8-95 1 steady operating conditions exist. 2 kinetic …...8-79 review problems 8-95 refrigerant-134a...

8-79

Review Problems 8-95 Refrigerant-134a is expanded adiabatically in an expansion valve. The work potential of R-134a at the inlet, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The properties of the refrigerant at the inlet and exit of the valve and at dead state are (Tables A-11 through A-13)

180 kPaR-134a 1.2 MPa

40°C

kJ/kg.K 0918.1kJ/kg 17.272

C20kPa 100

kJ/kg.K 42271.0kJ/kg 23.108

kPa 180

kJ/kg.K 39424.0kJ/kg 23.108

C40MPa 2.1

0

0

0

0

212

2

1

1

1

1

==

°==

=

===

==

°==

sh

TP

shh

P

sh

TP

The specific exergy of the refrigerant at the inlet and exit of the valve are kJ/kg 40.55=+−=−−−= kg.K1.0918)kJ/-K)(0.39424 273.15(20-kJ/kg)17.27223.108()( 010011 ssThhψ

kJ/kg 32.20kJ/kg.K 1.0918K)(0.42271 273.15(20kJ/kg)17.27223.108()( 020022 =−+−−=−−−= ssThhψ

(b) The exergy destruction is determined to be kJ/kg 8.34=+=−= /kg.K0.39424)kJ-K)(0.42271 273.15(20)( 120dest ssTx

(c) The second-law efficiency for this process may be determined from

0.794===kJ/kg 55.40kJ/kg 20.32

1

2II ψ

ψη

PROPRIETARY MATERIAL. © 2006 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.

8-80

8-96 Steam is accelerated in an adiabatic nozzle. The exit velocity, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Potential energy changes are negligible. Analysis (a) The properties of the steam at the inlet and exit of the turbine and at the dead state are (Tables A-4 through A-6)

kJ/kg.K 2678.0kJ/kg 54.75

0C18

kJ/kg.K 6753.6kJ/kg 9.2919

C250kPa 6.1

kJ/kg.K 4484.6kJ/kg 4.2978

C300MPa 5.3

0

00

2

2

2

2

1

1

1

1

==

=°=

==

°==

==

°==

sh

xT

sh

TP

sh

TP

Steam 3.5 MPa300°C

1.6 MPa 250°C Vel2

The exit velocity is determined from an energy balance on the nozzle

m/s 342.0=

+=

+

+=+

2

22

22

22

2

22

2

21

1

/sm 1000kJ/kg 1

2V

kJ/kg 9.2919/sm 1000

kJ/kg 12

m/s) (0kJ/kg 4.2978

22

V

Vh

Vh

(b) The rate of exergy destruction is the exergy decrease of the steam in the nozzle

kW 26.41=

−−

−+−

=

−−

−+−=

kJ/kg.K)4484.66753.6)(K 291(/sm 1000

kJ/kg 12

0m/s) (342kg2978.4)kJ/(2919.9kg/s) 4.0(

(2

22

2

120

21

22

12dest ssTVV

hhmX &&

(c) The exergy of the refrigerant at the inlet is

[ ]kW 72.441

kJ/kg.K)2678.04484.6)(K 291(0kJ/kg )54.75(2978.4kg/s) 4.0(

(2 010

21

011

=−−+−=

−−+−= ssTVhhmX &&

The second-law efficiency for this device may be defined as the exergy output divided by the exergy input:

0.940=−=−==kW 72.441

kW 41.26111

dest

1

2II X

XXX

&

&

&

&η


8-81

8-97 An electrical radiator is placed in a room and it is turned on for a period of time. The time period for which the heater was on, the exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constant specific heats. 3 The room is well-sealed. 4 Standard atmospheric pressure of 101.3 kPa is assumed. Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K (Table A-2). The properties of oil are given to be ρ = 950 kg/m3, coil = 2.2 kJ/kg.K. Analysis (a) The masses of air and oil are

kg 36.62K) 273K)(10/kgmkPa (0.287

)m kPa)(50 (101.33

3

1

1 =+⋅⋅

==RTP

maV

kg 50.28)m )(0.030kg/m (950 33oiloiloil === Vρm

An energy balance on the system can be used to determine time period for which the heater was kept on

