Solution manual fluid mechanics 7th edition chapter 9 PDF

Preview

Summary

Download Solution manual fluid mechanics 7th edition chapter 9 PDF

Description

Chapter 9  Compressible Flow 9.1 An ideal gas flows adiabatically through a duct. At section 1, p1  140 kPa, T1  260C, and V1  75 m/s. Farther downstream, p2  30 kPa and T2  207C. Calculate V2 in m/s and s2  s1 in J/(kgK) if the gas is (a) air, k  1.4, and (b) argon, k  1.67.

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:

1 1 1 m c pT  V2  constant  1005(260)  (75) 2  1005(207)  V 22 or V2  335 2 2 2 s

Ans.

 207  273   30  Meanwhile, s2  s1  cp ln(T2 /T1 )  R ln(p 2 /p1 )  1005ln  ,  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): m 1 1 1 c p T  V 2  518(260)  (75)2  518(207)  V22 , solve V2  246 2 2 2 s

 207  273   30  s2  s1  518ln   208ln    54  320  266 J/kg  K  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

Solution:

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

1 1 1 m c pT  V 2  1858(260)  (75) 2  1858(207)  V22, solve V2  450 2 2 2 s  207  273  30  s2  s1  1858 ln  461ln   195  710  515 J/kg  K   260  273  140

Ans. (a) Ans. (a)

(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

J ; kg

J kg

m 1 1 1 h  V 2  2.993E6  (75) 2  2.892E6  V22 , solve V2  443 s 2 2 2

at 140 kPa and 260 C, read s1  7915

Ans. (b)

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

Thus s2  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 200C 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  400   631 K  358C Ans. (a)  300 

1  2 ,  T2  T1 (p2 /p1 )  (200  273) 

Q  mc v T  (8)(650)(358  200)  8.2E5 J

Ans. (b)

J  358  273   1500 s2  s1  mcv 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/m3. At section D, the pressure is 28.2 kPa and the temperature is –19C. (a) Find the entropy change, if any. (b) Which way is the air flowing? Solution: Convert TD = -19+273 = 254 K. We need the temperature at section B:

pB 154, 000 Pa   472 K RB (287)(1.137 kg / m3) T p 472 154, 000 Then s B  sD  c p ln( B )  R ln( B )  (1005) ln( ) (287) ln( ) 254 28, 200 TD pD J or : s B  s D  623  487  +133 Ans.( a) kg  K

TB 

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

Ans.(b)

9.5 Steam enters a nozzle at 377C, 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 377C and 1.6 MPa, read h1  3.205E6 J/kg and s1  7153 J/kgK At saturation for s1  s2  7153, read p2  185 kPa, T2  118C, and h2  2.527E6 J/kg

1 1 1 m Then h  V2  3.205E6  (200) 2  2.527E6  V22, solve V2  1180 2 2 2 s

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 cp 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 ; cv = 980 J/kg-K ; k = cp/cv = 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 J 1.3(189)   819 ; k1 kg  K 1.3 1

cv | CO2 

R J 189   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 T2 = 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) V2, (d) p2, and (e) s2 – s1. 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

Then V 2 

 m 2 A2

kg

m kg )(0.0020 m2 )(122 .5 )  0.0429 s s m m 0.0429 kg/ s   806 Ans.( c) 3 2 s (0.266 kg / m )(0.002 m )

1  m  2  1 A1V1  (1.75 m

3

Ans.(b)

That’s pretty fast! Check a2 = (kRT2)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 )(281K )  21,450 Pa Ans (d ) Similarly , T1 

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

(e) Finally, with pressures and temperatures known, the entropy change follows from Eq. (9.8): T p 281 21450 J Ans.( e) s 2  s1  c p ln( 2 )  R ln( 2 )  1005 ln( )  287 ln( )   757  757  0 T1 p1 kg  K 597 300000 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 20C 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

dQ dWshaft dE CV   00  h atmm entering dt dt dt

or, after filling,

ECV,final  E CV,initial  h atmm entered, or: mc vT tank  mc pT atm Thus Ttank  (cp /cv )Tatm  (1.4)(20  273)  410 K  137C 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/m3. 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 Rgas and hence the gas exit temperature: J 54000 p  8314 , hence Texit    462   779 K Ans. (a) 18 kg K M R  462(0.15)

Solutions Manual  Fluid Mechanics, Seventh Edition

640

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

kg kg      0.15 3  (0.45m) 2 (1600m / s )  38 s m 4 

Ans. (b)

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

Thrust  e Ae Ve2  m

e



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, T1  288.16 K. At 12000 m standard, T2  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 300C and 1 atm, estimate the speed of sound of (a) nitrogen; (b) hydrogen; (c) helium; (d) steam; and (e) uranium hexafluoride 238UF6 (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,

