gas refrigeration cycles · pdf file11-38 11-56 [also solved by ees on enclosed cd] an...

11-36

Gas Refrigeration Cycles 11-49C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction. 11-50C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is

COPR,Stirling =−

11T TH L/

11-51C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature. 11-52C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin. 11-53C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling (h1 = h2) process. 11-54C By regeneration.

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.

11-37

11-55 An ideal-gas refrigeration cycle with air as the working fluid is considered. The maximum and minimum temperatures in the cycle, the COP, and the rate of refrigeration are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17),

T hP

T hP

r

r

1 1

1 3

250 050 7329

300191 386

1

3

= → ==

= → ==

250 K kJ / kg

300 K kJ / kg

..

..

T

4

3

QH · 2

1QRefrig ·

-23°C27°C

Thus,

( )( )

( )kJ/kg 97218

4620386131

kJ/kg 6034219872732903

4

min43

4

2

max21

2

34

12

.hTT..P

PPP

.hTT..P

PPP

rr

rr

===→=

==

===→===

K 219.0

K 342.2

s

(b) The COP of this ideal gas refrigeration cycle is determined from

COPRnet, in comp, in turb, out

= =−

qw

qw w

L L

where

q h h

w h hw h h

L = − = − == − = − == − = − =

1 4

2 1

3 4

250 05 218 97 3108342 60 250 05 92 5530019 218 97 8122

. . .

. . .. . .

kJ / kgkJ / kgkJ / kg

comp, in

turb, out

Thus, COPR =−

=31 08

92 55 8122.

. .2.74

(c) The rate of refrigeration is determined to be

( ) ( )( ) kJ/s2.49 kJ/kg 31.08kg/s 0.08refrig === LqmQ &&


11-38

11-56 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17),

T hP

T hP

r

r

1 1

1 3

285 2851411584

320 320 291 7375

1

3

= → ==

= → ==

K k

K k

..

..

J / kg

J / kg

T

4

3

QH · 2

1QRefrig·

47°C12°C

Thus,

( )

( )kJ/kg 76201

K8201347507375125050

kJ/kg 17452 K4450792515841

50250

4

43

4

2

21

2

34

12

.h

.T..PPPP

.h

.T..PPPP

rr

rr

==→=

==

==→=

==

s

Then the rate of refrigeration is

( ) ( ) ( )( ) kW 6.67=−=−== kJ/kg 201.76285.14kg/s 0.0841refrig hhmqmQ L &&&

(b) The net power input is determined from

& & &W W Wnet, in comp, in turb, out= −

where

( ) ( )( )( ) ( )( ) kW 9.48kJ/kg 201.76320.29kg/s 0.08

kW 13.36kJ/kg 285.14452.17kg/s 0.08

43outturb,

12incomp,

=−=−=

=−=−=

hhmWhhmW

&&

&&

Thus, & . .Wnet, in = − =13 36 9 48 3.88 kW

(c) The COP of this ideal gas refrigeration cycle is determined from

COP 6.67 kW3.88 kWR

net, in= = =

&

&Q

WL 1.72


11-39

11-57 EES Problem 11-56 is reconsidered. The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration, the net power input, and the COP are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input data" T[1] = 12 [C] P[1]= 50 [kPa] T[3] = 47 [C] P[3]=250 [kPa] m_dot=0.08 [kg/s] Eta_comp = 1.00 Eta_turb = 1.0 "Compressor anaysis" s[1]=ENTROPY(Air,T=T[1],P=P[1]) s2s=s[1] "For the ideal case the entropies are constant across the compressor" P[2] = P[3] s2s=ENTROPY(Air,T=Ts2,P=P[2])"Ts2 is the isentropic value of T[2] at compressor exit" Eta_comp = W_dot_comp_isen/W_dot_comp "compressor adiabatic efficiency, W_dot_comp > W_dot_comp_isen" m_dot*h[1] + W_dot_comp_isen = m_dot*hs2"SSSF First Law for the isentropic compressor, assuming: adiabatic, ke=pe=0, m_dot is the mass flow rate in kg/s" h[1]=ENTHALPY(Air,T=T[1]) hs2=ENTHALPY(Air,T=Ts2) m_dot*h[1] + W_dot_comp = m_dot*h[2]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[2]=ENTHALPY(Air,T=T[2]) s[2]=ENTROPY(Air,h=h[2],P=P[2]) "Heat Rejection Process 2-3, assumed SSSF constant pressure process" m_dot*h[2] + Q_dot_out = m_dot*h[3]"SSSF First Law for the heat exchanger, assuming W=0, ke=pe=0" h[3]=ENTHALPY(Air,T=T[3]) "Turbine analysis" s[3]=ENTROPY(Air,T=T[3],P=P[3]) s4s=s[3] "For the ideal case the entropies are constant across the turbine" P[4] = P[1] s4s=ENTROPY(Air,T=Ts4,P=P[4])"Ts4 is the isentropic value of T[4] at turbine exit" Eta_turb = W_dot_turb /W_dot_turb_isen "turbine adiabatic efficiency, W_dot_turb_isen > W_dot_turb" m_dot*h[3] = W_dot_turb_isen + m_dot*hs4"SSSF First Law for the isentropic turbine, assuming: adiabatic, ke=pe=0" hs4=ENTHALPY(Air,T=Ts4) m_dot*h[3] = W_dot_turb + m_dot*h[4]"SSSF First Law for the actual compressor, assuming: adiabatic, ke=pe=0" h[4]=ENTHALPY(Air,T=T[4]) s[4]=ENTROPY(Air,h=h[4],P=P[4]) "Refrigeration effect:" m_dot*h[4] + Q_dot_Refrig = m_dot*h[1] "Cycle analysis" W_dot_in_net=W_dot_comp-W_dot_turb"External work supplied to compressor" COP= Q_dot_Refrig/W_dot_in_net "The following is for plotting data only:" Ts[1]=Ts2 ss[1]=s2s Ts[2]=Ts4 ss[2]=s4s


11-40

COP ηcomp ηturb QRefrig

[kW] Winnet [kW]

0.6937 0.7 1 6.667 9.612 0.9229 0.8 1 6.667 7.224 1.242 0.9 1 6.667 5.368 1.717 1 1 6.667 3.882

0,7 0,75 0,8 0,85 0,9 0,95 10

1

2

3

4

5

6

7

ηcomp

QRefrig

[kW]

η turb

0.70.70.850.851.01.0

0,7 0,75 0,8 0,85 0,9 0,95 10

2

4

6

8

10

12

ηcomp

Win;net

[kW]

η turb

0.70.70.850.851.01.0

0,7 0,75 0,8 0,85 0,9 0,95 10

0,5

1

1,5

2

ηcomp

COP

η turb

0.70.70.850.851.01.0


11-41

11-58E An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17E),

T hP

T hP

r

r

1 1

1 3

500 119 4810590

580 138 661 7800

1

3

= → ==

= → ==

R B

R B

..

