ch06

6-1 Steam at 160 psia and 400oF enters a nozzle with a volumetric flow rate of 6615 cfm (cubic feet per

minute). The inlet area is 14.5 in2. If the steam leaves at 1500 ft/s at a pressure of 40 psia, find the exit

temperature.

Approach:Use the first law for an open system,

specialized for a nozzle. Find properties in

the steam tables.

Assumptions:1. Potential energy change is negligible.

2. The system operates in steady-state.

3. The nozzle is adiabatic.

1

o

1

160psia

400 F

PT

�

�

1 6615cfmV ��

2 40 psia

ft

P �

2 1500s

�V

Solution: From the first law, specialized for a nozzle,

2 2

1 21 2

2 2h h� � �

V V

The inlet velocity may be found from

� �

� �

3

11 2

2

2

ft 1min6615

min 60s ft1095

s1ft14.5in.

144in.

VA

� ��

� ��

�V

Taking the inlet enthalpy from Table B-12,

2 2 2

2 21 22 1 2

2

Btu 1 ft 1 Btu 1 lbf Btu1218 1095 1500 1197

lbm ft2 2 lbm 2 s 778ft lbf lbm32.2

s

h h

� ��

� � � � � � � ��

V V

2 2

BtuAt 40 psia, 1197 , from Table B-12,

lbmP h� �

o

2 320 FT � Answer

6 - 1

6-2 Oxygen at 220oF enters a well-insulated nozzle of inlet diameter 0.6 ft. The inlet velocity is 60 ft/s. The

oxygen leaves at 75oF, 10 psia. The exit area is 0.01767 ft2. Calculate the pressure at the inlet.


specialized for a nozzle. Apply conservation

of mass and the ideal gas law to determine

inlet pressure.




4. Specific heat is constant.

5. Oxygen behaves like an ideal gas under

these conditions.

o

1 220 F

ft

T �

160

s�V

o

2 75 FT �

210 psiaP �


2 2

1 21 2

2 2h h� � �

V V

Solving for and noting that, for an ideal gas with constant specific heat, 2V ph c T� � �

� � � �2 2

2 1 2 1 1 22 2 ph h c T T� � � � � �V V

1V

Using data for specific heat from Table B-8

� �2

2 2

Btu 778 lbf ft 32.17 lbm ft ft ft2 0.2215 220 75 R 60 1269

lbm R 1Btu 1lbf s s s

� �� V �

From conservation of mass,

1 2

1 1 1 2 2 2

m mA A� �

��

� �V V

Using the ideal gas law,

1 21 1 2 2

1 2

MP MPA ART RT

�V V

Solving for inlet pressure,

� ��

2 2 11 2 2

1 1 2

1269 0.01767 220 46010psia 16.8psia

60 75 4600.3

A TP PA T �

� � ��

VV

Answer

6 - 2

6-3 A well-insulated nozzle has an entrance area of 0.28 m2 and an exit area of 0.157 m2. Air enters at a velocity

of 65 m/s and leaves at 274 m/s. The exit pressure is 101 kPa and the exit temperature is 12�C. What is the

entrance pressure?


specialized for a nozzle. Apply conservation

of mass and the ideal gas law to determine

inlet pressure.





5. Air behaves like an ideal gas under these

conditions.

1

m65�V

s

2

o

2

101 kPa

12 C

PT

�

�

2

m274

s�V


2 2

1 21 2

2 2h h� � �

V V

For an ideal gas with constant specific heat, ,ph c T� � � therefore

� �2 2

2 11 2

2 2pc T T� � �

V V

Solving for T1, 2 2

2 11 2

2 p

T Tc�

� �V V

Specific heat depends on temperature, and should be evaluated at the average of inlet and outlet temperatures, but

the inlet temperature is unknown. As an approximation, evaluate the specific heat at the exit temperature, and

correct later if necessary. From Table A-8,

2

kJAt 273 12 285K, 1.004

kg KpT c� � � �

�

� � � �

� �

22 2

2

1

m274 65

s 285KkJ 1000J

2 1.004kg K 1kJ

T� ��

� ��

320 K�

From conservation of mass,

1 2

1 1 1 2 2 2

m mA A� �

��

� �V V

Using the ideal gas law,

1 21 1 2 2

1 2

MP MPA ART RT

�V V

Solving for inlet pressure,

2 2 11 2

1 1 2

A TP PA T

�VV

� �1

274 0.157 320101

65 0.28 285P � ��

�

268 kPa� � � � � �

� Answer

Comment:The entrance temperature of 320 K is very close to the exit temperature of 285 K. Specific heat does not vary

significantly between these two temperature and there is no need to iterate on specific heat.

6 - 3

6-4 Carbon monoxide enters a nozzle at 520 kPa, 100oC, with a velocity of 10 m/s. The gas exits at 120 kPa

and 500 m/s. Assuming no heat transfer and ideal gas behavior, find the exit temperature.


specialized for a nozzle.





5. Carbon monoxide behaves like an ideal

gas under these conditions.

1

o

1

520kPa

100 C

PT

�

�

1

m10

s�V

2 120 kPa

m

P �

2 500s

�V


2 2

1 21 2

2 2h h� � �

V V


� �2 2

2 11 2

2 2pc T T� � �

V V

Specific heat depends on temperature, and should be evaluated at the average of inlet and outlet temperatures, but

the outlet temperature is unknown. As an approximation, evaluate the specific heat at the inlet temperature, and

correct later if necessary. From Table A-8, at 1=100 273 373K,T � �kJ

1.045 .kg K

pc ��

2 2

1 22 1

2 p

T Tc�

� �V V

2 2

o

m m10 500

s s100 C

kJ 1000 J2 1.045

kg K 1 kJ

� � � ��

o

2 19.6 CT �� Answer

6 - 4

6-5 Low-velocity steam with negligible kinetic energy enters a nozzle at 320oC, 3 MPa. The steam leaves the

nozzle at 2 MPa with a velocity of 410 m/s. The mass flow rate is 0.37 kg/s. Determine

a. the exit state

b. the exit area


specialized for a nozzle. Find properties in

the steam tables.




o

1

1

320 C

3MPa

TP��

1 0�V

2 2 MPa

m

P �

2 410s

�V

Solution: a) From the first law, specialized for a nozzle,

2 2

1 21 2

2 2h h� � �

V V

Determining the inlet enthalpy by interpolation in Table A-12, 22 2

61 22 1

kJ 1000J 410m/s J kJ3042 2.96 10 2960

2 2 kg 1kJ 2 kg kgh h

� ��

V V ��

Interpolate in Table A-12 at 2 MPa and h = 2960 kJ/kg to find the final temperature of o

2 274 CT �

The exit state is superheated vapor at 2 MPa and 274 oC.

b) The mass flow rate is given by

2 2 2m A�� V

To determine the exit density, find the specific volume at the exit state by interpolation in Table A-12, 3

2 0.118m /kgv �

2 3 3

2

1 18.47

0.118m /kg mv� � � �

kg

Solving the mass flow rate equation for exit area,

4 2

2

2 23

kg0.37

s 1.07 10 mkg m

8.47 410m s

mA�

��

�V

Answer

6 - 5

6-6 Steam enters a diffuser at 250oC and 50 kPa and exits at 300oC and 150 kPa. The diameter at the entrance

is 0.25 m and the diameter at the exit is 0.5m. If the mass flow rate is 9.4 kg/s, find the heat transfer to the

surroundings.


eliminating work and potential energy.

Find properties in the steam tables.



o

1 250 CT �

150kPaP �

o

2 300 CT �

2150 kPaP �

Solution: a) From the first law,

2 2

2 2

cv i ecv cv i i i e e e

dE Q W m h gz m h gzdt

� � ��

� � �� V V �

�

Assuming steady conditions, one stream in and one stream out, no work, and no change in potential energy, the

first law becomes 2 2

1 21 20

2 2cvQ m h m h

� � ��

� � � �V V �

�

The area at the inlet is 2 2

2 211 1

0.250.0491m

2 2

DA r� � ��

The velocity at the inlet is, using data from Table A-12, 3

11 2

1

kg m9.4 4.82

s kg646m s

0.0491m

mvA

� ��

� ��

V

The exit area is 2

2

2

0.50.196m

2A � � ��

�

Obtaining specific volume by interpolating in Table A-12,

� � � �2

2

2

9.4 1.98 m95.0

0.196 s

mvA

� � ��

V

Rearranging the first law, 2 2

2 12 1

2 2cvQ m h h

� ��

�� V V

� � � �2 2

2 2

2 2

m m95 646

kg kJ 1000J kJ 1000Js s9.4 3073 2976s kg 1kJ 2 kg 1kJ 2

cvQ

� � ��

� � � ��

�

107,146 W 107.1kWcvQ � � � �� Answer

6 - 6

6-7 Air enters a diffuser at 50 kPa, 85oC with a velocity of 250 m/s. The exit pressure is atmospheric at 101

kPa. The exit temperature is 110oC. If the diameter at the inlet is 8 cm,

a. Find the exit velocity.

b. Find the diameter at the exit.

Assume constant specific heats.


eliminating work, heat, and potential

energy. Use ideal gas relations to

evaluate properties.



3. The diffuser is adiabatic.

4. Air behaves like an ideal gas under

these conditions.

