Chapter 15 Soln

8/10/2019 Chapter 15 Soln

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Updated June 2013

8e

SOLUTION MANUALCHAPTER 15

Fundamentals of

Thermodynamics

Borgnakke SonntagBorgnakke Sonntag


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Borgnakke and Sonntag

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In-Text Concept Questions


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15.a

Is stagnation temperature always higher than free stream temperature? Why?

Yes. Since kinetic energy can only be positive we have

h0= h

1+ V

12/2 > h

1

If it is a gas with constant heat capacity we getT

0= T

1+ V

12/2Cp


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15.b

By looking at Eq. 15.25, rank the speed of sound for a solid, a liquid, and a gas.

Speed of sound:

P

s= c2

For a solid and liquid phase the density varies only slightly withtemperature and constant s is also nearly constant T. We thus expect thederivative to be very high that is we need very large changes in P to givesmall changes in density.A gas is highly compressible so the formula reduces to Eq.15.28 whichgives modest values for the speed of sound.

15.c

Does speed of sound in an ideal gas depend on pressure? What about a real gas?

No. For an ideal gas the speed of sound is given by Eq.15.28

c = kRTand is only a function of temperature T.For a real gas we do not recover the simple expression above and there is adependency on P particularly in the dense gas region above the criticalpoint.

15.d

Can a convergent adiabatic nozzle produce a supersonic flow?

No. From Eq.15.33 and a nozzle so dP < 0 it is required to have dA > 0 toreach M > 1. A convergent nozzle will have M = 1 at the exit, which is the

smallest area. For lower back pressures there may be a shock standingoutside the exit plane.


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15.g

Which of the cases in Fig. 15.17 (a-h) have entropy generation and which do not?

a. There is no flow so sgen= 0.

b. Subsonic flow, reversible, so sgen= 0.

c. Limit for subsonic flow, reversible, so sgen= 0.d. The only supersonic reversible flow solution, so sgen= 0.

e. Supersonic reversible in nozzle sgen= 0, irreversible outside.

f. Supersonic reversible in nozzle sgen= 0, compression outside.

g. Shock stands at exit plane, sgen> 0 across shock.

h. Shock is located inside nozzle, sgen> 0 across shock.


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15.h

How does the stagnation temperature and pressure change in an adiabatic nozzleflow with an efficiency of less than 100%?

The stagnation temperature stays constant (energy eq.)

The stagnation pressure drops (s is generated, less kinetic energy).


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15.i

Table A.13 has a column for Poy/Poxwhy is there not one for Toy/Tox?

The stagnation pressure drops across the shock (irreversible flow) whereas thestagnation temperature is constant (energy equation).


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15.j

How high can a gas velocity (Mach number) be and still treat it as incompressibleflow within 2% error?

The relative error in the P versus kinetic energy, Eq.15.66, becomes

e = EA= 0.28314 = 0.02 M= Vco

=

Vco2 4 0.02


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Concept-Study Guide Problems


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15.1

Which temperature does a thermometer or thermocouple measure? Would youever need to make a correction to that?

Since the probe with the thermocouple in its tip is stationary relative to the

moving fluid it will measure something close to the stagnation temperature. If thatis high relative to the free stream temperature there will be significant heattransfer (convection and radiation) from the probe and it will measure a little less.For very high accuracy temperature measurements you must make somecorrections for these effects.


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15.3

Most compressors have a small diffuser at the exit to reduce the high gas velocitynear the rotating blades and increase the pressure in the exit flow. What does thisdo to the stagnation pressure?

For a reversible flow (ideal case) the stagnation pressure is constant. However, thereason it is done is to raise the pressure in a near reversible flow (diffuser) ratherthan let the flow reduce the peak velocities in a less reversible fashion whichwould lower the stagnation pressure.

15.4

A diffuser is a divergent nozzle used to reduce a flow velocity. Is there a limit forthe Mach number for it to work like this?

Yes, the flow must be subsonic. If the flow was supersonic then increasing theflow area would increase the velocity.


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15.5

Sketch the variation in V, T, P, andMfor a subsonic flow into a convergentnozzle withM = 1 at the exit plane?

V =M c =M A kRTEA= 2Cp(ToT)

Since we do not know the area versus length, we plot it versus mach numberM.

T, P and relative to the stagnation state is listed in Table A.12 and given ineqs.15.34-36. A small spread sheet (M step 0.1) did the calculations.

Only the first part 0 < M < 1 is the answer to this question.

The curves areplotted as thevariables:

T / To

/ oP / Po

V / 2CpTo

and for k = 1.40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Mach number

V

P

T


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15.6

Sketch the variation in V, T, P, andMfor a sonic (M = 1) flow into a divergentnozzle withM = 2 at the exit plane?

V =M c =M kRT = 2Cp(ToT)

Since we do not know the area versus length, we plot it versus mach numberM.

T, P and relative to the stagnation state is listed in Table A.12 and given ineqs.15.34-36.

Only the last part 1 < M < 2 is the answer to this question.

The curves areplotted as thevariables:

T / To

/ oP / Po

V / 2CpTo

and for k = 1.40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

Mach number

V

P

T


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15.7

Can any low enough backup pressure generate an isentropic supersonic flow?

No. Only one back pressure corresponds to a supersonic flow, which isthe exit pressure at state d in Figure 15.13. However a pressure lower than that

can give an isentropic flow in the nozzle, case e, with a drop in pressure outsidethe nozzle. This is irreversible leading to an increase in s and therefore notisentropic.

15.8

Is there any benefit to operate a nozzle choked?

Yes. Since the mass flow rate is constant (max value) between points cand d in Fig. 15.12 a small variation in the back pressure will not have anyinfluence. The nozzle then provides a constant mass flow rate free ofsurges up or down which is very useful for flow calibrations or other

measurements where a constant mass flow rate is essential.


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15.9

Can a shock be located upstream from the throat?

No. The flow adjust so M = 1 at the throat.


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15.10

The high velocity exit flow in Example 15.7 is at 183 K. Can that flow be used tocool a room?

Being that cold it sounds like it could. However when the flow enters a room it

eventually would have to slow down and then it has the stagnation temperature. Ifyou let the flow run over a surface there will be a boundary layer with zerovelocity at the surface and again there the temperature is close to the stagnationtemperature.


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15.11

A convergent-divergent nozzle is presented for an application that requires asupersonic exit flow. What features of the nozzle do you look at first?

You look at the cross section area change through the nozzle. At the throat M = 1

so in the divergent section the velocity increases and the ratio A/A* determineshow the flow changes. The exit area can then tell you what the exit mach numberwill be and if you can have a reversible flow or not.


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15.13

Suppose a convergent-divergent nozzle is operated as case h in Fig. 15.17. Whatkind of nozzle could have the same exit pressure but with a reversible flow?

A convergent nozzle, having subsonic flow everywhere assuming the

pressure ratio is higher than the critical.


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Stagnation Properties


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15.14

A stationary thermometer measures 80oC in an air flow that has a velocity of 200m/s. What is the actual flow temperature?

We assume that the thermometer measures the stagnation temperature, that is theprobe (bulb, thermistor or thermo-couple junction sits stationary).

h0= h

1+ V

12/2

1=

0 V

12/2Cp

1= 80oC

2000 1.004

2002J/kg-K

(m/s)2= 60oC


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15.15

Steam leaves a nozzle with a pressure of 500 kPa, a temperature of 350C, and avelocity of 250 m/s. What is the isentropic stagnation pressure and temperature?