..

tu / lbm

tu / lbm

T

4

3

QH · 2

1QRefrig ·

120°F40°F

Thus,

( )

( )Btu/lbm 14101

R 442359330780013010

Btu/lbm 68163R 9683177305901

1030

4

43

4

2

21

2

34

12

.h.T..P

PPP

.h.T..P

PPP

rr

rr

==→=

==

==→=

==

s


( ) ( ) ( )( ) Btu/s 9.17=−=−== Btu/lbm101.14119.48lbm/s 0.541refrig hhmqmQ L &&&



where

( ) ( )( )( ) ( )( ) Btu/s18.79 Btu/lbm101.14138.66lbm/s 0.5

Btu/s22.10Btu/lbm 119.48163.68lbm/s 0.5

43outturb,

12incomp,

=−=−=

=−=−=

hhmWhhmW

&&

&&

Thus, & . . .Wnet, in Btu / s= − = =2210 18 76 3 34 4.73 hp


COP 9.17 Btu / s3.34 Btu / sR

net, in= = =

&

&Q

WL 2.75


11-42

11-59 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible. Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17),

T hP

T hP

r

r

1 1

1 3

285 2851411584

320 320 291 7375

1

3

= → ==

= → ==

K k

K k

..

..

J / kg

J / kg

4

2

4s

3

QH ·

2

11QRefrig ·

T

47°C12°CThus,

( )

( )kJ/kg 76201

K8201347507375125050

kJ/kg 17452 K4450792515841

50250

4

43

4

2

21

2

34

12

.h

.T..PPPP

.h

.T..PPPP

s

srr

s

srr

==→=

==

==→=

==

s

Also,

( )( )(

kJ/kg 54219762012932085029320

433443

43

.....

hhhhhhhh

sTs

T

=−−=

−η−=→−−

=η)


( ) ( ) ( )( ) kW 5.25=−=−== kJ/kg 219.54285.14kg/s 0.0841refrig hhmqmQ L &&&



where

( ) ( )

( ) ( )[ ] ( )( ) ( )( ) kW 8.06kJ/kg 219.54320.29kg/s 0.08

kW 16.700.80/kJ/kg 285.14452.17kg/s 0.08

43outturb,

1212incomp,

=−=−=

=−=η−=−=

hhmW

/hhmhhmW Cs

&&

&&&

Thus, & . .Wnet, in = − =16 70 8 06 8.64 kW


COP 5.25 kW8.64 kWR

net, in= = =

&

&Q

WL 0.61


11-43

11-60 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).

4

2

4s

3

QH ·2s

11QRefrig ·

TAnalysis (a) From the isentropic relations,

( )( )( )

( )( ) K1.208

31K323

K2.4083K263

667.1/667.0k/1k

3

434

667.1/667.0k/1k

1

212

=

=

=

==

=

−

−

PP

TT

PP

TT

s

s

50°C

-10°C

sand

( ) ( )( )

( ) ( ) (K 5.444

80.0/2632.408263/

1.20832380.0323

121212

12

12

12

min

433443

43

43

43

=−+=−+=→

−−

=−−

=

==−−=−−=→

−−

=−−

=

Csss

C

sTss

T

TTTTTTTT

hhhh

TTTTT

TTTT

hhhh

ηη

ηηK 231.1

)

(b) The COP of this ideal gas refrigeration cycle is determined from

( ) ( )

( ) ( )

( ) ( ) 0.356=−−−

−=

−−−−

=

−−−−

=

−==

1.2313232635.4441.231263

COP

4312

41

4312

41

outturb,incomp,innet,R

TTTTTT

hhhhhh

wwq

wq LL

(c) The mass flow rate of helium is determined from

( ) ( )( ) kg/s 0.109=−⋅

=−

=−

==K 231.1263KkJ/kg 5.1926

kJ/s 18

41

refrig

41

refrigrefrig

TTcQ

hhQ

qQ

mpL

&&&&


11-44

11-61 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations,

( )( )( )

( )( ) min

4140k1k

4

545

4140k1k

1

212

K617341 K258

C122.3 K33954K 266

T.PPTT

.PPTT

././

././

=°−==

=

=

°===

=

−

−

C99.4

Qrege

QH · 2

4

6 QRefrig ·

5

-15°C

-7°C

3

1

T

(b) From an energy balance on the regenerator, 27°C

( ) ( 6143

outin

(steady) 0systemoutin 0

hhmhhmhmhm

EE

EEE

iiee −=−→=

=

=∆=−

∑∑ &&&&

&&

&&&

)sor,

or,

( ) ( )

( ) ( ) C49C15C27C74316

61436143

°−=°−+°−°−=+−=

−=−→−=−

TTTT

TTTTTTcmTTcm pp &&

Then the COP of this ideal gas refrigeration cycle is determined from

( ) ( )

( ) ( )( )

( )[ ] ( )[ ] 1.12=°−−−−°−−

°−−°−=

−−−−

=

−−−−

=

−==

C4.9915C73.122C4.99C49

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The mass flow rate is determined from

( ) ( ) ( )[ ] kg/s 0.237=°−−−°⋅

=−

=−

==C99.449CkJ/kg 1.005

kJ/s 12

56

refrig

56

refrigrefrig

TTcQ

hhQ

qQ

mpL

&&&&


11-45

11-62 An ideal-gas refrigeration cycle with air as the working fluid is considered. The lowest temperature that can be obtained by this cycle, the COP, and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