5. Specific heats are constant.

o

1

1

85 C

50kPa

TP��

1 250m/s�V

o

2 110 CT �

2101 kPaP �

Solution: a) From the first law, specialized for a diffuser

2 2

1 21 2

2 2h h� � �

V V


� �2 2

2 11 2

2 2pc T T� � �

V V

Evaluating specific heat at the average temperature of using data in Table A-8, o197 C 370KaveT � �

� � � � � �2

22

2 1 2 1 2

kJ 1000J m2 2 1.01 85 110 250

kg K 1kJ spc T T

� ��

V V

2

m110

s�V Answer

b) By conservation of mass

1 2m m��

1 1 1 2 2 2A A� ��V V

Using the ideal gas law, 2 2

1 1 21 2

1 22 2

PM D P M DRT RT

� ��

V V 2

2 21 21 1 2 2

1 2

P PD DT T

�V V

Solving for exit diameter,

1 2 12 1

2 1 2

50 110 273 2508 8.78cm�

101 85 273 110

P TD DP T

� ��

VV

Answer

6 - 7

6-8 Superheated steam enters a well-insulated diffuser at 14.7 psia, 320oF and 400 ft/s. The steam exits as

saturated vapor at a very low speed. Find the exit pressure and temperature.


eliminating work, heat, and potential energy.

Find properties in the steam tables.



3. The diffuser is adiabatic.

o

1

1

320 F

14.7 psia

TP��

1 400ft/s�V2 0�V

Solution: a) From the first law, specialized for a diffuser

2 2

1 21 2

2 2h h� � �

V V

Using values of enthalpy at the inlet from Table B-12 at 14.7 psia, 320oF,

� �2

2

2 2

2

ft Btu400 1Btu s lbm1202.1

ftlbm 225,037

s

h

� � � ��

� � �� !

Btu1205.3

lbm�

In Table B-10, for there are two possible solutions – one at 4401205.3Btu/lbm,gh � oF, and 381.2 psia and the

other at about 470oF, and 514.1 psia. Either solution could occur, depending on the exit pressure imposed. Note

also that the exit enthalpy is rather insensitive to pressure between these two values. Changing the exit pressure

has little effect on the exit state. Answer

6-9 Steam enters an adiabatic turbine at 0.8 MPa and 500oC. It exits at 0.05 MPa and 150oC. If the turbine

develops 24.5 MW of power, what is the mass flow rate?

Approach:Use the first law for an open system, eliminating heat, kinetic

energy and potential energy. Find properties in the steam tables.


2. Kinetic energy change is negligible.


4. The turbine is adiabatic.

Solution: Assuming an adiabatic turbine with negligible kinetic and

potential energy changes, the first law becomes

� �1 2W m h h� ��

o

1

1

500 C

0.8MPa

TP��

o

2

2

150 C

0.05MPa

TP

��

With values of enthalpy from Table A-12

� �3

1 2

24.5 10 kW34.98kg s

3480.6 2780.1kJ kg

Wmh h

��

� �

�� Answer

6 - 8

6-10 Air enters an adiabatic turbine at 900 K and 1000 kPa. The air exits at 400 K and 100 kPa with a velocity

of 30 m/s. Kinetic and potential energy changes are negligible. If the power delivered by the turbine is

1000 kW,

a. find the mass flow rate.

b. find the diameter of the duct at the exit.


energy and potential energy. Find properties using ideal gas

relations.





5. Air behaves like an ideal gas under these conditions.


Solution: a) Assuming an adiabatic turbine with negligible kinetic and


� �1 2W m h h� ��

1

1

900K

1000 kPa

TP��

2

2

2

400 K

100kPa

30m/s

TP

��V


� �1 2 1 2p

W Wmh h c T T

� ��

� ��

With values of specific heat from Table A-8 at the average temperature of 650 K,

� �1000kW kg

1.88kJ s

1.063 900 400 Kkg K

m � ��

�

� Answer

b) Exit area is related to velocity and mass flow rate through

22

2

mvA ��V

From the ideal gas law,

� �

� �

3

22

2

kJ8.314 400K

mkmol K1.15

kg kg28.97 100 kPa

kmol

RTvMP

� ��

Substituting values 3

2

2

kg m1.88 1.15

s kg0.0721m

m30

s

A

� ��

� ��

� �� 2

2

4 0.072140.303m

AD� �

� � � Answer

6 - 9

6-11 Saturated steam at 320oC enters a well-insulated turbine. The mass flow rate is 2 kg/s and the exit pressure

is 50 kPa. Determine the final state if the power produced is:

a. 100 kW

b. 400 kW









� �1 2W m h h� ��

o

1320 C

saturated steam

T �

2 50kPaP �

From Table A-10, the enthalpy of saturated steam at the initial temperature of 320 oC

1 2700kJ kgh �

2 1

Wh hm

� ��

� � 1000 W100 kW

kJ 1000 J 1 kW2700

kgkg 1 kJ2

s

� ��

� �6 J k

2.65 10 2650kg kg

� � �J

From Table A-11 at 50 kPa, we see that the exit enthalpy is greater than that of a saturated vapor; therefore, the

exit state is in the superheated region and, by interpolation in Table A-12, o

2 83.4 CT � Answer 2 50kPaP �

b) Recalculating the exit enthalpy for the higher power condition,

� �2

1000 W400 kW

kJ 1000 J 1 kW2700

kg 1 kJ 2 kg sh

� ��

� �6 J k

2.5 10 2500kg kg

� � �J

f

From Table A-11 at 50 kPa, we see that the exit enthalpy is between that of a saturated liquid and a saturated

vapor; therefore, the exit state is in the two-phase region and

� �2 f gh h x h h� � �

2 2500 340.50.937

2646 340.5

f

g f

h hx

h h� �

� � ��

Summarizing, the final state is two-phase with

2 50 kPaP �

0.937x� Answer

6 - 10

6-12 Superheated steam at 1.6 MPa, 600oC enters a well-insulated turbine. The exit pressure is 50 kPa. The

turbine produces 10 MW of power. If the exit pipe is 1.6 m in diameter and carries 11 kg/s of flow, find

the velocity at the exit. Neglect kinetic energy.







Solution: Assuming an adiabatic turbine with negligible kinetic and


� �1 2W m h h� ��

o

1

1

600 C

1.6 MPa

TP��

2 50kPaP �

With values of enthalpy from Table A-12 at 1.6 MPa, 600oC,

2 1

Wh hm

� ��

� � 1000kW10MW

1MWkJ kJ3693 2784

kg 11 kg/s kg

� ��

From Table A-12, at and . To find the exit velocity, 2 50kPaP � 2 2784 kJ/kg,h � o

2 150 CT �

2 2

2

Amv

�� V

Solving for velocity and using values of v2 from Table A-12 at ando

2 150 CT � 2 50 kPa,P �

3

22 2

2 2

kg m11 3.89

s kg m21.3

s1.6� m

2

mvA

� ��

� ��

�V Answer

6 - 11

6-13 Air at 550oC and 900 kPa is expanded through an adiabatic gas turbine to final conditions of 100 kPa and

300oC. The total power output desired is 1 MW. If the inlet velocity is 30 m/s, what should the inlet pipe

diameter be? Neglect kinetic and potential energy.



relations.









� �1 2W m h h� ��

o

1

1

1

550 C

900kPa

30m/s

TP��V

o

2

2

300 C

100kPa

TP

��


� �1 2 1 2p

W Wmh h c T T

� ��

� ��

With values of specific heat from Table A-8 at the average temperature of 700 K,

� � o

1000kW kg3.72

kJ s1.075 500 300 C

kg K

m � ��

�

�

b) Exit area is related to velocity and mass flow rate through

11

1

mvA ��V

From the ideal gas law,

� �

� �

3

11

1

kJ8.314 550 273 K

mkmol K0.262

kg kg28.97 900kPa

kmol

RTvMP

� � ��

Substituting values 3

2

1

kg m3.72 0.262

s kg0.0325m

m30

s

A

� ��

� ��

� �� 1

1

4 0.032540.203m

AD� �

� � � Answer

6 - 12

6-14 Air at 510�C and 450 kPa enters an ideal, adiabatic turbine. The exit pressure is 101 kPa. In steady state,

the turbine produces 50 kW of power. Find

a. the exit temperature. (Hint: use Eq. 2-56)

b. the mass flow rate.

Approach:Use Eq. 2-56 to find the final temperature. Then apply the first

law for an open system, eliminating heat, kinetic energy and

potential energy to obtain mass flow rate.







7. The process is quasi-static.

Solution: a) Since the turbine is ideal, the process is quasi-static. For an

adiabatic, quasi-static, process of an ideal gas with constant

specific heat (from Eq. 2-56),

o

1

1

510 C

450kPa

TP��

2 101kPaP �

1

2 2

1 1

kkT P

T P

�

� ��

From Table A-8, for air 1.4k �

� �1 1.4 1

1.42

2 1

1

101510 273

450

kkPT T

P

� �

� � � ��

o511 K 238 C� �

b) Assuming an adiabatic turbine with negligible kinetic and potential energy changes, the first law becomes

� �1 2W m h h� �� For an ideal gas with constant specific heat, ,ph c T� � � therefore

� �1 2 1 2p

W Wmh h c T T

� ��

� ��

With values of specific heat from Table A-8,

� � � �1 2

50 kW

kJ1.06 510 238 K

kg K

p

Wmc T T

� ��

��

�� kg

0.173s

� Answer

6 - 13

6-15 Saturated steam at 3 MPa enters a well-insulated turbine operating in steady state. The turbine produces

600 kW of power. The mass flow rate through the turbine is 84 kg/min and the exit quality is 0.93. Find

the exit temperature.


energy and potential energy. To find the final temperature, it is

necessary to iterate with data from the steam tables.







