lecture notes for Theory of Aerospace Propulsion, Second Edition Pasquale M. Sforza

LECTURE NOTES WELL ORGANIZED

THEORY OF AEROSPACE PROPULSION 1. Propulsion Principles and Engine Classification

P. M. SFORZA University of Florida

an idealized flow machine Ae A0 Ve

V0 p0 T0 • Steady, quasi‐one‐dimensional flow

pe=p0 Te

• Entrance and exit stations are pressure matched • No heat transfer across control volume • Friction outside the flow machine is negligible • Mass injected into the stream is negligible Theory of Aerospace Propulsion

2

quasi‐1‐d conservation equations Mass

Momentum

Energy

− ρ0 A0V0 + ρe AeVe = 0 m = ρ0 A0V0 = ρe AeVe = const

( − ρ0 A0V0 )V0 + ( ρe AeVe )Ve = F F = m (Ve − V0 ) = net force on fluid 1 2 2 ⎤ ⎡ ⎢( he − h0 ) + (Ve −V0 ) ⎥ = mQ + P m 2 ⎣ ⎦ Theory of Aerospace Propulsion

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zero heat addition, ΔQ=0 1 2 ⎡ 2 ⎤ m ⎢( he − h0 ) + (Ve − V0 ) ⎥ = P 2 ⎣ ⎦ he ≈ h0 because no heat is added to the flow

Force on fluid

1 m (Ve + V0 )(Ve − V0 ) = FVavg = P 2 P Power added P>0: thrust on machine F= Power removed P<0: drag on machine Vavg FV0 ηp = FVavg

Propulsive= Useful power Actual power efficiency Theory of Aerospace Propulsion

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propeller Motor= propeller blade motor housing motor

V0

0

1 Theory of Aerospace Propulsion

(a) Reciprocating piston engine (b) Electric motor (c) Gas turbine engine

e 5

ΔQ=0 and P>0: piston‐engine props

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ΔQ=0 and P>0: turboprops

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A modern large turboprop engine The Europrop International TP400‐D6 is a collaborative engine between Rolls‐Royce, MTU, Snecma and ITP. Power range in excess of 11,000shp TP400‐D6 is the most powerful Western turboprop ever Application: Airbus Military A400M

http://defence.rolls-royce.com/most-powerful-turboprop-engines/ Theory of Aerospace Propulsion

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fans and turbines Example of ΔQ=0 with P>0: Fans

Cincinnati Fan tube axial fan

Example of ΔQ=0 with P<0: Turbines

GE 1.5 MW wind turbine farm in Gatun, Spain Theory of Aerospace Propulsion

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turbopumps Example of DQ=0 with P>0: Turbopumps Even though Pratt & Whitney Rocketdyne's SSME weighs one‐seventh as much as a locomotive engine, its high‐pressure fuel pump alone delivers as much horsepower as 28 locomotives, while its high‐ pressure oxidizer pump delivers the equivalent horsepower for 11 more.

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propeller operation The propeller is speed‐limited: F/P =1/Vavg < 1/Vo 0.05

V0/Ve 0.9

0.04 F/P (kN/kW)

0.03

0.6

0.02

0.3

0.01 0 0

50

100

150

200

250

V0 (m/s) Theory of Aerospace Propulsion

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propulsive efficiency Propeller propulsive efficiency is high for Ve/Vo~1 1 0.8 0.6

ηp

0.4 0.2 0 0

0.2

0.4

0.6

0.8

1

V0/Ve

Propulsive efficiency ηp=V0/Vavg = 2/(1+Ve/V0) Theory of Aerospace Propulsion

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constant ΔQ but no net power: P=0 Energy eq.

Thrust

Overall efficiency

1 2 ⎡ 2 ⎤ m ⎢( he − h0 ) + (Ve − V0 ) ⎥ = m ΔQ 2 ⎣ ⎦ 1 m (Ve + V0 )(Ve − V0 ) = m ⎡⎣ ΔQ − ( he − h0 ) ⎤⎦ 2 waste heat m F= ⎡⎣ ΔQ − ( he − h0 ) ⎤⎦ Vavg FV0 V0 ⎡ ⎛ he − h0 ⎞ ⎤ = ηo = ⎢1 − ⎜ ⎥ = η pηth ⎟ m ΔQ Vavg ⎣ ⎝ ΔQ ⎠ ⎦ Theory of Aerospace Propulsion

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turbojet fuel lines inlet

compressor

combustor nozzle combustor

turbine 1

2

3

4

5

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14

turbojet with afterburner fuel lines inlet

afterburner

combustor

nozzle

combustor

1

2

3

4

5

5AB

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7

15

ΔQ>0 but P=0: the turbojet engine Pratt & Whitney J57 turbojet. World's first jet engine to develop 10,000 lbs. thrust. Powers F‐100 to supersonic flight.

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ramjet fuel lines inlet

combustor nozzle combustor

1

2, 3

4, 5

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17

ΔQ>0 with P=0: the ramjet Marquardt RJ43‐MA‐20‐B4 Ramjet powered the unmanned 42 ft long D‐ 21 (M=3, 90kft) after launching from an SR‐71aircraft

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early problems with jet aircraft Performance characteristics of early jet fighters were different than those of propeller‐driven aircraft with reciprocating engines Differences were due to the manner in which the thrust and power of turbojet engines vary with speed.

Reciprocating engines generate the same amount of power at takeoff as at high speeds, whereas the turbojet at the same altitude has nearly the same thrust at both high and low speeds.

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turbojets are weakly speed‐limited 0.03 prop takeoff 0.025 jet takeoff

0.02 1/Vavg 0.015 (s/m) 0.01

jet max prop max

0.005 0 0

50

100

150

200

250

300

V0 m/s)

F ~ 1/Vavg = 2/(Ve+V0); Propeller has Ve~V0 so F~1/V0 Turbojet has Ve~2.5V0,max so thrust decay is much slower Theory of Aerospace Propulsion

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jet vs. propeller comparison Acceleration of the jet on takeoff will be low and takeoff distance long (subscript denotes 25mph) Characteristic P25

Propeller 1600

Jet 167

1600 7200 1168 0.75 0.12

2740 2500 2500 0.25 0.25

P410 F25 F410 (F/W)25 (F/W)410

For the same drag area, the jet would be much faster than the propeller‐driven aircraft. (subscript denotes 410mph) Theory of Aerospace Propulsion

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specific fuel consumption Specific fuel consumption: “mass of fuel burned per hour for each unit of thrust produced”

m ΔQ = m f ηbQ f FV0 = η pηth m ΔQ ⎛ m ΔQ ⎞ ⎛ ⎞ V0 cj = =⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜ F ⎝ ηbQ f ⎠ ⎝ η pηth m ΔQ ⎠ Vavg V0 cj = = η pηthηbQ f ηthηbQ f m f

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specific fuel consumption (SI units) 0.18 M=1 0.15 0.12 cj 0.09 (kg/hr/N) 0.06 subsonic turbojets

supersonic turbojets

0.03 0 0

100

200

300

400

500

600

V0 (m/s)

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specific fuel consumption (English units) 2 M=1 1.5 cj (lbm/hr/lb)

1 0.5

subsonic turbojets

supersonic turbojets

0 0

500

1000

1500

2000

V0 (ft/s)

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early jet‐propeller combination Ryan Fireball FR‐1 jet‐propeller fighter

Leading edge inlets

F=1600 lbs

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modern jet‐propeller: the turbofan The special case of combined heat and power addition: the turbofan Ve,c, Te,c, Ae,c A0

cold hot

V0 p0 ρ0 T0

Ve,h, Te,h, Ae,h

cold

pe=p0 cold flow

Theory of Aerospace Propulsion

hot flow

26

the turbofan engine fan cowling

fan flow

fan flow

cold flow combustor

core flow

nozzle

hot flow

combustor

fan flow

cold flow

fuel line

fan 1

2 2.5

3

3F

4 5

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the turbofan advantage ⎛ m e , c ⎞ = m e , h (1 + β ) m = m 0 = m e , h + m e , c = m e , h ⎜ 1 + ⎟ ⎜ ⎟

m e ,h ⎠ ⎝ m m β F = Fh + Fc = Ve , h − V0 ) + Ve , c − V0 ) ( ( 1+ β 1+ β P Cold Fc = stream Vc , avg Hot stream Fh =

m Δ Qη th ⎡ Δ Q − ( he , c − h0 ) ⎤ = ⎦ (1 + β ) V Vh , avg ⎣ h , avg m e , h

ρ 0 A0η thη b ( f a ) Q f V0 D1 + ≈ + D2 F = 1+ β Vc , avg Vh , avg V0 P

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turbofan specific fuel consumption ( sfc )c Fc 1+ m f ,h + m f ,c ( sfc )h Fh sfc = = ( sfc )h F Fh + Fc

( sfc )c

bsfc ) V0 ( = 550η p ,c

1+

c

Fh

bsfc = power‐based specific fuel consumption of the fan, typically 0.5lbm/hp/hr

3600V0 ( sfc )h = 778η p ,hηthηb ( f a ) Q f

η p ,h ηthηb ( sfc )c = 7.43 ( bsfc ) <1 η p ,c β ( sfc )h Theory of Aerospace Propulsion

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the turbofan advantage η p ,h ηthηb ( sfc )c ≈ 7.43 ( bsfc ) η p ,c β ( sfc )h ( sfc )c 0.5 ( 0.5 )( 0.95 ) 1 ≈ 7.43 ( 0.5 ) ≈ <1 0.9 β β ( sfc )h Using practical numbers we see that (sfc)c /(sfc)h <1 for the turbofan, where β>1. Therefore the specific fuel consumption is always less than that of the turbojet

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low bypass turbofan engine Pratt & Whitney's first F100 military engine flew on July 27, 1972, on a twin‐engine F‐15 Eagle. It also powered the F‐16 Fighting Falcon.

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high bypass turbofan engine GE's commercial engines of the 1970s built on the technology of the TF39 military engine of the 1960s. By the 1980s, the CF6 family of engines was powering wide‐body aircraft, including the Boeing 747 and 767

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rockets The special case of Ao=0; the rocket

A0=0

nozzle

Ae

fuel at h=hi V0

oxidizer at hi 0

combustion chamber Theory of Aerospace Propulsion

Ve

e

33

rocket engine equations If A0=0 then m0=0 and the force on the fluid is

e F = mV

The energy equation may be rearranged to provide an equation for the force as a function of the heat addition

m F = ⎡⎣ ΔQ − ( he − hi ) ⎤⎦ = ρ e Ae ⎡⎣ ΔQ − ( he − hi ) ⎤⎦ Ve propulsive efficiency

FV0 V0 ηp = = FVe Ve Theory of Aerospace Propulsion

waste heat 34

specific impulse t

I sp = Airbreathing engines

I sp =

∫ Fdt 0 t

∫ mdt

F ≈ m

o

( f a )ηbQ f ηth

Rocket engines

I sp = Ve

Vavg

Isp is measured in units of velocity but is often given in units of time by dividing through by g, the acceleration of gravity Theory of Aerospace Propulsion

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specific impulse comparison Airbreathing engines vs. rocket engines 20000 M=10

M=5 16000 Isp (m/s)

LH2/Air

12000 HC/Air

8000 LH2/LOX 4000 RP-1/LOX

0 0

1000

2000

3000

4000

V0(m/s)

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liquid‐fueled rocket engine Designed for the Boeing Delta IV family launch vehicles, the Rocketdyne bell‐nozzle RS‐68 is a LH2‐LO2 booster engine Thrust Level: 745,000 lbf Weight: 14,560 lb Mixture Ratio:6.06 Isp:410 sec Chamber Pressure pc =1410 psia Expansion Ratio ε=21.5 Theory of Aerospace Propulsion

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solid propellant rocket Space Shuttle solid rocket booster motor (SRBM) produces 2.65 million pounds of thrust

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space propulsion systems with ΔQ>0 (a)

nuclear thermal rocket

nuclear reactor

propellant tank

nozzle

(b)

electric resistojet

propellant tank

nozzle resistance heater (c)

electric arcjet

propellant tank

nozzle electric arc

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space propulsion with electric acceleration electron ion accelerator emitter Electrostatic ion rocket

propellant tank

ion generator

(a)

electric arc

nozzle jxB force

Magnetoplasmadynamic (MPD) rocket

propellant tank (b) magnetic field coil Theory of Aerospace Propulsion

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Comparative fuel efficiencies of different propulsion systems

Theory of Aerospace Propulsion

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force field for airbreathing engines Ae A0 V0

Ve me=mf + m0

mf

p0 T0

pe=p0 Te 0

1 Theory of Aerospace Propulsion

7

e 42

force field on the control volume m 7V7 − m 0V0 = p0 A0 − p7 A7 + F1 + F2 F1

p0A0

F2 p7A7 m7V7

m0V0

pe=p0 ρe Te 0

1 Theory of Aerospace Propulsion

7

e 43

force field on the structure F + F4 − F2 = 0

0 + m f )V7 − mV 0 0 +( p7 − p0 ) A7 ⎤ − ⎡⎣ p0 ( A0 − A7 ) + F1 + F4 ⎤⎦ F = ⎡⎣( m ⎦ F

F4 F2

F= net thrust – nacelle drag Theory of Aerospace Propulsion

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grouping the forces F + F4 − F2 = 0 7 7 − mV 0 0 +( p7 − p0 ) A7 ⎤⎦ − ⎡⎣ p0 ( A0 − A7 ) + F1 + F4 ⎤⎦ F = ⎣⎡mV Net (nozzle) thrust

Nacelle Drag

pdAs pdA A

A+dA Theory of Aerospace Propulsion

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thrust and drag components Net thrust =

gross thrust

–

ram drag

⎡ w0 + w f ⎤ w0 Fn = ⎢ V7 + ( p7 − p0 ) A7 ⎥ − V0 ⎣ g ⎦ A1 g Nacelle drag F1 = ∫ pdA

D = p0 ( A0 − A7 ) + F1 + F4

A0 A7

F4 = ∫ pdA + D f

Additive, or inlet, drag

A1

A1

A7

A0

A1

D = ∫ ( p − p0 )dA + ∫ ( p − p0 ) dA + D f Theory of Aerospace Propulsion

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thrust components For preliminary studies it is generally assumed that wf/w0<<1 For airbreathing engines wf/w0~2%

Fn = m (Veff − V0 ) ≈ m 0 (Veff − V0 ) A7 Veff = V7 + ( p7 − p0 ) m 0 Consider the gross thrust and ram drag: 2 ⎡ ⎡ ⎤

Fg = ⎣( m0 + m f ) V7 + ( p7 − p0 ) A⎦ ≈ A7 ⎣ p7 (1 + γ 7 M 7 ) − p0 ⎤⎦ 2

Fr = m0V0 = γ 0 p0 A0 M 0 Theory of Aerospace Propulsion

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the nozzle as an energy transformer A nozzle can transform internal energy to kinetic energy. Consider a point on the exit plane, that is, station 7

⎛ ⎛ h7 ⎞ T7 ⎞ V7 = 2ht ,7 ⎜1 − ≈ 2c p ,7Tt ,7 ⎜1 − ⎜ h ⎟⎟ ⎜ T ⎟⎟ t ,7 ⎠ t ,7 ⎠ ⎝ ⎝

T7 ⎛ p7 ⎞ =⎜ ⎟⎟ ⎜ Tt ,7 ⎝ pt ,7 ⎠

γ 7 −1 γ7

V7

γ 7 −1 ⎡ ⎤ γ7 ⎛ ⎞ p V7 = 2c p ,7Tt ,7 ⎢⎢1 − ⎜ 7 ⎟ ⎥⎥ ⎜p ⎟ ⎢⎣ ⎝ t ,7 ⎠ ⎥⎦ Theory of Aerospace Propulsion

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total pressure ratio in the engine For a matched nozzle, i.e., one where p7=p0, we have γ 7 −1 ⎡ ⎤ γ ⎛ p0 pt ,0 ⎞ 7 ⎥ ⎢ V7 = 2c p ,7Tt ,7 ⎢1 − ⎜ ⎜ p p ⎟⎟ ⎥ ⎢⎣ ⎝ t ,0 t ,7 ⎠ ⎥⎦

pt ,0 pt ,7

=

pt ,0 pt ,2 pt ,3 pt ,4 pt ,5 pt ,2 pt ,3 pt ,4 pt ,5 pt ,7

Nozzle pressure recovery Turbine work removal Combustor pressure recovery Compressor work input Inlet pressure recovery Theory of Aerospace Propulsion

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Mach angle and the zone of silence Sound source moving supersonically

(M2‐1)1/2

1

μ

sound waves

M>1

μ

Sound source moving sonically

sound waves M=1

Mach angle μ=arcsin(1/M) Theory of Aerospace Propulsion

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nozzle types & related exit flow Mach lines μ<90o

M7>1

μ=90o

M7=1

Supersonic exhaust Exit flow can support pressure mismatch because external pressure information can’t get inside nozzle

Theory of Aerospace Propulsion

No real Mach lines

M7<1 Subsonic exhaust Exit must be pressure matched because pressure is felt everywhere 51

net thrust as a function of V7 7 maximum thrust

7

Fn saddle point p7=p0 M7>1

0 V0

V7=a7

Exit flow subsonic Theory of Aerospace Propulsion

V7 Exit flow supersonic

52

interpretation of maximum thrust Thrust loss – pressure too low Nozzle wall

Thrust loss ‐ high pressure contribution lost

Δlong p0 correct length – pressure matched

Δshort

p0 pressure

pressure along Theory of Aerospace Propulsion nozzle wall

53

Example: Pratt & Whitney J58 jet engine for the SR‐71 8,000gal/hr of JP‐7 (F/O=0.034) with Fg,=34,000lbs @200kts, M7=1 and TET=1,580F=Tt,7 (a)Veff=Fg/mf(1+O/F)= 2,400fps (b)Fn=Fg‐m0V0=29,400lbs (c)cj =mf/Fn=8,000(6.76lbm/gal)/29,400=1.84lbm/hr/lb (d)A0=m0/r0V0=17.4 ft2 (d1=4.7ft) (e)Puseful=FnV0=14,750 hp (11MW) (f) T7=Tt,7[1‐(γ7 ‐1)M72/2]‐1 =2,040R(0.86) =1,750R (use γ7=1.33) (g)V7=a7=(γ7RT7)1/2=2,000fps (h)A7=[m(1+γ7)(RT/γ7)1/2]7/p0 – Fg/p0 and A7=7.36ft2 (d7=3.1ft) (i) p7=(Fg/A7+p0)/(1+γ7M72)=1.37p0 Theory of Aerospace Propulsion

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Example: The German A‐4 Rocket (V‐2) The V‐2 burned about 8000kg of propellant (LOX and Alcohol, O/F=1.3) in 65s with Ve =2000m/s with a matched nozzle. Find: (a) F=mVe =(8000kg/65s)(2000m/s)=246kN (or 55.3klbs), (b) P=V0F= 0 on the pad and 2250kN/s (2.25MW or 660,000hp) at 2km/s, (c) Isp=F/mg=246kN/(8000kg/65s)(9.81m/s2) =203s, (d)ηo=FV0/wfhbQf=FV0/{(m/[1+O/F])ηbQf}= (492MNm/s)/{123kg/s(1)(27.2MJ/kg)/(4.3)}= 33.8%

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THEORY OF AEROSPACE PROPULSION 2. Quasi-One-Dimensional Flow Equations

P. M. SFORZA University of Florida

analysis of propulsion system components • Equations for flow through a combustor under the steady quasi‐one‐dimensional approximation • Applications extend to other propulsion subsystems, such as, nozzles, inlets, and turbomachinery cascades.

• The equations are simple to deal with, yet they are commonly acknowledged to be reasonably accurate • The equation set is particularly useful for preliminary design purposes.

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2

Q‐1D flow theory accounts for: • Changes in combustor cross‐sectional area • Chemical reactions • Variations in gas molecular weight and specific heat • Exchange of heat with the surroundings • Drag caused by internal bodies or solid particles in the flow • Losses due to friction on the combustor walls

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equation of state

p=

ρ RuT W

dp d ρ dT dW = + − p ρ T W

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speed of sound ⎛ ∂p ⎞ ⎡ ∂ γ RuT γ ⎤ a =⎜ ⎟ = ⎢ ( C ρ )⎥ = W ⎝ ∂ρ ⎠ s ⎣ ∂ρ ⎦ 2

da 1 ⎛ d γ dT dW ⎞ = ⎜ + − ⎟ a 2⎝ γ T W ⎠ Theory of Aerospace Propulsion

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Mach number

V M = a

dM dV dW d γ dT = 2 + − − 2 γ M V W T 2

2

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r

mass conservation G ˆ = ∫ ρVn dA =0 ∫ ρV ⋅ ndA A

z Vn

A

A(z)

In differential form for axisymmetric flow this becomes

∂ ∂ ( ρ rVr ) + ( ρ rVz ) = 0 ∂r ∂z

Integrating over the area A(z)=p[rw(z)] 2 leads to: rw ( z ) ⎡ ⎤ ⎡ drw Vr , w ⎤ ∂ − ⎢ 2π ∫ ρVz rdr ⎥ − ρVz , w rw ⎢ ⎥=0 ∂z ⎢⎣ ⎢⎣ dz Vz , w ⎥⎦ ⎥⎦ 0

mass flow

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mass conservation (continued) Wall boundary condition

drw Vr , w = dz Vz , w

w al l

dz

V

Vr drw

Vz

The mass conservation equation then reduces to rw ( z ) ⎡ ⎤ dm ∂ =0 ⎢ 2π ∫ ρVz rdr ⎥ = ∂z ⎢⎣ ⎥⎦ dz 0

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mass conservation (concluded) rw ( z )

m = 2π ∫ ρVz rdr = ρV π r = const 2 w

0

dm d ρ dV dA = + + =0 ρ m V A

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conservation of energy Rate of heat transfer in across the boundary

⎛ V ⎞ m ( dQ − dWk ) = m ⎜ dh + d ⎟ 2 ⎠ ⎝

Total energy change within boundary

2

Power taken out across the boundary Heat input rate ‐ Power output = Total energy change N

h = ∑ Yi hi i =1

ρi Yi = ρ

m = m o + m f T

hi = Δh + ∫ c p ,i dT 0 f ,i

Tr

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conservation of energy (continued) Enthalpy change

N

N

i =1

i =1

dh = ∑ Yi dhi + ∑ hi dYi

differential of species enthalpy

dhi = c p ,i dT N

specific heat of mixture

c p = ∑ Yi c p ,i i =1

temperature change

composition change N

change in enthalpy of mixture

dh = c p dT + ∑ hi dYi Theory of Aerospace Propulsion

i =1

11

conservation of energy (continued) dQ − dWk − dH dT dV 2 = + c pT T 2c p T across the control surface

within the control volume N

dH = ∑ hi dYi i =1

Ru c p − cv = W cp γ= cv

change in composition due to chemical reaction Theory of Aerospace Propulsion

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conservation of energy (continued) dQ − dWk − dH dT ⎛ γ − 1 ⎞ 2 dV = +⎜ ⎟M 2 c pT T ⎝ 2 ⎠ V N

general chemical reaction among N species

2

N

∑ C ′ A → ∑ C ′′ A i =1

i

i

i =1

i

i

N

1 N dH on a molar basis dH = ∑ h dY = H i dCi ∑ i i CW i =1 i =1 Theory of Aerospace Propulsion

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conservation of energy (concluded) The heat of reaction of a chemical reaction is given by T T N ⎡ ⎤ ⎡ ⎤ 0 0 ΔH (T ) = ∑ Ci′′ ⎢ ΔH f ,i + ∫ C p ,i dT ⎥ − ∑ Ci′ ⎢ ΔH f ,i + ∫ C p ,i dT ⎥ i =1 ⎢⎣ ⎥⎦ i =1 ⎢⎣ ⎥⎦ Tr Tr N

(

)

1 N ΔH ( T ) = Ci′′ − Ci′ H i ∑ CW i =1 To determine the heat of reaction we must determine the chemical composition of the flow.

1 N dH = H i dCi ∑ CW i =1

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species conservation equation ρi+dρi

ρi V

dρi/dt

V+dV A+dA

A dz The conservation of species equation is

∂ρi − ρiVA + ( ρi + d ρi )(V + dV )( A + dA) = Adz ∂t ‐(rate of species i into CV)+(rate of species i out of CV)= rate of production of species i in CV Theory of Aerospace Propulsion

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species conservation (cont.) ∂ρi d ( ρi AV ) = AVd ρi + ρi d ( AV ) = Adz ∂t global mass conservation equation :

d ( ρ AV ) = 0 = AVd ρ + ρ d ( AV ) ∂ρi = AVd ρi − AV ρi Adz ∂t ρ ∂ρi ρVdYi = dz ∂t dρ

Theory of Aerospace Propulsion

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normalizing the species equation introduce L and tc:

z z ′ = = O (1) L t τ= t=0 tc

L

t=τ

reaction completed

⎛τ f ⎞ 1 ⎛ ∂ρi ⎞ 1 ⎛ ∂ρi ⎞ τ f ⎜ ⎟ dz = ⎜ ⎟ dz ' = O ⎜ ⎟ ρV ⎝ ∂t ⎠ ρ ⎝ ∂τ ⎠ τ c ⎝ τc ⎠ Theory of Aerospace Propulsion

17

species conservation (concluded) Characteristic residence time of a fluid particle, L/V

τ f ⎛ 1 ∂ρi ⎞ dYi = ⎜ ⎟ dz′ τ c ⎝ ρ ∂τ ⎠ Characteristic time for reaction to be completed The quantity τf /τc is essentially the first Damkohler similarity parameter for reacting flows Theory of Aerospace Propulsion

18

characteristic times fast flow but slow reaction

slow flow but fast reaction

τf/τc>>1

τf/τc<<1

fluid particles move slowly through reaction zone

Fluid particles pass quickly through reaction zone

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19

Damkohler number limits τf/τc<<1: Here dYi ~0 because the residence time is very brief

compared to the reaction time, no appreciable reaction occurs, and dH=0.

τf/τc>>1: In the limit dYi is infinite because the reaction time

goes to zero and the reaction proceeds instantaneously to the equilibrium composition pertinent to the local temperature and pressure, and thus dH is determined by using the chemical equilibrium conditions.

Theory of Aerospace Propulsion

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Damkohler number limits For tf / tc>>1 the heat of reaction is N

(

)

1 ′′ ′ C C Hi ΔH ( T ) = − ∑ i i equil CW i =1 For τf/τc~O(1) and neither of the limits is applicable the reaction in the combustor is rate‐controlled and the actual chemical kinetics must be accounted for. This complicates the equation set and the solution techniques required.

Theory of Aerospace Propulsion

21

conservation of momentum G G G G dFs ∫A ρV ⋅ nˆ VdA = υ∫ ρ Fb + ∫A dA dA

(

)

p V

p+dp dFz

(drag force on fluid)

V+dV

τwdAw (wall shear force on fluid) isentropic flow

mdV = − Adp − τ w dAw − dFz Theory of Aerospace Propulsion

22

conservation of momentum Skin friction coefficient

Cf =

τw 1 ρV 2 2

Hydraulic diameter

dAw dz =4 A D

⎡ ⎤ 2 ⎥ dFz 1 dp 1 dV dz 2 2 ⎢ + γM + γ M ⎢ 4C f + =0 ⎥ 2 1 2 p 2 V D ⎢ γ pAM 2 ⎥ 2 ⎣ ⎦

Theory of Aerospace Propulsion

23

equations in standard form dA A dQ − dWk − dH c pT Independent variables

4C f dz D

+

dFz 1 γ pAM 2 2

dW W dγ

γ Theory of Aerospace Propulsion

24

equations in standard form (cont.) dM dV da dp dT d ρ , , , , , 2 M V a p T ρ 2

dependent variables:

Typical equation from influence coefficient chart

dp ⎡ γ M 2 ⎤ dA ⎡ γ M 2 ⎤ dQ − dWk − dH =⎢ −⎢ − 2⎥ 2⎥ p ⎣1 − M ⎦ A ⎣1 − M ⎦ c pT ⎡ ⎤ ⎡ γ M {1 + ( γ − 1) M } ⎤ ⎢ 4C f dz ⎥ ⎡ γ M 2 ⎤ dW dFz ⎢ ⎥⎢ + +⎢ ⎥ 2⎥ 2 1 − D M 1 2 2 (1 − M ) ⎢⎣ ⎥⎦ ⎢ ⎦ W γ pAM ⎥ ⎣ ⎣ ⎦ 2 2

2

Theory of Aerospace Propulsion

25

influence coefficients: constant W and γ

dM 2 M2

dT T

dρ

ρ dp p

dA A

dQ − dWk − dH c pT

⎡ γ −1 2 ⎤ ⎢1 + 2 M ⎥ −2 ⎢ ⎥ 2 1 − M ⎢ ⎥ ⎣ ⎦ (γ − 1) M 2 1− M 2

1+ γ M 2 1− γ M 2 1− γ M 2 1− M 2

− ( γ − 1) γ M 4

M2 1− M 2

−1 1− M 2

−γ M 2 2 (1 − M 2 )

γM2 1− M 2

−γ M 2 1− M 2

−γ M 2 ⎡⎣1 + ( γ − 1) M 2 ⎤⎦

area change

heat addition

4C f dz

+

dFz

1 γ pM 2 A 2 ⎡ γ −1 2 ⎤ γ M 2 ⎢1 + M ⎥ 2 ⎣ ⎦ 2 1− M D

2 (1 − M 2 )

Theory of Aerospace Propulsion

2 (1 − M 2 )

friction 26

gas flow with simple area change A(z) z

⎡ γ −1 2 ⎤ 1+ M ⎥ 2 ⎢ dM dA 2 = −2 ⎢ ⎥ 2 M ⎢ 1− M ⎥ A ⎣ ⎦

dp γ M dA = p 1− M 2 A 2

dT ( γ − 1) M dA = T 1− M 2 A 2

dρ

M 2 dA = ρ 1− M 2 A

Theory of Aerospace Propulsion

27

THEORY OF AEROSPACE PROPULSION 3A. Idealized Cycle Analysis Of Jet Propulsion Engines: Turbojets

P. M. SFORZA University of Florida

3.1 General jet engine cycle Ae A0 Ve

V0 p0 T0 0

Net thrust:

pe=p0 Te

fuel 1

7

e

⎛ Fn 2 2 A0 ⎞ = p7 (1 + γ 7 M 7 ) − p0 ⎜1 + γ 0 M 0 ⎟ A7 A7 ⎠ ⎝

3.1 General jet engine cycle Schematic plot of the Brayton‐ like cycle of an ideal jet engine Pratt&Whitney J57 Turbojet

h

4 constant pressure heat addition 3 2 isentropic 1 compression 0

5 isentropic 6 expansion

p=constant

7

back to p7=p0

p=constant=p0

combustor inlet

s Schematic diagram of typical turbojet engine showing major components and the standard numbering scheme for engine stations

fuel lines

nozzle

0

1

2

3

4

6 7

3.1 General jet engine cycle Generic turbojet cycle diagram for subsonic flight, M0<1 4 p5

h

p6

5 p3

6

3

p0 7

2

p2 0 p1

1

s

3.2 Ideal jet engine cycle analysis Idealized turbojet cycle in which all the compression and expansion processes are considered to be isentropic. p4=p3 p5

4

h

5 p3

p2

3

6

2

7

0 p1

1

s

p6 p0

3.3 Ideal turbojet in maximum power takeoff Inlet flow, stations 0‐2: Standard day sea level conditions: altitude z=0, p0=101.3kPa, T0=288K, and a0=340.3m/s.

The takeoff condition is taken as M0=0 (V0=0), i.e., the static thrust condition. The actual Mach number at aircraft rotation and subsequent lift‐off is low enough (M0<0.25, that is, V0<85m/s)) that the difference between static and stagnation conditions is small.