5

2

Qrege

27°C

QH · 2s

4

6 QRefrig ·

5

-15°C

-7°C

3

1

T

s

Analysis (a) The lowest temperature in the cycle occurs at the turbine exit. From the isentropic relations,

( )( )( )

( )( ) C99.4K 6173

41 K258

C122.3K 33954 K266

4140k1k

4

545

4140k1k

1

212

°−==

=

=

°===

=

−

−

.PPTT

.PPTT

././

s

././

s

and

( ) ( ) (( ))

( ) ( )( ) (C4165

750731227

4991580015

121212

12

12

12

min

544554

54

54

54

°=−−+−=η−+=→

−−

=−−

=η

=°−=−−−−−=−η−=→

−−

=−−

=η

.././TTTT

TTTT

hhhh

T..TTTT

TTTT

hhhh

Csss

C

sTss

TC82.5

)

)

(b) From an energy balance on the regenerator,

( ) ( 6143

outin(steady) 0

systemoutin 0

hhmhhmhmhm

EEEEE

iiee −=−→=

=→=∆=−

∑∑ &&&&

&&&&&

or,

or,

( ) ( )

( ) ( ) C49C15C27C74316

61436143

°−=°−+°−°−=+−=

−=−→−=−

TTTT

TTTTTTcmTTcm pp &&


( ) ( )

( ) ( )( )

( )[ ] ( )[ ] 0.32=°−−−−°−−

°−−°−=

−−−−

=

−−−−

=

−==

C5.8215C74.165C5.82C49

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The mass flow rate is determined from

( ) ( ) ( )[ ] kg/s 0.356=°−−−°⋅

=−

=−

==C82.549CkJ/kg 1.005

kJ/s 12

56

refrig

56

refrigrefrig

TTcQ

hhQ

qQ

mpL

&&&&


11-46

11-63 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2). Analysis (a) From the isentropic relations,

( )

( )( ) K 4.4325K 2.273 4.1/4.0k/1k

1

212 ==

=

−

PP

TsT

Turbine

RegeneratorHeat Exch.

Heat Exch.

QL .

2 1

Compressor

6

54

3

. QH

K 5.472

2.2732.2734.43280.0 2

2

12

12

12

12

=→−

−=

−−

=−−

=

TT

TTTT

hhhh ss

Cη

The temperature at state 4 can be determined by solving the following two equations simultaneously:

( ) 4.1/4.0

4

k/1k

4

545 5

1

=

=

−

TPP

TT s

ssT TT

Thhhh

54

4

54

54 2.19385.0

−−

=→−−

=η

Using EES, we obtain T4 = 281.3 K. An energy balance on the regenerator may be written as

5

2

Qregen

QH · 2s

4

6QRefrig ·

5s

3

1

s

T

or,

( ) ( )

K 3.2463.2812.3082.2734316

61436143

=+−=+−=

−=−→−=−

TTTT

TTTTTTcmTTcm pp &&

35°C

0°CThe effectiveness of the regenerator is

0.434=−−

=−−

=−−

=3.2462.3083.2812.308

63

43

63

43regen TT

TThhhh

ε -80°C

(b) The refrigeration load is

kW 21.36=−=−= K)2.19346.3kJ/kg.K)(2 5kg/s)(1.00 4.0()( 56 TTcmQ pL &&

(c) The turbine and compressor powers and the COP of the cycle are

kW 13.80K)2.27372.5kJ/kg.K)(4 5kg/s)(1.00 4.0()( 12inC, =−=−= TTcmW p&&

kW 43.35kJ/kg)2.19381.3kJ/kg.K)(2 5kg/s)(1.00 4.0()( 54outT, =−=−= TTcmW p&&

0.478=−

=−

==43.3513.80

36.21COPoutT,inC,innet, WW

QW

Q LL&&

&

&

&


11-47

(d) The simple gas refrigeration cycle analysis is as follows:

4

2

4s

3

QH ·

2

11QRefrig ·

( )

( ) K 6.19451K 2.3081 4.1/4.0k/1k

34 =

=

=

−

rTsT

T

K 6.2116.1942.308

2.30885.0 4

4

43

43 =→−

−=→

−−

= TT

TTTT

sTη 35°C

0°C

kW 24.74=−=

−=

kJ/kg)6.21173.2kJ/kg.K)(2 5kg/s)(1.00 4.0(

)( 41 TTcmQ pL &&

s

[ ] kW 32.41kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K) 5kg/s)(1.00 4.0(

)()( 4312innet,

=−−−=

−−−= TTcmTTcmW pp &&&

0.599===32.4174.24COP

innet,WQL&

&


11-48

Absorption Refrigeration Systems 11-64C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form. 11-65C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source. 11-66C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point. 11-67C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor. 11-68C The coefficient of performance of absorption refrigeration systems is defined as

geninpump,gen

R inputrequiredoutputdesiredCOP

QQ

WQQ LL ≅

+==

11-69C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving the pump. 11-70 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible. Analysis The maximum COP that this refrigeration system can have is

142268300

268 K403 K30011COP

0

0maxR, .

TTT

TT

L

L

s=

−

−=

−

−=

which is slightly greater than 2. Thus the claim is possible, but not probable. 11-71 The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined. Analysis The maximum COP that this refrigeration system can have is

2.64=

−

−=

−

−=

273298273

K 393K 29811COP

0

0maxR,

L

L

s TTT

TT


11-49

11-72 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis The maximum COP that this refrigeration system can have is

151243298

243K 403K 29811COP

0

0maxR, .