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

(c) helium: k  1.66, R  2077:

a  1.66(2077)(573)  1406 m/s Ans. (c) (d) steam: k  1.33, R  461: a  1.33(461)(573)  593 m/s Ans. (d) [NOTE: The EES “soundspeed” function would predict asteam = 586 m/s.] (e) For uranium hexafluoride, we need only to compute R from the molecular weight:

Chapter 9  Compressible Flow

(e)

238

UF6 : M  238  6(19)  352,  R 

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 20 C 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 n n  (B  1)( /a )  B; Bulk modulus K    n(B 1)pa ( /a ) , a   pa d

K/

We may then substitute numbers for water, with pa  101350 Pa and a  998 kg/m3: (a) at 1 atm: Kwater  7(3001)(101350)(1)7  2.129E9 Pa (21007 atm) (a)

speed of sound awater  K/  2.129E9/998 1460 m/s  1100  3000 (b) at 1100 atm:   998  3001 

Ans.

Ans. (a)

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  9000  3000  (c) at 9000 atm:   998   3001 

1/7

Ans. (b) 7

 1217

kg  1217  , 3 ; K  Ka  998  m

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 = p1,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 H2O, k = 1.33 and R = 461 m2/s2-K: aideal 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 p2  210 kPa , compute 2  DENSITY (Steam , p  210, s  s o )  0.9073 kg / m 3 At p1  190 kPa , compute 1  DENSITY (Steam , p  190, s  s o )  0.8404 kg / m 3 210000  190000 20000 p m2   298950 |s   0.9073 0.8404 0.0669 s2 m Ans .(c ) Finally, a differences  298950  547 s

Finite differences : a2 

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

P9.14 At 1 atm and 20C, the density of methyl alcohol is 49.4 lbm/ft3. At 300 atm, its density increases to 50.9 lbm/ft3. 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/ft3 = 790.7 kg/m3, and 50.9 lbm/ft3 = 814.7 kg/m3. Then use finite differences to approximate the formula: a methanol 

p  

(300  1)(101350 Pa) 3

(814.7  790.7) kg / m



m 30303650 Ans.  1124 s 24.0

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  ( B  1) ( ) n  B (1.19) po o 3 It looks like this, with  o = 790 kg/m3 from Table A.3. At 2000 atm,  887 kg/m . 3000 2500

ETHANOL

p , atm

2000 1500 1000 500

 , kg/m3

0 780

800

820

840

860

880

900

920

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

dp  a2  d

po

o

n ( B  1) (

 n1 ) o

887 kg/ m3 6  1)( ( 7 )( 1600 ) 790 kg / m 3 790 kg / m 3 At 1 atm, the speed of sound of ethanol is about 1200 m/s. Ethanol : a 

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

 pf 

2 liq Cliq  p

C   liqC liq

Ans . (b)

If liqCliq   C of air, then Vf  0 and pf  2p, 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 500C? 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

T o  T (1  0.2 Ma 2 )  ( 216 .66)[1  0.2(3.13) 2 ]  641 K  368 o C

Ans.(a )

(b) To have a front stagnation temperature of 500C = 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 p1 = 80 kPa. At section 2 downstream, the temperature has dropped to 0C. Find (a) the pressure, (b) velocity, and (c) Mach number at section 2. Solution: Assume k = 1.4 and, of course, convert T2 = 0C = 273 K. (b) The adiabatic energy equation will yield the new velocity: T1 

V12 V2 V22 (250) 2  To  330  361 K  T2  2  273  2c p 2(1005) 2c p 2(1005) Solve for V 2  421

m s

Ans .(b )

(a) To calculate p2, 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 p1

(

T2 T1

)k /(k 1) 

p2 273 3.5 )  0.515, or : p2  41 kPa ( 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



421m / s 421   1.27 331 m / s 1.4(287)(273)

Ans.( c)

9.21 CO2 expands isentropically through a duct from p1  125 kPa and T1  100C to p2  80 kPa and V2  325 m/s. Compute (a) T2; (b) Ma2; (c) To; (d) po; (e) V1; and (f) Ma1.

Chapter 9  Compressible Flow

647

Solution: For CO2, from Table A.4, take k  1.30 and R  189 J/kgK. Compute the specific heat: cp  kR/(k 1)  1.3(189)/(1.3 1)  819 J/kgK. The results follow in sequence:

(a) T2  T1 ( p2 /p1)( k1)/ k  (373 K)(80/125)(1.3 1)/1.3  336 K Ans. (a) (b) a2 

kRT2  (1.3)(189)(336)  288 m/s, Ma2  V 2 /a 2  325/288 1.13 Ans. (b)

 k 1 2   0.3  Ma2   (336)  1 (c) To1  To2  T2  1 (1.13)2   401 K   2 2   1.3/(1.3 1)

 k1  (d) po 1  po 2  p2 1  Ma22    2 (e) To1  ...