Solução

11-1

Chapter 11

REFRIGERATION CYCLES

The Reversed Carnot Cycle

11-1C Because the compression process involves the compression of a liquid-vapor mixture which

requires a compressor that will handle two phases, and the expansion process involves the expansion of

high-moisture content refrigerant.

11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The

coefficient of performance, the amount of heat absorbed from the refrigerated space, 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) Noting that TH = 30 C = 303 K and TL = Tsat @ 160 kPa = -15.60 C = 257.4 K, the COP of this

Carnot refrigerator is determined from

5.641K 4.257/K 303

1

1/

1COP CR,

LH TT

(b) From the refrigerant tables (Table A-11),

kJ/kg 58.93

kJ/kg 66.266

C30@4

C30@3

f

g

hh

hh

Thus,

and

kJ/kg 147.03kJ/kg 173.08K 303

K 257.4

kJ/kg 08.17358.9366.26643

HH

LL

L

H

L

H

H

qT

Tq

T

T

q

q

hhq

30 C

4 3

2 1

160 kPa

QL

QH

T

s

(c) The net work input is determined from

kJ/kg 26.0503.14708.173net LH qqw

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.

11-2

11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered.

The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net

work input are to be determined.


Analysis (a) Noting that TH = Tsat @ 90 psia = 72.78 F = 532.8 R and TL = Tsat @ 30 psia = 15.37 F = 475.4 R.

8.281R 475.4/R 532.8

1

1/

1COP CR,

LH TT

(b) Process 4-1 is isentropic, and thus

0.237418589.0

03793.008207.0

RBtu/lbm 0.08207

14525.005.007481.0

psia 30 @

1

1

psia 90 @ 441

fg

f

fgf

s

ssx

sxsss

(c) Remembering that on a T-s diagram the area enclosed

represents the net work, and s3 = sg @ 90 psia = 0.22006 Btu/lbm·R,

Btu/lbm7.92 RBtu/lbm 08207.022006.0)37.1578.72(43innet, ssTTw LH

Ideal and Actual Vapor-Compression Cycles

11-4C Yes; the throttling process is an internally irreversible process.

11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual

cycle.

11-6C No. Assuming the water is maintained at 10 C in the evaporator, the evaporator pressure will be

the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design

refrigeration or air-conditioning devices that involve such extremely low pressures.

11-7C Allowing a temperature difference of 10 C for effective heat transfer, the condensation temperature

of the refrigerant should be 25 C. The saturation pressure corresponding to 25 C is 0.67 MPa. Therefore,

the recommended pressure would be 0.7 MPa.

11-8C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the

reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the

latter cycle involves an irreversible process for which the process path is not known.

11-9C The cycle that involves saturated liquid at 30 C will have a higher COP because, judging from the

T-s diagram, it will require a smaller work input for the same refrigeration capacity.

QH

QL

4 3

21

T

s

11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature

of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.



11-3

11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of

the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical

maximum refrigeration load for the same power input to the compressor are to be determined.


Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)

0.479544

4

34

33

3

22

2

11

1

kJ/kg 23.111

kPa 60

kJ/kg 23.111

kJ/kg 23.111C42

kPa 1200

kJ/kg 16.295C65

kPa 1200

kJ/kg 03.230C34

kPa 60

xh

P

hh

hT

P

hT

P

hT

P

26 C Water

18 C

Win

QL

1.2 MPa

65 C

Expansion valve

Compressor

Evaporator

Condenser

42 C

QH

4

3 2

1 60 kPa

-34 C Using saturated liquid enthalpy at the given

temperature, for water we have (Table A-4)

kJ/kg 94.108

kJ/kg 47.75

C26 @2

C18 @1

fw

fw

hh

hh

(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor

kg/s 0455.0

g75.47)kJ/k94kg/s)(108. (0.25kJ/kg)23.11116.295(

)()( 1232

R

R

wwwR

m

m

hhmhhm

The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are

kW 367.8kJ/kg)23.11116kg/s)(295. 0455.0()( 32 hhmQ RH

kW 513.2kW 0.45kJ/kg)03.23016kg/s)(295. 0455.0()( in12in QhhmW R

kW 5.85513.2367.8inWQQ HL

(c) The COP of the refrigerator is determined from its

definition T

QH

QL

Win

·

2

·

·

4

3

2

1

2.33513.2

85.5COP

in

L

W

Q

(d) The reversible COP of the refrigerator for the

same temperature limits is

063.51)27330/()27318(

1

1/

1COPmax

LH TT

s

Then, the maximum refrigeration load becomes

kW 12.72kW) 513.2)(063.5(inmaxmaxL, WCOPQ



11-4

11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is

considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the

rate of heat rejection to the environment, and the COP are to be determined.


