thermofluids ch6

7/27/2019 Thermofluids ch6

1/52

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

P

T

=

=

1 6615cfmV =

2 40 psia

ft

P =

2 1500s

=V

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

2 2

1 2

1 22 2

h h+ = +V V

The inlet velocity may be found from

( )

( )

3

11 2

2

2

ft 1min

6615 min 60s ft1095

s1ft14.5in.

144in.

V

A

= = =

V

Taking the inlet enthalpy from Table B-12,

2 2 22 21 2

2 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


2/52

6-2 Oxygen at 220oF enters a well-insulated nozzle of inlet diameter 0.6 ft. The inlet velocity is 60 ft/s. Theoxygen leaves at 75oF, 10 psia. The exit area is 0.01767 ft2. Calculate the pressure at the inlet.

Approach:Use the first law for an open system,specialized for a nozzle. Apply conservation

of mass and the ideal gas law to determine

inlet pressure.



3. The nozzle is adiabatic.4. Specific heat is constant.

5. Oxygen behaves like an ideal gas under

these conditions.

o

1 220 Fft

T =

1 60s

=V

o2 75 FT =

2 10 psiaP =


2 2

1 21 2

2 2

h h+ = +V V

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

( ) ( )2 22 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 m

A 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 P

A T

+ = = = +

V

V Answer

6 - 2


3/52

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


specialized for a nozzle. Apply conservation

of mass and the ideal gas law to determineinlet pressure.

Assumptions:1. Potential energy change is negligible.2. The system operates in steady-state.


4. Specific heat is constant.

5. Air behaves like an ideal gas under theseconditions.

1

m65=V

s

2

o

2

101kPa

12 C

P

T

=

=

2

m274

s=V

Solution:

From the first law, specialized for a nozzle,2 21 2

1 22 2

h h+ = +V V

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

( )2 2

2 1

1 22 2

pc T T =

V V

Solving forT1,2 2

2 11 2

2 pT T

c

= +

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, andcorrect later if necessary. From Table A-8,

2

kJAt 273 12 285K, 1.004

kg Kp

T 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 m

A 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 P

A T=

V

V( )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


4/52

6-4 Carbon monoxide enters a nozzle at 520 kPa, 100oC, with a velocity of 10 m/s. The gas exits at 120 kPaand 500 m/s. Assuming no heat transfer and ideal gas behavior, find the exit temperature.

Approach:Use the first law for an open system,specialized for a nozzle.





5. Carbon monoxide behaves like an idealgas under these conditions.

1o

1

520kPa

100 C

P

T

=

=

1

m10

s=V

2 120 kPam

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, butthe 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=100 273 373 K,T + =kJ

1.045 .kg K

pc

2 2

1 22 1

2p

T Tc

= +

V V

2 2

o

m m10 500

s s100 C

kJ 1000 J2 1.045

kg K 1 kJ

= +

o

219.6 CT = Answer

6 - 4


5/52

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


2. The system operates in steady-state.3. The nozzle is adiabatic.

o

1

1

320 C

3MPa

T

P

=

=

1 0V

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,

22 261 2

2 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 ofo

2 274 CT =

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

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 32

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


6/52

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

1 50kPaP =

o

2 300 CT =

2 150 kPaP =

Solution:a) From the first law,

2 2

2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

V V

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

first law becomes2 2

1 21 20

2 2cvQ m h m h

= + + +

V V

The area at the inlet is2 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

mv

A

= = =

V

The exit area is2

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

mv

A= = =

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

= +

107,146W 107.1kWcvQ = = Answer

6 - 6


7/52

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

T

P

=

=

1 250m/s=V

o

2 110 CT =

2 101kPaP =

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

2 2

1 21 2

2 2h h+ = +

V V

For an ideal gas with constant specific heat, ,p

h c T = therefore

( )2 2

2 11 2

2 2p

c T T = V V

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

T = =

( ) ( ) ( )2

22

2 1 2 1 2

kJ 1000J m2 2 1.01 85 110 250

kg K 1kJ sp

c T T

= + = +

V V

2

m110

s=V

Answerb) 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 2

1 2

1 22 2

PM D P M D

RT RT

=

V V

2

2 21 21 1 2 2

1 2

P PD D

T 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 D

P T

+ = = +

V

V Answer

6 - 7


8/52

6-8 Superheated steam enters a well-insulated diffuser at 14.7 psia, 320oF and 400 ft/s. The steam exits assaturated vapor at a very low speed. Find the exit pressure and temperature.

