ThermoSolutions CHAPTER11

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. Thecoefficient 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 thisCarnot refrigerator is determined from

5.641K4.257/K303

11/

1COP CR,LH TT

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

kJ/kg58.93kJ/kg66.266

C30@4

C30@3

f

ghhhh

Thus,

and

kJ/kg147.03kJ/kg173.08K303K257.4

kJ/kg08.17358.9366.26643

HH

LL

L

H

L

H

H

qTT

qTT

qq

hhq

30 C4 3

21160 kPa

QL

QH

T

s

(c) The net work input is determined fromkJ/kg26.0503.14708.173net LH qqw

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 network input are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.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.

8.281R475.4/R532.8

11/

1COP CR,LH TT

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

0.237418589.0

03793.008207.0

RBtu/lbm0.08207

14525.005.007481.0

psia30@

11

psia90@441

fg

f

fgf

sss

x

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/lbm7.92RBtu/lbm08207.022006.0)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 actualcycle.

11-6C No. Assuming the water is maintained at 10 C in the evaporator, the evaporator pressure will bethe saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to designrefrigeration 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 temperatureof 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 thereversed 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 temperatureof 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 ofthe refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoreticalmaximum refrigeration load for the same power input to the compressor are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)

0.479544

4

34

33

3

22

2

11

1

kJ/kg23.111kPa60

kJ/kg23.111

kJ/kg23.111C42

kPa1200

kJ/kg16.295C65

kPa1200

kJ/kg03.230C34

kPa60

xhP

hh

hTP

hTP

hTP

26 C Water18 C

Win

QL

1.2 MPa 65 C

Expansionvalve

Compressor

Evaporator

Condenser42 C

QH

4

3 2

1 60 kPa -34 C

Using saturated liquid enthalpy at the giventemperature, for water we have (Table A-4)

kJ/kg94.108

kJ/kg47.75

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/s0455.0

g75.47)kJ/k94kg/s)(108.(0.25kJ/kg)23.11116.295()()( 1232

R

R

wwwR

m

mhhmhhm

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

kW367.8kJ/kg)23.11116kg/s)(295.0455.0()( 32 hhmQ RH

kW513.2kW0.45kJ/kg)03.23016kg/s)(295.0455.0()( in12in QhhmW R

kW5.85513.2367.8inWQQ HL

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

T

QH

QL

Win·

2

·

·4

3

2

1

2.33513.285.5COP

in

L

WQ

(d) The reversible COP of the refrigerator for thesame temperature limits is

063.51)27330/()27318(

11/

1COPmaxLH TT s

Then, the maximum refrigeration load becomes

kW12.72kW)513.2)(063.5(inmaxmaxL, WCOPQ

11-4

11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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.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, therefrigerant 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/kg82.88

kJ/kg82.88liquidsat.

MPa7.0

C95.34kJ/kg50.273MPa7.0

KkJ/kg94779.0kJ/kg97.236

vaporsat.kPa120

34

MPa7.0@33

2212

2

kPa120@1

kPa120@11

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL4s

·

Win·

·

0.7 MPa

4

32

10.12 MPa

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

s

andkW1.83

kW7.41

kJ/kg236.97273.50kg/s0.05

kJ/kg82.8897.236kg/s0.05

12in

41

hhmW

hhmQL

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

kW9.2383.141.7inWQQ LH

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

4.06kW1.83kW7.41COP

inR W

QL

11-5

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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.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, therefrigerant 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),



MPa9.0

C45.44kJ/kg93.278MPa9.0

KkJ/kg94779.0kJ/kg97.236

vaporsat.kPa120

34

MPa9.0@33

2212

2

kPa120@1

kPa120@11

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL4s

·

Win·

·

0.9 MPa

4

32

10.12 MPa

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from s

andkW2.10

kW6.77

kJ/kg236.97278.93kg/s0.05

kJ/kg61.10197.236kg/s0.05

12in

41

hhmW

hhmQL

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

kW8.8710.277.6inWQQ LH


3.23kW2.10kW6.77COP

inR W

QL

11-6

11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid isconsidered. 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.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.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/kg58.8248.2142802.049.22

2802.085503.0

09275.033230.0

kPa120@44

kPa120@

34

fgsfs

fg

fs

hxhh

sss

x T

QH

QL4s

·

Win·

·

0.7 MPa

4

32

10.12 MPa

Then, Q kW7.72kJ/kg82.58236.97kg/s0.0541 sL hhm

and 23.4kW1.83kW7.72COP

inR W

QL

Then the percentage increase in and COP becomesQ

4.2%

4.2%

06.406.423.4

COPCOP

COPinIncrease

41.741.772.7inIncrease

R

RR

L

LL Q

QQ s

11-15 [Also solved by EES on enclosed CD] An ideal vapor-compression refrigeration cycle withrefrigerant-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.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, therefrigerant 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),



MPa8.0

kJ/kg37.275MPa8.0

KkJ/kg94456.0kJ/kg16.239

vaporsat.kPa140

34

MPa8.0@33

212

2

kPa140@1

kPa140@11

hh

hhP

hss

P

sshhP

f

g

g

T

QH

QL

·

Win·

·

0.8 MPa

4

32

10.14 MPa

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

0.32208.212

08.2747.95

kPa140@

44

fg

f

hhh

xs

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

3.9716.23937.275

47.9516.239COP12

41

inR hh

hhwqL

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

kW1.2697.3

kW)60/300(COPR

inLQ

W

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_inCOP=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 1.554175 4.656 1.074250 6.315 0.7918325 8.388 0.5961400 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 300101

102

103

104

h [kJ/kg]

P [k

Pa]

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 1.002.2

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 4003

4

5

6

7

8

9

10

11

12

P[1] [kPa]

CO

Ppl

ot

100 150 200 250 300 350 4000.4

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 isconsidered. The quality of the refrigerant at the end of the throttling process, the COP, the power input tothe compressor, and the irreversibility rate 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),



MPa8.0

kJ/kg76.28185.0/16.23937.27516.239

/

kJ/kg37.275MPa8.0

KkJ/kg94456.0kJ/kg16.239

vapor.satkPa140

34

MPa8.0@33

121212

12

212

2

kPa140@1

kPa140@11

hh

hhP

hhhhhhhh

hss

P

sshhP

f

Css

C

ss

g

gT

2

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

kPa140@

44

fg

f

hhh

x

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

3.3716.23976.281

47.9516.239COP12

41

inR hh

hhwqL

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

kW1.4837.3kW5

COPinR

LQW

The exergy destruction associated with the compression process is determined from

120

0

0

surr120gen0destroyed ssmT

Tq

ssmTSTX

where

KkJ/kg96483.0kJ/kg76.281

MPa8.0

kg/s0348.0kJ/kg95.47239.16

kJ/s5

22

2

41

shP

hhQ

qQ

m L

L

L

Thus,

kW0.210KkJ/kg0.944560.96483kg/s0.0348K298destroyedX

11-11

11-18 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removalfrom 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.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13),


kJ/kg98.84C24MPa65.0

kJ/kg16.281MPa7.0


KkJ/kg97236.0kJ/kg36.246

C10MPa14.0

34

C24@33

3

212

2

22

2

1

1

1

1

hh

hhTP

hss

P

hTP

sh

TP

f

ss

s

T

QH

QL

0.7 MPa 50 C

0.14 MPa-10 C

Win·

2

·

·

s

0.65 MPa24 C

4

3

2s

10.15 MPa

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

and

kW5.06

kW19.4

kJ/kg246.36288.53kg/s0.12

kJ/kg84.98246.36kg/s0.12

12in

41

hhmW

hhmQL

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

82.5%36.24653.28836.24616.281

12

12

hhhh s

C


3.83kW5.06kW19.4COP

inR W

QL

11-12

11-19E An ice-making machine operates on the ideal vapor-compression refrigeration cycle, usingrefrigerant-134a as the working fluid. The power input to the ice machine is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant 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-12E and A-13E),

throttlingBtu/lbm39.33

Btu/lbm39.33liquidsat.

