, and M4. We know γ and M3 so we have Tt,4/Tt,3 = f(M4) and we can then relate all the ratios to m a , s / m a . Therefore we can likewise relate all the other ratios to the secondary airflow ratio. requested are shown in Figure E4.6.

11

The results

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

3000 constant pressure

2500 2000

constant area

T4 (K) 1500 1000 500 0 0

0.2

0.4

0.6

0.8

1

Secondary airflow ratio

(a) The temperature leaving the combustor and entering the turbine T4

1.06 1.04 1.02 1 0.98 pt,4 (MPa) 0.96 0.94 0.92 0.9 0.88 0.86

constant pressure

constant area

0

0.2

0.4

0.6

0.8

1

Secondary airflow ratio

(b) The stagnation pressure loss across the combustor is shown as the actual

stagnation pressure compared to the entrance stagnation pressure pt,3=1.052MPa

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1000 900 800 700 600 V4 (m/s) 500 400 300 200 100 0

constant area

constant pressure

0

0.2

0.4

0.6

0.8

1

Secondary airflow ratio

(c) The velocity leaving the combustor and entering the turbine V4 Figure E4.6 Combustor characteristics as a function of secondary airflow

E4.5 Hydrogen and fluorine make a powerful propellant combination. The space shuttle main engines use an H2-O2 propellant mix operating at a combustion chamber pressure of 200 atmospheres at a temperature of about 3500K. At this chamber pressure determine the combustion products and the adiabatic flame temperature for a mixture of H2 and F2 using 1 mole of each. Indicate the unknown quantities and the equation set being used to solve for them.

Solution

The chemical reaction is assumed to be H 2 + F2 → bHF HF + bH 2 H 2 + bH H + bF2 F2 + bF F

The stoichiometric reaction for these reactants is H 2 + F2 → 2 HF As a consequence, the stoichiometric number of product moles is bstoich=2. The atom conservation relations are as follows: H : 2 = bHF + 2bH 2 + bH F : 2 = bHF + 2bF2 + bF

(E4.10)

The equilibrium subreactions and the associated equilibrium constants in terms of the partial pressure species are given by

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 1 1 H 2 + F2 → HF : K p , HF = 2 2

pHF bHF = pH 2 pF2 bH 2 bF2

p 1 H 2 → H : K p,H = H = 2 pH 2

p bH b bH 2

1 p F2 → F : K p , F = F = 2 pF2

p bF b bF2

The equilibrium constants may be taken from the tables in Appendix G and are listed in Table E4.2 for four temperatures of interest.

Table E4.2 Equilibrium constants for HF, H, and F

T(K) Kp,HF

Kp,H

Kp,F

3500

17,057

0.59143

131.95

4100

4,162.0

1.8915

205.21

4500

2,005.4

3.4626

259.17

5000

950.82

6.4402

330.98

The total number of moles of product may be written as b = bHF + bH 2 + bH + bF2 + bF ≈ bHF + bH 2 + bH + bF

(E4.11)

The number of moles of F2 in the products, bF2, was neglected in the total number of moles because the magnitude of the equilibrium constant Kp,F suggests that bF2<<bF. Equations (E4.10), (E4.11), and the three equilibrium constants constitute six equations in six unknowns (the individual product moles bi and the total number of product moles) and one parameter (p) so that one may solve the set iteratively. Manipulating these equations leads to a cubic equation for x = bH 2 as follows: x 3 + c1 x 2 + c2 x + c3 = 0

The coefficients are given by c1 =

1 2

K p,F ⎞ b⎛ ⎜⎜ K p , H + 2 ⎟ p⎝ K p , HF ⎟⎠

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

c2 =

1 b ⎛ K p,H K p,F ⎞ ⎜ ⎟ 2 p ⎜⎝ K p , HF ⎟⎠ b ⎛ K p,F ⎞ ⎜ ⎟ p ⎜⎝ K p , HF ⎟⎠

c3 = −

Then for specified values of p and T one may solve for bH2. One may find the remaining product moles by solving the following set of equations sequentially:

(

(

)

b K p , H bH 2 p

bHF = 2 1 − bH 2 −

(

)

)

bH = 2 1 − bH 2 − bHF

bF = bHF − 2 Using the desired chamber pressure p and an assumed value for the total product moles b one may compute all the individual product moles bi and total number of moles b. The computed value of b is then compared to the assumed value. When the two values are sufficiently close the solution has been reached to within a desired degree of accuracy. In the case under consideration, where p=200 atmospheres, the equilibrium composition at various chamber temperatures are shown in Table E4.3.

Table E4.3 Equilibrium molar composition at p=200atm and various temperatures

Species bH2 bHF bH bF b

3500K 0.0070 1.9810 0.00496 0.01896 2.01192

4100K 0.0203 1.9323 0.0272 0.0677 2.0475

4500K 0.0341 1.8662 0.0655 0.1338 2.0996

5000K 0.0553 1.7302 0.1592 0.2698 2.2145

The difference between the enthalpy of the products, evaluated at an assumed value of the adiabatic flame temperature Tc’, and that of the reactants, which are injected at the temperature Tj, is defined as

( )

( ){ ( )

}

{

}

Qc Tc′ = ∑ bi Tc′ ⎡ H i Tc′ − H i0 ⎤ − ⎡⎣ H i (Tr ) − H i0 ⎤⎦ − ∑ ai ⎣⎡ H i (T j ) − H i0 ⎦⎤ − ⎡⎣ H i (Tr ) − H i0 ⎤⎦ ⎣⎢ ⎦⎥ i =1 i =1 N

N

Assuming that the fuel and oxidizer are injected at the standard temperature of 298.16K and using the tables presented in Appendix G this equation simplifies to

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

( )

4

( ) ( )

Qc Tc′ = ∑ bi Tc′ ⎡ H i Tc′ − H i ( 298.16 ) ⎤ ⎢⎣ ⎥⎦ i =1

The difference in the available heat of formation of the product and reactant species is

( )

( )

Qa Tc′ = ∑ bi Tc′ ΔH f ,i 0 − ∑ ai (T j ) ΔH f ,i 0 N

i =1

N

i =1

Because the heats of formation of both the fuel and the oxidizer are zero, the available heat equation simplifies to

( )

4

( )

Qa Tc′ = ∑ bi Tc′ ΔH 0f ,i i =1

Carrying out these calculations for the species listed in Table E4.3 leads to the results shown in Table E4.4 Table E4.4 Enthalpy difference and available heat of combustion

Tc' (K)

Qc (kJ)

Qa (kJ)

3500 4100 4500 5000

213 259 292 335

553 529 495 420

For the adiabatic flame temperature Tc we must have Qc=Qa and the results in Table E4.4 are plotted in Figure E4.7. Because the tables are limited to T=5000K an extrapolated solution is indicated on the figure as Tc=5300K. Product concentrations in Table E4.3 may also be extrapolated if the composition at Tc is desired.

600 500

Qa

Tc=5300K

400 Q (kJ) 300

Qc

200 p=200atm

100 0 3000

3500

4000

4500

5000

5500

6000

T (K)

Figure E4.7 Graphical determination of the adiabatic flame temperature

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.6 Turbopumps are used on the Space Shuttle Main Engines (SSME) to inject the hydrogen and oxygen propellants into the main combustion chamber. The fuel and oxidizer pumps are powered by turbines driven by combustion of some of the hydrogen and oxygen in a pre-burner. Find the adiabatic flame temperature in the pre-burner assuming that the molecular hydrogen and oxygen are injected at a temperature Tj = 298.16K and a pressure pc = 20atm and that they are injected in stoichiometric proportions, that is, two moles of H2 to every mole of O2.

Solution

The stoichiometric reaction of hydrogen and oxygen involves two moles of hydrogen for every mole of oxygen and yields only water: 1 H 2 + O2 → H 2O 2

But, even though the stoichiometric proportions of two moles of hydrogen for every mole of oxygen will balance the chemical reaction as shown, in general, there will be additional species produced by the reaction. We may suppose that, in general, the chemical reaction proceeds as follows: nH 2 + O2 → b1 H 2 + b2O2 + b3 H 2O + b4OH + b5O + b6 H The fuel to oxidizer ratio is O F=

m o noWo (1mole )( 32 g mol ) 16 = = = m f n f W f ( nmol )( 2 g mol ) n

The stoichiometric reaction is the case where n=2 and the equivalence ratio is therefore

φ=

( F O ) = n 16 = n ( F O )s 2 16 2

Recall that for φ <1 the mixture is fuel-lean, while for φ >1 the mixture is fuel-rich. We will continue the problem with the general value of n moles and then specialize the results to the desired case of n=2.

To determine the adiabatic flame temperature, we must first determine the equilibrium concentrations of the products for a given temperature. To do this we may proceed by as follows: (i). Atom conservation (1)H: 2n=2b1+2b3+b4+b6

(E4.12)

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (2)O: 2=2b2+b3+b4+b5

(E4.13)

(ii) Total moles (3) b=b1+b2+b3+b4+b5+b6

(E4.14)

(iii) Equilibrium sub-reactions and their associated equilibrium constants 1 H2 → H 2

K p,H = P

b6 b2

(E4.15)

1 (5) O2 → O 2

K p ,O = P

b5 b2

(E4.16)

1 1 (6) H 2 + O2 → OH 2 2

K p ,OH =

b4 b1b2

(E4.17)

1 (7) H 2 + O2 → H 2O 2

K p , H 2O =

b3 Pb1 b2

(E4.18)

(4)

Note that the parameter P =

p b

Solution scheme: (i) Pick a value for the total number of moles b, say, b’=b1+b2+b3+b4+b5+b6. A reasonable starting guess is the total number of reactant moles for the stoichiometric reaction, here, for example, we may guess b’ = 2. (ii) Pick a value for the adiabatic flame temperature Tc=Tc’ and input the associated Kp,i, (iii) Pick a value of b1 and calculate b6 from Equation (E4.15);

b6 =

K p , H b1 P

(iv) Dividing Equation (E4.17) by Equation (E4.18), substituting for b3 using Equation (E4.12), and solving for b4 yields b4 =

2n − 2b1 − b6

(

1 + 2 P b1 K p , H 2O K p ,OH

=

2n − 2b1 − b1 ( K p , H P )

) 1 + 2P b ( K

(v) We may now use Equation (E4.12) to calculate b3=0.5(2n-2b1-b4-b6) (vi) From Equation (E4.17) we calculate

18

1

p , H 2O

K p ,OH

)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 1⎛ b ⎞ b2 = ⎜ 4 ⎟ b4 ⎜⎝ K p ,OH ⎟⎠

2

(vii) From Equation (E4.16) we find

b5 =

K p ,O b2 P

(viii) Use Equation (E4.13) to assess the error in oxygen atoms: E=2b2+b3+b4+b5-2 (ix) Then check the error in total number of moles from Equation (E4.14) E1=b1+b2+b3+b4+b5+b6-b’ When both errors are as small as desired, the equilibrium concentration is defined.

The difference between the enthalpy of the products, evaluated at an assumed value of the adiabatic flame temperature Tc’, and that of the reactants, which are injected at the temperature Tj, is defined as

( )

( ){ ( )

}

{

}

Qc Tc′ = ∑ bi Tc′ ⎡ H i Tc′ − H i0 ⎤ − ⎡⎣ H i (Tr ) − H i0 ⎤⎦ − ∑ ai ⎡⎣ H i (T j ) − H i0 ⎤⎦ − ⎡⎣ H i (Tr ) − H i0 ⎤⎦ ⎣⎢ ⎦⎥ i =1 i =1 N

N

The difference in the available heat of formation of the product and reactant species is

( )

( )

Qa Tc′ = ∑ bi Tc′ ΔH f ,i 0 − ∑ ai (T j ) ΔH f ,i 0 N

1

N

1

For the adiabatic flame temperature Tc we must have Qc=Qav, therefore for each value of Tc’ for which the converged equilibrium composition has been calculated we must evaluate the error E2=Qc -Qav and choose a value of Tc’ which makes this error as small as desired. Interpolation results are shown in Figure E4.8. The selected solution is Tc=3500K for which the product concentrations are as shown in Table E4.5.

Table E4.5 Molar concentrations of the products at Tc=3500K Species H2 O2 H2O OH O H Total

b1 b2 b3 b4 b5 b6 b

Number of moles 0.3157 0.0938 1.4952 0.2648 0.0515 0.1134 2.3345

19

Mole fraction 0.135 0.040 0.640 0.113 0.022 0.049 1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

300 200

Tc=3500K

100 Qc'-Qa' (kJ)

0 -100 -200 -300 3000 3100 3200 3300 3400 3500 3600 3700 3800 3900 4000 Tc'(K)

Figure E4.8 Interpolation to find the adiabatic flame temperature

E4.7 For the propellant system comprised of CnH2n as fuel and O2 as oxidizer:

a1Cn H 2 n + a2O2 → products Determine (a) the stoichiometric fuel to oxidizer ratio (wf/wo)s and (b) the equilibrium composition of the products of reaction for the following conditions: p=20atm, T=3000K, equivalence ratio φ=1.2, and total mass of reactants equal to 1kg.

Solution

Assume that the reaction may be represented by. a1CnH2n+a2O2=b1CO2+b2CO+b3H2O+b4OH+b5H2+b6H+b7O2+b8O The chamber conditions are p=20atm, T=3000K, φ=1.2 and the total mass of reactants is1000g. This is related to the number of moles of reactants as follows: a1(12n+2n)+a2(32)=14na1+32a2=1000g (a) The stoichiometric reaction is given by v'1CnH2n+v'2O2=v"3CO2+v"4H2O For the stoichiometric reaction atom conservation yields C: nv'1=v”3, H: 2nv'1=2v"4 O: 2v'2=2v"3+v"4=2nv'1+nv'1=3nv'1

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The stoichiometric fuel to oxidizer ratio is found from the stoichiometric reaction, noting that ν2’=(3n/2)ν1’, as ⎛ m f ⎞ ν 1′ (12n + 2 ) 14nν 1′ 7 = = = 0.2917 ⎜ ⎟ = ν 2′ ( 32 ) 48nν 1′ 24 ⎝ m o ⎠ s The fuel to oxidizer ratio for the actual mixture of 1kg of reactants may be found by using the given value of the equivalence ratio, φ=1.2=6/5, to obtain m f ⎛ m ⎞ 6 7 7 =φ⎜ f ⎟ = = m o ⎝ m o ⎠ s 5 24 20 Then from the actual reactants we have m f n f W f a1 (14n ) 7 = = = m o noWo a2 ( 32 ) 20

Therefore na1=(7/20)(32/14)a2=(4/5)a2 and from the given mass of reactants 1000g= 14(4a2/5)+32a2=(216/5)a2 we find a2=23.15 a1=18.52/n Then the chemical reaction under consideration becomes (18.52/n)CnH2n+23.15O2=b1CO2+b2CO+b3H2O+b4OH+b5H2+b6H+b7O2+b8O The equations arising from the equilibrium of the sub-reactions for H, O, OH, and H2O are:

K p,H =

pH = pH 2

p b6 b b5

K p ,O =

pO = pO2

p b8 b b7

K p ,OH =

pOH b = 4 pH 2 pO2 b5b7

K p , H 2O =

pH 2 O pH 2 pO2

=

b b3 p b5 b7

Manipulating these equations and introducing P=p/b we find

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 2

⎛ b ⎞ b5 = P ⎜ 6 ⎟ ⎜K ⎟ ⎝ p,H ⎠

(E4.19) 2

⎛ b ⎞ b7 = P ⎜ 8 ⎟ ⎜K ⎟ ⎝ p ,O ⎠ ⎛ K p ,OH P ⎞ b4 = ⎜ bb ⎜ K K ⎟⎟ 6 8 ⎝ p , H p ,O ⎠ ⎛ K p , H 2O P 2 ⎞ 2 b3 = ⎜ 2 bb ⎜ K p , H K p ,O ⎟⎟ 6 8 ⎝ ⎠

(E4.20) (E4.21)

Atom conservation leads to C: na1=18.52=b1+b2 H: 2na1=37.04=2b3+b4+2b5+b6 O: 2a2=46.3=2b1+b2+b3+b4+2b7+b8 (b) Equilibrium composition: Substituting Equations (E4.19), (E4.20), and (E4.21) for b3, b4, and b5 in the H conservation equation leads to: 2 ( 7496 P 2b8 + 37.18 P ) b62 + ( 65.39 Pb8 + 1) b6 − 37.04 = 0

This is a quadratic equation for b6 with coefficients involving P and b8 and may be solved for particular values of those parameters. Then the oxygen concentration is b8=46.3-[2b1+b2+b3+b4+2b7] Values for b1 and b2 may be found by considering the CO2 equilibrium sub-reaction: 1 CO + O2 → CO2 2 The equilibrium constant for this reaction may be written in terms of the equilibrium

constants for CO2 and CO as K p ,1 =

pCO2 pCO pO2

=

K p ,CO2 K p ,CO

=

b1 b2 Pb7

Here the equilibrium constants for CO2 and CO refer to the equilibrium of the gaseous species when CO2 and CO are formed from oxygen and solid carbon and are tabulated in Appendix G. Substituting for b7 from the oxygen equilibrium sub-reaction and for b1 from the carbon atom conservation equation yields

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual K p ,CO2 K p ,CO

b1 K p ,O 18.52 − b2 K p ,O = b2 b8 Pb2 Pb8

=

Then the CO molar concentration is b2 =

18.52 ⎛ K p ,CO2 P ⎞ 1 + b8 ⎜ ⎟⎟ ⎜K K , , p CO p O ⎝ ⎠

Iteration of these equations is done on a spreadsheet and the final result is shown on the bar chart in Figure E4.9 and in Table E4.6 below.

18 16

T=3000K p=20atm b=37.88 moles

14 12 moles

10 8 6 4 2 0 CO2

CO

H2O

OH

H2

H

O2

Figure E4.9 Bar chart of product molar concentrations

Table E4.6 Molar concentrations of the products Species CO2 CO H2O OH H2 H O2 O total

Moles 10.31 8.212 16.00 0.8649 1.853 0.3072 0.2500 0.08154 37.88

23

O

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.8 Consider the gaseous propellant system described by the reaction

4H 2 + O2 → products Determine (a) the equivalence ratio φ and (b) the adiabatic flame temperature Tad. Note that this reaction is fuel rich and it should be possible to reduce the number of product species accordingly so as to simplify the calculation, if desired. Use p=20atm for the chamber pressure and assume the reactants are introduced as gases at a temperature of 298.16K. Solution

(a) The equivalence ratio is defined as

φ=

( f / a) ( f / a )stoich

Since this is a ratio we may use the molar concentrations and find φ =2 (b) To determine the adiabatic flame temperature, first consider the reaction to proceed as follows: 4H2+O2=b1H2+b2O2+b3H2O+b4OH+b5O+b6H 1. Equilibrium concentrations (i). Atom conservation H: 8=2b1+2b3+b4+b6

(E4.22)

O: 2=2b2+b3+b4+b5

(E4.23)

(ii) Total moles b=b1+b2+b3+b4+b5+b6

(E4.24)

(iii) Equilibrium sub-reactions Pb 1 H 2 → H : K p,H = 6 2 b1

(E4.25)

Pb 1 O2 → O : K p ,O = 5 2 b2

(E4.26)

b 1 1 H 2 + O2 → OH : K p ,OH = 4 2 2 b1b2

(E4.27)

b3 1 H 2 + O2 → H 2O : K p , H 2O = 2 Pb1 b2

(E4.28)

Note that here we have introduced a parameter P =

24

p b

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 2. Solution scheme: (i) Pick bguess=b, starting with, say, total number of reactant moles, b=5 (ii) Pick a value for Tc and input Kp,i =1-4 (iii) Pick a value of b1 and calculate b6 from Equation (E4.25) (iv) From Equation (E4.22) b3=(8-2b1-b4-b6)/2. Then, dividing Equation (E4.27) by Equation (E4.28) and substituting the result for b3 yields the equation for b4 b4 =

8 − 2b1 − b6

(

1 + 2 Pb1 K p , H 2O K p ,OH

=

8 − 2b1 − b1 ( K p , H P )

) 1 + 2Pb ( K 1

p , H 2O

K p ,OH

)

(v) From Equation (E4.22) we may now calculate b3= (8-2b1-b4-b6)/2 (vi) From Equation (E4.27) we find 1⎛ b ⎞ b2 = ⎜ 4 ⎟ b1 ⎜⎝ K p ,OH ⎟⎠

2

(vii) From Equation (4E.26) we find

b5 =

K p ,O b2 P

(viii) Check result for assumed value of b1 by using Equation (E4.23), which we haven't used yet, to find the error in hydrogen E=2b2+b3+b4+b5-2 (ix) Check error in total number of moles from Equation (E4.24) E1=b1+b2+b3+b4+b5+b6-bguess 3. Carry out calculations for various values of the adiabatic flame temperature and select the value which makes the error in total number of moles as small as desired.

Interpolation results are shown in Figures E4.10 and E4.11 below. The selected solution is Tc= 3054K for which the (interpolated) product concentrations are given in Table E4.7.

25

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Table E4.7 Product molar concentrations and mole fractions Species b1=H2 b2=O2 b3=H2O b4=OH b5=O b6=H total

100 80 60 40 Qc'-Qa' 20 (kJ) 0 -20 -40 -60 2900

bi (moles) 1.9654 0.00060924 1.9516 0.045398 0.0016443 0.12066 4.08529

Xi 0.48258 0.00015 0.47920 0.01115 0.00041 0.02963 1.0031

Tc=3054K

2950

3000

3050

3100

3150

3200

3250

3300

T (K)

Figure E4.10 Linear interpolation result for Tc

4.2 4.18 4.16 4.14 b 4.12 (moles) 4.1 4.08 4.06 4.04 2900

b=4.085

2950

3000

3050

3100

3150

3200

3250

T (K)

Figure E4.11 Interpolation for correct number of moles of product

26

3300

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.9 Consider the following reaction between solid carbon and gaseous oxygen in which all elemental carbon is consumed:

C ( c ) + O2 ( g ) → b1O2 ( g ) + b2CO ( g ) + b3CO2 ( g ) + b4O ( g ) Determine the adiabatic flame temperature for this reaction assuming the reactants are introduced at the reference temperature, 298.16K and a pressure of 20 atmospheres.

Solution

The atom conservation equations are as follows: C :1 = bCO + bCO2 O : 2 = 2bO2 + bCO + 2bCO2 + bO

(E4.29)

The total number of moles of product gas is given by b = b1 + b2 + b3 + b4

(E4.30)

We are told that there is no elemental carbon present in the products at equilibrium.

Therefore the equilibrium sub-reactions must involve only the four indicated product species. One acceptable sub-reactions is 1 CO + O2 ( g ) R CO2 2

The equilibrium constants for this reaction is

Kp =

pCO2 pCO pO2

=

pCO2

pO2

pO2

pCO

=

K p ,CO2 K p ,CO

Here the equilibrium constants for CO2 and CO refer to the equilibrium of the gaseous species when CO2 and CO are formed from oxygen and solid carbon and are tabulated in Appendix G. The other equilibrium sub-reaction is 1 O2 → O 2 We may combine the equilibrium constants for CO and CO2 to obtain

K p ,CO K p ,CO2

=

p bCO bO2 b bCO2

The equilibrium constant for oxygen is

27

(4E.31)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

K p ,o =

p bO b bO2

(4E.32)

We have 5 unknowns (the four bi and b) and upon assuming a temperature Tc’ and collecting the corresponding values for the Kp,i we may proceed to solve Equations (E4.29) through (E4.32) for the equilibrium composition at each value of the Tc’. The results are shown in Table E4.8. The energy balance between the available heat of formation Qc’ and the enthalpy of the products Qa’ for the above temperatures is shown in Figure E4.12 and the matching point defines the adiabatic flame temperature which is Tc=3756K Table E4.8 Product moles at various temperatures T (K) 1700 1900 2100 2300 2500 2700 2900 3100 3300 3500 3700 3900 4100

bo2

bo

bco

bco2

b

0.0004 0.0016 0.0047 0.0117 0.0249 0.0460 0.0770 0.1170 0.1615 0.2047 0.2390 0.2580 0.2580

1.961E-07 2.627E-06 2.135E-05 1.208E-04 5.169E-04 1.772E-03 5.123E-03 1.282E-02 2.828E-02 5.597E-02 1.005E-01 1.653E-01 2.506E-01

0.0008 0.0031 0.0094 0.0235 0.0502 0.0938 0.1591 0.2468 0.3513 0.4654 0.5785 0.6813 0.7666

0.9992 0.9969 0.9906 0.9765 0.9498 0.9062 0.8409 0.7532 0.6487 0.5346 0.4215 0.3187 0.2334

1.0004 1.0016 1.0047 1.0118 1.0254 1.0478 1.0821 1.1298 1.1898 1.2607 1.3395 1.4233 1.5086

400 Q a'

300

Tc=3756K

Q' 200 (kJ) Qc '

100

0 1700 1900 2100 2300 2500 2700 2900 3100 3300 3500 3700 3900 4100 T (K)

Figure E4.12 Interpolation for adiabatic flame temperature

28

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.10 Consider the reaction of 1 mole of fluorine (F2, gas) reacting with varying numbers of moles of hydrogen (H2, gas) injected at a temperature Tj=298.16K. Plot a graph of the variation of the adiabatic flame temperature Tc with the equivalence ratio for pc=6.8MPa. Solution

Consider the reaction to be defined as follows: nH 2 + F2 → b1 HF + b2 H 2 + b3 H + b4 F2 + b5 F

The fuel and oxidizer are assumed to be injected into the combustion chamber as gases at a temperature Tinj= 298.16K; furthermore all species formed are assumed to be gases. The atom conservation equations are H : 2n = b1 + 2b2 + b3 F : 2 = b1 + 2b4 + b5 The total number of moles of product formed is b = b1 + b2 + b3 + b4 + b5 There are three equations for 6 unknowns with the quantity x as a parameter. Therefore we need three more equations among the unknowns and these are the equations representing equilibrium of the sub-reactions: 1 1 H 2 + F2 → HF 2 2 1 H2 → H 2 1 F2 → F 2 From these equations we may form the equilibrium constants as follows: K p , HF =

b1 b2b4

K p,H =

b3 b2

p b

K p,F =

b5

p b

b4

29

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Using tabulated values of the Kp,i in Appendix G for various temperatures and choosing values for the number of moles of hydrogen injected we may solve for the corresponding equilibrium compositions.

The stoichiometric fuel to oxidizer ratio is

( F O )s =

m f n f W f (1 2 )( 2 ) = 0.1053 = = m o noWo (1 2 )(19 )

The equivalence ratio is therefore

φ=

( F O ) = ( 2n 19 ) = n ( F O )s ( 2 19 )

Note that the weight fraction f of hydrogen in the reactants is given by

f =

nWH 2 nWH 2 + WF2

=

n ⎛ n 2n ⎞ ≈ + ... ⎟ ⎜1 − 2n + 19 9.5 ⎝ 9.5 ⎠

Since n=O(1) we see that the weight fraction of hydrogen is approximately 10% of the equivalence ratio. The equivalence ratio here is φ= n and we choose three values: 0.5, 1.0, and 1.5, representing fuel-lean, stoichiometric, and fuel-rich mixtures. The equilibrium compositions for different temperatures are shown in the Table E4.9 below: The enthalpy of combustion at a specific combustion temperature Tc may be written as Qc = ∑ bi ,Tc ⎡( H i − H i0 ) − ( H i − H i0 ) ⎤ Tc Tinj ⎦ ⎥ ⎣⎢ i =1 N

The available heat of reaction is 0

Qavail = −∑ bi ,Tc ΔH 0f ,i i =1

When these two enthalpies are equal the assumed value of Tc is the actual adiabatic flame temperature. The resulting adiabatic flames temperatures are shown as a function of equivalence ratio in Figure E4.13 below.

30

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Table E4.9 Equilibrium composition for various temperatures and equivalence ratios T (K) 3500 3500 3500 3500 3900 3900 3900 3900 4300 4300 4300 4300 4700 4700 4700 4700 5000 5000 5000 5000

n 0.5 1.0 1.5 1.0 0.5 1.0 1.5 2.0 0.5 1.0 1.5 2.0 0.5 1.0 1.5 2.0 0.5 1.0 1.5 2.0

H2 0.0000 0.0090 0.4630 0.0090 0.0000 0.0188 0.4231 0.8716 0.0002 0.0326 0.3686 0.7690 0.0011 0.0489 0.3101 0.6424 0.0029 0.0607 0.2707 0.5446

HF 1.0000 1.9722 1.9956 1.9722 0.9989 1.9303 1.9830 1.9870 0.9936 1.8514 1.9476 1.9597 0.9747 1.7205 1.8647 1.8942 0.9391 1.5812 1.7530 1.8017

F 1.0000 0.0278 0.0044 0.0278 1.0011 0.0697 0.0170 0.0130 1.0064 0.1486 0.0524 0.0403 1.0253 0.2795 0.1353 0.1058 1.0609 0.4188 0.2470 0.1983

H 0.0000 0.0098 0.0783 0.0098 0.0011 0.0320 0.1708 0.2698 0.0063 0.0832 0.3151 0.5024 0.0255 0.1817 0.5151 0.8209 0.0610 0.2974 0.7057 1.1090

b 2.0500 2.0188 2.5413 2.0188 2.0011 2.0508 2.5939 3.1414 2.0064 2.1158 2.6837 3.2714 2.0266 2.2306 2.8252 3.4633 2.0640 2.3581 2.9763 3.6536

Note that adding fuel (H2) over and above that for a stoichiometric mixture will rapidly cause a drop-off in Tc after a value of φ =1.1 1500 1400 1300 1200 1100 Tc (K) 1000 900 800 700 600 0.4

0.6

0.8

1

1.2

1.4

1.6

Equivalence ratio φ

Figure E4.13 Adiabatic flame temperature as a function of equivalence ratio. The

weight fraction of hydrogen in the propellant is f=φ /(φ+9.5)

31

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E4.11 A turbojet engine combustor is fed with air from the compressor at a Mach number M3=0.3 and a stagnation temperature Tt,3=315C and discharges the combustion products to the turbine at a stagnation temperature Tt,4=838C. (a) For a burner efficiency ηb=90% and fuel heating value Qf=43.3MJ/kg find the overall fuel-air ratio f /a, assuming γ=1.33 and R=287J/kg. (b) If the combustor operates at constant pressure, determine the exit Mach number M4 Solution

(a) The one-dimensional energy equation across the combustor is m 0 c p 3Tt 3 + m f ηb Q f = ( m 0 + m f ) c p 4Tt 4

Assuming cp=γR/(γ−1) is constant we find the fuel-air ratio m f m a = 0.01561 (b) For operation at constant pressure the stagnation temperature ratio is given by

γ −1 2 Tt 4 M 32 1 + 2 M 4 = Tt 3 M 42 1 + γ − 1 M 2 3 2 Using the value Tt4/Tt3=1.889 and the assumption that γ is constant leads to M4=0.217

E4.12 Consider the high-performance propellant which uses hydrogen as the fuel and bromine as the oxidizer. The reaction may be taken in the following form:

H 2 + Br2 → b1 H 2 + b2 H + b3 HBr + b4 Br2 + b5 Br Determine the equilibrium composition at a temperature of 1500K and a pressure of 20atm. Some pertinent equilibrium constants at 1500K are given below:

K p,H =

p2 = 1.76 × 10−5 p1

K p , Br =

p5 = 0.298 p4

K p , HBr =

p3 p1 p4

= 163.0

Solution

Conservation of atoms may be expressed as follows

32

Theory of Aerospace Propulsion, Second Edition: Solutions Manual H: 2=2b1+b2+b3 Br: 2=b3+2b4+b5 There are six unknowns (the five bi and the total number of product moles, b) therefore we need three more equations. Equilibrium among the sub-reactions is expressed by 1 1 H 2 + Br2 → HBr 2 2 1 H2 → H 2 1 Br2 → Br 2

At the given temperature T=1500K the corresponding equilibrium constants are as follows: b3 = 163 b1b4

K p , HBr = K p,H =

p b2 = 1.76 × 10−5 b b1

K p , Br =

p b5 = 0.298 b b4

The equilibrium constant Kp,H<<1 so that it is reasonable to assume b2<<b1 and therefore form the atom conservation Equation for hydrogen we may write 2 = 2b1 + b2 + b3 ≈ 2b1 + b3 Eliminating b3 between the H and Br atom conservation equations yields 2b1 + b2 − b4 − b5 = 0 ≈ 2b1 − b4 − b5 We now have the following equations:

b1 = b4 + b5 = b4 + K p , Br K p , HBr =

b3 b1b4

=

2 (1 − b1 ) b1b4

=

b4 P

2 − b4 − K p , Br b4 P

( b4 2 ) ⎡⎣b4 + K p, Br b4 P ⎤⎦

(4E.33)

(4E.34)

We may now choose a value of P=p/b, insert Kp,HBr=163, and solve Equation (E4.34) for b4. Then we may find b1 from Equation (E4.33) and b2 and b5 from the equilibrium

33

Theory of Aerospace Propulsion, Second Edition: Solutions Manual constants for Kp,H and Kp,Br, respectively. Finally, we may solve for b3 from the hydrogen atom conservation equation. The total number of moles is used to calculate a new value of P which is compared to the initial guess. When the error is as small as desired the system is solved. The results for T=1500K and p=20atm is given in Table E4.10.

Table E4.10 Equilibrium composition for T=1500K and p=20atm Species H2 H HBr Br2 Br Total

moles 0.01164 2.06E-08 1.9767 0.01265 0.01063 2.01162

b1 b2 b3 b4 b5 b

E4.13 Compute the adiabatic flame temperature arising from the combustion of octane (C8H18) with air at a fuel to air ratio of 0.02 and where the initial temperature of the reactants is 811K. The lower heating value of octane is Qf=44.79MJ/kg at 25oC. Solution

The fuel to air ratio of 0.02 is much less than that for stoichiometric reaction, 0.0664, as was shown in chapter 4 in the text. The equivalence ratio

φ=

( f / a ) = 0.301 ( f / a )stoich

As pointed out in the text, this is a very lean mixture which would have very erratic burning, if it could be ignited at all. The fuel to air ratio is typical of the overall value, including both primary and secondary air, and we can determine what the adiabatic flame temperature should be, even though the actual reaction would be in two steps: burning in the primary zone and dilution in the secondary zone. In the current case, with so much extra air, we would expect a relatively low temperature and a good deal of extra oxygen in the products. So if we make that assumption we have the following reaction: 79 ⎞ ⎛ C8 H18 + α ⎜ O2 + N 2 ⎟ → b1CO2 + b2 H 2O + b3 N 2 + b4O2 21 ⎠ ⎝

34

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Here we see the usual stoichiometric products plus the additional oxygen. Assuming that the likely temperature is low enough to avoid dissociating the oxygen molecules this formula should hold. We must first find the number of moles of air that would give the correct fuel-air ratio of 0.02: f /a=

Wf C f Wa Ca

=

(1)(114 ) 79 ⎞ ⎛ α ⎜ 32 + 28 ⎟ 21 ⎠ ⎝

= 0.02

From this we see that α=41.5 moles. Then atom conservation yields C: 8=b1 H: 18=2b2 O: 2α=2b1+b2+2b4 N: 2(3.76)α=2b3 Now we find b1=8, b2=9, b3=156, b4=29, and the total number of moles b=202. The mole fractions are X1=0.0396, X2=0.0446, X3=0.772, and X4=0.144. At temperatures of about 2000K or less these results are accurate and insensitive to pressure (even to 100atm).

Carrying more species leads to difficulties in numerical error control and are unnecessary. Results may be checked with NASA’s on-line CEA (Chemical Equilibrium with Applications) service which does provide results for many more species. This composition is essentially constant in the range 1000K<T<2000K within which the expected answer lies. For the adiabatic flame temperature we must find the temperature Tc which satisfies the equation:

∑ ai ⎡⎣ΔH 0f ,i + ΔH i (T j ) − ΔH i (Tr )⎤⎦ − ∑ bi ⎡⎣ΔH 0f ,i + ΔH i (Tc ) − ΔH i (Tr )⎤⎦ = 0 3

4

i =1

i =1

Here the injection temperature Tj=811K. The first term is fixed and using the thermodynamic properties (see, for example, the THERMOBUILD routine in the on-line NASA CEA program for the properties of C8H18; the properties of the other species are available in Appendix G) for the species we have

(1molC8 H18 ) [ −224kJ / mol + 155kJ / mol − 0] + ( 41.5molAir ) [ −0.126kJ / mol + 15kJ / mol − 0] = 470kJ The second term is

35

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

(8molCO2 ) ⎡⎣ −393.7kJ / mol + ΔH (Tc ) − 9.37kJ / mol ⎤⎦ + ( 9molH 2O ) ⎡⎣ −241.9kJ / mol + ΔH (Tc ) − 9.91kJ / mol ⎤⎦ + (156molN 2 ) ⎡⎣ 0 + ΔH (Tc ) − 8.67 kJ / mol ⎤⎦ + ( 29molO2 ) ⎡⎣0 + ΔH (Tc ) − 8.67 kJ / mol ⎤⎦ Combining the terms leads to the full equation which may be written as

470kJ − 8ΔH CO2 (Tc ) − 9ΔH H 2O (Tc ) − 156ΔH N2 (Tc ) − 29ΔH O2 (Tc ) + 7095kJ = ΔQ Now all that remains is to enter the appropriate enthalpy values for an assumed Tc and iterate until the temperature is found which makes ΔQ=0, which is Tc=1183K, approximately, as shown in Figure E4.14.

Tc=1183K

1000 0

ΔQ (kJ)

-1000 -2000 -3000 1000

1100

1200

1300

1400

1500

T (K)

Figure E4.14 Iteration for the adiabatic flame temperature

36

1600

1700

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E5. NOZZLES FOR AIRBREATHING ENGINES

E5.1 A simple two-dimensional nozzle is shown in Figure E5.1. The area of the minimum section is Amin=0.1m2 and the area of the exit station is Ae=0.5m2 and the area varies linearly with axial distance z between the two locations. During a test, measurements indicate that the total pressure at the nozzle inlet pti=172.3kPa while the ambient pressure outside the nozzle pa=101.3kPa. The test engineer claims that a normal shock is standing midway between the minimum and exit stations, that is, at z/L=0.5. Assuming γ=1.4: (a) Determine if a normal shock is actually located at the station claimed. (b) If the shock is not located at that station is it upstream or downstream of the location stated by the test engineer. (c) Determine the exact location of the shock, if one is present at all. pa

r

Mi<1

0

pti

i

L

min

z

e

Figure E5.1 Nozzle for turbojet engine

Solution (a) If there is a shock in the nozzle the flow in the downstream portion of the nozzle will be subsonic and pe=pa. Halfway down the nozzle the area is 0.3m2 so that Ae/Amin=3.

Entering the supersonic portion of the compressible flow tables with this area ratio would show that the Mach number M0.5,up=2.64 so that if there were a shock there the Mach number just after the shock would be M0.5,down=0.50 and pt,0.5,down/pt0.5,up = pt,0.5,down/pt,I = 0.445. The area ratio behind the shock would be A0.5,down/A*=1.34. Then at z=L AL/A* = (AL/A0.5,down)(A0.5,down/A*) =(0.5/0.3)(1.34)=2.23. Therefore ML=0.27 and p7/pt7=0.951.

The exit pressure would then be p7=(p7/pt,7)(pt,0.5,down/pt,i)pt,i= 72.9kPa <101.3kPa. Therefore the shock cannot be at the station z/L=0.5. (b) Since the exit pressure is too low the shock must be located upstream of z/L=0.5.

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (c) One may select several other stations for the shock location and determine pe for each location and interpolate to find that the shock is at z/L=0.27.

E5.2 The contoured (Laval) nozzle shown in Figure E5.2 has a ratio of exit area to minimum area Ae/Amin=2 and exhausts into an ambient atmosphere where the static pressure is pa=pt /30, that is, one-thirtieth of the stagnation pressure pt in the nozzle flow.

For γ=1.4 determine the direction α of the free streamline leaving the exit plane of the nozzle. α >0 Free streamline

min

e

Figure E5.2 Contoured nozzle and unknown exit streamline deflections α

Solution The static to stagnation pressure ratio at the minimum section for sonic flow at the minimum section is −

γ

pmin ⎛ γ + 1 ⎞ γ −1 =⎜ = 0.5283 ⎟ pt ⎝ 2 ⎠ Since pa/pt=0.03333 <<pmin/pt it is reasonable to assume that the flow is supersonic in the nozzle. For an area ratio Ae/Amin=Ae /A*=2 the Mach number at the exit may be found from the isentropic flow tables to be Me=2.2. The corresponding pressure ratio pe/pt=0.0935. Therefore pe>pa and the nozzle is under-expanded and the free streamline leaving the nozzle must be deflected outward, that is, α>0.

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E5.3 The NASA Boeing F-18 Hornet Systems Research Aircraft was flown by NASA's Dryden (now Armstrong) Flight Research Center, Edwards, California. The F/A-18 former Navy fighter was modified to investigate new technologies in the areas of flight controls. A convergent nozzle like that shown in the photograph in the text is operated at standard sea level conditions with an exit pressure ratio p7/pt,5=0.55 and a nozzle entrance total temperature Tt,5=1450R (805.6K). If the exit area is 1 ft2(0.0929m2) and the flow in the nozzle flow is ideal, calculate the following, using γ=1.33: (a) exit velocity V7, (b) exit temperature T7, (c) the weight flow rate w 7 or mass flow rate m 7 , (d) the gross thrust Fg, (e) the gross thrust if the nozzle entrance total temperature is raised to Tt,5= 3000R (1667K), (f) the gross thrust if Tt,5 remains at 3000R (1667K) and the exit area is increased to1.3ft2(0.1208m2).

Solution (a) Note that the gas constant R=cp(γ-1)/γ =cp/4=1,716ft2/s2-R. Since the flow is ideal the stagnation temperature is constant (Tt,5=Tt,7) and the exit velocity is given by γ −1 ⎡ ⎤ γ ⎛ ⎞ p ⎢ ⎥ = 2 6,864 ft 2 s 2 ⋅ R 1, 450 R ⎡1 − 0.55 0.25 ⎤ 7 V7 = 2c pTt ,7 ⎢1 − ⎜ )⎣ ( ) ⎦ ⎟ ( )( ⎜p ⎟ ⎥ t ,7 ⎠ ⎥ ⎢⎣ ⎝ ⎦

V7=1,662ft/s (507m/s) (b) The exit temperature is −1

⎛ M 72 ⎞ ⎛ γ −1 2 ⎞ T7 = Tt ,7 ⎜1 + M 7 ⎟ = (1, 450 R ) ⎜1 + ⎟ 2 6 ⎠ ⎝ ⎠ ⎝

−1

We need the exit Mach number and since the nozzle flow is ideal we may use the stagnation pressure relation for isentropic flow −γ

−4

p7 ⎛ γ − 1 2 ⎞ γ −1 ⎛ M 72 ⎞ = ⎜1 + M 7 ⎟ = ⎜1 + ⎟ = 0.55 pt ,7 ⎝ 2 6 ⎠ ⎠ ⎝ M 7 = 6 ⎡( 0.55 ) ⎣

−1 4

− 1⎤ = 0.983 ⎦

Then we can substitute back to find T7=1,249R (649K) (c) The mass flow is m = ( ρ AV )7 and because we found that M7<1 the exit pressure p7 = p0 and therefore pt,7 = p0/0.55 = 3847lb/ft2 (184.1kPa) and we may write

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m = pt ,7 A7

γ +1

−

γ

⎛ γ − 1 2 ⎞ 2(γ −1) M 7 ⎜1 + M7 ⎟ RTt ,7 2 ⎝ ⎠

m = ( 32.2 lbm ⋅ ft lb ⋅ s )( 3847 lb ft )(1 ft ) 2

2

2

⎛ 0.9832 ⎞ 1.33 ( 0.983) ⎜1 + ⎟ 6 ⎠ (1716 ft s ⋅ R )(1450 R ) ⎝

m = 52.8 lbm s (23.9kg / s ) (d) The gross thrust Fg=wV7/g=(52.8lb/s)(1,662ft/s)/(32.2ft/s2)=2,730lb (12,100N) 7= Fg = mV

( 52.8 lbm s )(1, 662 ft s ) = 2, 730lb = 12,100 N

32.2 lbm ⋅ ft lb ⋅ s 2 (e) If the stagnation temperature is raised to 3,000R the gross thrust doesn’t change because the mass flow rate decreases in the same ratio as the exit velocity increases. We may write Fn=(ρAV)7V7 =ρ7A7a72M72 =γp7A7M72 and all these quantities are unchanged. (f) If the stagnation temperature is raised to 3,000R and A7 is increased to 1.3ft2 then the gross thrust increases accordingly to Fg= 3,550lb (15,790N)

E5.4 Two SR-71 aircraft have been used by NASA as test beds for high-speed and highaltitude aeronautical research. The aircraft, an SR-71A and an SR-71B pilot trainer aircraft, have been based at NASA's Dryden (now Armstrong) Flight Research Center, Edwards, California. The shock diamonds in the exhaust are caused by supersonic flow being generated by the exhaust nozzles. In a supersonic nozzle which has a converging diverging shape the flow can be accelerated to supersonic speed by passing through the critical point (M=1) at the minimum area formed by the converging section (A6=A*) and then through the diverging section where the area increases to the exit area A7. For such an ideal nozzle with γ=1.33, a minimum area A6=4ft2 (0.3716m2) and an expansion out to an exit area A7 at which the pressure p7 matches the ambient pressure p0=5psia (34.47kPa) at an altitude of 27000ft (8,230m) calculate: (a) the velocity, temperature, pressure, Mach number, and density at the minimum area station, (b) repeat the calculation for the exit station, (c) the gross thrust developed. Assume that the turbine exit conditions are Tt,5=2,040R (1133K) and pt,5=40psia (275.8kPa).

Solution (a) The throat conditions are as follows:

4

−3.5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual −1 ⎛T *⎞ 6 ⎛ γ −1 ⎞ T6 = T * = ⎜ T = 1+ ⎟ Tt ,5 = ( 2, 040 R ) = 1, 749 R = 972 K ⎜ T ⎟⎟ t ,5 ⎜⎝ 2 ⎠ 7 ⎝ t ,5 ⎠

V6 = V * = a* = γ RT * = 1.33 (1716 ft 2 s 2 ⋅ R )1, 749 R = 1,997 ft s = 609 m s −γ

⎛ p* ⎞ ⎛ γ − 1 ⎞ γ −1 2 2 p6 = p* = ⎜ pt ,5 = ⎜1 + ⎟ ⎟ pt ,5 = 0.539 ( 5, 760 lb ft ) = 3,104 lb ft = 148.6 kPa ⎜p ⎟ 2 ⎝ ⎠ ⎝ t ,5 ⎠

2 2 p * 3,104 lb ft ( 32.2 lbm ⋅ ft lb ⋅ s ) ρ6 = ρ * = = = 0.0333 lbm ft 3 = 0.533 kg m3 2 2 RT * (1716 ft s ⋅ R ) (1749 R )

(b) After noting that the stagnation pressure and stagnation temperature are constant throughout the nozzle and that the exit pressure matches the local atmospheric pressure the exit conditions are as follows: γ −1

γ −1

1

⎛ p ⎞γ ⎛ p ⎞γ T7 T ⎛ 5 ⎞4 = 7 = ⎜ 7 ⎟ = ⎜ 0 ⎟ = ⎜ ⎟ = 0.595 ⎜p ⎟ Tt ,7 Tt ,5 ⎜⎝ pt ,7 ⎟⎠ ⎝ 40 ⎠ ⎝ t ,5 ⎠ Then the exit temperature is T7 = 0.595(1240R) = 1,214R =674K The speed of sound at the exit is

a7 =

( γ RT )7 = 1.33 (1, 716 ft 2 s 2 ⋅ R ) (1, 214 R ) = 1, 665 ft s = 507 m s

The Mach number at the exit is M7 =

2 ⎛ Tt ,7 ⎞ − 1⎟ = 6 (1.681 − 1) = 2.02 ⎜ γ − 1 ⎝ T7 ⎠

And the exit speed is V7 = a7M7 =3.366ft/s=1,026m/s The density is ⎛ p ⎞ ( 720 lb ft )( 32.2 lbm ⋅ ft lb ⋅ s ) = 0.0111lbm ft 3 = 0.178 kg m3 ρ7 = ⎜ ⎟ = 2 2 ⎝ RT ⎠7 (1, 716 ft s ⋅ R ) (1, 214R ) 2

2

(c) The gross thrust is

Fg = m 7V7 + A7 ( p7 − p0 ) = m 7V7 = ( γ pAM 2 ) Then

Fg = 1.33 ( 720 lb ft 2 ) A7 ( 2.02 ) = 3,907 A7lb 2

5

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The area may be found from the mass conservation equation as follows:

⎛ ρ * ⎞⎛ V * ⎞ ⎛ 1.03 ×10−3 ⎞ ⎛ 1,997 ⎞ * 4 ft 2 ) = 7.1 ft 2 = 0.66m2 A7 = ⎜ A = ( ⎟⎜ ⎟ ⎜ ⎟ −4 ⎟ ⎜ ⎝ 3.46 ×10 ⎠ ⎝ 3,366 ⎠ ⎝ ρ7 ⎠⎝ V7 ⎠ Therefore Fg = 27,740lbs (123,400N)

E5.5 Design a converging diverging nozzle for a missile ramjet engine under the constraint of maximum thrust. The combustor delivers hot gas (assume γ = 1.33 = constant) to the nozzle at a stagnation pressure pt,5 = 81kPa through a minimum section of fixed area A6 = 0.0845 m2. The missile is to fly at M0 = 2.2 at altitudes between 12.2km and 18.3km. Provide an illustration of the nozzle contour, drawn to scale and showing all pertinent features, and explain these features in accompanying text. (a) Explain how your design ensures maximum thrust over the mission specification envelope, (b) Show a plot of the variation of gross thrust with altitude for your nozzle design, (c) How did you decide on the length of the nozzle and its contour? (d) Present a simplified design for a nozzle that will approximate the maximum gross thrust constraint, (d) Compare the gross thrust produced by your simplified design with that of your ideal design over the flight envelope.

Solution

(a) Matched pressure ensures maximum gross thrust over altitude range specified.

However, to achieve this, the nozzle exit area must increase as one ascends through that range. (b) and (d) Thrust is shown in Figure E5.3 as a function of altitude for the ideal nozzle and for three fixed geometry nozzles whose exit pressures match the ambient pressure at the altitudes noted. (c) and (d) Typical characteristic dimensions for throat radius of curvature and length are given in the text. Nozzle contours are usually determined by numerical approaches to the compressible flow equations. A smooth contour satisfying the characteristic dimensions mentioned above is provided by, for example, the following equation for nozzle radius r as a function of distance x along the nozzle axis: r=

1 1 ⎡ x 1 ⎤ ( r * + r7 ) + ( r7 − r *) sin ⎢π ⎛⎜ − ⎞⎟ ⎥ 2 2 ⎣ ⎝ L 2 ⎠⎦

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The fixed geometry nozzle for the best case of a match at 15km altitude is shown in the Figure E5.4. Nozzle details upstream of the throat are not particularly crucial but can be generated in a similar manner

20 19 18

10km

17 16 z (km) 15 14

matched exit pressure

20km

13 12

15km

11 10 4

5

6

7

8

9

10

Fg (kN)

Figure E5.3 Gross thrust variation with altitude for 3 fixed altitude designs and

one matched pressure design

25 nozzle radius r (cm)

20 15 10 5

flow deflection in degrees

0 0

20

40

60

80

100

nozzle axial station (cm)

Figure E5.4 Fixed geometry nozzle design for the best case of a match at 15km

altitude

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 5.6 Consider a jet engine that exhausts hydrocarbon-air combustion products (the overall fuel to air ratio here is f/a = 0.02) through a nozzle with stagnation conditions Tt,5=1500F (815.6C) and pt,5=30psia (205.8kPa). Use the one-dimensional adiabatic energy equation and the assumption of an isentropic expansion (assume γ=1.33) through the nozzle to develop equations that are used to generate a plot of the (dimensional) static and stagnation temperature and pressure variation as a function of Mach number through the nozzle, for Mach numbers in the range 0<M<3. Clearly state any assumptions you made in producing the plot.

Solution

The one-dimensional adiabatic energy equation is, for constant cp, given by the relation Tt= T + V2/2cp = constant. This can be manipulated so as to be in terms of Mach number:: Tt/T = [1+(γ-1)M2/2]. For isentropic flow in the nozzle the pressure is related to temperature by T =Cp(γ-1)/γ and therefore the stagnation pressure must also be constant in the nozzle. One may also cast this equation as follows: Tt/T = (pt/p)(γ-1)/γ. The graphs of the dimensional quantities are shown in Figure E5.5 below.

