Solution manual of Thermodynamics-Ch.11

PROPRIETARY MATERIAL. © 2008 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.

11-1

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

(c) The net work input is determined from

kJ/kg 26.05=−=−= 03.14708.173net LH qqw

s

T

QH

QL

160 kPa 1 2

3 430°C


11-2

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


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

( ) ( ) 8.28=−

=−

=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

s

T

QH

QL1 2

3 4


11-3

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.

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.


11-4

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


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

0.4795=⎭⎬⎫

==

==

=⎭⎬⎫

°==

=⎭⎬⎫

°==

=⎭⎬⎫

°−==

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

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

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

Then, the maximum refrigeration load becomes

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

60 kPa -34°C

1

23

4

QH42°C

Condenser

Evaporator

Compressor

Expansion valve

1.2 MPa 65°C

QL

Win

Water 18°C

26°C

QH

QL

1

2

3

4

s

T

·

·

2

Win ·


11-5

11-12 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the power requirement are to be determined.


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-11, A-12, and A-13),

)throttling( kJ/kg 32.107

kJ/kg 32.107 liquid sat.

MPa 1

kJ/kg 29.275 MPa 1

KkJ/kg 92927.0kJ/kg 77.252

vapor sat.

C4

34

MPa 1 @ 33

212

2

C4 @ 1

C4 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°

°

hh

hhP

hss

P

sshhT

f

g

g

The mass flow rate of the refrigerant is

kg/s 750.2kJ/kg 107.32)(252.77

kJ/s 400)(41

41 =−

=−

=⎯→⎯−=hh

QmhhmQ L

L

&&&&

The power requirement is

kW 61.93=−=−= kJ/kg 252.77)29kg/s)(275. 750.2()( 12in hhmW &&

The COP of the refrigerator is determined from its definition,

6.46===kW 61.93

kW 400COPin

R WQL&

&

QH

QL

4°C 1

2 3

4

1 MPa

s

T

·

Win ·

·

4s


11-6

11-13 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined.




kJ/kg 47.95 liquid sat.kPa 800

kJ/kg 90.269 kPa 800

KkJ/kg 92691.0kJ/kg 55.255

vapor sat.

kPa 400

34

kPa 800 @ 33

212

2

kPa 400 @ 1

kPa 400 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hh

hhP

hss

P

sshhP

f

g

g

The mass flow rate of the refrigerant is determined from

kg/s 0.06247=−

=−

=⎯→⎯−=kJ/kg )47.95(255.55

kJ/s 10)(41

41 hhQ

mhhmQ LL

&&&&


kW 0.8964=−=−= kJ/kg 255.55)90kg/s)(269. 06247.0()( 12in hhmW &&

QH

QL

0.4 MPa 1

2 3

4

0.8 MPa

s

T

·

Win·

·

4s


11-7

11-14E An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The mass flow rate of the refrigerant and the power requirement are to be determined.


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-11E, A-12E, and A-13E),

)throttling( Btu/lbm 869.37

Btu/lbm 869.37 liquid sat.psia 100

Btu/lbm 57.117 psia 100

RBtu/lbm 22660.0Btu/lbm 61.101

vapor sat.

F10

34

psia 100 @ 33

212

2

F10 @ 1

F10 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°−

°−

hh

hhP

hss

P

sshhT

f

g

g

The mass flow rate of the refrigerant is

lbm/h 376.5=−

=−

=⎯→⎯−=Btu/lbm )869.37(101.61

Btu/h 000,24)(41

41 hhQ

mhhmQ LL

&&&&


kW 1.761=⎟⎠⎞

⎜⎝⎛−=−=

Btu/h 3412.14kW 1Btu/lbm )61.101.57lbm/h)(117 5.376()( 12in hhmW &&

QH

QL

-10°F 1

2 3

4

100 psia

s

T

·

Win ·

·

4s


11-8

11-15E Problem 11-14E is to be repeated if ammonia is used as the refrigerant.

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

"Given" x[1]=1 T[1]=-10 [F] x[3]=0 P[3]=100 [psia] Q_dot_L=24000 [Btu/h] "Analysis" Fluid$='ammonia' "compressor" h[1]=enthalpy(Fluid$, T=T[1], x=x[1]) s[1]=entropy(Fluid$, T=T[1], x=x[1]) s[2]=s[1] P[2]=P[3] h[2]=enthalpy(Fluid$, P=P[2], s=s[2]) "expansion valve" h[3]=enthalpy(Fluid$, P=P[3], x=x[3]) h[4]=h[3] "cycle" m_dot_R=Q_dot_L/(h[1]-h[4]) W_dot_in=m_dot_R*(h[2]-h[1])*Convert(Btu/h, kW) Solution for ammonia Fluid$='ammonia' COP_R=5.847 h[1]=615.92 [Btu/lb_m] h[2]=701.99 [Btu/lb_m] h[3]=112.67 [Btu/lb_m] h[4]=112.67 [Btu/lb_m] m_dot_R=47.69 [lbm/h] P[2]=100 [psia] P[3]=100 [psia] Q_dot_L=24000 [Btu/h] s[1]=1.42220 [Btu/lb_m-R] s[2]=1.42220 [Btu/lb_m-R] T[1]=-10 [F] W_dot_in=1.203 [kW] x[1]=1 x[3]=0 Solution for R-134a Fluid$='R134a' COP_R=3.993 h[1]=101.62 [Btu/lb_m] h[2]=117.58 [Btu/lb_m] h[3]=37.87 [Btu/lb_m] h[4]=37.87 [Btu/lb_m] m_dot_R=376.5 [lbm/h] P[2]=100 [psia] P[3]=100 [psia] Q_dot_L=24000 [Btu/h] s[1]=0.22662 [Btu/lb_m-R] s[2]=0.22662 [Btu/lb_m-R] T[1]=-10 [F] W_dot_in=1.761 [kW] x[1]=1 x[3]=0


11-9

11-16 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined.


Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )

( )throttlingkJ/kg 82.88

kJ/kg 82.88liquid sat.

MPa 7.0

C95.34kJ/kg 50.273MPa 7.0

KkJ/kg 94779.0kJ/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

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

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&

&

QH

QL

0.121

2 3

4

0.7 MPa

s

T

·

Win ·

·

4s


11-10

11-17 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.



( )



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


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&

&

QH

QL

0.12 MPa1

2 3

4

0.9 MPa

s

T

·

Win ·

·

4s


11-11

11-18 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.


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

Then,

( ) ( )( ) kW 7.72kJ/kg 82.58236.97kg/s 0.0541 =−=−= sL hhmQ &&

and

23.4kW 1.83kW 7.72COP

inR ===

WQL&

&

Then the percentage increase in &Q and COP becomes

4.2%

4.2%

=−

=Δ

=

=−

=Δ

=

06.406.423.4

COPCOP

COPin Increase

41.741.772.7in Increase

R

RR

L

LL Q

QQ

&

&&

QH

QL

0.12 MPa 1

2 3

4

0.7 MPa

s

T

·

Win ·

·

4s


11-12

11-19 A refrigerator with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the isentropic efficiency of the compressor, and the COP of the refrigerator are to be determined.


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


kJ/kg 98.84C24MPa 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


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&

&

QH

QL

0.15 MP 1

2

3

4

0.65 MPa

s

T

·

·

2

Win ·

0.14 MPa

0.7 MPa50°C


11-13

11-20 An air conditioner operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system is to be determined.


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. The evaporating temperature will be 22-2=20°C. From the refrigerant tables (Tables A-11, A-12, and A-13),



MPa 1

kJ/kg 11.273 MPa 1

KkJ/kg 92234.0kJ/kg 59.261

vapor sat.

C20

34

MPa 1 @ 33

212

2

C20 @ 1

C20 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°

°

hh

hhP

hss

P

sshhT

f

g

g

The COP of the air conditioner is determined from its definition,

13.39=−−

===59.26111.27332.10759.261COP

12

41

inAC -hh

-hhwqL

QH

QL

20°C 1

2 3

4

1 MPa

s

T

·

Win ·

·

4s


11-14

11-21E A refrigerator operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The increase in the COP if the throttling process were replaced by an isentropic expansion is to be determined.


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-11E, A-12E, and A-13E),

expansion) c(isentropi Btu/lbm 80.59 F15



liquid sat.psia 300



vapor sat.

F20

434

4

34

psia 300 @ 3

psia 300 @ 33

212

2

F20 @ 1

F20 @ 11

=⎭⎬⎫

=°=

=≅

⋅====

⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°

°

s

f

f

g

g

hss

T

hh

sshhP

hss

P

sshhT

The COP of the refrigerator for the throttling case is

2.012=−−

===98.10568.125

339.6698.105COP12

41

inR -hh

-hhwqL

The COP of the refrigerator for the isentropic expansion case is

2.344=−−

===98.10568.12580.5998.105COP

12

41

inR -hh

-hhwq sL

The increase in the COP by isentropic expansion is 16.5%.

QH

QL

20°F 1

2 3

4

300 psia

s

T

·

Win ·

·

4s


11-15

11-22 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP of the system and the cooling load are to be determined.





kJ/kg 12.267 kPa 600

KkJ/kg 93766.0kJ/kg 51.244

vapor sat.

C10

34

kPa 600 @ 33

212

2

C01 @ 1

C01 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°−

°−

hh

hhP

hss

P

sshhT

f

g

g

The COP of the air conditioner is determined from its definition,

7.21=−−

===51.24412.267

51.8151.244COP12

41

inAC -hh

-hhwqL

The cooling load is

kW 1141=−=−= kJ/kg )51.8151kg/s)(244. 7()( 41 hhmQL &&

QH

QL

-10°C 1

2 3

4

600 kPa

s

T

·

Win ·

·

4s


11-16

11-23 A refrigerator with refrigerant-134a as the working fluid is considered. The power input to the compressor, the rate of heat removal from the refrigerated space, and the pressure drop and the rate of heat gain in the line between the evaporator and the compressor are to be determined.


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

( )

kJ/kg 33.239MPa 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

v

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

=−=−=

=−=−=Δ

hhmQ

PPP

&&

QH

QL

0.15 MPa1

2s

3

4

0.95 MPa

s

T

·

·

2

Win ·

0.14 MPa-10°C

1 MPa

-18.5°C


11-17

11-24 EES Problem 11-23 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.


"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]"


11-18

COPplot Win[kW] Qin [kW] ηc

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


11-19

11-25 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle. The mass flow rate of the refrigerant, the condenser pressure, and the COP of the refrigerator are to be determined.


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

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


kg/s 0.00727=−

=−

=kg236.97)kJ/(298.87

kW 45.0

12

in

hhW

m&

&

(c) The refrigeration load and the COP are

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

)( 41

=−=

−= hhmQL &&

2.43===kW 0.45kW 091.1COP

in

L

WQ&

&

QH

QL

120 kPa1

2 3

4

s

T

·

Win ·

·

4s

.

QH

60°C Condenser

Evaporator

Compressor

Expansion valve

120 kPa x=0.3 QL

Win

1

23

4

.

.


11-20

Selecting the Right Refrigerant

11-26C 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.

11-27C 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.

11-28C 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.

11-29 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.


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.

11-30 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.


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.


11-21

Heat Pump Systems

11-31C A heat pump system is more cost effective in Miami because of the low heating loads and high cooling loads at that location.

11-32C 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.


11-22

11-33 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.


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

The isentropic efficiency of the compressor is

0.670=−−

=−−

=87.24776.29187.24726.277

12

12

hhhh s

Cη

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

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

(c) The power input and the COP are

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

4.64===790.067.3COP

inWQH&

&

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

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

===

6.18=−−

=−−

=46.24425.273

47.9525.273COP12

32

hhhh

kW 3.20=−=−= kJ/kg)47.9525kg/s)(273. 018.0()( 32 hhmQH &&

QH

QL

0.2 MPa 1

2 3

4

0.8 MPa

s

T

·

Win ·

·

4s

Q

QL

1

2

3

4

s

T

·

·2

Win·

.

QH750 kPa

Condenser

Evaporator

Compressor

Expansion valve

800 kPa 55°C

QL

Win

1

2 3

4

.

.


11-23

11-34 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be determined.





kJ/kg 00.280 kPa 1200

KkJ/kg 93210.0kJ/kg 72.249

vapor sat.

kPa 280

34

kPa 1200 @ 33

212

2

kPa 280 @ 1

kPa 280 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hh

hhP

hss

P

sshhP

f

g

g


kg/s 0.6605kJ/kg )72.249(280.00

kJ/s 20)(12

in12in =

−=

−=⎯→⎯−=

hhW

mhhmW&

&&&

Then the rate of heat supplied to the evaporator is

kW 87.15=−=−= kJ/kg )77.11772kg/s)(249. 6605.0()( 41 hhmQL &&

The COP of the heat pump is determined from its definition,

5.36=−−

===72.24900.28077.11700.280COP

12

32

inHP -hh

-hhwqH

QH

QL

280 kPa 1

2 3

4

1.2 MPa

s

T

·

Win ·

·

4s


11-24

11-35 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of compressor irreversibilities on the COP of the cycle is to be determined.


Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),



kJ/kg 50.271 kPa 800

KkJ/kg 93210.0kJ/kg 72.249

vapor sat.

kPa 280

34

kPa 800 @ 33

212

2

kPa 280 @ 1

kPa 280 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hh

hhP

hss

P

sshhP

f

s

g

g

The actual enthalpy at the compressor exit is determined by using the compressor efficiency:

kJ/kg 34.27585.0

72.24950.27172.249C

1212

12

12C =

−+=

−+=⎯→⎯

−−

=η

ηhh

hhhhhh ss

The COPs of the heat pump for isentropic and irreversible compression cases are

8.082=−−

=−−

==72.24950.271

47.9550.271COP12

32

inideal HP, hh

hhwq

s

sH

7.021=−−

=−−

==72.24934.275

47.9534.275COP12

32

inactual HP, hh

hhwqH

The irreversible compressor decreases the COP by 13.1%.

QH

QL

-1.25°C 1

2s

3

4

0.8 MPa

s

T

·

Win ·

·

4s

2


11-25

11-36 A heat pump operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the COP of the cycle is to be determined.


Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 2°C. The saturation pressure of refrigerant at −1.25°C is 280 kPa. From the refrigerant tables (Tables A-11, A-12, and A-13),



kJ/kg 04.274 kPa 800

KkJ/kg 9403.0kJ/kg 96.251

C25.1225.1kPa 280

34

kPa 800 @ 33

212

2

1

1

1

1

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°=+−==

hh

hhP

hss

P

sh

TP

f

The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are

kJ/kg 50.271 kPa 800

KkJ/kg 93210.0kJ/kg 72.249

vapor sat.

kPa 280

212

2

kPa 280 @ 1

kPa 280 @ 11

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

s

g

g

hss

P

sshhP

The COPs of the heat pump for the two cases are

8.082=−−

=−−

==72.24950.271

47.9550.271COP12

32

inideal HP, hh

hhwq

s

sH

8.087=−−

=−−

==96.25104.274

47.9504.274COP12

32

inactual HP, hh

hhwqH

The effect of superheating on the COP is negligible.

QH

QL

-1.25°C 1

2

3

4

0.8 MPa

s

T

·

Win ·

·

4s


11-26

11-37E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of subcooling at the exit of the condenser on the power requirement is to be determined.


Analysis In this cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as subcooled liquid. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),


Btu/lbm 124.45 F100psia 160

F1005.95.1095.9


KkJ/kg 22188.0Btu/lbm 81.108

vapor sat.

psia 50

34

F100 @ 33

3

psia 160 @ sat3

212

2

psia 50 @ 1

psia 50 @ 11

=≅

=≅⎭⎬⎫

°==

°=−=−=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

°

hh

hhTP

TT

hss

P

sshhP

f

g

g

The states at the inlet and exit of the expansion valve when the refrigerant is saturated liquid at the condenser exit are



34

psia 160 @ 33

=≅

==⎭⎬⎫=

hh

hhP

f

The mass flow rate of the refrigerant in the ideal case is

lbm/h 0.1415Btu/lbm )519.48(119.19

Btu/h 000,100)(32

41 =−

=−

=⎯→⎯−=hh

QmhhmQ H

L

&&&&


kW 4.305=⎟⎠⎞

⎜⎝⎛−=−=


With subcooling, the mass flow rate and the power input are

lbm/h 1.1350Btu/lbm )124.45(119.19

Btu/h 000,100)(32

41 =−

=−

=⎯→⎯−=hh

QmhhmQ H

L

&&&&

kW 4.107=⎟⎠⎞

⎜⎝⎛−=−=


Subcooling decreases the power requirement by 4.6%.

QH

QL

50 psia 1

2 3

4

160 psia

s

T

·

Win·

·

4s


11-27

11-38E A heat pump operating on the vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The effect of superheating at the compressor inlet on the power requirement is to be determined.


Analysis In this cycle, the compression process is isentropic and leaves the condenser as saturated liquid at the condenser pressure. The refrigerant entering the compressor is superheated by 10°F. The saturation temperature of the refrigerant at 50 psia is 40.23°F. From the refrigerant tables (Tables A-11E, A-12E, and A-13E),




KkJ/kg 2262.0Btu/lbm 99.110

F2.501023.40psia 50

34

psia 160 @ 33

212

2

1

1

1

1

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°=+==

hh

hhP

hss

P

sh

TP

f

The states at the inlet and exit of the compressor when the refrigerant enters the compressor as a saturated vapor are


KkJ/kg 22188.0Btu/lbm 81.108

vapor sat.

psia 50

212

2

psia 50 @ 1

psia 50 @ 11

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hss

P

sshhP

g

g

The mass flow rate of the refrigerant in the ideal case is

lbm/h 0.1415Btu/lbm )519.48(119.19

Btu/h 000,100)(32

41 =−

=−

=⎯→⎯−=hh

QmhhmQ H

L

&&&&


kW 4.305=⎟⎠⎞

⎜⎝⎛−=−=


With superheating, the mass flow rate and the power input are

lbm/h 3.1366Btu/lbm )519.48(121.71

Btu/h 000,100)(32

41 =−

=−

=⎯→⎯−=hh

QmhhmQ H

L

&&&&

kW 4.293=⎟⎠⎞

⎜⎝⎛−=−=


Superheating decreases the power requirement slightly.

QH

QL

50 psia 1

2

3

4

160 psia

s

T

·

Win·

·

4s


11-28

11-39 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)

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 C40.52kPa 1400 @sat °=T

C59.48kJ 24.121

kPa 14003

3

3 °=⎭⎬⎫

==

ThP

(from EES)

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

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

&&

QH

QL 1

2

3

4

1.4 MPa

s

T

·

Win ·

·

4s

1.4 MPa s2 = s1

1

2 3

4

QH

20°C x=0.23

Condenser

Evaporator

Compressor Expansion valve

sat. vap.

Win

Water 50°C

40°C

.

.


11-29

Innovative Refrigeration Systems

11-40C 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.

11-41C Cascade refrigeration systems have higher COPs than the ordinary refrigeration systems operating between the same pressure limits.

11-42C 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.

11-43C We would favor the two-stage compression refrigeration system with a flash chamber since it is simpler, cheaper, and has better heat transfer characteristics.

11-44C Yes, by expanding the refrigerant in stages in several throttling devices.

11-45C To take advantage of the cooling effect by throttling from high pressures to low pressures.


11-30

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 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

The 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

(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

(steady) 0systemoutin

=−+=−+=

=

=

==−

∑∑h

hxhxh

hmhm

EE

EEE

iiee &&

&&

&&& Δ

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

1294incompII,incompI,in

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&

&

QL

0.14 MPa

1

2 5

8

0.5 MPa

s

T

·

41 MPa

6 9A

B 37


11-31

11-47 EES Problem 11-46 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"


11-32

h[1]=enthalpy(R134a,P=P[1],x=x1) "properties for state 1" 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


11-33

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

Qou

t [k

W]

CO

P

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


11-34

11-48 [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


0.2651=−

=−

=71.196

16.5532.10766

fg

f

hhh

x


( ) ( )( )( ) ( )( ) kJ/kg 84.25490.2552651.0188.2512651.0

11

0

9

26369

outin

(steady) 0systemoutin

=−+=−+=

=

=

==−

∑∑h

hxhxh

hmhm

EE

EEE

iiee &&

&&

&&& Δ

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


3.71===kW 9.10kW 33.80COP

innet,R W

QL&

&

QL

0.14 MPa

1

2 5

8

0.32 MPa

s

T

·

41 MPa

6 9

A

B3

7


11-35

11-49 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.


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

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 &&

(c) The power input and the COP are

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&

&

.

5

6 7

8

QH

Condenser

Evaporator

Compressor

Expansion valve

Win

1

2 3

4

Condenser

Evaporator

Compressor

Expansion valve

QL

Win

.

.

.


11-36

11-50 A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling rate of the high-temperature evaporator, the power required by the compressor, and the COP of the system are to be determined.


Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg 44.234 vapor sat.

C4.26

kJ/kg 45.250 vapor sat.

C0



C4.26 @ 77

C0 @ 55

364

kPa 800 @ 33

==⎭⎬⎫°−=

==⎭⎬⎫°=

=≅=

==⎭⎬⎫=

°−

°

g

g

f

hhT

hhT

hhh

hhP

The mass flow rate through the low-temperature evaporator is found by

kg/s 05757.0kJ/kg )47.95(234.44

kJ/s 8)(67

2672 =−

=−

=⎯→⎯−=hh

QmhhmQ L

L

&&&&

The mass flow rate through the warmer evaporator is then

kg/s 04243.005757.01.021 =−=−= mmm &&&

Applying an energy balance to the point in the system where the two evaporator streams are recombined gives

kJ/kg 23.2411.0

)44.234)(05757.0()45.250)(04243.0(7251117251 =

+=

+=⎯→⎯=+

mhmhm

hhmhmhm&

&&&&&

Then,

3

45

6

Compressor

Expansion valve1

2

Evaporator 27

Evaporator 1

Condenser

Expansion valve

Expansion valve

QH

QL

-26.4°C 1

2

3

6

800 kPa

s

T

·

Win·

·

4 5

7

0°C


11-37

kJ/kg 26.286 kPa 800

KkJ/kg 9789.0 kJ/kg 23.241

kPa 100

212

2

11

C26.4 @sat 1

=⎭⎬⎫

==

⋅=⎭⎬⎫

=≅= °−

hss

P

sh

PP

The cooling rate of the high-temperature evaporator is

kW 6.58=−=−= kJ/kg )47.95(250.45)kg/s 04243.0()( 451 hhmQL &&

The power input to the compressor is

kW 4.50=−=−= kJ/kg )23.24126kg/s)(286. 1.0()( 12in hhmW &&

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

3.24=+

==kW 4.50

kW 6.58)(8COP

inR W

QL&

&


11-38

11-51E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),

Btu/lbm 99.100 vapor sat.

psia 15


psia 30



psia 15 @ 77

psia 03 @ 55

364

psia 140 @ 33

==⎭⎬⎫=

==⎭⎬⎫=

=≅=

==⎭⎬⎫=

g

g

f

hhP

hhP

hhh

hhP

The mass flow rates through the high-temperature and low-temperature evaporators are found by

lbm/h 99.49Btu/lbm )304.45(105.32

Btu/h 3000)(45

1,14511, =

−=

−=⎯→⎯−=

hhQ

mhhmQ LL

&&&&

lbm/h 58.179Btu/lbm )304.45(100.99

Btu/h 000,10)(67

2,26722, =

−=

−=⎯→⎯−=

hhQ

mhhmQ LL

&&&&


Btu/lbm 93.10158.17999.49

)99.100)(58.179()32.105)(99.49()(21

725111217251 =

++

=++

=⎯→⎯+=+mm

hmhmhhmmhmhm&&

&&&&&&

Then,


RBtu/lbm 2293.0 Btu/lbm 93.101

psia 15

212

2

11

1

=⎭⎬⎫

==

⋅=⎭⎬⎫

==

hss

P

shP


kW 1.366=⎟⎠⎞

⎜⎝⎛−+=−+=

Btu/h 3412.14kW 1Btu/lbm )93.10124lbm/h(122. )58.17999.49())(( 1221in hhmmW &&&


2.79=⎟⎠⎞

⎜⎝⎛+

==Btu/h 3412.14

kW 1kW 1.366

Btu/h 3000)(10,000COP

inR W

QL&

&

3

45

6

Compressor

Expansion valve1

2

Evaporator 27

Evaporator 1

Condenser

Expansion valve

Expansion valve

QH

QL

15 psia 1

2

3

6

140 psia

s

T

·

Win·

·

45

7

30 psia


11-39

11-52E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The power required by the compressor and the COP of the system are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),


psia 15


psia 60



psia 15 @ 77

psia 06 @ 55

364

psia 140 @ 33

==⎭⎬⎫=

==⎭⎬⎫=

=≅=

==⎭⎬⎫=

g

g

f

hhP

hhP

hhh

hhP

The mass flow rates through the high-temperature and low-temperature evaporators are found by

lbm/h 7.740Btu/lbm )304.45(110.11

Btu/h 000,48)(45

1,14511, =

−=

−=⎯→⎯−=

hhQ

mhhmQ LL

&&&&

lbm/h 6.179Btu/lbm )304.45(100.99

Btu/h 000,10)(67

2,26722, =

−=

−=⎯→⎯−=

hhQ

mhhmQ LL

&&&&


Btu/lbm 33.10858.1797.740

)99.100)(58.179()11.110)(7.740()(21

725111217251 =

++

=++

=⎯→⎯+=+mm

hmhmhhmmhmhm&&

&&&&&&

Then,



psia 15

212

2

11

1

=⎭⎬⎫

==

⋅=⎭⎬⎫

==

hss

P

shP


kW 5.966=⎟⎠⎞

⎜⎝⎛−+=−+=

Btu/h 3412.14kW 1Btu/lbm )33.10845lbm/h(130. )6.1797.740())(( 1221in hhmmW &&&


2.85=⎟⎠⎞

⎜⎝⎛+

==Btu/h 3412.14

kW 1kW 5.966

Btu/h 000),10(48,000COP

inR W

QL&

&

3

45

6

Compressor

Expansion valve1

2

Evaporator I 7

Evaporator II

Condenser

Expansion valve

Expansion valve

QH

QL

15 psia 1

2

3

6

140 psia

s

T

·

Win·

·

45

7

60 psia


11-40

11-53 A two-stage compression refrigeration system with a separation unit is considered. The mass flow rate through the two compressors, the power used by the compressors, and the system’s COP are to be determined.



kJ/kg 74.252 kPa 200

KkJ/kg 96866.0kJ/kg 86.225

vapor sat.

