Chapter 9 Compressible Flow 9.1 An ideal gas flows adiabatically through a duct. At section 1, p1 140 kPa, T1 260C, and V1 75 m/...

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Fig. P9.1

Solution: (a) For air, take k 1.40, R 287 J/kg K, and cp 1005 J/kg K. The adiabatic steady-flow energy equation (9.23) is used to compute the downstream velocity:

m 1 1 1 cp T V 2 constant 1005(260) (75) 2 1005(207) V 22 or V2 335 2 2 2 s

Ans.

207 273 30 Meanwhile, s 2 s1 c pln(T2 /T1) R ln(p 2 /p1) 1005 ln 287 ln , 260 273 140

or s2 s1 105 442 337 J/kg K Ans. (a)

(b) For argon, take k 1.67, R 208 J/kg K, and cp 518 J/kg K. Repeat part (a): 1 1 1 m c pT V 2 518(260) (75)2 518(207) V22 , solve V2 246 2 2 2 s

207 273 30 208 ln 54 320 266 J/kg K s 2 s1 518 ln 260 273 140

Ans .

Ans. (b)

9.2 Solve Prob. 9.1 if the gas is steam. Use two approaches: (a) an ideal gas from Table A.4; and (b) real steam from EES or the steam tables [15].

Solutions Manual Fluid Mechanics, Seventh Edition

636

For steam, take k 1.33, R 461 J/kg K, and cp 1858 J/kg K. Then

Solution:

cp T

1 2 1 1 m V 1858(260) (75) 2 1858(207) V22, solve V2 450 2 2 2 s

Ans. (a)

207 273 30 461ln 195 710 515 J/kg K Ans. (a) s 2 s1 1858 ln 260 273 140

(b) For real steam, we look up each enthalpy and entropy in EES or the Steam Tables:

at 140 kPa and 260 C, read h1 2.993E6 at 30 kPa and 207 C, h2 2.892E6 Then

h

J kg

J ; kg

1 2 1 1 m V 2.993E6 (75)2 2.892E6 V22 , solve V 2 443 2 2 2 s

at 140 kPa and 260C, read s1 7915

Ans . (b)

J J , at 30 kPa and 207 C, s2 8427 kg K kg K

Thus s 2 s1 8427 7915 512 J/kg K

Ans. (b)

These are within 1.5% of the ideal gas estimates (a). Steam is nearly ideal in this range. 9.3 If 8 kg of oxygen in a closed tank at 200C and 300 kPa is heated until the pressure rises to 400 kPa, calculate (a) the new temperature; (b) the total heat transfer; and (c) the change in entropy. Solution:

For oxygen, take k 1.40, R 260 J/kg K, and cv 650 J/kg K. Then

1 2 , T2 T1 (p2 /p1 ) (200 273)

400 631 K 358 C Ans. (a) 300

Q mcv T (8)(650)(358 200) 8.2E5 J Ans. (b)

J 358 273 1500 s 2 s1 mc v ln(T2/T1) (8)(650) ln 200 273 K

Ans. (c)

Chapter 9 Compressible Flow

637

P9.4 Consider steady adiabatic airflow in a duct. At section B, the pressure is 154 kPa and the density is 1.137 kg/m 3. At section D, the pressure is 28.2 kPa and the temperature is –19C. (a) Find the entropy change, if any. (b) Which way is the air flowing? Solution: Convert T D = -19+273 = 254 K. We need the temperature at section B:

TB

pB 154, 000 Pa 472 K 3 R B (287)(1.137 kg / m )

T p 472 154, 000 Then s B s D c p ln( B ) R ln( B ) (1005) ln( ) (287) ln( ) TD pD 254 28, 200 or : s B s D 623 487 +133

J kg K

Ans.(a )

The entropy is higher at B. Therefore the (adiabatic) flow is from D to B.

Ans.(b)

9.5 Steam enters a nozzle at 377C, 1.6 MPa, and a steady speed of 200 m/s and accelerates isentropically until it exits at saturation conditions. Estimate the exit velocity and temperature. Solution:

At saturation conditions, steam is not ideal. Use EES or the Steam Tables:

At 377C and 1.6 MPa, read

h1 3.205E6 J/kg

and

s1 7153 J/kgK

At saturation for s1 s2 7153, read p2 185 kPa, T2 118C, and h 2 2.527E6 J/kg

Then h

m 1 2 1 1 V 3.205E6 (200) 2 2.527E6 V22, solve V2 1180 s 2 2 2

Ans.

This exit flow is supersonic, with a Mach number exceeding 2.0. We are assuming with this calculation that a (supersonic) shock wave does not form. P9.6 Use EES, other software, or the Gas Tables, to estimate c p and cv, their ratio, and their difference, for carbon dioxide at 800K and 100 kPa. Compare with estimates similar to Eqs. (9.4).

Solutions Manual Fluid Mechanics, Seventh Edition

638

Solution: The writer used EES, for example, cp = CP(CarbonDioxide,T=800,P=100) and obtained the following results at 800K and 100 kPa: cp = 1169 J/kg-K ; c v = 980 J/kg-K ; k = cp /c v = 1.19 ; cp – cv = 189 J/kg-K (=R)

The difference is still equal to the gas constant R, but the specific heats are about 50% higher

than would be estimated from Table A.4, which states (at room temperature), that k = 1.30 and R = 189 J/kg-K:

c p |CO2

kR 1.3(189 ) J ; 819 k 1 1.3 1 kg K

cv |CO2

R 189 J 630 k 1 1.3 1 kg K

So we give up a little accuracy by assuming constant specific heats if temperature changes are large.

P9.7 Air flows through a variable-area duct. At section 1, A1 = 20 cm2 , p1 = 300 kPa, 1 = 1.75 kg/m3 , and V1 = 122.5 m/s. At section 2, the area is exactly the same, but the density is much lower: 2 = 0.266 kg/m3, and T 2 = 281 K. There is no transfer of work or heat. Assume one-dimensional steady flow. (a) How can you reconcile these differences? (b) Find the mass flow at section 2. Calculate (c) V 2, (d) p2, and (e) s 2 – s 1. Hint: This problem requires the continuity equation. Solution: constant:

Part ( a) is too confusing, let’s try ( b, c, d, e ) first. (b) The mass flow must be

1 m 2 1 A1V1 (1.75 m Then V 2

m 2 A2

m kg ) 0.0429 s s m3 0.0429 kg/ s m 806 Ans.(c) s (0.266kg / m 3 )(0.002m 2 ) kg

)(0.0020 m 2 )(122 .5

Ans.(b )

That’s pretty fast! Check a2 = (kRT 2)1/2 = [1.4(287)(281)]1/2 = 336 m/s. Hence the Mach number at section 2 is Ma2 = V2/a2 = 806/336 = 2.40. The flow at section 2 is supersonic! (d) The pressure at section 2 is easy, since the density and temperature are given:

Chapter 9 Compressible Flow

639

p 2 2 RT 2 (0.266 kg / m 3)(287 m 2 / s 2 K )(281 K ) 21,450 Pa Ans( d )

Similarly , T1

p1 (300000 Pa) 597 K 2 3 R 1 ( 287 m / s2 K )(1.75 kg / m )

(e) Finally, with pressures and temperatures known, the entropy change follows from Eq. (9.8): T p 281 21450 J s2 s1 c p ln( 2 ) R ln( 2 ) 1005 ln( ) 287 ln( ) 757 757 0 Ans.(e) T1 p1 597 300000 kg K Ahah! Now I get it. (a) The flow is isentropic. Ans.(a) The stagnation properties, To = 605 K, po = 319 kPa, and o = 1.805 kg/m3 are constant in the flow from section 1 to section 2.

9.8 Atmospheric air at 20C enters and fills an insulated tank which is initially evacuated. Using a control-volume analysis from Eq. (3.63), compute the tank air temperature when it is full. Solution:

The energy equation during filling of the adiabatic tank is

dE dQ dWshaft 0 0 CV hatmmentering , or, after filling, dt dt dt

ECV,final ECV,initial hatm mentered , or: mcv Ttank mcp Tatm

Thus Ttank (cp /cv )Tatm (1.4)(20 273) 410 K 137C

Ans.

9.9 Liquid hydrogen and oxygen are burned in a combustion chamber and fed through a rocket nozzle which exhausts at 1600 m/s and exit pressure equal to ambient pressure of 54 kPa. The nozzle exit diameter is 45 cm, and the jet exit density is 0.15 kg/m 3. If the exhaust gas has a molecular weight of 18, estimate (a) the exit gas temperature; (b) the mass flow; and (c) the thrust generated by the rocket. Solution: R gas

(a) From Eq. (9.3), estimate R gas and hence the gas exit temperature:

J p 54000 8314 462 779 K Ans. (a) , hence Texit M 18 kgK R 462(0.15)

Solutions Manual Fluid Mechanics, Seventh Edition

640

(b) The mass flow follows from the exit velocity:

kg kg m AV 0.15 3 (0.45m)2 (1600m / s ) 38 m 4 s

Ans. (b)

(c) The thrust was derived in Problem 3.68. When pexit p ambient, we obtain

Thrust eAeVe2 mVe 38(1600) 61,100 N Ans. (c)

9.10 A certain aircraft flies at the same Mach number regardless of its altitude. Compared to its speed at 12000-m Standard Altitude, it flies 127 km/h faster at sea level. Determine its Mach number. Solution:

At sea level, T 1 288.16 K. At 12000 m standard, T 2 216.66 K. Then

a1 kRT1 1.4(287)(288.16) 340.3

m m ; a 2 kRT2 295.0 s s

Then Vplane Ma(a2 a1 ) Ma(340.3 295.0) Ma(45.3) [127 km/h] 35.27 m/s Solve for Ma

35.27 0. 0.78 78 Ans. 45.3

9.11 At 300C and 1 atm, estimate the speed of sound of (a) nitrogen; (b) hydrogen; (c) helium; (d) steam; and (e) uranium hexafluoride 238 UF6 (k 1.06). Solution: The gas constants are listed in Appendix Table A.4 for all but uranium gas (e): (a) nitrogen:

k 1.40, R 297, T 300 273 573 K:

a kRT 1.40(297)(573) 488 m/s Ans. (a)

(b) hydrogen: k 1.41, R 4124, (c) helium:

k 1.66, R 2077:

k 1.33, R 461:

a 1.41(4124)(573) 1825 m/s Ans. (b)

a 1.66(2077)(573) 1406 m/s Ans. (c)

a 1.33(461)(573) 593 m/s Ans. (d) [NOTE: The EES “soundspeed” function would predict asteam = 586 m/s.]

(d) steam:

(e) For uranium hexafluoride, we need only to compute R from the molecular weight:

Chapter 9 Compressible Flow

(e)

UF6 : M 238 6(19) 352, R

238

641

8314 23.62 m 2/s 2 K 352

then a 1.06(23.62)(573) 120 m/s Ans. (e)

9.12 Assume that water follows Eq. (1.19) with n 7 and B 3000. Compute the bulk modulus (in kPa) and the speed of sound (in m/s) at (a) 1 atm; and (b) 1100 atm (the deepest part of the ocean). (c) Compute the speed of sound at 20C and 9000 atm and compare with the measured value of 2650 m/s (A. H. Smith and A. W. Lawson, J. Chem. Phys., vol. 22, 1954, p. 351). Solution:

We may compute these values by differentiating Eq. (1.19) with k 1.0:

p dp (B 1)( / a )n B; Bulk modulus K n(B 1)pa ( /a )n , a K/ pa d

We may then substitute numbers for water, with p a 101350 Pa and a 998 kg/m3: (a) at 1 atm: (a)

Kwater 7(3001)(101350)(1) 7 2.129E9 Pa (21007 atm)

Ans.

speed of sound a water K/ 2.129E9/998 1460 m/s Ans. (a)

1100 3000 (b) at 1100 atm: 998 3001

1/7

998(1.0456) 1044 kg/m 3

K Katm (1.0456)7 (2.129E9)(1.3665) 2.91E9 Pa (28700 atm) Ans. (b)

a K/ 2.91E9/1044 1670 m/s Ans. (b)

9000 3000 (c) at 9000 atm: 998 3001

1/7

kg 1217 , 1217 3 ; K K a 998 m 7

or: K 8.51E9 Pa, a K/ 8.51E9/1217 2645 m/s (within 0.2%) Ans. (c)

P9.13 Consider steam at 500 K and 200 kPa. Estimate its speed of sound by three different methods: (a) using the handy new EES thermophysical function SOUNDSPEED(Steam, p = p 1,T = T1 ); (b) assuming an ideal gas from Table B.4; or (c) using finite differences for isentropic densities between 210 kPa and 190 kPa.

Solutions Manual Fluid Mechanics, Seventh Edition

642

Solution: (a) Enter EES and use the new function, setting units to kPa and degrees Kelvin: aEES SOUNDSPEED(Steam, p 200, T 500)

547

m s

Ans.(a )

(b) Ideal gas approximation: From Table B.4 for H2 O, k = 1.33 and R = 461 m2 /s2-K: a ideal gas

k RT

1.33(461)(500) 554

m s

Ans .(b )

This is 1.3% higher than EES, not bad. In this region, a better k would 1.30, not 1.33. (c) Using finite differences of density and pressure at the same entropy as the given state: Entropy level : so ENTROPY(Steam , p 200,T 500) 7.6168 kJ / kg K

At p 2 210kPa , compute 2 DENSITY (Steam , p 210, s so ) 0.9073 kg / m3 At p 1 190kPa , compute 1 DENSITY (Steam , p 190, s so ) 0.8404 kg / m3 210000 190000 20000 m2 p 298950 Finite differences : a | s 0.9073 0.8404 0.0669 s2 m Finally, a differences 298950 547 Ans .(c ) s 2

Part (c) is the same as the EES result, so maybe that’s how the new function works?

3 . Atis300 its P9.14 At 1 atm and 20C, the density of methyl alcohol">methyl alcohol 49.4atm, lbm/ft 3 density increases to 50.9 lbm/ft . Use this data to estimate the speed of sound. Comment on the possible uncertainty of this estimate.

Solution: For convenience, convert the density data to SI units: 49.4 lbm/ft 3 = 790.7 kg/m3, and 50.9 lbm/ft 3 = 814.7 kg/m 3. Then use finite differences to approximate the formula: a methanol

p

(300 1)(101350Pa ) 3 (814.7 790.7)kg / m

30303650 m 1124 Ans. 24.0 s

Chapter 9 Compressible Flow

643

Since it relies on a small difference between two large densities, the density measurement must be very accurate. For example, a 1% error in density might cause a 50% error in speed of sound.

P9.15 The pressure-density relation for ethanol is approximated by Eq. (1.19) with B = 1600 and n = 7. Use this relation to estimate the speed of sound of ethanol at a pressure of 2000 atmospheres.

Solution: Recall that Eq. (1.19) is a curve-fit equation of state for liquids:

p (1.19) ( B 1) ( ) n B po o It looks like this, with o = 790 kg/m 3 from Table A.3. At 2000 atm, 887 kg/m 3. 3000 2500 2000

ETHANOL

p , atm

1500 1000

, kg/m3

500 0

780

800

820

840

860

880

900

920

We see that the slope (or speed of sound squared) increases with pressure. Differentiate:

dp po a2 n ( B 1) ( )n 1 d o o

887 kg / m 3 6 Ethanol: a (7)(1600 1)( ) 790 kg / m 3 790 kg / m 3 At 1 atm, the speed of sound of ethanol is about 1200 m/s. 101350 Pa

1700

m s

Ans.

644

Solutions Manual Fluid Mechanics, Seventh Edition

9.16 A weak pressure wave (sound wave) p propagates through still air. Discuss the type of reflected pulse which occurs, and the boundary conditions which must be satisfied, when the wave strikes normal to, and is reflected from, (a) a solid wall; and (b) a free liquid surface.

Fig. P9.16

Solution: (a) When reflecting from a solid wall, the velocity to the wall must be zero, so the wall pressure rises to p 2 p to create a compression wave which cancels out the oncoming particle motion V. (b) When a compression wave strikes a liquid surface, it reflects and transmits to keep the particle velocity Vf and the pressure p pf the same across the liquid interface: Vf

2 C V ; C liqC liq

If liqC liq C of air, then

p f

2 liq Cliq p

C liqC liq

Ans. (b)

V f 0 and p f 2p, which is case (a) above.

9.17 A submarine at a depth of 800 m sends a sonar signal and receives the reflected wave back from a similar submerged object in 15 s. Using Prob. 9.12 as a guide, estimate the distance to the other object.

Chapter 9 Compressible Flow

Solution:

645

It probably makes little difference, but estimate a at a depth of 800 m:

at 800 m,

p 101350 1025(9.81)(800) 8.15E6 Pa 80.4 atm

p/pa 80.4 3001( /1025) 7 3000, solve 1029 kg/m3

a n(B 1)pa ( /a ) 7/ 7(3001)(101350)(1029/1025)7 /1029 1457 m/s

Hardly worth the trouble: One-way distance a t/2 1457(15/2) 10900 m. Ans. 9.18 Race cars at the Indianapolis Speedway average speeds of 185 mi/h. After determining the altitude of Indianapolis, find the Mach number of these cars and estimate whether compressibility might affect their aerodynamics. Solution: Rush to the Almanac and find that Indianapolis is at 220 m altitude, for which Table A.6 predicts that the standard speed of sound is 339.4 m/s 759 mi/h. Thus the Mach number is Maracer V/a 185 mph/759 mph 0.24 Ans.

This is less than 0.3, so the Indianapolis Speedway need not worry about compressibility.

P9.19 In 1976, the SR-71A Blackbird, flying at 20 km standard altitude, set the jet-powered aircraft speed record of 3326 km/h. (a) Estimate the temperature, in C, at its front stagnation point. (b) At what Mach number would it have a front stagnation-point temperature of 500C? Solution: At 20 km altitude, from Table A.6, T = 216.66K and a = 295.1 m/s. Convert the velocity from 3326 km/h to (3316)(1000)/(3600) = 924 m/s. Then Ma = V/a = 924/295.1 = 3.13. Compute

To T (1 0.2 Ma 2 ) ( 216.66)[1 0.2(3.13) 2 ] 641 K 368 oC

Ans.(a )

(b) To have a front stagnation temperature of 500C = 773 K, we could calculate

To 773 K ( 216.66)[1 0.2 Ma2 ] ,

solve for Ma 3.58

Ans.( b)

The SR-71A couldn’t fly that fast because of structural and heat transfer limitations.

Solutions Manual Fluid Mechanics, Seventh Edition

646

P9.20 Air flows isentropically in a channel. Properties at section 1 are V1 = 250 m/s, T1 = 330 K, and p 1 = 80 kPa. At section 2 downstream, the temperature has dropped to 0C. Find (a) the pressure, (b) velocity, and (c) Mach number at section 2. Solution: Assume k = 1.4 and, of course, convert T2 = 0C = 273 K. (b) The adiabatic energy equation will yield the new velocity: 2

2

2

V1 (250) 2 V V2 T1 To 330 361 K T2 2 273 2c p 2(1005) 2c p 2(1005) Solve for V2 421

m s

Ans .(b )

(a) To calculate p 2, we could go through the stagnation pressure (which is 110 kPa) or we could simply use the ideal gas temperature ratio, Eq. (9.9): p2 T p 273 3.5 ) 0.515, or : p2 41 kPa ( 2 )k /(k 1) 2 ( p1 T1 80 330

Ans.( a)

We have velocity and temperature at section 2, so we can easily calculate the Mach number:

Ma2

V2 a2

V2

kRT2

421 1.4(287)(273)

421m / s 1.27 331m / s

Ans .(c )

9.21 CO2 expands isentropically through a duct from p1 125 kPa and T1 100C to p2 80 kPa and V2 325 m/s. Compute (a) T 2; (b) Ma2; (c) T o; (d) po; (e) V 1; and (f) Ma 1.

Chapter 9 Compressible Flow

647

Solution: For CO 2, from Table A.4, take k 1.30 and R 189 J/kgK. Compute the specific heat: c p kR/(k 1) 1.3(189)/(1.3 1) 819 J/kgK. The results follow in sequence:

(a) T2 T1( p2 /p1 )( k1)/ k (373 K)(80/125)(1.31)/1.3 336 K Ans. (a)

(b) a 2 kRT 2 (1.3)(189)(336) 288 m/s, Ma 2 V 2/a 2 325/288 1.13 Ans . (b) k 1 2 0.3 (c) To1 To2 T2 1 Ma2 (336) 1 (1.13)2 401 K Ans. (c) 2 2 1.3/(1.3 k1 2 (d) po1 po 2 p2 1 Ma2 2

1)

0.3 (1.13) 2 (80) 1 2

1.3/0.3

171 kPa Ans. (d)

V12 V12 (e) To1 401 K T1 , Solve for V1 214 m/s 373 2c p 2(819)

Ans. (e)

(f) a 1 kRT 1 (1.3)(189)(373) 303 m/s, Ma 1 V 1/a 1 214/303 0.71 Ans. (f) 9.22 Given the pitot stagnation temperature and pressure and the static-pressure measurements in Fig. P9.22, estimate the air velocity V, assuming (a) incompressible flow and (b) compressible flow. Solution: Given p 80 kPa, po 120 kPa, and T 100°C 373 K. Then

o

po 120000 1.12 kg/m3 RTo 287(373)

(a) ‘Incompressible’:

o , V

2 p

Fig. P9.22

2(120000 80000) m 267 (7% low) Ans. (a) s 1.12

(b) Compressible: T To(p/p o)(k–1)/k 373(80/120) 0.4/1.4 332 K. Then T o 373 K T V2 /2cp 332 V2/[2(1005)], solve for V 286 m/s. Ans. (b)

Solutions Manual Fluid Mechanics, Seventh Edition

648

P9.23 A gas, assumed ideal, flows isentropically from point 1, where the velocity is negligible, the pressure is 200 kPa, and the temperature is 300 C, to point 2, where the pressure is 40 kPa. What is the Mach number Ma 2 if the gas is (a) air; (b) argon; or (c) CH 4? (d) Can you tell, without calculating, which gas will be the coldest at point 2? Solution: This is a standard exercise in using the isentropic-flow formulas. The term “negligible velocity” is code for stagnation conditions, hence p o = 200 kPa and To = 300C = 573 K. Work it out for the three different gases, using the ideal-gas isentropic-flow formulas: po p

(1

k 1 2 k/( k1) Ma ) 2

;

To k 1 2 (1 Ma ) T 2

For po /p = 200/40 = 5.0 and To = 573 K, we obtain (a) Air,

k = 1.40

;

(c) CH4,

k = 1.32

;

(b) Argon, k = 1.67

;

(d) The argon cools off the most

Ma 2 = 1.709

Ma 2 = 1.646

Ma 2 = 1.727

;

;

;

T2 = 362 K

Ans.(a)

T 2 = 388 K

Ans.(c)

T 2 = 301 K

Ans.(b)

Ans.(d). But, since both cp and V vary with k, the writer

was not smart enough to divine which was coolest without calculating the temperatures.

9.24 For low-speed (nearly incompressible) gas flow, the stagnation pressure can be computed from Bernoulli’s equation p0 p

1 V2 2

(a) For higher subsonic speeds, show that the isentropic relation (9.28a) can be expanded in a power series as follows: p0 p

1 2 k 1 V 2 1 Ma2 Ma4 2 24 4

(b) Suppose that a pitot-static tube in air measures the pressure difference p0 – p and uses the Bernoulli relation, with stagnation density, to estimate the gas velocity. At what Mach number will the error be 4 percent?

Chapter 9 Compressible Flow

Solution:

649

Expand the isentropic formula into a binomial series:

k/(k po k 1 2 Ma 1 p 2

k k 1 k 1 k k 1 Ma 2 Ma 2 1 1 k 1 2 k 1 2 k 1 2

1)

2

1

k k k(2 k) Ma 2 Ma4 Ma6 2 8 48

Use the ideal gas identity (1/2) V 2 (1/2)kp(Ma2) to obtain po p

(1/2) V 2

1

1 2 k Ma 2 Ma 4 4 24

Ans.

The error in the incompressible formula, 2p/oV 2, is 4% when

V o/ 1.04, 2 4 2(po p)/ 1 (1/4)Ma [(2 k)/24]Ma

o k 1 where Ma 2 1 2

1/(k 1)

For

k 1.4, solve this for 4% error at Ma 0.576 Ans.

9.25 If it is known that the air velocity in the duct is 750 ft/s, use that mercury manometer measurement in Fig. P9.25 to estimate the static pressure in the duct, in psia. Solution: Estimate the air specific weight in the manometer to be, say, 0.07 lbf/ft3. Then

Fig. P9.25

8 po pmeasured ( gmercury gair )h (846 0.07) ft 564 lbf/ft 2 12

Given T 100 F 560 R, a kRT 1.4(1717)(560) 1160 ft/s Then

Finally,

Ma V/a 750/1160 0.646

po p 564 [1 0.2(0.646)2 ]3.5 1 1.324 1 0.324 p p

Solve for

p static 1739 psf 12.1 lbf/in 2 (abs) Ans.