1200 Tt 1000 T (K) 800 600

T

400 0

0.5

1

1.5 M

(a)

8

2

2.5

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

250 pt 200 150 p (kPa) 100 50 p 0 0

0.5

1

1.5

2

2.5

3

M

(b) Figure E5.5 Variation of (a) temperature and (b) pressure as a function of Mach

number for isentropic nozzle flow E5.7 A turbojet engine with an exit area of 0.116m2 is operated at an altitude of 6.1km where p0 = 46kPa and T0 = 249K. Assume the nozzle is choked (M7 =1), the exit pressure is equal to the ambient pressure of the undisturbed atmosphere at the operating altitude, and the exhaust gases are characterized by γ =1.33 and R = 287m2/s2-K. Determine the following: (a) gross thrust produced (b) exit mass flow rate m 7 if the exit temperature T7=811K (c) ram drag at the flight speed V0=168m/s (assuming an overall fuel to air ratio of 0.02) (d) specific fuel consumption (e) propulsive efficiency (d) The thermal efficiency Solution

(a) With the simplification that the exit pressure is equal to the ambient pressure the gross thrust may be written as

Fg = ( ρ AV 2 ) = ( γ pAM 2 ) = (1.33)( 46, 000 N ) ( 0.116m2 ) (1) = 7, 097 N 7

7

(b) The mass flow at station 7 is ⎛ γ ⎞ m 7 = ⎜⎜ pAM ⎟ = ( 46, 000 N )(1.33)(1) RT ⎟⎠7 ⎝

1.33 = 12.76 kg s ( 287 m s 2 ⋅ K ) (811K ) 2

(c) Assuming an overall fuel to air ratio of 0.02 the exit mass flow is

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual f ⎞ ⎛ m 7 = m 0 + m f = m 0 ⎜1 + ⎟ = 1.02m 0 a⎠ ⎝ The ram drag at the flight speed V0=168m/s is Fr = m 0V0 =

m 7 12.76 kg s (168 m s ) = (168 m s ) = 2,102 N 1.02 1.02

(d) The specific fuel consumption is

cj =

m f g Fn

s ⎞⎛ kg ⎞⎛ m⎞ ⎛ f ⎞ ⎛ m 0 g ⎜ ⎟ ⎜ 3600 ⎟⎜12.51 ⎟⎜ 9.807 2 ⎟ ( 0.02 ) hr ⎠⎝ s ⎠⎝ s ⎠ ⎝a⎠=⎝ = = 1.769hr −1 7, 079 N − 2,102 N Fg − Fr

(e) The propulsive efficiency is ηp=V0/Vavg and Vavg=(V0+Ve)/2. Because the nozzle is matched Ve=V7=a7M7=a7(1). The exit sound speed is

a7 = γ 7 R7T7 = 1.33 ( 287 m2 s 2 ⋅ K ) ( 811K ) = 556 m s Then the propulsive efficiency is 2 (168 m s ) 2V0 = = 46.4% V0 + V7 (168 m s ) + ( 556 m s ) (f) The thermal efficiency is

ηp =

ηth = 1 −

c p ,7T7 − c p ,0T0 ⎛ f a ⎞ ηb ⎟ Q f ⎜ ⎝ 1+ f a ⎠

= 1−

(1.13 kJ kg ⋅ K ) 811K − (1.003 kJ kg ⋅ K ) 249 K = 21.6% ( 0.0196 )(1)( 43, 400 kJ kg )

E5.8 A turbojet engine with a nozzle exit radius of 19.23cm is operated with an exit pressure p7 = 62kPa at an altitude of 6.1km where p0 = 46kPa and T0 = – 24C. Assume the nozzle is choked (M7 = 1) and the exhaust gases are characterized by γ = 1.33 and R = 287J/kg-K. Determine the following: (a) The gross thrust Fg produced (b) The exit mass flow rate, if the exit temperature T7=538C (c) The ram drag Fr at the flight speed V0 = 168m/s (assuming an overall fuel to air ratio f/a = 0.02) (d) The specific fuel consumption cj Solution

(a) The gross thrust Fg = ⎣⎡ p7 (1 + γ 7 M 72 ) − p0 ⎦⎤ A7 = ⎣⎡62 kN m 2 (1 + 1.33 × 12 ) − 46 kN m 2 ⎦⎤ ( 0.116m 2 ) = 11.43kN

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) The exit mass flow ⎤ ⎡ γ ⎤ ⎡ 1.33 2 2 m 7 = ( ρ AV )7 = ⎢ pAM ⎥ ⎥ = ⎢( 62 kN m )( 0.116m ) (1) 287 m2 s 2 ⋅ K ( 811K ) ⎦⎥ RT ⎦ 7 ⎣⎢ ⎣ m 7 = 17.19 kg s −1

f ⎞ kg ⎞ m⎞ ⎛ ⎛ ⎛ (c) The ram drag Fr = m 0V0 = m 7 ⎜1 + ⎟ V0 = ⎜ 17.22 ⎟ ( 0.9804 ) ⎜ 168 ⎟ = 2,835 N a⎠ s ⎠ s⎠ ⎝ ⎝ ⎝

(d) Specific fuel consumption s ⎞⎛ kg ⎞ ⎛ f ⎞ ⎛ m 0 ⎜ ⎟ ⎜ 3600 ⎟ ⎜ 16.88 ⎟ ( 0.02 ) m f hr ⎠ ⎝ s ⎠ ⎝a⎠=⎝ = = 0.1413 kg hr N = 1.385 lbm hr lb cj = Fn Fg − Fr 11, 440 N − 2,835 N

E5.9 A turbojet engine has a variable exit area converging diverging nozzle which operates in the matched pressure mode and accepts the gas flow from the turbine at a stagnation temperature Tt,5 = 838C and a stagnation pressure pt,5 = 275.6kPa. The nozzle throat radius r6=10cm and γ=4/3 throughout the nozzle. (a) Calculate the exit Mach number M7 (b) Find the nozzle exit area A7 required to attain the maximum gross thrust Fg possible at an altitude of z=23 km, where p0 = 4.35kPa (c) Calculate the exit mass flow If the aircraft descends to an altitude of 8.3km where p0=34.5kPa and the exit area is adjusted so as to permit the nozzle to remain in the matched mode: (d) Calculate the exit Mach number M7 (e)Find the new nozzle exit area A7 required to attain the maximum gross thrust Fg possible (f) Calculate the new exit Mach number M7 Solution

(a)The maximum gross thrust in a converging-diverging (supersonic flow) nozzle occurs when p7=p0 so we know that here p7/pt,7 = p0/pt,5 = 4.35kPa/275.6kPa = 0.01578.

It is also known that for γ=4/3 the ratio pt,7/p7=[1+M72/6]4. From this we find M7=3.305. (b) For γ=4/3 and M6=1 we may use the area relation A7/A6=(1/M7)[(1+M72/6)/(7/6)]3.5 = 6.654 This yields A7=0.209m2 and r7=25.7cm. (c) The maximum mass flow through the throat is 11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m 6 = pt ,5 A6

γ

− ( γ +1)

⎡ γ + 1 ⎤ 2(γ −1) RTt ,5 ⎢⎣ 2 ⎥⎦

m 6 = ( 275.6 kN m 2 )( 0.0314m 2 )

4 ⎛7⎞ ⎜ ⎟ 2 2 3 ( 287 m s ⋅ K ) (1,111K ) ⎝ 6 ⎠

−3.5

= 10.32

kg s

(d) The maximum gross thrust at the new altitude occurs when p7=p0 so we know that now p7/pt,7 = p0/pt,5 = 34.5/275.6 = 0.1252. We also know that pt,7/p7=[1+M72/6]4 from which we find M7=2.02. (e) Again using the area relation A7/A6=(1/M7)[(1+M72/6)/(7/6)]3.5 = 1.775 which yields A7=0.0558m2 or r7=13.32cm. (f) The equation for the mass flow given in part (c) involves the stagnation properties and the throat area, neither of which have changed. Therefore the mass flow remains the same.

For information we note that with the same stagnation conditions in both cases the gross thrust for matched operation is Fg=p7A7(1+γM72) and at z=23km Fg=14.15kN while at z=8.3km Fg=12.59kN. However, it must be remembered that maintaining the given stagnation conditions at both altitudes requires that the aircraft be traveling at a higher Mach number at the higher altitude.

E5.10 At the entrance to a convergent turbojet nozzle the stagnation values of pressure and temperature are 138kPa and 560C and γ=4/3. (a) For a back pressure of 101.3kPa calculate the exit velocity for isentropic flow (b) If the exit area of the nozzle is 0.11m2 determine the mass flow rate and the thrust produced (c) If the back pressure in part (a) were reduced to the critical value find the values of exit pressure, exit velocity, mass flow rate, and thrust (d) If the back pressure were reduced further to 34.5kPa why would it be necessary to add a divergent segment to the nozzle to maintain isentropic flow conditions? (e) For this condition determine the exit velocity and mass flow rate (f) Calculate the pressure, temperature, and Mach number at the throat formed by the juncture of the converging and diverging sections Solution

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (a) A convergent nozzle has a maximum exit Mach number Me=1. For isentropic flow the exit static to stagnation pressure ratio is −

γ

pe ⎛ γ − 1 2 ⎞ γ −1 = ⎜1 + Me ⎟ 2 pt ⎝ ⎠ In this hot jet case γ =4/3 so pe/pt = [1+Me2/6]-4 and for pe/pt = 101.3/138 = 0.734 we have Me = 0.694. The temperature Te/Tt = [1+Me2/6]-1 and Tt = (560C+273) = 833K so that Te = 771K and the exit sound speed is ae = (γRT)e1/2 = [(4/3)(287)(771)]1/2 = 543 m/s thus Ve = 376m/s (b) The mass flow is

⎡ ⎤ ⎛ p ⎞ 101.3 kN m2 ⎥ ( 0.11m 2 ) ( 376 m s ) = 18.93 kg s m = ⎜ e ⎟ AeVe = ⎢ 2 2 ⎢⎣ ( 287 m s ⋅ K ) ( 771K ) ⎥⎦ ⎝ RTe ⎠ The thrust produced is Fg=(18.93kg/s)(376m/s)=7.118kN (c) If the back pressure were reduced to the critical pressure for the convergent nozzle the flow would be matched with Me =1. The critical pressure ratio p*/pt = (7/6)-4 = 0.539 and therefore the exit pressure pe = p* =74.38kPa. In the same fashion Te = T* = 833(7/6)-1 =714K. Therefore the exit sound speed is ae = [(4/3)(287m2/s2-K)(714K)]1/2 = 522.7m/s.

Since the exit flow is sonic ae = Ve = 522.7m/s. The mass flow is

⎡ ⎤ 74.38 kN m 2 ⎥ ( 0.11m2 ) ( 522.7 m s ) = 20.9 kg s m = ⎢ 2 2 ⎢⎣ ( 287 m s ⋅ K ) ( 714 K ) ⎥⎦ e = (20.9)(522.7)=10.92kN The gross thrust is Fg = mV

(d) If the back pressure is dropped to 34.5kPa the jet would be underexpanded because pe=74.5kPa. To get the exit pressure down to the back pressure the flow would have to expand out of the nozzle and shocks would ultimately appear in the exhaust plume rendering the flow there non-isentropic. To avoid this, a divergent section could be added to the nozzle to provide shock-free expansion to the required exit pressure. (e) With a divergent section with pe/pt = 34.5/138 = 0.25. Using the same approach as in part (a) the exit Mach number would be found to be Me = 1.58 and Te = 589K. Then ae = 474m/s and Ve = 749m/s. The mass flow would be the same as in part (c) because the throat is choked and that station was the exit station. If the stagnation pressure and 13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual temperature are fixed, as they are here, the mass flow remains fixed at 20.9kg/s as well.

Direct calculation would show that N Ae ⎛ m⎞ kg 2 m2 = 0.11 749 20.9 m = m ( ) ⎜ ⎟ A∗⎝ s⎠ s ⎛ m2 ⎞ 287 ⎜ ⎟ ( 589 K ) 2 s −K⎠ ⎝ 34,500

Using the area ratio found from the mass flow equation we may calculate Me=1.58 from the A/A* relation 7

Ae 1 ⎡ 6 + M e2 ⎤ 2 = ⎢ ⎥ = 1.23 A∗ Me ⎣ 7 ⎦ (f) The Mach number at the throat is M = 1 and the pressure p* and temperature T* there may be found from the isentropic relations p* = pt[1+1/6]-4= 138kPa(0.5398) = 74.49kPa and Te = Tt[1+1/6]-1 = 833K(0.8571) = 714K

E5.11 A blowdown supersonic tunnel, that is, one which operates by exhausting air from one receiver at high pressure through a contoured nozzle to another receiver at lower pressure, utilizes standard sea level air from the atmosphere as the high pressure receiver and a large vacuum vessel as the low pressure receiver. (a) For isentropic flow through the nozzle calculate the pressure, temperature, velocity, Mach number, and density at the nozzle throat. (b) If the cross-sectional area of the throat is 25.81cm2 and the exit pressure is 20.7kPa calculate the temperature, velocity, Mach number, density, and area at the nozzle exit. (c) Determine the mass flow through the nozzle Solution

(a) For isentropic flow with γ =1.4 the pressure ratio p/pt=[1+M2/5]–3.5 and the temperature ratio T/Tt = [1+Me2/5]–1. Αt the throat where M = 1 we have p*/pt = (6/5)-7/2 = 0.528 and T*/Tt = (6/5)-1 = 0.833. The standard sea level conditions are also the stagnation conditions and therefore we may calculate p* = 53.5kPa and T* = 240K. (b) The exit pressure ratio pe/pt=20.7kPa/101.3kPa=0.2043 so that Me = 1.694. The exit temperature ratio is Te/Tt = (1+Me2/5)-1 = 0.6353 and the stagnation temperature remains

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual constant at Tt=288K so that Te=182.9K. Then the speed of sound is ae = (γRT)e1/2 = 270.9m/s. The exit velocity Ve =1.694ae= 458.9m/s, the density is

ρe = (pe/RTe) = 20,700N/m2/[(287m2/s2-K)(182.9K)] = 0.3943kg/m3, The area ratio Ae/A* =(1/Me)[(1+Me2/5)/(6/5)]3

which, for Me=1.694, yields Ae/A* =

1.331 so that Ae = 34.37cm2. (c) The mass flow (ρAV) = [(0.3943kg/m3) (0.003437m2) (458.9m/s)] = 0.6213kg/s

E5.12 A contoured nozzle is to be designed to accelerate air from stagnation conditions of 345kPa and 889K to a Mach number of 2.0 at the nozzle exit at a flow rate of 14.6kg/s.

Assuming γ=4/3 calculate the pressure, temperature, velocity, area, and density at (a) The nozzle throat (b) The nozzle exit (c) The axial force on the nozzle due to the acceleration of the air Solution

(a) Because of the high stagnation temperature the ratio of specific heats is assumed to be

γ=4/3. At the throat the temperature ratio T/Tt = [1+M2/6]-1 is T*/Tt = 6/7 = 0.857 and therefore T* = 762K. The sound speed is a* = (γRT*)1/2 = 539m/s and since at the throat M=1 we have a*=V*=539m/s. The pressure ratio is p*/pt=(7/6)-4=0.54 and p*=186kPa. The density

ρ* = p*/RT* = (186,000N)/(287m2/s2-K)(762K) = 0.851kg/m3.

The throat area A* = (ρAV)/ρ*V*= (14.6kg/s)/(0.851kg/m3)(539m/s) = 0.0318m2. (b) For isentropic flow the static to stagnation pressure ratio is pe/pt=[1+(γ-1)Me2/2] –γ/(γ-1).

In this hot jet case we again assume γ =1.33 so pe/pt = [1+Me2/6]-4 = 0.13 so pe=44.9kPa. The temperature Te/Tt=[1+Me2/6]-1 = 0.6 with Tt=889K yields Te=533K and the exit sound speed is ae=(γRT)e1/2=[(1.33)(287m2/s2-K)(533K)]1/2=451 m/s Thus Ve = 902m/s. The exit density

ρe = pe/RTe = (44900N)/(287m2/s2-K)(533K) = 0.294kg/m3.

The exit area Ae = (ρAV)/ρeVe = 14.6/(0.29)(902) = 0.0558m2.

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual kg ⎞ ⎛ m⎞ ⎛ e = ⎜ 14.6 ⎟ ⎜ 902 ⎟ = 13.2kN (c) The axial force is the gross thrust Fg = mV s ⎠⎝ s⎠ ⎝

E5.13 A contoured nozzle with throat area of 323cm2 is operated with stagnation conditions of 276kPa and 1110K. It is designed to have the exit pressure match the ambient pressure of 34.5kPa. If γ =4/3 and the adiabatic expansion efficiency of the nozzle is 90%: (a) Calculate the pressure, temperature, velocity, Mach number, and density at the throat and the exit (b) Determine the exit area (c) Calculate the thrust developed (d) Determine the percent reduction in thrust due to the non-ideal flow in the nozzle..

Solution

First we notice that the nozzle does not operate isentropically but adiabatically, with an efficiency of expansion of 90%. We may observe the process on the T,s diagram in Figure E5.6, assuming that cp=constant. The energy equation states that the stagnation temperature is constant for adiabatic flow so that Tt,5 = Tt,7. In isentropic flow the stagnation pressure is also constant so that pt,5=pt,7 and expansion from pt,5 to the nozzle exit pressure p7 results in the ideal exit temperature T7’. However, in the process under consideration, the flow is not isentropic so expansion from pt5 to p7 results in an exit temperature T7>T7’ and consequently less kinetic energy in the flow. The efficiency of adiabatic expansion may be defined as the ratio of actual kinetic energy realized in the expansion process to the ideal kinetic energy achievable in an isentropic expansion.

Following the diagram on the T,s plane the efficiency may be written as follows: ⎛T ⎞ ⎛T ⎞ 1 − ⎜⎜ 7 ⎟⎟ 1 − ⎜⎜ 7 ⎟⎟ 2 Tt ,5 T −T ⎝ Tt ,5 ⎠ = V7 η n = t ,5 7 = ⎝ ⎠ = γ −1 Tt ,5 − T7′ V7′2 ⎛ T7′ ⎞ γ ⎛ ⎞ 1 − ⎜⎜ ⎟⎟ 1 − p7 ⎜⎜ ⎟⎟ ⎝ Tt ,5 ⎠ ⎝ pt ,5 ⎠

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

pt5 pt7 Tt5 = Tt7 p7

V72/2

2

V7’ /2 T7 T7’

S

Figure E5.6 Operation of the nozzle on the T-S plane

(a) At the throat the temperature ratio is given by the adiabatic relation T/Tt = [1+M2/6]-1 so that T*/Tt = 6/7 = 0.857 and therefore T*=951K. The sound speed is a*= (γRT*)1/2 = 603m/s and since at the throat M=1 we have a* = V* = 603m/s. However, the pressure ratio must be found using the efficiency equation in the form ⎛ p∗ ⎞ 1− ⎜ ⎜ p ⎟⎟ ⎝ t ,5 ⎠

γ −1 γ

⎛T ∗⎞ 1 − ⎜⎜ ⎟ Tt ,5 ⎟⎠ ⎝ =

ηn

This yields p*/pt,5=0.50 and therefore p*=138kPa. The density

ρ* = p*/RT* = (138,000N)/[(287 m2/s2-K)(951K)] = 0.506kg/m3 Carrying out the same process for the exit station yields p7 = pa = 34.5kPa

T7 =706K

M7 =1.86

V7 = 967m/s

ρ7 = 0.170kg/m3.

If the nozzle were ideal (ηn=100%) the following results would have been obtained: pt,7’=pt,5=276kPa

T7’=660K

M7’=2.02

V7’= 1,015m/s ρ7’=0.182kg/m3

(b) The mass flow is constant so (ρAV)* = (ρAV)7 and therefore A7 = (ρ*/ρ7)(V*/V7)A* = 0.0599m2 (c) The gross thrust for a matched nozzle is Fg = (γpAM2)7 =9.51kN. (d) If the nozzle were ideal the only difference in the gross thrust equation would be that the exit Mach number would be higher, 2.02 rather than 1.86 and the thrust would be

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Fg=11.22kN. The change in thrust is 9,510N-11,220N= -1,710N and the percent change in thrust is ΔFg /Fg,ideal= -15.2%

E5.14 A convergent nozzle exhausts to standard sea level conditions with a pressure ratio p6/pt5=0.55 and an inlet stagnation temperature Tt5=806K. If the exit diameter is 0.344m and the nozzle efficiency is 100% assume γ=1.30 and calculate the following: (a) exit velocity V6, (b) exit temperatureT6, (c) mass flow, (d) gross thrust Fg, (e) exit Mach number M6, (f) gross thrust if Tt5 is increased to 1667K, (g) gross thrust if the exit diameter is also increased to 0.392m. Note that for this nozzle, which is assumed to be adiabatic, the stagnation temperature Tt5 = Tt6 and since the nozzle is perfect the stagnation pressure pt5 = pt6 Solution

(a) For p6/pt5 = 0.55 the exit Mach number is M6 = 0.9931 and the temperature ratio is T6/Tt5 = 0.8711. Therefore the temperature T6 = 702.1K and the sound speed a6 = (γRT6)1/2 = 511.8m/s. Therefore the exit velocity V6 = 508.2m/s (b) As calculated in part (a) the exit temperature is T6=702.1K (c) The mass flow is given by 101.3 kN m 2 ) ( ⎛ p ⎞ m = ⎜ AV ⎟ = 0.09289m 2 ) ( 508.2 m s ) = 23.73 kg s ( 2 2 ⎝ RT ⎠6 ( 287 m s ⋅ K ) ( 702.1K ) (d) The gross thrust is

6 = ( 23.73 kg s )( 508.2 m s ) = 12.06kN Fg = mV (e) As calculated in part (a) the exit Mach number is M6=0.9931 (f) If we now increase the stagnation temperature to Tt5=1667K, the exit temperature increases to T6=1452K so the sound speed there increases to a6=736m/s. The exit Mach number M6 remains the same so that the exit velocity increases to V6=730.9m/s. As a result of the stagnation temperature increase the mass flow decreases to 101.3 kN m 2 ) ( ⎛ p ⎞ m = ⎜ AV ⎟ = 0.09289m 2 ) ( 730.9 m s ) = 16.50 kg s ( 2 2 ⎝ RT ⎠6 ( 287 m s ⋅ K ) (1452 K ) kg ⎞ ⎛ m⎞ ⎛ 6 = ⎜ 16.50 ⎟ ⎜ 730.9 ⎟ = 12.06kN . In other words, The gross thrust is given by Fg = mV s ⎠⎝ s⎠ ⎝

changing the stagnation temperature alone will not change the thrust because the gross thrust may also be written as Fg=ρAV2=γpAM2 and this is constant.

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (g) On the other hand, if A6 is increased the mass flow, and therefore the thrust, increases.

In this case Fg=1.3(101.3kN/m2)(0.1206m2) (0.99312)=15.66kN

5.15 Consider air (γ =1.4) flowing through a smoothly contoured Laval nozzle as shown in Figure E5.7. The pressure in the very large reservoir, region S, is p0=1 MPa. The pressure in region A, the unconfined ambient into which the nozzle issues, is pE=100 kPa. The area of the narrowest section is A1=0.15 m2 and that of the nozzle exit is A2=0.6 m2.

Determine the following: (a) the Mach number and pressure at station 1, M1 and p1, respectively (b) the Mach number and pressure at station 2, M2 and p2, respectively (c) the Mach number and pressure in the nozzle exit region 3, M3 and p3, respectively and sketch the flow pattern there Solution

(a) Because the pressure ratio pE/p0=10 it is safe to assume that the throat is choked and M1=1. For γ=1.4 the static to stagnation pressure ratio is p/pt=[1+M2/5]−3.5 and at station 1 p1/p0=p*/p0= [6/5]−3.5 =0.5283 and therefore p1=528.3kPa (b) The mass flow is

m = ρ AV =

p γ γ RT AM = pAM RT RT

The mass flow is constant and the flow is isentropic so we may write p2 A2 M 2 p1 A1M 1

T1 ⎛ p2 ⎞ =⎜ ⎟ T2 ⎝ p1 ⎠

67

⎛ A2 ⎞ ⎛ M 2 ⎞ ⎜ ⎟⎜ ⎟ =1 ⎝ A1 ⎠ ⎝ M 1 ⎠

The pressure ratio between stations 1 and 2 is given by p1 ⎛ p1 ⎞ ⎛ p0 ⎞ ⎛ 5 + M 22 ⎞ = ⎜ ⎟⎜ ⎟ = ⎜ ⎟ p2 ⎝ p0 ⎠ ⎝ p2 ⎠ ⎝ 5 + M 12 ⎠

3.5

⎛ 5 + M 22 ⎞ =⎜ ⎟ ⎝ 6 ⎠

3.5

Using the given area ratio of 4, p1=528.3kPa and M1=1 in the previous two equations leads to M2=2.94 and p2=29.8 kPa (c) M and p in region 3 are the same as at the exit, i.e., station 2, except near the nozzle lip where a slip surface emerges due to an oblique shock emanating from the lip and tilted up toward the nozzle axis as shown in Figure E5.8. If one goes through more analysis one may show that since the shock must provide a pressure ratio of p2/pE=29.8/100=0.298 so

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual the Mach number in region 3 normal to this shock is approximately M3,n=1.74 and the shock angle to the axis of the nozzle is arcsin(1.74/2.94)=36.28o.

A

3

S

A 2

1

Figure E5.7 Schematic diagram of nozzle A

p=100kPa

p=29.3 kPa

3

shock slipstream A 2

Figure E5.8 Pressure variation near the exit of the nozzle

E5.16 Consider the axisymmetric contoured converging-diverging nozzle shown in Figure E5.9. The nozzle is attached at station 1 to a large air tank kept at a constant pressure of 1725 kPa and a constant temperature of 677K. The nozzle exhausts, at station 2, into the free atmosphere which is at a pressure and temperature of 101 kPa and 300K, respectively. The nozzle contour is designed to produce an exit Mach number and pressure of 2.5 and 101 kPA, respectively. (a) Determine the exit diameter, d2, and the mass flow rate, m , of the nozzle, (b) Assuming that the pressure in the air tank is doubled

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual to 3450 kPa while the temperature in the tank is maintained at 677K what is the velocity vector leaving the exit plane at point A in the diagram? (c) Assume now that the supply pressure in the air tank is drastically reduced to 182.4 kPa while the temperature remains at 677K. A probe at the exit plane, station 2, measures the stagnation pressure to be pt,2=117.2 kPa, suggesting the presence of a shock in the nozzle. Determine the diameter, ds, inside the nozzle at which the shock is located and the Mach number at the exit plane, M2.

Solution

(a) The area ratio A2/A* corresponding to a design Mach number M2=2.5 may be found from Equation (5.25). Assuming γ=1.4 we could also have used the compressible flow tables in Appendix A. The result is A2/A* = 2.637 and for sonic flow at the throat Amin=A* and dmin=d*. Using A2/A* = (d2/d*)2 we find d2=4.34 cm. and A* = (πdmin2/4) = 5.536x104

m2. We then use Equation (5.23) to find the mass flow as follows:

(1.725 ×10 N m )( 5.536 ×10 m ) m = 6

−4

2

2

677 K

1.4 −3 (1.2 ) = 1.483 kg s 2 2 287 m s ⋅ K

pt,1 pt,2 y, j A

2.67 cm

Mx My x, i

1

minimum

2

Figure E5.9 Schematic diagram of a nozzle attached to a large air reservoir at station 1 and exhausting into the atmosphere. Mx and My refer to the upstream and downstream locations of a shock in the nozzle as do pt,1 and pt,2

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

(b) Doubling the stagnation pressure also doubles the static pressure at the exit plane of the nozzle to p2=202 kPa, which is greater than the ambient pressure of the surroundings, and therefore the nozzle is under-expanded. The Mach number at the exit plane, M2, is still 2.5 and the Prandtl-Meyer angle there is still ν2=39.12o, as shown in Equation (A.22) or in the tables in Appendix A. However, the pressure on the streamline leaving at point A must match the back pressure pb=101 kPa and yet have the new stagnation pressure pt=3450kPa. Therefore, the ratio pt/pb=34.16 and this corresponds to MA =2.96,

νΑ=48.98o, and Tt/TA=2.752. Then TA=246K and aA=314.8 m/s, yielding VA=931.74 m/s.

The streamline at A is therefore tilted up from the horizontal by an angle Δν =νA-ν2 = 48.98o-39.12o = 9.86o. Then VAcosνA=918 m/s and VAsinνA=159.6 m/s so that the vector velocity at point A is V = (918 m/s)i + (159.6 m/s)j (c)The shock separates two regions of constant stagnation pressure with pt,1 = 182.4kPa being constant up to the location of the shock as illustrated in Figure E5.9 and pt,2 = 117.8kPa from the shock and on downstream. The stagnation pressure ratio across the shock is pt,2/pt,1 = (117.8/182.4) = 0.6453, and from the shock relations of Equation (A.11) or the compressible flow tables of Appendix A this ratio corresponds to a Mach number ahead of the shock Mx = 2.16. The corresponding area ratio is Ax/Ax* = 1.935 = (ds/d*)2, so that the shock is located at the axial station where ds=3.72 cm. Using the shock relations of Equation (A.11) or the compressible flow tables of Section A.4, the Mach number just downstream of the shock (a location denoted by the subscript y) is found to be My=0.5525 and the corresponding area ratio is Ay/Ay*=1.255. The Mach number at the nozzle exit, station 2, may be found by noting that the area ratio at station 2 is A2/Ay* = (Ay/Ay*)(A2/Ay) = (1.255)(4.34m/3.72m)2 =1.4. This area ratio corresponds to a local exit Mach number M2=0.47. Note that, as a check, we find the ratio p2/pt,2=101.3/117.8=0.8593 corresponds to the exit Mach number M2=0.47.

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

E5.17 Consider a flow exhausting into a large reservoir that is held at a back pressure p0=101 kPa. For frictionless flow find the velocity leaving the nozzle lip at point A, V=Vxi + Vyj. A detailed view of the conditions at the exit appears in Figure E5.10.

M=2.02 pt=315.6 kPa Tt= 300K

Me=2.02

θ VA

shock waves

Mn,e

δ

A p0

A

Figure E5.10 Schematic diagram of supersonic flow exiting a two-dimensional nozzle

Solution (i) The exit plane pressure in the nozzle at M=2.02 is p= (pt/p)-1pt =315.6/8.071=39.1 kPa.

The nozzle is therefore over-expanded and the shock waves shown are necessary to bring the pressure on the streamline leaving the nozzle with velocity VA up to the back pressure level of p0=101 kPa. (ii) The Mach number normal to the shock at the exit, Mn,e=2.02sinθ, where θ is the angle the shock makes with the horizontal at the point A of the nozzle lip. The pressure across the shock is px/py=p0/pe=101/39.1=2.565, the temperature ratio T0/Te=1.34, and the normal component of the Mach number behind the shock is Mn,0=0.684. These results correspond to a shock angle θ = 49o. The graphs or equations for shock angle vs. flow deflection in Appendix A show that at M=2.02 and θ = 49o, the flow deflection δ=17.86o. (iii) Then the downstream Mach number MA=Mn,0/sin(θ−δ)=0.684/.517=1.323 The downstream temperature is Tb=1.34Tt(Te/Tt) =1.34(300)/1.816=221K The sound speed

23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ae=[1.4(287)(221)]1/2=298 m/s Therefore Vb=abMb=395 m/s. (iv) Then the velocity components are Vx=Vbcos δ =375.6 m/s and Vy=Vbsin δ=121.0 m/s and the velocity vector leaving the lip of the nozzle is VA=375.6i+121j

E5.18 A sonic nozzle exit (station 6) with p6=0.94atm is shown in Figure E5.11 exhausting into a quiescent environment with pressure p0=0.5atm. Determine the path of the streamline exiting the lip of the duct from station 3 to station 4 and the Mach number on that streamline.

M7=1 p7=0.94atm

nozzle wall

Me, pe p0=0.5 atm 6

φ exiting streamline

Figure E5.11 Schematic diagram of unmatched sonic nozzle Solution

Assuming frictionless flow the stagnation pressure is constant along the exiting streamline. The pressure ratio p6/pt=0.5283 for sonic flow of air with γ =1.4 so the stagnation pressure must be pt=1.78atm. The pressure along the exiting streamline must match the ambient pressure (pe=p0) so the Mach number may be found from the ratio pe/pt,e=0.5/1.78=0.281. In the isentropic flow tables in Appendix A we find that this pressure ratio corresponds to Me=1.48. From Prandtl-Meyer flow tables or Equation (A.22) we find the Prandtl-Meyer angle at M3=1 is ν3 =0 and at Me =1.48 it is νe =11.31.

24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This leads to a turn angle φ=νe – ν3=11.31o and is directed downward as shown in Figure E5.11.

E5.19 A converging nozzle whose exit area is A7 = (2)-1/2m2 is operated with air as the working fluid at a stagnation temperature of 300K. The nozzle is passing the maximum mass flow and has matched pressure at the exit while exiting into an ambient atmosphere whose pressure is p0=141kPa. Determine the gross thrust Fg produced.

Solution Fg = m7V7 = ρ7 A7V72 = ρ7 A7 a72 M 72 Fg = ρ7 A7

γ p7 2 M 7 = γ p7 A7 M 72 ρ7

1 ⎛ 1 2⎞ m2 Fg = (1.4 )(141kPa ) ⎜ m ⎟ (1) = 1.4 (141, 000 N / m 2 ) 1.41 ⎝ 2 ⎠ Fg = 140, 000 N = 140kN

E5.20 A converging-diverging nozzle is tested in a laboratory and found to produce a maximum thrust of 100kN at a mass flow of 100kg/s. What is the exit velocity V7?

Solution

The maximum gross thrust occurs for a matched nozzle, for which kg ⎞ ⎛ 7 = ⎜ 100 ⎟ V7 = 100kN Fg ,max = mV s ⎠ ⎝ 100, 000 N m = 1, 000 V7 = kg s 100 s

E5.22 A converging-diverging nozzle operating at a stagnation pressure of 16atm and a stagnation temperature of 1200K discharges into the atmosphere at standard sea level conditions. Experimental readings taken at the exit station and the minimum area station show the temperature ratio to be T7/T6= 6/10. Determine the exit Mach number and the nozzle area ratio A7/A6.

Solution

25

Theory of Aerospace Propulsion, Second Edition: Solutions Manual M 62 6 1+ T7 T7 Tt ,6 5 = 5 = 6 = = M2 M 2 10 T6 Tt ,7 T6 1+ 7 1+ 7 5 5 2 M 1+ 7 = 2 5 M7 = 5

(a)

3

⎛ M 72 ⎞ 3 3 1+ ⎟ A7 ⎛ M 6 ⎞ ⎜ 1 ⎛5+5⎞ 1 ⎛ 10 ⎞ 5 ⎟ = =⎜ ⎟⎜ ⎜ ⎟ = ⎜ ⎟ A6 ⎝ M 7 ⎠ ⎜ M 62 ⎟ 5 ⎝ 5 +1 ⎠ 5⎝ 6 ⎠ + 1 ⎜ ⎟ 5 ⎠ ⎝

(b)

3

A7 1 ⎛5⎞ = ⎜ ⎟ = 2.07 A6 5 ⎝3⎠

E5.23 Consider the axisymmetric contoured converging-diverging nozzle like that shown in Figure 5.1 of the text. The nozzle is attached at station 5 to a large air tank kept at a constant pressure of 1725 kPa and a constant temperature of 677K. The nozzle exhausts, at station 7, into the free atmosphere which is at a pressure and temperature of 101 kPa and 300K, respectively. The nozzle contour is designed to produce an exit Mach number and pressure of 2.5 and 101 kPa, respectively. Determine the exit diameter, d7, and the mass flow rate, m , of the nozzle if γ=1.4.

Solution

The area ratio A7/A* corresponding to a design Mach number M7=2.5 may be found from Equation (5.25). Assuming γ=1.4 we could also have used the compressible flow tables in Section A.4 in the appendix. The result is A7/A* = 2.637 and for sonic flow at the throat Amin=A* and dmin=d*. Using A7/A* = (d7/d*)2 we find d7=4.34 cm. and A* = (πdmin2/4) = 5.536x10-4m2. We then use Equation (5.23) to find the mass flow as follows:

(1.725 ×10 N m )( 5.536 ×10 m ) m = 6

2

−4

2

677 K

26

1.4 −3 (1.2 ) = 1.483 kg s 2 2 287 m s ⋅ K

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E6. INLETS FOR AIRBREATHING ENGINES E6.1 A simple conically diverging inlet with a cone half-angle of 10o is operating at the maximum mass flow condition at a flight speed of M0=2. (a) Sketch the inlet and any shock waves, (b) find the captured stream tube area ratio A1/A0 and (c) calculate the stagnation pressure recovery.

Solution (a) The flow field for the inlet is shown in Figure E6.1.

10o M0=2

M2<1

A0

Figure E6.1 Simple conically diverging inlet operating at maximum mass flow

(b) For maximum mass flow the normal shock will be at the inlet lip as shown in Figure E6.1. In that case the captured streamtube area is equal to the inlet area or A1/A0 =1 (c) The pressure recovery for γ =1.4 may be obtained by using Equation (A.11) to yield 7 2

5 2

7

5

⎛ 6M ⎞ ⎛ ⎞ ⎛ 24 ⎞ 2 ⎛ 6 ⎞ 2 6 =⎜ 2 ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ = 0.7209 pt ,0 ⎝ M 0 + 5 ⎠ ⎝ 7 M 02 − 1 ⎠ ⎝ 9 ⎠ ⎝ 27 ⎠ pt ,2

2 0

E6.2 An internal compression inlet is started and is operating in the ideal mode with the normal shock positioned exactly at the minimum section as shown in Figure E6.2. The Mach number behind the shock is Mmin=(3)-1/2 and it is desired that the Mach number at the end of the inlet be M2=1/3.Determine the required area ratio A2/Amin.

Solution From the given information we may form the ratio

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

A2 ⎛ A2 ⎞ ⎛ A* ⎞ = ⎜ ⎟⎜ ⎟ Amin ⎝ A* ⎠ ⎝ Amin ⎠

M0>1

Mmin

M2

Figure E6.2 An internal compression inlet with the shock fixed at the minimum section Using Equation (5.25) for the critical area ratio for isentropic flow in a nozzle with γ =1.4 we find 3

⎛ M 22 ⎞ 3 + 1 ⎟ A2 ⎛ M min ⎞ ⎜ 5 ⎟ = ⎛ 1/ 3 ⎞ ⎛ 5 + 1/ 9 ⎞ =⎜ ⎜ ⎜⎜ ⎟⎟ ⎜ ⎟ ⎟ 2 M min Amin ⎝ M 2 ⎠ ⎜ ⎟ ⎝ 1/ 3 ⎠ ⎝ 5 + 1/ 3 ⎠ + 1 ⎜ ⎟ 5 ⎠ ⎝ 3

3

A2 ⎛ 46 ⎞ ⎛ 23 ⎞ = 3 ⎜ ⎟ = 3 ⎜ ⎟ = 1.524 Amin ⎝ 48 ⎠ ⎝ 24 ⎠

E6.3 Consider the internal compression inlet shown in Figure E6.3. (a) For an area ratio A1/Amin=1.26 determine the starting Mach number Ms and the corresponding values of pt2/pt1and (A0/A1)max (b) For a flight Mach number M0=2.5 find the corresponding values of pt2/pt1and (A0/A1)max (c) For a flight Mach number M0=0.25 find the corresponding values of pt2/pt1and (A0/A1)max Solution (a) For the starting Mach number Ms we must have the normal shock at the lip of the inlet and the Mach number of the minimum section Mmin=1. The area ratio behind the lip shock with Mach number M1,downstream=M1,x<1 is A1,x/Amin = A1/A* =1.26. Entering the subsonic flow portion of the isentropic flow tables in Appendix A with this area ratio shows that M1,x=0.55. Entering the supersonic portion of the isentropic flow tables with this value downstream of the normal shock at the lip shows that M1,x=M0=Ms=2.18. To 2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual get the maximum pressure recovery we must position the shock at the minimum station.

Then we consider that the flow is compressed isentropically from A1 to Amin. We know that at the lip M1=2.18 and A1/A*=1.97 so that at the minimum station, just before the shock that is positioned there, the area ratio Amin/A* = (A1/A*)(Amin/A1) = 1.97/1.26 =1.56.

The Mach number corresponding to this area ratio is Mmin= 1.90 and the corresponding pressure recovery is pt,2/pt,1 = pt,2/pt,0 = 0.767. Since the shock is at the lip of the inlet and the flow at the minimum section is sonic we have (A0/A1)max=1.

M0 p0

0

1 minimum

2

Figure E6.3 Converging diverging internal compression inlet

(b) At a flight Mach number M0=2.5 we may again adjust the shock to be positioned at the minimum section to maintain the maximum possible mass flow through the inlet.

Once more we consider that the flow is compressed isentropically from A1 to Amin. We know that at the lip M1 =2.5 and A1/A* =2.637 so that at the minimum station, just before the shock that is positioned there, the area ratio Amin/A* = (A1/A*)(Amin/A1) = 2.637/1.26 = 2.093. The Mach number corresponding to this area ratio is Mmin = 2.25 and the normal shock pressure recovery is pt,2/pt,1 = pt,2/pt,0= 0.606. As before, since the shock is swallowed the maximum area ratio is (A0/A1)max=1. (c) At subsonic Mach numbers like M0 = 0.25 the pressure recovery is pt,2/pt,1 = pt,2/pt,0 = 1. However the captured streamtube area may be found by recalling that for maximum mass flow through the inlet Mmin=1. Entering the subsonic portion of the isentropic flow tables with M0=0.25 shows that A0/A*=2.403. Therefore (A0/A1)max= (A0/A*)/(A1/A*) = 2.403/1.26 = 1.91.

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E6.4. Consider a converging-diverging internal compression inlet as shown in the Figure E6.3. (a) For an area ratio A1/Amin=1.30 determine the starting Mach number Ms and the corresponding values of pt2/pt1 and (A0/A1)max, (b) For a flight Mach number M0=3 find the corresponding values of pt2/pt1 and (A0/A1)max (c) For a flight Mach number M0=2 find the corresponding values of pt2/pt1 and (A0/A1)max, (d) For a flight Mach number M0=0.2 find the corresponding values of pt2/pt1and (A0/A1)max Solution (a) The starting Mach number Ms = 2.43. When the shock is positioned at the minimum section for maximum stagnation pressure recovery the Mach number there is Mmin= 2.14.

This corresponds to a pressure recovery pt,2/pt,1 = pt,2/pt,0 = 0.656 and a maximum captured streamtube area ratio (A0/A1)max=1 (b) For M0=3>Ms the shock may be swallowed and at the minimum section Mmin = 2.74, the stagnation pressure ratio across the shock pt,2/pt,1 = pt,2/pt,0 = 0.415, and (A0/A1)max =1 (c) For M0 = 2<Ms the shock cannot be swallowed but remains expelled with pt,1/pt,0 = 0.721, M1,x=0.5774 and A1,x/A*=1.218. Note that pt,2/pt,1 =1 because the flow behind the expelled shock is isentropic and therefore the total pressure recovery pt,2/pt,0 = (pt,2/pt,1)( pt,1/pt,0)=0.721. This leads to (A0/A1)max = 1.218/1.3 = 0.936 (d) For M0 =0.2<1<Ms, there is no shock and pt,2/pt,1 = pt,2/pt,0 = 1 with (A0/A1)max= 2.96/1.3 =2.28

E6.5. Consider an internal compression inlet of the type shown in Figure E6.3. (a) For an area ratio A1/Amin=1.1265 determine the starting Mach number Ms and the corresponding values of pt2/pt1 and (A0/A1)max, (b) For a flight Mach number M0=2 find the corresponding values of pt2/pt1and (A0/A1)max (c) For a flight Mach number M0=1.25 find the corresponding values of pt2/pt1and (A0/A1)max (d) For a flight Mach number M0=0.75 find the corresponding values of pt2/pt1 and (A0/A1)max Solution (a) The starting Mach number Ms = 1.63. When the shock is positioned at the minimum section for maximum stagnation pressure recovery the Mach number there is Mmin=1.43.

This corresponds to a pressure recovery pt,2/pt,1= pt,2/pt,0 = 0.950 and a maximum captured streamtube area ratio (A0/A1)max=1 (b) For M0 = 2>Ms the shock may be swallowed and at the minimum section Mmin =1.85, the stagnation pressure ratio across the shock pt,2/pt,1 = pt,2/pt,0 = 0.790, and (A0/A1)max =1;

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (c) For M0 =1.25<Ms the shock cannot be swallowed but remains expelled with pt,1/pt,0 = 0.987. Behind the shock M1,x=0.813, and (A1,x/A*) =1.033. Note that pt,2/pt,1 =1 because the flow behind the expelled shock is isentropic and therefore the total pressure recovery pt,2/pt,0 = (pt,2/pt,1)( pt,1/pt,0)=0.987. The maximum captured streamtube area is (A0/A1)max= 1.033/1.1265=0.916 (d) M0 = 0.75<1<Ms and therefore there is no shock. As a result, pt,1/pt,0= pt,2/pt,0 = 1 and (A0/A1)max=1.0624/1.1265=0.943

E6.6. Consider the internal compression inlet shown in Figure E6.3. (a) For an area ratio A1/Amin=1.23 determine the starting Mach number Ms and the corresponding values of pt2/pt1and (A0/A1)max (b) For a flight Mach number M0=2.5 find the corresponding values of pt2/pt1and (A0/A1)max (c) For a flight Mach number M0=1.5 find the corresponding values of pt2/pt1and (A0/A1)max Solution (a) The starting Mach number Ms = 2.04. When the shock is positioned at the minimum section for maximum stagnation pressure recovery the Mach number there is Mmin=1.78.

This corresponds to a pressure recovery pt,2/pt,1= pt,2/pt,0 = 0.822 and a maximum captured streamtube area ratio (A0/A1)max=1; (b) For M0 = 2.5>Ms the shock may be swallowed and at the minimum section Mmin = 2.28, the stagnation pressure across the shock pt,2/pt,1 = pt,2/pt,0 = 0.594, and (A0/A1)max= 1. (c) For M0 =1.5<Ms the shock cannot be swallowed but remains expelled with pt,1/pt,0 = 0.93. Behind the shock M1,x=0.704, and (A1,x/A*)=1.09. Note that pt,2/pt,1 =1 because the flow behind the expelled shock is isentropic and therefore the total pressure recovery pt,2/pt,0 = (pt,2/pt,1)( pt,1/pt,0)=0.93. The maximum captured streamtube area is (A0/A1)max= 1.09/1.23=0.886 E6.7. The North American Aviation F-100 Super Sabre had a wing area of 400ft2, a wing sweepback angle of 45 degrees, a taper ratio of 0.25, and an airfoil-section thickness ratio of about 7 percent. The oblong nose inlet shown in the photograph in the text provides an immediate recognition feature of the F-100 series of aircraft. As compared with a circular inlet, the oblong design (A1=6ft2) provides better pilot visibility over the nose and, since the duct passes under the pilot's seat, the vertical dimension of the fuselage is reduced at this location. The Pratt & Whitney J57-P-21A engine (16,000 pounds thrust with afterburner) propelled the F-100 to a maximum speed of M=1.39 at

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 35,000ft altitude. Under this condition, and assuming the 1976 U.S. standard atmosphere properties apply, determine (a) the free stream stagnation pressure, pt,0, (b) the free stream stagnation temperature, Tt,0, (c) the maximum pressure recovery of the inlet, pt,2/pt,0, (d) the static pressure at the end of the inlet, p2, assuming that the inlet duct expands smoothly to A2/A1=5/3, (e) the weight flow through the inlet, (f) assuming the gross thrust to be the full afterburning thrust quoted determine the drag coefficient cD=D/qS where q is the dynamic pressure and S the wing area.

Solution Because inlet flows are not at very high temperatures we assume throughout that γ =1.4.

The atmospheric conditions are p0 = 499.4lb/ft2, T0 = 394.1R, ρ0 = 0.02376lbm/ft3, a0 = 973.1ft/s, q0 = 0.7p0M02 = 675 lb/ft2. (a) pt,0 = p0(1+M02/5)3.5 = 1567lb/ft2(0.74atm) (b)Tt,0 = T0(1+M02/5) = 546.4R (c) The normal shock pressure recovery is ⎡ 6 M 02 ⎤ =⎢ ⎥ pt ,0 ⎣ M 02 + 5 ⎦ pt ,2

3.5

⎡ 6 ⎤ ⎢ ⎥ 2 ⎣ 7 M 0 − 1⎦

2.5

= 0.961

(d) The area ratio 3

A2 M 1 ⎛ 5 + M 22 ⎞ 5 = ⎜ ⎟ = A1 M 2 ⎝ 5 + M 12 ⎠ 3

The Mach number behind the normal shock at M0 =1.39 and it is given by M1 =

M 02 + 5 = 0.743 7 M 02 − 1

Using this in the area ratio relation we find M2=0.35. Because the stagnation pressure behind the normal shock remains constant throughout the rest of the inlet the static to stagnation pressure ratio p2/pt,2 =p2/pt,1=(1+ M22/5)-3.5=0.919. The pressure recovery is the stagnation pressure across the normal shock recovery so that (pt,1/pt,0)=0.961 and therefore the pressure at station 2 is ⎛ p ⎞⎛ p ⎞ p2 = ⎜ 2 ⎟ ⎜ t ,1 ⎟ pt ,0 = 0.919 ( 0.961) (1567 lb ft 2 ) = 1383 lb ft 2 ⎜ p ⎟⎜ p ⎟ ⎝ t ,1 ⎠ ⎝ t ,0 ⎠

(e) The mass flow through the inlet is m = ( ρ AV )0 = ( 0.02376 lbm ft 3 )( 6 ft 2 ) ( 973.1 ft s )(1.39 ) = 194 lbm s

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (f) The net thrust Fn = Fg – Fr =16,000lb – Fr Fr = m 0V0 =

(194 lbm s )(1,353 ft s ) = 8,155lb 32.174 lbm ⋅ ft lb ⋅ s 2

Fn= 16,000lb-8,155lb=7,844lb The drag coefficient is cD =

D Fn 7,844lb = = = 0.029 qS qS ( 675.3 lb ft 2 )( 400 ft 2 )

E6.8. The McDonnell F4 Phantom, a supersonic all-weather fighter-bomber had a maximum speed of Mach 2.23 at 40,000ft altitude. It was powered by two GE J79-GE-17 turbojets producing 17,880lbs of thrust each. The inlet used on the Phantom is a “Dtype” inlet of area 6.5ft2 mounted on either side of the fuselage and containing variable angle ramps and a boundary layer splitter plate to divert low speed boundary layer flow around the inlet, rather than into it. Some of these details can be seen in the photograph in the text. The inlet can be considered to have nominally two-dimensional flow entering it for the supersonic flight condition shown in Figure E6.4. The configuration shown has one ramp deflected up 10o in order to position the oblique shock generated by the ramp such that it impinges directly on the cowl lip. The back pressure is adjusted so that the supersonic flow in region “a” shocks down through a normal shock wave, as shown on the diagram and passes into region “b” and then on through the duct to the exit station 2.

For the conditions upstream of the inlet shown on Figure E6.4 calculate the following: (a) the angle of the oblique shock wave θ, (b) the Mach number, pressure, stagnation pressure, temperature, and stagnation temperature in region “a”, (c) the weight flow rate through the inlet, (d) the Mach number, pressure, stagnation pressure, temperature, and stagnation temperature in region “b”, (e) the stagnation pressure at inlet station 2 and the pressure recovery of the inlet, (f) the ratio of the area at station 2 to the area at the location of the normal shock that would bring the Mach number at station 2 to M2=0.35, (g) the ram drag of the inlet, Fr Solution (a) The shock wave angle may be found from the oblique shock wave graph from NACA TR 1135 shown in Appendix A or found from the implicit equation relating the flow deflection angle and the upstream Mach number. For a flow deflection δ=10o and an upstream Mach number M0=2 the shock wave angle θ =39.3o.

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1

2

cowl lip M0 =2.0 pt,0 =43.5psia Tt,0 =778R b

a

θ

inlet ramp angle = 10o

fuselage splitter plate

diverted boundary layer flow

Figure E6.4 Schematic diagram of F4 inlet (b) M0sinθ = 1.267 = M0,n is the normal component of the Mach number entering the oblique shock and there •

The stagnation pressure ratio may be read from the tables in Appendix A or calculated from Equation (A.11) in the text: ⎛ 6M 0,2 n ⎞ =⎜ ⎟ pt ,0 ⎜⎝ M 0,2 n + 5 ⎟⎠

pt , a •

3.5

⎛ ⎞ 6 ⎜⎜ ⎟⎟ 2 ⎝ 7 M 0, n − 1 ⎠

2.5

= 0.985

The free stream stagnation pressure is given in Figure E6.4 as pt,0=43.5psia and therefore pt,a = 42.85psia

•

The free stream stagnation temperature is given in Figure E6.4 as Tt,0 = 778R and because the flow is adiabatic Tt,a=778R

•

Equation (6.21) in the text gives the following relation: M 02 sin 2 θ + 5 M sin (θ − δ ) = = 0.6452 7 M 02 sin 2 θ − 1 2 a

2

Knowing θ and δ we find Ma=1.64.

