15. REFRIGERATION

15.1 An ideal single-stage vapor compression refrigeration cycle uses R-22 as the working fluid. The condensing temperature is 110 F (43 C) and the evaporating temperature is 40 F (4.5 C). The system produces 10 tons (35.2 kW) of cooling effect. Determine the (a) coefficient of performance, (b) refrigerating efficiency, (c) hp/ton (kW input per kW of refrigeration), (d) mass flow rate of the displacement, (e) theoretical input to the compressor in hp (kW), and (f) theoretical piston displacement of the compressor in cfm (m3/s).

Solution:

English Units From Chart 4 and Table A-3a At 110 F, psiaPP 14.24141 == At 40 F, psiaPP 28.8332 ==

lbmBtuii 717.4221 == lbmBtui 191.1083 = , ( )FlbmBtuss == 2204.043 , lbmftv 33 6561.0=

lbmBtui 74.1194 =

(a) Coefficient of Performance

67.5191.10874.119717.42191.108

34

23=

=

=

iiiiCOP

(b) Refrigerating Efficiency

c

R COPCOP

=

138.740110

67.45940=

+=

=

ec

ec TT

TCOP

794.0138.767.5

==R

(c) hp/ton

15. REFRIGERATION

Equation 15-3

tonhpCOPton

hp 8325.067.572.472.4

===

(d) mass flow rate of refrigerant

( )23 iimqe = &&

( )( ) hrlbm

iiq

m e 1833717.42191.108

000,121023

=

=

=

&&

(e) theoretical input to the compressor in hp

( )( ) hptontonhphp 325.8108325.0 ==

(f) theoretical piston displacement

Equation 15-10

0.13 ==PD

vmv

&

( )( )cfmvmPD 044.20

606561.01833

3 === &

SI units From Chart 4 and Table A-3b At 43 C, MPaPP 65.141 == At 4.5 C, kPaPP 4.57532 ==

kgkJii 72.25321 == kgkJi 47.4063 = , ( )KkgkJss == 7436.143 , kgmv 33 04091.0= kgkJi 63.4324 =

(a) Coefficient of Performance

84.547.40663.43272.25347.406

34

23=

=

=

iiiiCOP

(b) Refrigerating Efficiency

c

R COPCOP

=

212.75.44315.2735.4

=

+=

=

ec

ec TT

TCOP

810.0212.784.5

==R

15. REFRIGERATION

(c) hp/ton

Equation 15-4

tonkWCOPton

kW 6027.084.552.352.3

===

(d) mass flow rate of refrigerant

( )23 iimqe = &&

( ) skgiiq

m e 2305.072.25347.406

2.3523

=

=

=

&&

(e) theoretical input to the compressor in hp

( )( ) kWtontonkWkW 027.6106027.0 ==

(f) theoretical piston displacement

Equation 15-10

0.13 ==PD

vmv

&

( )( ) smvmPD 33 0094.004091.02305.0 === &

15.2 A vapor compression refrigeration cycle uses R-22 and follows the theoretical single-stage cycle. The condensing temperature is 48 C, and the evaporating temperature is 18 C. The power input to the cycle is 2.5 kW, and the mass flow rate of refrigerant is 0.05 kg/s. Determine (a) the heat rejected from the condenser, (b) the coefficient of performance, (c) the enthalpy at the compressor exit, and (d) the refrigerating efficiency.

Solution:

From Chart 4 and Table A-3b

15. REFRIGERATION

At 48 C, MPaPP 8555.141 == At -18 C, kPaPP 77.26432 ==

kgkJii 51.26021 == kgkJi 81.3973 = , ( )KkgkJss == 7787.143 , kgmv 33 08615.0= kgkJi 81.4474 =

(a) heat rejected from the condenser

skgm 05.0=& ( ) ( ) kWiimqc 365.951.26081.44705.014 === &&

(b) coefficient of performance

746.281.39781.44751.26081.397

34

23=

=

=

iiiiCOP

(c) enthalpy at the compressor exit

kgkJi 81.4474 =

(d) refrigerating efficiency

c

R COPCOP

=

( ) 866.3184815.27318

=

+=

=

ec

ec TT

TCOP

710.0866.3746.2

==R

15.3 An R-134a system is arranged as shown in Fig. 15-34. Compression is isentropic. Assume frictionless flow. Find the system hp/ton.