[ ] [ ][ ] [

min 34s 2038 ==∆°−°+°−°=∆−

−+−=∆−

tt

TTmcTTmctQW a

C)10C)(50kJ/kg. kg)(2.2 50.28(C)10C)(20kJ/kg. kg)(0.718 36.62(kW) 35.08.1()()()( oil1212outin v

&&

]

Room

Radiator

Q10°C

(b) The pressure of the air at the final state is

kPa 9.104m 50

K) 273K)(20/kgmkPa kg)(0.287 (62.363

32

2 =+⋅⋅

==V

aaa

RTmP

The amount of heat transfer to the surroundings is kJ 5.713s) kJ/s)(2038 (0.35outout ==∆= tQQ &

The entropy generation is the sum of the entropy changes of air, oil, and the surroundings

kJ/K 5548.1kPa 101.3kPa 104.9kJ/kg.K)ln (0.287

K 273)(10K 273)(20

kJ/kg.K)ln (1.005kg) (62.36

lnln1

2

1

2

=

−

++

=

−=∆

PP

RTT

cmS pa

kJ/K 2893.8K 273)(10K 273)(50kJ/kg.K)ln kg)(2.2 (28.50ln

1

2oil =

++

==∆TT

mcS

kJ/K 521.2K 273)(10

kJ 713.5

surr

outsurr =

+==∆

TQ

S

kJ/K 365.12521.22893.85548.1surroilagen =++=∆+∆+∆= SSSS

The exergy destruction is determined from kJ 3500=+== kJ/K) K)(12.365 273(10gen0dest STX

(c) The second-law efficiency may be defined in this case as the ratio of the exergy recovered to the exergy input. That is,

[ ][ ] kJ 729.7kJ/K) K)(1.5548 27310(C10)-C)(20kJ/kg. (0.718kg) (62.36

)( 0122,

=+−°°=

∆−−= aa STTTcmX v

[ ][ ] kJ 13.162kJ/K) K)(8.2893 27310(C10)-C)(50kJ/kg. (2.2kg) (28.50

)( 0122,oil

=+−°°=

∆−−= aSTTTCmX

4.6%0.046 ==+

=∆

+==

s) kJ/s)(2038 (1.8kJ 162.13)(7.729

in

oil,2,2

supplied

recovered

tW

XXXX a

II &η


8-82

8-98 Hot exhaust gases leaving an internal combustion engine is to be used to obtain saturated steam in an adiabatic heat exchanger. The rate at which the steam is obtained, the rate of exergy destruction, and the second-law efficiency are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air properties are used for exhaust gases. 4 Pressure drops in the heat exchanger are negligible. Properties The gas constant of air is R = 0.287 kJkg.K. The specific heat of air at the average temperature of exhaust gases (650 K) is cp = 1.063 kJ/kg.K (Table A-2). Analysis (a) We denote the inlet and exit states of exhaust gases by (1) and (2) and that of the water by (3) and (4). The properties of water are (Table A-4)

350°CHeat Exchanger

Sat. vap.200°C

Water20°C

Exh. gas400°C 150 kPa

kJ/kg.K 4302.6kJ/kg 0.2792

1C200

kJ/kg.K 29649.0kJ/kg 91.83

0C20

4

4

4

4

3

3

3

3

==

=°=

==

=°=

sh

xT

sh

xT

An energy balance on the heat exchanger gives

kg/s 0.01570=−=°−°

−=−+=+

w

w

wpa

wawa

mm

hhmTTcmhmhmhmhm

&

&

&&

&&&&

kJ/kg)91.830.2792(C)350400(C)kJ/kg 3kg/s)(1.06 8.0(

)()( 3421

4231

(b) The specific exergy changes of each stream as it flows in the heat exchanger is

