AE 2010 Thermodynamics (Problem Set #4 Mass Conservation Control Volume Analysis) NEW UPDATE Georgia

AE 2010 Thermodynamics (Problem Set #4: Mass Conservation: Control Volume Analysis) NEW UPDATE Georgia Institute Of Technology • Always indicate any assumptions you make. If you use any results or equations from the class notes or text in you solutions, please note and reference them (but you better be sure they are applicable to the problem at hand).

• Show all your work, no credit for just answers. When applicable, try to solve the problem algebraically first. Only use numbers/values in the final steps of your solution – and be sure to include units when you insert numbers.

• If the problem statement is given in ENGLISH units, the answer must also be in English units; if the problem statement is in SI units, the answer must be in SI units.

1. Low Speed Wind Tunnel A low speed, open-return wind tunnel is being designed with a fan at the inlet, and a test section having a rectangular cross-section of 40in.  42in. The designers want to achieve a 55 mph wind speed, with the air in the test-section nominally at 72 ºF and 0.95 atm. How much air (mass flow rate) will the fan that runs the wind tunnel have to supply to meet these conditions? If the pressure of the air at the exit of the wind tunnel was 0.2 psi lower than the pressure in the test section, what would be the air velocity at the exit, assuming the air temperature remained at 72 ºF? 2. Steady-State Aircraft Engine In a jet engine, air enters from the front, fuel is added farther downstream in the engine, and the fuel and air burn. Thrust is produced when the gases leave the engine through a nozzle at high velocity. Consider a case where a jet engine is being tested on the ground. It is operating at steady-state and burning 0.25 kg/s of jet fuel. The velocity (v) and temperature (T) profiles of the gases exiting the round nozzle have been measured, and are found to depend only on the radial distance (r) from the center of the nozzle – they are not dependent on the azimuthal position around the nozzle. The measured data can be described approximated by the following expressions: r ()  r 2 and v r = 420.m s − 105m s   T (r) = 1350K − 675K R R where R is the radius of the nozzle exit (=0.25 m for our engine). In addition, the pressure and molecular weight of the exiting gas are 1.0 bar and 28.1 (and these values are uniform at the exit). From the measured profiles, determine the fuel-air ratio of the engine, i.e., the mass flow rate of fuel entering the engine divided by the mass flow rate of air entering the engine.

Solution for Problem Set #4: Mass Conserv.: Control Volume Analysis Problem 1. Low Speed Wind Tunnel Given: Wind tunnel with test section conditions shown in figure.

Test Section

m fan

55 mph 72 F 0.95 atm 40”

test

Find: a) mass flow rate supplied by fan b) air velocity at tunnel exit

Exit

42” pexit=0.95 atm-0.2psi

Assume: 1) air is ideal gas at room T and p conditions, 2) steady flow 3) uniform flow 4) exit has same area as test section Analysis: a) Mass Flowrate Assuming steady conditions,

0

(

0=

+ mfan − m test

uniform flow in test section

mfan = mtest = test vtest Atest

Using air is ideal gas

mfan =

mfan =

)

ptest vtest Atest RTtest

0.95atm (40in)(42in) 1 ft 2 55mph 3 60mph 144in 2 0.7302 ft atm lbmol (460 + 72)R lbmol R 28.8lbm

m fan = 0.0704

lbm (80.7 ft s)11.7 ft2 = 66.3lb m s 3 ft

b) vexit

mexit = mtest

At exit Or

exit vexit Aexit = test vtest Atest

Same T’s

vexit =

p ptest RTtest vtest = test v test pexit pexit RTexit

vexit =

0.95atm atm 0.95atm − 0.2 psi 14.7 psi

55mph = 55.8mph = 81.8 ft s

Implications: Large wind tunnels require lots of air, but if the numbers here are accurate, the fan running them does not need to produce much of a pressure increase.