Figure 15-34 Schematic for Problem 15-3.

15. REFRIGERATION

Solution:

From Chart 3 and Table A-2a At 100 F, psiaPP 83.13841 == At -10 F, psiaPP 626.1632 ==

lbmBtui 943.441 = lbmBtui 542.1013 =

At 100 F lbmBtuii 978.3721 ==

3311 = iiii 542.101978.37943.44 3 = i

lbmBtui 507.1083 = at 16.626 psia

( )FlbmBtuss == 2612.043 , lbmftv 33 5842.3= lbmBtui 22.1344 =

743.2507.10822.134978.37507.108

34

23=

=

=

iiiiCOP

Equation 15-3

tonhpCOPton

hp 721.1743.272.472.4

===

15.4 Consider a reciprocating compressor operating with R-134a. Refrigerant enters the cylinder at 20 psia (138 kPa) and 20 F (-7 C), but leaves the evaporator saturated at 0.5 F (-18 C). The vapor is discharged from the cylinder at 180 psia (1.24 MPa). Compute the volumetric efficiency for (a) a clearance factor of 0.03; (b) a clearance factor of 0.15; (c) compare the mass flow rates for parts (a) and (b); and (d) compare the power input to the compressor in parts (a) and (b).

15. REFRIGERATION

Solution:

English Units

At 3, 0.5 F saturated, Table A-2a, R134a lbmftv 33 1362.2=

At b, 20 psia, 20 F, Chart 3 lbmftvb 3413.2=

psiaPb 20= At c, 180 psia

psiaPc 180= Equation 15-8

b

n

b

cv

v

v

PPCC 3

1

1

+=

30.1=n for R-134a

(a) 03.0=C

7679.0413.2

1362.220

18003.003.0130.1

1

=

+=v

(b) 15.0=C

2983.0413.2

1362.220

18015.015.0130.1

1

=

+=v

(c) Equation 15-10

PDvm

v3&

=

Assume PD same as Example 15-4

15. REFRIGERATION

For Part (a)

( )( )( )( )( ) minlbmv

PDm v 59.3

17281362.21725107679.0

3

===

&

For Part (b)

( )( )( )( )( ) minlbmv

PDm v 394.1

17281362.21725102983.0

3

===

&

Mass flow rate of part (a) is larger than part (b)

(d) Equation 15-12 and Equation 15-13, 0.1=m

( )( )

=

11

1n

n

b

cbb P

PvP

n

nmW &&

For Part (a)

( ) ( ) ( )( )( )( )

=

120

180413.2144180130.1

30.159.330.1

130.1

W&

hpminlbfftW 47.19553,642 ==&

For Part (b)

( ) ( ) ( )( )( )( )

=

120

180413.2144180130.1

30.1394.130.1

130.1

W&

hpminlbfftW 56.7504,249 ==&

Power input flow rate of part (a) is larger than part (b)

SI Units

At 3, -18 C saturated, Table A-2b, R134a kgmv 33 13597.0=

At b, 138 kPaa, -7 C, Chart 3 kgmvb

31503.0= kPaaPb 138=

At c, 1.24 MPaa kPaaPc 1240=

Equation 15-8

b

n

b

cv

v

v

PPCC 3

1

1

+=

30.1=n for R-134a

15. REFRIGERATION

(a) 03.0=C

7849.01503.0

13597.0138

124003.003.0130.1

1

=

+=v

(b) 15.0=C

3057.01503.0

13597.0138124015.015.01

30.11

=

+=v

(c) Equation 15-10

PDvm

v3&

=

Assume PD same as Example 15-4 ( )( )

( ) ( )( ) smPD3

3 00471.0601728281.3172510

==

For Part (a)

( )( )( ) skgv

PDm v 0272.0

13597.000471.07849.0

3

===

&

For Part (b)

( )( )( ) skgv

PDm v 0106.0

13597.000471.03057.0

3

===

&

Mass flow rate of part (a) is larger than part (b)

(d) Equation 15-12 and Equation 15-13, 0.1=m

( )( )

=

11

1n

n

b

cbb P

PvP

n

nmW &&

For Part (a)

( ) ( ) ( )( )( )

=

1138

12401503.01240130.1

30.10272.030.1

130.1

W&

kWW 49.14=&

For Part (b)

( ) ( ) ( )( )( )