Solutions Manual Fluid Mechanics, Seventh Edition

650

9.26 Show that for isentropic flow of a perfect gas if a pitot-static probe measures p0 , p, and T0, the gas velocity can be calculated from p ( k 1)/ k V 2 2 cp T0 1 p 0

What would be a source of error if a shock wave were formed in front of the probe? Solution:

Assuming isentropic flow past the probe,

T To (p/p o ) (k1)/k

p (k 1)/k V2 , solve V2 2cp To 1 To p 2c p o

Ans.

If there is a shock wave formed in front of the probe, this formula will yield the air velocity inside the shock wave, because the probe measures p o2 inside the shock. The stagnation pressure in the outer stream is greater, as is the velocity outside the shock.

P9.27 A pitot tube, mounted on an airplane flying at 8000 m standard altitude, reads a stagnation pressure of 57 kPa. Estimate (a) the velocity in mi/h, and (b) the Mach number. Solution: We assume that the static pressure is the standard atmosphere pressure at 8000 m, which from Table B.6 is 35,581 Pa. Then the isentropic pressure formula will yield the Mach number: po 57000 1.602 (1 0.2 Ma 2 ) 3.5 , p 35581

Solve Ma 0.85 Ans.(b)

Gratifyingly, the speed of sound at 8000 m is given right in Table B.6: a = 308 m/s. Then

V (Ma )( a ) (0.85)(308 m / s ) 262 m / s 586 mi / h

Ans.(a )

9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea-level standard air through a converging nozzle of throat diameter 3 cm. Estimate (a) the mass flow rate; and (b) the Mach number at the throat.

Chapter 9 Compressible Flow

651

Solution: For sea-level air take To 288 K, o 1.225 kg/m 3, and po 101350 Pa. The pressure ratio is given, and we can assume isentropic flow with k 1.4:

p e 60000 1 0.2 Ma2e po 101350

3.5

, solve Mae 0.899 Ans. (b)

We can then solve for exit temperature, density, and velocity, finally mass flow:

e o [1 0.2(0.899)2 ] 2.5 0.842

kg

m

, Te 3

pe 60000 248 K R e 287(0.842)

m s kg Finally, m e AeVe (0.842) (0.03) 2(284) 0.169 s 4 Ve Mae ae 0.899[1.4(287)(248)]1/2 284

Ans. (a)

9.29 Steam from a large tank, where T 400C and p 1 MPa, expands isentropically through a small nozzle until, at a section of 2-cm diameter, the pressure is 500 kPa. Using the Steam Tables, estimate (a) the temperature; (b) the velocity; and (c) the mass flow at this section. Is the flow subsonic? Solution: “Large tank” is code for stagnation values, thus To 400C and po 1 MPa. This problem involves dogwork in the tables and well illustrates why we use the ideal-gas law so readily. Using k 1.33 for steam, we find the flow is slightly supersonic:

Ideal-gas simplification:

1.33 1 1 2 Ma Ans.(a)

1.33 2 0.33

p o 1000 2.0 p 500

Solve Ma 1.07

,

That was quick. Instead, for more accuracy, use EES, assuming constant entropy: At To 400C and p o 1 MPa,

read so 7465

J kgK

and h o 3.264E6

Then, at p 0.5 MPa, assuming s so , read T 302°C 575 K Ans. (a) Also read h 3.067E6 J/kg

and 1.908 kg/m3.

J kg

Solutions Manual Fluid Mechanics, Seventh Edition

652

With h and ho known, the velocity follows from the adiabatic energy equation:

J h V /2 h o , or 3.067E6 V /2 3.264E6 kg m Solve V 627 Ans. (b) s 2

2

m2 or 2 , s

The speed of sound is now included in EES, thanks, EES! a SOUNDSPEED(steam , p 500,s 7.465) 583 Then

Ma V/a

m s

627 1. 1.075 075 Ans . (a) (slightly supersonic) 583

Finally, the mass flow is computed from the density and velocity:

m A V

(1.908kg /m 3 ) ( / 4)(0.02m )2 (627 m / s ) 0.376kg / s

Ans .(c )

We could have done nearly as well (2%) by simply assuming an ideal gas with k 1.33.

P9.30 When does the incompressible-flow assumption begin to fail for pressures? Construct a graph of po/p for incompressible flow of air as compared to Eq. (9.28a). Neglect gravity. Plot both versus Mach number for 0 Ma 0.6 and decide for yourself where the deviation is too great. Solution: The Bernoulli incompressible equation can be converted to Mach number form:

po V 2 kV2 k V2 k p o |incompressible p V , or : 1 1 1 1 Ma 2 2 p 2p 2 kRT 2 a2 2 po (k 1) Compare with |compressible [1 Ma2 ]k /( k 1) p 2 The two formulas are compared in the chart below. The difference becomes visible (but less 2

than 1%) at Ma = 0.3 (the usual criterion) but is still small (<2%) at Ma = 0.5.

Chapter 9 Compressible Flow

653

1.25

Incompressible

1.20

Compressible

1.15

po/p

1.10 1.05 1.00

k = 1.40 0

0.1

0.2

0.3

Ma

0.4

0.5

0.6

9.31 Air flows adiabatically through a duct. At one section, V1 400 ft/s, T 1 200F, and p1 35 psia, while farther downstream V2 1100 ft/s and p2 18 psia. Compute (a) Ma2; (b) U max; and (c) p o2/po1. Solution: (a) Begin by computing the stagnation temperature, which is constant (adiabatic): V2 (400)2 V2 (200 460) 673 R T2 To1 To2 T1 2c p 2(6010) 2c p

(1100)2 Then T2 673 573 R, 2(6010) V 1100 100 Ma2 2 0.938 Ans. (a) a2 1173 1.4(1717)(573)

(b) Umax 2cp To 2(6010)(673) 2840 ft/s Ans. (b)

(c) We need Ma1 V1 /a1 400/ 1.4(1717)(200 460) 400/1260 0.318

then po1 p1 1 0.2Ma12

3.5

1.072p1 37.53 psia

p o2 31.74 0.846 Ans . (c) p o1 37.53 ______________________________________________________________________ and po2 p2 1 0.2Ma22

3.5

1.763p2 31.74 psia,

Solutions Manual Fluid Mechanics, Seventh Edition

654

9.32 The large compressed-air tank in Fig. P9.32 exhausts from a nozzle at an exit velocity of 235 m/s. The mercury manometer reads h 30 cm. Assuming isentropic flow, compute the pressure (a) in the tank and (b) in the atmosphere. (c) What is the exit Mach number? Solution: The tank temperature To 30C 303 K. Then the exit jet temperature is

Fig. P9.32

Ve2 (235) 2 235 Te To 303 276 K, Mae 0.706 Ans. (c) 2c p 2(1005) 1.4(287)(276) 3.5 p Then tank 1 0.2Ma2e 1.395 and ptank pe ( mercury tank )gh pe

Guess tank 1.6 kg/m 3, p o p e (13550 1.6)(9.81)(0.30) 39900 Pa

Solve the above two simultaneously for pe 101 kPa and ptank 140.8 kPa Ans. (a, b) 9.33 Air flows isentropically from a reservoir, where p 300 kPa and T 500 K, to section 1 in a duct, where A 1 0.2 m2 and V1 550 m/s. Compute (a) Ma1; (b) T 1; (c) p1 ; (d) m; and (e) A*. Is the flow choked?

Solution:

Use the energy equation to calculate T1 and then get the Mach number: V 21 (550) 2 500 350 K T1 To 2c p 2(1005)

Then a 1 1.4(287)(350) 375 m/s, Ma 1 V1/a 1

Ans . (b)

550 1.47 Ans. (a) 375

The flow must be choked in order to produce supersonic flow in the duct. Answer.

p 1 p o 1 0.2

Ma21

3.5

300 /[1 0.2(1.47)2 ]3.5 86 kPa Ans. (c)

Chapter 9 Compressible Flow

1

p1 86000 kg kg 0.854 3 , m AV (0.854)(0.2)(550) 94 s RT1 287(350) m

Finally,

Ans . (d)

A 1 (1 0.2Ma2 )3 1.155 if Ma 1.47, A* Ma 1.728 A*

P9.34

655

0.2 0.173 m 2 Ans. (e) 1.155

Carbon dioxide, in a large tank at 100C and 151 kPa, exhausts to a 1-atm

environment through a converging nozzle whose throat area is 5 cm2 . Using isentropic ideal-gas theory, calculate (a) the exit temperature, and (b) the mass flow.

Solution: For CO2, from Table A.4, take R = 189 m2/s2-K and k = 1.30. Find the exit Mach number from the exit pressure ratio (guessing that the exit flow is subsonic, so pexit = patm ):

po 151000Pa k 1 )Mae2 ]k /(k 1) [1 0.15Mae2 ]1.3/ 0.3 , solve Mae 0.802 [1 ( p 101350Pa 2

Then Te

To 373 K 2 k1 2 [1 0.15* (0.802) ] [1 ( ) Mae ] 2

340 K 67 oC

Ans.(a)

To compute the mass flow, either find Ve and e or use the mass-flow function, Eq. (9.47).

o

151000 kg 2.14 kg ; e 2.14 1.58 (189 )(373 ) m3 [1 0.15(0.802 )2 ]1/ 0.3 m3

V e Ma e a e Ma e kRT e (0.802 ) 1.3(189 )(340 ) 232 m / s

Finally, m e Ae Ve (1.58)(0.0005 m2 )(232 ) 0.183 kg / s Ans.( b)

With pressures known, an alternate method would use the mass-flow function, Eq. (9.47):

Solutions Manual Fluid Mechanics, Seventh Edition

656

m RT o A po

m (189 )(373 ) (0.0005 )(151000 )

2k p p ( ) 2 / k[1 ( )( k 1) / k ] k 1 po po

2(1.3) 101350 2 /1.3 101350 0.3 /1.3 ( ) [1 ( ) ] , Solve 0.3 151000 151000

0.183 m

kg s

Ans.( b)

The mass-flow function is a bit messy, but it gives the mass flow directly.

9.35 Helium, at To 400 K, enters a nozzle isentropically. At section 1, where A1 0.1 m2, a pitot-static arrangement (see Fig. P9.25) measures stagnation pressure of 150 kPa and static pressure of 123 kPa. Estimate (a) Ma 1; (b) mass flow; (c) T1; and (d) A*. Solution: For helium, from Table A.4, take k 1.66 and R 2077 J/kgK. (a) The local pressure ratio is given, hence we can estimate the Mach number: po 150 1.66 1 Ma 21 1 p1 123 2

1.66/(1.66 1)

, solve for Ma1 0.50 Ans. (a)

Use this Mach number to estimate local temperature, density, velocity, and mass flow: T1

1

To 400 370 K 2 1 (k 1)Ma1 /2 1 0.33(0.50)2

Ans. (c)

p1 123000 kg 0.160 3 RT1 2077(370) m

m s kg Finally, m 1A1V1 (0.160)(0.1)(565) 9.03 s Finally, A* can be computed from Eq. (9.44), using k 1.66: V1 Ma1 a1 0.50[1.66(2077)(370)]1/2 565

A1 1 1 0.33 Ma12 A* Ma1 (1.66 1)/2

(1/2)(2.66)/(0.66)

Ans. (b)

1.323, A* 0.0756 m2

Ans. (d)

9.36 An air tank of volume 1.5 m 3 is at 800 kPa and 20C when it begins exhausting through a converging nozzle to sea-level conditions. The throat area is 0.75 cm2. Estimate (a) the initial mass flow; (b) the time to blow down to 500 kPa; and (c) the time when the nozzle ceases being choked.

Chapter 9 Compressible Flow

657

Solution: For sea level, pambient 101.35 kPa 0.528ptank , hence the flow is choked until the tank pressure drops to pambient /0.528 192 kPa. (a) We obtain minitial m max 0.6847

po A*

RTo

0.6847

800000(0.75 E4 m 2 ) 287(293)

0.142

kg s

Ans. (a)

(b) For a control volume surrounding the tank, a mass balance gives

d p A* dpo m 0.6847 o , separate the variables: (o ) dt RTo dt RTo

A* RTo p (t ) t e 0.00993t exp 0.6847 p (0)

At 500 kPa, we obtain

At choking (192 kPa),

until p(t) drops to 192 kPa

500/800 exp(–0.00993t),

192/800 exp(–0.00993t),

or t 47 s Ans. (b)

or t 144 s Ans. (c)

9.37 Make an exact control volume analysis of the blowdown process in Fig. P9.37, assuming an insulated tank with negligible kinetic and potential energy. Assume critical flow at the exit and show that both p o and T o decrease during blowdown. Set up firstorder differential equations for po (t) and To(t) and reduce and solve as far as you can.

Solution:

Fig. P9.37

For a CV around the tank, write the mass and the energy equations:

mass:

d d po (o ) m, or dt dt RTo

energy:

B

po 0.6847A* , where B To R

dQ dW d p 0 o cv To mcp To dt dt dt RTo

Solutions Manual Fluid Mechanics, Seventh Edition

658

We may rearrange and combine these to give a single differential equation for T o:

dTo 0.6847 CTo3/ 2 , where C (k 1)A* R , or dt

1 1 Integrate: To (t) Ct To (0) 2

2

dTo C dt 3/2 To

Ans.

With To(t) known, we could go back and solve the mass relation for po(t), but in fact that is not necessary. We simply use the isentropic-flow assumption: po (t) po (0)

k/(k 1)

To (t) To (0)

1 1 CTo1/2 (0)t 2

2k/(k 1)

0.6847(k 1)A* R C

Ans.

Clearly, tank pressure also decreases with time as the tank blows down.

9.38 Prob. 9.37 makes an ideal senior project or combined laboratory and computer problem, as described in Ref. 30, sec. 8.6. In Bober and Kenyon’s lab experiment, the tank had a volume of 0.0352 ft3 and was initially filled with air at 50 lb/in2 gage and 72F. Atmospheric pressure was 14.5 lb/in2 absolute, and the nozzle exit diameter was 0.05 in. After 2 s of blowdown, the measured tank pressure was 20 lb/in2 gage and the tank temperature was –5F. Compare these values with the theoretical analysis of Prob. 9.37. Solution:

Use the formulas derived in Prob. 9.37 above, with the given data:

To (0) 72 460 532 R,

0.6847( /4)(0.05/12) 2(1.4 1) 1717 1 “C” 0.0044 0.0352 ft3 s R1/2

1 Then To (t) To (0) 1 (0.0044) 532t 2

2

532/[1 0.0507t]2

Similarly, p o p o(0)[To/To(0)] k/(k 1) (50 14.5 psia)/[1 0.0507t] 7

Some numerical predictions from these two formulas are as follows: t, sec: To, R:

po , psia:

0 532.0 64.5

0.5 506.0 54.1

1.0 481.9 45.6

1.5 459.5 38.6

2.0 438.6R

32.8 psia

At t 2 sec, the tank temperature is 438.6R –21.4F, compared to –5F measured.

Chapter 9 Compressible Flow

659

At t 2 sec, the tank pressure is 32.8 psia 18.3 psig, compared to 20 psig measured. The discrepancy is probably due to heat transfer through the tank walls warming the air. 9.39 Consider isentropic flow in a channel of varying area, between sections 1 and 2. Given Ma1 2.0, we desire that V 2/V1 equal 1.2. Estimate (a) Ma 2 and (b) A2/A 1. (c) Sketch what this channel looks like, for example, does it converge or diverge? Is there a throat? Solution:

This is a problem in iteration, ideally suited for EES. Algebraically, 1/2

2 V2 Ma2 a2 Ma2 a o 1 0.2 Ma 2 1.2, given that Ma1 2.0 V1 Ma1 a1 Ma1 a 1 0.2 Ma 2 1/2 o 1 For adiabatic flow, a o is constant and cancels. Introducing Ma1 2.0, we have to solve Ma 2/[1 0.2Ma 22]1/2 1.789. By iteration, the solution is: Ma 2 2.98 Ans. (a)

Then

A2 A2 / A* 4.1547 (Table B.1) 2.46 A / A * 1.6875 A1 1

There is no throat between, a supersonic expansion. Ans. (c)

Ans. (b)

(1)

(2) Supersonic

P9.40 Steam, in a tank at 300 kPa and 600 K, discharges isentropically to a low-pressure atmosphere through a converging nozzle with exit area 5 cm2. (a) Using an ideal gas approximation from Table B.4, estimate the mass flow. (b) Without actual calculations, indicate how you would use real properties of steam, from EES, to find the mass flow. Solution: The code words “low-pressure atmosphere” mean that the flow is choked at the exit. For steam, from Table B.4, assume k = 1.33 and R = 461 m2/s 2-K. Then we are at maximum mass flow:

2 0.5(k 1) /(k 1) p o A * k 1.33 : mmax [k 1/ 2 ( ) ] k 1 RT o (300000 Pa )(0.0005 m 2 ) [0.6726] (461 J / kg K )(600 K )

0.192

kg s

Ans.(a )

Solutions Manual Fluid Mechanics, Seventh Edition

660

(b) Using EES for real steam, we don’t have these nice power-law formulas, but we can use the energy equation and the continuity equation and the fact that the entropy is constant: Energy : h

1 2 V h o ; Continuity : m A V constant ; s s o constant 2

First establish so from the given p o and T o. Then guess a velocity V, perhaps starting at 300 m/s, compute h from energy, then compute = (h, so) from the EES thermophysical functions. This enables us to calculate the (guessed) mass flow = A V. Is it a maximum? Probably not. Keep changing V until you reach a maximum mass flow. The final result obtained by the writer is At

V 569 m / s and 0.683 kg / m 3 , m mmax 0.191

kg s

Ans.(b)

The perfect-gas result is very accurate. Steam is nearly ideal in this superheat region. Meanwhile, you could also use the new EES thermophysical function SOUNDSPEED for this problem, monitoring a for each guessed velocity V and computing the Mach number V/a. you would find that, at V = 559 m/s, the mass flow is maximum, and Ma = 1.000 (choked).

9.41 Air, with a stagnation pressure of 100 kPa, flows through the nozzle in Fig. P9.41, which is 2 m long and has an area variation approximated by A 20 20 x 10 x 2

with A in cm2 and x in m. It is desired to plot the complete family of isentropic pressures p(x) in this nozzle, for the range of inlet pressures 1 p(0) 100 kPa. Indicate those inlet pressures which are not physically possible and discuss briefly. If your computer has an online graphics routine, plot at least 15 pressure profiles; otherwise just hit the highlights and explain.

Chapter 9 Compressible Flow

661

Fig. P9.41

Solution: There is a subsonic entrance region of high pressure and a supersonic entrance region of low pressure, both of which are bounded by a sonic (critical) throat, and both of which have a ratio A x 0 /A* 2.0. From Table B.1 or Eq. (9.44), we find these two conditions to be bounded by a) subsonic entrance: A/A* 2.0, Mae 0.306, pe 0.9371po 93.71 kPa b) supersonic entrance:

A/A* 2.0, Ma e 2.197, p e 0.09396p o 9.396 kPa

Thus no isentropic flow can exist between entrance pressures 9.396 p e 93.71 kPa. The complete family of isentropic pressure curves is shown in the graph on the following page. They are not easy to find, because we have to convert implicitly fro m area ratio to Mach number.

Solutions Manual Fluid Mechanics, Seventh Edition

662

9.42 A bicycle tire is filled with air at 169.12 kPa (abs) and 30C. The valve breaks, and air exhausts into the atmosphere of 100 kPa (abs) and 20C. The valve exit is 2-mmdiameter and is the smallest area in the system. Assuming one-dimensional isentropic flow, (a) find the initial Mach number, velocity, and temperature at the exit plane. (b) Find the initial mass flow rate. (c) Estimate the exit velocity using the incompressible Bernoulli equation. How well does this estimate agree with part (a)? Solution:

(a) Flow is not choked, because the pressure ratio is less than 1.89:

po 169.12 1 0.2 Mae2 p 100

3.5

, solve Mae 0.90;

Read Te 0.8606To 261 K

Ve Mae ae (0.90) 1.4(287)(261) 0.90(324) 291

m s

Ans. (a)

kg s

Ans . (b)

(b) Evaluate the exit density at Ma 0.90 and thence the mass flow:

e

pe 100000 kg 1.335 3 , RTe 287(261) m

Then m e AeVe (1.335)

4

(0.002)2 (291) 0.00122

Chapter 9 Compressible Flow

663

(c) Assume o tire, for how would we know exit if we didn’t use compressibleflow theory? Then the incompressible Bernoulli relation predicts

o Ve ,inc

2 p

o

po 169120 kg 1.945 3 RTo 287(303) m

2(169120 100000) m 267 s 1.945

Ans. (c)

This is 8% lower than the “exact” estimate in part (a).

P9.43 Air flows isentropically through a variable-area duct. At section 1, A 1 = 20 cm2, p 1 = 300 kPa, 1 = 1.75 kg/m3, and Ma 1 = 0.25. At section 2, the area is exactly the same, but the flow is much faster. Compute (a) V2; (b) Ma 2; and (c) T2, and (d) the mass flow. (e) Is there a sonic throat between sections 1 and 2? If so, find its area. Solution: If the areas are the same but the velocities different, there must be a sonic throat in be-tween. (e) We can find the throat area A* right away. For k = 1.40, from Eq. (9.45), Ma1 = 0.25,

A1 20 cm 2 1 (1 0.2Ma12 )3 2.4027 , solve A * Athroat 8.32 cm 2 Ans.(e) A* A* Ma1 1.728

(d) Compute V 1 and then we can find the mass flow:

T1

p1 300000 597 K ; V1 Ma1 a1 0.25 1.4(287 )(597 ) 123 m / s R 1 287 (1.75)

1 A1V1 (1.75 Then m

kg

m

3

)(0.0020 m 2 )(123

m kg ) 0.43 s s

Ans.(d )

We also need To T1 (1 0.2Ma12 ) (597 K )[1 0.2(0.25)2 ] 604K Now go over to section 2 and compute those properties. We have the same area ratio:

Solutions Manual Fluid Mechanics, Seventh Edition

664

A2 also 2.4027 ; Table B.1 : Ma2 2.400 Ans.( b) A* T2 To /(1 0.2 Ma22 ) (604 K ) /[1 0.2(2.400) 2] 281 K Ans.(c) m V2 Ma2 kRT2 (2.400 ) 1.4( 287 )(281) 806 Ans.(a ) s

9.44 In Prob. 3.34 we knew nothing about compressible flow at the time so merely assumed exit conditions p2 and T2 and computed V2 as an application of the continuity equation. Suppose that the throat diameter is 3 in. For the given stagnation conditions in the rocket chamber in Fig. P3.34 and assuming k 1.4 and a molecular weight of 26, compute the actual exit velocity, pressure, and temp-erature according to one-dimensional theory. If pa 14.7 lbf/in2 absolute, compute the thrust from the analysis of Prob. 3.68. This thrust is entirely independent of the stagna-tion temperature (check this by changing To to 2000°R if you like). Why?

Fig. P3.34

Solution: If M 26, then R gas 49720/26 1912 ftlbf/slugR. Assuming choked flow in the throat (to produce a supersonic exit), the exit area ratio yields the exit Mach number: A e D e 5.5 3.361, whence Eq. 9.45 (for k 1.4) predicts Mae 2.757 A* D* 3.0 2

2

Then isentropic pe 400/[1 0.2(2.757) 2]3.5 15.7 psia Te 4000 R/[1 0.2(2.757) ] 1587R 2

Ans.

Ans.

Chapter 9 Compressible Flow

665

Then Ve Ma e kRTe 2.757 1.4(1912)(1587) 5680 ft/s

Ans.

We also need e p e/RTe (15.7 144)/[1912(1587)] 0.000747 slug/ft 3 From Prob. 3.68, Thrust F A e eVe2 (p e p a) , 2 5.5 or: F [0.000747(5680)2 (15.7 14.7) 144] 4000 lbf Ans. 4 12

Thrust is independent of To because e 1/To and Ve To, so T o cancels out.