psia80

Btu/lbm00.115psia80

RBtu/lbm22567.0Btu/lbm73.102

vaporsat.psia20

34

psia08@33

212

2

psia20@1

psia20@11

hh

hhP

hss

P

sshhP

f

g

g T

QH

QL

·

Win·

·

80 psia

4

32

120 psia

sThe cooling load of this refrigerator is

Btu/s0.7042Btu/lbm169lbm/s15/3600iceice hmQL

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

and

hp0.176Btu/s0.7068

hp1Btu/lbm102.73115.00lbm/s0.01016

lbm/s0.01016Btu/lbm33.39102.73

Btu/s0.7042

12in

41

hhmW

hhQm

R

LR

11-13

11-20 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to thecompressor, 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.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

kJ/kg33.239MPa14165.0

vaporsat.C5.18



kJ/kg20.289MPa0.1

/kgm14605.0KkJ/kg97236.0

kJ/kg36.246

C10kPa140

5

55

34

C30@33

3

212

2

31

1

1

1

1

hPT

hh

hhTP

hss

P

sh

TP

f

ss

vT

QH

QL -18.5 C

1 MPa

0.14 MPa-10 C

Win·

2

·

·

s

0.95 MPa30 C

4

3

2s

10.15 MPa

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

kW1.8878.0/kJ/kg246.36289.20kg/s0.03423/

kg/s0.03423/kgm0.14605/sm0.3/60

12in

3

3

1

1

Cs hhmW

mvV

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

kW4.99kJ/kg93.58239.33kg/s0.0342345 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/kg239.33246.36kg/s0.03423

14065.141

51gain

15

hhmQ

PPP

11-14

11-21 EES Problem 11-20 is reconsidered. The effects of the compressor isentropic efficiency and thecompressor 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=0.78Fluid$='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_inCOP=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 0.62.381 0.6985 1.663 0.72.721 0.6112 1.663 0.83.062 0.5433 1.663 0.93.402 0.4889 1.663 1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10

1

2

3

4

5

6

7

8

9

c

Win

V1 m3/min1.01.0 0.50.5 0.10.1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10

0.5

1

1.5

2

2.5

3

3.5

4

c

CO

Ppl

ot V1 m3/min1.01.0 0.50.5 0.10.1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10

3.6

7.2

10.8

14.4

18

c

Qin

[kW

] 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 COPof the refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13)

4

3 2

1

Win.

120 kPa x=0.3

Expansionvalve

Compressor

Evaporator

Condenser60 C

QH

.

.QL

kJ/kg97.236 vap.)(sat.1

kPa120

kJ/kg87.298C60

kPa8.671

liq.)(sat.0kJ/kg83.86

kJ/kg83.8630.0

kPa120

11

41

22

2

32

33

3

43

44

4

hx

PP

hTP

PP

Pxh

hh

hxP

kPa671.8

The mass flow rate of the refrigerant is determined from T

QH

QL4s

·

Win·

·4

32

10.12 MPa

kg/s0.00727kg236.97)kJ/(298.87

kW45.0

12

in

hhW

m

(c) The refrigeration load and the COP are

Q kW091.1kJ/kg)83.8697kg/s)(236.0727.0()( 41 hhmL

2.43kW0.45kW091.1COP

in

L

WQ

s

11-17

Selecting the Right Refrigerant

11-23C The desirable characteristics of a refrigerant are to have an evaporator pressure which is above theatmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above thetemperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic,noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes themass flow rate) and, of course, being available at low cost.

11-24C The minimum pressure that the refrigerant needs to be compressed to is the saturation pressure ofthe refrigerant at 30 C, which is 0.771 MPa. At lower pressures, the refrigerant will have to condense at temperatures lower than the temperature of the surroundings, which cannot happen.

11-25C Allowing a temperature difference of 10 C for effective heat transfer, the evaporation temperatureof the refrigerant should be -20 C. The saturation pressure corresponding to -20 C is 0.133 MPa.Therefore, the recommended pressure would be 0.12 MPa.

11-26 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a isconsidered. Reasonable pressures for the evaporator and the condenser are to be selected.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis Allowing a temperature difference of 10 C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be -20 C and 35 C, respectively. The saturationpressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.133 MPa and 0.888 MPa, respectively.

11-27 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered.Reasonable pressures for the evaporator and the condenser are to be selected.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis Allowing a temperature difference of 10 C for effective heat transfer, the evaporation and condensation temperatures of the refrigerant should be 0 C and 32 C, respectively. The saturationpressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommendedevaporator and condenser pressures are 0.293 MPa and 0.816 MPa, respectively.

Heat Pump Systems

11-28C A heat pump system is more cost effective in Miami because of the low heating loads and highcooling loads at that location.

11-29C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps havehigher COPs than the air-source systems because the temperature of water is higher than the temperature of air in winter.

11-18

11-30E A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a isconsidered. The power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant 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-12E and A-13E),

throttlingBtu/lbm79.41

Btu/lbm79.41liquidsat.

psia120

Btu/lbm62.116psia120

RBtu/lbm22188.0Btu/lbm81.108

vaporsat.psia50

34

psia120@33

212

2

psia50@1

psia50@11

hh

hhP

hss

P

sshhP

f

g

g

QH

QL

House·

Win·

·

120 psia

4

32

150 psia

T

The mass flow rate of the refrigerant and the power input to the compressor are determined from s

and

Btu/s0.7068=hp1sinceBtu/s.7381Btu/lbm108.81116.62kg/s0.2227

lbm/s0.2227Btu/lbm41.79116.62Btu/s060,000/360

12in

32

hp2.46hhmW

hhQ

qQ

m H

H

H

The electrical power required without the heat pump is

Thus,

kW0.7457=hp1sincekW75.1546.258.23

hp.5823Btu/s0.7068

hp1Btu/s060,000/360

insaved

hp21.1WWW

QW

e

He

11-19

11-31 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is considered.The power input to the heat pump is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, therefrigerant 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),



MPa4.1

kJ/kg54.282MPa4.1

KkJ/kg93006.0kJ/kg88.251

vaporsat.kPa320

34

MPa4.1@33

212

2

kPa320@1

kPa320@11

hh

hhP

hss

P

sshhP

f

g

g

QH

QL

House·

Win·

·

1.4 MPa

4

32

10.32 MPa

T

The heating load of this heat pump is determined from

kW.0515C1545CkJ/kg4.18kg/s0.12water12 TTcmQH s

and

Then,

kW2.97kJ/kg251.88282.54kg/s0.09688

kg/s0.09688kJ/kg127.22282.54

kJ/s15.05

12in

32

hhmW

hhQ

qQm

R

H

H

HR

11-20

11-32 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source. The power input to the heat pump, the rate of heat absorption from the water, and theincrease in electric power input if an electric resistance heater is used instead of a heat pump are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

QH

QL

1 MPa

Water, 8 C

60 CHouse

0 C

Win·

2

·

·

s

4

330 C

10.28 MPa

T


kJ/kg58.93C30MPa0.1


kJ/kg83.250C0kPa280

34

C30@33

3

22

2

11

1

hh

hhTP

hTP

hTP

f

The mass flow rate of the refrigerant is

kg/s0.08341kJ/kg93.58293.38

kJ/s0060,000/3,6

32 hhQ

qQm H

H

HR

Then the power input to the compressor becomes

kW3.55kJ/kg250.83293.38kg/s0.0834112in hhmW

(b) The rate of hat absorption from the water is

kW13.12kJ/kg93.58250.83kg/s0.0834141 hhmQL

(c) The electrical power required without the heat pump is

Thus,

kW13.1255.367.16

kW16.67kJ/s3600/000,60

inincrease WWW

QW

e

He

11-21

11-33 EES Problem 11-32 is reconsidered. The effect of the compressor isentropic efficiency on the power input to the compressor and the electric power saved by using a heat pump rather than electric resistanceheating is to be investigated.Analysis The problem is solved using EES, and the solution is given below.