Stagnation enthalpy from energy equation and values from steam tables B.1.3

h0= h

1+ V

12/2 = 3167.7 kJ/kg +

25022000

(m/s)2J/kJ

= 3198.4 kJ/kg

s0= s

1= 7.6329 kJ/kg K

It can be linearly interpolated from the printed tables

Computer software: (ho, s

o) T

o= 365C, P

o= 556 kPa


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15.17

An object from space enters the earths upper atmosphere at 5 kPa, 100 K, with arelative velocity of 2500 m/s or more . Estimate the object`s surface temperature.

ho1

- h1= V

12/2 = 25002/2000 = 3125 kJ/kg

ho1

= h1+ 2000 = 100 + 3125 = 3225 kJ/kg => T =2767 K

The value for h1from ideal gas table A.7 was estimated since the lowest T in the

table is 200 K.


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15.18

The products of combustion of a jet engine leave the engine with a velocity

relative to the plane of 500 m/s, a temperature of 525C, and a pressure of 75 kPa.Assuming that k=1.32, Cp=1.15 kJ/kg K for the products, determine thestagnation pressure and temperature of the products relative to the airplane.

Energy Eq.: ho1

- h1= V

12/2 = 5002/2000 = 125 kJ/kg

To1

- T1= (h

o1- h

1)/Cp= 125/1.15 = 108.7 K

To1

= 525 + 273.15 + 108.7 = 906.8 K

Isentropic process relates to the stagnation pressure

Po1

= P1(T

o1/T

1)k/(k-1)= 75(906.8/798.15)4.125= 127 kPa


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15.19

Steam is flowing to a nozzle with a pressure of 400 kPa. The stagnation pressure

and temperature are measured to be 600 kPa and 350oC, respectively. What arethe flow velocity and temperature?

Stagnation state Table B.1.3: ho1= 3165.66 kJ/kg, so1= 7.5463 kJ/kg K

State 1: 400 kPa, s1= s

o1= 7.5463 kJ/kg K

T1= 250 + (300 250)

7.5463 - 7.3788

7.5661 - 7.3788= 294.7oC

h1= 2964.16 +

7.5463 - 7.3788

7.5661 - 7.3788(3066.75 2964.16) = 3055.9 kJ/kg

Energy equation gives

V12/2 = h

o1- h

1= 3165.66 3055.9 = 109.76 kJ/kg

V1= 2 (ho1- h1) = (2 109.76 1000) J/kg = 468.5 m/s


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15.20

A meteorite melts and burn up at temperatures of 3000 K. If it hits air at 5 kPa, 50K how high a velocity should it have to experience such a temperature?

Assume we have a stagnation T = 3000 K

h1+ V12

/2 = hstagn.Use table A.7, h

stagn.= 3525.36 kJ/kg, h

1= 50 kJ/kg

V12/2 = 3525.36 50 = 3475.4 kJ/kg (remember convert to J/kg = m2/s2)

V1

= 2 3475.4 1000 J/kg = 2636 m/s


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15.21

Air leaves a compressor in a pipe with a stagnation temperature and pressure of

150C, 300 kPa, and a velocity of 125 m/s. The pipe has a cross-sectional area of0.02 m2. Determine the static temperature and pressure and the mass flow rate.

ho1

- h1= V

12/2 = 1252/2000 = 7.8125 kJ/kg

To1

- T1= (h

o1- h

1)/C

p= 7.8125/1.004 = 7.8 K

T1= T

o1- T = 150 - 7.8 = 142.2 C = 415.4 K

P1= P

o1(T

1/T

o1)k/(k-1)= 300 kPa (415.4/423.15)3.5= 281 kPa

m.

= AV=AV

v=

P1AV

1

RT1

=281.2(0.02)(125)

0.287(415.4)= 5.9 kg/s


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15.22

I drive down the highway at 110 km/h on a day with 25C, 101.3 kPa. I put myhand, cross sectional area 0.01 m2, flat out the window. What is the force on myhand and what temperature do I feel?

The air stagnates on the hand surface : h1

+ V1

2/2 = hstagn.

Use constant heat capacity

Tstagn.

= T1

+

E/2

CpE= 25 + EEA

0.5 110A2A(1000/3600)A2AE1004E

A= 25.465CV

12

Assume a reversible adiabatic compression

Pstagn.

= P1(T

stagn./T

1)Ak/(k-1) EA= 101.3 kPa (298.615/298.15)A3.5E

= 101.85 kPa


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15.23

A stagnation pressure of 110 kPa is measured for an airflow where the pressure is

100 kPa and 20C in the approach flow. What is the incomming velocity?

Assume a reversible adiabatic compression

To1= T1(Po1/P1)A(k-1)/kE

A = 293.15 K (A110100EA)A0.2857E

A = 301.24 K

VA12

AEE/2 = ho1

- h1= C

p(T

o1- T

1) = 1.004 (301.24 293.15) = 8.1246 kJ/kg

V1

= A 2 8.1246 1000EA= 127.5 m/s

1

2

cb

To the left a Pitot tube, blue inner tube measures stagnation pressure and yellowouter tube with holes in it measures static pressure. To the right is a stagnationpoint on a wall relative to the free stream flow at state 1.


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15.24

A 4-cm inner diameter pipe has an inlet flow of 10 kg/s water at 20AoEAC, 200 kPa.

After a 90 degree bend as shown in Fig. P15.24, the exit flow is at 20AoEAC, 190 kPa.Neglect gravitational effects and find the anchoring forces F

AxE

A

and FAy

E

A

.

D = 0.04 m = A4E

ADA2EA= 0.001257 mA2 EA

Vavg

= Am.

E

AE

A= A10 0.001002

0.001257EA= 7.971 m/s

Now we can do the x and y direction momentum equations for steady flowand the same magnitude of the velocity, but different directions

X-dir: 0 = Am.E

AVavg 1

+ Fx

Am.E

A 0 + (P1

Po

) A

Y-dir: 0 = Am.

E

A 0 + Fy Am

.E

A (Vavg 2

) + (P2 P

o) A

Fx= Am

.EAV

avg 1 (P

1 P

o) A

= 10 7.97 100 0.001257 1000 = 205 N

Fy= Am

.EAV

avg 2 (P

2 P

o) A

= 10 7.97 90 0.001257 1000 = 193 N

x

y

F

Fx

y


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15.25

A jet engine receives a flow of 150 m/s air at 75 kPa, 5C across an area of 0.6 mA2EAwith an exit flow at 450 m/s, 75 kPa, 800 K. Find the mass flow rate and thrust.

Am.E

A= AV; ideal gas = P/RT

Am.

E

A= (P/RT)AV = (A75

0.287 278.15EA) 0.6 150 = 0.9395 0.6 150

= 84.555 kg/s

Fnet

= Am.E

A(Vex

- Vin

) = 84.555 (450 - 150) = 25 367 N

Inlet High P Low P exit

cb

Fnet

The shaft must have axial load bearings to transmit thrust to aircraft.


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15.26

How large a force must be applied to a squirt gun to have 0.1 kg/s water flow outat 20 m/s? What pressure inside the chamber is needed?