Approach:Use the first law for an open system,eliminating work, heat, and potential energy.

Find properties in the steam tables.



3. The diffuser is adiabatic.

o

1

1

320 F14.7psia

TP

==

1 400ft/s=V2 0V

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,g

h = 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 turbinedevelops 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.3. The system operates in steady-state.

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

T

P

=

=

o

2

2

150 C

0.05MPa

T

P

=

=

With values of enthalpy from Table A-12

( )

3

1 2

24.5 10 kW34.98kg s

3480.6 2780.1kJ kg

Wm

h h

= = =

Answer

6 - 8


9/52

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

Assumptions:1. Potential energy change is negligible.2. Kinetic energy change is negligible.


4. The turbine is adiabatic.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

1000kPa

T

P

=

=

2

2

2

400K

100kPa

30m/s

T

P

=

=

=V


h c T = therefore

( )1 2 1 2p

W Wm

h h c T T = =

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

( )

1000kW kg1.88

kJ s1.063 900 400 K

kg 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

RTv

MP

= = =

Substituting values3

2

2

kg m1.88 1.15

s kg0.0721m

m30

s

A

= =

( )( )2

2

4 0.072140.303m

AD

= = = Answer

6 - 9


10/52

6-11 Saturated steam at 320oC enters a well-insulated turbine. The mass flow rate is 2 kg/s and the exit pressureis 50 kPa. Determine the final state if the power produced is:

a. 100 kW

b. 400 kW




2. Kinetic energy change is negligible.

3. The system operates in steady-state.4. The turbine is adiabatic.



( )1 2W m h h=

o

1 320 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 h

m=

( )1000 W

100 kWkJ 1000 J 1 kW

2700kgkg 1 kJ

2s

=

6 J k2.65 10 2650

kg 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 kW2700kg 1 kJ 2 kg s

h

=

6 J k2.5 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


11/52

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






4. The turbine is adiabatic.

Solution:Assuming an adiabatic turbine with negligible kinetic and


( )1 2W m h h=

o

1

1

600 C

1.6MPa

T

P

=

=

2 50kPaP =

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

2 1

Wh h

m=

( )

1000kW

10MW 1MWkJ kJ3693 2784

kg 11 kg/s kg

= =

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

2 150 CT

2 2

2

Am

v=

V

Solving for velocity and using values ofv2 from Table A-12 at ando

2 150 CT 2 50kPa,P = 3

22 2

2 2

kg m11 3.89

s kg m21.3

s1.6 m

2

mv

A

= = =

V Answer

6 - 11


12/52

6-13 Air at 550oC and 900 kPa is expanded through an adiabatic gas turbine to final conditions of 100 kPa and300oC. 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.



3. The system operates in steady-state.4. The turbine is adiabatic.5. Air behaves like an ideal gas under these conditions.




( )1 2W m h h=

o

1

1

1

550 C

900kPa

30m/s

T

P

=

=

=V

o

2

2

300 C

100kPa

T

P

=

=


( )1 2 1 2p

W Wm

h h c T T = =

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

( ) o1000kW kg

3.72kJ s

1.075 500 300 Ckg K

m = =

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

1

1

1

mvA =

V

From the ideal gas law,

( )

( )

3

11

1

kJ8.314 550 273 K

mkmol K0.262

kg kg28.97 900kPa

kmol

RTv

MP

+ = = =

Substituting values3

2

1

kg m3.72 0.262

s kg0.0325m

m30

s

A

= =

( )( )1

1

4 0.03254

0.203m

AD

= = =

Answer

6 - 12


13/52

6-14 Air at 510C 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.



4. The turbine is adiabatic.5. Air behaves like an ideal gas under these conditions.6. Specific heat is constant.

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

T

P

=

=

2 101kPaP =

1

2 2

1 1

k

kT P

T P

=

From Table A-8, for air 1.4k=

( )

1 1.4 1

1.42

2 1

1

101510 273

450

k

kPT 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 Wm

h 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

Wm

c T T= =

kg

0.173s

= Answer

6 - 13


14/52

6-15 Saturated steam at 3 MPa enters a well-insulated turbine operating in steady state. The turbine produces600 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.



3. The system operates in steady-state.4. The turbine is adiabatic.



( )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 3MPa is

1 2804 kJ kggh h= =

2 1

Wh h

m= +

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= + f104.9 0.93 2547 2442= + ( )( ) 2376 kJ kg=

This is very close to the calculated value of 2375, therefore2 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


15/52

6-16 In a 3-hp compressor, carbon dioxide flowing at 0.023 lbm/s is compressed to 120 psia. The gas enters at60oF 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.