kJ/kg.K 08206.0K 273)(400K 273)(350kJ/kg.K)ln 3kg/s)(1.06 (0.8ln

1

2 −=++

==∆TT

cs pa

kJ/kg 106.29kJ/kg.K) 6K)(-0.0820 273(20C400)-C)(350kJ/kg. 063.1(

)( 012

−=+−°°=

∆−−=∆ apa sTTTcψ

kJ/kg 913.910)kJ/kg.K29649.0K)(6.4302 273(20kJ/kg)91.830.2792(

)( 34034

=−+−−=

−−−=∆ ssThhwψ

The exergy destruction is determined from an exergy balance on the heat exchanger to be

or

kW 8.98=

−=+=∆+∆=−

dest

dest kW 98.8kJ/kg )913.910)(kg/s 01570.0(kJ/kg) 106kg/s)(-29. 8.0(

X

mmX wwaa

&

&&& ψψ

(c) The second-law efficiency for a heat exchanger may be defined as the exergy increase of the cold fluid divided by the exergy decrease of the hot fluid. That is,

0.614=−

=∆−∆

=kJ/kg) 106kg/s)(-29. 8.0(

kJ/kg) 913.910)(kg/s 01570.0(II

aa

ww

mm

ψψη

&

&


8-83

8-99 The inner and outer surfaces of a brick wall are maintained at specified temperatures. The rate of exergy destruction is to be determined. Assumptions 1 Steady operating conditions exist since the surface temperatures of the wall remain constant at the specified values. 2 The environment temperature is given to be T0 = 0°C. Analysis We take the wall to be the system, which is a closed system. Under steady conditions, the rate form of the entropy balance for the wall simplifies to

30

Q Brick Wall

{

W/K0.166 0K 278

W900K 293

W900

0

0

wallgen,wallgen,

wallgen,outb,

out

inb,

in

entropy of change of Rate

0system

generation entropy of Rate

gen

mass andheat by ansferentropy trnet of Rate

outin

=→=+−

=+−

=∆=+−

SS

STQ

TQ

SSSS

&&

&&&

43421&&

43421&&

5°C20°C

The exergy destroyed during a process can be determined from an exergy balance or directly from its definition , gen0destroyed STX =

W45.3=== ) W/K166.0(K) 273(gen0destroyed STX &&

8-100 A 1000-W iron is left on the iron board with its base exposed to air. The rate of exergy destruction in steady operation is to be determined. Assumptions Steady operating conditions exist. Analysis The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the iron and its immediate surroundings so that the boundary temperature of the extended system is 20°C at all times. It gives

{

0

0

genoutb,

out


0system


gen


outin

=+−

=∆=+−

STQ

SSSS

&&

43421&&

43421&&

Therefore,

W/K3.413K 293 W1000

0outb,

outgen ====

TQ

TQ

S&&

&


W1000=== ) W/K413.3(K) 293(gen0destroyed STX &&

Discussion The rate of entropy generation within the iron can be determined by performing an entropy balance on the iron alone (it gives 2.21 W/K). Therefore, about one-third of the entropy generation and thus exergy destruction occurs within the iron. The rest occurs in the air surrounding the iron as the temperature drops from 150°C to 20°C without serving any useful purpose.