TTT

TT

L

L

s=

−

−=

−

−=

Thus, ( )( ) kJ/h 105.75 5×=×== kJ/h 10515.1COP 5genmaxR,maxL, QQ &&

11-73E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined. Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is ( )( ) Btu/h 100.55 5×=== Btu/h10550COP 5

genR .QQL&&

11-74 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The rate at which the steam condenses, the power input to the reversible refrigerator, and the second law efficiency of an actual chiller are to be determined. Properties The enthalpy of vaporization of water at 200C is hfg = 1939.8 kJ/kg (Table A-4). Analysis (a) The thermal efficiency of the reversible heat engine is

370.0K )15.273200(

K )15.27325(11 0revth, =

++

−=−=sT

Tη

The COP of the reversible refrigerator is

52.7K )15.27310()15.27325(

K )15.27310(COP0

revR, =+−−+

+−=

−=

L

L

TTT

T0

TL

Rev.Ref.

Ts

T0

Rev. HE

The COP of the reversible absorption refrigerator is 78.2)52.7)(370.0(COPCOP revR,revth,revabs, ===η

The heat input to the reversible heat engine is

kW 911.72.78

kW 22COP revabs,

in === LQQ

&&

Then, the rate at which the steam condenses becomes

kg/s 0.00408===kJ/kg 1939.8kJ/s 911.7in

fgs h

Qm

&&

(b) The power input to the refrigerator is equal to the power output from the heat engine kW 2.93==== )kW 191.7)(370.0(inrevth,HEout,Rin, QWW &&& η

(c) The second-law efficiency of an actual absorption chiller with a COP of 0.7 is

0.252===2.78

7.0COPCOP

revabs,

actualIIη


11-50

Special Topic: Thermoelectric Power Generation and Refrigeration Systems 11-75C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit. 11-76C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect. 11-77C No. 11-78C No. 11-79C Yes. 11-80C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages. 11-81C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals. 11-82C The efficiency of a thermoelectric generator is limited by the Carnot efficiency because a thermoelectric generator fits into the definition of a heat engine with electrons serving as the working fluid. 11-83E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined. Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,

%3.31=−=−==R800R550

11Carnotth,maxth,H

L

TT

ηη

11-84 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined. Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits,

Thus, ( ) ( ) ( )

W 12.1

10.72

===

=−

=−

==

72.10 W130

COP

1K 268/K 2931

1/1COPCOP

maxminin,

CarnotR,max

L

LH

QW

TT

&&


11-51

11-85 A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR,

COPCOP

180 W0.15R

inin

R= → = = =

&

&&

&QW

W QL L 1200 W

11-86E A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined. Analysis The required power input is determined from the definition of COPR,

hp 3.14=Btu/min 33.310.15Btu/min 20

COPCOP

Rin

inR ===→= LL Q

WWQ &

&&

&

11-87 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/L and cp = 4.18 kJ/kg·°C (Table A-3). Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks,

W6.71=kW 00671.0s 360012

kJ 290

kJ 290=C3)-C)(25kJ/kg kg)(4.18 15.3(kg 3.15=L) 0kg/L)(0.35 1(9

coolingcooling

cooling

=×

=∆

=

°°⋅=∆=×==

tQ

Q

TmcQm

&

Vρ

The electric power consumed by the refrigerator is

W36=A) V)(3 12(in == IW V&

Then the COP of the refrigerator becomes

COP W36 W

cooling

in= = = ≈&

&. .

QW

6 71 0 200.186


11-52

11-88E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min. The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and the electric power drawn from the battery of the car are to be determined. Assumptions Heat transfer through the walls of the refrigerator is negligible. Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0 Btu/lbm.°F (Table A-3E). Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks,

W36.2=

×=

∆=

°°⋅=∆=

Btu 1J 1055

s 6015Btu 84.30

Btu 30.84=F38)-F)(78Btu/lbm lbm)(1.0 771.0(

coolingcooling

cooling

tQ

Q

TmcQ p

&

(b) The average heating rate of the refrigerator is simply the rate of increase of the energy content of the canned drinks,

W49.7=

×=

∆=

°°⋅=∆=

Btu 1J 1055

s 6015Btu 4.42

Btu 42.4=F75)-F)(130Btu/lbm lbm)(1.0 771.0(

heatingheating

heating

tQ

Q

TmcQ p

&

(c) The electric power drawn from the car battery during cooling and heating is

W441

W181

.1.2

W7.49COP

2.112.01COPCOP

0.2 W2.36

COP

heating

heatingheatingin,

coolingheating

cooling

coolingcoolingin,

===

=+=+=

===

QW

QW

&&

&&

11-89 The maximum power a thermoelectric generator can produce is to be determined. Analysis The maximum thermal efficiency this thermoelectric generator can have is

1420K 353K 30311maxth, .

TT

H

L =−=−=η

Thus,

kW 39.4==== kJ/h 142,000kJ/h) (0.142)(106inmaxth,maxout, QW && η


11-53

Review Problems 11-90 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP, the condenser and evaporator pressures, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The COP of this refrigeration cycle is determined from

( ) ( ) ( ) 5.06=−

=−

=1K 253/K 303

11/

1COP CR,LH TT

(b) The condenser and evaporative pressures are (Table A-11)

kPa 770.64kPa 132.82

====

°

°−

C03@satcond

C20@satevap

PPPP

(c) The net work input is determined from

( ) ( )( )( ) ( )( ) kJ/kg 82.19591.21280.049.25

kJ/kg 43.5791.21215.049.25

C20@22

C20@11

=+=+=

=+=+=

°−

°−

fgf

fgf

hxhhhxhh

-20°C

30°C

q L

4 3

2 1

T

s

kJ/kg 27.35===

=−=−=

06.5kJ/kg 138.4

COP

kJ/kg4.13843.5782.195

Rinnet,

12

L

Lq

w

hhq

11-91 A large refrigeration plant that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant, the power input to the compressor, and the mass flow rate of the cooling water are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )

( )throttling kJ/kg 82.88

kJ/kg 82.88liquid sat.