Isentropic flow through the inlet pt,2=pt,0 Tt,2=Tt,0 0

1

2

3.3 Ideal turbojet in maximum power takeoff Compressor flow, stations 2‐3: For ideal compression the stagnation temperature at the compressor exit is ⎛ pt ,3 ⎞ Tt ,3 = 288 ⎜ ⎜ p ⎟⎟ ⎝ t ,2 ⎠

0.286

Stagnation pressure and temperature at the compressor exit at takeoff for various compressor pressure ratios: pt,3/pt,2

pt,3 (MPa)

Tt,3 (K)

5

0.5065

456.7

10

1.013

556.8

20

2.026

678.9

30

3.039

762.3

3.3 Ideal turbojet in maximum power takeoff Combustor flow, stations 3‐4

3

m f ⎛ m f ⎞ ηb Q f ⎜1 + ⎟ c pTt ,4 − c pTt ,3 = m 0 ⎝ m 0 ⎠ m f Tt ,4 − Tt ,3 f = = m 0 a ( 43.4 ×103ηb c p ) − Tt 4

4

Fuel‐air ratio f/a for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,2:

Tt,3 (K)

1200K

1300K

1400K

1500K

1600K

1700K

5

457

0.0176

0.0200

0.0224

0.0249

0.0273

0.0298

10

557

0.0152

0.0176

0.0200

0.0225

0.0249

0.0274

20

679

0.0123

0.0147

0.0171

0.0196

0.0220

0.0244

30

762

0.0103

0.0127

0.0151

0.0176

0.0200

0.0224

3.3 Ideal turbojet in maximum power takeoff Turbine flow, stations 4‐5

Compressor‐turbine matching m 0 c p (Tt ,3 − Tt ,2 ) = ( m 0 + m f ) c p (Tt ,4 − γTt ,5 ) In terms of turbine pressure ratio this becomes

⎡ ⎤ γ −1 γ −1 ⎧ ⎫ pt ,5 ⎢ Tt ,2 1 ⎪⎛ pt ,3 ⎞ γ ⎪⎥ = ⎢1 − ⎟⎟ − 1⎬⎥ ⎨⎜⎜ f ⎞ ⎪⎝ pt ,2 ⎠ pt ,4 ⎢ Tt ,4 ⎛ ⎥ ⎪ 1 + ⎜ ⎟ ⎭⎥ ⎢⎣ a ⎠⎩ ⎝ ⎦

Turbine exit stagnation pressures pt ,5 (kPa) for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

301.3

314.8

326.8

337.4

346.9

355.5

10

423.7

457.8

488.6

516.4

541.7

564.7

20

521.0

592.7

659.5

721.6

779.2

832.8

30

535.5

636.9

734.1

826.6

914.2

999.7

3.3 Ideal turbojet in maximum power takeoff

4

5

For ideal expansion ⎛ pt ,4 ⎞ ⎟⎟ through the turbine: Tt ,5 = Tt ,4 ⎜⎜ ⎝ pt ,5 ⎠

− ( γ −1)

Turbine exit stagnation temperatures Tt 5 (K) for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

1034

1135

1235

1336

1436

1536

10

935.2

1036

1137

1237

1338

1438

20

813.8

914.7

1016

1117

1217

1318

30

730.4

831.5

932.6

1034

1135

1237

γ

3.3 Ideal turbojet in maximum power takeoff Nozzle flow, stations 5‐7: The typical nozzle is a converging duct so that stations 6 and 7 are coincident. Maximum thrust occurs when the exit station is choked, that is, M7=1. The nozzle is ideal so that pt,5=pt,6=pt,7.

Thus the pressure at the exit station p7 may be found using the isentropic flow relation with the condition M7=1 and the values for pt,5 from a previous slide.

Nozzle exit pressures p7 (kPa) for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,

4

5

6,7

2

1200K 1300K 1400K 1500K 1600K 1700K

5

159.2

166.3

172.6

178.3

183.3

187.8

10

223.8

241.8

258.1

272.8

286.2

298.3

20

275.3

313.1

348.4

381.2

411.7

440.0

30

282.9

336.5

387.8

436.7

483.0

528.2

3.3 Ideal turbojet in maximum power takeoff

4

5

6,7

We may calculate the nozzle exit temperature T7 using the isentropic flow relation with the condition M7=1 and the values for the turbine exit stagnation temperature from a previous slide. The results for T7 are shown in the table on the following slide. The exit velocity can then be found from the relation V=aM, where the sound speed

(

a7 = γ RT7 = 20.04 m s ⋅ K

) T

7

Here γR=1.4(287m2/s2‐K) and the results for V7 are also shown on the next slide.

3.3 Ideal turbojet in maximum power takeoff Nozzle exit temperatures T7 (K) for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off

4

5

6,7

pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

5

861.9

945.6

1029.2 1112.9 1196.6 1280.2

10

779.3

863.2

947.0

1030.9 1114.8 1198.6

20

678.1

762.2

846.3

930.4

1014.5 1098.6

30

608.6

692.9

777.1

861.4

945.7

Nozzle exit velocities V7 (m/s) for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

588.3

616.2

642.9

668.5

693.2

717.0

10

559.4

588.8

616.7

643.4

669.1

693.8

20

521.9

553.3

583.0

611.3

638.3

664.2

30

494.4

527.5

558.7

588.2

616.3

643.4

1700K

1030.8

3.3 Ideal turbojet in maximum power takeoff Net thrust

⎛ f⎞ Fn = m 0 ⎜1+ ⎟V7 − m 0V0 + A7 ( p7 − p0 ) ⎝ a⎠

Mass f ⎞ ⎛

m0 ⎜ 1 + ⎟ = m 7 = ρ 7V7 A7 conservation a⎠ ⎝ m 0 ⎛ f ⎞ 1 + ρ7V7 ⎜⎝ a ⎟⎠

Exit area

A7 =

Specific thrust

Fn ⎛ p0 ⎞ ⎤ 1 ⎛ f ⎞ ⎡ = ⎜ 1 + ⎟ V7 ⎢1 + 1 − ⎟ ⎥ − V0 2 ⎜

m0 ⎝ a ⎠ ⎣ γ M7 ⎝ p7 ⎠ ⎦

3.3 Ideal turbojet in maximum power takeoff Net thrust

⎛ f⎞ Fn = m 0 ⎜1+ ⎟V7 − m 0V0 + A7 ( p7 − p0 ) ⎝ a⎠

Mass f ⎞ ⎛

m0 ⎜ 1 + ⎟ = m 7 = ρ 7V7 A7 conservation a⎠ ⎝ m 0 ⎛ f ⎞ 1 + ρ7V7 ⎜⎝ a ⎟⎠

Exit area

A7 =

Specific thrust

Fn ⎛ p0 ⎞ ⎤ 1 ⎛ f ⎞ ⎡ = ⎜ 1 + ⎟ V7 ⎢1 + 1 − ⎟ ⎥ − V0 2 ⎜

m0 ⎝ a ⎠ ⎣ γ M7 ⎝ p7 ⎠ ⎦

3.3 Ideal turbojet in maximum power takeoff 1100

Tt,4 (K) 1700 1600

1000

Fn/m (m/s)

1500 1400

900 800

1300 1200

TJ, sonic nozzle Mo=0, z=0

700 600 0

5

10

15

20

25

30

35

pt,3/pt,2

The specific thrust, which has the units of velocity, gives a useful view of the net thrust produced per unit mass flow of air passing through the engine without specifying the size of the engine.

3.3 Ideal turbojet in maximum power takeoff Specific fuel consumption

⎛ f ⎞ ⎛ Fn ⎞ cj = = ⎜ ⎟⎜ ⎟ Fn ⎝ a ⎠ ⎝ m 0 ⎠ m f

−1

1.1 TJ, sonic nozzle Mo=0, z=0km

1 0.9

Tt,4 (K) 1700 1600 1500 1400 1300 1200

0.8 cj (lbm/hr/lb) 0.7 0.6 0.5 0.4 0

5

10

15

20

pt,3/pt,2

25

30

35

3.4 Ideal turbojet in subsonic cruise in the stratosphere Inlet flow, stations 0‐2: In subsonic flight, M0<1, the inlet of the ideal turbojet operates isentropically so that the stagnation values of pressure and temperature delivered to the compressor face, station 2, are equal to those in the free stream, pt,2=pt,0 and Tt,2=Tt,0.

Using the isentropic relations the stagnation conditions at station 2 are readily calculated to be pt,2=pt,0=40.4kPa and Tt,2=Tt,0=252K.

0

1

2

3.4 Ideal turbojet in subsonic cruise in the stratosphere Compressor flow, stations 2‐3: For a given compressor pressure ratio the compressor exit stagnation temperature is given by ⎛ pt ,3 ⎞ Tt ,3 = 252 ⎜ ⎜ p ⎟⎟ ⎝ t ,2 ⎠ 2

0.286

3

Stagnation pressure and temperature at the compressor exit in M0=0.8 cruise at 10km altitude pt,3/pt,2

pt,3 (kPa)

Tt,3 (K)

5

202.0

399.3

10

404.0

486.9

20

808.0

593.6

30

1212

666.5

3.4 Ideal turbojet in subsonic cruise in the stratosphere Combustor flow, stations 3‐4

3

m f ⎛ m f ⎞ ηb Q f ⎜1 + ⎟ c pTt ,4 − c pTt ,3 = m 0 ⎝ m 0 ⎠ m f Tt ,4 − Tt ,3 f = = m 0 a ( 43.4 ×103ηb c p ) − Tt 4

4

Fuel‐air ratio f/a for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

0.01898

0.02140

0.02383

0.02628

0.02873

0.03120

10

0.01691

0.01932

0.02175

0.02419

0.02664

0.02910

20

0.01438

0.01679

0.01921

0.02164

0.02409

0.02654

30

0.01265

0.01506

0.01747

0.01990

0.02234

0.02479

3.4 Ideal turbojet in subsonic cruise in the stratosphere Turbine flow, stations 4‐5

Compressor‐turbine matching m 0 c p (Tt ,3 − Tt ,2 ) = ( m 0 + m f ) c p (Tt ,4 − γTt ,5 ) In terms of turbine pressure ratio this becomes

⎡ ⎤ γ −1 γ −1 ⎧ ⎫ pt ,5 ⎢ Tt ,2 1 ⎪⎛ pt ,3 ⎞ γ ⎪⎥ = ⎢1 − ⎟⎟ − 1⎬⎥ ⎨⎜⎜ f ⎞ ⎪⎝ pt ,2 ⎠ pt ,4 ⎢ Tt ,4 ⎛ ⎥ ⎪ 1 + ⎜ ⎟ ⎭⎥ ⎢⎣ a ⎠⎩ ⎝ ⎦

Turbine exit stagnation pressures pt ,5 (kPa) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

128.9

133.8

138.2

142.0

145.5

148.5

10

191.2

204.1

215.7

226.0

235.4

243.9

20

255.3

283.9

310.2

334.3

356.6

377.0

30

281.6

324.0

363.9

401.1

435.9

468.3

3.4 Ideal turbojet in subsonic cruise in the stratosphere

4

5

For ideal expansion ⎛ pt ,4 ⎞ ⎟⎟ through the turbine: Tt ,5 = Tt ,4 ⎜⎜ ⎝ pt ,5 ⎠

− ( γ −1)

γ

Turbine exit stagnation temperatures Tt 5 (K) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

1055

1156

1256

1356

1457

1557

10

969.0

1070

1170

1271

1371

1472

20

863.2

964.0

1065

1166

1266

1367

30

790.6

891.6

992.5

1093

1194

1295

3.4 Ideal turbojet in subsonic cruise in the stratosphere Nozzle flow, stations 5‐7: The typical nozzle is a converging duct so that stations 6 and 7 are coincident. Maximum thrust occurs when the exit station is choked, that is, M7=1. The nozzle is ideal so that pt,5=pt,6=pt,7.

Thus the pressure at the exit station p7 may be found using the isentropic flow relation with the condition M7=1 and the values for pt,5 from a previous slide.

Nozzle exit pressures p7 (kPa) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude

4

5

6,7

pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

68.09

70.70

73.00

75.03

76.85

78.47

10

101.02

107.8

113.9

119.4

124.4

128.9

20

134.9

150.0

163.9

176.6

188.4

199.2

30

148.8

171.2

192.2

211.9

230.3

247.4

3.4 Ideal turbojet in subsonic cruise in the stratosphere

4

5

6,7

We may calculate the nozzle exit temperature T7 using the isentropic flow relation with the condition M7=1 and the values for the turbine exit stagnation temperature from a previous slide. The results for T7 are shown in the table on the following slide. The exit velocity can then be found from the relation V=aM, where the sound speed

(

a7 = γ RT7 = 20.04 m s ⋅ K

) T

7

Here γR=1.4(287m2/s2‐K) and the results for V7 are also shown on the next slide.

3.4 Ideal turbojet in subsonic cruise in the stratosphere

Nozzle exit temperatures T7 (K) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude

4

5

6,7

pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

879.5

963.1

1047

1130

1214

1298

10

807.5

891.3

975.1

1059

1143

1226

20

719.4

803.3

887.3

971.3

1055

1139

30

658.8

743.0

827.1

911.2

995.4

1080

Nozzle exit velocities V7 (m/s) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

593.1

620.7

647.1

672.4

696.8

720.4

10

568.3

597.1

624.5

650.8

676.1

700.4

20

536.4

566.9

595.8

623.3

649.7

675.1

30

513.4

545.1

575.2

603.7

631.0

657.1

3.4 Ideal turbojet in subsonic cruise in the stratosphere Variable area nozzle: If A7 could be increased over A6 thereby accelerating the nozzle flow to lower pressures such that a matched condition, p7=p0, could be achieved, the nozzle exit Mach number, temperature, and velocity would change. Noting that for an isentropic nozzle pt,7=pt,5 and for a matched exit p7=p0, and keeping γ=1.4 for this ideal study, we may find M7 shown in the table on the next slide by using the isentropic relation ⎛ M ⎞ = = ⎜1 + ⎟ 5 ⎠ p7 p0 ⎝

pt ,7

5

6

7

pt ,5

2 7

3.5

3.4 Ideal turbojet in subsonic cruise in the stratosphere Matched nozzle exit Mach numbers M7 for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

1.685

1.710

1.731

1.749

1.765

1.779

10

1.940

1.982

2.017

2.047

2.073

2.096

20

2.122

2.190

2.246

2.294

2.335

2.370

30

2.183

2.272

2.346

2.408

2.461

2.506

Matched nozzle exit velocities V7 (m/s) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=0.8 cruise at 10km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

875.8

925.0

971.7

1016

1059

1100

10

913.1

970.9

1025

1077

1127

1174

20

904.6

971.4

1034

1093

1149

1202

30

878.0

951.1

1019

1083

1143

1200

3.4 Ideal turbojet in subsonic cruise in the stratosphere Fn ⎛ p0 ⎞ ⎤ 1 ⎛ f ⎞ ⎡ = ⎜ 1 + ⎟ V7 ⎢1 + 1 − ⎟ ⎥ − V0 2 ⎜ m 0 ⎝ a ⎠ ⎣ γ M7 ⎝ p7 ⎠ ⎦

Sonic nozzle 900

Tt,4 (K) 1700 1600

850 800

FFn n/m m 0

1500

750 700

( m s ) 650 (m/s)

1400 1300

600 TJ, sonic nozzle Mo=0.8, z=10km

550 500 0

5

10

1200 15

20

pt,3/pt,2

25

30

35

3.4 Ideal turbojet in subsonic cruise in the stratosphere ⎛ f ⎞ ⎛ Fn ⎞ cj = = ⎜ ⎟⎜ ⎟ Fn ⎝ a ⎠ ⎝ m 0 ⎠ m f

Sonic nozzle

−1

1.3 TJ, sonic nozzle Mo=0.8, z=10km

1.2 1.1 cj (lbm/lb/hr)

Tt,4 (K) 1700 1600 1500 1400 1300 1200

1 0.9 0.8 0.7 0

5

10

15

20

pt,3/pt,2

25

30

35

3.4 Ideal turbojet in subsonic cruise in the stratosphere Matched nozzle specific thrust compared to sonic nozzle specific thrust 1.18 TJ at M0=0.8 and z=10km

(Fn/m)m/(Fn/m)s

1.16

Tt4 (K) 1700 1600 1500 1400 1300 1200

1.14 1.12 1.1 1.08 1.06 0

5

10

15

20

pt,3/pt,2

25

30

35

3.4 Ideal turbojet in subsonic cruise in the stratosphere Matched nozzle cj compared to sonic nozzle cj 0.96 TJ at M0=0.8 and z=10km

0.94 0.92

Tt,4 (K) 1200 1300 1400 1500 1600 1700

c j, m/c j,s 0.9 0.88 0.86 0

5

10

15

20

pt,3/pt,2

25

30

35

3.5 Ideal turbojet in supersonic cruise in the stratosphere Schematic diagram of an afterburning turbojet engine combustor

fuel lines

inlet afterburner

1

2

3

4

5

5b

6

nozzle

7

The basic cruise state is standard day stratospheric conditions, that is, altitude z=15km, pressure p0=12.1kPa, temperature T0=217K, and speed of sound a0=295m/s. The typical cruise flight condition is taken here as M0=2 (V0=590m/s).

3.5 Ideal turbojet in supersonic cruise in the stratosphere p4=p3

Tt,4=Tt,5

4

5b

5

T p3

3 p5b=p5 2

p2 p1=p0

0,1

s

6 p6

afterburner 7

p0

3.5 Ideal turbojet in supersonic cruise in the stratosphere Inlet flow, stations 0‐2 Although M >1 the inlet of the ideal 0

turbojet is still considered to operate isentropically so that pt,2=pt,0 and Tt,2=Tt,0 Using isentropic flow relations for the given supersonic cruise conditions pt,2=pt,0=94.7kPa and Tt,2=Tt,0=391K. 0 Weak shock

1

2

We will show, in Chapter 6, that a well‐ designed inlet can deliver a pressure recovery of more than 90% at M0=2.

3.5 Ideal turbojet in supersonic cruise in the stratosphere Compressor flow, stations 2‐3 Stagnation pressure and temperature at the compressor exit in M0=2 cruise at 15km altitude

2

3

⎛ pt ,3 ⎞ Tt ,3 = 391⎜ ⎜ p ⎟⎟ ⎝ t ,2 ⎠

0.286

pt,3/pt,2

pt,3 (kPa)

Tt,3 (K)

5

474

620

10

947

755

20

1894

921

30

2841

1034

The stagnation pressure being delivered to the combustor is much higher in supersonic cruise than in subsonic cruise even though the ambient atmospheric pressure at z=15km is only one‐half that at z=10km. This is due to the higher ram pressure at supersonic speed.

3.5 Ideal turbojet in supersonic cruise in the stratosphere

Combustor flow, stations 3‐4

m f ⎛ m f ⎞ ηb Q f ⎜1 + ⎟ c pTt ,4 − c pTt ,3 = m 0 ⎝ m 0 ⎠ m f Tt ,4 − Tt ,3 f = = m 0 a ( 43.4 ×103ηb c p ) − Tt 4

3

4

Fuel‐air ratio f/a for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=2 cruise at 15km altitude pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

0.01375

0.01616

0.01858

0.02101

0.02345

0.02591

10

0.01054

0.01294

0.01535

0.01777

0.02020

0.02265

20

0.00661

0.00900

0.01140

0.01382

0.01624

0.01868

30

0.00392

0.00631

0.008707

0.01112

0.01353

0.01596

Note that Tt,3 in supersonic cruise is higher than in subsonic cruise. As a result f/a is lower in supersonic cruise because less heat can be added to get to the limiting turbine inlet stagnation temperature.

3.5 Ideal turbojet in supersonic cruise in the stratosphere γ

Turbine flow, stations 4‐5

4

⎡ ⎤ γ −1 γ −1 ⎧ ⎫⎥ γ ⎛ ⎞ pt ,5 ⎢ Tt ,2 p 1 ⎪ t ,3 ⎪ = ⎢1 − − 1 ⎟⎟ ⎨⎜⎜ ⎬⎥ f ⎞ ⎪⎝ pt ,2 ⎠ pt ,4 ⎢ Tt ,4 ⎛ ⎥ ⎪ + 1 ⎜ ⎟ ⎭⎥ ⎢⎣ a ⎠⎩ ⎝ ⎦

Turbine exit stagnation pressures pt,5 (kPa) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=2 cruise at 15km altitude

5

pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

228.6

243.5

256.9

268.9

279.8

289.6

10

271.1

304.7

335.8

364.6

391.2

415.7

20

250.8

309.4

367.0

422.7

476.2

527.2

30

196.3

266.0

338.4

411.7

484.6

556.2

Note that pt,5 in supersonic cruise is considerably larger than in subsonic cruise. Even though p0 at 15km is half that at 10km altitude, supersonic ram pressure increases more than make up for that reduction.

3.5 Ideal turbojet in supersonic cruise in the stratosphere Turbine flow, stations 4‐5 −( γ −1) For ideal expansion ⎛ pt ,4 ⎞ γ through the turbine: Tt ,5 = Tt ,4 ⎜⎜ p ⎟⎟ ⎝

4

t ,5

⎠

Turbine exit stagnation temperatures Tt,5 (K) for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=2 cruise at 15km altitude

5

pt,3/pt,2

1200K

1300K

1400K

1500K

1600K

1700K

5

974.3

1074

1175

1276

1376

1477

10

839.1

939.9

1041

1142

1243

1343

20

673.1

774.3

875.5

976.8

1078

1179

30

558.8

660.3

761.8

863.3

964.8

1066

3.5 Ideal turbojet in supersonic cruise in the stratosphere Afterburner flow, stations 5‐5b The turbine exit flow is still quite rich in oxygen and therefore fuel may be injected into the duct leading to the nozzle throat and burned, again at constant pressure. The afterburner section acts like a large constant pressure combustor and the relatively low temperature gas coming from the turbine exit may be boosted to higher levels, Tt,5b, which permits the generation of higher exit velocity V7 and thereby increases the thrust capability of the engine. fuel line

afterburner

4

5

5b

6

nozzle

7

3.5 Ideal turbojet in supersonic cruise in the stratosphere Afterburner flow, stations 5‐5b p4=p3 Tt,4=Tt,5

4

5b

5

T

3 p3

p5b=p5

2 p2 p1=p0

0,1

s

6 p6

afterburner 7

p0

3.5 Ideal turbojet in supersonic cruise in the stratosphere Afterburner flow, stations 5‐5b: Some of the oxygen in the air entering the engine has been consumed in the combustor and no longer available for combustion. A simple approximation assumes that the stagnation temperature rise achievable is linearly proportional to a function of the fuel to air ratio used in the core engine φ (f/a) as given by

f a) ( K= = = 1− Tt ,5bi − Tt 5 Q f ( f a )stoich Tt ,5b − Tt 5

Qeff

For Jet A the stoichiometric fuel to air ratio may be taken as (f/a)stoich=0.0678, as described in Chapter 4, which leads to

⎡ ⎛ f ⎞⎤ Qeff = ⎢1 − 14.7 ⎜ ⎟ ⎥ Q f ⎝ a ⎠⎦ ⎣

3.5 Ideal turbojet in supersonic cruise in the stratosphere Afterburner flow, stations 5‐5b: ⎡ ⎛ f ⎞⎤ 1 + ⎜ ⎟ ⎥ (Tt ,5b − Tt ,5 ) ⎢

⎛ mf ⎞ ⎛ f ⎞ ⎝ a ⎠⎦ ⎣ ⎜ ⎟ =⎜ ⎟ = ⎛ f ⎞⎤ ⎝ a ⎠ AB ⎝ m0 ⎠ AB 3 ⎡ 43.4 × 10 ⎢1 − 14.7 ⎜ ⎟ ⎥ − Tt ,5b ⎝ a ⎠⎦ ⎣ Afterburner fuel‐air ratio (f/a)AB for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=2 cruise at 15km altitude pt3/pt2

1200K

1300K

1400K

1500K

1600K

1700K

5

0.006844 0.007196 0.007588 0.007697 0.008524 0.009088

10

0.01028

0.01077

0.01132

0.01152

0.01262

0.01338

20

0.01396

0.01459

0.01528

0.01560

0.01687

0.01781

30

0.01622

0.01691

0.01766

0.01809

0.01941

0.02043

3.5 Ideal turbojet in supersonic cruise in the stratosphere Nozzle flow, stations 5b–7 The nozzle considered for supersonic cruise must be of the variable‐area converging‐diverging duct type so that it may operate in the matched pressure mode p7=p0, thereby assuring the maximum thrust achievable. Since the stagnation pressure is constant throughout the nozzle, pt,5b=pt,6=pt,7. Keeping γ=1.4 for this ideal study, we find M7 by using the isentropic relation Matched nozzle exit Mach numbers M7 for various compressor pressure ratios and turbine inlet stagnation temperatures in M0=2 cruise at 15km altitude pt3/pt2

1200K

1300K

1400K

1500K

1600K

1700K

5

2.566

2.608

2.642

2.672

2.698

2.720

10

2.677

2.754

2.817

2.871

2.917

2.959

20

2.627

2.764

2.875

2.969

3.048

3.117

30

2.469

2.665

2.822

2.951

3.060

3.153

3.5 Ideal turbojet in supersonic cruise in the stratosphere Nozzle flow, stations 5b–7 Note that the exit Mach numbers lie in the range 2.56<M7<3.15 which require area ratios in the range 2.8<A7/A6<4.8 in order to produce the matched exit flow. The exit velocity V7 in the matched case may be found by first using the isentropic relation to find T7 from M7 and Tt,7=Tt,5b and then using the Mach number definition M7=V7/a7.

3.5 Ideal turbojet in supersonic cruise in the stratosphere Fn ⎡ ⎛ f ⎞ ⎛ f ⎞ ⎤ ⎡ p0 ⎞ ⎤ 1 ⎛ = ⎢1 + ⎜ ⎟ + ⎜ ⎟ ⎥ V7 ⎢1 + 1 − ⎟ ⎥ − V0 2 ⎜ m 0 ⎣ ⎝ a ⎠ ⎝ a ⎠ AB ⎦ ⎣ γ M 7 ⎝ p7 ⎠ ⎦ 1000 Tt,4 (K) 1700 1600

900 800

1500 1400

Fn/m 700 (m/s)

1300 600 1200

ABTJ: Tt,7=Tt,4 M0=2, z=15km

500 400 0

5

10

15

20

pt,3/pt,2

25

30

35

3.5 Ideal turbojet in supersonic cruise in the stratosphere Ratio of specific thrust with and without afterburning as a function of compressor pressure ratio and turbine inlet temperature 3.4

Tt,4 (K) ABTJ compared to TJ M0=2, z=15km

(Fn/m)AB /(Fn/m)

3

1200

2.6 1300 2.2 1400 1500 1600 1700

1.8 1.4 1 0

5

10

15

20 pt,3/pt,2

25

30

35

3.5 Ideal turbojet in supersonic cruise in the stratosphere ( f a ) + ( f a ) AB c j = 3600 g ( Fn m 0 )

1.40 1700

Tt,4 (K)

ABTJ: Tt,7=Tt,4 M0=2, z=15km

1.36 1600 1.32 cj 1.28 (lbm/hr/lb)

1500 1400

1.24

1300 1200

1.20 1.16 0

5

10

15

20

pt,3/pt,2

25

30

35

3.5 Ideal turbojet in supersonic cruise in the stratosphere Ratio of specific fuel consumption with and without afterburning as a function of compressor pressure ratio and turbine inlet temperature 1.9 Tt4 (K) 1200

ABTJ compared to TJ M0=2, z=15km

1.8 1.7 1.6

1300 1400 1500 1600 1700

c j,AB/c j 1.5 1.4 1.3 1.2 1.1 0

5

10

15

20

pt,3/pt,2

25

30

35

THEORY OF AEROSPACE PROPULSION 3.B Idealized Cycle Analysis Of Jet Propulsion Engines: Turbofans

P. M. SFORZA University of Florida

3.7 Ideal turbofan in maximum power takeoff Schematic diagram of turbofan engine fan cowling fan flow fan flow

cold flow combustor

core flow

nozzle

combustor fan flow

fuel line

hot flow cold flow

fan 1

2

3

3F

4 5

6,7

3.7 Ideal turbofan in maximum power takeoff Standard day sea level conditions: altitude z=0, p0=101.3kPa, T0=288K, and a0=340.3m/s.

The takeoff condition is taken as M0=0 (V0=0), i.e., the static thrust condition. The actual Mach number at aircraft rotation and subsequent lift‐off is low enough (M0<0.25, that is, V0<85m/s)) that the difference between static and stagnation conditions is small.

Isentropic flow through the inlet pt,2=pt,0 Tt,2=Tt,0 0

1

2

3.7 Ideal turbofan in maximum power takeoff Compressor flow, stations 2‐3: For ideal compression the stagnation temperature at the compressor exit is ⎛ pt ,3 ⎞ Tt ,3 = 288 ⎜ ⎜ p ⎟⎟ ⎝ t ,2 ⎠

0.286

The stagnation pressure and temperature at the compressor exit at takeoff for various compressor pressure ratios: pt,3/pt,2

pt,3 (MPa)

Tt,3 (K)

5

0.5065

456.7

10

1.013

556.8

20

2.026

678.9

30

3.039

762.3

3.7 Ideal turbofan in maximum power takeoff Fan flow, stations 2‐3F: The fan provides an increase in stagnation pressure from the inlet value as given by the fan pressure ratio pt,3F/pt,2.

Current engines operate in the range 1.5<pt,3F/pt,2<3. We will use only the lower value of 1.5, which is typical of commercial engines, for comparison purposes. The fan exit stagnation temperature is ⎛ pt ,3 F ⎞ =⎜ ⎟⎟ ⎜ Tt ,2 ⎝ pt ,2 ⎠

Tt ,3 F

γ −1 γ

p3 F ⎛ γ − 1 2 ⎞ M 3F ⎟ = ⎜1 + 2 pt ,3 F ⎝ ⎠

−γ γ −1

For p3,F=p0 use this eq. to find M3,F<1 For choked exit M3,F=1 use this eq. to find p3,F

pt,3F/pt,2

pt,3F(kPa)

Tt,3F (K)

M3F

p3F (kPa)

T3F (K)

V3F(m/s)

1.5

152

323

0.784

101

288

267

2.0

203

351

1.000

107

293

343

2.5

253

374

1.000

134

312

354

3.7 Ideal turbofan in maximum power takeoff The flow passing through the combustor is the same for the turbojet and turbofan cases because we are considering the same compressor pressure ratios and the same turbine entrance total temperatures.

Therefore the results for the fuel air ratio as a function of compressor pressure ratio and turbine inlet stagnation temperature, as given previously, are the same for both the turbojet and turbofan cases.

3.7 Ideal turbofan in maximum power takeoff combustor

3

m f ⎛ m f ⎞ ηb Q f ⎜1 + ⎟ c pTt ,4 − c pTt ,3 = m 0 ⎝ m 0 ⎠ m f Tt ,4 − Tt ,3 f = = m 0 a ( 43.4 ×103ηb c p ) − Tt 4

4

Fuel‐air ratio f/a for various compressor pressure ratios and turbine inlet stagnation temperatures at take‐off pt,3/pt,2:

Tt,3 (K)

1200K

1300K

1400K

1500K

1600K

1700K

5

457

0.0176

0.0200

0.0224

0.0249

0.0273

0.0298

10

557

0.0152

0.0176

0.0200

0.0225

0.0249

0.0274

20

679

0.0123

0.0147

0.0171

0.0196

0.0220

0.0244

30

762

0.0103

0.0127

0.0151

0.0176

0.0200

0.0224

3.7 Ideal turbofan in maximum power takeoff The matching condition to be met is that the turbine power produced equal the sum of the required compressor and fan power, which may be written as follows:

m F c p (Tt ,3 F − Tt ,2 ) + m C c p (Tt ,3 − Tt ,2 ) = m C c p (Tt ,4 − Tt ,5 )

β c p (Tt ,3 F − Tt ,2 ) + c p (Tt ,3 − Tt ,2 ) = c p (Tt ,4 − Tt ,5 ) ⎡⎛ p ⎞0.286 ⎤ ⎡⎛ p ⎞ 0.286 ⎤ T ⎡ ⎛ p ⎞ −0.286 ⎤ ⎥ β ⎢⎜⎜ t ,3 F ⎟⎟ − 1⎥ + ⎢⎜ t ,3 ⎟ − 1⎥ = t ,4 ⎢1 − ⎜ t ,4 ⎟ ⎢⎝ pt ,2 ⎠ ⎥ ⎢⎜⎝ pt ,2 ⎟⎠ ⎥ Tt ,2 ⎢ ⎜⎝ pt ,5 ⎟⎠ ⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ T ⎧ ⎛p ⎞ t ,2 ⎪ t ,3 F ⎢ = 1− β ⎟⎟ ⎨ ⎜⎜ pt ,5 ⎢ Tt ,4 ⎪ ⎝ pt ,2 ⎠ ⎩ ⎣

pt ,4

0.286

⎛ pt ,3 ⎞ +⎜ ⎜ p ⎟⎟ ⎝ t ,2 ⎠

0.286

⎫⎪⎤ − ( β + 1) ⎬⎥ ⎪⎭⎥⎦

−3.5

3.7 Ideal turbofan in maximum power takeoff

Therefore, for every selected value of the turbine inlet temperature the required turbine pressure ratio depends on three independent parameters so that a large number of solutions are possible.

Using the equation at the bottom of the previous slide we find the results for pt5 as shown in Table 3.30 in the textbook for the selected limiting turbine inlet stagnation temperatures and several bypass and compressor pressure ratios.

The turbine exit stagnation temperature may then be found from the −( γ −1) isentropic relation ⎛p ⎞ γ Tt ,5 = Tt ,4 ⎜ t ,4 ⎟ ⎜p ⎟ ⎝ t ,5 ⎠

The corresponding results for Tt,5 for the different compressor pressure ratios and limiting turbine inlet stagnation temperatures are shown in Table 3.31 of the textbook.

3.7 Ideal turbofan in maximum power takeoff

3.8 Ideal turbofan in high subsonic cruise in the stratosphere Low bypass ratio internal turbofan with internal mixing of fan and core flows fuel lines

fan cowling inlet

cold fan flow

fan flow combustor

hot core flow mixing zone nozzle

combustor fan flow

core flow fan flow 1 2

3

4

5

5F

6

7

3.8 Ideal turbofan in high subsonic cruise in the stratosphere Mach number M0<1 the inlet of the ideal turbojet operates isentropically so that the stagnation values of pressure and temperature delivered to the compressor face, station 2, are equal to those in the free stream, pt,2=pt,0 and Tt,2=Tt,0.

Using the isentropic relations the stagnation conditions at station 2 are readily calculated to be pt,2=pt,0=40.4kPa and Tt,2=Tt,0=252K.