� �1 2W m h h� ��

1 3MPa

saturated steam

P �

2 0.93x �

From Table A-11, the enthalpy of saturated steam at the initial pressure of 3 MPa is

1 2804 kJ kggh h� �

2 1

Wh hm

��

600 kW kJ2804

kg 1min kg84

min 60 s

��

kJ2375

kg�

The exit state is in the two-phase region, but neither temperature nor pressure is known. The exit temperature is

found by trial and error. Begin by assuming a value for , then use2T 2x to compute with data from Table A-10. If

, the iteration is complete, if not, select a new value of and recompute To begin, assume

From Table A-10,

2h

2 2375kJ/kgh � 2T 2.h

2 25 C.T � �

� �2 2f gh h x h h� � � f

104.9 0.93 2547 2442� � � � �� 2376 kJ kg�

This is very close to the calculated value of 2375, therefore

2 25 CT � � Answer

Comment:In this example, we selected the correct result immediately. In reality, it would be necessary to try several

temperatures before zeroing in on the correct value.

6 - 14

6-16 In a 3-hp compressor, carbon dioxide flowing at 0.023 lbm/s is compressed to 120 psia. The gas enters at

60oF and 14.7 psia. The inlet and outlet pipes have the same diameter. Find the final temperature and the

volumetric flow rate at the exit (in ft3/min). Assume constant specific heat at 100oF.



relations.




4. The compressor is adiabatic.

5. Carbon dioxide behaves like an ideal gas under these

conditions.


Solution: Assuming an adiabatic compressor with negligible kinetic and


� �1 2W m h h� ��

o

1

1

60 F

14.7 psia

TP��

2 120psiaP �


� �1 2

2 1

p

p

W mc T T

WT Tmc

� �

� �

� ��

With data from Table B-8

� �o o

2

Btu2544

h3hp1hp

60 F 533 Flbm Btu 3600s

0.023 0.195s lbm R 1h

T

� ��

� � � ��

� ��

Answer

From conservation of mass

2 2 21 2

2 2

V V P Mm mv RT

� � ��

� �

� �

� �

3

3

1 22

2

lbm psia ft 60s0.023 10.73 533 + 460 R

s lbmol R 1min ft2.78

lbm min120 psia 44.01

lbmol

m RTVP M

� ��

� ��

�� Answer

Comments:If more accuracy is desired, the calculation should be repeated with specific heat evaluated at the average of the

inlet and outlet temperatures, i.e. at (60+533)/2 or 296.5 F. The outlet temperature was not known at the beginning

of the calculation, so specific heat at a temperature near the inlet temperature was used.

6 - 15

6-17 A well-insulated compressor is used to raise saturated R-134a vapor at a pressure of 360 kPa to a final

pressure of 900 kPa. The compressor operates in steady state with a power input of 850 W. If the flow rate

is 0.038 kg/s, what is the final temperature?


energy and potential energy.




4. The compressor is adiabatic.

Solution: Assuming an adiabatic compressor with negligible kinetic and


� �2 1W m h h� ��

1 360kPa

saturated vapor

P �

2 900kPaP �

2 1

Wh hm

� ��

From Table A-15 at 360 kPa, 1 250.6 kJ/kggh h� �

2

850 W kJ250.6

kg kg0.039

s

h � �kJ

272kg

�

From Table A-16, at P2 = 900 kPa and h2 = 271.6, T2 � 40 �C Answer

6 - 16

6-18 Air flowing at 0.5 m3/min enters a compressor at 101 kPa and 25oC. The air exits at 600 kPa and 300oC.

During this process, 250 W of heat are lost to the environment. What is the required power input?

Approach:Use the first law for an open system, eliminating kinetic energy and

potential energy. Find properties using ideal gas relations.






Solution: From the first law for an open system

2 2

2 2



� � � ��

� �� V V

Assuming steady conditions, one stream in and one stream out, no

change in kinetic or potential energy, the first law becomes

� �1 20 Q W m h h� � � ��

o

1

1

3

25 C

101kPa

0.5m /min

TP

V

��

��

2 120psiaP �


� �1 2pW Q mc T T� � ��

Using data in Table A-1,

� �

� �

3

31 1 1

1 1

m kg 1min0.5 101kPa 28.97

min kmol 60s kg9.84 10

kJ s8.314 25 273 K

kmol K

V V PMmv RT

�

� � � ��

� � ��

� ��

The average temperature of the air is

o25 300162 C 435K

2aveT �

� � �

Using specific heat values from Table A-8 interpolated at Tave

� �3 okg J250 W 9.84 10 1018 25 300 C 3000 W 3kW

s kg KW � � �� Answer

6 - 17

6-19 Refrigerant-134a enters a compressor at 0oF and 10 psia with a volumetric flow rate of 15 ft3/min. The

refrigerant exits at 70 psia and 140oF. If the power input is 2 hp, find the rate of heat transfer in Btu/h.

Approach:Use the first law for an open system, eliminating kinetic and

potential energy.




Solution: From the first law, neglecting kinetic and potential energy

� �1 20 Q W m h h� � � ��

The mass flow rate is

1

Vmv

��

�

o

1

1

3

1

0 F

10psia

15ft /min

TP

V

��

��

o

2

2

140 F

70psia

TP

��

Taking the inlet specific volume, from TableB-16, 1,v

3

3

ft15

lbmmin 3.19ft min

4.703lbm

m � ��

2 1( )Q W m h h� � �� Again using Table B-16,

Btu3.412

746 W lbm Btu 60minh( 2hp) 3.19 129.1 102.91 hp 1W min lbm 1 h

Q

� ��

� � ��

� ��

Btu

76.0h

Q � �� Answer

6 - 18

6-20 A pump is used to raise the pressure of a stream of water from 10 kPa to 0.7 MPa. The temperature of the

water is the same at inlet and outlet and equal to 20oC. The velocity also does not change across the pump.

If the mass flow rate is 14 kg/s, what power is needed to drive the pump? Assume frictionless flow and no

significant elevation change.

Approach:Use the equation for pump work in the form 1 2( )W mv P P� �� .




4. The pump is ideal.

5. Water is incompressible.

Solution: For a frictionless pump with no elevation changes:

1 2( )W mv P P� �� so, using the density of water at 20oC from Table A-6,

1 2P PW m��

��

� �

3

10kPa 700 kPakg 1000Pa14 9677 W 9.68 kW

kgs 1kPa998.2

m

� � ��

Answer

6-21 A 2-hp pump is used to raise the pressure of saturated liquid water at 5 psia to a higher value. Assume the

velocity is constant, the water is incompressible, and the flow is frictionless. If the mass flow rate is 6

lbm/sec, find the final pressure.








1 2( )W mv P P� ��

2 1

WP Pmv

� ��

From Table B-11, vf = 0.0164 ft3/lbm at P = 5 psia. Substituting values

� �2

2

2 2 3

ft lbf550

1fts2hp1 hp 144in.

lbf5

in. lbm ft6 0.0164

s lbm

P

��

� � � ��

��

� �

2 82.6 psiaP � Answer

6 - 19

6-22 Water is pumped at 12 m/s through a pipe of diameter 1.2 cm. The inlet pressure is 30 kPa. If the pump

delivers 6 kW, find the final pressure. Assume frictionless, incompressible flow with no elevation or

velocity changes.








1 21 2

( )( )

m P PW mv P P��

� � ��

The mass flow rate is

m A�� V2

3

kg m 0.012m kg998 12 1.35

m s 2��

� � � s

2 1

WP Pm�

� ��

� �3

1000 W kg6kW 998

1000Pa 1 kW m30kPa

kg1 kPa1.35

s

� �� 64.45 10 Pa 4.45MPa� � �

��

2 4.45MPaP � Answer

6-23 A 1-hp pump delivers oil at a rate of 10 lbm/s through a pipe 0.75 in. in diameter. There is no elevation

change between inlet and exit, no velocity change, and no oil temperature change. The oil density is 56

lbm/ft3. Find the pressure rise across the pump.






5. The oil is incompressible.


1 2( )W mv P P� ��

1 2

W WP Pmv m

��

� ��

� �2

3 2

1 2

ft lbf550

lbm 1fts56 1hpft 1 hp 144in.

lbm10

s

P P

��

� �� 21.4psia� Answer

6 - 20

6-24 An architect needs to pump 2.3 lbm/s of water to the top of the Empire State building, which is about 1000

ft high. Assume water at 45 psia is available at the base of the building. What is the power of the pump

needed, in hp, if the flow is assumed frictionless? The velocity of the water is constant.

Approach:Apply the first law for an open system, dropping terms for

transients, heat, and kinetic energy. Replace the enthalpy

difference with � �2 1 2 1 .h h v P P� � �

Assumptions:1. The system is adiabatic and isothermal.





