numerical vpc ceoncept

11-1

Chapter 11 REFRIGERATION CYCLES

The Reversed Carnot Cycle 11-1C Because the compression process involves the compression of a liquid-vapor mixture which requires a compressor that will handle two phases, and the expansion process involves the expansion of high-moisture content refrigerant. 11-2 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) Noting that TH = 30°C = 303 K and TL = Tsat @ 160 kPa = -15.60°C = 257.4 K, the COP of this Carnot refrigerator is determined from

( ) ( ) 5.64=−

=−

=1K 4.257/K 303

11/

1COP CR,LH TT

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

kJ/kg 58.93kJ/kg 66.266

C30@4

C30@3====

°

°

f

ghhhh

Thus,

and

( ) kJ/kg 147.03=

==→=

=−=−=

kJ/kg 173.08K 303K 257.4

kJ/kg 08.17358.9366.26643

HH

LL

L

H

L

H

H

qTT

qTT

qq

hhq

30°C 4 3

2 1160 kPa

QL

QH

T

s

(c) The net work input is determined from kJ/kg 26.05=−=−= 03.14708.173net LH qqw

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

11-2

11-3E A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the quality at the beginning of the heat-absorption process, and the net work input are to be determined. 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.

( ) ( ) 8.28=−

=−

=1R 475.4/R 532.8

11/

1COP CR,LH TT

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

( ) ( )( )

0.2374=−

=

−=

⋅=

+=+==

18589.003793.008207.0

RBtu/lbm 0.08207

14525.005.007481.0

psia 30 @

11

psia 90 @ 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/lbm7.92 RBtu/lbm 08207.022006.0)37.1578.72(43innet, =⋅−−=−−= ssTTw LH

Ideal and Actual Vapor-Compression Cycles 11-4C Yes; the throttling process is an internally irreversible process. 11-5C To make the ideal vapor-compression refrigeration cycle more closely approximate the actual cycle. 11-6C No. Assuming the water is maintained at 10°C in the evaporator, the evaporator pressure will be the saturation pressure corresponding to this pressure, which is 1.2 kPa. It is not practical to design refrigeration or air-conditioning devices that involve such extremely low pressures. 11-7C Allowing a temperature difference of 10°C for effective heat transfer, the condensation temperature of the refrigerant should be 25°C. The saturation pressure corresponding to 25°C is 0.67 MPa. Therefore, the recommended pressure would be 0.7 MPa. 11-8C The area enclosed by the cyclic curve on a T-s diagram represents the net work input for the reversed Carnot cycle, but not so for the ideal vapor-compression refrigeration cycle. This is because the latter cycle involves an irreversible process for which the process path is not known. 11-9C The cycle that involves saturated liquid at 30°C will have a higher COP because, judging from the T-s diagram, it will require a smaller work input for the same refrigeration capacity.

QH

QL

4 3

21

T

s

11-10C The minimum temperature that the refrigerant can be cooled to before throttling is the temperature of the sink (the cooling medium) since heat is transferred from the refrigerant to the cooling medium.


11-3

11-11 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical maximum refrigeration load for the same power input to the compressor are to be determined. 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)

0.4795=

==

==

=

°==

=

°==

=

°−==

44

4

34

33

3

22

2

11

1

kJ/kg 23.111kPa 60

kJ/kg 23.111

kJ/kg 23.111C42

kPa 1200

kJ/kg 16.295C65

kPa 1200

kJ/kg 03.230C34

kPa 60

xhP

hh

hTP

hTP

hTP

26°C Water 18°C

Win

QL

1.2 MPa 65°C

Expansion valve

Compressor

Evaporator

Condenser 42°C

QH

4

3 2

1 60 kPa -34°C

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

kJ/kg 94.108

kJ/kg 47.75

C26 @2

C18 @1

==

==

°

°

fw

fw

hh

hh

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

kg/s 0455.0

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

=→

−=−−=−

R

R

wwwR

m

mhhmhhm

&

&

&&

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

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

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

kW 5.85=−=−= 513.2367.8inWQQ HL&&&

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

T

QH

QL

Win·

2

·

·

4

3

2

1

2.33===513.285.5COP

in

L

WQ&

&

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

063.51)27330/()27318(

11/

1COPmax =−+−+

=−

=LH TT

s

Then, the maximum refrigeration load becomes

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


11-4

11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )

( )throttlingkJ/kg 82.88

kJ/kg 82.88liquid sat.