3.8 Ideal turbofan in high subsonic cruise in the stratosphere

3.9 Ideal internal turbofan in supersonic cruise in the stratosphere Schematic diagram of an afterburning internal turbofan engine fuel lines

fan cowling inlet

fan flow

afterburner

combustor core flow

nozzle

combustor fan flow

core flow fan flow 1 2

fuel injection 3F

3

4

5

5b

6

7

The basic cruise state is standard day stratospheric conditions, that is, altitude z=15km, pressure p0=12.1kPa, temperature T0=217K, and speed of sound a0=295m/s. The typical cruise flight condition is taken here as M0=2 (V0=590m/s).

3.9 Ideal internal turbofan in supersonic cruise in the stratosphere pt,4=pt,3 Tt,4=Tt,5b

4

5b p6 5

T 3

pt,3

p5

core flow pt,3F=pt,5 pt,2

6

afterburner heating of core p0 and fan flow

fan flow

7

2 1 isentropic inlet compression

pt,1=pt,0 0

s

Afterburner operation shown on a turbofan cycle diagram

3.9 Ideal turbojet in supersonic cruise in the stratosphere Although M0>1 the inlet of the ideal turbojet is still considered to operate isentropically so that the stagnation values of pressure and temperature delivered to the compressor face are equal to those in the free stream, i.e., pt,2=pt,0 and Tt,2=Tt,0 Using isentropic flow relations for the given supersonic cruise conditions the stagnation conditions at station 2 are readily calculated to be pt,2=pt,0=94.7kPa and Tt,2=Tt,0=391K. We will show, in Chapter 6, that a well‐designed inlet can deliver a pressure recovery of more than 90% at M0=2.

3.10 Ideal scramjet in high supersonic cruise in the stratosphere M0>>1 inlet/forebody fuel lines

inlet shock waves

0

nozzle/afterbody

combustor

1

2

3

4 5

6

7

Schematic diagram of a scramjet engine showing integrated forebody and plug‐type nozzle afterbody

3.10 Ideal scramjet in high supersonic cruise in the stratosphere scramjet inlet

Full scale model of the X-43A mounted upside down on a strut for testing in NASA Langley’s 8 foot (2.4 m), high-temperature wind tunnel.

THEORY OF AEROSPACE PROPULSION 3C. Idealized Cycle Analysis Of Jet Propulsion Engines: Ramjets and Scramjets

P. M. SFORZA University of Florida

3.6 Ideal ramjet in high supersonic cruise in the stratosphere Schematic diagram of an ideal ramjet for high supersonic cruise fuel line inlet shocks combustor

inlet

1

Flight conditions

nozzle

2, 3

4, 5

6

7

M0

z (km)

p0 (kpa)

T0 (K)

q0 (kPa) pt0 (kPa)

Tt0 (K)

2

15

12.1

217

33.9

95

391

3

20

5.52

217

34.8

203

608

4

24

2.97

221

33.3

451

928

3.6 Ideal ramjet in supersonic cruise in the stratosphere Although M0>1 the inlet of the ideal turbojet is still considered to operate isentropically so that pt,2=pt,0 and Tt,2=Tt,0

Weak shocks

1

2

Using isentropic flow relations for the given supersonic cruise conditions pt,2=pt,0=94.7kPa and Tt,2=Tt,0=391K. We will show, in Chapter 6, that a well‐ designed inlet can deliver relatively high stagnation pressure recovery

3.10 Ideal scramjet in high supersonic cruise in the stratosphere M0>>1 inlet/forebody fuel lines

inlet shock waves

0

nozzle/afterbody

combustor

1

2

3

4 5

6

7

Schematic diagram of a scramjet engine showing integrated forebody and plug‐type nozzle afterbody

3.10 Ideal scramjet in high supersonic cruise in the stratosphere scramjet inlet

Full scale model of the X-43A mounted upside down on a strut for testing in NASA Langley’s 8 foot (2.4 m), high-temperature wind tunnel.

THEORY OF AEROSPACE PROPULSION 4. Combustion Chambers for Airbreathing Engines

P. M. SFORZA University of Florida

combustion chambers for airbreathing engines A good combustor should: •Provide full combustion with minimum pressure loss •Operate without significant accumulation of deposits •Ignite the fuel readily •Give reliable service over long periods of time

Theory of Aerospace Propulsion

2

combustion chambers for airbreathing engines Good combustor design is based on fundamentals: Proper mixture ratio and the three “Ts” of good combustion •Temperature of the reactants •Turbulence for good mixing •Time for mixing and combustion to go to completion

Theory of Aerospace Propulsion

3

constant area combustor Equations developed from the influence coefficient table for no change in area, heat transfer, power, friction, or drag

⎡1 + γ M ⎤ ⎡ dH dW ⎤ d γ dM = −⎢ + ⎥− 2 2 ⎥⎢ M ⎣ 1 − M ⎦ ⎢⎣ c pT W ⎥⎦ γ 2

2

4 3

1 ⎡ dH dW ⎤ = + ⎥ 2 ⎢ ρ 1 − M ⎢⎣ c pT W ⎥⎦

dρ

dp γ M ⎡ dH dW ⎤ = + ⎥ 2 ⎢ p 1 − M ⎢⎣ c pT W ⎥⎦ 2

Theory of Aerospace Propulsion

4

constant area combustor

Tt,4

Modeling the heat release as a change in Tt

dH = cp4Tt4 - cp3Tt3

4

Tt,3

3

dH d ( c pTt ) ⎛ dTt dc p ⎞ Tt − = =⎜ + ⎟ ⎜ T ⎟T c pT c pT c p ⎠ ⎝ t Theory of Aerospace Propulsion

5

constant area combustor Tt,4

Combining equations yields

4 3

d (γ M ) dp =− 2 p 1+ γ M 2

ρ3 γ 4 M = ρ4 γ 3 M

2 4 2 3

Tt,3T t,3

integrate

(1 + γ M ) (1 + γ M ) 3

2 3

4

2 4

Eq. of state

p3 1 + γ 4 M = p4 1 + γ 3 M

2 4 2 3

2 1 + γ M T4 γ 4 M ( 3 3 ) W4 = T3 γ 3 M (1 + γ 4 M 42 ) W3

Theory of Aerospace Propulsion

2 4 2 3

6

constant area combustor The stagnation pressure change is γ4

⎡ γ 4 − 1 2 ⎤ γ 4 −1 M4 ⎥ 2 ⎢1 + pt ,4 1 + γ 3 M 3 ⎣ 2 ⎦ = γ3 pt ,3 1 + γ 4 M 42 ⎡ γ 3 − 1 2 ⎤ γ 3 −1 ⎢⎣1 + 2 M 3 ⎥⎦

4 3

ρV = constant V4/V3=ρ3/ρ4

The stagnation temperature change is

⎡ γ 4 −1 2 ⎤ Tt ,4 Tt ,4 T4 T3 γ 4 M 42W4 ⎡1 + γ 3 M 32 ⎤ ⎢1 + 2 M 4 ⎥ = = ⎥ ⎢ 2 2⎥ ⎢ − γ 1 Tt ,3 T4 T3 Tt ,3 γ 3 M 3 W3 ⎣1 + γ 4 M 4 ⎦ ⎢1 + 3 M 2 ⎥ 3 2 ⎣ ⎦ 2

Theory of Aerospace Propulsion

7

constant pressure combustor Same assumptions as before except that now area can change but pressure cannot

dp γ M ⎛ dA dH dW ⎞ = + + =0 ⎟ 2 ⎜ p 1 − M ⎜⎝ A c pT W ⎟⎠

p4=p3

2

4

p3

3

⎡ γ −1 2 ⎤ 1+ M ⎥ 2 2 ⎡ ⎢ ⎡ dM dA 1 + γ M ⎤ dH dW ⎤ d γ 2 = −2 ⎢ −⎢ + ⎥− ⎥ 2 2 2 ⎥⎢ M ⎢ 1− M ⎥ A ⎣ 1 − M ⎦ ⎢⎣ c pT W ⎥⎦ γ ⎣ ⎦

d (γ M ) 2

γM

2

dA =− A

integrating

A3 γ 4M = A4 γ 3M

Theory of Aerospace Propulsion

2 4 2 3

8

constant pressure combustor Here we show that V=constant in a constant pressure combustor

V4=V3 4

V3

3

dV 1 ⎛ dA dH dW ⎞ 1 dp =− − − =− =0 ⎟ 2 ⎜ 2 V 1 − M ⎜⎝ A c pT W ⎟⎠ γM p 2

dp 1 2 dV + γM =0 2 p 2 V Theory of Aerospace Propulsion

9

constant pressure combustor Final properties are known if initial properties are given

ρ3 A4 = ρ 4 A3

T3 A3 W3 = T4 A4 W4 p3 V3 =1= p4 V4 M3 γ 4 A4 = γ 3 A3 M4

γ4

⎡ γ 4 − 1 2 ⎤ γ 4 −1 + M 1 4 ⎥ pt ,4 ⎢⎣ 2 ⎦ = γ3 pt ,3 ⎡ γ 3 − 1 2 ⎤ γ 3 −1 ⎢⎣1 + 2 M 3 ⎥⎦ 2 1+

γ 4 −1

M 42

W4 γ 3 M 3 2 = Tt ,3 W3 γ 4 M 42 1 + γ 3 − 1 M 2 3 2

Tt ,4

Theory of Aerospace Propulsion

10

combustor configuration Fuel injector

Secondary air for cooling

From compressor

To turbine

diffuser

primary zone

secondary zone

Theory of Aerospace Propulsion

11

annular‐type combustor

Theory of Aerospace Propulsion

12

double‐annular type combustor

Theory of Aerospace Propulsion

13

Fuels for airbreathing engines Paraffin hydrocarbon fuel burning in air

79 ⎤ 79 ⎡ Cn H 2 n + 2 + α ⎢O2 + N 2 ⎥ → ACO2 + BH 2O + α N2 21 ⎦ 21 ⎣

conservation of atoms:

C→n= A H → 2n + 2 = 2 B O → 2α = 2 A + B

Theory of Aerospace Propulsion

14

Fuels for airbreathing engines 3n + 1 ⎡ 79 ⎤ 3n + 1 ⎛ 79 ⎞ Cn H 2 n + 2 + O2 + N 2 ⎥ → nCO2 + ( n + 1) H 2O + ⎜ ⎟ N2 ⎢ 2 ⎣ 21 ⎦ 2 ⎝ 21 ⎠

3n + 1 α= 2

The molar fuel to air ratio

⎛ wf ⎞ ⎜ ⎟ ⎝ wa ⎠ stoich

∑W C (12 ) n + (1)( 2n + 2 ) 14n + 2 = = = ∑W C 3n + 1 ⎡⎢( 32 ) + 79 ( 28)⎤⎥ 68.67 ( 3n + 1) i

i

i

i

fuel

air

2

⎣

21

Theory of Aerospace Propulsion

⎦

15

Fuels for airbreathing engines Octane: C8H18 ⎛ 79 ⎞ ⎛ 79 ⎞ C8 H18 + 12.5O2 + 12.5 ⎜ ⎟ N 2 → 8CO2 + 9 H 2O + 12.5 ⎜ ⎟ N 2 ⎝ 21 ⎠ ⎝ 21 ⎠

Fuel to air ratio by weight

wf wa

Theory of Aerospace Propulsion

= 0.066

16

Fuels for airbreathing engines For hydrocarbon fuels of this type : •Combustion possible in the range 0.04 < f/a < 0.25 •Maximum reaction rate occurs for f/a =0.073 •Complete combustion occurs for f/a =0.066 •Combustion is slow and erratic for f/a <0.055 •Equivalence ratio >1: fuel‐rich, CO, H2O ⎛ m ⎞ f

⎜ ⎟ ma ⎠ f a) ( ⎝ φ= = ⎛ m f ⎞ ( f a )stoich ⎜ ⎟ ⎝ ma ⎠ stoich

=1: stoichiometric <1: fuel‐lean; CO2, O2

Theory of Aerospace Propulsion

17

Fuels for airbreathing engines Fuel

(f/a)stoich Oxides

Qf (MJ/kg)1 Qf,u (MJ/l)1 Tc (K)2

Jet A

0.0678

CO2, H2O

43.4

34.7

2,295

Hydrogen 0.0292

H2O

120.24

8.42

2,431

Methane 0.0582

CO2, H2O

50

21.2

2.246

Ethanol

0.111

CO2, H2O

27.23

21.62

2,214

Ammonia 0.165

NO, H2O

18.55

11.41

2,084

(1) lower heating value is implied (2) Tc= adiabatic flame temperature Theory of Aerospace Propulsion

18

combustor heat release Total rate of heat addition

⎡⎛ f ⎞ ⎤ mdH = m4 ht ,4 − m3 ht ,3 = m3 ⎢⎜1 + ⎟ c p ,4Tt ,4 − c p ,3Tt ,3 ⎥ a⎠ ⎝ ⎣ ⎦ m f

m3

m4

m 3c p ,3Tt ,3 + m f c p , f Tt , f + m f ηbQ f = ( m 3 + m f ) c p ,4Tt ,4 3

3’

4

⎡ f c p , f Tt , f ⎤ f ⎞ ⎛ m 3c p ,3Tt ,3 ⎢1 + ⎥ + m f ηbQ f = m 3 ⎜1 + ⎟ c p ,4Tt ,4 a⎠ ⎝ ⎢⎣ a c p ,3Tt ,3 ⎥⎦ Theory of Aerospace Propulsion

19

heat addition in the combustor Tt,0=(1+M2/5)T0 = Tt,3~Tt,0(pt,3/pt,2).286 =

17C (takeoff) 115C (M=2) 405C (takeoff) 235C (M=2)

Cp =

Fuel to air ratio f/a~0.02

•1.94kJ/kgK (JetA liquid at 15C)

Fuel temperature Tt,f<rTt,0

• 1.51kJ/kgK (Jet A vapor at 25C)

r=recovery factor~0.8

• 1.00 kJ/kgK (Air at 25C) • 1.14kJ/kg (Air at 750C)

m 3 (c p ,4Tt ,4 − c p ,3Tt ,3 ) ≈ m f ηb Q f ≈ m 3c p (Tt ,4 − Tt ,3 ) Theory of Aerospace Propulsion

20

combustor efficiency m + m ) h − m h − m h ( η = 3

b

f

t ,4

3 t ,3

f

t, f

m f Q f

For f/a <<1 the combustor efficiency becomes

m f m 4 ht ,4 − ht ,3 − ht , f ht ,4 − ht ,3 c p (Tt ,4 − Tt ,3 ) m 3 m 3 ≈ ≈ ηb = m f m f m f Qf Qf Qf m 3 m 3 m 3 Theory of Aerospace Propulsion

21

combustor efficiency 100

95

ηb (%) 90

85 0 12

4

8

pt,3Tt,3/V3 x10‐6 (N‐s‐K/m3)

combustor parameter pt,3Tt,3/ V3 is a function of T,time, and mixing Theory of Aerospace Propulsion

22

combustor efficiency 100

Sea level

90 ηb (%) 80

40 kft 20 kft

70

70

80

90

100 N(% rpm) γ

pt ,3 =

⎛ γ − 1 2 ⎞ γ −1 p0 = p0 ⎜1 + M0 ⎟ 2 p0 pt ,2 pt ,0 ⎝ ⎠

pt ,3 pt ,2 pt ,0 pt ,2 pt ,0

pt ,3 pt ,2

Theory of Aerospace Propulsion

23

Apportioning the airflow

10%

82%

w3

w4

8% 10% Burning total 28% (primary air)

Dilution total 72% (secondary air)

Theory of Aerospace Propulsion

24

effect of cooling on exit temperature 80

Typical operating range

Cooling air (%)

p3 =6 atm T3 =590K

40

Stoichiometric 2450K

0 1300

1900

Theory of Aerospace Propulsion

2500 T4 (K)

25

curve fit for specific heat of air Curve fits for T<2000K are: Cp(kJ/kg‐K)=0.94+0.23(T/1000) for f/a=0.02 Cp(KJ/kg‐K)=0.94+0.2(T/1000) for f/a=0 (shown below) 2.50 (c) f/a=0

2.00 cp (kJ/kg-K)

p=20atm 3atm

1.50 1atm 1.00 0.50 0.00 500

1000

1500

2000

2500

3000

Temperature T (K)

Theory of Aerospace Propulsion

26

example: constant area combustor fuel Qf =44MJ/kg Tt,3=488K M3=0.35 3

4

What is maximum f/a?

m 3 (c p ,4Tt ,4 − c p ,3Tt ,3 ) ≈ m f ηb Q f ≈ m 3c p (Tt ,4 − Tt ,3 )

⎡ γ 4 −1 2 ⎤ Tt ,4 Tt ,4 T4 T3 γ 4 M 42W4 ⎡1 + γ 3 M 32 ⎤ ⎢1 + 2 M 4 ⎥ = = ⎥ ⎢ 2 2⎥ ⎢ γ 1 − Tt ,3 T4 T3 Tt ,3 γ 3 M 3 W3 ⎣1 + γ 4 M 4 ⎦ ⎢1 + 3 M 2 ⎥ 3 ⎣ ⎦ 2 2

Theory of Aerospace Propulsion

27

example: constant area combustor Assume that γ3=γ4=γ=4/3 and note that

2 3 4 M 6 M + + Tt ,4 − Tt ,3 M ( ) ( 4 ) = −1 2 2 Tt ,3 M ( 3 + 4M 2 ) ( 6 + M 3 ) 4 2 4 2 3

2 2 3

The maximum heat addition is that which makes M4=1. Setting M4=1 and M3=0.35 yields ΔTt,max=1.327(488)=648K and Tt4,max=1136K. Then

ΔH max = m 3c p (Tt ,4max − Tt ,3 ) = m f ηbQ f ⎛ kJ ⎞ 1.15 648K ) ( ⎜ ⎟ c p (Tt ,4max − Tt ,3 ) ⎝ ⎛ m f ⎞ kg ⋅ K ⎠ = = 0.0178 ⎜ ⎟ = ηb Q f ⎛ MJ ⎞ ⎝ m 3 ⎠ max 0.95 ⎜ 44 ⎟ kg ⎝ ⎠ Theory of Aerospace Propulsion

28

THEORY OF AEROSPACE PROPULSION 4A. Equilibrium Chemistry

P. M. SFORZA University of Florida

general chemical reactions Reactants

Products

n

n

∑ν ′ M → ∑ν ′′M j =1

j

j

j =1

j

j

Here νj’ and νj’’ are the stoichiometric coefficients of the reactants and products for the species Mj, j=1, 2, 3,.., n, where n is the total number of species involved.

Theory of Aerospace Propulsion

2

criteria for chemical equilibrium dS=0 for equilibrium at constant p and H dG=0 for equilibrium at constant p and T

ΔG = ∑ (ν ′′j −ν ′j )G j n

j =1

Gibbs function

G = H − TS n ⎡ ν ′′j −ν ′j ⎤ 0 ΔG = ∑ (ν ′′j − ν ′j ) G j + RT ⎢ ∑ ln ( p j ) ⎥ j =1 ⎣ j =1 ⎦ n

Theory of Aerospace Propulsion

3

minimum Gibbs function ΔG − ΔG = RT ∑ ln ( p j ) n

ν ′′j −ν ′j

o

j =1

Then for equilibrium ΔG=0 and ν ′′j −ν ′j

ΔG = − RT ln ∏ ( p j ,e ) n

o

j =1

= − RT ln K p

The equilibrium constant Theory of Aerospace Propulsion

4

manipulating the Gibbs function Using the definition of the Gibbs function

ΔH − T ΔS = − RT ln K p o

o

And using the combined 1st and 2nd laws

dH = TdS + Vdp

d ⎛G ⎞ H ⎜ ⎟=− 2 dT ⎝ T ⎠ T o

o

Theory of Aerospace Propulsion

5

Van’t Hoff equation d ⎛ ΔG o ⎞ d ln K p ) ( ⎜ ⎟ = −R dT ⎝ T ⎠ dT The Van’t Hoff equation

ΔH d ln K p ) = − ( R ⎛1⎞ d⎜ ⎟ ⎝T ⎠

o

permits the experimental determination of the heat release in the reaction carried out at 1 atmosphere from the equilibrium constant Kp calculated from the measured equilibrium composition Theory of Aerospace Propulsion

6

general chemical reactions nk

n

∑a A + ∑a A ↔ ∑b B i =1

i

i,g

k =1

mq

m

k

k ,c

j =1

j

j,g

+ ∑ bq Bq ,c q =1

Minimizing G leads to the following equilibrium constant for the gaseous species

∏( p m

Kp =

j =1 n

B j ,e

)

∏( p ) i =1

bj

ai

Ai ,e

Theory of Aerospace Propulsion

7

equilibrium among sub‐reactions Requirement for equilibrium composition:

all the sub‐reactions are simultaneously in equilibrium Thus we will deal with the equilibrium constants for these reactions:

K p = ∏ ( p j ,e ) n

ν ′′j −ν j

′

j =1

Theory of Aerospace Propulsion

8

homogeneous case: direct solution Hydrogen‐Fluorine reaction

H 2 + F2 → a1′′HF + a2′′H 2 + a3′′H + a4′′F2 + a5′′F H : 2 = a1′′ + 2a2′′ + a3′′ F : 2 = a1′′ + 2a4′′ + a5′′

Atom conservation

5

Total number of moles Dalton’s law

a = ∑ ai′′ i =1 n

p = ∑ pi i =1

Theory of Aerospace Propulsion

pi ai′′ Xi = = p a′′ Mole fraction of products 9

sub‐reactions in the H‐F system overall reaction: H + F → HF

K p , HF p1 a1′′ ⎛ a′′ ⎞ K p ,1 = = = ⎜ ⎟ p3 p5 a3′′a5′′ ⎝ p ⎠ K p , F K p , H sub‐reaction 1: 2 H → H 2

p2 a2′′ ⎛ a′′ ⎞ 1 K p ,2 = 2 = 2 ⎜ ⎟ = 2 p3 a3′′ ⎝ p ⎠ K p , H sub‐reaction 2:

2 F → F2 p4 a4′′ ⎛ a′′ ⎞ 1 K p ,4 = 2 = 2 ⎜ ⎟ = 2 p5 a5′′ ⎝ p ⎠ K p , F Theory of Aerospace Propulsion

10

reducing the set of unknowns There are six equations and six unknowns. This set may be solved for the equilibrium composition at a given pressure and temperature. We may simplify the set if we can estimate that some unknowns may be neglected.

3000K

4500K

Log10Kp,1

3.7832

0.3249

Log10Kp,2

1.6069

‐1.0731

Log10Kp,4

‐3.8207

‐4.8158

Note: Kp,4<<Kp,1, Kp,2 Theory of Aerospace Propulsion

11

reduced equations: H‐F system From the Kp values we see a4’’<<a5’’2 so F2 is negligible Eliminating a3”, a5”, and a” from this set leaves the following

a1′′( 4 − a1′′ − a2′′ ) pK p , HF = ( 2 − a1′′)( 2 − a1′′ − 2a2′′ ) a1′′( 2 − a1′′ − 2a2′′ ) = K p,H2 a2′′ ( 2 − a1′′)

K p , HF

For given pressure (in atm) and temperature one may find the concentrations of products Theory of Aerospace Propulsion

12

homogeneous case : trial & error The hydrogen‐oxygen system

H 2 + O2 → a1′′O2 + a2′′H 2 + a3′′O + a4′′H + a5′′OH + a6′′H 2O

Atom conservation

H : 2 = 2a2′′ + a4′′ + a5′′ + 2a6′′ O : 2 = 2a1′′ + a3′′ + a5′′ + a6′′

Theory of Aerospace Propulsion

13

the equilibrium sub‐reactions p3 1 O2 → O : K p ,O = K p ,1 = 2 p1 p4 1 H 2 → H : K p , H = K p ,2 = 2 p2 1 1 H 2 + O2 → OH : K p ,OH = K p ,3 = 2 2

p5 p1 p2

p6 1 H 2 + O2 → H 2O : K p , H 2O = K p ,4 = 2 p2 p1

Theory of Aerospace Propulsion

14

manipulating the equations pV = a′′RT

Equation of state

6

p = ∑ pi

Dalton’s law

ai′′ pi = a′′ p

Mole fraction identity

i =1

Recast atom conservation

p 2 = 2 p2 + p4 + p5 + 2 p6 a′′ p 2 = 2 p1 + p3 + p5 + p 6 a′′

Theory of Aerospace Propulsion

15

the equilibrium equations Put the partial pressures of the products in terms of the reactants

p3 = K p ,O p1 p4 = K p , H

p2

p5 = K p ,OH

p1 p2

p6 = K p , H 2O p2 p1 Substituting into an atom conservation equation

⎡ p ⎤ ⎢⎣ 2 a′′ − 2 p2 − K p , H p2 ⎥⎦ p1 = 2 K p , H 2O p2 + K p ,OH p2 Theory of Aerospace Propulsion

16

iteration equation p ⎡ p 2 = ⎢ 2 − 2 p2 − K p , H a′′ ⎣ a′′

⎤ K p , H 2O + K p ,OH p2 + K p ,O p2 ⎥ ⎦ 2 K p , H 2O p2 + K p ,OH p2

⎡ p ⎤ ⎢ 2 a′′ − 2 p2 − K p , H p2 ⎥ +2 ⎢ ⎥ ⎢ 2 K p , H 2O p2 + K p ,OH p2 ⎥ ⎣ ⎦

2

a” is unknown and must be guessed to calculate p2 and then all the other pressures are calculated until finally a new value for a” is determined and compared to the initial guess.

Theory of Aerospace Propulsion

17

assessing neglected species Which possible product species should be included? ‐Use an “educated guess” based on previous experience ‐Include all possible species

Once a choice is made it is tacitly assumed that the neglected species are unimportant energetically (though they may actually be important kinetically)

If there is some doubt as to the importance of a neglected product one can rather readily check Theory of Aerospace Propulsion

18

example: dissociation of an air‐like gas Consider an air‐like mixture (atmospheric air would have 3.76 moles of N2 rather than 4) 4N2 + O2

a1”N2 + a2”O2 + a3”NO + a4” O

Assume p=0.01atm and T=4000K which corresponds to M0=12 at an altitude of 250,000ft Atom conservation: N: 8=2a1” + a3” O: 2= 2a2”+ a3” + a4” Total moles: a”=a1” + a2” + a3” + a4”

Theory of Aerospace Propulsion

19

example: air‐like gas (continued) p4 1 p a4′′ O2 → O : K p ,O = = 2 a′′ a2′′ p2 1 1 N 2 + O2 : K p , NO = 2 2

p3 a3′′ = p1 p2 a1′′a2′′

Solving the 5 equations for 5 unknowns yields a1” = 3.985

a2” = 2.77 x 10‐3

a3” = 3.13 x 10‐2

a4” = 1.963

Here we have assumed that little atomic nitrogen is formed, that is, aN” << ai” of all other species. Is that correct?

Theory of Aerospace Propulsion

20

example: air‐like gas (continued) Consider atomic nitrogen

pN 1 p a′′N = N2 → N : K p,N = 2 a′′ a1′′ p1 Under the given conditions Kp,N=5.5x10‐2 Thus aN” ~(5)1/2 Therefore neglecting N leads to an error in composition. The calculation must be repeated, this time including N. However, notice that O2 may now be safely ignored. Thus the complexity of the problem hasn’t increased

Theory of Aerospace Propulsion

21

adiabatic flame temperature (AFT) Find the temperature resulting from the combustion of fuel and oxidizer under given pressure conditions in the combustor.

Consider the following assumptions: •Velocities in the combustor are small enough so kinetic energy contribution may be neglected •The pressure in the combustor is constant •The combustion products behave like perfect gases •The combustion process is adiabatic; all the heat released goes into the products Theory of Aerospace Propulsion

22

AFT and the first law The first law of thermodynamics for isobaric combustion is dQ=dH‐Vdp=dH, For an adiabatic combustion chamber dQ=dH=0. Then, for the general chemical reaction N

N

∑ a′A →∑ a′′A i =1

i

i

i =1

i

i

the change in enthalpy is

ΔH = ∑ ai′′H i (Tc ) − ∑ ai′H i (T j ) = 0 N

i =1

Theory of Aerospace Propulsion

23

formulation of the AFT The enthalpy change may be expanded to N

∑ ai′′⎡⎣ ΔH i =1

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

0 f ,i

i =1

ΔHf,i0 is the molar heat of formation of species i at p=1 atm and T=Tr ΔHi(T) = Hi(T) – Hi(0) Tr = reference temperature, 298.16K Tc = adiabatic flame temperature in degrees K Tj = injection temperature of the reactants in degrees K Theory of Aerospace Propulsion

24

simplifications: gas phase reactions • If the injection temperature equals the reference temperature, Tj = Tr N

N

i =1

i =1

0 0 ′′ ′ ⎡ ⎤ a H H T H T a H Δ + Δ − Δ − Δ ( ) ( ) ∑ i ⎣ f ,i ∑ i f ,i = 0 c r ⎦

• If the reactants are injected at Tj=Tr and they are in their reference state N

0 ′′⎡ a Δ H ∑ i ⎣ f ,i + ΔH (Tc ) − ΔH (Tr )⎦⎤ = 0 i =1

• If the reactants are in a liquid state the latent heat must be taken into account (see the discussion in the text) Theory of Aerospace Propulsion

25

The AFT process diagram enthalpy of reactants=enthalpy of products

H 3

1

5

Tj

Tc

Hj 2 Tr

4

Theory of Aerospace Propulsion

26

steps in the AFT process 1. Propellants enter the combustion chamber at Tj with

H j = ∑ ai′ΔH i (T j ) N

i =1

2. Propellants are brought to the reference state N

H j − ∑ ai′ΔH i (Tr ) i =1

and this level of enthalpy is arbitrarily chosen as the abscissa in the process chart 3. Propellants are brought to the reference composition, that N is, 0

∑ a′ΔH i =1

i

f ,i

is added to the results of step 2 at T=Tr Theory of Aerospace Propulsion

27

steps in the AFT process (concl.)

4. Propellants are transformed into products of reaction at T=Tr, that is, the quantity 0 ′′⎡ Δ a H ∑ i ⎣ f ,i − ΔH i (Tr )⎤⎦

is subtracted from the results of step 3 5. Products are brought to the final temperature Tc such that ΔH=H(5) – H(1) = 0, that is, N

∑ a′′H (T ) i =1

i

i

c

is added to the result of step 4 Theory of Aerospace Propulsion

28

an AFT calculation method Define an available heat of reaction: N

Qavail = ∑ [ ai′ − ai′′] ΔH 0f ,i i =1

Define a heat of combustion at an assumed combustion temperature T’c: N

Qcomb (Tc′) = ∑ ai′′(Tc′) ⎡⎣ ΔH i (Tc′) − ΔH i (Tr ) ⎤⎦ − H j i =1

where

H j = ∑ ai′ ⎡⎣ ΔH i (T j ) − ΔH i (Tr ) ⎤⎦ N

i =1

Theory of Aerospace Propulsion

29

an AFT calculation (concluded) 1. Assume a combustion temperature T=Tc’ and calculate the equilibrium composition in the combustor at Tc’ and p 2. Use the equilibrium composition of the products as calculated in the previous step to determine Qavail(Tc’) 3. Calculate the quantity Qcomb(Tc’) 4. If Qavail(Tc’) > Qcomb(Tc’) then the adiabatic flame temperature Tad > Tc’ and vice versa .