C40



C1.10



kJ/kg 24.273 kPa 800

KkJ/kg 93773.0kJ/kg 46.244

vapor sat.

C1.10

878

C10.1 @sat 8

C40 @ 7

C40 @ 77

56

C1.10 @ 55

34

kPa 800@ 33

212

2

C1.10 @ 1

C1.10 @ 11

=⎭⎬⎫

===

⋅====

⎭⎬⎫°−=

=≅

==⎭⎬⎫°−=

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°−

°−

°−

°−

°−

°−

hssPP

sshhT

hh

hhT

hh

hhP

hss

P

sshhT

g

g

f

f

g

g

The mass flow rate through the evaporator is determined from

kg/s 0.1601=−

=−

=⎯→⎯−=kJ/kg )43.38(225.86

kJ/s 30)(67

6676 hhQ

mhhmQ LL

&&&&

An energy balance on the separator gives

kg/s 0.2303=−−

=−−

=⎯→⎯−=−47.9546.24443.3874.252)1601.0()()(

41

5862412586 hh

hhmmhhmhhm &&&&

The total power input to the compressors is

kW 10.93=

−+−=−+−=

kJ/kg )46.24424kg/s)(273. 2303.0(kJ/kg )86.22574kg/s)(252. 1601.0()()( 122786in hhmhhmW &&&


2.74===kW 10.93

kW 30COPin

R WQL&

&

3

4 5

6

Compressor

Expansion valve

1

2

Evaporator7

Separator

Condenser

Expansion valve

Compressor

8

QL

-40°C 7

8 3

6

s

T

·

20.8 MPa

4 15

-10.1°C


11-41

11-54E A two-stage compression refrigeration system with a separation unit is considered. The cooling load and the system’s COP are to be determined.


Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),



vapor sat.

psia 60







vapor sat.

psia 120

878

8

psia 60 @ 7

psia 60 @ 77

56

psia 120 @ 55

34

psia 300 @ 33

212

2

psia 120 @ 1

psia 120 @ 11

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

=≅

==⎭⎬⎫=

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hss

P

sshhP

hh

hhP

hh

hhP

hss

P

sshhP

g

g

f

f

g

g


6641

5862412586 5258.1

339.6616.115787.4128.116)()( mm

hhhh

mmhhmhhm &&&&&& =−−

=−−

=⎯→⎯−=−

The total power input to the compressors is given by

)(5258.1)(

)()(

126786

122786in

hhmhhmhhmhhmW

−+−=−+−=&&

&&&

Solving for 6m& ,

lbm/h 4681Btu/lbm )16.11506.123(5258.1)11.110(116.28

Btu/h )14.341225()(5258.1)( 1278

in6 =

−+−×

=−+−

=hhhh

Wm

&&

The cooling effect produced by this system is then

Btu/h 319,800=−=−= Btu/lbm )787.41.11lbm/h)(110 4681()( 676 hhmQL &&


3.75===kW 25

kW 412.14)(319,800/3COP

inR W

QL&

&

3

4 5

6

Compressor

Expansion valve

1

2

Evaporator7

Separator

Condenser

Expansion valve

Compressor

8

QL

60 psia7

8 3

6

s

T

·

2300 psia

4 15120 psia


11-42

11-55 A two-stage cascade refrigeration system is considered. Each stage operates on the ideal vapor-compression cycle with upper cycle using water and lower cycle using refrigerant-134a as the working fluids. The mass flow rate of R-134a and water in their respective cycles and the overall COP of this system 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 From the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13),



kJ/kg 59.267 kPa 400

KkJ/kg 96866.0kJ/kg 86.225

vapor sat.

C40



MPa 6.1

kJ/kg 4.5083 MPa 6.1

KkJ/kg 0249.9kJ/kg 1.2510

vapor sat.

C5

78

kPa 040 @ 77

656

6

C40 @ 5

C40 @ 55

34

MPa 6.1 @ 33

212

2

C5 @ 1

C5 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°−

°−

°

°

hh

hhP

hss

P

sshhT

hh

hhP

hss

P

sshhT

f

g

g

f

g

g

The mass flow rate of R-134a is determined from

kg/s 0.1235=−

=−

=⎯→⎯−=kJ/kg )94.63(225.86

kJ/s 20)(85

85 hhQ

mhhmQ LRRL

&&&&

An energy balance on the heat exchanger gives the mass flow rate of water

kg/s 0.01523=−−

=−−

=⎯→⎯

−=−

44.8581.251094.6359.267kg/s) 1235.0(

)()(

41

76

4176

hhhh

mm

hhmhhm

Rw

wR

&&

&&


kJ/s 35.44

kJ/kg )1.2510.4kg/s)(5083 01523.0(kJ/kg )86.22559kg/s)(267. 1235.0()()( 1256in

=−+−=

−+−= hhmhhmW wR &&&


0.451===kJ/s 44.35

kJ/s 20COPin

R WQL&

&

QL -40°C 5

6 3

8

s

T

·

2 1.6 MPa

4 1 7

5°C 400 kPa


11-43

11-56 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.


Analysis From Prob. 11-55 and the water and refrigerant tables (Tables A-4, A-5, A-6, A-11, A-12, and A-13),

K 303C30K 303C30K 243C30

kJ/kg 0.4225kJ/kg 92.161

kg/s 01751.0kg/s 1235.0

KkJ/kg 27423.0KkJ/kg 24757.0

KkJ/kg 96866.0KkJ/kg 0869.3KkJ/kg 3435.2

KkJ/kg 0249.9

0

32

85

8

7

65

4

3

21

=°==°==°−=

=−==−=

==

⋅=⋅=

⋅==⋅=⋅=

⋅==

TTT

hhqhhq

mmss

ssss

ss

H

L

H

L

w

R&

&

The exergy destruction during a process of a stream from an inlet state to exit state is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie

Application of this equation for each process of the cycle gives

[ ][ ] kJ/s 52.4)96866.024757.0)(1235.0()0869.30249.9)(01751.0()303(

)()(

kJ/s 05.1243

92.16127423.096866.0)303)(1235.0(

kJ/s 00.1)24757.027423.0)(303)(1235.0()(

kJ/s 94.3)3435.20869.3)(303)(01751.0()(

kJ/s 53.38303

0.42250249.93435.2)K 303)(01751.0(

67R41w0exchheat destroyed,

850R85 destroyed,

780R78 destroyed,

340w34 destroyed,

23023 destroyed,

=−+−=

−+−=

=⎟⎠⎞

⎜⎝⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=−=−=

=−=−=

=⎟⎠⎞

⎜⎝⎛ +−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

ssmssmTX

Tq

ssTmX

ssTmX

ssTmX

Tq

ssTmX

L

L

H

Hw

&&&

&&

&&

&&

&&

For isentropic processes, the exergy destruction is zero:

0

0

56 destroyed,

12 destroyed,

=

=

X

X&

&

Note that heat is absorbed from a reservoir at -30°C (243 K) and rejected to a reservoir at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly. The greatest exergy destruction occurs in the condenser.

QL-40°C 5

6 3

8

s

T

·

21.6 MPa

4 17

5°C 400 kPa


11-44

Gas Refrigeration Cycles

11-57C The ideal gas refrigeration cycle is identical to the Brayton cycle, except it operates in the reversed direction.

11-58C The reversed Stirling cycle is identical to the Stirling cycle, except it operates in the reversed direction. Remembering that the Stirling cycle is a totally reversible cycle, the reversed Stirling cycle is also totally reversible, and thus its COP is

COPR,Stirling =−

11T TH L/

11-59C In the ideal gas refrigeration cycle, the heat absorption and the heat rejection processes occur at constant pressure instead of at constant temperature.

11-60C In aircraft cooling, the atmospheric air is compressed by a compressor, cooled by the surrounding air, and expanded in a turbine. The cool air leaving the turbine is then directly routed to the cabin.

11-61C No; because h = h(T) for ideal gases, and the temperature of air will not drop during a throttling (h1 = h2) process.

11-62C By regeneration.


11-45

11-63 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.

Analysis (a) We assume both the turbine and the compressor to be isentropic, the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17),

T hP

T hP

r

r

1 1

1 3

285 2851411584

320 320 291 7375

1

3

= ⎯ →⎯ ==

= ⎯ →⎯ ==

K kJ / kg

K kJ / kg

..

..

Thus,

( )

( )kJ/kg 76201

K8201347507375125050

kJ/kg 17452 K4450792515841

50250

4

43

4

2

21

2

34

12

.h

.T..PPPP

.h

.T..PPPP

rr

rr

==⎯→⎯=⎟

⎠⎞

⎜⎝⎛==

==⎯→⎯=⎟

⎠⎞

⎜⎝⎛==

Then the rate of refrigeration is

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

(b) The net power input is determined from

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

where

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

kW 13.36kJ/kg 285.14452.17kg/s 0.08

43outturb,

12incomp,

=−=−=

=−=−=

hhmWhhmW

&&

&&

Thus,

& . .Wnet, in = − =13 36 9 48 3.88 kW

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

COP 6.67 kW3.88 kWR

net, in= = =

&

&Q

WL 1.72

s

T

1

2 QH

47°C 12°C

3

4

·

QRefrig ·


11-46

11-64 EES Problem 11-63 is reconsidered. The effects of compressor and turbine isentropic efficiencies on the rate of refrigeration, the net power input, and the COP are to be investigated. Analysis The problem is solved using EES, and the solution is given below.

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


11-47

COP ηcomp ηturb QRefrig

[kW] Winnet [kW]

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

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

1

2

3

4

5

6

7

ηcomp

QR

efrig

[kW

]

η turb

0.70.70.850.851.01.0

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

2

4

6

8

10

12

ηcomp

Win

;net

[kW

]

η turb

0.70.70.850.851.01.0

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

0,5

1

1,5

2

ηcomp

CO

P

η turb

0.70.70.850.851.01.0


11-48

11-65 [Also solved by EES on enclosed CD] An ideal-gas refrigeration cycle with air as the working fluid is considered. The rate of refrigeration, the net power input, and the COP are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with variable specific heats. 3 Kinetic and potential energy changes are negligible.

Analysis (a) We assume the turbine inlet temperature to be the temperature of the surroundings, and the compressor inlet temperature to be the temperature of the refrigerated space. From the air table (Table A-17),

T hP

T hP

r

r

1 1

1 3

285 2851411584

320 320 291 7375

1

3

= ⎯ →⎯ ==

= ⎯ →⎯ ==

K kJ / kg

K kJ / kg

..

..

Thus,

( )

( )kJ/kg 76201

K8201347507375125050

kJ/kg 17452 K4450792515841

50250

4

43

4

2

21

2

34

12

.h

.T..PPPP

.h

.T..PPPP

s

srr

s

srr

==⎯→⎯=⎟

⎠⎞

⎜⎝⎛==

==⎯→⎯=⎟

⎠⎞

⎜⎝⎛==

Also,

( )( )( )

kJ/kg 54219762012932085029320

433443

43

.....

hhhhhhhh

sTs

T

=−−=

−η−=⎯→⎯−−

=η

Then the rate of refrigeration is

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

(b) The net power input is determined from

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

where

( ) ( )

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

kW 16.700.80/kJ/kg 285.14452.17kg/s 0.08

43outturb,

1212incomp,

=−=−=

=−=η−=−=

hhmW

/hhmhhmW Cs

&&

&&&

Thus,

& . .Wnet, in = − =16 70 8 06 8.64 kW

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

COP 5.25 kW8.64 kWR

net, in= = =

&

&Q

WL 0.61

s

T

1

2QH

47°C12°C

3

4s

·

QRefrig ·

2

1 4


11-49

11-66 A gas refrigeration cycle with helium as the working fluid is considered. The minimum temperature in the cycle, the COP, and the mass flow rate of the helium are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.

Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).

Analysis (a) From the isentropic relations,

( )( )( )

( )( ) K1.208

31K323

K2.4083K263

667.1/667.0k/1k

3

434

667.1/667.0k/1k

1

212

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

PP

TT

PP

TT

s

s

and

( ) ( )( )

( ) ( ) ( )K 5.444

80.0/2632.408263/

1.20832380.0323

121212

12

12

12

min

433443

43

43

43

=−+=−+=⎯→⎯

−−

=−−

=

==−−=−−=⎯→⎯

−−

=−−

=

Csss

C

sTss

T

TTTTTTTT

hhhh

TTTTT

TTTT

hhhh

ηη

ηηK 231.1

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

( ) ( )

( ) ( )

( ) ( ) 0.356=−−−

−=

−−−−

=

−−−−

=

−==

1.2313232635.4441.231263

COP

4312

41

4312

41

outturb,incomp,innet,R

TTTTTT

hhhhhh

wwq

wq LL

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

( ) ( )( ) kg/s 0.109=−⋅

=−

=−

==K 231.1263KkJ/kg 5.1926

kJ/s 18

41

refrig

41

refrigrefrig

TTcQ

hhQ

qQ

mpL

&&&&

s

T

1

2QH

50°C-10°C

3

4s

·

QRefrig ·

2

1 4


11-50

11-67 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with constant specific heats.

Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2).

Analysis (a) From the isentropic relations,

( )

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

1

212 ==⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

PP

TT s

K 5.472

2.2732.2734.43280.0 2

2

12

12

12

12

=⎯→⎯−

−=

−−

=−−

=

TT

TTTT

hhhh ss

Cη

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

( ) 4.1/4.0

4

k/1k

4

545 5

1⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

TPP

TT s

ssT TT

Thhhh

54

4

54

54 2.19385.0

−−

=→−−

=η

Using EES, we obtain T4 = 281.3 K.

An energy balance on the regenerator may be written as

or,

( ) ( )

K 3.2463.2812.3082.2734316

61436143

=+−=+−=

−=−⎯→⎯−=−

TTTT

TTTTTTcmTTcm pp &&

The effectiveness of the regenerator is

0.434=−−

=−−

=−−

=3.2462.3083.2812.308

63

43

63

43regen TT

TThhhh

ε

(b) The refrigeration load is

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

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

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

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

0.478=−

=−

==43.3513.80

36.21COPoutT,inC,innet, WW

QW

Q LL&&

&

&

&

s

T

1

3

0°C

56

4

2sQH ·

QRefrig ·

35°C

Qrege

2

5-80°C

.

3

45

6

QH

Compressor

12

QL

Heat Exch.

Heat Exch.

Regenerator

Turbine

.


11-51

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

( )

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

34 =⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

−

rTT s

K 6.2116.1942.308

2.30885.0 4

4

43

43 =⎯→⎯−

−=⎯→⎯

−−

= TT

TTTT

sTη

kW 24.74=

−=

−=

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

)( 41 TTcmQ pL &&

[ ]kW 32.41

kJ/kg)6.211(308.2)2.273(472.5kJ/kg.K) 5kg/s)(1.00 4.0(

)()( 4312innet,

=−−−=

−−−= TTcmTTcmW pp &&&

0.599===32.4174.24COP

innet,WQL&

&

s

T

1

2QH

35°C 0°C

3

4s

·

QRefrig ·

2

1 4


11-52

11-68E An ideal gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined.

Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible.

Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).

Analysis From the isentropic relations,

R 8.37641R) 560(

R 7.668R)(4) 450(

4.1/4.0/)1(

3

434

4.1/4.0/)1(

1

212

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

kk

kk

PP

TT

PP

TT

The COP of this ideal gas refrigeration cycle is determined from

2.06=−−−

−=

−−−−

=

−−−−

=

−==

)8.376560()4507.668(8.376450)()(

)()(

COP

4312

41

4312

41


TTTTTT

hhhhhh

wwq

wq LL

s

T

1

2 QH

100°F-10°F

3

4

·

QRefrig ·


11-53

11-69E An gas refrigeration cycle with air as the working fluid has a compression ratio of 4. The COP of the cycle is to be determined.


Properties The properties of air at room temperature are cp = 0.240 Btu/lbm·R and k = 1.4 (Table A-2Ea).


R 9.402psia 19psia 6R) 560(

R 7.668R)(4) 450(

4.1/4.0/)1(

3

434

4.1/4.0/)1(

1

212

=⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

kk

s

kk

s

PP

TT

PP

TT

and

R 1.470)87.0/()4507.668(450/)(

R 3.412)9.402560)(94.0(560)(

121212

12

12

12

433443

43

43

43

=−+=−+=⎯→⎯

−−

=−−

=

=−−=−−=⎯→⎯

−−

=−−

=

Csss

C

sTss

T

TTTTTTTT

hhhh

TTTTTTTT

hhhh

ηη

ηη

The COP of this gas refrigeration cycle is determined from

0.364=−−−

−=

−−−−

=

−−−−

=

−==

)3.412560()4504.701(3.412450)()(

)()(

COP

4312

41

4312

41


TTTTTT

hhhhhh

wwq

wq LL

s

T

1

2s QH

100°F -10°F

3

4s

·

QRefrig ·

2

1 4


11-54

11-70 An ideal gas refrigeration cycle with air as the working fluid provides 15 kW of cooling. The mass flow rate of air and the rates of heat addition and rejection are to be determined.


Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a).


K 3.191kPa 500kPa 100K) 303(

K 1.464kPa 100kPa 500K) 293(

4.1/4.0/)1(

3

434

4.1/4.0/)1(

1

212

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

−

kk

kk

PP

TT

PP

TT

The mass flow rate of the air is determined from

kg/s 0.1468=−⋅

=−

=⎯→⎯−=K )3.191K)(293kJ/kg (1.005

kJ/s 15)(

)(41

Refrig41Refrig TTc

QmTTcmQ

pp

&&&&

The rate of heat addition to the cycle is the same as the rate of cooling,

kW 15== Refrigin QQ &&

The rate of heat rejection from the cycle is

kW 23.8=−⋅=−= K)303K)(464.1kJ/kg 5kg/s)(1.00 1468.0()( 32 TTcmQ pH &&

s

T

1

2 QH

30°C20°C

3

4

·

QRefrig ·


11-55

11-71 An ideal gas refrigeration cycle with air as the working fluid is considered. The minimum pressure ratio for this system to operate properly is to be determined.



Analysis An energy balance on process 4-1 gives

K 1.248

KkJ/kg 1.005kJ/kg 20

K 268

)(

Refrig14

41Refrig

=⋅

−=−=

−=

p

p

cq

TT

TTcq

The minimum temperature at the turbine inlet would be the same as that to which the heat is rejected. That is,

K 2933 =T

Then the minimum pressure ratio is determined from the isentropic relation to be

1.79=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

− 4.0/4.1)1/(

4

3

4

3

K 248.1K 293

kk

TT

PP

s

T

1

2 QH

20°C -5°C

3

4

·

QRefrig ·


11-56

11-72 An ideal gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined.




K 2.128161K) 283(

K 5.420K)(4) 283(

K 9.378K)(4) 255(

4.1/4.0/)1(

5

656

4.1/4.0/)1(

3

434

4.1/4.0/)1(

1

212

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

kk

kk

kk

PP

TT

PP

TT

PP

TT


1.19=−−−+−

−=

−−−+−−

=

−−−+−−

=

−==

)2.128283()2835.420()2559.378(2.128255

)()()(

)()()(

COP

653412

61

653412

61


TTTTTTTT

hhhhhhhh

wwq

wq LL


kg/s 0.163=−⋅

=−

=⎯→⎯−=K )2.128K)(255kJ/kg (1.005

kJ/s )3600/000,75()(

)(61

Refrig61Refrig TTc

QmTTcmQ

pp

&&&&

10°C

s

T

1

2

3

4

5

6

-18°C

3

4 5

6

. Q

1

2

Turbine

Q .

Heat exch.

Heat exch.

. Q

Heat exch.


11-57

11-73 A gas refrigeration cycle with with two stages of compression with intercooling using air as the working fluid is considered. The COP of this system and the mass flow rate of air are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas with constant specific heats. 3 Kinetic and potential energy changes are negligible. Properties The properties of air at room temperature are cp = 1.005 kJ/kg·K and k = 1.4 (Table A-2a). Analysis From the isentropic relations,

K 2.128161K) 283(

K 5.420K)(4) 283(

K 9.378K)(4) 255(

4.1/4.0/)1(

5

656

4.1/4.0/)1(

3

434

4.1/4.0/)1(

1

212

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

kk

s

kk

s

kk

s

PP

TT

PP

TT

PP

TT

and

K 9.135)2.128283)(95.0(283)(

K 8.44485.0/)2835.420(283/)(

K 8.40085.0/)2559.378(255/)(

655665

65

65

65

343434

34

34

34

121212

12

12

12

=−−=−−=⎯→⎯−−

=−−

=

=−+=−+=⎯→⎯−−

=−−

=

=−+=−+=⎯→⎯−−

=−−

=

sTss

T

Csss

C

Csss

C

TTTTTTTT

hhhh

TTTTTTTT

hhhh

TTTTTTTT

hhhh

ηη

ηη

ηη


0.742=−−−+−

−=

−−−+−−

=

−−−+−−

=

−==

)9.135283()2838.444()2558.400(9.135255

)()()(

)()()(

COP

653412

61

653412

61


TTTTTTTT

hhhhhhhh

wwq

wq LL


kg/s 0.174=−⋅

=−

=⎯→⎯−=K )9.135K)(255kJ/kg (1.005

kJ/s )3600/000,75()(

)(61

Refrig61Refrig TTc

QmTTcmQ

pp

&&&&

10°C

s

T

1

2

3

4

5

6

-18°C

2s4s

6s

3

4 5

6

. Q

1

2

Turbine

Q .

Heat exch.

Heat exch.

. Q

Heat exch.


11-58

Absorption Refrigeration Systems

11-74C Absorption refrigeration is the kind of refrigeration that involves the absorption of the refrigerant during part of the cycle. In absorption refrigeration cycles, the refrigerant is compressed in the liquid phase instead of in the vapor form.

11-75C The main advantage of absorption refrigeration is its being economical in the presence of an inexpensive heat source. Its disadvantages include being expensive, complex, and requiring an external heat source.

11-76C In absorption refrigeration, water can be used as the refrigerant in air conditioning applications since the temperature of water never needs to fall below the freezing point.

11-77C The fluid in the absorber is cooled to maximize the refrigerant content of the liquid; the fluid in the generator is heated to maximize the refrigerant content of the vapor.

11-78C The coefficient of performance of absorption refrigeration systems is defined as

geninpump,gen

R inputrequiredoutputdesiredCOP

QQ

WQQ LL ≅

+==

11-79C The rectifier separates the water from NH3 and returns it to the generator. The regenerator transfers some heat from the water-rich solution leaving the generator to the NH3-rich solution leaving the pump.


11-59

11-80 The COP of an absorption refrigeration system that operates at specified conditions is given. It is to be determined whether the given COP value is possible.

Analysis The maximum COP that this refrigeration system can have is

142268300

268 K403 K30011COP

0

0maxR, .

TTT

TT

L

L

s=⎟

⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

which is slightly greater than 2. Thus the claim is possible, but not probable.

11-81 The conditions at which an absorption refrigeration system operates are specified. The maximum COP this absorption refrigeration system can have is to be determined.


2.64=⎟⎠⎞

⎜⎝⎛

−⎟⎠⎞

⎜⎝⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

273298273

K 393K 29811COP

0

0maxR,

L

L

s TTT

TT

11-82 The conditions at which an absorption refrigeration system operates are specified. The maximum rate at which this system can remove heat from the refrigerated space is to be determined.


151243298

243K 403K 29811COP

0

0maxR, .

TTT

TT

L

L

s=⎟

⎠⎞

⎜⎝⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

Thus,

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

11-83E The conditions at which an absorption refrigeration system operates are specified. The COP is also given. The maximum rate at which this system can remove heat from the refrigerated space is to be determined.

Analysis For a COP = 0.55, the rate at which this system can remove heat from the refrigerated space is

( )( ) Btu/h 100.55 5×=== Btu/h10550COP 5genR .QQL&&


11-60

11-84 A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The rate at which the steam condenses, the power input to the reversible refrigerator, and the second law efficiency of an actual chiller are to be determined.

Properties The enthalpy of vaporization of water at 200°C is hfg = 1939.8 kJ/kg (Table A-4).

Analysis (a) The thermal efficiency of the reversible heat engine is

370.0K )15.273200(

K )15.27325(11 0revth, =

++

−=−=sT

Tη

The COP of the reversible refrigerator is

52.7K )15.27310()15.27325(

K )15.27310(COP0

revR, =+−−+

+−=

−=

L

L

TTT

The COP of the reversible absorption refrigerator is

78.2)52.7)(370.0(COPCOP revR,revth,revabs, ===η

The heat input to the reversible heat engine is

kW 911.72.78

kW 22COP revabs,

in === LQQ

&&

Then, the rate at which the steam condenses becomes

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

fgs h

Qm

&&

(b) The power input to the refrigerator is equal to the power output from the heat engine

kW 2.93==== )kW 191.7)(370.0(inrevth,HEout,Rin, QWW &&& η

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

0.252===2.78

7.0COPCOP

revabs,

actualIIη

Rev. HE

T0

Ts

Rev.Ref.

TL

T0


11-61

Special Topic: Thermoelectric Power Generation and Refrigeration Systems

11-85C The circuit that incorporates both thermal and electrical effects is called a thermoelectric circuit.

11-86C When two wires made from different metals joined at both ends (junctions) forming a closed circuit and one of the joints is heated, a current flows continuously in the circuit. This is called the Seebeck effect. When a small current is passed through the junction of two dissimilar wires, the junction is cooled. This is called the Peltier effect.

11-87C No.

11-88C No.

11-89C Yes.

11-90C When a thermoelectric circuit is broken, the current will cease to flow, and we can measure the voltage generated in the circuit by a voltmeter. The voltage generated is a function of the temperature difference, and the temperature can be measured by simply measuring voltages.