3. The system operates in steady-state.4. The compressor is adiabatic.5. Carbon dioxide behaves like an ideal gas under these

conditions.6. Specific heat is constant.

Solution:Assuming an adiabatic compressor with negligible kinetic and


( )1 2W m h h=

o

1

1

60 F

14.7psia

T

P

=

=

2 120psiaP =


( )1 2

2 1

p

p

W m c T T

WT T

m c

=

=

With data from Table B-8

( )

o o

2

Btu2544

h3hp1hp

60 F 533 F

lbm Btu 3600s0.023 0.195s lbm R 1h

T

= =

Answer

From conservation of mass

2 2 21 2

2 2

V V P M m m

v RT= = =

( )

( )

3

3

1 22

2

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

s lbmol R 1min ft2.78

lbm min120psia 44.01

lbmol

m RTV

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

6 - 15


16/52

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

m= +

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


17/52

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.




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


Solution:From the first law for an open system

2 2

2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

V V

Assuming steady conditions, one stream in and one stream out, nochange 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

T

P

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 PM m

v RT

= = = =

+

The average temperature of the air is

o25 300162 C 435K

2ave

T+

= = =

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

( )3 okg J

250 W 9.84 10 1018 25 300 C 3000 W 3kWs kg K

W = + = =

Answer

6 - 17


18/52

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

Vm

v=

o

1

1

3

1

0 F

10psia

15ft /min

T

P

V

=

=

=

o

2

2

140 F

70psia

T

P

=

=

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

3

3

ft15lbmmin 3.19

ft min4.703

lbm

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

= +

Btu76.0

h

Q = Answer

6 - 18


19/52

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 thewater 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= .

Assumptions:

1. Potential energy change is negligible.2. Kinetic energy change is negligible.


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 P

W m

=

( )

3

10kPa 700kPakg 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 thevelocity is constant, the water is incompressible, and the flow is frictionless. If the mass flow rate is 6lbm/sec, find the final pressure.

Approach:


Assumptions:

1. Potential energy change is negligible.


4. The pump is ideal.

5. Water is incompressible.


1 2( )W mv P P=

2 1

WP P

mv

=

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

( )2

2

2 2 3

ft lbf 550

1fts2hp1 hp 144in.

lbf5

in. lbm ft6 0.0164

s lbm

P

=

2 82.6 psiaP = Answer

6 - 19


20/52

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

velocity changes.

Approach:







1 21 2

( )( )

m P PW mv P P

= =

The mass flow rate is

m A= V

2

3kg m 0.012m kg998 12 1.35m s 2

= = s

2 1

WP P

m

=

( )

3

1000 W kg6kW 998

1000Pa 1 kW m30kPa

kg1 kPa1.35

s

64.45 10 Pa 4.45 MPa= =

=

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 elevationchange 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.

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

.




4. The pump is ideal.5. The oil is incompressible.


1 2( )W mv P P=

1 2

W WP P

mv m

= =

( )2

3 2

1 2

ft lbf 550

lbm 1fts56 1hpft 1 hp 144in.

lbm10

s

P P

= 21.4psia= Answer

6 - 20


21/52

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

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

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 2 1 2P P

W 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 hp2.3lbm lbf fts

62.1 550ft 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


22/52

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

P

T

=

=

=V2 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

From conservation of mass1 2

1 1 2 2 2

m m

A 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 RT

PM P M =V V

Solving for exit velocity,

22 1

1

m 400 m3.6 14.4

s 100 s

P

P

= = =

V V Answer

6 - 22


23/52

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





o

1 24 CT =

saturated liquid2 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 ofx2 is too high, so a different value ofT2 is chosen. By trial and error, the final value ofT2 is found to

beo

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


24/52

6-27 Saturated liquid R-134a at 80oF undergoes a throttling process. The pressure decreases to of its originalvalue. Find the exit quality.





o

1 80 FT =

saturated liquid2 1 / 4P P=

Solution:From the first law applied to a throttle,

1 2h h=

From Table B-14,

If theno

80 FsatT = 101.4psiasatP =

1 1

BtuAt 101.4, 37.3

lbmfP h h= = =

12 101 25.2psia

4 4PP = = =

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


25/52

6-28 A supply line contains a two-phase mixture of steam and water at 240C. To determine the quality of themixture, 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 125C, what is the quality of the mixture in themain 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 = 240CkJ kJ