•

The static temperature Ta = Tt,a(1+Ma2/5)-1 = 778/(1+2.69/5) = 505.8R

•

The static pressure pa = pt,0(pt,a/pt,0) (1+Ma2/5)-3.5 = (43.5psia)(0.985)(0.222) = 9.51psia

(c) The mass flow is

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m 0 = p0 A0 M 0

γ RT0

= pt ,0 A0 M 0

γ

⎛ γ −1 2 ⎞ M0 ⎟ ⎜1 + RTt ,0 ⎝ 2 ⎠

−3

−3

⎛ 22 ⎞ ⎛ ⎛ lb ⎞ 1.4 lbm ⋅ ft ⎞ m 0 = ⎜ 6, 264 2 ⎟ ( 6.5 ft 2 ) ( 2 ) ⎜1 + ⎟ ⎜ 32.17 ⎟ 2 ft ⎠ 5⎠ ⎝ lb ⋅ s 2 ⎠ ⎛ ft ⎞ ⎝ ⎝ ⎜ 1716 2 ⎟ ( 778R ) s ⋅R ⎠ ⎝ lbm m 0 = 460 s (d) The Mach number entering the normal shock is Ma=1.64 and therefore •

stagnation pressure: pt,b/pt,a=[6 Ma2/( Ma2+5)]3.5[6/(7Ma2-1)]2.5 = 0.88 and therefore pt,b = 37.7psia

•

stagnation temperature: Tt,b=Tt,0=778R because the flow is adiabatic

•

Mb2 = [(Ma)2+5]/[7(Ma)2-1] = 0.431 so Mb = 0.657

•

Tb = Ttb(1+Mb2/5)-1 = 778/(1+0.431/5) =716R

•

pb = pt,b (1+Mb2/5)-3.5 = 37.7(0.748) = 28.23psia

(e) The stagnation pressure at station 2 is pt,2 = pt,b = 37.7psia because the inlet flow behind the normal shock is assumed to be isentropic. The stagnation pressure recovery is therefore pt,2/pt,0 = 37.7/43.5 = 0.867 (f) The area ratio A2/Ab=(Mb/M2)(1+ M22/5)3(1+ Mb2/5)-3 and we know that Mb=0.657 and M2 is required to be 0.35 so A2/Ab=1.575 (g) The ram drag for one inlet under the stipulated flight conditions where the local sound speed a0=1019ft/s is Fr = m 0V0 =

( 460 lbm s )(1019 ft s )( 2 ) = 29,140lb 32.17 lbm ⋅ ft s 2 ⋅ lb

Note that this is larger than the available maximum gross thrust of 17,880 lbs from one engine. This means that the F4 cannot fly at the given conditions, which correspond to an altitude of 24,000ft. The aircraft must be above 30,500 ft to be able to fly at Mach 2.

Flight tests suggest that the drag coefficient is around 0.040 to 0.045 at altitudes above 35,000 ft. For M0=2 at a 45,000 ft altitude with a nominal drag coefficient of 0.0425 and a wing planform area S=530ft2 the total aircraft drag is ⎛1 ⎞ D = cD q0 S = cD ⎜ γ p0 M 02 ⎟ S = ( 0.0425 )( 0.7 ) ( 309 lb ft 2 ) ( 4 ) ( 530 ft 2 ) = 19, 490lb ⎝2 ⎠

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The ram drag for one inlet at this altitude, where a0=968ft/s, would be

Fr = m 0V0 = ( 309 lb ft 2 )( 6.5 ft 2 ) ( 2 )

1.4 ( 968 ft s )( 2 ) = 11, 240lb (1716 ft s 2 ⋅ R ) ( 390R ) 2

The net thrust with both engines is Fn = 2[(17,888lb)-(11,240lb)] = 13,270lbs, which is still considerably less than the estimated drag. The uncertainty in the calculation here is primarily in the inlet area, which is estimated from small-scale drawings and photographs, and the actual value of the drag coefficient. Careful evaluation of more detailed drawings suggest that the area of 6.5ft2 quoted in the problem statement is much too large and a value for A1=4.7ft2 is more accurate. This inlet area leads to a ram drag of 8,130lbs for one inlet so that the net thrust becomes Fn = 2(17,888-8,130) = 19,520lbs, which now is approximately equal to the total drag, as it should be for equilibrium flight.

Furthermore, the mass flow with this smaller inlet area would be about 187lbm/s which is close to the actual mass flow of 170lbm/s for the J79 engine.

E6.9. The fuselage mounted two-shock D-inlet shown in Figure E6.5 has a height h=0.193m and a width w =0.577m. It is operated at M0=2.5 at an altitude where p0=68.91kPa and T0=250K. The pressure recovery inside the inlet from station y to station 2 is pt,2/pt,y=0.9. For the case of maximum mass flow through the inlet and assuming two-dimensional flow calculate (a) α, (b) M1, (c) p1, (d) T1, (e) the mass flow, (f) A0/A1, (g) My, (h) py, (i) Ty, (j) pt2, and (k) the ram drag Fr.

cowl

2

M0 α

1

y

h

fuselage

Figure E6.5 Fuselage mounted D-inlet with two-shock compression

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Solution (a) Entering the compressible flow chart in Appendix A with a Mach number of 2.5 and a flow deflection of 10o we find α=32o, (b) Using the charts we may find the Mach number behind the oblique shock we may use the oblique shock relation M 12 sin 2 (θ − δ ) = 0.1403M 12 =

M 02 sin 2 θ + 5 = 0.5985 7 M 02 sin 2 θ − 1

From this we find M1=2.066 (c) The normal component of M0 on the oblique shock is M0,n = M0sinα =1.32 and for this normal shock Mach number the pressure ratio p1/p0 =1.866 so that p1 =128.6kPa. The stagnation pressure ratio behind the oblique shock is therefore pt,1=(pt,1/pt,0)pt,0=0.976pt,0.

From the isentropic relations for γ =1.4 we know that pt,0=p0[1+M02/5]3.5=1.177MPa so that pt,1=1.149MPa (d) Similarly, the temperature ratio T1/T0=1.204 so that T1=301K (e) The maximum mass flow occurs when the free stream flow is fully captured in the manner shown in Figure E6.5 so that captured streamtube has an area given by A0 = hw

= 84.5kg/s. = 0.111m2, m = ρ0A0a0M0 = p0A0M0[γ/T0R]1/2 so that m (f) The flow area at station 1 is A1 = wh(cos10)[1-(tan10)/(tanα)] = 0.707wh so that the area ratio A0/A1=1.414. Thus the flow area at the lip of the cowl A1= 0.707A0 = 0.0785m2. (g) The flow behind the oblique shock passes through a normal shock and becomes subsonic. The Mach number behind the normal shock may be found by entering the compressible flow tables with M1 = 2.066 and reading off the value My = 0.567 (h) The pressure behind the normal shock is py = (py/p1)p1 where the tables show py/p1 = 4.81 so that py = 618.6kPa. Then of course pt,y = (pt,y/pt,1)pt,1 = 0.69pt,1 = 793kPa (i) In the same fashion we find Ty/T1 = 1.741 so that Ty = 524K (j) The total pressure at station 2 is pt,2 = (pt,2/pt,y)pt,y = 0.9pt,y and pt,2=714kPa. Note that the flow may be considered to be adiabatic so that the stagnation temperature is constant throughout at Tt,0 = T0[1+M02/5] = 562.5K (k) The ram drag Fr= γp0M02A0= 66.93kN

11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E6.10. Derive the expression for the kinetic energy efficiency of an inlet for constant thermal properties may be given by the following equation: −

γ −1

⎛ pt 0 ⎞ γ −1 ⎜ ⎟ pt 2 ⎠ ⎝ ηke = 1 − γ −1 2 M0 2 Solution

The definition of ηke is the ratio of the kinetic energy at the end of the inlet (after isentropic re-expansion to ambient pressure p0) to the kinetic energy in the free stream, which is given by

ηke =

V2′2 2 V02 2

(E6.1)

This may be written as follows: h0′ −1 h0′ − h0 h0 η KE = 1 − = 1− ht 0 ht 0 − h0 −1 h0

(E6.2)

The process and the variables are shown on the h-s diagram in Figure E6.6. Then, under the assumption that cp is constant we have for the denominator of the fraction on the right hand side of Equation (E6.2) the following: ht 0 T γ −1 2 −1 = t0 −1 = M0 2 h0 T0

(E6.3)

The numerator of the fraction in Equation (E6.2) involves h0 and h0’ and both are evaluated at the same pressure, that is, p0. For constant cp the ratio of the two may be written as γ −1

T0′ ⎛ p0 ⎞ γ γ −1 ⎜ ⎟ p ⎛ ⎞ ′ h0 Tt 0 ⎝ t 2 ⎠ pt 0 γ = = = ⎜ ⎟ γ −1 h0 T0 pt 2 ⎠ ⎝ γ ⎛ p ⎞ Tt 0 ⎜ 0 ⎟ ⎝ pt 0 ⎠

12

(E6.4)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

h

pt0 pt2 2

ht0

2

p0 V2’ /2

V02/2 h0’ h0

0’ 0

S

Figure E6.6 A typical h-s diagram for inlet flow

Substituting Equations (E6.3) and (E6.4) into Equation (E6.2) yields γ −1

⎛ pt 0 ⎞ γ ⎜ ⎟ −1 pt 2 ⎠ ⎝ η KE = 1 − γ −1 2 M0 2

One might also have noted that along the line p0 = constant Tds = dh and therefore Δs =

cp[ ln(T0’/T0)]. However, Equation (6.14) in the text shows that Δs = -R[ln(pt2/pt0)].

Combining these two results for the entropy change yields ln(T0’/T0)= -(R/cp)[ln(pt2/pt0)] and since R/cp=(γ -1)/γ the same result for Equation (E6.4) is obtained and the derivation is complete.

E6.11. Derive an expression for the contraction ratio Amin/A1 for a started internal contraction inlet as a function of Mach number M0 and present a plot of the result. Prove that as M0 grows without bound the limiting value of the contraction ratio is 1.666 for air where γ=1.4. Indicate this value on your graph.

Solution

There are several ways to approach this problem. To prove the limiting value case one may consider that in order for the inlet to start the shock must be brought to the lip of the

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual inlet and the subsequent contraction must be suitable for bringing the subsonic flow behind that shock up to M=1 at the minimum section. For a constant value of γ the shock relations in Appendix A show, in Equation (A.4), that the Mach number behind a normal shock may be written as M1 =

(γ − 1) M 02 + 2 2γ M 02 − ( γ − 1)

Here we are using the subscript 1 to denote conditions behind the shock standing at the lip, that is, station 1. Taking the limit of this equation as the flight Mach number M0 grows large, we see that M1 becomes independent of M0 and approaches a constant value given by M1 =

γ −1 2γ

For γ=7/5=1.4 M1 approaches 7-1/2=0.378 and A/A* in Equation (5.25) becomes 3

A1 ⎛6⎞ = 7 ⎜ ⎟ = 1.667 A1 * ⎝7⎠

To plot the contraction ratio we first note that across the normal shock at the lip of the inlet pt ,0 A0 * = pt ,1 A1* = pt ,1 Amin

Here the critical area behind the shock A1* is equal to the minimum area Amin where we expect the Mach number to become unity. At the same time, with the shock at the lip of the inlet we also have A1=A0. Then the contraction ratio A p A1 A1 = = 0 t ,1 p Amin A0 * pt ,0 t ,0 A0 * pt ,1

Using Equations (A.8) and (5.25) this becomes, for γ=1.4 2 3 ⎞ 36 M 02 A1 1 ⎛ 5 + M0 ⎞ ⎛ ⎜ ⎟ = ⎜ ⎟ Amin M 0 ⎝ 6 ⎠ ⎜ ( 7 M 02 − 1)( M 02 + 5 ) ⎟ ⎝ ⎠

7/2

⎛ 7 M 02 − 1 ⎞ ⎟ ⎜ 6 ⎠ ⎝

The inverse of this equation is plotted in Figure E6.7 along with the related pressure recovery.

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1.0 0.8 Amin/A0 0.6 pt,2/pt,0 0.4 0.2 0.0 1

2

3

4

5

6

M0

Figure E6.7 The contraction ratio for a started internal compression inlet E6.12 A supersonic inlet is being designed for an aircraft using the configuration shown in Figure E6.8, i.e., a 3 shock inlet (2 oblique shocks and one normal shock). Assuming that the flow is two-dimensional, (a) find the proper length of the δ =10o wedge, xs, such that the flow picture is as shown, with no additional shocks appearing; (b) find the pressure and temperature in each of the 4 regions; (c) find the stagnation pressure recovery pt,4/pt,1; and (d) briefly describe how might you redesign the wedge portion of the inlet so as to improve the stagnation pressure recovery, pt,4/pt,1.

2

3

4 0.5 m

M1=3 P1=10 kPa T1=225 K

10o

xs

Figure E6.8 Three shock inlet schematic diagram Solution

(a)First we find the orientation of the first shock which emanates from the leading edge of the 10o wedge. The pertinent dimensions and angles are defined in Figure E6.9.

Entering the oblique shock graph in Appendix A with the Mach number M1=3 and wedge 15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual angle δ=10o we find the shock angle to be approximately θ2 =27.4o.The normal component of the free stream Mach number is M1,n=M1sinθ2=1.38.

3 θ2 M1=3 p1=10 kPa T1=225 K

δ=10o

2 δ=10o

θ3

4 0.5 m

y

z xs

Figure E6.9 Schematic diagram of three shock inlet showing pertinent

Using this value we can enter the normal shock tables in Appendix A to find the conditions behind the shock in region 2: M2,n=0.7483, p2/p1=2.055, T2/T1=1.242, and

pt,2/pt,1=0.9630. Then the Mach number in region 2 is M2=M2,n/sin(θ2-δ)=2.5. To enter region 3, the flow at M2=2.5 goes through another compressive turn of 10oand the oblique shock chart shows that θ3=32o. Then the Mach number normal to and upstream of the second shock is M2,n=2.5sinθ3=1.32. For this Mach number we have p3/p2 =1.866,

T3/T2=1.204, and pt,3/pt,2=0.9758 with the normal component of the Mach number behind the shock as M3,n=0.7760. Then the Mach number in region 3 is M3=M3,n/sin(θ3-δ)=2.07.

The conditions across the last, normal, shock wave with an upstream Mach number

M3=2.07 are M4=0.5659, p4/p3 =4.832, T4/T3=1.745, and pt,3/pt,2=0.6882. To get the second shock on the shoulder with no other waves we must have, using the terminology on the sketch, the following:

z = 0.5cot θ 2

y = ( 0.5 − xs tan δ ) cot (θ3 − δ ) xs = y + z = ( 0.5 − xs tan δ ) cot (θ3 − δ ) + 0.5cot θ 2 From this equation we find xs=1.53m (b) Using the property ratios we have already calculated we may find the pressure and temperature in each of the 4 regions and they will be constant in each of the regions.

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual p2 = (p2/p1)p1= (2.055)(10kPa)=20.55kPa T2 = (T2/T1)T1= (1.242)(225K) = 279.5K p3 = (p3/p2)p2= (1.866)(20.55kPa)=38.35kPa T3 = (T3/T2)T2= (1.204)(279.5K) = 336.5K p4 = (p4/p3)p3= (4.832)(38.35kPa)=185.3kPa T4 = (T4/T3)T3= (1.745)(336.5K) = 587.2K (c) The stagnation pressure ratio across a shock depends upon the normal component of Mach number and is given by Equation (A.11). It is often more convenient to instead use the normal shock tables in Appendix A. We have calculated the stagnation pressure ratios across all the shocks so that the overall pressure recovery is ⎛ p ⎞⎛ p ⎞⎛ p ⎞ = ⎜ t ,4 ⎟ ⎜ t ,3 ⎟ ⎜ t ,2 ⎟ = ( 0.6882 )( 0.9758 )( 0.9630 ) = 0.6467 pt ,1 ⎜⎝ pt ,3 ⎟⎠ ⎜⎝ pt ,2 ⎟⎠ ⎜⎝ pt ,1 ⎟⎠

pt ,4

(d) The MIL-E-5008B specification suggests that the inlet pressure recovery should be

pt ,4 pt ,1

= 1 − 0.075 ( M 1 − 1)

1.35

= 0.808

The current design is almost 20% below the suggested value. It is obvious that the Mach number ahead of the terminal normal shock is the major contributor to the pressure recovery. Reducing that Mach number will increase the pressure recovery and one modification that would help would be to alter the wedge as shown in Figure E6.10. The introduction of another oblique shock reduces the Mach number without large losses in stagnation pressure.

3 M1=3 P1=10 kPa T1=225 K

4 0.5 m

2 10o

xs

Figure E6.10 Schematic diagram of a four shock inlet that is likely to perform

better than the original three shock inlet

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E.6.13 A conventional axisymmetric ramjet with a simple converging-diverging inlet shown in Figure E6.11 is flying at a supersonic speed M0 at an altitude, z0=50,000ft where p0=1.69psia and T0=390R. The ratio of specific heats may be assumed to be constant at γ =1.4, the heating value of the fuel is Qf=18,500Btu/lb, and the burner efficiency is ηb=095. For the conditions listed for the three cases listed in Table E6.1 carry out the following investigations: Case 1 If the inlet is operated at maximum pressure recovery, find (a) inlet minimum area Amin, (b) inlet area A1, (c) nozzle throat area A6, (d) net thrust Fn if A7/A3=1.25, and (e) net thrust Fn,max, that is, net thrust for p7=p0.

Case 2 If the inlet is operated at maximum pressure recovery, find (a) the net thrust Fn and (b) the specific fuel consumption cj.

Case 3 In this case the only change from Case 2 is that stagnation temperature at the exit of the constant area burner is now 3500R instead of 2500R. Table E6.1 Three cases for Problem E6.13

Case M0 Ms 1 2 3

M3 A1/Amin A3 (ft2) 2.2 2.2 0.3 1.0 2.5 1.26 2.5 1.26

1

min

A6 (ft2) 1.0 1.0

2,3

A7/A6 M5 A5 (ft2) 0.7 1.0 2.0 2.0

4,5

p0 (psia) 1.69 1.69

6

T0 (R) 390 390

Tt,4 (R) 2500 3500

7

Figure E6.11 Cross-section of an axisymmetric ramjet showing the various stations. Note that the absence of a compressor causes stations 2 and 3 to be equivalent while absence of a turbine makes stations 4 and 5 the same.

Solution

Case 1: (a) If the inlet operates at maximum pressure recovery, then Ms=M0. At the starting Mach number Ms=2.2 the shock is standing at A1 and the Mach number behind the shock at A1 is M1,y=0.5471 while the corresponding A1/A*=1.259=A1/Amin, the inlet contraction ratio. For maximum pressure recovery the shock is drawn inside the inlet to stand at the inlet minimum section Amin. The inlet contraction ratio just determined

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual yields a Mach number at the minimum section of Mmin=1.932 with a pressure recovery

pt,3/pt,0=0.753 and a Mach number behind the shock Mmin,y=0.590. Then, given that M3=0.3 we find ⎛ A ∗ ⎞ 1.200 Amin ⎛ A ⎞ =⎜ = 0.59 ⎜ ⎟= ⎟ 2.035 A3 ⎝ A ∗ ⎠ M min ′ ⎝ A3 ⎠ Then, since A3=1ft2, we find the minimum area to be Amin=0.59ft2. (b) We may find the inlet area from A1=(A1/A*)Amin = 1.259Amin = 1.259(0.59ft2) = 0.743ft2. (c) In the nozzle, we are given that M5=0.7 and we know that the nozzle should be choked so that M6=1, therefore A5/A6=A5/A*=1.0944. Then the nozzle throat area

A6=A5/1.0944=0.914ft2. (d) For the situation where the area ratio (A7/A3) is stipulated to be 1.25 we may find the nozzle expansion ratio A7/A6=(A7/A3)(A3/A6)=1.25(1ft2/0.914ft2)=1.368=A7/A*. Therefore

M7=1.73, A7=1.25ft2, and p7/pt,7=0.194. In the constant area combustor we have p5 1 + γ M 32 1.126 = = = 0.668 p3 1 + γ M 52 1.686 pt ,5 pt ,3

=

pt ,5 p5 p3 p5 p3 pt ,3

=

1 ( 0.668)( 0.940 ) = 0.87 0.721

The nozzle flow is assumed to be isentropic and therefore the stagnation pressure is constant,

pt,5=pt,7.

The

free

stream

stagnation

pressure

is

pt,0=(pt,0/p0)p0=

(1/0.0935)(1.69psia) =18.07psia = 2603psf. Then we may find the nozzle exit pressure as pt ,7 = pt ,5 =

⎛ lb ⎞ lb pt ,0 = 0.87 ( 0.753) ⎜ 2603 2 ⎟ = 1705 2 pt ,3 pt ,0 ft ⎠ ft ⎝

pt ,5 pt ,3

The nozzle exit static pressure p7=0.194(1705psf)=331psf=2.297psia. Thus, for this condition the nozzle is under-expanded and the thrust is

Fn = A7 ⎡⎣ p7 (1 + γ M 72 ) − p0 ⎤⎦ − γ p0 A0 M 02 = 618lb (e) If the nozzle exit area were variable we could expand the flow further until p7=p0. In this case p7/pt,7=(1.69x144)/1705=0.142 so that M7=1.93, A7/A*=1.593 and A7=1.46ft2.

The net thrust is then simply

Fn,max = γ p7 A7 M 72 − γ p0 A0 M 02 = 628lb

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Note that in this case the net thrust for a matched nozzle is not much greater than the net thrust for the fixed nozzle with A7/A3=1.25.

Case 2: (a) For maximum pressure recovery the shock is drawn inside the inlet to stand at the inlet minimum section Amin. The given inlet contraction ratio A1/Amin=1.26 while the area ratio at the free stream Mach number M0=2.5 is A1/A*=2.637. Therefore the critical area ratio at the inlet minimum section is

Amin ⎛ Amin ⎞ ⎛ A1 ⎞ 2.637 =⎜ = 2.092 ⎟⎜ ⎟= A ∗ ⎝ A1 ⎠ ⎝ A ∗ ⎠ 1.26 This area ratio yields a Mach number at the minimum section of Mmin=2.25 with a pressure recovery pt,3/pt,0=0.6055 and a Mach number behind the shock Mmin,y=0.5406.

The free stream stagnation pressure is pt,0=p0(pt,0/p0)=1.69psia/(0.05853)=28.87psia. The total temperature is constant so that Tt,3=Tt,0. The free stream stagnation temperature is

Tt,0=T0(Tt,0/T0)=390R/(0.4440)=878R. The fuel to air ratio is given by the combustor energy balance

T −T 2500 R − 878R f = t ,4 t ,3 = = 0.0223 18,500 Btu lbm Qf a − 2500 R ηb − Tt ,4 0.95 0.24 Btu lbm ⋅ R cp In order to find the fuel flow rate we may first determine the total weight flow rate which is given by the choked flow at the nozzle throat:

w 6 = g

pt ,6 A6 Tt ,6

−

γ +1

γ ⎛ γ + 1 ⎞ 2(γ −1) ⎜ ⎟ R⎝ 2 ⎠

=

pt ,6 (1 ft 2 )( 32.17 ft s 2 ) 2500 R

1.4 −3 (1.2 ) 2 2 1716 ft s ⋅ R

w 6 = ( 0.0106 ft 2 s ) pt ,6 Then all that is necessary is the stagnation pressure at the nozzle throat which may be written as pt ,6 =

pt ,6 pt ,5 pt ,3 pt ,5 pt ,3 pt ,0

pt ,0 = (1)(1)( 0.6055 )( 28.87 psia ) = 17.48 psia

Note that we assume that the stagnation pressure loss in the constant area burner is negligible, though in practice it is generally at least several percent. Then the weight flow through the throat is w 6 = ( 0.0106 ft 2 s )(17.48 lb in 2 )(144 in 2 ft 2 ) = 26.68 lb s and the

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual fuel flow is 0.0223(26.68lb/s) = 0.595lb/s. The mass flow through the nozzle is constant so we may write the net thrust as Fn =

w 7 26.68 lb s V − V0 ) + ( 2 ft 2 ) ( p7 − 243.4 lb ft 2 ) (V7 − V0 ) + A7 ( p7 − p0 ) = 2 ( 7 32.17 ft s g

Thus we need to know V0, V7, and p7. In the free stream the velocity is V0 = a0 M 0 = 2.5 (1.4 )1716 ft 2 s 2 ⋅ R ( 390 R ) = 2419 ft s The nozzle expansion ratio was given as A7/A6=2.0 so the exit Mach number is M7=2.2 and p7/pt,7=0.09352, while T7/Tt,7=0.5081. The nozzle stagnation conditions are pt,7 = pt,5 = 17.48psia and Tt,7 = Tt,5 =2500R so p7=1.635psia and T7=1270R. Then a7=1747ft/s and V7=3843ft/s. Finally Fn = ( 0.829 lb ⋅ s ft )( 3843 ft s − 2419 ft s ) + ( 2 ft 2 )( 235.4 lb ft 2 − 243.4 lb ft 2 ) Fn = 1180lb − 16lb = 1164lb

Note that the nozzle is almost exactly matched in this case. (b) The specific fuel consumption is then cj =

w f Fn

=

0.59

lb s × 3600 s hr = 1.82hr −1 1164lb

Case 3: In this case the only change from Case 2 is that stagnation temperature at the exit of the constant area burner is now 3500R instead of 2500R. (a) The fuel to air ratio is given by the combustor energy balance

T −T 3500 R − 878R f = t ,4 t ,3 = = 0.0376 Qf 18,500 Btu lbm a − 3500 R ηb − Tt ,4 0.95 0.24 Btu lbm ⋅ R cp In order to find the fuel flow rate we may first determine the total weight flow which is given by the choked flow at the nozzle throat: − ( γ +1)

w 6 = g

pt ,6 A6 Tt ,6

γ ⎛ γ + 1 ⎞ 2(γ −1) ⎜ ⎟ R⎝ 2 ⎠

=

pt ,6 (1 ft 2 )( 32.17 ft s 2 ) 3500 R

1.4 −3 (1.2 ) 2 2 1716 ft s ⋅ R

w 6 = ( 0.00899 ft 2 s ) pt ,6

Note that we assume that the stagnation pressure loss in the constant area burner is negligible, though in practice it is generally at least several percent. Then the weight flow

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual through the nozzle throat is w 6 = ( 0.00899 ft 2 s )(17.48 lb in 2 )(144 in 2 ft 2 ) = 22.63 lb s and the fuel flow is 0.0376(22.55lb/s) = 0.847lb/s. The mass flow through the nozzle is constant so we may write the net thrust as Fn =

w 7 22.55 lb s V − V0 ) + ( 2 ft 2 )( p7 − 243.4 lb ft 2 ) (V7 − V0 ) + A7 ( p7 − p0 ) = 2 ( 7 32.17 ft s g

The nozzle expansion ratio was given as A7/A6=2.0 so the exit Mach number is M7=2.2 and p7/pt,7=0.09352, while T7/Tt,7=0.5081. The nozzle stagnation conditions are pt,7 = pt,5 = 17.48psia and Tt,7 = Tt,5 =3500R so p7=1.635psia and T7=1778R. Then a7=2067ft/s and V7=4547ft/s. Finally

⎛ lb − s ⎞ ⎛ ft ft ⎞ lb lb ⎞ 2 ⎛ Fn = ⎜ 0.703 ⎟ ⎜ 4547 − 2420 ⎟ + ( 2 ft ) ⎜ 235.4 2 − 243.4 2 ⎟ ft ⎠ ⎝ s s⎠ ft ft ⎠ ⎝ ⎝ Fn = 1496lb − 16lb = 1480lb (b) The specific fuel consumption is then cj =

w f Fn

=

0.847 lb s × 3600 s h = 2.06hr −1 1480lb

We see that the increase in burner exit temperature increases the thrust by about 27% but the increase in specific fuel consumption is only about 10%.

E6.14 Air may be drawn from the local atmosphere, which is at standard temperature and pressure, into a very large vacuum tank through either of two smooth inlets shown in Figure E6.12. The area distribution for each case is listed in the figure by (a) and (b). A normal shock is detected at area A3 =3. What is the pressure in the tank?

Solution

(a) Assuming that the flow through the diverging inlet duct shown as case (a) is frictionless and isentropic we expect the area A1=1=A* and therefore A3/A*=3/1=3. Then we may use isentropic flow tables or equations to determine that the Mach number at station 3, just ahead of the point where the shock is located, is M3 =2.64. Across the shock at this Mach number the stagnation pressure ratio is 0.445. The stagnation pressure just upstream of the shock is equal to the atmospheric pressure, pt,1=1atm, and is constant right up to the shock at station 3. The lower value of stagnation pressure, just downstream

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual of the shock, is constant all the way into the tank. Because the air in the large tank is essentially at rest, this is also the static pressure of the tank. Therefore, ptank = pt,3 = (0.445)(1atm) = 0.445atm. (a) A1 = 1 A2 = 2 A3 = 3 A4 = 4

A1

A4

A4

tank

tank

(b) A1 = 3 A2 = 2 A3 = 3 A4 = 4

A3 A2

A2 A3

A1

Figure E6.12 Schematic diagram of a large reservoir tank which may accept air through either a diverging inlet (a) or a converging-diverging inlet (b). The crosssectional area distribution of each inlet is noted on the diagram

(b) When the inlet is a converging-diverging inlet with Amin=A2=2, then A3/A*=3/2=1.5.

Again we may use the isentropic flow tables or equations to determine that the Mach number at station 3, just ahead of the point where the shock is located, is M3 =1.86. At this Mach number the stagnation pressure ratio across the shock is 0.785. The upstream stagnation pressure is equal to the atmospheric pressure, pt,1=1 atm, and is constant up to the shock at station 3, and the stagnation pressure behind the shock is constant all the way into the tank, and since the large tank is essentially quiescent, this is the pressure of the tank. Therefore, ptank = pt,3 = (0.785)(1atm) = 0.785atm.

E6.15 The measurement of free stream stagnation and static pressure is crucial to the air data system of an aircraft, as described in Appendix F. The Pitot-static tube is a simple and rugged instrument used for measuring the Mach number M1 of a supersonic air stream (γ =7/5). Calibration of this probe has shown that the pressure coefficient at the static holes is 0.1. Considering the measured pressures indicated on the diagram in Figure E6.13, calculate the free stream Mach number M1.

23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

M1 20.7kPa static holes 138kPa

Figure E6.13 Pitot-static tube with measurements at the Pitot port and the static port indicated on the diagram Solution

The pressure in the stagnation tube of the probe is the stagnation pressure behind a normal shock. For γ=7/5, the so-called Rayleigh Pitot equation is ⎛ 6 M 12 ⎞ =⎜ ⎟ p1 ⎝ 5 ⎠

pt ,2

7 2

⎛ ⎞ 6 ⎜ ⎟ 2 ⎝ 7M1 − 1 ⎠

52

=

138 p1

The pressure measured at the static taps is defined by p − p1 2 ⎛ 20.7 ⎞ Cp = 2 = − 1⎟ = 0.1 ⎜ γ M 12 ⎝ p1 q1 ⎠ Eliminating p1 between the two expressions yields ⎛ 6 M 12 ⎞ ⎜ ⎟ ⎝ 5 ⎠

72

⎛ ⎞ 6 ⎜ ⎟ 2 ⎝ 7M1 −1 ⎠

52

⎛ 7 M 12 ⎞ = 6.667 ⎜1 + 0.1 ⎟ 10 ⎠ ⎝

This equation may be solved by simple iteration as follows; M1 1.7 2.7 2.75 2.8

pt,2/p1 [r.h.s of the equation] 8.015 10.069 10.212 10.325

pt,2/p1 [l.h.s of the equation] 4.223 9.863 10.196 10.569

From the two columns it is seen that the value M1=2.75 is close enough to satisfy our equations.

24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7. TURBOMACHINERY E7.1 An axial flow compressor for a jet engine is operating on a test stand under standard sea level atmospheric conditions. The pressure ratio of the compressor is pt,3 /pt,2=10 and it processes a mass flow rate of 45.3kg/s. The dimensions of the compressor are shown in Figure E7.1. Assuming that all processes are isentropic and p1=97kPa determine the force F experienced by the load cell on the test stand, the power P required to drive the compressor, and the temperature leaving the compressor T3.

1

2

3

38.1cm

25.4cm

33cm

Load cell Test stand

Figure E7.1 Schematic diagram of axial flow compressor on test stand

Solution (a) The areas A1= πr12 = 0.456m2 and A3 = π(r3a2-r3,b2) = 0.139m2. Because the test stand is stationary the stagnation pressure and temperature are the standard day values pt,1==101.kPa and Tt,1=288K. We are given p1=97kPa so we know that p/pt,1=0.9576 and using this in the isentropic relation yields M1=0.25. The static temperature T1=288K(1+.252/5)-1=284.4K. The speed of sound a1=(γRT1)1/2 = 338m/s. Therefore the velocity at station 1 is V1=a1M1=84.5m/s. Conservation of mass requires that

ρAV=constant=45.3kg/s. This may be written as p1 A1M 1

γ RT1

= p3 A3 M 3

1

γ RT3

= 45.3 kg s

Theory of Aerospace Propulsion, Second Edition: Solutions Manual We note that pt,1=pt,2 and Tt,1=Tt,2 so that the above equation becomes 12

p1 pt ,2 ⎛ pt ,2 ⎞ M 1 ⎛ T3 Tt ,3 ⎞ ⎜ ⎟ ⎜ ⎟ p3 pt ,3 ⎜⎝ pt ,3 ⎟⎠ M 3 ⎜⎝ T1 Tt ,1 ⎟⎠

12

⎛ Tt ,3 ⎞ ⎜⎜ ⎟⎟ ⎝ Tt ,2 ⎠

=

A3 = 0.3048 A1

Because all processes are said to be isentropic we know that p~ργ~T(γ-1)/γ and if γ=7/5 then p~ρ7/5~T2/7 and the equation becomes ⎛1⎞ ⎜ ⎟ ⎝ 10 ⎠

67

−3

M 1 ⎛ 5 + M 12 ⎞ ⎜ ⎟ = 0.3048 M 3 ⎝ 5 + M 32 ⎠

Using M1=0.25 in the equation above we find M3=0.11. The stagnation pressure pt,3=10pt,2=1013kPa and p3=1013kPa(1+0.112/5)-3.5=1004.5kPa. The stagnation pressure ratio is Tt,3/Tt,2=(pt,3/pt,2)2/7=1.9307 and therefore Tt,3=1.9307(288K)=556K and T3=556K(1+0.112/5)-1=554.6K. Then a3=472.1m/s and V3=51.9m/s. Applying the conservation of momentum to the fluid in a control volume between stations 1 and 3 yields

p1 A1 + ( −m 1 ) V1 + F − p3 A3 + m 3V3 = 0 Then solving for F yields

F = p3 A3 − p1 A1 + m (V1 − V3 ) 2 2 kg ⎞ ⎡⎛ m⎞ ⎛ m⎞ ⎤ ⎛ F = (1004.5kPa ) ( 0.139m ) − ( 97kPa ) ( 0.456m ) + ⎜ 45.3 ⎟ ⎢⎜ 51.9 ⎟ − ⎜ 84.5 ⎟ ⎥ s ⎠ ⎣⎢⎝ s⎠ ⎝ s ⎠ ⎦⎥ ⎝ 2

2

F = 95.39kN − 1.48kN = 93.91kN This force on the fluid is acting to the right so the force on the load cell is Fcell=93.91kN to the left; therefore it is a thrust force. If we form the difference of the impulse function at stations 1 and 5 we find I 3 − I1 = p3 A3 (1 + γ M 32 ) − p1 A1 (1 + γ M 12 ) 2 2 I 3 − I1 = (1004.5kPa ) ( 0.139m 2 ) ⎡1 + 1.4 ( 0.11) ⎤ − ( 97 kPa ) ( 0.456m 2 ) ⎡1 + 1.4 ( 0.25 ) ⎤ ⎣ ⎦ ⎣ ⎦ I 3 − I1 = 141.39 N − 48.1N = 93.29 N Thus the force F = I3-I2 = 98.85 is the same, within the accuracy of the calculations, as

the force obtained by using the control volume approach. (b) The energy equation for adiabatic flow through the compressor is

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 1 1 h1 + V12 + WC = h3 + V32 2 2

Assuming cp=1.004kJ/kg/K the work done by the compressor is 2 2 1 2 1 ⎡⎛ m ⎞ ⎛ m⎞ ⎤ 2 WC = ( h3 − h1 ) + (V3 − V1 ) = c p (T3 − T1 ) + ⎢⎜ 51 ⎟ − ⎜ 84.5 ⎟ ⎥ 2 2 ⎢⎣⎝ s ⎠ ⎝ s ⎠ ⎥⎦

⎛ m 2 ⎞ ⎛ kJ / kg ⎞ kJ WC = (1.004 kJ kg ⋅ K )( 554.6 K − 284.5K ) − ⎜ 2270 2 ⎟ ⎜ 3 2 2 ⎟ = 268.9 s ⎠ ⎝ 10 m / s ⎠ kg ⎝ C = 12.18MW Then the power expended is P = mW

(c) The temperature after isentropic compression was already determined in part (a) as T3=554.7K.

E7.2 A turbojet engine is operated under standard sea level conditions and the following measurements are made at the compressor exit: stagnation temperature equals 216C and stagnation pressure equals 482kPa. The airflow through the compressor is determined to be 27.2kg/s. Determine: (a) the work input required per unit mass of air, (b) the power required, (c) the adiabatic compression efficiency of the compressor, and (d) the torque exerted on the fluid at a rotational speed of 12,000rpm.

Solution

(a) The engine is being run up on the ground so M0=0 and therefore pt,2=101.3kPa and Tt,2=288K. Assuming a constant specific heat for air, cp=1.004kJ/kg-K, the work required by the compressor is Wc=cp(Tt,3 – Tt,2)=(1.004kJ/kg-K)(489K – 288K)=201.8kJ/kg (b) The power required is c = 27.2kg/s(201.8kJ/kg) = 5.489MW P = mW

(c) Assuming that γ =1.4 the adiabatic efficiency of compression is

ηc = [(pt,3/pt,2)(γ-1)/γ – 1][(Tt,3/Tt,2) – 1]-1 = [(482/101.3)2/7 – 1][(489/288) – 1]-1 = 0.805 ηc=80.5% (d) The torque exerted on the fluid is Tq = P /ω = (5,489kN-m/s)/[2π(12,000rpm)/60s-1] = 4.368kN-m

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7.3 A compressor operates with a stagnation pressure ratio of 4 and an inlet stagnation temperature of 5C. If the exit stagnation temperature is 171C, what is the adiabatic efficiency of compression?

Solution

Assuming a constant value for γ =1.4, the adiabatic efficiency of compression may be written as

ηc = [(pt,3/pt,2) (γ-1)/γ – 1][(Tt,3/Tt,2) – 1]-1 = [(4)2/7 – 1][(444/278) – 1]-1 = 0.814 or 81.4%

E7.4 Axial flow compressors are characterized by flow passages whose radial coordinates change little over the length of the machine while centrifugal compressors have blade passages that start near the axis of rotation and increase substantially in radius by the time the exit of the machine is reached. Consider standard sea level air entering a centrifugal flow compressor with the following conditions: r2 =6in, c2 =300ft/s, and α2 =70o. The air leaves the rotor with r3 =18in, c3 =1200ft/s, and α3 =25o.

The rotational speed of the compressor is 9000rpm and it processes 32.17lbm/s of air.

Carry out the following: (a) draw the combined entrance and exit velocity diagrams to scale, (b) determine the entrance and exit blade angles β2 and β3, (c) determine the torque required to drive the compressor, (d) determine the horsepower required, (e) find the ideal pressure head, (f) determine the contribution to the pressure rise provided by the external, internal, and centrifugal effects, (g) determine the whirl velocity Δcu, (h) determine the ideal pressure ratio of the compressor Solution

(a)The velocity triangles for the centrifugal compressor are shown in Figure E7.2.

c3=1200ft/s

w2=464ft/s c2=300ft/s

α2=70o β2=37o

α3=25o

β3=57.3o

u2=471ft/s

w3=603ft/s

u3=1414ft/s

Figure E7.2 Velocity diagrams for the centrifugal compressor of exercise E7.4

(b) From the diagram we can find c2,a = c2sinα2 = (300ft/s)sin70 = 281.9ft/s and c3,a = c3sinα3 = (1200ft/s)sin25 = 507.1ft/s. Similarly, the tangential components of the absolute velocities are c2,u = c2cosα2 = (300ft/s)cos70 = 102.6ft/s

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual c3,u = c3cosα2 = (1200ft/s)cos25 = 1088ft/s.

The linear velocities of rotation are u2 = (2πN/60)r2 = (942.5s-1)(0.5ft) = 471.2ft/s u3 = (2πN/60)r3 = (942.5s-1)(1.5ft) =1414ft/s.

The relative speeds are w2=[(u2-c2,u)2+c2,a2]1/2=[(471.2-102.6)2+281.92]1/2=464.0ft/s w3=[(u3-c3,u)2+c3,a2]1/2=[(1414-1088)2+507.12]1/2=602.8ft/s The rotor blade angles are

β2 = arcsin(c2,a/w2) = arcsin(281.9/464) = 37.41o β3 = arcsin(c3,a/w3) = arcsin(507.1/602.8) = 57.27o (c)The required torque Tq = m (r3c3,u-r2c2,u) Tq= [(32.17lbm/s)/(32.17lbm-ft/s2-lb)][(1.5ft)(1088ft/s)-(0.5ft)(102.6ft/s)] Tq= 1581ft-lbs (d) The power required by the compressor is P = Tqω = (1581ft-lbs)(942.5s-1)=1.49x106ft-lb/s(778ft-lb/Btu)-1=1,915Btu/s P=(1.49x106ft-lb/s)/(550ft-lb/s/hp) = 2709hp (e) The ideal pressure head is written in terms of the velocity components as follows: H= (1/2g)[(c32-c22)+ (u32-u22)+ (w22-w32)] H= (2x32.17ft/s2)-1[(135.0x104+177.7x104-14.81x104)ft2/s2] H= 46,310ft-lb/lb (=1.67atm or 24.6psia for air at standard conditions) (f) The contribution to the ideal pressure head due to external effect = (1/2g)(c32-c22) = 20,980ft (45.3%) internal effect = (1/2g)(w22-w32) = -2,301ft (-5%) centrifugal effect = (1/2g)(u32-u22) = 27,620ft (59.7%) (g)Whirl velocity Δcu=c3,u – c2,u = 1088ft/s-102.6ft/s = 985.4ft/s (h) The pressure ratio pt,3/pt,2 = [(ηcWc/cpTt,2)+1]γ/(γ−1) and the work done is WC =

P 1,915 Btu s = = 59.53 Btu lbm m 32.17 lbm s

The stagnation temperature Tt2=T2+c22/2cp= 520R+(300ft/s)2/2(6006ft2/s2-R)=527.5R The ideal pressure ratio is

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ⎡ ηW ⎤ = ⎢1 + c C ⎥ pt ,2 ⎣⎢ c pTt ,2 ⎦⎥ pt ,3

γ ( γ −1)

⎡ (1)( 59.54 Btu lbm ) ⎤ = ⎢1 + ⎥ ⎣ ( 0.24 Btu lbm ⋅ R )( 527.5 R ) ⎦

3.5

= 3.844

E7.5 Repeat problem 4, changing only the entrance radius to r2=r3=18in, so that the machine now acts as an axial flow compressor.

Solution

(a) The velocity triangles for the axial compressor are shown in Figure E7.3.

w2=1341ft/s c3=1200ft/s

c2=300ft/s w3=603ft/s α2=70o

α3=25o

β3=57.3o

β2=12.2o u2=1414ft/s

u3=1414ft/s

Figure E7.3 Velocity diagram for the axial compressor of exercise E7.5

(b) From the diagram we can find c2,a = c2sinα2 = (300ft/s)sin70 = 281.9ft/s and c3,a = c3sinα3 = (1200ft/s)sin25 = 507.1ft/s. Similarly, the tangential components of the absolute velocities are c2,u = c2cosα2 = (300ft/s)cos70 = 102.6ft/s c3 u = c3cosα2 = (1200ft/s)cos25 = 1088ft/s.

The linear speeds of rotation are u2 = (2πN/60)r2 = (942.5s-1)(1.5ft) = 1414ft/s u3 = (2πN/60)r3 = (942.5s-1)(1.5ft) = 1414ft/s.

The relative speeds are w2 = [(u2-c2,u)2+c2,a2]1/2 = [(1414-102.6)2+281.92]1/2 =1341ft/s w3 = [(u3-c3,u)2+c3,a2]1/2 = [(1414-1088)2+507.12]1/2 = 602.8ft/s The rotor blade angles are

β2 = arcsin(c2,a/w2) = arcsin(281.9/1341) = 12.14o and β3 = arcsin(c3,a/w3) = arcsin(507.1/602.8) = 57.27o 6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (c)The required torque Tq = m (r3c3,u-r2c2,u) = Tq=[(32.17lb/s)/(32.17lbm-ft/s2-lb)][(1.5ft)(1088ft/s)-(1.5ft)(102.6ft/s)] Tq=1,478ft-lb (d) The power required is P = Tqω = 1478ft-lb(942.5s-1)=1.393x106ft-lb/s=1.393x106ft-lb/s/(778ft-lb/Btu) P=1,790Btu/s P= (1.393x106ft-lb/s)/(550ft-lb/s/hp) = 2,533hp (e) The ideal pressure head is given by H = (1/2g)[(c32-c22)+ (u32-u22)+ (w22-w32)] H= (2x32.2ft/s2)-1[(135.0x104 ft2/s2) + 0 + (143.5x104 ft2/s2)] H = 43,284ft-lb/lb (=1.565atm or 23psia for air at standard conditions) (f) Pressure head due to external effect = (1/2g)(c32-c22) = 20,980ft or 48.5% internal effect = (1/2g)(w22-w32) = 22,302ft or 51.5% centrifugal effect = (1/2g)(u32-u22) = 0 ft (g) Whirl velocity Δcu = c3,u – c2,u = 1088-102.6 = 985.4ft/s (h) Ideal pressure ratio pt,3/pt,2 = [(ηcWc/cpTt,2)+1]γ/(γ−1)and the work done is WC = ⎡ ηW ⎤ = ⎢1 + c C ⎥ pt ,2 ⎢⎣ c pTt ,2 ⎥⎦ pt ,3

P 1790 Btu s = = 55.64 Btu lbm m 32.17 lbm s

γ ( γ −1)

⎡ (1)( 55.64 Btu lbm ) ⎤ = ⎢1 + ⎥ ⎣ ( 0.24 Btu lbm ⋅ R )( 527.5 R ) ⎦

3.5

= 3.58

E7.6 A compressor rotor is operated at 10,000rpm on a test stand where the atmospheric conditions are those of a standard sea level day. The compressor has an axial inlet with a hub diameter of 13.35cm and an eye diameter of 25.40cm. The compressor has a radial exhaust and the rotor tip diameter is 50.80cm. If the average inlet Mach number is 0.7 determine: (a) the mass flow through the compressor, (b) the power required to drive the compressor, (c) the total temperature at the compressor exit, (d) the portion of the required power due to the centrifugal effect, (e) the overall pressure ratio of the machine, if the adiabatic efficiency of compression ηc=85%.

Solution

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual A schematic diagram of the compressor is shown in Figure E7.4. (a) The mass flow through the compressor (ρAca)2. The Mach number at the inlet is M2=0.7 so c2a=a2M2=0.7(γRT2)1/2=0.7[(1.4)(287m2/s2-K)(T2)]1/2.

The compressor is operating on a laboratory test stand so the stagnation pressure and the stagnation temperature entering the compressor assume the standard sea level values.

Thus Tt,2 = 288K and pt,2 =101.3kPa. The static pressure T2 = Tt,2[1+M22/5]-1 = (288K)[1+ (0.7)2/5]-1 = 262.3K. Similarly p2 = pt,2[1+M22/5]-3.5 = (101.3kPa)[1+(0.7)2/5]-3.5 = 73.02kPa Then c2a = 0.7[(1.4)(287m2/s2-K)(262.3K)]1/2 = 227.2m/s The density is obtained from the equation of state ρ = p/RT so that

ρ2 = (73,030N/m2)/[(287 m2/s2-K)(262.3K)] = 0.9694kg/m3 The flow area is A2 = π(reye2 – rhub2) = π[(25.4cm2)-(13.35cm2)]10-4/4 = 0.03667m2 Then the mass flow is

m = (0.97kg/m3)(0.03667m2)(227.2m/s) =8.077kg/s (b)A schematic diagram of the compressor is shown in Figure E7.4. The power required is P = m ( u3 c3u − u2 c2u ) . However, c2u=0 because the flow enters the compressor axially and u3 =c3u because the flow leaves the impeller in the radial direction. The linear speed of rotation at the exit is u3 = (2πN/60) r3 = (2π)(10,000/60)[(0.5)(0.508m)] = 265.4m/s.

Therefore the power 32 = (8.077kg/s) (265.4m/s)2 = 568.9kW P = mu

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual c2a,hub w2hub c2a,eye

axial entry reye

w2eye

w2eye w2,hub

rhub

c2a

u2,hub

u2,eye u2,eye

α2,eye

α2hub

blade leading edge twists radially from α2hub to α2eye

w3=w3r c3

u3 Since w3 is radial, c3u = u3

Figure E7.4 Centrifugal compressor showing hub and eye details

(c) The stagnation temperature at the exit may be found from WC =cp(Tt,3 – Tt,2) or

Tt ,3 = Tt ,2 +

WC 568.9kN ⋅ m P = Tt ,2 + = 288K + = 358.4 K p cp mc (8.077 kg s )(1.004 kN ⋅ m kg ⋅ K )

(d) At the exit u3= 265.4m/s as already calculated. At the eye station of the exit u2,eye = (2πN/60)r2 = (2π)(10,000/60)[(0.5)(0.254m)] =133m/s.

The linear speed of rotation decreases as the hub is approached so we’ll use the eye value just calculated to give a conservative estimate of the fraction of the power arising from the centrifugal effect. Then Pcentrifugal P

1 1 2 2 m ( u32 − u22 ) ( 8.077 kg s ) ⎡⎣( 265.4 m s ) − (133 m s ) ⎤⎦ =2 =2 = 0.374 571.7 kJ s P

(e) The pressure ratio of the machine is ⎡ ⎛T ⎞⎤ = ⎢1 + ηc ⎜ t ,3 − 1⎟ ⎥ ⎜T ⎟ pt ,2 ⎢⎣ ⎝ t ,2 ⎠ ⎥⎦ pt ,3

γ ( γ −1)

⎡ ⎛ 358.4 K ⎞ ⎤ = ⎢1 + 0.85 ⎜ − 1⎟ ⎥ ⎝ 288K ⎠⎦ ⎣

9

3.5

= 1.936

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7.7 Carry out problem E7.6 for the case where instead of the engine being operated on a test stand it is operating in an aircraft flying at M0=0.7 at sea level where the atmospheric conditions are those of a standard day.

Solution

A schematic diagram of the compressor was shown previously in Figure E7.4. (a) The mass flow through the compressor is (ρAca)2. At the inlet M2 = 0.7 so c2a = a2M2 = 0.7(γRT2)1/2 = 0.7[(1.4)(287m2/s2-K)(T2)]1/2.

The compressor is operating in flight so the static pressure and the static temperature entering the compressor assume the standard sea level values. Thus T2 = 288K and p2 =101.3kPa. With M2 = 0.7, the stagnation temperature and pressure are Tt,2 =T2[1+0.72/5]=316.2K pt,2 = p2[1+0.72/5]3.5=140.5kPa Then c2a=M2a2=0.7[(1.4)(287m2/s2-K)(288K)]1/2=238.1m/s The flow area is A2 = π(reye2-rhub2) = 3.14[(25.4cm2)-(13.35cm2)]10-4/4 = 0.03667m2 The density is the standard sea level value ρ2=1.225kg/m3 so the mass flow is m = ( ρ ca A )2 = (1.225 kg m3 ) ( 238.1 m s ) ( 0.03667 m 2 ) = 10.7 kg s

(b) The power required is P = m ( u3 c3u − u2 c2u ) . However, c2u=0 because the flow enters the compressor axially and u3 =c3u because the flow leaves the impeller in the radial direction. The linear speed of rotation at the exit is u3 = (2πN/60)r3 = (2π)(10,000/60)[0.5)(0.508m)] = 265.4m/s.

Therefore the power 32 = (10.67kg/s) (265.4m/s)2 = 753.6kW P = mu

(c) The stagnation temperature at the exit may be found from Wc = cp(Tt,3 – Tt,2) or

Tt ,3 = 316.2 K +

753.6 kN ⋅ m s = 386.3K 10.7 kg s (1.004 kJ kg ⋅ K )

(d) At the exit u3= 266m/s as already calculated. At the eye station of the exit u2,eye = (2πN/60)r2 = (2)(3.14)(10,000/60)[(0.5)(0.254m)] =133m/s.

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The linear speed of rotation decreases as the hub is approached so we’ll use the eye value just calculated to give a conservative estimate of the fraction of the power arising from the centrifugal effect. Then Pcentrifugal P

1 1 2 2 m ( u32 − u22 ) (10.7 kg s ) ⎡⎣( 265.4 m s ) − (133 m s ) ⎤⎦ =2 =2 = 0.374 753.6 kJ s P

(e) The pressure ratio of the machine is ⎡ ⎛ 386.3K ⎞ ⎤ = ⎢1 + 0.85 ⎜ − 1⎟ ⎥ pt ,2 ⎣ ⎝ 316.2 K ⎠ ⎦ pt ,3

3.5

= 1.829

E7.8 A radial bladed centrifugal compressor designed with zero pre-whirl (axial entry) for a turbojet engine is shown in Figure E7.5 and has the following data: Slip angle=14o Impeller tip radius=26.3cm Impeller eye radius=15cm Impeller hub radius=7cm Impeller tip width=4.4cm Number of vanes=29 Diffuser inner radius=30.5cm Diffuser outer radius=40.6cm N=16,750rpm The compressor operates with an axial inlet Mach number of 0.8 at maximum rpm at the impeller inlet as the aircraft moves through the atmosphere at standard sea level conditions. (a) Determine the linear speed of rotation at the impeller tip, eye, and hub (b) Determine the inlet flow area allowing for 5% blockage by vane thickness (c) Sketch the inlet blade design showing proper diameters and blade angles at the hub, eye, and at the station midway between the two. (d) Calculate the airflow rate (e) If τ=0.94 draw the exit velocity diagram to scale and indicate all magnitudes (f) If ηc=80% determine p3, pt,3, T3, and Tt,3 (g) What is the entrance angle for the fixed diffuser vanes? (Note that angular momentum of the fluid is conserved in the annulus between the impeller tip and the diffuser entrance (h) What is the diffuser entrance Mach number? (i) What is the pressure ratio of the compressor pt,3/pt,2? (j) What is the torque and power required to drive the impeller?