Solution:

The first law for an open system is 2 2

2 2



� � ��

� � �� V V �

�

Apply this between station 1 and station 2 on the figure above, neglecting kinetic energy and assuming steady,

adiabatic operation, to get

� � � �1 2 1 20 W m h h mg z z� � � � � ��

For an incompressible liquid undergoing an isothermal process

� �2 1 2 1h h v P P� � �

Substituting this into the first law and noting that 1v ��

� �1 21 2

P PW m mg z z�

� ��

��

Using values for the density of water from Table B-6 and being very careful with units,

� �2

2 2

3

lbf 144 in.45 14.7

in. 1 ftlbm 1 hp2.3

lbm lbf fts62.1 550

ft s

W

� ��

� � ��

� �2

2

lbm ft 1 lbf 1 hp2.3 32.17 1000ft

lbm ft lbf fts s32.17 550

s s

� � ��

��

� �

3.89 hpW � �� Answer

Comment:The actual pump chosen should have more horsepower than this because, in reality, there are some frictional

effects.

6 - 21

6-25 Air at 150oC, 40 kPa is throttled to 100 kPa. The inlet velocity is 3.6 m/s. Find the exit velocity.

Approach:Use the first law specialized for a throttle.




4. Air may be considered an ideal gas under these conditions.

5. The throttle is adiabatic.

6. The inlet and exit pipe have the same diameter.


1

o

1

1

40kPa

150 C

3.6m/s

PT

�

��V

2 100kPaP �

Solution: For a throttle with no heat transfer, no change in kinetic or potential energy, and no work, the first law reduces to

2 10 h h� �


� �2 1 10 pc T T T T� � " � 2


1 2

1 1 2 2 2

m mA A� ��

�

� �V V

Assuming the diameter of the inlet and exit pipes are the same and using the ideal gas law,

1 21 2

1 2

RT RTPM P M

�V V

Solving for exit velocity,

22 1

1

m 400 m3.6 14.4

s 100 s

PP

� � � ��

V V Answer

6 - 22

6-26 Saturated liquid R-134a at 24oC is throttled until the final quality is 0.116. Find the final temperature and

pressure.






o

1 24 CT �saturated liquid

2 0.116x �

Solution: From Table A-14 at 24oC

1 =82.9 kJ/kgfh h�From the first law applied to a throttle,

1 2h h�This problem must be solved iteratively, since neither the final temperature nor the final pressure is known. First

assume a final temperature and determine the corresponding quality, x2. For example, assume T2 = 0oC. Then,

from Table A-14

2

2

82.9 50.00.167

247 50.0

f

g f

h hx

h h� �

� � ��

This value of x2 is too high, so a different value of T2 is chosen. By trial and error, the final value of T2 is found to

be o

2 8 CT �

To verify this result, calculate the final quality as

2

2

82.9 60.70.116

252 60.7

f

g f

h hx

h h� �

� � ��

The final pressure is the saturation pressure at 8oC, which is given in Table A-14 as

2 0.388MPaP � Answer

6 - 23

6-27 Saturated liquid R-134a at 80oF undergoes a throttling process. The pressure decreases to ¼ of its original

value. Find the exit quality.






o

1 80 FT �saturated liquid

2 1 / 4P P�

Solution: From the first law applied to a throttle,

1 2h h�From Table B-14,

If then o80 FsatT � 101.4psiasatP �

1 1

BtuAt 101.4, 37.3

lbmfP h h� � �

12

10125.2 psia

4 4

PP � � �

To find the quality at the exit, we need to interpolate in Table B-14. From Table B-14

T P hf hg

5 23.8 13.14 102.5

10 26.7 14.66 103.2

By interpolation at 25.2 psia, hf =13.9 and hg =102.8. The quality is

2

2

25.2 13.90.127

102.8 13.9

f

g f

h hx

h h� �

� � ��

Answer

6 - 24

6-28 A supply line contains a two-phase mixture of steam and water at 240�C. To determine the quality of the

mixture, a throttling calorimeter is used. In this device, a small sample of the two-phase mixture is bled off

from the line and expanded through a throttling valve to atmospheric pressure. If the temperature on the

downstream side of the throttling valve is measured to be 125�C, what is the quality of the mixture in the

main steam line?






Solution: From the first law applied to a throttle,

1 2h h�From Table A-12 (superheated vapor),

2

kJ2726

kgh �

From Table A-10 at T1 = 240�C

kJ kJ1037 2804

kg kgf gh h� �

� �1 2f g fh h x h h h� � � �

1 2726 10370.956

2804 1037

f

g f

h hx

h h� �

� � ��

Answer

6 - 25

6-29 In a heat pump, R-134a is throttled through an expansion coil, which is a long copper tube of small

diameter. The tube is bent in a coil both to fit in a compact space and to provide a large pressure drop. The

refrigerant enters as saturated liquid at 5oC with a flow rate of 0.025 kg/s and exits as a two-phase mixture

at a pressure of 200 kPa. The wall of the coil may be assumed to be at the average temperature of the inlet

and outlet. Heat is exchanged by natural convection and radiation from the outer surface of the coil with a

combined heat transfer coefficient of 6 W/m2·oC to the surroundings at 20oC. The expansion coil has an

outside diameter of 8mm and a length of 2.2 m. Calculate the quality at the exit state.

Approach:Use the first law specialized for an open system, eliminating kinetic and

potential energy changes, work, and the transient term. Calculate the

heat transfer from the convection rate equation.




4. The heat transfer coefficient is uniform over the surface of the coil

and independent of temperature.

Solution: From the first law for an open system,

2 2

2 2



� � ��

� � �� V V �

�

Assuming no change in kinetic or potential energy and steady conditions with no work, the first law becomes

� �1 20 Q m h h� � ��

Interpolating in Table A-14 gives the enthalpy of saturated liquid at 5oC as 1 56.7 kJ/kg.h � The heat lost from the

surface of the coil is

� �ave surrQ hA T T� ��

The saturation temperature at the exit condition of 200 kPa is, from Table A-15, The average

temperature of the outside of the coil is

o

2 10.1 C.T � �

� � o1 25 10.1

2.55 C2 2

aveT TT

� ��

The surface area of the coil is

� � � � 21m8mm 2.2m 0.0553m

1000mmA DL� �

� ��

�

Therefore, the rate of heat loss by convection/radiation is

� � � �� 2 o

2 o

W6 0.0553m 2.55 20 C 7.48W

m Cave surrQ hA T T � ��

�

Solving the first law for exit enthalpy,

2 1

7.48W kJ 1000J56.7 56,396J 56.4kJ

kg kg 1kJ0.025

s

Qh hm

� ��

� �

�

�

With values at the exit pressure of 200 kPa from Table A-15, the exit quality is

2 56.4 36.840.0956

241.3 36.84

f

g f

h hx

h h� �

� � ��

Answer

Comment:The exit enthalpy was very close to the inlet enthalpy, with heat loss having little influence on the final enthalpy.

This is often the case; therefore, throttles are usually assumed to be adiabatic.

6 - 26

6-30 One way to produce saturated liquid water is to mix subcooled liquid water with steam. In the tank below,

40 kg/s of subcooled liquid water enters at 15oC and 50 kPa. Superheated steam enters at 200oC and 50

kPa. What mass flow rate is required so that the exit stream is saturated liquid water at 50 kPa? Assume

the tank is well-insulated.

Approach:Apply conservation of mass and the first law

for an open system.




4. The tank is adiabatic.

Solution: From conservation of mass

1 2m m m� �� 3

From the first law for an open system 2 2

2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming an adiabatic tank and steady operation, the first

law becomes

1 1 2 2 3 30 m h m h m h� � ��

For state 1, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 15oC,

which is, from Table A-10,

1

kJ63

kgh �

From Table A-12 at 200oC and 50 kPa,

2

kJ2877

kgh �

From Table A-11, the enthalpy of saturated liquid at 50 kPa is

3

kJ340.5

kgh �

Eliminating in the first law by using conservation of mass, 3m�

� ��

1 1 2 2 1 2 3

1 1 3 2 2 3

0

0

m h m h m m h

m h h m h h

� � � �

� � � �

� � � �

� � �

Solving for 2m�

� � � �

� �

1 1 3

2

2 3

kg kJ40 63 340.5

s kg4.38kg s

kJ2877 340.5

kg

m h hm

h h

� ��

� � �

�� Answer

6 - 27

6-31 In a desuperheater, superheated steam is converted to saturated steam by spraying liquid water into the

steam. Using data on the figure, calculate the mass flow rate of liquid water.

Approach:Apply conservation of mass and the first

law for an open system.




4. The desuperheater is adiabatic.

5. Pressure is constant during the

process.


1 2m m m� �� 3


2 2



� � ��

� � �� V V �

�


law becomes

1 1 2 2 3 30 m h m h m h� � ��

From Table A-12, the enthalpy of superheated steam at 250oC and 2 MPa is

1 2902kJ kgh �



2 125.8 kJ kgh �

The desuperheater is effectively a mixing chamber, and the pressure is the same for all three streams. The

enthalpy of saturated vapor at 2 MPa, is, from Table A-11,

3 2799 kJ kgh �

Eliminating in the first law by using conservation of mass, 3m�

� ��

1 1 2 2 1 2 3

1 1 3 2 2 3

0

0

m h m h m m h

m h h m h h

� � � �

� � � �

� � � �

� � �

Solving for 2m�

� � � �

� �

1 1 3

2

2 3

kg kJ0.3 2902 2799

s kg0.0116 kg s

kJ125.8 2799

kg

m h hm

h h

� ��

� � �

�� Answer

6 - 28

6-32 A laundry requires a stream of 8 kg/sec of hot water at 40oC. To obtain this supply, liquid water at 20oC is

mixed in an adiabatic chamber with saturated steam. All three process streams are at 100 kPa. What are

the required mass flow rates of the two inlet streams?