MPa 7.0

C95.34kJ/kg 50.273MPa 7.0

KkJ/kg 94779.0kJ/kg 97.236

vaporsat.kPa 120

34

MPa 7.0 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

=≅

===

°==

==

⋅====

=

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL 4s

·

Win·

·

0.7 MPa

4

32

1 0.12 MPa

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

s

and ( ) ( )( )

( ) ( )( ) kW 1.83

kW 7.41

=−=−=

=−=−=

kJ/kg 236.97273.50kg/s 0.05

kJ/kg 82.8897.236kg/s 0.05

12in

41

hhmW

hhmQL

&&

&&

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

kW 9.23=+=+= 83.141.7inWQQ LH&&&

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

4.06===kW 1.83kW 7.41COP

inR W

QL&

&


11-5

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )



MPa 9.0

C45.44kJ/kg 93.278MPa 9.0

KkJ/kg 94779.0kJ/kg 97.236

vaporsat.kPa 120

34

MPa 9.0 @ 33

2212

2

kPa 120 @ 1

kPa 120 @ 11

=≅

===

°==

==

⋅====

=

hh

hhP

Thss

P

sshhP

f

g

g

T

QH

QL 4s

·

Win·

·

0.9 MPa

4

32

1 0.12 MPa

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

and ( ) ( )( )

( ) ( )( ) kW2.10

kW 6.77

kJ/kg 236.97278.93kg/s 0.05

kJ/kg 61.10197.236kg/s 0.05

12in

41

=−=−=

=−=−=

hhmW

hhmQL

&&

&&

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

kW 8.87=+=+= 10.277.6inWQQ LH&&&


3.23===kW 2.10kW 6.77COP

inR W

QL&

&


11-6

11-14 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The throttling valve in the cycle is replaced by an isentropic turbine. The percentage increase in the COP and in the rate of heat removal from the refrigerated space due to this replacement are to be determined. 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/kg 58.8248.2142802.049.22

2802.085503.0

09275.033230.0

kPa 120 @44

kPa 120 @

34

=+=+=

=−

=

−=

fgsfs

fg

fs

hxhh

sss

x T

QH

QL 4s

·

Win ·

·

0.7 MPa

4

32

10.12 MPa

Then, Q ( ) ( )( ) kW 7.72kJ/kg 82.58236.97kg/s 0.0541 =−=−= sL hhm&&

and 23.4kW 1.83kW 7.72COP

inR ===

WQL&

&

Then the percentage increase in and COP becomes &Q

4.2%

4.2%

=−

=∆

=

=−

=∆

=

06.406.423.4

COPCOP

COPin Increase

41.741.772.7in Increase

R

RR

L

LL Q

QQ

&

&&

s

11-15 [Also solved by EES on enclosed CD] An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The quality of the refrigerant at the end of the throttling process, the COP, and the power input to the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),



MPa 8.0

kJ/kg 37.275MPa 8.0

KkJ/kg 94456.0kJ/kg 16.239

vapor sat.kPa 140

34

MPa 8.0 @ 33

212

2

kPa 140 @ 1

kPa 140 @ 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

0.322=−

=

−=

08.21208.2747.95

kPa 140 @

44

fg

f

hhh

x s

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

3.97=−−

=−−

==16.23937.275

47.9516.239COP12

41

inR hh

hhwqL

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

kW 1.26===97.3

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_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c

P1 [kPa] COPplot Win [kW] 100 3.216 1.554 175 4.656 1.074 250 6.315 0.7918 325 8.388 0.5961 400 11.15 0.4483


11-8

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

-25

0

25

50

75

100

125

s [kJ/kg-K]

T[C]

800 kPa

140 kPa

R134a

T-s diagram for η = 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 η = 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 η = 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 is considered. The quality of the refrigerant at the end of the throttling process, the COP, the power input to the compressor, and the irreversibility rate associated with the compression process are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis (a) The refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )( )



MPa 8.0

kJ/kg 76.28185.0/16.23937.27516.239

/

kJ/kg 37.275MPa 8.0

KkJ/kg 94456.0kJ/kg 16.239

vapor.sat kPa 140

34

MPa 8.0 @ 33

121212

12

212

2

kPa 140 @1

kPa 140 @ 11

=≅

===

=−+=

−+=→−−

=

=

==

⋅====

=

hh

hhP

hhhhhhhh

hss

P

sshhP

f

Css

C

ss

g

g

ηη ( )