11-91C The performance of thermoelectric refrigerators improves considerably when semiconductors are used instead of metals.

11-92C 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.


11-62

11-93E A thermoelectric generator that operates at specified conditions is considered. The maximum thermal efficiency this thermoelectric generator can have is to be determined.

Analysis The maximum thermal efficiency of this thermoelectric generator is the Carnot efficiency,

%3.31=−=−==R800R550

11Carnotth,maxth,H

L

TT

ηη

11-94 A thermoelectric refrigerator that operates at specified conditions is considered. The maximum COP this thermoelectric refrigerator can have and the minimum required power input are to be determined.

Analysis The maximum COP of this thermoelectric refrigerator is the COP of a Carnot refrigerator operating between the same temperature limits,

Thus, ( ) ( ) ( )

W 12.1

10.72

===

=−

=−

==

72.10 W130

COP

1K 268/K 2931

1/1COPCOP

maxminin,

CarnotR,max

L

LH

QW

TT

&&

11-95 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

11-96E A thermoelectric cooler that operates at specified conditions with a given COP is considered. The required power input to the thermoelectric cooler is to be determined.

Analysis The required power input is determined from the definition of COPR,

hp 3.14=Btu/min 33.310.15Btu/min 20

COPCOP

Rin

inR ===⎯→⎯= LL Q

WWQ &

&&

&


11-63

11-97 A thermoelectric refrigerator powered by a car battery cools 9 canned drinks in 12 h. The average COP of this refrigerator is to be determined.

Assumptions Heat transfer through the walls of the refrigerator is negligible.

Properties The properties of canned drinks are the same as those of water at room temperature, ρ = 1 kg/L and cp = 4.18 kJ/kg·°C (Table A-3).

Analysis The cooling rate of the refrigerator is simply the rate of decrease of the energy of the canned drinks,

W6.71=kW 00671.0s 360012

kJ 290

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

coolingcooling

cooling

=×

=Δ

=

°°⋅=Δ=×==

tQ

Q

TmcQm

&

Vρ

The electric power consumed by the refrigerator is

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

Then the COP of the refrigerator becomes

COP W36 W

cooling

in= = = ≈&

&. .

Q

W

6 71 0 200.186


11-64

11-98E A thermoelectric cooler is said to cool a 12-oz drink or to heat a cup of coffee in about 15 min. The average rate of heat removal from the drink, the average rate of heat supply to the coffee, and the electric power drawn from the battery of the car are to be determined.

Assumptions Heat transfer through the walls of the refrigerator is negligible.

Properties The properties of canned drinks are the same as those of water at room temperature, cp = 1.0 Btu/lbm.°F (Table A-3E).

Analysis (a) The average cooling rate of the refrigerator is simply the rate of decrease of the energy content of the canned drinks,

W36.2=⎟

⎠⎞

⎜⎝⎛

×=

Δ=

°°⋅=Δ=

Btu 1J 1055

s 6015Btu 84.30

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

coolingcooling

cooling

tQ

Q

TmcQ p

&

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

W49.7=⎟

⎠⎞

⎜⎝⎛

×=

Δ=

°°⋅=Δ=

Btu 1J 1055

s 6015Btu 4.42

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

heatingheating

heating

tQ

Q

TmcQ p

&

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

W441

W181

.1.2

W7.49COP

2.112.01COPCOP

0.2 W2.36

COP

heating

heatingheatingin,

coolingheating

cooling

coolingcoolingin,

===

=+=+=

===

QW

QW

&&

&&

11-99 The maximum power a thermoelectric generator can produce is to be determined.

Analysis The maximum thermal efficiency this thermoelectric generator can have is

1420K 353K 30311maxth, .

TT

H

L =−=−=η

Thus,

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


11-65

Review Problems

11-100 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.


Analysis (a) The COP of this refrigeration cycle is determined from

( ) ( ) ( ) 5.06=−

=−

=1K 253/K 303

11/

1COP CR,LH TT

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

kPa 770.64kPa 132.82

====

°

°−

C03@satcond

C20@satevap

PPPP

(c) The net work input is determined from

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

kJ/kg 43.5791.21215.049.25

C20@22

C20@11

=+=+=

=+=+=

°−

°−

fgf

fgf

hxhhhxhh

kJ/kg 27.35===

=−=−=

06.5kJ/kg 138.4

COP

kJ/kg4.13843.5782.195

Rinnet,

12

L

Lq

w

hhq

s

T

qL 1 2

3 430°C

-20°C


11-66

11-101 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a as the working fluid is used to heat a house. The rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the COP of this heat pump are to be determined.



( )throttling kJ/kg 61.101


MPa 9.0

kJ/kg 75.275MPa 9.0

/kgm 099867.0KkJ/kg 93773.0

kJ/kg 46.244

vaporsat.kPa 200

34

MPa 0.9 @ 33

212

2

3kPa 200 @ 1

kPa 200 @ 1

kPa 200 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

==

⋅====

⎭⎬⎫=

hh

hhP

hss

P

sshh

P

f

g

g

g

vv

The rate of heat supply to the house is determined from

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

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

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

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

5.57=−−

=−−

==46.24475.27561.10175.275COP

12

32

inR hh

hhwqL

11-102 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

=q

wL

where

( )( )

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

66816

11

with1

hhhhxwww

hhh

xhhxqfg

fL

−+−−=+=

−=−−=

QH

QL

200 kPa 1

2 3

4

0.9 MPa

s

T

·

Win ·

·

House


11-67

11-103 A two-stage compression refrigeration system using refrigerant-134a as the working fluid is considered. The fraction of the refrigerant that evaporates as it is throttled to the flash chamber, the amount of heat removed from the refrigerated space, the compressor work, and the COP are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 The flashing chamber is adiabatic.

Analysis (a) The enthalpies of the refrigerant at several states are determined from the refrigerant tables to be (Tables A-11, A-12, and A-13)

kJ/kg 94.63kJ/kg 47.95

kJ/kg 58.260

,kJ/kg 94.63,kJ/kg 47.95

,kJ/kg 55.255,kJ/kg 16.239

8

6

2

7

5

3

1

=

=

=

=

===

hh

h

hhhh


0.1646=−

=−

=62.191

94.6347.9566

fg

f

hhh

x


( ) ( )

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

0

9

26369

outin(steady) 0

systemoutin

=−+=−+=

=

=→==−

∑∑h

hxhxh

hmhm

EEEEE

iiee &&

&&&&& Δ

KkJ/kg 94168.0kJ/kg 75.259

MPa 4.09

9

9 ⋅=⎭⎬⎫

==

shP

Also,

kJ/kg 47.274KkJ/kg 94168.0

MPa 8.04

94

4 =⎭⎬⎫

⋅===

hss

P

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

( )( ) ( )( )

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

kJ/kg 146.4

=−+−−=−+−−=+=

=−−=−−=

kJ/kg 259.75274.47kJ/kg 239.16260.580.1646111

kJ/kg 63.94239.161646.011


816

hhhhxwww

hhxqL


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

inR w

qL

qL

0.14 MPa

1

2 5

8

0.4 MPa

s

T

40.8 MPa

6 9A

B 37


11-68

11-104 A refrigerator operating on a vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy loss is to be determined.


Analysis In this cycle, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13),

KkJ/kg 34605.0 kJ/kg 82.88

C10


KkJ/kg 33230.0kJ/kg 82.88

liquid sat.kPa 700

kJ/kg 38.270 kPa 700

KkJ/kg 93766.0kJ/kg 51.244

vapor sat.

C10

44

4

34

kPa 700 @ 3

kPa 700 @ 33

212

2

C10 @ 1

C10 @ 11

⋅=⎭⎬⎫

=°−=

=≅

⋅====

⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°−

°−

shT

hh

sshhP

hss

P

sshhT

f

f

s

g

g


kJ/kg 95.27485.0

51.24438.27051.244C

1212

12

12C =

−+=

−+=⎯→⎯

−−

=η

ηhh

hhhhhh ss

and kJ/kg 95252.0 kJ/kg 95.274

kPa 7002

2

2 =⎭⎬⎫

==

shP

The heat added in the evaporator and that rejected in the condenser are

kJ/kg 186.13kJ/kg )82.8895.274(

kJ/kg 155.69kJ/kg )82.88244.51(

32

41

=−=−==−=−=

hhqhhq

H

L


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


kJ/kg 6.29=⎟⎠

⎞⎜⎝

⎛ −−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−=−=

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⋅−=−=

K 273kJ/kg 69.155

34605.093766.0)K 295(

kJ/kg 06.4KkJ/kg )33230.034605.0)(K 295()(

kJ/kg 17.3K 295kJ/kg 13.186

95252.033230.0)K 295(

kJ/kg 38.4KkJ/kg )93766.095252.0)(K 295()(

41041 destroyed,

34034 destroyed,

23023 destroyed,

12012 destroyed,

L

L

H

H

Tq

ssTx

ssTxTq

ssTx

ssTx

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

QH

QL

-10°C 1

2s

3

4

0.7 MPa

s

T

·

Win ·

·

4s

2


11-69




KkJ/kg 33146.0 kJ/kg 98.84

C10


KkJ/kg 31958.0kJ/kg 98.84

C247.27.26

7.2kPa 700

kJ/kg 38.270 kPa 700

KkJ/kg 93766.0kJ/kg 51.244

vapor sat.

C10

44

4

34

C24 @ 3

C24 @ 3kPa 700 @sat 3

3

212

2

C10 @ 1

C10 @ 11

⋅=⎭⎬⎫

=°−=

=≅

⋅=≅=≅

⎪⎭

⎪⎬

⎫

°=−=−=

=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°

°

°−

°−

shT

hh

sshh

TTP

hss

P

sshhT

f

f

s

g

g


kJ/kg 95.27485.0

51.24438.27051.244C

1212

12

12C =

−+=

−+=⎯→⎯

−−

=η

ηhh

hhhhhh ss

and kJ/kg 95252.0 kJ/kg 95.274

kPa 7002

2

2 =⎭⎬⎫

==

shP


kJ/kg 189.97kJ/kg )98.8495.274(

kJ/kg 159.53kJ/kg )98.84244.51(

32

41

=−=−==−=−=

hhqhhq

H

L


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


kJ/kg 6.44=⎟⎠

⎞⎜⎝

⎛ −−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−=−=

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⋅−=−=

K 273kJ/kg 53.159

33146.093766.0)K 295(

kJ/kg 50.3KkJ/kg )31958.033146.0)(K 295()(

kJ/kg 25.3K 295kJ/kg 97.189

95252.031958.0)K 295(

kJ/kg 38.4KkJ/kg )93766.095252.0)(K 295()(

41041 destroyed,

34034 destroyed,

23023 destroyed,

12012 destroyed,

L

L

H

H

Tq

ssTx

ssTxTq

ssTx

ssTx

The greatest exergy destruction occurs in the evaporator. Note that heat is absorbed from freezing water at 0°C (273 K) and rejected to the ambient air at 22°C (295 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

QH

QL

-10°C 1

2s

3

4

0.7 MPa

s

T

·

Win ·

· 2


11-70



Analysis In this cycle, the refrigerant leaves the condenser as saturated liquid at the condenser pressure. The compression process is not isentropic. From the refrigerant tables (Tables A-11, A-12, and A-13),

KkJ/kg 4988.05089.0

kJ/kg 77.117C37


KkJ/kg 42441.0kJ/kg 77.117

liquid sat.