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),

throttlingkJ/kg 82.88

kJ/kg 82.88liquid sat.

MPa 7.0

C95.34kJ/kg 50.273MPa 7.0

KkJ/kg 94779.0

kJ/kg 97.236

vaporsat.

kPa 120

34

MPa 7.0 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

hh

hhP

Thss

P

ss

hhP

f

g

g

T

QH

QL 4s

·

Win

·

·

0.7 MPa

4

3

2

1

0.12 MPa

Then the rate of heat removal from the refrigerated

space and the power input to the compressor are

determined from s

and

kW 1.83

kW 7.41

kJ/kg 236.97273.50kg/s 0.05

kJ/kg 82.8897.236kg/s 0.05

12in

41

hhmW

hhmQL

(b) The rate of heat rejection to the environment is determined from

kW 9.2383.141.7inWQQ LH

(c) The COP of the refrigerator is determined from its definition,

4.06kW 1.83

kW 7.41COP

in

RW

QL



11-5


considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the

rate of heat rejection to the environment, and the COP are to be determined.







MPa 9.0

C45.44kJ/kg 93.278MPa 9.0

KkJ/kg 94779.0

kJ/kg 97.236

vaporsat.

kPa 120

34

MPa 9.0 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

hh

hhP

Thss

P

ss

hhP

f

g

g

T

QH

QL 4s

·

Win

·

·

0.9 MPa

4

3

2

1

0.12 MPa

Then the rate of heat removal from the

refrigerated space and the power input to the

compressor are determined from s

and

kW2.10

kW 6.77

kJ/kg 236.97278.93kg/s 0.05

kJ/kg 61.10197.236kg/s 0.05

12in

41

hhmW

hhmQL

(b) The rate of heat rejection to the environment is determined from

kW 8.8710.277.6inWQQ LH


3.23kW 2.10

kW 6.77COP

in

RW

QL



11-6


considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase

in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be

determined.


Analysis If the throttling valve in the previous problem is replaced by an isentropic turbine, we would have

s4s = s3 = sf @ 0.7 MPa = 0.33230 kJ/kg·K, and the enthalpy at the turbine exit would be

kJ/kg 58.8248.2142802.049.22

2802.085503.0

09275.033230.0

kPa 120 @44

kPa 120 @

3

4

fgsfs

fg

f

s

hxhh

s

ssx

T

QH

QL 4s

·

Win ·

·

0.7 MPa

4

3

2

1

0.12 MPa

Then, Q kW 7.72kJ/kg 82.58236.97kg/s 0.0541 sL hhm

and 23.4kW 1.83

kW 7.72COP

inR

W

QL

Then the percentage increase in and COP becomes Q

4.2%

4.2%

06.4

06.423.4

COP

COPCOPin Increase

41.7

41.772.7in Increase

R

RR

L

LL

Q

QQ

s

11-15 [Also solved by EES on enclosed CD] An ideal vapor-compression refrigeration cycle with

refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the end of the

throttling process, the COP, and the power input to the compressor are to be determined.







MPa 8.0

kJ/kg 37.275MPa 8.0

KkJ/kg 94456.0

kJ/kg 16.239

vapor sat.

kPa 140

34

MPa 8.0 @ 33

212

2

kPa 140 @ 1

kPa 140 @ 11

hh

hhP

hss

P

ss

hhP

f

g

g

T

QH

QL

·

Win

·

·

0.8 MPa

4

3

2

1

0.14 MPaThe quality of the refrigerant at the end of the throttling process is

0.32208.212

08.2747.95

kPa 140 @

4

4fg

f

h

hhx

s

(b) The COP of the refrigerator is determined from its definition,

3.9716.23937.275

47.9516.239COP

12

41

inR

hh

hh

w

qL

(c) The power input to the compressor is determined from

kW 1.2697.3

kW)60/300(

COPRin

LQW



11-7

11-16 EES Problem 11-15 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]=140 [kPa]} {P[2] = 800 [kPa] Fluid$='R134a' Eta_c=1.0 "Compressor isentropic efficiency" Q_dot_in=300/60 "[kJ/s]"} "Compressor" x[1]=1 "assume inlet to be saturated vapor" h[1]=enthalpy(Fluid$,P=P[1],x=x[1]) T[1]=temperature(Fluid$,h=h[1],P=P[1]) "properties for state 1" s[1]=entropy(Fluid$,T=T[1],x=x[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" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" P[3]=P[2] "neglect pressure drops across condenser" T[3]=temperature(Fluid$,h=h[3],P=P[3]) "properties for state 3" h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,T=T[3],x=0) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "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] "neglect pressure drop across evaporator" 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