T2

QH · 2s

Win·

3 0.8 MPa

0.14 MPa1

QL ·4

s

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

0.322=−

=

−=

08.21208.2747.95

kPa 140 @

44

fg

f

hhh

x

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

3.37=−−

=−−

==16.23976.281

47.9516.239COP12

41

inR hh

hhwqL

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

kW 1.48===37.3kW 5

COPinR

LQW

&&

The exergy destruction associated with the compression process is determined from

( )120

0

0

surr120gen0destroyed ssmT

Tq

ssmTSTX −=

+−== &&&&

where

( )

KkJ/kg 96483.0kJ/kg 76.281

MPa 8.0

kg/s 0348.0kJ/kg 95.47239.16

kJ/s 5

22

2

41

⋅=

==

=−

=−

==

shP

hhQ

qQ

m L

L

L&&

&

Thus,

( )( )( ) kW 0.210=⋅−= KkJ/kg 0.944560.96483kg/s 0.0348K 298destroyedX&


11-11

11-18 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined. 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/kg 98.84C24MPa 65.0

kJ/kg 16.281MPa 7.0

kJ/kg 53.288C50MPa 7.0

KkJ/kg 97236.0kJ/kg 36.246

C10MPa 14.0

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

1 0.15 MPa

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

and ( ) ( )( )

( ) ( )( ) kW 5.06

kW19.4

=−=−=

=−=−=

kJ/kg 246.36288.53kg/s 0.12

kJ/kg 84.98246.36kg/s 0.12

12in

41

hhmW

hhmQL

&&

&&

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

82.5%=−−

=−−

=36.24653.28836.24616.281

12

12

hhhh s

Cη


3.83===kW 5.06kW 19.4COP

inR W

QL&

&


11-12

11-19E An ice-making machine operates on the ideal vapor-compression refrigeration cycle, using refrigerant-134a as the working fluid. The power input to the ice machine is to be determined. 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, 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-12E and A-13E),

( )throttlingBtu/lbm 39.33

Btu/lbm 39.33liquid sat.

psia 80

Btu/lbm 00.115psia 80

RBtu/lbm 22567.0Btu/lbm 73.102

vapor sat.psia 20

34

psia 08 @ 33

212

2

psia 20 @ 1

psia 20 @ 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/s 0.7042Btu/lbm 169lbm/s 15/3600iceice ==∆= hmQL &&

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

and ( )

( ) ( )( ) hp 0.176=

−=−=

=−

=−

=

Btu/s 0.7068hp 1Btu/lbm 102.73115.00lbm/s 0.01016

lbm/s 0.01016Btu/lbm 33.39102.73

Btu/s 0.7042

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 the compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat gain in the line between the evaporator and the compressor are to be determined. 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/kg 33.239MPa 14165.0