8-84

8-101 The heating of a passive solar house at night is to be assisted by solar heated water. The amount of heating this water will provide to the house at night and the exergy destruction during this heat transfer process are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the glass containers themselves is negligible relative to the energy stored in water. 3 The house is maintained at 22°C at all times. 4 The outdoor temperature is given to be 5°C. Properties The density and specific heat of water at room temperature are ρ = 997 kg/m3 and c = 4.18 kJ/kg·°C (Table A-3). Analysis The total mass of water is

( )( ) kg 95.348m 0.350kg/m 997 33 === Vρwm

water45°C

22°C

The amount of heat this water storage system can provide is determined from an energy balance on the 350-L water storage system

water12systemout

energies etc. potential, kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

)( TTmcUQ

EEE

−=∆=−

∆=−4342143421

Substituting, kJ 33,548=C22)-C)(45kJ/kg kg)(4.18 (348.95out °°⋅=Q

The entropy generated during this process is determined by applying the entropy balance on an extended system that includes the water and its immediate surroundings so that the boundary temperature of the extended system is the environment temperature at all times. It gives

{

watergenoutb,

out

entropyin Change

system

generationEntropy

gen

mass andheat by ansferentropy trNet

outin

SSTQ

SSSS

∆=+−

∆=+−4342143421

Substituting,

( )( )

KkJ 215.4K 295

kJ 33,548K 318K 295

lnKkJ/kg 4.18kg 348.95

lnroom

out

water1

2

outb,

outwatergen

/=

+⋅=

+

=+∆=

TQ

TT

mcTQ

SS


kJ 1172=== )kJ/K 215.4(K) 278(gen0destroyed STX


8-85

8-102 The inner and outer surfaces of a window glass are maintained at specified temperatures. The amount of heat loss and the amount of exergy destruction in 5 h are to be determined Assumptions Steady operating conditions exist since the surface temperatures of the glass remain constant at the specified values. Analysis We take the glass to be the system, which is a closed system. The amount of heat loss is determined from Jk 57,600=×=∆= s) 3600kJ/s)(5 (3.2tQQ &

Under steady conditions, the rate form of the entropy balance for the glass simplifies to

{

W/K 0.2868 0K 276W 3200

K 283W 3200

0

0

glassgen,wallgen,

glassgen,outb,

out

inb,

in


0system


gen


outin

=→=+−

=+−

=∆=+−

SS

STQ

TQ

SSSS

&&

&&&

43421&&

43421&&

Glass

3°C10°C Then the amount of entropy generation over a period of 5 h becomes J/K 5162s) 3600K)(5/W 2868.(0glassgen,glassgen, =×== tSS ∆&



Discussion The total entropy generated during this process can be determined by applying the entropy balance on an extended system that includes the glass and its immediate surroundings on both sides so that the boundary temperature of the extended system is the room temperature on one side and the environment temperature on the other side at all times. Using this value of entropy generation will give the total exergy destroyed during the process, including the temperature gradient zones on both sides of the window. 8-103 Heat is transferred steadily to boiling water in the pan through its bottom. The inner and outer surface temperatures of the bottom of the pan are given. The rate of exergy destruction within the bottom plate is to be determined. Assumptions Steady operating conditions exist since the surface temperatures of the pan remain constant at the specified values. Analysis We take the bottom of the pan to be the system, which is a closed system. Under steady conditions, the rate form of the entropy balance for this system can be expressed as

{

W/K 0.00561 0K 377W 800

K 378W 800

0

0

systemgen,systemgen,

systemgen,outb,

out

inb,

in


0system


gen


outin

=→=+−

=+−

=∆=+−

SS

STQ

TQ

SSSS

&&

&&&

43421&&

43421&&

800 W

104°C


105°C



8-86

8-104 A elevation, base area, and the depth of a crater lake are given. The maximum amount of electricity that can be generated by a hydroelectric power plant is to be determined. Assumptions The evaporation of water from the lake is negligible. Analysis The exergy or work potential of the water is the potential energy it possesses relative to the ground level, mgh = PE =Exergy

d

z

12 m

140 m

Therefore,

( )kWh 109.55 4×=

−×

×=

−==

====

∫∫ ∫ ∫

2222

2243

21

22

s/m 1000kJ/kg 1

s 3600h 1)m 140(m) 152(

)m/s 81.9)(m 102)(kg/m 1000(5.0

2/)(

)(Exergy

2

1

zzAgzdzAg

AdzgzdmgzdPEPE

z

zρρ

ρ

8-105E The 2nd-law efficiency of a refrigerator and the refrigeration rate are given. The power input to the refrigerator is to be determined. Analysis From the definition of the second law efficiency, the COP of the refrigerator is determined to be