MPa 2.1

kJ/kg 11.298 MPa 2.1

KkJ/kg 9865.0kJ/kg 09.233

C30737kPa 60

4

4

4

4

34

MPa 1.2 @ 3

MPa 1.2 @ 33

212

2

1

1

1

C37 @sat 1

⋅==

⎭⎬⎫

=°−=

=≅

⋅====

⎭⎬⎫=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°−=+−=== °−

sx

hT

hh

sshhP

hss

P

sh

TPP

f

f

s


kJ/kg 33.30590.0

09.23311.29809.233C

1212

12

12C =

−+=

−+=⎯→⎯

−−

=η

ηhh

hhhhhh ss

and KkJ/kg 0075.1 Btu/lbm 33.305

MPa 2.12

2

2 ⋅=⎭⎬⎫

==

shP


kJ/kg 56.187kJ/kg )77.11733.305(kJ/kg 32.115kJ/kg )77.11709.233(

32

41

=−=−==−=−=

hhqhhq

H

L


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


kJ/kg 57.1K 273)34(

kJ/kg 32.1154988.09865.0)K 303(

KkJ/kg )42441.04988.0)(K 303()(

kJ/kg 88.10K 303kJ/kg 56.187

0075.142441.0)K 303(

kJ/kg 36.6KkJ/kg )9865.00075.1)(K 303()(

41041 destroyed,

34034 destroyed,

23023 destroyed,

12012 destroyed,

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−=−=

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⋅−=−=

L

L

H

H

Tq

ssTx

ssTxTq

ssTx

ssTx

kJ/kg 22.54

The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

QH

QL

-37°C 1

2s

3

4

1.2 MPa

s

T

·

Win ·

·

4s

2


11-71




KkJ/kg 4585.04665.0

kJ/kg 26.108C37


KkJ/kg 39486.0kJ/kg 26.108

C403.63.46

3.6MPa 2.1

kJ/kg 11.298 MPa 2.1

KkJ/kg 9865.0kJ/kg 09.233

C30737kPa 60

4

4

4

4

34

C40 @ 3

C40 @ 3MPa 1.2 @sat 3

3

212

2

1

1

1

C37 @sat 1

⋅==

⎭⎬⎫

=°−=

=≅

⋅=≅=≅

⎪⎭

⎪⎬

⎫

°=−=−=

=

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°−=+−===

°

°

°−

sx

hT

hh

sshh

TTP

hss

P

sh

TPP

f

f

s


kJ/kg 33.30590.0

09.23311.29809.233C

1212

12

12C =

−+=

−+=⎯→⎯

−−

=η

ηhh

hhhhhh ss

and KkJ/kg 0075.1 Btu/lbm 33.305

MPa 2.12

2

2 ⋅=⎭⎬⎫

==

shP


kJ/kg 07.197kJ/kg )26.10833.305(

kJ/kg 83.124kJ/kg )26.10809.233(

32

41

=−=−==−=−=

hhqhhq

H

L


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


kJ/kg 73.1K 273)34(

kJ/kg 83.1244585.09865.0)K 303(

KkJ/kg )39486.04585.0)(K 303()(

kJ/kg 44.11K 303kJ/kg 07.197

0075.139486.0)K 303(

kJ/kg 36.6KkJ/kg )9865.00075.1)(K 303()(

41041 destroyed,

34034 destroyed,

23023 destroyed,

12012 destroyed,

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−−=⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−=−=

=⎟⎠

⎞⎜⎝

⎛ +−=⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

=⋅−=−=

L

L

H

H

Tq

ssTx

ssTxTq

ssTx

ssTx

kJ/kg 19.28

The greatest exergy destruction occurs in the expansion valve. Note that heat is absorbed from fruits at -34°C (239 K) and rejected to the ambient air at 30°C (303 K), which is also taken as the dead state temperature. Alternatively, one may use the standard 25°C (298 K) as the dead state temperature, and perform the calculations accordingly.

QH

QL

-37°C 1

2s

3

4

1.2 MPa

s

T

·

Win ·

· 2


11-72

11-108E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The cooling load of both evaporators per unit of flow through the compressor and the COP of the system are to be determined.


Analysis From the refrigerant tables (Tables A-11E, A-12E, and A-13E),


F5.29


F30



F5.29 @ 77

F30 @ 55

364

psia 160 @ 33

==⎭⎬⎫°−=

==⎭⎬⎫°=

=≅=

==⎭⎬⎫=

°−

°

g

g

f

hhT

hhT

hhh

hhP

For a unit mass flowing through the compressor, the fraction of mass flowing through Evaporator II is denoted by x and that through Evaporator I is y (y = 1-x). From the cooling loads specification,

)(2)(

2

6745

2 vape,1 vape,

hhyhhx

QQ LL

−=−

= &&

where

yx −= 1

Combining these results and solving for y gives

3698.0)519.4840.107()519.4868.98(2

519.4840.107)()(2 4567

45 =−+−

−=

−+−−

=hhhh

hhy

Then,

6302.03698.011 =−=−= yx


3

45

6

Compressor

Expansion valve1

2

Evaporator 27

Evaporator 1

Condenser

Expansion valve

Expansion valve

QH

QL

-29.5°F 1

2

3

6

160 psia

s

T

·

Win·

·

45

7

30°F


11-73

Btu/lbm 18.1041

)68.98)(3698.0()40.107)(6302.0(1

751175 =

+=

+=⎯→⎯=+

yhxhhhyhxh

Then,



psia 10

212

2

11

F29.5 @sat 1

=⎭⎬⎫

==

⋅=⎭⎬⎫

=≅= °−

hss

P

sh

PP

The cooling load of both evaporators per unit mass through the compressor is

Btu/lbm 55.66=

−+−=−+−=

Btu/lbm )519.48(98.68)3698.0(Btu/lbm )519.48(107.40)6302.0()()( 6745 hhyhhxqL

The work input to the compressor is

Btu/lbm 26.96Btu/lbm )18.104(131.1412in =−=−= hhw


2.06===Btu/lbm 26.96Btu/lbm 55.66COP

inR w

qL


11-74

11-109E A two-evaporator compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.


Analysis From Prob. 11-107E and the refrigerant tables (Tables A-11E, A-12E, and A-13E),

Btu/lbm 621.82Btu/lbm 161.50Btu/lbm 881.58

3698.016302.0

RBtu/lbm 22948.0RBtu/lbm 11286.0RBtu/lbm 22260.0RBtu/lbm 10238.0RBtu/lbm 09774.0

RBtu/lbm 2418.0

6767,

4545,

7

6

5

4

3

21

==−==−=

=−==

⋅=⋅=⋅=⋅=⋅=

⋅==

H

L

L

qhhqhhq

xyx

sssss

ss


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie

Application of this equation for each process of the cycle gives the exergy destructions per unit mass flowing through the compressor:

[ ] Btu/lbm 3.18=−−=

−−=

=⎟⎠⎞

⎜⎝⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⎟⎠⎞

⎜⎝⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−×+×=

−+=

=⎟⎠⎞

⎜⎝⎛ +−=⎟⎟

⎠

⎞⎜⎜⎝

⎛+−=

)11286.0)(3698.0()22260.0)(6302.0(2418.0)R 555(

)(

Btu/lbm 54.0R 440

Btu/lbm 161.5011286.022948.0)R 555)(3698.0(

Btu/lbm 44.0R 495

Btu/lbm 881.5810238.022260.0)R 555)(6302.0(

Btu/lbm 73.4RBtu/lbm )09774.011286.03698.010238.06302.0)(R 555(

)(

Btu/lbm 2.67R 555

Btu/lbm 621.822418.009774.0)R 555(

6510mixing destroyed,

67,67067 destroyed,

45,45045 destroyed,

3640346 destroyed,

23023 destroyed,

ysxssTX

Tq

ssyTx

Tq

ssxTx

sysxsTxTq

ssTx

L

L

L

L

H

H

&


012 destroyed, =X&

The greatest exergy destruction occurs during the mixing process. Note that heat is absorbed in evaporator 2 from a reservoir at -20°F (440 R), in evaporator 1 from a reservoir at 35°F (495 R), and rejected to a reservoir at 95°F (555 R), which is also taken as the dead state temperature.

QH

QL

-29.5°F 1

2

3

6

160 psia

s

T

·

Win ·

·

45

7

30°F


11-75

11-110 A two-stage compression refrigeration system with a separation unit is considered. The rate of cooling and the power requirement are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. Analysis From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg 51.264 kPa 400

KkJ/kg 95813.0kJ/kg 91.230

vapor sat.

C32


kJ/kg 94.63 liquid sat.C9.8



kJ/kg 49.281 kPa 1400

KkJ/kg 92691.0kJ/kg 55.255

vapor sat.

C9.8

878

C8.9 @sat 8

C32 @ 7

C32 @ 77

56

C9.8 @ 55

34

kPa 0014 @ 33

212

2

C9.8 @ 1

C9.8 @ 11

=⎭⎬⎫

===

⋅====

⎭⎬⎫°−=

=≅

==⎭⎬⎫°=

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°=

°

°−

°−

°

°

°

hssPP

sshhT

hh

hhT

hh

hhP

hss

P

sshhT

g

g

f

f

g

g


kg/s 280.194.6351.26422.12755.255kg/s) 2()()(

58

4126412586 =

−−

=−−

=⎯→⎯−=−hhhh

mmhhmhhm &&&&

The rate of cooling produced by this system is then

kJ/s 213.7=−=−= kJ/kg )94.6391kg/s)(230. 280.1()( 676 hhmQL &&


kW 94.89=−+−=

−+−=kJ/kg )55.25549kg/s)(281. 2(kJ/kg )91.23051kg/s)(264. 280.1(

)()( 122786in hhmhhmW &&&

3

45

6

Compressor

Expansion valve

1

2

Evaporator 7

Separator

Condenser

Expansion valve

Compressor

8

QL

-32°C 7

8 3

6

s

T

·

2 1.4 MPa

4 1 5

8.9°C


11-76

11-111 A two-stage vapor-compression refrigeration system with refrigerant-134a as the working fluid is considered. The process with the greatest exergy destruction is to be determined.


Analysis From Prob. 11-109 and the refrigerant tables (Tables A-11, A-12, and A-13),

K 29827325K 25527318kJ/kg 27.154kJ/kg 97.166

kg/s 280.1kg/s 2

KkJ/kg 95813.0KkJ/kg 2658.0KkJ/kg 24761.0

KkJ/kg 4720.0KkJ/kg 45315.0

KkJ/kg 92691.0

0

32

67

lower

upper

87

6

5

4

3

21

=+===+−=

=−==−=

==

⋅==⋅=⋅=

⋅=⋅=

⋅==

TTT

hhqhhq

mm

ssssss

ss

H

L

H

L

&

&


⎟⎟⎠

⎞⎜⎜⎝

⎛+−−==

sink

out

source

in0gen0dest T

qT

qssTsTx ie


[ ][ ] kW 11.0)4720.092691.0)(kg/s 2()95813.024761.0)(kg/s 280.1()K 298(

)()(

kW 32.14K 255kJ/kg 97.1662658.095813.0)K 298)(kg/s 280.1(

kW 94.6KkJ/kg )24761.02658.0)(K 298)(kg/s 280.1()(

kW 23.11KkJ/kg )45315.04720.0)(K 298)(kg/s 2()(

K 298kJ/kg 27.15492691.045315.0)K 298)(kg/s 2(

41upper85lower0separator destroyed,

670lower67 destroyed,

560lower56 destroyed,

340upper34 destroyed,

230upper23 destroyed,

=−+−=

−−−=

=⎟⎠

⎞⎜⎝

⎛ −−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

=⋅−=−=

=⋅−=−=

=⎟⎠

⎞⎜⎝

⎛ +−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−=

ssmssmTX

Tq

ssTmX

ssTmX

ssTmX

Tq

ssTmX

L

L

H

H

&&&

&&

&&

&&

&&

kW 26.18


0

0

78 destroyed,

12 destroyed,

=

=

X

X&

&

Note that heat is absorbed from a reservoir at 0°F (460 R) and rejected to the standard ambient air at 77°F (537 R), which is also taken as the dead state temperature. The greatest exergy destruction occurs during the condensation process.

QL

-32°C7

8 3

6

s

T

·

2 1.4 MPa

4 1 5

8.9°C


11-77

11-112 A regenerative gas refrigeration cycle with helium as the working fluid is considered. The temperature of the helium at the turbine inlet, the COP of the cycle, and the net power input required are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Helium is an ideal gas with constant specific heats at room temperature. 3 Kinetic and potential energy changes are negligible.

Properties The properties of helium are cp = 5.1926 kJ/kg·K and k = 1.667 (Table A-2).

Analysis (a) The temperature of the helium at the turbine inlet is determined from an energy balance on the regenerator,

( ) ( )6143

outin

(steady) 0systemoutin 0

hhmhhmhmhm

EE

EEE

iiee −=−⎯→⎯=

=

=Δ=−

∑∑ &&&&

&&

&&&

or,

Thus,

( ) ( )

( ) ( ) K 278C25C10C206134

61436143

=°=°−+°−−°=+−=

−=−⎯→⎯−=−

C5TTTT

TTTTTTcmTTcm pp &&

(b) From the isentropic relations,

( )( )( )

( )( ) C93.9K 1179

31 K278

C135.2 K24083 K263

66716670k1k

4

545

66716670k1k

1

212

°−==⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

°===⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

.PPTT

.PPTT

././

././

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

( ) ( )

( ) ( )( )

( )[ ] ( )[ ] 1.49=°−−−°−−

°−−°−=

−−−−

=

−−−−

=−

==

C93.95C10135.2C93.9C25

COP

5412

56

5412

56


TTTTTT

hhhhhh

wwq

wq LL

(c) The net power input is determined from

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

( )( ) ( )[ ] ( )[ ]( )kW 108.2=

−−−−−°⋅=

−−−=

−−−=−=

93.9510135.2CkJ/kg 5.1926kg/s 0.455412

5412outturb,incomp,innet,

TTTTcm

hhhhmWWW

p&

&&&&

s

T

1

3

-10°C

-25°C 56

4

2QH ·

QRefrig ·

20°C

Qregen ·


11-78

11-113 An absorption refrigeration system operating at specified conditions is considered. The minimum rate of heat supply required is to be determined.


259.1263298

263K358K298

11COP0

0maxR, =⎟

⎠⎞

⎜⎝⎛

−⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

L

L

s TTT

TT

Thus,

kW53.9 1.259

kW 12COP maxR,

mingen, === LQQ

&&

11-114 EES Problem 11-113 is reconsidered. The effect of the source temperature on the minimum rate of heat supply is to be investigated.