vapor sat.C5.18

throttlingkJ/kg 58.93

kJ/kg 58.93C30MPa 95.0

kJ/kg 20.289MPa 0.1

/kgm 14605.0KkJ/kg 97236.0

kJ/kg 36.246

C10kPa 140

5

55

34

C 30 @ 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

1 0.15 MPa

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

( ) ( ) ( )[ ] ( ) kW1.88 78.0/kJ/kg 246.36289.20kg/s 0.03423/

kg/s 0.03423/kgm 0.14605/sm 0.3/60

12in

3

3

1

1

=−=−=

===

Cs hhmW

m

η&&

&&

vV

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

( ) ( )( ) kW 4.99=−=−= kJ/kg 93.58239.33kg/s 0.0342345 hhmQL &&

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

and

( ) ( )( ) kW0.241

1.65

kJ/kg 239.33246.36kg/s 0.03423

14065.141

51gain

15

=−=−=

=−=−=∆

hhmQ

PPP

&&


11-14

11-21 EES Problem 11-20 is reconsidered. The effects of the compressor isentropic efficiency and the compressor inlet volume flow rate on the power input and the rate of refrigeration are to be investigated. Analysis The problem is solved using EES, and the solution is given below. "Input Data" "T[5]=-18.5 [C] P[1]=140 [kPa] T[1] = -10 [C]} V_dot[1]=0.1 [m^3/min] P[2] = 1000 [kPa] P[3]=950 [kPa] T[3] = 30 [C] Eta_c=0.78 Fluid$='R134a'" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],T=T[1]) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],T=T[1]) v[1]=volume(Fluid$,P=P[1],T=T[1])"[m^3/kg]" m_dot=V_dot[1]/v[1]*convert(m^3/min,m^3/s)"[kg/s]" h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" Wc=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+Wc=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) W_dot_c=m_dot*Wc "Condenser" h[3]=enthalpy(Fluid$,P=P[3],T=T[3]) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],T=T[3]) h[2]=q_out+h[3] "energy balance on condenser" Q_dot_out=m_dot*q_out "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]=pressure(Fluid$,T=T[5],x=0)"pressure=Psat at evaporator exit temp." P[5] = P[4] h[5]=enthalpy(Fluid$,T=T[5],x=1) "properties for state 5" q_in + h[4]=h[5] "energy balance on evaporator" Q_dot_in=m_dot*q_in COP=Q_dot_in/W_dot_c "definition of COP" COP_plot = COP W_dot_in = W_dot_c Q_dot_line5to1=m_dot*(h[1]-h[5])"[kW]"


11-15

COPplot Win

[kW] Qin

[kW] ηc

[kW] 2.041 0.8149 1.663 0.6 2.381 0.6985 1.663 0.7 2.721 0.6112 1.663 0.8 3.062 0.5433 1.663 0.9 3.402 0.4889 1.663 1

0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10

1

2

3

4

5

6

7

8

9

η 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

η 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

η 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 COP of 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

Expansion valve

Compressor

Evaporator

Condenser 60°C

QH

.

.QL

kJ/kg 97.236 vap.)(sat. 1

kPa 120

kJ/kg 87.298C60

kPa 8.671

liq.) (sat. 0kJ/kg 83.86

kJ/kg 83.8630.0

kPa 120

11

41

22

2

32

33

3

43

44

4

=

===

=

°==

=

=

==

=

=

==

hx

PP

hTP

PP

Pxh

hh

hxP

kPa 671.8

The mass flow rate of the refrigerant is determined from T

QH

QL 4s

·

Win ·

·4

32

1 0.12 MPa

kg/s 0.00727=−

=−

=kg236.97)kJ/(298.87

kW 45.0

12

in

hhW

m&

&

(c) The refrigeration load and the COP are

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

2.43===kW 0.45kW 091.1COP

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 the atmospheric pressure, and a condenser pressure which corresponds to a saturation temperature above the temperature of the cooling medium. Other desirable characteristics of a refrigerant include being nontoxic, noncorrosive, nonflammable, chemically stable, having a high enthalpy of vaporization (minimizes the mass 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 of the 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 temperature of 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 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 -20°C and 35°C, respectively. The saturation pressures corresponding to these temperatures are 0.133 MPa and 0.888 MPa. Therefore, the recommended evaporator 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 saturation pressures corresponding to these temperatures are 0.293 MPa and 0.816 MPa. Therefore, the recommended evaporator 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 high cooling loads at that location. 11-29C A water-source heat pump extracts heat from water instead of air. Water-source heat pumps have higher 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 is considered. 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, 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-12E and A-13E),

( )throttlingBtu/lbm 79.41

Btu/lbm 79.41liquid sat.

psia 120

Btu/lbm 62.116psia 120

RBtu/lbm 22188.0Btu/lbm 81.108

vapor sat.psia 50

34

psia 120 @ 33

212

2

psia 50 @ 1

psia 50 @ 11

=≅

===

=

==

⋅====

=

hh

hhP

hss

P

sshhP

f

g

g

QH

QL

House ·

Win·

·

120 psia

4

32

1 50 psia

T

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

and ( )

( ) ( )( )Btu/s 0.7068 = hp 1 sinceBtu/s .7381

Btu/lbm 108.81116.62kg/s 0.2227

lbm/s 0.2227Btu/lbm 41.79116.62Btu/s 060,000/360

12in

32

hp 2.46==−=−=

=−

=−

==

hhmW

hhQ

qQ

m H

H

H

&&

&&&

The electrical power required without the heat pump is

Thus,

( )

kW 0.7457 = hp 1 sincekW 75.1546.258.23

hp .5823Btu/s 0.7068

hp 1Btu/s 060,000/360

insaved

==−=−=

=

==

hp 21.1WWW

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, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )throttling kJ/kg 22.127