57.5375.1245.0COPCOP

COPCOP

375.121495/535

11/

1COP

revR,IIRrevR,

R

revR,

=×==→=

=−

=−

=

ηη II

LH TT

η II = 0.45

200 Btu/min

R

75°F

35°F

Thus the power input is

hp 0.85=

==Btu/min 42.41hp 1

57.5Btu/min 020

COPRin

LQW

&&


8-87

8-106 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a closed system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR. Assumptions Kinetic and potential changes are negligible. Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as

Energy balance: (1) RR QQUUWUUWQQEEE ++−=→−=−+→∆=− 021120systemoutin

Entropy balance: 0

012gensystemgenoutin )(

TQ

TQ

SSSSSSSR

R −+

−+−=→∆=+− (2)

Solving for Q0 from (2) and substituting in (1) yields

QR

SourceTR

System

gen00

21021 1)()( STTT

QSSTUUWR

R −

−−−−−=

The useful work relation for a closed system is obtained from

)(1)()( 120gen0

021021

surr

VV −−−

−−−−−=

−=

PSTTT

QSSTUU

WWW

RR

u

Then the reversible work relation is obtained by substituting Sgen = 0,

−−−+−−−=

RR T

TQPSSTUUW 0

21021021rev 1)()()( VV

A positive result for Wrev indicates work output, and a negative result work input. Also, the QR is a positive quantity for heat transfer to the system, and a negative quantity for heat transfer from the system.


8-88

8-107 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a steady-flow system that exchanges heat with surroundings at T0 at a rate of Q as well as a heat reservoir

at temperature T

&0

R in the amount Q . &R

Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as

Energy balance: outinsystemoutin EEEEE &&&&& =→∆=−

)2

()2

(22

0 ii

iiee

eeR gzV

hmgzV

hmWQQ ++∑−++∑=−+ &&&&&

System

or Ree

eeii

ii QQgzV

hmgzV

hmW &&&&& ++++∑−++∑= 0

22)

2()

2( (1)

Entropy balance: inoutgensystemgenoutin SSSSSSS &&&&&&& −=→∆=+−

0

0gen T

QTQsmsmSR

Riiee

−+

−+∑−∑=

&&&& (2)

Solving for from (2) and substituting in (1) yields &Q0

−−−−++∑−−++∑=

RRee

eeeii

iii T

TQSTsTgz

VhmsTgz

VhmW 0

gen00

2

0

21)

2()

2( &&&&&


−−−++∑−−++∑=

RRee

eeeii

iii T

TQsTgz

VhmsTgz

VhmW 0

0

2

0

2

rev 1)2

()2

( &&&&



8-89

8-108 Writing energy and entropy balances, a relation for the reversible work is to be obtained for a uniform-flow system that exchanges heat with surroundings at T0 in the amount of Q0 as well as a heat reservoir at temperature TR in the amount QR. Assumptions Kinetic and potential changes are negligible. Analysis We take the direction of heat transfers to be to the system (heat input) and the direction of work transfer to be from the system (work output). The result obtained is still general since quantities wit opposite directions can be handled the same way by using negative signs. The energy and entropy balances for this stationary closed system can be expressed as Energy balance: systemoutin EEE ∆=−

cvii

iiee

eeR UUgzV

hmgzV

hmWQQ )()2

()2

( 12

22

0 −+++∑−++∑=−+

or, Rcvee

eeii

ii QQUUgzV

hmgzV

hmW ++−−++∑−++∑= 012

22)()

2()

2( (1)

Entropy balance: systemgenoutin SSSS ∆=+−

SourceTR

Q

System

0

012gen )(

TQ

TQsmsmSSSR

Riieecv

−+

−+∑−∑+−= (2)