P9.45 Air flows isentropically from a large tank through a variable-area nozzle. At section 1, where A1 = 12 cm2, conditions are p1 = 20.5 kPa, T1 = 232 K, and V 1 = 733 m/s. Find (a) the mass flow; (b) Ma1; (c) the throat area, if any; and (d) the pressure and temperature in the tank. Solution: We have enough information at section 1 to find the mass flow:

1

p1 20500 kg kg ; m 1 A1V1 (0.308)(0.0012)(733) 0.271 0.308 RT1 (287)(232) s m3

And the information at section 1 also yields the local Mach number: a1

kRT1 1.4(287)(232) 305

m V 733m / s 2.40 Ans.(b ) ; Ma1 1 s a1 305 m / s The local flow is supersonic, so there definitely is a sonic throat:

Ma1 2.40:

A1 12cm 2 2.403 , A * 5.0 cm 2 A* 2.403

Ans.(c)

Finally, the tank (stagnation) conditions are calculated by the isentropic ratios: po Ma1 2.40 : [1 0.2(2.4) 2 ]3.5 14.6 , po 300 kPa 20.5 kPa

To [1 0.2(2.4) 2 ] 2.15, To 500 K 232 K

Ans.(d )

Ans.( a)

Solutions Manual Fluid Mechanics, Seventh Edition

666

P9.46

A one-dimensional isentropic airflow has the following properties at one section where

the area is 53 cm 2: p = 12 kPa, = 0.182 kg/m 3, and V = 760 m/s. Determine (a) the throat area; (b) the stagnation temperature; and (c) the mass flow.

Solution: We already have what we need to compute the mass flow:

m AV (0.182 kg / m 3 )(0.0053 m 2 )(760 m / s )

0.733 kg / s

Ans.(c )

Now we need the Mach number at this section:

T

p 12000 Pa 230 K R (0.182 kg / m 3 )( 287 m 2 / s 2 K )

Ma

V a

V 760 760 2.50 304 kRT 1.4(287 )(230 ) The flow at this station is supersonic; therefore a sonic throat exists. We may now calculate

T o T (1 0.2Ma 2 ) (230 K )[1 0.2(2.50) 2 ]

518 K

Ans.(b )

A (1 0.2 Ma2 ) 3 [1 0.2(2.50) 2 ]3 0.0053 2.64 ; A * 0.0020 m2 Ans.( a) A* 1.728 Ma 1.728(2.50) 2.64

9.47 In wind-tunnel testing near Mach 1, a small area decrease caused by model blockage can be important. Let the test section area be 1 m 2 and unblocked conditions are Ma 1.1 and T 20°C. What model area will first cause the test section to choke? If the model cross-section is 0.004 sq.m., what % change in test-section velocity results? Solution:

First evaluate the unblocked test conditions:

T 293K, a kRT 1.4(287)(293) 343

m m , V (1.1)(343) 377 s s

A [1 0.2(1.1)2 ]3 Also, 1.007925 , or A* 1.0/1.007925 0.99214 m2 A* 1.728(1.1) (unblocked)

Chapter 9 Compressible Flow

667

If A is blocked by 0.004 m 2, then Anew 1.0 0.004 0.996 m2, and now

A new 0.996 1.00389, solve Eq. (9.45) for Ma(blocked) 1.0696 A* 0.99214

m Ans. s Thus a 0.4% decrease in test section area has caused a 2.1% decrease in test velocity. Same To 364 K, new T 296 K, new a 345 m/s, new V Ma(a) 369

9.48 A force F 1100 N pushes a piston of diameter 12 cm through an insulated cylinder containing air at 20°C, as in Fig. P9.48. The exit diameter is 3 mm, and pa 1 atm. Estimate (a) V e, (b) Vp, and (c) me .

Fig. P9.48

Solution:

First find the pressure inside the large cylinder: pp

F 1100 1 atm 101350 198600 1.96 atm A ( /4)(0.12)2

Since this is greater than (1/0.5283) atm, the small cylinder is choked, and thus Vexit

2k 2(1.4) RTo (287)(293) 313 m/s k 1 1.4 1

Ans . (a)

V piston ( e /p)(Ae /Ap)Ve (0.6339)(0.003/0.12) 2(313) 0.124 m/s Finally, m m max 0.6847

(198600)( /4)(0.003) 2 287(293)

0.00331

kg s

Ans. (b) Ans. (c)

The mass flow increases with F, but the piston velocity and exit velocity are independent of F if the exit flow is choked.

Solutions Manual Fluid Mechanics, Seventh Edition

668

9.49 Consider the venturi nozzle of Fig. 6.40c, with D 5 cm and d 3 cm. Air stagnation temperature is 300 K, and the upstream velocity V 1 72 m/s. If the throat pressure is 124 kPa, estimate, with isentropic flow theory, (a) p1; (b) Ma 2; and (c) the mass flow. Solution: Given one-dimensional isentropic flow of air. The problem looks sticky— sparse, scattered information, implying laborious iteration. But the energy equation yields V1 and Ma1: V 12 (72 m/s)2 To T1 , solve for T1 297.4K 300K T1 2 cp 2(1005J/kgK)

V1

Ma1

72

kRT1

1.4(287)(297.4)

72 m/s 0.208 346 m/s

Area-ratio calculations will then yield A* and Ma2 and then po and p1: A1 ( /4)(0.05m) A* A* 2

1 0.2 Ma12

Solve

1.728 Ma1

1 0.2 Ma22

po p2 1 0.2 Ma 22

p1 po 1 0.2 Ma12

3.5

3.5

[1 0.2(0.208)2 ]3 2.85, 1.728(0.208)

A* 0.0006886 m 2

A 2 ( /4)(0.03 m) A* 0.0006886 m2 1.728Ma 2 2

3

3

1.027, Solve Ma 2 0.831 Ans. (b)

(124 kPa)[1 0.2(0.831)2 ]3.5 195 kPa

(195 kPa)/[1 0.2(0.208)2 ]3.5 189 kPa Ans. (a)

The mass flow follows from any of several formulas. For example:

p 189000 kg 2 m 1 A1V1 1 A1V1 (0.05) (72) 0.313 s RT1 287(297.4) 4

Ans . (c)

P9.50 Carbon dioxide is stored in a tank at 100 kPa and 330 K. It discharges to a second tank through a converging nozzle whose exit area is 5 cm 2. What is the initial mass flow rate if the second tank has a pressure of (a) 70 kPa, or (b) 40 kPa?

Chapter 9 Compressible Flow

669

Solution: For CO2 , from Table B.4, R = 189 m2/s 2-K and k = 1.30. Calculate the pressure p* if the flow is choked at the exit:

po (k 1) (1.3 1) [1 (1.0)2 ]k /( k 1) [1 (1.0)2 ]1.3/ 0.3 1.8324 p* 2 2 100 kPa Hence, for this case, p * 54.6 kPa 1.8324 Thus, for case (a), p b = 70, the nozzle is not choked. For case (b), p b = 40, it is choked. (a) We could plow through the various exit properties and finally find the mass flow: pe 70 kPa ; Mae 0.756 ; e 1.22

m m Maa ae 207 ; V e s s m3 m e Ae Ve (1.22)(0.0005)(207) 0.126 kg / s Ans( a) kg

; ae 273

Or we could do it all in one step with the handy mass flow function, Eq.(9.47), p/po = 0.546: m RT o Ae po

2(1.3) (0.546)2 /1.3 [1 (0.546)0.3/1.3 ] 0.629 1.3 1 (0.0005)(100, 000) kg Hence m 0.629 0.126 s (189)(330)

Ans.( a)

(b) For an outside pressure of only 40 kPa < 54.6 kPa, the exit is choked:

m mmax [k1/ 2 (

2 0.5( k1) /( k1) po A * ) ] k 1 RTo

[0.6673]

(100,000)(0.0005) (189)(330)

0.134

kg s

Ans.(b)

P9.51 The scramjet engine of Fig. 9.30 is supersonic throughout. A sketch is shown in Fig. C9.8. Test the following design. The flow enters at Ma = 7 and air properties for 10,000 m altitude. Inlet area is 1 m2, the minimum area is 0.1 m2, and the exit area is 0.8 m2 . If there is no combustion, (a) will the flow still be supersonic in the throat? Also, determine (b) the exit Mach number, (c) exit velocity, and (d) exit pressure.

Solution: From Table B.6 at 10,000 m, read p 1 = 26416 Pa, T1 = 223.16 K, and 1 = 0.4125 kg/m3. Establish area ratio and stagnation conditions at the inlet, section 1:

Solutions Manual Fluid Mechanics, Seventh Edition

670

Ma1 7:

A1 104.1 ; po 1 (26416)[1 0.2(7) 2 ]3.5 1.094E8 Pa A* To 1 (223.16)[1 0.2(7)2 ] 2410 K

Now, assuming isentropic flow (no combustion), work your way through the area ratios:

A2 0.1m2 : A2 / A * (104.1)(0.1m2 /1.0m2 ) 10.41 , for which Mathroat 3.97 Ans.(a ) A4 0.8 m2 : A4 / A * (104.1)(0.8 m2 /1m2 ) 83.28, for which Ma4 6.655

Ans.(b)

T4 2410 K /[1 0.2(6.655)2 ] 244 K , V4 Maa a4 (6.655) 1.4(287)(244) 2080 m / s Ans.(c) p4 1.094E8 Pa /[{1 0.2(6.655)2 }3.5 ] 36,400 Pa

Ans.(d )

9.52 A converging-diverging nozzle exits smoothly to sea-level standard atmosphere. It is supplied by a 40-m3 tank initially at 800 kPa and 100 C. Assuming isentropic flow, estimate(a) the throat area; and (b) the tank pressure after 10 sec of operation. The exit area is 10 cm 2. Solution: The phrase “exits smoothly” means that exit pressure atmospheric pressure, which is 101 kPa. Then the pressure ratio specifies the exit Mach number: po /pexit

3.5 800 1 0.2 Ma2e , solve for Maexit 2.01 101

Thus Ae /A* 1.695 and A* (10 cm 2)/1.695 5.9 cm 2 Ans .(a) Further, m mmax 0.6847(800000)(0.00059)

287(373) 0.99 kg/s

The initial mass in the tank is quite large because of large volume and high pressure:

o

po 800000 kg 7.47 3 , thus m tank,t 0 (7.47)(40) 299 kg RTo 287(373) m

After 10 sec, blowing down at 0.99 kgs, we have about 299 10 289 kg left in the tank. The pressure will drop to about 800(289299) 773 kPa. Ans. (b).

Chapter 9 Compressible Flow

671

9.53 Air flows steadily from a reservoir at 20C through a nozzle of exit area 20 cm2 and strikes a vertical plate as in Fig. P9.53. The flow is subsonic throughout. A force of 135 N is required to hold the plate stationary. Compute (a) Ve , (b) Mae , and (c) p 0 if pa 101 kPa.

Fig. P9.53

Solution: Assume pe 1 atm. For a control volume surrounding the plate, we deduce that F 135N e Ve2 Ae kpe Mae2 Ae 1.4(101350)(0.002 m2 )Ma2e , or Mae 0.69 Ans. (b)

Te 293/[1 0.2(0.69)2 ] 268 K, ae 1.4(287)(268) 328 Ve ae Mae (328)(0.69) 226 m/s Ans . (a)

Finally, ptank po 101350[1 0.2(0.69) 2]3.5 139000 Pa P9.54

m , thus s

Ans. (c)

The airflow in Prob. P9.46 undergoes a normal shock just past the section where

data was given. Determine the (a) Mach number, (b) pressure, and (c) velocity just

downstream of the shock. [Recall the given data: the area is 53 cm 2, p = 12 kPa, = 0.182 kg/m3, and V = 760 m/s.]

Solution: In Prob. P9.46 we found that the local Mach number at the section was 2.5. Table B.2: Ma2 = 0.513

Ans.(a) ;

p2/p1 = 7.125 , hence p2 = 7.125*12000 = 85500 Pa

V1/V2 = 2/ 1 = 3.333

hence

V2 = (760)/(3.333) =

Ans.(b) 228 m/s

Ans.(c)

The shock results were quite easy – finding the upstream Mach number was the hard part.

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672

9.55 Air, supplied by a reservoir at 450 kPa, flows through a converging-diverging nozzle whose throat area is 12 cm2. A normal shock stands where A1 20 cm2. (a) Compute the pressure just downstream of this shock. Still farther downstream, where A 3 30 cm2, estimate (b) p3; (c) A 3*; and (d) Ma 3. Solution:

If a shock forms, the throat must be choked (sonic). Use the area ratio at (1):

A1 20 450 1.67, or Ma1 1.985, whence p1 59 kPa 2 3.5 A* 12 [1 0.2(1.985) ]

Then, across the shock,

p2 2.8(1.985)2 0.4 4.43, p1 2.4

p 2 4.43(59) 261 kPa

Across the shock, at Ma1 1.985, At A 3 30 cm2 , Finally, po2

A*2 A*1

Ans. (a)

1.374, A *2 1.374(12) 16.5 cm2

Ans.(c)

A3 30 1.82, whence subsonic Ma 3 0.34 Ans. (d) 16.5 * A2

p o1 328 328 kPa, p3 303 kPa 1.374 [1 0.2(0.34)2 ]3.5

Ans. (b)

9.56 Air from a reservoir at 20C and 500 kPa flows through a duct and forms a normal shock downstream of a throat of area 10 cm2. By an odd coincidence it is found that the stagnation pressure downstream of this shock exactly equals the throat pressure. What is the area where the shock wave stands? Solution:

If a shock forms, the throat is sonic, A* 10 cm 2. Now p*1 0.5283po1 0.5283(500) 264 kPa po2

Then

also

po2 264 0.5283: Table B.2, read Ma 1 2.43 po1 500

[1 0.2(2.43) 2] 3.0 So A1 /A* 2.47, or A1 (at shock) 2.47(10) 24.7 cm 2 1 1.728(2.43)

Ans.

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673

9.57 Air flows from a tank through a nozzle into the standard atmosphere, as in Fig. P9.57. A normal shock stands in the exit of the nozzle, as shown. Estimate (a) the tank pressure; and (b) the mass flow. Solution: The throat must be sonic, and the area ratio at the shock gives the Mach number:

Fig. P9.57

1 0.2Ma12 14 1.4 A1 /A* , solve Ma1 1.76 upstream of the shock 10 1.728Ma1 3

2.8(1.76)2 0.4 101350 Then p 2/p 1 shock 3.46, p 2 1 atm, p 1 29289 Pa 2.4 3.46

Thus ptank po1 29289[1 0.2(1.76) 2]3.5 159100 Pa

Ans. (a)

Given that To 100C 373 K and a critical throat area of 10 cm 2, we obtain

m m max 0.6847p oA*

RTo 0.6847(159100)(0.001) 0.333

kg s

287(373)

Ans. (b)

P9.58 The data of Prob. P9.4 represent conditions before and after a normal shock wave. If the velocity at section B is 238 m/s, what are (a) the velocity at D; (b) the stagnation temperature, (c) the stagnation pressure at D, and (d) the Mach number at section D? (D)

p = 28.2 kPa T = -19C = 254 K

p = 154 kPa = 1.137 kg/m^3 V = 238m/s

(B)

Solution: Assume k = 1.4. From Prob. P9.4, we found that section D is upstream. The pressure ratio will give us the upstream Mach number MaD: pB 154,000Pa 5.46 ; Table B.1 or Eq.(9.54) : MaD 2.20 Ans .(d ) pD 28, 200 Pa

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674

With the upstream Mach number known, we can find the stagnation pressure there:

p0 D pB (1 0.2 MaD2 )3.5 (28.2 kPa )[1 0.2(2.20)2 ]3.5 300 kPa Ans.(c ) The downstream stagnation pressure (not asked) is much less, 189 kPa. Similarly, given TD = -19C = 254 K, we can calculate the stagnation temperature (good for both sides):

T0 D TD (1 0.2 Ma2D ) (254 K)[1 0.2(2.20)2 ] 500 K T0 B Ans.( b) Finally, we could compute V D from the ratio across the shock, Eq. (9.58), 2 VD ( k 1) MaD V 2.4(2.20) 2 m Ans.(a) 2.95 D , VD 702 2 2 VB 238 s (k 1)Ma D 2 (0.4)(2.20) 2

Alternately, V D = Ma D (kRTD) 1/2 = (2.20)[1.4(287)(254)]1/2 = (2.20)(319) = 702 m/s. 9.59 Air, at stagnation conditions of 450 K and 250 kPa, flows through a nozzle. At section 1, where the area 15 cm 2, there is a normal shock wave. If the mass flow is 0.4 kgs, estimate (a) the Mach number; and (b) the stagnation pressure just downstream of the shock. Solution:

If there is a shock wave, then the mass flow is maximum:

mmax 0.4 Then

kg p A* 250000 A* 0.6847 o 0.6847 , solve A* 0.000840 m 2 s RTo 287(450) A1 0.0015 1.786 Table B.1: Read Ma1,upstream 2.067 A* 0.00084

Finally, from Table B.2, read Ma 1, downstream 0.566 Also, Table B.2:

Ans. (a)

po2 0.690, p0,downstream 0.690(250) 172 kPa Ans . (b) po1

9.60 When a pitot tube such as Fig. (6.30) is placed in a supersonic flow, a normal shock will stand in front of the probe. Suppose the probe reads po 190 kPa and p 150 kPa. If

Chapter 9 Compressible Flow

675

the stagnation temperature is 400 K, estimate the (supersonic) Mach number and velocity upstream of the shock.

Fig. P9.60

Solution:

We can immediately find Ma inside the shock: po2 /p2

190 1.267 1 0.2Ma22 150

Then, across the shock, T1

Ma12

3.5

, solve Ma2 0.591

0.4(0.591)2 2 , solve Ma 1 1.92 2.8(0.591)2 0.4

Ans.

400 230 K, a1 1.4(287)(230) 304 m/s, [1 0.2(1.92) 2]

V1 Ma1a1 (1.92)(304) 585 m/s

Ans.

P9.61 Air flows from a large tank, where T = 376 K and p = 360 kPa, to a design condition where the pressure is 9800 Pa. The mass flow is 0.9 kg/s. However, there is a normal shock in the exit plane just after this condition is reached. Estimate (a) the throat area; and, just downstream of the shock, (b) the Mach number, (c) the temperature, and (d) the pressure. Solution: The low design pressure definitely indicates a supersonic condition:

p 9800 0.0272 (1 0.2 Ma12 )3.5 , solve for po 360000

m max 0.6847 Chok ed : m

po A * RTo

0.6847

(360000 ) A * 287 (376 )

Ma1

3.00

, Solve A * 0.0012 m 2 Ans.(a )

Find T1 in front of the shock and then use the normal shock conditions:

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676

T1 To /(1 0.2 Ma12 ) (376 K ) /[1 0.2(3.0) 2 ] 134 K Table B.2, Ma1 3.0 : Ma2 0.475 Ans.( b)

T 2 / T1 2.679 , T 2 (2.679 )(134 K ) 360 K Ans.(c) p 2 / p1 10.333 , p 2 (10.333 )(9800 ) 101,300 Pa Ans.(d ) The problem worked out so that the external pressure was 1 atm.

9.62 An atomic explosion propagates into still air at 14.7 psia and 520R. The pressure just inside the shock is 5000 psia. Assuming k 1.4, what are the speed C of the shock and the velocity V just inside the shock? Solution:

The pressure ratio tells us the Mach number of the shock motion:

5000 2.8Ma 21 0.4 340 p 2 /p1 , solve for Ma1 17.1 14.7 2.4 ft a1 1.4(1717)(520) 1118 ft/s, V1 C 17.1(1118) 19100 s

Ans. (a)

We then compute the velocity ratio across the shock and thence the relative motion inside:

0.4(17.1) 2 2 0.1695, V2 0.1695(19100) 3240 ft/s V2 /V1 2 2.4(17.1) Then Vinside C V2 19100 3240 15900 ft/s Ans. (b)

Chapter 9 Compressible Flow

677

9.63 Sea-level standard air is sucked into a vacuum tank through a nozzle, as in Fig. P9.63. A normal shock stands where the nozzle area is 2 cm 2, as shown. Estimate (a) the pressure in the tank; and (b) the mass flow. Solution: The flow at the exit section (“3”) Fig. P9.63 is subsonic (after a shock) therefore must equal the tank pressure. Work our way to 1 and 2 at the shock and thence to 3 in the exit: 101350 p01 101350 Pa, A1 /A* 2.0, thus Ma1 2.1972, p1 9520 Pa [1 0.2(2.2)2 ]3.5 p2 2.8(2.2) 2 0.4 5.47, p2 5.47(9520) 52030 Pa p1 2.4 Also compute A* /A* 1.59, or A* 1.59 cm2 2

1

2

Also compute po2 1013501.59 63800 Pa. Finally compute A 3 /A * 2 3/1.59 1.89, 2 3.5 read Ma3 0.327, whence p 3 63800[1 0.2(0.327) ] 59200 Pa. Ans. (a). With T o 288 K, the (critical) mass flow 0.6847p oA*(RTo ) 0.0241 kg s. Ans. (b)

P9.64Air, from a reservoir at 350 K and 500 kPa, flows through a converging-diverging nozzle. The throat area is 3 cm 2. A normal shock appears, for which the downstream Mach number is 0.6405. (a) What is the area where the shock appears? Calculate ( b) the pressure and (c) the temperature downstream of the shock. Solution: This Ma 2 = 0.6405 occurs right in Table B.2, and we read Ma1 = 1.70. (a) Hence, for isentropic flow up to that point, from Eq. (9.45), A 1 1 0.2(1.70)2 ]3 ( 1.3376 , Ashock (1.3376)(3cm 2 ) 4.0 cm 2 Ans.(a ) )[ A* 1.70 1.728

(b) Calculate pressure before and after the shock:

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678

Read Table B.1:

p1 0.2026 , p1 (0.2026)(500 kPa) 101.3 kPa po

Read Table B.2 at Ma 1 1.7 :

p2 3.205; p 2 (3.205)(101.3) 325 kPa Ans.( b) p1

(c) Do the same thing as (b) to determine temperature downstream of the shock:

Read Table B.1:

T1 0.6337 , p1 (0.6337)(350 K ) 222 K To

Read Table B.2 at Ma1 1.7 :

T2 1.4583; T2 (1.4583)(222) 323 K Ans.( c) T1

9.65 Air flows through a convergingdiverging nozzle between two large reservoirs, as in Fig. P9.65. A mercury manometer reads h 15 cm. Estimate the downstream reservoir pressure. Is there a shock wave in the flow? If so, does it stand in the exit plane or farther upstream? Solution: The manometer reads the pressure drop between throat and exit tank:

Fig. P9.65

pthroat ptank#2 (mercury air )gh (13550 0)(9.81)(0.15) 19940 Pa

The lowest possible pthroat p* 0.5283(300) 158.5 kPa, for which pe 138.5 kPa

But this pe is much lower than would occur in the duct for isentropic subsonic flow.

We can check also to see if isentropic supersonic flow is a possibility: With A e A* 3.0, the exit Mach number would be 2.64, corresponding to pe 0.047p o 14 kPa (?). This is much too low, so that case fails also. Suppose we had supersonic flow with a normal shock wave in the exit plane:

ptank#2 2.8(2.64)2 0.4 A e /A* 3.0, Ma e 2.64, p e 14 kPa, 7.95, pe 2.4

or: ptank#2 7.95(14) 113 kPa, compared to ptank (manometer reading) 138.5 kPa

Chapter 9 Compressible Flow

679

This doesn’t match either, the flow expanded too much before the shock wave. Therefore the correct answer is: a normal shock wave upstream of the exit plane. Ans. 9.66 In Prob. 9.65 what would be the mercury manometer reading if the nozzle were operating exactly at supersonic “design” conditions? Solution: We worked out this idealized isentropic-flow condition in Prob. 9.65: Fig. P9.65

Ae 3.0, Mae 2.64, pe 14 kPa ptank #2 ; p* pthroat 158.5 kPa A* Then p t pe 158500 14200 144300 (13550 0)(9.81)h,

Design flow:

solve h 1.09 m Ans.

P9.67 A supply tank at 500 kPa and 400 K feeds air to a converging diverging nozzle whose throat area is 9 cm2. The exit area is 46 cm 2. State the conditions in the nozzle if the pressure outside the exit plane is (a) 400 kPa; (b) 120 kPa; and (c) 9 kPa. (d) In each of these cases, find the mass flow. Solution: For reference purposes, find the design (isentropic supersonic) condition for this nozzle:

Ae 46 cm 2 5.111 ; Table B.1 : Read |design 2 A* 9 cm p exit , design

po

(1 0.2Ma e2 )3.5

500 ,000 Pa

[1 0.2(3.20) ]

Madesign 3.20

2 3.5

10140 Pa

This immediately establishes what happens in case (c): poutside = 9000 Pa < pdesign, hence supersonic expansion outside the exit (Case I, Fig. 9.12) Now determine the outside pressure if there is a normal shock in the exit plane:

Ma1 3.20; TableB.2 : p2 / p1 11.78 , hence p2 (11.78)(10140 ) 119,500 Pa

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680

This is close enough to case (b), a back pressure of 120 kPa, that we can say there is a normal shock in the exit. (case F, Fig. 9.12) Ans.(b) Finally, check the situation if the nozzle downstream of the throat is subsonic-isentropic:

Ae 46 cm2 5.111 ; Table B.1 : Read |subsonic 2 A* 9 cm

p exit ,subsonic

po

Masubsonic 0.114

500 ,000 Pa

(1 0.2 Mae2 )3.5 [1 0.2(0.114 )2 ]3.5 This is greater than case (a), a postulated back pressure of 450,000 Pa.