MPa 7.0

C95.34 kJ/kg 50.273MPa 7.0

KkJ/kg 94779.0kJ/kg 97.236

vapor sat.kPa 120

34

MPa 0.7 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

=≅

===

°==

==

⋅====

=

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL

·

Win·

·

0.7 MPa

4

32

1 0.12 MPa

The mass flow rate of the refrigerant is determined from

( ) kg/s 0.675=−

=−

=kJ/kg 88.82236.97

kJ/s 100

41 hhQ

m L&

& s

(b) The power input to the compressor is ( ) ( )( ) kW 24.7=−=−= kJ/kg 236.97273.50kg/s 0.67512in hhmW && (c) The mass flow rate of the cooling water is determined from

( ) ( )( )

( )( ) kg/s 3.73=°°⋅

=∆

=

=−=−=

C8CkJ/kg 4.18kJ/s 124.7

)(

kW 124.7kJ/kg 88.82273.50kg/s 0.675

watercooling

32

TcQ

m

hhmQ

p

H

H&

&

&&


11-54

11-92 EES Problem 11-91 is reconsidered. The effect of evaporator pressure on the COP and the power input is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" "P[1]=120 [kPa]" P[2] = 700 [kPa] Q_dot_in= 100 [kW] DELTAT_cw = 8 [C] C_P_cw = 4.18 [kJ/kg-K] Fluid$='R134a' Eta_c=1.0 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" {h[2]=enthalpy(Fluid$,P=P[2],T=T[2]) } T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" Q_dot_out=m_dot*Qout "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" Q_dot_in=m_dot*Q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c m_dot_cw*C_P_cw*DELTAT_cw = Q_dot_out


11-55

P1 [kPa] COP WC [kW]

120 4,056 24,66 150 4,743 21,09 180 5,475 18,27 210 6,27 15,95 240 7,146 13,99 270 8,126 12,31 300 9,235 10,83 330 10,51 9,517 360 11,99 8,34 390 13,75 7,274

100 150 200 250 300 350 400

4

8

12

16

5

9

13

17

21

25

P[1] [kPa]

COP

Wc

[kW]


11-56

11-93 A large refrigeration plant operates on the vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of the refrigerant, the power input to the compressor, the mass flow rate of the cooling water, and the rate of exergy destruction associated with the compression process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )



MPa 7.0

C95.34 kJ/kg 50.273MPa 7.0

KkJ/kg 94779.0kJ/kg 97.236

vapor sat.kPa 120

34

MPa 0.7 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

=≅

===

°==

==

⋅====

=

hh

hhP

Thss

P

sshhP

f

sss

g

g

T

QH

QL

·

Win·

·

0.7 MPa

4

32

1 0.12 MPa

The mass flow rate of the refrigerant is determined from s

( ) kg/s 0.675=−

=−

=kJ/kg 88.82236.97

kJ/s 100

41 hhQ

m L&

&

(b) The actual enthalpy at the compressor exit is

( ) ( ) (kJ/kg 285.67

75.0/97.23650.27397.236/121212

12

=−+=−+=→

−−

= Css

C hhhhhhhh

ηη )

Thus, ( ) ( )( ) kW 32.9=−=−= kJ/kg 236.97285.67kg/s 0.67512in hhmW &&

(c) The mass flow rate of the cooling water is determined from

and ( ) ( )( )

( )( ) kg/s 3.97=°°⋅

=∆

=

=−=−=

C8CkJ/kg 4.18kJ/s 132.9

)(

kW 132.9kJ/kg 88.82285.67kg/s 0.675

watercooling

32

TcQ

m

hhmQ

p

H

H

&&

&&

The exergy destruction associated with this adiabatic compression process is determined from

& & & ( )X T S T m sdestroyed gen= = −0 0 2 s1

where

KkJ/kg 98655.0kJ/kg 67.285

MPa 7.02

2

2 ⋅=

==

shP

Thus,

( )( )( ) kW 7.80=⋅−= KkJ/kg 0.947790.98655kg/s 0.675K 298destroyedX&


11-57

11-94 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),



MPa 9.0

kJ/kg 75.275MPa 9.0

/kgm 099867.0KkJ/kg 93773.0

kJ/kg 46.244

vaporsat.kPa 200

34

MPa 0.9 @ 33

212

2

3kPa 200 @ 1

kPa 200 @ 1

kPa 200 @ 11

=≅

===

=

==

==

⋅====

=

hh

hhP

hss

P

sshh

P

f

g

g

g

vvQH

QL

House ·

Win ·

·

0.9 MPa

4

32

1200 kPa

T

s The rate of heat supply to the house is determined from

( ) ( )( ) kW 55.73=−=−= kJ/kg .61101275.75kg/s 0.3232 hhmQH &&

(b) The volume flow rate of the refrigerant at the compressor inlet is

( )( ) /sm 0.0320 3=== /kgm 0.099867kg/s 0.32 311 vV m&&

(c) The COP of t his heat pump is determined from

5.57=−−

=−−

==46.24475.27561.10175.275COP

12

32

inR hh

hhwqL

11-95 A relation for the COP of the two-stage refrigeration system with a flash chamber shown in Fig. 11-12 is to be derived. Analysis The coefficient of performance is determined from

COPRin

=qw

L

where

( )( )

( )( ) ( )( 94126incompII,incompI,in

66816

11

with1

hhhhxwww

hhh

xhhxqfg

fL

−+−−=+=

−=−−=

)


11-58

11-96 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flashing chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13)

kJ/kg 94.63kJ/kg 47.95

kJ/kg 58.260

,kJ/kg 94.63,kJ/kg 47.95

,kJ/kg 55.255,kJ/kg 16.239

8

6

2

7

5

3

1

==

=

====

hh

h

hhhh

T

qL

73B

A9 6

0.8 MPa 4

0.4 MPa

8

5 2

1

0.14 MPaThe fraction of the refrigerant that evaporates as it is throttled to the flash chamber is simply the quality at state 6,

0.1646=−

=−

=62.191

94.6347.9566

fg

f

hhh

x

s (b) The enthalpy at state 9 is determined from an energy balance on the mixing chamber:

( ) ( )

( )( ) ( )( ) kJ/kg 75.25958.2601646.0155.2551646.011

0

9

26369

outin(steady) 0

systemoutin

=−+=−+=

=

=→==−

∑∑h

hxhxh

hmhm

EEEEE

iiee &&

&&&&& ∆

KkJ/kg 94168.0kJ/kg 75.259

MPa 4.09

9

9 ⋅=

==

shP

Also, kJ/kg 47.274KkJ/kg 94168.0

MPa 8.04

94

4 =

⋅===

hss

P

Then the amount of heat removed from the refrigerated space and the compressor work input per unit mass of refrigerant flowing through the condenser are