P1 [kPa] COPplot Win [kW] 100 3.216 1.554

175 4.656 1.074

250 6.315 0.7918

325 8.388 0.5961

400 11.15 0.4483



11-8

0,0 0,2 0,4 0,6 0,8 1,0 1,2-50

-25

0

25

50

75

100

125

s [kJ/kg-K]

T[C]

800 kPa

140 kPa

R134a

T-s diagram for = 1.0

0 50 100 150 200 250 300

101

102

103

104

h [kJ/kg]

P [

kP

a]

31.33 C

-18.8 C

R134a

P-h diagram for = 1.0

0,0 0,2 0,4 0,6 0,8 1,0 1,2-50

-25

0

25

50

75

100

125

s [kJ/kg-K]

T[C]

800 kPa

140 kPa

R134a

T-s diagram for = 0.6



11-9

0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00

2.2

2.4

2.6

2.8

3.0

3.2

3.4

3.6

3.8

4.0

Compressor efficiency

CO

P

COP vs Compressor Efficiency for R134a

100 150 200 250 300 350 400

3

4

5

6

7

8

9

10

11

12

P[1] [kPa]

CO

Pp

lot

100 150 200 250 300 350 400

0.4

0.6

0.8

1

1.2

1.4

1.6

P[1] [kPa]

Win



11-10

11-17 A nonideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is

considered. The quality of the refrigerant at the end of the throttling process, the COP, the power input to

the compressor, and the irreversibility rate associated with the compression process are to be determined.


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 8.0

kJ/kg 76.281

85.0/16.23937.27516.239

/

kJ/kg 37.275MPa 8.0

KkJ/kg 94456.0

kJ/kg 16.239

vapor.sat

kPa 140

34

MPa 8.0 @ 33

121212

12

212

2

kPa 140 @1

kPa 140 @ 11

hh

hhP

hhhhhh

hh

hss

P

ss

hhP

f

Css

C

ss

g

g

T2

QH · 2s

Win

·

3 0.8 MPa

0.14 MPa1

QL ·4

s

The quality of the refrigerant at the end of the throttling process is

0.32208.212

08.2747.95

kPa 140 @

4

4fg

f

h

hhx

(b) The COP of the refrigerator is determined from its definition,

3.3716.23976.281

47.9516.239COP

12

41

inR

hh

hh

w

qL

(c) The power input to the compressor is determined from

kW 1.4837.3

kW 5

COPin

R

LQW

The exergy destruction associated with the compression process is determined from

120

0

0

surr120gen0destroyed ssmT

T

qssmTSTX

where

KkJ/kg 96483.0kJ/kg 76.281

MPa 8.0

kg/s 0348.0kJ/kg 95.47239.16

kJ/s 5

22

2

41

sh

P

hh

Q

q

Qm L

L

L

Thus,

kW 0.210KkJ/kg 0.944560.96483kg/s 0.0348K 298destroyedX



11-11

11-18 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal

from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor,

and the COP of the refrigerator are to be determined.


Analysis (a) From the refrigerant tables (Tables A-12 and A-13),


kJ/kg 98.84C24

MPa 65.0

kJ/kg 16.281MPa 7.0

kJ/kg 53.288C50

MPa 7.0

KkJ/kg 97236.0

kJ/kg 36.246

C10

MPa 14.0

34

C24 @ 33

3

212

2

22

2

1

1

1

1

hh

hhT

P

hss

P

hT

P

s

h

T

P

f

ss

s

T

QH

QL

0.7 MPa

50 C

0.14 MPa

-10 C

Win

·

2

·

·

s

0.65 MPa

24 C

4

3

2s

1 0.15 MPa

Then the rate of heat removal from the refrigerated space and

the power input to the compressor are determined from

and

kW 5.06

kW19.4

kJ/kg 246.36288.53kg/s 0.12

kJ/kg 84.98246.36kg/s 0.12

12in

41

hhmW

hhmQL

(b) The adiabatic efficiency of the compressor is determined from

82.5%36.24653.288

36.24616.281

12

12

hh

hh sC


3.83kW 5.06

kW 19.4COP

in

RW

QL



11-12

11-19E An ice-making machine operates on the ideal vapor-compression refrigeration cycle, using

refrigerant-134a as the working fluid. The power input to the ice machine is to be determined.


Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the


as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12E and A-13E),

throttlingBtu/lbm 39.33

Btu/lbm 39.33liquid sat.

psia 80

Btu/lbm 00.115psia 80

RBtu/lbm 22567.0

Btu/lbm 73.102

vapor sat.

psia 20

34

psia 08 @ 33

212

2

psia 20 @ 1

psia 20 @ 11

hh

hhP

hss

P

ss

hhP

f

g

g

T

QH

QL

·

Win

·

·

80 psia

4

3

2

120 psia

sThe cooling load of this refrigerator is

Btu/s 0.7042Btu/lbm 169lbm/s 15/3600iceice hmQL

Then the mass flow rate of the refrigerant and the power input become

and

hp 0.176Btu/s 0.7068

hp 1Btu/lbm 102.73115.00lbm/s 0.01016

lbm/s 0.01016Btu/lbm 33.39102.73

Btu/s 0.7042

12in

41

hhmW

hh

Qm

R

LR



11-13

11-20 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the

compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat

gain in the line between the evaporator and the compressor are to be determined.


Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

kJ/kg 33.239

MPa 14165.0

vapor sat.

C5.18


kJ/kg 58.93C30

MPa 95.0

kJ/kg 20.289MPa 0.1

/kgm 14605.0

KkJ/kg 97236.0

kJ/kg 36.246

C10

kPa 140

5

55

34

C 30 @ 33

3

212

2

31

1

1

1

1

h

PT

hh

hhT

P

hss

P

s

h

T

P

f

ss

v

T

QH

QL -18.5 C

1 MPa

0.14 MPa

-10 C

Win

·

2

·

·

s

0.95 MPa

30 C

4

3

2s

1 0.15 MPa

Then the mass flow rate of the refrigerant and the power input becomes

kW1.88 78.0/kJ/kg 246.36289.20kg/s 0.03423/

kg/s 0.03423/kgm 0.14605

/sm 0.3/60

12in

3

3

1

1

Cs hhmW

mv

V

(b) The rate of heat removal from the refrigerated space is

kW 4.99kJ/kg 93.58239.33kg/s 0.0342345 hhmQL

(c) The pressure drop and the heat gain in the line between the evaporator and the compressor are

and

kW0.241

1.65

kJ/kg 239.33246.36kg/s 0.03423

14065.141

51gain

15

hhmQ

PPP



11-14

11-21 EES Problem 11-20 is reconsidered. The effects of the compressor isentropic efficiency and the

compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated.

Analysis The problem is solved using EES, and the solution is given below.

"Input Data" "T[5]=-18.5 [C] P[1]=140 [kPa] T[1] = -10 [C]} V_dot[1]=0.1 [m^3/min] P[2] = 1000 [kPa] P[3]=950 [kPa] T[3] = 30 [C] Eta_c=0.78 Fluid$='R134a'" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],T=T[1]) v[1]=volume(Fluid$,P=P[1],T=T[1])"[m^3/kg]" m_dot=V_dot[1]/v[1]*convert(m^3/min,m^3/s)"[kg/s]" 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" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],T=T[3]) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,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]=pressure(Fluid$,T=T[5],x=0)"pressure=Psat at evaporator exit temp." P[5] = P[4] h[5]=enthalpy(Fluid$,T=T[5],x=1) "properties for state 5" q_in + h[4]=h[5] "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 Q_dot_line5to1=m_dot*(h[1]-h[5])"[kW]"



11-15

COPplot Win [kW]

Qin [kW]

c [kW]

2.041 0.8149 1.663 0.6

2.381 0.6985 1.663 0.7

2.721 0.6112 1.663 0.8

3.062 0.5433 1.663 0.9

3.402 0.4889 1.663 1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0

1

2

3

4

5

6

7

8

9

c

Win

V1 m3/min

1.01.0 0.50.5 0.10.1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0

0.5

1

1.5

2

2.5

3

3.5

4

c

CO

Pp

lot

V1 m3/min

1.01.0 0.50.5 0.10.1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

0

3.6

7.2

10.8

14.4

18

c

Qin

[k

W] 1.0 1.0 0.50.5 0.10.1

V1 m3/min



11-16

11-22 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-

compression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, and the COP

of the refrigerator are to be determined.


Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13)

4

32

1

Win .

120 kPa x=0.3

Expansion valve

Compressor

Evaporator

Condenser

60 C

QH

.

.QL

kJ/kg 97.236 vap.)(sat. 1

kPa 120

kJ/kg 87.298C60

kPa 8.671

liq.) (sat. 0

kJ/kg 83.86

kJ/kg 83.8630.0

kPa 120

11

41

22

2

32

33

3

43

44

4

hx

PP

hT

P

PP

Px

h

hh

hx

P

kPa 671.8

The mass flow rate of the refrigerant is

determined from T

QH

QL 4s

·

Win ·

·4

3

2

1

0.12 MPa

kg/s 0.00727kg236.97)kJ/(298.87

kW 45.0

12

in

hh

Wm

(c) The refrigeration load and the COP are

Q kW 091.1kJ/kg)83.8697kg/s)(236. 0727.0()( 41 hhmL

2.43kW 0.45

kW 091.1COP

in

L

W

Q

s



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