"Input Data""Input Data is supplied in the diagram window""P[1]=280 [kPa] T[1] = 0 [C] P[2] = 1000 [kPa] T[3] = 30 [C] Q_dot_out = 60000 [kJ/h] Eta_c=1.0Fluid$='R134a'""Use ETA_c = 0.623 to obtain T[2] = 60C"

"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])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$,P=P[3],T=T[3]) "properties for state 3"s[3]=entropy(Fluid$,P=P[3],T=T[3])h[2]=Qout+h[3] "energy balance on condenser"Q_dot_out*convert(kJ/h,kJ/s)=m_dot*Qout


"Evaporator"P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator"Q_dot_in=m_dot*Q_inCOP=Q_dot_out*convert(kJ/h,kJ/s)/W_dot_c "definition of COP"COP_plot = COP W_dot_in = W_dot_c E_dot_saved = Q_dot_out*convert(kJ/h,kJ/s) - W_dot_c"[kW]"

11-22

Win [kW] c Esaved

3.671 0.6 133.249 0.7 13.422.914 0.8 13.752.641 0.9 14.032.415 1 14.25

0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 12,4

2,6

2,8

3

3,2

3,4

3,6

3,8

c

Win

0,6 0,65 0,7 0,75 0,8 0,85 0,9 0,95 112,75

13,05

13,35

13,65

13,95

14,25

c

Esaved

[kW]

11-23

11-34 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP andthe rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compressioncycle between the same pressure limits are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13)

26.277kPa800

kJ/kg9506.0kJ/kg87.247

C)409.10(kPa200

C09.10kJ/kg91.87

kJ/kg91.87C)306.29(

kPa750

C06.29

kJ/kg76.291C55kPa800

212

2

1

1

1

1

kPasat@200

34

33

3

kPasat@7503

22

2

shss

P

sh

TP

Thh

hTP

TT

hTP

4

3 2

1

Win

.

800 kPa 55 C

Expansionvalve

Compressor

Evaporator

Condenser

750 kPa QH.

.QL

The isentropic efficiency of the compressor is T

QH

QL

Win·

2

·

·4

3

2

1

0.67087.24776.29187.24726.277

12

12

hhhh s

C

(b) The rate of heat supplied to the room is

Q kW3.67kJ/kg)91.8776kg/s)(291.018.0()( 32 hhmH

(c) The power input and the COP are

W kW790.0kJ/kg)87.24776kg/s)(291.018.0()( 12in hhm

s4.64

790.067.3COP

inWQH

(d) The ideal vapor-compression cycle analysis of the cycle is as follows: T

QH

QL4s

·

Win·

·

0.8 MPa

4

32

10.2 MPa

kJ/kg.K9377.0kJ/kg46.244

kPa200@1

kPa200@1

g

g

sshh

kJ/kg25.273MPa800

212

2 hss

P

34

kPa800@3 kJ/kg47.95hhhh f

s6.18

46.24425.27347.9525.273COP

12

32

hhhh

kW3.20kJ/kg)47.9525kg/s)(273.018.0()( 32 hhmQH

11-24

11-35 A geothermal heat pump is considered. The degrees of subcooling done on the refrigerant in thecondenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimumpower input 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)

40 CWater50 C

Win.

sat. vap.

Expansionvalve Compressor

Evaporator

Condenser

20 Cx=0.23

QH.

4

32

1

1.4 MPa s2 = s1

.QL

kJ/kg00.280kPa1400

kJ/kg9223.0kJ/kg59.261

vap.)(sat.1kPa1.572

kJ/kg24.121kPa1.572

23.0C20

212

2

1

1

1

1

43

4

4

4

4

hss

P

sh

xP

hhhP

xT

From the steam tables (Table A-4)

kJ/kg53.167

kJ/kg34.209

C40@2

C50@1

fw

fw

hh

hh

The saturation temperature at the condenser pressure of 1400 kPa and the actual temperatureat the condenser outlet are

T

QH

QL4s

·

Win·

·

1.4MPa

4

3

2

1

C40.52kPa1400@satT

C59.48kJ24.121

kPa14003

3

3 ThP

(from EES)

Then, the degrees of subcooling isC3.8159.4840.523satsubcool TTT

(b) The rate of heat absorbed from the geothermal water in the evaporator is

s

kW718.2kJ/kg)53.16734kg/s)(209.065.0()( 21 wwwL hhmQThis heat is absorbed by the refrigerant in the evaporator

kg/s0.01936)kJ/kg24.121(261.59

kW718.2

41 hhQ

m LR

(c) The power input to the compressor, the heating load and the COP are kW6564.0kJ/kg)59.26100kg/s)(280.01936.0()( out12in QhhmW R

kW3.074kJ/kg)24.12100kg/s)(280.01936.0()( 32 hhmQ RH

4.68kW0.6564

kW074.3COPin

H

WQ

(d) The reversible COP of the cycle is

92.12)27350/()27325(1

1/1

1COPrevHL TT

The corresponding minimum power input is

kW0.23892.12kW074.3

COPrevminin,

HQW

11-25

Innovative Refrigeration Systems

11-36C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, twoor more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, butthey have higher COP's, they can incorporate two or more different refrigerants, and they can achievemuch lower temperatures.

11-37C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operatingbetween the same pressure limits.

11-38C The saturation pressure of refrigerant-134a at -32 C is 77 kPa, which is below the atmosphericpressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration systemwith a different refrigerant at the bottoming cycle is recommended in this case.

11-39C We would favor the two-stage compression refrigeration system with a flash chamber since it issimpler, cheaper, and has better heat transfer characteristics.

11-40C Yes, by expanding the refrigerant in stages in several throttling devices.

11-41C To take advantage of the cooling effect by throttling from high pressures to low pressures.