"Input Data:" T_L = -10 [C] T_0 = 25 [C] T_s = 85 [C] Q_dot_L = 8 [kW] "The maximum COP that this refrigeration system can have is:" COP_R_max = (1-(T_0+273)/(T_s+273))*((T_L+273)/(T_0 - T_L)) "The minimum rate of heat supply is:" Q_dot_gen_min = Q_dot_L/COP_R_max

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

50 90 130 170 210 2502

4

6

8

10

12

14

Ts [C]

Qge

n;m

in [

kW]


11-79

11-115 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


11-80

11-116 A regenerative gas refrigeration cycle using air as the working fluid is considered. The effectiveness of the regenerator, the rate of heat removal from the refrigerated space, the COP of the cycle, and the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3 Air is an ideal gas with variable specific heats.

Analysis (a) For this problem, we use the properties of air from EES:

kJ/kg 50.433kPa 500

kJ/kg.K 6110.5C0

kPa 100kJ/kg 40.273C0

212

2

11

1

11

=⎭⎬⎫

==

=⎭⎬⎫

°==

=⎯→⎯°=

shss

P

sTP

hT

kJ/kg 63.308C35

kJ/kg 52.47340.273

40.27350.43380.0

33

2

2

12

12

=⎯→⎯°=

=−

−=

−−

=

hT

hh

hhhh s

Cη

For the turbine inlet and exit we have

kJ/kg 45.193C80 55 =⎯→⎯°−= hT

sT hh

hhhT

54

54

44 ?

−−

=

=⎯→⎯=

η

=⎭⎬⎫

==

=⎭⎬⎫

==

=⎭⎬⎫

°==

shss

P

sTP

sTP

545

5

44

4

11

1

kPa 500

?kPa 500

kJ/kg.K 6110.5C0

kPa 100

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

T4 = 281.8 K, h4 = 282.08 kJ/kg

An energy balance on the regenerator gives

kJ/kg 85.24608.28263.30840.2734316 =+−=+−= hhhh

The effectiveness of the regenerator is determined from

0.430=−−

=−−

=85.24663.30808.28263.308

63

43regen hh

hhε

s

T

1

3

0°C

5s6

4

2s QH·

QRefrig ·

35°C

Qregen

2

5-80°C

.

3

45

6

QH

Compressor

12

QL

Heat Exch.

Heat Exch.

Regenerator

Turbine

.


11-81

(b) The refrigeration load is

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

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

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

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

0.479=−

=−

==45.3505.80

36.21COPoutT,inC,innet, WW

QW

Q LL&&

&

&

&

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

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

3

2

1

===

hhh

kJ/kg 2704.5C35kPa 500

33

3 =⎭⎬⎫

°==

sTP

kJ/kg.K 52.194kPa 100

434

1 =⎭⎬⎫

==

shss

P

kJ/kg 64.21152.19463.308

63.30885.0 4

4

43

43 =⎯→⎯−

−=⎯→⎯

−−

= hh

hhhh

sTη

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

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

0.599===25.4170.24COP

innet,WQL&

&

s

T

1

2QH

35°C 0°C

3

4s

·

QRefrig ·

2

14


11-82

11-117 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it.

Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air.

Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes

COP COPair-cond heat pump= − = − =1 2 2 1 12. .

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

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

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


11-83

11-118 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined.



kPa 200kJ/kg 46.244

vapor sat.

C1.10


kJ/kg 32.79 kPa 800

C203.113.31

3.11


C10.1 @sat 6

C1.10 @ 66

45

C20 @ 4

4

kPa 800 @sat 4

kPa 800 @ 33

====

⎭⎬⎫°−=

=≅

=≅⎪⎭

⎪⎬

⎫

=°=−=

−=

==⎭⎬⎫=

°−

°−

°

PPhhT

hh

hhP

TT

hhP

g

f

f

An energy balance on the heat exchanger gives

kJ/kg 61.26046.24432.7947.95)()( 64314361 =+−=+−=⎯→⎯−=− hhhhhhmhhm &&

Then,

kJ/kg 17.292 kPa 800

KkJ/kg 9970.0 kJ/kg 61.260

kPa 200

212

2

11

1

=⎭⎬⎫

==

⋅=⎭⎬⎫

==

hss

P

shP


5.23=−−

=−−

==61.26017.292

32.7946.244COP12

56

inR hh

hhwqL

3

4

5

6

Compressor

Throttle valve

1

2

Heat exchanger

Condenser

Evaporator

QH

QL

-10.1°C 1

2

3

5

800 kPa

s

T

·

Win ·

·

6

4


11-84

11-119 An innovative vapor-compression refrigeration system with a heat exchanger is considered. The system’s COP is to be determined.



kPa 200kJ/kg 46.244

vapor sat.

C1.10


kJ/kg 43.65 kPa 800

C103.213.31

3.21


C10.1 @sat 6

C1.10 @ 66

45

C10 @ 4

4

kPa 800 @sat 4

kPa 800 @ 33

====

⎭⎬⎫°−=

=≅

=≅⎪⎭

⎪⎬

⎫

=°=−=

−=

==⎭⎬⎫=

°−

°−

°

PPhhT

hh

hhP

TT

hhP

g

f

f

An energy balance on the heat exchanger gives

kJ/kg 50.27446.24443.6547.95)()( 64314361 =+−=+−=⎯→⎯−=− hhhhhhmhhm &&

Then,

kJ/kg 28.308 kPa 800

KkJ/kg 0449.1 kJ/kg 50.274

kPa 200

212

2

11

1

=⎭⎬⎫

==

⋅=⎭⎬⎫

==

hss

P

shP


5.30=−−

=−−

==50.27428.30843.6546.244COP

12

56

inR hh

hhwqL

3

4

5

6

Compressor

Throttle valve

1

2

Heat exchanger

Condenser

Evaporator

QH

QL

200 kPa 1

2

3

5

800 kPa

s

T

·

Win ·

·

6

4


11-85

11-120 An ideal gas refrigeration cycle with with three stages of compression with intercooling using air as the working fluid is considered. The COP of this system is to be determined.




K 5.72555

1K) 288(

K 1.456K)(5) 288(

K 7.400K)(5) 253(

4.1/4.0/)1(

7

878

4.1/4.0/)1(

3

4364

4.1/4.0/)1(

1

212

=⎟⎠⎞

⎜⎝⎛

××=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

==⎟⎟⎠

⎞⎜⎜⎝

⎛==

==⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−

−

kk

kk

kk

PP

TT

PP

TTT

PP

TT


0.673=−−−+−

−=

−−−+−−

=

−−−+−+−−

=

−==

)5.72288()2881.456(2)2537.400(5.72253

)()(2)(

)()()()(

COP

873412

81

87563412

81


TTTTTTTT

hhhhhhhhhh

wwq

wq LL

15°C

s

T

1

2

3

4

5

6

-20°C

7

8


11-86

11-121 An ideal vapor-compression refrigeration cycle with refrigerant-22 as the working fluid is considered. The evaporator is located inside the air handler of building. The hardware and the T-s diagram for this heat pump application are to be sketched. The COP of the unit and the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler are to be determined.


Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant-22 data from the problem statement,

)throttling( kJ/kg 101

kJ/kg 101 liquid sat.kPa 1728

kJ/kg 7.283 kPa 1728

KkJ/kg 9344.0kJ/kg 1.248

vapor sat.

C5

34

kPa 1728 @ 33

212

2

C5 @ 1

C5 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫°−=

°−

°−

hh

hhP

hss

P

sshhT

f

g

g

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

5.13=−−

=−−

==1.2487.283

1017.283COP12

32

inHP hh

hhwqH

(c) An energy balance on the condenser gives

TcTcmhhmQ pa

apaRH Δ=Δ=−=

vV&

&&& )( 32

Rearranging, we obtain the ratio of volume flow rate of air entering the air handler to mass flow rate of R-22 through the air handler

R22/s) kgair/min)/( (m 462 3=

=

⋅

−=

Δ−

=

R22/s) air/s)/(kg (m 699.7

)K 20)(K)kJ/kg 005.1)(/kgm 847.0/1(kJ/kg )1017.283(

)/1(3

332

Tchh

m pR

a

v

V&

&

Note that the specific volume of air is obtained from ideal gas equation taking the pressure of air to be 101 kPa and using the room temperature of air (25°C = 298 K) to be 0.847 m3/kg.

QH

QL

-5°C 1

2 3

4

45°C

s

T

·

Win ·

·

4s1

2 3

4

-5°C

Condenser

Evaporator

Compressor Expansion

valve

sat. vap.QL

Win

Air

45°C


11-87

11-122 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. Cooling water flows through the water jacket surrounding the condenser. To produce ice, potable water is supplied to the chiller section of the refrigeration cycle. The hardware and the T-s diagram for this refrigerant-ice making system are to be sketched. The mass flow rates of the refrigerant and the potable water are to be determined.


Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant-134a data from the problem statement,



kJ/kg 07.284 kPa 1200

KkJ/kg 94456.0kJ/kg 16.239

vapor sat.

kPa 140

34

kPa 1200 @ 33

212

2

kPa 140 @ 1

kPa 140 @ 11

=≅

==⎭⎬⎫=

=⎭⎬⎫

==

⋅====

⎭⎬⎫=

hh

hhP

hss

P

sshhT

f

g

g

(b) An energy balance on the condenser gives

TcmhhmQ pwRH Δ=−= &&& )( 32

Solving for the mass flow rate of the refrigerant

kg/s 50.3=−

⋅=

−

Δ=

kJ/kg)77.11707.284()K 10)(K)kJ/kg 18.4)(kg/s 200(

32 hhTcm

m pwR

&&

(c) An energy balance on the evaporator gives

ifwRL hmhhmQ &&& =−= )( 41

Solving for the mass flow rate of the potable water

kg/s 18.3=−

=−

=kJ/kg 333

kJ/kg)77.11716.239)(kg/s 3.50()( 41

if

Rw h

hhmm

&&

QH

QL

140 kPa 1

2 3

4

1.2 MPa

s

T

·

Win ·

·

4s1

2 3

4

140 kPa

Condenser

Evaporator

Compressor Expansion

valve

sat. vap.

Win

Potable water

1.2 MPa

Water 200 kg/s


11-88

11-123 A vortex tube receives compressed air at 500 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law.

Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist.

Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cpT.

Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.

(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system & &E Ein out= for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed as

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

+=

+=+=

&&&

&&&

Canceling cp and solving for T3 gives

K 307.3=

×−=

−=

75.027825.0300

75.025.0 21

3TT

T

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

(c) The entropy balance for this steady flow system & & &S S Sin out gen− + = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−+−=−+−=

+−+=−+=

−=

1

3

1

3

1

2

1

2

1312

133122

1323322113322

inoutgen

lnln75.0lnln25.0

)(75.0)(25.0)()(

)(

PP

RTT

cPP

RTT

c

ssssssmssm

smmsmsmsmsmsm

SSS

pp

&&

&&&&&&&

&&&

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

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

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

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

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

=

⎟⎠⎞

⎜⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ −=S&

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

1

3 2

Compressed air

Cold air Warm air


11-89

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

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

)()( c1c1cooling

=

−=−= TTcmhhmQ pcc &&&

kW 1.1571kPa 100kPa 500

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

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

−=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−

− kk

PP

kRTm

Wη

&&

Then the COP of the vortex refrigerator becomes

0350.kW 1.157kW 5.53COP

in comp,

cooling ===W

Q&

&

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

COP 278 K KCarnot =

−=

−=

T

T TL

H L ( )300 27812.6

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


11-90

11-124 A vortex tube receives compressed air at 600 kPa and 300 K, and supplies 25 percent of it as cold air and the rest as hot air. The COP of the vortex tube is to be compared to that of a reversed Brayton cycle for the same pressure ratio; the exit temperature of the hot fluid stream and the COP are to be determined; and it is to be shown if this process violates the second law.

Assumptions 1 The vortex tube is adiabatic. 2 Air is an ideal gas with constant specific heats at room temperature. 3 Steady operating conditions exist.

Properties The gas constant of air is 0.287 kJ/kg.K (Table A-1). The specific heat of air at room temperature is cp = 1.005 kJ/kg.K (Table A-2). The enthalpy of air at absolute temperature T can be expressed in terms of specific heats as h = cp T.

Analysis (a) The COP of the vortex tube is much lower than the COP of a reversed Brayton cycle of the same pressure ratio since the vortex tube involves vortices, which are highly irreversible. Owing to this irreversibility, the minimum temperature that can be obtained by the vortex tube is not as low as the one that can be obtained by the revered Brayton cycle.

(b) We take the vortex tube as the system. This is a steady flow system with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow entropy balance equation for this system & &E Ein out= for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed as

321

332211

332211

75.025.01 TcTcTc

TcmTcmTcmhmhmhm

ppp

ppp

+=

+=+=

&&&

&&&

Canceling cp and solving for T3 gives

K 307.3=×−

=

−=

75.027825.0300

75.025.0 21

3TT

T

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

(c) The entropy balance for this steady flow system & & &S S Sin out gen− + = 0 can be expressed as with one inlet and two exits, and it involves no heat or work interactions. Then the steady-flow energy balance equation for this system for a unit mass flow rate at the inlet ( &m1 kg / s)= 1 can be expressed

⎟⎟⎠

⎞⎜⎜⎝

⎛−+⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

−+−=−+−=

+−+=−+=

−=

1

3

1

3

1

2

1

2

1312

133122

1323322113322

inoutgen

lnln75.0lnln25.0

)(75.0)(25.0)()(

)(

PP

RTT

cPP

RTT

c

ssssssmssm

smmsmsmsmsmsm

SSS

pp

&&

&&&&&&&

&&&

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

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

K300 K3307kJ/kg.K)ln 0051(750

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

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

.