When Qavail(Tc’) = Qcomb(Tc’) then Tc’ = Tc. A detailed example is carried out in the text Theory of Aerospace Propulsion

30

THEORY OF AEROSPACE PROPULSION Chapter 5. Nozzles for Airbreathing Engines

P. M. SFORZA University of Florida

nozzle characteristics turbine exit

throat

5

6

nozzle exit

dV d ρ dA =− − V A ρ

Theory of Aerospace Propulsion

7

2

pressure field along the nozzle dp ⎡ γ M 2 ⎤ dA ⎡ γ M 2 ⎤ dQ − dWk − dH =⎢ + ⎢− 2⎥ p ⎣1 − M ⎦ A ⎣ 1 − M 2 ⎥⎦ c pT ⎛ ⎞ 2 ⎡ ⎤⎜ ⎟ + − M 1 1 γ ⎡ ⎤ dW ( ) dF dz M γ 2 z ⎥ ⎜ 4C f + ⎢ −γ M + +⎢ ⎟ 2⎥ 2 1 − D M 1 γ 2 (1 − M ) ⎦⎥ ⎜ ⎢⎣ ⎦ W γ pM 2 A ⎟ ⎣ ⎝ ⎠ 2 2

Friction

Drag

4CfL/D

Fz

Cf=.005

cD qS cD S = = 1 qA A 2 γ pAM 2 CD=0.1

Theory of Aerospace Propulsion

3

nozzle flow with simple area change dp γ M 2 dA = p 1− M 2 A

ds dT γ − 1 dp = − =0 cp T γ p

2 1 γ − M ( ) dA

dT = 2 1− M T

2

dM = −2 2 M

A

1+

γ −1

M

2 2 1− M

Isentropic flow

2

dA A

Theory of Aerospace Propulsion

p~ργ~Tγ/(γ−1)

4

pressure distribution in the nozzle γ dp 2 dM =− p ⎛ γ −1 2 ⎞ 2 ⎜1 + M ⎟ 2 ⎝ ⎠ p

p=p* when M=1 −

γ

⎡ 2 ⎛ γ − 1 2 ⎞ ⎤ γ −1 =⎢ M ⎟⎥ ⎜1 + 2 p * ⎣γ +1⎝ ⎠⎦

integrating

−

γ

⎡ γ − 1 2 ⎤ γ −1 M7 ⎥ p7 ⎢1 + 2 =⎢ ⎥ 1 γ − p5 ⎢1 + M 52 ⎥ 2 ⎣ ⎦

p=pt when M=0 −

γ

p ⎛ γ − 1 2 ⎞ γ −1 = ⎜1 + M ⎟ pt ⎝ 2 ⎠

Theory of Aerospace Propulsion

5

nozzle temperature distribution Adiabatic flow along a streamline

1 2 c p T + V = c p Tt = const 2 −1 or ⎡ γ −1 2 ⎤ 1 + M 7 ⎥ ⎢ T 7 2 Tt γ −1 2 = ⎢ γ −1 ⎥ = 1+ M T5 ⎢1 + 2 T 2 M5 ⎥ ⎣ 2 ⎦

Theory of Aerospace Propulsion

6

mass flow through the nozzle Mass conservation equation

p m = ρ AV = AaM RT Substituting in stagnation values and speed of sound:

pt A γ ⎛ γ − 1 2 ⎞ m = M ⎜1 + M ⎟ 2 Tt R ⎝ ⎠ Theory of Aerospace Propulsion

γ +1 − 2( γ −1)

7

nozzle maximum mass flow ⎡ ⎤ 2 ⎢ ⎥ ⎡ ⎤ Γ − 1 dm M Γ Γ 2 = Kf ⎢1 + ( γ − 1) M = Kf ⎢ ⎥ =0 ⎥ − 1 γ dM f 2 ⎣ ⎦ ⎢1 + M ⎥ ⎣ ⎦ 2

d m d ⎡ m m M Γ ⎤ = + ( γ − 1) ⎢ ⎥ 2 dM dM ⎣ M M f ⎦ 2

2

⎡ d 2 m ⎤ 2 = −m <0 ⎢ dM 2 ⎥ γ +1 ⎣ ⎦ M =1 Theory of Aerospace Propulsion

M=1 critical Mach number

mass flow maximized 8

area distribution in the nozzle dM 2 = −2 2 M

1+

γ −1

M2

2 2 1− M

dA A

1 ⎛ γ −1 2 ⎞ A = C ⎜1 + M ⎟ M⎝ 2 ⎠

γ +1 2( γ −1)

Constant of integration Theory of Aerospace Propulsion

9

isentropic flow with simple A change p

Note that the pressure is double‐valued at a given A

Subsonic branch

M=1

p*

A5

A6

A7

Supersonic branch

A*

A7

A5

A

Theory of Aerospace Propulsion

10

nozzle critical area 2 ⎡ ⎤ γ 1 M + ( ) dA A =0 when M=1 = − ⎢1 − ⎥ dM M ⎣ 2 + ( γ − 1) M 2 ⎦

⎡ ⎤ 2 2 ⎢ ( γ − 1) M (1 − M ) ⎥ d2A A = 1+ ⎢ ⎥ 2 γ 1 − γ 1 − dM 2 2⎛ 2⎞ ⎢ ⎥ M + 1 M ⎜1 + M ⎟ 2 ⎦ 2 ⎝ ⎠⎣

⎡ d2A ⎤ 2A = >0 ⎢ dM 2 ⎥ ⎣ ⎦ M =1 γ + 1

Critical area A* is a minimum

Theory of Aerospace Propulsion

11

maximum mass flow in nozzle pt A * γ ⎛ γ + 1 ⎞ m max = m * = ⎜ ⎟ Tt R ⎝ 2 ⎠

−

γ +1 2( γ −1)

pt A * m * ≈ ( 0.192γ + 0.417 ) RTt approximation

Theory of Aerospace Propulsion

12

nozzle area ratio ⎡ γ −1 2 ⎤ 1+ M ⎥ 1 ⎢ A 2 = ⎢ ⎥ γ + 1 A* M ⎢ ⎥ ⎣ ⎦ 2

γ +1 2( γ −1)

Subsonic branch

Supersonic branch

1

M

A/A* 1

A 1 M << 1: → A* M

0

⎛ γ −1 ⎞ A →⎜ M >> 1: ⎟ A* ⎝ γ +1⎠

γ +1 2( γ −1)

M

2 γ −1

Theory of Aerospace Propulsion

13

pressure in a convergent‐divergent nozzle 5

1 p/pt

Note: γ =1.4

7

Subsonic flow

0.528 no solutions

range of pexit

shock in nozzle

continuous supersonic flow 0

Z sta.5

sta.6

sta.7 matched nozzle

Theory of Aerospace Propulsion

14

conditions for maximum thrust The gross thrust = Fg = m 7V7 + A7 ( p7 − p0 ) Taking the first derivative and seeking the stationary point yields 1

∂Fg

∂V7 ∂p7 ∂A7 = m 7 + A7 + ( p7 − p0 ) =0 ∂p7 ∂p7 ∂p7 ∂p7 mdV = − Adp

from momentum conservation

Finally we find

∂Fg

So either p7=p0 or A ⎛ ∂A ⎞ = ( p7 − p0 ) ⎜ ⎟ = 0 doesn’t vary with p ∂p7 ⎝ ∂p ⎠7 (which is when M=1) Theory of Aerospace Propulsion

15

maximum thrust (continued) We consider the 2nd derivative to seek a maximum

⎛ ∂ Fg ⎞ ⎛ ∂ 2 A ⎞ ⎛ ∂A ⎞ ⎜⎜ 2 ⎟⎟ = ( p7 − p0 ) ⎜ 2 ⎟ + ⎜ ⎟ ⎝ ∂p ⎠7 ⎝ ∂p ⎠7 ⎝ ∂p ⎠7 2

For a matched nozzle, p7=p0

⎛ ∂ 2 Fg ⎞ ⎛ ∂A ⎞ =⎜ ⎟ ⎜⎜ 2 ⎟⎟ ⎝ ∂p ⎠ p7 = p0 ⎝ ∂p ⎠7

>0, for M7>1Fg=maximum =0, for M=1 Fg=saddle pt.

Note that for subsonic exit flow p7=p0 so F = m 7V7 and therefore there is no stationary point Theory of Aerospace Propulsion

16

Mach angle and the zone of silence Sound source moving supersonically

(M2‐1)1/2

1

μ

sound waves

M>1

μ

Sound source moving sonically

sound waves M=1

Mach angle μ=arcsin(1/M) Theory of Aerospace Propulsion

17

nozzle types & related exit flow Mach lines μ=90o

μ<90o

M7>1

M7=1

Exit flow can support pressure mismatch because external pressure information can’t get inside nozzle

Theory of Aerospace Propulsion

No real Mach lines exist

M7<1 Exit must be pressure matched because pressure is felt everywhere 18

nozzle maximum thrust 1.8 1.6

locus of maxima

1.4 1.2 Fg/ptA* 1 0.8

M7=4

3

2.5

2 1.75 1.5

0.6 0.4 0.001

1.0 0.01

0.1

0.8 0.6 1

p0/pt

Gross thrust coefficient for various back pressures and exit M7 Theory of Aerospace Propulsion

19

sample problem: nozzle pressure mismatch effects Nozzle with pt=2.76MPa and Aexit/Amin=5.16 was designed for h=30.5km but launch takes place at sea level (p=101.3kPa). Find Flaunch/F30.5km USAF AEDC Rocket Test Cell

⎡ γ −1 2 ⎤ 1+ M ⎥ 1 ⎢ A 2 = ⎢ ⎥ γ + 1 A* M ⎢ ⎥ ⎣ ⎦ 2

γ +1 2( γ −1)

=5.16

Assuming γ=1.3 we may iterate to find Mexit=3. Then F=A7[p7(γM72+1)‐p0] and Flaunch=110.6A7 while F30.5km=125.2A7 Therefore Flaunch/F30.5km=0.884, a 12% reduction in thrust from the design value.

Theory of Aerospace Propulsion

20

nozzle design •

Specify Mach number or pressure distribution

• Determine nozzle area distribution as f(Z) • Determine nozzle radius or height as a f(Z) • Wall angle α at exit affects thrust: λ=Fg /mV7= sinα/α for 2‐D nozzle and (1+cosα)/2 for round nozzle • Check that throat radius of curvature ~ d* to 2d* • Length of nozzle may be three or more throat diameters Theory of Aerospace Propulsion

21

normal shock inside nozzle

p pdown Shock jump p* pup

As

Subsonic branch Shock M=1 Supersonic branch A*As A7

A5

A5

A6

A7

A

Theory of Aerospace Propulsion

22

shock waves and the exit pressure 7

5

1 p/pt

0.70 shock inside

0.528 continuous supersonic flow

shock at exit

0.28

0 Sta.5

Sta. 6

Sta.7

Z

Design exit Mach number is M7=3 Theory of Aerospace Propulsion

23

locating a shock in a nozzle m max = pt ,up A

* up

Turbine exit

γ ⎛ γ +1⎞

⎜ ⎟ RTt ⎝ 2 ⎠

−

γ +1 2( γ −1)

= pt ,down A

* down

γ ⎛ γ +1⎞

⎜ ⎟ RTt ⎝ 2 ⎠

−

γ +1 2( γ −1)

Nozzle exit

Throat

pt,7<pt,5 Tt,7=Tt,5 Mup 5

Mdown<1

6

Z’

pe given 7

z

* * pt ,up Aup = pt ,down Adown Theory of Aerospace Propulsion

24

locating a shock in the nozzle ⎛ pt ,7 ⎞ ⎜⎜ ⎟⎟ ⎝ p,5 ⎠

γ −1 γ

1 ⎛ pe ⎞ = ⎜ ⎟⎟ ⎜ 2 ⎝ pt ,5 ⎠

γ −1 γ

1 ± 2

2 ⎛ γ +1⎞ G= ⎜ ⎟ γ −1 ⎝ 2 ⎠

⎛ pe ⎞ ⎜⎜ ⎟⎟ ⎝ pt ,5 ⎠

2( γ −1)

γ

γ +1 2( γ −1)

⎡ ⎤ γ 2 A6 ⎛ pe ⎞ ⎥ ⎢ +⎢ ⎜⎜ ⎟⎟ ⎥ G A7 ⎝ pt ,5 ⎠ ⎢⎣ ⎥⎦ −2

• Knowing geometry A(x) and pe/pt,5 calculate pt,7/pt,5 •From shock tables find Mx upstream of shock •From Mx find Ax/A*=Ax/A6 •From geometry locate the station Ax/A6 Theory of Aerospace Propulsion

25

2

chart for locating a shock in nozzle 0.8 0.7 0.6

A6/A7=0.5

0.5

0.4 0.3

pt,7/pt,5 0.4 0.3 0.2 0.1 0

γ=1.33

0.2 0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

pe/pt,5

For any value of pe/pt,5 and A6/A7 one can find pt,7/pt,5 and then from normal shock tables or equations one can find Mx and A*/A. Only the physically valid portion of the solution curves are shown Theory of Aerospace Propulsion

26

shock wave angle vs. flow deflection (γ=1.4) 90

δ θ

θ

M=3 20o 0

M=oo 0

δ

34o

Theory of Aerospace Propulsion

46o 27

over‐expanded nozzle 7

pe/pt,5=0.056 slip surface θ=27.3

M7=3 p7/pt,5=.02722

10o

θr= 22 centerline M=2.5 pe/pt,e =(pe/p7)(p7/pt,5)(pt,5/pt,e) = (2.055)(.027022)(.963)=.058

Theory of Aerospace Propulsion

28

underexpanded nozzle pe/pt,5=0.01

slip surface

7

θ=49.76o

M7=3 p7/pt,5=.02722 ν=49.76o

11.69o

θr= 73.10

centerline

M=3.69 pe/pt,e =.01

Theory of Aerospace Propulsion

29

nozzle geometry requirements subsonic

supersonic

Ae/A6=1

Ae/A6=2.6

Theory of Aerospace Propulsion

30

nozzle regimes for fighter aircraft 4000

Full A/B γ=1.28

Tt,5 (oR)

γ=1.32

2000

γ=1.38

Part A/B M=1.4 Mil

1000

M=0.9 0

pt,5/p7 4

Theory of Aerospace Propulsion

6

8

31

J47 exhaust nozzle

Courtesy NASA

Theory of Aerospace Propulsion

32

F‐15 with 2‐D vectoring nozzles

Courtesy NASA Theory of Aerospace Propulsion

33

2‐D variable area nozzle Courtesy NASA

Theory of Aerospace Propulsion

34

2‐D variable area nozzle

Courtesy NASA

Theory of Aerospace Propulsion

35

variable‐geometry thrust vectoring nozzles

Boeing F-15B used by NASA for research on advanced controls technology.

Theory of Aerospace Propulsion

36

THEORY OF AEROSPACE PROPULSION Chapter 6. Inlets for Airbreathing Engines

P. M. SFORZA University of Florida

inlet Requirements •Duct air to the engine with low drag and total pressure loss •Diffuse flow to high pressure and low Mach number •Minimize flow distortion in flow to engine •Handle range of mass flow •Low weight, small size, and mechanically simple

Theory of Aerospace Propulsion

2

inlets set up the flow for the compressor inlet

compressor

combustor turbine

20p0

nozzle

2V0

V0 p0 Theory of Aerospace Propulsion

3

subsonic inlets 0

1

2

M0<1

Theory of Aerospace Propulsion

4

subsonic inlet γ −1 2 ⎞ ⎛ 1 + M0 ⎟ ⎛ A0 ⎞ M1 ⎜ 2 = ⎜ ⎟ ⎜ ⎟ − γ 1 M0 ⎜ ⎝ A1 ⎠ m ax M 12 ⎟⎟ ⎜1+ ⎝ 2 ⎠

γ +1 2 ( γ −1)

=

1 ⎡ 2 ⎛ γ − 1 2 ⎞⎤ M 0 ⎟⎥ 1 + ⎜ ⎢ 2 M 0 ⎣γ + 1⎝ ⎠⎦

(A0/A1)max 4

γ +1 2 (γ −1)

For γ =1.4

⎛ A0 ⎞ 1 ⎡ 5 ⎛ M ⎞⎤ ⎢ ⎜1 + ⎜ ⎟ = ⎟⎥ 5 ⎠⎦ ⎝ A1 ⎠ max M 0 ⎣ 6 ⎝ 2 0

2

0

0

.25

.5

.75

1

M0

Theory of Aerospace Propulsion

5

3

captured streamtube: subsonic inlet M0~0

M0~1

M0=0.5

Theory of Aerospace Propulsion

6

subsonic inlet with M0>1 streamlines don’t know about shock wave so are straight

streamlines must bend away from axis, so A0<A1 Theory of Aerospace Propulsion

7

normal shock inlet with different p2 A0/A1=1.02 M0=1.14 A0/A1=0.91

Shock wave moves further away from inlet lip as the back pressure (p2) increases.

Mass flow through the inlet decreases and captured streamtube diameter decreases

A0/A1=0.81 Theory of Aerospace Propulsion

8

normal Shock equations 1

ρ1u1 = ρ 2u2

2

p1 + ρ u = p2 + ρ u 2 1 1 2 1

2 2 2

2 2

u u h1 + = h2 + 2 2 ρ1 u2 =ε = ρ2 u1

Theory of Aerospace Propulsion

normal shock wave

9

jump conditions (γ=constant) p2 2γ = 1+ M 12 − 1) ( γ +1 p1

⎧⎪ (γ − 1) M + 2 ⎫⎪ M 2 = u2 / a2 = ⎨ ⎬ 2 ⎪⎩ 2γ M 1 − (γ − 1) ⎪⎭ 2 1

1/ 2

ρ 2 u1 q1 γ +1 = = = ρ1 u2 q2 (γ − 1) + 2 / M 12 2 2 ⎡ ⎤ ⎡ ⎤ 2 γ M − γ − 1 γ − 1 M ( ) ( ) T2 ⎛ a2 ⎞ 1 1 + 2⎦ ⎣ ⎦ ⎣ =⎜ ⎟ = 2 T1 ⎝ a1 ⎠ (γ + 1) M 12 2

pt ,2 pt ,1

=

ρ

t ,2

ρt ,1

=e

−

Δs R

γ

⎡ ( γ + 1) M 12 ⎤ γ −1 ⎡ ⎤ γ +1 =⎢ ⎥ ⎢ ⎥ 2 2 ⎣ ( γ − 1) M 1 + 2 ⎦ ⎣ 2γ M 1 − ( γ − 1) ⎦ Theory of Aerospace Propulsion

1 γ −1

10

max. mass flow in subsonic inlet, M0>1 pt,2/pt,0 A0=A1

1.0 0.8

M0>1

0.6

M<1

0.4 0.2 0 0

1

2

3

4

5 M0

1 Theory of Aerospace Propulsion

2 11

subsonic inlets at M0>1 M>M0 1

A0/A1

shock moving inside shock at lip

back pressure dropping

pitot tube 0

0 (pt,2/pt,0)/(pt,2/pt,0)normal shock 1 Theory of Aerospace Propulsion

12

internal compression inlet A0=A1

M0=Ms>1

Amin

⎛ A0* ⎞ ⎛ A0 ⎞ ⎛ A0* ⎞ A1 ⎛ A1 ⎞ =⎜ *⎟ ⎜ ⎟=⎜ *⎟ ⎜ ⎟ Amin ⎝ A0 ⎠ M = M ⎝ Amin ⎠ ⎝ A0 ⎠ M = M ⎝ Amin ⎠ s

Theory of Aerospace Propulsion

s

13

internal compression inlets for M0>1

shock expelled

M=1

shock swallowed

M>1 M<1

M=1 M<1

M<M0

Theory of Aerospace Propulsion

M<1

inlet started

shock adjusted

14

critical areas across a shock Recall that the maximum mass flow rate in a streamtube is

pt A * γ ⎛ γ + 1 ⎞ m max = m * = ⎜ ⎟ Tt R ⎝ 2 ⎠

γ +1 − 2( γ −1)

A pt ,0 = Amin pt , y * o

Ms

My

pt,0

pt,y

Tt,0

Tt,y

Shock wave pt,0>pt,y

pt , y ⎛ A ⎞ A1 = ⎜ ⎟ Amin pt ,0 ⎝ A * ⎠ M = M s Theory of Aerospace Propulsion

Tt,0=Tt,y γ=constant W=constant 15

internal compression inlet with Ms=2 (a) For M0<1 3 (A0/A1)max

* ⎛ A0 ⎞ A0 A0 ⎜ ⎟ = ⎝ A1 ⎠max A * A1

Ms=2

2 1

But Amin=A0*

0

⎛ A0 ⎞ A0 Amin ⎜ ⎟ = * ⎝ A1 ⎠ max A0 A1

0

1

M0

2

3

Theory of Aerospace Propulsion

16

internal compression inlets * ⎛ A0 ⎞ A0 A0 A0 pt , y Amin = * ⎜ ⎟ = * ⎝ A1 ⎠ max A0 A1 A0 pt ,0 A1

(b) For Ms>M0>1 3 (A0/A1)max

Ms=2

2

For M0>Ms

⎛ A0 ⎞ ⎜ ⎟ =1 ⎝ A1 ⎠ max

1

0 0

1

M0

2

3

Theory of Aerospace Propulsion

17

normal shock vs. internal compression pt,2/pt,0 1.0 0.8

Internal compression inlet

0.6

normal shock inlet

0.4 0.2 0 0

1

2

3

4

Theory of Aerospace Propulsion

5 M0 18

internal compression: off design pt,2/pt,0=0.72

pt,2/pt,0=0.83

M0=Ms=2 M0=2

M=1

M0=2 M=1.75 M<1

M0=3

M=0.65

M0=3 M=2.8

M0=3>Ms

pt,2/pt,0=0.33 Theory of Aerospace Propulsion

M<1

pt,2/pt,0=0.39 19

internal compression inlet at Ms>M0>1 and M0<1 A0<A1 M0=1.5

pt2/pt,0=0.93 (A0/A1)max=0.90

A0>A1 M0=0.5

pt2/pt,0=1 (A0/A1)max=1.1 Theory of Aerospace Propulsion

20

example: internal compression inlet M0

A2/A1=2

p0

Amin/A1=0.85 0

1

min

2

Determine: (a) The starting Mach number Ms for the inlet; (b) Assuming that the pressure in the free stream at the altitude of flight is p0=68.9kPa find the corresponding pressure p2 and the capture stream tube area ratio A0/A1 for the maximum stagnation pressure recovery when the flight Mach number is equal to the starting Mach number, M0=Ms Theory of Aerospace Propulsion

21

example: internal compression inlet (cont.) (a) Ms is that value of M0 for which the normal shock wave is located at station 1 and Mmin=1. Under those conditions the area ratio A1/Amin=1/0.85=1.176=A/A*. This corresponds to a Mach number of 0.61 just behind the shock at the lip. The upstream Mach number Ms=1.83 and the pressure recovery is 0.799

M0=Ms=1.83

pt2/pt0=0.799

M=0.61 Mmin=1

0

1

min

2

Theory of Aerospace Propulsion

22

example: internal compression inlet (cont.) (b) M0=Ms=1.83 and the corresponding value of A1/A* =1.472. Then Amin/A*=(Amin/A1)(A1/A*)=(0.85)(1.472)= 1.251 and Mmin=1.6 which yields a normal shock pressure recovery of 0.895

M0=Ms=1.83 Mmin=1.60 pt,0/p0=6.017

Amin/A*=1.1179

M2=0.227

M=0.668

A2/A*=2.63 pt2/pt0=0.895

0

1

min

2

Behind the shock the critical area A* is constant and we find A2/A*=((A2/A1)(A1/Amin)(Amin/A*) A2/A*=(2)(1.176)(1.1179)= 2.63 This corresponds to M2=2.27 Theory of Aerospace Propulsion

23

example: internal compression inlet (conc.)

Then the corresponding static to stagnation pressure ratio at station 2 is p2/pt,2=0.965. Then the pressure at station 2 may be found from p2=( p2/pt,2)( pt,2/pt,0)( pt,0/p0)p0 =(0.965)(0.895)(6.017)(68.9kPa)=358kPa

Theory of Aerospace Propulsion

24

additive drag The momentum equation for the control volume below is

− ( ρ0 A0V0 ) V0 + ( ρ1 AV 1 1 ) V1 = p0 A0 − p1 A1 + pavg ( A1 − A0 ) pavg

p1

p0

Additive drag in external compression inlets

A0 0

1

Theory of Aerospace Propulsion

25

additive drag (cont.)

Assuming pavg=constant yields

⎛ pavg ⎞ F1 = ∫ ( p − p0 ) dA = ⎜ − 1⎟ p0 ( A1 − A0 ) ⎝ p0 ⎠ A0 A1

Casting this in the form of an inlet drag coefficient we have:

A0 ⎞ ⎛ 1− ⎜ ⎟ p ⎛ ⎞ F1 A1 avg ⎟ Δc D = =⎜ − 1⎟ ⎜ 1 1 p 2 2 ⎟ ⎜ 0 ⎝ ⎠ γ p0 M 0 A1 γ M0 ⎟ ⎜ 2 ⎝2 ⎠ Theory of Aerospace Propulsion

26

additive drag (concl.)

Taking pavg equal to the pressure behind the shock, an optimistic assumption, we get

7 ( M − 1) 2γ M − ( γ − 1) p −1 ≈ −1 = p0 6 γ +1 2 0

2 0

~0.5

~ 0.15

~0.125

5⎛ 1 ⎞⎛ A0 ⎞ −1 ΔcD = ⎜1 − 2 ⎟⎜1 − ⎟ ≈ O (10 ) A1 ⎠ 3 ⎝ M0 ⎠⎝ Theory of Aerospace Propulsion

27

isentropic compression Mach waves μ=arcsin(1/M)

μ

waves coalesce

Isentropic compression pt=constant

M0 compression ramp weak shock waves

Near isentropic compression

finite turns

Δpt~(Δδ)3 Theory of Aerospace Propulsion

28

external compression inlets

two‐shock inlet

three‐shock inlet shock wave angle θ

1<M2<M1 δ

M1>1

flow deflection angle

oblique shock wave Theory of Aerospace Propulsion

29

oblique shocks (γ=constant) p2 2γ = 1+ M 12 sin 2 θ − 1) ( γ +1 p1

2 2 M γ + 1 sin θ ( ) ρ2 1 = = ρ1 u2,n ( γ − 1) M 12 sin 2 θ + 2

u1,n

4 2 2 2 2 2 γ + 1 M sin θ − 4 M sin θ − 1 γ M sin ( ) 1 ( 1 )( 1 θ + 1) 2

M 22 =

⎡⎣ 2γ M 12 sin 2 θ − ( γ − 1) ⎤⎦ ⎡⎣( γ − 1) M 12 sin 2 θ + 2 ⎤⎦

2 2 2 2 ⎡ ⎤ ⎡ 2 γ M sin θ − γ − 1 γ − 1 M sin θ + 2 ⎦⎤ ( ) ⎦ ⎣( ) 1 T2 ⎛ a2 ⎞ 1 ⎣ =⎜ ⎟ = 2 2 2 T1 ⎝ a1 ⎠ γ 1 M sin θ + ( ) 1 2

pt ,2 pt ,1

=

ρ

t ,2

ρt ,1

=e

−

Δs R

γ

⎡ ( γ + 1) M 12 sin 2 θ ⎤ γ −1 ⎡ ⎤ γ +1 =⎢ ⎥ ⎢ ⎥ 2 2 2 2 ⎣ ( γ − 1) M 1 sin θ + 2 ⎦ ⎣ 2γ M 1 sin θ − ( γ − 1) ⎦ Theory of Aerospace Propulsion

30

1 γ −1

shock wave angle vs. flow deflection M1

90

δ θ

strong shock region

θ M1=2 30o

M1 =3

20o

M1 =oo 0

34o δ 23o The curves are given by (for γ=1.4) 0

46o

1 γ +1 tan δ 6 tan δ 2 2 = sin θ − = sin θ − 2 M1 2 tan δ + cot θ 5 tan δ + cot θ Theory of Aerospace Propulsion

31

example: 3‐shock external compression inlet Find the pressure recovery for this 3 shock inlet at h=35kft where p0=3.44psia, T0=390R, M0=2.5, and for which: pt,0 = p0(1+M02/5 )3.5 =58.78psia and Tt,0=T0(1+M02/5 )=878R

c b 10o

a 10o

The regions a, b, and c may be treated independently since they are related only by the jump conditions across the 3 shock waves

Theory of Aerospace Propulsion

32

example : 3‐shock external compression inlet (cont.)

From oblique shock chart (or equations) find θa=32o

M sin θ a + 5 M sin (θ a − δ a ) = 7 M 02 sin 2 θ a − 1 2 a

2

2 0

2

cowl cowl lip angle c

M0=2.5

a

b ramps

We have M0,n=M0sinθa=1.32 so that pt,a/pt,0=0.975 and therefore pt,a=57.31psia. From charts or the equation above Ma=2.07. Then pa/pt,a= (1+Ma2/5)‐3.5 =0.115 which leads to pa=.115pt,a=6.59psia Theory of Aerospace Propulsion

33

example : 3‐shock external compression inlet (cont.)

We continue this process in the other shock cells: From oblique shock chart (or equations) θb=38o and Masinθb=1.27 Then Mb=1.71 from the equation on the previous slide and pb/pt,b=0.20 Since Masinθb=1.27, then ptb/pt,a=0.984 and pt,b/pt,0=(pt,b/pt,a)(pt,a/pt,0)=(.984)(.975)=0.959 Thus pt,b=56.4psia and pb=11.3psia

Theory of Aerospace Propulsion

34

example: 3‐shock external compression inlet (cont.)

Across the normal shock between b and c, using Mb=1.71 in the normal shock relations, yields ptc/ptb=0.852 and therefore pt,c/pt,0=0.852(0.959) =0.817. The Mach number Mc=0.638 and therefore pc=0.76pt,c= 36.05psia. A normal shock inlet would have given pt,c/pt,0=0.499. The pressure distribution is shown below: 40

c

3-shock inlet

p (psia) normal shock inlet

20

0

b 0

a distance along inlet wall Theory of Aerospace Propulsion

35

off‐design operation and additive drag

Subcritical flow

Supercritical flow

Schlieren flow visualization at M0=2 Theory of Aerospace Propulsion

36

operation at different M0

M0=1.49

M0=2.44 Theory of Aerospace Propulsion

37

operation at different M0

M0=2.73

M0=3.01

Theory of Aerospace Propulsion

38

aircraft Inlets F‐86 Pitot inlet

F‐8 Chin inlet (also F‐ 16)

F‐101 wing root inlet

F‐11F side inlet

Theory of Aerospace Propulsion

39

aircraft inlets

F‐4 side or “D” inlet Theory of Aerospace Propulsion

40

aircraft inlets

The F‐22 Raptor’s inlets have a splitter‐type diverter which is clearly evident in this photograph (USAF) Theory of Aerospace Propulsion

41

aircraft inlets

The inlets of the F‐35 Lightning II have integrated Bump‐type boundary layer diverters to steer boundary layer air away from the inlet entries Theory of Aerospace Propulsion

42

external compression inlets F‐14

Theory of Aerospace Propulsion

43

mixed compression inlets •Reduced cowl drag •Increased gust sensitivity •Increased unstart distortion

external compression

internal compression

Theory of Aerospace Propulsion

44

mixed‐compression Inlet

Theory of Aerospace Propulsion

45

mixed compression inlets SR‐71 Blackbird

Theory of Aerospace Propulsion

46

Lockheed SR‐71A inlets

Theory of Aerospace Propulsion

47

Mach 3.5 inlet model schematic

RL=cowl lip radius =24.86cm Theory of Aerospace Propulsion

48

THEORY OF AEROSPACE PROPULSION Chapter 7. Turbomachinery

P. M. SFORZA University of Florida

thermodynamics of a turbomachine 1 2 1 2 h2 + V2 + Qext + Wk = h3 + V3 2 2 1 2⎞ ⎛ 1 2⎞ ⎛ Wc = ⎜ h3 + V3 ⎟ − ⎜ h2 + V2 ⎟ = ht ,3 − ht ,2 2 ⎠ ⎝ 2 ⎠ ⎝ Q

Wt = ht ,4 − ht ,5

h3, V3

h2, V2

Wk

2 Theory of Aerospace Propulsion

3 2

efficiency of adiabatic compression ht p3 ht,3

p2

h’t,3 ht,2

adiabatic‐friction process

isentropic process

s2 s3

ηad ,c =

ht′,3 − ht ,2 ht ,3 − ht ,2

s

=

Tt′,3 − Tt ,2 Tt ,3 − Tt ,2

Theory of Aerospace Propulsion

3

work required for compression isentropic relation

⎛ pt ,3 ⎞ =⎜ ⎟⎟ ⎜ Tt ,2 ⎝ pt ,2 ⎠ Tt′,3

γ −1 γ

⎛ pt ,3 ⎞ ⎜⎜ ⎟⎟ pt ,2 ⎠ ⎝ ηad ,c = Tt ,3 Tt ,2

compressor work

⎡ c p ,2Tt ,2 ⎢⎛ pt ,3 ⎞ Wc = ⎜⎜ ⎟⎟ ⎢ ηad ,c ⎝ pt ,2 ⎠ ⎢⎣ Theory of Aerospace Propulsion

γ 2 −1 γ2

γ 2 −1 γ2

−1

−1

⎤ ⎥ − 1⎥ ⎥⎦ 4

adiabatic efficiency of expansion ht

ht ,4 − ht ,5 ηad ,e = ht ,4 − ht′,5

pt,4 pt,

ht,4

5

ht,5 h’t,5

isentropic process

adiabatic‐ friction process

s4 s5

turbine work output

s

1−

ηad ,e =

Tt ,5 Tt ,4

⎛ pt ,5 ⎞ 1− ⎜ ⎜ p ⎟⎟ ⎝ t ,4 ⎠

γ 4 −1 γ4

γ 4 −1 ⎡ ⎤ γ ⎛ pt ,5 ⎞ 4 ⎥ ⎢ Wt = ηad ,e c p ,4Tt ,4 ⎢1 − ⎜ ⎜ p ⎟⎟ ⎥ ⎢⎣ ⎝ t ,4 ⎠ ⎥⎦ Theory of Aerospace Propulsion

5

turbojet cycle general numbering system for the turbojet engine

fuel lines

inlet combustor compressor

nozzle

combustor turbine

4 h 3

p4=p3

5 6

p3 2 p2 p1

7

1

2

3

4

5

6

7

p5

p6 p0

thermodynamic cycle for the turbojet engine operating with a matched nozzle and constant pressure combustor

1 s Theory of Aerospace Propulsion

6

velocity components for a rotor c2r c2u r

Inertial velocity components c2a

ω absolute velocity c Relative wind seen by blade w c blade speed u

u

c=u+w Theory of Aerospace Propulsion

7

velocity in rotor and stator passages w2

stator

u

c2

c2

u

stator

w3

rotor

c3 u

Theory of Aerospace Propulsion

c3

8

forces and torque acting on the fluid Fa = m ( c3,a − c2,a )