11-26

11-42 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through thelower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, andthe COP of this cascade refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compressionrefrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressoras a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturatedliquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from therefrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg47.95kJ/kg91.269

kJ/kg94.63kJ/kg58.260

kJ/kg,47.95,kJ/kg55.255

,kJ/kg94.63,kJ/kg16.239

8

6

4

2

7

5

3

1

hhhh

hhhh

T

The mass flow rate of the refrigerant through thelower cycle is determined from an energy balance on the heat exchanger:

kg/s0.1954kg/s0.2494.6358.26047.9555.255

0

32

85

3285

outin

(steady)0systemoutin

AB

BA

iiee

mhhhh

m

hhmhhm

hmhm

EE

EEE

3

QL

B

A 58

0.8 MPa

6

·

0.4 MPa

4

7 2

1

0.14 MPa

s

(b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:

kW7.63

kW34.24

kJ/kg239.16260.58kg/s0.1954kJ/kg255.55269.91kg/s0.24

kJ/kg63.94239.16kg/s0.1954

1256incompII,incompI,in

41

hhmhhmWWW

hhmQ

BA

BL

(c) The COP of this refrigeration system is determined from its definition,

494.kW7.63kW34.24COP

innet,R W

QL

11-27

11-43 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid. The mass flow rate of refrigerant through thelower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, andthe COP of this cascade refrigerator are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compressionrefrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressoras a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturatedliquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from therefrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg47.95kJ/kg66.268

kJ/kg54.77kJ/kg34.267

kJ/kg,47.95,kJ/kg92.260

,kJ/kg54.77,kJ/kg16.239

8

6

4

2

7

5

3

1

hhhh

hhhh

T

The mass flow rate of the refrigerant through thelower cycle is determined from an energy balance on the heat exchanger:

kg/s0.2092kg/s24.054.7734.26747.9592.260

0

32

85

3285

outin


AB

BA

iiee

mhhhh

m

hhmhhmhmhm

EE

EEE

QL

3B

A 58

0.8 MPa

6

·

0.55 MPa

4

7 2

1

0.14 MPa

s

(b) The rate of heat removed by a cascade cycle is the rate of heat absorption in the evaporator of the lowest stage. The power input to a cascade cycle is the sum of the power inputs to all of the compressors:

kW7.75

kW33.81

kJ/kg16.23934.267kg/s2092.0kJ/kg92.26066.268kg/s24.0

kJ/kg54.7716.239kg/s2092.0

1256incompII,compI,inin

41

hhmhhmWWW

hhmQ

BA

BL

(c) The COP of this refrigeration system is determined from its definition,

4.36kW75.7kW81.33COP

innet,R W

QL

11-28

11-44 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system withrefrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it isthrottled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg33.73kJ/kg32.107

kJ/kg31.265

,kJ/kg33.73,kJ/kg32.107,kJ/kg30.259

kJ/kg,16.239

8

6

2

7

5

3

1

hh

h

hhhh T

QL

73B

A96

1 MPa 4

·

0.5 MPa

8

5 2

1

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

0.182898.185

33.7332.10766

fg

f

hhh

x s

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

kJ/kg21.26431.2651828.0130.2591828.011

0

9

26369

outin


hhxhxh

hmhm

EE

EEE

iiee

also,

kJ/kg97.278KkJ/kg94083.0

MPa14

94

4 hss

P

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

kW.039kJ/kg239.16265.31kg/s0.2043kJ/kg264.21278.97kg/s0.25

kJ/kg73.33239.16kg/s0.2043

kg/s0.2043kg/s0.251828.011


81

6

hhmhhmWWW

hhmQ

mxm

BA

BL

AB

kW33.88

(c) The coefficient of performance is determined from

3.75kW9.03kW33.88COP

innet,R W

QL

11-29

11-45 EES Problem 11-44 is reconsidered. The effects of the various refrigerants in EES data bank for compressor efficiencies of 80, 90, and 100 percent is to be investigated.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Input Data""P[1]=140 [kPa] P[4] = 1000 [kPa]P[6]=500 [kPa] Eta_compB =1.0Eta_compA =1.0"m_dot_A=0.25 [kg/s]

"High Pressure Compressor A"P[9]=P[6]h4s=enthalpy(R134a,P=P[4],s=s[9]) "State 4s is the isentropic value of state 4"h[9]+w_compAs=h4s "energy balance on isentropic compressor"w_compA=w_compAs/Eta_compA"definition of compressor isentropic efficiency"h[9]+w_compA=h[4] "energy balance on real compressor-assumed adiabatic"s[4]=entropy(R134a,h=h[4],P=P[4]) "properties for state 4"T[4]=temperature(R134a,h=h[4],P=P[4])W_dot_compA=m_dot_A*w_compA

"Condenser"P[5]=P[4] "neglect pressure drops across condenser"T[5]=temperature(R134a,P=P[5],x=0) "properties for state 5, assumes sat. liq. at cond. exit"h[5]=enthalpy(R134a,T=T[5],x=0) "properties for state 5"s[5]=entropy(R134a,T=T[5],x=0)h[4]=q_out+h[5] "energy balance on condenser"Q_dot_out = m_dot_A*q_out

"Throttle Valve A"h[6]=h[5] "energy balance on throttle - isenthalpic"x6=quality(R134a,h=h[6],P=P[6]) "properties for state 6"s[6]=entropy(R134a,h=h[6],P=P[6])T[6]=temperature(R134a,h=h[6],P=P[6])

"Flash Chamber"m_dot_B = (1-x6) * m_dot_A P[7] = P[6] h[7]=enthalpy(R134a, P=P[7], x=0) s[7]=entropy(R134a,h=h[7],P=P[7])T[7]=temperature(R134a,h=h[7],P=P[7])

"Mixing Chamber"x6*m_dot_A*h[3] + m_dot_B*h[2] =(x6* m_dot_A + m_dot_B)*h[9] P[3] = P[6]h[3]=enthalpy(R134a, P=P[3], x=1) "properties for state 3"s[3]=entropy(R134a,P=P[3],x=1)T[3]=temperature(R134a,P=P[3],x=x1)s[9]=entropy(R134a,h=h[9],P=P[9]) "properties for state 9"T[9]=temperature(R134a,h=h[9],P=P[9])

"Low Pressure Compressor B"x1=1 "assume flow to compressor inlet to be saturated vapor"h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1"

11-30

T[1]=temperature(R134a,P=P[1], x=x1) s[1]=entropy(R134a,P=P[1],x=x1)P[2]=P[6]h2s=enthalpy(R134a,P=P[2],s=s[1]) " state 2s is isentropic state at comp. exit"h[1]+w_compBs=h2s "energy balance on isentropic compressor"w_compB=w_compBs/Eta_compB"definition of compressor isentropic efficiency"h[1]+w_compB=h[2] "energy balance on real compressor-assumed adiabatic"s[2]=entropy(R134a,h=h[2],P=P[2]) "properties for state 2"T[2]=temperature(R134a,h=h[2],P=P[2])W_dot_compB=m_dot_B*w_compB

"Throttle Valve B"h[8]=h[7] "energy balance on throttle - isenthalpic"x8=quality(R134a,h=h[8],P=P[8]) "properties for state 8"s[8]=entropy(R134a,h=h[8],P=P[8])T[8]=temperature(R134a,h=h[8],P=P[8])

"Evaporator"P[8]=P[1] "neglect pressure drop across evaporator"q_in + h[8]=h[1] "energy balance on evaporator"Q_dot_in=m_dot_B*q_in

"Cycle Statistics"W_dot_in_total = W_dot_compA + W_dot_compB COP=Q_dot_in/W_dot_in_total "definition of COP"

compA compB Qout COP0,8 0,8 45,32 2,963

0,8333 0,8333 44,83 3,0940,8667 0,8667 44,39 3,225

0,9 0,9 43,97 3,3570,9333 0,9333 43,59 3,4880,9667 0,9667 43,24 3,619

1 1 42,91 3,751

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] 1000 kPa

500 kPa

140 kPa

R134a

1

2

39

45

67

8

11-31

0,8 0,84 0,88 0,92 0,96 142,5

43

43,5

44

44,5

45

45,5

2,9

3

3,1

3,2

3,3

3,4

3,5

3,6

3,7

3,8

comp

Qout

[kW]