11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

impeller shroud

impeller wheel 40.6

4.4

30.5 15 7

hub

26.3 eye diffuser

Figure E7.5 Sketch of centrifugal compressor. All dimensions in cm

Solution

(a) The linear speed of rotation u = ωr = (2πN/60)r =1754r, therefore utip= 461.3m/s ueye= 263.1m/s uhub=122.8m/s (b) The flow area A2 = π(reye2 – rhub2) – Ablades. Thus A2 = 0.0525m2 (c) Blade design is shown in Figure E7.6. For clarity the velocity diagram for the point midway between the hub and eye is omitted. It should be clear how to construct it. Note that c2a=M2aa =0.8[1.4(287m2/s2-K)(288K)]1/2 = 272.1m/s

c2a=272m/s

β3=30.5o

hub

βeye=46.0o βhub=65.7o

eye

Figure E7.6 Blade angles at hub, eye, and tip

(d) Airflow rate = ρc2aA2 = (1.225kg/m3)(272.1m/s)(0.0525m2) = 17.5kg/s

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (e) The slip coefficient τ = c3u/u3 = 0.94. From part (a) u3 = 461.3m/s so c3u = 433.6m/s.

The slip angle is given as 14o and the velocity diagram is shown in Figure E7.7. The velocity triangle yields [w3r / (u3-c3u)] = tan76o = 4.01 and u3-c3u = u3-0.94 u3 = 0.06 u3.

Therefore w3r = 0.06(461.3m/s)(4.01) = 111m/s; note that w3r=c3r. The velocity c3 = [c3,u2+w3,r2]1/2 = 447.6m/s. Then w3 = w3r/sin76o = 114.3m/s

w3=114m/s

c3=447m/s

c3r=111m/s 14o

α3=14.4o

0

100

200

u3=461m/s c3u=433m/s 300

400

500

(m/s)

Figure E7.7 Velocity diagram for τ = 0.94

(f) The work done per unit mass

Wc = ( u3c3u − u2 c2u ) Because the flow enters the compressor axially, c2u=0 and Wc= (u3c3u) = (461.3m/s)(433.6m/s) = 2x105m2/s2.

But the work per unit mass is also given by Wc = cp(Tt,3 – Tt,2) . Because the compressor is moving through the air at M0=0.8 the stagnation temperature Tt,2 = T2[1+M02/5] = 288K(1+ 0.82/5) =324.9K. Then, using cp=1.03kJ/kg-K Tt,3 = [(2x105m2/s2)/(1.03kJ/kg-K)]+324.9K and Tt,3=519.1K. The total pressure ratio is given by pt,3/pt,2 ={1+ηc[(Tt,3/Tt,2) -1]}γ/(γ-1) pt,3/pt,2 = {1+0.8[(519.1K/324.9K) -1]}3.5 = 3.93 The stagnation pressure entering the compressor is pt,2 = p2[1+(γ-1)M22/2] γ / (γ -1) = (101.3kPa)[1+0.82/5]3.5=154.4kPa Then pt,3=3.93pt,2=606.8kPa.

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The static temperature may be found from the adiabatic energy equation for a streamline leaving the compressor: cpT3+c32/2 = cpTt,3. Then T3= 519.1K- (447.6m/s)2/(2)(1.030kJ/kg-K) = 421.8K. Therefore the sound speed a3 = (γRT3)1/2 = [1.4(287m2/s2-K)(421.8K)1/2 = 411.7m/s.

The Mach number M3 = c3/a3 = (447.6m/s)/(411.7m/s) =1.087. Then the static pressure is given by p3 = pt,3[1+ (γ –1)M32/2]-γ/(γ-1) = (606.8kPa)[1+1.0872/5]-3.5 = 288.7kPa. For reference, the density

ρ3 = p3/RT3 = (288,700N/m2)/[(287J/kg-K)(421.8K)] = 2.384kg/m3. (g) The flow leaves the impeller with angular momentum per unit mass of r3c3u = (0.263m)(433.6m/s) = 114m2/s Because no work is done on the fluid as it proceeds to the diffuser entrance, the angular momentum per unit mass remains constant and the tangential component of absolute velocity at the diffuser entrance is cu,ent=(c3ur3)/rent. This becomes cu,ent = (114m2/s)/0.305m =373.8m/s.

The conservation of mass requires that (ρAcr)3 = (ρAcr)ent and therefore, assuming that the height of the passage, h, is constant across the gap, we find cent , r = c3,r

2π hr3 ρ3 ρ 0.263m ρ3 = (111 m s ) = ( 95.7 m s ) 3 2π hrent ρ ent 0.305m ρ ent ρ ent

Because there should be little, if any, density change across the gap, and the ratio of radial to tangential absolute velocity is small, a reasonable approximation is the density is constant and therefore cent,r ~95.7m/s.We know that 2 ent

c

=c

2 u ,ent

⎡ ⎛ c ⎞2 ⎤ ⎡ ⎛ 95.7 ⎞ 2 ⎛ ρ ⎞ 2 ⎤ r , ent 2 2 3 ⎢1 + ⎜ ⎟⎟ ⎥ = cu ,ent ⎢1 + ⎜ ⎟ ⎥ ≈ 1.066cu ,ent ⎟ ⎜ ⎜ ρ 373.8 c ⎢ ⎥ ⎠ ⎝ ent ⎠ ⎦⎥ ⎣⎢ ⎝ ⎣ ⎝ u ,ent ⎠ ⎦

Therefore cent ~1.032 cu,ent =1.032(373.8m/s)=385.8m/s. There is essentially no difference between the diffuser blade entry angle and the impeller blade exit angle, as can be seen in the diagrams shown in Figure E7.8.

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

cent=386m/s

cr,ent~95.7m/s

αent=14.4o

Diffuser entrance

cu,ent =374m/s

c3=447.6m/s

14o

α3=14.4o

Impeller exit

0

100

200

w3=114m/s u3=461.3m/s c3u=433.6m/s

300

400

500

(m/s)

Figure E7.8 Velocity diagram for impeller exit and diffuser entrance

(h) The energy equation across the gap is cpT3+ c32/2 = cpTent+ cent2/2 so that, assuming cp = 1.03kJ/kg-K, the temperature at the entrance to the diffuser is

( 447.6 m s ) − ( 385.8 m s ) = 446.8K c2 − c2 Tent = T + 3 ent = 421.8K + 2c p 2 (1, 030 m 2 s 2 ⋅ K ) 2

2

The speed of sound is

ant = 1.4 ( 287 m2 s 2 ⋅ K ) ( 446.8K ) = 423.7 K The Mach number entering the diffuser is thus Ment = cent / aent = 385.8m/s/423.7m/s, or Ment = 0.911. (i) The pressure ratio across the machine is pt,3/pt,2 since there is assumed to be no friction in the diffuser and the total pressure remains constant through it. Therefore, from part (f), we have pt,3/pt,2=606.8kPa/154.4kPa = 3.93.

3 c3u = (17.5kg/s)(461.3m/s)(433.6m/s) = 3.50MW and the torque (j) The power is P = mu

is Tq = P/ω = (3.50x106N-m/s)/[2π(16,750)/60s-1] = 1,994N-m

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7.9 Redo problem E7.8 at a reduced rotational speed N=6,750rpm. Discuss major differences between the two cases.

Solution

(a) The linear speed of rotation u=ωr=2πNr/60=706.9r, therefore utip=185.9m/s ueye=106m/s uhub=49.5m/s (b) The flow area A2=π(reye2-rhub2)-Ablades. Thus A2=0.0525m2 (c) Blade design is shown in Figure E7.9. For clarity the velocity diagram for the point midway between the hub and eye is omitted. It should be clear how to construct it. Note that c2a=M2aa =0.8[1.4(287m2/s2-K)(288K)]1/2 = 272.1m/s

c2a= 272m/s

hub eye

βeye=68.8o

βhub=79.7

β3=55.8o

Figure E7.9 Blade angles at hub, eye, and tip

(d) Airflow rate ρc2aA2=(1.225kg/m3)(272.1m/s)(0.0525m2) =17.5kg/s (e) The slip coefficient τ = c3u/u3 = 0.94. From part (a) u3=185.9m/s so c3u=174.7m/s. The velocity diagram appears in Figure E7.10. The velocity triangle yields [w3r/(u3-c3u)] = tan76o = 4.01 and u3-c3u = u3-0.94 u3 = 0.06 u3. Therefore w3r = 0.06(185.9m/s)(4.01) = 44.7m/s and c3 = [c3,u2+w3,r2]1/2 =180.3m/s. Then w3 = w3r/sin76o = 46.06m/s

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

c3r=44.7m/s w3=46.06m/s

c3=180.3m/s

14o

α3=14.4o

0

40

80

u3=185.9m/s c3u=174.7m/s

120

160

200

(m/s)

Figure E7.10 Velocity diagram for τ = 0.94

(f) The work done per unit mass

Wc = ( u3c3u − u2 c2u ) Because the flow enters the compressor axially, c2u=0 and Wc= (u3c3u) = (185.9m/s)(174.7m/s)=32,480m2/s2.

But the work per unit mass is also given by Wc = cp(Tt,3 – Tt,2) = (1.03kg/kJ-K)(Tt,3 – Tt,2).

Because the compressor is moving through the air at M0=0.8, Tt,2 = T2[1+ (γ − 1)M02/2] = 288K[(1+ 0.82/5] =324.9K. Then Tt,3 = [(3.248x104m2/s2)/(1030m2/s2-K)]+324.9K =356.4K. The total pressure ratio is given by pt,3/pt,2 ={1+ηc[(Tt,3/Tt,2) -1]}γ/(γ-1) so that pt,3/pt,2 = {1+0.8[(356.4/324.9) -1]}3.5 = 1.298.

The stagnation pressure entering the compressor is pt,2 = p2[1+ (γ –1)M22/2] γ/(γ-1) = (101.3kPa)[1+0.82/5]3.5 = 154.4kPa Thus pt,3=1.298(154.4kPa)=200.4kPa. The static temperature may be found from the adiabatic energy equation for a streamline leaving the compressor: cpT3+c32/2 = cpTt,3. Then T3= 356.4K- (180.3m/s)2/(2)(1030m2/s2-K) =340.6K. Therefore the sound speed a3 = (γRT3)1/2 = [1.40(287m2/s2-K)(340.6K)]1/2 = 369.9m/s.

The Mach number M3 = c3/a3 = (180.3m/s)/(369.9m/s) =0.487.

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Then the static pressure is given by p3 = pt,3[1+(γ –1)M32/2]-γ/(γ-1) = (200.4kPa)[1+0.4872/5]-3.5 = 170.3kPa. For reference, the density

ρ3 = p3/RT3 = (170,300N/m2)/(287J/kg-K)(340.6K) = 1.742kg/m3. (g) The flow leaves the impeller with angular momentum per unit mass of r3c3u = (0.263m)(174.7m/s) = 45.95m2/s. Since no work is done on the fluid as it proceeds to the diffuser entrance, the angular momentum per unit mass remains constant and the tangential component of absolute velocity at the diffuser entrance is cent,u =c3u(r3/rent).

This becomes cu,ent = (174.7m/s)(0.263m/0.305m) =150.7m/s. The conservation of mass requires that (ρAcr)3 = (ρAcr)ent and therefore cent , r = c3, r

2π hr3 ρ3 ρ 0.263m ρ3 = 44.7 m s = 38.5 m s 3 2π hrent ρ ent 0.305m ρ ent ρ ent

Because there should be little, if any, density change across the gap, and the ratio of radial to tangential absolute velocity is small, a reasonable approximation is the density is constant and therefore cent,r ~38.5m/s.We know that ⎡ ⎛ c ⎞2 ⎤ ⎡ ⎛ 38.5 ⎞ 2 ⎛ ρ ⎞ 2 ⎤ r , ent 2 2 3 c = c ⎢1 + ⎜ ⎟ ⎥ = cu ,ent ⎢1 + ⎜ ⎟ ⎥ ≈ 1.065cu ,ent ⎟ ⎜ ⎜ ⎟ c ⎢ ⎥ ⎢⎣ ⎝ 150.7 ⎠ ⎝ ρent ⎠ ⎥⎦ ⎣ ⎝ u ,ent ⎠ ⎦ Therefore cent ~1.031 cu,ent =1.031(150.7m/s)=155.4m/s. Here we see no difference 2 ent

2 u , ent

between the diffuser blade angle and the impeller exit angle, as can be seen in the results shown in Figure E7.11. (h) The energy equation across the gap, assuming cp = constant, is cpT3+ c32/2 = cpTent+ cent2/2 so that Tent = T3 + (c32 - cent2)/2cp = 342K+(180.32-155.42)/2(1030m2/s2-K) = 346K The speed of sound is aent = [1.4(287m2/s2-K)(346K)]1/2 = 373m/s.

The Mach number entering the diffuser is thus Ment = cent / aent = 155.4m/s/373m/s, or Ment = 0.417. (i) The pressure ratio across the machine is pt,3/pt,2 since there is assumed to be no friction in the diffuser. From part (f) pt,3/pt,2=200.4kPa/154.4kPa = 1.297.

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

cent=155.4m/s

cr,ent~38.5m/s

αent=14.3o

Diffuser entrance

cu,ent =150.7m/s

w3=46.06m/s

c3=180.3m/s

o

14

α3=14.3o

Impeller exit

0

40

80

u3=185.9m/s c3u=174.7m/s 120

160

200 (m/s)

Figure E7.11 Velocity diagram for impeller exit and diffuser entrance 3 c3u = (17.5kg/s)(185.9m/s)(174.7m/s) = 0.568MW and the (j) The power is P = mu

torque is Tq = P/ω = (0.568x106N-m/s)/[2π(6,750)/60s-1] = 803.6N-m The obvious difference between operating this compressor at reduced rotational speed compared to that of the previous problem is that the power expended is proportional to the square of that speed while the pressure rise is roughly directly proportional to that speed. However, more importantly, the impeller blade width was given as 4.4cm and that is consistent with the mass flow at 16,750rpm while at the lower speed the impeller width would have to be about three times larger to pass the same mass flow. The low speed case would be appropriate for a mass flow of about 5.4kg/s.

E7.10 Consider a staged axial flow compressor with the general configuration shown in Figure E7.12. Only the rotor of the first stage is shown and it accepts air from the inlet guide vanes with the following conditions: p2=96.2kPa, T2=289K. The rotor mean line has r=30.5cm, α2=30o, β2=10o, and u2=427m/s. (a) Draw the velocity diagram of the entry to the rotor and find w2, c2, c2a and then find pt,2, Tt2, and M2, where the last 3 items are based upon the absolute velocity c2. (b) Determine N, the rotational speed of the machine in rpm. (c) If the stagnation pressure ratio across the rotor is1.15 and ηc=85%, find c3u-c2u. (d) Draw the exit velocity diagram to scale and find w3, c3, β3, α3, assuming c2a=c3a. (e) Find the blade angle of the first stator passage at the median radius r of the rotor.

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (f) Determine the airflow rate for the compressor if the blade height is 12.7cm assuming that 5% of the flow passage is blocked by blade leading edges. (g) Calculate the required torque, power, and work per unit mass for this first stage. (h) Determine the power required to drive a compressor using of 13 stages like the first stage. (i) Calculate the static pressure rise achieved in the one stage shown. (j) Determine the percent reaction of the one stage shown w2

inlet guide vane

c2a

c2

c2

u

30.5cm

rotor u stator

w3 c 3a

12.7cm

first stage

c3

c3 u

Figure E7.12 Sketch of a stage of an axial flow compressor Solution

(a) A velocity diagram may be drawn to scale using the data given and Figure E7.13 as a guide. Velocities from velocity diagram: w2 = 332.3m/s, c2 = 115.3m/s, c2a = 57.7m/s.

From the given value of the temperature T2 = 289K the sound speed is found to be a2= 340.8m/s. Then from the definition of the Mach number M2 = c2/a2 =0.338. The given pressure p2 = 96.2kPa may be used to find the stagnation pressure from pt,2

=

p2[1+M22/5]3.5 = 104.1kPa. Similarly, the stagnation temperature may be found from Tt,2 = T2[1+M22/5] = 295.6K.

w2

β2

w3

ca u

u

c2

c3

α2 c3,u-c2,u

Figure E7.13 Velocity triangle

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) The rotational speed ω=u/r=(427m/s)/0.305m=1400s-1. Then N=60ω/2π and therefore N = 13,369rpm (c) For the given value of pressure ratio and efficiency WC = (cpTt,2/ηc)[(pt,3/pt,,2)0.286-1] = 14,180N-m/kg = u(c3u-c2u), Therefore (c3u-c2u) = (14,180N-m/kg)/(427m/s)=33.2m/s (d) Note that c2,u=c2cos30=99.85m/s, therefore c3,u=33.2m/s+c2,u=33.18m/s+99.85m/s=133.1m/s.

We continue in this vein and find α3=tan-1(ca/c3,u)=23.45o and β3=11.1o, w3=299.7m/s, and c3=144.9m/s. (e) The blade angle of stator is α3=23.45o (f) The airflow rate depends on

ρ2=p2/RT2=(96,200N/m2)/[(287kJ/kg-K)(289K)=1.159kg/m3 ρ2A2c2a=(1.159kg/m3)π[(0.305m+0.0635m)2–(0.305m–0.0635m)2](0.95)(57.7m/s) = 15.46kg/s

( c3u − c2u ) =(15.46kg/s)(0.305m)(33.2m/s)=156.5N-m, the power (g)The torque Tq = mr Pc= ωTq = 219kW and the work per unit mass of fluid Wc =

P = 14.17kJ/kg m

(h) To find the power required for 13 identical stages of pressure ratio =1.15 we calculate the overall pressure ratio to be pt,3/pt,2 = (1.15)13 = 6.153. Then the overall efficiency is

η ad ,c =

(p

t ,3

pt ,2 )

(γ −1) γ

−1 n

⎧⎪ ⎫⎪ (γ −1) nγ 1 ⎡ − 1⎤ ⎬ − 1 ⎨1 + i ⎢( pt ,3 pt ,2 ) ⎦⎥ ⎪⎭ ⎪⎩ η ad ,c ⎣

=

( 6.153)

0.286

−1 13

1 ⎡ 0.286 13 ⎧ ⎫ − 1⎤ ⎬ − 1 ( 6.153) ⎨1 + ⎣ ⎦ ⎩ 0.85 ⎭

= 0.8122

The overall work is γ ⎡ ⎤ c pTt ,2 ⎢⎛ pt ,3 ⎞ γ −1 ⎥ (1 kJ kg ⋅ K )( 295.6 K ) ⎡( 6.153)0.286 − 1⎤ = 248 kJ kg − = 1 Wc = ⎜ ⎟ ⎥ ⎣ ⎦ 0.8122 ηad ,c ⎢⎜⎝ pt ,2 ⎟⎠ ⎢⎣ ⎥⎦

Then the overall power is Pc = (15.46kg/s)(248kJ/kg) = 3.834MW (i) The stagnation pressure rise in the stage: pt,3-pt,2 = pt,2[(pt,3/pt,2)-1] = 104.1(1.15-1)kPa= 15.62kPa.

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Static pressure rise across stage: p3-p2. The work result may be used to find Tt,3 = 309.7K and the energy equation along a streamline yields T3 = 299.3K leading to M3 = 0.416, showing a small increase in the Mach number across the rotor. Then p3 = 106.3kPa and therefore the static pressure rise is p3-p2 =106.3kPa - 96.2kPa= 10.1kPa. (j)The degree of reaction is defined based on incompressible flow as rinc = (w22 – w32)/[(c32 – c22)+ (w22 – w32)] = 73.3%.

Although the density is almost constant in this problem, the flow is still considered compressible and the compressible flow relations have been used. If we instead use r = (p3-p2)/(pt,3-pt,2)=10.2kPa/15.64kPa = 65.2%

E7.11 A 50% reaction stage of an axial flow turbine shown in Figure E7.14 has blades 12.7cm in height and a linear speed of rotation u=366m/s at the mean line of the rotor where the diameter is d=76.2cm. The stagnation temperature and stagnation pressure entering the stator from the combustor are Tt=1144K and pt=687kPa. The angle for the stator exit absolute velocity is α4=25o. (a) Draw 2 possible combined velocity diagrams for this stage. The diagrams should be neat, executed with a straight edge or by computer, of reasonable size, clearly marked, and shown to scale. The combined velocity diagram is one which shows the inlet and outlet velocities referenced to a single u velocity. For example, the general velocity diagrams shown in the figure may be superimposed to form a combined velocity diagram. (b) Calculate the work per unit mass Wt for the case where the turbine is extracting the maximum work per unit of entering kinetic energy 2Wt/c42 (c) Determine the blade efficiency ηb, which is defined as ηb=u/c4, for the case of maximum work described in part (b) and draw the corresponding combined velocity diagram. (d) Plot the distribution of stagnation and static values of pressure and temperature, as well as the Mach number, through the stage as a function of x, starting from the entrance to the stator blades and ending at the exit of the rotor blades. (e) If the blades block 5% of the flow area determine the mass flow rate through the turbine (f) Determine the power developed by the turbine

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual axial flow from combustor exit

c4

stator

w4

α4

stage

u

38.1cm

c4 u

c5

u

rotor

stator

w5

12.7cm

Figure E7.14 Sketch of an axial flow turbine Solution

(a) Since only u and α4 are given there are a wide variety of possible velocity diagrams for a 50% reaction axial turbine stage and these are shown in Figure E7.15.

Δcu Δcu Δcu

c5

c5

c5

w4 w4 w4

β4

α4=25o 0

100

200

300

400 u (m/s)

Figure E7.15 Possible velocity diagrams for fixed values u=366m/s and α4=25o.

Values of β4 are shown as red dash-dot lines for β4<π/2, in blue dashed lines for

β4=π/2, and in solid black lines for β4>π/2, along with the corresponding velocities c4 and w4. (b) Note that the work done per unit mass by flow turning, as expressed by the product of u and the difference in u-components of velocity Δcu=c4u – c5u is also given by

23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Wt = uΔcu = u(c4u – c5u) Using Figure E7.15 this equation for the work done may also be written as Wt = (1/2)[(c42-c52) – (w42-w52)]

(E7.1)

Then the work extracted, for a 50% reaction axial turbine, using the basic definition where c4=w5 and c5=w4, is given by either of the two following equalities: Wt = (c42-w42) or Wt = u(c4cosα4 – w4cosβ4)

(E7.2)

Note that cosβ4 = -cos(π − β4) so that the work extracted continually grows as β4 increases, and that w4cosβ4 = u-c4cosα4 so that Equation (E7.2) may be written as Wt = u(2c4cosα4-u) = c42[2(u/c4)cosα4 – (u/c4)2]

(E7.3)

Then Equation (E7.3) may be expressed non-dimensionally as Wt/c42 = 2ηbcosα4 – ηb2

(E7.4)

This ratio of the work to twice the kinetic energy entering the rotor is expressed as a function of what may be called the blade efficiency ηb = u/c4 and it has a maximum value when ηb = cosα4 and therefore when u = c4cosα4

(E7.5)

That is, when the absolute velocity leaving the blade is purely axial, as shown by the blue dashed lines in Figure E7.15, the blade efficiency is a maximum. Under this maximum blade efficiency condition the work extracted per unit mass is Wt,max.η = c42cos2α4 = u2

(E7.6)

The velocity c4=404m/s and the axial velocity component c4a=c4sin(25)=171m/s. For fixed values of u and α4 the work extracted per unit mass continues to grow linearly with c4, and therefore with the mass flow, which is proportional to the axial component of velocity c4a=c4sinα4. If we choose to consider the case of the maximum blade efficiency, then the work extracted is Wt,max η =(366m/s)2=134kJ/kg

(E7.7)

(c) To determine the pressure and temperature profiles through the blade passages we must consider the flow to be compressible. The variation of the stagnation temperature and the stagnation pressure (red and blue solid lines, respectively) and the static 24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual temperature and static pressure (red and blue dashed lines, respectively) are shown in Figure E7.16.

M=0.21

1137K 1144K

665kPa 687kPa

c4 1074

w4

α4

1144

531

0.631

u

687kPa

0.267

u 0.649 1016

1029

w5 c5

959K

0.27

427

449kPa

337kPa

449kPa

u M=0.667

1029K

Figure E7.16 The variation of the stagnation temperature and the stagnation

pressure (solid lines) and the static temperature and static pressure (dashed lines) through the turbine passages. Also shown is the variation of the absolute Mach number entering and leaving the rotor (solid lines) and the relative Mach number entering and leaving the passage (dashed line)

We assume that there are no friction losses and no heat transfer through the passages so that the stagnation pressure and temperature are constant through the stator passages. We further assume that cp = 1.16kJ/kg-K and γ = 4/3 and that both are constant throughout the passages. We are given that the stagnation temperature entering the rotor is Tt,4 =1144K and in the rotor passage the work extracted per unit mass, which we have calculated already as 134kJ/kg, is given by Wt = cp(Tt,4-Tt,5) = (1.16kJ/kg-K)(1144K-Tt,5) =134kJ/kg We find then that the stagnation temperature leaving the rotor is Tt,5=1029K. The stagnation temperature variation through the passages is shown in Figure E7.16. The static temperature (and the corresponding sound speeds) entering and leaving the rotor may be found by applying the energy equation along a streamline for adiabatic flow as follows: T4 = Tt4 – c42/2cp = 1144K – (404m/s)2/[2(1160m2/s2-K)] =1074K 25

Theory of Aerospace Propulsion, Second Edition: Solutions Manual a4 = (γRT4)1/2 =(1.33*287m2s2-K*1074K)=640m/s

M4 = (404m/s)/(640m/s) =0.631 M4’ = (171m/s)/(640m/s) = 0.267 (relative to rotor at rotor entrance) T5 = Tt5 – c52/2cp = 1029K – (171m/s)2/[2(1160m2/s2-K)] =1016K a5 = (γRT5)1/2 = (1.33*287m2s2-K*1016K) =622m/s

M5 = (171m/s)/(622m/s)=0.275 M5’= (404m/s)/(622m/s)= 0.649 (relative to rotor at rotor exit) Knowing the sound speed at any point we may find the corresponding Mach numbers from M=c/a. Then the isentropic relations between static and stagnation properties may be used to find the static pressures and densities. At the entrance to the rotor the stagnation pressure is given as 687kPa and the absolute Mach number has been calculated to be 0.631 so the static pressure at the rotor entrance is p4 = pt4[1+ (γ-1)M42/2] - γ / (γ-1) = 687kPa[1+M42/6]-4 = 531kPa The density may be found from the equation of state

ρ4=p4/RT4=(531kPa)/[(0.287kJ/kg-K)(1074K)] =1.724kg/m3 Note that the conservation of mass requires (ρAca)4 = (ρAca)5. We have assumed that ca is constant, in which case the density must be constant through the stage if the entrance and exit areas of the rotor passage are taken as equal. We must consider this situation in greater detail.

The stagnation pressure at the rotor exit may be found from the work done according to the following equation: Wt = cpTt,4ηad,e[1-(pt,5/pt,4)(γ-1)/γ]=(1.16kJ/kg)(1144K)ηad,e[1-(pt,5/pt,4)-1/4] This leads to the result (pt,5/pt,4)=[1-0.1009/ηad,e]4

26

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The adiabatic efficiency of expansion for the rotor is not specified, but in keeping with this ideal analysis we set ηad,e=100%. Then we find that the exit stagnation pressure is pt,5=448.8kPa. The corresponding static pressure is p5=pt,5(1+M52/6)-4=448.8kPa(1+0.2752/6)-4=426.9kPa Then using the state equation to calculate the density leaving the rotor we obtain

ρ5=p5/RT5 = (426,900N/m2)/[(0.287kJ/kg-K)(1016K)] =1.464kg/m3 Note that the density ratio is ρ4/ρ5=(1.724kg/m3)/(1.464kg/m3)=1.178. Thus the density drops by almost 18% through the rotor. This result, coupled with the mass conservation requirement, indicates that A5/A4=1.178 in order to keep ca constant. (d) The area of the turbine flow path is, accounting for 5% blockage due to blade width, given by A = 0.95π(ro2 – ri2) = 0.95π[(0.381+0.0635)2 – (0.381-0.0635)2] = 0.289m2 The mass flow through the turbine is then

ρ4A4ca4 =ρ4(0.289m2)ca= (1.724kg/m3)(0.289m2)(171m/s) = 85.2 kg/s (e) The power developed by the turbine is P= (134kJ/kg)(85.2kg/s)=11.42MW Note: To find the flow conditions at the entrance to the inlet guide vanes one may make

use of the fact that one knows the mass flow, the area, and the stagnation conditions at that station so that using m = pt A

γ

γ −1

⎛ ⎞ M ⎜1 + M2⎟ 2 RTt ⎝ ⎠

−

γ +1 2( γ −1)

Solving this equation yields M=0.22. Then using the usual isentropic flow equations we find pentry=665kPa and Tentry=1137K. Similarly at the exit of the stator we use c4=404m/s and find Texit=Tt,4-c42/2cp=959K, aexit=605m/s, and Mexit=0.667. Then pexit= pt,5(1+Mexit2/6)-4=337kPa This yields a pressure drop in the rotor of 104kPa and a pressure drop in the stator of 90kPa which yields a degree of reaction r = (104kPa)/(104kPa+90kPa)=0.536 This result is very close to the original assumption of r=0.5.

27

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7.12 A turbine blade receives hot gas from a stator with an outlet angle of 70o as shown in Figure E7.17. The blade is designed to have 3o incidence at the root when the linear speed of rotation at the root is 213.4 m/s. The absolute speed leaving the stator is 548.6m/s and the radius of the turbine disc at the root. (a) Find the rotor blade angle β4 for a blade that is untwisted over its entire length (b) Find the incidence at the tip for this untwisted blade (c) Find the stator outlet angle α4 and the rotor blade inlet angle β4 at the root and at the tip under the following set of assumptions: (i) The original conditions pertain only to the root station (ii) There is free vortex flow in the gap between the stator and the rotor, which implies that ucu = constant through the gap and, likewise, the axial component ca=constant. (iii) The incidence at all points along the leading edge of the blade is 3o (iv) The rotor outlet absolute velocity c4 is 231.6m/s and is in the axial direction (v) The deviation is 5o at all points along the trailing edge of the stator blade

stator

70o stage

50.8cm

c4

rotor

u stator -i 10.16cm Note that i is the incidence and δ is the deviation. These angles are measured with respect to the tangent to the camber line at the leading and trailing edges, respectively.

The curvature of the blade as shown is purely illustrative and the angles are exaggerated for clarity.

i

+δ

Figure E7.17 Sketch of an axial flow turbine stage

28

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Solution

The solution is given in the details appearing on Figure E7.18.

Root velocities

c4=548.6m/s

Tip velocities

c4=548.6m/s

w4=407m/s

α4=20o

β4=144.2o

α4=20o

β4=148.3o

u=213.4m/s

u=256.1m/s

i =3o

Rotor blade root leading edge

w4=407m/s

Rotor blade tip leading edge

i =-1.1o

boundary layer edges

Free vortex flow in the gap between stator and rotor: ucu=constant, therefore c3tip=(rroot/rtip)c3root=414.3m/s and c2=474.7m/s

c4,root

α2= 20+5 deviation o

c4,tip

29o Stator tip trailing edge δ =5o 5o

w4,root

c4,a

w4,tip

39o

56o

24o Rotor root leading edge i =3o

42o

Rotor tip leading 59o edge i =3o

Figure E7.18 Diagram of blade properties for the axial turbine stage

29

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E7.13 An axial flow turbine blade 10.2cm in height is attached to a 50.8cm radius turbine wheel as shown in Figure E7.19. The blade is designed to receive air from a stator that has an outlet angle α4=20o. When the linear speed of rotation at the root of the blade (that is, at the point where the blade is attached to the wheel) is 213m/s the angle of incidence of the chord line of the blade there is i=3o. The absolute velocity from the stator outlet is constant at c4=549m/s.

Consulting the sketch below and assuming incompressible flow: (a) Find the stagger angle γ of the blade chord at the root (b) If the blade is untwisted throughout its height find the angle of incidence at the tip of the blade under the given conditions (c) Find the work performed per unit mass of fluid processed if the rotor blade provides an exit absolute velocity that is purely axial.

w4

blade tip blade root

w4

-i +i

stator 20O

50.8cm

stage

chord line c4

rotor

u

γ 10.2cm

Definition of incidence and stagger angles

Figure E7.19 Sketch of an axial flow turbine blade

Solution

(a) Velocity diagrams are shown in Figure E7.20. The axial velocity component is ca = c4sinα4 = 549m/s(sin20) =188m/s The component c4u=c4cosα4=549cos20=516m/s At the blade root the angle

π −β4,root = arctan[ca/(c4u-uroot)] = arctan[188/(516-213)] = 31.8o.

The stagger angle at the root is γ = (π −β4) + i = 31.8+3 = 34.8o

30

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

c4=549m/s ca=188m/s α4=20o

31.8o

uroot= 213m/s

α4=20o

35.9o

utip=256m/s

i =3o

i = -1.1o

34.8o

34.8o

root

tip

Figure E7.20 Velocity diagrams constructed for the turbine blade

(b) The linear speed of rotation at the tip is utip = uroot (rtip / rroot) = (213m/s)(50.8+10.2)/50.8 = 256m/s Then the angle

π − β4,tip = arctan[ca/(c4u-utip)] = arctan[188/(516-256)] = 35.9o The incidence at the tip for an untwisted blade is itip = γ−(π−β4,tip) = 34.8o-35.9o = -1.1o (c) The work per unit mass is W = umeanΔcu = 0.5(utip+uroot)(c4u-c5u) =0.5(213+256)(516-0) W = 121,000m2/s2 or 121kJ/kg

E7.14. An axial flow compressor stage is designed for 50% reaction, a rotational speed of 13,760rpm, and a mean radius of 25.4cm. Carry out all analyses at the mean radius location and assume the flow is incompressible in this problem. (a) If the rotor blade relative velocity exits at an angle β3 =70o and ca=152m/s, draw the velocity diagram for the entrance and the exit of the rotor at the mean radius approximately to scale and determine all appropriate angles and velocities and note them directly on the diagram. (b) If the airflow rate is 36.2kg/s find the torque and the power required and the ideal total head produced.

Solution

The stage is a 50% (symmetric) stage so β3 = α2 and β2 = α3 while w3 = c2 and w2 = c3.

31

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The rotational speed is

ω = 2πN/60 = 2π(13,760)/60 =1441s-1 The linear speed of rotation at the mean radius is u = ωrm= (1441)(0.254m) = 366m/s.

The axial velocity ca=w3sin70 or w3 = (152m/s)(sin70)-1=162m/s and thus c2 =162m/s The velocity component c3u = u -w3cos70 = 366m/s-(162m/s)(0.342) = 311m/s The absolute velocity c3 = (c3u2+c3a2)1/2 = (3112+1522)1/2 = 346m/s The velocity component c2u = c2cos70 = 162m/s(cos70) = (162m/s)(0.3420) =55.4m/s (a)

w3=162m/s

c3=346m/s

ca=152m/s

ca=152m/s 70o

70o

26o

w3=346m/s

c2=162m/s

u=366m/s

26o u=366m/s

(b) (i) The torque is Tq= m rm(c3u-c2u) = (36.2kg/s)(0.254m)(311m/s – 55.4m/s) = 2,350N-m (ii) The power required is P = Tqω = (2,350N-m)(1,441s-1) = 3,387kW (iii) The ideal total head H is given by WC /g or H=

P 3.387 ×106 N ⋅ m s = = 9,540m

mg ( 36.2 kg s ) ( 9.807 m s 2 )

32

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E8. BLADE ELEMENT ANALYSIS FOR AXIAL FLOW TURBOMACHINES

E8.1 (a) Derive the Kutta-Joukowski law for subsonic compressible flow; (b) Determine the velocity components and show the appearance of the Prandtl-Glauert factor (1-M2)1/2 in the results; (b) Explain how the lifting characteristics of a blade are affected by the compressibility effect.

Solution Consider a circular contour C drawn around a two-dimensional lifting body in a steady frictionless compressible flow, as shown in Figure E8.1

v

Γ L

u

r

V0

D

θ

Contour C

Figure E8.1 Circular contour around a body showing a polar coordinate system

(a) The Kutta-Joukowski Law (i) Circulation: The circulation Γ is defined by the contour integral 2π G G Γ = − v∫ V ⋅ rdθ = − ∫ vrdθ C

(E8.1)

0

(ii) Velocity components: If the contour radius r is allowed to increase, the flow at the contour more and more closely resembles the sum of a uniform flow and a point vortex for which the velocity components u and v may be represented by

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ∞ ⎛ ⎞ An (θ ) ⎟ ∑ ⎜ ⎛ A (θ ) ⎞ ⎟ → lim u = V0 ⎜ cos θ + 1 u = V0 ⎜ cos θ + n =1 n ⎟ r ⎠ r ⎜ ⎟ r →∞ ⎝ ⎜ ⎟ ⎝ ⎠

(E8.2)

∞ ⎛ ⎞ Bn (θ ) ⎟ ∑ ⎜ B (θ ) ⎞ ⎛ ⎟ → lim v = −V0 ⎜ sin θ + 1 v = −V0 ⎜ sin θ + n =1 n ⎟ r ⎠ r ⎜ ⎟ r →∞ ⎝ ⎜ ⎟ ⎝ ⎠

(E8.3)

The circulation is then a constant given by 2π B1 (θ ) ⎤ ⎡ Γ = − ∫ ⎢ −V0 sin θ − V0 ⎥rdθ = ∫ V0 B1 dθ r ⎦ 0 ⎣ 0 2π

(E8.4)

(iii) Mass conservation: Applying the conservation of mass equation to the control volume enclosed by the contour leads to the following result: 2π

G

2π

2π

2π

2π

0

0

0

0

ˆ θ = ∫ ρ urdθ = ∫ ρV0 cos θ rdθ + ∫ ρV0 A1 dθ = ρV0 ∫ A1 dθ = 0 (E8.5) ∫ ρV ⋅ nrd 0

Therefore A1 must be zero or even in θ.

G (iv) Irrotationality: The irrotationality condition ∇ × V = 0 is given in the polar

coordinate system by

V dA ∂ ∂ ( rv ) − ( u ) = −V0 sin θ − V0 ( − sin θ ) + 0 1 = 0 r dθ ∂r ∂θ

(E8.6)

This equation shows that A1=0. Then the velocity components, to O(r-2) are given by u = V0 cos θ B ⎞ ⎛ v = −V0 ⎜ sin θ + 1 ⎟ r ⎠ ⎝ (vi) Continuity: Likewise, the continuity equation is

B ⎤⎞ ∂ ∂ ∂ ∂ ⎛ ⎡ ( ρ ur ) + ( ρ v ) = ( ρV0 [ r cos θ ]) + ⎜ − ρV0 ⎢sin θ + 1 ⎥ ⎟ = 0 ∂r ∂θ ∂r ∂θ ⎝ r ⎦⎠ ⎣

ρV0 cos θ + rV0 cos θ

B ⎞ ∂ρ ∂ρ 1 dB1 ⎛ =0 − ρV0 cos θ − ρV0 − V0 ⎜ sin θ + 1 ⎟ r dθ r ⎠ ∂θ ∂r ⎝

2

(E8.7)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual We may solve for B1 as follows: B

dB1 ⎡ 2 1 ∂ρ 1 ∂ρ ⎤ = ⎢ r cos θ − ( r sin θ + B1 ) dθ ⎣ ρ ∂r ρ ∂θ ⎦⎥

(E8.8)

(vii) Density and pressure fields: It is necessary to find the behavior of the density

ρ(r,θ). The Bernoulli equation for compressible flow is

γ

p

γ −1 ρ

+

1 2 γ p0 1 2 + V u + v2 ) = ( 2 γ − 1 ρ0 2 0

(E8.9)

Then, for isentropic flow where p ~ ργ and noting that the sound speed a2 = γ p/ρ, we find that

γ γ −1

C (ρ

γ −1

2 2 2 ⎤ 1γp ⎡ ⎤ B1 ⎞ 1 2 ⎡⎛ u ⎞ ⎛ v ⎞ 2 ⎛ 2 0 − ρ0 ) = V0 ⎢⎜ ⎟ + ⎜ ⎟ − 1⎥ = M 0 ⎢( cos θ ) + ⎜ sin θ + ⎟ − 1⎥ 2 ⎢⎝ V0 ⎠ ⎝ V0 ⎠ 2 ρ0 r ⎠ ⎝ ⎣⎢ ⎦⎥ ⎣ ⎦⎥ γ −1 2 ⎤ ⎛ ρ ⎞ B1 γ −1 2 ⎡ 2 ⎛ B1 ⎞ 2 1 M cos θ sin θ 2 sin θ = + + + + − 1⎥ ⎜ ⎟ 0 ⎢ ⎜ ⎟ 2 r ⎝ r ⎠ ⎢⎣ ⎥⎦ ⎝ ρ0 ⎠

γ −1

Then, collecting terms, we have 1

⎛ ρ ⎞ ⎡ γ −1 2 γ − 1 B12 ⎤ γ −1 1 M B sin θ = + + ( ) ⎜ ⎟ ⎢ ⎥ 0 1 r 2 r2 ⎦ ⎝ ρ0 ⎠ ⎣ As we let the radius of the contour r become increasingly large we may neglect terms of O(1/r2) and higher and expand the result in a binomial series the density ratio is approximately given by M 02 ρ = 1− ( B1 sin θ ) ρ0 r

(E8.10)

The pressure may be found directly from the density through the isentropic relation which yields

γ M 02 p = 1− ( B1 sin θ ) p0 r

(E8.11)

(viii) The density derivatives: We now need the derivatives of the density for Equation (8.10) which are as follows:

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual M 02 ( B1 sin θ ) r 2 ∂ρ r2 ∂ ⎛ ρ ⎞ = ≅ M 02 ( B1 sin θ ) ⎜ ⎟= ρ ∂r ⎛ ρ ⎞ ∂r ⎝ ρ0 ⎠ M 02 1− ( B1 sin θ ) ⎜ ⎟ r ρ ⎝ 0⎠ M 02 ⎛ dB ⎞ − B1 cos θ − 1 sin θ ⎟ ⎜ 1 ∂ρ 1 ∂ ⎛ ρ ⎞ r ⎝ dθ ⎠ = ⎜ ⎟= 2 ρ ∂θ ⎛ ρ ⎞ ∂θ ⎝ ρ 0 ⎠ M 1 − 0 ( B1 sin θ ) ⎜ ⎟ r ⎝ ρ0 ⎠

(E8.12)

(E8.13)

Then Equation (E8.13) becomes dB 1 ⎛ ∂ρ ⎞ M 02 ⎛ ⎞ − B1 cos θ − 1 sin θ ⎟ ⎜ ⎟≅ ⎜ r ⎝ dθ ρ ⎝ ∂θ ⎠ ⎠

(E8.14)

Using Equations (E8.12) and (E8.14) in Equation (E8.8) and remembering that terms of O(1/r) are to be neglected with respect to terms of O(1), yields

dB1 1 − M 02 sin 2 θ ) = M 02 ( 2 B1 sin θ cos θ ) ( dθ

(E8.15)

(ix) Solving for B1: We may rearrange Equation (E8.15) as follows: B

d ( M 02 sin 2 θ ) dB1 2 M 02 sin θ cos θ d θ = = B1 1 − M 02 sin 2 θ 1 − M 02 sin 2 θ

(E8.16)

Integrating Equation (E8.16) yields B1 =

K 1 − M 02 sin 2 θ

(E8.17)

Here K is a constant of integration, which we may evaluate by using the circulation Equation (E8.1) since we now know that

⎛ ⎞ B ⎞ K 1 ⎛ v = −V0 ⎜ sin θ + 1 ⎟ = −V0 ⎜ sin θ + ⎟ 2 2 r ⎠ r 1 − M 0 sin θ ⎠ ⎝ ⎝ Then Equation (E8.1) becomes 2π

2π

0

0

2π

dθ 1 − M 02 sin 2 θ 0

Γ = − ∫ vrdθ = V0 r ∫ sin θ dθ + V0 K ∫

(E8.18)

The first term on the right hand side of Equation (E8.18) is zero and we are left with

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

2π

Γ = V0 K

(E8.19)

1 − M 02

Then the constant K is given by K=

Γ 1 − M 02 2π V0

(E8.20)

When M0=0, that is, for incompressible flow, this term becomes K inc =

Γ 2π V0

(E8.21)

(x) Lift and drag: The force acting on the two-dimensional body may be determined by applying the conservation of momentum equation to the control volume defined by the contour C. It is convenient to choose the lift and drag components per unit span, which are given by 2π

L = − ∫ ⎡⎣ p sin θ + ρ u ( u cos θ + v sin θ ) ⎤⎦rdθ 0

2π

D = − ∫ ⎡⎣ p cos θ + ρ v ( u cos θ − v sin θ ) ⎤⎦rdθ 0

Breaking these equations down to their elements yields 2π

2π

2π

0

0

0

2 2 2 2 ∫ p sin θ rdθ = ∫ ( p0 r sin θ − γ p0 M 0 B1 s in θ )dθ = −γ p0 M 0 ∫ B1 sin θ dθ

2π

2π

2π

0

0

0

2 2 ∫ p cos θ rdθ = ∫ ( p0 r cos θ − γ p0 M 0 B1 sin θ cos θ )dθ = −γ p0 M 0 ∫ B1 sin θ cos θ dθ 2π

2π

0

0

2 2 ∫ ρ u ( u sin θ + v cos θ )rdθ = −V0 ∫ ρ B1 cos θ dθ

2π

2π

0

0

2 ∫ ρ u ( u cos θ − v sin θ )rdθ = V0 ∫ ρ B1 sin θ cos θ dθ

Since to O(1/r) the density ρ = ρ0 and p = p0 so that γ p0M02=ρ0V02 and then the lift and drag per unit span are given by 2π

2π

0

0

L = − ∫ ⎡⎣ − ρ 0V02 ( B1 sin 2 θ + B1 cos 2 θ ) ⎤⎦dθ = ρ 0V02 ∫ B1 dθ = ρ 0V0 Γ

5

(E8.22)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 2π

D = − ∫ ⎡⎣ − ρ 0V02 ( B1 sin θ cos θ − B1 sin θ cos θ ) ⎤⎦dθ = 0

(E8.23)

0

Therefore the Kutta-Joukowski law regarding the lift, L=ρVΓ, holds for compressible flow while the drag D=0. The relationship between the incompressible lift and the incompressible lift is

L Γ 1 = = Linc Γinc 1 − M 02

(E8.24)

The behavior of the reciprocal of the Prandtl-Glauert factor of β = 1 − M 02 Equation (E8.24) is shown in Figure E8.2.

5 4 1/β

3 2 1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

M0

Figure E8.2 The reciprocal of the Prandtl-Glauert factor β is shown as a function

of Mach number (b)The velocity components are therefore u = V0 cos θ v = −V0 sin θ −

1 − M 02 Γ 2π r 1 − M 02 sin 2 θ

The velocity components in Cartesian coordinates are as follows:

1 − M 02 sin θ B1 Γ Vx = V0 + V0 sin θ = V0 + r 2π r ( M 02 sin 2 θ − 1)

6

(E8.25)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual 2 B1 Γ 1 − M 0 cos θ Vy = −V0 cos θ = − 2π r M 02 sin 2 θ − 1 r

(E8.26)

(c) Relationship of compressibility to lifting capability of blades: The blade lift is manifested as a pressure force integrated over the entire blade surface. As shown in Equation (E8.24) the lift increases in compressible flow as the free stream Mach number increases. The non-dimensional pressure coefficient is generally used to represent the pressure field and using Equation (E8.1) we find Cp =

1 − M 02 sin θ ⎛ p ⎞ p − p0 p0 B1 Γ (E8.27) = − = − = − 1 2 sin 2 θ ⎜ ⎟ 1 1 r 2π rV0 M 02 sin 2 θ − 1 2 2 ⎝ p0 ⎠ ρ 0V0 γ p0 M 0 2 2

Comparing Equation (E8.27) with Equation (E8.26) we see that

⎛ V − V0 ⎞ C p = −2 ⎜ x ⎟ ⎝ V0 ⎠

(E8.28)

This shows that the pressure coefficient is the negative of twice the perturbation velocity component in the x-direction, the result found for linearized compressible flow in Section 8.12 of the text.

E8.2 Consider an infinite circular cylinder rotating at constant angular speed ω. The cylinder is divided into sectors of included angle 2α which are filled with an inviscid fluid: (a) Show that the fluid in each sector circulates in the opposite sense to that of ω, (b) Calculate the flow in a sector of included angle 60o and show streamlines and velocity vectors. Note that Figure 7.15 of Chapter 7 shows such a cylinder with α=π/4.

Solution

(a) We consider a spinning cylinder of radius a, as shown in Figure E8.3. The segment of included angle 2α is divided into two equal parts as shown. The flow in the segment is inviscid so that the circulation is zero and, by Helmholtz’s theorem it remains so.The velocity in the flow is given by the sum of the velocity relative to the segment and the velocity due to rotation of the segment: G G G G V = u +ω×r The circulation around the contour C taken as positive in the ω direction (counterclockwise) is then given by

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual G G a G G G G Γ = v∫ V ⋅ ds = ∫ ( u + ω × r ) ⋅ ds = 0 C

0

α

θ

a

α=0

−α ω

Figure E8.3 Schematic diagram of a spinning cylinder with segment of included

angle 2α. The segment is filled with frictionless fluid which rotates in a sense opposite to that of the cylinder.

This requires that a

α

a

0

−α

0

G G G G G G G G G G G v∫ u ⋅ ds = −∫ (ω × r ) ⋅ dr − ∫ (ω × r ) ⋅ rdθ − ∫ (ω × r ) ⋅ ( −dr ) C

Then α

α

−α

−α

G G G G G 2 2 v∫ u ⋅ ds = − ∫ (ω × r ) ⋅ rdθ = − ∫ ω r dθ = −2αω r C

Since the integration around the contour was taken in the sense of –ω, the direction of the velocity u must be in the opposite direction and therefore the fluid in the segment rotates in the sense opposite to that of the spinning cylinder. (b) To calculate the flow we may follow the approach by Lamb (Hydrodynamics, Dover, 1945) who introduces the stream and potential functions, respectively, as follows:

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ( 2 n +1)π

( 2n + 1) πθ 1 ⎛ cos 2θ ⎞ ⎛ r ⎞ 2α cos − 1⎟ + ∑ A2 n +1 ⎜ ⎟ ψ = ωr 2 ⎜ 2 2α ⎝ cos 2α ⎠ ⎝a⎠ 1 ⎛ sin 2θ ⎞ ⎛r⎞ φ = − ωr 2 ⎜ − 1⎟ − ∑ A2 n +1 ⎜ ⎟ 2 ⎝ cos 2α ⎠ ⎝a⎠

( 2 n +1)π 2α

sin

( 2n + 1) πθ 2α

The coefficient A2 n +1 = ( −1)

n +1

⎡

⎤ 1 2 1 − + ⎥ ⎣ ( 2n + 1) π − 4α ( 2n + 1) π ( 2n + 1) π + 4α ⎦

ωa2 ⎢

The non-dimensional velocity components in the radial and tangential directions, respectively, are given by ⎡ sin 2θ ( 2n + 1) πθ ⎤ u 1 1 ∂ψ π ( 2 n +1) 2πα − 2 =− =η ⎢ + ∑ A2′n +1 ( 2n + 1) η sin ⎥ (E8.29) ωa ω a r ∂θ 2α 2α ⎣ cos 2α ⎦ ⎡ cos 2θ ( 2n + 1) πθ ⎤ π ( 2 n +1) 2πα − 2 v 1 ∂ψ = =η ⎢ − 1 + ∑ A2′n +1 ( 2n + 1) η cos ⎥ (E8.30) ω a ω a ∂r α α α cos 2 2 2 ⎣ ⎦

Here we have introduced the non-dimensional radial coordinate η=r/a and the coefficient

A2′n +1 =

A2 n +1 ω a2

(E8.31)

The non-dimensional stream function may be written as ⎡ cos 2θ ψ 1 2 η = − + ∑ A2′n +1η ⎢ 2 ωa ⎢⎣ 2 cos 2α 2

( 2 n +1)π 2α

−2

cos

( 2n + 1) πθ ⎤ 2α

⎥ ⎥⎦

(E8.32)

Now Equations (E8.29), (E8.30), and (E8.31) are functions only of η and θ with α appearing as a parameter so that the velocity components and the stream function may be calculated. Because the walls of the segment must be a streamline of the flow we may set the stream function to zero along them. For the stipulated case where 2a=60o this leads to the expression

ψ 1 ⎡ ⎤ 3 2 n +1 − 2 = η 2 ⎢cos 2θ − + ∑ A2′n +1η ( ) cos ( 2n + 1) 3θ ⎥ 2 2 ωa ⎣ ⎦ The coefficient

A2′n +1 = ( −1)

n +1

8

1

π ( 2n + 1) ⎡9 ( 2n + 1)2 − 4 ⎤ ⎣

9

⎦

Theory of Aerospace Propulsion, Second Edition: Solutions Manual It is clear that for n=0 A1’=-8/5π while for n=1 this coefficient is very small so that the non-dimensional stream function becomes

ψ ⎡ 1 ⎛ 8 8 ⎞⎤ η 7 cos 9θ − ... ⎟ ⎥ = η 2 ⎢cos 2θ − + ⎜ − η cos 3θ + 2 2 ⎝ 5π 231π ωa ⎠⎦ ⎣ Because η <1 the second and subsequent terms in the series are negligible and the stream function may be accurately calculated using just the first term. The resulting flow pattern for ψ /ωa2 = 0.3 and 0.6 is shown for 0< θ <30o in Figure E8.4 because the flow is symmetrical about θ = 0.