=

1138

12401503.01240130.1

30.10106.030.1

130.1

W&

kWW 65.5=&

15. REFRIGERATION

Power input flow rate of part (a) is larger than part (b)

15.5 Consider a four-cylinder, 3 in. bore by 4 in. stroke, 800 rpm, single-acting compressor for use with R-134a. The proposed operating condition for the compressor is 100 F condensing temperature and 40 F evaporating temperature. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid, that vapor will leave the evaporator at a temperature of 45 F, and that vapor will enter the compressor at a temperature of 55 F. Assume a compressor volumetric efficiency of 70 percent and frictionless flow. Calculate the refrigeration capacity in tons.

Solution: ( ) ( ) ( ) minftPD 3

2

36.528001728

434

4 =

=

pi

70.0=v

Entering the compressor at 3, 55 F psiaP 732.543 = at 45 F saturated, Table A-2a

Chart 3, lbmftv 33 894.0=

Equation 15-10, PD

vmv

3=

&

( )( ) ( ) hrlbmv

PDm v 246060

894.036.5270.0

3

===

&

Table A-2a At 1, 100 F, lbmBtuii 538.3921 == At 3, 45 F, lbmBtui 6.1083 = Refrigeration capacity in tons

( ) ( )( ) tonshrBtuiimqe 16.14893,169538.396.108246023 ==== &&

15. REFRIGERATION

15.6 Consider the compressor of Fig. 15-7. (a) Construct the pressure-enthapy diagram for a condensing temperature of 130 F (54 C) and an evaporating temperature of 45 F (7 C). (b) What is the heat transfer rate in the evaporator and the power input? (c) Suppose that the load on the evaporator decreases to 110,000 Btu/hr (32 kW), and find the evaporating temperature and power input. Assume that the condensing temperature remains constant.

Solution:

(a) Pressure-enthalpy diagram

(b) From Fig. 15-7

Condensing Temperature = 130 F Evaporator Temperature = 45 F Then, Capacity = 145,000 Btu/hr Power Input = 14.7 kW

(c) From Fig. 15-7

Condensing Temperature = 130 F Capacity = 110,000 Btu/hr Then, Evaporator Temperature = 32.5 F Power Input = 13.3 kW

15.7 Refer to Problem 15-6a, and sketch the capacity and power curves for 130 F (54 C) from Fig. 15-7. (a) Sketch the capacity curve for the evaporator, assuming its capacity is proportional to the evaporating temperature in the ratio 4000 Btu/(hr-F) (2.1 kW/C). (b) The evaporator load decreases to 130,000 Btu/hr (38 kW) while the condensing temperature decreases to 115 F (46 C). Sketch the new evaporator-capacity curve. (c) Suppose in part (a) that the evaporator operating conditions remain fixed but the condensing temperature increases to 145 F (63 C). What will the capacity and evaporator temperature be?

15. REFRIGERATION

Solution: Use English Units only

(a)

(b)

15. REFRIGERATION

(c)

From Fig. 15-7 Evaporator Temperature = 20 F Condenser Temperature = 145 F Then Capacity = 70,000 Btu/hr Evaporator Temperature = 20 F

15.8 A refrigeration system using the compressor described in Fig. 15-7 is designed to operate with a condensing temperature of 115 F with an evaporating temperature of 50 F when the outdoor ambient is 95 F. After the system is put into operation, the evaporator pressure is measured to be 69 psia,and the power to the compressor is 12 kW. Superheat and subcooling are assumed to be as given in Fig. 15-7, and the ambient is about 95 F. (a) Estimate the condensing and evaporating temperatures, (b) compare the actual and expected performance, and (c) suggest what might be done to obtain the design conditions.

Solution: Figure 15-7 Condensing Temperature = 115 F Evaporating Temperature = 50 F Then Capacity = 175,000 Btu/hr Power Input = 13.7 kW

At 69 psia, saturation temperature = 29.5 F, Table A-3a.

15. REFRIGERATION

(a) Evaporating Temperature = 29.5 F

Power input = 12 kW Then: Condensing Temperature = 115 F Evaporating Temperature = 29.5 F

(b) Actual capacity = 115,000 Btu/hr

Expected capacity 175,000 Btu/hr Actual capacity < expected capacity

(c) Increase the evaporator pressure to 98.8 psia for evaporator temperature of 50 F.