Fu = m ( c3,u − c2,u ) Fr = m ( c3, r − c2, r )

C3,u r3 r2 C2,u

Tq = m ( r3c3,u − r2 c2,u ) Theory of Aerospace Propulsion

9

work depends on change in whirl P = Tqω = mω ( r3c3,u − r2 c2,u )

P = m ( u3c3,u − u2 c2,u ) P 1 Wc = = ( u3c3,u − u2 c2,u ) mg g 1 Wt = ( u4 c4,u − u5c5,u ) g Theory of Aerospace Propulsion

10

the velocity triangle

relative velocity to an observer on the rotor w

v

absolute velocity in laboratory coordinates c

β

α cu

u linear velocity of rotation of the rotor tangential component of the absolute velocity Theory of Aerospace Propulsion

11

velocity component relations G G G u+w=c

c =c +v 2

2 u

2

w = ( u − cu ) + v 2

2

2

c − w = −u + 2ucu 2

2

2

1 2 2 2 ucu = ( c + u − w ) 2 Theory of Aerospace Propulsion

12

general ideal total head equation Note: the pressure “head” is the work done per unit weight flow such that the units of W are simply those of length, i.e., ft or m

Centrifugal effect: increasing pressure by increasing the radius of exit

P 1 2 2 2 2 2 2 ⎡ ⎤ = − + − + − Wc = c c u u w w ( ) ( ) ( ) 3 2 3 2 2 3 ⎦ mg 2 g ⎣ External effect: increasing pressure by diffusing c3 outside the rotor, i.e., in an external device

Internal effect: increasing pressure by diffusing w2 inside the rotating passage

Theory of Aerospace Propulsion

13

THEORY OF AEROSPACE PROPULSION 7A. Centrifugal Compressors

P. M. SFORZA University of Florida

centrifugal compressor Array of identical blades spaced at an angular pitch of ωp ω

r3 r2 ωp c2

w2 u2

w3 c3 u3

Theory of Aerospace Propulsion

2

energy equation in the compressor 1 2⎞ 1 2⎞ ⎛ ⎛ = m ⎜ h3 + c3 ⎟ m ⎜ h2 + c2 ⎟ + mW 2 ⎠ 2 ⎠ ⎝ ⎝ p3 1 2 p2 1 2 e2 + + c2 + W = e3 + + c3 ρ2 2 ρ3 2

h = e + p/ρ

For liquids, as well as for gases undergoing a small pressure rise, e ~ constant (small T change) and therefore ρ ~ constant and the equation becomes

pt 3 − pt 2 = ρW Then

1 1 2 2 p3 − p2 = ρ ( u3 − u2 ) + ρ ( w22 − w32 ) 2 2 Theory of Aerospace Propulsion

For liquid pumps or small gas pressure rise, e.g., fans 3

power transfer in the compressor P = ωTq = m ω ( r3c3u − r2c2u ) = m ( u3c3u − u2 c2u ) Velocity component relation:

c + u − w = 2ucu 2

2

2

In these equations no assumption about compressibility has been made so far For constant density flow, the stagnation pressure change across the cascade is

1 2 2 pt 3 − pt 2 = ( p3 − p2 ) + ρ ( c3 − c2 ) 2 Theory of Aerospace Propulsion

4

compressibility effect In a gas compressor the ideal process is an isentropic compression where p ~ργ and thus 1/ γ

ρ3 ⎛ p3 ⎞ =⎜ ⎟ ρ 2 ⎝ p2 ⎠

We may now calculate the density and temperature changes for an ideal compression

Theory of Aerospace Propulsion

5

single‐stage isentropic compression An axial flow compressor is limited to small pressure rise in a single stage because of boundary layer separation on the blades 1.5

γ=1.4

1.4

density ratio ρ3/ρ2

axial flow compressor range

1.3 1.2

temperature ratio Tt,3/Tt,2 1.1 1 1

1.1

1.2

pt,3/pt,2

1.3

Theory of Aerospace Propulsion

1.4

1.5

6

single‐stage isentropic compression The centrifugal compressor is not limited to a small pressure rise because of the powerful centrifugal effect in the blade passages 3.5

γ=1.4

ρ3/ρ2

centrifugal compressor range

3 2.5

axial compressor range

2

Tt,3/Tt,2

1.5 1 1

1.5

2

2.5pt,3/pt,2 3

3.5

Theory of Aerospace Propulsion

4

4.5

5

7

Mach number effect on pressure 30

Cp-Cp,inc (%)

25 20 15 10 5 0 0

0.2

0.4

0.6

0.8

1

1.2

Mach number

pt − p∞ pt − p∞ 2 ⎡⎛ pt ⎞ ⎤ C p ,t = = = − 1⎥ ⎟ 2 ⎢⎜ 1 q∞ γ p∞ M ∞2 γ M ∞ ⎣⎝ p∞ ⎠ ⎦ 2 Theory of Aerospace Propulsion

8

centrifugal compressors u3 c2

b2

c3

w2

w3

u2 b3 ω

P = ωTq = m ω ( r3c3u − r2c2u ) = m ( u3c3u − u2c2u ) Theory of Aerospace Propulsion

9

Velocity diagrams

w2

c2

c2r c3

β2

α2

w3 c3r

u2

β3

α3 u3 Theory of Aerospace Propulsion

10

typical compressor design Axial entry (no pre‐whirl):c2u=0 ω w2eye w2hub c2a =const u2eye

Radial exit: c3u=u3 u3 c3 w3 Theory of Aerospace Propulsion

11

entry blade twist for no pre‐whirl To keep the entry speed ca (yellow arrows) constant the blade angles must change because u (red arrows) increases as r increases

Theory of Aerospace Propulsion

12

mass flow considerations C3,r

The mass flow is constant

h3

m = ρ 2 c2 a (π r22 ) = ρ3c3r (2π r3 h3 ) r2 Then the

r3

C2,a compressible

c3r ρ 2 r r = = c2 a ρ3 2r3 h3 2r3 h3 2 2

axial entrance

2 2

radial exit

incompressible

Theory of Aerospace Propulsion

13

radial exit with no pre‐whirl P = ωTq = m ω ( r3c3u − r2c2u ) = m ( u3c3u − u2c2u ) No pre‐whirl

c2 = c2 r

Total head

w3 = w3r

c3u = u3

c2u = 0 Power required

Radial exit

2

P = mu3

P u32 Wc = =

mg g Theory of Aerospace Propulsion

14

general exit with no pre‐whirl c3u 3u u3 = mu

P = mc u3 2 3

slip angle

c3 u3

slip coefficient τ

β3 w3 blade tip edge of rotor wheel r3 Theory of Aerospace Propulsion

15

head rise for zero pre‐whirl tan β 3 τ= tan α 3 + tan β 3

2 2 u c u P = 3 3u = 3 τ Wc =

mg g u3 g

c3

c3

c3 β3 inc

cr3

u3 Forward curved blade Theory of Aerospace Propulsion

16

work required for compression isentropic relation

⎛ pt ,3 ⎞ =⎜ ⎟⎟ ⎜ Tt ,2 ⎝ pt ,2 ⎠ Tt′,3

γ −1 γ

⎛ pt ,3 ⎞ ⎜⎜ ⎟⎟ pt ,2 ⎠ ⎝ ηad ,c = Tt ,3 Tt ,2

compressor work

⎡ c p ,2Tt ,2 ⎢⎛ pt ,3 ⎞ Wc = ⎜⎜ ⎟⎟ ⎢ ηad ,c ⎝ pt ,2 ⎠ ⎢⎣ Theory of Aerospace Propulsion

γ 2 −1 γ2

γ 2 −1 γ2

−1

−1

⎤ ⎥ − 1⎥ ⎥⎦ 17

THEORY OF AEROSPACE PROPULSION 7B. Axial Flow Compressors

P. M. SFORZA University of Florida

linear, or 2‐D, Cascade

l

w3 c3,a s

w2 c2,a

c3

u

u c2

Theory of Aerospace Propulsion

2

circulation about a single blade consider a stationary cascade:

B

B

A

A

(

)

ˆj ⋅ dyjˆ = c3u s V ⋅ dl = c 3 u ∫ ∫ B

l

s C

A y,j

D D

D

(

(

ˆj ⋅ − dyjˆ V ⋅ dl = c u 2 ∫ ∫

C

) ) = −c s 2u

x,i

C

Theory of Aerospace Propulsion

3

circulation about a single blade C

−Γ =

A

∫ V ⋅ dl = c s + ∫ V ⋅ dl − c s + ∫ V ⋅ dl 3u

contour

2u

B

−Γ = ( c3u − c2u ) s

D

These terms cancel each other

The circulation is proportional to the amount the flow is turned Theory of Aerospace Propulsion

4

the tangential force component

Fu = m ( c3u − c2u )

m Γ Fu = − s

m = ρ ca sh =constant

Fu = − ρ ca hΓ

Theory of Aerospace Propulsion

5

the axial force component Fa = m ( c3a − c2 a )

m = ρ 2 c2 a sh = ρ3c3a sh For incompressible flows ca=constant Thus there is no net external axial force acting on the fluid, that is, Fa=0 for incompressible flow

Theory of Aerospace Propulsion

6

compressibility effect In a gas compressor the ideal process is an isentropic compression where p ~ργ and thus 1/ γ

c3a ρ 2 ⎛ p2 ⎞ = =⎜ ⎟ c2 a ρ3 ⎝ p3 ⎠

⎛ p3 ⎞ =⎜ ⎟ ⎝ p2 ⎠

−1/ γ

Note: For a pressure ratio of 1.2 and γ=1.4, c3a/c2a=0.88 The pressure rise causes a density rise, in turn requiring a 12% decrease in the axial velocity component.

Theory of Aerospace Propulsion

7

forces on the control volume B

l

p(x,y)

Fu’

C

Fa’

La p2

p3

Lu D

s

A y

p(x,y) x Theory of Aerospace Propulsion

8

forces in the control volume Fu = Fu′ + ∫ − p ( x, y )nˆ ⋅ ˆjdA A

ˆ Fa = Fa′ + ( p2 − p3 ) sh + ∫ − p ( x, y ) nˆ ⋅ idA A

Lu = − Fu′ = − m ( c3u − c2u ) ⎛ ⎞ ρ 3 3a ⎜1 − ⎟ La = − Fa′ = ( p2 − p3 ) sh − mc ⎝ ρ2 ⎠ Theory of Aerospace Propulsion

9

incompressible flow in the cascade Bernoulli’s equation

1 1 2 2 2 2 p2 + ρ ( c2u + c2 a ) = p3 + ρ ( c3u + c3u ) 2 2 1 2 2 p2 − p3 = ρ ( c3u − c2u ) 2

Theory of Aerospace Propulsion

10

the lift forces in terms of circulation Lu = −m ( c3u − c2u ) = ρ ca hΓ 1 2 2 La = ( p3 − p2 ) sh = ρ ( c3u − c2u ) sh 2

c − c = ( c3u + c2u )( c3u − c2u ) 2 3u

2 2u

1 La = − ρ ( c3u + c2u ) hΓ = − ρ cˆu hΓ 2 Theory of Aerospace Propulsion

11

forces and equivalent velocity field La α L

ĉ

Lu

ˆc = cˆu2 + ca2 α

ˆc u= cˆu2 + ca2

ca Theory of Aerospace Propulsion

12

compressible flow in the cascade Bernoulli’s equation for compressible flow

p3 γ − 1 c γ −1 c + = + ρ2 γ 2 ρ3 γ 2 p2

2 2

2 3

Momentum equation along a streamline

dp + ρ cdc = 0 Theory of Aerospace Propulsion

13

compressible flow result

Then

1 1 2 2 ρ cdc = d ( ρ c ) − c d ρ 2 2 1 1 2 2 dp + d ( ρ c ) − c d ρ = 0 2 2

The integrated momentum equation is then

1 ˆ2 p3 − p2 = − ρˆ cˆu ( c3u − c2u ) − ρˆ cˆa ( c3a − c2 a ) + M ( p3 − p2 ) 2 Theory of Aerospace Propulsion

14

momentum equation terms defined 1 cˆu = ( c3u + c2u ) 2 1 cˆa = ( c3a + c2 a ) 2 ρ3c32 − ρ c22 ρˆ = 2 2 c3 − c2 3

∫ M dp 2

2 ˆ M =

2

( p3 − p2 )

Theory of Aerospace Propulsion

15

Axial component of the lift force −1

−1

⎛ 1 ˆ 2⎞ ˆ ⎛ 1 ˆ 2⎞ ˆ La = ⎜ 1 − M ⎟ ρ cˆu ( c3u − c2u ) sh = − ⎜ 1 − M ⎟ ρ cˆu Γh ⎝ 2 ⎠ ⎝ 2 ⎠

For incompressible flow M=0 and ρ2=ρ3 the previous result is recovered:

1 La = − ρ ( c3u + c2u ) Γh = − ρ cˆu Γh 2

Theory of Aerospace Propulsion

16

velocity triangles view stator rotor stator w2

stator

u

c2

c2

u

stator

w3

rotor

c3 u

c3 Theory of Aerospace Propulsion

17

combined velocity diagrams

w2

c2,a

c2

c3,a

c3 w3 β2 u2

β3 u3

α2

α3 Δcu

c2,u c3,u

Theory of Aerospace Propulsion

18

pressure rise in axial flow stages Incompressible flow:

1 p3,i − p2,i = ρ ( w2,2 i − w3,2 i ) 2

Pressure rise in the rotor:

2i

i-1

stator u

rotor

3i 2i+1

i

low subsonic speeds, M<0.6

stator rotor i+1

3i+1 Pressure rise in the stator:

1 2 2 p2,i +1 − p3,i = ρ ( c3,i − c2,i +1 ) 2

Theory of Aerospace Propulsion

19

pressure rise across rotor + stator 1 p3,i − p2,i = ρ ( w2,2 i − w3,2 i ) 2

1 2 2 p2,i +1 − p3,i = ρ ( c3,i − c2,i +1 ) 2

across rotor

across stator

1 2 2 2 2 ⎡ p2,i +1 − p2,i = ρ ⎣( c3,i − c2,i +1 ) + ( w2i − w3i ) ⎤⎦ 2 Theory of Aerospace Propulsion

20

static and stagnation pressure rise across an axial stage Static pressure rise across stage was shown to be

1 p2,i +1 − p2,i = ρ ⎡⎣( c3,2 i − c2,2 i +1 ) + ( w22i − w32i ) ⎤⎦ 2 Stagnation pressure constant in stator ( no work done, no losses)

1 2 pt 3,i = pt 2,i +1 = p2,i +1 + ρ c2,i +1 2 Theory of Aerospace Propulsion

21

static and stagnation pressure rise across an axial stage (cont.)

Stagnation pressure rises across stage because work done on fluid raises pt in the rotor

1 2 2 2 2 ⎡ pt 3,i − pt 2,i = ρ ⎣( c3,i − c2,i ) + ( w2,i − w3.i ) ⎤⎦ 2 Static pressure rise across stage

1 p2,i +1 − p2,i = pt 3,i − pt 2,i − ρ ( c2,2 i +1 − c2,2 i ) 2 Theory of Aerospace Propulsion

22

work and stagnation pressure work per unit mass in [(N‐m/s)/kg/s)]=[J/kg]

Wc = P m work per unit weight in [(ft‐lb/s)/(lb/s)]=[ft‐lb/lb]=[ft] or[m]

Wc = P mg pt ,2 ⎛ pt ,3 ⎞ − 1⎟ Wc = ⎜⎜ ρ g ⎝ pt ,2 ⎟⎠

Incompressible flow

γ 2 −1 ⎡ ⎤ c p ,2Tt ,2 ⎢⎛ pt ,3 ⎞ γ 2 ⎥ − 1⎥ Wc = ⎜⎜ ⎟⎟ ⎢ ηad ,c ⎝ pt ,2 ⎠ ⎢⎣ ⎥⎦ Theory of Aerospace Propulsion

Compressible flow

23

work required in different processes

Wcomp/Wincomp

1.2 1.1

ηad,c =90%

1 0.9

ηad,c =100%

0.8 0.7 0.6 1

1.1

1.2

1.3

1.4

1.5

pressure ratio pt,3 /pt,2

Theory of Aerospace Propulsion

24

work in lightly loaded stages neglect ⎛ pt 3 ⎞ ⎜ ⎟ ⎝ pt 2 ⎠

γ −1 γ

⎛ Δpt ⎞ = ⎜1 + ⎟ pt 2 ⎠ ⎝

γ −1 γ

2

γ − 1 ⎛ Δpt ⎞ γ − 1 1 ⎛ γ − 1 ⎞ ⎛ Δpt ⎞ = 1+ − 1⎟ ⎜ ⎜ ⎟+ ⎟ + ... ⎜ γ ⎝ pt 2 ⎠ γ 2 ⎝ γ ⎠ ⎝ pt 2 ⎠

γ −1 ⎡ ⎤ γ c pTt 2 ⎢⎛ pt 3 ⎞ c pTt 2 ⎡ γ − 1 ⎛ Δpt ⎞ ⎤ ⎥ Wc = ⎢ ⎜ ⎟ − 1⎥ ≈ ⎜ ⎟⎥ ⎢ ηc ⎝ pt 2 ⎠ ηc ⎣ γ ⎝ pt 2 ⎠ ⎦ ⎢⎣ ⎥⎦

c pTt 2 R ⎛ pt 3 ⎞ pt 2 ⎛ pt 3 ⎞ Wc ≈ − 1⎟ = − 1⎟ ⎜ ⎜ ηc c p ⎝ pt 2 ⎠ ηc ρt 2 ⎝ pt 2 ⎠ This is like incompressible flow but with an average density Theory of Aerospace Propulsion

25

degree of reaction Stagnation pressure rise

1 pt 3 − pt 2 = ( p3 − p2 ) + ρ ( c32 − c22 ) 2 Static pressure rise

1 p3 − p2 = ρ ( w22 − w32 ) 2 Degree of reaction= ratio of static to total pressure rise

p3 − p2 ( w22 − w32 ) r= = 2 2 2 2 pt 3 − pt 2 ( w2 − w3 ) + (c3 − c2 ) Theory of Aerospace Propulsion

26

100% reaction stage All the pressure rise occurs in the rotor and stator blade is symmetric

stator

w2

χρ

c2 ca u

c3

stator

rotor

χs

w3

u

c3 u

c2

Theory of Aerospace Propulsion

27

50% reaction : symmetric sections Half the pressure rise in the rotor and half in the stator and both blades look alike

w2

stator

c2

u

χr

rotor

c3

stator w3

c3 u

c4

Theory of Aerospace Propulsion

χs

28

0% reaction stage: the impulse stage w2

stator

c2 u

χr

c3

rotor

w3

χs

c3 u c2

All the pressure rise occurs in the stator and the rotor is symmetric Theory of Aerospace Propulsion

29

axial outlet stage: c2>c3 and %r>100 Whirl velocity is removed from the exit flow stator w2

c2 u

χr rotor w3

c3 u

exit passage c3=ca

Theory of Aerospace Propulsion

30

axial pressure variation through the different types of stages

p2

100% stator

ex

ax ia l

ic r t e m m sy

dp/dx is a minimum for the symmetric stage

it

impulse

p3

hydraulic turbines use the impulse stage

Swirl removed by axial exit x stage

rotor

Theory of Aerospace Propulsion

31

stages necessary for fixed pressure gradient in each stage stage 1

stage 2

stage 3

stage 4

ex it ax ia l

10 0%

sy

mm etr ic

p3

p2 x

Theory of Aerospace Propulsion

32

staging in turbomachines conditions on pressure chain rule for pressure rise

(1)

( 2)

( n)

pt(3) = pt(2+1) i

n

(i )

pt 3 pt 3 pt 3 pt 3 pt 3 = (1) ( 2) ... ( n ) = ∏ ( i ) pt 2 pt 2 pt 2 pt 2 i =1 pt 2

i

(1)

pt 2 = pt 2 ptn3 = pt 3

special case: pressure rise is constant across each stage

(i ) ⎞ ⎛ pt 3 pt 3 = ⎜⎜ ( i ) ⎟⎟ pt 2 ⎝ pt 2 ⎠

n

stator

Theory of Aerospace Propulsion

rotor

stator

pit3=pi+1t2 ith stage 33

pt,3 /pt,2

effect of staging on overall pressure ratio 40 35 30 25 20 15 10 5 0

Typical Range

(pt3/pt2)(i)= 1.2 1.15 1.10 0

5

10

15

20

no. of stages

Theory of Aerospace Propulsion

34

staging in turbomachines (cont.) stage efficiency

stage temperature ratio γ −1 γ

⎛ pt(3i ) ⎞ ⎜ (i ) ⎟ − 1 pt 2 ⎠ (i ) ⎝ η ad ,c = Tt (3i ) −1 (i ) Tt 2

γ −1 ⎡ ⎤ i) (i ) ( Tt 3 1 ⎢⎛ pt 3 ⎞ γ ⎥ = 1 + ( i ) ⎜⎜ ( i ) ⎟⎟ − 1 (i ) ⎥ Tt 2 ηad ,c ⎢⎝ pt 2 ⎠ ⎢⎣ ⎥⎦

overall temperature ratio γ −1 ⎡ ⎧ ⎫⎤ i) ( n γ Tt 3 1 ⎪⎛ pt 3 ⎞ ⎪⎥ ⎢ = ∏ ⎢1 + i ⎨⎜⎜ (i ) ⎟⎟ − 1⎬⎥ ηad ,c ⎪⎝ pt 2 ⎠ Tt 2 i =1 ⎪⎥ ⎢⎣ ⎩ ⎭⎦ Theory of Aerospace Propulsion

35

efficiency of adiabatic compression γ −1 γ

⎛ pt 3 ⎞ ⎜ ⎟ −1 pt 2 ⎠ ⎝ η ad ,c = γ −1 ⎛ ⎡ ⎧ ⎫ ⎤⎞ i) ( n γ ⎜ ⎢ 1 ⎪⎛ pt 3 ⎞ ⎪ ⎥ ⎟ ⎜ ∏ ⎢1 + (i ) ⎨⎜ ( i ) ⎟ ⎬ − 1⎥ ⎟ − 1 ⎜ i =1 ⎢ η ad ,c ⎩⎪⎝ pt 2 ⎠ ⎭⎪ ⎥ ⎟ ⎦⎠ ⎝ ⎣

Overall efficiency of adiabatic compression for n stages of varying stage efficiency

But, for constant pressure rise in each stage (i )

pt 3 ⎛ pt 3 ⎞ =⎜ ⎟ (i ) pt 2 ⎝ pt 2 ⎠

1 n

ηad ,c =

⎛ pt 3 ⎞ ⎜ ⎟ p ⎝ t2 ⎠

γ −1 γ

⎛ ⎡ ⎜1 + 1 ⎢⎛ pt 3 ⎞ ⎜ η ( i ) ⎢⎜ p ⎟ ad , c ⎝ t 2 ⎠ ⎜ ⎢⎣ ⎝

Theory of Aerospace Propulsion

−1

γ −1 nγ

n

⎤⎞ − 1⎥ ⎟⎟ − 1 ⎥ ⎥⎦ ⎟⎠ 36

variation of work and efficiency with number of stages Stage efficiency η(i)ad,c = constant

Wc the compressor work

ηad,c the efficiency of adiabatic compression

Overall pressure ratio pt3/pt2=constant

Number of stages

Theory of Aerospace Propulsion

37

variation of work and efficiency with number of stages overall efficiency ηad,c=constant

η(i) ad,c the stage efficiency of adiabatic compression

overall pressure ratio pt3/pt2=constant

Wc the compressor work number of stages Theory of Aerospace Propulsion

38

Cruise dynamic pressure Dynamic pressure q=1/2γp0M02 100

q=250psf 500psf

Altitude (kft)

80 60

1000psf

40

General flight regime for aircraft

20 0 0

0.5

1

1.5

2

2.5

3

3.5

4

Mach number

Theory of Aerospace Propulsion

39

compressor exit conditions For pt,3/pt,2=20 and flight at q~500psf 2500

γ =1.4 ηc =.85

2000

Pt,3 (kPa)

1500

Tt,3 (K) 1000 500 0 0

0.5

1

1.5

2

2.5

3

3.5

Mach number

Note: We assume γ=1.4 but actually γ~1.37, 1.34, and 1.3 at M=1, 2, and 3, respectively Theory of Aerospace Propulsion

40

effect of γ on compressor exit conditions 2500

Pt3 (kPa)

ηc=0.85

2000

Tt3(K), γ=1.4 γ=1.3

1500 1000 500 0 0

0.5

1

1.5

2

2.5

3

3.5

Mach number

Theory of Aerospace Propulsion

41

compressor power required The mass flow to the compressor is m0=(ρAV)0=[pM(γ/RT)1/2]0 The power per unit area is P/A=Wcm/A0

P ower (M W/m ^2)

70

γ=1.4

60 50

γ=1.3

40 30

ηc=0.85

20 10 0 0

0.5

1

1.5

2

2.5

3

Mach number Theory of Aerospace Propulsion

42

THEORY OF AEROSPACE PROPULSION 7C. Axial Turbines and Performance Maps

P. M. SFORZA University of Florida Theory of Aerospace Propulsion

1

blade element forces w, relative velocity L, lift

β4

β2

β2

i Fu, turning force

β2 D, drag

Fa, axial force chord line

u, turbine

u, compressor γ

stagger angle

Theory of Aerospace Propulsion

2

airfoil and angle relationships w β4

cL,max

β2

cL sinβ4=sinβ2 cosβ4= - cosβ2

cD 1 0 i0

0

is

i

0

π β

Airfoil lift and drag curves Theory of Aerospace Propulsion

3

turning force & blade aerodynamics Fu = Lsin β2 + Dcos β2

Fu = Lsin (π − β4 ) + Dcos (π − β4 ) 1 2 Fu = cL ρSb w ( sin β2 + ε cos β2 ) 2 1 2 Fu = cL ρSb w ( sin β4 − ε cos β4 ) 2

aerodynamic efficiency ε=cD/cL

turning force

uFu ω rFu u = Wt = m ( c5,u − c4,u ) = mg mg mg Theory of Aerospace Propulsion

4

turbine work: aerodynamic analysis Introducing the aerodynamic coefficients

ωr

1 2 Wt = cL ρ 4 ( nb Sb ) w4 ( sin β 4 − ε cos β 4 ) mg 2

cA w4 = sin β 4

m = ρ 4 c A A4

c4 β4

w4

cA

u

1 − ε cot β 4 2π N 1 r Wt = mg ) cL ( nb Sb ) 2 2 ( 60 g ρ 4 A4 2 sin β 4 Theory of Aerospace Propulsion

5

the corrected flow variables π rnb Sb δ 4 ⎛ N ⎞ ⎛ m θ 4 ⎞ cL Wt = 1 − ε cot β 4 ) ⎜ ⎟⎜ ⎟ ( 2 60 gA4 ρ 4 ⎜⎝ θ 4 ⎟⎠ ⎜⎝ δ 4 ⎟⎠ sin β 4 −1

pref ⎛ p4 ⎞⎛ T4 ⎞ δ 4 p4 = ρ4 = ⎜⎜ ⎟⎜ ⎟⎟ ⎟⎜ RT4 RTref ⎝ pt ,4 ⎠⎝ Tt ,4 ⎠ θ 4

Tt θ= Tref pt δ= pref

state equation

−1 ⎡ π dnb Sb RTref ⎛ T4 ⎞ ⎛ p4 ⎞ ⎤ N m θ 4 cL Wt =⎢ (1 − ε cot β 4 ) ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎥ 2 θ 4 ⎢120 gA4 pref ⎝ Tt ,4 ⎠ ⎝ pt ,4 ⎠ ⎥ θ 4 δ 4 sin β 4 ⎣ ⎦

Theory of Aerospace Propulsion

6

aerodynamic operating regimes of turbines and compressors compressor turbine operating operating range range cL

cL sin β

i0 0

is

i c

γ-i=π-β4 u

γ

0

π/2

m θ4

δ4

w

N Theory of Aerospace Propulsion

θ4

π

β

increasing

=constant 7

turbine performance map Wt

θ4

Wt N

θ4

increasing

θ4

N m θ4

=K

θ4

δ4

cL sin β 4

β4 = π − γ + i

m θ4

0

maximum Theory of Aerospace Propulsion

δ4 8

mass flow through turbine m=

m θ4

δ4

pt ,4 A4 RTt ,4

=

f (γ 4 , M 4 ) =

pref RTref

pref RTref

A4 f ( γ , M )

δ4 f (γ , M 4 ) θ4 The corrected mass flow is maximum when entrance to turbine is choked

From the thermodynamic analysis γ −1 ⎡ ⎤ γ ⎛ pt ,5 ⎞ ⎥ ⎢ Wt = c p ,4 (Tref θ 4 )ηad ,t ⎢1 − ⎜ ⎜ p ⎟⎟ ⎥ ⎢⎣ ⎝ t ,4 ⎠ ⎥⎦ Theory of Aerospace Propulsion

9

turbine performance map pt ,4 pt ,5 N/θ41/2 increasing

N

θ4 m θ4 0

w’θ41/2/δ4

maximum

⎡ K N ⎛ cL ⎞ m θ 4 ⎤ ≈ ⎢1 − ⎥ ⎜ ⎟ pt ,5 ⎢⎣ η ad ,t c p ,4Tref θ 4 ⎝ sin β 4 ⎠ δ 4 ⎥⎦

pt ,4

Theory of Aerospace Propulsion

−γ γ −1

δ4

γ=1.33 -γ/(γ−1)= -4 10

representative turbine map: pressure 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

Theory of Aerospace Propulsion

11

representative turbine map: efficiency 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

1.9

2

2.1

2.2

2.3

2.4

pt,4/pt,5

Theory of Aerospace Propulsion

12

compressor performance weight flow increasing

weight flow increasing

w

cL

c u

sinβ2 1

γ 0

β

π/2

Theory of Aerospace Propulsion

For fixed c2 direction increased weight flow increases β2 and decreases i 13

compressor map Wc

θ2

or

Surge or stall line (cL maximum) Operating line

pt ,3

ηadc =constant

pt ,2

N increasing

θ2

m θ2

δ2 Theory of Aerospace Propulsion

14

representative compressor map 12 11 10 9 8 7 pt,3/pt,2 6 5 4 3 2 1

110% compressor surge line

100%

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

Theory of Aerospace Propulsion

15

dimensional analysis Dimensional analysis also yields the corrected flow variables:

pt ,out = f ( pt ,in , Tt ,in , m, γ R, ω, μ , d )

8 primary variables and 4 dimensions (L, t, M, T) leads to 4 dimensionless groupings: Π1 = pt ,out pt ,in Π2 =

m γ RTt ,in pt ,in d 2

∼

ω

m θin

δ in d 2

ωd Nd Π3 = ∼ γ RTin θin

d m

m ρ Aca ρ ca d Π4 = = = Re d ∼ μd μd μ Theory of Aerospace Propulsion

16

dimensional analysis The last dimensionless product may be put in the following form: −( γ −1) ⎡ ⎤ ⎛ ⎞ 2γ pref γ R ⎢ δ ⎛ γ −1 2⎞ ⎥ d⎜ M in ⎟ Π4 = ⎟ M in ⎜ 1 + ⎥ 2 φ θ ⎜⎝ μref Tref ⎟⎠ ⎢ ⎝ ⎠ ⎢⎣ ⎥⎦ Reynolds number index (where φ=μ/μref)

For a given compressor or turbine the characteristic diameter d is a common factor therefore

⎛ m θ2 N δ2 ⎞ = f⎜ , , ⎟ ⎜ ⎟ δ pt ,2 θ φ θ 2 2 2 ⎠ ⎝ 2 ⎛ m θ4 N pt ,5 δ4 ⎞ = f⎜ , , ⎟ ⎜ ⎟ pt ,4 δ θ φ θ 4 4 4 4 ⎠ ⎝ pt ,3

Theory of Aerospace Propulsion

pressure ratios have a weak dependence on RNI 17

HW E7.11: 50% stage turbine A 50% reaction stage axial flow turbine with u=366m/s at the mean line of the rotor where d=76.2cm has Tt=1144K and pt=687kPa entering the stator from the combustor The blade angle α4=25o.

Axial flow from combustor exit

stator 38.1cm

stator c

5

12.7cm

c4 α4 u

u

w4 c4

stage

rotor u

w5

Theory of Aerospace Propulsion

18

HW E7.11: velocity diagram Δcu Δcu Δcu c5

c5

c5

w4 w4

β4

α4=25o 0

100

w4

200

300

400 u (m/s)

W=uΔcu=u(c4u – c5u)=(1/2)[(c42-c52) – (w42-w52)] Theory of Aerospace Propulsion

19

HW E7.11: work from turbine Because c4=w5 and c5=w4 either equality holds: W = (c42‐w42) = u(c4cosα4-w4cosβ4) And since w4cosβ4=u‐c4cosα4 W = u(2c4cosα4‐u) = c42[2(u/c4)cosα4 – (u/c4)2] W/c42 = 2ηbcosα4 – ηb2 Where the blade efficiency: ηb=u/c4 Theory of Aerospace Propulsion

20

HW E7.11: Temperature leaving rotor Maximum efficiency occurs when ηb=cosα4 or u= c4cosα4 Wmax. blade eff. =c42cos2α4 = u2 Wmax. blade eff. =(366m/s)2=134kJ/kg W=cp(Tt4‐Tt5)= (1.16kJ/kg‐K)(1144K‐Tt5)=134kJ/kg Tt5=1029K

Theory of Aerospace Propulsion

21

HW E7.11: Flow properties T4 = Tt4 – c42/2cp = 1144K – (404m/s)2/[2(1160m2/s2‐K)] =1074K a4 = (γRT4)1/2 =(1.33*287m2/s2‐K*1074K)=640m/s M4= (404m/s)/(640m/s) =0.631 M4’ =(171m/s)/(640m/s) = 0.267 (relative to rotor at rotor exit) T5 = Tt5 – c52/2cp = 1029K – (171m/s)2/[2(1160m2/s2‐K)] =1016K a5 = (γRT5)1/2 = (1.33*287m2/s2‐K*1016K)=622m/s M5=(171m/s)/(622m/s)=0.275 M5’=(404m/s)/(622m/s)= 0.649 (relative to rotor at rotor exit) Theory of Aerospace Propulsion

22

HW E7.11: Flow properties (cont.)