MPa 4.1

kJ/kg 54.282MPa 4.1

KkJ/kg 93006.0kJ/kg 88.251

vapor sat.kPa 320

34

MPa 4.1 @ 33

212

2

kPa 320 @ 1

kPa 320@ 11

=≅

===

=

==

⋅====

=

hh

hhP

hss

P

sshhP

f

g

g

QH

QL

House ·

Win·

·

1.4 MPa

4

32

1 0.32 MPa

T

The heating load of this heat pump is determined from

( )[ ]

( )( )( ) kW .0515C1545CkJ/kg 4.18kg/s 0.12water12

=°−°⋅=

−= TTcmQH && s

and

Then, ( )

( ) ( )( ) kW 2.97=−=−=

=−

=−

==

kJ/kg 251.88282.54kg/s 0.09688

kg/s 0.09688kJ/kg 127.22282.54

kJ/s 15.05

12in

32

hhmW

hhQ

qQ

m

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 the increase 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

1 0.28 MPa

T

( )throttling kJ/kg 58.93

kJ/kg 58.93C30MPa 0.1

kJ/kg 38.293C60MPa 0.1

kJ/kg 83.250C0kPa 280

34

C30 @ 33

3

22

2

11

1

=≅

=≅

°==

=

°==

=

°==

°

hh

hhTP

hTP

hTP

f

The mass flow rate of the refrigerant is

( ) kg/s 0.08341kJ/kg 93.58293.38

kJ/s 0060,000/3,6

32=

−=

−==

hhQ

qQm H

H

HR

&&&

Then the power input to the compressor becomes

( ) ( )( ) kW 3.55=−=−= kJ/kg 250.83293.38kg/s 0.0834112in hhmW &&

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

( ) ( )( ) kW 13.12=−=−= kJ/kg 93.58250.83kg/s 0.0834141 hhmQL &&

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

Thus,

kW 13.12=−=−=

===

55.367.16

kW 16.67kJ/s 3600/000,60

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 resistance heating 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=1.0 Fluid$='R134a'" "Use ETA_c = 0.623 to obtain T[2] = 60C" "Compressor" h[1]=enthalpy(Fluid$,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$,P=P[3],T=T[3]) h[2]=Qout+h[3] "energy balance on condenser" Q_dot_out*convert(kJ/h,kJ/s)=m_dot*Qout "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" Q_dot_in=m_dot*Q_in COP=Q_dot_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] ηc Esaved

3.671 0.6 13 3.249 0.7 13.42 2.914 0.8 13.75 2.641 0.9 14.03 2.415 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

ηc

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

ηc

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 and the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression 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 of refrigerant-134a are (Tables A-11 through A-13)

26.277kPa 800

kJ/kg 9506.0kJ/kg 87.247

C)409.10(kPa 200

C09.10kJ/kg 91.87

kJ/kg 91.87C)306.29(

kPa 750

C06.29

kJ/kg 76.291C55kPa 800

212

2

1

1

1

1

kPa sat@200

34

33

3

kPa sat@7503

22

2

=

==

==

°+−==

°−===

=

°−==

°==

=

°==

shss

P

sh

TP

Thh

hTP

TT

hTP

4

3 2

1

Win .

800 kPa 55°C

Expansion valve

Compressor

Evaporator

Condenser

750 kPa QH.

.QL

The isentropic efficiency of the compressor is T

QH

QL

Win·

2

·

·

4

3

2

1

0.670=−−

=−−

=87.24776.29187.24726.277

12

12

hhhh s

Cη

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

Q kW 3.67=−=−= kJ/kg)91.8776kg/s)(291. 018.0()( 32 hhmH &&

(c) The power input and the COP are

W kW 790.0kJ/kg)87.24776kg/s)(291. 018.0()( 12in =−=−= hhm&&

s 4.64===

790.067.3COP

inWQH&

&

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

QH

QL 4s

·

Win ·

·

0.8 MPa

4

32

1 0.2 MPa

kJ/kg.K 9377.0kJ/kg 46.244

kPa 200 @1

kPa 200 @1

====

g

g

sshh

kJ/kg 25.273MPa 800

212

2 =

==

hss

P

34

kPa 800 @3 kJ/kg 47.95hhhh f

===

s 6.18=

−−

=−−

=46.24425.273

47.9525.273COP12

32

hhhh

kW 3.20=−=−= kJ/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 the condenser, the mass flow rate of the refrigerant, the heating load, the COP of the heat pump, the minimum power 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°C Water50°C

Win .

sat. vap.