1037 2804kg kg

f 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


26/52

6-29 In a heat pump, R-134a is throttled through an expansion coil, which is a long copper tube of smalldiameter. 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/m2oC to the surroundings at 20oC. The expansion coil has anoutside 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

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

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 56.7kJ/kg.h = The heat lost from the

surface of the coil is

( )ave surr Q 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 2 5 10.1 2.55 C2 2

aveT TT + += = =

The surface area of the coil is

( ) ( ) 21m

8mm 2.2m 0.0553m1000mm

A DL

= = =

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

( ) ( )( )2 o2 oW

6 0.0553m 2.55 20 C 7.48Wm C

ave surr Q 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 h

m

= + = + = =

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


27/52

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.



3. The system operates in steady-state.4. The tank is adiabatic.

Solution:From conservation of mass

1 2m m m+ = 3

From the first law for an open system2 2

2 2

cv i e

cv cv i i i e e e

dE

Q W m h gz m h gzdt

= + + + + +

V V

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


28/52

6-31 In a desuperheater, superheated steam is converted to saturated steam by spraying liquid water into thesteam. Using data on the figure, calculate the mass flow rate of liquid water.

Approach:Apply conservation of mass and the firstlaw for an open system.




4. The desuperheater is adiabatic.5. Pressure is constant during the

process.


1 2m m m+ = 3


2 2

cv i e

cv cv i i i e e e

dE

Q W m h gz m h gzdt

= + + + + +

V V


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


29/52

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

cv i e

cv cv i i i e e e

dE

Q W m h gz m h gzdt

= + + + + +

V V


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 83.96kJ 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,

3 167.57kJ kgh =

Using conservation of mass to eliminate in the first law1m

( )3 2 1 2 2 3

3 1 2 1 2 2 3 3

0

0

m m h m h m h

m 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


30/52

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

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

V V

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

1 20 Q mh mh= +

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

( ) ( )1 251.4 0.88 2610 251.4 2327 kJ kgf g fh h x h h= + = + = For state 2, the enthalpy of the subcooled liquid may be approximated by the enthalpy of saturated liquid at 55oC,which is, from Table A-10,

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 stube

m A

= = =

V

The total mass flow rate for all 20 tubes is

( )kg

20 20 0.512 10.24s

tubem m= = =

From the first law

( ) ( )2 1kg kJ

10.24 230 2327 21, 478kWs kg

Q m h h= = = Answer

6 - 30


31/52

6-34 Saturated steam at 120oF is condensed in a tube, as shown. Cooling water at 50oF flows in crossflow overthe exterior of the pipe, giving a heat transfer coefficient of 200 Btu/hft2oF. Find the exit quality.

Approach:Use the thermal resistance analogy todetermine 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 thecondensing steam temperature.

Solution:Apply the thermal resistance analogy tofind the heat removed from the pipe.

Assuming that the condensing heat transfer

coefficient is very large, the interior wall ofthe 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

TQ

R

= = =


2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

V V

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

becomes

1 2

0 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 h

m

= + = + =

Using data in Table A-10,

2 735 880.63

1113.5 88

f

g f

h hx

h h

= = =

Answer

6 - 31


32/52

6-35 Superheated R-134a enters a well-insulated heat exchanger at 0.7 MPa, 70oC. It exits as saturated liquid at0.7 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.


3. The system operates in steady-state.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 V

v v

= = = = =


2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

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 a

m 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


33/52

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


2 2

2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

V V

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

p

h hm m

c T T

=

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

( )

( ) o

kJ

254.03 66.18kg kgkg5 93.5

s skJ1.005 25 15 C

kg K

am = =

Answer

6 - 33


34/52

6-37 Superheated steam at 5 psia and 200F is condensed in a heat exchanger. The steam flows at 39 lbm/s andexits as saturated liquid. Cooling water at 45F is used to condense the steam. The water and steam are notmixed in the heat exchanger, but enter and leave as separate streams. If the maximum allowable water

temperature rise is 15F and the maximum allowable water velocity is 11 ft/s, what is the diameter of thepipe which carries water to the heat exchanger?





4. The condenser is adiabatic.5. Pressure is constant during the process.

6. The cooling water is incompressible.7. The specific heat of the water is constant.


2 2

2 2

cv i e

cv cv i i i e e edE Q W m h gz m h gzdt

= + + + + +

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, and1m 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

=

lbm

2646s

=

To find the diameter

3m A= V2

2

D

=

V

32m

D

=

V

3

lbm2646

s2ft lbm

11 62.4s ft

=

2.21ft= Answer

6 - 34


35/52

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




3. The system operates in steady-state.4. The condenser is adiabatic.5. Pressure is constant during the process.

6. The river water is incompressible.7. The specific heat of the river water is

constant.