Approach:Apply conservation of mass and the first law

for an open system.




4. The mixing chamber is adiabatic.

5. Pressure is constant during the process.


1 2m m m� �� 3


2 2



� � ��

� � �� V V �

�


law becomes

1 1 2 2 3 30 m h m h m h� � ��

From Table A-11, for saturated steam at 100 kPa,

1 2675.5kJ kgh �



2 83.96 kJ kgh �

For state 3, the enthalpy of the subcooled liquid is approximated by the enthalpy of saturated liquid at 40oC, which

is, from Table A-10,

3 167.57 kJ kgh �

Using conservation of mass to eliminate in the first law 1m�

� �3 2 1 2 2 3

3 1 2 1 2 2 3 3

0

0

m m h m h m hm h m h m h m h

� � � �

� � � �

� � � ��

3

Gathering terms

� � � �

� ��

� �

� �

2 1 2 3 1 3

3 1 3

2

1 2

kg kJ8 2675.5 167.57

s kg7.74kg s

kJ2675.5 83.96

kg

m h h m h h

m h hm

h h

� � �

��

� � ��

� �

��

Answer

1 28 0.26kg sm m� � �� Answer

6 - 29

6-33 Steam with a quality of 0.88 and a pressure of 20 kPa enters a condenser. The steam flow is divided

equally among 20 tubes 2.1-cm in diameter which run in parallel through the condenser. The same amount

of heat is removed from each tube. Liquid water exits each tube with a velocity of 1.5 m/s and a

temperature of 55oC. Find the total amount of heat removed from the entire condenser.

Approach:Apply the first law for an open system,

eliminating all terms except heat and

enthalpy change.




4. Pressure is constant in each tube.


2 2

2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming steady operation with no work, the first law

becomes

1 20 Q mh mh� � ��

Using data from Table A-11 for saturated steam at 20 kPa,

� � � �1251.4 0.88 2610 251.4 2327 kJ kgf g fh h x h h� � � � � � �



2 230kJ kgh �

The mass flow rate in each tube is (with values of density from Table A-6) 2

2 2 2 3

kg m 2.1cm 1m kg985.7 1.5 0.512

m s 2 100cm stubem A� �

� � ��

� V

The total mass flow rate for all 20 tubes is

� � kg20 20 0.512 10.24

stubem m� � ��

From the first law

� � � �2 1

kg kJ10.24 230 2327 21,478kW

s kgQ m h h� � � � � �� Answer

6 - 30

6-34 Saturated steam at 120oF is condensed in a tube, as shown. Cooling water at 50oF flows in crossflow over

the exterior of the pipe, giving a heat transfer coefficient of 200 Btu/h·ft2·oF. Find the exit quality.

Approach:Use the thermal resistance analogy to

determine the rate of heat transfer. Apply

the first law for an open system,

eliminating all terms except heat and

enthalpy change.




4. Pressure is constant in each tube.

5. The interior of the tube wall is at the

condensing steam temperature.

Solution: Apply the thermal resistance analogy to

find the heat removed from the pipe.

Assuming that the condensing heat transfer

coefficient is very large, the interior wall of

the tube is at the temperature of the

saturated steam. The resistance to

conduction in the tube wall is

� �1 o

o

3.25ln

Btu30.000212

Btu h F2 2ft 30

h ft F

R�

� ��

��

The resistance to convection on the exterior of the tube is

� � � �2 o

2 o

1 B0.00147

h FBtu 1ft200 2� 3.25in. 2ft

h ft F 12in.

R � �tu

��

The two resistances add in series. The total heat transferred is

� �

� �

o

o

120 50 F Btu41,640

Btu h0.000212 + 0.00147

h Ftot

TQR

��

�

�


2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming steady operation with no work, the first law

becomes

1 20 Q mh mh� � ��

Solving for exit enthalpy and using data in Table A-10,

2 1

Btu41,640

Btu Btuh 1113.5 735lbm lbm lbm

110h

Qh hm

��

�

Using data in Table A-10,

2 735 880.63

1113.5 88

f

g f

h hx

h h� �

� � ��

Answer

6 - 31

6-35 Superheated R-134a enters a well-insulated heat exchanger at 0.7 MPa, 70oC. It exits as saturated liquid at

0.7 MPa with a volumetric flow rate of 6000 cm3/min. The R-134a exchanges heat with an air flow, which

enters at 18oC at a mass flow rate of 195 kg/min. Find the exit air temperature.

Approach:Apply the first law for an open

system.

Assumptions:1. Potential energy change is

negligible.



4. The heat exchanger is adiabatic.

5. Pressure is constant during the

process.

6. Air may be considered an ideal gas

under these conditions.

7. The specific heat of air is constant.

Solution: Using data in Table A-15 for a saturated liquid at 0.7 MPa, the mass flow rate of the refrigerant is

33

3

cm 1m6000

min 100cm kg7.2

m min0.0008328

kg

r rr r r

r f

V Vm Vv v

�

� ��

� ��


2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming an adiabatic heat exchanger in steady operation,

the first law becomes

� � � �1 2 3 40 r am h h m h h� � � ��


� � � �1 2 3 40 r a pm h h m c T T� � � ��

Using data from Tables A-16, A-15, and A-8,

� � � �1 2 o

4 3

kg kJ7.2 307 86.78

min kg18 26.1 C

kg kJ195 1

min kg K

r

a p

m h hT T

m c

� � ��

��

Answer

6 - 32

6-36 R-134a flows through the evaporator of a refrigeration cycle at a rate of 5 kg/s. The R-134a enters as

saturated liquid and leaves as saturated vapor at 12oC. Air at 25oC enters the shell side of the heat

exchanger. If the air leaves at 15oC, what mass flow rate of air is required?

Approach:Apply the first law for an open system.




4. The evaporator is adiabatic.


6. Air may be considered an ideal gas under

these conditions.

7. The specific heat of air is constant.


2 2

2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming an adiabatic evaporator in steady operation, the

first law becomes

� � � �1 2 3 40 r am h h m h h� � � ��


� � � �1 2 3 40 r a pm h h m c T T� � � ��

Solving for the mass flow rate of the air,

� �

2 1

3 4

a rp

h hm mc T T

��

��

With values of enthalpy from Table A-14 and specific heat from Table A-8,

� �

� � o

kJ254.03 66.18

kg kgkg5 93.5

s skJ1.005 25 15 C

kg K

am�

� ��

� Answer

6 - 33

6-37 Superheated steam at 5 psia and 200�F is condensed in a heat exchanger. The steam flows at 39 lbm/s and

exits as saturated liquid. Cooling water at 45�F is used to condense the steam. The water and steam are not

mixed in the heat exchanger, but enter and leave as separate streams. If the maximum allowable water

temperature rise is 15�F and the maximum allowable water velocity is 11 ft/s, what is the diameter of the

pipe which carries water to the heat exchanger?





4. The condenser is adiabatic.


6. The cooling water is incompressible.

7. The specific heat of the water is constant.

Solution:


2 2



� � ��

� � �� V V �

�

2

Neglecting changes in kinetic and potential energy, and assuming an adiabatic condenser in steady operation, the

first law becomes

1 1 3 3 2 2 4 4m h m h m h m h� � ��

Since the streams do not mix, and 1m m�� 3 4;m m�� therefore

� � � �1 1 2 3 4 3m h h m h h� � ��

Assuming water is an ideal fluid,

� � � �1 1 2 3 4 3pm h h m c T T� � ��

� ��

1 2

3 1

4 3p

h hm m

c T T�

��

� �

With h1 from Table B-12, h2 from Table B-11, and cp from Table B-6,

� �

� �3

o

o

Btu1148 130.2

lbm lbm39Btus

1 15 Flbm F

m��

� � �

� lbm2646

s�

To find the diameter

3m A�� V2

2

D� � � ��

�V

32mD��

��

V3

lbm2646

s2ft lbm

11 62.4s ft

��

� ��

2.21 ft� Answer

6 - 34

6-38 A two-phase mixture of steam and water with a quality of 0.93 and a pressure of 5 psia enters a condenser

at 14.3 lbm/s. The mixture exits as saturated liquid. River water at 45oF is fed to the condenser through a

large pipe. The exit temperature of the river water is 70oF less than the exit temperature of the other

stream. If the maximum allowable average velocity in the pipe carrying river water is 15 ft/s, calculate the

pipe diameter.





4. The condenser is adiabatic.


6. The river water is incompressible.

7. The specific heat of the river water is

constant.