F = Ad mV

dtE

A= Am.E

AV = 0.1 20 kg m/sA2EA= 2 N

Eq.15.21: vP = 0.5 VA2E

P = 0.5 VA

2EA

/v = 0.5 20A

2EA

/ 0.001= 200 000 Pa = 200 kPa


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15.27

A jet engine at takeoff has air at 20Ao EAC, 100 kPa coming at 25 m/s through the 1.0m diameter inlet. The exit flow is at 1200 K, 100 kPa, through the exit nozzle of0.4 m diameter. Neglect the fuel flow rate and find the net force (thrust) on theengine.

1= A

4E

ADA2EA= 0.7854 mA2EA; 2= A

4E

ADA2EA= 0.1257 mA2E

v1= A

RT

PEA= A

0.287 293.15100E

A= 0.8409 mA3EA/kg; v2= 3.444 mA3EA/kg

Am.

E

A= AV/v = 1V

1/v

1= A

0.7854 250.8409E

A= 48.0 kg/s

V2=

m.

v2A

2

= A48.0 3.444

0.1257EA= 1315 m/s

Now we can do the x direction momentum equation for steady flow and the samemass flow rate in and out

X-dir: 0 =A

m.E

A

V1+ F

x+ (P

1 P

o) A

1

A

m.E

A

V2 (P

2 P

o) A

2

Fx= Am

.EAV

1 (P

1 P

o) A

1+ Am

.EAV

2+ (P

2 P

o) A

2

=A

m

.E

A

(V2 V1) 0 + 0 = 48 (1315 25) = 61 920 N

Inlet High P Low P exit

cb

Fnet


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15.28

A water turbine using nozzles is located at the bottom of Hoover Dam 175 mbelow the surface of Lake Mead. The water enters the nozzles at a stagnationpressure corresponding to the column of water above it minus 20% due to losses.

The temperature is 15C and the water leaves at standard atmospheric pressure. If

the flow through the nozzle is reversible and adiabatic, determine the velocity andkinetic energy per kilogram of water leaving the nozzle.

The static pressure at the 175 m depth is:

P = gZ = AgZ

vEA= A

9.807 1750.001001 1000EA= 1714.5 kPa

Pac

= 0.8 P = 1371.6 kPa

Use Bernoullis equation: vP = VAex2

AE E/2

Vex

= A 2vPEA

Vex

= A 2 0.001001 1000 1371.6EA= 62.4 m/s

VAex2

AE E/2 = vP = 1.373 kJ/kg


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15.29

A water cannon sprays 1 kg/s liquid water at a velocity of 100 m/s horizontallyout from a nozzle. It is driven by a pump that receives the water from a tank at

15C, 100 kPa. Neglect elevation differences and the kinetic energy of the waterflow in the pump and hose to the nozzle. Find the nozzle exit area, the required

pressure out of the pump and the horizontal force needed to hold the cannon.

Am.E

A= AV= AV/v A = Am. E

Av/V= 1A0.001001

100EA= 1.010A-5EAmA2 EA

A

W.E

A

p=

A

m.E

A

wp

=A

m.E

A

v(Pex

- Pin

) =A

m.E

A

VA

ex2

AE E

/2

Pex

= Pin

+ VAex2

AE E/2v = 100 + 100A2 EA/2 1000 0.001 = 150 kPa

F = Am.

E

AVex

= 1 100 = 100 N


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15.30

An irrigation pump takes water from a lake and discharges it through a nozzle asshown in Fig. P15.30. At the pump exit the pressure is 900 kPa, and the

temperature is 20C. The nozzle is located 15 m above the pump and theatmospheric pressure is 100 kPa. Assuming reversible flow through the system

determine the velocity of the water leaving the nozzle.

Assume we can neglect kinetic energy in the pipe in and out of the pump.

Incompressible flow so Bernoulli's equation applies (V1V

2V

30)

Bernoulli: v(P3 P

2) + (VA

32

AEE VA22

AEE)/2 + g(Z3 Z

2)= 0

P3= P

2

g(Z3- Z

2)

v= 900 A

9.807 151000 0.001002EA= 753 kPa

Bernoulli: VA

42

AEE

/2 = v(P3 P

4)

V4= 2v(P

3- P

4) = A 2 0.001002 653 1000EA = 36.2 m/s


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15.31

A water tower on a farm holds 1 m3liquid water at 20C, 100 kPa in a tank on topof a 5 m tall tower. A pipe leads to the ground level with a tap that can open a 1.5cm diameter hole. Neglect friction and pipe losses, and estimate the time it willtake to empty the tank for water.

Incompressible flow so we can use Bernoulli Equation.

PAeEA= Pi; Vi= 0; Ze= 0; Zi= H

VAe2

AEE/2 = gZi VAeEA= A 2gZEA = A 2 9.807 5EA = 9.9 m/s

Am.E

A= AVAeEA= AVAeEA/v = m/t

m = V/v; A = DA2EA/4 = 0.015A2EA/ 4 = 1.77 10A-4EAmA2 EA

t = mv/AVAeEA= V/AVAeE

t =EA

1

1.77 10A-4A9.9EA

= 571.6 sec = 9.53 min


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15.32

Find the speed of sound for air at 100 kPa at the two temperatures 0C and 30C.Repeat the answer for carbon dioxide and argon gases.

From eq. 15.28 we have

c0= A kRTEA= A 1.4 0.287 273.15 1000EA= 331 m/s

c30

= A 1.4 0.287 303.15 1000EA= 349 m/s

For Carbon Dioxide: R = 0.1889 kJ/kg K, k = 1.289

c0= A 1.289 0.1889 273.15 1000EA= 257.9 m/s

c30

= A 1.289 0.1889 303.15 1000EA= 271.7 m/s

For Argon: R = 0.2081 kJ/kg K, k = 1.667

c0= A 1.667 0.2081 273.15 1000EA= 307.8 m/s

c30

= A 1.667 0.2081 303.15 1000EA= 324.3 m/s


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15.35

Use the CATT3 software to solve the previous problem.

From Eq. 15.25: cA2EA= (APE

A) AsEA= vA2 EA(APv

E

A)AsE

Superheated vapor water at 400AoEAC, 6000 kPa

CATT3: v = 0.04739 mA3 EA/kg, s = 6.541 kJ/kg K

At P = 6200 kPa & s = 6.541 kJ/kg K: T = 405.1Ao EAC, v = 0.0462 mA3 EA/kg


cA2EA= -(0.04739)A2EA(A 6.2 - 5.80.0462 - 0.04866E

A

)A

MJ

kgEA

= 0.36517 10A6EAmA2 EA/sA2E

=> c = 604 m/s

From Table A.8: Cp= A1338.56 - 1235.3

50EA

= 2.0652 kJ/kg K

Cv= Cp R = 2.0652 0.4615 = 1.6037 kJ/kg K

k = Cp/Cv= 1.288; R = 0.4615 kJ/kg K (from A.5)

Now do the speed of sound from Eq.15.28

c = A kRTEA= A 1.288 0.4615 673.15 1000EA= 632.6 m/s


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15.36

If the sound of thunder is heard 5 seconds after the lightning is seen and the

weather is 20C. How far away is the lightning taking place?

The sound travels with the speed of sound in air (ideal gas). Use theformula in Eq.15.28

L = c t = A kRTEA t = A 1.4 0.287 293.15 1000EA5 = 1716 m

For every 3seconds after thelightning thesound travelsabout 1 km.