495,000 Pa

Thus case (a) will result in a normal shock just downstream of the throat. (Case D, Fig. 9.12) In all three cases, the throat is choked, hence the mass flow is maximum:

m max 0.6847 m

po A * RTo

0.6847

(500000 )(0.0009 ) (287 )( 400 K )

0.909

kg Ans.( a, b, c) s

9.68 Air in a tank at 120 kPa and 300 K exhausts to the atmosphere through a 5-cm2-throat converging nozzle at a rate of 0.12 kgs. What is the atmospheric pressure? What is the maximum mass flow possible at low atmospheric pressure? Solution:

Let us answer the second question first, to see where 0.12 kgs stands:

mmax

0.6847po A* RTo

0.140 kg/s

0.6847(120000)(0.0005) 287(300)

Ans. (b) (if patm 63 kPa)

So the given mass flow is about 86% of maximum and p atm 63 kPa. We could just go at it, guess the exit pressure and iterating, or we could express it more elegantly:

m AV

o

A Ma kR (1 0.2Ma2 )2.5

To

1 0.2Ma2

Const Ma , (1 0.2Ma2 )3

where Const 0.2419 in SI units. If m 0.12 kg/s, we thus solve for Ma:

Ma 0.496(1 0.2Ma2 )3

to obtain Ma 0.62, p atm 92.6 kPa

Ans.

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681

9.69 With reference to Prob. 3.68, show that the thrust of a rocket engine exhausting into a vacuum is given by F

p0 Ae 1 k Ma2e

k 1 2 1 Ma e 2

k /(k 1)

where Ae exit area Mae exit Mach number p0 stagnation pressure in combustion chamber Note that stagnation temperature does not enter into the thrust. Solution:

In a vacuum, p atm 0, the solution to Prob. 3.68 is

F eA eVe2 A e(p e 0) A e p e eVe2 ,

but e Ve2 kpe Mae2, hence F p eA e 1 kMa e2 For isentropic flow, pe po

k 1 2 Ma e 1 2

k/(k 1)

, F

p oA e 1 kMa 2e

k 1 2 1 2 Mae

k/(k 1)

Ans.

P9.70 Air, with p o = 500 kPa and To = 600 K, flows through a converging-diverging nozzle. The exit area is 51.2 cm 2, and the mass flow is 0.825 kg/s. What is the highest possible back pressure that will still maintain supersonic flow inside the diverging section? Solution: Naturally assume isentropic flow with k = 1.40. If the flow is to be supersonic, there must be a choked throat. Find its area:

m m max 0.6847

po A * RTo

0.6847

(500000) A * (287)(600)

0.825

kg s

Solve for A * Athroat 0.0010 m2 10 cm2

With A* known, we find the desired supersonic Mach number in the exit plane:

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682

Ae 51.2 cm2 5.12 ; Table B.1 or Eq.(9.48) : Mae 3.2 A* 10cm 2 pe

po

[1 0.2(Mae2 )]3.5

500, 000

[1 0.2(3.2) ]

2 3.5

10,100 Pa

If 10,100 Pa were the back pressure, that would be the design condition. The highest possible back pressure would cause a normal shock in the exit plane and still allow supersonic flow inside:

Normal shock : Ma1 3.2 ; Table B.2 : Ma 2 0.4643 , Then p 2 pb,max

p2 11.78 p1 (11.78)(10100) 119,000 Pa Ans .

P9.71 A converging-diverging nozzle has a throat area of 10 cm 2 and an exit area of 20 cm2. It is supplied by an air tank at 250 kPa and 350 K. (a) What is the design pressure at the exit? At one operating condition, the exit properties are pe = 183 kPa, Te = 340 K, and Ve = 144 m/s. (b) Can this condition be explained by a normal shock inside the nozzle? (c) If so, at what Mach number does the normal shock occur? [Hint: Use the change in A* to locate, if necessary, this position.] Solution: First find the design conditions and also the actual Mach number at the exit:

Ae 20 cm2 2.0 ; 2 A * 10 cm

p design

po

(1 0.2 Ma2e ) 3.5

Actual Mae

Ve

Madesign 2.1972 2.20

Table B.1 or EES :

250,000

[1 0.2(2.1972 ) ] 144 m / s

2 3.5

23,500 Pa

0.390

Ans.( a)

(subsonic ) kRTe 1.4( 287 )(340K ) So the actual exit flow is nowhere near design conditions. If we were so careful as to check for a normal shock in the exit plane, we would find an exit pressure of 128 kPa, much less than the given value of 183 kPa. Then yes, there is a normal shock inside the nozzle. Ans.(b) (c) To find where the shock is located, calculate the stagnation pressure and/or the area A*2 for the given exit conditions.

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683

po 2 pe (1 0.2Mae2 )3.5 (183 kPa)[1 0.2(0.389 )2 ]3.5 (183)(1.110) 203 .1 kPa At Ma e 0.389

Ae

1.627

20 cm 2

; Solve

A *2 12.3 cm 2

A2* A2* Either one of these will tell us the Mach number where the normal shock occurred:

po 2 203.1 kPa 0.8124 ; po 1 250 kPa A*2 A1*

12.3 cm2 10.0 cm

2

1.23

TableB.2 or EES : Ma 1 1.80

; TableB.2 or EES : Ma 1 1.80

Ans.(c) Ans.(c)

9.72 A large tank at 500 K and 165 kPa feeds air to a converging nozzle. The back pressure outside the nozzle exit is sea-level standard. What is the appropriate exit diameter if the desired mass flow is 72 kg/h?

Solution: Given To 500 K and po 165 kPa. The pressure ratio across the nozzle is (101.35 kPa)/(165 kPa) 0.614 0.528. Therefore the flow is not choked but instead exits at a high subsonic Mach number, with pthroat p atm 101.35 kPa. Equation (9.47) is handy:

m A

RTo po

2k p k 1 po

2/k

p (k 1)/k (72/3600) 287(500) 4.59E5 m 2 1 p A 165000 A o

Solve for A exit 6.82E 5 m2

4

2(1.4) 0.4

101 165

2/1.4

101 0.4/1.4 1 0.673 165

D 2exit , Solve D exit 0.0093 m Ans .

The exit Mach number is approximately 0.86. 9.73 Air flows isentropically in a converging-diverging nozzle with a throat area of 3 cm 2. At section 1, the pressure is 101 kPa, T 1 300 K, and V1 868 ms. (a) Is the nozzle choked? Determine (b) A1 ; and (c) the mass flow. Suppose, without changing stagnation conditions of A 1, the flexible throat is reduced to 2 cm2. Assuming shock-free flow, will

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684

there be any changes in the gas properties at section 1? If so, calculate the new p1 , V1, and T1 and explain. Solution: Ma1

Check the Mach number. If choked, calculate the mass flow:

V1 868 2.50 Supersonic: the nozzle is choked. Ans. (a) A1 1.4(287)(300)

po 101[1 0.2(2.50) 2]3.5 1726 kPa; To 300[1 0.2(2.50) 2] 675 K At Ma1 2.50,

m m max 0.6847

A1 2.64 A1 2.64(3) 7.91 cm2 A*

Ans. (b)

kg po A* 1726000(0.0003) 0.6847 0.805 s RTo 287(675)

Ans. (c)

If po and T o are unchanged and the throat (A*) is reduced from 3.0 to 2.0 cm 2, the mass flow is cut by one-third and, if A1 remains the same (7.91 cm2), the area ratio changes and the Mach number will change at section 1:

New

A1 7.91 3.955; Table B.1: read Ma1, new 2.928 2.0 * Anew

Since the Mach number changes, all properties at section 1 change: p1, new

1726000 52300 Pa 2 3.5 [1 0.2(2.928) ]

T1, new

675 249 K [1 0.2(2.928) 2]

V1, new 2.928 1.4(287)(249) 926

m s

A practical question might be: Does the new, reduced throat shape avoid flow separation and shock waves? P9.74 Use your strategic ideas, from part (b) of Prob. P9.40, to actually carry out the calculations for mass flow of steam, with po = 300 kPa and T o = 600 K, discharging through a converging nozzle of choked exit area 5 cm 2. Take advantage of the new EES thermodynamic function SPEEDSOUND(Steam, p = p1, s = s1).

Chapter 9 Compressible Flow

685

Solution: Using EES for real steam, we don’t have these nice power-law formulas, but we can use the energy equation and the continuity equation and the fact that the entropy is constant: Energy : h

1 2 V h o ; Continuity : m A V constant ; s s o constant 2

First establish so and ho from the given po and To:

s o Entropy(Steam , p 300,T 600) 7.7952kJ /(kg K ).

ho Enthalpy(Steam, p 300,T 600) 3123.8 kJ / kg Then guess an exit velocity V, which we know, from Prob. 9.40a, lies somewhere between 500 and 600 m/s. Compute h from energy, then compute =(h, so) from the EES thermophysical functions. This enables us to calculate the (guessed) mass flow = A V. Is it a maximum? Probably not. Keep changing V until you reach a maximum mass flow and a Mach number of 1.0. The final result obtained by the writer is At the exit: Ma 1.0, V 559 m / s , 0.683kg / m3 , m mmax 0.191

Here is a plot of the exit mass flow versus velocity, from an EES Table: 0.191

mdot [kg/s]

0.1905

0.19

maximum flow = 0.191 kg/s at V = 559 m/s

0.1895

0.189

0.1885

0.188 500

520

540

Ve [m/s]

560

580

600

kg s

Ans.

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686

And here is a plot of the exit Mach number versus velocity, from an EES Table. Each time we changed V, we calculated a new SOUNDSPEED and a new Ma = V/a. 1.1 1.075 1.05

Mach

1.025

-----------------------------------------------------------

1 0.975 0.95

Ma = 1.0 at V = 559 m/s

0.925 0.9 0.875 500

520

540

Ve [m/s]

560

9.75 A double-tank system in Fig. P9.75 has two identical converging nozzles of 1-in 2 throat area. Tank 1 is very large, and tank 2 is small enough to be in steady-flow equilibrium with the jet from tank 1. Nozzle flow is isentropic, but entropy changes between 1 and 3 due to jet dissipation in tank 2. Compute the mass flow. (If you give up, Ref. 14, pp. 288–290, has a good discussion.)

580

600

Fig. P9.75

Solution: We know that 1V1 2V 2 from continuity. Since p atm is so low, we may assume that the second nozzle is choked, but the first nozzle is probably not choked. We may guess values of p 2 and compare the computed values of flow through each nozzle: Assume 2nd nozzle choked: 3 V3

Guess p2 80 psia:

100 1 0.2Ma21 80

0.6847p 2 (RTo)

3.5

with To 520 R constant

or Ma1 0.574, then 1 0.0138

slug ft 3

Chapter 9 Compressible Flow

687

and V1 621 ft/s, or 1V1 8.54 slug/sft 2, whereas 3V3 8.35 slug/s ft Guess p2 81 psia: Ma1 0.557, 1V1 8.38

slug sft2

and 3 V3 8.45

Interpolate to: p2 80.8 psia, Ma1 0.561, m AV 0.0585

slug s

2

slug sft2 Ans .

9.76 A large reservoir at 20C and 800 kPa is used to fill a small insulated tank through a converging-diverging nozzle with 1-cm2 throat area and 1.66-cm 2 exit area. The small tank has a volume of 1 m 3 and is initially at 20C and 100 kPa. Estimate the elapsed time when (a) shock waves begin to appear inside the nozzle; and (b) the mass flow begins to drop below its maximum value. Solution:

During this entire time the nozzle is choked, so let’s compute the mass flow: me mmax

0.6847 po A* 0.6847(800000)(0.0001) 0.189 kg/s RT o 287(293)

Meanwhile, a control volume around the small tank reveals a linear pressure rise with time: me

d d dp (mtank ) ( tank ) tank dt dt dt RTo

Carrying out the numbers gives

, or:

dptank RTo m dt

dp 287(293)(0.189) Pa 15900 constant dt 1.0 s

We are assuming, for simplicity, that the tank stagnation temperature remains at 293 K. Shock waves move into the nozzle when the tank pressure rises above what would occur if the nozzle exit plane were to have a normal shock: If

Ae 1.66 800000 105400 Pa. , then Mae 1.98, p1 A* 1.0 [1 0.2(1.98)2 ]3.5

p 2 2.8(1.98)2 0.4 After the shock, 4.41, p2 ptank 4.41(105.4) 465 kPa p1 2.4

Above this tank pressure, the shock wave moves into the nozzle. The time lapse is t shocks in nozzle

p tank 465000 100000 Pa 23 sec Ans . (a) dp/dt 15900 Pa/s

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Assuming the tank pressure rises smoothly and the shocks do not cause any instability or anything, the nozzle ceases to be choked when ptank rises above a subsonic isentropic exit: Ae 800000 1.66, than Ma e (subsonic) 0.380, p e 724300 Pa A* [1 0.2(0.38)2 ]3.5

Then t choking stops

p 724300 100000 39 sec dp/dt 15900

Ans. (b)

9.77 A perfect gas (not air) expands isentropically through a supersonic nozzle with an exit area 5 times its throat area. The exit Mach number is 3.8. What is the specific heat ratio of the gas? What might this gas be? If p o 300 kPa, what is the exit pressure of the gas? Solution:

We must iterate the area-ratio formula, Eq. (9.44), for k:

Ma e 3.8,

k 1 2 2(k 1)

k 1 1 (3.8) Ae 1 2 5 A* 3.8 (k 1)/2

,

Solve for k 1.667

Monatomic gas: could be helium or argon.

With k known, pe

300 kPa

1 [1.667 1] / 2 (3.8)

2 1.667/0.667

Ans. (a)

Ans. (b) 3.7 kPa

Ans. (c)

9.78 The orientation of a hole can make a difference. Consider holes A and B in Fig. P9.78, which are identical but reversed. For the given air properties on either side, compute the mass flow through each hole and explain the difference. Solution:

Case B is a converging nozzle

Fig. P9.78

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689

with p2p1 100150 0.667 0.528, therefore case B is not choked. Case A is choked at the entrance and expands to a (subsonic) pressure of 100 kPa, which we may check from a subsonic calculation. The results are: Nozzle B:

p2 293 0.667, read Ma2 0.784, Te 261 K, 2 po [1 0.2(0.784) ] 100000 kg m m 1.34 3 , ae 324 , Ve 254 , e 287(261) m s s kg m e A eVe 0.0068 Ans. (B) s

Nozzle A: m mmax

kg 0.6847(150000)(0.0002) (5% more) Ans. (A) 0.071 s 287(293)

9.79 A large reservoir at 600 K supplies air flow through a converging-diverging nozzle with a throat area of 2 cm 2. A normal shock wave forms at a section of area 6 cm 2. Just downstream of this shock, the pressure is 150 kPa. Calculate (a) the pressure in the throat; (b) the mass flow; and (c) the pressure in the reservoir.

Solution: The throat is choked, and just upstream of the shock is a supersonic flow at an area ratio A/A* (6 cm2)/(2 cm2) 3.0. From Table B.1 estimate Ma1 2.64. That is,

A1 3.0 A*

1 0.2 Ma12

1.728 Ma1

3

, Solve Ma1 2.637

(a, c) The pressure ratio across the shock is given by Eq. (9.55) or Table B.2:

p2 150 kPa 1 2 kMa12 k 1 p1 p1 k 1 1 [2(1.4)(2.637)2 0.4] 7.95, or p1 18.9 kPa 2.4

p tank p o p1 1 0.2Ma12

3.5

(18.9)[1 0.2(2.637)2 ]3.5 399 kPa

Ans . (c)

At the throat, p p* 0.5283po (0.5283)(399 kPa) 211 kPa Ans . (a)

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690

(b) To avoid bothering with density and velocity, Eq. (9.46b) is handy for choked flow.

mmax

0.6847 po A* RTo

0.6847(399000 Pa)(0.0002 m 2 ) (287m /s K)(600 K) 2

2

0.132 kg/s Ans. (b)

9.80 A sea-level automobile tire is initially at 32 lbfin2 gage pressure and 75F. When it is punctured with a hole which resembles a converging nozzle, its pressure drops to 15 lbfin2 gage in 12 min. Estimate the size of the hole, in thousandths of an inch. Solution: The volume of the tire is 2.5 ft2 . With patm 14.7 psi, the absolute pressure drops from 46.7 psia to 29.7 psia, both of which are sufficient to cause a choked exit. A theory for isothermal blowdown of a choked tank was given in Prob. 9.36:

A* RTo p tank p(0)exp 0.6847 t ,

A* 1717(535) or: 29.7 46.7exp 0.6847 12 60 , 2.5

solve A* = 2.4E-6 ft2 (d 0.00175 ft 21 thousandths of an inch)

Ans.

P9.81Air, at po = 160 lbf/in2 and To = 300F, flows isentropically through a convergingdiverging nozzle. At section 1, where A1 = 288 in2, the velocity is V 1 = 2068 ft/s. Calculate (a) Ma1; (b) A*; (c) p1; and (d) the mass flow, in slug/s. Solution: That is a high velocity, 2068 ft/s, even higher than the stagnation speed of sound, ao = (kRTo)1/2 = 1351 ft/s. So the flow at section 1 is supersonic. We need to find Ma 1 such that

V1 Ma1 a1 Ma1 kRT1 Ma1

kRT o

1 0.2( Ma12 )

Ma1

1.4(1716)(300 460) 1 0.2( Ma12 )

You could iterate, or EES would find the result in a flash: Ma1 = 2.10 (b, c, d) The remaining properties follow from the Mach number:

Ans.(a)

2068

ft s

Chapter 9 Compressible Flow

691

A1 288 in 2 (b) Ma1 2.10 : Table B.1: 1.837 , A * 157 in 2 1.089 ft 2 Ans.(b) A* 1.837 po 160 psia lbf 17.5 psia 2520 (c) p1 Ans.(c ) [1 0.2( Ma12 )]3.5 [1 0.2(2.1) 2 ] 3.5 ft2 (d ) m m max ( 0.6847)

(160x144 psf )(1.089 ft 2 )

(1716 ft / s R)(760 R) 2

2

o

o

15.04

slug s

Ans.( d )

9.82 Air at 500 K flows through a converging-diverging nozzle with throat area of 1 cm 2 and exit area of 2.7 cm2. When the mass flow is 182.2 kgh, a pitot-static probe placed in the exit plane reads p o 250.6 kPa and p 240.1 kPa. Estimate the exit velocity. Is there a normal shock wave in the duct? If so, compute the Mach number just downstream of this shock. Solution: These numbers just don’t add up to a purely isentropic flow. For example, po p 250.6240.1 yields Ma 0.248, whereas AA* 2.7 gives Ma 0.221. If the mass flow is maximum, we can estimate the upstream stagnation pressure: ?

m m max

182.2 ? p (0.0001) 0.6847 o1 3600 287(500)

if p o1 280 kPa

This doesn’t check with the measured value of 250.6 kPa, nor does an isentropic choked subsonic expansion lead to p exit 240.1—it gives 271 kPa instead. We conclude that there is a normal shock wave in the duct before the exit plane, reducing po: Normal shock:

p o2 250.6 0.895 if Ma1 1.60 and Ma 2 0.67 p o1 280.0

This checks with A*2 1.12 cm 2 , Te

500

[1 0.2(0.248) 2]

Ans. (b)

Ae 2.42, Ma e 0.248 (as above) A* 2

494 K, a e kRTe 445

m m , V e Ma ea e 110 s s

9.83 When operating at design conditions (smooth exit to sea-level pressure), a rocket engine has a thrust of 1 million lbf. The chamber pressure and temperature are 600 lbfin2 absolute and 4000R, respectively. The exhaust gases approximate k 1.38

692

Solutions Manual Fluid Mechanics, Seventh Edition

with a molecular weight of 26. Estimate (a) the exit Mach number and (b) the throat diameter. Solution: “Design conditions” mean isentropic expansion to pe 14.7 psia 2116 lbfft 2: po 600 1.38 1 2 40.8 1 Ma e pe 14.7 2

1.38/0.38

, solve for Ma e 3.06

Ans. (a)

From Prob. 3.68, if pe pa ,

F e Ae Ve2 kpe Ae Ma2e 1.38(2116)Ae(3.06) 2 1E6 lbf, solve A e 36.6 ft 2

Assuming an isentropic expansion to Mae 3.06, we can compute the throat area: 2.38 2(0.38)

A e 36.6 1 2 0.38(3.06) A* A* 3.06 1.38 1 2

4.65, or A* =

Solve for throat diameter D* 3.2 ft

36.6 7.87 ft 2 D*2 4.65 4

Ans. (b)

9.84 Air flows through a duct as in Fig. P9.84, where A1 24 cm2, A 2 18 cm2, and A3 32 cm 2. A normal shock stands at section 2. Compute (a) the mass flow, (b) the Mach number, and (c) the stagnation pressure at section 3. Solution: We have enough information at section 1 to compute the mass flow: Fig. P9.84

a1 1.4(287)(30 273) 349 m/s, V1 2.5(349) 872

m p kg , 1 1 0.46 3 s RT1 m

Then m e Ae Ve 0.46(0.0024)(872) 0.96 kg/s

Ans. (a)

Chapter 9 Compressible Flow

693

Now move isentropically from 1 to 2 upstream of the shock and thence across to 3:

Ma1 2.5,

A1 24 A 2 18 2.64, A1* 9.1 cm 2, and 1.98 * 2.64 * 9.1 A1 A1

Read Ma2,upstream 2.18, po1 po2 40[1 0.2(2.5)2 ]3.5 683 kPa, across the

shock,

A* 3 1.57, A*3 14.3 cm2 , A*2

A3 2.24 sub , Ma 3 0.27 Ans. (b) * A3

Finally, go back and get the stagnation pressure ratio across the shock: at Ma2 2.18,

po3 0.637, po3 0.637(683) 435 kPa p o2

Ans. (c)

P9.85 A typical carbon dioxide tank for a paintball gun holds about 12 oz of liquid CO 2. The tank is filled no more than one-third with liquid, which, at room temperature, maintains the gaseous phase at about 850 psia. (a) If a valve is opened that simulates a converging nozzle with an exit diameter of 0.050 in, what mass flow and exit velocity result? (b) Repeat the calculation for helium. Solution: For CO2, from Table A.4, R = 189 J/kg-K and k = 1.30. By “room temperature” we assume To = 293 K. Convert po = 853 psia = 5.86 MPa. Convert D exit = 0.050 in = 0.00127 m. Assume an outside pressure of 1 atm, which ensures that the flow through the nozzle will be choked. The maximum mass flow results, Eq. (9.46a), and V exit = V*, Eq. (9.33):

0.6673 CO2 , k 1.30 : m and

po A * RT o

0.6673

V exit V *

(5.86 E6 Pa)( / 4)(0.00127 m) 2 (189 )(293)

2k RT o k 1

0.021

2(1.3) m (189 )( 293) 250 (1.3 1) s

kg Ans .(a ) s

Ans.(a )

(b) For helium, from Table A.4, R = 2077 J/kg-K and k = 1.66. Then

Helium, k 1.66 : m 0.6673 and

p oA * RTo

V exit V *

0.6673

(5.86 E 6 Pa )( / 4)(0.00127 m ) 2

2k RT o k 1

(2077 )(293)

0.0069

2(1.66) m (2077 )(293) 872 (1.66 1) s

kg Ans.(b ) s

Ans.(b )

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694

9.86 Air enters a 3-cm diameter pipe 15 m long at V1 73 ms, p1 550 kPa, and T1 60C. The friction factor is 0.018. Compute V2, p 2, T2, and p02 at the end of the pipe. How much additional pipe length would cause the exit flow to be sonic? Solution:

First compute the inlet Mach number and then get (fLD) 1:

Then

(fL/D2 14.53 – (0.018)(15)/(0.03) 5.53, read Ma 2 0.295

m 73 fL 0.20, read 14.53, , Ma1 D s 366 for which p/p* 5.4554, T/T* 1.1905, V/V* 0.2182, and p o /p*o 2.9635 a1 1.4(287)(60 273) 366

At this new Ma 2, read pp* 3.682, TT* 1.179, V/V* 0.320, po /p*o 2.067. Then V2 V1

m V2 /V* 0.320 73 107 V1 /V* s 0.218

Ans. (a)

3.682 p 2 550 371 kPa Ans. (b) 5.455 1.179 T2 333 330 K Ans. (c) 1.190

Now we need po1 to get po2 :

2.067 394 kPa p o1 550[1 0.2(0.2)2 ]3.5 566 kPa, so p o2 566 2.964

The extra distance we need to choke the exit to sonic speed is (fLD)2 5.53. That is, L 5.53

D 0.03 5.53 9.2 m Ans. f 0.018

P9.87 Problem C6.9 gives data for a proposed Alaska-to-Canada natural gas (assume CH4) pipeline. If the design flow rate is 890 kg/s and the entrance conditions are 2500 lbf/in 2 and 140F, determine the maximum length of adiabatic pipe before choking occurs.