Solving for Q0 from (2) and substituting in (1) yields

[ ]

−−−−−−+

−++∑−−++∑=

RRgencv

eee

eeiii

ii

TT

QSTSSTUU

sTgzV

hmsTgzV

hmW

0021021

0

2

0

2

1)()(

)2

()2

(

me

The useful work relation for a closed system is obtained from

[ ] )(1)()(

)2

()2

(

1200

gen021021

0

2

0

2

surr

VV −−

−−−−−−+

−++∑−−++∑=−=

PTT

QSTSSTUU

sTgzV

hmsTgzV

hmWWW

RRcv

eee

eeiii

iiu


[ ]

−−−+−−−+

−++∑−−++∑=

RRcv

eee

eeiii

ii

TT

QPSSTUU

sTgzV

hmsTgzV

hmW

021021021

0

2

0

2

rev

1)()()(

)2

()2

(

VV



8-90

8-109 An electric resistance heater is immersed in water. The time it will take for the electric heater to raise the water temperature to a specified temperature, the minimum work input, and the exergy destroyed during this process are to be determined. Assumptions 1 Water is an incompressible substance with constant specific heats. 2 The energy stored in the container itself and the heater is negligible. 3 Heat loss from the container is negligible. 4 The environment temperature is given to be T0 = 20°C. Properties The specific heat of water at room temperature is c = 4.18 kJ/kg·°C (Table A-3). Analysis Taking the water in the container as the system, which is a closed system, the energy balance can be expressed as

water12ine,

waterine,


system


outin

)(

)(

TTmctW

UW

EEE

−=∆

∆=

∆=−

&

4342143421

Heater

Water 40 kg

Substituting, (800 J/s)∆t = (40 kg)(4180 J/kg·°C)(80 - 20)°C Solving for ∆t gives ∆t = 12,544 s = 209.1 min = 3.484 h

Again we take the water in the tank to be the system. Noting that no heat or mass crosses the boundaries of this system and the energy and entropy contents of the heater are negligible, the entropy balance for it can be expressed as

{

watergen

entropyin Change

system

generationEntropy

gen


outin

0 SS

SSSS

∆=+

∆=+−4342143421

Therefore, the entropy generated during this process is

( )( ) kJ/K31.18 K 293K 353

ln KkJ/kg 4.184kg 40ln1

2watergen =⋅==∆=

TT

mcSS



The actual work input for this process is

kJ 042,10=s) 52kJ/s)(12,5 8.0(inact,inact, =∆= tWW &

Then the reversible (or minimum required )work input becomes kJ 906=−=−= 9136042,10destroyedinact,inrev, XWW


8-91

8-110 A hot water pipe at a specified temperature is losing heat to the surrounding air at a specified rate. The rate at which the work potential is wasted during this process is to be determined. Assumptions Steady operating conditions exist. Analysis We take the air in the vicinity of the pipe (excluding the pipe) as our system, which is a closed system.. The system extends from the outer surface of the pipe to a distance at which the temperature drops to the surroundings temperature. In steady operation, the rate form of the entropy balance for this system can be expressed as

{

W/K 0.0344 0K 278

W 45K 353

W 45

0

0

systemgen,systemgen,

systemgen,outb,

out

inb,

in


0system


gen


outin

=→=+−

=+−

=∆=+−

SS

STQ

TQ

SSSS

&&

&&&

43421&&

43421&&

80°C

L = 10 m

D = 5 cm

Q Air, 5°C




8-92

8-111 Two rigid tanks that contain water at different states are connected by a valve. The valve is opened and steam flows from tank A to tank B until the pressure in tank A drops to a specified value. Tank B loses heat to the surroundings. The final temperature in each tank and the work potential wasted during this process are to be determined. Assumptions 1 Tank A is insulated and thus heat transfer is negligible. 2 The water that remains in tank A undergoes a reversible adiabatic process. 3 The thermal energy stored in the tanks themselves is negligible. 4 The system is stationary and thus kinetic and potential energy changes are negligible. 5 There are no work interactions. Analysis (a) The steam in tank A undergoes a reversible, adiabatic process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6),