Solution:


2 2



� � ��

� � �� V V �

�

2

Neglecting changes in kinetic and potential energy, and assuming an adiabatic condenser in steady operation, the

first law becomes

1 1 3 3 2 2 4 4m h m h m h m h� � ��


� � � �1 1 2 3 4 3m h h m h h� � ��

Assuming water is an ideal fluid,

� � � �1 1 2 3 4 3pm h h m c T T� � ��

� ��

1 2

3 1

4 3p

h hm m

c T T�

��

� �

State 2 is a saturated liquid at 5 psia, since pressure is constant across the condenser. From Table B-11 o

2 162.2 FT �

4 2 70T T� � o92.2 F�

State 1 is a two-phase mixture at 5 psia. With data from Table B-11,

� � � �1

Btu130.2 0.93 1131 130.2 1061

lbmf g fh h x h h� � � � � � �

The enthalpy of state 2 is

2h � 130.2 Btu lbmfh �

Using the value of specific heat from Table B-6 at the average river water temperature of 69 oF

� �

� �

� �

� �1 1 4

3o3 4

o

lbm Btu14.3 1061 130.2

s lbm

Btu1.00 92.2 45 F

lbm F

m h hm

h h

� � ��

�� 282 lbm s�

2

32

Dm A� � � � ��

� V V

34mD� �

��V

3

lbm4 282

s

lbm ft62.2 15

ft s

D�

� ��

� ��

0.620 ft

� � �

� Answer

6 - 35

6-39 A heat exchanger is used to cool engine oil. The specific heat of the oil is 0.6 Btu/lbm·oF. Using data on

the figure below, find the exit temperature of the air.





4. The heat exchanger is adiabatic.


6. The oil is incompressible.

7. The specific heat is constant.


conditions.

Solution:


2 2



� � ��

� � �� V V �

�

2

Neglecting changes in kinetic and potential energy, and assuming an adiabatic heat exchanger in steady operation,

the first law becomes

1 1 3 3 2 2 4 4m h m h m h m h� � ��


� � � �1 1 2 3 4 3m h h m h h� � ��

Assuming the oil is an ideal liquid and the air is an ideal gas,

� � � �1 , 1 2 3 , 4 3p oil p airm c T T m c T T� � ��

� � � � o

o1 , 1 2 o

4 3

3 ,

lbm Btu3.8 0.6 160 90 F

s lbm F50 85 F

lbm Btu19 0.24

s lbm R

p oil

p air

m c T TT T

m c

� ��

��

Answer

6 - 36

6-40 Saturated liquid Refrigerant-134a at 36oC is throttled to –8oC. The refrigerant then enters an evaporator

and exits as saturated vapor. The evaporator is used to cool liquid water from 20oC to 10oC. If the mass

flow rate of refrigerant is 0.013 kg/s, what is the mass flow rate of the water?

Approach:Apply the first law for an open system to each

component.




4. The throttle and evaporator are adiabatic.

5. Pressure is constant across the evaporator.

6. The water is incompressible.

7. The specific heat of the water is constant.

Solution:

Taking a control volume around the throttle, and applying the first law,

1h h� 2

Using data in Table A-14

1

kJ100.3

kgfh h� �

From Table A-14 at –8oC, hf = 39.54 kJ/kg and hg = 242.5 kJ/kg. Since h2 = 100.3 kJ/kg falls between these two

values, state 2 is a two-phase mixture. Taking a control volume around the evaporator and applying the first law 2 2

2 2



� � ��

� � �� V V �

�

Neglecting changes in kinetic and potential energy, and assuming an adiabatic evaporator in steady operation, the

first law becomes

2 2 4 4 3 3 5 5m h m h m h m h� � ��

The water and R-134a do not mix, therefore

2 3 Rm m m� �� 4 5 Wm m m� ��

� � ��

� �

�2 3 4 5

3 2

4 5

0 R w

Rw

m h h m h h

m h hm

h h

� � � �

��

�

� �

��

Assuming the water is an ideal liquid

� ��

3 2

4 5

Rw

p

m h hm

c T T�

��

��

The saturated vapor at state 3 is at the same pressure as state 2. We have already concluded that state 2 is a two-

phase mixture, so its saturation temperature is –8oC. During a constant pressure process entirely in the two-phase

region, temperature remains constant. Therefore, state 3 is a saturated vapor at –8oC. From Table A-14

3 242.5kJ/kggh h� �

2 1 100.3kJ/kgh h� �

Using values of specific heat at the average water temperature of 15 oC,

� �� o

0.013kg s 242.5 100.3 kJ kg kg0.0441

4.187 kJ kg K 20 10 C swm

��

� �� Answer

6 - 37

6-41 In a flash chamber, a pressurized liquid is throttled to a lower pressure where it becomes a two-phase

mixture. The saturated liquid and vapor streams are removed in separate lines. In the figure below, liquid

R-134a at 10�F and 30 psia is throttled to 12 psia. If 21.6 lbm/h of saturated vapor exits the flash chamber,

what is the inlet flow rate? Assume the flash chamber is adiabatic.

Approach:Apply conservation of mass and the first law for an

open system to the combination of the two

components.




4. The throttle and flash chamber are adiabatic.

5. Pressure is constant in the flash chamber.

Solution: In this problem, the easiest approach is to define a

control volume around the combination of the two

components. You could also analyze each

component separately, but that would take more

effort and produce the same final result. Applying

conservation of mass to the control volume shown,

1 2 3m m m� ��

From the first law with no kinetic or potential

energy change, and no heat transfer or work,

1 1 2 2 3 3m h m h m h� ��

State 1 is a compressed liquid. Approximate the

enthalpy of the liquid as the enthalpy of the

saturated liquid at the same temperature. From

Table B-14

h1 = hf at 10�F = 14.66 Btu/lbm

From Table B-15, the enthalpy of states 2 and 3 is

h2 = hg at 5 psi = 93.79 Btu/lbm

h3 = hf at 5 psi = –3.73 Btu/lbm

Substituting the given value of mass flow rate into conservation of mass produces

1 3

lbm21.6

hm m� ��

Substituting and the values of enthalpy into the first law, 1m�

� �3 3

lbm Btu lbm Btu lbm Btu21.6 14.66 21.6 93.79 3.73

h lbm h lbm h lbmm m� � � ��

� � � � � �

�

Solving for gives 3m�

3

lbm92.9

hm ��


1

lbm lbm21.6 92.9

h hm � �� lbm

114.5h

� Answer

6 - 38

6-42 Saturated liquid water at 40 kPa enters a 140 kW pump. The output of the pump is fed into a boiler where

heat is added at a rate of 302 MW. There is negligible pressure drop across the boiler. If the mass flow

rate of water is 70 kg/s, determine the boiler pressure and the state at the exit of the boiler.

Approach:Apply to the pump to

determine the exit pressure form the pump.

This is equal to the pressure at the exit of the

boiler, since there is negligible pressure drop

across the boiler. Draw a control volume

around both components and apply the first law

to determine exit temperature from the boiler.

� 1 2W mv P P� ��




4. The pump is adiabatic and ideal.

5. Pressure is constant across the boiler.

6. The water is incompressible.

Solution:

For the pump, power is related to pressure rise by

� �1 2cvW mv P P� ��

Solving for exit pressure and using the specific volume of saturated liquid water at from Table A-11, 40kPa

2 1

cvWP Pmv

� ��

� �3

1000 W140kW

1000Pa 1 kW40kPa

1 kPa kg m70 0.001027

s kg

61.99 10 Pa 2.0MPa� � �

� ��

� �

Pressure is constant across the boiler, so

3 2 2.0 MPaP P� � Answer To find the state at the exit of the boiler, draw a

control volume around both components. The first

law for this control volume is (assuming steady-state,

and no kinetic or potential energy changes)

0 cv cv i i e eQ W m h m h� � � ��

� �1 30 cv cvQ W m h h� � � ��

3 1

cv cvQ Wh hm�

� ��

302,000kW ( 140)kW kJ

318kg kg

70s

� ��

3

kJ4634

kgh �

The exit state of the boiler has a pressure

of and an enthalpy of From Table

A-12, this is superheated vapor with a temperature of

2MPa 4634kJ/kg.

o1000 CT � Answer

6 - 39

6-43 Air at 2000 R enters the turbine of a turbojet engine. The turbine is well insulated and produces 100 Btu of

work per pound mass of air flowing through the engine. Upon exiting the turbine, the air enters the inlet of

an insulated nozzle at 20 ft/s. The air leaves the nozzle at 2800 ft/s through an exit flow area of 0.6 ft2.

The pressure at the nozzle exit is 10 lbf/in2. What is the mass flow rate of air through the engine in lbm/s?

Approach:Draw a control volume around each component

and apply the first law.


2. Kinetic energy change in the turbine is

negligible.


4. The nozzle and turbine are adiabatic.


conditions.

6. The specific heat of the air is constant.

Solution:

Begin by constructing a control volume around the turbine. Assuming an adiabatic turbine with negligible kinetic

and potential energy changes, the first law becomes

� �1 2W m h h� �� For an ideal gas with constant specific heat, ,ph c T� � � therefore

� �1 2pW mc T T� ��

Specific heat depends on the average air temperature in the turbine, however, the exit temperature is unknown. To

make further progress, use cp of air at the inlet temperature of (about ) and correct the calculation

later if necessary. Solving for T2000R o1500 F

2 and using values of cp from Table B-8 at 1500�F,

2 1

100 Btu lbm2000R 1638R

Btu0.276

lbm R

cv

p

WT Tmc

� � � � �

�

�

�

Apply the first law for an open system to a control volume around the nozzle: 22

322 2 3

2 2

cvcv cv


� ��

� �� VV

3

Assuming an adiabatic nozzle, no work, no change in potential energy and steady conditions, 22

322 30

2 2m h m h

� ��

� �� VV


� �2 2

2 3

2 30

2pc T T �

� � �V V

2 2

2 3

3 22 p

T Tc�

� �V V

The specific heat should be evaluated at the average temperature in the nozzle; however, the exit temperature is

unknown. The nozzle inlet temperature is 1638 R or 1178�F and we expect the outlet temperature to be lower.