Chapter 9 Compressible Flow

695

Solution: For CH4 , from Table A.4, R = 518 m2/s 2-K, k = 1.32, and = 1.03E-5 kg/m-s. Convert to SI units: p1 = 2500 psi = 1.72E7 Pa, D = 52 in = 1.321 m, T1 = 140F = 333 K. From the ideal gas law and the known mass flow, find the entrance density and pressure:

1 V1

p1 1.724E 7 99.9 kg / m3 RT1 (518)(333) m

1 ( / 4)D 2

890

(99.9)( / 4)(1.321)2

6.50 m / s

We can now find the Reynolds number and the smooth-wall friction factor:

ReD

(99.9)(6.50)(1.321) VD 8.33E7 1.03E 5

Smooth wall, Eq.(6.48) :

1 2.51 ) yields 2.0 log 10 ( f ReD f

f 0.00608

Find the entrance Mach number:

Ma1

V1

kRT1

6.50 1.32(518)(333)

6.50 0.0136 477.2

Now apply Eq. (9.66) for compressible adiabatic flow with friction, for k = 1.32: f L * 1 Ma2 k 1 ( k 1) Ma2 ln[ ] 4074 for k 1.32 and Ma 0.0136 2 D 2 k 2 ( k 1) Ma k Ma 2 This yields L*, the pipe length which will cause choking:

L* (

fL * D 1.321 ) ) 885, 000 m 550 miles (4074)( D f 0.00608

Ans.

P9.88 Air flows adiabatically through a 2-cm-diameter pipe. Conditions at section 2 are p2 = 100 kPa, T2 = 15°C, and V2 = 170 m/s. The average friction factor is 0.024. At section 1, which is 55 meters upstream, find (a) the mass flow; (b) p1; and (c) p o1.

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696

Solution: The mass flow is the same at sections 1 and 2, so use section 2 to calculate To find things at section 1, we need compressible pipe-flow theory. It happens that the given data simulates the situation in Examples 9.10 and 9.11 in the text. At section 2,

2

p2 100,000 kg kg AV (1.21) (0.02) 2 (170) 0.0646 Ans.(a ) 1.21 3 ; m RT 2 287 (288K ) 4 s m

a2

kRT2 1.4(287 )(288 K) 340 m / s ;

V 2 170 m / s 0.50 a 2 340 m / s fL fL fL f L 0.024(55m ) 1.0691 Table B.3 : |2 1.0691 ; Then |1 |2 D D D D 0.02m TableB.3 : Read Ma1 0.100 1.0691 66.0 67.0691

p1 10 .9435 p* p / p* 10.9435 (100k Pa) 512 kPa Then p1 p2 1 p2 / p * 2.1381

Table B.3 :

p2 2.1381 p*

Ma2

,

Ans.( b)

p o 1 p 1[1 0.2Ma12 ]3.5 512[1 0.2(0.1)2 ]3.5 516 k Pa Ans .(c )

We need the pressure ratios to finish the problem, and then calculate p o1 from the Mach number

9.89 Carbon dioxide flows through an insulated pipe 25 m long and 8 cm in diameter. The friction factor is 0.025. At the entrance, p 300 kPa and T 400 K. The mass flow is 1.5 kg/s. Estimate the pressure drop by (a) compressible; and (b) incompressible (Sect. 6.6) flow theory. (c) For what pipe length will the exit flow be choked?

Chapter 9 Compressible Flow

697

Solution: For CO2, from Table A.4, take k 1.30 and R 189 J/kgK. Tough calculation, no appendix tables for CO 2, should probably use EES. Find inlet density, velocity, Mach number:

1

p1 300000 Pa kg 3.97 3 RT 1 (189J/kg K)(400 K) m

m 1.5 Ma1

kg kg 1A1V1 3.97 3 (0.08 m)2 V1 , Solve for V1 75.2 m/s s m 4

V1 75.2m/s 75.2 m/s 0.240 2 2 kRT1 313.5m/s (1.3)(189m /s K)(400 K)

Between section 1 (inlet) and section 2 (exit), the change in (f L /D) equals (0.025)(25 m)/ (0.08 m) 7.813. We have to find the correct exit Mach number from this change: 1 Ma 2 k 1 (k 1)Ma 2 fL*/ D ln 2 kMa 2 k 2 (k 1)Ma 2

For k 1.3 and Ma1 0.240 compute ( fL*/D)1 10.190

Then ( fL*/D)2 10.190 7.813 2.377 for what Mach number? Then iterate (or use EES) to the exit value Ma2 = 0.408

Now compute p1/p* = 4.452 and

p2/p* 2.600

Then p2 p1( p2 /p*)/( p1 /p*) (300 kPa)(2.600)/(4.452) 175 kPa

The desired compressible pressure drop 300 175 125kPa

Ans. (a)

(b) The incompressible flow theory (Chap. 6) simply predicts that 3.97 kg/m3 fL 1 2 p inc V1 (7.813) D 2 2

2 m 75.2 88000 Pa 88 kPa s

Ans. (b)

The incompressible estimate is 30% low. Finally, the inlet value of (fL /D) tells us the maximum possible pipe length for choking at the exit:

Lmax

fL * D D 1 f

0.08 m (10.19) 32.6 m Ans. (c) 0.025

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698

P9.90 Air flows through a rough pipe 120 ft long and 3 inches in diameter. Entrance conditions are p = 90 lbf/in2 , T = 68F, and V = 225 ft/s. The flow chokes at the end of the pipe. (a) What is the average friction factor? (b) What is the pressure at the end of the pipe? Solution: Find the Mach number at the entrance: a1

k R T1

Ma1

1.4(1716 ft 2 / s2 o R)(460 68 o R) 1126 ft / s

V1 225 ft / s 0.200 a1 1126 ft / s

Excellent, this Mach number is right in Table B.3.

f (

fL D 0.25 ft ) ) (14.5333)( D L 120 ft

Read fL*/D = 14.5333. Then

0.0303

Ans.( a)

Also, in Table B.3, we can read the pressure ratio:

Ma 0.20 ,

p 90 psia lbf 5.4554 , hence p * 16.5 Ans .(b ) p* 5.4554 in 2

9.91 Air flows steadily from a tank through the pipe in Fig. P9.91. There is a converging nozzle on the end. If the mass flow is 3 kgs and the flow is choked, estimate (a) the Mach number at section 1; and (b) the pressure in the tank.

Fig. P9.91

Chapter 9 Compressible Flow

699

Solution: For adiabatic flow, T* constant To 1.2 3731.2 311 K. The flow chokes in the small exit nozzle, D 5 cm. Then we estimate Ma2 from isentropic theory:

A 2 6 cm 1.44, read Ma 2 (subsonic) 0.45, for which fL/D2 1.52, A* 5 cm p2 /p* 2.388, po2 /po* 1.449, 2 /* 2.070, T2 /T* 1.153 or T2 359 K 2

Given m 3

kg p2 2 2 A2 V2 (0.06) (0.45) 1.4(287)(359), s 287(359) 4

Solve for p2 640 kPa. Then p* 640/2.388 268 kPa At section 1,

fL fL fL 0.025(9) 2 1.52 5.27, read Ma1 0.30 D D D 0.06 for which p1 /p* 3.6, or p1 3.6(268) 965 kPa.

Assuming isentropic flow in the inlet nozzle,

ptank 965[1 0.2(0.30)2 ]3.5 1030 kPa

P9.92

Ans . (a)

Ans. (b)

Air enters a 5- cm-diameter pipe at 380 kPa, 3.3 kg/m 3, and 120 m/s. The friction

factor is 0.017. Find the pipe length for which the velocity (a) doubles; (b) triples; and (c) quadruples.

Solution: First find the conditions at the entrance, which we will call section 1:

T1

p1 380000 401 K ; Ma1 R 1 287 (3.3)

Table B.3 or EES :

V1 / V * 0.3245

,

V1 kRT1

120 0.299 1.4(287 )( 401)

fL / D 5.353

(a) Since V* is constant, we simply double (V 1/V*) and find the new Mach number:

V2 0.6490 ; Table B.3 : Ma2 0.6144 , fL / D | 2 0.4367 V* D fL fL 0.05 Pipe length L ( |1 |2 ) (5.353 0.4367 ) 14.46 m Ans.(a ) f D D 0.017

( a) Double

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700

(b) Again, since V* is constant, we simply triple (V1/V*) and find the new Mach number:

V2 fL / D | 2 0.0013 0. 9735 ; Table B.3 : Ma 2 0.9684 , V* D fL fL 0.05 Pipe length L ( |1 |3 ) (5.353 0.0013 ) 15.74 m Ans.(b ) f D D 0.017

( b) Triple

(c) We are already at choking, it is impossible to quadruple the velocity for this flow . Ans.(c) 9.93 Air flows adiabatically in a 3-cm-diameter duct with f 0.015. At the entrance, V 950 ms and T 250 K. How far down the tube will (a) the Mach number be 1.8; and (b) the flow be choked? Solution:

Ma1

(a) Find the entrance Mach number and its value of fLd:

950

1.4(287)(250) read f

3.00; Table B.3: read f

L1 0.5222; at Ma 2 1.8, D

L2 L 0.2419; f 0.5222 0.2419 0.28, D D

0.28(0.03) 0.56 m Ans. (a) 0.015 (b) To go all the way to choking requires the full change L

fL1D 0.5222,

or:

Lchoke (0.5222)(0.03)/(0.015) 1.04 m

Ans. (b)

9.94 Compressible pipe flow with friction, Sec. 9.7, assumes constant stagnation enthalpy and mass flow but variable momentum. Such a flow is often called Fanno flow, and a line representing all possible property changes on a temperature-entropy chart is called a Fanno line. Assuming an ideal gas with k 1.4 and the data of Prob. 9.86, draw a Fanno line for a range of velocities from very low (Ma <<1) to very high (Ma >>1). Comment on the meaning of the maximum-entropy point on this curve.

Chapter 9 Compressible Flow

701

Solution: Recall from Prob. 9.86 that, at Section 1 of the pipe, V1 73 ms, p1 550 kPa, and T1 60C 333 K, with f 0.018. We can then easily compute Ma1 0.20, 1 5.76 kgm3, Vmax 822 ms, and T o 336 K. Our basic algebraic equations are: V2 V2 Energy: T To , or: T 336 K 2(1005) 2c p

Continuity: V 1 V1 5.76(73), or: 420/V

Entropy: s 718 ln(T/333) 287 ln( /5.76)

(a) (b) (c)

We simply let V vary from, say, 10 ms to 800 ms, compute from (b) and T from (a) and s from (c), then plot T versus s. [We have arbitrarily set s 0 at state 1.] The result of this exercise forms the Fanno Line for this flow, shown on the next page. Some Mach numbers are listed, subsonic on the top, supersonic on the bottom, and exactly sonic at the right-hand (maximum-entropy) side. Ans.

Solutions Manual Fluid Mechanics, Seventh Edition

702

9.95 Helium (Table A.4) enters a 5-cm-diameter pipe at p 1 550 kPa, V 1 312 ms, and T1 40C. The friction factor is 0.025. If the flow is choked, determine (a) the length of the duct and (b) the exit pressure. Solution: For helium, take k 1.66 and R 2077 JkgK. We have no tables for k 1.66, have to do our best anyway. Compute the Mach number at section 1:

a1 (1.66)(2077)(40 273) 1039 m/s, Ma 1 V1/a 1 Eq. 9.66:

312 0.300 1039

(2.66)(0.3)2 fL 1 (0.3)2 2.66 ln 4.37 at section 1 D 1.66(0.3)2 2(1.66) 2 (0.66)(0.3)2

Choked: fL/D2 0, L 4.37D/f 4.37(0.05)/(0.025) 8.7 m Ans. (a)

1 2.66 Also, p1/p* 0.3 2 0.66(0.3)2

1/2

3.79, p exit p*

550 145 kPa 3.79

Ans.(b)

9.96 Methane (CH4) flows through an insulated 15-cm-diameter pipe with f 0.023. Entrance conditions are 600 kPa, 100C, and a mass flow of 5 kg/s. What lengths of pipe will (a) choke the flow; (b) raise the velocity by 50%; (c) decrease the pressure by 50%?

Chapter 9 Compressible Flow

703

Solution: For methane (CH 4), from Table A.4, take k 1.32 and R 518 J/kgK. Tough calculation, no appendix tables for methane, should probably use EES. Find inlet density, velocity, Mach number:

1

p1 600000 Pa kg 3.11 3 RT 1 (518 J/kg K)(373K) m

m 5 kg/s 1A1V1 (3.11 kg/m 3) (0.15 m) 2V1, solve for V1 91.1 m/s 4 V 91.1 m/s 0.180 a1 kRT1 1.32(518)(373) 505 m/s, Ma1 1 a1 505 m/s

Now we have to work out the pipe-friction relations, Eqs. (9.66) and (9.68), for k 1.32. We need fL*/D, V/V*, and p/p* at the inlet, Ma 0.18:

fL* 1 Ma 2 k 1 ( k 1) Ma 2 ln 19.63 at Ma1 0.18 and k 1.32 2 2 D 2 k kMa 2 ( k 1) Ma

D 0.15 m 128 m Solve L*choking 19.63 19.63 0.023 f (b) First find V/V* and increase it by 50% to find the new condition:

Ans. (a)

V1 k 1 1/ 2 ] Ma1 [ 0.1938 if k 1.32, Ma 0.18 2 V* 2 (k 1)Ma1

V2 2(0.1938) 0.3876 ; Solve for Ma2 0.2715, fL* / D 7.35 V* 0.15m Velocity increases 50% when L (19.63 7.35) * 80 m Ans.( b) 0.023 Then

(c) Now find p/p* and decrease it by 50% to find the required pipe length:

p 1 k 1 p* Ma 2 (k 1)Ma 2

1/2

Decrease 50% to:

5.954 at Ma1 0.18 and

k 1.32

p/ p* 2.977 Solve for: Ma2 0.358, fL */ D 3.46

Solve: L* (19.63 3.46)

D 0.15m 16.17 105 m 0.023 f

Ans. (c)

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704

9.97 By making a few algebraic substitutions, show that Eq. (9.74), or the relation in Prob. 9.96, may be written in the density form

2 k fL 2 ln 1 2 k 1 D

12 22 * 2

Why is this formula awkward if one is trying to solve for the mass flow when the pressures are given at sections 1 and 2? Solution: This much less laborious algebraic derivation is left as a student exercise. There are two awkward bits: (1) we don’t know 1 and 2; and (2) we don’t know * either, and preliminary computations are necessary. 9.98 Compressible laminar flow, f 64Re, may occur in capillary tubes. Consider air, at stagnation conditions of 100C and 200 kPa, entering a tube 3 cm long and 0.1 mm in diameter. If the receiver pressure is near vacuum, estimate (a) the average Reynolds number, (b) the Mach number at the entrance, and (c) the mass flow in kgh. Solution: The pipe is choked, “receiver pressure near vacuum,” so L L* and we need only to correctly guess the inlet Mach number and iterate until the Table B.3 value of (fLD) matches the actual value, with f 64Re from laminar pipe theory. Since Re VD and V is constant due to mass conservation, Re varies only due to the change in with temperature (from about 2.1E5 in the entrance to 1.9E5 kgms at the exit). We assume avg 2.0E5 kgms. Try Ma1 from 0.1 to 0.2 and find 0.12 to be the best estimate: Ma1 0.12

Ans.(b) Table B.3: (fL/D)1 45.4, also compute

T1 372 K, V1 46 m/s, 1 1.85

kg 1.85(46)(0.0001) 430 Ans. (a) Re , avg m3 2.0E5 Then flaminar 64 /430 0.15, f(L/D) 0.15(300) 45.0 (close enough for me!)

The mass flow is m 1A1V1 1.85( /4)(0.0001)2 (46) 6.74E7

kg Ans. (c) s (0.00243 kg/h)

Chapter 9 Compressible Flow

705

9.99 A compressor forces air through a smooth pipe 20 m long and 4 cm in diameter, as in Fig. P9.99. The air leaves at 101 kPa and 200C. The compressor data for pressure rise versus mass flow are shown in the figure. Using the Moody chart to estimate f , compute the resulting mass flow. Solution: The compressor performance is approximate by the parabolic relation

Fig. P9.99

p compressor 250 1563 m 2, with p in kPa and m in kg/s

We must match this to the pressure drop due to friction in the pipe. For preliminaries, compute e pe RTe ) 0.744 kgm 3, and ae (kRTe ) 436 ms. Guess the mass flow:

m 0.2 kg/s, then Re ?

Ve

4m 4(0.2) 255000, f Moody 0.015 D (0.04)(2.5E 5)

Re 255000(2.5E5) m 214 fL 214 , Mae 0.491, read 1.15, e D 0.744(0.04) s 436 D e also read p e/p* 2.18, p*

101000 46300 Pa 2.18

Then, at the pipe entrance (Sect. 1), we may compute fLD and find the pressure there:

fL 1 1.15 0.015(20) 8.65, read Ma 1 0.248, p1 4.39, D 0.04 p* p2 4.39(46300) or p2 203 kPa, or ppipe 203 101 102 kPa whereas pcomp 187 kPa (m too small)

We increase the mass flow until p pipe pcompressor. The final converged result is:

m 0.256

kg m , Re 326000, f 0.0142, Ve 274 , Ma e 0.628, read s s

fL/De 0.39, p e/p* 1.68, p* 60.1 kPa, add f L/D 7.1 to get fL/D 1 7.49,

read Ma1 0.263, p1 /p* 4.14, p2 249 kPa, ppipe 148 kPa pcompressor (OK)

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706

For these operating conditions, the approximate flow rate is

0.256 kgs. Ans.

P9.100 Air flows adiabatically through a pipe 10 cm in diameter and 66 m long. Entrance conditions are p 1 = 550 kPa, T1 = 350 K, and V1 = 75 m/s. The average friction factor is 0.022. (a) Is the pipe flow choked at the exit? (b) What is the exit pressure? (c) At what distance down the pipe is the velocity 184 m/s? Solution: Find the inlet Mach number from the given data:

Ma1

V1

kRT1

75 m/ s 1.4(287)(350)

Excellent, this Mach number is right in Table B.3. at the end of the pipe, x = 66 meters:

75 0.20 375

Read fL*/D = 14.5333. Then find (fL/D)

fL (0.022)(66 m) |exit 14.5333 14.5333 14.52 0.0133 0 D 0.1 m

As close as we can calculate, Yes, it is choked at the exit.

Ans.(a)

(b) The exit then must be at (sonic) pressure p*. Calculate this from the inlet pressure:

Ma1 0.2:

p 550 kPa 5.4554 , p1 550 kPa , thus p * 101 kPa Ans.(b) p* 5.4554 (c) Go back and find V/V* at the inlet. Then we can move downstream to the new velocity: Ma1 0.2 : V / V * 0.2182 ; thus V * 75 m/ s / 0.2182 344 m / s Downstream, V2 / V * (184 m/ s) / 344 m/ s 0.5349 ; Table B.3: Read Ma2 0.50; then

fL / D 1.0691 , f L / D 14.5333 1.0691 13.46; L (13.46)(0.1) / .022 61.2 m Ans.(c )

Chapter 9 Compressible Flow

707

9.101 How do the compressible-pipe-flow formulas behave for small pressure drops? Let air at 20C enter a tube of diameter 1 cm and length 3 m. If f 0.028 with p 1 102 kPa and p2 100 kPa, estimate the mass flow in kgh for (a) isothermal flow, (b) adiabatic flow, and (c) incompressible flow (Chap. 6) at the entrance density. Solution: For a pressure change of only 2%, all three estimates are nearly the same. Begin by noting that fLD 0.028(3.00.01) 8.4, and 1 102000[287(293)] 1.213 kgm 3. Take these estimates in order:

(102000)2 (100000)2 p21 p22 m (a) Isothermal: 569 A [ 2 ln( / )] 287(293)[8.4 2 ln(102/100)] RT fL/D p1 p2 2

Then m/A 23.9, m isothermal 23.9( /4)(0.01)2 0.00187

(b) Adiabatic: Given To 293 K, ao

kg s

Ans. (a)

kRTo 343 m/s, use Eqs. 9.74 and 9.75:

V1 (343) 2 [1 (0.9803)2 ] 2 0.9803, V1 388, or V1 19.7 m/s Converges to V2 1.4(8.4) 2.4 ln(1.02)

Then m 1AV1 1.213( /4)(0.01)2 (19.7) 0.00188 kg/s

Ans. (b)

(c) Incompressible: p (fL/D)( /2)V2 , or 2000 (8.4)(1.213/2)V2 , or V 19.8 m/s. Then mincompressible AV 1.213( /4)(0.01)2 (19.8) 0.00189 kg/s. Ans. (c)

9.102 Air at 550 kPa and 100C enters a smooth 1-m-long pipe and then passes through a second smooth pipe to a 30-kPa reservoir, as in Fig. P9.102. Using the Moody chart to compute f, estimate the mass flow through this system. Is the flow choked?

Fig. P9.102

Solution: Label the pipes “A” and “B” as shown. Given (L/D)A 20 and (L/D)B 40. Label the relevant sections 1, 2, 3, 4 as shown. With po1/pe 550/30 18.3, these short

Solutions Manual Fluid Mechanics, Seventh Edition

708

pipes are sure to be choked, with an exit pressure p 4 much larger than 30 kPa. One way is to guess Ma1 and work your way through to section 4 to require Ma 4 1.0 (choked). Take a constant average viscosity 2.2E5 kg/ms. Assume isentropic expansion to section 1 from the reservoir, frictional flow through pipe A, isentropic expansion from 2 to 3, and a second frictional flow through pipe B to section 4. The correct solution is Ma 1 0.18: 373 K 450000 Pa 371 K, p 440000 Pa, 1 2 2 3.5 1 0.2(0.18) [1 0.2(0.18) ] p 440000 kg m VD 1 1 4.14 3 , V1 Ma1 kRT1 69.5 , Re1 1 1 A 653,000 RT1 287(371) m s Ma1 0.18, T1

Moody chart: f A 0.0125; Table B.3:

fL 1 18.54, fL 2 18.54 0.0125(20) 18.29 D D

Pipe A is so short that the Mach number hardly changes. At (fL/D)2 18.29, read Ma 2 0.181. Now, at Ma2 0.181, determine A 2/A* 3.26, hence A3/A* (3.26)(3/5)2 1.17, read Ma3 0.613 and (fL/D) 3 0.442. Stop to calculate 3 3.49 kg/m 3, V 3 229 m/s, ReB 1.09E6, from the Moody chart, f B 0.0115. Then (fL/D)4 0.442 – 0.0115(40) 0.018. (?) This last value should have been exactly (fL/D)4 0 if the exit Mach number is 1.0. But we were close. The mass flow follows from the conditions at section 1: kg kg m m 1A1V1 4.14 3 (0.05 m)2 69.5 0.565 s m 4 s

Ans.

EES can barely improve upon this: Ma1 0.1792, yielding a mass flow of 0.5616 kg/s. The exit pressure is p4 201 kPa, far larger than the receiving reservoir pressure of 30 kPa.

9.103 Natural gas, with k 1.3 and a molecular weight of 16, is to be pumped through 100 km of 81-cm-diameter pipeline. The downstream pressure is 150 kPa. If the gas enters at 60C, the mass flow is 20 kgs, and f 0.024 , estimate the required entrance pressure for (a) isothermal flow and (b) adiabatic flow.

Chapter 9 Compressible Flow

Solution:

709

The gas constant is R gas 831416 520 JkgK. First use Eq. 9.73:

20 p21 p32 m (a) Isothermal: 2 A ( /4)(0.81) RT[fL/D 2 ln(p1/p 2 )] 2

2

p 21 (150000) 2 , 520(333)[2963 2 ln(p1 /150000)] solve for p1 892 kPa

Ans. (a)

Part (a) indicates a low inlet Mach number, 0.02, so Te To, a e ao 475 m/s. Then use Eqs. (9.74) and (9.75) —the latter simply indicates that the bracket [] 1.000. Then

V1 p 2 V2 p1

and

V12

a2o [1 (V1 /V2 )2 ] (475)2[1 (V1/V2) 2 ] , k(fL/D) (k 1)ln(V2/V1) 1.3(2963) 2.3ln(V2/V1)

plus 1V1 2V2 m/A 38.81 kg/s m 2. Solve for V1 7.54 p1 1RT1 5.15(520)(333) 892 kPa

m kg , 1 5.15 3 , s m

9.104 A tank of oxygen (Table A.4) at 20C is to supply an astronaut through an umbilical tube 12 m long and 1.5 cm in diameter. The exit pressure in the tube is 40 kPa. If the desired mass flow is 90 kg/h and f 0.025, what should be the air pressure in the tank?