( )( )( )( )( )( )

( ) ( )( )( )( )

KkJ/kg 7.7100kJ/kg 2731.4

/kgm 1.1989

C250kPa 200

kJ/kg 2125.9kJ/kg 1982.17895.011.561/kgm 0.47850001073.060582.07895.0001073.0

7895.03200.5

6717.18717.5

mixture sat.

kPa 300

KkJ/kg 5.87171191.58.07765.1kJ/kg 2163.39.19488.022.604

/kgm 0.37015001084.046242.08.0001084.0

8.0kPa 400

,1

,1

3,1

1

1

,2,2

3,2,2

,2,2

300@,2

12

2

1,1

1,1

31,1

1

1

⋅===

°==

=+=+==−+=+=

=−

=−

=

°==

==

⋅=+=+==+=+=

=−+=+=

==

B

B

B

fgAfA

fgAfA

fg

fAA

kPasatA

fgfA

fgfA

fgfA

su

TP

uxuux

sss

x

TT

ssP

sxssuxuu

x

xP

v

vvv

vvv

:BTank

: ATank

C133.52

900 kJ

×B

m = 3 kg steam

T = 250°C P = 200 kPa

A V = 0.2 m3

steam P = 400 kPa

x = 0.8

The initial and the final masses in tank A are

and

kg 0.4180/kgm0.479

m0.2

kg 0.5403/kgm 0.37015

m 0.2

3

3

,2,2

3

3

,1,1

===

===

A

AA

A

AA

m

m

v

V

v

V

Thus, 0.540 - 0.418 = 0.122 kg of mass flows into tank B. Then, m mB B2 1 0122 3 0122, , . .= − = + = 3.122 kg

The final specific volume of steam in tank B is determined from

( ) ( )( )

/kgm 1.152m 3.122

/kgm 1.1989kg 3 33

3

,2

11

,2,2 ====

B

B

B

BB m

vmm

vV

We take the entire contents of both tanks as the system, which is a closed system. The energy balance for this stationary closed system can be expressed as

( ) ( )BA

BA

umumumumQWUUUQ

EEE

11221122out

out


system


outin

0)=PE=KE (since )()(−+−=−

=∆+∆=∆=−

∆=−4342143421

Substituting,


8-93

( )( ) ( )( ){ } ( ) ( )( ){ }

kJ/kg 2425.94.27313122.33.21635403.09.2125418.0900

,2

,2

=−+−=−

B

B

uu

Thus,

KkJ/kg .97726kJ/kg 2425.9

/kgm 1.152

,2

,2

,2

3,2

⋅=°=

=

=

B

B

B

B

sT

u

C110.1v

(b) The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes both tanks and their immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives

{

BAgensurrb,

out

entropyin Change

system

generationEntropy

gen


outin

SSSTQ

SSSS

∆+∆=+−

∆=+−4342143421

Rearranging and substituting, the total entropy generated during this process is determined to be

( ) ( )

( )( ) ( )( ){ } ( )( ) ( )( ){ }kJ/K 234.1

K 273kJ 900

7100.739772.6122.38717.55403.08717.5418.0

surrb,

out11221122

surrb,

outgen

=

+−+−=

+−+−=+∆+∆=TQ

smsmsmsmTQ

SSS BABA

The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition , gen0destroyed STX =