0.5 0.45 0.4 0.35 0.3 y/a 0.25 0.2 0.15 0.1 0.05 0

ψ/ωa^2=0

cylinder

0.03

segment wall

0.06

0

0.2

0.4

0.6

0.8

1

1.2

x/a

Figure E8.4 The stream non-dimensional stream function contours inside a 60o

segment of a spinning cylinder is shown. Only half the segment is shown because the flow is symmetrical about θ=0. The numbers indicate values of ψ /ωa2

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E8.3 Using the equation for the efficiency of adiabatic compression, Equation (8.73), with the assumption that ε=εs=εr show that contours of constant efficiency may be depicted as circles on a φ,ri plot. Show the contours for ε=0.04 and efficiencies of 88%, 90%, and 92%, and indicate the value and location of the maximum efficiency on the same plot.

Solution

The general equation for the efficiency of adiabatic compression, Equation (8.73) in the text, is ⎡ r − φε r 1 − ri − φε s ⎤ η ad ,c = φ ⎢ i + ⎥ ⎣ φ + ε r ri φ + ε s (1 − ri ) ⎦

We may expand this general expression for ηad,c and after neglecting products of εr and εs we obtain

η ad ,c =

φ − ( ε r + ε s ) (φ 2 + ri 2 − ri ) φ + ε r ri + ε s (1 − ri )

Setting εr=εs=ε in this equation and rearranging terms we obtain the following equation which describes circles in the φ,ri plane: ⎛ ⎝

1⎞

2

(φ − φ0 ) + ⎜ ri − ⎟ = R 2 2 2

⎠

The quantities

φ0 =

1 − ηad ,c 4ε

R 2 = φ02 +

1 − 2ηad ,c 4

The centers of the circles lie on the line ri=0.5 and at the points φ=φ0 which are determined by the magnitude of the efficiency and the drag to lift ratio of the blades. The radii of the circles are given by the quantity R and depend on the same quantities. Now we have a relationship between ri and φ for fixed values of ηad,c and ε which appear as circles of constant efficiency of 88%, 90%, and 92% for ε=0.04 in Figure E8.5. The maximum efficiency of 92.6% is also shown. This is obtained by setting the radius R=0 and solving for ηad,c.

11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1 (ηad,c)max=92.6% 0.75

ri 0.5 92%

90%

88%

0.25

0 0

0.25

0.5

0.75

1

1.25

1.5

φ

Figure E8.5 Circles of constant efficiency ηad,c for εr=εs=ε=0.04 as a

function of flow rate parameter φ and the theoretical degree of reaction ri.

E8.4 A compressor stage processing air under standard sea level conditions is designed for ideal degree of reaction r*=50%. The rotor speed is u=460m/s, the axial velocity component ca=200m/s, and the rotor blade entrance angle is 35o. The blade chord length l=5cm, the blade pitch distance s=7cm, the blade height is h=12.7cm and the hub radius ri=30cm.The aerodynamic efficiency of the blade in the design operating range is ε=0.008. (a) Draw the combined entrance and exit velocity diagram to scale. (b) Find the slip coefficient τ and the flow rate coefficient φ. (c) Find the force components Fu, Fa, and the resultant force F. (d) Find the lift and drag components L and D. (e) Draw a scaled vector diagram showing the forces found in parts (b) and (c) (f) Determine the solidity of the compressor (g) Find the torque and power required per blade (h Determine the pressure rise in the rotor

Solution

(a) The combined velocity diagram is shown in Figure E8.6. The ideal degree of reaction r*=1/2, therefore

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

r∗ =

1 1 w2,u + w3,u wm ,u = = u u 2 2

This shows that wm,u=230m/s. (b) The slip coefficient τ and the flow rate coefficient φ are given as follows:

τ=

w2,u − w3,u u

φ=

=

286 − 174 = 0.243 460

ca 200 = = 0.435 u 460

(c)The blade area is lh=(5cm)(12.7cm)=63.5cm2=6.35x10-3m2. Then the force components and the resultant force are given by

Fu = ρ u 2φτ A = (1.225kg / m3 ) ( 460m / s ) ( 0.435)( 0.243) ( 6.35 × 10−3 m2 ) = 174 N 2

Fa = Fu

1 − ε tan β m 1 − 0.008 tan 41 = (174 N ) = 197 N ε + tan β m 0.008 + tan 41

F = Fu2 + Fa2 = 263N (d) The lift and drag components are given by

L=

Fu 174 N = = 262.8 N sin β m + ε cos β m sin 41 + 0.008cos 41

D=

Fu − L sin β m 174 N − 262.8 N ( sin 41) = = 2.10 N cos β m cos 41

Here we kept an additional significant figure in L so that the aerodynamic efficiency

ε =D/L=0.008 will be maintained. We had been keeping three significant figures and can easily show that if we rounded off L to 263N the aerodynamic efficiency ε =D/L=0.0073.

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

w2=349m/s

c2

c3

ca=200m/s wm w3 β2=35 βm=41 β3=48.6 L

u=460m/s

D

Figure E8.6 Combined velocity diagram and schematic of blade camber line

(e) A scaled diagram of the forces, except for the drag, is shown in Figure E8.7. The drag force is exaggerated in scale here because it would not be visible on the scale of the other forces. It is clear that though the lift is large and the drag small, the axial and tangential forces on the blade are large and of the same order. (f) The solidity is given by σ = l /s = 5cm/7cm=0.714

(g) The power required to turn the blade is P = uFu = (460m/s)(174N) = 80kW and the torque required is Tq = P/ω = rmFu = (0.30m+0.127m/2)(174N) = 63.2N-m. The rotational speed based on the pitch radius rm (60s/min)(1265/s) /2π =12,080.

(h) The pressure rise in the rotor is given by ⎛ r ∗ −εφ ⎞ 2 ⎛ 0.5 − 0.008 × 0.435 ⎞ 3 Δp = ρ u 2τφ ⎜ ⎟ ⎟ = (1.225kg / m ) ( 460m / s ) ( 0.243)( 0.435 ) ⎜ ⎝ 0.435 + 0.008 × 0.5 ⎠ ⎝φ +εr ∗⎠ Δp = 30.99kPa

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

wm=349m/s

Fu=174N

Fa=197N L=263N

D=2.1N (not to scale)

βm=41o

Figure E8.7 The forces acting on the blade are shown to scale, except for the

drag, which is exaggerated in scale here because it would not be visible on the scale of the other forces.

E8.5 Compressor characteristics may be known for a specific design point, typically sea level static conditions, and performance is desired for other, off-design, flight conditions.

A semi-empirical equation has been developed which provides quite accurate estimates of the compressor pressure ratio pt3/pt2 at a given altitude and speed when the value is known at some other flight condition, both being evaluated at the same rpm N. The empiricism arises from the following ad hoc definition of the compressor pressure ratio equation: pt ,3 ⎡ u ( c3,u − c2,u ) ⎤ = ⎢1 + ⎥ ηad ,c c pT2 ⎦⎥ pt ,2 ⎣⎢

Note that in this empirical definition the temperature in the denominator in the equation above is not Tt2 as arises in the proper formulation of the energy equation but instead the average stagnation pressure given by γ −1 ⎧ ⎡ ⎤⎫ ⎛ ⎞ 1 1 ⎢ pt ,3 γ ⎪ ⎥⎪ T2 = (Tt ,2 + Tt ,3 ) = Tt 2 ⎨1 + ⎜ ⎟ − 1⎥ ⎬ ⎢ ⎜ ⎟ 2 ⎪ 2ηad ,c ⎢⎝ pt ,2 ⎠ ⎥⎦ ⎪⎭ ⎣ ⎩ The basis for the remainder of the development is that the kinematic work term u(c3,u – c2,u) is independent of the compressor inlet temperature Tt,2, and of course Tt,2 varies with altitude and Mach number. Consider two conditions: subscript a, where the pressure ratio and temperature are specified and b, where the flight conditions are specified but the compression ratio must be found.

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (a) Use the equations above, the correct basic equation for the stagnation temperature ratio (Tt,3/Tt,2), and the condition that u(c3,u – c2,u) is a constant to find a relation for (pt3/pt2)b. (b) For the case where ηad,c=100% plot a graph which shows the variation of (pt3/pt2)b as a function of (pt3/pt2)a with the temperature ratio (Tt,2b/Tt,2a) as a parameter. Consider case a to be the known case which is evaluated at standard sea level static conditions and case b to cover flight in the atmosphere between sea level and 11km and Mach numbers between 0 and 1. (c) For a compressor with a stagnation pressure ratio of 20 under standard sea level static conditions determine the its value at an altitude of 11km and M0=0.85 for ηad,c=100% and 85% and discuss the difference between the two answers Solution

(a) The work required by the compressor is given thermodynamically as Wc=ht,3 – ht,2 while dynamically, evaluating it at the mean, or pitch, radius it is Wc = u(c3,u – c2,u). Then using the adiabatic efficiency of compression ηad,c=η, the compressor stagnation pressure ratio pt,3/pt,2 = R, and assuming a constant average cp andγ through the compressor we may write c p Tt ,2 ⎛ γ γ−1 ⎞ ⎜ R − 1⎟ = u Δcu ⎟ η ⎜⎝ ⎠ If we assume that the dynamical form of the work equation is independent of the entering stagnation temperature then the suggested empirical pressure ratio of the compressor under conditions a and b are given, respectively, by γ

⎡η ( u Δcu ) ⎤ γ −1 Ra = ⎢ + 1⎥ ⎣⎢ c p Tt ,2 a ⎦⎥

⎡η ( u Δcu ) ⎤ Rb = ⎢ + 1⎥ ⎢⎣ c p Tt ,2b ⎥⎦

γ γ −1

Then, assuming that the efficiency of adiabatic compression, which depends primarily on fluid dynamic behavior, remains constant, the pressure ratio under condition b becomes, ⎡ η c p Tt ,2 a ⎛ Rb = ⎢ ⎜ Ra ⎢⎣ c p Tt ,2b η ⎝⎜

γ −1 γ

⎞ ⎤ − 1⎟ + 1⎥ ⎟ ⎠ ⎦⎥

16

γ γ −1

⎡ Tt ,2 a ⎛ = ⎢1 + ⎜ Ra ⎢⎣ Tt ,2b ⎜⎝

γ −1 γ

⎞⎤ − 1⎟ ⎥ ⎟⎥ ⎠⎦

γ γ −1

(E8.33)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual So now the compressor pressure ratio at condition b is known in terms of the given compressor pressure ratio at condition a in terms of the average stagnation temperatures.

These average temperatures may be written as follows: γ −1 ⎧⎪ ⎤ ⎫⎪ 1 ⎡ γ Tt ,2 a = Tt ,2 a ⎨1 + R 1 − ⎢ a ⎥⎬ ⎥⎦ ⎪⎭ ⎪⎩ 2η ⎢⎣

(E8.34)

γ −1 ⎧⎪ ⎤ ⎫⎪ 1 ⎡ γ Tt ,2b = Tt ,2b ⎨1 + R 1 − ⎢ b ⎥⎬ ⎥⎦ ⎭⎪ ⎩⎪ 2η ⎢⎣

(E8.35)

Combining Equations (E8.33), (E8.34), and (E8.35) we obtain the following equation for the compressor pressure ratio at the new condition b: γ −1 γ −1 ⎛ γ γ−1 ⎞ ⎡ ⎞ ⎤ Tt ,2 a ⎛ γ γ−1 ⎞ ⎡ ⎞⎤ 1 ⎛ γ 1 ⎛ γ R R R R 1 1 1 1 1 − + − = − + ⎢ ⎥ ⎢ ⎜ b ⎟ ⎜ b ⎟ ⎜ a ⎟ ⎜ a − 1⎟ ⎥ ⎟ ⎜ ⎟ ⎢ 2η ⎜ ⎟ ⎥ Tt ,2b ⎜ ⎟ ⎢ 2η ⎜ ⎝ ⎠⎣ ⎝ ⎠⎦ ⎝ ⎠⎣ ⎝ ⎠ ⎥⎦

(E8.36)

(b) If we assume γ =1.4 and η =100%, Equation (E8.36) becomes

(R

0.286 b

T0, s.l . ⎡ 1 ⎤ ⎡ 1 ⎤ Ra0.286 − 1) ⎢1 + ( Ra0.286 − 1) ⎥ − 1) ⎢1 + ( Rb0.286 − 1) ⎥ = ( 2 ⎣ 2 ⎦ T0 (1 + 0.2 M 0 ) ⎣ 2 ⎦

This may be simplified to and directly solved to yield ⎡ Ra0.572 − 1 ⎤ ⎥ Rb = ⎢1 + ⎢⎣ θ 0 (1 + 0.2M 02 ) ⎥⎦

1.748

(E8.37)

Here the quantity θ0 is the standard atmosphere temperature ratio, T0/T0,s.l. and M0 is the flight Mach number. With the altitude z measured in kilometers the temperature ratio is

θ0=1-0.02253z for 0<z<11km and θ0=0.752 for 11km<z<20km. A representative plot of the compressor pressure ratio R as a function of Mach number and altitude is shown in Figure E8.8. The figure is based on initial values Ra=10, 20, and 30 and flight condition pairs M0, z that are consistent for practical flight: M0=0.2, 0.4, 0.6, 0.8, and 0.9 at the respective altitudes of 0, 4, 8, 12, and 14km.

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

50

z=0

4km

8km

12km 14km

45

30

40 35 pt,3/pt,2 30

20

25 20 15

10

10 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

M0

Figure E8.8 Compressor pressure ratio as a function of Mach number and altitude

for three compressors with sea level static pressure ratios of 10, 20, and 30

(c) For a compressor with a stagnation pressure ratio of 20 under standard sea level static conditions we may determine the value at an altitude of 11km and M0=0.85 for

ηad,c=100% by applying Equation (E8.37). The result is Rb=24.85. However, if the efficiency η = 85% we must use Equation (E8.36) which yields a quadratic equation whose solution 24.81, is within the accuracy used. Thus, it is clear that the efficiency, which is assumed to be constant, doesn’t materially affect the compressor ratio at different flight speeds and altitudes at the same engine rpm N. Recall that the efficiency of adiabatic compression is determined largely by boundary layer fluid dynamics and therefore will be subject to Reynolds number effects which are dependent upon altitude and speed though it has also been indicated that under normal operating conditions the Reynolds number effects are small. The change in compressor pressure ratio with a change in inlet stagnation temperature can be explained by turning to the thermodynamic T,s plane as shown in Figure E8.9.

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

pt,3b pt,3a pt,3c

T

Tt,3c

pt,2a

Tt,3a Tt3,b Tt,2c Tt,2a Tt,2b

a

s

Figure E8.9 Constant work compression processes on a T, s plane for different

starting stagnation temperatures

For constant work, (Tt,3 – Tt,2) = constant, starting at the same stagnation pressure, a reduction in starting stagnation temperature, like Tt2b, will lead to a higher output pressure, thereby increasing the compressor pressure ratio. The converse is true for a higher starting stagnation temperature, like Tt,2c. The basic effect is due to the fact that on a T, s plane the constant pressure lines diverge as entropy is increased. This effect remains even for efficiency less than 100%.

E8.6 Another consideration for off-design performance calculations concerns the flow rate through the engine at flight conditions different from those at the design point.

Considering the compressor to act like a constant volume pump over the range of flight conditions as long as the engine rpm N is maintained constant has been found to be a quite accurate assumption. The mass flow entering the compressor may be written in terms of the axial component of the axial velocity and the flow area as follows: m = ρ 2 c2,a A2 = ρ 2υ

The assumption of constant volume flow rate is seen to be equivalent to requiring c2,a=constant. Then the mass flow ratio G=

m b ρ 2,b p2,b T2,a = = m a ρ 2,a p2,a T2,b

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Assume, as in the previous problem, that case a is the design condition which is evaluated at standard sea level static conditions and case b covers flight in the atmosphere between sea level and 11km and Mach numbers between 0 and 1. (a) Derive an equation to estimate the mass flow ratio G in terms of the case b flight conditions and describe any further assumptions made in arriving at the final result. (b) Plot a graph showing the variation of G with as a function of flight Mach number with altitude as a parameter. (c) For a compressor processing a mass flow of 82kg/s at sea level static conditions determine the mass flow at an altitude of 11km and M0=0.85.

Solution

(a) Using the perfect gas equation of state we may write the mass flow ratio as G=

m b ⎛ p2b ⎞ ⎛ pt ,0 a ⎞ ⎛ pt ,0b ⎞ ⎛ T2 a ⎞ ⎛ Tt ,0b ⎞ ⎛ Tt ,0 a ⎞ =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ ⎟⎜ ⎟⎜ m a ⎜⎝ pt ,0b ⎟⎠ ⎝ p2 a ⎠ ⎜⎝ pt ,0 a ⎟⎠ ⎜⎝ Tt ,0 a ⎟⎠ ⎝ T2b ⎠ ⎜⎝ Tt ,0b ⎟⎠

For the conditions given the stagnation pressure and temperature remain constant from station 0 to station 2 so that we may rewrite this equation as follows: ⎛ p ⎞ ⎛ pt ,2 a ⎞ ⎡⎛ p0b ⎞⎛ pt ,0b ⎞ ⎛ p0 a ⎞ ⎤ ⎛ T2 a ⎞ ⎛ Tt ,2b ⎞ ⎡⎛ T0 a ⎞⎛ Tt ,0 a ⎞ ⎛ T0b ⎞ ⎤ G = ⎜ 2b ⎟ ⎜ ⎜ p ⎟ p ⎟ ⎢⎜ p ⎟⎜ p ⎟ ⎜⎜ p ⎟⎟ ⎥ ⎜⎜ T ⎟⎟ ⎜ T ⎟ ⎢⎜ T ⎟⎜ T ⎟ ⎜⎜ T ⎟⎟ ⎥ ⎝ t ,2b ⎠ ⎝ 2 a ⎠ ⎣⎢⎝ 0 a ⎠⎝ 0b ⎠ ⎝ t ,0 a ⎠ ⎦⎥ ⎝ t ,2 a ⎠ ⎝ 2b ⎠ ⎣⎢⎝ 0b ⎠⎝ 0 a ⎠ ⎝ t ,0b ⎠ ⎥⎦ Using the isentropic relations between static and stagnation conditions and assuming

γ =7/5, we obtain 5

−

5

5

−

5

⎛ p ⎞⎛ T ⎞⎛ 5 + M 02b ⎞ 2 ⎛ 5 + M 22b ⎞ 2 σ 0b ⎛ 5 + M 02b ⎞ 2 ⎛ 5 + M 22b ⎞ 2 G = ⎜ 0b ⎟⎜ 0 a ⎟⎜ = ⎟ (E8.38) ⎜ ⎟ ⎜ 2 ⎟ ⎜ 2 ⎟ σ 0 a ⎝ 5 + M 02a ⎠ ⎝ 5 + M 22a ⎠ ⎝ p0 a ⎠⎝ T0b ⎠⎝ 5 + M oa ⎠ ⎝ 5 + M 2 a ⎠ Here σ 0 = ρ0/ρsl =σ 0(z) is the atmospheric density ratio which is a function of altitude z.

The Mach numbers at the compressor face, M2a and M2b, are generally around 0.5. More importantly, the ratio of M2a to M2b is only on the order of the ratio of the square root of the temperature ratio at the compressor face. It is easy to verify that the last term on the right hand side of Equation (E8.38) is very close to unity. Then, if the reference case a is sea level static conditions, Equation (E8.38) becomes simply 5

⎛ M 2 ⎞2 G = σ 0b ⎜1 + 0b ⎟ 5 ⎠ ⎝

20

(E8.39)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The mass flow ratio is depicted in Figure E8.10 as a function of M0 with altitude as a parameter in order to illustrate the general behavior.

1.8

z=0

1.6 1.4 1.2 G

4

1 0.8

8

0.6 12 14

0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

1.2

M0

Figure E8.10 Variation of the mass flow parameter G as a function of flight

Mach number for a range of altitudes between 0 and 14km

(c) For a compressor processing a mass flow of 82kg/s at sea level static conditions the mass flow at an altitude of 11km and M0=0.85 may be readily calculated from Equation (E8.39) using σ0b=0.37479 to yield G=0.425.

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E9. AIRBREATHING ENGINE PERFORMANCE AND COMPONENT INTEGRATION E9.1 Consider an engine with the compressor and turbine performance maps given in Figures E9.1- E9.3. The exhaust nozzle minimum section has a diameter d6=8.55in and the flight condition considered is takeoff at M0=0.25 and sea level standard day conditions. Assuming that the turbine inlet temperature is Tt4=3400R determine the engine operating rpm N (in % of the design value), the free stream capture area A0, and the net thrust produced. The nozzle may be assumed to be choked with A7=A6. The gas properties may be assumed as follows: γ2=7/5, γ4=4/3, and R=1716ft2/s2-R. 2.4 N=110% 100% 90% 80%

2.2 2

70%

pt,4/pt,5 1.8 1.6 1.4 1.2 12.4

12.6

12.8

13

13.2

13.4

13.6

13.8

m4 in lbm/s

Figure E9.1 Turbine pressure ratio as a function of corrected mass flow m4 = m θ 4 δ 4

0.95 N=100%

0.9

90%

ηt 0.85

110% 80%

0.8 70% 0.75 1.2

1.3

1.4

1.5

1.6

1.7

1.8 pt,4/pt,5

Figure E9.2 Turbine efficiency map

1

1.9

2

2.1

2.2

2.3

2.4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

12 11 10 9 8 7 pt,3/pt,2 6 5 4 3 2 1

110% 100%

compressor surge line

η=87%

90%

80% 85%

80% N=70%

80% exit choke line

8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 m2 in lbm/s

Figure E9.3 Compressor pressure ratio as a function of corrected mass flow m2 = m θ 2 δ 2 Solution The mass flow at the choked nozzle exit, station 6, is p δ m = ρ6 A6V6 = 6 gA6 γ 6 RT6 = 6 gA6 pref RT6 θ6

γ6 ⎛

⎜1 + RTref ⎝

γ 6 −1 2

− ( γ 6 +1)

⎞ 2(γ 6 −1) M 62 ⎟ ⎠

Then the corrected weight flow at station 6, for γ6 = 4/3, pref = 2116lb/ft2, Tref = 520R, A6 = 0.399ft2, and M6 =1 is m θ 6 δ 6 =19.33 lbm s The nozzle may be assumed to be ideal so that pt,5 = pt,6 and Tt,5 = Tt,6 and therefore m θ5 δ 5 = m θ 6 δ 6 = 19.33 lb s The corrected weight flow at station 5 is related to that at station 4 by m θ5

δ5

=

m θ 4

Tt ,5 pt ,4

δ4

Tt ,4 pt ,5

The turbine map of Figure E9.1 gives the pressure ratio of the turbine pt,4/pt,5 as a function of the corrected weight flow at station 4 and the turbine efficiency ηt which is related to the turbine pressure and temperature ratio by

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual γ −1 − 4 ⎤ ⎡ γ4 ⎛ ⎞ p ⎥ = 1 − ηt ⎢⎢1 − ⎜ t ,4 ⎟ ⎥ ⎜ ⎟ Tt ,4 p ⎢⎣ ⎝ t ,5 ⎠ ⎥⎦

Tt ,5

Then, entering Figure E9.1 with a value of corrected weight flow, one may read off corresponding values of pt,4/pt,5 and ηt for each engine speed N and calculate the corrected weight flow at station 5 using the following:

m θ5

δ5

⎡ ⎛ p ⎞ −0.25 ⎤ m θ 4 ⎛ pt ,4 ⎞ t ,4 = ⎜⎜ ⎟⎟ 1 − ηt ⎢1 − ⎜⎜ ⎟ ⎥ δ 4 ⎝ pt ,5 ⎠ ⎢ ⎝ pt ,5 ⎟⎠ ⎥ ⎣ ⎦

For convenience we write the corrected mass flow at any station i as follows: mi = m θi δ i This permits the construction of a plot relating the corrected weight flows at stations 4 and 5 as shown in Figure E9.4. The calculated value of the corrected weight flow at station 5 is m5 = m θ5 / δ 5 = 19.33 lbm s and is shown on Figure E9.4 as a dashed line.

Where that line crosses the curves for the different corrected rpm of the turbine defines the values of m4 = m θ 4 / δ 4 required.

20 19.9 19.8 19.7 19.6 m5 19.5 19.4 19.3 19.2 19.1 19

N=100%

90%

80%

required m5

13

13.05

13.1

13.15

13.2

13.25

13.3

13.35

13.4

m4

Figure E9.4 Correlation of the corrected mass flows m4 and m5 at stations 4 and

5, respectively. The dashed line is the required value of m5

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual For each of these 3 required values of m4 shown in Figure E9.4 we may enter the turbine map of Figures E9.1 and E9.2 and extract the corresponding values of pt,4/pt,5 and ηt and calculate the work done by the turbine at each of the 3 values of N using Wt = c p ,4Tt ,4ηt ⎡1 − ( pt ,4 pt ,5 ) ⎢⎣

− ( γ 4 −1) γ 4

⎤ = η ( 938.4 Btu lbm ) ⎡1 − ( p p )−0.25 ⎤ t t ,4 t ,5 ⎥⎦ ⎣ ⎦

The results for turbine work are shown in Table E9.1.

Table E9.1 Turbine work m4 pt,4/pt,3 ηt (lbm/s) 13.03 1.55 0.75 13.22 1.54 0.87 13.31 1.52 0.88

N (%) 100 90 80

Wt (Btu/lbm) 73.04 83.54 82.07

The relationship between the corrected mass flow at station 4, m4, and that at station 2, m2, is given by m2 = m4

pt ,4 pt ,3

Tt ,2

pt ,3 pt ,2

Tt ,4

Rearranging this equation and assuming no stagnation pressure loss in the combustor, that is, pt,4=pt,3, we may write pt ,3 pt ,2

=

Tt ,4 Tt ,2 m4

⎛ 2.541 ⎞ m2 = ⎜ ⎟ m2 ≈ 0.193m2 ⎝ m4 ⎠

From Figure E9.4 we see that for each value of N there is a required value of m4 and for these values the slope of the preceding equation equals varies less than 1% from the value shown so that this is an essentially unique straight line that defines the required compressor pressure ratio to match the mass flow at both stations. We may superimpose this matching line on the compressor map of Figure E9.3 and this is presented in Figure E9.5. The turbine work shown in Table E.1 must equal the work done by the compressor which is given by c p ,2Tt ,2 ⎡⎛ pt ,3 ⎞ ⎢⎜ Wc = ⎟ ηc ⎢⎜⎝ p,2 ⎟⎠ ⎣

(γ 2 −1) γ

⎤ 123.36 Btu lbm ⎡⎛ p ⎞0.286 ⎤ ⎢⎜ t ,3 ⎟ − 1⎥ − 1⎥ ≈ ηc ⎥ ⎢⎜⎝ pt ,2 ⎟⎠ ⎥ ⎣ ⎦ ⎦

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

12 11 10 9 8 7 pt3/pt2 6 5 4 3 2 1

110% 100%

compressor surge line

η=87%

90%

80% 85% 80% exit choke line

N=70%

required pt,3/pt,2 line

8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 m2 (lbm/s) 80%

Figure E9.5 The matching line for consistent mass flow is shown on the

compressor map as a function of m2, the corrected mass flow at station 2 and the matching point is shown on that line as a solid symbol

Using the compressor map the compressor work may be estimated at several values of N along the matching line and the results are shown in Table E9.2

Table E9.2 Compressor work

N(%)

pt,3/pt,2

ηc

Wc

100 90 80

6 5 4

0.86 0.85 0.82

98.35 86.90 74.98

Within the accuracy of the graphs used, the matching point is at approximately N = 84% with m2 = m θ 2 δ 2 ≈ 24.6 lbm s , Wc=84Btu/lbm, and a compressor pressure ratio pt,3/pt,2 = 4.75, as indicated by the solid symbol on Figure E9.5.. Because pt,2/pt,0 ~1 at M0 = 0.25 and T0 = Tref and p0 = pref, the corrected weight flow is given by m θ 2

δ2

= m

Tt ,2 pref Tref pt ,2

= m 1 + 0.2 M 02 (1 + 0.2 M 02 )

Then

5

−3.5

= 0.963m = 24.6 lbm s

Theory of Aerospace Propulsion, Second Edition: Solutions Manual m = 25.54 lbm s = ρ 0 A0V0 = ( 0.0765 lbm ft 3 ) A0 ( 0.25 )(1117 ft s ) = 21.36 A0

Therefore the free stream capture area is A0 = 25.54/21.36 = 1.196 ft2. The nozzle is choked so the exit Mach number M6=1 and the exit velocity is

V6 = a6 = γ 6 R (T6 Tt ,6 ) Tt ,6 =

( 4 3) (1716 ft 2 s 2 ⋅ R ) ( 6 7 ) Tt ,6

The exit stagnation temperature is ⎛ ⎛T ⎞ Wt ⎞ Tt ,6 = Tt ,5 = ⎜ t ,5 ⎟ T,4 = ⎜1 − T ⎜T ⎟ ⎜ c T ⎟⎟ t ,4 t ,4 p ,4 t ,4 ⎝ ⎠ ⎝ ⎠ ⎡ ⎤ 84 Btu lbm Tt ,6 = ⎢1 − ⎥ 3400 R = 3096 R ⎣ ( 0.276 Btu lbm ⋅ R )( 3400 R ) ⎦

Then V6 = 2,464ft/s. The exit pressure may be obtained from the following chain rule: ⎛ p ⎞⎛ p ⎞⎛ p ⎞⎛ p ⎞⎛ p ⎞⎛ p ⎞ ⎛ p ⎞ p = ⎜ 6 ⎟⎜ t ,6 ⎟⎜ t ,5 ⎟⎜ t ,4 ⎟⎜ t ,3 ⎟⎜ t ,2 ⎟ ⎜ t ,0 ⎟ p0 ⎜ p ⎟⎜ p ⎟⎜ p ⎟⎜ p ⎟⎜ p ⎟⎜ p ⎟ p ⎝ t ,6 ⎠⎝ t ,5 ⎠⎝ t ,4 ⎠⎝ t ,3 ⎠⎝ t ,2 ⎠⎝ t ,0 ⎠ ⎝ 0 ⎠ −4

3.5 −1 ⎛7⎞ p = ⎜ ⎟ (1)(1.52 ) (1)( 4.75 )(1) (1 + 0.252 5 ) ( 2116 lb ft 2 ) = 3, 727lb ⎝6⎠ The net thrust is given by

Fn = m (V6 − V0 ) + A6 ( p6 − p0 ) 25.54 lbm s ( 2, 464 ft s − 279 ft s ) + ( 0.399 ft 2 ) ( 3727lb − 2116lb ) 2 32.17 ft ⋅ lbm s ⋅ lb Fn = 2,377lb

Fn =

E9.2 Consider an engine with the compressor and turbine performance maps given in the Figures E9.6 and E9.7. The exhaust nozzle minimum section has a diameter d6=41.4cm and the flight condition is cruise at M0=0.8 at an altitude of 5km. Assuming that the turbine inlet temperature Tt4=1000C determine the corrected rpm N (in % of the design value) and the free stream capture area A0. The nozzle may be assumed to be choked with A7=A6. The gas properties may be assumed as follows: γ2=1.40, cp,2=1.0kJ/kg-K, γ4=1.33, cp,4=1.16kJ/kg-K and R=287m2/s2-K. The reference pressure and temperature are 101.3kPa and 288K, respectively. The pressure and temperature at 10km altitude are 54.0kPa and 255K, respectively.

Solution

The mass flow at the choked nozzle exit, station 6, is

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m = ρ6 A6V6 =

p6 δ A6 γ 6 RT6 = 6 A6 pref RT6 θ6

γ6 ⎛

⎜1 + RTref ⎝

γ 6 −1

− ( γ 6 +1)

⎞ 2(γ 6 −1) M 62 ⎟ 2 ⎠

2.8 2.6 2.4 2.2 2 pt,4/pt,5 1.8 1.6 1.4 1.2 1 15

15.4

15.8

16.2

16.6

17

17.4

17.8

18.2

m4 in kg/s

Figure E9.6 Turbine pressure ratio as a function of corrected mass flow m4 = m θ 4 δ 4 . The lines for corrected speed in the range 80%< N<100% fall

closely upon one another so that the curve shown is for all those values of corrected N. In addition, the turbine efficiency may be taken as ηt=90% for all corrected speeds N. 10 9

surge line

95%

8 90%

pt3/pt2 7 6

85%

5

N=80%

4 32

36

40

44

48

52

56

60

m2 in kg/s

Figure E9.7 Compressor pressure ratio as a function of corrected mass flow m2 = m θ 2 δ 2 . The compressor efficiency may be taken to be constant at

ηc=85% for all corrected speeds N shown.

Then the corrected mass flow at station 6, for γ6 = 1.33, pref,=101.3kPa, Tref,=288K, A6 = 0.785(0.414m)2=0.1346m2, and M6 =1 is

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m θ 6

δ6

= 31.9kg / s

The nozzle may be assumed to be ideal so that pt,5 = pt,6 and Tt,5 = Tt,6 and therefore

m θ5

δ5

=

m θ 6

δ6

= 31.9kg / s

The corrected mass flow at station 5 is related to that at station 4 by m θ5

δ5

=

m θ 4

Tt ,5 pt ,4

δ4

Tt ,4 pt ,5

The turbine map in Figure E9.6 gives the pressure ratio of the turbine pt,4/pt,5 as a function of the corrected mass flow at station 4 and the turbine efficiency ηt which is related to the turbine pressure and temperature ratio by γ −1 − 4 ⎤ ⎡ γ4 ⎛ ⎞ Tt ,5 p ⎥ = 1 − ηt ⎢⎢1 − ⎜ t ,4 ⎟ ⎥ ⎜ ⎟ Tt ,4 p ⎢⎣ ⎝ t ,5 ⎠ ⎥⎦

Then, entering Figure E9.6 with arbitrarily chosen values of corrected mass flow, read off corresponding values of pt,4/pt,5 and ηt (here considered constant at 90%) for each engine speed N and calculate the corresponding corrected weight flow at station 5 using the following: m θ5

δ5

⎡ ⎛ p ⎞ −0.248 ⎤ m θ 4 ⎛ pt ,4 ⎞ t ,4 ⎥ = ⎜ ⎟ 1 − ηt ⎢1 − ⎜⎜ ⎟ p δ 4 ⎝⎜ pt ,5 ⎠⎟ ⎢ ⎝ t ,5 ⎠⎟ ⎥ ⎣ ⎦

This permits the construction of a plot relating the corrected weight flows at stations 4 and 5 as shown in Figure E9.8. The required corrected mass flow at station 4 is thus found to be m4=17.23kg/s. Then the corresponding turbine pressure ratio may be determined from the turbine performance map, as shown in Figure E9.9.

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

40 36 32 m5 28 (kg/s) 24 20 16 15

15.4

15.8

16.2

16.6

17

17.4

17.8

m4 (kg/s)

Figure E9.8 Plot of corrected mass flow at station 4 as a function of corrected

mass flow at station 5.

2.8 2.6 2.4 2.2 2 pt,4/pt,5 1.8 1.6 1.4 1.2 1 15

15.4

15.8

16.2

16.6

17

17.4

17.8

18.2

m4 (kg/s)

Figure E9.9 Turbine performance plot showing the required turbine pressure ratio

Assuming no change in stagnation pressure across the combustor the corrected mass flow at station 2 (where γ2=1.4) is

m θ 2

δ2

=

m θ 4

δ4

(

255 K 1 + 0.2 ( 0.8 ) Tt 2 pt 4 = 17.23 Tt 4 pt 2 1273K

2

) p = 8.2 p t3

t3

pt 2

pt 2

This line is plotted on the compressor map, as shown in Figure E9.10 and where the compressor work equals the turbine work we have the matching point.

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

10 9

surge line

8.2(pt,3/pt,2)

8 pt3/pt2 7 6

90%

95%

85%

5 N=80%

4 32

36

40

44

48

52

56

60

m2 (kg/s)

Figure E9.10 Compressor map with compressor pressure ratio line showing

matching point for corrected mass flow m2 and N The relation between the corrected turbine work and the corrected compressor work is Wt

θ4

=

Wc Tt ,2 Wc 288 W = = 0.226 c θ 2 Tt ,4 θ 2 1273 θ2

The turbine work is γ −1 − 4 ⎤ ⎡ γ4 ⎛ ⎞ p ⎥ = 47.5 = c pTref ηt ⎢⎢1 − ⎜ t ,4 ⎟ ⎥ ⎜ ⎟ p θ4 ⎢⎣ ⎝ t ,5 ⎠ ⎥⎦

Wt

Then Wc/θ2 =210kJ/kg, and we know that γ2

⎡ ηc Wc ⎤ γ 2 −1 = ⎢1 + ⎥ pt ,2 ⎢⎣ c p ,2Tref θ 2 ⎥⎦ pt ,3

This yields pt,3/pt,2 = 5.41 at an rpm of N = 85% and this corresponds to a corrected mass flow of 8.2(5.41) = 44.4kg/s and a compressor work Wc = 210kJ/kg(Tt,2/Tref) or Wc = 210kJ/kg. The corrected mass flow in the free stream is m θ 0

δ0

=

m θ 2 pt ,2

Tt ,0

δ2

Tt ,2

pt ,0

At the flight Mach number M0 = 0.8 we expect the pressure recovery to be perfect and the stagnation temperature to remain constant so that the corrected mass flow at station 0 is equal to that at station 2. Then the actual captured mass flow is

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

m = ( 44.4kg / s )

pt ,0 pref

−

1

1

⎛ p ⎞ ⎛ p ⎞ ⎛ T ⎞ 2 ⎛ T ⎞2 = ( 44.4kg / s ) ⎜ t ,0 ⎟ ⎜ 0 ⎟ ⎜ t ,0 ⎟ ⎜ ref ⎟ ⎜ ⎟ Tt ,0 ⎝ p0 ⎠ ⎝ pref ⎠ ⎝ T0 ⎠ ⎝ T0 ⎠

Tref

The reference and local atmospheric pressure and temperature were given and the static to stagnation pressure and temperature ratios are given by the isentropic relations yielding 1

⎛ 54kPa ⎞⎛ 288 ⎞ 2 m = ( 44.4kg / s ) (1 + 0.2M ) ⎜ ⎟⎜ ⎟ = 36.1 kg s ⎝ 101.3kPa ⎠⎝ 255 ⎠ 2 3 0

The free stream capture area may be found from the mass flow relation A0 =

m m m = = p0 ρ 0V0 γ RT0 M 0 p0 M 0 RT0

RT0

γ

= 0.191m 2

E9.3 Assume that a turbine performance map may be characterized by the following equation for the corrected mass flow as a function of the turbine pressure ratio: m θ 4

δ4

⎛p ⎞ = 15 ⎜ t ,4 ⎟ ⎜p ⎟ ⎝ t ,5 ⎠

0.2

This relation may be considered to apply with constant turbine efficiency ηt=90% for all % corrected rpm N. The corresponding compressor performance map is that shown previously in Figure E9.3. The compressor efficiency may also be considered constant at ηc=85%. Assume that, for a certain nozzle minimum area, the corrected mass flow at station 5 is given (in kg/s) by m θ5 = 31.9 δ5 For the case where Tt,4=1300C, find the matching point (pt,3/pt,2, %N, and Wc) for the turbine-compressor unit if the aircraft is traveling at M0=0.8 at an altitude of 5km (p0=54kPa and T0=255K). It may be assumed that the stagnation pressure across the combustor is constant (pt,3/pt,4=1).

Solution

(a) The corrected mass flow at station 5 is related to that at station 4 by m θ5

δ5

= 31.9kg / s =

11

m θ 4

δ4

Tt 5 pt 4 Tt 4 pt 5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual We know the following: m θ 4

δ4

1/5

⎛p ⎞ = 15 ⎜ t 4 ⎟ ⎝ pt 5 ⎠

γ −1 ⎧ − 4 ⎤⎫ ⎡ γ4 ⎛ ⎞ p ⎪ ⎥⎪ = ⎨1 − ηt ⎢1 − ⎜ t 4 ⎟ ⎢ ⎥⎬ Tt 4 pt 5 ⎠ ⎝ ⎪ ⎪ ⎢⎣ ⎦⎥ ⎭ ⎩

Tt5

Then ⎛p ⎞ 31.9 = 15 ⎜ t 4 ⎟ ⎝ pt 5 ⎠

65

⎧⎪ ⎡ ⎛ p ⎞−0.248 ⎤ ⎫⎪ t4 ⎥⎬ ⎨1 − 0.9 ⎢1 − ⎜ ⎟ p ⎢⎣ ⎝ t 5 ⎠ ⎥⎦ ⎪⎭ ⎪⎩

(E9.1)

Choosing (pt,4/pt,5) = 1, 2, and 3 yields, for the right hand side of Equation (E9.1), the values 17, 31.9, and 49.6kg/s. Obviously, the correct answer is (pt,4/pt,5) =2. This then yields the corrected mass flow, in kg/s, at station 4 as

m θ 4

δ4

= 17.23

(b) Assuming no change in stagnation pressure across the combustor the corrected mass flow at station 2 (where γ=1.4) is

m θ 2

δ2

=

m θ 4

Tt ,2 pt ,4

δ4

Tt ,4 pt ,2

= 17.23

(

255K 1 + 0.2 ( 0.8 ) 1573K

2

) p = 7.37 p t ,3

t ,3

pt ,2

pt ,2

This line is plotted on the compressor map, as shown in Figure E9.11 and where the compressor work equals the turbine work we have the matching point. The relation between the corrected turbine work and the corrected compressor work is Wt

θ4

=

Wc Tt ,2 Wc 288 W = = 0.183 c θ 2 Tt ,4 θ 2 1573 θ2

The turbine work is γ −1 − ⎡ ⎤ γ ⎛ ⎞ p Wt t ,4 ⎢ ⎥ = c pTref ηt ⎢1 − ⎜ ⎟⎟ ⎥ = 47.5 ⎜ θ4 p ⎢⎣ ⎝ t ,5 ⎠ ⎥⎦

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

10 7.367(pt,3/pt,2) 9

surge line

8 pt3/pt2 7 6

90%

95%

85%

5 N=80%

4 32

36

40

44

48

52

56

60

m2 (kg/s)

Figure 9.11 Compressor map with compressor pressure ratio line and an arrow

showing the matching point Then Wc/θ2 =261kJ/kg, and we know that γ

⎡ η Wc ⎤ γ −1 = ⎢1 + c ⎥ pt ,2 ⎢⎣ c p Tref θ 2 ⎥⎦ pt ,3

This yields pt,3/pt,2 = 7.31 at an rpm of N = 95% and this corresponds to a corrected mass flow of 7.37(7.31) = 53.9kg/s and a compressor work Wc = 260kJ/kg(Tt,2/Tref) or Wc = 260kJ/kg.

E9.4 A mini-turbojet (11in diameter by 19in long; weight under 30lb) operates at standard sea level conditions passing 2.2lbm/s of air. The axial turbine has a pressure ratio of 4.2 at a rotational speed of 60,000rpm and the turbine inlet temperature Tt4=1750F. Assume that the efficiency of the centrifugal compressor is ηc=85%, the efficiency of the axial turbine ηt=95%, the burner efficiency ηb=90%, and the heating value of the fuel HV=18,900Btu/lb. Determine (a) The compressor pressure ratio pt,3/pt,2 (b) The static thrust Fn (c) The specific fuel consumption cj Solution

(a) The work done by the turbine is assumed to be equal to the work required by the compressor and this may be expressed as follows:

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual γ 4 −1 γ 2 −1 ⎡ ⎤ ⎡ ⎤ γ4 γ2 ⎛ ⎞ ⎛ ⎞ c T p p p ,2 t ,2 ⎢ ⎢ ⎥ ⎥ t ,5 t ,3 ηt c p ,4Tt ,4 ⎢1 − ⎜⎜ ⎟⎟ ⎥ = ⎜⎜ ⎟⎟ − 1⎥ ⎢ ηc ⎝ pt ,2 ⎠ p ⎢⎣ ⎝ t ,4 ⎠ ⎥⎦ ⎢⎣ ⎥⎦

Here it is assumed that the average values of the specific heats in the turbine section and the compressor section are given respectively by cp,4 = 0.276Btu/lb and cp,2 = 0.24Btu/lb while the average specific heat ratios are γ4 =1.33 and γ2 =1.4. For the stipulated turbine pressure ratio of 4.2 we find the compressor pressure ratio pt,3/pt,2 =54.3. (b) The static thrust is equal to the gross thrust which is given by

6 + A6 ( p6 − p0 ) Fg = mV We assume a choked nozzle with exit conditions p6 and A6, neither of which are known at this point. We may determine the stagnation pressure in the nozzle as follows: pt ,5 =

pt ,5 pt ,4 pt ,3 pt ,2 pt ,0 pt ,4 pt ,3 pt ,2 pt ,0 p0

p0

Assuming that the stagnation pressure loss through the burner is negligible this becomes ⎛ lb ⎞ lb ⎛ 1 ⎞ pt ,5 = ⎜ ⎟ (1)(15.3)(1)(1) ⎜ 2116 2 ⎟ = 7, 708 2 = 53.53 psia ft ⎠ ft ⎝ 4.2 ⎠ ⎝ Thus the pressure ratio across the nozzle is p0/pt,5 =0.274 and indeed a simple converging nozzle will be choked. Assuming that losses in the nozzle are negligible and that M6=1 we have γ6

p6 p p ∗ ⎛ 2 ⎞ γ 6 −1 = 6 = =⎜ = 0.540 ⎟ pt ,6 pt ,5 pt ,6 ⎝ γ 6 + 1 ⎠ Here we have assumed that in the nozzle γ6 = γ 4 =1.33 and therefore p6 = 4,162lb/ft2 = 28.9psia. Similarly for the temperature we have

T6 T 2 T* = 6 = = = 0.857 Tt ,6 Tt ,5 Tt ,5 γ 6 + 1 The turbine exit stagnation temperature may be found using the work equation to obtain γ 4 −1 ⎧ ⎡ ⎤⎫ γ4 ⎛ ⎞ p ⎪ ⎢ ⎥ ⎪ = 1581R t ,5 Tt ,6 = Tt ,5 = Tt ,4 ⎨1 − ηt ⎢1 − ⎜ ⎟ ⎜ p ⎟ ⎥⎬ ⎪ ⎝ t ,4 ⎠ ⎥ ⎪ ⎢ ⎣ ⎦⎭ ⎩

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Then the static temperature at the nozzle exit is T6 =1,355R. The exit velocity of the choked nozzle equals the sound speed V7 = a7 = (γ 7RT7)1/2 =1,759ft/s. The area at the nozzle exit may be found from the mass conservation equation: A7 =

m m m = = p6 ρ 6V6 γ 6 RT6 M 6 p6 M 6 RT6

RT6

γ6

= 0.0217 ft 2

The diameter d7=1.99in and the static thrust for the choked nozzle is 2.2 lbm s 1759 ft s + 0.0217 ft 2 ( 4,162 lb ft 2 − 2116 lb ft 2 ) 32.17 ft ⋅ lbm s 2 ⋅ lb Fg = 120.3lb + 44.5lb = 164.8lb Fg =

(c) The mass flow of fuel consumed may be found by using the energy balance across the combustor which yields m f c p ,4Tt ,4 − c p ,3Tt ,3 ( 0.276 Btu lbm ⋅ R × 2210 R ) − ( 0.24 Btu lbm ⋅ R × Tt ,3 ) = = m a ηb Q f − c p ,4Tt ,4 0.9 × 18,900 Btu lbm − ( 0.276 Btu lbm ⋅ R × 2210 R ) The compressor exit stagnation temperature is found from the turbine work equation as γ 4 −1 ⎡ ⎤ c p ,4Tt ,4 ⎢ ⎛ pt ,5 ⎞ γ 4 ⎥ 1− ⎜ Tt ,3 = Tt ,2 + ηt = 1243R ⎟ c p ,2 ⎢ ⎜⎝ pt ,4 ⎟⎠ ⎥ ⎢⎣ ⎥⎦

Then the fuel flow rate is m f = ( 2.2 lbm s )

m f = ( 2.2 lbm s )( 0.019 ) = 0.0418 lbm s m a

The static specific fuel consumption is cj =

3600m f Fn

=

3600 ( 0.0418 lbm s ) = 0.913 lbm hr lb (164.8lb )

Practical considerations The pressure ratios of both the turbine and the compressor are high if single stage compressors and turbines are considered, particularly for the small-scale engine treated here. In an engine with single stages the more likely case would have the compressor pressure ratio equal to 4.2 which would lead to the more reasonable pressure ratio across a single stage axial flow turbine of pt,4/pt,5=1.347. Following through with the previous analysis leads to a turbine exit stagnation pressure pt,5=6,597lb/ft2=45.8psia and a turbine

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual exit stagnation temperature Tt,5=2,061R. The static conditions would be p5= 3563lb/ft2 = 24.74psia and T6 = 1766R leading to V6 = 2,008ft/s. Then the static thrust developed is Fg=179lb and the specific fuel consumption cj=1.196lbm/hr/lb. Practical values for a real engine with the given dimensions and conditions on flow rate and combustor temperature are Fg=125lb and cj=1.2lbm/hr/lb. The variation of thrust and specific fuel consumption with compressor pressure ratio for this engine are shown in Figures E9.12 and E9.13.

200 160 120 Fg (lb) 80 40 0 0

2

4

6

8

10

12

14

16

pt,3/pt,2

Figure E9.12 The variation of thrust with compressor pressure ratio

1.8 1.6 1.4 1.2 1 cj (lbm/hr/lb) 0.8 0.6 0.4 0.2 0 0

2

4

6

8

10

12

14

16

pt,3/pt,2

Figure E9.13 The variation of specific fuel consumption with compressor

pressure ratio

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E9.5 The large (29m long and a take-off mass around 29,000kg) Soviet DBR 123 unmanned reconnaissance vehicle sketched in Figure E9.14 was powered by the KR-15 jet engine and was in service between the mid-1960s and the late 1980s.The vehicle was ground launched using two solid propellant rockets which were jettisoned after burnout.

The DBR 123 could achieve a range of 3,000km flying at around 22km altitude at speed in excess of Mach 2. The maximum speed attainable was around M0=2.55 and it carried a fuel load of 19,000 liters. Assuming that the lift to drag ratio could be represented reasonably by the equation L/D=4(M0+3)/M0: (a) Estimate the specific fuel consumption at a constant flight speed of M0=2.25 at the given altitude (b) Estimate the thrust of the KR-15 engine (c) Estimate the inlet area A1 and the nozzle exhaust area A7 assuming matched operation

Figure E9.14 Sketch of Soviet DBR 123 unmanned reconnaissance vehicle

Solution

(a) The basic Breguet equation for R, the range covered in cruise at constant M, cj, and L/D is given in Equation (F.7) in Appendix F as

R=−

aM ⎛ L ⎞ ⎛ mF ⎞ ⎟ ⎜ ⎟ ln ⎜1 − cj ⎝ D ⎠ ⎝ m1 ⎠

The quantities mF and m1 denote the mass of fuel used and the mass of the vehicle at the start of cruise, respectively. Any units may be used but they must be such that the units of range are consistent with the units of a/cj, where a is the local speed of sound and cj is the specific fuel consumption. At an altitude of 22km the atmospheric pressure and temperature are 4.046kPa and 218.6K, respectively. The sound speed is therefore a0 = 296.4m/s. Assuming flight at constant speed in the stratosphere, where a is approximately constant, the flight is also at constant Mach number and the specific fuel consumption based on the range equation, for R=3,000km, becomes

cj = −

( 296.4 m s ) M 0 L ln ⎛1 − mF ⎞ 3 ×106 m

17

D

⎜ ⎝

⎟ m1 ⎠

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This equation suggests that a high L/D means high specific fuel consumption. However, it really expresses the fact that at higher values of L/D a reduced fuel fraction mF/m1 may be used for a given mission range. The fuel capacity of the DBR-123 is given as 19,000 liters and because jet fuel has a mass density of about 0.8kg/l the total mass of the fuel is mF=15,200kg. Because the vehicle is launched to speed by expendable rockets the mass at the start of cruise may be taken as the take-off mass mto~m1=29,000kg and therefore the ratio mF/m1=0.5241. Therefore, a reasonable estimate of the specific fuel consumption is given by c j = 7.334 × 10−5 s −1M 0

L L = 0.264hr −1M 0 D D

The rule of thumb for L/D was given in the problem statement so we find that

⎡ k ( M 0 + 3) ⎤ −1 c j ≈ 0.264hr −1M 0 ⎢ ⎥ = 0.264hr ⎡⎣ k ( M 0 + 3) ⎤⎦ M 0 ⎣ ⎦

(E9.2)

The rule of thumb quoted in the problem statement has k=4. As shown by Sforza (Manned Spacecraft Design Principles, Elsevier, 2015) this rule of thumb becomes increasingly optimistic as the Mach number increases beyond 2. Available data suggest that the coefficient should probably lie between 2.5 < k < 3.3. For example, the XB-70A Valkyrie supersonic bomber had a lift to drag ratio range of 5.9<L/D<7.5 between 1.65<M0<3 and the SR-71 Blackbird had an even lower L/D of around 5 around M=3.

The sparse information available concerning the Tumansky KR-15 engine used in the DBR-123, which was an expendable variant of the engine used in the MiG-25 Foxbat, an aircraft capable of Mach 3.2, quotes a value of cj = 2.75lbm/hr/lb for static thrust under full afterburner conditions. A more detailed article (written in the Czech language) is available online (www.leteckemotory.cz/motory/r-15/-44k ) and it has some information on the R-15B-300, a later version of the KR-15, used on the MiG-25. The net thrust and specific fuel consumption for that engine at an altitude of 20km and M0=2.25 are quoted as Fn=43.4kN and cj = 2.4lbm/hr/lb, respectively. As a comparison, the Pratt & Whitney TF30 used on the Grumman F-14A shows Fn=35.6kN and cj=2.82lbm/hr/lb at M0=2.25 and 18.3km altitude. The estimated result for the specific fuel consumption number will be discussed in more detail subsequently.