2 2

2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

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-11o

2 162.2 FT =

4 2 70T T= o

92.2 F=

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

( ) ( )1Btu

130.2 0.93 1131 130.2 1061lbm

f 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 lbmBtu

1.00 92.2 45 Flbm F

m h hmh 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


36/52

6-39 A heat exchanger is used to cool engine oil. The specific heat of the oil is 0.6 Btu/lbmoF. Using data onthe figure below, find the exit temperature of the air.


Assumptions:

1. Potential energy change is negligible.2. Kinetic energy change is negligible.

3. The system operates in steady-state.4. The heat exchanger is adiabatic.5. Pressure is constant during the process.

6. The oil is incompressible.7. The specific heat is constant.

8. Air behaves like an ideal gas under these

conditions.


2 2

2 2

cv i e

cv cv i i i e e e

dE

Q W m h gz m h gzdt

= + + + + +

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 0.24s lbm R

p oil

p air

m c T T T T

m c

= + = + =

Answer

6 - 36


37/52

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



3. The system operates in steady-state.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

kgf

h 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 law2 2

2 2

cv i e

cv cv i i i e e e

dEQ W m h gz m h gz

dt

= + + + + +

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

R

w

m h h m h h

m h hm

h h

= +

=

Assuming the water is an ideal liquid

( )

( )3 2

4 5

R

w

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-phaseregion, 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,

( )( )

( )( ) o0.013 kg s 242.5 100.3 kJ kg kg

0.04414.187 kJ kg K 20 10 C s

wm

= =

Answer

6 - 37


38/52

6-41 In a flash chamber, a pressurized liquid is throttled to a lower pressure where it becomes a two-phasemixture. The saturated liquid and vapor streams are removed in separate lines. In the figure below, liquid

R-134a at 10F 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 twocomponents.




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 acontrol volume around the combination of the two

components. You could also analyze eachcomponent 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 theenthalpy of the liquid as the enthalpy of the

saturated liquid at the same temperature. From

Table B-14

h1 = hfat 10F = 14.66 Btu/lbm

From Table B-15, the enthalpy of states 2 and 3 ish2 = hg at 5 psi = 93.79 Btu/lbm

h3 = hfat 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 3lbm Btu lbm Btu lbm Btu

21.6 14.66 21.6 93.79 3.73h lbm h lbm h lbm

m m

+ = +

Solving for gives3m

3

lbm92.9

hm =

From conservation of mass

1

lbm lbm21.6 92.9

h hm = +

lbm114.5

h= Answer

6 - 38


39/52

6-42 Saturated liquid water at 40 kPa enters a 140 kW pump. The output of the pump is fed into a boiler whereheat 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 theboiler, since there is negligible pressure drop

across the boiler. Draw a control volume

around both components and apply the first lawto 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 1cvWP P

mv=

( )

3

1000W140kW

1000Pa 1 kW40kPa

1 kPa kg m70 0.001027

s kg

61.99 10 Pa 2.0 MPa=

=

Pressure is constant across the boiler, so

3 2 2.0MPaP P= = Answer

To find the state at the exit of the boiler, draw acontrol 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 1cv cvQ Wh h

m

= +

302,000kW ( 140)kW kJ318

kg kg70

s

= +

3kJ4634kg

h =

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


40/52

6-43 Air at 2000 R enters the turbine of a turbojet engine. The turbine is well insulated and produces 100 Btu ofwork 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.3. The system operates in steady-state.

4. The nozzle and turbine are adiabatic.5. Air behaves like an ideal gas under these

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=


h 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 forT

2000Ro

1500 F

2 and using values ofcp from Table B-8 at 1500F,

2 1

100Btu lbm2000R 1638R

Btu0.276

lbm R

cv

p

WT T

m c= = =

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

322 2 3

2 2

cv

cv cv

dEQ W m h gz m h gz

dt

= + + + + +

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


h c T = therefore

( )2 2

2 32 30

2p

c 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 1178F and we expect the outlet temperature to be lower.For simplicity, we assume an average temperature of 1000F and will correct the calculation later if necessary.Using data from Table B-8,