Solution: For oxygen, from Table A.4, take k 1.40 and R 260 J/kgK. Given To 293 K and fL/D (0.025)(12 m)/(0.015 m) 20. Use isothermal flow, Eq. (9.73), as a first estimate:

p12 p22 m 90/3600 kg/s 2 A RT [ fL /D 2 ln(p1 / p2 )] ( /4)(0.015 m) 2

2

p12 (40000) 2 Solve for p1 192 kPa (260)(293)[20 2 ln( p1/40000)]

Solutions Manual Fluid Mechanics, Seventh Edition

710

This is a very good estimate of p1, but we really need adiabatic flow, Eqs. (9.66) and (9.68a). First estimate the entrance Mach number from p 1 and T 1 To :

1

p1 192000 kg 2.52 3 , RTo 260(293) m

m

90 kg 1 AV1 (2.52) (0.015m) 2 V1 3600 s 4

solve V1 56 m/s, a1 kRTo 1.4(260)(293) 327 m/s, Ma1

56 0.17 327

We can guess Ma 1 around 0.17, find (fL*/D)1, subtract (fL/D) 20, find the new Mach number and p*, thence back up to obtain p1. Iterate to convergence. For example: Ma 1 0.17, (fL*/D)1 21.12, p1/p* 6.43, (fL*/D)2 21.12 – 20 1.12, compute

Ma 2 0.49, p2/p* 2.16, p* 40000/2.16 18500 Pa, p 1 6.43(18500) 119000 Pa, T1 To/[1 0.2(0.17) 2] 291 K, 1 1.57 kg/m3 , mass flow 1AV1 55 kg/h

The mass flow is too low, so try Ma 1 a little higher. The iteration is remarkably sensitive to Mach number because the correct exit flow is close to sonic. The final converged solution is Ma1 0.1738, Ma2 0.7792, p1 189.4 kPa

Ans.

This problem is clearly well suited to EES, which converges rapidly to the final pressure.

P9.105 Modify Prob. P9.87 as follows. The pipeline will not be allowed to choke. It will have pumping stations about every two hundred miles. (a) Find the length of pipe for which the pressure has dropped to 2000 lbf/in 2. (b) What is the temperature at that point? Solution: From Prob. P9.87, we found the following data. For CH4, from Table A.4, R = 518 m2/s2 -K, k = 1.32, and = 1.03E-5 kg/m-s. Convert to SI units: p1 = 2500 psi = 1.72E7 Pa, D = 52 in = 1.321 m, T 1 = 140F = 333 K. From the ideal gas law and the known mass flow (890 kg/s), find the entrance density and pressure, Reynolds number, and smooth-wall friction factor:

1 99.9

kg 3

m

; V1 6.50

m ; Re D 8.33E 7 ; s

f

smooth

Then we found the entrance Mach number and thus the factor (fL*/D):

0.00608

Chapter 9 Compressible Flow

Ma1

V1 6.5 0.0136 ; k 1.32 ; Eq .(9.66) : a1 477

711

f L* 4074 D

The present problem involves pressures and temperatures, so we need to calculate p/p* and T/T*: Ma1 0.0136; k 1.32, Eqs.(9.68) :

p1 T 79.06 , 1 1.16 p* T*

(a) When the pressure p2 drops to 2000 psia = 1.39E6 Pa, the new pressure ratio will be

p2 p p 2000 2 1 ( )(79.06) 63.25 ; Eq. (9.68a ) yields Ma2 0.0170 p* p1 p * 2500

fL * fL * f L ) 2605, hence ( ) 4074 2605 1469 D D D (1469)(1.321) Finally, p2 2000 psia at L 319,000 m 198 mi Ans.( a) 0.00608

At this Ma2 , (

(b) At this same Ma2 = 0.0170, compute T 2/T* also equals 1.16, same as T1/T*. Therefore the temperature has hardly changed at all: T2

*P9.106

333K = 140F.

Ans.(b)

Air, from a 3 cubic meter tank initially at 300 kPa and 200C, blows down

adiabatically through a smooth pipe 1 cm in diameter and 2.5 m long. Estimate the time required to reduce the tank pressure to 200 kPa. temperature and f 0.020.

t = 0: 200C 300 kPa 3 m3

For simplicity, assume constant tank

(1

(2)

Fig. P9.106

p a = 100 kPa

Solutions Manual Fluid Mechanics, Seventh Edition

712

Solution: We know that p2 at the exit is 100 kPa, and we are given f = 0.020, thanks! For the given L and D, we can immediately calculate fL/D = (0.02)(2.5m/0.01m) = 5.0. Suggested procedure: Calculate the mass flow at p tank = 300 and 200 kPa and

average these two. Guess Ma 2 at the exit, calculate fL/D 2, increase upstream by 5.0 to fL/D 1, find Ma 1 and p1 when p o1 = ptank, estimate 1, V 1, and the mass flow. To start,

guess sonic flow at the exit, Ma2 = 1.0, fL/D 2 = 0. Then fL/D1 = 5.0, for which Ma1 =

0.3066 (Table B.3), p 1/p* = 3.54, hence p1 = 354 kPa and po1 = 378 kPa, too high! So the exit Mach numbers are less than 1.0. Iteration (with EES) leads to the following results that converge to a tank pressure of 300 kPa:

Ma2 0.814 ; p1

p2 p fL fL | 2 0.0605 ; | 1 5.0605 ; Ma1 0.3052 ; 1 3.5564 1.2646 ; p* D D p*

p1 / p * 3.5564 ( p 2) (100 ) 281 .2 kPa ; po 1 (281 .2)[1 0.2(0.3052 ) 2 ] 3.5 300 kPa p2 / p* 1.2646

Repeat these (laborious) calculations for a tank pressure of 200 kPa:

Ma2 0.536 ; p1

p2 fL fL p | 2 0.8113 ; | 1 5.8113 ; Ma1 0.2898 ; 1 3.7491 1.9874 ; p* D D p*

p1 / p * 3.7491 ( p2) (100 ) 188 .6 kPa ; p o1 (188 .6)[1 0.2(0.2898 ) 2 ]3.5 200 kPa p2 / p* 1.9874

Now, with inlet conditions known, find V1 and 1 and thus the two mass flows:

m kg 2 ; m AV ( 2 . 110 ) ( 0 . 01 ) ( 131 . 8 ) 0 . 0219 1 1 3 s 4 s m kg m kg 200 kPa : 1 1.413 125 .3 ; m 1AV1 (1.413) (0.01) 2 (125 .3) 0.0139 ; V 1 s 4 s m3 Average mass flow (0.0219 0.0139 ) / 2 0.0179 kg/ s

po1 300 kPa : 1 2.110 p o1

kg

;V1 131 .8

Tank at 300 kPa: m = [po /RTo ](Vol) = [(300000/287/473)kg/m3](3m 3) = (2.21)(3) = 6.63 kg Tank at 200 kPa: m = [po /RTo ](Vol) = [(200000/287/473)kg/m3](3m 3) = (473)(3) = 4.42 kg

Chapter 9 Compressible Flow

Time to blow down from 300 to 200 kPa = (6.63-4.42 kg)/(0.0179 kg/s)

713

91 s

Ans.

9.107 A fuel-air mixture, assumed equivalent to air, enters a duct combustion chamber at V1 104 m/s and T1 300 K. What amount of heat addition in kJ/kg will cause the exit flow to be choked? What will be the exit Mach number and temperature if 504 kJ/kg is added during combustion? Solution:

Evaluate stagnation temperature and initial Mach number:

V12 (104)2 104 To T1 300 305 K; Ma1 0.30 2c p 2(1005) 1.4(287)(300)

305 Table B.4: To/To* 0.3469, hence To* 880 K 0.3469 Thus q choke c p To,max 1005(880 305) 5.78E5 J/kg 578 kJ/kg Ans. (a)

A heat addition of 504 kJ/kg is (just barely) less than maximum, should nearly choke:

To2 To1

q 540000 305 842 K, cp 1005

Finally, without using Table B.4,

To2

T* o

842 0.957, Ma2 0.78 880

Ans. (b)

T2 842 /[1 0.2(0.78)2 ] 751 K Ans. (c)

9.108 What happens to the inlet flow of Prob. 9.107 if the combustion yields 1500 kJ/kg heat addition and po1 and To1 remain the same? How much is the mass flow reduced? Solution:

The flow will choke down to a lower mass flow such that To2 = To*:

To2 T*o 305

1500000 T 305 1798 K, thus o1 0.17, Ma1,new 0.198 1005 1798 T* o

(m/A)new 1V1 1a1Ma1 oa o Ma1 /[1 0.2Ma12 ]3

if po1 , To1, o1 are the same.

m new 0.198 1 0.2(0.30)2 Then 0.68 (about 32% less flow) Ans . m old 0.30 1 0.2(0.198)2

Solutions Manual Fluid Mechanics, Seventh Edition

714

9.109 A jet engine at 7000-m altitude takes in 45 kg/s of air and adds 550 kJ/kg in the combustion chamber. The chamber cross section is 0.5 m2, and the air enters the chamber at 80 kPa and 5°C. After combustion the air expands through an isentropic converging nozzle to exit at atmospheric pressure. Estimate (a) the nozzle throat diameter, (b) the nozzle exit velocity, and (c) the thrust produced by the engine.

Fig. P9.109

Solution:

1

At 700-m altitude, pa 41043 Pa, T a 242.66 K to use as exit conditions.

p1 80000 kg kg 1.00 45 AV 1.00(0.5)V1 , V1 90 m/s , m RT1 287(278) m3 s

Ma1

90 0.27, 1.4(287)(278)

Table B.4: To1/T*o 0.29,

To1 278 (90)2 /[2(1005)] 282 K,

To* 282/0.29 973 K

550000 T 829 829 K, thus o2 0.85, read Ma2 0.63 1005 973 T*o also read p1 /p* 2.18, p2 /p* 1.54, p2 80(1.54/2.18) 57 kPa,

Add heat: To2 282

po2 p 2 1 0.2 Ma 22

3.5

2 3.5 57[1 0.2(0.63) ] 74 kPa

With data now known at section 2, expand isentropically to the atmosphere:

3.5 pe 41043 T T 0.72 1 0.2Ma 2e , solve Ma e 0.70, e e 0.910 , po2 57000 To2 829

Solve Te 755 K, e pe /RTe 0.189 kg/m 3, a e kRTe 551 m/s , Ve Ma ea e 385 m/s

m 45 0.189(385)

Ans. (b)

D 2e, solve D e 0.89 m Ans. (a)

4 Finally, if pe patm , from Prob. 3.68, Fthrust mVe 45(385) 17300 N Ans. (c)

Chapter 9 Compressible Flow

715

9.110 Compressible pipe flow with heat addition, Sec. 9.8, assumes constant momentum (p V 2) and constant mass flow but variable stagnation enthalpy. Such a flow is often called Rayleigh flow, and a line representing all possible property changes on an temperature-entropy chart is called a Rayleigh line. Assuming air passing through the flow state p1 548 kPa, T 1 588 K, V1 266 m/s, and A 1 m2, draw a Rayleigh curve of the flow for a range of velocities from very low (Ma 1) to very high (Ma 1). Comment on the meaning of the maximum-entropy point on this curve.

Solution:

First evaluate the Mach number and density at the reference state: p 548000 kg 3.25 3 ; Ma RT 287(588) m

Our basic algebraic equations are then:

V

kRT

266

1.4(287)(588)

Momentum:

p V2 548000 3.25(266)2 778000

Entropy:

s 718 ln(T/588) 287 ln( /3.25)

Continuity:

V 3.25(266), or:

864/V

0.55

(a)

(b)

(c)

We simply let V vary from, say, 10 m/s to 800 m/s, compute from (b), p from (a), T p/ T, and s from (c), then plot T versus s. [We have arbitrarily set s 0 at state 1.]

The result of this exercise forms the Rayleigh Line for this flow, shown below. Some Mach numbers are listed, subsonic on the top, supersonic on the bottom, and exactly sonic at the right-hand (maximum-entropy) side. Ans.

Solutions Manual Fluid Mechanics, Seventh Edition

716

*9.111 Add to your Rayleigh line of Prob. 9.110 a Fanno line (see Prob. 9.94) for stagnation enthalpy equal to the value associated with state 1 in Prob. 9.110. The two curves will intersect at state 1, which is subsonic, and also at a certain state 2, which is supersonic. Interpret these two cases vis-a-vis Table B.2. Solution:

For T1 588 K and V 1 266 m/s, the stagnation temperature is

V2 (266)2 V2 To T1 588 623 K ; Elsewhere, T 623 2c p 2(1005) 2(1005)

Also, from Prob. 9.110, 864/V (b )

(d)

T and s 728 ln 287 ln (c) 588 3.25

Chapter 9 Compressible Flow

717

By varying V and computing (T, , s) from (b, c, d), we plot the Fanno line and add it to the previous Rayleigh line. The composite graph is as follows:

The subsonic intersection is state 1, Ma1 0.55, T2 588 K, and the supersonic intersection is at Ma 2 2.20, where, for example, T 2 316 K. Also, s 1 s2. These two points thus correspond to the two sides of a normal shock wave, where “2” is the supersonic upstream and “1” the subsonic downstream condition. We may check these results in Table B-2, where, at Ma 2.20, the temperature ratio across the shock is 1.857—for our calculations, this ratio is 588 K/316 K 1.86 (agreement would be perfect if we kept more significant figures). Shock flow satisfies all the four equations of Rayleigh and Fanno flow combined—continuity, momentum, energy, and the equation of state. 9.112 Air enters a duct subsonically at section 1 at 1.2 kg/s. When 650 kW of heat is added, the flow chokes at the exit at p 2 95 kPa and T2 700 K. Assuming frictionless heat addition, estimate (a) the velocity; and (b) the stagnation pressure at section 1. Solution: Then

Since the exit is choked, p 2 p* and T2 T* and, of course, Ma 2 1.0.

V2 V* kRT* 1.4(287)(700) 530 m/s, and q Q/m

650 kJ 542 1.2 kg

542000 Also, T* 301 K o 1.2T2 1.2(700) 840 K; hence To1 840 1005

Solutions Manual Fluid Mechanics, Seventh Edition

718

301 0.358; Table B.4: read Ma1 0.306 , read V1 /V* 0.199 Then To1 /To* 840

So V1 530(0.199) 105 m/s

Ans. (a)

Also read p o1/p*o 1.196, where p*o p2 /0.5283 180 kPa, Hence p o1 1.196(180) 215 kPa Ans. (b) 9.113 Air enters a constant-area duct at p1 90 kPa, V 1 520 m/s, and T1 558°C. It is then cooled with negligible friction until it exists at p 2 160 kPa. Estimate (a) V2; (b) T2 ; and (c) the total amount of cooling in kJ/kg. We have enough information to estimate the inlet Ma 1 and go from there: m 520 p 0.90, read 1 1.1246, a1 1.4(287)(558 273) 578 , Ma1 s 578 p*

Solution:

or p*

90 p 160 80.0 kPa, whence 2 2.00, read Ma2 0.38, 1.1246 p* 80 read T2 /T* 0.575, V2 /V* 0.287, To2 /T*o 0.493

We have to back off to section 1 to determine the critical (*) values of T, V, T o:

Ma1 0.9, T1 /T* 1.0245, T*

558 273 811 K, T2 0.575(811) 466 K Ans. (b) 1.0245

520 571 m/s, so V2 0.287(571) 164 m/s Ans. (a) 0.911 966 973 K To1 /To* 0.9921, where To1 T1 V12/2cp 966 K, To* 0.9921 Finally, To2 0.493(973) 480 K, kJ Ans. (c) q cooling cp To 1.005(966 480) 489 kg

also, V1 /V* 0.911, V*

Chapter 9 Compressible Flow

719

P9.114 The scramjet of Fig. 9.30 operates with supersonic flow throughout. An idealized sketch is in Fig. C9.8, at the end of this chapter. Assume that the heat addition of 500 kJ/kg, between sections 2 and 3, is frictionless and at constant area of 0.2 m 2. Given Ma2 = 4.0, p2 = 260 kPa, and T2 = 420 K. Assume airflow at k = 1.40. At the combustion section exit, find (a) Ma3, (b) p 3, and (c) T3 . Solution: Use frictionless heat-addition theory to get from section 2 to section 3:

Ma2 4.0 : To*

To 2 To*

0.5891 , To2 (420 K )[1 0.2(4.0) 2] 1764 K

1764 T 420 2994 K ; 2 0.1683 , T * 2496 K 0.5891 T* 0.1683 p2 260 kPa Q 500000 0.1026 , p * 2534 kPa ; To 3 To 2 1764 2262 K p* 0.1026 cp 1005

Now find the new stagnation temperature ratio and hence the properties at section 3: To3 To*

2262 0.7552 , thus Ma3 2.205 2994

Ans.( a) ;

p3 0.3074 , p 3 (0.3074)(2534 kPa) 779 kPa p*

Ans.(b)

T3 0.4594 , T3 (0.4594)(2496 K ) 1146 K T*

Ans.( c)

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720

P9.115 Air enters a 5-cm-diameter pipe at 380 kPa, 3.3 kg/m3, and 120 m/s. Assume frictionless flow with heat addition. Find the amount of heat addition for which the velocity (a) doubles; (b) triples; and (d) quadruples.

Solution: First find the conditions at the entrance, which we will call section 1:

T1

p1 380000 401 K ; Ma1 R 1 287 (3.3)

Table B.4 or EES :

V1 / V * 0.1905

,

V1

kRT1

120 0.299 1.4(287 )( 401)

To1 / To* 0.3448

Then To1 T1 (1 0.2 Ma12 ) (401 K)[1 0.2(0.299 )^2] 408 K And

To* To1 /(To1 / To* ) 408 K /(0.3448 ) 1184 K

Since V 1 is less than one-fifth of V*, it looks like we can make it up to quadruple velocity:

( a) V2 / V * 2(0.1905 ) 0.3810 ; Table B.4 : Ma2 0.452 , To 2 / To* 0.6168

To 2 1184 (0.6168 ) 731 K ; q c p (To2 To1 ) 1005 ( 731 408) 324 ,000 J / kg Ans.(a )

( b) V3 / V * 3(0.1905 ) 0.5715 ; Table B.4 : Ma3 0.598 , To3 / To* 0.8164 To3 1184 (0.8164 ) 967 K ; q cp (To3 To1 ) 1005 (967 408) 561,000 J / kg Ans.(b )

( c) V4 / V * 4(0.1905 ) 0.7620 ; Table B.4 : Ma4 0.756 , To4 / To* 0.9434 T o4 1184 (0.9434 ) 1117 K ; q c p(T o4 To1 ) 1005 (1117 408) 713,000 J / kg Ans.(c)

Note that we could even quintuple, going up to 5.25V 1 before choking due to heat addition.

Chapter 9 Compressible Flow

721

9.116 An observer at sea level does not hear an aircraft flying at 12000 ft standard altitude until it is 5 statute miles past her. Estimate the aircraft speed in ft/sec.

Fig. P9.116

Solution:

The average temperature over this range is 498°R, hence

a kRT 1.4(1717)(498) 1094 ft/s, and tan or: 24.4, Ma plane csc 2.42, Vplane

12000 0.455, 26400 ft Ans. 2.42(1094) 2640 s

P9.117 A tiny scratch in the side of a supersonic wind tunnel creates a very weak wave

of angle 17, as shown in Fig. P9.117, after which a normal shock occurs. The air temperature in region (1) is 250 K. Estimate the temperature in region (2). shock 1

2

17o

Fig. P9.117 Solution: The weak wave is a Mach wave, hence the Mach number in region 1 is Ma1 = 1/sin(17) = 3.42. The normal shock relations, Table B.2 or Eq. (9.58), thus give the temperature ratio: Ma1 3.42 : T2 /T1 3.2069 Thus T2 3.2069 (250K ) 800 K Ans.

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Solutions Manual Fluid Mechanics, Seventh Edition

9.118 A particle moving at uniform velocity in sea-level standard air creates the two disturbance spheres shown in Fig. P9.118. Compute the particle velocity and Mach number. Solution: If point “a” represents t 0 units, the particle reaches point “b” in 8 3 5 units. But the distance from a to b is only 3 units. Therefore the (subsonic) Mach number is

Ma

Fig. P9.118

V t 3 units 0.6 Ans. (a) a t 5 units

Vparticle Ma(a) 0.6 1.4(287)(288 K) 204 m/s Ans. (b)

Chapter 9 Compressible Flow

723

9.119 The particle in Fig. P9.119 is moving supersonically in sea-level standard air. From the two disturbance spheres shown, compute the particle (a) Mach number; (b) velocity; and (c) Mach angle.

Fig. P9.119

Solution: If point “a” represents t 0 units, the particle reaches point “b” in 8 – 3 5 units. But the distance from a to b is 8 8 3 19 units. Therefore the Mach number is

Ma

V t 19 units 3.8 at 5 units

Ans. (a)

1 15.3 Ans. (c) 3.8

wave sin 1

Vparticle Ma(a) 3.8 1.4(287)(288 K) 1290 m/s

Ans. (b)

9.120 The particle in Fig. P9.120 is moving in sea-level standard air. From the two disturbance spheres shown, estimate (a) the position of the particle at this instant; and (b) the temperature in °C at the front stagnation point of the particle. Fig. P9.120 Solution: Given sea-level temperature 288 K. If point “a” represents t 0 units, the particle reaches point “b” in 6 – 3 3 units. But the distance from a to b is 6 units. Therefore the particle Mach number is

Ma

6 2.0, Tstagnation To 288[1 0.2(2.0) 2 ] 518 K 3

Ans. (b)

sin 1 30 , and the particle is at point “P” at 6 meters ahead of “b.” Ans. (a) 2 1

9.121 A thermistor probe, in the shape of a needle parallel to the flow, reads a static temperature of –25°C when inserted in the stream. A conical disturbance of half-angle 17° is formed. Estimate (a) the Mach number; (b) the velocity; and (c) the stagnation temperature of the stream.

Fig. P9.121

Solutions Manual Fluid Mechanics, Seventh Edition

724

Solution: If the needle is “very thin,” it reads the stream static temperature, T –25°C 248 K. We are given the Mach angle, 17°, so everything else follows readily: Ma csc csc17 3.42

Ans. (a)

To 248[1 0.2(3.42) 2 ] 828 K 555C

Ans. (c)

U Ma a 3.42 1.4(287)(248) 3.42(316) 1080 m/s

Ans. (b)

9.122 Supersonic air takes a 5° compression turn, as in Fig. P9.122. Compute the downstream pressure and Mach number and wave angle, and compare with smalldisturbance theory. Solution: From Fig. 9.23, 25°, and we can iterate Eq. (9.86) to a closer estimate:

Ma1 3.0, 5 , compute 23.133 , p2 /p1 1.454, p 2 145.4 Pa

From Eq. 9.83f, compute Ma 2 2.750

Fig. P9.122

Ans. (a, b, c) Exact oblique shock theory.

This is a small deflection. The linear theory of Eqs. 9.88 and 9.89 is reasonably accurate: (k 1) tan 1 0.0557 4 cos 3 2 p 1.4(3) tan 5 0.39, 22.9 (1% low ) 2 p1 3 1

sin 1 19.47 , sin sin 3 1

or: linear

0.389

p 2 100(1.39) 139 kPa (4% low )

P9.123 The 10 deflection in Ex. 9.17 caused a final Mach number of 1.641 and a pressure ratio of 1.707. Compare this with the case of the flow passing through two 5 deflections. Comment on the results and why they might be higher or lower in the second case.

Chapter 9 Compressible Flow

Solution: The sketch shows both the intermediate and the final states. The final Mach number is slightly Ma= higher, Ma = 1.649. The final 1.821 Ma= p = 1.315 pressure is slightly lower, 1.706. 2.0 They are close to the 10 case because p = 1.0 5 ten degrees is small, nearly isentropic. Comment: The closer the flow is to isentropic, the higher the final Mach number and the lower the pressure.

725

Ma= 1.649 p = 1.706 5

9.124 When a sea-level air flow approaches a ramp of angle 20, an oblique shock wave forms as in Figure P9.124. Calculate (a) Ma1; (b) p2; (c) T2 ; and (d) V 2.