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) The net thrust is a maximum at the beginning of cruise, because the vehicle mass is greatest at that point, and is given by W m g 29, 000kg ( 9.807 m s ) 284.4kN Fn ,1 = D = 1 = 1 = = L D L D L D L D 2

(E9.3)

At constant L/D the thrust decreases throughout the constant speed cruise as the weight of the aircraft decreases and the aircraft slowly ascends. The specific fuel consumption may also be calculated on the basis of the net thrust according to cj =

m f Fn

=

m 0 ( f a ) ρ 0 A0 a0 M 0 ( f a ) = Fn Fn

( 0.0645 kg m ) ( 296.4 m s ) A M ( f a ) c = 3

0

j

(E9.4)

0

Fn

We don’t know the fuel to air ratio f/a, but it is likely to fall between rather narrow limits.

For a non-afterburning engine f /a ~ 0.02 and for afterburning engines f /a > 0.4 so it is reasonable to assume that in this case a typical value would be f /a= 0.05. The area A0 is discussed next in part (c). (c) The gross thrust produced by the engine is Fg = Fn + Fr. The ram drag depends upon the inlet area, assuming no inlet spillage, Fr =γ0p0A0M02, and we may assume that the captured streamtube area is approximately equal to the engine frontal area, A0=A2=0.785d22=0.785*(1.45m)2=1.65m2.

Discussion of results The rule of thumb L/D=k(M0+3)/M0 was used in the net thrust Equation (E9.3) for values of k=3 and 4 and the results are shown in Figure E9.15. Also shown on Figure E9.15 are the reported values of Fn for the R-15 at M0=2.25 and 20km altitude and for the Pratt & Whitney TF30 at M0=2.25 and 18km altitude. The results suggest that k=3 is a more appropriate coefficient for the empirical L/D equation. The results for specific fuel consumption cj based on the Breguet Equation (E9.2) and on Equation (E9.4) using f/a=0.05 are shown in Figure E9.16. Also shown are the reported levels of cj for the R-15 at M0=2.25 and 20km altitude and for the TF30 at M0=2.25 and 18km altitude. We see that the calculated results for Fn and cj are quite reasonable if k=3 is used. The Breguet equation estimate for cj is about one-third higher than that made with Equation (E9.4) and this is likely due to uncertainty in the fuel weight and gross weight information. Because 19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual the original KR-15 engine was an expendable engine it is reasonable to expect the specific fuel consumption to be higher than in the more refined and reusable R15B-300.

50 45 40 35 30 Fn (kN) 25 20 15 10 5 0

R-15-B TF30

k=3 k=4

1.5

1.75

2

2.25

2.5

2.75

3

M0

Figure E9.15 Fn as a function of M0 for k=3 and 4 calculated with Equation

(E9.3). Also shown are reported values of Fn for the R-15 (M0=2.25, z=20km) and for the Pratt & Whitney TF30 (M0=2.25, z=18km)

6 5.5 5 4.5 cj 4 (lbm/h/lb) 3.5 3 2.5 2

k=4

Breguet eq.

k=3 k=4

f/a=0.5

k=3

TF30 R-15B

1.5

1.75

2

2.25

2.5

2.75

3

M0

Figure E9.16 Specific fuel consumption cj as a function of M0 for k=3 and 4

based on the Breguet Equation (E9.2) and on Equation (E9.4), assuming f/a=0.5.

Also shown are the reported cj for the R-15 (M0=2.25, z=20km) and for the TF30 (M0=2.25, z=18km)

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E9.6 In 1965, GE won the contract to develop engines for the USAF's Lockheed C-5A 'Galaxy' transport, the first of the jumbo jets. The GE TF39 engines introduced the high bypass turbofan. Before the TF39, bypass ratios of turbofan engines were less than twoto-one. The TF39 showed that bypass ratios of 8 were possible, far more than the bypass ratio of 2 that was common at the time. This innovation resulted in specific fuel consumption as much as 25% less than previous other available engines. In the late 60's, Lockheed and Douglas entered the commercial wide-body airliner market with their trijet offerings. GE introduced a new engine design, the CF6-50, modeled after the TF39 for use on the Douglas DC-10 and Lockheed L-1011. Consider a generic layout of the CF6-50 as shown in Figure E9.17, and assuming that the engine is operating in a ground-based test cell, determine: (a) the power developed by the low pressure and high pressure turbines, (b) the thrust developed by the engine for the case of ideal expansion of the fan flow and the core flow, (c) the specific fuel consumption 1

2

3A

3B

3

4

4A

5

7

Vf V6 =3 outer driveshaft inner driveshaft

Figure E9.17 Fan pressure ratio pt,3A/pt,2=1.72, low pressure compressor pressure

ratio pt3B,/pt2=1.69, and overall pressure ratio pt,3/pt,2=29.4. Compressor and turbine efficiencies ηc=ηt=0.9 and burner efficiency ηb=0.95. In the compressor

γ=1.4 and in the turbine and beyond γ=1.33. The temperature Tt,4=1360C and the heating value Qf = 44MJ/kg. The fan flow rate m f =540kg/s and the core flow rate m c =124kg/s Solution

(a) The power generated by the low-pressure turbine must equal the power required by the fan and the low-pressure compressor, or 21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Plpt = Pf + Plpc = m f W f + m cWlpc

Here the specific work is in kJ/kg and the mass flow is in kg/s. The compressor work in general is given by Wc = (cpTt,initial/ηc)Xc where Xc = compressor pressure function = [(pt,outl/pt,in) (γ−1)/γ -1]. In the cold section of the engine we will assume γ =1.4 and cp =1kJ/kg-K. The overall pressure ratio is given as 29.4, the fan pressure ratio is given as 1.72, and the low pressure compressor pressure ratio is given as 1.69. The information in the caption of Figure E9.17 may be interpreted as given so that the high pressure compressor pressure ratio must be 29.4/1.69=17.4. The overall pressure ratio is fixed and in some engines the fan section feeds the low pressure compressor so that one may consider the low pressure compressor pressure ratio to be 1.69x1.72=2.907 and the high pressure compressor ratio to be 29.4/(1.72x1.69)=10.11. In such a case the distribution of temperatures through the compressor sections is somewhat different from what we will find but the overall results would be essentially equivalent. (i) Fan: The fan work and the fan exit temperature are given by Wf = (1kJ/kg)(289K/0.9)[(1.72)0.286-1] = 53.88kJ/kg Tt,3a = (Wf /cp) + Tt,2 = (53.88kJ/kg)/(1kJ/kg-K)+289K = 342.9K.

(ii) Low pressure compressor: The low pressure compressor work and the power required from the low pressure turbine are as follows: Wlpc = (1kJ/kg)(289K/0.9)[(1.69)0.286-1] = 51.99kJ/kg Plpt = (540kg/s)(53.88kJ/kg)+(124kg/s)(51.99kJ/kg) = 35.54MW (iii) High pressure compressor: The work required to drive the high pressure compressor is Whpc = (1kJ/kg)(Tt,3b/0.9)[(pt,3/pt,3b )0.286-1].

Here the high pressure compressor has a pressure rise and entrance temperature given by pt,3/pt,3b = 29.4/1.69 = 17.40 Tt,3b = (Wlpc /cp)+Tt,2 = (51.99kJ/kg/1kJ/kg-K) +289 = 341K. The work done by the high pressure compressor and its exit stagnation temperature is

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Whpc = (1kJ/kg)(341K/0.9)[(17.4 )0.286-1] = 478.8kJ/kg Tt,3 = (478.8kJ/kg/1kJ/kg-K) + 341K =819.8K The power that must be delivered by the high pressure turbine is Phpt = (124kg/s)(478.8kJ/kg) = 59.37MW (b) Gross thrust: The gross thrust generated by the engine is the sum of the fan thrust and the nozzle thrust or Fg = Fg , f + Fg , n = m f V f + m cV7

(i) Fan nozzle: If we assume that the fan nozzle exit pressure equals the ambient pressure, and the stagnation pressure through the fan nozzle is constant, then the static-tostagnation pressure ratio at the fan nozzle exit p3a/pt,3a=p0/pt,3a. Note that since the flow entering the fan comes from the stationary atmosphere the stagnation pressure entering the fan is pt,2 = p0 = 101.3kPa, standard sea level pressure. Then p0/pt,3a = 1/1.72 = 0.5814 and from isentropic flow equations or tables one finds that at the fan nozzle exit MF=0.915. Similarly, the temperature at the fan nozzle exit is T3a = Tt,3a[1+(γ-1)Mf2/2]-1 = 342.9K/(1+0.1693) = 293.7K. The corresponding speed of sound and fan nozzle exit velocity is af = (γRT3a)1/2 = [(1.4)(287m2/s2K)(293.7K)]1/2 =343.5m/s.

Vf = M3aa3a = 0.915(343.5m/s) = 314.3m/s The fan thrust is Fg,F = (524kg/s)(314.3m/s) = 164.7kN (ii) Turbine analysis: In the hot section of the engine we assume γ =1.33 and cp =1.16kJ/kg-K. The static-to-stagnation pressure ratio of the core nozzle p7/pt,7 = p7/pt,5 = p0/pt,5 when the nozzle flow is assumed isentropic and the expansion ideal, that is, when pt,7=pt,5 and p7=p0. The turbine work is given by Wt = cpTt,inηtXt where Xt = turbine pressure function = [1 - (pout / pin )(γ−1)/γ] Thus the work of the high pressure turbine with Tin = Tt,4 =1633K (specified) is Whpt = 478.8kJ/kg = (1.16kJ/kg-K)(1633K)(0.9) Xt. This requires the turbine pressure function to be Xt = 0.2808 which corresponds to pt,4/pt,4a = 3.775. The temperature leaving the high pressure turbine is 23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Tt,4a = Tt,4 - Whpt/(1.16kJ/kg-K) = 1633K-413K =1220K. The low-pressure turbine specific work is Wlpt =

Plpt 35.54MW = = 286.6 kJ kg m c 124 kg s

Wlpt = (1.16kJ/kg-K)(Tt4a)(0.9)Xt = (1.16kJ/kg-K)(1220K)(0.9)Xt. We must then have Xt = 0.2250 which corresponds to pt,4a/pt,5 = 2.794. The temperature leaving the high pressure turbine is Tt,5 = Tt,4a - Whpt/(1.16kJ/kg-K) = 1220K-247 = 973K (iii) Core nozzle flow: If the core flow is assumed to exit the nozzle in a matched pressure condition, then p7 = p0 = pt,2 = 101.3kPa. We may find the exit Mach number from the isentropic relation p7/pt,7 = [1+ (γ-1)M72/2]-γ/(γ-1)=[1+0.165M72]-4.03.

Since the nozzle flow may be assumed to be isentropic we have pt,7 = pt,5 = (pt,5/pt4a)(pt,4a/pt,4)(pt4/pt,3)(pt,3/pt,2)pt,2 pt,7 = [(2.794)(3.775)]-1(1)(29.4)pt,2=2.787pt,2 Thus the core nozzle pressure ratio is pt,2/pt,7 = p7/pt,7 = 0.3588. Note that we have assumed that there is no stagnation pressure drop across the combustor. Then, since p7/pt,7 = [1+0.165M72]-4.03 we find that M7 = 1.324. The temperature ratio at the nozzle exit is T7/Tt,7 =T7/Tt,5= Tt,5 [1+0.165M72]-1 = 0.7755 Therefore the temperature, sound speed, and velocity at the nozzle exit are as follows: T7 = 754.6K a7 = (γRT)71/2 = 536.7m/s V7 = 1.324a7=710.6m/s.

The core nozzle thrust for ideal expansion is therefore kg ⎞ ⎛ m⎞ ⎛ Fgc = m cV7 = ⎜ 124 ⎟ ⎜ 710.6 ⎟ = 88.11kN s ⎠⎝ s⎠ ⎝

The total thrust Fg = 164.7+88.11 = 252.8kN (56,830lbs) 24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This result is 10% higher than the factory rating of Fg = 226.8kN (51,000lb). The higher value is related primarily to the relatively high values specified for the efficiency of the high pressure compressor and the turbine entry temperature. (iv) Choked nozzle exit: If instead we assume that the flow at station 7 is choked, then M7 =1 and the exit temperature and exit velocity are given by T7 = Tt,7/(1+0.165) = Tt,5/1.165 = 835K V7 = a7=(γRT)71/2 = 564.6m/s.

The exit pressure may be calculated as follows: p7 = pt,7(1+0.165)-4.03 = pt,5(1.165)-4.03 =2.787pt,2(0.5404)=1.506pt,2 The gross thrust of the core nozzle becomes Fg = m 7V7 + A7 ( p7 − p0 ) = (124 kg s )( 564.6 m s ) + ( A7 m 2 )( pt ,2 N m 2 ) (1.506 − 1)

In order to find A7 use the mass flow relation with pt,2=p0=101.3kPa: A7 =

124 kg s ( 287 m 2 s 2 ⋅ K ) ( 835 K ) m 7 m 7 = = = 0.3449m 2 2 p ρ7V7 7 (152, 600 N m ) ( 564.6 m s ) V7 RT7

This yields a core nozzle thrust Fg,choked = 87.78kN. The total thrust with a choked core nozzle is therefore Fg = 164.7kN+87.7kN = 252.4kN (56,740lbs). This value is still 10% higher than the standard factory rating. The difference in calculated and actual values is therefore not due to assumptions regarding the core nozzle expansion process, but to the factors discussed previously. (c) The specific fuel consumption may be found by considering the heat release in the combustor. Assuming the higher value for cp in the combustor because the compressor exhaust is hot, we have the following approximation: m f c p ,4 (Tt ,4 − Tt ,3 ) (1.16 kJ kg ⋅ K )(1633K − 819.8K ) = = = 0.0236 m a ηb HV − c p ,4Tt ,4 0.95 × 44 × 103 kJ kg − (1.16 kJ kg ⋅ K × 1633K ) This yields a fuel flow rate of 2.93kg/s. The specific fuel consumption

cj =

3600m f g Fn

s ⎞⎛ kg ⎞ ⎛ ⎜ 3600 ⎟⎜ 2.93 ⎟ hr ⎠⎝ s ⎠ =⎝ = 0.0417 kg hr N = 0.409 lbm hr lb 252,800 N 25

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The factory rating is 0.38lbm/hr/lb and the calculated value is therefore 7.6% higher than the standard factory rating. Though this seems close we must recall that the calculated thrust level in the denominator is already 10% too high. Therefore the calculated fuel flow rate is much too high. The burner efficiency at 95% is quite reasonable so that the inaccuracy likely lies in the combustor temperature difference (Tt,4-Tt,3) being too large.

This means that either Tt,4 is too large or Tt,3 is too small, or both.

Operationally, keeping Tt,4 below its maximum value extends engine life and for the CF650 Tt,4 is generally kept at 1577K or less. The efficiencies used in this problem are somewhat on the optimistic side and reduced high pressure compressor efficiency will result in a larger value of Tt,3 for the same work requirement. (d) Performance under modified operating conditions: Let us recalculate the properties in the hot section of the engine using the reduced turbine inlet temperature of 1577K, rather than 1633K, and further assume that the high pressure compressor efficiency is 0.85 rather than 0.9. We again assume γ =1.33 and cp=1.16kJ/kg-K. Because of the reduced efficiency of the high pressure compressor the work it requires is Whpt = 478.8kJ/kg(.9/.85) = 507kJ/kg The temperature leaving the high pressure compressor and entering the combustor is now Tt,3=Whpc/(1.kJ/kg-K)+Tt,3b=507K+341K=848K The work of the high pressure turbine must also satisfy the following equation: Whpt = 507kJ/kg=(1.16kJ/kg-K)(1577K)(0.85)Xt. This requires the turbine pressure function to be Xt = 0.3260 and therefore pt,4/pt,4a = 4.905. The temperature leaving the high pressure turbine becomes Tt,4a = Tt,4 - Whpt/(1.16kJ/kg-K) = 1577K-437K =1140K. The low-pressure turbine specific work remains the same at Wlpt =

Plpt m c

=

35.54MW kJ = 286.6 kg kg 124 s

But with the higher value for the low pressure turbine entry temperature, Tt,4a=1140K

26

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Wlpt = 286.6kJ/kg = (1.16kJ/kg-K)(Tt4a)(0.9)Xt = (1.16kJ/kg-K)(1140K)(0.9)Xt. This leads to Xt = 0.2408 and pt,4a/pt,5 = 3.035. The temperature entering the nozzle is given by Tt,5 = Tt,4a –Wlpt/(1.16kJ/kg-K) = 1140K-247 = 893K If the flow is assumed to exit the nozzle at a matched pressure condition, then p7 = p0 = pt,2 =101.3kPa. Then we may find the exit Mach number from the isentropic relation p7/pt,7=[1+0.5(γ-1)M72]-γ/(γ-1)=[1+0.165M72]-4.03.

Since the nozzle flow may be assumed to be isentropic we have pt,7 =pt,5=(pt,5/pt4a)(pt,4a/pt,4)(pt4/pt,3)(pt,3/pt,2)pt,2 pt,7 =[(3.035)(4.905)]-1(1)(29.4)pt,2=1.974pt,2 Thus the core nozzle pressure ratio is pt,2/pt,7 = p7/pt,7 = 0.506 and this leads to a nozzle exit Mach number M7 = 1.06. The temperature, sound speed and velocity at the exit is T7=Tt,5(1+0.165M72)-1=754K a7=536.5m/s V7=1.06a7=574m/s.

The gross thrust of the core nozzle becomes kg ⎞⎛ m⎞ ⎛ Fg = m 7V7 = ⎜ 124 ⎟⎜ 574 ⎟ = 71.18kN s ⎠⎝ s⎠ ⎝

The total thrust is therefore Fg =165.3kN+71.18kN = 235.9kN (53,000lbs). This value is now just 3.9% higher than the standard factory rating.

Then the specific fuel consumption may be found from c p ,4 (Tt ,4 − Tt ,3 ) m f = = m a ηb HV − c p ,4Tt ,4

(1.16 kJ kg ⋅ K )(1577 K − 848K ) ⎞ kJ ⎛ kJ − ⎜ 1.16 × 1577 K ⎟ 0.95 × 44 × 10 kg ⎝ kg − K ⎠

= 0.0212

3

This yields a fuel flow rate of 2.623kg/s. The specific fuel consumption

cj =

3600m f g Fn

s ⎞⎛ kg ⎞ ⎛ ⎜ 3600 ⎟⎜ 2.623 ⎟ hr ⎠⎝ s ⎠ =⎝ = 0.04 kg hr N = 0.393 lbm hr lb 235,900 N

27

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This value is now only 3.4% higher than the quoted value.

We see from these results that there is a good deal of sensitivity to the operational features of the engine. The change of the high pressure compressor efficiency to 85% from 90% and the 3.4% reduction of the high pressure turbine inlet temperature made a substantial difference in bringing the calculated performance of the engine more in line with actual performance.

E9.7 The Northrop Grumman MQM-74A “Chukar” target drone had a neatly tapered cigar-shaped fuselage, straight mid-mounted wings, an underslung jet engine with the intake under the wings, and a conventional tail configuration with the tailplanes set in an inverted “V” configuration. It was powered by a Williams International WR24-6 turbojet engine and was launched by RATO booster from the ground or a ship. Engine characteristics: N=61,000 rpm, centrifugal compressor, axial flow turbine type with pressure ratio of 4.2; length: 0.57m, maximum diameter: 0.27m; dry mass of engine: 13.6 kg, air flow: 1kg/s, turbine entry temperature: 955C. For sea level static conditions and assuming pt,4/pt,3=0.9, ηc=75%, ηt=90%, ηb=90%, and Qf=44MJ/kg, find (a) the compressor pressure ratio pt,3/pt,2, (b) the thrust, and (c) the specific fuel consumption Solution

(a) Here it is assumed that the average values of the specific heats in the turbine section and the compressor section are cp,4 = 1.155kJ/kg-K and cp,2 = 1.005kJ/kg-K while the average specific heat ratios are γ4 =1.33 and γ2 =1.4. Then, given the pressure ratio of 4.2, noting that Tt,4 =1228K, Tt,2 = 288K, and using the work equivalence for the turbine and the compressor, we have the following equation for the compressor pressure ratio: γ2

γ 4 −1 ⎧ ⎡ ⎤ ⎫γ 2 −1 c p ,4Tt ,4 ⎢ ⎛ pt ,5 ⎞ γ 4 ⎥ ⎪ pt ,3 ⎪ = ⎨1 + ηcηt 1− ⎜ ⎟ ⎬ pt ,2 ⎪ c p ,2Tt ,2 ⎢ ⎜⎝ pt ,4 ⎟⎠ ⎥ ⎪ ⎢⎣ ⎥⎦ ⎭ ⎩ 3.5

0.248 ⎧⎪ 1.155 kJ kg ⋅ K )(1228 K ) ⎡ ⎛ 1 ⎞ ⎤ ⎫⎪ ( = ⎨1 + ( 0.75 )( 0.9 ) ⎢1 − ⎥⎬ pt ,2 ⎪⎩ (1.005 kJ kg ⋅ K )( 288K ) ⎣⎢ ⎜⎝ 4.2 ⎟⎠ ⎦⎥ ⎪⎭

pt ,3

= 11.1

(b) The gross thrust is Fg = ρ7A7V72+A7(p7-p0); note that the ram drag is zero because we are considering the case of static thrust. However, A7 is unknown and to find it we first assume that the nozzle exit is choked (M7=1) and that the flow through the nozzle is

28

Theory of Aerospace Propulsion, Second Edition: Solutions Manual isentropic. Assuming that in the nozzle γ =1.33 then (p/pt)7 = p*/pt,5 = [2/(γ+1)]γ/(γ-1) = 0.540. In the same manner we find (T/Tt)7 = 0.858. The stagnation temperature in the nozzle is constant so Tt,7 = Tt,5 and γ 4 −1 ⎧ ⎡ ⎤⎫ 0.248 γ4 ⎛ ⎞ p ⎪ ⎢ ⎥ ⎪ = 1228 K ⎪⎧1 − .9 ⎡1 − ⎛ 1 ⎞ ⎤ ⎪⎫ = 897 K t ,5 Tt ,5 = Tt ,4 ⎨1 − ηt ⎢1 − ⎜ ⎢ ⎜ ⎨ ⎟ ⎥⎬ ⎜ p ⎟⎟ ⎥ ⎬ ⎝ 4.2 ⎠ ⎥⎦ ⎭⎪ ⎢ t ,4 ⎠ ⎪ ⎪ ⎪ ⎝ ⎣ ⎩ ⎢⎣ ⎥⎦ ⎭ ⎩

Thus the exit temperature is given by T7 = 0.858(897K) = 770K. The exit stagnation pressure is ⎛ p ⎞⎛ p ⎞ ⎛ p ⎞ ⎛ 1 ⎞ pt ,7 = pt ,5 = ⎜ t ,5 ⎟⎜ t ,4 ⎟ ⎜ t ,3 ⎟ pt ,2 = ⎜ ⎟ ( 0.9 )(11.1)(101.3kPa ) = 241kPa ⎜ p ⎟⎜ p ⎟ ⎜ p ⎟ ⎝ 4.2 ⎠ ⎝ t ,4 ⎠⎝ t ,3 ⎠ ⎝ t ,2 ⎠

Then p7 = 130.1kPa and p7-p0 = 28.8kPa. Note that the stagnation pressure pt,2 = pt,0 = p0 and p0 =101.3kPa which is the standard sea level value. Now we may find the exit area since ρ7A7V7 = 1kg/s:

A7 =

1 kg s 1 kg s = ρ7 a7 M 7 p7 M 7

RT7

γ7

= 3.13 × 10−3 m2 = 31.3cm 2

The exit (inside) diameter is calculated to be d7=6.31cm, while the actual (outside) diameter, scaled from a schematic diagram, is around 8cm. For static thrust conditions Fg = m7V7 + A7 ( p7 − p0 ) = (1 kg s )( 541.7 m s ) + 0.00313m 2 ( 28.8 kN m 2 ) = 632 N

The quoted static sea level thrust for this engine is 538N. (c) The fuel weight flow rate is found from the heat balance equation ⎛ ⎞ ⎛ ⎞ kJ kJ 1.155 × 1228 K ⎟ − ⎜1.005 × Tt ,3 ⎟ ⎜ m f c p ,4Tt ,4 − c p ,3Tt ,3 ⎝ kg − K kg − K ⎠ ⎝ ⎠ = = m a ηb Q f − c p ,4Tt ,4 ⎛ ⎞ kJ kJ 0.9 × 44 × 103 − 1.155 × 1228 K ⎟ kg ⎜⎝ kg − K ⎠

The temperature Tt,3 is found from the compressor work which is equal to the turbine work so that kJ γ 4 −1 ⎡ ⎤ 1.155 γ 4 ⎛ ⎞ c p ,4Tt ,4 ⎢ p kg ⋅ K 1 − ⎜ t ,5 ⎟ ⎥⎥ = 288 K + 0.9 Tt ,3 = Tt ,2 + ηt (1228K )( 0.299 ) = 668K ⎢ ⎜ ⎟ kJ c p ,2 pt ,4 ⎠ ⎝ 1.005 ⎢⎣ ⎥⎦ kg ⋅ K

29

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The fuel-air ratio is therefore f / a=0.0196 and the fuel mass flow rate is 0.0196kg/s. Then the specific fuel consumption becomes

cj =

3600m f g Fn

s ⎞⎛ kg ⎞ ⎛ ⎜ 3600 ⎟⎜ 0.0196 ⎟ hr ⎠⎝ s ⎠ =⎝ = 0.112 kg s N = 1.094 lbm hr lb 632 N

Small turbojets of this type have specific fuel consumption in the range of 1.0lbm/h/lb to 1.3lbm/h/lb.

Note: This engine is often reported as having a turbine pressure ratio of 4.2 indicated in the problem statement but this is erroneous. As pointed out in the solution to problem E9.4, it is actually the compressor ratio which is 4.2 and this is borne out by Leyes and Fleming (The History of North American Small Gas Turbine Aircraft Engines of North America, Smithsonian Institution, Washington, DC, 1999). However, both the thrust and the specific fuel consumption are relatively insensitive to the pressure ratios as can be seen in the results for this engine shown in Figures E9.18 and E9.19. The relationship between the turbine pressure ratio and the compressor pressure ratio is shown in Figure E9.20. The important feature is that for single stage centrifugal compressors the maximum practical pressure ratio is about 6 and lower values are more likely for smallscale compressors. This compression ratio requires a turbine pressure ratio of about 2 which is achievable in a single-stage axial flow turbine.

700 600 500 Fg (kN)

400 300 200 100 0 0

2

4

6

8

10

12

pt,3/pt,2

Figure E9.18 Static thrust Fg as a function of compressor pressure ratio

30

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

2 1.6 cj (lbm/h/lb)

1.2 0.8 0.4 0 0

2

4

6

8

10

12

pt,3/pt,2

Figure E9.19 Specific fuel consumption cj as a function of compressor pressure

ratio 5 4 practical range

3 pt,4/pt,5 2 1 0 0

2

4

6

8

10

12

pt,3/pt,2

Figure E9.20 Variation of the required turbine pressure ratio with the compressor pressure ratio

E9.8 Consider the matching of the various components of the turbojet engine with a compressor ratio pt,3/pt,2=15 in two cases: (i) where the compressor and turbine are ideal, ηc=ηt=1, and (ii) where ηc=0.85 and ηt=0.95. (a) Determine the pressure and temperature ratios across the compressor-turbine combination, pt,5/pt,2 and Tt,5/Tt,2, respectively; (b) Determine the fuel-air ratio f/a assuming flight at M0=2 at an altitude z=50,00ft where the free stream temperature T0=390K, with Qf=18,900Btu/lb, the stagnation temperature entering the combustor Tt,4=4Tt,2 and a combustor efficiency ηb=0.95; (c) Determine the net thrust Fn and the specific fuel consumption cj for inlet pressure recoveries pt2/pt0=1.0 and 0.9, respectively if M2=0.3 and A2=1.0ft2 with the nozzle operating in the matched mode; (d) Briefly compare the results for the cases of different compressor and turbine efficiencies.

31

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Solution

(a) The basic matching condition is that the compressor power required is equal to the turbine power supplied. If the mass flow through the engine is constant the matching condition reduces to the specific work required for the compressor being equal to the specific work produced by the turbine. This equality may be written as follows: γ c −1 γ −1 − t ⎤ ⎡ ⎤ ⎡ γt ⎛ ⎞ c p ,cTt ,2 ⎢⎛ pt ,3 ⎞ γ c p ⎥ = η c T ⎢1 − t ,4 ⎥ − 1 ⎜ ⎟ ⎜ ⎟ t p t t , ,4 ⎢ ⎥ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ p ηc ⎝ pt ,2 ⎠ ⎢⎣ ⎥⎦ ⎢⎣ ⎝ t ,5 ⎠ ⎥⎦

Solving for turbine pressure and temperature ratio yields γt

γ c −1 ⎧ ⎡ ⎤ ⎫γ t −1 pt ,5 ⎪ c p ,cTt ,2 1 ⎢⎛ pt ,3 ⎞ γ c ⎥⎪ = ⎨1 − ⎜ ⎟ − 1⎥ ⎬ ⎢ ⎜ ⎟ pt ,4 ⎪ c p ,tTt ,4 ηcηt ⎝ pt ,2 ⎠ ⎢⎣ ⎥⎦ ⎪⎭ ⎩ γ t −1 ⎡ ⎤ γt ⎛ ⎞ p ⎢ ⎥ t ,5 = 1 − ηt ⎢1 − ⎜ ⎟ ⎥ ⎜ ⎟ Tt ,4 p ⎢⎣ ⎝ t ,4 ⎠ ⎥⎦

Tt ,5

Then pt ,5 pt ,2

pt ,5 pt ,4 pt ,3

=

Tt ,5 Tt ,2

pt ,4 pt ,3 pt ,2 Tt ,5 Tt ,4

=

Tt ,4 Tt ,2

(b) Now, for Tt,4=4Tt,2 and assuming no pressure drop across the combustor, the pressure ratio becomes p ⎧ ⎫ 0.24 1 ⎡ 0.286 15 ) = t ,3 ⎨1 − − 1⎤ ⎬ ( ⎦ pt ,2 pt ,2 ⎩ 0.276 4ηcηt ⎣ ⎭ pt ,5

4.03

⎡ 1.1695 ⎤ = 15 ⎢1 − ⎥ ⎣ 4.6ηcηt ⎦

4.03

⎡ 0.2542 ⎤ = 15 ⎢1 − ηcηt ⎦⎥ ⎣

Likewise, the temperature ratio becomes Tt ,5

T = 4 t ,5 Tt ,2 Tt ,4 The fuel-to-air ratio is obtained from the energy balance across the combustor as

32

4.03

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Btu Btu ⎛ ⎞ ⎛ ⎞ 0.276 × 4Tt ,2 ⎟ − ⎜ 0.24 × Tt ,3 ⎟ m f c p ,4Tt ,4 − c p ,3Tt ,3 ⎜⎝ lbm ⋅ R lbm ⋅ R ⎠ ⎠ ⎝ = = Btu ⎛ Btu m a ηbQ f − c p ,4Tt ,4 ⎞ 0.95 ×18,900 − ⎜ 0.276 × 4Tt ,2 ⎟ lbm ⎝ lbm ⋅ R ⎠ Assuming that across the compressor cp=cp,c=0.24Btu/lbm-R and across the turbine cp=cp,t=0.276Btu/lbm-R, the compressor exit temperature may be found from the matching condition Wc = Wt. In that case cp,c(Tt,3 - Tt,2) = cp,t(Tt,4-Tt,5) so that 0.248 ⎡ ⎤ ⎛ Tt ,5 ⎞ 0.276 ⎡ ⎛ pt ,5 ⎞ ⎤ ⎥ ⎢ ⎢ ⎥ 1 η Tt ,3 = Tt ,2 + Tt ,4 ⎜1 − = Tt ,2 1 + 6 − ⎟ ⎜ ⎟ t ⎜ T ⎟ ⎢ 0.24 ⎢ ⎜⎝ pt ,4 ⎟⎠ ⎥ ⎥ c p ,c t ,4 ⎠ ⎝ ⎣ ⎦⎦ ⎣

c p ,t

At a flight Mach number M0 = 2 an adiabatic inlet would have Tt,2 = Tt,0 = (1+M02/5)T0 so that Tt,2 =702R. Then Btu Btu ⎛ ⎞ ⎛ ⎞ 0.276 × 4Tt ,2 ⎟ − ⎜ 0.24 × Tt ,3 ⎟ ⎜ m f ⎝ lbm ⋅ R lbm ⋅ R ⎠ ⎝ ⎠ = Btu Btu

ma ⎛ ⎞ 0.95 × 18,900 − ⎜ 0.276 × 4Tt ,2 ⎟ lbm ⎝ lbm ⋅ R ⎠ ⎧ ⎡ ⎛ p ⎞0.248 ⎤ ⎫⎪ Btu Btu ⎪ − 0.24 × 702 ⎨1 + 4.6ηt ⎢1 − ⎜⎜ t ,5 ⎟⎟ ⎥ ⎬ 775 lbm ⎢ ⎝ pt ,2 ⎠ ⎥ ⎪ lbm ⎪⎩ m f ⎣ ⎦⎭ = Btu Btu m a − 775 17,955 lb lb 0.248 ⎡ ⎛p ⎞ ⎤ 606.5 − 775ηt ⎢1 − ⎜⎜ t ,5 ⎟⎟ ⎥ ⎡ ⎛ p ⎞0.248 ⎤ ⎢ ⎝ pt ,4 ⎠ ⎥ m f ⎣ ⎦ = = 0.0353 − 0.04511ηt ⎢1 − ⎜ t ,5 ⎟ ⎥ 17,180 m a ⎢ ⎜⎝ pt ,4 ⎟⎠ ⎥ ⎣ ⎦

Results for the fuel-air ratios are presented in TableE9.3.

Table E9.3 Results for various engine properties

pt,5/pt,4 Tt,5/Tt,4 pt,5/pt,2 Tt,5/Tt,2

ηc=1

ηc=0.85

ηt=1

ηt=0.95

0.3067

0.2179

0.7459

0.7011

4.600

3.269

2.984

2.804

33

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Tt,2 (R ) Tt,4 ( R) Tt,3 ( R) f /a

702.0

702.0

2808

2808

1523 0.02384

1667 0.02182

(c) The stagnation pressure entering the nozzle is pt,5 and the nozzle is assumed to be ideal and adapted to the external pressure so that the net thrust is Fn = m 7V7 − m 0V0 The area and Mach number at the compressor entrance has been given as A2=1ft2 and M2=0.3 so we may find the free stream mass flow from the mass conservation equation.

We assume that the inlet is adiabatic so Tt,2 = Tt,0, and that the entering air is perfect with

γ =1.4 and R =1716ft2/s2-R so that the mass flow becomes m 0 = m 2 = ρ 2V2 A2 =

⎛ p ⎞ 1 p2 γ RT2 M 2 A2 = pt ,2 ⎜⎜ 2 ⎟⎟ RT2 ⎝ pt ,2 ⎠ Tt ,2

Tt ,2

γ

T2

R

M 2 A2

The isentropic relations yield Tt ,2

p2 pt ,2

T2

= (1 + 0.2M 22 )

−3.5+ 0.5

= 0.9479

The stagnation conditions may be determined as follows: pt ,2 Tt ,2

= p0

pt ,0 pt ,2 p0 pt ,0

3.5 − 0.5 p pt ,2 p pt ,2 1 T0 1 + 0.2 M 02 ) = 0 = 5.832 0 ( T0 Tt ,0 T0 pt ,0 T0 pt ,0

At the altitude z=50,000ft the pressure and temperature are p0=243.6lb/ft2 and T0=390R so that the mass flow becomes lb ⎞ ⎛ 243.6 2 ⎟ ⎜ pt ,2 lbm 1.4 ft ⎛ pt ,2 ⎞ 2 ⎟⎜ 1 0.3 18.8 t = m 0 = 0.9479 ⎜ 5.832 f ( ) ( ) ⎟ ft 2 pt ,0 s 390 R ⎟ ⎜⎝ pt ,0 ⎟⎠ ⎜ 1716 2 ⎜ ⎟ s −R ⎝ ⎠ Then since a0=968ft/s the flight velocity is 1,936ft/s and therefore the ram drag is ⎛ pt ,2 ⎞ Fr = 1131⎜ lb ⎜ p ⎟⎟ ⎝ t ,0 ⎠

The mass flow and the weight flow at the nozzle exit are given by

34

Theory of Aerospace Propulsion, Second Edition: Solutions Manual ⎛ p ⎞⎛ f ⎞ f ⎞ lbm ⎛ m 7 = ⎜1 + ⎟ m 0 = 0.5843 ⎜ t ,2 ⎟ ⎜ 1 + ⎟ ⎜ p ⎟⎝ a⎠ s a⎠ ⎝ ⎝ t ,0 ⎠ ⎛ p ⎞⎛ f ⎞ lb w 7 = 18.8 ⎜ t ,2 ⎟ ⎜1 + ⎟ ⎜ p ⎟⎝ a⎠ s ⎝ t ,0 ⎠

To find the nozzle exit velocity for the matched, ideal nozzle we note that the nozzle exit stagnation pressure pt,7 = pt,5 and p7 = p0. Therefore the nozzle pressure ratio is

⎞ ⎛ 0.2542 ⎞ −3.5 ⎛ p p7 p p p p p p = 0 = 0 t ,2 = 0 t ,0 t ,2 = (1 + 0.2M 02 ) ⎜ t ,0 ⎟15 ⎜1 − ⎟ ⎜p ⎟ ηcηt ⎠ pt ,7 pt ,5 pt ,2 pt ,5 pt ,0 pt ,2 pt ,5 ⎝ t ,2 ⎠ ⎝

4.03

Here we used the pressure ratio pt,5/pt,2 found in the solution to part (b) of this problem.

We may find the Mach number from the isentropic relation 1 ⎡⎛ p7 ⎞ ⎢⎜ M7 = ⎟ 0.165 ⎢⎝⎜ pt ,7 ⎠⎟ ⎣

−0.248

⎤ − 1⎥ ⎥ ⎦

Then the static temperature, sound speed, and exit velocity follow directly: −1

−1

T7 = Tt ,7 ⎡⎣1 + 0.165M 72 ⎤⎦ = Tt ,5 ⎡⎣1 + 0.165M 02 ⎤⎦ = Tt ,0

Tt ,5

⎡1 + 0.165M 02 ⎤⎦ Tt ,2 ⎣

−1

a7 = γ 7 RT7 = 2402T7 = 49.01 T7 ft s V7 = a7 M 7 = 49.01M 7 T7 The net thrust for an adapted nozzle (p7=p0) is given by ⎡⎛ f ⎞ ⎤ Fn = m 7V7 − m 0V0 = m 0 ⎢⎜1 + ⎟ V7 − a0 M 0 ⎥ a⎠ ⎣⎝ ⎦ The specific fuel consumption is cj =

3600m f g Fn

=

f 3600m 0 g a Fn

All the information is now available and the results are shown in Figures E9.21 and E9.22. (d) The results shown in Figures 9.21 and 9.22 illustrate the importance of high component efficiencies. Note that in this example, the effect of imperfect inlet pressure recovery is about the same as that of imperfect compressor and turbine efficiencies (columns 2 and 3 in Figures 9.21 and 9.22). Considering the reasonable values of inlet

35

Theory of Aerospace Propulsion, Second Edition: Solutions Manual pressure recovery and turbomachinery efficiencies used here shows a reduction in thrust of about 22% and an increase in specific fuel consumption by about 6% compared to the ideal situation of 100% efficiency of the inlet, compressor, and turbine.

2000 1600

ηc=1 ηt=1 IPR=1

ηc=0.85, ηt=0.95 IPR=1

ηc=1 ηt=1 IPR=0.9

1

2

3

ηc=0.85 ηt=0.95 IPR=0.9

1200 Fn (lb) 800 400 0 4

Figure E9.21 Net thrust for different efficiencies and inlet pressure recoveries

(IPR) 1.6 1.5 1.4

ηc=1 ηt=1 IPR=1

ηc=0.85, ηt=0.95 IPR=1

ηc=1 ηt=1 IPR=0.9

ηc=0.85 ηt=0.95 IPR=0.9

cj 1.3 (lbm/hr/lb) 1.2 1.1 1 1

2

3

4

Figure E9.22 Specific fuel consumption for different efficiencies and inlet

pressure recoveries (IPR)

E9.9 A ramjet-powered missile flies at a Mach number M0= 3 at an altitude of 26km, where the ambient pressure and temperature are p0=0.028 atmospheres and T0=-50C. Subsonic combustion takes place in the constant area region between stations 3 and 4 and results in increasing the stagnation temperature of the hot gas. The stagnation pressure recovery of the inlet, pt,3/pt,0, is that corresponding to a normal shock standing

36

Theory of Aerospace Propulsion, Second Edition: Solutions Manual at the lip of the inlet, station 1, as shown in Figure E9.23; this corresponds to maximum mass flow entering the engine. The increase in stagnation temperature due to combustion is Tt,4-Tt,3= 600C. The combustor diameter d=25cm and the inlet area ratio A2/A3=0.68.

Assuming R=0.286 kJ/kg-K and γ = 1.4 throughout, and neglecting the (small) mass of the added fuel, provide the following information: (a) For the maximum mass flow condition shown, what is the area of the nozzle minimum section, A6 (b) For maximum thrust the nozzle must be “matched”, that is, p7 = p6. What is the exit area required to achieve this condition? (c) What is the engine thrust? (d) Carefully sketch the distribution of stagnation pressure and temperature from station 0 to station 7 and explain how engine thrust could be increased shock wave fuel

combustor

inlet

0 1

nozzle

2, 3

4, 5

6

7

Figure E9.23 Schematic diagram of ramjet engine Solution

(a) The conditions at station 1 (where γ=1.4) may be found from the given information as follows: pt,0=p0(1+M02/5)3.5=(2.836kPa)(1+32/5)3.5=104.17kPa Tt,0=T0(1+M02/5)=(223K)(1+32/5)=624.4K pt,1/pt,0=0.3283 (from compressible flow tables in Appendix A) M1=0.4752 (from compressible flow tables in Appendix A) pt,1= (pt,1/pt,0)(pt,0)=(0.3283)(104.17kPa)=34.2kPa p1=pt,1(1+M12/5)-3.5=(34.2kPa)(1+0.47522/5)-3.5=29.3kPa Tt,1=Tt,0=624.4K (adiabatic flow through the normal shock)

37

Theory of Aerospace Propulsion, Second Edition: Solutions Manual T1=Tt,1(1+M12/5)-1=(624.4K)(1+0.47522/5)-1=597.4K A1/A3=0.68 A3=0.785d32=0.785(.25m)2=0.04906m2 d3=25cm A1=0.68(0.04906m2)=0.03336m2 d1=20.61cm The mass flow ρAV =constant (when neglecting mass of fuel added) so we use the information above to determine the mass flow at station 1:

m = p1 A1M 1

γ RT1

= 1.327 kg s

For maximum mass flow M6=1 and A6=A* so

m = pt ,6 A6

⎛ p6 ⎞ Tt ,6 = 1.327 kg s ⎜ ⎟ RTt ,6 ⎜⎝ pt ,6 ⎟⎠ T6

γ

We assume that the nozzle flow from stations 4 to 7 is isentropic so Tt,6 = Tt,4 = Tt,3+600K. Similarly we assume the inlet flow from stations 1 to 3 is also isentropic, so Tt,3= Tt,1=(Tt,1/T1)T1 = 2.8(223K) = 624.4K. Therefore Tt,4 =1224.4K. We know the static to stagnation ratios at station 6 from the isentropic flow tables because M6=1, so all we need now is pt,6, and because the nozzle flow is isentropic pt,6=pt,4. Using the relations for onedimensional flow with heat addition presented in Chapter 2 allows us to connect stations 3 and 4, in between which the heat is added. For that, we need M3, which decreases isentropically within the diffuser from the value behind the normal shock at the lip of the inlet. We found M1=0.4752 from the normal shock tables for M0=3 and then found A1/A*=1.39 which corresponds to M1. Then A3/A*= (A3/A1)(A1/A*) = (0.68)-1(1.39)=2.04 This corresponds to M3 = 0.3. Going back to the equations for constant area combustors with heat addition in Chapter 4 of the text we find that for M3=0.3 the ratios Tt,3/Tt* = 0.3469 and pt,3 /pt* =1.199. Using the temperature ratio we obtain Tt,4/Tt*= (Tt,3/Tt*)(Tt,4/Tt,3)=(0.3469)(1224.4/624.4)=0.6795 Therefore M4=0.492. This corresponds to pt,4/pt*=1.117 and pt,4/pt* =(pt,3/pt*)(pt,4/pt,3)=(1.199) (pt,4/pt,3)= (1.117)

38

Theory of Aerospace Propulsion, Second Edition: Solutions Manual From this we find (pt,4/pt,3)=0.9323 and because pt,3=pt,1=34.2kPa we calculate pt,4=31.89 kPa. Using these values in our mass flow equation evaluated at station 6 allows us to find A6=0.036m2 and d6=21.43cm (b) For a matched nozzle p7 = (p7/pt,7)pt,7 = (p7/pt,7)pt,4 = p0 = 2.836 kPa so that (p7/pt,7) = 2.836kPa/31.89kPa=0.0883 and therefore M7 =2.234. From Equation (5.25) in the text we find that M7 corresponds to A7/A* = 2.066 but A6 = A* so A7 = 2.066A6 = 0.0744 m2 and d7=30.79cm (c) The thrust for a matched nozzle is just the change in momentum across the engine.

Since we may assume that the added weight of fuel is negligible the thrust is simply ⎡ T ⎤ ⎡⎛ a ⎞ ⎤ 0 ⎢⎜ 7 ⎟ M 7 − M 0 ⎥ = m γ 0 RT0 ⎢ 7 M 7 − M 0 ⎥ Fn = m (V7 − V0 ) = ma ⎣⎝ a0 ⎠ ⎦ ⎣⎢ T1 ⎦⎥

Now we need to know T7 in order to complete the equation for Fn. Because the nozzle exit Mach number M7=2.224, the static to stagnation temperature ratio there is T7/Tt,7 = 0.5006 and Tt,7 = Tt,4 =1224.4K so that we find T7 = 612.9K and Fn = 279.3 N. (d) The distribution of stagnation temperature and pressure are shown in Figure E9.24 superimposed on a notional cross-section of the engine based on the calculated diameters.

An improvement in stagnation pressure recovery through the inlet and/or an increase in the rise in total temperature would result in greater thrust. Increasing inlet pressure recovery is an active research and development field and a normal shock inlet would not be used for this M0 = 3 flight. Using an external compression inlet would yield much higher stagnation pressure recovery, and therefore thrust. Increasing the temperature, which could be done here, would ultimately run into practical problems with the structural integrity of materials at elevated temperatures. For example, consider using the standard MIL-5008B pressure recovery for this case: pt ,2 pt ,0 = 1 − 0.075 ( M 0 − 1)

1.35

= 0.8083

This increases the stagnation pressure in the inlet by a factor of 2.46 which makes it possible to have a much smaller nozzle throat with more than a 50% increase in thrust.

39

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

160 10xengine radius 120 Tt(K)/10

80

pt (kPa)

40 0 -0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x/L

Figure E9.24 Variation of stagnation pressure and temperature through the

engine is shown superimposed on a notional engine cross-section

40

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E10. PROPELLERS E10.1 A pair of propellers is flying in tandem at a separation distance sufficient to eliminate mutual interference, as shown schematically in Figure E10.1. The rear propeller is of a diameter that it just contains the slipstream of the front propeller. If that from propeller has an efficiency of 90% and both propellers provide equal thrust, calculate (a) the efficiency of the combined unit and (b) the efficiency of the rear propeller.

V0

V0

Figure E10.1 Two propellers flying at constant speed V0 and therefore fixed separation

Solution (a) We assume that the propellers may be replaced by actuator discs and the velocity at the forward and aft discs is denoted by V1 and V2, respectively, while the velocity in the slipstream of the first disc is Vs,1. The efficiency of the first propeller is

η1 =

FV 9 1 0 = FV 10 1 1

The velocity V1 at the first disc is the average of the flight and slipstream velocities so that V1 =

10 1 V0 = (V0 + Vs ,1 ) 9 2

Solving this equation yields the slipstream velocity Vs ,1 = equation is

ρV1 A1 = ρV2 A2

1

11 V0 . The conservation of mass 9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Noting that the mass flow is constant and that the thrust of each disc is the same the momentum conservation equation applied to each disc yields F = ρ AV 1 1 (Vs ,1 − V0 ) = ρ AV 1 1 (Vs ,2 − Vs ,1 )

This permits solving for the slipstream velocity of disc 2 as follows: Vs ,2 = 2Vs ,1 − V0 = 2

11 13 V0 − V0 = V0 9 9

The efficiency of the combined unit is

η=

2 FV0 2 2 2 9 = = = = FV1 + FV2 V1 V2 10 1 ⎛ Vs ,1 Vs ,2 ⎞ 10 1 ⎛ 11 13 ⎞ 11 + + ⎜ + ⎟ + ⎜ + ⎟ V0 V0 9 2 ⎝ V0 V0 ⎠ 9 2 ⎝ 9 9 ⎠

Therefore the combined unit efficiency is about 81.8% (b) Then the efficiency of the aft disc may be found, but it must be recalled that both propellers are flying at the same speed V0, as indicated above. Then the efficiency is V0 ⎛9⎞ 2⎜ ⎟ Vs ,1 V0 3 11 = = ⎝ ⎠ = η2 = 1 V 13 4 Vs ,1 + Vs ,2 ) 1 + s ,2 V0 ( 2 Vs ,1 1 + 9 11 V0 9 2

Therefore it is more efficient to have the propellers in parallel than in tandem, for then the combined unit would have an efficiency of 90%. On the other hand, the rear propeller in a tandem unit would have a smaller diameter than the forward propeller, d2/d1=0.913.

E10.2 A three-bladed propeller with AF =140, diameter of 15ft, and design lift coefficient of CL,d =0.5 is driven by a free turbine engine at standard sea level conditions.

The engine delivers 2000hp to the propeller at a free turbine rotational speed of 15,000rpm through a transmission with a gear reduction of 12:1. The power delivered decreases linearly to 1800hp as the rotational speed is decreased to 10,000rpm. Using Figure 10.16 plot the variation of the static thrust delivered by the propeller as a function of propeller rpm. For CPc<0.12 you may use the relation CTc/CPc=0.62CPc-0.5.

Solution The general power coefficient is given by

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual CP =

P ρ n3 d 5

The typical propeller manufacturer’s chart uses a more specific equation with English units: BHP the power in horsepower, σ the density ratio ρ / ρs.l., N the propeller speed in rpm, and d the propeller diameter in feet. In that case the power coefficient is BHP 2000 CP ,c = 3 5 ⎛ N ⎞ ⎛d ⎞ σ⎜ ⎟ ⎜ ⎟ ⎝ 1000 ⎠ ⎝ 10 ⎠

The thrust is given by F=

CT P C p nd

The power delivered by the engine may be expressed by the following linear equation where Ns is the engine shaft speed: ⎛ N ⎞ BHP = 1800 + 40 ⎜ s − 10 ⎟ ⎝ 1000 ⎠ The propeller is driven through a 12:1 reduction gearbox so that N=Ns/12. Then the power coefficient is ⎛ 12 N ⎞ 0.9 + 0.02 ⎜ − 10 ⎟ ⎝ 1000 ⎠ CP ,c = 3 5 ⎛ N ⎞ ⎛ 15 ⎞ σ⎜ ⎟ ⎜ ⎟ ⎝ 1000 ⎠ ⎝ 10 ⎠

The curve of CT,c/CP,c as a function of CP,c (Figure 10.16 of the text) is shown here as Figure E10.2 and can be used to determine CT,c/CP,c as a function of propeller speed N. Then the static thrust (in lb) may be found from F=

CT ,c 33, 000 BHP CP ,c

Nd

The results for static thrust as a function of propeller speed are shown in Figure E10.3

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

2.4 CL,i 0.7 0.5 0.3

2.2 2.0 1.8 CT/CP 1.6 1.4 1.2 1.0 0.8 0.04

0.08

0.12

0.16

0.2

0.24

0.28

0.32

CP

Figure E10.2 Static thrust chart for a 3-bladed propeller with AF =140 and various design lift coefficients.

11000 10000 9000 Fs (lb) 8000 7000 6000 5000 800

900

1000

1100

1200

N (rpm)

Figure E10.3 Static thrust shown as a function of propeller speed

4

1300

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E10.3 Consider the propeller chart for a three-bladed propeller shown in Figure E10.4.

The speed power coefficient is Cs=J/Cp1/5 which is a coefficient independent of the propeller diameter. A propeller must be selected for an aircraft that will cruise at V=200kts at an altitude of 3km. The engine provides 1175hp (877kW) at a shaft speed of 1105rpm. Determine the diameter, the efficiency, the blade pitch angle at 3/4 radius, and the thrust.

1

η

0.8

γ=15

0.6

30

0.4 45 0.2 0 0.5

1

1.5

2

2.5

2

2.5

Cs

2.0 1.6 1.2

γ=45

0.8

30 15

J

0.4 0.0 0.5

1

1.5 Cs

Figure E10.4 Charts of efficiency vs. speed coefficient (upper) and advance ratio vs. speed coefficient (lower) for a three-bladed propeller with various 3/4 radius blade pitch angles γ shown in degrees

Solution For convenience, Figure E10.2 of the text is shown here as Figure E10.4. At an altitude of 3km the density is ρ0=0.001765lb-s2/ft4 and the speed V0=200kts=230.3mph=337.9ft/s.