15.9 Saturated R-22 at 45 F (7 C) enters the compressor of a single-stage system. Discharge pressure is 275 psia (1.90 Mpa). Suction valve pressure drop is 2 psi (13.8 kPa). Discharge valve pressure drop is 4 psi (27.6 kPa). Assume the vapor is superheated 10 F (5.6 C) in the cylinder during the intake stroke. Piston clearance is 5 percent. Determine the (a) volumetric efficiency, (b) pumping capacity in lbm/min (kg/s) for 20 ft3/min (9.44 L/s) piston displacement, and (c) horsepower (kW) requirement if the mechanical efficiency is 80 percent.

Solution:

English Units psiaP 2754 =

psiaPPP dc 279427544 =+=+== At 45 F, Table A-3a,

psiaP 791.903 = psiaPPb 791.882791.9024 ===

Table A-3a, lbmftv 33 6029.0= At Point b, Ftb 551045 =+=

15. REFRIGERATION

Chart 4, psiaPb 791.88= , Ftb 55= Then, lbmftvb 36386.0=

16.1=n for R22 05.0=C

(a) Equation 15-8

b

n

b

cv

v

v

PPCC 3

1

1

+=

8646.06386.06029.0

791.8827905.005.01

16.11

=

+=v

(b) Equation 15-10

( )( )minlbm

v

PDm v 68.28

6029.0208646.0

3

===

&

(c) Equation 15-12 and 15-13

( )( )

=

11

1n

n

b

cbb

m PP

vPn

nmW&

&

( ) ( )( )( )( )

=

1791.88

2796386.0144791.88116.1

16.180.068.28 16.1

116.1

W&

hpminlbftW 0.11054,363 ==&

SI Units kPaMPaP 190090.14 ==

kPaPPP dc 6.19276.2719006.274 =+=+== At 7 F, Table A-3b,

kPaP 94.6213 = kPaPPb 14.60808.1394.62124 ===

Table A-3b, kgmv 33 03791.0= At Point b, Ctb 6.126.57 =+= Chart 4, kPaPb 14.608= , Ftb 55= Then, kgmvb

304013.0= 16.1=n for R22 05.0=C

(a) Equation 15-8

15. REFRIGERATION

b

n

b

cv

v

v

PPCC 3

1

1

+=

8642.004013.003791.0

14.6086.192705.005.01

16.11

=

+=v

(b) Equation 15-10

( )( )( )( ) skgv

PDm v 2152.0

100003791.044.98642.0

3

===

&

(c) Equation 15-12 and 15-13

( )( )

=

11

1n

n

b

cbb

m PP

vPn

nmW&

&

( ) ( )( )( )

=

114.6086.192704013.014.608

116.116.1

80.02152.0 16.1

116.1

W&

kWW 2.8=& 15.10 Consider the single-stage vapor-compression cycle shown in Fig. 15-35.

Design conditions using R-134a are: Lq& = 30,000 Btu/hr 3P = 200 psia 1P = 60 psia saturated 43 PP = 2 psi 2P = 55 psia C = 0.04 2T = 60 F m = 0.90 PD = 9.4 cfm

(a) Determine W& , Hq& , 12m& , and sketch the cycle on a P-i diagram. If the load Lq& decreases to 24,000 Btu/hr and the system comes to equilibrium with 2P =

50 psia and 2T = 50 F , (b) determine W& , Hq& , m& , and locate the cycle on a P-i diagram.

Figure 15-35 Schematic for Problem 15-10.

15. REFRIGERATION

Solution:

(a)

hrBtuqL 000,24=& 2P = 50 psia and 2T = 50 F

Equation 15-8

b

n

b

cv

v

v

PPCC 3

1

1

+=

+=

n

v PPCC

1

2

31

899.05020004.004.01

1.11

=

+=v

Equation 15-10

2v

PDm v

=&

at 50 psia, 50 F, R-134a lbmftv 32 975.0= ( )( )

minlbmm 6673.8975.0

4.9899.0==&

( )( )

=

11

1n

n

b

cbb

m PP

vPn

nmW&

&

( )( )

=

11

1

2

322

nn

m PP

vPn

nmW&

&

( ) ( )( )( )( )