For simplicity, we assume an average temperature of 1000�F and will correct the calculation later if necessary.

Using data from Table B-8,

� �2 2 2 2

3

2

20 - 2800 ft s 1Btu 1lbf1638R 1043R

ft lbm778ft lbfBtu 32.22 0.263slbm R

T

� ��

� ��

�

From the ideal gas law:

6 - 40

� �

� �

2

2

33

323

ft lbf 1ft1545 1043R

lbmol R 144in.38.6ft lbm

lbm28.97 10lbf in

lbmol

RTvMP

� ��

� ��

The mass flow rate may now be calculated as

� �� 2

3 3

3

3

2800ft s 0.6ft43.5lbm s

38.6ft lbm

Amv

� � �� V Answer

Comments:For greater accuracy, you could recalculate the mass flow rate with improved values of specific heat, based on the

calculated temperatures. The specific heat is not a strong function of temperature over the range considered, so

this may not be necessary.

6 - 41

6-44 A well-insulated, rigid tank of volume 0.7 m3 is initially evacuated. The tank develops a leak and

atmospheric air at 20oC, 100 kPa enters. Eventually the air in the tank reaches a pressure of 100 kPa. Find

the final temperature.

Approach:Apply the first law for an open system and integrate over

time. Note that the temperature of the air leaking into the

tank is constant with time, so the enthalpy of the air is

also a constant.




4. Air may be considered to be an ideal gas.


Solution:

From the first law for an open system with no change in kinetic or potential energy

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

The tank is adiabatic. No work is done and nothing leaves the tank, so the first law reduces to

cvi i

dU m hdt

� �

Integrating over time

cv i idU m h dt�# # �For an ideal gas, enthalpy depends only on temperature. The air entering the tank is at the same temperature

throughout the process, so enthalpy is constant and may be removed from the integral.

2 1 i i iU U h m dt h m� � �# � 2

i

where m2 is the mass in the tank at the end of the process. At the beginning of the process, the tank is evacuated,

therefore 1 0.U � 2 2 2im u h m� 2 i iu h u Pv� � �

For an ideal gas with constant specific heat, vu c T� � �

2

( ) iv i i

RTc T T PvM

� � �

Solving for 2T

2i

iv

RTT TMc

� �� kJ

8.314 20 273 Kkmol K

(20 273)Kkg kJ

28.97 0.718kmol kg K

� � ��

�

2 410KT � o137 C� Answer

6 - 42

6-45 Helium at 150oF and 40 psia is contained in a rigid, well-insulated tank of volume 5 ft3. A valve is cracked

open and the helium slowly flows from the tank until the pressure drops to 20 psia. During this process, the

helium in the tank is maintained at 150oF with an electric resistance heater. Find

a. the mass of helium withdrawn.

b. the energy input to the heater.

Approach:Use a mass balance and the ideal gas law to find the mass of

helium withdrawn. Use the first law for an open system to find

the energy input to the heater.




4. Helium may be considered to be an ideal gas.


Solution:

a) From a mass balance

1 2em m m� �where is the mass of helium exiting. Using the ideal gas law, em

1 1 2 2

1 2

ePMV P MVm

RT RT� � 1

1 2( )

MV P PRT

� ��

� �

3

3

lbm4 5ft

lbmol(40 20)psia

psia ft10.73 150 460 R

lbmol R

� ��

� ��

0.061lbm�

b) From the first law with no kinetic or potential energy

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

The tank is adiabatic and nothing is entering, therefore

cvcv e e

dU W m hdt

� � ��

Integrating over time,

cv cv e edU W dt m h dt� � �# # #� �

Since the temperature of the helium in the tank does not change, is constant and eh

2 1

dcv e eU U W h m t� � � � # � 2 2 1 1 cv e em u m u W h m� � � �

2 2 1 1 ( )cv e e e em u m u W u P v m� � � � �

� �� 2 2 1 1 1 2cv e e em u m u W u P v m m� � � � � �Since u is only a function of temperature for an ideal gas and the temperature is constant

1 2 eu u u u� � �

Therefore

� �1 20 cv e eW P v m m� � � �

� �1 2

ecv

RTW m m e em RTM

� � M

� � �

� � � �ft lbf0.061lbm 1545 150 460 R

1Btulbm R

lbm 778ft lbf4

lbmol

��

18.5BtucvW � � Answer

6 - 43

6-46 A residential hot water heater initially contains water at 140oF. Someone turns on a shower and draws

water from the tank at a rate of 0.2 lbm/s. Cold make-up water at 50oF is added to the tank at the same rate.

A burner supplies 5472 Btu/h of heat. The water tank, which is a cylinder of diameter 1.8 ft, is filled to a

height of 4 ft. How long will it be before the exiting water reaches 100oF? Assume a well-mixed tank.

Approach:Apply the first law for an open system. Use

ideal liquid relations to rewrite enthalpy and

internal energy in terms of temperature.

Separate variables and integrate.



3. Water is an ideal liquid.

4. The tank is well-mixed and all tank contents

are at the same temperature.

Solution:

The first law for an open system with one stream in and one stream out, neglecting kinetic and potential energy

changes, is

cvcv i i e e

dU Q W m h m hdt

� � � ��

The entering and exiting mass flow rates are equal and no work is done, therefore

� � � �cv cvcv i e

d m uQ m h h

dt� � ��

The mass in the control volume does not change with time. For an ideal liquid .ph c T� � � With these ideas, the

first law becomes

� �cvcv cv p i e

dum Q mc Tdt

� � �� T

The tank is well-mixed, so the temperature, T, of the water in the tank equals the temperature of the exiting

stream. The internal energy of the water in the tank is related to its temperature by

cv v pdu c dT c dT� �

since, for an ideal liquid, . The first law takes the form v pc c�

� �cv p cv p idTm c Q mc T Tdt

� � ��

Separating variables and integrating from initial temperature T1 to final temperature T2

� �2 2

1 0

T tcv p

Tcv p i

m c dTdt

Q mc T T�

� �# #� �

� �let cv p i

p

Q mc T Td mc dT$

$

� � �

� �

� ��

2

12cv p

p

dm c tmc

$

$

$$

� �# �

2

1 2lncvm tm

$$$� �

�

� �2 1ln lncvm tm

$ $� �� 2�

22

1

lncvm tm

$$

� ��

6 - 44

� ��

2

2

1

lncv p icv

cv p i

Q mc T Tm tm Q mc T T

� ��

� ��

Find cvm

� � � �

� � � �

� � � ��

2

3 2

o

o

2

1.862lbm ft � ft 4ft 631lbm

2

1h Btu5472Btu h + 0.2 lbm s 1 50 100 F

2600s lbm F631 lbmln

10.2lbm s5472 + 0.2 1 50 140

3600

cvm V

t

� � ��

� � � � � ��

��

2 2096s 35mint � � Answer

6 - 45

6-47 In an industrial process, two streams are mixed in a tank and a single stream exits. Both streams may be

assumed to have the properties of water. The volume of fluid in the tank is constant. A paddle wheel stirs

the tank contents, doing work Initially, the water in the tank is at temperature T.W� 1. At time t = 0, stream

A at temperature TA enters with mass flow rate , and stream B enters at TAm� B with rate Bm� . The quantities

TA, TB, , and Am� Bm� are all constant with time. Assuming a well-mixed tank, derive a formula for the time,

t2, at which the tank water temperature is T2. The tank is well insulated.

Approach:Apply the first law for an open system. Use

ideal liquid relations to rewrite enthalpy and

internal energy in terms of temperature.

Separate variables and integrate.



3. The liquid is ideal.

4. The tank is well-mixed and all tank contents

are at the same temperature.


6. All fluids have the properties of water.

Solution:

The first law for an open system neglecting kinetic and potential energy changes, is

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

Since the tank is well-insulated

� �cv cvA A B B C C

d m uW m h m h m h

dt� � � � ��

The mass of liquid in the tank is constant, and conservation of mass requires that ,C A Bm m m� �� therefore

� �

� � � �

cvcv A A B B A B C

cvcv A A C B B C

dum W m h m h m m hdt

dum W m h h m hdt

� � � � � �

� � � � � �

� � � � �

� � � h

The liquid in the tank is ideal and has a temperature T. The exiting stream is also at T, because the tank is well-

mixed. With these considerations, the first law becomes

� � � �

� �

cv p A p A B p B

cv p A p A B p B A p B p

dTm c W m c T T m c T TdtdTm c W m c T m c T m c m c Tdt

� � � � � �

� � � � � �

� � �

� � � � �

A p A B p B A B

cv p cv

W m c T m c T m mdT Tdt m c m

� � � � ��

�

� � � � �

Define K1 and K2 so that

1 2

dT K K Tdt

� �

Separating variables and integrating

2

1 01 2

T

T

dT dtK K T

��#

2t

#

1 2

let

K K T$ � �

2d K dT$ � �

6 - 46

2

1

2

1

2

2

2

2

2 12

2 1 2 1 2 1

1ln

1 1ln ln

d tK

tK

2 2K K TtK K K K T

$

$

$$

$$

$

$$

� �

� �

� ��

#

� �

� �

2

2

1

ln

cvA A B B A B

pcv

cvA BA A B B A B

p

W m T m T m m TcmtWm m m T m T m m Tc

� � � � � ��

��

��

�� Answer

6 - 47

6-48 A well insulated tank of volume 0.035 m3 is initially evacuated. A valve is opened, and the tank is charged

with superheated steam from a supply line at 600 kPa, 500oC. The valve is closed when the pressure

reaches 300 kPa. How much mass enters?