( )( ) ( )( )

( )( ) ( )( )( )( ) ( ) kJ/kg 32.6

kJ/kg 146.4

=−+−−=−+−−=+=

=−−=−−=

kJ/kg 259.75274.47kJ/kg 239.16260.580.1646111

kJ/kg 63.94239.161646.011

94126incompII,incompI,in

816

hhhhxwww

hhxqL

(c) The coefficient of performance is determined from

4.49===kJ/kg 32.6kJ/kg 146.4COP

inR w

qL

11-97 An aircraft on the ground is to be cooled by a gas refrigeration cycle operating with air on an open cycle. The temperature of the air leaving the turbine is to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The specific heat ratio of air at room temperature is k = 1.4 (Table A-2). Analysis Assuming the turbine to be isentropic, the air temperature at the turbine exit is determined from


11-59

( )

( ) C9.0°−==

=

=

−

K264kPa 250kPa 100 K343

4140k1k

3

434

././

PPTT


11-60

11-98 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined. Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible. Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2). Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator,

Qregen ·

QH ·2

4

6 QRefrig ·

5-25°C

-10°C

3

1

T

( ) ( 6143

outin

(steady) 0systemoutin 0

hhmhhmhmhm

EE

EEE

iiee −=−→=

=

=∆=−

∑∑ &&&&

&&

&&&

)

20°C

or,

Thus,

( ) ( )

( ) ( ) K 278C25C10C206134

61436143

=°=°−+°−−°=+−=

−=−→−=−

C5TTTT

TTTTTTcmTTcm pp &&

s

(b) From the isentropic relations,

( )( )( )

( )( ) C93.9K 1179

31 K278

C135.2 K24083 K263

66716670k1k

4

545

66716670k1k

1

212

°−==

=

=

°===

=

−

−

.PPTT

.PPTT

././

././


( ) ( )

( ) ( )( )

( )[ ] ( )[ ] 1.49=°−−−°−−

°−−°−=

−−−−

=

−−−−

=−

==

C93.95C10135.2C93.9C25

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The net power input is determined from

( ) ( )[ ] ( ) ( )[ ]

( )( ) ( )[ ] ( )[ ]( ) kW 108.2=−−−−−°⋅=

−−−=−−−=−=

93.9510135.2CkJ/kg 5.1926kg/s 0.4554125412outturb,incomp,innet, TTTTcmhhhhmWWW p&&&&&

11-99 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined. Analysis The maximum COP that this refrigeration system can have is

259.1263298

263K358K298

11COP0

0maxR, =

−

−=

−

−=

L

L

s TTT

TT

Thus, kW53.9 1.259

kW 12COP maxR,

mingen, === LQQ

&&


11-61

11-100 EES Problem 11-99 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data:" T_L = -10 [C] T_0 = 25 [C] T_s = 85 [C] Q_dot_L = 8 [kW] "The maximum COP that this refrigeration system can have is:" COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L)) "The minimum rate of heat supply is:" Q_dot_gen_min = Q_dot_L/COP_R_max

Qgenmin [kW] Ts [C] 13.76 50 8.996 65 6.833 80 5.295 100 4.237 125 3.603 150 2.878 200 2.475 250

50 90 130 170 210 2502

4

6

8

10

12

14

Ts [C]

Qgen;min

[kW]


11-62

11-101 A house is cooled adequately by a 3.5 ton air-conditioning unit. The rate of heat gain of the house when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis Noting that 1 ton of refrigeration is equivalent to a cooling rate of 211 kJ/min, the rate of heat gain of the house in steady operation is simply equal to the cooling rate of the air-conditioning system,

& & ( .Q Qheat gain cooling ton)(211 kJ / min) = 738.5 kJ / min == = 3 5 44,310 kJ / h

11-102 A room is cooled adequately by a 5000 Btu/h window air-conditioning unit. The rate of heat gain of the room when the air-conditioner is running continuously is to be determined. Assumptions 1 The heat gain includes heat transfer through the walls and the roof, infiltration heat gain, solar heat gain, internal heat gain, etc. 2 Steady operating conditions exist. Analysis The rate of heat gain of the room in steady operation is simply equal to the cooling rate of the air-conditioning system,

& &Q Qheat gain cooling= = 5,000 Btu / h


11-63

11-103 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES:

kJ/kg 50.433kPa 500

kJ/kg.K 6110.5C0

kPa 100kJ/kg 40.273C0

212

2

11

1

11

=

==

=

°==

=→°=

shss

P

sTP

hT

Turbine

RegeneratorHeat Exch.

Heat Exch.

QL.

2 1

Compressor

6

54

3

. QH

kJ/kg 63.308C35

kJ/kg 52.47340.273

40.27350.43380.0

33

2

2

12

12

=→°=

=−

−=

−−

=

hT

hh

hhhh s

Cη

For the turbine inlet and exit we have

kJ/kg 45.193C80 55 =→°−= hT

sT hh

hhhT

54

54

44 ?