The propeller advance ratio is given by 5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

V = J= nd

ft s = 18.35 1 d 18.42 ( d ) s 337.9

The power coefficient is BHP 1175 0.5875 105 6.583 × 104 2000 2000 = = = CP = 3 5 3 5 0.8925 d 5 d5 ⎛ N ⎞ ⎛d ⎞ ⎛ 1175 ⎞ ⎛ d ⎞ σ⎜ 0.7423 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 1000 ⎠ ⎝ 10 ⎠ ⎝ 1105 ⎠ ⎝ 10 ⎠

Then the speed coefficient is Cs =

J 18.35 d = = 1.995 1/5 d 9.198 CP

From Figure E10.4 we may construct a plot of J and η as a function of the blade pitch angle at 3/4 radius β for Cs=1.995, as shown in Figure E10.5. We may choose the design point for J at the maximum efficiency. The associated propeller diameter d as a function of β is shown in Figure E10.6. The choice of d=14ft has maximum efficiency but since the fall-off in efficiency is slow it is possible to select a somewhat smaller propeller, which might reduce cost.

1.6 1.4

J

1.2

design points

1 0.8 η

0.6 0.4 0

5

10

15

20

25

30

35

40

45

50

pitch at 3/4 radius γ (deg)

Figure E10.5 The advance ratio J and the propeller efficiency η as a function of

the blade pitch angle at 3/4 radius β

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Theory of Aerospace Propulsion, Second Edition: Solutions Manual

24 22 20 d (ft)

design point

18 16 14 12 10 15

20

25

30

35

40

45

pitch at 3/4 radius b (deg)

Figure E10.6 The blade diameter as a function of the blade pitch angle at 3/4

radius β

E10.4 A figure of merit for propeller static thrust is Φ=Pi/P, the ratio of the ideal power to the actual power used in producing a given thrust F. (a)Derive the following expression for the figure of merit: CT3/ 2 Φ = 0.798 CP

(b) Rearrange the equation for Φ, using the definitions of the coefficients to form an expression which would be useful in preliminary estimation of rotor diameter of a helicopter of known weight and engine power. (c) The Boeing V-22 Osprey has a maximum mass of 27,400kg and is powered by 2 Rolls-Royce Allison engines each providing 4590kW of power to 11.6m diameter blades. Estimate the required rotor diameter at sea level and compare it to the actual value. Discuss the effect of altitude on this result.

Solution

(a) The expression for power is given in Equation (10.43) in the text as P = CP ρ n3 d 5

The corresponding expression for thrust, also from Equation (10.43), is F = CT ρ n 2 d 4

The ideal power for a propeller under static conditions is given in Equation (10.14) in the text as

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

F3 2ρ A

Pi = Then the ratio of the ideal to actual power is Φ=

Pi CT3/2 ρ 3/2 n3 d 6 1 2 CT3/2 = = π CP P CP ρ n3 d 5 2ρ A

(b) Taking the ratio of the thrust coefficient to the power coefficient we obtain CT F ρ n3 d 5 Fn = = d CP ρ n 2 d 4 P P Then, using the result from (a) we find the figure of merit to be Φ=

2 ⎛ Fnd ⎞ 2 ⎛ Fnd ⎞ ⎛ F 1 ⎞ 2F 3 1 C = = ⎜ ⎟ π ⎜⎝ P ⎟⎠ T π ⎜⎝ P ⎟⎠ ⎜⎝ ρ nd 2 ⎟⎠ πρ P 2 d

Now we have a relationship between the figure of merit, which must lie in the range 0<Φ<1, and the diameter of the propeller. Using the information given on the vehicle mass and the power an estimate of the propeller diameter may be made using a reasonable value for Φ, generally in the range of 0.7<Φ<0.8. (c) If we apply the result of part (b) to the V-22 we find the result shown in Figure E10.7.

The minimum propeller diameter at sea level is about 10m but that would have to increase by about 1.5m to accommodate operation from high altitude locations. Note that there are two rotors, each supporting half the mass of the V-22. The actual diameter of the V-22 rotors is 11.6m.

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1 0.9 0.8 0.7 0.6 Φ 0.5 0.4 0.3 0.2 0.1 0

design region

z=3km z=0

0

5

10

15

20

25

30

35

40

d (m)

Figure E10.7 Estimate of minimum propeller diameter for the V-22.

E10.5 Consider the use of 3-bladed, AF=140, 4.72m diameter propellers for a VTOL application. Two integrated design lift coefficient choices are possible for the proposed airplane: CLd=0.3 or 0.5. In takeoff (V0=0) the engine supplies 4030kW at a propeller speed of 1230rpm and in cruise (V0=250kts=128.7m/s) the engine supplies 750kW at a propeller speed of 900rpm. Determine (a) the static thrust at takeoff for each proposed propeller and the figure of merit for each, (b) The efficiency ηp in cruise of the propeller with CL,d=0.5, (c) Using these results and the information that the CL,d=0.3 case yields an efficiency in cruise of ηp=0.9 discuss the design implications of the magnitude of the integrated lift coefficient CL,i.

Solution

(a) The power coefficient used in propeller charts is given in Section 10.6.4 of the text, in English units, as CP ,c =

BHP 2000

σ ( N 1000 ) ( d 10 ) 3

5

In the present problem P=4030kW=5400hp and d=4.72m=15.5ft. Assuming takeoff under standard sea level conditions where σ=1 we obtain C P ,c =

2.7

(1)(1.23) (1.55 ) 3

5

= 0.162

From the propeller chart shown in Figure 10.16 of the text we estimate the following values for CT/CP: CT/CP=1.44 for CL,i=0.5

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual CT/CP=1.33 for CL,i=0.3 The static thrust is given, in lbs, in Section 10.6.4 as F=

CT ⎛ BHP ⎞ CT ⎛ 5, 400 ⎞ C ( 33, 000 ) = 9,347 T ⎜ ⎟ ( 33, 000 ) = ⎜ ⎟ CP ⎝ Nd ⎠ CP ⎝ 1, 230 ×15.5 ⎠ CP

Thus, the static thrust is F=13,460lb=59.87kN for CL.i=0.5 F=12,432lb=55.29kN for CL,i=0.3 The figure of merit is 2 CT3 2 2CP ⎛ CT ⎞ Φ= = ⎜ ⎟ π CP π ⎝ CP ⎠

32

⎛C ⎞ = 0.321⎜ T ⎟ ⎝ CP ⎠

32

Then Φ=0.555 for CL,i=0.5 Φ=0.492 for CL,i=0.3 (b) In the case of forward flight P=750kW=1005hp and N=900rpm. Assuming level flight under standard sea level conditions where σ=1 we obtain, using the equation for CP from part (a), we find CP =

1, 005

( 2, 000 )( 0.9 ) (1.55 ) 3

5

= 0.077

Then the advance ratio is J = 101.4

( 250kts ) = 1.817 V = 101.4 Nd ( 900 )(15.5 ft )

Entering Figure 10.15 of the text with CP=0.077 and J=1.817 we estimate η=84% at a blade pitch angle γ3/4=35o. (c) It is stated that for an integrated design lift coefficient in forward flight of CL,i=0.3 the efficiency η=0.9, 7% greater than that for Cl,i=0.5. This shows the difficulty in designing for vertical flight, for which a high value of CL,I is favorable, and designing for horizontal flight, where a low value of CL,I is desirable. A reasonable compromise must be found by the designer. A propeller with variable camber of its constituent airfoil elements would be

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual a great asset in this regard, but variable geometry solutions often end up being too heavy and not sufficiently reliable.

E10.6 Estimate the static thrust and rpm of a 3m diameter propeller with an input power of 746kW as a function of the number of blades. Assume that the propeller is limited to a tip Mach number of 0.8 and that the propeller operates at standard sea level conditions.

Discuss the results in terms of the blade loading (the fraction of the thrust carried by each blade) and its relationship to the drag losses incurred.

Solution

Using the equation for estimating CP given in Section 10.6.2 we have

⎛ 6.44 ×10−4 ⎞ 746 ⎛ 6.44 × 10−4 ⎞ Pa = 0.104 = CP = ⎜ ⎟⎟ ⎟ 3 2 ⎜⎜ 3 2 32 1 1 ( )( ) 0.8 3 ( ) ⎝ σθ ⎠ M tip d ⎝ ⎠ The estimate for the figure of merit from the same section is Φ = 0.85 − 18 B −3 2CP

Using this result we may find the static thrust Fs from the equation given in Section 10.6.2:

⎡⎛ 2 CT3 2 ⎞ ⎤ Fs = 124 ⎢⎜⎜ ⎟⎟ σ Ps ,a d ⎥ ⎢⎣⎝ π CP ⎠ ⎥⎦

23

= 124 ⎡⎣Φ σ Ps ,a d ⎤⎦

23

The results are shown in Figure E10.8 where it is seen that the static thrust produced is seen to drop off rapidly for B<4.

20 18 16 14 12 Fs (kN) 10 8 6 4 2 0 2

3

4

5

6

7

B

Figure E10.8 Static thrust as a function of B, the number of blades

11

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Theory of Aerospace Propulsion, Second Edition: Solutions Manual

The blade loading and the figure of merit are shown in Figure E10.9 where it is seen that the thrust carried by each blade is a maximum at B=3 and the figure of merit slowly increases with the number of blades.

5.0 4.0 Fs/B (kN) 3.0 2.0

Φ

1.0 0.0 2

3

4

5

6

7

8

B

Figure E10.9 Blade loading and figure of merit as a function of the number of

blades

Reducing the loading on each blade makes it easier for the boundary layer to stay attached thereby reducing the drag and improving performance. The propeller speed may be found by using the specified tip Mach number as follows: N=

⎞ θ M tip 60ω 30 ⎛ utip ⎞ 30 ⎛ γ RTslθ = ⎜ = ⎜ = 1, 732rpm M tip ⎟ = 6497 ⎟ ⎟ 2π π ⎜⎝ rtip ⎟⎠ π ⎜⎝ d 2 d ⎠

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11. LIQUID PROPELLANT ROCKET MOTORS

E11.1 An H2-O2 rocket engine is under development in an experimental facility and certain estimates of performance and geometry must be made: (a) A combustion chamber and associated propellant feed system has been designed, built, and tested. During the test program it was found that for a combustion chamber pressure of 20 atmospheres and an injection temperature of 298.16K the H2-O2 reaction occurs stoichiometrically. Under these conditions calculate the temperature in the combustion chamber. (b) A rocket nozzle was fitted to the combustion chamber and the complete unit operated on a special test stand. The nozzle exit plane was monitored and the pressure there was found to be equal to that of the ambient atmosphere in the test chamber while the temperature was measured to be 1700K. Determine the specific impulse of the engine under these conditions. (c) The static thrust of the rocket engine, as measured on the test stand, was found to be 100,000lbs. Calculate the mass flow of propellant required and the diameter of the nozzle throat. (d) The area ratio, exit-to-throat, of the nozzle on this engine was 84.66. Find the ambient pressure in the test facility.

Solution (a) The stoichiometric reaction is 1 H 2 + O2 → H 2O 2

As written, this reaction has all the reactants completely combined into one product species. Though chemically balanced, it is not at all clear that the reaction, as shown, can be in equilibrium at the given chamber pressure of 20 atmospheres. However, let us assume that is the case as a start. The general change in enthalpy of the species between injection and complete reaction in an adiabatic process is Qc − Qa = ∑ bi H i (Tc ) − ∑ ai H i (T j ) N

N

i =1

i =1

(E11.1)

For the reaction of current interest we may write this out as

{

Qc − Qa = (1mol ) H O ⎡⎢( H c − H 298 ) H O − ( H ref − H 298 ) ⎤⎥ + ΔH f , H 2O H 2O ⎦ 2 2 ⎣

{

} ⎛1 ⎞ − ⎜ mole ⎟ {⎣⎡( H − H ) − ( H − H ) ⎦⎤ + ΔH } = 0 ⎝2 ⎠ − (1mole ) H ⎡⎣( H j − H 298 ) − ( H ref − H 298 ) ⎤⎦ + ΔH f , H 2 2 H2 j

298

ref

O2

1

298

O2

f ,O2

} (E11.2)

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The injection temperature Tj is given as being equal to the reference temperature, Tref = 298.16K, making (Hj -H298)=(Href -H298)=0. Furthermore, the heats of formation ΔHf of molecular hydrogen and oxygen are zero, so the change in enthalpy is given by the first term alone in Equation (E11.2). Using the tables in Appendix G we have Qc − Qa = ( H c − H 298 ) H O − 241.8kJ = 0

(E11.3)

2

Therefore the adiabatic flame temperature is that temperature for which Equation (E11.3) is satisfied, such that (Hc-H298)H2O = 241.8kJ/mol and the enthalpy tables in Appendix G show the temperature to be Tc=4929K. (b) The one-dimensional adiabatic energy equation may be written as

Ve = 2 ( ht − he ) = 2 ⎡⎣Wc−1 ( H c − H 298 ) − We−1 ( H e − H 298 ) ⎤⎦

(E11.4)

Assuming that the stoichiometric reaction is satisfied and that only H2O is present in the chamber so that Tc=4929K and further, that throughout the nozzle the composition does not change, then at the specified value of Te = 1700K the exit velocity would be

Ve =

2 ( 241.8 kJ mol − 57.87 kJ mol ) (1000 kg ⋅ m2 s 2 ⋅ kJ ) 0.018 kg mol

Ve = 4,520 m s The specific impulse may be given as Isp=4,520m/s or Isp=Ve/g=461s. e for a (c) The measured thrust F = 100,000lbs = 444,800N and F = F = mV

matched nozzle and therefore the mass flow is given by

m = F Ve = ( 444,800 N ) ( 4,520 m s ) = 98.41 kg s The throat area is At =

m ( Ru Wc ) Tc pc Γ

(E11.5)

Water at 4929K has Cp=60.93J/mol-K so that the quantity (γc -1)/γc= Ru/Cp,c=0.135 and therefore we estimate that γc =1.16. More accurate calculations, may indicate lower values, say, γc=1.125. However, the quantity Γ is not very sensitive to the value of γ.

Using the approximate expression Γ =0.192γ+0.417 with γ=1.16 we find Γ= 0.640 while for γc=1.125 the quantity Γ=0.633, difference of about 1.1%. Therefore we will use Γ=0.640, Tc = 4929K, pc = 2.026MPa, and

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Rc =

Ru 8.31451 kJ kmol ⋅ K 1000 kg ⋅ m 2 s 2 ⋅ kJ = 461.9 m 2 s 2 ⋅ K = 18 kg kmol Wc

Substituting these values into Equation (E11.5) yields At = 0.1145m2. The diameter of the throat is therefore dt = 0.38m (d) We find pe by using the mass flow and the specified area ratio:

m = ρe AeVe = ρe ( A A* ) A*Ve = ρe ( 84.66 ) ( 0.1145m2 ) ( 4,520 m s ) = 98.41 kg s e

From this equation we find ρe=0.002246kg/m3 and substituting this into the state equation, noting that at the exit We=18kg/kmol, we find pe = ρ e ( Ru We ) Te = ( 0.002246 kg m3 )( 461.9 m s 2 ⋅ K ) (1, 700 K ) = 1.76kPa

This is a pressure of about 13.2mmHg which is readily achieved in a vacuum chamber.

At Te=1700K the ratio of specific heats is γe=1.2 and the exit Mach number is given by

Ve

Me =

γ e ( Ru We ) Te

4520 m s

=

1.2 ( 461.9 m 2 s 2 ⋅ K ) (1700 K )

= 4.66

The exit static pressure may be also be obtained from the isentropic relation −γ

e pe ⎛ γ e −1 2 ⎞ = 2026kPa ⎜1 + pe = pc Me ⎟ 2 pc ⎝ ⎠

(γ e −1)

= 2026 ⎣⎡1 + 0.1( 4.662 ) ⎦⎤ = 1.99kPa −6

The value of the static pressure obtained using the isentropic relation (which is quite sensitive to γ) with the estimated local value at the exit of γe =(1-Ru/Cp,e)-1=1.2 is about 12% higher than the value obtained from the state equation which doesn’t depend strongly on the value of γ.

We may now check to see if the flow was in chemical equilibrium throughout the nozzle by checking if the entropy remained constant. The entropy equation is 6

⎡ X i ,c

X i ,e

⎤

∑ ⎢ W ( S − R ln X ) − W ln p − W ( S − R ln X ) − W ln p ⎥ = 0 (E11.6) i =1

⎣

Ru

i ,c

c

u

i ,c

c

Ru

i ,e

c

e

u

i ,e

e

e

⎦

The sum of the first two terms in Equation (11.6) is the chamber entropy and that of the last two terms is the exit enthalpy. It is specified that Te=1700K but the pressure pe was calculated from the other information given. We have all the information required by Equation (E11.5) and carrying out the calculation we find that the entropy difference is not zero at Te=1700K, but instead at a temperature about 1740K, as indicated in Figure

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11.1, about 2.35% high. The entropy at the exit for Te=1700K is less than that in the combustion chamber which is not physically acceptable although the difference is small.

However, a more important difficulty appears when we examine more closely the applicability of our assumptions. Equilibrium reaction of a stoichiometric mixture of H2 and O2 was the subject of Exercise E4.6 and the calculated mole fractions Xi are shown as in Figure E11.2 and the molar concentrations bi are listed in Table E11.1.

0.08 0.06 0.04 0.02 0 Sc-Se -0.02 (kJ/kg-K) -0.04 -0.06 Te=1740K -0.08 -0.1 -0.12 1700 1710 1720 1730 1740 1750 1760 1770 1780 1790 1800 Te (K)

Figure E11.1 Entropy difference between the combustion chamber and the nozzle

exit shown as a function of exit temperature for the stoichiometric reaction

1 0.9 0.8 0.7 0.6 Xi 0.5 0.4 0.3 0.2 0.1 0 3000

H2O H O

3200

3400

3600

3800

4000

4200

4400

4600

4800

5000

T (K)

Figure E11.2 The equilibrium mole fractions Xi are shown as a function of

temperature for a stoichiometric mixture of H2 and O2 at p=20atm

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual From the molar concentrations and the Tables in Appendix G for ΔHf,i0 and H(T)-H(298) one may iterate to find that the temperature which satisfies Equation (E11.1) is approximately Tc=3500K. From Figure E11.2 we see that even at 3000K the H2O comprises only about 85% of the total number of moles produced by the reaction. Indeed, the H2O mole fraction decays almost linearly with temperature until it only constitutes several percent of the total mixture at 5000K. Table 11.1 Equilibrium molar concentrations bi for a stoichiometric mixture of H2 and O2 at p=20atm T (K) 3000 3200 3500 4000 5000

b1(H2) 0.1310 0.2175 0.3157 0.5593 0.5073

b2(O2) 0.04595 0.0717 0.09384 0.1603 0.1131

b3(H2O) 1.8191 1.6841 1.4952 0.9867 0.1392

b4(OH) 0.08126 0.1472 0.2648 0.4585 0.5107

b5(O) 0.00834 0.02341 0.05151 0.2345 1.1262

b6(H) 0.01849 0.04952 0.1134 0.4495 2.1962

b 2.1041 2.1935 2.3345 2.8488 4.5927

Carrying out an equilibrium flow calculation starting at Tc=3500K and ending the expansion at the specified temperature Te=1700K shows that the flow would be isentropic only if the flow could expand to pe=0.58kPa, about one-third the value computed from the mass flow calculation. Thus the area ratio is about one-third the required value and would have to increase to about 250, like that of the Pratt & Whitney RL10-B which is rated at about 25,000lb thrust.

The problem, as stated, is over-determined because Te and pe were specified independently. An isentropic expansion on a T-s plot goes straight down from Tc to a specified Te which lies on one particular pe=constant line. An adiabatic expansion starting from the same point must have ds>0, so for it to end on the specified Te it could only do so on a different constant pressure line where pe’< pe. On the other hand, if that adiabatic expansion were to end on a specified constant pressure line pe it could only do so at a different temperature Te’>Te.

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11.2 A rocket engine which uses H2 as the fuel and O2 as the oxidizer is being designed to produce 20,000lb of thrust. The engine gas properties are assumed to be as follows: γ=1.25 and W=9.80lbm/lbm-mol; the universal gas constant Ru=49,720ft2/s2-R. The chamber pressure pc=300psia and the nozzle exit pressure pe=14.7psia. The engine is to consume 60lbm/s of propellant and the nozzle is to operate in the matched mode. Under these conditions determine the following: (a) Nozzle thrust coefficient cF and specific impulse (b) Throat diameter dt (c) Nozzle exit diameter de, velocity Ve, and temperature Te (d) Chamber temperature Tc (e) The fuel mass flow and the oxidizer mass flow e for the case where the chamber temperature (in degrees K) depends upon the number of moles of H2 and O2 in the following fashion: nH Tc = 4200 − 300 2 nO2 Solution

(a) Assuming adiabatic flow throughout the nozzle so that the stagnation enthalpy ht = hc = constant, perfect operation of the nozzle, and a constant average specific heat cp so that Tt = Tc = constant, the energy equation yields the following expression for exit velocity ⎛ T ⎞ Ve = 2c p Tc ⎜1 − e ⎟ ⎝ Tc ⎠ Similarly, the mass flow through the assumed ideal nozzle with constant average thermodynamic properties and γ=1.25 as given is as follows: −

(γ +1)

p A ∗ γ ⎛ γ + 1 ⎞ 2(γ −1) ⎛ s ⋅ R1/2 ⎞ pc A ∗ 0.00925 m = c = ⎜ ⎟ ⎜ ⎟ R⎝ 2 ⎠ ft ⎠ Tc Tc ⎝ Here the gas constant has been taken as R=

Ru 1545.36 ft ⋅ lb mol ⋅ R × 32.174 lbm ⋅ ft s 2 ⋅ lb = = 5074 ft 2 s 2 ⋅ R 9.8 lbm mol W

The nozzle flow is critical at the throat so that A* is the throat area and because the engine operates in the matched mode the thrust is ⎛ T ⎞ e = 0.00925 pc A ∗ 2c p ⎜1 − e ⎟ F = mV ⎝ Tc ⎠ Then the thrust coefficient is given by

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

cF =

⎛ Te ⎞ ⎛ F s − R1/ 2 ⎞ = ⎜ 0.00925 ⎟ 2c p ⎜1 − ⎟ pc A ∗ ⎝ ft ⎠ ⎝ Tc ⎠

For ideal flow in the nozzle the isentropic relation yields γ −1

Te ⎛ pe ⎞ γ ⎛ 14.7 psia ⎞ =⎜ ⎟ =⎜ ⎟ Tc ⎝ pc ⎠ ⎝ 300 psia ⎠

0.2

= 0.547

The average specific heat may be found from the average molecular weight and average specific heat ratio according to cp =

γ

Ru = 5 ( 5074 ft 2 s 2 ⋅ R ) = 25,370 ft 2 s 2 ⋅ R γ −1 W

Substituting these values into the equation for the thrust coefficient yields cF = 1.402.

Using the given mass flow of propellants, the specific impulse is 20, 000lb ( 32.174 ft ⋅ lbm s ⋅ lb ) F Ip = = = 333s

mg 60 lbm s ( 32.174 ft s 2 ) 2

(b) The throat diameter may be found by rearranging the equation for the thrust coefficient as follows: A∗ =

F 20, 000lb = = 47.55in 2 = 0.330 ft 2 cF pc 1.402 ( 300 lb in 2 )

The throat diameter is thus 0.649ft, or 7.78in. (c) The nozzle exit velocity is just Ve=gIsp=10,710ft/s while the exit area is given by Ae =

0.00925 pc A ∗ RTe m = pe ρeVe Tc

1 γ −1 ⎡ ⎤ 2c pTc ⎢1 − ( pe pc ) γ ⎥ ⎣ ⎦

γ −1 0.00925 A ∗ R pc Te ⎡ ⎤ γ Ae = 1 p p − ( ) e c ⎢ ⎥⎦ pe Tc ⎣ 2c p

−1 2

Substituting all the known values leads to the following result for exit area:

Ae = 0.2084 ( 0.333 ft 2 ) ( pe pc )

−1 γ

⎡1 − ( p p )(γ −1) γ ⎤ e c ⎣ ⎦

−1 2

= 1.151 ft 2

The area ratio ε = Ae/A* =3.457 which corresponds to an exit Mach number Me=2.57, as does the given nozzle pressure ratio pe /pc = 0.049. The exit temperature may be found from the Mach number as follows:

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Me =

Ve 10, 710 ft s 10, 710 ft s = = = 2.57 ae γ RTe 1.25 ( 5074 ft 2 s 2 ⋅ R ) Te

This leads to an exit temperature Te = 2738R. (d) The chamber temperature may be found from ⎛ γ −1 2 ⎞ Tc = Te ⎜ 1 + M e ⎟ = ( 2738 R ) (1 + 0.125 × 2.57 2 ) = 4999 R 2 ⎝ ⎠

(e) The chamber temperature Tc = 4999R = 2777K and is said to be related to the number of moles of the fuel and oxidizer as follows:

Tc = 4200 − 300

nH 2 nO2

= 2777 K

Then the mass fraction of fuel to oxidizer is

m H 2 m O2

=

nH 2 WH 2 nO2 WO2

= 4.743

2 = 0.296 32

The sum of the mass flows of hydrogen and oxygen is 60lbm/s so the mass flow of oxygen is 46.3lbm/s and that of hydrogen is 13.7lbm/s. The stoichiometric ratio of hydrogen to oxygen is 2 (1 mole of hydrogen for every 0.5 mole of oxygen) so that the equivalence ratio for this fuel-oxygen mixture is φ = 4.743/2 = 2.37, a fairly rich mixture and the oxidizer to fuel ratio is O/F=5.97, a value typical of practical rocket engines.

E11.3 A liquid propellant rocket motor using a JP-4/RFNA propellant combination must produce 19.6kN of thrust at sea level. The characteristics of the combustion products are combustion chamber temperature Tc=3000K, average specific heat ratio γ=1.2, molecular weight W=27, and chamber pressure pc=19.34 atmospheres. Calculate the following: (a) the throat and exit areas, At and Ae, for matched nozzle exit flow at sea level assuming a nozzle efficiency ηn= 95%, (b) the characteristic velocity c*, the propellant mass flow rate, and the specific impulse of the engine at sea level, (c) the thrust developed at an altitude of 11.5km where the pressure is p0=0.2 atmospheres, (d) for comparison purposes calculate the thrust and exit area at this altitude for matched nozzle exit operation.

8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Solution

(a) For a nozzle efficiency of ηn = 95% the nozzle velocity coefficient cV = ηn1/2 = 0.975 and with a molecular weight W = 27 the gas constant R =Ru/W = 8.3145/27 = 0.308kJ/kgK=308m2/s2-K. Then, with γ =1.2, the specific heat cp = γR/(γ-1) = 6R = 1.85kJ/kg-K and Γ =0.649. The thrust coefficient is given in Equation (11.17) of the text and is reproduced here for convenience in Equation (E11.7). The current problem stipulates that the nozzle is matched so that pe = po = 1atm = 101.3kPa and therefore the pressure ratio pe/pc = 1/19.34 = 0.0517.

1/ 2

γ −1 ⎧ ⎡ ⎤⎫ F ⎪ 2γ ⎢ ⎛ pe ⎞ γ ⎥ ⎪ cF = 1− = λ cd cV Γ ⎨ ⎢ ⎜ ⎟ ⎥⎬ pc At ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎥ ⎪ ⎣ ⎦⎭ ⎩

+

Ae ⎛ pe p0 ⎞ − ⎟ ⎜ At ⎝ pc pc ⎠

(E11.7)

The nozzle exit angle is not given so it is usual to assume a perfectly contoured nozzle providing parallel flow at the nozzle exit and therefore λ =1. Likewise, without a specification of the discharge coefficient it is usual to assume it to be unity also, that is, cd =1. With this information the thrust coefficient is 16 cF = (1)(1)( 0.975 )( 0.649 ) 12 ⎡1 − ( 0.0517 ) ⎤ = 1.368 ⎣ ⎦

The throat area is therefore At = 19,600N/(19.34x101,300N/m2)(1.368) = 73.3x10-4 m2 or At = 73.3 cm2. With a discharge coefficient cd =1 the mass flow through the throat is 2 2 pc At Γ (19.34 × 101,300 N m )( 0.00733m ) ( 0.649 ) m = = = 9.69 kg s RTc ( 308 m2 s 2 ⋅ K ) ( 3000 K )

This mass flow passes through the exit of the nozzle where m = ρe AeVe . In addition, the measured thrust together with this actual mass flow yields the specific impulse for the matched nozzle as follows: Ip =

F Ve = = 206.3s

mg g

Therefore Ve = 2023m/s and we now need only the exit density to find the exit area. The flow is not isentropic since ηn = 0.95, but it is reasonable to assume it is adiabatic so that the energy equation along a streamline from the combustion chamber to the exit plane may be written as follows: cpTe+Ve2/2 = cp,cTc or, assuming a constant value for cp,

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

( 2023 m s ) = 1,893K Ve2 = 3, 000 K − 2c p 2 (1,848 J kg ⋅ K ) 2

Te = Tc −

From the equation of state pe 101,300 N m 2 ρe = = = 0.174 kg m3 2 2 RTe ( 308 m s ⋅ K ) (1893K ) The exit area is Ae =

m 9.69 kg s = = 0.0275m 2 = 275cm 2 ρ eVe ( 0.174 kg m3 ) ( 2023 m s )

(b) The characteristic velocity c* =

RTc = Γ

( 308 m s ⋅ K ) ( 3000K ) 2

2

0.649

= 1, 481 m s

We have already determined the mass flow to be 9.69kg/s and the specific impulse to be 206.3s. (c) The thrust developed by the engine at an altitude of 11.5km where p0 = 0.2atm is e + Ae ( pe − p0 ) = 19.6kN + ( 0.0275m 2 ) (1atm − 0.2atm ) (101,300 N m 2 ) = 21.83kN F = mV

(d) If the nozzle were matched at an altitude of 11.5 km the pressure ratio would become pe/pc= 0.2/19.34 = 0.01034 and the thrust coefficient would change to 16 cF = (1)(1)( 0.975 )( 0.649 ) 12 ⎡1 − ( 0.01034 ) ⎤ = 1.601 ⎣ ⎦

The thrust would then be F = cF pc At = (1.601) (19.34 × 101,300 N m 2 )( 0.00733m 2 ) = 22.99kN

The pressure ratio and area ratio are related to the Mach number by the following equations: −γ

pe ⎛ γ − 1 2 ⎞ γ −1 Me ⎟ = ⎜1 + pc ⎝ 2 ⎠ γ +1

2 Ae 1 ⎛ 2 + ( γ − 1) M e ⎞ 2(γ −1) = ⎜ ⎟ At M e ⎝ γ +1 ⎠

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Using the pressure ratio of 0.01034 in the first equation yields Me =3.38. Using that Mach number in the second equation yields Ae/At =11.57. With the throat area At = 73.3cm2 the exit area required for matched nozzle operation at 11.2km altitude is Ae = 848cm2, a large increase over that for matched operation at sea level where Ae=247.7m2.

E11.4 The Rocketdyne RS-X engine was a developmental outgrowth of the RS-27 engine used on Delta launchers. It is an RP-1/LOX engine that develops an Isp=264s at sea level and an Isp,vac =299 sec in space. The engine O/F=2.25 and the chamber pressure of pc=775psia feeds a nozzle with an expansion ratio ε=10. The engine weighs approximately 4500lbs and is 134in long and 72.6in wide (nozzle exit diameter). The nozzle has an angular divergence α=12o at the exit plane. The combustion chamber temperature may be taken as Tc=6400R while the molecular weight and ratio of specific heats are considered constants at the values 21.9 and 1.135, respectively. Experiments have shown that the nozzle discharge and velocity coefficients are cd=0.05 and cV= 0.903, respectively. Under these conditions and assuming sea level static test conditions determine: (a) the exit and throat areas Ae and At, (b) the mass flow of the propellants, (c) the thrust produced F, (d) the thrust coefficient cF, (e) the nozzle exit Mach number Me, (f) the nozzle exit pressure pe, (g) the nozzle exit velocity Ve, (h) the effective exhaust velocity Ve, (i) the characteristic velocity c*.

Solution

(a) The area ratio ε and nozzle exit diameter are given so the exit and throat areas are Ae = 0.785(72.6in/12in/ft)2 = 28.73ft2 At = Ae/ε= 28.73ft2/10 = 2.873ft2, The throat diameter is therefore dt = 22.96in. (b) The ideal mass flow of propellant comes from Equation (11.2) in the text and is given below. The gas constant is R = Ru/W = 1545ft-lb/mol-R/21.9lbm/mol =70.55ft-lb/lbm= 2270ft2/s2-R so the ideal mass flow is 2 2 2 2 pc At Γ Wc ( 775 lb in × 144 in ft )( 2.873 ft ) ( 0.64 ) m i = = ( 32.174 ft ⋅ lbm s 2 ⋅ lb ) 2 2 Ru Tc ( 2270 ft s ⋅ R ) ( 6400R )

m i = 1731lbm s

Therefore the actual mass flow is m = cd m i = 0.95 (1, 731lbm s ) = 1, 645 lbm s (c) The thrust may be obtained from the sea level value of specific impulse F ( 32.174 ft ⋅ lbm s 2 ⋅ lb ) F I sp = = = 264s

mg (1645 lbm s ) ( 32.174 ft s 2 )

11

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The thrust at sea level F=434,280 lbs (d) The thrust coefficient cF = F / pcA t= (434,280lb)/(111,600 lb/ft2)(2.873ft2) =1.354 (e) The ideal nozzle exit Mach number may be found from the area ratio according to the equation:

⎡ 2 ⎛ γ − 1 2 ⎞⎤ A* = 0.1 = M e ⎢ M e ⎟⎥ ⎜1 + Ae 2 ⎠⎦ ⎣γ +1⎝

−( γ +1) 2( γ −1)

Using the given value γ =1.135 one finds Me = 3.10. (f) The value for the ideal pressure ratio pe/pc is given by −γ

pe ⎛ γ − 1 2 ⎞ γ −1 Me ⎟ = ⎜1 + 2 pc ⎝ ⎠ The ideal expansion process takes us from the chamber pressure to the same exit pressure as the actual process, the ideal pressure and the actual pressure are equivalent and we find pe / pc= 0.0149 and pe = 11.58psia=1668lb/ft2. (g) The ideal exit velocity Ve ,i = γ e RTe ,i M e and the ideal exit temperature is

Te ⎛ γ − 1 2 ⎞ = ⎜1 + Me ⎟ 2 Tc ⎝ ⎠

−1

For the values determined thus far the ideal exit temperature Te,i = 0.6065Tc = 3882R. This temperature will be lower than the actual temperature. The speed of sound is (γeRTe,i)1/2 = 3163 ft/s and Ve,i=9805ft/s. The stagnation temperature Tc = Te + Ve2/2cp = Te,i+ Ve,i2/2cp and since Ve =ηn1/2Ve,i = cVVe,i = 0.903(9805ft/s), we have Ve = 8854ft/s. (h) The effective exhaust velocity is Ae ( pe − p0 ) m 28.73 ft 2 )( 32.174 ft ⋅ lbm s 2 ⋅ lb ) ( Veff = 8854 ft s + 1668 lb ft 2 − 2116 lb ft 2 ) = 8602 ft s ( (1645 lbm s ) Veff = Ve +

(i) The characteristic velocity is c∗ =

RTc = Γ

12

( 2270 ft s ⋅ R ) ( 6400 R ) 2

2

0.64

= 5956 ft s

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11.5 An RP-1/LOX propellant combination (O/F=2.28) is used in a rocket motor producing a thrust F=49.1kN at sea level. The chamber pressure pc=20 atmospheres and the combustion products have the following characteristics: the combustion chamber temperature may be taken as Tc=3430K while the molecular weight and ratio of specific heats are considered constants at the values 21.98 and 1.226, respectively and are assumed constant throughout the expansion. Note that this last assumption is approximate and that the actual values will change somewhat as the flow proceeds down the nozzle and the pressure and temperature drop. Determine: (a) the nozzle throat and exit areas for matched nozzle exit flow at sea level conditions, (b) the characteristic velocity and the specific impulse at sea level, (c) the evolution of p, T, M, and V along the nozzle whose coordinates are given in Table E11.2 below in terms of axial distance x/rt and corresponding radial distance r/rt. The origin is at the throat and the exit radius is not given since it is to be calculated. Using the final dimensional results also draw the nozzle.

Table E11.2 Nozzle Coordinates

x/rt -2.60 -2.0 -1.5 -1.0 -0.5 0 0.5 1.0 1.5 2.0 3.15 r/rt 2.65 2.5 2.1 1.65 1.25 1 1.15 1.35 1.5 1.7 calc. (d) the residence time of a fluid particle in the nozzle.

Solution

(a) Assuming cd, cV, and λ are all equal to unity we have At =F/cFpc and for an ideal nozzle matched at pe/pc =1/20 and with Γ = 0.652 we have 1/ 2

γ −1 ⎧ ⎡ ⎤⎫ F ⎪ 2γ ⎢ ⎛ pe ⎞ γ ⎥ ⎪ cF = 1− = λ cd cV Γ ⎨ ⎢ ⎜ ⎟ ⎥⎬ pc At ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎥ ⎪ ⎣ ⎦⎭ ⎩

+

Ae ⎛ pe p0 ⎞ ⎜ − ⎟ =1.4 At ⎝ pc pt ⎠

From this equation we find the throat area At =

49,100 N = 0.0173m 2 2 (1.4 ) ( 20 ×101,300 N m )

From the mass conservation equation we may find the exit area 2

−1 pc At Γ 2 ⎛ ρc ⎞ At Γ 2 ⎛ ρc ⎞ m m 2 ⎡ pc At Γ ⎤ Ae = ρ c p A = =⎢ = ⎡ ⎤ ( ) ⎥ ⎜ ⎟= ⎜ ⎟ ρeVe ρe F ⎢⎣ RTc ⎥⎦ ⎣ e F c t ⎦ RTc ρc cF ⎝ ρe ⎠ cF ⎝ ρ e ⎠

Because the nozzle flow is assumed to be ideal (ρc/ρe) = (pc/pe)1/γ and therefore 1γ

A Γ2 ⎛ p ⎞ Ae = t ⎜ c ⎟ cF ⎝ pe ⎠

( 0.0173m ) ( 0.652 ) 20 = 2

1.4

13

2

( )

1 1.226

= 0.0605m 2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) The characteristic velocity c* =

8.31451 kJ kg ⋅ mol ⋅ K ( 3, 430 K ) 3 2 2 RT 1 = (10 m s ⋅ kJ ) = 1, 747 m s Γ 0.652 21.98 kg kg ⋅ mol

The specific impulse at sea level is c * cF (1747 m s )(1.4 ) I sp = = = 249.4s 9.807 m s 2 g (c) The evolution of p, T, M, and V along the nozzle as a function of Mach number is shown in Table E11.3 and as a function of axial distance in Table E11.4. The variation of the ratios of flow properties to their critical values as a function of Mach number is shown in Figure E11.3. The nozzle contour is shown as the variation of radius with axial distance in Figure E11.4. The variation of Mach number as a function of axial distance in the nozzle is shown in Figure E11.5. The variation of pressure as a function of axial distance in the nozzle is shown in Figure E11.6. The variation of temperature as a function of axial distance in the nozzle is shown in Figure E11.7 and the variation of velocity as a function of axial distance in the nozzle is shown in Figure E11.8. (d)Numerically integrating dx/V to find the residence time t yields t ~ 1ms.

3.0 A/A* 2.5 V/a* 2.0 1.5 1.0

T/T*

0.5 p/p* 0.0 0.0

0.5

1.0

1.5

2.0

2.5

3.0

M

Figure E11.3 Variation of the ratios of flow properties to their critical values as a

function of Mach number

14

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Table E11.3 Evolution of flow properties along the nozzle as a function of M

M

A/A*

p/p*

T/T*

V/a*

r (cm)

p (Mpa)

T (K)

V (m/s)

0.2

3.017362

1.744259

1.107992

0.210522

12.92747

1.971012

3414.831

251.5743

0.4

1.611674

1.621883

1.093234

0.418231

9.447969

1.832727

3369.348

499.7865

0.6

1.197147

1.439761

1.069493

0.620498

8.142796

1.62693

3296.177

741.4948

0.8

1.040565

1.223845

1.037936

0.815033

7.591619

1.382945

3198.92

973.9648

1

1

1

1

1

7.442175

1.13

3082

1195

1.2

1.033464

0.788921

0.957238

1.174063

7.565671

0.891481

2950.208

1403.005

1.4

1.129272

0.603779

0.91119

1.336388

7.908592

0.68227

2808.287

1596.983

1.6

1.289293

0.450404

0.863273

1.486599

8.450368

0.508957

2660.606

1776.486

1.8

1.524128

0.32901

0.814716

1.624709

9.18778

0.371781

2510.955

1941.528

2

1.852111

0.236356

0.766529

1.751033

10.12822

0.267083

2362.442

2092.484

2.2

2.299947

0.167638

0.719494

1.866106

11.28649

0.189431

2217.481

2229.997

2.4

2.90435

0.117796

0.674186

1.970612

12.68307

0.133109

2077.841

2354.881

2.6

3.71448

0.082251

0.630995

2.065316

14.34329

0.092944

1944.728

2468.052

0.1

5.935085

1.776542

1.111744

0.105439

18.13065

2.007492

3426.394

125.9999

0.3

2.067916

1.691984

1.101795

0.314899

10.70203

1.911942

3395.731

376.3046

0.5

1.354028

1.536741

1.082422

0.520197

8.659915

1.736517

3336.023

621.636

0.7

1.099466

1.334337

1.054606

0.718858

7.803523

1.507801

3250.297

859.0356

0.9

1.009473

1.111457

1.01967

0.908808

7.47734

1.255947

3142.622

1086.026

1.1

1.008628

0.891857

0.979124

1.088458

7.474211

1.007799

3017.661

1300.707

1.3

1.073691

0.692587

0.934532

1.256726

7.711514

0.782623

2880.229

1501.787

1.5

1.200775

0.523001

0.887383

1.413015

8.155128

0.590991

2734.914

1688.553

1.7

1.396396

0.385856

0.839006

1.557153

8.794357

0.436017

2585.816

1860.798

1.9

1.675018

0.279373

0.790522

1.689315

9.631847

0.315691

2436.39

2018.731

2.1

2.059012

0.199327

0.742827

1.809936

10.67897

0.225239

2289.393

2162.873

2.3

2.579831

0.14066

0.696596

1.919633

11.95351

0.158946

2146.909

2293.962

2.7

4.216238

0.068624

0.610274

2.109242

15.28138

0.077545

1880.865

2520.544

2.5

3.280057

0.098493

0.652308

2.019139

13.47847

0.111297

2010.412

2412.871

15

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Table E11.4 Flow development as a function of axial distance x/rt

r/rt

x(cm)

r(cm)

A/At

M

p (Mpa)

T(K)

V(m/s)

-52

53

-19.34

19.7

7.02

0

2.020

3430

0

-40

50

-14.88

18.6

6.25

0.09

2.010

3427

113.4

-30

42

-11.16

15.6

4.41

0.15

1.992

3421

188.9

-20

33

-7.44

12.3

2.72

0.23

1.956

3409

289.1

-10

25

-3.72

9.30

1.56

0.4

1.833

3369

499.8

0

20

0

7.44

1.00

1

1.130

3082

1195

10

23

3.72

8.56

1.323

1.65

0.4714

2623

1819

20

27

7.44

10.04

1.823

2

0.2671

2362

2092

30

30

11.16

11.16

2.250

2.2

0.1894

2217

2230

40

34

14.88

12.65

2.890

2.4

0.1331

2077

2355

63

37

23.44

13.88

3.478

2.54

0.1035

1983

2435

25 20 15 r (cm) 10 5 0 -20

-15

-10

-5

0

5

10

15

20

25

x(cm)

Figure E11.4 Nozzle contour shown as radius vs. axial distance 3.0 2.5 2.0 M 1.5 1.0 0.5 0.0 -20

-15

-10

-5

0

5

10

15

20

25

x (cm)

Figure E11.5 Variation of Mach number as a function of axial distance

16

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

2.4 2 1.6 p (Mpa) 1.2 0.8 0.4 0 -20

-15

-10

-5

0

5

10

15

20

25

x (cm)

Figure E11.6 Variation of pressure as a function of axial distance

3500 3250 3000 2750 T (K) 2500 2250 2000 1750 1500 -20

-15

-10

-5

0

5

10

15

20

25

x (cm)

Figure E11.7 Variation of temperature as a function of axial distance

3000 2500 2000 V (m/s) 1500 1000 500 0 -20

-15

-10

-5

0

5

10

15

20

25

x (cm)

Figure E11.8 Variation of velocity as a function of axial distance in the nozzle

17

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11.6 Consider the gaseous propellant system described by the reaction

4H 2 + O2 → products Note that this reaction is fuel rich and it should be possible to reduce the number of product species accordingly so as to simplify the calculation, if desired. Use p=20atm for the chamber pressure. Determine (a) the equivalence ratio φ (b) the adiabatic flame temperature Tad. If a combustor operating under these conditions is part of a rocket engine operating on a static test stand and the nozzle is matched at an exit pressure of pe=p0=1atm calculate the following for the case of frozen flow in the nozzle and for the case of equilibrium flow in the nozzle: (c) The exit temperature Te (d) The mixture molecular weight at the exit We,mix (e) The exit velocity Ve (f) The specific impulse Isp (g) The exit area Ae Solution to the chemical reaction part of the problem (same as problem 4.8)

(a) The equivalence ratio is given by

( F O ) = ( 4mol × 2 g mol )(1mol × 32 g mol ) = 2 φ= ( F O )stoich (1mol × 2 g mol )( 0.5mol × 32 g mol )−1 −1

Thus this is a fuel-rich propellant mix. The oxidizer to fuel ratio for the present case is O/F=4 and that for a stoichiometric mixture is (O/F)stoich=8. Since this is a ratio we may use the molar concentrations and find φ =2

(b) To determine the adiabatic flame temperature, first consider the reaction to proceed as follows: 4H 2 + O2 → b1 H 2 + b2O2 + b3 H 2O + b4OH + b5O + b6 H 1. Equilibrium concentrations (i). Atom conservation (1) H: 8=2b1+2b3+b4+b6

(E11.8)

(2) O: 2=2b2+b3+b4+b5

(E11.9)

(ii) Total moles (3) b =b1+b2+b3+b4+b5+b6

(E11.10)

18

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (iii) Equilibrium sub-reactions (4) 0.5H2=H:

Kp2=Pb6/b11/2

(5) 0.5O2=O:

Kp1=Pb5/b21/2

(E11.11) (E11.12) 1/2

(6) 0.5H2+0.5O2=OH:

Kp3=b4/(b1b2)

(7) H2+0.5O2=H2O:

Kp4=(1/P)b3/b1b21/2

(E11.13) (E11.14)

Note: P=(p/b)1/2 2. Solution scheme: (i) Pick bguess=b, starting with total number of reactant moles, b=5 (ii) Pick a value for Tc and input Kpi with i=1-4 (iii) Pick a value of b1 and calculate b6 from Equation 4 (iv) From Equation (E11.8) b3=0.5(8-2b1-b4-b6). Then dividing Equation (E11.13)

by Equation (E11.14) and substituting the result for b3 yields the equation for b4 b4= (8-2b1-b6)(Kp3/Kp4)[2Pb11/2+(Kp3/Kp4)]-1. (v) From Equation (E11.8) we may now calculate b3=0.5(8-2b1-b4-b6) (vi) From Equation (E11.13) we find b2 = (b4/Kp3) 2/b1 (vii) From Equation (E11.12) we find b5=Kp1b21/2/P (viii) Check result for assumed value of b1 by using Equation (E11.9), which we haven't used yet to find the error in hydrogen E=2b2+b3+b4+b5-2 (ix) Check error in total number of moles from Equation (E11.10) E1=b1+b2+b3+b4+b5+b6-bguess 3. Carry out calculations for various values of the adiabatic flame temperature and select the value which makes the error in total number of moles as small as desired.

Interpolation results are shown in Figures 11.9 and 11.10 below. The selected solution is Tc= 3075K for which the molar concentrations and mole fractions are as follows: b1=1.958, b2=0.645x10-4, b3=1.958, b4=0.0385, b5=1.841x10-3, b6=0.129, b=4.085 X1=0.4793, X2=1.579x10-5, X3=0.4793, X4=9.424x10-3, X5=4.506x10-4, X6=0.03158

19

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

20 15 Tc=3075K

10 5

Qc-Qa

0 -5 -10 3000

3050

3100

3150

3200

3250

T (K)

Error in no. of moles

Figure E11.9 Linear interpolation result and actual Tc

0.06 0.04 0.02 0 -0.02

match at 4.067

-0.04 4

4.01

4.02

4.03

4.04

4.05

4.06

4.07

4.08

4.09

4.1

total moles

Figure E11.10 Interpolation for correct number of moles of product

Solution to nozzle flow part of the problem

For both frozen flow and equilibrium flow in the nozzle the entropy of the mixture is a constant if the flow is frictionless and no heat transfer to the walls occurs. This is the basic condition which permits the determination of the entire flow field in both cases.

(i)Frozen Flow (a) Nozzle Exit Temperature: The entropy relation between the combustion chamber and the exit plane is

∑ X ( S − S ) = R ln p N

i =1

pc

i

Tc

Te

e

20

Theory of Aerospace Propulsion, Second Edition: Solutions Manual With the conditions pc=20atm and pe=1atm, and noting that the molecular weight of the mixture is 9.62 which yields R=8.31451J/mol-K the entropy equation becomes

∑ X ( S − S ) − 24.908 = 0 N

i =1

i

Tc

Te

(E11.15)

The chamber temperature and composition was given in item 3 in part (b) of this problem and that composition is frozen so Xi = constant. One may then assume various values of Te until one is found which satisfies Equation (E11.15). Carrying this process out leads to the result Te=1700K (b) Nozzle Exit Mixture Molecular Weight: The mixture molecular weight is given by N

Wmix = ∑ X iWi i =1

Using the results given in item 3 in part (b) of this problem we find Wmix = 9.62kg/kg-mol (c) Nozzle Exit Velocity: The one-dimensional energy equation for adiabatic frictionless flow is 1 hc = he + Ve2 2

In frozen flow the composition is unchanged so we may solve for the exit velocity in terms of the molar change of enthalpy between the chamber and the exit as follows:

Ve = 2

Hc − He Wmix

The change in enthalpy is given by N

H c − H e = ∑ X i ⎡⎣ H i (Tc ) − H i (Te ) ⎤⎦ i =1

The temperatures Tc and Te are known, as are the (constant) mole fractions Xi, and we calculate that Hc – Hi = 58.57MJ/kg-mol and therefore Ve =3460m/s (d) Nozzle Specific Impulse: Assuming a matched nozzle issuing parallel flow at the exit we know that the specific impulse Isp = Ve/g so that Isp=352s (e) Nozzle Expansion Ratio: For this isentropic flow we may calculate the expansion ratio using the exit Mach number Me and the effective isentropic exponent γeff which is defined by the relation

21

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

Te ⎛ pe ⎞ =⎜ ⎟ Tc ⎝ pc ⎠

γ eff −1 γ eff

This equation yields the result γeff =1.246. Then the exit Mach number is given by Ve/ae and ae = (γeffRTe)1/2 and R=Ru/Wmix =(8.3145kJ/mol-K)/(9.62kg/kJ-mol) = 0.864kJ/kg-K = 864m2/s2-K. This yields ae =1,353m/s and Me =2.56. The area ratio equation is γ +1

⎡ γ − 1 2 ⎤ 2(γ −1) M ⎥ 1+ A 1 ⎢ 2 = ⎢ γ +1 ⎥ A* M ⎢ ⎥ 2 ⎣ ⎦ Using the calculated values for Me and γeff we find Ae/A*=3.42. Note that this is a small area ratio for a rocket nozzle and this is because the pressure ratio across the nozzle is quite modest. (f) Nozzle Thrust Coefficient: This coefficient is defined by cF = F/pcAt where At = A* is the throat area. We have no information on the size of the engine or the thrust but we have shown that it may also be written as cF = Ve/c* where c* is the characteristic exhaust velocity which is given by

c* =

1 Γ eff

RTc

γ mix

Here the quantity γ eff +1

⎛ 2 ⎞ 2(γ eff −1) Γ eff = ⎜ ⎜ γ + 1 ⎟⎟ ⎝ eff ⎠

Using the information generated thus far we find Γeff = 0.589 and c* = 2479m/s so that cF = c*/Ve=1.396. (ii) Equilibrium Flow (a) Nozzle Exit Temperature: The entropy in the combustion chamber is equal to that at the exit plane in this case as well and may be expressed by the following relation: ⎡

⎤ ⎡ ⎤ W X i ,Te ST0e − R ln X i ,Te ⎥ = R ⎢ ln pc − mix ,c ln pe ⎥ (E11.16) Wmix ,e mix ,e ⎦⎥ ⎣⎢ ⎦⎥

∑ ⎢⎢ X ( S − R ln X ) − W ⎣

i ,Tc

0 Tc

Wmix ,c

i ,Tc

(

)

22

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The adiabatic flame temperature is the same as calculated in the case of frozen flow, that is, Tc = 3075K. With the conditions pc=20atm and pe=1atm, and noting that R =8.31451 J/mol-K the entropy in Equation (E11.16) may be solved for iteratively. One may choose an exit temperature, calculate the equilibrium composition at the exit station, and use those values in Equation (11.9) until it is satisfied to some degree of accuracy. The correct temperature is found to lie between 1800K and 1900K and linear interpolation yields Te =1890K, a value 11% higher than that found for frozen flow. (b) Nozzle Exit Mixture Molecular Weight: The mixture molecular weight at any station along the nozzle is given by N

Wmix = ∑ X iWi i =1

Here we know that in the combustion chamber Wmix,c = 9.62 and carrying out an equilibrium chemistry calculation at the exit temperature and pressure we find Wmix,e = 10.012 (c) Nozzle Exit Velocity: The one-dimensional energy equation for adiabatic frictionless flow is 1 hc = he + Ve2 2

In equilibrium flow the composition is varies along the nozzle so we may solve for the exit velocity in terms of the mass change of enthalpy between the chamber and the exit.