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15.37

Find the speed of sound for carbon dioxide at 2500 kPa, 60AoEAC using either thetables or the CATT3 software (same procedure as in Problem 15.34) and comparethat with Eq.15.28.

From Eq. 15.25: cA2EA= (APE

A) AsEA= vA2 EA(AP

vE

A)AsE

Superheated carbon dioxide at 60A

o EA

C, 2500 kPa

CATT3: v = 0.02291 mA3 EA/kg, s = 1.521 kJ/kg K


At P = 2400 kPa & s = 1.521 kJ/kg K: T = 56.9A

o EA

C, v = 0.02365 mA

3EA

/kg

cA

2EA

= (0.02291)A

2EA

(A

2.6 - 2.4

0.02221 - 0.02365E

A

)A

MJ

kgE

A

= 7.8983 10A

4EA

mA

2EA

/sA

2E

=> c = 270 m/s

From Table A.5: k = Cp/Cv= 1.289; R = 0.1889 kJ/kg K


c = A kRTEA= A 1.289 0.1889 333.15 1000EA= 284.8 m/s


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15.38

A jet flies at an altitude of 12 km where the air is at -40AoEAC, 45 kPa with a velocityof 1000 km/h. Find the Mach number and the stagnation temperature on the nose.

From Table A.5: k = Cp/Cv= 1.4; R = 0.287 kJ/kg K


c = A kRTEA= A 1.4 0.287 233.15 1000EA= 306 m/s

V= 1000 km/h = 1000 (km/h)1000 (m/km) /3600 (s/h)= 278 m/s

M = V/ c = 278/306 = 0.908

h0= h1+ VA12

AEE

/2 T0= T1+ VA12

AEE

/ 2Cp= 40 + A2782

E2 1004EA= -1.5AoE

A

C


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15.39

The speed of sound in liquid water at 25AoEAC is about 1500 m/s. Find the stagnation

pressure and temperature for aM= 0.1 flow at 25AoEAC, 100 kPa. Is it possible to geta significant mach number flow of liquid water?

V =Mc = 0.1 1500 = 150 m/s

h0= h1+ VA12

AEE/2

Bernoulli Eq.: P = VA12

AEE/2v = A1502

E2 0.001EA= 11.25 10A6E

APa = 11.25 MPa

P0= P1+ P = 100 + 11 250 = 11 350 kPa

T0= T1+ VA12

AEE

/ 2Cp= 25 + A1502

E2 4180EA= 27.7Ao E

A

C

Remark: Notice the very high pressure. To get a higher velocity you need a higherpressure to accelerate the fluid, that is not feasible for any large flow rate.


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15.40

Steam flowing at 15 m/s 1800 kPa, 300Ao EAC expands to 1600 kPa in a convergingnozzle. Find the exit velocity and area ratio Ae/ Ai.

Solve the problem with the steam tables.

Inlet state: vi= 0.14021 mA3E

A/kg, hi= 3029.21 kJ/kg, si= 6.8226 kJ/kg-K

Exit state: (Pe,se= si) ve= 0.15371 mA3 E

A

/kg, he= 3000.995 kJ/kg

Energy Eq.: VAi2

AEE/ 2 + hi= VAe2

AEE/ 2 + he ; VAe2

AEE= VAi2

AEE+ 2(hihe)

Ve= A 15 15 + 2000(3029.213000.995)EA= 238 m/sSame mass flow rate so

Ae/Ai= (ve/vi)(Vi/Ve) = A0.15371

0.14021EA A

15

238EA= 0.06909

If we solved as ideal gas with constant specific heat we get (k = 1.327)

Te= Ti(Pe/Pi)A(k-1)/kE

A= 573.15 (1600/1800)A0.2464 E

A= 556.76 K

Ve= A V i2

AE+ 2Cp(TiTe)E= A 15 15 + 2 1872(573.15 556.76)E= 248 m/s

Ae/Ai= (ve/vi)(Vi/Ve) = (Pi/Pe)A1/kE

A(Vi/Ve) = A

1800

1600

0.7536EA A

15

248EA= 0.0661


51/127



15.41

A convergent nozzle has a minimum area of 0.1 mA

2 EA

and receives air at 175 kPa,1000 K flowing with 100 m/s. What is the back pressure that will produce themaximum flow rate and find that flow rate?

APA

*AEEA

PoE= (A 2k+1E

A)Ak

k-1E

A= 0.528 Critical Pressure Ratio

Find Po:

h0= h1+ VA12

AEE/2 = 1046.22 + 100A2EA/2000 = 1051.22 kJ/kg

T0= Ti+ 4.4 = 1004.4 K from table A.7

P0= Pi(T0/Ti)Ak/(k-1)

E

A= 175 (1004.4/1000)A3.5EA= 177.71 kPa

The mass flow rate comes from the throat properties

PA

*EA

= 0.528 Po= 0.528 177.71 = 93.83 kPa

TA*EA= 0.8333 To= 836.97 K

A*EA= EEAPA*A

E

RTA

*AE

A= A93.83

0.287 836.97EA= 0.3906 kg/mA3E

V= c = EA kRTA* AEA = A 1.4 1000 0.287 836.97EA= 579.9 m/s

Am.E

A= AV= 0.3906 0.1 579.9 = 22.65 kg/s


52/127


53/127



15.43

To what pressure should the steam in problem 15.40 expand to reach Mach one?Use constant specific heats to solve.

Find stagnation properties

T0= T1+ VA

12

AEE

/2Cp= 573.15 K + 15A

2E

A

(m/s)A

2E

A

/(2 1872 J/kgK) = 573.21 KP0= P1(T0/T1)A

k/(k-1) EA= 1800 kPa (573.21/573.15)A

4.058E

A= 1800.765 kPa

From Eq.15.35 we get ( k = 1.327)

P = P0[1 + Ak - 1

2E

AM A2EA]Ak/(k1)E

A= 1800.765 kPa [1 + 0.1635]A4.058E

A= 974 kPa


54/127



15.44

A jet plane travels through the air with a speed of 1000 km/h at an altitude of 6

km, where the pressure is 40 kPa and the temperature is12C. Consider the inletdiffuser of the engine where air leaves with a velocity of 100 m/s. Determine thepressure and temperature leaving the diffuser, and the ratio of inlet to exit area of

the diffuser, assuming the flow to be reversible and adiabatic.

V= 1000 km/h = 277.8 m/s, v1= RT/P = 0.287 261.15/40 = 1.874 mA3EA/kg

h1= 261.48 kJ/kg,

ho1

= 261.48 + 277.8A2 EA/2000 = 300.07 kJ/kg

To1

= 299.7 K,

Po1

= P1(T

o1/T

1)Ak/(k-1) EA= 40 (299.7/261.15)A3.5EA= 64.766 kPa

h2= 300.07 - 100A2E

A/2000 = 295.07 kJ/kg T2= 294.7 K,P

2= P

o1(T

2/T

o1)Ak/(k-1) EA= 64.766 (294.7/299.7)A3.5EA= 61 kPa

v2= RT2/P2= 0.287kJ/kg-K 294.7 K/61 kPa = 1.386 mA

3EA/kg

A1/A

2= (v

1/v

2)(V

2/V

1) = (1.874/1.386)(100/277.8) =0.487


55/127


56/127


57/127



15.47

An air flow at 600 kPa, 600 K,M= 0.3 flows into a convergent-divergent nozzlewith M= 1 at the throat. Assume a reversible flow with an exit area twice thethroat area and find the exit pressure and temperature for subsonic exit flow toexist.