Expansion valve Compressor

Evaporator

Condenser

20°C x=0.23

QH.

4

32

1

1.4 MPa s2 = s1

.QL

kJ/kg 00.280kPa 1400

kJ/kg 9223.0kJ/kg 59.261

vap.)(sat. 1kPa 1.572

kJ/kg 24.121kPa 1.572

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/kg 53.167

kJ/kg 34.209

C40 @ 2

C50 @ 1

==

==

°

°

fw

fw

hh

hh

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

T

QH

QL 4s

·

Win ·

·

1.4MPa

4

3

2

1

C40.52kPa 1400 @sat °=T

C59.48kJ 24.121

kPa 14003

3

3 °=

==

ThP

(from EES)

Then, the degrees of subcooling is C3.81°=−=−=∆ 59.4840.523satsubcool TTT (b) The rate of heat absorbed from the geothermal water in the evaporator is

s

kW 718.2kJ/kg)53.16734kg/s)(209. 065.0()( 21 =−=−= wwwL hhmQ && This heat is absorbed by the refrigerant in the evaporator

kg/s 0.01936=−

=−

=)kJ/kg24.121(261.59

kW 718.2

41 hhQ

m LR

&&

(c) The power input to the compressor, the heating load and the COP are kW 6564.0kJ/kg)59.26100kg/s)(280. 01936.0()( out12in =−=+−= QhhmW R

&&&

kW 3.074=−=−= kJ/kg)24.12100kg/s)(280. 01936.0()( 32 hhmQ RH &&

4.68===kW 0.6564

kW 074.3COPin

H

WQ&

&

(d) The reversible COP of the cycle is

92.12)27350/()27325(1

1/1

1COPrev =++−

=−

=HL TT

The corresponding minimum power input is

kW 0.238===92.12kW 074.3

COPrevminin,

HQW

&&


11-25

Innovative Refrigeration Systems 11-36C Performing the refrigeration in stages is called cascade refrigeration. In cascade refrigeration, two or more refrigeration cycles operate in series. Cascade refrigerators are more complex and expensive, but they have higher COP's, they can incorporate two or more different refrigerants, and they can achieve much lower temperatures. 11-37C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits. 11-38C The saturation pressure of refrigerant-134a at -32°C is 77 kPa, which is below the atmospheric pressure. In reality a pressure below this value should be used. Therefore, a cascade refrigeration system with 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 is simpler, 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 the lower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, and the COP of this cascade refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compression refrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressor as a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturated liquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg 47.95kJ/kg 91.269

kJ/kg 94.63kJ/kg 58.260

kJ/kg, 47.95,kJ/kg 55.255

,kJ/kg 94.63,kJ/kg 16.239

8

6

4

2

7

5

3

1

====

====

hhhh

hhhh

T

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

( ) ( )

( ) kg/s 0.1954=−−

=−−

=

−=−

=

=

==−

∑∑

kg/s 0.2494.6358.26047.9555.255

0

32

85

3285

outin

(steady) 0systemoutin

AB

BA

iiee

mhhhh

m

hhmhhm

hmhm

EE

EEE

&&

&&

&&

&&

&&& ∆

3

QL

B

A 5 8

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:

( ) ( )( )( ) ( )

( )( ) ( )( )kW 7.63

kW 34.24

=−+−=

−+−=+=

=−=−=

kJ/kg 239.16260.58kg/s 0.1954kJ/kg 255.55269.91kg/s 0.24

kJ/kg 63.94239.16kg/s 0.1954

1256incompII,incompI,in

41

hhmhhmWWW

hhmQ

BA

BL

&&&&&

&&

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

494.kW 7.63kW 34.24COP

innet,R ===

WQL&

&


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 the lower cycle, the rate of heat removal from the refrigerated space, the power input to the compressor, and the COP of this cascade refrigerator are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The heat exchanger is adiabatic. Analysis (a) Each stage of the cascade refrigeration cycle is said to operate on the ideal vapor compression refrigeration cycle. Thus the compression process is isentropic, and the refrigerant enters the compressor as a saturated vapor at the evaporator pressure. Also, the refrigerant leaves the condenser as a saturated liquid at the condenser pressure. The enthalpies of the refrigerant at all 8 states are determined from the refrigerant tables (Tables A-11, A-12, and A-13) to be

kJ/kg 47.95kJ/kg 66.268

kJ/kg 54.77kJ/kg 34.267

kJ/kg, 47.95,kJ/kg 92.260

,kJ/kg 54.77,kJ/kg 16.239

8

6

4

2

7

5

3

1

====

====

hhhh

hhhh

T

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

( ) ( )