COP

200 300 400 500 600 700 800 900 1000 11003.45

3.50

3.55

3.60

3.65

3.70

3.75

3.80

P[6] [kPa]

CO

P

COP vs Flash Chamber Pressure, P6

11-32

11-46 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system withrefrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it isthrottled to the flash chamber, the rate of heat removed from the refrigerated space, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3The flash chamber is adiabatic. Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg16.55kJ/kg32.107

kJ/kg90.255

,kJ/kg16.55,kJ/kg32.107,kJ/kg88.251

kJ/kg,16.239

8

6

2

7

5

3

1

hh

h

hhhh

T

QL

73B

A96

1 MPa 4

·

0.32 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,

0.265171.196

16.5532.10766

fg

f

hhh

x

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

kJ/kg84.25490.2552651.0188.2512651.011

0

9

26369

outin


hhxhxh

hmhm

EE

EEE

iiee

and

KkJ/kg94074.0kJ/kg84.254

MPa32.09

9

9 shP

also, kJ/kg94.278KkJ/kg94074.0

MPa14

94

4 hss

P

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

kW9.10kJ/kg239.16255.90kg/s0.1837kJ/kg254.84278.94kg/s0.25

kJ/kg55.16239.16kg/s0.1837

kg/s0.1837kg/s0.250.265111


81

6

hhmhhmWWW

hhmQ

mxm

BA

BL

AB

kW33.80


3.71kW9.10kW33.80COP

innet,R W

QL

11-33

11-47 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant throughthe upper cycle, the rate of heat removal from the refrigerated space, and the COP of the refrigerator are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):

kJ/kg.K9377.0kJ/kg46.244

kPa200@1

kPa200@1

g

g

sshh

Win.Expansion

valveCompressor

Evaporator

Condenser

4

32

1

Win

.Expansionvalve

Compressor

Evaporator

Condenser

QH.

8

76

5

.QL

kJ/kg30.263kPa500

212

2sh

ssP

kJ/kg01.26846.244

46.24430.26380.0 22

12

12

hh

hhhh s

C

kJ/kg33.73kJ/kg33.73

34

kPa500@3

hhhh f

kJ/kg.K9269.0kJ/kg55.255

kPa400@5

kPa004@5

g

g

sshh

kJ/kg33.278kPa1200

656

6sh

ssP

kJ/kg02.28455.255

55.25533.27880.0 66

56

56

hh

hhhh s

C

kJ/kg77.117kJ/kg77.117

78

kPa1200@7

hhhh f

The mass flow rate of the refrigerant through the upper cycle is determined from an energy balance on theheat exchanger

kg/s0.212AA

BA

mm

hhmhhm

kJ/kg)33.7301kg/s)(268.15.0(kJ/kg)77.117.55255(

)()( 3285

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

kW25.67kJ/kg)33.7346kg/s)(244.15.0()( 41 hhmQ BL


kW566.9kJ/kg)46.24401kg/s)(268.212.0(kJ/kg)55.25502kg/s)(284.15.0()()( 1256in hhmhhmW BA

2.68566.9

67.25COPin

L

WQ

11-34

11-48 A two-stage cascade refrigeration cycle with a flash chamber is considered. The mass flow rate ofthe refrigerant through the high-pressure compressor, the rate of heat removal from the refrigerated space, the COP of the refrigerator, and the rate of heat removal and the COP if this refrigerator operated on a single-stage cycle between the same pressure limits are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.Analysis (a) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):

kJ/kg.K9377.0kJ/kg46.244

kPa200@1

kPa200@1

g

g

sshh

7

Flashchamber

Expansionvalve Low-press.

Compressor

Evaporator8

9

2

1

Expansionvalve High-press.

Compressor

Condenser

QH

.

6

54

3

.QL

kJ/kg09.261kPa450

212

2sh

ssP

kJ/kg24.26546.244

46.24409.26180.0 22

12

12

hh

hhhh s

C


kJ/kg77.117kJ/kg77.117

kJ/kg53.257

78

kPa450@7

56

kPa1200@5

kPa450@3

hhhhhhhh

hh

f

f

g

2594.0kPa450

kJ/kg77.1176

6

6 xPh

The mass flow rate of the refrigerant through thehigh pressure compressor is determined from amass balance on the flash chamber

kg/s0.20250.2594-1

kg/s15.01 6

7

xm

m

Also, kg/s05255.015.02025.073 mmm

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

kJ/kg24.263kJ/kg)53kg/s)(257.(0.05255kJ/kg)24kg/s)(265.(0.15kg/s)2025.0( 99

33279

hh

hmhmhm

Then,

kJ/kg9451.0kJ/kg24.263

kPa4509

9

9 shP

kJ/kg27.284kPa1200

494

4sh

ssP

kJ/kg53.28924.263

24.26327.28480.0 44

94

94

hh

hhhh s

C

11-35

The rate of heat removal from the refrigerated space is

kW26.35kJ/kg)81.6846kg/s)(244.15.0()( 817 hhmQL


kW442.8kJ/kg)24.26353kg/s)(289.2025.0(kJ/kg)46.24424kg/s)(265.15.0()()( 94127in hhmhhmW

3.12442.8

35.26COPin

L

WQ

(d) If this refrigerator operated on a single-stage cycle between the same pressure limits, we would have

kJ/kg.K9377.0kJ/kg46.244

kPa200@1

kPa200@1

g

g

sshh

T

QH

QL

Win·

2

·

·4

3

2

1

kJ/kg84.281kPa1200

212

2sh

ssP

kJ/kg19.29146.244

46.24484.28180.0 22

12

12

hh

hhhh s

C

s

kJ/kg77.117

kJ/kg77.117

34

kPa1200@3

hh

hh f

kW25.66kJ/kg)77.11746kg/s)(244.2025.0()( 41 hhmQL

kW465.9kJ/kg)46.24419kg/s)(291.2025.0()( 12in hhmW

2.71465.9

66.25COPin

L

WQ

11-36

Gas Refrigeration Cycles

11-49C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reverseddirection.

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,Stirling1

1T 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 surroundingair, 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.

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. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperatureto 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/kg9721846203861

31

kJ/kg6034219872732903

4

min43

4

2

max21

2

34

12

.hTT..P

PPP

.hTT..P

PPP

rr

rr

K219.0

K342.2s

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

COPRnet, in comp, in turb, out

qw

qw w

L L

whereq 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, COPR31 08

92 55 8122.