To find these properties we need the stagnation properties from the inlet state

From Table A.12: Mi= 0.3: Pi/Po= 0.93947, Ti/To= 0.98232

Po= 600 kPa/ 0.93947 = 638.66 kPa, T

o= 600 K/ 0.98232 = 610.8 K

This flow is case c in Figure 15.13. From Table A.12: AE/AA*EA= 2

PE/P

o= 0.9360, T

E/T

o= 0.98127

PE

= 0.9360 Po

= 0.936 638.66 kPa = 597.8 kPa

TE= 0.98127 T

o= 0.98127 610.8 K = 599.4 K


58/127



15.48

Air at 150 kPa, 290 K expands to the atmosphere at 100 kPa through a convergent

nozzle with exit area of 0.01 m2. Assume an ideal nozzle. What is the percenterror in mass flow rate if the flow is assumed incompressible?

Te= Ti(PePi)

A

k-1

k E

A

= 258.28 K

VAe2

AEE/2 = hihe= Cp(Ti- Te) = 1.004 (290 - 258.28) = 31.83 kJ/kg

Ve= 252.3 m/s; ve=RTePe

= A0.287 258.28

100EA= 0.7412 mA3EA/kg

Am.E

A= AVe/ ve= A0.01 252.3

0.7413E

A= 3.4 kg/s

Incompressible Flow: vi= RT/P = 0.287 290/150 = 0.55487 m3/kg

VAe2 AEE/2 = v P = vi(Pi- Pe) = 0.55487 (150 - 100) = 27.74 kJ/kg

=> Ve= 235 m/s => Am.E

A

= AVe/vi= 0.01 235 / 0.55487 = 4.23 kg/s

m.

incompressible

m.

compressible

= A4.23

3.4E

A= 1.25 about 25% overestimation.


59/127



15.49

Find the exit pressure and temperature for supersonic exit flow to exist in thenozzle flow of Problem 15.47.

We assume a reversible as the possibility which is case d in Figure 15.13.

To find these properties we need the stagnation properties from the inlet stateFrom Table A.12: Mi= 0.3: Pi/Po= 0.93947, Ti/To= 0.98232

Po= 600 kPa/ 0.93947 = 638.66 kPa, T

o= 600 K/ 0.98232 = 610.8 K

From Table A.12: AE/AA*EA= 2, P

E/P

o= 0.09352, T

E/T

o= 0.50813

PE= 0.09352 P

o= 0.09352 638.66 = 59.7 kPa

TE= 0.50813 T

o= 0.50813 610.8 = 310.4 K

This is significant lower P and T, but then we also have M = 2.2


60/127



15.50

Air is expanded in a nozzle from a stagnation state of 2 MPa, 600 K to a

backpressure of 1.9 MPa. If the exit cross-sectional area is 0.003 mA2EA, find themass flow rate.

This corresponds to case c and is a reversible flow.

PE/P

ox= 1.9/2.0 = 0.95 Table A.12: M

E= 0.268

TE= (T/T

o)ET

o= 0.9854 600 = 591.2 K

cE= kRT

E= A 1.4 1000 0.287 591.2EA= 487.4 m/s

VE= M

Ec

E= 0.268 487.4 = 130.6 m/s

vE = RT/P = 0.287 kJ/kg-K 591.2 K/1900 kPa = 0.0893 mA3 EA/kg

A

m

.EA

= AEVE/vE= 0.003 mA

2EA

130.6 (m/s)/0.0893 mA

3EA

/kg = 4.387 kg/s


61/127



15.51

A 1-m3insulated tank contains air at 1 MPa, 560 K. The tank is now dischargedthrough a small convergent nozzle to the atmosphere at 100 kPa. The nozzle has

an exit area of 2 105m2.

a. Find the initial mass flow rate out of the tank.

b. Find the mass flow rate when half the mass has been discharged.

a. The back pressure ratio:

PB/P

o1= 100/1000 = 0.1 < (PA*EA/P

o)crit= 0.5283

so the initial flow is choked with the maximum possible flow rate.

ME= 1 ; P

E= 0.5283 1000 = 528.3 kPa

TE= TA*EA= 0.8333 560 K = 466.7 K

VE= c = A kRT* EA= A 1.4 1000 0.287 466.7EA= 433 m/s

vE= RTA*E

A/PE= 0.287 kJ/kg-K 466.7 K/528.3 kPa = 0.2535 mA3E

A/kg

Am.E

A

1= AV

E/v

E= 2105mA2EA 433 (m/s)/0.2535 mA3EA/kg = 0.0342 kg/s

b. The initial mass is

m1= P

1V/RT

1= 1000 1/(0.287 560) = 6.222 kg

with a mass at state 2 as m2= m

1/2 = 3.111 kg.

Assume an adiabatic reversible expansion of the mass that remains in the tank.

P2= P

1(v

1/v

2)AkE

A= 100 0.5A1.4EA= 378.9 kPa

T2= T

1(v

1/v

2)Ak-1E

A= 560 0.5A0.4EA= 424 K

The pressure ratio is still less than critical and the flow thus choked.P

B/P

o2= 100/378.9 = 0.264 < (PA*EA/P

o)crit

ME= 1 ; P

E= 0.5283 378.9 = 200.2 kPa

TE= T

A

*EA

= 0.8333 424 = 353.7 K

VE= c = A kRT* EA= A 1.4 1000 0.287 353.7EA= 377 m/s

Am.

E

A

2= AV

EP

E/RT

E= A

210-5(377)(200.2)E0.287(353.7)E

A= 0.0149 kg/s

We could also have found the two mass flow rates from Eq. 15.42 withoutfinding the exit plane conditions.


62/127



15.52

A convergent-divergent nozzle has a throat diameter of 0.05 m and an exitdiameter of 0.1 m. The inlet stagnation state is 500 kPa, 500 K. Find the backpressure that will lead to the maximum possible flow rate and the mass flow ratefor three different gases as: air; hydrogen or carbon dioxide.

There is a maximum possible flow when M = 1 at the throat,

TA

*EA

=A

2

k+1EA

To; P

A

*EA

= Po(

A

2

k+1EA

)A

k

k-1E

A

; A

*EA

= o(

A

2

k+1EA

)A

1

k-1 EA

Am.E

A= A*EAAA*EAV = A*EAAA*EAc = PA*EAAA*EAA k/RT*EAA

A

*EA

= DA

2EA

/4 = 0.001963 mA

2EA

k TA*EA PA*EA c A*EA Am.E

Aa) 1.400 416.7 264.1 448.2 2.209 1.944b) 1.409 415.1 263.4 1704.5 0.154 0.515c) 1.289 436.9 273.9 348.9 3.318 2.273

AE/A

A

*EA

= (DE/D

A

*EA

)A

2EA

= 4. There are 2 possible solutions corresponding to points c and d

in Fig. 15.13 and Fig. 15.17. For these we haveSubsonic solution Supersonic solutionM

E P

E/P

o M

E P

E/P

o

a) 0.1466 0.985 2.940 0.0298b) 0.1464 0.985 2.956 0.0293c) 0.1483 0.986 2.757 0.0367

PB= PE0.985 500 = 492.5 kPa all cases point c

a) PB= P

E= 0.0298 500 = 14.9 kPa, point d

b) PB= P

E= 0.0293 500 = 14.65 kPa, point d

c) PB= P

E= 0.0367 500 = 18.35 kPa, point d


63/127



15.53

Air is expanded in a nozzle from a stagnation state of 2 MPa, 600 K, to a staticpressure of 200 kPa. The mass flow rate through the nozzle is 5 kg/s. Assume theflow is reversible and adiabatic and determine the throat and exit areas for thenozzle.