Solution: For sea-level air, take p1 101.35 kPa, T 1 288.16 K, and 1 1.2255 Fig. P9.12 kg/m3 . (a) The approach Mach number is determined by the specified angles, 60 and 20. From Fig. 9.23 we read that Ma 1 is slightly less than 2.0. More accurately, use Eq. (9.86): tan

2 cot Ma12 sin 2 1 Ma12( k

cos2 ) 2

for 20 and 60

Iterate, or use EES, to find that Ma1 1.87. Ans. (a)

(b, c) With Ma1 known, use Eqs. (9.83a, c) to find p2 and T 2:

p2 1 2 kMa12 sin 2 k 1 2.893, p 2 2.893(101.35 kPa) 293 kPa Ans. (b) p1 k 1

T2 2 kMa12 sin 2 k 1 2 2 2 ( k 1) Ma1 sin 1.401, T1 ( k 1)2 Ma12 sin2 T2 1.401(288.16 K) 404 K

Ans. (c)

(d) Finally, to find V 2, first find V1 from the approach Mach number, then use Eq. (9.83b):

Solutions Manual Fluid Mechanics, Seventh Edition

726

a1 kRT1 1.4(287)(288.16) 340

m m , V1 a1 Ma1 (340)(1.87) 636 s s

V2 cos cos 60 m 0.653, V2 0.653(636 m/s) 415 s V1 cos ( ) cos 40

Ans. (d)

P9.125 We saw in the text that, for k = 1.40, the maximum possible deflection caused

by an oblique shock wave occurs at infinite approach Mach number and is max = 45.58. Assuming an ideal gas, what is max for (a) argon; and (b) carbon dioxide?

Solution: In the limit as Ma1 , the normal-velocity ratio across a shock wave, Eq. (9.58), is

V1 V2

k 1 k 1

It is this limiting ratio that we use to calculate max for the two gases above, with Eq. (9.85).

max |Ma tan 1 ( r ) tan 1 (

1

) ,

where r

r The results, using Table B.4 to determine k, are as follows:

k 1 k 1

(a) Argon, k = 1.67, r = 3.985: max = 63.392 - 26.608 = 36.78 (b) CO2 , k = 1.30, r = 7.667:

max = 70.142 - 19.858 = 50.28

Ans.(a)

Ans.(b)

9.126 Consider airflow at Ma1 2.2. Calculate, to two decimal places, (a) the deflection angle for which the downstream flow is sonic; and (b) the maximum deflection angle.

Chapter 9 Compressible Flow

727

Solution: We are near the peak of the (invisible) curve for Ma 1 2.2 in Fig. 9.23. The wave angles are 65, which we guess for finding the sonic downstream condition: Ma1 2.2, guess 65 , and using Eq. 9.86, compute 26.1 and Ma 2 0.92 (not quite sonic)

Converges to Ma 2 1.000 when 61.9 and 25.9

Maximum deflection occurs at 64.6 and max 26.1

9.127 Do the Mach waves upstream of an oblique-shock wave intersect with the shock? Assuming supersonic downstream flow, do the downstream Mach waves intersect the shock? Show that for small deflections the shock-wave angle lies halfway between 1 and 2 for any Mach number. Solution:

Ans. (a)

Ans . (b)

Fig. P9.127

Yes, Mach waves both upstream and downstream will intersect the shock:

(k 1)Ma12 Linear theory: 1 2 4 a1

(k 1)Ma12 and 2 2 4 Ma 1 1

Thus, to first order (small deflection), the shock wave angle will lie halfway between 1 and ( 2 ), as sketched in the figure above. 9.128 Air flows past a two-dimensional wedge-nosed body as in Fig. P9.128. Determine the wedge half-angle for which the horizontal component of the total pressure force on the nose is 35 kN/m of depth into the paper. Solution: Regardless of the wedge angle , the horizontal force equals the pressure inside the shock times the projected vertical area of the nose:

Fig. P9.128

Solutions Manual Fluid Mechanics, Seventh Edition

728

Fhoriz p 2A vert proj p 2(0.12 m)(1.0 m) 35000 N, or p 2 291700 Pa Eq. (9.83a):

p2 291700 1 2 2.917 0.4 , solve Ma1 sin 1.63 2.8Ma1n p1 100000 2.4

2.81 1 Use Eq. (9.86) to compute wedge 15. 5.5 5 Ans . or sin1 1.63 / 3.0 32.8

9.129 Air flows at supersonic speed toward a compression ramp, as in Fig. P9.129. A scratch on the wall at a creates a wave of 30 angle, while the oblique shock has a 50 angle. What is (a) the ramp angle ; and (b) the wave angle caused by a scratch at b?

Fig. P9.129

Solution: The two “scratches” cause Mach waves which are directly related to Mach No.: 1 30°, Ma 1 csc 30° 2.0, 50°, Eq. 9.86 yields 18.13 Ans. (a)

Then Ma 2n 0.690 Ma 2sin(50 18.13°),

1 Ma2 1.307, sin 1 49.9 1.307

Ans. (b)

P9.130 A supersonic airflow, at a temperature of 300 K, strikes a wedge and is deflected 12. If the resulting shock wave is attached, and the temperature after the shock is 450 K, (a) estimate the approach Mach number and wave angle. ( b) Why are there two solutions? Solution: We search through the oblique-shock solutions until we

find a = 12 deflection for which T2 /T1 = 450K/300K = 1.50.

(a, b) There are two solutions, weak and strong:

Ma 1

strong shock

weak shock 12 o

Fig. P9.130

Chapter 9 Compressible Flow

Ma1 = 4.69 , Ma1 = 1.78 ,

= 22 = 80

729

(weak shock)

(strong shock) Answers (a)

9.131 The following formula has been suggested as an alternate to Eq. (9.86) to relate upstream Mach number to the oblique shock wave angle and turning angle : sin2

1 (k 1)sin sin 2 cos( ) Ma12

Can you prove or disprove this relation? If not, try a few numerical values and compare with the results from Eq. (9.86). Solution: The formula is quite correct and serves as an interesting alternative to Eq. (9.86). Notice that one can immediately solve for Ma1 if and are known, which would have been a great help in Prob. 9.124. For details of the proof, see page 371 of R. M. Olson, Essentials of Engineering Fluid Mechanics, 4 th ed., Harper and Row, New York, 1980. 9.132 Air flows at Ma 3 and p 10 psia toward a wedge of 16° angle at zero incidence, as in Fig. P9.132. (a) If the pointed edge is forward, what is the pressure at point A? If the blunt edge is forward, what is the pressure at point B? Solution:

For Ma 3, 8°, Eq. 9.86:

25.61 ,

2.8(3sin 25.61 )2 0.4 1.80, p A /p1 2.4

p A 18.0 psia Ans. (a)

Fig. P9.132

730

Solutions Manual Fluid Mechanics, Seventh Edition

(b) A normal shock forms, and p B po2 inside the shock. Given p o1 p1/0.0272 367 psia, Table B.2, Ma 3: p o2/po1 0.3283, hence p o2 0.3283(367) 121 psia. Ans. (b) 9.133 Air flows supersonically toward the double-wedge system in the figure. The (x,y) coordinates of the tips are given. Both wedges have 15° deflection angles. The shock wave of the forward wedge strikes the tip of the aft wedge. What is the freestream Mach number?

Fig. P9.133

Solution: However tricky the shock reflection might be at the upper (aft) wedge, the problem is solved by knowing the shock angle at the lower (forward) wedge:

tan 1(1.0) 45°, 15°, Eq. (9.86) yields Ma 2.01 Ans.

9.134 When an oblique shock strikes a solid wall, it reflects as a shock of sufficient strength to cause the exit flow Ma3 to be parallel to the wall, as in Fig. P9.134. For airflow with Ma1 2.5 and p1 100 kPa, compute Ma3, p3, and the angle .

Fig. P9.134

Solution: With 1 40°, we can compute the first shock deflection, which then must turn back the same amount through the second shock:

Ma1n 2.5 sin 40° 1.607; Eq. (9.86): 1 17.68°, Ma 2n 0.666, Ma 2 1.754,

Also 2 17.68°, solve 2 60.45°, Ma 2n 1.526, Ma 3n 0.692 Ma 2sin( 2 2 ), Finally Ma 3 1.02

Ans. (a) p 2/p1 2.85, p 2 285 kPa, p3 /p2 2.55.

Keep going: p3 2.55(285) 727 kPa Ans. (b)

Finally, 2 2 60.56 17.68 42.8 Ans. (c)

Chapter 9 Compressible Flow

731

9.135 A bend in the bottom of a supersonic duct flow induces a shock wave which reflects from the upper wall, as in Fig. P9.135. Compute the Mach number and pressure in region 3. Solution:

Given 10°, find state 2:

Ma1 3.0, 1 10°, Eq. 9.86 predicts 1 27.38°,

Ma1n 1.380, Ma 2n 0.748,

Fig. P9.135

Ma2 2.505, 2 1 10°, 2 31.80°, Ma 2n 1.32, Ma 3n 0.776, Ma3 2.09 Ans. Meanwhile, p 2 /p1 2.054, or p 2 205.4 kPa,

and p3 /p2 1.866, p3 1.866(205.4) 383 kPa 9.136 Figure P9.136 is a special application of Prob. 9.135. With careful design, one can orient the bend on the lower wall so that the reflected wave is exactly canceled by the return bend, as shown. This is a method of reducing the Mach number in a channel (a supersonic diffuser). If the bend angle is 10°, find (a) the downstream width h and (b) the downstream Mach number. Assume a weak shock wave.

Ans.

Fig. P9.136

Solution: The important thing is to find the angle between the second shock and the upper wall, as shown in the figure. With initial deflection 10°, proceed forward to “3”:

Ma1 3.5, 1 10°, compute 1 24.384°, Ma 2 2.904, 2 1 10°, 2 28.096°,

upper wall 2 2 28.096 10 18.096°, Ma3 2.427 Ans. (b)

Length CB in the figure (1 m)/sin(24.384°) 2.422 m, angle ACB 14.384°, angle CBA 2 28.096°, by the law of sines, AB/sin(14.384°) 2.422/sin(28.096°) or the length AB 1.278 m. Finally, duct width h 1.278sin(18.096°) 0.40 m. Ans. (a) The horizontal distance from one lower corner to the next is 3.42 m. The length of CB is 3.47 m. Thus the shocks are not drawn to scale in the figure.

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Solutions Manual Fluid Mechanics, Seventh Edition

9.137 A 6 half-angle wedge creates the reflected shock system in Fig. P9.137. If Ma3 2.5, find (a) Ma1; and (b) the angle .

Fig. P9.137

Solution: (a) We have to go backward from region 3 to regions 2 and 1, using Eq. (9.86). In both cases the turning angle is 6. EES is of course excellent for this task, otherwise the iteration will be laborious. The results are:

Ma2 2.775, 2 25.66°, Ma 1 3.084, 1 23.36° Ans. (a)

Since the wall is horizontal, it is clear from the geometry of Fig. P9.137 that

180° 25.66° 23.36° 130.98 Ans. (b)

9.138 The supersonic nozzle of Fig. P9.138 is overexpanded (case G of Fig. 9.12) with A e /At 3.0 and a stagnation pressure of 350 kPa. If the jet edge makes a 4° angle with the nozzle centerline, what is the back pressure pr in kPa? Solution: The nozzle is clearly choked because there are shock waves downstream. Thus

Fig. P9.138

Ae 3.0, read Mae Ma1 2.64, pe 350/[1 0.2(2.64) 2 ]3.5 16.5 kPa A* 4°, Eq. 9.86 gives 2 25.3°, Ma1n 2.64 sin 25.3° 1.125, p2 /p1 1.311 Thus p2 preceiver 1.311(16.5) 21.7 kPa Ans.

Chapter 9 Compressible Flow

9.139 Airflow at Ma 2.2 takes a compression turn of 12° and then another turn of angle in Fig. P9.139. What is the maximum value of for the second shock to be attached? Will the two shocks intersect for any less than max ? Solution:

733

Fig. P9.139

First get the conditions in section (2) and then iterate for max:

Ma1 2.2, 12°, Eq. 9.86: 1 37.87°, Ma1n 1.351, Ma2n 0.762 Ma2 sin( ) or: Ma 2 1.745. For this Mach number, estimate max 18° from Fig. 9.23 Iterate, by trial and error, find max 18.02

Ans.

NOTE: The two shocks 1 and 2 always intersect for any 2 max . Ans. 9.140 The solution to Prob. 9.122 is Ma2 2.750 and p2 145.5 kPa. Compare these results with an isentropic compression turn of 5°, using Prandtl-Meyer theory. Solution: Find for Ma 3 and subtract 5°: Ma1 3.0, Table B.5: 1 49.76°,

2 49.76 5 44.76°, read Ma 2 2.753 3.5

Fig. P9.122

p 2 /po 1 0.2(2.753) 2 Then p 2 p1 145.4 kPa Ans. 100 2 p /p 1 0.2(3.0) 1 o This is almost identical to the shock wave result, because a 5° turn is nearly isentropic. 9.141 Supersonic airflow takes a 5° expansion turn, as in Fig. P9.141. Compute the downstream Mach number and pressure and compare with small-disturbance theory. Solution:

Find for Ma 3 and add 5°:

Ma1 3.0, Table B.5: 1 49.76°, 2 49.76 5 54.76°, Read Ma 2 3.274 Ans.

Fig. P9.141

734

Solutions Manual Fluid Mechanics, Seventh Edition 3.5

p 2 /p o 1 0.2(3.274) 2 Then p 2 p 1 66.7 kPa Ans. 100 2 p /p 1 0.2(3.0) 1 o The linear theory is not especially accurate because even a 5° turn is slightly nonlinear: k Ma2 1.4(3)2 p Eq. 9.89: tan 2 tan( 5) 0.390, 2 2 p Ma 1 3 1 p 100(1 0.390) 61 kPa

Ans. (9% low)

9.142 A supersonic airflow at Ma1 3.2 and p1 50 kPa undergoes a compression shock followed by an isentropic expansion turn. The flow deflection is 30° for each turn. Compute Ma2 and p 2 if (a) the shock is followed by the expansion and (b) the expansion is followed by the shock. Solution: The solution is given in the form of the two sketches below. A shock wave with a 30° turn is a hugely non-isentropic flow, so the final conditions are nowhere near the original and they do not agree with each other either.

Fig. P9.142

9.143 Airflow at Ma 1 3.2 passes through a 25° oblique-shock deflection. What isentropic expansion turn is required to bring the flow back to (a) Ma1 and (b) p1?

Fig. P9.143

Chapter 9 Compressible Flow

Solution:

735

First work out state (2):

Ma1 3.2, 1 42.56°, Ma2 1.83, p2 5.30 units, 2 21.59°

(a) Ma3 3.2 means 2 53.47°, or 53.47 21.59 31.9 Ans . (a)

(b) p3 1 unit means Ma3 2.906, 3 47.90°, 47.90 21.59 26.3 Ans. (b)

P9.144 The 10 deflection in Ex. 9.17 caused the Mach number to drop to 1.64. (a) What turn angle will create a Prandtl-Meyer fan and bring the Mach number back up to 2.0? (b) What will be the final pressure?

2

1

Ma 1 = 2 p 1 = 10 psia

2

3

Fig. P9.144

10

Solution: Recall from Ex. 9.17 that the shock wave caused Ma 2 = 1.64 and p 2 = 17.07 psia. First calculate the (reduced) stagnation pressure in section 2, plus the Prandtl-Meyer angle 2: Table B.5 or Eq.(9.99) : Ma2 1.64 ; p2 17.07 psia ; 2 16.04 o Then po 2 p2 (1 0.2Ma22 ) 3.5 (17.07)[1 0.2(1.64) 2] 3.5 77.0 psia

Now find the Prandtl-Meyer angle in section 3, knowing that Ma3 = 2.0: Ma3 2.0 :

Table B.5 :

3 26.38o

Thus 3 2 26.38o 16.04o

10.34o Ans.( a)

p o3 p o2 77psia : p3 (77) /[1 0.2(2.0)2 ]3.5 9.84 psia Ans .(b )

The final wall is just slightly down from horizontal, and the final pressure is just slightly reduced from the original value of 10 psia. The 10 deflection angle is almost isentropic.

Solutions Manual Fluid Mechanics, Seventh Edition

736

9.145 Air at Ma 1 2.0 and p1 100 kPa undergoes an isentropic expansion to a downstream pressure of 50 kPa. What is the desired turn angle in degrees? Solution: This is a real ‘quickie’ compared to what we have been doing for the past few problems. Isentropic expansion to a new pressure specifies the downstream Mach number: 3.5 p o p1 1 0.2 Ma21 100[1 0.2(2)2 ]3.5 782 kPa 50 0.0639, read Ma 2 2.44, read 2 37.79°, p 2 /p o 782 while 1 26.38°, 37.79 26.38 11.4 Ans. 9.146 Air flows supersonically over a surface which changes direction twice, as in Fig. P9.146. Calculate (a) Ma 2; and (b) p 3.

Fig. P9.146

Solution: (a) At the initial condition Ma1 2.0, from Table B.5 read 1 26.38. The first turn is 10, so 2 26.38 10 36.38. From Table B.5 read Ma 2 2.38. For more accuracy, use Eq. (9.99) to obtain Ma 2 2385. Ans. (a)

(b) The second turn is 12, so 3 36.38 12 48.38. From Table B.5 read Ma3 2.93. For more accuracy, use Eq. (9.99) to obtain Ma2 2.9296. (Not worth the extra effort.) To find pressures, we need the stagnation pressure, which is constant:

p o p1 1 0.2 Ma12

Then

p3 p o 1 0.2 Ma32

3.5

3.5

(200 kPa)[1 0.2(2.0) 2 ]3.5 1565 kPa

(1565)/[1 0.2(2.9296)2 ]3.5 47.4 kPa Ans. (b)

Chapter 9 Compressible Flow

737

9.147 A converging-diverging nozzle with a 4:1 exit-area ratio and p0 500 kPa operates in an underexpanded condition (case 1 of Fig. 9.12) as in Fig. P9.147. The receiver pressure is pa 10 kPa, which is less than the exit pressure, so that expansion waves form outside the exit. For the given conditions, what will the Mach number Ma 2 and the angle of the edge of the jet be? Assume k 1.4 as usual. Solution: Get the Mach number in the exit and then execute a Prandtl-Meyer expansion:

Fig. P9.147

Ae 4.0, read Mae 2.94, Table B.5: 1 48.59°, po1 po2 500 kPa A* 500 po /p2 50, read Ma2 3.21 Ans . (a) Read 2 53.61 , 10 53.61 48.59 5.02°, or see figure above 90 85.0 Ans . (b)

9.148 Air flows supersonically over a circular-arc surface as in Fig. P9.148. Estimate (a) the Mach number Ma2 and (b) the pressure p2 as the flow leaves the circular surface.

Fig. P9.148

Solutions Manual Fluid Mechanics, Seventh Edition

738

Solution: (a) At the initial condition Ma1 2.5, from Table B.5 read 1 39.12. (a) Circular arc or not, the turn angle is 32, so 2 39.12 32 71.12. From Table B.5 read Ma 2 4.44. For more accuracy, use Eq. (9.99) to obtain Ma2 4.437. Ans. (a) (b) To find p2, first find the stagnation pressure:

1 0.2 Ma

po p1 1 0.2 Ma12

Then

p2 po

2 2

3.5

3.5

(150 kPa)[1 0.2(2.5) 2 ] 3.5 2563 kPa

(2563) /[1 0.2(4.437)2 ]3.5 9.6 kPa

Ans. (b)

P9.149 Air flows at Ma = 3.0 past a doubly symmetric diamond airfoil whose front and rear included angles are both 24°. For zero angle of attack, compute the drag coefficient with shock-expansion theory and compare with Ackeret theory. Solution: The airfoil and its front and rear pressures are shown below. Ma 2 = 2.406 p2 = 2.3404 p 2 = 36.892 o

Ma = 3.0

Ma 3 = 3.653 p 3 = 0.0099 p 3 = 60.892 o

24

p

24

Fig. P9.149

The foil thickness is C[tan(12)] = 0.2126C. The drag coefficient is:

CD

Drag

(k

/ 2)Ma2 p bC

( 2.3404 p 0.0099 p ) b(0.2126 C) 2

(1.4 / 2))(3.0) p bC

0.0787 Ans.

From Ackeret theory, Eq. (9.107), the drag coefficient prediction is somewhat lower: C D,Ackeret

dy [ 2 ( ) 2avg ] 2 dx Ma 1 4

4

3 1 2

[02 tan2 (12o )] 0.0639

Ans.

Chapter 9 Compressible Flow

739

Ackeret (small-disturbance) theory is 19% low because the airfoil is too thick (21%). 9.150 A flat plate airfoil with chord C 1.2 m is to have a lift of 30 kN/m when flying at 5000-m standard altitude with U 641 m/s. Using Ackeret theory, estimate (a) the angle of attack; and (b) the drag force in N/m. Solution: Ma

At 5000 m, 0.7361 kg/m3, T 256 K, and p 54008 Pa. Compute Ma: 641

1.4(287)(256)

2.00, CL

30000

(1/2)(0.7361)(641)2 (1.2)(1)

Solve for 0.0716 rad 4.10

0.1653

Ans.

4

(2) 2 1

With angle of attack known, Ackeret theory simply predicts that

D L tan 30000tan(4.10°) 2150 N/m Ans. (b)

9.151 Air flows at Ma 2.5 past a halfwedge airfoil whose angles are 4, as in Fig. P9.151. Compute the lift and drag coefficients at equal to (a) 0; and (b) 6. Solution:

Let’s use Ackeret theory here:

(a) 0°: CL 0; CD

Fig. P9.151

2 1 2 0 (tan 4 0) 0.00427 2 2 (2.5) 1 4

Ans. (a)

[A complete shock-expansion calculation gives C L –0.0065, CD 0.00428.] (b) 6° 0.105 rad: CL CD

4(0.105) 0.183,

4 [(0.105)2 (1/2)(tan2 4° 0)] 0.0234

Ans. (b)

[A complete shock-expansion calculation gives C L 0.179, CD 0.0219.]

740

Solutions Manual Fluid Mechanics, Seventh Edition

P9.152 The X-43 model A scramjet aircraft in Fig. 9.30 is small, W = 3000 lbf, and unmanned, only 12.33 ft long and 5.5 ft wide. The aerodynamics of a slender arrowhead-shaped hypersonic vehicle is beyond our scope. Instead, let us assume it is a flat plate airfoil of area 2.0 m2. Let Ma = 7 at 12,000 m standard altitude. Estimate the drag, by shock-expansion theory. [HINT: Use Ackeret theory to estimate the angle of attack.] Solution: From Table B.6 at 12,000 m, = 1 p1 = 19312 Pa and T1 = 216.66 K. Ma 1 = 7 Knowing W = 3000 lbf = 13345 N, 2 we can estimate C L = W/[0.5kpMa1 ] = (13345N)/[0.5(1.4)(19312)(7)2(2.0)] = 0.0101. Use Ackeret theory, Eq. (9.104), to estimate :

p 3 = 16209 Pa

p 2 = 22902 Pa

C L = 0.0101 = 4/(Ma 2 – 1) 1/2 , solve for 1.0 degree.

Assume = 1 degree, as in the figure. From shock theory, Eqs. (9.86) and (9.83a), calculate Ma2 = 5.582, =8.848, p2 /p = 1.186, p2= 1.186(19312) = 22902 Pa The Prandtl-Meyer expansion is on the top, with a turning angle of 1 degree. From Eq. (9.99), Ma1 7 , 1 90.97 o , 2 90.97 1o 91.97o , read Ma3 7.195 po1 (19312 Pa)[1 0.2(7) 2 ]3.5 7.995E7 Pa po3 p3 (7.995E7 Pa) /[1 0.2(7.195) 2 ]3.5 16212 Pa

The (hypothesized) wing area is 2.0 m2. The total force, lift, and drag of the flat plate is thus

F ( p2 p3 )A p (22902 16209 Pa)(2.0 m 2 ) 13386 N 3009 lbf Lift Drag F sin(1o ) (13386 N )(0.01745) 234 N 53 lbf

Ans. (close enough!)

We could iterate, but we are already within 0.3% of the correct lift. Ackeret theory was surprisingly accurate for such a high Mach number, because of the very small deflection.

Chapter 9 Compressible Flow

741

9.153 A supersonic transport has a mass of 65 Mg and cruises at 11-km standard altitude at a Mach number of 2.25. If the angle of attack is 2 and its wings can be approximated by flat plates, estimate (a) the required wing area; and (b) the thrust. Solution:

At 11 km (Table B.6), take p 22612 Pa. (a) Use linearized theory:

CL

4

M 2 1

4(2 /180) 2.25 2 1

0.0693

A 115 m2

W

k pMa 2 A 2 Ans . (a)

65000(9.81) , 0.7(22612)(2.25)2 A

(b) According to linearized (Ackeret) theory, if there is no thickness drag, then

Drag Lift , or Drag Thrust 65000(9.81)(2 /180) 22,300 N Ans. (b)

9.154 A symmetric supersonic airfoil has its upper and lower surfaces defined by a sine waveshape: t x y sin 2 C where t is the maximum thickness, which occurs at x C/2. Use Ackeret theory to derive an expression for the drag coefficient at zero angle of attack. Compare your result with Ackeret theory for a symmetric double-wedge airfoil of the same thickness. Solution:

Evaluate the mean-square surface slope and then use Eq. (9.107):

t

1 x 2 y cos , y 2C C0 C

C

2 t t 2 x cos dx 2C C 8 C 2

2 2 t 2 1 At 0, C D , 0 2 2 2 2 8 C Ma 1

4

2 or: C D Ma2 1 8 4

t 2 C

Ans.