−−

=

=→=

η

5

2

Qregen

QH· 2s

4

6QRefrig ·

5s

3

1

s

T

35°C

=

==

=

==

=

°==

shss

P

sTP

sTP

545

2

44

4

11

5

kPa 500

?kPa 500

kJ/kg.K 6110.5C0

kPa 100

0°C

-80°C

We can determine the temperature at the turbine inlet from EES using the above relations. A hand solution would require a trial-error approach. T4 = 281.8 K, h4 = 282.08 kJ/kg An energy balance on the regenerator gives kJ/kg 85.24608.28263.30840.2734316 =+−=+ hh−= hh

The effectiveness of the regenerator is determined from

0.430=−−

=−−

=85.24663.30808.28263.308

63

43regen hh

hhε

(b) The refrigeration load is

kW 21.36=−=−= kJ/kg)45.19385kg/s)(246. 4.0()( 56 hhmQL &&


11-64

(c) The turbine and compressor powers and the COP of the cycle are

kW 05.80kJ/kg)40.27352kg/s)(473. 4.0()( 12inC, =−=−= hhmW &&

kW 45.35kJ/kg)45.19308kg/s)(282. 4.0()( 54outT, =−=−= hhmW &&

0.479=−

=−

==45.3505.80

36.21COPoutT,inC,innet, WW

QW

Q LL&&

&

&

&

(d) The simple gas refrigeration cycle analysis is as follows:

kJ/kg 63.308kJ/kg 52.473kJ/kg 40.273

3

2

1

===

hhh

4

2

4s

3

QH ·

2

11QRefrig ·

T

35°C kJ/kg 2704.5

C35kPa 500

33

3 =

°==

sTP

0°C

kJ/kg.K 52.194kPa 100

434

1 =

==

shss

P s

kJ/kg 64.21152.19463.308

63.30885.0 4

4

43

43 =→−

−=→

−−

= hh

hhhh

sTη

kW 24.70=−=−= kJ/kg)64.21140kg/s)(273. 4.0()( 41 hhmQL &&

[ ] kW 25.41kJ/kg)64.211(308.63)40.273(473.52kg/s) 4.0()()( 4312innet, =−−−=−−−= hhmhhmW &&&

0.599===25.4170.24COP

innet,WQL&

&


11-65

11-104 An air-conditioner with refrigerant-134a as the refrigerant is considered. The temperature of the refrigerant at the compressor exit, the rate of heat generated by the people in the room, the COP of the air-conditioner, and the minimum volume flow rate of the refrigerant at the compressor inlet are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) From the refrigerant-134a tables (Tables A-11 through A-13)

kJ/kg 77.117

kJ/kg 77.117

39.277kPa 1200

kJ/kg 9240.0kJ/kg 04112.0

kJ/kg 30.259

1kPa 500

34

kPa @12003

212

2

1

1

1

1

1

==

==

=

==

===

==

hh

hh

hss

Ps

h

xP

f

s

v

.

34°C

26°C

4

3 2

1

Win

1200 kPa

Expansion valve

Compressor

Condenser

500 kPa

. QH

. QL

Evaporator kJ/kg 42.283

30.25930.25939.27775.0 2

2

12

12

=→−

−=

−−

=

hh

hhhh s

Cη

C54.5°=

==

22

2

kJ/kg 42.283kPa 1200

ThP

(b) The mass flow rate of the refrigerant is T

QH

QL

Win ·

2

·

·4

3

2s

1

kg/s 04053.0/kgm 0.04112

s 60min 1

L 1000m 1L/min) 100(

3

3

1

1 =

==v

V&&m

The refrigeration load is

kW 737.5kJ/kg)77.11730kg/s)(259. 04053.0(

)( 41

=−=−= hhmQL &&

which is the total heat removed from the room. Then, the rate of heat generated by the people in the room is determined from

s

kW 0.67=−−=−−= kW )9.060/250737.5(equipheatpeople QQQQ L&&&&

(c) The power input and the COP are kW 9774.0kJ/kg)30.25942kg/s)(283. 04053.0()( 12in =−=−= hhmW &&

5.87===9774.0737.5COP

in

L

WQ&

&

(d) The reversible COP of the cycle is

38.371)27326/()27334(

11/

1COPrev =−++

=−

=LH TT

The corresponding minimum power input is

kW 1535.038.37kW 737.5

COPrev

Lminin, ===

QW

&&

The minimum mass and volume flow rates are

kg/s 006366.0kJ/kg)30.259(283.42

kW 1535.0

12

minin,min =

−=

−=

hhW

m&

&

L/min 15.7==== /s)m 0002617.0(/kg)m 112kg/s)(0.04 006366.0( 331minmin1, vV m&&


11-66

11-105 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it. Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air. Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes COP COPair-cond heat pump= − = −1 2 2 1 12. .=

Then the rate of cooling (heat absorption from the air) becomes

& & ( . )( .Q Wincooling air-cond= COP kW) kW = 8640 kJ / h= =12 2 2 4

since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.


11-67

11-106 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cpT. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle. (b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system

for a unit mass flow rate at the inlet & &E Ein out= ( &m1 kg / s)= 1 can be expressed as

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

+=

+=+=

&&&

&&&

Warm air Cold air

Compressed air

2 3

1

Canceling cp and solving for T3 gives

K 307.3=×−

=

−=

75.027825.0300

75.025.0 21

3TT

T

Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K.

(c) The entropy balance for this steady flow system can be expressed as with one inlet

and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system for a unit mass flow rate at the inlet

& & &S S Sin out gen− + = 0

( &m1 kg / s)= 1 can be expressed

−+

−=

−+−=−+−=

+−+=−+=

−=

1

3

1

3

1

2

1

2

1312

133122

1323322113322

inoutgen

lnln75.0lnln25.0

)(75.0)(25.0)()(

)(

PP

RTT

cPP

RTT

c

ssssssmssm

smmsmsmsmsmsm

SSS

pp

&&

&&&&&&&

&&&

Substituting the known quantities, the rate of entropy generation is determined to be

0>kW/K 461.0kPa 500kPa 100ln kJ/kg.K) 287.0(

K 300K 3.307ln kJ/kg.K) 005.1(75.0

kPa 500kPa 100ln kJ/kg.K) 287.0(

K 300K 278ln kJ/kg.K) 005.1(25.0gen

=

−+

−=S&

which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.


11-68

(d) For a unit mass flow rate at the inlet ( &m1 kg / s)= 1 , the cooling rate and the power input to the compressor are determined to

kW 5.53=278)K-00kJ/kg.K)(3 5kg/s)(1.00 25.0(

)()( c1c1cooling

=

−=−= TTcmhhmQ pcc &&&

kW 1.1571kPa 100kPa 500

80.0)14.1(K) 00kJ/kg.K)(3 7kg/s)(0.28 1(

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

=

−

−=

−

−

=

−

− kk

PP

kRTm

Wη

&&

Then the COP of the vortex refrigerator becomes

0350.kW 1.157kW 5.53COP

in comp,

cooling ===W

Q&

&

The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is

COP 278 K KCarnot =

−=

−=

TT T

L

H L ( )300 27812.6

Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.