PA*EA= PoA

2

k+1E AA

k

k-1E

= 2 0.5283 = 1.056 MPa

TA*EA= To A

2

k+1E

A= 600 0.8333 = 500 K

vA*EA= RTA*EA/PA*EA= 0.287 500/1056= 0.1359 mA3 EA/kg

The critical speed of sound is

cA*EA= A kRT*EA= A 1.4 1000 0.287 500EA= 448.2 m/s

AA*EA= Am.E

AvA*EA/cA*EA= 5 0.1359/448.2 = 0.00152 mA2EA

P2/P

o= 200/2000 = 0.1 MA* AE

2E= 1.701 = V

2/cA*E

We used the column in Table A.12 with mach no. based on throat speed of sound.

V2= 1.701 448.2 = 762.4 m/s

T2= T

o(T

2/T

o) = 600 0.5176 = 310.56 K

v2= RT

2/P

2= 0.287 kJ/kg-K 310.56 K/200 kPa = 0.4456 mA3 EA/kg

A2= Am

.EAv

2/V

2= 5 kg/s 0.4456 (mA3 EA/kg) / 762.4 m/s = 0.00292 mA2 EA

Velocity

DensityArea

Mach #

2.0 MPa 0.2 MPaP


64/127


65/127



15.55

What is the exit pressure that will allow a reversible subsonic exit flow in theprevious problem?

This flow is case c in Fig.15.17 (and c in Fig. 15.13) the only reversible subsonic

flow with M = 1 at the throat.

AE/AA*EA= 2 see Table A.12 (M< 1)

PE/P

o= 0.9360, T

E/T

o= 0.98127

PE= 0.9360 P

o= 0.936 1200= 1123 kPa

( TE= 0.98127 To= 0.98127 600 = 588.8 K )


66/127



15.56

A flow of helium flows at 500 kPa, 500 K with 100 m/s into a convergent-divergent nozzle. Find the throat pressure and temperature for reversible flow andM= 1 at the throat.

We need to find the stagnation properties first ( k = 1.667 )

To= T1+ VA12

AEE/2Cp= 500 + 100A2E

A/(2 5193) = 500.963 K

Po= P1(T0/T1)Ak/(k-1) E

A

= 500 (500.963/500)A

2.5EA

= 502.41 kPa

From the analysis we get Eqs.15.37-38

PA*EA= Po A

2

k + 1

k/(k-1)E

A= 502.41 A

2

1.667 + 1

2.5E

A= 244.7 kPa

TA*EA= ToA

2

k + 1E

A= 500.963 A2

1.667 + 1E

A= 375.7 K


67/127



15.57

Assume the same tank and conditions as in Problem 15.51. After some flow outthe nozzle flow changes to become subsonic. Find the mass in the tank and themass flow rate out at that instant.

The initial mass ism

1= P

1V/RT

1= 1000 1/(0.287 560) = 6.222 kg

The flow changes to subsonic when the pressure ratio reaches critical.

PB/P

o3= 0.5283 P

o3= 189.3 kPa

v1/v

3= (P

o3/P

1)A1/k

E

A= (189.3/1000)A0.7143 EA= 0.3046

m3= m

1v

1/v

3= 6.222 0.3046 = 1.895 kg

At this point the tank temperatures is (assuming isentropic expansion)

T3= T

1(v

1/v

3)Ak-1EA= 560 0.3046A0.4EA= 348 K

PE

= PB

= 100 kPa ; ME

= 1

TE= 0.8333 348 = 290 K ; V

E= kRT

E= 341.4 m/s

Am.

E

A

3= AV

EP

E/RT

E= A

210-5(341.4)(100)E0.287(290)E

A= 0.0082 kg/s

We could have used Eq.15.42 with Po= Po3= 189.3 kPa, To= T3= 348 K

P

AIR

ecb

g


68/127



15.58

A given convergent nozzle operates so it is choked with stagnation inlet flowproperties of 400 kPa, 400 K. To increase the flow, a reversible adiabaticcompressor is added before the nozzle to increase the stagnation flow pressure to

500 kPa. What happens to the flow rate?

Since the nozzle is choked the mass flow rate is given by Eq.15.42. Thecompressor changes the stagnation pressure and temperature.

Isentropic: Po new= Porp and To new= ToA( )rpk-1

k AE

E

Po new/ To new= A( )rpk+1

2k AE

E

[Po/ To]so the mass flow rate is multiplied with the factor

A( )rpk+1

2k AE

E= A

500

400

2.4

2.8E

A= 1.21


69/127



15.59

A 1-m3uninsulated tank contains air at 1 MPa, 560 K. The tank is nowdischarged through a small convergent nozzle to the atmosphere at 100 kPa whileheat transfer from some source keeps the air temperature in the tank at 560 K. The

nozzle has an exit area of 2 105m2.

a. Find the initial mass flow rate out of the tank.

b. Find the mass flow rate when half the mass has been discharged.a. The back pressure ratio:

PB/P

o1= 100/1000 = 0.1 < (PA*EA/P

o)crit= 0.5283

so the initial flow is choked with the maximum possible flow rate given inEq.15.42

Am.E

A= AA*EAPo k/RTo[ Ak + 1

2EA]A

(k+1)/2(k1)E

= 2 1051 106A

1.4/(0.287 560 1000)EA

1.2A

2.4/0.8E

= 0.034 kg/s

b. The initial mass is

m1= P

1V/RT

1= 1000 1/(0.287 560) = 6.222 kg

with a mass at state 2 as m2= m

1/2 = 3.111 kg.

P2= P

1/2 = 500 kPa ; T

2= T

1; P

B/P

2= 100/500 = 0.2 < (PA*EA/P

o)crit

The flow is choked and the mass flow rate is similar to the previously statedone but with half the stagnation pressure (same stagnation temperature):

A

m

.EA

2= AA

*E

A

Po k/RTo[A

k + 1

2EA

]A

(k+1)/2(k1)EA

=A

m

.EA

1/ 2 = 0.017 kg/s

P

AIR

ecb


70/127



15.60

Assume the same tank and conditions as in Problem 15.59. After some flow outthe nozzle flow changes to become subsonic. Find the mass in the tank and themass flow rate out at that instant.

The initial mass ism1= P

1V/RT

1= 1000 1/(0.287 560) = 6.222 kg

Flow changes to subsonic when the pressure ratio reaches critical.P

B/P

o3= 0.5283 ; P

3= P

o3= P

B/0.5283 = 100/0.5283 = 189.3 kPa

m3= m

1P

3/P

1= 1.178 kg;

T3= T

1

TE= TA*EA= 0.8333 560 = 466.7 K, P

E= P

B= 100 kPa

VE= c = A kRT* EA= A 1.4 1000 0.287 466.7EA= 433 m/s

A

m

.EA

3= AVEPE/RTE=A

210-5(433)(100)E0.287(466.7)E

A

= 0.00646 kg/sFlow is critical so we could have used Eq.15.42 with Po= 189.3 kPa, To= 560 K


71/127



Normal Shocks


72/127



15.61

The products of combustion, use air, enter a convergent nozzle of a jet engine at a

total pressure of 125 kPa, and a total temperature of 650C. The atmosphericpressure is 45 kPa and the flow is adiabatic, with a rate of 25 kg/s. Determine theexit area of the nozzle.