. .2.74

(c) The rate of refrigeration is determined to be

kJ/s2.49kJ/kg31.08kg/s0.08refrig LqmQ

11-38

11-56 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluidis 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. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperatureto 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/kg76201 K82013475073751

25050

kJ/kg17452 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

kW6.67kJ/kg201.76285.14kg/s0.0841refrig hhmqmQ L

(b) The net power input is determined from

W W Wnet, in comp, in turb, out

where

kW9.48kJ/kg201.76320.29kg/s0.08kW13.36kJ/kg285.14452.17kg/s0.08

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

QW

L 1.72

11-39

11-57 EES Problem 11-56 is reconsidered. The effects of compressor and turbine isentropic efficiencies onthe 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]=Ts2ss[1]=s2sTs[2]=Ts4ss[2]=s4s

11-40

COP comp turb QRefrig[kW]

Winnet[kW]

0.6937 0.7 1 6.667 9.6120.9229 0.8 1 6.667 7.2241.242 0.9 1 6.667 5.3681.717 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

comp

QRefrig

[kW]

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

comp

Win;net

[kW]

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

comp

COP

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. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperatureto 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/lbm14101R44235933078001

3010

Btu/lbm68163R9683177305901

1030

4

43

4

2

21

2

34

12

.h.T..P

PPP

.h.T..P

PPP

rr

rr s


Btu/s9.17 Btu/lbm101.14119.48lbm/s0.541refrig hhmqmQ L



where

Btu/s18.79 Btu/lbm101.14138.66lbm/s0.5 Btu/s22.10Btu/lbm119.48163.68lbm/s0.5

43outturb,

12incomp,

hhmWhhmW

Thus, . . .Wnet, in Btu / s2210 18 76 3 34 4.73 hp


COP 9.17 Btu / s3.34 Btu / sR

net, in

QW

L 2.75

11-42

11-59 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluidis 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. 3Kinetic and potential energy changes are negligible.Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and thecompressor 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/kg76201 K82013475073751

25050

kJ/kg17452 K4450792515841

50250

4

43

4

2

21

2

34

12

.h

.T..PPPP

.h

.T..PPPP

s

srr

s

srrs

Also,

kJ/kg54219762012932085029320

433443

43

.....

hhhhhhhh

sTs

T


kW5.25kJ/kg219.54285.14kg/s0.0841refrig hhmqmQ L



where

kW8.06kJ/kg219.54320.29kg/s0.08

kW16.700.80/kJ/kg285.14452.17kg/s0.08

43outturb,

1212incomp,

hhmW

/hhmhhmW Cs

Thus, . .Wnet, in 16 70 8 06 8.64 kW


COP 5.25 kW8.64 kWR

net, in

QW

L 0.61

11-43

11-60 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperaturein 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. 3Kinetic and potential energy changes are negligible.Properties The properties of helium are cp = 5.1926 kJ/kg·Kand k = 1.667 (Table A-2).

4

2

4s

3

QH·

2s

11QRefrig·

TAnalysis (a) From the isentropic relations,

K1.20831K323

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

K5.44480.0/2632.408263/

1.20832380.0323

121212

12

12

12

min

433443

43

43

43

Csss

C

sTss

T

TTTTTTTT

hhhh

TTTTT

TTTT

hhhh

K231.1

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

0.3561.2313232635.444

1.231263

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/s0.109K231.1263KkJ/kg5.1926

kJ/s18

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 temperaturethat 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. 3Kinetic 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 K33954K266

T.PPTT

.PPTT

././

././

C99.4

Qrege

QH· 2

4

6QRefrig·

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

1.12C4.9915C73.122

C4.99C49

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The mass flow rate is determined from

kg/s0.237C99.449CkJ/kg1.005

kJ/s12

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 temperaturethat 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. 3Kinetic and potential energy changes are negligible.Properties The properties of air at room temperatureare cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

5

2

Qrege

27 C

QH· 2s

4

6QRefrig·

5

-15 C

-7 C

3

1

T

s

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

C99.4K617341 K258

C122.3K33954 K266

4140k1k

4

545

4140k1k

1

212

.PPTT

.PPTT

././

s

././

s

and

C4165750731227

4991580015

121212

12

12

12

min

544554

54

54

54

.././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


0.32C5.8215C74.165

C5.82C49

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The mass flow rate is determined from

kg/s0.356C82.549CkJ/kg1.005

kJ/s12

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. 3Air 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,

K4.4325K2.273 4.1/4.0k/1k

1

212 P

PTsT

Turbine

RegeneratorHeatExch.

HeatExch.

QL.

21

Compressor

6

54

3

.QH

K5.4722.273

2.2734.43280.0 22

12

12

12

12

TT

TTTT

hhhh ss

C

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

4.1/4.0

4

k/1k

4

545 5

1TPP

TT s

ssT TT

Thhhh

54

4

54

54 2.19385.0

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,K3.2463.2812.3082.2734316

61436143

TTTT

TTTTTTcmTTcm pp

35 C

0 CThe effectiveness of the regenerator is

0.4343.2462.3083.2812.308

63

43

63

43regen TT

TThhhh

-80 C

(b) The refrigeration load is

kW21.36K)2.19346.3kJ/kg.K)(25kg/s)(1.004.0()( 56 TTcmQ pL

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

kW13.80K)2.27372.5kJ/kg.K)(45kg/s)(1.004.0()( 12inC, TTcmW p

kW43.35kJ/kg)2.19381.3kJ/kg.K)(25kg/s)(1.004.0()( 54outT, TTcmW p

0.47843.3513.80

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·

K6.19451K2.3081 4.1/4.0k/1k

34 rTsT

T

K6.2116.1942.308

2.30885.0 4

4

43

43 TT

TTTT

sT

35 C 0 C

kW24.74kJ/kg)6.21173.2kJ/kg.K)(25kg/s)(1.004.0(

)( 41 TTcmQ pL

s

kW32.41kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K)5kg/s)(1.004.0(

)()( 4312innet, TTcmTTcmW pp

0.59932.4174.24COP

innet,WQL

11-48

Absorption Refrigeration Systems

11-64C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerantduring part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquidphase 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 externalheat source.

11-66C In absorption refrigeration, water can be used as the refrigerant in air conditioning applicationssince 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 inthe 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,genR inputrequired

outputdesiredCOPQQ

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 thepump.

11-70 The COP of an absorption refrigeration system that operates at specified conditions is given. It is tobe 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 maximumCOP this absorption refrigeration system can have is to be determined.Analysis The maximum COP that this refrigeration system can have is

2.64273298

273K393K29811COP

0

0maxR,

L

L

s TTT

TT

11-49

11-72 The conditions at which an absorption refrigeration system operates are specified. The maximumrate 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

243K403K29811COP

0

0maxR, .

TTT

TT

L

L

s

Thus, kJ/h105.75 5kJ/h10515.1COP 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/h100.55 5 Btu/h10550COP 5genR .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 lawefficiency 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,

sTT

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,

The heat input to the reversible heat engine is

kW911.72.78

kW22COP revabs,

inLQ

Q

Then, the rate at which the steam condenses becomes

kg/s0.00408kJ/kg1939.8kJ/s911.7in

fgs h

Qm

(b) The power input to the refrigerator is equal to the power output from the heat enginekW2.93)kW191.7)(370.0(inrevth,HEout,Rin, QWW

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

0.2522.78

7.0COPCOP

revabs,

actualII

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 closedcircuit and one of the joints is heated, a current flows continuously in the circuit. This is called theSeebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction iscooled. 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 temperaturedifference, 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 maximumthermal efficiency this thermoelectric generator can have is to be determined.Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,

%3.31R800R550

11Carnotth,maxth,H

L

TT

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,

W12.1

10.72

72.10 W130

COP

1K268/K2931

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. Therequired power input to the thermoelectric cooler is to be determined.Analysis The required power input is determined from the definition of COPR,

hp3.14=Btu/min33.310.15Btu/min20

COPCOP

Rin

inR

LL QW

WQ

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, = 1 kg/Land 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 canneddrinks,

W6.71=kW00671.0s360012

kJ290

kJ290=C3)-C)(25kJ/kgkg)(4.1815.3(kg3.15=L)0kg/L)(0.351(9

coolingcooling

cooling

tQ

Q

TmcQm V

The electric power consumed by the refrigerator is

W36=A)V)(312(in IW V

Then the COP of the refrigerator becomes

COP W36 W

cooling

in

. .Q

W6 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 theelectric 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,

W36.2Btu1

J1055s6015

Btu84.30

Btu30.84=F38)-F)(78Btu/lbmlbm)(1.0771.0(

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,

W49.7Btu1

J1055s6015

Btu4.42

Btu42.4=F75)-F)(130Btu/lbmlbm)(1.0771.0(

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

1420K353K30311maxth, .