8-94

8-112E A cylinder initially filled with helium gas at a specified state is compressed polytropically to a specified temperature and pressure. The actual work consumed and the minimum useful work input needed are to be determined. Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 The compression or expansion process is quasi-equilibrium. 5 The environment temperature is 70°F. Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R (Table A-1E). The specific heats of helium are cv = 0.753 and cv = 1.25 Btu/lbm.R (Table A-2E). Analysis (a) Helium at specified conditions can be treated as an ideal gas. The mass of helium is

lbm 0.264)R 530)(R/lbmftpsia 2.6805(

)ft 15)(psia 25(3

3

1

11 =⋅⋅

==RTP

mV

Q

HELIUM 15 ft3

PVn = const

The exponent n and the boundary work for this polytropic process are determined to be

539.1682.715

2570

ft 7.682)ft 15()psia 70)(R 530()psia 25)(R 760(

2

1

1

21122

331

2

1

1

22

2

22

1

11

=→

=

→

=

→=

===→=

nPP

PP

PP

TT

TP

TP

nnnn

V

VVV

VVVV

Then the boundary work for this polytropic process can be determined from

( )

( )( )( )Btu 55.9

1.5391R530760RBtu/lbm 0.4961lbm 0.264

111211222

1inb,

=−

−⋅−=

−−

−=−−

−=−= ∫ nTTmR

nPP

dPWVV

V

Also,

Thus, Btu 36.0=−=−=

=

⋅−−=−−=

9.199.55

Btu 9.19ftpsia 5.4039

Btu 1ft5)12psia)(7.68 7.14()(

insurr,inb,inu,

33

120insurr,

WWW

PW VV

(b) We take the helium in the cylinder as the system, which is a closed system. Taking the direction of heat transfer to be from the cylinder, the energy balance for this stationary closed system can be expressed as

)()(

)(

12inb,out

inb,12out

12inb,out


system


outin

TTmcWQWuumQuumUWQ

EEE

−−=

−−=−

−=∆=+−

∆=−

v

4342143421

Substituting, ( )( )( ) Btu 10.2R530760RBtu/lbm 0.753lbm 0.264Btu 55.9out =−⋅−=Q

The total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the cylinder and its immediate surroundings so that the boundary temperature of the extended system is the temperature of the surroundings at all times. It gives


8-95

{

sysgensurrb,

out

entropyin Change

system

generationEntropy

gen


outin

SSTQ

SSSS

∆=+−

∆=+−4342143421

where the entropy change of helium is

RBtu 01590psia 25psia 70

ln)RBtu/lbm 0.4961(R 530R 760

ln)RBtu/lbm 1.25()lbm 0.264(

lnln1

2

1

2avg,heliumsys

/.−=

⋅−⋅=

−=∆=∆

PP

RTT

cmSS p

Rearranging and substituting, the total entropy generated during this process is determined to be

Btu/R 0033450R 530

Btu 10.2Btu/R) 0159.0(

0

outheliumgen .=+−=+∆=

TQ

SS

The work potential wasted is equivalent to the exergy destroyed during a process, which can be determined from an exergy balance or directly from its definition , gen0destroyed STX =

Btu 1.77)Btu/R 003345.0(R) 530(gen0destroyed === STX

The minimum work with which this process could be accomplished is the reversible work input, Wrev, in. which can be determined directly from Btu 34.23=−=−= 77.10.36destroyedinact,inrev, XWW

Discussion The reversible work input, which represents the minimum work input Wrev,in in this case can be determined from the exergy balance by setting the exergy destruction term equal to zero,

12inrev,

exergyin Change

system

ndestructioExergy

e)(reversibl 0destroyed

mass and work,heat,by nsferexergy traNet outin XXWXXXX −=→∆=−−

43421444 3444 2143421

Substituting the closed system exergy relation, the reversible work input during this process is determined to be

Btu 34.24=⋅−

−°−⋅=−+−−−=

]ftpsia Btu/5.4039[ft)15682.7psia)( (14.7+

Btu/R) R)(-0.0159 530(F70)R)(300Btu/lbm lbm)(0.753 264.0()()()(

33

12012012rev VVPSSTUUW


8-95 1 steady operating conditions exist. 2 kinetic …...8-79 review problems 8-95 refrigerant-134a...

Documents