The isentropic relation yields p4 =pt4[1+0.5(γ‐1)M42] -γ/(γ-1) =685kPa[1+M42/6] ‐4 = 532kPa The equation of state yields

ρ4=p4/RT4=(532kPa)/[(0.287kJ/kg‐K)(1074K)] =1.73kg/m3 The conservation of mass requires m=ρ4Aca4 = ρ5Aca5

Theory of Aerospace Propulsion

23

HW E7.11: mass flow and power (cont.) p5 =ρ5RT5=(1.73kg/m3)(0.287kJ/kg‐K)(1016K) =504kPa pt5 =p5[1+0.5(γ‐1)M52] γ/(γ-1) =504kPa[1+M52/6] 4 = 529kPa Flow area A=0.95π(ro2‐ri2)=0.95π[(0.381+0.0635)2‐ (0.381‐0.0635)2]=0.304m2 Mass flow through the turbine m=ρ4Aca4 =ρ4(0.304m2)ca=(1.73kg/m3)(0.304m2)(171m/s) m= 85.2 kg/s Power developed by the turbine is P=mW = (134kJ/kg)(85.3kg/s)=11.42MW Theory of Aerospace Propulsion

24

HW E7.11: Flow in blade passage (concl.)

1137K

1074K

1016K

Tt

pt

T

p

M=0.21

1144K

c4 w4 α4 u

1144K

1029K

c5 959K

1029K

w5

c4 0.0267 u M=0.27

685kPa

M=0.631 532kPa 0.649 504kPa

685kPa

529kPa

u M=0.667 476kPa 529kPa

Theory of Aerospace Propulsion

25

THEORY OF AEROSPACE PROPULSION 8. Blade Element Theory

P. M. SFORZA University of Florida

quasi‐two‐dimensional flow in an infinitesimal annular cylinder dr

dr r

Theory of Aerospace Propulsion

2

force field on blade element dL

chord line w m

dFa

dF

αm dFu

w2

dD

β2

βm Theory of Aerospace Propulsion

3

circulation‐based lift Frictionless flow in straight cascades gave the following result:

L = L + L = ρ cˆΓh 2 u

2 a

ˆc = cˆu2 + ca2 1 cˆu = ( c3u + c2u ) 2

Theory of Aerospace Propulsion

4

combined velocity diagram

w2

w m w3 ca

β2

βm β3

w2u

wmu

c2

c3 α2

w3u

α3

wmu=(w3u+w2u)/2

Theory of Aerospace Propulsion

5

force field geometry dL

chord line w m

dFa

dF ε

dFu

βm

αm

w2

dD

β2

βm Theory of Aerospace Propulsion

6

turning and axial forces from lift and drag forces dFu = dL sin β m + dD cos β m dFa = dL cos β m − dD sin β m ⎛1 2 ⎞ dL = cl ⎜ ρ wm ⎟ ldr ⎝2 ⎠ ⎛1 2 ⎞ dD = cd ⎜ ρ wm ⎟ ldr ⎝2 ⎠ Theory of Aerospace Propulsion

7

momentum & blade theory compared Blade element theory

⎡ ⎤ cd ⎛1 2 ⎞ dFu = cl ⎜ ρ wm ⎟ ldr ⎢sin β m + cos β m ⎥ cl ⎝2 ⎠ ⎣ ⎦

⎡ ⎤ cd ⎛1 2 ⎞ dFa = cl ⎜ ρ wm ⎟ ldr ⎢cos β m − sin β m ⎥ cl ⎝2 ⎠ ⎣ ⎦ Cascade momentum analysis for frictionless flow

dFu = ( c3u − c2u ) ρ ca sdr dFa = ( p3 − p2 ) sdr Theory of Aerospace Propulsion

8

combination of momentum and blade element analyses ⎛1 2 ⎞ cl ⎜ ρ wm ⎟ l 2 ⎝ ⎠ ( c3u − c2u ) = [sin β m + ε cos β m ] ρ ca s ⎛1 2 ⎞ cl ⎜ ρ wm ⎟ l 2 ⎝ ⎠ ( p3 − p2 ) = [cos β m − ε sin β m ] s

Theory of Aerospace Propulsion

9

power using blade element results ⎛1 2 ⎞ ( c3u − c2u ) = ucl ⎜ ρ wm ⎟ ldr [sin β m + ε cos β m ] dP = mu ⎝2 ⎠ Assuming a constant average density for the flow in the stage yields

l u ⎛1 2 ⎞ pt 3 − pt 2 = cl ⎜ ρ wm ⎟ [sin β m + ε cos β m ] s ca ⎝ 2 ⎠

Theory of Aerospace Propulsion

10

degree of reaction p3 − p2 ca cos β m − ε sin β m r= = pt 3 − pt 2 u sin β m + ε cos β m 1 − ε tan β m r = φ cot β m 1 + ε cot β m The parameter φ =ca/u is typically used as a non‐ dimensional measure of mass flow through the stage Theory of Aerospace Propulsion

11

r and the mass flow function φ 1 1 w3u + w2u ) w3u + ( u − c2u ) ) ( ( wmu 2 = =2 cot β m = ca ca ca ⎞ 1⎛1 cot β m = ⎜ + cot β3 − cot α 2 ⎟ 2⎝φ ⎠

1 r = ⎡⎣1 + φ ( cot β 3 − cot α 2 ) ⎤⎦ 2 1 r = (1 + φ B ) 2

where

B = cot β 3 − cot α 2

Theory of Aerospace Propulsion

12

influence of the trailing edge angles u w3

β3 u

rotor

c3

stator w2

α2

c2

u Theory of Aerospace Propulsion

13

influence of the trailing edge angles (cont.) u w3

β3 u

rotor

c3

1 r = ⎡⎣1 + φ ( cot β 3 − cot α 2 ) ⎤⎦ 2 w2

α2

constant=B

φ =ca/u determines r

c2

u

Theory of Aerospace Propulsion

14

non‐dimensional pressure coefficient ψ pt 3 − pt 2 = ρ u ( c3u − c2u ) = ρ u [1 − φ B1 ] 2

where

B1 = cot β3 + cot α 2

The pressure coefficient is

pt 3 − pt 2 ψ= = 2 (1 − φ B1 ) 1 2 ρu 2 Theory of Aerospace Propulsion

15

pressure coefficient vs. mass flow coefficient 3

B1<0

2

B1=0

ψ

Typical range for lightly loaded stage

1

compressors

0 0

fans

B1>0

φdesign

φ

Theory of Aerospace Propulsion

16

non‐dimensional combined velocity diagrams

w2/u wm/u w3/u ca/u =φ β2

βm

c2/u

β3

cm/u α2

c3/u αm

1

α3 1

τ r*

r*=rfrictionless=φcotβm Theory of Aerospace Propulsion

17

rotor and stator forces βm+εr

dDr

dLs

dLr

w2/u β2

wm/u w3/u βm

ca/u= φ

c2/u

β3

dDs

αm+εs

cm/u

c3/u

α2 1

αm

α3 1

τ r*

Theory of Aerospace Propulsion

18

the deflection coefficient τ From the combined velocity diagram we see that

c2u + w2u = u c3u + w3u = u Adding or subtracting yields

c3u − c2u = w2u − w3u

( c3u + c2u ) + ( w3u + w2u ) = 2u c3u − c2u w2u − w3u = τ= u u Theory of Aerospace Propulsion

19

the theoretical degree of reaction r* w2u + w3u )( w2u − w3u ) ( w22 − w32 r* = 2 = 2 2 2 w2 − w3 + c3 − c2 ( w2u + w3u )( w2u − w3u ) + ( c3u + cu )( c3u − c2u )

1 w2u + w3u r* = u 2

Theory of Aerospace Propulsion

20

rotor and stator forces dFr ,u = ( ρ ca sr dr )( w3u − w2u ) = ρ u 2τφ ( sr dr ) dFs ,u = ( ρ ca ss dr )( c3u − c2u ) = ρ u 2τφ ( ss dr ) dFr ,a = Δpr ( sr dr ) dFs ,a = Δps ( ss dr )

dFr ,a = dFr ,u cot ( β m + ε r ) dFs ,a = dFs ,a cot (α m + ε s ) Theory of Aerospace Propulsion

21

rotor and stator forces (cont.)

1 − ε r tan β m dFr ,a = dFr ,u ε r + tan β m 1 − ε s tan α m dFs ,a = dFs ,u ε s + tan α m

tan β m = The mean angles are

tan α m =

Theory of Aerospace Propulsion

φ r*

φ

1− r * 22

pressure change across rotor and stator including friction ⎛ r * −ε rφ ⎞ Δpr = ρ u τφ ⎜ ⎟ ⎝ φ + εrr * ⎠ 2

⎡ 1 − r * −ε s ⎤ Δps = ρ u τφ ⎢ ⎥ ⎣ φ + ε s (1 − r *) ⎦ 2

⎡ r * −ε rφ 1 − r * −ε s ⎤ Δp = Δpr + Δps = ρ u τφ ⎢ + ⎥ ⎣ φ + ε r r * φ + ε s (1 − r *) ⎦ 2

Theory of Aerospace Propulsion

23

power required The power required based on the turning force exerted by the rotor may be expressed as follows:

P = uFr ,u = ρ u τφ ( sr h ) 3

For a constant density flow

⎛ Δp ⎞ ⎛ Δp ⎞ P=⎜ ⎟ m r = ⎜ ⎟ ρ uφ sr h ⎝ ρ ⎠ ⎝ ρ ⎠

Theory of Aerospace Propulsion

24

efficiency of compression The ratio of theoretical power required to actual power required is

P * Δp Δp ηc = = = 2 P Δp * ρ u τ The efficiency of compression is then given by

⎡ r * −φε r 1 − r * −φε s ⎤ + ηc = φ ⎢ ⎥ ⎣ φ + ε r r * φ + ε s (1 − r *) ⎦ Theory of Aerospace Propulsion

25

maximum efficiency ⎛ εr ⎞ ropt * ≈ ⎜ 1 + ⎟ ⎝ εs ⎠

−1

Assuming εr=εs=ε leads to the following:

1 − 2εφ ηc = 2φ ε + 2φ

1 φopt = 2

(

)

1− ε 1+ ε − ε ≈ 2 2

ηmax = 1 + 2ε − 2ε 1 + ε ≈ 1 − 2ε (1 − ε ) 2

2

Theory of Aerospace Propulsion

26

contour plot of efficiency 1.0 r* ηc=0.926 0.92 0.91 0.90 0.88

0.5

0 0

0.5

φ

1.0

Theory of Aerospace Propulsion

27

THEORY OF PROPULSION 8A. Compressibility and Flow Losses

P. M. SFORZA University of Florida

the Euler equations ∂wx ∂wx 1 ∂p wx + wy =− ∂x ∂y ρ ∂x momentum ∂wy ∂wy 1 ∂p wx + wy =− ρ ∂y ∂x ∂y

continuity

∂ ∂ ( ρ wx ) + ( ρ wy ) = 0 ∂x ∂y

Theory of Aerospace Propulsion

2

combined Euler and continuity equations a =γ 2

p

ρ

∂p ∂ 2 ∂ρ γ γ −1 ∂ρ = ( C ρ ) = Cγρ =a ∂x ∂x ∂x ∂x ∂p 2 ∂ρ =a ∂y ∂y 2 ∂ w ⎛ ⎞ ⎛ ∂wx ∂wy ⎞ ∂ w ∂ w ∂ w a y 2 2 x x wx + wx wy ⎜ + = ⎜ + ⎟ + wy ⎟ ρ ⎝ ∂x ∂x ∂y ⎠ ∂y ∂y ⎠ ⎝ ∂x

Theory of Aerospace Propulsion

3

the perturbation potential y w i

wm

w’ x

wm

∂ϕ ∂ϕ (1 − M ) ∂x 2 + ∂y 2 = 0 wx = wm + w′x ∂ϕ ′ w = x ′ wy = wy ∂x ∂ϕ w′x , w′y << wm w′y = ∂ y Theory of Aerospace Propulsion 4 2

2 m

2

transform to Laplace’s equation Transform the coordinates

Introduce a new dependent variable

xi = x yi = y 1 − M

2 m

φi=Λφ

To obtain Laplace’s equation

∂ 2φi ∂ 2φi + 2 =0 2 ∂xi ∂yi Theory of Aerospace Propulsion

5

relationships among the variables w′x ,i = Λw′x

incompressible

compressible

Λ

w′y ,i =

1− M

2 m

w′y

The angle of incidence

ii ≈ tan ii =

w′y wm + w′x

≈

w′y wm

For geometrically similar airfoils Therefore Theory of Aerospace Propulsion

=

Λ 1 − M m2

i

i = ii Λ = 1 − M m2 6

the perturbation velocities and Cp compressible

w′x =

1 1 − M m2

w′x ,i

incompressible

w′y = w′y ,i The pressure coefficient in linearized flow

p − pm w′x Cp = ≈ −2 1 wm ρ m wm2 2

C p ,i = −2

Theory of Aerospace Propulsion

w′x ,i wm

incompressible 7

the Prandtl‐Glauert rule The relation between the compressible and incompressible pressure coefficient is given by

Cp =

1 1− M

2 m

Then blade spacing is given by

And solidity by

cl =

C p ,i s=

1 1− M

1 1− M

2 m

2 m

cl ,i

si

l 2 li σ = = 1 − M m = 1 − M m2 σ i s si Theory of Aerospace Propulsion

8

velocity distributions over airfoils (v/V)2

(v/V)2

x/l Theory of Aerospace Propulsion

9

critical Mach number as f(cL) The critical Mach number is the free stream value of M for which the Mach number at some point on the surface first reaches unity Mcr 0.7 0.65 Mcr for NACA 65‐415

0

0.2

0.55

Theory of Aerospace Propulsion

cl 10

limitation of the Prandtl‐Glauert rule Cp =

1 1− M

2 m

C p ,i

is unbounded as Mm

0

The full plane inviscid subsonic flow potential equation is 2 2 2 ⎡ ⎤ ⎡ 2 ⎛ ∂φ ⎞ 2 ⎤ ∂ 2φ ⎛ ⎞ ∂ ∂ φ φ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ∂ φ 2 + ⎢a − ⎜ ⎟ ⎥ 2 = 0 ⎢a − ⎜ ⎟ ⎥ 2 − 2 ⎜ ⎟ ⎜ ⎟ ⎝ ∂x ⎠ ⎥⎦ ∂x ⎝ ∂x ⎠ ⎝ ∂y ⎠ ∂x∂y ⎢⎣ ⎝ ∂y ⎠ ⎥⎦ ∂y ⎢⎣

Karman and Tsien solved this equation using Chaplygin’s hodograph method and an approximate equation of state

⎛1 1 ⎞ p − pm = ρ a ⎜ − ⎟ ⎝ ρ ρm ⎠ 2 m m

Theory of Aerospace Propulsion

11

the Karman‐Tsien correction for Cp It is still necessary that Mm2τ <<1, that is, thin airfoils 2 ⎡ ⎤ C M p ,i 2 m ⎥ C p = C p ,i ⎢ 1 − M m + 2 ⎢⎣ 1 + 1 − M m 2 ⎥⎦

−1

The Karman‐Tsien correction is smaller in magnitude than that given by the Prandtl‐Glauert correction

Theory of Aerospace Propulsion

12

boundary layer effects Curved streamlines initiate secondary flows

Pitch line streamline

R

Theory of Aerospace Propulsion

∂p ρ w2 = ∂r r

13

gap leakage losses Outer casing

Blade gap

δ

Pressure side of Blade height blade

Gap leakage vortex

h Blade chord l

Linear speed of rotation u Theory of Aerospace Propulsion

14

curved streamline effects casing wall boundary layer streamline

w2,tip mean pitch line

w2,mean

disc boundary layer streamline

w2,root

Theory of Aerospace Propulsion

15

curved flow losses Curved blade passage

Viscous stream‐ lines

tip

Inviscid stream‐ lines

n, coordinate normal to the blade surface

ω

Near‐wall (top and bottom) boundary Secondary layer flows forced by flow centrifugal pressure vortices Theory of Aerospace Propulsion gradient

16

cross‐flows in the blade passage Outer casing

Blade gap δ

Pressure side

Secondary flow vortices

Blade height h

Blade chord l

Linear speed of rotation u Theory of Aerospace Propulsion

17

boundary layer regions Separation region

Turbulent boundary layer Transition

Laminar boundary layer

turbulent wake

Theory of Aerospace Propulsion

18

blade loading Blade element theory for axial compressors gives the blade loading:

σ cL = 2sin β m ( cot β 2 − cot β3 )(1 − ε cot β m ) τ deflection coefficient cot β 2 − cot β3 = = φ mass flow coefficient Boundary layer separation limits the amount of turn allowable, so

σ cL ≤ (σ cL )max Theory of Aerospace Propulsion

19

correlation for thin blades For (t/l)max~10% experiments suggest 2

⎛ sin β 3 ⎞ 3 cL ⎜ ≤ 0.675 ⎟ ⎝ sin β m ⎠ 6 − σ Which, in turn, leads to

sin 2 β 3 ≤ 0.1125σ ( 6 − σ ) ( cot β 2 − cot β3 ) sin β m Δβ=β2 - β3 results obtained from experiment show that this correlation is conservative for values of σ ~O(1) This suggests that there should be a basis for this in the physics of the flow field.

Theory of Aerospace Propulsion

20

blade pressure distributions Δβ=18.6

suction side w2 β

o

w3 2

β2

β3

pressure side

u u 1 Sp 0 Sp -1

Ss

Ss -2 -3 -3.1(1-x/l)

-3.1(1-x/l)

-4

0

0.2

0.4

0.6

Theory of Aerospace Propulsion

0.8

1 x/l

21

lift on the blade l

L = h ∫ ( pl − pu )dx 0

Pressure coefficient of the pressure, or lower, surface

Pressure coefficient of the suction, or upper surface

pu − p3 Ss = 1 2 ρ w3 2

pl − p3 Sp = 1 ρ w32 2 1 ⎛ x⎞ 2 L = ρ w3 hl ∫ ( S p − S s )d ⎜ ⎟ 2 ⎝l⎠ 0 1

Theory of Aerospace Propulsion

22

the blade lift coefficient Lower (pressure) surface

Upper (suction) surface

⎛ dx ⎞ S s ≈ K ⎜1 − ⎟ l ⎠ ⎝

l

∫ S dx ≈ 0 p

0

1 ⎛ x⎞ ⎛ x⎞ 1 ⎛ 1 ⎞ 2 2 L ≈ ρ w3 hl ∫ K ⎜ 1 − ⎟ d ⎜ ⎟ = ρ w3 hl ⎜ − K ⎟ 2 ⎝ l⎠ ⎝l⎠ 2 ⎝ 2 ⎠ 0 1

1 ⎛ w3 ⎞ =− K⎜ cL = ⎟ 1 2 ⎝ wm ⎠ ρ wm2 hl 2 L

Theory of Aerospace Propulsion

2

23

modeled lift coefficient 1 ⎛ w3 ⎞ =− K⎜ cL = ⎟ 1 2 ⎝ wm ⎠ ρ wm2 hl 2 L

2

Neglecting the drag (ε<<1) yields

sin 2 β3 1 =− σK ( cot β 2 − cot β3 ) sin β m 4

Theory of Aerospace Propulsion

24

comparing flow turning predictions Modeled flow

sin 2 β 3 = 0.775 ( cot β 2 − cot β3 ) sin β m

Empirical correlation

sin 2 β 3 = 0.5625 ( cot β 2 − cot β3 ) sin β m

Experimental result

sin 2 β3 = 0.781 ( cot β 2 − cot β3 ) sin β m

Theory of Aerospace Propulsion

25

experimental drag profile 0.15 operating point Δβ=0.8(Δβ)stall

0.10

ε 0.05

Δβ=13.8o

Δβ=18.6o stall

0 5

10

15 20 β3-β2 (degrees)

Theory of Aerospace Propulsion

25 26

turbine expansion processes Frictional and heat transfer losses reduce the energy of the flow pt,4 Tt Tt,4 real

Tt,5 Tt,5 Tt,5’

isentropic

pt,5

adiabatic

s Theory of Aerospace Propulsion

27

heat transfer in the blade passage S

S

l

l

0

0

0

0

Pt = ∫ ρ 4 ca ,4 c p ,4Tt ,4 dy − ∫ ρ5ca ,5c p ,5Tt ,5 dy − ∫ q pressure dx − ∫ qsuction dx

S

pt,4

T(y)

p ,5Tt ,5 mc

p ,4Tt ,4 mc

pt,5

Theory of Aerospace Propulsion

28

thermal boundary layer ⎛ ∂T ⎞ qw = − k w ⎜ ⎟ ⎝ ∂y ⎠ w

Heat transfer to the wall: y

δ

T(y,0)

T(y,t), for large t

Heat transfer into the wall

Zero heat transfer

0 0

Tw,i

Te

Theory of Aerospace Propulsion

Tw,ad

T 29

adiabatic wall temperature If pe, Te, and Tw are constant and Prandtl number Pr=μcp/k=1:

1 2 1 2 T+ w = Te + we 2c p 2c p T ( y ) = Te +

1 ⎡⎣ue2 − u 2 ( y ) ⎤⎦ 2c p

⎡ ⎛ ∂u ⎞ ⎤ ⎛ ∂T ⎞ ⎜ ⎟ = ⎢u ⎜ ⎟ ⎥ = 0 ⎝ ∂y ⎠ w ⎣ ⎝ ∂y ⎠ ⎦ w

1 2 Tw,ad = Te + ue 2c p

stagnation enthalpy is constant

temperature distribution

wall heat transfer is zero

adiabatic wall temperature

Theory of Aerospace Propulsion

30

for non‐unity Prandtl number μ ρ

ν Pr = = cpk α ρ

=

Momentum diffusivity Thermal diffusivity

γ −1 2 ⎞ ⎛ Tw,ad = Te ⎜ 1 + r Me ⎟ 2 ⎝ ⎠

r ≈ Pr =recovery factor Theory of Aerospace Propulsion

31

turbine blade cooling: external view tip cap holes

film cooling holes

trailing edge slots blade platform

cooling air Theory of Aerospace Propulsion

32

turbine blade cooling: internal view film cooling of tip cap blade leading edge

pin‐fin cooling of aft region

impingement cooling on inner leading edge

porous blade trailing edge

blade platform cooling air

cooling air

Theory of Aerospace Propulsion

internally ribbed cooling surfaces 33

THEORY OF AEROSPACE PROPULSION 9. Airbreathing Engine performance and Component Integration

P. M. SFORZA University of Florida

example: basic matching fuel lines

Given: M0=2.0, Z=50kft (15.24km) pt,2/pt,0=0.9, A2=1ft2 , M2=0.3 pt,3/pt,2=15,hc=0.85, ηb= ηt=0.95, Qf =18,900Btu/lbm (44MJ/kg) Find: Fn and cj for various Tt,4

inlet compressor

combustor

nozzle

combustor turbine

1

2

3

4

5

6

7

pt,2=243.6lb/ft2(1+22/5)3.5=1906lb/ft2 Tt,2 =390R(1+22/5)=702R cp,c=0.24Btu/lb‐R(1kJ/kg‐K); γc=1.4 ⎛ pt ,3 ⎞ ⎜⎜ ⎟⎟ p t ,2 ⎠ ηc = ⎝ Tt ,3 Tt ,2

γ c −1 γc

−1

cp,t=0.276Btu/lb‐R(1.2kJ/kg‐K); γt=1.33 ηt =

−1

T 1 − t ,5 Tt ,4 ⎛ pt ,5 ⎞ 1 − ⎜⎜ ⎟⎟ p ⎝ t ,4 ⎠

γ t −1 γt

Solve to find Tt,3=1667R Theory of Aerospace Propulsion

2

example: basic matching (cont.)

The matching condition is

Wc = c p ,c (Tt ,3 − Tt ,2 ) = Wt = c p ,t (Tt ,4 − Tt ,5 ) Assume Tt,4 /Tt,2 =3 which means Tt,4 =2106R (1170K) and further assume pt,4/pt,3=1. The mass flow is m = ( ρ AV )2 =

γc

pt ,2 A2 M 2

(1 + M / 5) 2 2

3

RTt ,2

= 3.4 lbm s

The exit velocity for a matched ideal nozzle (p7=p0 and pt7=pt,5) is (γ t −1) γ t ⎤ ⎡ V7 = 2c p ,tTt ,7 1 − ( p0 pt ,5 ) ⎢⎣ ⎥⎦ Theory of Aerospace Propulsion

3

example: basic matching (concl.)

From the matching condition we find γt

γ t −1 γ c −1 γ t −1 ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ γ γ t ⎛ pt ,5 ⎞ ⎥ Tt ,5 pt ,5 ⎪ c p ,cTt ,2 1 ⎢⎛ pt ,3 ⎞ c ⎢ ⎪ ⎥ = 1 − ηt ⎢1 − ⎜ = ⎨1 − ⎟⎟ ⎥ ⎜⎜ ⎟⎟ − 1⎥ ⎬ and ⎜ ⎢ Tt ,4 pt ,4 ⎠ pt ,4 ⎪ c p ,tTt ,4 ηcηt ⎝ pt ,2 ⎠ ⎝ ⎪ ⎢⎣ ⎥⎦ ⎢ ⎥ ⎣ ⎦ ⎩ ⎭ We use these equations to find pt,5= 3316lb/ft2 and Tt,5= 1277R. Then we can calculate: 3.4 lbm s Fn = m (V7 − V0 ) = ( 2899 ft s − 1936 ft s ) = 101.8lb 2 32.17 ft ⋅ lbm s ⋅ lb −1

⎤ ηb Q f m 0 ( 3600 s h ) ⎡ = − 1⎥ = 0.82 lbm h lb cj = ⎢ Fn Fn ⎢⎣ c p ,b (Tt ,4 − Tt ,3 ) ⎥⎦ m f

Note: The thrust is linearly proportional to A2 while cj is independent of A2 Theory of Aerospace Propulsion

4

turbojet component matching: 1 fuel lines inlet

Given: M0, altitude, pt,2/pt,0, Tt,4, and A6

combustor

compressor

nozzle

combustor

Find: N, A0, A7

turbine

Corrected mass flow

1

2

3

4

5

6

7

m θ5 = constant through nozzle ~ A 6

δ5 Relation between corrected mass flows across turbine

m θ5

δ5

m θ 4 pt ,4 = δ 4 pt ,5

Tt ,5 Tt ,4

m θ5

δ5

ht ,4 − ht ,5 m θ 4 pt ,4 1− = δ 4 pt ,5 c pTref θ 4

Theory of Aerospace Propulsion

5

correlation of corrected weight flow Picking values of m θ4 for various

pt,4/pt,5

δ4

N

N

values of

θ4

efficiency and pressure ratio

θ 4 yields a turbine

ηt decreasing

m θ 4

δ4 Use maps to make a corrected mass flow correlation

m θ 4

δ4

N

found

θ4 Chosen rpm

known

Theory of Aerospace Propulsion

increasing

m θ5

δ5

6

matching condition m θ 2

δ2

pt ,3 m θ 4 pt ,4 pt ,3 Tt ,2 = =C δ 4 pt ,3 pt ,4 Tt ,4 pt ,2

known from previous slide assume constant (typically close to 1.0)

known from given conditions known from previous slide

Matching condition: Δht,t = Δht,c Δht,c/θ2 = (Δht,t/θ4)(Tt4/Tt2)

on compressor map

Δhtc/θ2 = cpTrefηadc-1[(pt3/pt2)(γ-1)/γ-1] Theory of Aerospace Propulsion

7

compressor map pt ,3

m θ 2

stall line

δ2

=C

pt ,3 pt ,2

pt ,2 Each point yields pt,3/pt,2 and η from which Δhc/θ2 may be calculated

current

N

θ2

m θ 2

δ2 Theory of Aerospace Propulsion

8

matching compressor and turbine m θ 2

δ2

pt ,3

=C

pt ,3 pt ,2

pt ,2

Δht,c/θ2 matching point: this is the required N

m θ 2

δ2 Theory of Aerospace Propulsion

9

capture area and nozzle exit area m θ 0

δ0

m θ 2 pt 2 = δ 2 pt 0

m θ 0

δ0

inlet pressure recovery

Tt 0 Tt 2

= A0 M 0 pref

Tt0/Tt2=1 (adiabatic flow)

γ0 ⎛

γ −1

⎞ M ⎟ ⎜1 + 2 RTref ⎝ ⎠ 2 0

γ 0 +1 2( γ 0 −1)

Required area for captured streamtube γ 6 +1 ⎡ ⎤ − γ 1 A7 ⎢ γ 6 − 1 ⎛ γ 6 + 1 ⎞ 6 ⎥ = A6 ⎢ 2 ⎜⎝ 2 ⎟⎠ ⎥ ⎣ ⎦

1 2

γ 6 +1 2 ⎡ ⎤ γ γ ⎢⎛ p0 ⎞ 6 − ⎛ p0 ⎞ 6 ⎥ ⎜ ⎟ ⎥ ⎢⎜ pt 7 ⎟ pt 7 ⎠ ⎝ ⎠ ⎝ ⎢⎣ ⎥⎦

−

1 2

Required nozzle exit area Theory of Aerospace Propulsion

10

sizing the engine fuel lines

inlet

M0

A6

combustor

A0

p0

Amin

nozzle

A7

combustor

T0 1

3

inlet performance given

1 pt2/pt0 0

2

4

5

6

7

Tt4 1

A0/Amin 1

M

0

Theory of Aerospace Propulsion

1

M

11

turbojet component matching: 2 Given: M0, altitude, pt,2/pt,0, m and N

fuel lines

inlet combustor

nozzle

combustor

Find: Tt4, A6, A7 1

2

3

4

5

6

7

Calculate corrected mass flows, rpm, and free stream capture area:

⎛ γ −1 2 ⎞ Tt ,2 = Tt ,0 = T0 ⎜1 + M0 ⎟ 2 ⎝ ⎠

m θ 0 m θ 2 ,

δ0

A0 =

δ2

m RTref θ 0

pref δ 0 f ( γ , M 0 )

γ

⎛ γ − 1 2 ⎞ γ −1 pt ,2 = p0 ⎜1 + pt ,2 = pt ,0 M0 ⎟ 2 pt ,0 ⎝ ⎠ pt ,0 pt ,2

N

θ2 Theory of Aerospace Propulsion

12

compressor map surge or stall line (cL maximum)

pt ,3 operating line

pt ,2

ηadc =constant

N

θ2

increasing

m θ 2 Find pt3/pt2 and ηadc Theory of Aerospace Propulsion

δ2 13

corrected mass flow Δhtc/θ2 = cpTrefηadc-1[(pt3/pt2)(γ-1)/γ ‐ 1] known from compressor map Tt3/Tt2 = 1+ηadc-1[(pt3/pt2)(γ-1)/γ ‐1] Determine corrected mass flow

m θ 3

δ3

m θ 2 pt ,2 = δ 2 pt ,3

Tt ,3 Tt ,2

Theory of Aerospace Propulsion

14

mass flow and work for turbine Across the (constant pressure) combustor assume an exit temperature Tt4 and find M4 from the combustor equations: Tt4/Tt3 = f(γ,M3, M4) and pt4/pt3 = g(γ,M3,M4)

corrected mass flow corrected turbine work corrected rpm

m θ 4

δ4

m θ3 pt ,3 = δ 3 pt ,4

Tt ,4 Tt ,3

Δht

Δht Tt ,2 = θ4 θ 2 Tt ,4 N Tt ,2 = θ4 θ 2 Tt ,4

N

Theory of Aerospace Propulsion

15

matching condition Δht/θ4 locus of results for various Tt4 matching point

m θ 4

δ4 Theory of Aerospace Propulsion

16

turbine performance map pt ,4 pt ,5

N N/θθ41/2 increasing 4

m θ 4 0

w’θ41/2/δ4

δ4 maximum

⎡ K N ⎛ cL ⎞ m θ 4 ⎤ ≈ ⎢1 − ⎥ ⎜ ⎟ pt ,5 ⎢⎣ η ad ,t c p ,4Tref θ 4 ⎝ sin β 4 ⎠ δ 4 ⎥⎦

pt ,4

Theory of Aerospace Propulsion

−γ γ −1

γ =1.33 -γ/(γ−1)= ‐4 17

nozzle minimum and exit areas The stagnation temperature ratio is obtained from Tt5/Tt4 = 1‐ (ΔWt/θ4)/cpTref Corrected weight flow entering nozzle