TT

H

L

Thus,

kW39.4kJ/h142,000kJ/h)(0.142)(106inmaxth,maxout, QW

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

5.061K253/K303

11/

1COP CR,LH TT

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

kPa770.64kPa132.82

C03@satcond

C20@satevap

PPPP

(c) The net work input is determined from

kJ/kg82.19591.21280.049.25kJ/kg43.5791.21215.049.25

C20@22

C20@11

fgf

fgf

hxhhhxhh -20 C

30 C

qL

4 3

21

T

s

kJ/kg27.3506.5kJ/kg138.4

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-134aas the working fluid is considered. The mass flow rate of the refrigerant, the power input to thecompressor, 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, therefrigerant 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),



MPa7.0

C95.34kJ/kg50.273MPa7.0

KkJ/kg94779.0kJ/kg97.236

vaporsat.kPa120

34

MPa0.7@33

2212

2

kPa120@1

kPa120@11

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL

·

Win·

·

0.7 MPa

4

32

10.12 MPa

The mass flow rate of the refrigerant is determined from

kg/s0.675kJ/kg88.82236.97

kJ/s100

41 hhQ

m Ls

(b) The power input to the compressor iskW24.7kJ/kg236.97273.50kg/s0.67512in hhmW

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

kg/s3.73C8CkJ/kg4.18

kJ/s124.7)(

kW124.7kJ/kg88.82273.50kg/s0.675

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


"Evaporator"P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator"Q_dot_in=m_dot*Q_inCOP=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 24,66150 4,743 21,09180 5,475 18,27210 6,27 15,95240 7,146 13,99270 8,126 12,31300 9,235 10,83330 10,51 9,517360 11,99 8,34390 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 theworking fluid. The mass flow rate of the refrigerant, the power input to the compressor, the mass flow rateof the cooling water, and the rate of exergy destruction associated with the compression process are to bedetermined.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-12and A-13),



MPa7.0

C95.34kJ/kg50.273MPa7.0

KkJ/kg94779.0kJ/kg97.236

vaporsat.kPa120

34

MPa0.7@33

2212

2

kPa120@1

kPa120@11

hh

hhP

Thss

P

sshhP

f

sss

g

g

T

QH

QL

·

Win·

·

0.7 MPa

4

32

10.12 MPa

The mass flow rate of the refrigerant is determined froms

kg/s0.675kJ/kg88.82236.97

kJ/s100

41 hhQ

m L

(b) The actual enthalpy at the compressor exit is

kJ/kg285.6775.0/97.23650.27397.236/1212

12

12Cs

sC hhhh

hhhh

Thus, kW32.9kJ/kg236.97285.67kg/s0.67512in hhmW

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

and

kg/s3.97C8CkJ/kg4.18

kJ/s132.9)(

kW132.9kJ/kg88.82285.67kg/s0.675

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 gen0 0 2 s1

where

KkJ/kg98655.0kJ/kg67.285

MPa7.02

2

2 shP

Thus,

kW7.80KkJ/kg0.947790.98655kg/s0.675K298destroyedX

11-57

11-94 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as theworking fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of therefrigerant 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, therefrigerant 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),



MPa9.0

kJ/kg75.275MPa9.0

/kgm099867.0KkJ/kg93773.0

kJ/kg46.244

vaporsat.kPa200

34

MPa0.9@33

212

2

3kPa200@1

kPa200@1

kPa200@11

hh

hhP

hss

P

sshh

P

f

g

g

g

vvQH

QL

House·

Win·

·

0.9 MPa

4

32

1200 kPa

T

sThe rate of heat supply to the house is determined from

kW55.73kJ/kg.61101275.75kg/s0.3232 hhmQH

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

/sm0.0320 3/kgm0.099867kg/s0.32 311 vV m

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

5.5746.24475.27561.10175.275COP

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


66816

11

with1

hhhhxwww

hhh

xhhxqfg

fL

11-58

11-96 A two-stage compression refrigeration system using refrigerant-134a as the working fluid isconsidered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amountof 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. 3The 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/kg58.260

,kJ/kg94.63,kJ/kg47.95

,kJ/kg55.255,kJ/kg16.239

8

6

2

7

5

3

1

hh

h

hhhh

T

qL

73B

A96

0.8 MPa 4

0.4 MPa

8

5 2

1

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

0.164662.191

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/kg75.25958.2601646.0155.2551646.011

0

9

26369

outin(steady)0

systemoutin

hhxhxh

hmhm

EEEEE

iiee

KkJ/kg94168.0kJ/kg75.259

MPa4.09

9

9 shP

Also, kJ/kg47.274KkJ/kg94168.0

MPa8.04

94

4 hss

P

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

kJ/kg32.6

kJ/kg146.4

kJ/kg259.75274.47kJ/kg239.16260.580.1646111

kJ/kg63.94239.161646.011


816

hhhhxwww

hhxqL


4.49kJ/kg32.6kJ/kg146.4COP

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 roomtemperature. 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 K264kPa250kPa100 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 tobe 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 theregenerator,

Qregen·

QH·

2

4

6QRefrig·

5-25 C

-10 C

3

1

T

6143

outin

(steady)0systemoutin 0

hhmhhmhmhm

EE

EEE

iiee

20 C

or,

Thus,K278C25C10C206134

61436143

C5TTTT

TTTTTTcmTTcm pp

s

(b) From the isentropic relations,

C93.9K117931 K278

C135.2 K24083 K263

66716670k1k

4

545

66716670k1k

1

212

.PPTT

.PPTT

././

././


1.49C93.95C10135.2

C93.9C25

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The net power input is determined from

kW108.293.9510135.2CkJ/kg5.1926kg/s0.4554125412outturb,incomp,innet, TTTTcmhhhhmWWW p

11-99 An absorption refrigeration system operating at specified conditions is considered. The minimumrate 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, kW53.91.259

kW12COP maxR,

mingen,LQ

Q

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] 13.76 508.996 656.833 805.295 1004.237 1253.603 1502.878 2002.475 250

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 heatgain 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 tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Air is an ideal gas with variable specific heats. Analysis (a) For this problem, we use the properties of air from EES:

kJ/kg50.433kPa500

kJ/kg.K6110.5C0

kPa100kJ/kg40.273C0

212

2

11

1

11

shss

P

sTP

hT

Turbine

RegeneratorHeatExch.

HeatExch.

QL.

21

Compressor

6

54

3

.QH

kJ/kg63.308C35

kJ/kg52.47340.273

40.27350.43380.0

33

2

2

12

12

hT

hh

hhhh s

C

For the turbine inlet and exit we have

kJ/kg45.193C80 55 hT

sT hh

hhhT

54

54

44 ?

5

2

Qregen

QH· 2s

4

6QRefrig·

5s

3

1

s

T

35 C

shss

P

sTP

sTP

545

2

44

4

11

5

kPa500

?kPa500

kJ/kg.K6110.5C0

kPa100 0 C

-80 C

We can determine the temperature at the turbine inlet from EES using the above relations. A hand solutionwould require a trial-error approach.