The critical pressure Table 15.1: Pcrit= P2= 1250.5283 = 66kPa > Pamb

The flow is then choked with M = 1 at the exit and a pressure drop outside of thenozzle.

T2= 923.150.8333 = 769.3 K

V2= c2= A 1.4 1000 0.287 769.3EA= 556 m/s

v2= RT2/P2= 0.287 769.3/66 = 3.3453 mA3E

A/kg

A2=A

m

.EA

v2/ V2= 25 3.3453/556 = 0.1504 mA

2E

A


73/127



15.62

Redo the previous problem for a mixture with k = 1.3 and molecular mass of 31.

The critical pressure Table 15.1: Pcrit= P2= 1250.5457 = 68.2kPa > Pamb

The flow is then choked. T2= 923.150.8696 = 802.8 KThe gas constant is R = 8.31451 / 31 = 0.2682 kJ/kg-K

V2= c2= A 1.3 1000 0.2682 802.8EA= 529 m/s

v2= RT2/P2= 0.2682802.8/68.2 = 3.157 mA3E

A

/kg

A2= Am.E

A

v2/ V2= 25 3.157/529 = 0.149 mA2

E

A


74/127



15.63

At what Mach number will the normal shock occur in the nozzle of Problem15.52 flowing with air if the back pressure is halfway between the pressures atcand din Fig. 15.17?

First find the two pressures that will give exit at c and d. See solution to 15.52

a)

AE/AA*EA= (D

E/DA*EA)A

2 EA= 4

For case c: PE= 492.5 kPa For case d: P

E= 14.9 kPa

Actual case: PE= (492.5 + 14.9)/2 = 253.7 kPa

Assume Mx= 2.4 My= 0.5231 ; Poy/Pox= 0.54015

Ax/AA* AE

xE= 2.4031 ; A

x/AA*AE

yE= 1.298

Work backwards to get PE:

AE/AA* AE

yE= (A

E/AA* AE

xE) (A

x/AA* AE

yE) / (A

x/AA*AE

xE) = 4 1.298/2.4031 = 2.1605

ME= 0.2807 ; P

E/P

oy= 0.94675

PE=(PE/Poy) (Poy/Pox) Pox=0.94675 0.54015 500 = 255.7kPa

Repeat if Mx= 2.5 PE= 233.8 kPa

Interpolate to match the desired pressure => Mx= 2.41


75/127



15.64

Consider the nozzle of Problem 15.53 and determine what back pressure willcause a normal shock to stand in the exit plane of the nozzle. This is case g in Fig.15.17. What is the mass flow rate under these conditions?

We assume reversible flow up to the shockTable A.12: P

E/P

o= 200/2000 = 0.1 ; M

E= 2.1591 = M

x

Shock functions Table A.13: My= 0.5529 ; P

y/P

x= 5.275

PB= P

y= 5.275 P

x= 5.275 200 = 1055 kPa

Am.

E

A= 5 kg/s same as in Problem 15.53 since M = 1 at throat.


76/127


77/127



15.66

How much entropy per kg flow is generated in the shock in Example 15.9?

The change in entropy is

sgen= sy- sx= Cpln

Ty

Tx R ln

Py

Px

= 1.004 ln 1.32 0.287 ln 2.4583

= 0.27874 0.25815 =0.0206 kJ/kg K

Notice that this function could have been tabulated also as

sgen/R = Ak

k - 1EAln

TyTx

lnPyPx


78/127



15.67

Consider the diffuser of a supersonic aircraft flying atM=1.4 at such an altitudethat the temperature is 20C, and the atmospheric pressure is 50 kPa. Considertwo possible ways in which the diffuser might operate, and for each case calculatethe throat area required for a flow of 50 kg/s.

a. The diffuser operates as reversible adiabatic with subsonic exit velocity.

b. A normal shock stands at the entrance to the diffuser. Except for the normalshock the flow is reversible and adiabatic, and the exit velocity is subsonic.This is shown in Fig. P15.67.

a. Assume a convergent-divergent diffuser withM=1 at the throat. Rememberthat for the supersonic flow the convergent front will reduce the velocity.

Relate the inlet state to the stagnation state state table A.12 withM=1.4:

P/Po= 0.31424 => Po= 50 kPa / 0.31424 = 159.11 kPa

T/To= 0.71839 => To= 253.15 K / 0.71839 = 352.385 K

The maximum flow rate for chocked flow from Eq.15.42

Am.E

A/AA*EA= Po k/RTo[ Ak + 1

2E

A]A(k+1)/2(k1)E

= 159.11 103A 1.4/(0.287 253.15 1000)EA1.2A2.4/0.8

E

A

= 342.59 kg/smA

2E

AA*EA= Am.E

A/342.59 kg/smA2EA= 0.1459 mA2EA

b. Across the shock we have, Mx= 1.4 so from Table A.13

My= 0.7397 ; Poy/Pox= 0.95819 ;

The stagnation properties after the shock are:

Toy

= Tox

= 253.15 / 0.71839 = 352.385 K (as before)

Poy= Pox0.95819 = (50 kPa / 0.31424)0.95819 = 152.46 kPa

Only the stagnation pressure changed due to the shock so Eq.15.42 gives

A

m.E

A

/AA

*EA

= 0.95819 342.59 kg/smA

2EA

= 328.266 kg/smA

2E

AA*EA= Am.E

A/328.266 kg/smA2EA= 0.1523 mA2EA


79/127



15.68

A flow into a normal shock in air has a total pressure 400 kPa, stagnationtemperature of 600 K andMx= 1.2. Find the upstream temperatureTx, the

specific entropy generation in the shock and the downstream velocity.

From Table A.12: Mx= 1.2 has Tx/To= 0.7764, Px/Po= 0.41238

Tx= 0.7764 To= 0.7764 600 K = 465.84 K

From Table A.13: Ty/Tx= 1.128, Py/Px= 1.5133, My= 0.84217


sgen= sy- sx= CplnTyTx

R lnPyPx

= 1.004 ln 1.128 - 0.287 ln 1.5133

= 0.12093 0.11890 =0.00203 kJ/kg K

From the shock relations we had

Ty= 1.128 Tx= 1.128 465.84 K = 525.47 K

Vy= Mycy= My kRTy= 0.84217 A 1.4 0.287 525.47 1000EA

= 0.84217 459.49 = 387 m/s


80/127



15.69

Consider the nozzle in problem 15.42 flowing helium. What should thebackpressure be for a normal shock to stand at the exit plane? This is case g inFig.15.17. What is the exit velocity after the shock?

Reversible flow up to the shock with M = 1 at the throat.