The change in enthalpy is given by

(

)

(

)

Hc He 1 N ⎡ 1 N ⎡ 0 ⎤ − = Δ + − − X H H H X i ,Te ΔH 0f ,i + H Te − H Tref ⎤ ∑ ∑ i ,Tc f ,i Tc Tref ⎣ ⎦ ⎣ ⎦ Wc We Wc i =0 We i =1 The temperatures Tc and Te are known, as are the mole fractions Xi , and we calculate that hc – hi = 61.86MJ/kg and therefore Ve = 3517m/s (d) Nozzle Specific Impulse: Assuming a matched nozzle issuing parallel flow at the exit we know that the specific impulse Isp = Ve/g so that Isp = 359s.

23

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (e) Nozzle Expansion Ratio: For this isentropic flow we may calculate the expansion ratio using the exit Mach number Me and the effective isentropic exponent γeff which is defined by the relation Te ⎛ pe ⎞ =⎜ ⎟ Tc ⎝ pc ⎠

γ eff −1 γ eff

This equation yields the result γeff = 1.193. The exit Mach number is given by Ve/ae where ae= (γeffReTe)1/2 and R=Ru/Wmix,e=(8.3145kJ/kg-mol-K)/(10.012kg/kg-mol) = 831m2/s2-K. This yields ae=1369m/s and Me = 2.57. The area ratio equation is γ +1

⎡ γ − 1 2 ⎤ 2(γ −1) M ⎥ 1+ A 1 ⎢ 2 = ⎢ γ +1 ⎥ A* M ⎢ ⎥ 2 ⎣ ⎦ Using the calculated values for Me and γeff we find Ae/A*=3.79. Note again that this is a small area ratio for a rocket nozzle and this is because the pressure ratio across the nozzle is quite modest. (f) Nozzle Thrust Coefficient: This coefficient is defined by cF = F/pcAt where At = A* is the throat area. We have no information on the size of the engine or the thrust but we have shown that it may also be written as cF = Ve/c* where c* is the characteristic exhaust velocity which is given by

c* =

1 Γ eff

RTc

γ mix

Here the quantity γ eff +1

⎛ 2 ⎞ 2(γ eff −1) Γ eff = ⎜ ⎜ γ + 1 ⎟⎟ ⎝ eff ⎠

Using the information generated thus far we find Γeff = 0.592 and c* = 2472m/s so that cF =Ve/c*== 1.423 which is almost 2% greater than the frozen flow value.

The equilibrium flow case yields better results than the frozen flow case because some recombination occurs along the nozzle releasing some energy and raising the exit temperature and therefore the exit velocity.

24

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E11.7 Consider a rocket motor operating with a chamber pressure and temperature of pc=2.07MPa and Tc=2860K, respectively, and a nozzle which produces the optimum thrust of 1334kN when exhausting into the atmosphere under standard sea level conditions. The exhaust gases are characterized by a mean molecular weight and isentropic exponent W=21.87kg/kg-mol and γ =1.229, respectively. Determine the following properties of the rocket motor: (a) Thrust coefficient cF (b) Characteristic exhaust velocity c* (c) Nozzle minimum area At (d) Nozzle exit area Ae Solution

(a) Assuming ideal operation, the thrust coefficient is given by 1

γ −1 ⎧ ⎡ ⎤⎫2 γ ⎛ ⎞ p 2 γ ⎪ ⎢1 − e ⎥ ⎪ − Ae ⎡⎛ pe ⎞ − ⎛ p0 ⎞ ⎤ cF = Γ ⎨ ⎢⎜ ⎟ ⎜ ⎟ ⎥ ⎜ ⎟ ⎢ ⎥⎬ At ⎣⎝ pc ⎠ ⎝ pc ⎠ ⎦ ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎥ ⎪ ⎣ ⎦⎭ ⎩

The nozzle is said to produce optimum thrust so we may further assume that the nozzle is matched with pe = p0 so that the last term in the equation above is zero. We also know the nozzle pressure ratio pe / pc = 0.04834 and γ +1

⎛ 2 ⎞ γ −1 Γ= γ⎜ ⎟ ⎝ γ +1⎠ We find then that Γ=0.654 and the thrust coefficient cF =1.407. (b) The characteristic velocity is given by

RTc

c* =

Γ

The gas constant R=Ru/W = (8.31451kJ/kg-mol)/(21.87kg/kg-mol)=0.3802kJ/kg-K= 380.2m2/s2-K and therefore the characteristic velocity c* =1594m/s. (c) The definition of the thrust coefficient is cF =

F pc At

We may solve for At because we know all the other terms in this definition and find the minimum, or throat, area At=0.4573m2. (d) The exit Mach number Me may be found from the isentropic flow relation

25

Theory of Aerospace Propulsion, Second Edition: Solutions Manual γ

pc ⎡ γ − 1 2 ⎤ γ −1 = 1+ Me ⎥ 2 pe ⎢⎣ ⎦ Then the exit area may be found from the nozzle area ratio equation γ +1

Ae 1 ⎡ 2 ⎛ γ − 1 2 ⎞ ⎤ 2(γ −1) M e ⎟⎥ = ⎜1 + At M e ⎢⎣ γ + 1 ⎝ 2 ⎠⎦

This yields Me=2.567 and therefore Ae=1.621m2

E11.8 A rocket nozzle operating with a stagnation pressure pt=2.76MPa and an area ratio Ae/Amin=11 was designed for optimal operation at an altitude z=10km, but launch takes place at sea level (p=101.3kPa). Find the ratio of thrust at launch to that at the optimum altitude F(0)/F(10) and the value of γ at the nozzle exit.

Solution

The thrust is given by F = Ae ⎡⎣(1 + γ e M e2 ) pe − p0 ⎤⎦ If the engine is optimized for z=10km the thrust there would be

F (10 ) = Aeγ e pe M e2 At launch z=0 and the thrust is

F ( 0 ) = Aeγ e pe M e2 + Ae ( pe − p0 ) The ratio of the two thrusts is F ( 0) 1 − ( p0 pe ) = 1+ γ e M e2 F (10 )

At z=10km the pressure ratio pe/p0=0.26153 and pe=26.49kPa. Therefore the nozzle exit pressure ratio for adapted operation is pe/pt=(26.49kPa)/(2760kPa)=0.0096 and the ratio p0/pe=(0.26153)-1=3.824. The exit Mach number is given by the isentropic relation −γ

pe ⎛ γ e − 1 2 ⎞ e Me ⎟ = ⎜1 + 2 pt ⎝ ⎠

(γ e −1)

The ratio of exit area to minimum (throat) area is given by

26

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

γ e −1

⎡ ⎤ M e2 ⎥ 1+ Ae 1 ⎢ 2 = ⎢ γ e +1 ⎥ Amin M e ⎢ ⎥ ⎣ 2 ⎦

γ e +1 2( γ e −1)

Here the exit area ratio is given as Ae/Amin=11 and the nozzle pressure ratio is pe/pt=0.0096 but both γe and Me are unknown. For the given nozzle pressure ratio we may find Me=Me(γe) and use that in the equation for the exit area ratio to yield the area ratio as a function of γe as shown in Figure E11.11.

16 14 12

Ae/At

10 8 6

Me

4 2 1.1

1.15

1.2

1.25

1.3

1.35

1.4

γe

Figure E11.11 The exit Mach number and area ratio for a nozzle with exit

pressure matched at z=10km is shown as a function of γ. The data symbols indicate the specific results for the case considered.

For Ae/Amin =11 the ratio of specific heats corresponding to a nozzle adapted to 10km altitude is shown in Figure E11.11 to be γe=1.25, for which Me=3.5. The ratio of launch thrust to that at z=10km is shown as a function of γe in Figure E11.12. For γe=1.25 the ratio F(0)/F(10)=0.816 showing that the nozzle exit pressure imbalance leads to a 18.5% reduction in thrust from the design value.

27

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

0.86 0.84 0.82 F(0)/F(10) 0.8 0.78 0.76 0.74 1.1

1.15

1.2

1.25

1.3

1.35

1.4

γe

Figure E11.12 The variation of launch (z=0) thrust to that at z=10km is shown as

a function of γe. The data symbol indicates the specific result for the case considered.

28

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E12. SOLID PROPELLANT ROCKET MOTORS

E12.1 Calculate the dimensions of an end-burning solid propellant grain needed to pressurize the tanks of a 19.62kN thrust liquid rocket engine. The gas generator is required to produce 3.31MPa for 30 seconds. The liquid propellants are RFNA (1520kg/m3) and a hydrocarbon fuel (1050kg/m3). The mixture ratio F/O=0.3 and the mass flow rate of propellant is 10.7kg/s. The characteristics of the double base plateau burning solid propellant are: burning rate r=0.406cm/s, combustion chamber temperature Tc=2000K, molecular weight of the combustion gases W=21.2kg/kg-mol, ratio of specific heats γ=1.28 and solid propellant density ρp=1560kg/m3.

Solution A schematic diagram of the gas pressurization system is shown in Figure E12.1.

RP-1 Fuel

d

r

L

RFNA oxidizer

Figure E12.1 Schematic diagram of gas generator for pressurizing liquid propellant tanks

The gas generator will produce a mass flow of gas that must be sufficient to maintain a constant pressure on a volume υ that continually increases with time. It is clear from the arrow shown on the burning surface of the end burning grain and those on the free surfaces of the liquid propellants that the volume between those surfaces will continually increase. The total mass flow of the propellants is

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual m = m O + m F = m O ⎡⎣1 + ( F O ) ⎤⎦

Then the volume flow of oxidizer is

υ O =

m O

ρO

=

1 10.7 kg s m = = 5.414 × 10−3 m3 s 3 ρO 1 + ( F O ) 1520 kg m 1 + 0.3 1

Likewise, the volume flow of fuel is

υ F =

m F

ρF

=

0.3 (10.7 kg s ) 1 m ( F O ) 1 = = 2.352 × 10−3 m3 s 3 ρ F 1 + ( F O ) 1050 kg m 1 + 0.3

The recession rate of the burning end grain also results in an increase in system volume according to the relation

υ g =

Ab L t

The time rate of change of the total volume filled with gas at any time t is therefore

υ ( t ) = (υ O + υ F + υ g ) = 7.766 × 10 −3 + Ab r During the burning period the pressure in the gas must remain constant at pc =3,310kPa. The integral form of the mass conservation equation is

G ∂ ˆ =0 ρ dυ + ∫ ρV ⋅ ndA ∫ ∂t υ A In the present case there is no net efflux of mass across the bounding surface A because this surface is entirely liquid or solid. Therefore the mass conservation equation becomes dυ p ∂ ∂ ∂ ∂ ρ dυ = ∫ ρ p dυ + ∫ ρ c dυ = ρ p + ( ρcυc ) = 0 ∫ dt ∂t υ ∂t υ p ∂t υ c ∂t Here υp is the volume of solid propellant consumed and we note that the solid propellant density ρp is constant. We also assume that the density of the gas produced ρc is not a function of the volume υc containing the gas generated. Then ∂ρ ∂υ ∂ρ ∂ ( ρcυc ) = υc c + ρc c = υc c + ρ c ( 7.766 ×10−3 + Ab r ) ∂t ∂t ∂t ∂t

Our mass conservation equation becomes

ρ p ( −rAb ) +

υ p dpc RTc dt

+ ρc ( 7.766 × 10−3 + Ab r ) = 0

2

Theory of Aerospace Propulsion, Second Edition: Solutions Manual If the pressure is to remain constant we must have p ( ρ − ρ ) A r = 7.766 ×10 RT −3

p

c

c

b

c

Noting that ρc<< ρp the equation is reduced to

Ab = ( 7.766 × 10−3 m3 s )

( 3310 kN m ) 2

(1560kg )

8.3145 kJ kmol ⋅ K ( 2000 K )( 0.00406 m s ) 21.2 kg kmol

Ab = 5.174 × 10−3 m 2 Thus the diameter of the grain is 8.12cm and the length of the grain is m⎞ ⎛ L = rtb = ⎜ 0.00406 ⎟ ( 30 s ) = 0.122m s⎠ ⎝

E12.2 The Space Shuttle’s integrated solid rocket boosters use a propellant mixture of ammonium perchlorate (NH4ClO4), polybutadiene acrylonitrile copolymer (PBAN), aluminum powder, and iron oxide catalyst. The characteristics of the propellant are given as follows: burn rate r=0.434 in/s at 1000psia, burn rate exponent n=0.35, propellant density ρp=0.063lb/in3, temperature coefficient of pressure (πp)k=0.11(%/F), characteristic exhaust velocity c*=5053ft/s, combustion temperature Tc=6,092F, effective ratio of specific heats γc=1.138 and γe=1.147. The nozzle throat diameter is 53.86in, the exit diameter is 152.6in, and the nozzle produces an average thrust of 2.59 million pounds at an ambient temperature of 60F. The average chamber pressure is 662psia during operation. For operation at sea level liftoff determine: (a) specific impulse, (b) chamber molecular weight, (c) exit Mach number, (d) exit pressure, (e) exit velocity, (f) exit temperature, (g) percent change in average thrust if the ambient temperature drops to 32F.

Solution F c ∗ and the throat (a) From Equation (11.19) of the text the thrust is given by F = mc

area is At = π(53.86in)2/4 = 2278in2 so the specific impulse is c * cF c * F 5, 053 ft s 2.59 ×106 lb I sp = = = = 269.7 s g g pc At 32.174 ft 2 s ( 662 lb in 2 )( 2, 278in 2 ) (b) The chamber ratio of specific heats is specified as γc =1.138 and from Equation (11.3) of the text we find Γc = 0.636. We may then rearrange the equation for the characteristic velocity to evaluate 3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

( 5, 053 ft s × 0.636 ) = 1576 ft 2 s 2 ⋅ R Ru ( c * Γ ) = = Wc Tc 6,552 R 2

Rc =

2

Therefore the chamber molecular weight is Wc = Ru/R = 49720/1576 =31.55. (c) From Equation (5.25) of the text the nozzle area ratio is given by γ +1

⎡ ⎤ 2(γ −1) γ +1 ⎢ ⎥ A* 2 ⎥ = Me ⎢ Ae ⎢ ⎛1 + γ − 1 M 2 ⎞ ⎥ e ⎟ ⎢⎣ ⎜⎝ 2 ⎠ ⎥⎦ From the given information γe=1.147 and A*/Ae = (53.86in/152.6in)2 = 0.1246 the area ratio equation may be solved to yield Me = 3. (d) The (average) exit pressure may be found from the isentropic relation using γe =1.147: ⎛ γ −1 2 ⎞ pe = pc ⎜ 1 + e Me ⎟ 2 ⎝ ⎠

− γ e ( γ e −1)

= ( 662 lb in 2 )(1 + 0.0735 × 32 )

7.803

= 12.6 lb in 2

(e) The exit velocity may be found by using Equation (11.21) of the text: I sp =

⎡ ⎛ 14.7 lb in 2 ⎞ ⎤ Ve ⎡ ⎛ p0 ⎞ 1 ⎤ Ve 1 + − = = 270 s 1 1 1 + ⎜1 − ⎢ ⎜ ⎟ 2⎥ 2 ⎢ 2 ⎟ 2⎥ g ⎣ ⎝ pe ⎠ γ e M e ⎦ 32.174 ft s ⎣ ⎝ 12.6 lb in ⎠ 1.147 × 3 ⎦

Substituting the known quantities yields Ve = 8,830ft/s (f) The exit temperature may be found in a similar fashion: Te = Tc[1+(γ-1)Me2/2]-1 and with Tc = 6,552R the exit temperature is therefore Te = 3,943R −1

−1 ⎛ γ −1 2 ⎞ Te = Tc ⎜ 1 + M e ⎟ = ( 6,552 R ) (1 + 0.0735 × 32 ) = 3,943R 2 ⎝ ⎠

(g) The percent change in thrust due to a drop in ambient temperature of 32F may be found using the temperature sensitivity coefficient defined by Equation (12.17) of the text which shows ⎛

c

⎞

⎝

i

⎠k

(π ) = p1 ⎜ dp ⎟ = 0.0011/ F dT p k

c

(E12.1)

This describes the change in chamber pressure as a function of the change in temperature of the propellant for a fixed value of k, the ratio of burner area to throat area.

4

Theory of Aerospace Propulsion, Second Edition: Solutions Manual The thrust F = cFpcAt and, because the throat area At is fixed, dF/F=dpc/pc + dcF/cF. Furthermore, because the area ratio Ae/A* is also fixed, the actual pressure ratio pe/pc is likewise fixed, independent of the change in the chamber pressure. Therefore the only change in the thrust coefficient, as given by Equation (11.17) in the text, is due to the change in actual exit pressure through the nozzle pressure mismatch term and thus p ⎞ ⎤ −1 ⎛ A ⎞ ⎛ p ⎞ 1 ⎛ A ⎞ ⎛ p ⎞ dp dcF 1 ⎡A ⎛ p = d ⎢ e ⎜ e − 0 ⎟⎥ = ⎜ e ⎟ d ⎜ 0 ⎟ = ⎜ e ⎟ ⎜ 0 ⎟ c cF cF ⎣ At ⎝ pc pc ⎠ ⎦ cF ⎝ At ⎠ ⎝ pc ⎠ cF ⎝ At ⎠ ⎝ pc ⎠ pc

Using Equation (E12.1) in the above equation yields p A 1 ⎛ dpc ⎞ p0 Ae dcF = 0 e (π p )k dTi ⎜ ⎟ dTi = cF cF pc At pc ⎝ dTi ⎠ k cF pc At For the solid rocket motor under investigation we have the actual values involved, which are as follows: 14.7 lb in 2 (18, 280in 2 ) p0 Ae = = 0.1038 cF pc At 2.59 ×106 lb Then dF dpc dcF ⎡ 1 ⎛ A ⎞ ⎛ p ⎞⎤ = + = ⎢1 + ⎜ e ⎟ ⎜ 0 ⎟ ⎥ (π p ) dTi k F pc cF ⎣ cF ⎝ At ⎠ ⎝ pc ⎠ ⎦ dF = 1.1038 (π p ) dTi = 1.1038 (.0011F −1 ) ( 32 F − 60 F ) = −0.034 k F

Thus ΔF/F = -3.4% and though this modern solid propellant has relatively small temperature sensitivity, the actual thrust loss is still 88,060lb.

E12.3 An end-burning grain solid propellant rocket is required to have the following characteristics at a normal ambient temperature of Tio=15C: sea level thrust F=4.91kN and a burn time of 15s. The characteristics of the propellant, a potassium perchlorateasphalt combination (76.5%KClO4 and 23.5% asphalt and oil) are as follows (all pressures are expressed in atmospheres): • Burning rate (cm/s) r={0.102exp[0.0015(Ti-15)]}p0.745 • Characteristic velocity (cm/s) c*={105,100exp[0.00021(Ti-15)]}p0.015 • Specific heat ratio γ=1.27 • Molecular weight W=30 • Propellant density ρp=1770kg/m3 • Chamber pressure pc=135.5 atm

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Determine (a) the propellant grain dimensions Ab and L, (b) the specific impulse, (c) the chamber temperature,(d) the influence of a 1% variation in the burning rate of the propellant on the thrust and the chamber pressure, (e) the motor performance at Ti=55C and -25C.

Solution

(a) The throat area At = F/pccF. Assuming ideal nozzle with exit pressure matched to the ambient we may simplify Equation (11.17) of the text, noting that γ =1.27 and therefore Γ = 0.662: 1/2

γ −1 ⎧ ⎡ ⎤⎫ ⎪ 2γ ⎢ ⎛ pe ⎞ γ ⎥ ⎪ 1− cF = λ cd cV Γ ⎨ ⎢ ⎜ ⎟ ⎥⎬ ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎥ ⎪ ⎣ ⎦⎭ ⎩

cF = (1)(1)(1)( 0.662 )

⎡ ⎛ 1atm ⎞0.2126 ⎤ 9.407 ⎢1 − ⎜ ⎥ = 1.634 ⎟ ⎣⎢ ⎝ 135.5atm ⎠ ⎦⎥

Then At =

F 4.91kN = = 2.189 × 10−4 m 2 2 cF pc 1.634 (135.5atm ) (101.3 kN m ⋅ atm )

This corresponds to a throat diameter dt=1.67cm. We may use our equilibrium chamber pressure Equation (12.15) of the text, which is ⎡A ⎤ pc = ⎢ b a ρ p c *⎥ ⎣ At ⎦

1 1− n

Then pc1− n =

Ab p p a ρ p c* = cn = c At pc r a

We may rewrite this as pc = Krρpc* where K = Ab/At. Then

(135.5atm ) (101.3 kN m 2 ⋅ atm ) 3kN pc 2 −4 Ab = At K = At = ( 2.189 × 10 m ) = r ρ pc * r ρ pc * r ρ pc * At an initial temperature Ti=15C we have

{

}

r = 0.102 exp ⎡⎣ 0.015 (Ti − 15 ) ⎤⎦ pc0.745 = 0.102 (135.5atm )

{

}

0.745

= 3.953 cm s

c* = 1.05 × 105 exp ⎡⎣0.00021(Ti − 15 ) ⎤⎦ pc0.015 = 1.05 ×105 (135.5atm )

6

0.015

= 1,130 m s

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Therefore Ab = 3kN/(0.03953m/s)(1131m/s)(1770kg/m3) = 0.0379m2. Then the length of the end burning grain is just L = rtb-o = (3.953cm/s)(15s) = 59.30cm (b) The specific impulse is Isp = cFc*/g = (1.634*1131m/s)/9.81m/s2 = 188.4s (c) The chamber temperature may be obtained from the characteristic velocity equation c* = (RTc )1/2/Γ or Tc =(c*Γ)2/ R =(1131m/s)(0.662)2[(8315J/mol-K/30kg/mol)]-1 = 2023K (d) The thrust F = cFpcAt so the change in thrust is

dF dcF dpc dAt ⎛ Ae p0 ⎞ dpc = + + = ⎜1 + ⎟ F cF pc At ⎝ At pc ⎠ pc Note that the throat area At=constant (as is Ae/At) and the thrust coefficient, assuming, with no loss in generality, λ=cv=1 is

cF = c F , i +

Ae ⎛ pe p0 ⎞ Ae pe Ae p0 − ⎜ − ⎟ = cF , i + At ⎝ pc pc ⎠ At pc At pc

A p ⎛ −dp ⎞ A p dp dcF = − e 0 ⎜ 2c ⎟ = e 0 cF At ⎝ pc ⎠ At pc pc Furthermore, from our equation for the equilibrium chamber pressure we have dpc/pc = da/a + ndpc/pc + dc/c* since the propellant density and K, the ratio of burning area to throat area are constant. Then the change in burning rate is da/a = (1-n)dpc/pc – dc*/c*.

Since c*~ pcq our equation becomes da/a=(1-n-q)dpc/pc and therefore dF ⎛ Ap ⎞ −1 da = ⎜1 + ⎟ (1 − n − q ) a F ⎝ Ap ⎠ The exponents given in the problem are n=0.745 and q=0.015. Thus

dF ⎛ Ae p0 ⎞ ⎛ da ⎞ = ⎜1 + ⎟ ⎜ 4.167 ⎟ a ⎠ F ⎝ At pc ⎠ ⎝ Therefore a 1% change in burning rate yields dF/F =0.04167[1+(Ae/At)(p0/pc)] change in thrust. This constitutes an increase (or decrease) in the thrust of at least 4.167%. (e) The motor performance at two different temperatures requires determination of the new burning rates and characteristic velocities. Using our definition pc=Krρpc* where K=Ab/At we find that Ab = AtK = Atpc/( r ρp c* )= (2.191x10-4m2)(135.5x101kN/m2)/r ρp c* =3kN/rρp c*

7

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (i) Ti = 55C we have •

r={0.102exp[0.0015(Ti-15)]}p0.745 = {0.102*exp[(0.0015*40)]}(pc)0.745 = 0.1083(pc)0.745cm/s = 0.001083 (pc)0.745 m/s

•

c* ={105,100exp[0.00021(Ti-15)]}p0.015 =1060 (pc)0.015m/s

The equilibrium pressure pc=Krρpc* and we know that K = 0.0379/0.000219 or K =173.1.

Then pc = (173.1*1770kg/m3)[ 0.001083(pc)0.745m/s]( 1060 (pc)0.015m/s). This may be written as pc1-0.745-0.015 = pc0.24 = (351.7kPa)/(101.3kPa/atm)=3.472, so pc=178.8atm and that means c* =1145m/s. The thrust is then ⎡ A ⎛p p ⎞⎤ F = ⎢cF ,i + e ⎜ e − 0 ⎟ ⎥ (178.8atm × 101.3 kPa atm ) ( 2.191× 10−4 m 2 ) At ⎝ pc pc ⎠ ⎦ ⎣ ⎡ ⎤ A F = ⎢1.634 + e ( 0.00738 − 0.00559 ) ⎥ ( 3.9684kN ) At ⎣ ⎦ To find the area ratio we may use the pressure ratio and Me=3.69. The area ratio

pe ⎛ γ − 1 2 ⎞ = ⎜1 + Me ⎟ 2 pc ⎝ ⎠

− γ ( γ −1)

= (1 + 0.135M e2 )

−4.704

From this equation we find and Me=3.69. The area ratio ⎡ ⎤ γ +1 ⎢ ⎥ A* 2 ⎥ = Me ⎢ Ae ⎢ ⎛1 + γ − 1 M 2 ⎞ ⎥ e ⎟ ⎢⎣ ⎜⎝ 2 ⎠ ⎥⎦

γ +1 2( γ −1)

1.135 ⎡ ⎤ = 3.69 ⎢ 2⎥ ⎣1 + 0.135 × 3.69 ⎦

4.204

= 0.0783

Thus Ae = At/0.0783 = (2.191x10-4)/0.0783 = 0.0028m2 and the thrust is F = 3.968[1.634+(12.77)(0.00738-0.00559)] = 6.57kN The thrust increase is dF/F=33.8%. (ii) At Ti=-25C we have •

r ={0.102exp[0.0015(Ti-15)]}p0.745 = {0.102*exp[-(0.0015*40)]}(pc)0.745 = 0.0961(pc)0.745cm/s = 0.000961 (pc)0.745 m/s

•

c* = {105,100exp[0.00021(Ti-15)]}p0.015 =1042 (pc)0.015m/s

Following the method used previously the new chamber pressure is pc = (173.1*1770kg/m3)[ 0.000961(pc)0.745m/s]( 1042 (pc)0.015m/s) 8

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This may be written as pc0.24 = (306.8kPa)/(101.3kPa/atm)=3.029, so pc = 101.3atm and that means c* =1117m/s. Because pe/pc = 1/135.5 = 0.00738 = constant and Ae/At=12.77 the thrust becomes F = [1.634+(Ae/At) (pe/pc -p0/pc )](101.3atm)(101.3kPa/atm)(2.191x10-4m2 ) F = [1.634+(Ae/At)(0.00738 – 1/101.3)](2.247kN) F = [1.634-0.00249(12.77)](2.247kN) The thrust is therefore F=3.60kN. The thrust decrease is dF/F=-26.78%.

Note: This older solid propellant has a value for (πp)k around 0.8 %C-1 or around 1.5%F-1,

which is more than an order of magnitude greater than that of the propellant in the Space Shuttle SRB.

E12.4 Consider an end burning grain solid rocket motor with the following characteristics: ρp=1770kg/m3, Tp=15C, c*=1,051p0.015in m/s (Note that in some printings of the textbook the coefficient was mistakenly given as 1.051) when p is in atmospheres, n=0.745, a=0.10 when p is in atmospheres. The thrust developed is F=4900N at a chamber pressure pc=13.74MPa over a burning time tb=15s where the combustion gases have γ=1.27 and a molecular weight W=30. Determine: (a) the chamber temperature Tc, (b) the thrust coefficient of the nozzle cF, (c) the throat area At, (d) the burning area Ab, (e) the length L of the grain, and (e) the specific impulse Isp.

Solution

(a) The characteristic velocity is given as c∗ = 1, 051 pc

⎛ 13, 740kPa ⎞ = 1, 051⎜ ⎟ ⎝ 101.3 kPa atm ⎠

0.015

0.015

= 1,131

m s

The chamber temperature is related to c* by the equation −1

c∗ =

γ +1 ⎡ ⎤ 2( γ −1) RTc ⎛ ⎞ 2 1 8.31451 kJ kmol ⋅ K ⎢ ⎥ Tc = RTc γ ⎜ = ⎟ ⎢ ⎝ γ +1⎠ ⎥ 0.6618 30 kg kmol Γ ⎣⎢ ⎦⎥

( 277.2 m s ⋅ K ) T 2

c* =

2

c

0.6618

Then the chamber temperature Tc = 2,021K. (b) The thrust coefficient

9

Theory of Aerospace Propulsion, Second Edition: Solutions Manual γ ⎡ ⎤ γ −1

⎛ ⎞ mV V p 1 F e e e ⎢ 2c p Tc 1 − ⎜ ⎟ ⎥ cF = = = = ⎢ ⎝ pc ⎠ ⎥ pc At pc At c ∗ c ∗ ⎢⎣ ⎥⎦

The specific heat is cp =

γ

Ru 8.315 kJ kmol ⋅ K = 4.704 = 1.304 kJ kg ⋅ K γ −1 W 30 kg kmol

The nozzle pressure ratio is pe /pc = 0.007373 so that the thrust coefficient is cF = 1.634. (c) The mass flow may be found from the thrust relation

m =

F F 4900 N = = = 2.651 kg s Ve cF c ∗ 1.634 (1131 m s )

Then the throat area may be found from At =

∗ 2.651 kg s (1131 m s ) F mc = = = 2.182 ×10−4 m 2 = 2.182cm 2 2 cF pc pc 13, 740, 000 N m

(d) The burning area is

Ab =

pc At Γ r ρ p RTc

(13.74 ×10 N m )( 2.182 ×10 m ) ( 0.6618) 6

=

2

−4

2

0.745 ⎡ ⎤ ⎛ 13, 740kPa ⎞ 0.001 m s ⎥ (1770kg ) 277.2 kJ kg ⋅ K ( 2021K ) ⎢ ⎜ ⎟ ⎝ 101.3kPa ⎠ ⎢⎣ ⎥⎦

Ab = 0.0386m 2 (e) The length of the grain is L= rb tb = (0.03878m/s)(15s) = 0.5817m. (f) The specific impulse is I sp =

F 4900 N = = 188.5s

mg 2.651 kg s ( 9.807 m s 2 )

10

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E12.5 Consider the cylindrical grain solid rocket motor with constant initial port area Ap as shown in Figure E12.2. Assume that (i) the flow between station 2 and the throat is isentropic, (ii) the burning rate is constant along the length of the grain, that is, the effects of erosive burning are balanced by the effect of longitudinal pressure drop, (iii) the flow in the chamber is in equilibrium, and the effect of the cap end is negligible.

Determine the pressure at the throat as a function of time pc(t).

Ap

At

Rc

L

1

2

Figure E12.2 Solid rocket motor with cylindrical grain

Solution

The differential equation describing the pressure in the chamber is

ρ p RTc Ab r RTc At Γ dpc pc − =0 + dt υc υc The initial port radius is R0 and the volume of the chamber varies with time as given by 2

υc = π L ( R0 + rt )

2

⎛ ⎛ rt ⎞ rt ⎞ = π R L ⎜1 + ⎟ = υc ,0 ⎜ 1 + ⎟ R0 ⎠ R0 ⎠ ⎝ ⎝

2

2 0

Using Rb to denote the radius of the burning surface, the burning surface area is given by

⎛ ⎛ rt ⎞ rt ⎞ Ab = 2π Rb L = 2π R0 L ⎜1 + ⎟ = υc ,0 ⎜1 + ⎟ ⎝ R0 ⎠ ⎝ R0 ⎠ Then the pressure equation becomes

dpc + dt

RTc At Γ ⎛ rt ⎞ υc ,0 ⎜ 1 + ⎟ R0 ⎠ ⎝

⎡

⎛

⎣

⎝

ρ p RTc r ⎢ 2π R0 L ⎜ 1 + 2

pc −

⎛ rt ⎞ υc ,0 ⎜1 + ⎟ R0 ⎠ ⎝

11

2

rt ⎞ ⎤ ⎟⎥ R0 ⎠ ⎦

=0

Theory of Aerospace Propulsion, Second Edition: Solutions Manual This equation may be rewritten as 2 ρ p RTc r ⎛ RTc At Γ ⎛ rt ⎞ dpc rt ⎞ pc − + ⎜1 + ⎟ ⎜1 + ⎟=0 R0 ⎠ dt 2π R0 L R0 R0 ⎠ ⎝ ⎝ 2

This is an ordinary differential equation with initial condition pc(0)=pc,0. The solution is ⎡ ⎛ a ⎞ ⎛ a ⎞ ⎛ a ⎞ ⎛ a ⎞⎤ pc − pc ,0 = 2 ρ p c ∗2 Γ 2 ⎢ exp ⎜ ⎟ Ei ⎜ − ⎟ − exp ⎜ ⎟ Ei ⎜ ⎟⎥ + R R R rt 0 0 0 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ R0 + rt ⎠ ⎦ ⎣

Here Ei is the exponential integral and the dimensionless coefficient a/R0 is

RTc At Γ c ∗ At Γ 2 a = = R0 π R0 Lr π R0 Lr The order of magnitude of the dimensionless coefficient a/R0 may be evaluated as follows: O (103 m s ) O (10−4 m 2 ) O (10−1 ) ⎛ a ⎞ O⎜ ⎟ = = O (102 ) m R ⎛ ⎞ ⎝ 0 ⎠ O (1) O (10−2 m ) O (100 m ) O 10−2 ⎜ ⎟ s⎠ ⎝ The order of magnitude of the dimensional coefficient

kg ⎞ ⎛ m2 ⎞ N ⎞ ⎛ ⎛ O ( 2 ρ p c ∗2 Γ 2 ) = O (1) O ⎜103 3 ⎟ O ⎜106 2 ⎟ O (10−1 ) = O ⎜108 2 ⎟ = O (103 atm ) m ⎠ ⎝ s ⎠ m ⎠ ⎝ ⎝ Then, using the typical orders of magnitude discussed, we may assume, for illustrative purposes, that a/R0=100 and 2ρpc*2Γ2=1000atm. The solution for the chamber pressure as a function of time is shown in Figure E12.3 where it is seen to be almost linear with the normalized instantaneous port radius R/R0=1 + rt/R0 and therefore with time. Since R0 and r are both O(10-2), the ratio r/R0=O(1) and the abscissa in Figure E12.3 is therefore on the order of time in seconds. Note that the pressure rises continually and does not drive toward an equilibrium chamber pressure because that rise permits the continually increasing mass flow to be passed through the fixed area throat, that is, this flow will not choke.

12

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

40 30

pc-pc,0 20 (atm) 10 0 1

2

3

4

5

port radius ratio R/R0

Figure E12.3 The variation of chamber pressure with the normalized port radius

13

Theory of Aerospace Propulsion, Second Edition: Solutions Manual E13. SPACE PROPULSION

E13.1 An electrostatic thruster produces 0.5N thrust for 3 hours and consumes a total of 0.5 kg using a power of 1kW. (a) Find the specific impulse Isp of the thruster, (b) find η’, the efficiency of the propulsive jet Solution (a) The specific impulse is I sp =

F =

mg

0.5 N 0.5kg 9.807 m s 2 ) ( 3hr × 3600 s hr

= 1,101s

(b) The efficiency of the propulsive jet is 1 2 2 Pe 2 mVe 1 FVe 1 0.5 N (1101s ) ( 9.807 m s ) η′ = = = = = 0.675 P P 2 P 2 4000 N ⋅ m s

E13.2 A spacecraft with a solar electric power plant with a specific mass α=15kg/kW is considered for a space mission that requires 4.8 months. (a) Estimate the optimum specific impulse to carry out this mission, (b) Assuming that the efficiency of the propulsive jet η’=50% determine the maximum thrust that can be developed if the solar array power delivered to the engine is 25kW, (c) Find the total impulse delivered over the entire mission, (d) Calculate the mass of the powerplant and of the propellant required for the mission, (d) estimate the exhaust velocity using Equation (13.13) and compare it to the value obtained for this case and discuss the difference, if any.

Solution (a) Using the estimate for optimal exhaust velocity given as Equation (14.13) in the text we find

I sp =

tp Ve 1 1000 = g g α

The planned mission time is 4.8months or 146 days so the propellant burn time tp= 1.261x107seconds. Then, with α =15kg/kW the estimated optimum specific impulse is Isp=2,957s.

1

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (b) The thrust F is directly proportional to the propulsive jet efficiency η’ and the powered delivered to the engine ηP and inversely proportional to the jet exhaust velocity; the relation is given by F = 2000η ′

(η P ) = 2000 0.5 ( ) Ve

25kW = 0.862 N 29, 000 m s

(c) The total impulse delivered is tp

I = ∫ Fdt = Ft p = ( 0.862 N ) (1.261× 107 s ) = 1.087 × 107 N ⋅ s 0

(d) The mass of the powerplant is meng = α(ηP) = (15kg/kW)(25kW) = 375kg. The mass of propellant may be obtained from the flow rate tp

m p = ∫ m p dt = m p t p = 0

F I 1.087 × 107 N − s tp = = = 374.8kg m gI sp gI sp 9.807 2 ( 2,957 s ) s

E13.3 A resistojet is to be employed on a satellite for attitude control in low earth orbit.

It will use gaseous nitrogen from the orbit modification propellant pressurization system.

The thrust level required is 100mN and the propellant feed is available at 1MPa. If the heater is limited to a maximum temperature of 500C determine (a) the specific impulse, (b) the mass flow of nitrogen, (c) the mass of nitrogen used if the design calls for a total delivered impulse of 104N-s, and (d) the nozzle dimensions for a practical device.

Solution (a) The thrust is given by

e + Ae ( pe − p0 ) F = λ mV The resistojets will be used in the vacuum conditions of space so the exit pressure p0=0 and the thrust becomes e + pe Ae = λ mV e+ Fvac = λ mV

The mass flow through the nozzle is

2

pe Ae pc At pc At

Theory of Aerospace Propulsion, Second Edition: Solutions Manual − ( γ +1)

pA m = cd c t Tc

γ ⎛ γ + 1 ⎞ 2(γ −1) ⎜ ⎟ R⎝ 2 ⎠

= cd

pc At Γ RTc

Therefore we may rewrite the vacuum thrust as ⎡ p A m RTc p A RTc ⎤ e+ e e = m ⎢λVe + e e Fvac = λ mV ⎥ pc At cd Γ pc At cd Γ ⎥⎦ ⎢⎣ The vacuum specific impulse is therefore

I sp ,vac =

Fvac λVe pe Ae RTc = +

mg g pc At cd g Γ

The exhaust velocity is cvVe,i where cv2=ηn so that it may be written as ( γ −1) γ ⎤ ⎡ ⎛ p ⎞(γ −1) γ ⎤ γ RTc ⎡ ⎛ pe ⎞ e ⎥ = 2η n ⎢1 − ⎜ ⎟ ⎥ 2c pTc ⎢1 − ⎜ ⎟ γ − 1 ⎢ ⎝ pc ⎠ ⎢⎣ ⎝ pc ⎠ ⎥⎦ ⎥⎦ ⎣

Ve = cvVe,i = cv

Here we have used the perfect gas identity cp=γR/(γ-1). Using γ=1.4, for which Γ=0.6847, and the given temperature of 500C=773K the vacuum specific impulse becomes ⎡ ⎡ ⎛ p ⎞0.286 ⎤ 0.552 p A ⎤ e e ⎥ ⎢ I sp ,vac = (129.2 s ) λ η n ⎢1 − ⎜ e ⎟ ⎥ + ⎢ ⎥ p c p ⎢⎣ ⎝ c ⎠ ⎥⎦ d c At ⎢⎣ ⎥⎦

The practical operational factors λ,ηn, and cd are generally specified based on laboratory testing so the only unknown is the nozzle pressure ratio pc /pe which depends on the nozzle design, i.e., the nozzle expansion ratio ε=Ae/At. These two ratios are related by pe Ae ⎛ 2 ⎞ = Me ⎜ ⎟ pc At ⎝ γ +1⎠

(γ +1) 2(γ −1)

⎛ γ −1 2 ⎞ Me ⎟ ⎜1 + 2 ⎝ ⎠

−1 2

⎛ γ −1 2 ⎞ = 0.5787 M e ⎜1 + Me ⎟ 2 ⎝ ⎠

−1 2

Therefore, by choosing a pressure ratio one defines an exit Mach number according to pe ⎛ γ − 1 2 ⎞ = ⎜1 + Me ⎟ pc ⎝ 2 ⎠

− γ ( γ −1)

⎛ M2 ⎞ = ⎜1 + e ⎟ 5 ⎠ ⎝

−3.5

From the Mach number one may find the area ratio from the equation relating pe/pc and Ae/At given previously. The pressure ratio and expansion ratios are shown in Figure E13.1.

3

Theory of Aerospace Propulsion, Second Edition: Solutions Manual

250

Ae/At

200 pc/pe

150 100 50 0 2

2.5

3

3.5

4

4.5

Me

Figure E13.1 Variation of nozzle expansion ratio and pressure ratio with Mach number for γ =1.4 As can be seen in Figure E13.1, the nozzle pressure and expansion ratios are quite close so that the ratio peAe/pcAt is close to unity. A good approximation is pe Ae ≈ 0.7 M e0.35 pc At

Therefore selecting λ,ηn, and cd and specifying the nozzle pressure ratio pc /pe one may calculate the vacuum specific impulse.

(b) Having determined the vacuum specific impulse the mass flow may be determined from m =

Fvac 0.1N 0.0102kg = = 2 gI sp ,vac ( 9.807 m s ) I sp ,vac I sp ,vac

(c) Then the total mass of propellant required is determined by the total impulse that must be delivered

I = Ftbo = ( 0.1N ) tbo = 104 N ⋅ s Thus the burnout time tbo=105s=27.78h and the propellant mass required is

bo = m p = mt

4

1, 020kg ⋅ s I sp ,vac

Theory of Aerospace Propulsion, Second Edition: Solutions Manual (d) The nozzle dimensions may be found by solving for the throat area in the equation for mass flow as follows: m RTc m At = = cd pc Γ cd

( 296.9 m s ⋅ K ) ( 733K ) m = ( 6.813 × 10 m N ⋅ s ) c (10 N m ) ( 0.6847 ) 2

2

−4

6

3

2

d

The exit area is found from the nozzle expansion ratio corresponding to the selected pressure ratio so that Ae=(Ae/At)At. The pressure ratio of the nozzle depends upon the area ratio selected, as do the throat and exit diameters. The length L of the nozzle and the correction factor λ both depend upon the expansion angle α of the nozzle. The discharge coefficient depends mainly on nozzle scale while the nozzle efficiency depends upon the nozzle shape and both depend on the Reynolds number. We may select some typical values for all these factors for illustrative purposes.

Then, taking Ae/A*=100 leads to pe/pc=0.011 for γ=1.4. A typical expansion angle of the nozzle is 15o which yields λ =0.983. The nozzle efficiency and discharge coefficient will be taken nominally as ηn=0.9 and cd=0.9. With these assumptions the vacuum specific impulse 0.552 ⎡ ⎤ 1.104 ⎥ = 181.9 s I sp ,vac = (129.2 s ) ⎢0.983 0.9 (1 − 0.0110.286 ) + 0.9 ⎣ ⎦

Then the mass flow is

m =

0.102kg 0.102kg = = 5.37 ×10−4 kg s I sp ,vac 189.9s

The total mass of propellant carried is mp =

1020kg ⋅ s = 5.37 kg 189.9 s

The throat area is At = ( 6.813 × 10−4 m3 N ⋅ s )

m 5.37 × 10−4 kg s = ( 6.813 × 10−4 m3 N ⋅ s ) = 4.065 × 10−7 m 2 cd 0.9

5

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Then the throat diameter dt =7.2mm, the exit diameter de = 7.2cm, and the nozzle length is L=14.78cm. The small scale of these space propulsion resistojets is typical, see, for example, Morren,m W.E., Whalen, M.V., and Sowey, J.S., “Performance and Endurance Tests of a Laboratory Model Multi-propellant Resistojet, Journal of Propulsion and Power, Vol. 6, No. 1, January, 1990, pp. 18-27.

E13.4 Determine the operating time required for an ion engine with a specific impulse Isp=10,000s to accelerate a vehicle through a velocity increment ΔV=5km/s if the thrust to weight ratio of the vehicle is 10-4. If an MPD engine with Isp=1000s is substituted with the vehicle thrust to weight ratio now equal to 10-3 how long would the acceleration take? Estimate the propellant mass needed in each case.

Solution The thrust is equal to the product of the mass flow of propellant and the exhaust velocity of the propellant jet and is also equal to the product of the mass of the vehicle and its acceleration:

F = m pVe = mV The rocket equation relates the change in velocity imparted to a space vehicle to the rocket’s specific impulse and overall mass ratio and is obtained from the above equation as V = Ve

m p m

= −Ve

m m = − gI sp m m

Here we note that the rate of consumption of propellant is equal to the negative of the rate of change of the mass of the vehicle. For a constant specific impulse we may integrate the equation to obtain ⎛m ⎞ mp ⎞ ⎛ ΔV = gI sp ln ⎜ initial ⎟ = − gI sp ln ⎜ 1 − ⎟ ⎜m ⎟ ⎝ minitial ⎠ ⎝ final ⎠

The burn time is often more important for low thrust space propulsion systems so we instead solve our equation for elapsed time dt =

m dV F

6

Theory of Aerospace Propulsion, Second Edition: Solutions Manual Then by integrating we obtain t final − tinitial = Δt = ∫

gI sp minitial gI sp m p m m⎛ dm ⎞ dV = ∫ ⎜ − gI sp m final − minitial ) = − ( ⎟=− F F⎝ m ⎠ F F minitial

By using our previous result for ΔV we obtain Δt =

I sp F minitial g

⎡ ⎛ ΔV ⎞ ⎤ ⎢1 − exp ⎜⎜ − ⎟⎟ ⎥ ⎢⎣ ⎝ gI sp ⎠ ⎥⎦

For the ion rocket with a thrust to initial weight ratio F/minitialg = 10-4 our equation becomes ⎡ ⎛ 509.8 ⎞ ⎤ Δt = I sp × 104 ⎢1 − exp ⎜ − ⎜ I ⎟⎟ ⎥⎥ sp ⎝ ⎠⎦ ⎣⎢ Then for the ion rocket with Isp=104s we find Δt=1,381hr=57.5days. For the MPD rocket with a thrust to initial weight ratio F/minitialg = 10-3 our equation becomes ⎡ ⎛ 509.8 ⎞ ⎤ Δt = I sp × 103 ⎢1 − exp ⎜ − ⎜ I ⎟⎟ ⎥⎥ sp ⎝ ⎠⎦ ⎣⎢ The burn time for the MPD rocket, with Isp=1000s, is Δt =110.9hr = 4.623days.

The mass fraction of propellant required is given by the rocket equation as ⎛ ΔV ⎞ ⎛ 509.8 ⎞ = 1 − exp ⎜ − = 1 − exp ⎜ − ⎟ ⎜ gI ⎟ ⎜ I ⎟⎟ mi sp ⎠ sp ⎝ ⎝ ⎠

mp

Therefore the propellant mass fraction m/minitial =0.0497 for the ion rocket and 0.399 for the MPD rocket. The propellant mass required in each case can then be determined from the initial mass of the spacecraft

E13.5 A nuclear thermal rocket using hydrogen as the propellant has a specific impulse Isp,NUKE=900s and a liquid rocket using LH2-LOX as the propellant has a specific impulse Isp,LOX=450s when both use the same nozzle and operate under the same ambient pressure conditions with the nozzle matched. The liquid rocket engine operates with a chamber temperature Tc=3600K. Calculate a reasonable estimate of the chamber temperature at which the nuclear rocket is operating.

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Theory of Aerospace Propulsion, Second Edition: Solutions Manual Solution The specific impulse is given by I sp =

c RTc cF Ru c pA c pA F = F c t = F c t = F = pc At mg mg Γg Γg Γg RTc

I sp , NUKE I sp , LOX

=

Tc , NUKE Wc , LOX Tc , LOX Wc , NUKE

≈

Tc , NUKE 18 Tc , LOX 2

=3

Tc , NUKE Tc , LOX

Tc Wc

=2

Therefore the chamber temperature for the nuclear rocket is 2

4 ⎛2⎞ Tc , NUKE = ⎜ ⎟ Tc , LOX = 3600 K = 1600 K 9 ⎝3⎠

E13.6 A fast trip to Mars, including landing and return, in a spacecraft with initial and final mass Mi and Mf, respectively, requires a spacecraft velocity increment ⎛M ⎞ m ΔV = gI sp ln ⎜ i ⎟ = 40, 000 ⎜M ⎟ s ⎝ f ⎠ The ideal mission would require 22,000m/s, about half that of the fast mission.(a) For both missions compare the required propellant mass fraction Mp /Mi required for chemical, solid-core nuclear, gas-core nuclear, and electric propulsion devices having Isp=450s, 1000s, 2000s, and 10,000s, respectively.

Solution The mass of propellant is the difference between the initial and final mass of the spacecraft so the propellant mass fraction required is ⎡ ΔV ⎤ = 1 − exp ⎢ − ⎥ Mi ⎢⎣ gI sp ⎥⎦

Mp

The results for the two missions and four propulsion systems are shown in Figure E13.2.

The higher the specific impulse the more practical the spacecraft will be. Of course this assumes that the thrust to weight ratio will be sufficiently large to be able to provide sufficient acceleration for a fast trip. For total transfer times on the order of a year or more the low thrust, high specific impulse propulsion system will have a smaller initial mass, while for shorter times the opposite is true.

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Theory of Aerospace Propulsion, Second Edition: Solutions Manual

1 0.9 0.8 0.7 0.6 mp/m0 0.5 0.4 0.3 0.2 0.1 0

Fast Mars landing and

Ideal path Mars landing and

0

1000 2000 3000 4000 5000 6000 7000 8000 9000 1000 0 Isp (s)

Figure E13.2 Propellant mass fraction required for two round trip Mars missions as a function of propulsion system specific impulse.

E13.7 A nuclear ramjet like that shown in Figure 13.28 is considered for flight in the stratosphere of the earth’s atmosphere. The reactor directly heats the air passing through it and the reactor is kept at a constant temperature Tr in the range 1500K<Tr<1900K. Show that the optimum specific net thrust Fn m 0 occurs for flight Mach numbers near M0=3. It may be assumed that the exhaust nozzle is always matched so that p7=p0 and that the stagnation pressure ratio across the reactor pt,ri/pt,re =2 and remains essentially constant.

Solution For the case of no mass addition to the stream, the specific thrust for a matched nozzle whose exit station is denoted by the subscript 7 is

Fn = Ve − V0 = 2c p (Tt ,7 − T7 ) − γ RT0 M 0 m 0 The air taken in through the inlet passes through the reactor is assumed to exit the reactor with the temperature Tr. For a matched nozzle the area ratio of the nozzle is changed to change the exit Mach number so that the exit pressure is matched to the local ambient pressure. Assuming constant γ the specific thrust may be approximated as

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Theory of Aerospace Propulsion, Second Edition: Solutions Manual γ −1 ⎡ ⎤ ⎛ p7 ⎞ γ ⎥ Fn ⎢ = 2c p Tr ⎢1 − ⎜ − γ RT0 M 0 ⎜ p ⎟⎟ ⎥ m 0 t ,7 ⎠ ⎝ ⎢⎣ ⎥⎦

Everything is specified except for the nozzle pressure ratio. The nozzle is matched so p7=p0 and the nozzle pressure ratio may be expressed as follows: γ

− pt ,ri p7 p pt ,0 pt ,ri ⎛ γ − 1 2 ⎞ γ −1 ⎡ 1.35 = 0 = ⎜1 + 1 − 0.075 ( M 0 − 1) ⎤ M0 ⎟ ⎣ ⎦p 2 pt ,7 pt ,0 pt ,ri pt , re ⎝ ⎠ t , re

Here we have assumed that the nozzle is ideal so that the exit stagnation pressure pt,7 is equal to the reactor exit stagnation pressure pt,re and that the ratio of the reactor entry stagnation pressure pt,ri to the free stream stagnation pressure is given by the MIL-E5008B standard pressure recovery relation. The stagnation pressure ratio across the reactor pt,ri/pt,re =2 and remains essentially constant. The specific thrust of the nuclear ramjet may be calculated as a function of M0 using γ =1.4 and cp =1000J/kg-K. Furthermore, in the stratosphere the temperature T0=216K and is constant. The results for two reactor exit temperatures are shown in Figure E13.3.

700 600

Tr=1900K

500 400 Fn/(dm/dt)0

Tr=1500K

300 200 100 0 0

1

2

3

4

5

M0

Figure E13.3 The variation of the specific thrust Fn m 0 with flight Mach number for a nuclear ramjet. The air temperature leaving the reactor is shown as Tr.

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