T4 = 281.8 K, h4 = 282.08 kJ/kgAn energy balance on the regenerator gives

kJ/kg85.24608.28263.30840.2734316 hhhh

The effectiveness of the regenerator is determined from

0.43085.24663.30808.28263.308

63

43regen hh

hh

(b) The refrigeration load is

kW21.36kJ/kg)45.19385kg/s)(246.4.0()( 56 hhmQL

11-64

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

kW05.80kJ/kg)40.27352kg/s)(473.4.0()( 12inC, hhmW

kW45.35kJ/kg)45.19308kg/s)(282.4.0()( 54outT, hhmW

0.47945.3505.80

36.21COPoutT,inC,innet, WW

QW

Q LL

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

kJ/kg63.308kJ/kg52.473kJ/kg40.273

3

2

1

hhh

4

2

4s

3

QH·

2

11QRefrig·

T

35 CkJ/kg2704.5

C35kPa500

33

3 sTP

0 C

kJ/kg.K52.194kPa100

434

1sh

ssP

s

kJ/kg64.21152.19463.308

63.30885.0 4

4

43

43 hh

hhhh

sT

kW24.70kJ/kg)64.21140kg/s)(273.4.0()( 41 hhmQL

kW25.41kJ/kg)64.211(308.63)40.273(473.52kg/s)4.0()()( 4312innet, hhmhhmW

0.59925.4170.24COP

innet,WQL

11-65

11-104 An air-conditioner with refrigerant-134a as the refrigerant is considered. The temperature of therefrigerant 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/kg77.117

kJ/kg77.117

39.277kPa1200

kJ/kg9240.0kJ/kg04112.0

kJ/kg30.259

1kPa500

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

Expansionvalve

Compressor

Condenser

500 kPa

.QH

.QL

EvaporatorkJ/kg42.283

30.25930.25939.27775.0 2

2

12

12

hh

hhhh s

C

C54.522

2

kJ/kg42.283kPa1200

ThP

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

QH

QL

Win·

2

·

·4

3

2s

1

kg/s04053.0/kgm0.04112

s60min1

L1000m1L/min)100(

3

3

1

1

v

Vm

The refrigeration load is

kW737.5kJ/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

kW0.67kW)9.060/250737.5(equipheatpeople QQQQ L

(c) The power input and the COP are kW9774.0kJ/kg)30.25942kg/s)(283.04053.0()( 12in hhmW

5.879774.0737.5COP

in

L

WQ

(d) The reversible COP of the cycle is

38.371)27326/()27334(

11/

1COPrevLH TT

The corresponding minimum power input is

kW1535.038.37kW737.5

COPrev

Lminin,

QW

The minimum mass and volume flow rates are

kg/s006366.0kJ/kg)30.259(283.42

kW1535.0

12

minin,min hh

Wm

L/min15.7/s)m0002617.0(/kg)m112kg/s)(0.04006366.0( 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 airor room air.Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning systembecomes

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 / h12 2 2 4

since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since itscooling 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 roomtemperature. 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 roomtemperature 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 thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat 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 itinvolves no heat or work interactions. Then the steady-flow energy balance equation for this system

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

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

Warm air Cold air

Compressedair

2 3

1

Canceling cp and solving for T3 gives

K307.375.0

27825.030075.025.0 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 inletand two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equationfor 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/K461.0kPa500kPa100lnkJ/kg.K)287.0(

K300K3.307lnkJ/kg.K)005.1(75.0

kPa500kPa100lnkJ/kg.K)287.0(

K300K278lnkJ/kg.K)005.1(25.0genS

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 thecompressor are determined to

kW5.53=278)K-00kJ/kg.K)(35kg/s)(1.0025.0(

)()( c1c1cooling TTcmhhmQ pcc

kW1.1571kPa100kPa500

80.0)14.1(K)00kJ/kg.K)(37kg/s)(0.281(

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

kk

PP

kRTm

W

Then the COP of the vortex refrigerator becomes

0350.kW1.157kW5.53COP

incomp,

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 Carnotrefrigerator 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 roomtemperature. 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 roomtemperature 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 thisirreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the onethat 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 itinvolves no heat or work interactions. Then the steady-flow entropy balance equation for this system

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

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

Warm air Cold air

Compressedair

2 3

1

Canceling cp and solving for T3 gives

K307.375.0

27825.030075.025.0 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 inletand two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equationfor 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/K5130kPa600kPa100kJ/kg.K)ln2870(

K300 K3307kJ/kg.K)ln0051(750

kPa600kPa100kJ/kg.K)ln2870(

K300 K278kJ/kg.K)ln0051(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 thecompressor are determined to

kW5.53=278)K-00kJ/kg.K)(35kg/s)(1.0025.0(

)()( c1c1cooling TTcmhhmQ pcc

kW9.1791kPa100kPa600

80.0)14.1(K)00kJ/kg.K)(37kg/s)(0.281(

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

kk

PP

kRTm

W

Then the COP of the vortex refrigerator becomes

0310.kW9.179kW5.53COP

incomp,

cooling

W

Q

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

12.6K)278300(

K278COPCarnotLH

L

TTT

Discussion Note that the COP of the vortex refrigerator is a small fraction of the COP of a Carnotrefrigerator 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 refrigerationcycle 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 c P1 [kPa]1.851 0.7 1002.863 0.7 2004.014 0.7 3005.462 0.7 4007.424 0.7 500

100 150 200 250 300 350 400 450 5000

2

4

6

8

10

P[1] [kPa]

COP

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 refrigerationcycle 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

comp

1.01.00.70.7

COP c P2 [kPa]4.935 0.7 4003.04 0.7 650

2.258 0.7 9001.803 0.7 11501.492 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 fluidexecuted 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 cycleis(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 numericalvalues).

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)+273TL=TEMPERATURE(R134a,x=0,P=P2)+273q_H=h_fgCOP=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


TH=20+273TL=-5+273Q_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 theworking fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from therefrigerated 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


P1=120 "kPa"P2=800 "kPa"P3=P2P4=P1s2=s1Q_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 theworking fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerantis 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


P1=320 "kPa"P2=1200 "kPa"P3=P2P4=P1s2=s1m=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) "Usingh_fg at P1"

11-114 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates betweenthe pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phaseat 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


P1=120 "kPa"P2=1000 "kPa"P3=P2P4=P1h1=ENTHALPY(R134a,x=1,P=P1)h3=ENTHALPY(R134a,x=0,P=P3)h4=h3x4=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-134aas the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performanceof this heat pump is(a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9

Answer (e) 5.9


P1=320 "kPa"P2=1200 "kPa"P3=P2

11-76

P4=P1s2=s1h1=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=h3COP_HP=qH/WinWin=h2-h1qH=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 limitsof 80 kPa and 280 kPa. Air is cooled to 35 C before entering the turbine. The lowest temperature of thiscycle is(a) –58 C (b) -26 C (c) 0 C (d) 11 C (e) 24 C

Answer (a) –58 C


k=1.4P1= 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 thecompressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before itenters 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


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k=1.667Cp=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 geothermalsource 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


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 kPaand 50 C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of thisrefrigerator is(a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4

Answer (d) 3.2


P1=160 "kPa"P2=800 "kPa"T2=50 "C"P3=P2P4=P1h1=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=h3COP_R=qL/WinWin=h2-h1qL=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

ThermoSolutions CHAPTER11

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