Find A6 from

m θ5

Nozzle exit area

δ5

= A6 pref

A7 = A6

m θ5

δ5

γ

⎛ γ +1⎞ ⎜ ⎟ RTref ⎝ 2 ⎠

γ +1 ⎛ γ +1⎞

⎜ ⎟ 2 ⎝ 2 ⎠

m θ 4 pt ,4 = δ 4 pt ,5

Tt ,5 Tt ,4

γ +1 2( γ −1)

γ +1 2( γ −1)

Theory of Aerospace Propulsion

2

⎛ p7 ⎞ γ ⎛ p7 ⎞ ⎜⎜ ⎟⎟ − ⎜⎜ ⎟⎟ p p ⎝ t ,7 ⎠ ⎝ t ,7 ⎠

γ +1 γ

18

representative compressor map 12 110% 11 100% 10 compressor surge line 9 8 η=87% 90% 80% 7 pt3/pt2 6 85% 80% 5 4 N/θ2^0.5=70% 80% 3 exit choke line 2 1 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40

m θ 2 δ 2 in lbm/s Note: x-axis in Figure 9.12 of text should be corrected to this form Theory of Aerospace Propulsion

19

representative turbine map: pressure 2.4 2.2

N/θ4^0.5=110% 100% 90% 80%

2

70%

pt4/pt5 1.8 1.6 1.4 1.2 44.8 45.2 45.6 46 46.4 46.8 47.2 47.6 48 48.4 48.8 49.2 49.6 . mθ4^0.5/ δ4 (lbm/s)

Theory of Aerospace Propulsion

20

representative turbine map: efficiency 0.95 N/θ^0.5=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

1.9

2

2.1

2.2

2.3

2.4

pt4/pt5

Theory of Aerospace Propulsion

21

sizing the engine fuel lines

inlet

M0

A0

p0

combustor

Amin

A6 nozzle A7

combustor

T0 1

2

3

Inlet performance given

1 pt2/pt0

4

5

6

7

1

A0/Amin

0 1

M

0

Theory of Aerospace Propulsion

1

M

22

inlet‐engine matching M0>1

A0

A1

Amin

For isentropic flow and γ=1.4 m θ

δ

For adiabatic flow (Tt=constant):

kg / s ⎞ AM ⎛ = ⎜ 415.8 2 ⎟ =const 2 3 m ⎝ ⎠⎛ M ⎞ ⎜1 + ⎟ 5 ⎝ ⎠ ⎛ m θ ⎞ m θ 0 pt ,0 ⎜⎜ ⎟⎟ = δ 0 pt ,min ⎝ δ ⎠ min

Theory of Aerospace Propulsion

23

captured streamtube area m θ 2

δ2 compressor requirement

−1

m θ 0 ⎛ pt ,2 ⎞ A0 M 0 = ⎜⎜ ⎟⎟ = 416.8 2 3 δ 0 ⎝ pt ,0 ⎠ ⎛ M ⎞ pt ,2 ⎜1 + ⎟ 5 ⎝ ⎠ pt ,0

captured streamtube mass flow

inlet pressure recovery

Assume Mil‐E‐5008B pressure recovery: (pt2/pt0)std = 1‐0.075(M0‐1)1.35 Then the captured area ‐ mass flow relation is:

A0 =

3

⎡1 − 0.075 ( M 0 − 1)1.35 ⎤ ⎛⎜ 1 + M ⎞⎟ ⎣ ⎦⎝ 5 ⎠ m θ 2 2 0

416.8M 0 Theory of Aerospace Propulsion

δ2

24

captured streamtube area needed 0.012 0.01 0.008 A0' 0.006 0.004 0.002 0.2

0.6

1

1.4

1.8

2.2

2.6

3

M0

A0′ =

A0 m θ 2 δ 2

captured streamtube area per unit corrected mass flow Theory of Aerospace Propulsion

25

inlet requirements A0 =(A0/A1)(A1/Amin)Amin captured streamtube

M>1

contraction ratio

minimum area

M>1

M<1

M0>1

M<1

M0>1

Subcritical: inlet flow influenced by engine operation

Supercritical: inlet flow independent of engine operation

Theory of Aerospace Propulsion

26

captured streamtube area as f(M0) 0.012 optimum 4 shock

0.01

MIL standard 0.008

internal compression

A0' 0.006 0.004 0.002 0

0.5

1

1.5

2

2.5

3

M0

Inlet operating below maximum pressure recovery will cause Fn and sfc to suffer. Excess mass flow must be spilled: (a) A0 decrease, (b) internal bleed, or (c) spillage and additive drag Theory of Aerospace Propulsion 27

effect of shock location in inlet fuel lines

inlet combustor

nozzle combustor

1

2

3

4

M

1

1

pt2/pt0

0 x

min xa xb

x

xmin xa xb

5

6

7

x

Theory of Aerospace Propulsion

28

nozzle conditions & shock location Turbine‐fixed Burner fixed

pt 5 pt 4 pt 3 pt 2 pt 5 = pt 0 pt 4 pt 3 pt 2 pt 0 Compressor –fixed Inlet‐fixed Minimum area total pressure

Flight condition‐fixed

pt 2 pt 6 = pt 5 ~ pt 0

Theory of Aerospace Propulsion

29

pressure at nozzle minimum area pt,5 = (pt,5/pt,4) (pt,4/pt,3) (pt,3/pt,2) (pt,2/pt,0) pt,0 f(M0)

fixed pt,6 = pt,5 ~ pt,2/pt,0 A6

pt,6

To stay choked

X Theory of Aerospace Propulsion

X 30

if A6 is fixed M~pt,6A6/T61/2

T6

X Tt,5= (Tt,5/Tt,4)Tt,4

Therefore Tt,4 must drop

Theory of Aerospace Propulsion

31

effect of shock location on A6 and pt,6 fuel lines

inlet

combustor

nozzle x

combustor

1

2

3

4

5

6

7

pt6

A6

0 xmin xa xb

x

(a)

0 xmin xa xb

Theory of Aerospace Propulsion

x 32

effect of shock location on Tt,6 and pt,6 fuel lines

inlet

combustor

nozzle x

combustor

1

2

3

4

Tt6

5

6

Tt6~(Tt5/Tt4)Tt4~Tt4

pt6

0

7

turbine inlet temperature

0 xmin xa xb

x

xmin xa xb x

Theory of Aerospace Propulsion

33

THEORY OF AEROSPACE PROPULSION Chapter 10. Propellers

P. M. SFORZA University of Florida

momentum theory for propellers •The thrust is uniformly distributed over the disc •No rotation is imparted to the flow by the actuator disc •Streamtube entering and leaving defines the flow distinctly •Pressure far ahead and behind matches the ambient value streamtube V0

thrust Ve actuator disc

Theory of Aerospace Propulsion

2

control volume for actuator disc p

control volume

V0

p+Δp V0

Ve F r

R disc of area A inflow along horizontal boundaries Theory of Aerospace Propulsion

3

conservation of mass in the C.V. G ˆ =0 ∫∫ ρV ⋅ ndS S

− ρV0π R 2 + ρV0π ( R 2 − r 2 ) + ρVeπ r 2 − ρ Q = 0 Q = π r 2 (Ve − V0 ) Volume flow into the C.V. through the horizontal boundaries of the C.V.

Theory of Aerospace Propulsion

4

momentum conservation in the C.V. G G ∫∫ ρV ⋅ nˆ VdS = Ffluid

(

)

S

− ρV π R − ρV0Q + ρV π ( R − r ) + ρV π r = Ffluid 2 0

2

2 0

2

2

2 e

2

Ffluid = − ρπ r (V0Ve − V − V + V ) = − Fdisc 2

2 0

2 e

2 0

Fdisc = ρπ r Ve (Ve − V0 ) 2

momentum into C.V.

momentum out of C.V. force on the fluid

Theory of Aerospace Propulsion

5

Bernoulli’s eqn.‐ up and downstream disc

1 1 2 2 p0 + ρV0 = p + ρV1 Up: V1 2 2 p p+Δp 1 1 2 2 Down: pe + ρVe = p + Δp + ρV1 2 2 1 1 2 2 Pressure Δp = ρ (Ve − V0 ) = ρ (Ve − V0 )(Ve + V0 ) jump 2 2 Continuity requires that And therefore

π r2Ve=AV1 1 V1 = (V0 + Ve ) 2

Theory of Aerospace Propulsion

6

velocity induced by actuator disc V’=V1‐V0 is called the induced velocity and the thrust is

F = ρπ r 2Ve (Ve − V0 ) = ρ AV1 ⎡⎣ 2 (V1 − V0 ) ⎤⎦ F = 2 ρ AV ′ (V0 + V ′ )

Thrust = (mass through disc) (overall change in velocity)

Theory of Aerospace Propulsion

7

conservation of energy in the C.V. G G G 1 G G ∫∫S ρV ⋅ nˆ 2 V ⋅VdS = F ⋅V = Pfluid

(

)

2 2 ⎡ ⎛ ⎞ 1 r r Q 2⎤ 2 3 3 3 ρπ R ⎢ −V0 + ⎜1 − 2 ⎟ V0 + 2 Ve − 2 V0 ⎥ = Pfluid πR 2 R ⎝ R ⎠ ⎣ ⎦

The power absorbed by the fluid is

1 ⎡1 ⎤ 2 2 Pfluid = ρ AV1 (Ve − V0 ) = ⎡⎣ ρ AV1 (Ve − V0 ) ⎤⎦ ⎢ (Ve + V0 ) ⎥ 2 ⎣2 ⎦ Pfluid = FVavg Theory of Aerospace Propulsion

8

ideal propulsive efficiency In terms of the induced velocity the power absorbed is

⎛ V′⎞ P = FV0 ⎜1 + ⎟ ⎝ V0 ⎠ power required to keep V=V0 The ideal propulsive efficiency is then

Pi ηi = = P

1 V′ 1+ V0

Theory of Aerospace Propulsion

9

ideal propulsive efficiency Operating range 1.0 V0 V’

0.8

ηi

V1=Vavg=V0+V’ 0.4

0.2 0

0

0.1

0.2

0.3

Theory of Aerospace Propulsion

0.4

V’/V0 10

thrust coefficient ⎡ V ′ ⎛ V ′ ⎞⎤ 1 2 F = 2 ρ AV ′ (V0 + V ′ ) = ρV0 A ⎢ 4 ⎜1 + ⎟ ⎥ 2 ⎣ V0 ⎝ V0 ⎠ ⎦ The thrust coefficient, which is dimensionless, is defined as Disc loading

F A

V′⎛ V′⎞ CT = = 4 ⎜1 + ⎟ 1 V V 2 0 0 ⎠ ⎝ ρV0 2

V′ 1 = V0 2

( 1 + C − 1) T

Theory of Aerospace Propulsion

11

power coefficient ⎛ V′⎞ 1 P C P ,i = = CT ⎜ 1 + ⎟ = CT 1 + 1 + CT V0 ⎠ 2 ⎛1 2⎞ ⎝ ρV0 A ⎜ V0 ⎟ ⎝2 ⎠

(

)

NOTE: For a statically thrusting propeller V0=0 and the non‐ dimensional coefficients don’t apply. Instead,

Vs′ =

Fs 2ρ A

Ps =

Fs3 2ρ A

Theory of Aerospace Propulsion

12

thrust variation with flight speed Equating the power to the propeller for the moving and static cases yields:

⎛ F 1 2F ⎞ = FV0 ⎜⎜ 1 + ⎟⎟ 2ρ A 2 ρA ⎠ ⎝ 3 s

This equation may be put in the following form: 2 ⎡ V0 F⎤ 1 F V0 1= +4 ⎥ ⎢ + 2 Vs′ Fs ⎥⎦ 2 Fs ⎢⎣Vs′

Theory of Aerospace Propulsion

13

thrust variation with flight speed Note that thrust drops as speed increases 1.0 F/Fstatic V’static=(Fstatic/2ρA)1/2

0 0

V0/V’static Theory of Aerospace Propulsion

1 14

force and velocity on blade element dF = dL cos (φ + α i ) − dD sin (φ + α i )

dL

φ+αi

φ

αi

dF

V’

Ve

φ

VR

V0

ωr

dD

γ

Induced angle Axis of rotation

Chord line Theory of Aerospace Propulsion

15

the blade element for a propeller 1 dF = BcL ( cdr ) ρVe2 ⎡⎣ cos (φ + α i ) − ε sin (φ + α i ) ⎤⎦ 2 dr r

C(r) Axis of rotation

Blade element

Theory of Aerospace Propulsion

16

thrust and power for a propeller R 1 F = ρ B ∫ ccLVe2 cos (φ + α i ) ⎡⎣1 − ε tan (φ + α i ) ⎤⎦ dr 0 2

R 1 P = ρ B ∫ ccLω rVe2 sin (φ + α i ) ⎡⎣1 + ε cot (φ + α i ) ⎤⎦ dr 0 2

The induced angle αi depends on the induced velocity V’

Theory of Aerospace Propulsion

17

propeller characteristics Blade activity factor (AF) is a measure of solidity and therefore power absorption capability 3

10 c ⎛ r ⎞ ⎛ r ⎞ AF = ⎜ ⎟ d⎜ ⎟ ∫ 16 0 D ⎝ R ⎠ ⎝ R ⎠ 5 1

c(r)

r

constant chord blade

Bc −3 B ( AF ) σ= = 1.28 × 10 πR π

Typical range:

R

100<AF<150 Theory of Aerospace Propulsion

18

geometric pitch Advance ratio J= V0 /ωD

γ

Chord line

ωr

Axis of rotation Advance during 1 revolution

γ = blade pitch angle γ

2π r tanγ

2πr In‐plane distance moved in 1 revolution Theory of Aerospace Propulsion

19

coarse pitch operation V0

L R

ωr

F γ D

axis of rotation chord line

Theory of Aerospace Propulsion

20

fine pitch operation

L F

R

γ

ωr

V0

D

axis of rotation chord line Theory of Aerospace Propulsion

21

the two‐speed propeller

η

1

Coarse pitch Fine pitch 0 0

Advance ratio J Theory of Aerospace Propulsion

22

the constant speed propeller γ is varied to keep ω constant at best engine speed 1

η

γ1

γ2

γ3

γ4

0 0

Advance ratio J Theory of Aerospace Propulsion

23

blade pitch control Coarse (high) pitch for cruise

Fine (high) pitch for takeoff

Use of pitch control.

Credits - NASA

Theory of Aerospace Propulsion

24

contra‐rotating propellers

Theory of Aerospace Propulsion

25

contra‐rotating propellers

Theory of Aerospace Propulsion

26

nondimensional blade parameters Advance ratio; J=V/nD

Thrust coefficient: CT=F/ρn2D4 Torque coefficient: Cq=Tq/rn2D5 Power coefficient: CP=P/ρn3D5 Efficiency: η=JCT/CP Theory of Aerospace Propulsion

27

propeller performance map Power coefficient CP=P/ρω3D5, J=V0/nD, and γ3/4=pitch angle at3R/4 0.50 0.45 0.40 0.35 0.30 CP 0.25 0.20 0.15 0.10 0.05 0.00

γ 3/4=50

40 35

η=89%

30

88%

85%

25

0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3

J Theory of Aerospace Propulsion

28

radial variation of blade properties Thickness to chord ratio, section lift coefficient and pitch angle (degrees/50) 0.9 0.8 0.7 0.6

Cl

0.5

γ /50

0.4 t/c

0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

r/R

Theory of Aerospace Propulsion

29

twist of propeller blades VR,tip ωrtip

VR,hub

ωrhub

prop blade tip

airfoil sections prop hub

prop root Theory of Aerospace Propulsion

30

ducted propeller 0

ducted propeller disc of area A

e Ve

V0 Ae A0

duct wall

propeller streamtube

p pte p0

pt0 p0

p2 p1

Δp x

Theory of Aerospace Propulsion

31

ducted propeller

V0

Theory of Aerospace Propulsion

32

turboprop with free turbine

propeller

free turbine nozzle

gearbox

Outer driveshaft

Inner driveshaft 1

2

3

Theory of Aerospace Propulsion

4

5 5FT

6

33

free turbine cycle diagram pt,4 4

pt,5

h 5

pt,5FT 5FT 3

p6t,6=p0

2 1,0

s Theory of Aerospace Propulsion

34

advanced turboprops

Lockheed Martin C‐130J Super Hercules showing the Rolls‐Royce AE2100D3 engines which produce 3458kW each and drive the GE‐Dowty R391 4.1m diameter 6‐bladed composite propellers Theory of Aerospace Propulsion

35

geared turbofan VFAN

V6

outer driveshaft inner driveshaft

fan speed reduction gearbox

Theory of Aerospace Propulsion

36

geared turbofan Reduction gear ratio G as a function of fan, rotor, or prop diameter 16 turboprops

β 12 G 8 geared turbofans

4

open rotors

0 0

1

2

3

4

5

6

7

RF/RT

Theory of Aerospace Propulsion

37

open rotor or unducted fan swept fan blades inlet

fuel line

nozzle

outer driveshaft inner driveshaft

1

2

3

4

5

5B1

5B2

6

gearbox to reverse rotation

Theory of Aerospace Propulsion

38

open rotor or unducted fan

The GE 36 Unducted Fan engine was flight tested on Boeing B737 and McDonnell Douglas MD80 single‐aisle aircraft in the 1980s Theory of Aerospace Propulsion

39

THEORY OF AEROSPACE PROPULSION Chapter 11. Liquid Propellant Rocket Motors

P. M. SFORZA University of Florida

features of liquid propellant rockets high pressure nitrogen gas pressure regulator

fuel tank oxidizer tank

fuel tank

oxidizer tank vent turbo‐ pumps turbine

combustion chamber Combustion chambers are under high pressure, up to 200 atm in the SSME

nozzle

The nozzle is the major factor in performance.

Regenerative cooling protects nozzle and prepares the fuel for injection

cooled nozzle

(a) Pressure‐fed system

(b) Turbopump‐fed system

Theory of Aerospace Propulsion

2

performance coefficients for nozzles Nominal performance estimates are typically based on quasi‐one‐ dimensional flow and require correction for operational use

m cd = m i

= discharge coefficient =

Actual flow rate Ideal flow rate

0.97 < cd <1.15.

Ve cV = Ve,i

= nozzle coefficient=

Actual exit velocity Ideal exit velocity

0.88 < cV < 0.92 Theory of Aerospace Propulsion

3

nozzle expansion process mo mf Enthalpy, h pc

pe

hc actual

Ve2/2 he

Ve,i2/2

ideal

Ve cV = = ηn Ve ,i

hc − he ηn = hc − he′

he ’

Δs Theory of Aerospace Propulsion

Entropy, s 4

nozzle geometry expansion wall injector

Ru

subsonic flow

Rd

transonic flow

supersonic flow

0.6<Ru/Rt<1.5 SSME: Rt=10.3in Theory of Aerospace Propulsion

5

thrust coefficient

λ = (1+cosα)/2

Thrust

e + Ae ( pe − p0 ) F = λ mV ⎧ ⎡ ⎪ 2γ ⎢ ⎛ pe ⎞ F = λ cd cV Γ ⎨ 1− ⎜ ⎟ ⎢ ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎣ ⎩ Thrust coefficient

γ −1 γ

α 1/ 2

⎤⎫ ⎥⎪ ⎥⎬ ⎥⎦ ⎪⎭

⎧ ⎡ F ⎪ 2γ ⎢ ⎛ pe ⎞ cF = 1− ⎜ ⎟ = λ cd cV Γ ⎨ ⎢ pc At ⎪ γ − 1 ⎢ ⎝ pc ⎠ ⎣ ⎩

pc At + Ae ( pe − po )

γ −1 γ

1/ 2

⎤⎫ ⎥⎪ ⎥⎬ ⎥⎦ ⎪⎭

Theory of Aerospace Propulsion

Ae ⎛ pe p0 ⎞ + ⎜ − ⎟ At ⎝ pc pc ⎠ 6

ideal vacuum thrust coefficient launch engines

space engines

2.2 CF,vac,i

γ=1.1 1.2 1.4

1.8 ε=20

ε=100

ε=5

1.4

p0 c F = c F , vac − ε pc 1.0

10

100 pc/pe 1,000 Theory of Aerospace Propulsion

10,000 7

effect of γ on exit velocity 4.0 Ve/[RuTc/Wc]1/ 2

limit

3.0

limit

γ=1.2 γ=1.3 2.0 10

100

pc/pe 1,000

Theory of Aerospace Propulsion

10,000 8

characteristic velocity propellant performance

nozzle performance

* cF F = cF pc At = mc

c* = ac/Γγ1/2 =characteristic velocity~Tc1/2 ⎛ γ −1 ⎞ Γ = γ ⎜1 + ⎟ 2 ⎝ ⎠

−

γ +1 2( γ −1)

=0.192γ+0.417

ac=(γRuTc/Wc)1/2 =chamber sound speed

Theory of Aerospace Propulsion

Tc/Wc should be large for best performance

9

specific impulse F Ve Ae I sp = = + ( pe − po ) m e g g

Definition of specific impulse

Ve ⎡ ⎛ p0 ⎞ 1 ⎤ I sp = ⎢1 + ⎜ 1 − ⎟ 2⎥ g ⎣ ⎝ pe ⎠ γ e M e ⎦ Ae 1 ⎡ 2 ⎛ γ − 1 2 ⎞⎤ ε= = M e ⎟⎥ ⎜1 + ⎢ 2 At M e ⎣ γ + 1 ⎝ ⎠⎦

Specific impulse depends on Ve and pe γ +1 2( γ −1)

Nozzle area ratio for gas with constant γ

γ +1

⎛ γ + 1 ⎞ 2 γ −1 M ≈⎜ ⎟ ε ⎝ γ −1 ⎠ 2 e

Limit for Me>>1

Theory of Aerospace Propulsion

10

expansion ratio ε vs. Me for various γ space engines 250

γ = 1.1 1.2 1.3

1.4

Area Ratio,

= Ae/At

200 150 100

launch engines

50 0 0

2

4

6

8

10

Exit Mach Number, Me

Theory of Aerospace Propulsion

11

altitude variation of specific impulse Ve ⎛ 1 ⎞ I sp ,vac = ⎜1 + 2 ⎟ g ⎝ γ Me ⎠

When p0=0

I sp

p0 1 1 p0 = 1− ≈ 1− ≤1 2 2 I sp ,vac γ M e pe 1 + γ M e pe

⎛ z⎞ exp ⎜ − ⎟ = 1− I sp ,vac pc ⎝ zs ⎠ I sp

εΓ′ γ

1 ⎛ γ − 1 ⎞ γ −1 ⎛ γ + 1 ⎞ Γ′ = ⎜ γ ⎝ 2 ⎟⎠ ⎜⎝ γ − 1 ⎟⎠

Exponential atmosphere

γ +1 2( γ −1)

Theory of Aerospace Propulsion

12

altitude variation of specific impulse When p0=0

Ve ⎛ 1 ⎞ I sp ,vac = ⎜1 + 2 ⎟ g ⎝ γ Me ⎠

I sp

p0 1 1 p0 = 1− ≈ 1− ≤1 2 2 I sp ,vac γ M e pe 1 + γ M e pe

Theory of Aerospace Propulsion

13

altitude variation of specific impulse High ε

3.5

Low ε

3 2.5

ε= 28.4

2

Isp/aγ 1.5

116.3 228.1

1 0.5 0 0

50

100

150

200

250

altitude (kft)

Fixed γ=1.2 and chamber pressure pc=200 atm for different expansion ratios, ε. The specific impulse is normalized to the product of γ and chamber sound speed a.

Theory of Aerospace Propulsion

14

comparing expansion ratios 1. ε=10, Me=3.92, pe=7.33 psia, Te=1227R: 2. ε=25, Me=5.00, pe=1.89 psia, Te=833R: 230 220 210

Isp

200

Nozzle 1, ε =10 Nozzle 2, ε =25

190 180 170 160 150 0

50

100

150

200

250

altitude (kft) Theory of Aerospace Propulsion

15

two‐position bell nozzle

Position 1 Low altitude Low ε

Position 2 High altitude High ε

Theory of Aerospace Propulsion

16

characteristics of propellants Type

Propellant

Isp,vac F (s) (lbs)

Cold gas

N2, NH3, He, Freon

50‐ 75

Liquid mono‐ propellant

H2O2 N2H4

150 200

Liquid Bipropellant

RP-1/O2

270‐ 1 to 360 106

1

Liquid Bipropellant

H2/O2

360‐ 1 to 450 106

1.26 very high performance

cryogenic, complicated

Liquid Bipropellant

UMDH/N2O4

270‐ 1 to 340 106

1.14 storable, good

complicated

Liquid Bipropellant

N2H4/F2

425

1.1

advantages

disadvantages

0.01 to 50

simple reliable low cost

low performance, high weight

0.01‐ 0.1

1.46 simple reliable low cost 1

low performance, higher weight

1 to 106

SG

high performance

complicated

performance very high performance

Theory of Aerospace Propulsion

toxic, dangerous, complicated

17

performance of fuel‐oxidizer combinations Tc (R) γ

Oxidizer Fuel

O/F

c* (fps)

LOX

LH2

LOX

RP‐1

LF2

LH2

4.0 5375 1.26 7985 φ~1.3 2.45 6605 1.24 5915 φ~1.4 8.0 7130 1.33 8355

LF2

RP‐1

2.8

7940 1.22 6390

Isp Isp (equili‐ (frozen) brium) 391

388

301

286

410

398

326

304

These values are for pc = 1000 psi and po = 14.7 psi Theory of Aerospace Propulsion

18

cryogenic propellants Name

Type SSME

Boiling Density Molecular ΔHv kJ/kg point, K kg/m3 weight

Liquid Oxygen LOX

oxidizer 90

114

32.00

213

Liquid Hydrogen LH2 Liquid Methane LCH4 Liquid Fluorine LF2

fuel

20.3

69.5

2.01

450

fuel

111.5

162

16.03

578

oxidizer 84.7

151

38.00

172

Theory of Aerospace Propulsion

19

hypergolic propellants Name

Type Formula

1011 1450

32.05 92.02

F O

N2H4 N2O4

F

CH3NHNH2

360

221

880

46.03

F

(CH3)2NNH2

336

221

790

60.04

O

HNO3 +13% N2O4+3%H2O

216

358

1550

63

Orbiter OMS and RCS Hydrazine Nitrogen Tetroxide Monomethyl‐ hydrazine(MM) Unsymmetrical‐ dimethylhydra‐ zine(UMDH) Red fuming nitric acid (RFNA)

W ρ (kg/m3)

Boil Freeze point point (K) (K) 386 274 294 262

Theory of Aerospace Propulsion

20

expansion ratio

expansion ratios for liquid rockets LH2‐LOX

90 80 70 60 50 40 30 20 10 0

CH4/LOX RP‐1/LOX

0

100

200

300

400

500

specific impulse Isp (sec)

Theory of Aerospace Propulsion

21

chamber pressures in liquid rockets

600

LH2/LOX

500 400

Isp(sec) 300

CH4/LOX

200

RP‐1/LOX

100 0 0

500

1000

1500

2000

2500

3000

3500

chamber pressure pc (psia) Theory of Aerospace Propulsion

22

the rocket equation and mass fraction F=(dm/dt)Ve

⎛ dm p ⎞ ⎛ dV ⎞ ⎜ ⎟ Ve − mg = m ⎜ ⎟ ⎝ dt ⎠ ⎝ dt ⎠ m = m p + me

dm p dm =− dt dt dm Ve − gdt = dV − m m0 Vb −o = Ve ln − gtb −o V at burn-out me mg

Vb −o = ln Ve

tb − o 1 − m p I sp 1− m0 Theory of Aerospace Propulsion

propellant mass fraction 23

nozzle weight is important For high V,b‐o mp/m0

1. But m0=mp + me, so me is important also

me=mstructure + mengine + mpayload nozzle, powerhead, TPS tankage, feed systems, etc

F‐1 engine

Nozzle weight~ length

Ve

short long Theory of Aerospace Propulsion

24

weight of LH2/LOX rocket engines

Predicted Weight (lbs)

F/W=10.5F0.144 : economy of scale – large rockets weigh less per pound of thrust 9000 8000 7000 6000 5000 4000 3000 2000 1000 0

LH2/LOX W=0.095F0.856

0

2000

4000

6000

8000

Actual Engine Weight (lbs) Theory of Aerospace Propulsion

25

weight of RP‐1/LOX rocket engines

Predicted Weight (lbs)

F/W=14.3F0.144 : economy of scale – large rockets weigh less per pound of thrust 20000

RP‐1/LOX 15000

W=0.07F0.856

10000 5000 0 0

5000

10000

15000

20000

Actual Engine Weight (lbs) Theory of Aerospace Propulsion

26

propellant injectors O

O

O

F

F

F

O

O

O

F

F

F

(a) showerhead

(b) impinging jet

Theory of Aerospace Propulsion

(c) splash plate

27

thrust vectoring s=yδ/cosθ

gimbal axis

0

actuator y

θ yδ

nozzle

δ Theory of Aerospace Propulsion

28

liquid propellant tanks Propellant tanks are subject to both static and dynamic loads.

The main static loads are as follows: •Hydrostatic pressure – due to the state of the fluid contained •Overpressure‐ sometimes used to avoid cavitation in the fuel pumps •Payload weight – the tanks generally also serve as structural members The dynamic loads are as follows: •Inertia loads – due to vehicle acceleration •Control loads – due to thrust vectoring induced bending •Aerodynamic loads – due to vehicle geometry and wind shear Theory of Aerospace Propulsion

29

tank configuration inter‐stage structure

oxidizer pressurant source

fuel turbopump

thrust mount nozzle

oxidizer tank

fuel tank

powerhead

outer skin structure

fuel pressurant source

oxidizer turbopump

aft skirt structure

General layout of propellant tank system for a large lower stage liquid bipropellant rocket Theory of Aerospace Propulsion

30

fuel tank stress analysis Hoop stress in cylindrical tank p(2RdL)

pπR2

t

dL

R

σH tdL

σH tdL

pπR2 2πRσAt

2πRσt Maximum stress at equator of spherical tank

Axial stress in spherically capped cylindrical tank Theory of Aerospace Propulsion

31

allowable wall thickness for tanks Spherical tank

2π Rsσ s t = π Rs2 p 1 p ts = Rs 2 σs

Spherically capped cylindrical tank

Rc σH = p tc 1 Rc 1 σA = p = σH 2 tc 2 tc = Rc

p

σ

L

= 2ts

Theory of Aerospace Propulsion

32

volume and weight of tanks 4 Vs = π Rs3 3 Ws = 2π R ρ 3 s

p

Spherical tank

σ

Specific yield stress

4 3 2⎞ 2 3⎛ L Vc = π Rc + π Rc ( L − 2 Rc ) = π Rc ⎜ − ⎟ 3 ⎝ Rc 3 ⎠ ⎛ L ⎞ p Ws = 2π R ⎜ − 1⎟ ρ ⎝ Rc ⎠ σ 3 c

Spherically capped cylindrical tank

Theory of Aerospace Propulsion

33

ultimate strength of tank materials 750

50 psia tank

σ/ρ (10‐3 lb‐in)

steel

500

Steel t=0.0167 in Aluminum t=0.0385 in

250 Aluminum 0 0

500

Theory of Aerospace Propulsion

1000 T (F)

34

comparison of equal volume tanks L −1 Wc 4 Rc = Ws 3 L − 2 Rc 3 L/Rc=2

L 2< <∞ Rc Wc 4 < 1< Ws 3 L/Rc=8

Theory of Aerospace Propulsion

35

weight and cross‐section of tanks of equal volume 1.4 1.2

Wc/Ws

1 0.8 0.6 0.4

Rc/Rs

0.2 0 0

2

4

6

8

10

L/Rc Theory of Aerospace Propulsion

36

dynamic load effects Spherical and L/Rc=8 tank of equal volume Sphere has greater drag

Acceleration causes an increase in hydrostatic pressure proportional to the load factor n Acceleration=ng

Rs~2Rc Drags~4Dragc

Cylinder has Increased hydrostatic pressure

Theory of Aerospace Propulsion

37

effect of attachments The method of securing the tanks and the associated fittings must be considered t

Theory of Aerospace Propulsion

38

weight with additional effects Simple pressure vessel

Wc/Ws

With stage attachments and dynamic loads

1

Optimal range

With stage attachments

0 0

2

4

6

8

Theory of Aerospace Propulsion

10

L/Rc

39

buckling of long thin‐wall cylinders Empirical relation for compressive buckling stress 1.6

⎛ t ⎞ ⎛σ ⎞ ⎛t⎞ = 9 ⎜ ⎟ + 0.16 ⎜ ⎟ ⎜ ⎟ ⎝ E ⎠initial ⎝L⎠ ⎝ Rc ⎠

1.3

For steel with E=3x107psi, L=2R=100in, and t=0.0167in, σ= 822psi For aluminum, σ=1019psi To improve performance use: •Rings and stringers •Internal pressurization Theory of Aerospace Propulsion

40

pressurization to reduce buckling Gas at pressure p = p′ + ρ p ngx p’ 0

p′ + ρ p ngx t t p = = = 2 Rc D 2σ 2σ L

x

Wc = ρ ∫ π Dtdx 0

Liquid propellant with density ρp Weight of cylindrical section Theory of Aerospace Propulsion

41

weight of cylinders and end caps Integration yields

2 ⎛ ⎞ ρ ngL ′ L p p 2 + Wc = π D ρ ⎜ ⎟⎟ ⎜2σ 4σ ⎠ ⎝

p′ = 4σ D ts ,lower p′ + ρ p ngL = 4σ D ρ p ngL ⎞ 1 2 3 ⎛ p′ Ws = ρπ D ( ts ,upper + ts ,lower ) = ρπ D ⎜ + ⎟ 2 8σ ⎠ ⎝ 4σ ts ,upper

Theory of Aerospace Propulsion

42

tank to propellant weight ratio propellant type

internal tank pressure (psia)

propellant specific gravity

Wtank/Wpropellant

liquid

50

1

0.005

solid

400

1

0.04

liquid

50

0.3

0.0167

solid

1000

1.8

0.056

Theory of Aerospace Propulsion

43

propellant feed systems gas generator

oxidizer

fuel

pump

Gas generator cycle

turbines

fuel

pump

oxidizer pump

pump

turbine exhaust

nozzle

turbine

Expander cycle

nozzle

circulating fuel Theory of Aerospace Propulsion

44

propellant feed systems (concl.) oxidizer

fuel

turbopumps

nozzle

The staged combustion cycle propellant feed system. The scale of the pre‐burners is exaggerated for clarity

Theory of Aerospace Propulsion

45

exercise E11.5 An RP‐1/LOX propellant combination (O/F=2.28) in a rocket motor produces F=49.1kN at s. l. with pc=20 atm, Tc=3430K. W= 21.98 and γ =1.226 are assumed constant throughout the expansion. (Note that the actual values will change somewhat as the flow proceeds down the nozzle and the pressure and temperature drop) Determine: (a) Ae and At for matched nozzle exit flow at sea level conditions, (b) c* and Isp at sea level, (c) the evolution of p, T, M, and V along the nozzle. Using the final dimensional results also draw the nozzle, (d) Also find the residence time of a fluid particle in the nozzle.

