hw 9 solutions -chapter 7

8/11/2019 HW 9 Solutions -Chapter 7

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HW- Chapter 7- Solutions

(7 th edition)

33,36,44,51,56

7-33 A rigid tank is divided into two equal parts by a partition. One part is filled with compressedliquid water while the other side is evacuated. The partition is removed and water expands into theentire tank. The entropy change of the water during this process is to be determined.

Analysis The properties of the water are (Table A-4)

K kJ/kg0.8313

/kgm0.001017

C60

kPa004

C60@1

3C60@1

1

1

f

f

s sT

P v v

Noting that

/kgm0.002034001017.022 312 v v

K kJ/kg0278.16430.60002524.00261.1

0002524.0001026.0993.3

001026.0002034.0

/kgm340020.0

kPa40

22

22

32

2

fg f

fg

f

s x s s

x P v

v v

v

Then the entropy change of the water becomes

kJ/K0.492K kJ/kg0.83131.0278kg2.512 s smS

7-36 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The wateris heated electrically at constant pressure. The entropy change of the water during this process is to bedetermined.

Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Thecompression or expansion process is quasi-equilibrium.

Analysis From the steam tables (Tables A-4 through A-6),

K kJ/kg1.4337

kJ/kg467.13

/kgm0.001053

.

kPa150

kPa150@1

kPa150@1

3kPa150@1

1

f

f

f

s s

hhliquid sat

P v v

Also, kg4.75/kgm0.001053

m0.0053

3

1v

V m

We take the contents of the cylinder as the system. This is a closed system since no mass enters orleaves. The energy balance for this stationary closed system can be expressed as

)( 12ine,

out b,ine,

energiesetc. potential, kinetic,internal,inChange

system

massandwork,heat, bynsfer energy tra Net

outin

hhmW

U W W

E E E

2.5 kg

compressed liquid

Vacuum

H2O

150 kPa2200 kJ


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since U + W b = H during a constant pressure quasi-equilibrium process. Solving for h 2,

kJ/kg33.930kg4.75kJ2200

13.467ine,12 mW

hh

Thus,

K kJ/kg6384.27894.52081.04337.1

2081.00.2226

13.46733.930

kJ/kg33.930

kPa150

22

22

2

2

fg f

fg

f

s x s sh

hh xh

P

Then the entropy change of the water becomes

kJ/K 5.72K kJ/kg1.43372.6384kg4.7512 s smS 7-44 A cylinder is initially filled with saturated water vapor at a specified temperature. Heat istransferred to the steam, and it expands in a reversible and isothermal manner until the pressure dropsto a specified value. The heat transfer and the work output for this process are to be determined.

Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulated

and thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible and isothermal.

Analysis From the steam tables (Tables A-4 through A-6),

K kJ/kg6.8177

kJ/kg2631.1kPa800

K kJ/kg6.4302

kJ/kg2594.2

.

C200

2

2

12

2

C200@1

C200@11

s

u

T T

P

s s

uu

vapor sat

T

g

g

The heat transfer for this reversible isothermal process can bedetermined from

kJ 219.9K kJ/kg)6.43026.8177)(kg1.2)(K 473(12 s sTmS T Q We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. Theenergy balance for this closed system can be expressed as

)(

)(

12inout b,

12out b,in

energiesetc. potential, kinetic,internal,inChange

system

massandwork,heat, bynsfer energy tra Net

outin

uumQW

uumU W Q

E E E

Substituting, the work done during this process is determined to be

kJ 175.6kJ/kg)2594.22631.1)(kg1.2(kJ9.219out b,W

7-51 Steam is expanded in an isentropic turbine. The work produced is to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 The process isisentropic (i.e., reversible-adiabatic).

H2O

200 C

sat. vaporQ


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Analysis There is one inlet and two exits. We take the turbine as the system, which is a control volumesince mass crosses the boundary. The energy balance for this steady-flow system can be expressed inthe rate form as

outin

energiesetc. potential, kinetic,internal,inchangeof Rate

(steady) 0system

massandwork,heat, by

nsfer energy tranetof Rate

outin 0

E E

E E E

332211out

out332211

hmhmhmW

W hmhmhm

From a mass balance,

kg/s75.4)kg/s5)(95.0(95.0

kg/s25.0)kg/s5)(05.0(05.0

13

12

mm

mm

Noting that the expansion process is isentropic, theenthalpies at three states are determined as follows:

6)-A(TableK kJ/kg6953.7

kJ/kg4.2682

C100

kPa50

3

3

3

3 s

h

T

P

6)-A(Table kJ/kg3.3979 KkJ/kg6953.7

MPa41

31

1 h s s

P

6)-A(Table kJ/kg1.3309 KkJ/kg6953.7

kPa7002

32

2 h s s

P

Substituting,

kW6328

kJ/kg).4kg/s)(268275.4(kJ/kg).1kg/s)(330925.0(kJ/kg).3kg/s)(39795(332211out

hmhmhmW

7-56 Refrigerant-134a is compressed in an adiabatic compressor reversibly. The process is to besketched on the T-s diagram and the volume flow rate at the inlet is to be determined.

Assumptions The process is steady.

Analysis (b) Noting that the process is isentropic(constant entropy) the inlet and exit states are

obtained from R-134a tables (Tables A-12 and A-13) as follows:

/kgm06360.0

K kJ/kg9301.0kJ/kg88.251

1

kPa320

31

1

1

1

1

v

sh

x

P

kJ/kg35.279 KkJ/kg9301.0

kPa12002

12

2 h s s

P

T

s

3

1

4 MPa

0.7 MPa

2

Steam

4 MPa

50 kPa

700 kPa

T

s

1

21200

320


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We take the compressor as the system, which is a control volume since mass crosses the boundary.Noting that one fluid stream enters and leaves the compressor, the energy balance for this steady-flowsystem can be expressed in the rate form as

)(

0) peke(since

0

12in

2in1

outin

energiesetc. potential, kinetic,internal,inchangeof Rate

(steady) 0system

massandwork,heat, bynsfer energy tranetof Rate

outin

hhmW

QhmW hm

E E

E E E

Solving for the mass flow rate and substituting,

kg/s640.3)kJ/kg88.251(279.35

kW100

12

in

hhW

m

The volume flow rate at the inlet is then,

/sm0.232 3)/kgm06360.0)(kg/s640.3( 311 v V m

hw 9 solutions -chapter 7

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