Approach:Use the first law for an open system. Integrate the equation,

recognizing that the enthalpy entering is constant with time.

Properties are available in the steam tables.




4. The state of steam in the supply line is constant.

Solution: The first law for an open system neglecting kinetic and potential energy changes, is

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

The tank is well-insulated, no mass exits, and no work is done, therefore

cvi i

dU m hdt

� �

� �

� �

cv cvi i

cv cv i i

d m um h

dtd m u m h dt

�

�# #

�

�

The state of the steam entering remains the same throughout the process, so hi is a constant. If m2 is the mass in

the tank at the end of the process and m1 is the mass at the start

� �2 2 1 1 2 1i i im u m u h m dt h m m� � � �# �

Since the tank is initially evacuated, and 1 0,m �

2 2 2

2

i

i

m u h mu h

�

�

From Table A-12, 3483kJ kg ,ih � therefore

2 3483kJ kgu �

At the final pressure of 300 kPa and 2 3483kJ kgu �

3

2 1.5m kgv �

Therefore the mass in the tank at the end of the process is 3

2 3

2

0.035m0.0233kg

m1.5

kg

Vmv

� � � Answer

6 - 48

6-49 A well-insulated piston-cylinder assembly contains 0.06 kg of R-134a at –15oC with a quality of 0.92. A

supply line introduces superheated R-134a at 10oC, 200 kPa into the cylinder. Assuming the pressure in the

cylinder is constant, calculate the volume just when all the liquid has evaporated.

Approach:Use the first law for an open system. Integrate the equation,

recognizing that the enthalpy entering is constant with time.



3. The piston-cylinder assembly is adiabatic.

4. The state of R-134a in the supply line is constant.

5. The pressure in the cylinder is constant.

6. The expansion is quasi-static.

Solution: From conservation of mass,

2 1im m m� �

where m2 is the mass in the tank at the end of the process, m1 is the mass at the start, and mi is the total mass

introduced during the process. The first law for an open system neglecting kinetic and potential energy changes, is

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

The tank is well-insulated and no mass exits, therefore

cvcv i i

dU W m hdt

� � ��

� �

� �

cv cvcv i i

cv cv cv i i

d m uW m h

dtd m u W dt m h dt

� � �

� � �# # #

� �

� �

The state of the R-134a entering remains the same throughout the process, so hi is a constant.

2 2 1 1 cv i i cv i im u m u W h m dt W h m� � � � � � �# �

Noting that this is a quasi-static expansion at constant pressure, and substituting conservation of mass

� � �2 2 1 1 2 1 2 1im u m u P V V h m m� � � � � � �

� � � �2 2 1 1 2 2 1 1 2 1im u m u P m v m v h m m� � � � � �

� � � �2 2 2 2 1 1 1 1 2 1im u Pm v m u Pm v h m m� � � � �

Using the definition of enthalpy, and recognizing that ,h u Pv� � 1 2 ,P P P� �

� �2 2 1 1 2 1im h m h h m m� � �

Solving for m2,

� �1 11 1 12

2 2

ii

i i

m h hm h m hmh h h h

��

� �

The enthalpy of the R-134a in the cylinder at the start of the process is

� � � �1

kJ30.6 0.92 238.3 30.6 222

kgf g fh h x h h� � � � � � �

where hf and hg are found by interpolation in Table A-14 at –15oC. The contents of the cylinder remain at the

same pressure throughout the process and the R-134a changes from a two-phase mixture to a saturated vapor. A

constant pressure process in the two-phase region is also a constant temperature process, so the final enthalpy is

the enthalpy of saturated vapor at –15oC, that is 2 238.3kJ/kg.h � The enthalpy in the supply line, hi, may be

found in Table A-16. The final mass may now be calculated as

6 - 49

� � � ��

� �1 1

2

2

kJ0.06kg 222 259

kg0.106kg

kJ238 259

kg

i

i

m h hm

h h

��

� � ��

The final volume, using data for specific volume interpolated in Table A-14, is

� �3

3

2 2 2

m0.120 0.106kg 0.0127 m

kgV v m

� ��

� Answer

6 - 50

6-50 The pressure inside a pot is maintained at an elevated level by a steel bob which rests on an open tube of

inside diameter 0.5 cm. The bob, which has a mass of 0.401 kg, jiggles whenever the pressure in the pot is

high enough to displace it, and steam is released. Heat is added to the bottom of the pot at a rate of 900 W.

Heat is lost from the sides and the top of the pot by natural convection with a heat transfer coefficient of 3.9

W/m2·K. The pot has a height of 0.154 m and a diameter of 0.256 m. The ambient is at 20�C. Assume

conduction resistance through the pot sides and top is very small, and that there is no air in the pot (only

water and steam). The pot is half filled with water when the bob first lifts.

a. Find the temperature inside the pot.

b. Find the net rate of heat addition to the pot.

c. Find the initial mass of the two-phase mixture in the pot.

d. Find the fraction of the pot which is filled with liquid after one hour.

Approach:Use the first law for an open system.

Integrate the equation, recognizing that

the enthalpy entering is constant with

time. Properties are available in the

steam tables.

Assumptions:1. Potential energy change is

negligible.


3. The pot contains only steam and

water.

4. The conduction resistance is small.

5. The convection resistance on the

inside of the pot is small.

6. The work done in displacing the bob

is negligible.

Solution:

a) Define m as the mass of the bob and r as the radius of the small tube that the bob rests on. By definition,

FPA

�� 2

22

2

m0.401kg 9.81

s200,300Pa

0.25� m

100

mgr�

� ��

� ��

Therefore, throughout the process, the pressure in the pot is 200.3 kPa.P � The saturation temperature at this

pressure is, from Table A-11, T = 120�C. Answer

b) The conduction resistance through the walls and lid of the pot is assumed to be small. Also the convective

heat transfer coefficient on the inside is assumed to be very large, so that the inside convective resistance is small.

Therefore, the outside of the pot is at the saturation temperature of the water. The net heat into the control volume

is 2

2cv burner burner

DQ Q hA T Q h DL T� ��

��

� � � � � �2

2 o

2

W 0.256900 W 3.9 0.256 0.154 m 120 20 C

m K 2cvQ � �

� � � � ��

831.4 WcvQ �� Answer c) The volume of the pot is

� �2

2

DV L� � ��

� �2

0.2560.154

2� � ��

30.00793 m�

The initial volumes of liquid and vapor are

6 - 51

/ 2 and / 2f gV V V V� �

From Table A-11, Therefore 3 30.00106m / kg and 0.8857 m / kg at 200 kPa.f gv v� �

1 3.74 kgf g

f g

V Vm

v v� � � Answer

d) The first law for an open system neglecting kinetic and potential energy changes, is

cvcv cv i i e e

dU Q W m h m hdt

� � � ��

No mass enters the pot and no work is done, therefore

cvcv e e

dU Q m hdt

� ��

� �

� �

cv cvcv e e

cv cv cv e e

d m uQ m h

dtd m u Q dt m h dt

� �

� �# # #

� �

� �

The state of the steam leaving remains the same throughout the process, so he is a constant. If m2 is the mass in the

pot at the end of the process and m1 is the mass at the start

2 2 1 1 cv e e cv e em u m u Q t h m dt Q t h m� � � � � � �#� ��

where me is the total mass that escapes. Using conservation of mass,

�2 2 1 1 2 1cv em u m u Q t h m m� � � � �� Eq. (1)

To find u1, we need the initial quality of the mixture in the pot. Use 3 3

1

1

0.00793m m0.00212

3.74kg kg

Vvm

� � �

With data from Table A-11 at 200 kPa,

1

1

0.00212 0.00106

0.8857 0.00106

f

g f

v vx

v v� �

� ��

0.001197�

� �1 1f g fu u x u u� � � � �504 0.001197 2529 504 507 kJ/kg� � � �

Let y be the fraction of the liquid at the end by volume so that

� �2

1

f g

y VyVmv v

�� Eq. (2)

Additional equations for the final state are

2 /v V m� 2

�f

�f

Eq. (3)

�2 2f gv v x v v� � � Eq. (4)

�2 2f gu u x u u� � � Eq. (5)

Equations (1) through (5) are five equations with five unknowns, i.e., m2, v2, x2, u2, and y. It is not a linear system

of equations because Eq. (3) is non-linear. Therefore, the solution is iterative. One approach is:

Assume m2Find v2 from Eq. (3)

Find x2 from Eq. (4)

Find u2 from Eq. (5)

Check: Is Eq. (1) satisfied? If not, adjust m2. Once m2 is known, find y from Eq. (2). The final results are:

2

2

2

3

2

2.38 kg

509.8kJ/kg

0.002566

0.003326 m /kg

0.32

muxvy

��

��

Therefore, 32% of the pot is filled with water at the end of the process. Answer

6 - 52

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