11-69

11-107 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law. Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist. Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cp T. Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle. (b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system

for a unit mass flow rate at the inlet & &E Ein out= ( &m1 kg / s)= 1 can be expressed as

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

+=

+=+=

&&&

&&&

Warm air Cold air

Compressed air

2 3

1

Canceling cp and solving for T3 gives

K 307.3=×−

=

−=

75.027825.0300

75.025.0 21

3TT

T

Therefore, the hot air stream will leave the vortex tube at an average temperature of 307.3 K.

(c) The entropy balance for this steady flow system can be expressed as with one inlet

and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system for a unit mass flow rate at the inlet

& & &S S Sin out gen− + = 0

( &m1 kg / s)= 1 can be expressed

−+

−=

−+−=−+−=

+−+=−+=

−=

1

3

1

3

1

2

1

2

1312

133122

1323322113322

inoutgen

lnln75.0lnln25.0

)(75.0)(25.0)()(

)(

PP

RTT

cPP

RTT

c

ssssssmssm

smmsmsmsmsmsm

SSS

pp

&&

&&&&&&&

&&&

Substituting the known quantities, the rate of entropy generation is determined to be

0>kW/K 5130kPa 600kPa 100kJ/kg.K)ln 2870(

K300 K3307kJ/kg.K)ln 0051(750

kPa 600kPa 100kJ/kg.K)ln 2870(

K300 K278kJ/kg.K)ln 0051(250gen

.

....

...S

=

−+

−=&

which is a positive quantity. Therefore, this process satisfies the 2nd law of thermodynamics.


11-70

(d) For a unit mass flow rate at the inlet ( &m1 kg / s)= 1 , the cooling rate and the power input to the compressor are determined to

kW5.53=278)K-00kJ/kg.K)(3 5kg/s)(1.00 25.0(

)()( c1c1cooling

=

−=−= TTcmhhmQ pcc &&&

kW 9.1791kPa 100kPa 600

80.0)14.1(K) 00kJ/kg.K)(3 7kg/s)(0.28 1(

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

=

−

−=

−

−

=

−

− kk

PP

kRTm

Wη

&&

Then the COP of the vortex refrigerator becomes

0310.kW 9.179kW 5.53COP

in comp,

cooling ===W

Q&

&

The COP of a Carnot refrigerator operating between the same temperature limits of 300 K and 278 K is

12.6=−

=−

=K )278300(

K 278COPCarnotLH

L

TTT

Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnot refrigerator operating between the same temperature limits.


11-71

11-108 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=100 [kPa] P[2] = 1000 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"

COP ηc P1 [kPa] 1.851 0.7 100 2.863 0.7 200 4.014 0.7 300 5.462 0.7 400 7.424 0.7 500

100 150 200 250 300 350 400 450 5000

2

4

6

8

10

P[1] [kPa]

COP

η comp

1.01.00.70.7


11-72

11-109 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" P[1]=120 [kPa] P[2] = 400 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"

400 600 800 1000 1200 14000

1

2

3

4

5

6

7

8

P[2] [kPa]

COP

η comp

1.01.00.70.7

COP ηc P2 [kPa] 4.935 0.7 400 3.04 0.7 650

2.258 0.7 900 1.803 0.7 1150 1.492 0.7 1400


11-73

Fundamentals of Engineering (FE) Exam Problems

11-110 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg (d) 144 kJ/kg (e) 275 kJ/kg Answer (a) 28 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=800 "kPa" P2=140 "kPa" h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP "Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation" 11-111 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it to an environment at 20°C. The minimum required power input is (a) 30 W (b) 33 W (c) 56 W (d) 124 W (e) 350 W Answer (b) 33 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=20+273 TL=-5+273 Q_L=0.35 "kJ/s" COP_max=TL/(TH-TL) w_min=Q_L/COP_max "Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"


11-74

11-112 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is (a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s (d) 0.28 kg/s (e) 0.81 kg/s Answer (c) 0.23 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=800 "kPa" P3=P2 P4=P1 s2=s1 Q_refrig=32 "kJ/s" m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" 11-113 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is (a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kW Answer (d) 31 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 m=0.193 "kg/s" Q_supply=m*(h2-h3) "kJ/s" h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2)


11-75

h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1" 11-114 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) 0.35 Answer (b) 0.60 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=1000 "kPa" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3) 11-115 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performance of this heat pump is (a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9 Answer (e) 5.9 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2


11-76

P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" 11-116 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of this cycle is (a) –58°C (b) -26°C (c) 0°C (d) 11°C (e) 24°C Answer (a) –58°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 P1= 80 "kPa" P2=280 "kPa" T3=35+273 "K" "Mimimum temperature is the turbine exit temperature" T4=T3*(P1/P2)^((k-1)/k) - 273 "Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent" 11-117 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is (a) 9.3 kW (b) 27.6 kW (c) 48.8 kW (d) 93.5 kW (e) 119 kW Answer (b) 27.6 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values).


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k=1.667 Cp=5.1926 "kJ/kg.K" P1= 100 "kPa" T1=-10+273 "K" P2=250 "kPa" T3=20+273 "K" m=0.2 "kg/s" "Mimimum temperature is the turbine exit temperature" T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) W_netin=m*Cp*((T2-T1)-(T3-T4)) "Some Wrong Solutions with Common Mistakes:" W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*Cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB-1)/kB); kB=1.4 "Using air properties" W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" 11-118 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermal source at 140°C. The minimum rate of heat supply required is (a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s (d) 61 kJ/s (e) 150 kJ/s Answer (c) 30 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=20+273 "K" Q_refrig=150 "kJ/s" To=35+273 "K" Ts=140+273 "K" COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max "Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"


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11-119 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is (a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4 Answer (d) 3.2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=160 "kPa" P2=800 "kPa" T2=50 "C" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression" 11-120 ··· 11-129 Design and Essay Problems


gas refrigeration cycles · pdf file11-38 11-56 [also solved by ees on enclosed cd] an...

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