Problem 2. Steady-State Aircraft Engine Given: Engine operating steadily with air and fuel entering as shown in picture, and gas exiting with profiles only dependent on radius r, as m given by. fuel =0.25 kg/s

2   r   v e (r ) = 1054.00 −    m s R    

r

pe=1.0bar, MWe=28.1 r=R=0.25m

  r  mair Te (r ) = 6752.00 −  K  R  

r=0 Te

Find: fuel-air ratio of engine Assume: Exhaust is ideal/thermally perfect gas Analysis:

0=

dmCV + mouts dt

− m ins = me − (m air + m fuel )

mair = me − mfuel . We can find the exit mass flow rate from R → me =  e (ve  nˆ)dA =  v2rdr

with the density given by e = pe RTe

0

pe

or me = 2

R



ve

rdr

R MWe 0 Te Using the profiles given in the problem (with vc105 m/s and Tc675 K)

me = 2π

  r 2    r 2  vc 4 −    4 −    R p v R   R     R   pe e c rdr = 2π rdr R MWe Tc 0   r  R MWe 0   r  Tc 2 −   2 −   R      R 

Simplifying the integral,

  r   r  R 2 +   2 −   rdr = 2π e c pe vc R 2r +  r 2 dr p v m = 2π R R     e R MW T  R R MW T    r      e c 0 e c 0 2 −     R  and integrating,

R

pe vc 2 vc r 2 + r 3  = 2π pe v c R 2 1+ 1 8π m = 2π = R   e R MWe Tc  3R 0 R MWe Tc  3 3 R MWe Tc pe

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me

dA=2rdr (perimeter  thickness)

0, steady

Use mass conservation So

ve

Introducing the numbers

m = e

So

8π

(

5 )105m s  1 m  = 8π 10 (28.1)(105) 48 8314(675) 28.1 kg kmol 675K  4  2

105)Pa (

3 8314 J kmolK me = 27.5 kg s

Therefore

m fuel m air

=

m fuel m e − m fuel

=

0.25kg s = 0.00917 (27.5 − 0.25)kg s

Implications: 1) If the numbers are realistic, jet engines use a lot more air than fuel ~100 more by mass in this case 2) the velocities and temperatures at the exit of a jet engine are high

Problem 3. Unsteady Rocket Given: Rocket combustion chamber with given flow rate of reactants into chamber and pressure dependent exit flowrate (shown in figure)

p,T = 2600K , MW = 9.42 m1 = 45.0

kg

V = 1.20m3

s

m 2 = 8.2 10−6 p

Find: a) products if p=4.40 MPa b) dm/dt in chamber if p=4.40 MPa c) pressure for chamber to achieve steady state Assume: products are ideal(thermally perfect) gases Analysis: a) Product Density

 prod =

p 4.40 106 Pa = = RT (8314 9.42)J kgK(2600K )

b) dmCV/dt

dmCV + m2 dt

From Mass Conservation

m1 =

or

dmCV = m1 − m2 = m1 − const  p dt

dmCV kg kg = 45.0 − 8.2010−6 (4.4 106 Pa) dt s Pas dmCV = (45.0 − 36.1) kg = 8.9 kg dt s s

So at 4.40 MPa

So the mass inside the combustion chamber is increasing.

dmCV = m1 − m2 dt kg   −6 8.20 10 or  p = m1 m2 = m1  sPa  

0=

c) For steady-state, we have or so

p = 45.0

kg s

8.20 10−6

kg = 5.49MPa = 54.9bar Pas

which is also ~54 atm or ~800 psia.

Implications: 1) With the exit flow rate dependent on the chamber pressure, at p<psteady, the mass entering the chamber exceeds the mass leaving, so the pressure will increase until the steady pressure is achieved.

2) Since the product density is ~2 kg/m3 and the volume is 1.2 m3, there are about 2.4 kg of products in the combustion chamber at 44 bar. Thus it won’t take long for the pressure to increase if it is increasing at a rate of ~9 kg/s.

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kg s Pa

3. Unsteady Rocket Propellants enter a rocket combustion chamber (whose volume is 1.20 m 3) at a constant rate of 45.0 kg/s. They burn and their temperature rises to 2600 K, and the molecular weight of the burned products is 9.42. The nozzle downstream of the combustion chamber limits the exiting mass flowrate to 8.2010-6p kg/(Pasec), where p is the pressure in the combustion chamber. a) If the combustion chamber is at a pressure of 44.0 bar, what is the density of the combustion products in the combustion chamber? You can assume that the products are ideal gases at these high temperature conditions. b) For the same combustion chamber of 44.0 bar, what is the rate of change of mass in the combustion chamber? c) At what combustion chamber pressure will the rocket achieve steady-state operation?

This study source was downloaded by 100000900134793 from CourseHero.com on 12-03-2025 13:58:12 GMT -06:00

This study source was downloaded by 100000900134793 from CourseHero.com on 12-03-2025 13:58:12 GMT -06:00

This study source was downloaded by 100000900134793 from CourseHero.com on 12-03-2025 13:58:12 GMT -06:00