Meanwhile, for a double-wedge of the same thickness t, from Prob. 9.152, CD,double-wedge

t (19% less) Ans . 2 Ma 1 C 4

2

2

Solutions Manual Fluid Mechanics, Seventh Edition

742

P9.155 The F-35 airplane in Fig. 9.29 has a wingspan of 10 m and a wing area of 41.8 m2. It cruises at about 10 km altitude with a gross weight of about 200 kN. At that altitude, the engine develops a thrust of about 50 kN. Assume the wing has a symmetric diamond airfoil with a thickness of 8%, and accounts for all lift and drag. Estimate the cruise Mach number of the airplane. For extra credit, explain why there are two solutions. Solution:

At 10 km standard altitude, from Table B.6, = 0.4125 kg/m3, and

atmospheric pressure is p = 26416 Pa. We need to match lift and drag to the given weight/thrust data:

Lift W 200,000 N C L

k Ma 2 p A , where C L 2

k Drag Thrust 50,000 N C D Ma 2 p A , where C D 2

4

Ma 1 4 2

, in radians

dy [ 2 ( )2avg ] dx Ma2 1

For a symmetric 8% diamond airfoil, as in Fig. E9.21, the slope everywhere is |dy/dx| = 0.08, so the numbers for lift and drag are as follows:

W 200,000 N T 50,000 N

4

1.4 )Ma 2 ( 26416Pa )(41.8m 2 ) Ma2 1 2 4 1.4 [ 2 (0.08)2 ] ( ) Ma2 ( 26416 Pa)(41.8 m 2 ) 2 Ma 2 1 (

to be solved for and Ma. It was a bit of a surprise to the writer that there are two solutions to this system, both quite realistic:

(1) High velocity, low lift coefficient: = 0.02895 rad (1.66), Ma = 1.90

(2) Low velocity, high lift coefficient: = 0.02895 rad (1.66), Ma = 1.176

Ans.(1) Ans.(2)

NOTE: The true Mach number of the F-35A would include drag and lift of the fuselage and tail. EXTRA CREDIT: Why are there two solutions? Because the lift and drag vary as

Ma 2/(Ma2 -1)1/2 , which, for a given , comes down and then back up again, like this:

Chapter 9 Compressible Flow

250000

743

= 0.02895 radians

225000

Two solutions for 200 kN

Lift, N

200000

Fig. P9.155

175000 150000

1

1.2

1.4

Ma

1.6

1.8

2

To match both lift and drag requires = 0.02895 radians, then solving the lift as a quadratic equation yields the two solutions, Ma = 1.176 and 1.900. COMMENT: Ma = 1.176 is close to the transonic regime, for which Ackeret theory falters. 9.156 A thin circular-arc airfoil is shown in Fig. P9.156. The leading edge is parallel to the free stream. Using linearized (smallturning-angle) supersonic-flow theory, derive a formula for the lift and drag coefficient for this orientation, and compare with Ackeret-theory results for an angle of attack tan –1(h/L). Solution: For the (x,y) coordinate system shown, the formula for the plate surface is

yfoil R2 x2 R h,

L2 h 2 where R , and 2h

dy x dx R2 x2

Fig. P9.156

Solutions Manual Fluid Mechanics, Seventh Edition

744

If h << L (small disturbances), R <

dy kMa 2 p ; Thus p lower p upper B , where B 2 p dx Ma 1 Lift

foil

2Bx

R x 2

2

p

p total dA foil 2Bp x(R 2 x 2 ) 1/2 cos b dx, L 0

where cos 1 (dy/dx)2

1/2

Carry out this integration, assuming that h << L, R << L, chord length C L, to obtain

Lift

2kMa2 Ma2

1

p bh, or C L

But h/L radians, therefore CL 4

Lift

(k/2)Ma 2p bL

4

h Ma2 1 L

Ma 2 1 , which is exactly Ackeret theory.

Similarly, Drag 2Bp x(R2 x2 )1/2sin b dx 2kMa2 Ma2 1 0 L

Ans.(a)

or: C D 4/ Ma 2 1 (h/L) 2 1

1 1 2 C L 1 3 3

1/ 2

p bh 4h 3L

Ans. (b)

This is exactly the same as Ackeret theory. The extra term “1/3” is the “thickness -drag” contribution (actually the camber slope contribution) from Eqs. (9.106, 107) of the text.

P9.157

The Ackeret airfoil theory of Eqs. (9.104-5) is meant for moderate supersonic

speeds, 1.2 < Ma < 4. How does it fare for hypersonic speeds? To illustrate, calculate

(a) CL and (b) CD for a flat-plate airfoil at = 5, with Ma = 8.0, using shock-expansion theory, and compare with Ackeret theory. Comment.

Chapter 9 Compressible Flow

745

Solution: Like Example 9.19, we need only calculate a (steep) shock wave on the

bottom and a (steep) expansion wave fan on the top. The results are shown in the figure below.

expansion waves

= 5

Ma 3 = 9.412 p 3 = 0.3442 p

Ma =8.0, p

= 10.85

Fig. P9.157

shock

p 2 = 2.4771 p

If the airfoil has length C and span b, the force on the plate is F = (p 2 - p3)bC. Then we compute Lift = F cos and Drag = F sin . The two force coefficients are: CL CD

(2.4771 p 0.3442 p )bC cos 5 o 2 ( k / 2) p Ma bC

(2.4771 p 0.3442 p )bC sin 5 o (k

/ 2) p Ma 2 bC

2.125 (1.4 / 2)(8.0) 0.186 (1.4 / 2)(8.0)

2

0.0474

2

0.00415

Ans.( a) Ans.(b )

If we apply Ackeret theory to the same data, we obtain, much more simply, CL , Ackeret C D , Ackeret

4

Ma2 1 2 4

Ma 2

1

4 (5 / 180 ) 8 1 2

4 (5 / 180 )2 8 1 2

0.0440 0.00384

Ans.(a ) 7% low Ans.( b) 8% low

Even at hypersonic Mach numbers, Ackeret theory is not bad – and temptingly simple! The theory is helped here by the fact that the angle of attack and foil thickness are small.

746

Solutions Manual Fluid Mechanics, Seventh Edition

FUNDAMENTALS OF ENGINEERING EXAM PROBLEMS: Answers In the following problems, assume one-dimensional flow of ideal air, R 287 J/(kgK) and k 1.4. FE9.1 For steady isentropic flow, if the absolute temperature increases 50%, by what ratio does the static pressure increase? (a) 1.12 (b) 1.22 (c) 2.25 (d) 2.76 (e) 4.13 FE9.2 For steady isentropic flow, if the density doubles, by what ratio does the static pressure increase? (a) 1.22 (b) 1.32 (c) 1.44 (d) 2.64 (e) 5.66 FE9.3 A large tank, at 500 K and 200 kPa, supplies isentropic air flow to a nozzle. At section 1, the pressure is only 120 kPa. What is the Mach number at this section? (a) 0.63 (b) 0.78 (c) 0.89 (d) 1.00 (e) 1.83 FE9.4 In Prob. FE9.3 what is the temperature at section 1? (a) 300 K (b) 408 K (c) 417 K (d) 432 K (e) 500 K FE9.5 In Prob. FE9.3, if the area at section 1 is 0.15 m2, what is the mass flow? (a) 38.1 kg/s (b) 53.6 kg/s (c) 57.8 kg/s (d) 67.8 kg/s (e) 77.2 kg/s FE9.6 For steady isentropic flow, what is the maximum possible mass flow through the duct in Fig. FE9.6? (To 400 K, po 300 kPa) (a) 9.5 kg/s (b) 15.1 kg/s (c) 26.2 kg/s (d) 30.3 kg/s (e) 52.4 kg/s

FE9.7

Fig. FE9.6

If the exit Mach number in Fig. FE9.6 is 2.2, what is the exit area? (a) 0.10 m 2 (b) 0.12 m2 (c) 0.15 m 2 (d) 0.18 m2 (e) 0.22 m 2 FE9.8 If there are no shock waves and the pressure at one duct section in Fig. FE9.6 is 55.5 kPa, what is the velocity at that section? (a) 166 m/s (b) 232 m/s (c) 554 m/s (d) 706 m/s (e) 774 m/s FE9.9 If, in Fig. FE9.6, there is a normal shock wave at a section where the area is 0.07 m2, what is the air density just upstream of that shock? (a) 0.48 kg/m3 (b) 0.78 kg/m3 (c) 1.35 kg/m 3 (d) 1.61 kg/m3 (e) 2.61 kg/m3 FE9.10 In Prob. FE9.9, what is the Mach number just downstream of the shock wave? (a) 0.42 (b) 0.55 (c) 0.63 (d) 1.00 (e) 1.76

Chapter 9 Compressible Flow

747

COMPREHENSIVE PROBLEMS C9.1 The converging-diverging nozzle in the figure has a design Mach number of 2.0 at the exit plane for isentropic flow from tank a to b. (a) Find the exit area A e and back pressure p b which will allow design conditions. (b) The back pressure grows as tank b fills with air, until a normal shock wave appear in the exit plane. At what back pressure does this occur? (c) If tank b remains at constant T 20°C, how long will it take for the flow to go from condition (a) to condition (b)?

Fig. C9.1

Solution:

(a) Compute the isentropic pressure ratio and area ratio for Ma 2.0:

Mae 2.0: Table B.1:

pe 0.1278, pe 0.1278(1E6) 128,000 Pa po

Ae 1.6875, Ae 1.6875(0.07) 0.118 m2 A* (b) Compute the pressure ratio across a shock for Ma 2.0: Mae 2.0: Table B.2:

Ans. (a)

Ans. (a)

pb 4.5, pb 4.5(128000) 575,000 Pa pe

Ans. (b)

(c) Compute the (constant) mass flow and the mass needed to fill the tank: p A* (1E6)(0.07) kg 126.5 constant m mmax 0.6847 o 0.6847 s RTo 287(500)

128000 575000 2 (100m ) 531.9 kg m tank ( final initial ) tank 287(293) 287(293) Since mass flow is constant, the time required is simply the ratio of these two: t required

m 531.9 kg 4.2 sec Ans. (c) m 126.5 kg/s

Solutions Manual Fluid Mechanics, Seventh Edition

748

C9.2 Two large air tanks, one at 400 K and 300 kPa and the other at 300 K and 100 kPa, are connected by a straight tube 6 m long and 5 cm in diameter. The average friction factor is 0.0225. Assuming adiabatic flow, estimate the mass flow through the tube.

Solution: The higher-pressure tank denotes the inlet stagnation conditions, p o 300 kPa and T o 400 K. The flow will be subsonic, but we have no idea whether it is choked. Assume that the tube exit pressure equals the receiver pressure, 100 kPa. We must iterate—an ideal job for EES! We do know (f L/D): f

L 6.0 0.0225 2.70 D 0.05

If p2 100 kPa, we must ensure that the inlet Mach number is just sufficient that the inlet stagnation pressure po1 300 kPa: Guess Ma2, back off (fL/D) 2.70, find Ma1, check p*, p 1, and p o1. Example:

Ma2 1.0, p * p2 100 kPa, Read

Ma1 0.380,

f

L*2 L* 0, f 1 2.70, D D

p1 2.84 Then p*

p1 2.84(100) 284 kPa,

p1 [1 0.2(0.380)2 ] 3.5, solve po1 314 kPa 300 po1

So we back off and try values of Ma2 1.0 and proceed until the inlet matches. The solution (performed by the author using EES) is Ma2 0.962, Ma1 0.380, p2 p* 100 kPa, p1 271.5 kPa, T1 389K, kg kg m Ans. 1 2.434 3 , V1 150 , m 1 D 2V1 0.718 s 4 s m

So the tube is nearly, but not quite, choked.

C9.3 Fig. C9.3 shows the exit of a converging-diverging nozzle, where an oblique shock pattern is formed. In the exit plane, which has an area of 15 cm2 , the air pressure is 16 kPa and the temperature is 250 K. Just outside the exit shock, which makes an angle of 50° with the exit plane, the temperature is 430 K. Estimate (a) the mass flow; (b) the throat area; (c) the turning angle of the exit flow; and, in the tank supplying the air, (d) the pressure and (e) the temperature.

Chapter 9 Compressible Flow

749

Fig. C9.3

Solution: We know the temperature ratio and the shock wave angle, so we can muddle through oblique-shock-wave theory to find the shock conditions: T2 430 1.72; 40 T1 250

Iterate Eqns.(9.83) or use EES!

Solution: Maexit 3.17; 1 0.223 A throat 3.00 cm 2 V1 Ma1 a1 1006

m , s

m 1 A1V1 0.336

kg s

kg A1 15 ; 4.99 , m 3 A* A throat Ans. (b)

Ans. (a) shock 22.88

Ans. (c)

With the exit Mach number known, it is easy to compute stagnation conditions:

po, tank 16[1 0.2(3.17) 2] 3.5 760kPa

To, tank 250[1 0.2(3.17) 2] 753K

Ans.(d)

Ans. (e)

C9.4 The properties of a dense gas (high pressure and low temperature) are often approximated by van der Waals’ equation state [Refs. 17 and 18]:

p

RT 2 a1 1 b1

2 2 ft lb4 27 R Tc 9E 5 where a1 for air 64 p c slug 2

ft 3 RTc and b 1 for air 0.65 slug 8pc Find an analytic expression for the speed of sound of a van der Waals gas. Assuming k = 1.4, compute the speed of sound of air, in ft/s, at –100F and 20 atm, for (a) a

Solutions Manual Fluid Mechanics, Seventh Edition

750

perfect gas, and (b) a van der Waals gas. What percentage higher density does the van der Waals relation predict? NOTE: 1 atm = 2116 lbf/ft3 . Solution: For air, take R = 1716 ft-lbf/slugR. First evaluate the densities, T = 360R:

20 2116 lbf / ft 2 p slug Ideal gas : 0.0684 RT (1716 ft lbf / slug o R )(360o R ) ft 3 van der Waals : 20 2116

slug (1716)(360) 9 E5 2 , Solve for 0.0724 3 1 0.65 ft

The van der Waals relation predicts air density 5.7% higher. (a) For an ideal gas,

aideal

kRT (1.4)(1716)(360) 930ft/s

Ans.(a)

(b) For the van der Waals gas, a2 k(

RTb1 RT p 2 a1 ] )T k[ 1 b1 (1 b1 )2

Plug in the numbers : avan der Waals 877 ft/s

Ans .(b )

_____________________________________________________________________ C9.5 Consider one-dimensional steady flow of a non-ideal gas, steam, in a converging nozzle. Stagnation conditions are p o 100 kPa and T o 200C. The nozzle exit diameter is 2 cm. If the nozzle exit pressure is 70 kPa, (a) calculate the mass flow and the exit temperature for real steam, from the Steam Tables or using EES. (As a first estimate, assume steam to be an ideal gas from Table A.4.) Is the flow choked? (b) Find the nozzle exit pressure and mass flow for which the steam flow is choked, using EES or the Steam Tables. Solution: (a) For steam as an ideal gas, from Table A.4, k 1.33 and R 461 J/kgK. First use this approximation to find the exit Mach number: p 70 kPa 0.33 2 Mae 1 po 100 kPa 2

1.33/0.33

, solve for Maexit 0.75 Flow is not choked

Chapter 9 Compressible Flow

751

For real steam, we use EES. The nice Power-law ideal-gas formulas, Eqs. (9.26–9.28), are invalid, but the energy equation (9.22) is valid, and the nozzle flow is isentropic. First evaluate h o ENTHALPY(steam, p 100, T 473) 2875 kJ/kgK s o ENTROPY(steam, p 100, T 473) 7.833 kJ/kgK

Then use the energy equation with the known pressure at the exit:

Ve2 ho 2875 he with he ENTHALPY( steam, p 70, s 7.833) 2(1000) EES returns the result he 2800 kJ/kg and Ve 385 m/s

The specific request was for the exit temperature and the mass flow:

Te TEMPERATURE(steam, p 70, s 7.833) 434 K Ans. (a)

e DENSITY(steam, p 70, s 7.833) 0.351 kg/m3

mass flow eAe V e (0.351 kg/m3 )(/4)(0.02 m) 2(385 m/s) 0.0425 kg/s

Ans. (a)

EES now has a speed of sound function, so we can compute

ae = SOUNDSPEED(steam, p 70, s 7.833) 512 m/s

Hence the exit Mach number is Mae = (385 m/s)/(512 m/s) = 0.75 The agreement with perfect-gas theory is excellent.

(b) We are asked to determine pe for which the flow is choked, using EES. Ideal gas theory for k 1.33 predicts from Eq. (9.32) that p*/po 0.54. For EES, real steam, we use the same procedure as in part (a) above and reduce pexit gradually until the mass flow is a maximum. The final result is pe = 54.2 kPa: pe /po 0.542, Mae 1.00, mmax 0.04517 kg/s Ans . (b)

C9.6 Extend Prob. C9.5 as follows. Let the nozzle be converging-diverging, with an exit diameter of 3 cm. Assume isentropic flow. (a) Find the exit Mach number, pressure, and temperature for an ideal gas, from Table A.4. Does the mass flow agree with the value of 0.0452 kg/s in Prob. C9.5? (b) Investigate, briefly, the use of EES for this problem and explain why part (a) is unrealistic and poor convergence of EES is obtained. [HINT: Study the pressure and temperature state predicted by part (a).]

Solutions Manual Fluid Mechanics, Seventh Edition

752

Solution:

(a) For steam as an ideal gas, from Table A.4, k 1.33 and R 461 J/kgK. 0.5( k 1) /( k 1)

2 Aexit ( /4)(0.03) 1 1 0.5(k 1)Ma e 2.25 A* ( /4)(0.02) 2 Ma e 0.5(k 1) 2

Solve for Ma e 2.27

for k 1.33

Ans. (a)

For the exit pressure, temperature, and mass flow, use the ideal Power-law relations:

Te To 1 0.5( k 1) Ma2e 473/[1 0.165(2.27)2 ] 256 K

Ans. (a)

k/( k 1)

pe po 1 0.5( k 1)Mae2 100/[1 0.165(2.27) 2 ]4.03 8.4 kPa Ans. (a) p kg m e e 0.0713 3 ; Ve Mae ae 898 ; m e AeVe 0.0452 kg/s Ans . (a) RTe m s

The mass flow does equal the choked-flow value from Prob. C9.5, as expected.

(b) Recall from Prob. C9.5 that po 100 kPa and To 200C, which corresponds for real steam (EES) to h o 2875 kJ/kgK and so 7.833 kJ/kg. We proceed through the nozzle, using the energy equation, h o h V2/2, plus the condition of constant entropy. We also know that the flow is choked at 0.0452 kg/s with a throat diameter of 2 cm. The results are, for real steam, Ma e 2.22; p e 10.2 kPa; Te 319 K; e 0.0725 kg/m3; Quality 96%

The ideal-gas theory is still reasonably accurate at this point, but it is unrealistic, since the real steam has entered the two-phase (wet) region. C9.7 Professor Gordon Holloway and his student, Jason Bettle, of the University of New Brunswick, obtained the following tabulated data for blow -down air flow through a converging-diverging nozzle similar in shape to Fig. P3.22. The supply tank pressure and temperature were 29 psig and 74 F, respectively. Atmospheric pressure was 14.7 psia. Wall pressures and centerline stagnation pressures were measured in the expansion section, which was a frustrum of a cone. The nozzle throat is at x 0.

Chapter 9 Compressible Flow

x (cm):

Diameter (cm): pwall (psig): pstagnation (psig):

0

1.00 7.7 29

1.5

1.098 –2.6 26.5

3

1.195 –4.9 22.5

4.5

1.293 –7.3 18

753

6

1.390 –6.5 16.5

7.5

1.488 –10.4 14

9

1.585 –7.4 10

Use the stagnation pressure data to estimate the local Mach number. Compare the measured Mach numbers and wall pressures with the predictions of one-dimensional theory. For x 9 cm, the stagnation pressure data was not thought by Holloway and Bettle to be a valid measure of Mach number. What is the probable reason? Solution: From the cone’s diameters we can determine A/A* and compute theoretical Mach numbers and pressures from Table B.1. From the measured stagnation pressures we can compute measured (supersonic) Mach numbers, because a normal shock forms in front of the probe. The ratio p o2/p o1 from Eq. (9.58) or Table B.2 is used to estimate the Mach number. x (cm): Ma-theory:

pw-theory (psia):

0 1.00 23.1

1.5 1.54 11.2

3 1.79 7.72

4.5 1.99 5.68

6 2.16 4.36

7.5 2.31 3.44

9 2.45 2.77

The comparison of measured and theoretical Mach number is shown in the graph below.

Problem C9.7

Solutions Manual Fluid Mechanics, Seventh Edition

754

The comparison of measured and theoretical static pressure is shown in the graph below.

Holloway and Bettle discounted the data for x 9 cm, which gave Mach numbers and pressures widely divergent from theory. It is probably that a normal shock formed in the duct. C9.8 Engineers call the supersonic combustion, in the scramjet of Fig. 9.30, almost miraculous, “like lighting a match in a hurricane”. Figure C9.8 is a crude idealization of the engine. Air enters, burns fuel in the narrow section, then exits, all at supersonic speeds. There are no shock waves. Assume areas of 1 m 2 at sections 1 and 4 and 0.2 m 2 at sections 2 and 3. Let the entrance conditions be Ma1 = 6, at 10,000 m standard altitude. Assume isentropic flow from 1 to 2, frictionless heat transfer from 2 to 3 with Q = 500 kJ/kg, and isentropic flow from 3 to 4. Calculate the exit conditions and the thrust produced. Combustion

Ma > 1

Ma > 1 1

2

Fig. C9.8

3 4

Chapter 9 Compressible Flow

755

Solution: From Table B.6 at 10,000 m, p1 = 26416 Pa, T1 = 223.16 K, 1 = 0.4125 kg/m 3, and a1 = 299.5 m/s. Calculate the inlet velocity and mass flow: V1 Ma1 a1 (6.0)(299.5) 1797 m/ s ,

m 1 A1 V1 (0.4125)(1.0)(1797) 741 kg / s

Now find conditions at section 2, assuming one-dimensional isentropic steady flow: 2 2 3.5 To1 To 2 (223.16)[1 0.2(6) ] 1830 K ; po1 po 2 (26416)[1 0.2(6) ] 4.17E7 Pa

A1 A2 0.2 m2 At Ma1 6.0 , ) 10.64 for which Ma2 3.991 53.18 ; then (53.18)( 2 A* A* 1.0 m 1830 K 4.171E7 Pa 437 K ; p2 T2 278,000 Pa [1 0.2(3.991)2 ] [1 0.2(3.991)2 ]3.5

Use frictionless heat-addition theory to get from section 2 to section 3:

Ma2 3.991:

To 2 *

To

0.5895 , To* 3104 K ;

T2 0.169 , T * 2587 K T*

p2 Q 500000 0.103 , p * 2.698E6 Pa ; To 3 To 2 1830 2327 K p* cp 1005 To3 To*

2327 p 0.7497 , thus Ma3 2.238 ; 3 0.2996 , p3 808, 500 Pa 3104 p*

Now move isentropically from section 3 to section 4 (staying supersonic, of course):

Solutions Manual Fluid Mechanics, Seventh Edition

756

p o3 p o4 p3 [1 0.2(Ma32 )]3.5 (808500)[1 0.2(2.238)2 ]3.5 9.17E6 Pa

Ma3 2.2376 , p4

4

A3

A*3

9.17E6 Pa

2.073 , then

[1 0.2(3.963)2 ]3.5

A4

A*3

(2.073)(

63, 480 Pa ; T4

1.0 m 2 0.2m

2

2327 K

) 10.37 and Ma4 3.963

[1 0.2(3.963)2 ]

562 K

p4 kg m 0.394 3 ; V4 Ma4 kRT4 (3.963)(475) 1883 RT4 s m

Check the mass flow = (1883)(1.0)(0.394) =741 kg/s, Yes. For a control volume around the scramjet, we find that the thrust is due to both velocity changes and pressure changes: Thrust F m( V4 V1 ) A1 ( p4 p1 ) (741

kg m )(1883 1797 ) (1.0 m2 )(63480 26416 Pa ) s s 64150 37060 101, 200 N 22, 750 lbf

Ans.

Excellent force, but the combustion chamber has to withstand high pressures (8 atm) and high temperatures (2600 K). For this simple one-dimensional model, the thrust goes up linearly with Q.