( ) kg/s 0.2092=−−

=−−

=

−=−=

=

==−

∑∑

kg/s 24.054.7734.26747.9592.260

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:

( ) ( )( )( ) ( )

( )( ) ( )( )kW 7.75

kW 33.81

=−+−=

−+−=+=

=−=−=

kJ/kg 16.23934.267kg/s 2092.0kJ/kg 92.26066.268kg/s 24.0

kJ/kg 54.7716.239kg/s 2092.0

1256incompII,compI,inin

41

hhmhhmWWW

hhmQ

BA

BL

&&&&&

&&

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

4.36===kW 75.7kW 81.33COP

innet,R W

QL&

&


11-28

11-44 [Also solved by EES on enclosed CD] A two-stage compression refrigeration system with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled 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. 3 The 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/kg 33.73kJ/kg 32.107

kJ/kg 31.265

,kJ/kg 33.73,kJ/kg 32.107,kJ/kg 30.259

kJ/kg, 16.239

8

6

2

7

5

3

1

==

=

====

hh

h

hhhh T

QL

73B

A9 6

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 the quality at state 6,

0.1828=−

=−

=98.185

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/kg 21.26431.2651828.0130.2591828.0

11

0

9

26369

outin


=−+=−+=

=

=

==−

∑∑h

hxhxh

hmhm

EE

EEE

iiee &&

&&

&&& ∆

also,

kJ/kg 97.278KkJ/kg 94083.0

MPa 14

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/kg 239.16265.31kg/s 0.2043kJ/kg 264.21278.97kg/s 0.25

kJ/kg 73.33239.16kg/s 0.2043

kg/s 0.2043kg/s 0.251828.011


81

6

=−+−=

−+−=+=

=−=−=

=−=−=

hhmhhmWWW

hhmQ

mxm

BA

BL

AB

&&&&&

&&

&&

kW 33.88

(c) The coefficient of performance is determined from

3.75===kW 9.03kW 33.88COP

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 =1.0 Eta_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"

ηcompA ηcompB Qout COP 0,8 0,8 45,32 2,963

0,8333 0,8333 44,83 3,094 0,8667 0,8667 44,39 3,225

0,9 0,9 43,97 3,357 0,9333 0,9333 43,59 3,488 0,9667 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 1

42,5

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

ηcomp

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 with refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled 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. 3 The 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/kg 16.55kJ/kg 32.107

kJ/kg 90.255

,kJ/kg 16.55,kJ/kg 32.107,kJ/kg 88.251

kJ/kg, 16.239

8

6

2

7

5

3

1

==

=

====

hh

h

hhhh

T

QL

73 B

A9 6

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,

0.2651=−

=−

=71.196

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/kg 84.25490.2552651.0188.2512651.0

11

0

9

26369

outin


=−+=−+=

=

=

==−

∑∑h

hxhxh

hmhm

EE

EEE

iiee &&

&&

&&& ∆

and

KkJ/kg 94074.0kJ/kg 84.254

MPa 32.09

9

9 ⋅=

==

shP

also, kJ/kg 94.278KkJ/kg 94074.0

MPa 14

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 9.10kJ/kg 239.16255.90kg/s 0.1837kJ/kg 254.84278.94kg/s 0.25

kJ/kg 55.16239.16kg/s 0.1837

kg/s 0.1837kg/s 0.250.265111


81

6

=−+−=

−+−=+=

=−=−=

=−=−=

hhmhhmWWW

hhmQ

mxm

BA

BL

AB

&&&&&

&&

&&

kW 33.80

(c) The coefficient of performance is determined from

3.71===kW 9.10kW 33.80COP

innet,R W

QL&

&


11-33

11-47 A two-stage cascade refrigeration cycle is considered. The mass flow rate of the refrigerant through the upper cycle, the rate of heat removal from the refrigerated space, 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) The properties are to be obtained from the refrigerant tables (Tables A-11 through A-13):

kJ/kg.K 9377.0kJ/kg 46.244

kPa 200 @1

kPa 200 @1

====

g

g

sshh

Win .Expansion

valve Compressor

Evaporator

Condenser

4

32

1

Win .Expansion

valve Compressor

Evaporator

Condenser

QH.