( )2 2 2 23

2

20 - 2800 ft s 1Btu 1lbf 1638R 1043R

ft lbm778ft lbfBtu 32.22 0.263slbmR

T

= +

=

From the ideal gas law:

6 - 40


41/52

( )

( )

2

2

33

323

ft lbf 1ft1545 1043R

lbmol R 144in.38.6ft lbm

lbm28.97 10lbf in

lbmol

RTv

MP

= = =

The mass flow rate may now be calculated as

( ) ( )23 3

3

3

2800 ft s 0.6 ft43.5lbm s

38.6ft lbm

Am

v

= = =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


42/52

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


3. The tank is adiabatic.

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

5. The specific heat is constant.

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

cv

cv cv i i e e

dUQ W m h m h

dt

= +

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

cv

i i

dUm h

dt=

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 temperaturethroughout 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,v

u c T =

2( )i

v i i

RTc T T Pv

M = =

Solving for 2T

2i

i

v

RTT T

Mc= +

( )kJ

8.314 20 273 K kmol K

(20 273)K kg kJ

28.97 0.718kmol kg K

+ = +

+

2 410KT =o

137 C= Answer

6 - 42


43/52

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

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

e

P MV P MV m

RT RT= 1 1 2( )

MVP P

RT=

( )

( )

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

cv

cv cv i i e e

dUQ W m h m h

dt= +

The tank is adiabatic and nothing is entering, therefore

cv

cv e e

dUW m h

dt

=

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 andeh

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 2e

cv

RTW m m e e

m RT

M=

M=

( ) ( )ft lbf

0.061lbm 1545 150 460 R 1Btulbm R

lbm 778ft lbf4

lbmol

+ =

18.5BtucvW = Answer

6 - 43


44/52

6-46 A residential hot water heater initially contains water at 140oF. Someone turns on a shower and drawswater 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 contentsare at the same temperature.

Solution:The first law for an open system with one stream in and one stream out, neglecting kinetic and potential energychanges, is

cv

cv i i e e

dUQ W m h m h

dt= +

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

( )( )cv cv cv 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 edu

m Q mc T dt

= + 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 formv pc c

( )cv p cv p idT

m c Q mc T T dt

= +

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 T

d mc dT

= +

=

2

12cv p

p

dm c t

mc

=

2

1 2lncv

mt

m

=

( )2 1ln lncvm t

m

2=

22

1

lncvm

tm

=

6 - 44


45/52

( )

( )2

2

1

lncv p icv

cv p i

Q mc T T mt

m Q mc T T

+ = +

Find cvm

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

2

3 2

o

o

2

1.862lbm ft ft 4ft 631lbm

2

1h Btu5472Btu h + 0.2lbm s 1 50 100 F2600s lbm F631 lbm

ln10.2lbm s

5472 + 0.2 1 50 1403600

cvm V

t

= = =

=

2 2096s 35mint = = Answer

6 - 45


46/52

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

cv

cv cv i i e e

dUQ W m h m h

dt= +

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

( )

( ) ( )

cv

cv A A B B A B C

cv

cv A A C B B C

dum W m h m h m m h

dtdu

m 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 T

dt

dTm c W m c T m c T m c m c T

dt

= + +

= + + +

A p A B p B A B

cv p cv

W m c T m c T m mdTT

dt m c m

+ + +=

Define K1 and K2 so that

1 2

dTK K T

dt=

Separating variables and integrating

2

1 01 2

T

T

dTdt

K K T=

2t

1 2

let

K K T =

2d K dT =

6 - 46


47/52

2

1

2

1

2

2

2

2

2 12

2 1 2 1 2 1

1ln

1 1ln ln

dt

K

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

Wm T m T m m T

cmt

Wm mm T m T m m T

c

+ + +

= + + + +

Answer

6 - 47


48/52

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



3. The tank is adiabatic.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

cv

cv cv i i e e

dUQ W m h m h

dt= +

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

cv

i idU m h

dt=

( )

( )

cv cv

i i

cv cv i i

d m um h

dt

d m u m h dt

=

=

The state of the steam entering remains the same throughout the process, so hi is a constant. Ifm2 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, and1 0,m =

2 2 2

2

i

i

m u h m

u 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 is3

2 3

2

0.035m0.0233kg

m1.5

kg

Vm

v= = = Answer

6 - 48


49/52

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

cv cv i i e e

dUQ W m h m h

dt= +

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

cv

cv i i

dUW m h

dt= +

( )

( )

cv cv

cv i i

cv cv cv i i

d m uW m h

dt

d 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 substitutin

thermofluids ch6

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