Theory of Aerospace Propulsion

46

exercise E11.5 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.

(a)Assuming cd = cV = l =1 we have At=F/cFpc and for an ideal nozzle matched at pe/pc=1/20 and with Γ=0.65 we have 1/ 2 γ −1 ⎧ ⎡ ⎤⎫ 0 γ ⎛ ⎞ ⎛ ⎞ p A p p F 2 γ ⎪ e 0 ⎢1 − e ⎥⎪ + e cF = = λ cd cV Γ ⎨ − ⎬ ⎜ ⎟ =1.4 ⎢ ⎜⎝ pc ⎟⎠ ⎥ γ 1 pc At A p p − t ⎝ c t ⎠ ⎪ ⎪ ⎢ ⎥ ⎣ ⎦⎭ ⎩ Therefore At=174cm2. The exit area is given by

pc At Γ ) ( pc At Γ 2 m m = = = Ae = ρeVe ρ e F ρ e RTc ( cF pc At ) ρ e RTc cF 2

2

Theory of Aerospace Propulsion

47

exercise E11.5 (a) For ideal nozzle flow (ρc/ρe)=(pe/pc)1/γ therefore Ae=605cm2. (b) The characteristic velocity c*=(RTc)1/2Γ-1= 1752m/s and the specific impulse at sea level is Isp = c*cF/g=250s (c) the evolution of p, T, M, and V along the nozzle 25

r (cm)

20 15 10 5 0 -20

-10

0

10

20

30

x(cm) Theory of Aerospace Propulsion

48

exercise E11.5

2.5 2 1.5

p (Mpa) 1 0.5 0 -20

-10

0

10

20

30

x (cm)

Theory of Aerospace Propulsion

49

exercise E11.5

4000 3500 3000 2500

T (K) 2000 1500 1000 500 0 -20

-10

0

10

20

30

x (cm)

Theory of Aerospace Propulsion

50

exercise E11.5

3 2.5 2

M 1.5 1 0.5 0 -20

-10

0

10

20

30

x (cm)

Theory of Aerospace Propulsion

51

exercise E11.5 3000 2500 2000

V (m/s) 1500 1000 500 0 -20

-10

0

10

20

30

x (cm)

(d) Integrating dx/V to find the residence time t yields t~1ms Theory of Aerospace Propulsion

52

THEORY OF AEROSPACE PROPULSION 12. Solid Propellant Rocket Motors

P. M. SFORZA University of Florida

solid rocket motor components

Motor casing

Igniter

Burning surface Throat

Nozzle Solid propellant Combustion chamber

Theory of Aerospace Propulsion

2

attributes of solid propellant rockets Advantages: •Simplicity – no moving parts, no injection system, no need to be fueled prior to launch •Reliability – high because of low part count and no moving parts •Payload ratio – high because of lack of subsystems

Disadvantages: •Performance – lower than liquid propellants •Controllability – no accurate thrust modulation or shut‐off •Temperature sensitivity Theory of Aerospace Propulsion

3

fundamental parameter: burning rate The burning rate is the recession rate of the surface r=Δz/Δt (cm/s) original surface

r

Δz

r = ap

n c

Surface after time Δt pc = chamber pressure n = propellant characteristic independent of propellant temperature Tp a = temperature dependent function which is characteristic of the propellant Theory of Aerospace Propulsion

4

burning rate varies with temperature The hotter the propellant the faster the burning rate

Limited range of pc log r

Tp increasing

log r = log a + n log pc log pc Theory of Aerospace Propulsion

5

the burning rate equation A = ao ⎡⎣τ (Tp ,i − Tp ,i o ) ⎤⎦ a= T1 − Tp A T1 ao Top,i Tp,i

τ

= constant = self‐ignition temperature = constant = a at Tp = Tpi o = reference temperature, say 15oC = initial temperature of propellant = constant

Theory of Aerospace Propulsion

6

solid propellant grains • Homogeneous or double base propellant: molecule contains both fuel and oxidizer (e.g., nitroglycerin and nitrocellulose on approximately equal amounts).

• Heterogeneous or composite propellant: a combination of oxidizer (e.g., 70 to 90% ammonium perchlorate) mixed in a fuel binder (e.g., 10 to 30% polyethylene) with a metal (e.g., 10 to 20% aluminum) for additional energy release Theory of Aerospace Propulsion

7

composite propellant performance Propellant2

rp

Tc

(kg/m3) (K)

Wc

c*

Te

(m/s)

(K)

We

Isp,max3 (s)

0.7NH4ClO4, 0.3(CH2)x

1460

1312 18.41 1190

883

20.97 200

0.8NH4ClO4, 0.2(CH2)x

1590

2036 20.43 1406

924

20.99 221

0.9NH4ClO4, 0.1(CH2)x

1750

3013 26.13 1513

1615 26.85 252

0.81NH4ClO4, 0.09(CH2)x, 0.1Al

1820

3433 28.54 1546

2147 29.78 261

0.72NH4ClO4, 0.08(CH2)x, 0.2Al

1890

3784 31.17 1552

2490 33.21 266

1Siegel and Schieler (1964), 2mass ratios, 3expansion from p =68atm to c

Theory of Aerospace Propulsion

pe=1atm 8

propellant grain configurations STS ISRM has 11‐pointed star transitioning into cylindrical grain

Cruciform grain

Star grain

Cylindrical grain

Theory of Aerospace Propulsion

End‐burning grain

9

propellant grain and burning rate plateau‐burning grain log r

mesa‐burning grain log r

log pc

log pc Theory of Aerospace Propulsion

10

combustion chamber volume An end‐burning grain is shown for illustration Propellant of density ρp

Propellant burning area Ab

Surface recession rate r

Throat area At

Combustion chamber volume υc(t)

Theory of Aerospace Propulsion

11

mass conservation in the motor ∂ ˆ =0 ρ dυ + ∫ ρV ⋅ ndA ∫ ∂t υ A

Conservation of mass equation

dυ p ∂ ∂ ∂ + ρ cυc ρ p dυ + ∫ ρc dυ = ρ p ∫ dt ∂t υ p ∂t υc ∂t

pc At

ˆ =m = Γ ∫ ρV ⋅ ndA RT t

A

c

⎛ 2 ⎞ Γ= γ⎜ ⎟ 1 γ + ⎝ ⎠

L(t) Theory of Aerospace Propulsion

γ +1 γ −1

12

mass conservation (cont.)

The propellant volume υp=AbL(t),

∂υ p

∂υc dL = Ab = − rAb = − dt ∂t ∂t

The mass conservation then becomes

Chamber volume increases as propellant volume decreases

pc ∂ − ρ p rAb + ( ρ cυc ) + Γ=0 ∂t RTc

∂ρ c ∂υc ∂ρ c υc + ρc = υc + ρ c rAb ∂t ∂t ∂t Theory of Aerospace Propulsion

13

pressure variation Combining terms transforms the mass conservation equation into

υc ∂pc

( ρ − ρ ) rA + RT ∂t + RT Γ = 0 c

p

pc A

b

c

c

This assumes Tc=constant and leads to

⎤ dpc ⎡ RTc ⎤ n ⎡ RTc At Γ ⎥ pc = 0 + ⎢( ρ c − ρ p ) aAb ⎥ pc + ⎢ dt ⎣ υc ⎦ ⎢⎣ υc ⎥⎦ t

t

υc ( t ) = υ0 + ∫ rAb dt = υ0 + ∫ Ab ap dt n c

0

Theory of Aerospace Propulsion

0

14

steady‐state operation The transient requires solution of an integro‐differential equation. For the steady state dpc/dt=0 and the equilibrium value of pc is

⎡ Ab pc = ⎢ ⎢⎣ At

RTc Γ

⎤ a ( ρ p − ρc )⎥ ⎥⎦ c* =

1 1− n

RTc Γ

⎡ A a ( ρ p − ρc ) ⎤ b ⎢ pc = c *⎥ 1− n ⎢⎣ At ⎥⎦

1 1− n

Theory of Aerospace Propulsion

Typical values of n: 0.4 <n <0.85 1.67 < (1‐n)‐1 <6.67 15

gas generation rate for different n Mass flow rate

m ∼ pcn for n<1 due to gas generation

Choked throat, m ∼ pcn Due to gas leaving through throat

Operating point:

min = mout

m ∼ pcn for n>1 due to gas generation Chamber pressure, pc Theory of Aerospace Propulsion

16

pressure stability of a solid rocket motor Mass flow rate

m ∼ pcn for n >1: unstable m ∼ pcn for n<1: stable

Δm > 0

Δm < 0

operating point

Δp chamber pressure pc Theory of Aerospace Propulsion

17

solid rocket motor thrust Practical stable propellants have 0.4<n<0.85 The exponent in the pc equation is then 1.667<(1‐n)‐1<6.667 .

⎡ Ab ⎤ pc = ⎢ a ρ p c *⎥ ⎣ At ⎦

1 1− n

F = cF pc At

Theory of Aerospace Propulsion

18

burning to throat area ratio fixes thrust Ab/At is often given the symbol K

Ti increasing log Ab/At

n=0.4 Ti increasing n=0.7 log pc Theory of Aerospace Propulsion

19

temperature sensitivity coefficients 1 ⎛ dpc ⎞ (π p )k = p ⎜ dT ⎟ c ⎝ i ⎠k

The important design parameter is the ratio of burning area to throat area k=Ab/At and 100<k<2000

1 ⎛ dr ⎞ 1 =τ (π r ) p = ⎜ ⎟ = r ⎝ dTi ⎠ p Tign − Ti

1 ⎛ dr ⎞ (π r )k = ⎜ ⎟ r ⎝ dTi ⎠k

1 (π p )k = (π r )k = 1 − n (π r ) p Theory of Aerospace Propulsion

20

transient operation: start‐up υc

dpc dt = 2 * Γ c At pc ( ρ p c*kapcn −1 − 1) The igniter produces an initial pressure rise so rapid that υ~υc,i, c*=ceq* and is constant, bringing the pressure to a level pc,ign which is high enough for the grain to light and burn. Then

1− n ⎡ ⎤ ρ ac * k p − 1 p c ,ign t= ⎢ 2 1− n ⎥ 1 − n Γ c * At ⎢⎣ ρ p ac * k − pc ⎥⎦

υ c ,i

Theory of Aerospace Propulsion

21

transient operation: burn‐out The pressure drop off at burn‐out may be determined by assuming T~const which leads to

dpc pc At c * Γ 2 =− υc dt Integrating this equation yields

pc Γ 2 c * At =− ( t − tb − o ) pc ,eq υ final

Theory of Aerospace Propulsion

22

Space Transportation System ISRB

Theory of Aerospace Propulsion

23

STS ISRB Components

Theory of Aerospace Propulsion

24

Space Transportation System ISRB

SRB Thrust (lbs)

3 million lbs 100 sec 2 million lbs

Mission elapsed time (sec) Theory of Aerospace Propulsion

25

Shuttle SRB propellant composition Ammonium perchlorate (NH4ClO4) – 70% Polybutadiene acrylonitrile (PBAN) – 14% Aluminum powder – 16% Iron oxidizer powder – 0.07% (catalyst) Propellant mass fraction =88% Ammonium perchlorate produces HCl in the combustion products and forms a white cloud when exhauseted into even mildly humid air.

PBAN is a polymeric rubber‐based binder that alkso serves as fuel Aluminum is added to enhance thrust (HV=32 MJ/kg) Theory of Aerospace Propulsion

26

example: tubular grain design Requirements: •Sea level thrust F=4900N with a burn‐out time tb‐o=0.5s, •Chamber pressure pc=9.81MPa with an internal radius ri =3.2cm Propellant: Double‐base JPN (Nitrocellulose and nitroglycerin) •Burning rate r=(2.17x10‐9) p0.69 m/s @15C, (p in Pa) •Combustion temperature Tc=3125K •Specific heat ratio γ=1.215 and molecular weight W=26.4 •Propellant density rp=1620kg/m3 Design the grain dimensions

Theory of Aerospace Propulsion

27

Ro

example: tubular grain design (cont.) Γ=0.651

Ri

cF,i= 1.632 dc=6.4cm

(a) Find nozzle throat area: consider a matched nozzle with pe=p0=101kPa. Further assume cv=95% and find At=F/cvpccF,I=4900/(0.95*9.81x106*1.632)=3.22x10‐4m2 (b) Find burning area: c*=(RTc)1/2/Γ=(3125*8315/26.4)1/2/0.651 c*=1523m/s. Eq. (12.15) says Ab/At=pc1‐n/aρpc*=pc/rρpc* Ab/At=(9.81x106N/m2 )/(0.0213m/s*1620kg/m3 *1523m/s)=186.8 Ab=0.06m2 (c) Find grain dimensions: tb‐o=0.5s and r=.0213m/s so the grain thickness (plus 5% allowance) is ΔR=2(1.05)rtb‐o=2.24cm=Ro‐Ri Then since a common rule is RoRi/(Rc2‐Ro2)=1, i.e., Ap/Ab = const we have Ro=2.89cm and therefore Ri=0.65cm Theory of Aerospace Propulsion

28

example: tubular grain design (concl.) (c) The length of the grain is determined from 2πL(Ro+Ri)=Ab and L =(600cm2)/2π(2.89cm+0.65cm)=26.98cm

30cm

7cm

7.4cm

6.5cm

36cm (d) Exit area Ae/At=(Γ2/cF)(pe/pc)‐1/γ=(0.6512/1.632)(101.3/9815)0.823 Ae=11.2At=3.61x10‐3m2=36.1cm2 and de=6.78cm Theory of Aerospace Propulsion

29

comment on exercise E12.3(a) ⎡ Ab ⎤ pc = ⎢ a ( ρ p − ρ c ) c *⎥ ⎣ At ⎦

1 1− n

1− n c

p

Ab pc pc * aρ pc = n = = At pc r a

Assuming that ρc<<ρp we may rewrite this as follows: pc=Krρpc* where K=Ab/At. Then Ab=AtK=Atpc/(r ρp c*)=(2.191x10‐4m2) (135.5x101kN/m2)/rρp c* Ab=(3kN)/rρp c*.

At Ti=15C the burning rate is r={0.102exp[0.0015(Ti‐15)]}p0.745 r=0.102(135.5)0.745=3.953cm/s=0.03953m/s c*={105,100exp[0.00021(Ti‐15)]}p0.015 =11.31x104cm/s=1131m/s Theory of Aerospace Propulsion

30

hybrid rocket motors solid fuel

fuel surface nozzle

liquid oxidizer throttling valve

injector

Hybrid rockets occupy a middle ground between liquid propellant and solid propellant rockets: they use a solid fuel but a liquid oxidizer in the manner illustrated schematically above Theory of Aerospace Propulsion

31

hybrid rocket motors centerline gas flow

oxidizer spray

combustion layer vaporized fuel layer solid fuel

Hybrid rocket motors are easier to handle, like solid rockets, yet more controllable, like liquid rockets. Although hybrid rockets are designed to combine the best attributes of pure solid or pure liquid rockets, there are still hurdles to overcome.

Theory of Aerospace Propulsion

32

THEORY OF AEROSPACE PROPULSION Chapter 13. Space Propulsion

P. M. SFORZA University of Florida

spacecraft mass breakdown Initial mass payload mass propellant mass propulsion system mass structural mass

M 0 = M pay + M pro + M eng + M str 1 = μ pay + μ pro + μeng + μ str Theory of Aerospace Propulsion

2

rocket thrust, impulse, and power Thrust (N)

F = m pVe

Impulse (s)

Fdt V F ∫ I= = = ∫ m g dt m g g

e

p

Power in exhaust (kW)

0

p

0

0

⎛ 1 2 ⎞ g 0 FI Pe = m p ⎜ Ve ⎟ = ⎝ 2 ⎠ 2000 Theory of Aerospace Propulsion

3

specific impulse and specific power Propulsion system mass (kg) Specific mass (kg/kW) conversion efficiency prime power (kW)

M eng = αη P Power in exhaust (kW) Jet conversion efficiency

Pe = η ' (η P ) Theory of Aerospace Propulsion

4

thrust to weight ratio The propulsion system is characterized by its thrust to weight ratio

F F 2000η ' = = 2 Weng M eng g 0 g0α I F 1 ∼ Weng α I Theory of Aerospace Propulsion

5

mass ratios and rocket burn time =O(1) because μpl, μs <<1

Ft p g 0α I F 1 = μ pay + + + μ str 2000η ' M 0 g 0 IM 0

⎛ 1 − μeng ⎞ α I Then the rocket burn time is t ≈ ⎜⎜ ⎟⎟ p 20 ⎝ μeng ⎠ 2

Theory of Aerospace Propulsion

6

propulsion systems characteristics 105 Type II Nuclear electric

104 Type I

I (s)

Gas core fission

103

102

Solar electric

Solid core fission Chemical F/W0=10 1 10‐3

10‐2

10‐1

10‐1 1 α (kg/kW) Theory of Aerospace Propulsion

10‐2 10

10‐3 102 7

electric propulsion “The acceleration of gases for propulsion by electric heating and/or by electric and magnetic body forces.”

The general classes of systems for electric propulsion are: •Electrostatic propulsion devices •Electrothermal propulsion devices •Electromagnetic propulsion devices.

Theory of Aerospace Propulsion

8

electrothermal rockets Electrothermal rocket engines are very similar in principle to chemical and nuclear thermal rockets, differing only in using electrical heating to raise the temperature of the propellant prior to accelerating it in a nozzle.

•Electric resistance heating, as in the “resistojet” •Heating of the propellant by a high energy arc discharge passing through it, as in the “arc jet” •Heating by passing radio frequency (RF) electromagnetic waves through the propellant to heat it.

Fairly high thrust (F) at reasonably high Isp, but thermal limitations are the same as in chemical and nuclear thermal rocket engines.

Theory of Aerospace Propulsion

9

resistojet Type

Propellant Energy

Isp,vac (s) Thrust Density (N) (g/cc) 150‐700 0.005‐0.5 0.28, 0.60, 1.0, 0.019

Resisto N2, NH3, N2O4, H2 ‐jet

Resistive η =0.9

valve

heat exchanger

electric power supply

Theory of Aerospace Propulsion

nozzle

10

resistojet The SSTL* Low Power Resistojet is designed for low‐cost small satellite applications such as orbit correction and station keeping.

It can be used to augment Isp achieved from a compressed gas or a liquefied gaspropulsion system. m=65g φ14mm

56mm

*Surrey Satellite Technologies Ltd Theory of Aerospace Propulsion

11

resistojet • Propellant: Nitrogen, Xenon,butane & most gases • Thrust: up to 100 mN • Feed pressure: up to 10 bar • Operation temperature to 500°C • Redundant heaters • K type thermocouple to monitor chamber temperature • Supplied with heat shield • Measured performance Xenon – 55 sec N2 – 99 sec Butane – 100 sec

Theory of Aerospace Propulsion

12

resistojet Enthalpy of hydrogen and the effect of equilibrium dissociation 300

Velocity in nozzle:

atomic

h (MJ/kg)

p = 10‐6

10‐4

10‐2

1 atm

u = [2(ht‐h)]1/2 ht=const

200

heating (dissociation)

cooling, (recombination)

100

molecular 0

0

1000

2000

3000 T(K)

4000

Theory of Aerospace Propulsion

5000 13

resistojet Effect of nozzle length on recombination L1

L2

Ve

Ve

Damkohler number: τf / τc=(L/Ve)/τc Longer nozzle gives more time for recombination to occur.

Downside is that there is more frictional loss Theory of Aerospace Propulsion

14

arcjet Type

Propell‐ Energy ant Resisto N2, Resistive jet NH3, heating N2O4, η=0.9 H2 Arcjet NH3, Arc heating H2, η=0.3 N2H4

Isp,vac (s) 150‐ 700

Thrust (N) 0.005‐0.5

Density (g/cc) 0.28, 0.60, 1.0, 0.019

450‐ 1500

0.05‐5

0.60, 0.019 1.0

Theory of Aerospace Propulsion

15

electrothermal : arcjet propellant tank

electric power supply

pump positive electrode

~

Arc accelerated plasma

negative electrode propellant cooled chamber wall radiator

circulating nozzle wall coolant pump Theory of Aerospace Propulsion

16

arcjet nozzle block temperatures

anode attachment zone

propellant flow

2300K

1800K 1900K 2000K

100K

Theory of Aerospace Propulsion

2200K

17

space experiment: ESEX arcjet Rocket Research, now Aerojet General, manufactured this arcjet using ammonia as the propellant to produce F=2.00 N (0.40 lbf) with Isp: 800 sec.

ESEX was the first High Power Electric Propulsion flight operating on 26 kW to the thruster. At the time (1999) this was the highest power subsystem ever flown in space. The ammonia fuel was fed at 0.25mg/sec steady state. Design life of the engine was 1500 hours and 400 on/off cycles.

Theory of Aerospace Propulsion

18

electrostatic rockets The temperature limitations of electrothermal rockets may be avoided if the acceleration of the propellant is achieved by electric body forces. The ion rocket accomplishes this by using •An ion source to produce a stream of positively charged particles •A negatively charged grid electrode to electrostatically accelerate the ions •An electron source to neutralize the accelerated ions Thus there is no physical nozzle or pressure chamber and the only temperature limitations are on the ion source device.

Theory of Aerospace Propulsion

19

ion rocket capabilities electric field E=dV/dx

V=V0 x=0

V=0 x=xa Ua

Ion stream

Ion source

ion mass flow thrust

accelerating grid

m = N a MU a A F M 8 ⎛ V0 ⎞ = mU a = j U a = ε 0 ⎜ ⎟ 9 ⎝ xa ⎠ A q Theory of Aerospace Propulsion

2

20

ion rocket capabilities ion jet velocity

power required

electric field

q U a = 2 V0 M 5 2 0 2 a

P 1 FU a 4 q V = = ε0 2 9 A 2 A M x

4 V0 ⎛ dV ⎞ Ea = − ⎜ ⎟ = ⎝ dx ⎠a 3 xa

Max<107volts/m

F 1 = ε 0 Ea2 ≈ 440 N / m 2 A 2

Limiting thrust

Theory of Aerospace Propulsion

21

electrostatic: Ion Rocket in space propellant line

ion source accelerating electrode neutralizer: electron emitter battery ions electrons Theory of Aerospace Propulsion

22

ion rocket Type

Propellant Energy

Isp,vac (s) Thrust (N) 150‐ 0.005 ‐ 700 0.5 450‐ 0.05 ‐ 5 1500

Resisto N2, NH3, N2O4, H2 jet Arcjet NH3, H2, N‐2H4

Resistive η=0.9 Arc heat η=0.3

Ion

Electrostatic 2000‐ 6000 η=0.75

Hg, A,Xe, Cs

Theory of Aerospace Propulsion

5x10‐6 – 0.5

Density (g/cc) 0.28, 0.60, 1.0, 0.019 0.60, 0.019 1.0 13.5, 0.44,2.73, 1.87

23

Deep Space 1: Xenon Ion Rocket DS1 is the first spacecraft to use ion propulsion as the primary propulsion system (NSTAR). A solar array provided power and the rocket used up to 2.5kW The xenon ions travel up to 35 km/s (Isp=3200 to 1900s) DS 1, using less than 74 kg of xenon, and produces F=92mN (0.3oz) It thrusted for 678 days, far longer than any propulsion system had ever been operated Theory of Aerospace Propulsion

24

Theory of Aerospace Propulsion

25

Interplanetary mission performance

Theory of Aerospace Propulsion

26

electromagnetic rockets Type

Propellan t Resisto N2, NH3, N2O4, H2 jet Arcjet NH3, H2, N‐2H4 Ion Hg, A,Xe, Cs

MHD

A

Pulsed Teflon plasma

Energy source Resistive η=0.9 Arc heat η=0.3 Electrostati c η=0.75 Magnetic

Isp,vac (s)

Thrust (N) 150‐700 0.005 – 0.5 450‐1500 0.05 – 5

2000

Magnetic

1500

2000‐ 6000

Theory of Aerospace Propulsion

Density (g/cc) 0.28, 0.60, 1.0, 0.019 0.60, 0.019 1.0 5x10‐6 – 13.5, 0.5 0.44,2.73, 1.87 25 – 0.44 200 5x10‐6 – 2.2 5x10‐3 27

electromagnetic: MHD engine propellant tank electric power supply

pump

positive electrode magnetic field coils cathode

~

electric current negative electrode

accelerating plasma

Arc

anode

J

JXB

Electromagnetic force (J X B) B Theory of Aerospace Propulsion

28

pulsed plasma thruster

Courtesy NASA GRC

Theory of Aerospace Propulsion

29

pulsed plasma thruster

Courtesy NASA GRC

Theory of Aerospace Propulsion

30

nuclear propulsion historical background In May, 1961, after 4 months in office President Kennedy addressed Congress to outline major goals of his administration. He set 4 National Space Goals: •Manned Moon Landing •Development of communications •Weather satellites •Greatly increased funding for the ROVER nuclear rocket The reason for the last was to “give promise of some day providing a means for even more exciting and ambitious explorations of space, perhaps beyond the moon, perhaps to the very end of the solar system itself.”

Theory of Aerospace Propulsion

31

nuclear propulsion background (cont.)

Nuclear thermal propulsion (NTP) is an attractive space propulsion option: •Reduced mass requirements in orbit from which to stage high ΔV missions •Reasonable interplanetary trip times •Potential for usage of in situ propellant resources After extensive planning and research leading to ground testing of systems NTR development was canceled in 1972: •Nuclear test Ban Treaty signed •NASA cancels major Lunar/Planetary exploration efforts New interest in space exploration has rekindled interest in NTP and nuclear electric propulsion (NEP) Theory of Aerospace Propulsion

32

exhaust velocity advantage Tc Ve ~ Wc H2 ‐ O2

Tc= 3400oK

Wc = 11

Nuclear

Tc= 2,500oK

Wc = 2

Advantage

Ve,chemical

1 ≈ Ve,nuclear 2

Theory of Aerospace Propulsion

33

typical NTP rocket configuration

Theory of Aerospace Propulsion

34

solid core reactor radiation shield (B, Fe, Pb) reflector (C, BeO, D2O) moderator (Be, H2O, LiH) fuel (C, W, cermet)

hot gas flow

movable control rods (B, Cd) Theory of Aerospace Propulsion

35

NTP Experience 20 reactors were designed, built, and tested from 1953‐1973 ($1.4B).

The Demonstrated performance included: •Power (MWt) – 1100 (NRX) to 4100 (Phoebus‐2A •Thrust (klb) – 55 (NRX) to 210 Phoebus‐2A •Fuel temperature (K) – 2750 (PEWEE) •Specific Impulse (s) – 850 (vac, PEWEE) •Burn endurance – 1 to 2 hours •Start/Stop cycles – 28 automatic cycles (XE)

Theory of Aerospace Propulsion

36

candidate NTP propellants Propellant CO2

CH4

H2

NH3 H2O

Source Martian atmosphere Martian frost Earth Asteroids, Phobos/Deimos Earth Outer planets Lunar polar ice, Lunar Silane NEO asteroids Earth, Outer planets Earth, Outer planets Lunar ice, Martian ice Planetary moons Theory of Aerospace Propulsion

Isp potential (s) 160‐380

460‐670

800‐1200

350‐700 160‐240 37

the KIWI experimental nuclear reactor

Theory of Aerospace Propulsion

38

fuel rods for solid core reactors external surfaces coated with ZrC fuel rod (graphite‐UC‐ ZrC composite)

Support elements, shown here as the central feature, are surrounded by fuel elements (coolant passages are omitted from fuel rods here for clarity).

Inconel tube graphite sleeve

coolant channels pyrolitic graphite thermal insulation

fuel rods Theory of Aerospace Propulsion

ZrH moderator 39

NTP vs. chemical rockets Parameter

Nuclear Thermal Rocket

Chemical Rocket

Isp Performance

Solid core >800s Gas core >1500s

250 ‐ 450s

Thrust to weight ratio

2<F/W<5

F/W>50

Propellant

Monopropellant: H2, other gases

Bipropellant: LOX‐LH2, LOX‐ Hydrocarbon, etc.

Chamber pressure

2MPa to 7MPa

3.5MPa to 245MPa

Chamber temperature

2500K to 3200K

2700K to 3800K

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particle bed reactors high density graphite

outer porous tube

ZrC coating packed bed of fuel particles UC2

Hot H2

hexagonal moderator block low density graphite inner porous tube (a) coated fuel particle

(b) fuel element cross‐section Theory of Aerospace Propulsion

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particle bed reactors cold H2 in

hexagonal moderator block cold H2 in

array of fuel elements in moderator blocks

Hot H2 out hot H2 out (a) fuel element assembly

to nozzle (b) particle bed reactor

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the MITEE nuclear powerplant MITEE (MIniature ReacTor EnginE); Ref.: Acta Astronautica, Vol. 44, No.2‐4, 1999) Derivative of Ultra‐Light Particle Bed Reactor (PBR) developed for SDIO (1985‐1993; ~ $200M) MITEE is lighter and more compact than a standard PBR. MITEE can be adapted for a ramjet application: Replace H2 by ambient atmosphere from inlet Reduce outlet temperature from 3000 K to 1500 K Reduce power density from 10 to 2 MW/liter Increase operating life from ~1 hour to months Theory of Aerospace Propulsion

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properties of planetary atmospheres γ

(km)

Composition Molecular R weight (m2/s2‐K)

Venus

6,052.8

N2, CO2

30

286

1.4

735 0.542

Earth

6,378.1

N2, O2

29

287

1.4

185 0.273

Mars

3,397

CO2, N2, A

44

190

1.3

186 0.214

Jupiter

71,492

H2, He

2

4170

1.4

137 0.894

Titan

2,575

N2, CH4

27

309

1.4

92

0.199

Saturn

60,268

H2, He

2

4170

1.4

‐

‐

Planet

Radius

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T (K) a

(km/s)

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exploration of planetary atmospheres Interest in long‐term study of planetary atmospheres revealed a need for a space probe that had the following characteristics: •3‐D mobility for atmospheric mapping •Capability to explore for many months •Simple, reliable, small mass and size •Scalable to different planets A nuclear ramjet can use the planetary atmosphere as the propellant and carry out long term missions flying at supersonic speeds. Data can be relayed through a companion data satellite orbiting the planet Theory of Aerospace Propulsion

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Jupiter Flyer internal layout

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Project Pluto: TORY IIC Nuclear Ramjet

This ramjet was planned to power a supersonic low altitude missile (SLAM) and a prototype engine was fired for 5 minutes in 1964. The project was canceled in favor of ICBMs whose development was proving to be increasingly successful.

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gas core nuclear reactors H2 and UF6 gas injected through tangential slots to form a vortex reflector material casing nozzle

hot H2‐UF6 plasma vortex

The thermo‐structural problems are mitigated but some radioactive material will be exhausted through the nozzle making this approach applicable only to interplanetary missions Theory of Aerospace Propulsion

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gas core nuclear rocket higher allowable temperature yields Isp ~ 2000s simulated UF6 fuel chamber nozzle throat

simulated propellant flow

fuel containment region

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nuclear electric propulsion (NEP) For deep space missions solar power becomes problematic and interest in nuclear reactors for providing power for electric propulsion systems. NASA has studied such approaches including the Prometheus system shown here

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Xenon ion rocket for Prometheus NEP

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