....

...S

=

⎟⎠⎞

⎜⎝⎛ −+

⎟⎠⎞

⎜⎝⎛ −=&

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

1

3 2

Compressed air

Cold air Warm air


11-91

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

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

)()( c1c1cooling

=

−=−= TTcmhhmQ pcc &&&

kW 9.1791kPa 100kPa 600

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

1)1(

4.1/)14.1(

/)1(

0

1

comp

00incomp,

=⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛

−=

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛−

=

−

− kk

PP

kRTm

Wη

&&

Then the COP of the vortex refrigerator becomes

0310.kW 9.179kW 5.53COP

in comp,

cooling ===W

Q&

&

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

12.6=−

=−

=K )278300(

K 278COPCarnotLH

L

TTT

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


11-92

11-125 EES The effect of the evaporator pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated.


"Input Data" P[1]=100 [kPa] P[2] = 1000 [kPa] Fluid$='R134a' Eta_c=0.7 "Compressor isentropic efficiency" "Compressor" h[1]=enthalpy(Fluid$,P=P[1],x=1) "properties for state 1" s[1]=entropy(Fluid$,P=P[1],x=1) T[1]=temperature(Fluid$,h=h[1],P=P[1]) h2s=enthalpy(Fluid$,P=P[2],s=s[1]) "Identifies state 2s as isentropic" h[1]+Wcs=h2s "energy balance on isentropic compressor" W_c=Wcs/Eta_c"definition of compressor isentropic efficiency" h[1]+W_c=h[2] "energy balance on real compressor-assumed adiabatic" s[2]=entropy(Fluid$,h=h[2],P=P[2]) "properties for state 2" T[2]=temperature(Fluid$,h=h[2],P=P[2]) "Condenser" P[3] = P[2] h[3]=enthalpy(Fluid$,P=P[3],x=0) "properties for state 3" s[3]=entropy(Fluid$,P=P[3],x=0) h[2]=Qout+h[3] "energy balance on condenser" "Throttle Valve" h[4]=h[3] "energy balance on throttle - isenthalpic" x[4]=quality(Fluid$,h=h[4],P=P[4]) "properties for state 4" s[4]=entropy(Fluid$,h=h[4],P=P[4]) T[4]=temperature(Fluid$,h=h[4],P=P[4]) "Evaporator" P[4]= P[1] Q_in + h[4]=h[1] "energy balance on evaporator" "Coefficient of Performance:" COP=Q_in/W_c "definition of COP"

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

100 150 200 250 300 350 400 450 5000

2

4

6

8

10

P[1] [kPa]

CO

P

η comp

1.01.00.70.7


11-93

11-126 EES The effect of the condenser pressure on the COP of an ideal vapor-compression refrigeration cycle with R-134a as the working fluid is to be investigated.


"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"

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

2.258 0.7 900 1.803 0.7 1150 1.492 0.7 1400

400 600 800 1000 1200 14000

1

2

3

4

5

6

7

8

P[2] [kPa]

CO

P

η comp

1.01.00.70.7


11-94

Fundamentals of Engineering (FE) Exam Problems 11-127 Consider a heat pump that operates on the reversed Carnot cycle with R-134a as the working fluid executed under the saturation dome between the pressure limits of 140 kPa and 800 kPa. R-134a changes from saturated vapor to saturated liquid during the heat rejection process. The net work input for this cycle is (a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg (d) 144 kJ/kg (e) 275 kJ/kg Answer (a) 28 kJ/kg Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=800 "kPa" P2=140 "kPa" h_fg=ENTHALPY(R134a,x=1,P=P1)-ENTHALPY(R134a,x=0,P=P1) TH=TEMPERATURE(R134a,x=0,P=P1)+273 TL=TEMPERATURE(R134a,x=0,P=P2)+273 q_H=h_fg COP=TH/(TH-TL) w_net=q_H/COP "Some Wrong Solutions with Common Mistakes:" W1_work = q_H/COP1; COP1=TL/(TH-TL) "Using COP of regrigerator" W2_work = q_H/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = h_fg3/COP; h_fg3= ENTHALPY(R134a,x=1,P=P2)-ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2" W4_work = q_H*TL/TH "Using the wrong relation"


11-95

11-128 A refrigerator removes heat from a refrigerated space at –5°C at a rate of 0.35 kJ/s and rejects it to an environment at 20°C. The minimum required power input is (a) 30 W (b) 33 W (c) 56 W (d) 124 W (e) 350 W Answer (b) 33 W Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TH=20+273 TL=-5+273 Q_L=0.35 "kJ/s" COP_max=TL/(TH-TL) w_min=Q_L/COP_max "Some Wrong Solutions with Common Mistakes:" W1_work = Q_L/COP1; COP1=TH/(TH-TL) "Using COP of heat pump" W2_work = Q_L/COP2; COP2=(TH-273)/(TH-TL) "Using C instead of K" W3_work = Q_L*TL/TH "Using the wrong relation" W4_work = Q_L "Taking the rate of refrigeration as power input"


11-96

11-129 A refrigerator operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 120 kPa and 800 kPa. If the rate of heat removal from the refrigerated space is 32 kJ/s, the mass flow rate of the refrigerant is (a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s (d) 0.28 kg/s (e) 0.81 kg/s Answer (c) 0.23 kg/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=800 "kPa" P3=P2 P4=P1 s2=s1 Q_refrig=32 "kJ/s" m=Q_refrig/(h1-h4) h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_mass = Q_refrig/(h2-h1) "Using wrong enthalpies, for W_in" W2_mass = Q_refrig/(h2-h3) "Using wrong enthalpies, for Q_H" W3_mass = Q_refrig/(h1-h44); h44=ENTHALPY(R134a,x=0,P=P4) "Using wrong enthalpy h4 (at P4)" W4_mass = Q_refrig/h_fg; h_fg=ENTHALPY(R134a,x=1,P=P2) - ENTHALPY(R134a,x=0,P=P2) "Using h_fg at P2"


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11-130 A heat pump operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. If the mass flow rate of the refrigerant is 0.193 kg/s, the rate of heat supply by the heat pump to the heated space is (a) 3.3 kW (b) 23 kW (c) 26 kW (d) 31 kW (e) 45 kW Answer (d) 31 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 m=0.193 "kg/s" Q_supply=m*(h2-h3) "kJ/s" h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 "Some Wrong Solutions with Common Mistakes:" W1_Qh = m*(h2-h1) "Using wrong enthalpies, for W_in" W2_Qh = m*(h1-h4) "Using wrong enthalpies, for Q_L" W3_Qh = m*(h22-h4); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)" W4_Qh = m*h_fg; h_fg=ENTHALPY(R134a,x=1,P=P1) - ENTHALPY(R134a,x=0,P=P1) "Using h_fg at P1"


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11-131 An ideal vapor compression refrigeration cycle with R-134a as the working fluid operates between the pressure limits of 120 kPa and 1000 kPa. The mass fraction of the refrigerant that is in the liquid phase at the inlet of the evaporator is (a) 0.65 (b) 0.60 (c) 0.40 (d) 0.55 (e) 0.35 Answer (b) 0.60 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=120 "kPa" P2=1000 "kPa" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 x4=QUALITY(R134a,h=h4,P=P4) liquid=1-x4 "Some Wrong Solutions with Common Mistakes:" W1_liquid = x4 "Taking quality as liquid content" W2_liquid = 0 "Assuming superheated vapor" W3_liquid = 1-x4s; x4s=QUALITY(R134a,s=s3,P=P4) "Assuming isentropic expansion" s3=ENTROPY(R134a,x=0,P=P3)


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11-132 Consider a heat pump that operates on the ideal vapor compression refrigeration cycle with R-134a as the working fluid between the pressure limits of 0.32 MPa and 1.2 MPa. The coefficient of performance of this heat pump is (a) 0.17 (b) 1.2 (c) 3.1 (d) 4.9 (e) 5.9 Answer (e) 5.9 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=320 "kPa" P2=1200 "kPa" P3=P2 P4=P1 s2=s1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,s=s2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_HP=qH/Win Win=h2-h1 qH=h2-h3 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h1-h4)/(h2-h1) "COP of refrigerator" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h22-h3)/(h22-h1); h22=ENTHALPY(R134a,x=1,P=P2) "Using wrong enthalpy h2 (hg at P2)"


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11-133 An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 kPa and 280 kPa. Air is cooled to 35°C before entering the turbine. The lowest temperature of this cycle is (a) –58°C (b) -26°C (c) 0°C (d) 11°C (e) 24°C Answer (a) –58°C Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.4 P1= 80 "kPa" P2=280 "kPa" T3=35+273 "K" "Mimimum temperature is the turbine exit temperature" T4=T3*(P1/P2)^((k-1)/k) - 273 "Some Wrong Solutions with Common Mistakes:" W1_Tmin = (T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K" W2_Tmin = T3*(P1/P2)^((k-1)) - 273 "Using wrong exponent" W3_Tmin = T3*(P1/P2)^k - 273 "Using wrong exponent"


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11-134 Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 100 kPa and –10°C and is compressed to 250 kPa. Helium is then cooled to 20°C before it enters the turbine. For a mass flow rate of 0.2 kg/s, the net power input required is (a) 9.3 kW (b) 27.6 kW (c) 48.8 kW (d) 93.5 kW (e) 119 kW Answer (b) 27.6 kW Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). k=1.667 Cp=5.1926 "kJ/kg.K" P1= 100 "kPa" T1=-10+273 "K" P2=250 "kPa" T3=20+273 "K" m=0.2 "kg/s" "Mimimum temperature is the turbine exit temperature" T2=T1*(P2/P1)^((k-1)/k) T4=T3*(P1/P2)^((k-1)/k) W_netin=m*Cp*((T2-T1)-(T3-T4)) "Some Wrong Solutions with Common Mistakes:" W1_Win = m*Cp*((T22-T1)-(T3-T44)); T22=T1*P2/P1; T44=T3*P1/P2 "Using wrong relations for temps" W2_Win = m*Cp*(T2-T1) "Ignoring turbine work" W3_Win=m*1.005*((T2B-T1)-(T3-T4B)); T2B=T1*(P2/P1)^((kB-1)/kB); T4B=T3*(P1/P2)^((kB-1)/kB); kB=1.4 "Using air properties" W4_Win=m*Cp*((T2A-(T1-273))-(T3-273-T4A)); T2A=(T1-273)*(P2/P1)^((k-1)/k); T4A=(T3-273)*(P1/P2)^((k-1)/k) "Using C instead of K"


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11-135 An absorption air-conditioning system is to remove heat from the conditioned space at 20°C at a rate of 150 kJ/s while operating in an environment at 35°C. Heat is to be supplied from a geothermal source at 140°C. The minimum rate of heat supply required is (a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s (d) 61 kJ/s (e) 150 kJ/s Answer (c) 30 kJ/s Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). TL=20+273 "K" Q_refrig=150 "kJ/s" To=35+273 "K" Ts=140+273 "K" COP_max=(1-To/Ts)*(TL/(To-TL)) Q_in=Q_refrig/COP_max "Some Wrong Solutions with Common Mistakes:" W1_Qin = Q_refrig "Taking COP = 1" W2_Qin = Q_refrig/COP2; COP2=TL/(Ts-TL) "Wrong COP expression" W3_Qin = Q_refrig/COP3; COP3=(1-To/Ts)*(Ts/(To-TL)) "Wrong COP expression, COP_HP" W4_Qin = Q_refrig*COP_max "Multiplying by COP instead of dividing"


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11-136 Consider a refrigerator that operates on the vapor compression refrigeration cycle with R-134a as the working fluid. The refrigerant enters the compressor as saturated vapor at 160 kPa, and exits at 800 kPa and 50°C, and leaves the condenser as saturated liquid at 800 kPa. The coefficient of performance of this refrigerator is (a) 2.6 (b) 1.0 (c) 4.2 (d) 3.2 (e) 4.4 Answer (d) 3.2 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). P1=160 "kPa" P2=800 "kPa" T2=50 "C" P3=P2 P4=P1 h1=ENTHALPY(R134a,x=1,P=P1) s1=ENTROPY(R134a,x=1,P=P1) h2=ENTHALPY(R134a,T=T2,P=P2) h3=ENTHALPY(R134a,x=0,P=P3) h4=h3 COP_R=qL/Win Win=h2-h1 qL=h1-h4 "Some Wrong Solutions with Common Mistakes:" W1_COP = (h2-h3)/(h2-h1) "COP of heat pump" W2_COP = (h1-h4)/(h2-h3) "Using wrong enthalpies, QL/QH" W3_COP = (h1-h4)/(h2s-h1); h2s=ENTHALPY(R134a,s=s1,P=P2) "Assuming isentropic compression" 11-137 ··· 11-145 Design and Essay Problems

Solution manual of Thermodynamics-Ch.11

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Transcript of Solution manual of Thermodynamics-Ch.11