Px o

= Po, T

x o= T

o, A

E/AA*EA= 175 / 100 = 1.75

As we do not have a Table 12 made for helium ( k = 5/3 = 1.667) we need to usethe formulas. The area to M relation is given in Eq.15.43 so

A /AA*EA= 1.75 = (1/Mx) [ A

2

k + 1(1 + (1/3)M

2

xE) A]A2E

A= [3 + M2

x]A

2EA/16 M

x

There are two solutions and we need the one with Mx> 1, see fig 15.10 we should

expect a solution around Mx= 2. With trial and error on

3 + M2x= 16 1.75 Mx which gives Mx= 2.2033

Now we can do the normal shock from Eq.15.53

Mx= 2.2033 M2

y= A

2.20332+ 3

E5 2.20332- 1EA= 0.3375,

The temperature ratio from Eq. 15.49 and the Txfrom Eq. 15.34:

Ty= T

xA1 + 0.5(k-1) M

2

Ex

1 + 0.5(k-1) M2

yE

A=T

x o

1 + 0.5(k-1) M2

y

= A375

1 + 0.3330.3375EA= 337.1

K

Vy= Mycy= My kRTy= A 0.3375EA A 1.667 2.0771 337.1 1000EA

= 0.58095 1080.4 = 627.6 m/s

A*

M Mx y

AE


81/127



15.70

Find the specific entropy generation in the shock of the previous Problem.

Reversible flow up to the shock with M = 1 at the throat.

Px o= Po, Tx o= To, AE/AA

*EA

= 175 / 100 = 1.75

Table A.12: ME= M

x= 2.042, P

x/P

o x= 0.12, T

x/ T

o x= 0.5454

Now we can do the normal shock from Table A.13

Mx= 2.042 M

y= 0.5704, P

y/P

x= 4.6984, T

y/T

x= 1.7219


sgen= sy- sx= CplnTyTx

R lnPyPx

= 1.004 ln 1.7219 - 0.287 ln 4.6984

= 0.5456 0.44405 = 0.1015 kJ/kg K


82/127



Nozzles, Diffusers, and Orifices


83/127



15.71

Steam at 600 kPa, 300C is fed to a set of convergent nozzles in a steam turbine.The total nozzle exit area is 0.005 m2and they have a discharge coefficient of0.94. The mass flow rate should be estimated from the measurement of thepressure drop across the nozzles, which is measured to be 200 kPa. Determine the

mass flow rate.

Inlet B.1.3 hi= 3061.6 kJ/kg, s

i= 7.3724 kJ/kg K

Exit: (Pe, s

e,s) P

e= P

i 200 = 400 kPa, s

e,s= s

i= 7.3724 kJ/kg K

he,s

= 2961 kJ/kg and ve,s

= 0.5932 mA3EA/kg,

Ve,s

= A 2 1000(3061.6 - 2961)EA= 448.55 m/s

Am.E

A

s

= AVe,s

/ve,s

= 0.005 448.55/0.5932 = 3.781 kg/s

Am. E

A

a= C

DAm.E

A

s= 0.94 3.781 = 3.554 kg/s


84/127



15.72

Air enters a diffuser with a velocity of 200 m/s, a static pressure of 70 kPa, and a

temperature of 6C. The velocity leaving the diffuser is 60 m/s and the staticpressure at the diffuser exit is 80 kPa. Determine the static temperature at thediffuser exit and the diffuser efficiency. Compare the stagnation pressures at the

inlet and the exit.

Stagnation T at the inlet

To1

= T1+ V

A

12

AEE

/2Cp

= 267.15 + 200A

2EA

/(2000 1.004) = 287.1 K

Energy Eq. gives the same stagnation T at exit

To2

= To1

T2= T

o2- V

A

22

AEE

/2Cp

= 287.1 - 60A

2EA

/(2000 1.004) = 285.3 K

To1

- T1

T1

= Ak-1

kEAP

o1- P

1

P1

Po1

- P1= 18.25 P

o1= 88.3 kPa

To2- T2T

2

= Ak - 1

kEAPo2- P2

P2

Po2

- P2= 1.77 P

o2= 81.8 kPa

Tex

s = T1(Po2/P1)Ak-1/kE

A= 267.15 1.0454 = 279.3 K

D

=T

ex

s - T1T

o1- T

1

= A279.3 - 267.15

287.1 - 267.15E

A= 0.608


85/127



15.73

Repeat Problem 15.44 assuming a diffuser efficiency of 80%.

V= 1000 km/h = 277.8 m/s, v1= RT/P = 0.287 261.15/40 = 1.874 mA3EA/kg

h1= 261.48 kJ/kg,

ho1

= 261.48 + 277.8A2 EA/2000 = 300.07 kJ/kg

To1

= 299.7 K,

Po1

= P1(T

o1/T

1)

A

k/(k-1) EA

= 40 (299.7/261.15)A

3.5EA

= 64.766 kPa

Same as problem 15.44, except

D

= 0.80. We thus have from 15.44

h3- h

1

ho1- h1=

h3- 261.48

300.07 - 261.48= 0.8

h3= 292.35 kJ/kg, T

3= 291.9 K

Po2

= P3= P

1(

3/

1)

A

k/(k-1) EA

= 40 (291.9/261.15)A

3.5EA

= 59.06 kPa

To2

= To1

= 299.7 K

h2= 300.07 - 100A2

E

A/2000 = 295.07 kJ/kg T2= 294.7 K,

P2= Po2(T2/To1)Ak/(k-1) E

A= 59.06 (294.7/299.7)A3.5EA= 55.68 kPa

v2= RT2/P2= 0.287 294.7/55.68 = 1.519 mA

3E

A

/kgA1/A2= (v1/v2)(V2/V1) = (1.874/1.519) (100/277.8) =0.444

h

01

02

3

1

2

s


86/127



15.74

A sharp-edged orifice is used to measure the flow of air in a pipe. The pipediameter is 100 mm and the diameter of the orifice is 25 mm. Upstream of the

orifice, the absolute pressure is 150 kPa and the temperature is 35C. The pressuredrop across the orifice is 15 kPa, and the coefficient of discharge is 0.62.

Determine the mass flow rate in the pipeline.

T = Ti

A

k-1

k E A

PP

i

= 308.15 A

0.4

1.4EA

A

15

150EA

= 8.8 K

vi= RT

i/P

i= 0.5896 m

A

3 EA

/kg

Pe= 135 kPa, T

e= 299.35 K, v

e= 0.6364 mA3EA/kg

Am.E

A

i= Am

.EA

e V

i/ V

e = (D

e/D

i)A2 E

Avi/v

e= 0.0579

hi- h

e= VA

e2

AEE(1 - 0.0579A2 EA)/2 = Cp

(Ti- T

e)

Ve s

= A 2 1000 1.004 8.8/(1 - 0.0579)2EA= 133.1 m/s

Am.E

A= CDAV/v = 0.62 (/4) (0.025)A2 E

A133.1 / 0.6364 = 0.06365 kg/s


87/127



15.75

A critical nozzle is used for the accurate measurement of the flow rate of air.

Exhaust from a car engine is diluted with air so its temperature is 50C at a totalpressure of 100 kPa. It flows through the nozzle with throat area of 700 mm2bysuction from a blower. Find the needed suction pressure that will lead to critical

flow in the nozzle and the mass flow rate.

PA*EA= 0.5283 Po= 52.83 kPa, TA*EA= 0.8333 T

o= 269.3 K

vA*EA= RTA*EA/PA*EA= 0.287 269.3/52.83 = 1.463 mA3EA/kg

cA*EA= A kRT*EA= A 1.4 1000 0.287 269.3EA= 328.9 m/s

Am.E

A= AcA*EA/vA*EA= 700 10A-6EA 328.9/1.463 = 0.157 kg/s

8/10/2019 Chapter

Chapter 15 Soln

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