8

76

5

.QL

kJ/kg 30.263kPa 500

212

2 =

==

shss

P

kJ/kg 01.268

46.24446.24430.26380.0 2

2

12

12

=→−

−=

−−

=

hh

hhhh s

Cη

kJ/kg 33.73

kJ/kg 33.73

34

kPa 500 @3

====

hhhh f

kJ/kg.K 9269.0kJ/kg 55.255

kPa 400 @5

kPa 004 @5

====

g

g

sshh

kJ/kg 33.278kPa 1200

656

6 =

==

shss

P

kJ/kg 02.284

55.25555.25533.27880.0 6

6

56

56

=→−

−=

−−

=

hh

hhhh s

Cη

kJ/kg 77.117

kJ/kg 77.117

78

kPa 1200 @7

====

hhhh f

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

kg/s 0.212=→−=−

−=−

AA

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

kW 25.67=−=−= kJ/kg)33.7346kg/s)(244. 15.0()( 41 hhmQ BL &&


kW 566.9kJ/kg)46.24401kg/s)(268. 212.0(kJ/kg)55.25502kg/s)(284. 15.0()()( 1256in

=−+−=−+−= hhmhhmW BA &&&

2.68===566.9

67.25COPin

L

WQ&

&


11-34

11-48 A two-stage cascade refrigeration cycle with a flash chamber is considered. The mass flow rate of the 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.K 9377.0kJ/kg 46.244

kPa 200 @1

kPa 200 @1

====

g

g

sshh

7

Flash chamber

Expansion valve Low-press.

Compressor

Evaporator 8

9

2

1

Expansion valve High-press.

Compressor

Condenser

QH .

6

54

3

. QL

kJ/kg 09.261kPa 450

212

2 =

==

shss

P

kJ/kg 24.265

46.24446.24409.26180.0 2

2

12

12

=→−

−=

−−

=

hh

hhhh s

Cη

kJ/kg 81.68kJ/kg 81.68

kJ/kg 77.117kJ/kg 77.117

kJ/kg 53.257

78

kPa 450 @7

56

kPa 1200 @5

kPa 450 @3

====

====

==

hhhhhhhh

hh

f

f

g

2594.0kPa 450

kJ/kg 77.1176

6

6 =

==

xPh

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

kg/s 0.2025==−

=0.2594-1

kg/s 15.01 6

7

xm

m&

&

Also, kg/s 05255.015.02025.073 =−=−= mmm &&&

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

kJ/kg 24.263kJ/kg) 53kg/s)(257. (0.05255kJ/kg) 24kg/s)(265. (0.15kg/s) 2025.0( 99

33279

=→+=

+=

hh

hmhmhm &&&

Then,

kJ/kg 9451.0kJ/kg 24.263

kPa 4509

9

9 =

==

shP

kJ/kg 27.284kPa 1200

494

4 =

==

shss

P

kJ/kg 53.289

24.26324.26327.28480.0 4

4

94

94

=→−

−=

−−

=

hh

hhhh s

Cη


11-35

The rate of heat removal from the refrigerated space is

kW 26.35=−=−= kJ/kg)81.6846kg/s)(244. 15.0()( 817 hhmQL &&


kW 442.8kJ/kg)24.26353kg/s)(289. 2025.0(kJ/kg)46.24424kg/s)(265. 15.0()()( 94127in

=−+−=−+−= hhmhhmW &&&

3.12===442.8

35.26COPin

L

WQ&

&

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

kJ/kg.K 9377.0kJ/kg 46.244

kPa 200 @1

kPa 200 @1

====

g

g

sshh

T

QH

QL

Win ·

2

·

·

4

3

2

1

kJ/kg 84.281kPa 1200

212

2 =

==

shss

P

kJ/kg 19.29146.244

46.24484.28180.0 22

12

12

=→−

−=

−−

=

hh

hhhh s

Cη

s

kJ/kg 77.117

kJ/kg 77.117

34

kPa 1200 @3

==

==

hh

hh f

kW 25.66=−=−= kJ/kg)77.11746kg/s)(244. 2025.0()( 41 hhmQL &&

kW 465.9kJ/kg)46.24419kg/s)(291. 2025.0()( 12in =−=−= hhmW &&

2.71===465.9

66.25COPin

L

WQ&

&


numerical vpc ceoncept

Environment