=

150200975.014450

11.11.1

90.06673.8 1.1

11.1

W&

hpminlbftW 03.3882,99 ==&

15. REFRIGERATION

psiaPP 1802200234 ===

( )23 iimJWm

= &

&

At 50 psia, 50 F lbmBtui 02.1112 = ( )( )( )( ) lbmBtuiJm

Wi m 35.12402.1117786673.8882,9990.0

23 =+=+=&

&

hrBtuqL 000,24=& ( )12 iimqL = &&

( )( ) lbmBtumqii L 87.64

606673.8000,2402.11121 ===

&

&

lbmBtuii 87.6414 == ( ) ( )( )( )87.6435.124606673.843 == iimqH &&

hrBtuqH 932,30=&

(b)

hrBtuqL 000,30=& 2P = 55 psia and 2T = 60 F

Equation 15-8

b

n

b

cv

v

v

PPCC 3

1

1

+=

+=

n

v PPCC

1

2

31

9107.05520004.004.01

1.11

=

+=v

Equation 15-10

2v

PDm v

=&

15. REFRIGERATION

at 55 psia, 60 F, R-134a lbmftv 32 90165.0= ( )( )

minlbmm 4944.990165.0

4.99107.0==&

( )( )

=

11

1n

n

b

cbb

m PP

vPn

nmW&

&

( )( )

=

11

1

2

322

nn

m PP

vPn

nmW&

&

( ) ( )( )( )( )

=

15520090165.014455

11.11.1

90.04944.9 1.1

11.1

W&

hpminlbftW 13.3191,103 ==& psiaPP 1802200234 ===

( )23 iimJWm

= &

&

At 55 psia, 60 F lbmBtui 83.1122 = ( )( )( )( ) lbmBtuiJm

Wi m 40.12583.1127784944.9191,10390.0

23 =+=+=&

&

hrBtuqL 000,30=& ( )12 iimqL = &&

( )( ) lbmBtumqii L 17.60

604944.9000,3083.11221 ===

&

&

lbmBtuii 17.6014 == ( ) ( )( )( )17.6040.125604944.943 == iimqH &&

hrBtuqH 159,37=&

15.11 Consider an ordinary single-stage vapor-compression air-conditioning system. Because of clogged filters the air flow over the evaporator is gradually reduced to a very low level. Explain how the evaporator and compressor will be affected if the system continues to operate.

Answer:

Air flow over the evaporator when reduced to a very low level will indicate that the compressor is not loaded and also indicates that the evaporator is not transferring the expected quantity of heat to the refrigerant.

15.12 A vapor-compression cycle is subject to short periods of very light load; it is not practical to shut the system down. During these periods of light load, moisture condenses from the air flowing over the evaporator and freezes. Suggest a modification to the system to prevent this condition.

15. REFRIGERATION

Answer:

The system can be modified by reducing the suction pressure and evaporator temperature.

15.13 A vapor-compression cycle is subject to occasional overload that leads to the tripping of circuit breakers. Explain how the system can be modified to prevent compressor overload without shutting the system off.

Answer:

Add an evaporator pressure regulator to maintain a relatively constant minimum pressure in the evaporator. Because most of the evaporator surface is subjected to two-phase refrigerant a constant minimum temperature will also be maintained. Evaporator pressure is sensed internally and balanced by spring loaded diagphragm. When evaporator pressure falls below a set value, the valve will close, restricting the flow of refrigerant so that evaporator pressure will rise, therefore prevent freezing.

15.14 A saturated liquid aqua-ammonia solution at 220 F and 200 psia is throttled to a pressure of 10 psia. Find (a) the temperature after the throttling process, and (b) the relative portions of liquid and vapor in the mixture after throttling.

Solution:

(a) Temperature = Ft 4.622 =

15. REFRIGERATION

(b)

3.02 =fx , 976.02 =gx , 368.02 =x

mixturelbmvaporlbm

xx

xx

m

m

fg

fv 1006.03.0976.03.0368.0

22

22=

=

=

&

&

mixturelbmvaporlbm

m

m

m

mvf 8994.01006.011 ===&

&

&

&

15.15 A solution of ammonia and water at 180 F, 100 psia, and with a concentration of 0.25 lbm ammonia per lbm of solution is heated at constant pressure to a temperature of 280 F. The vapor is then separated from the liquid and cooled to a saturated liquid at 100 psia. What are the temperature and concentration of the saturated liquid?

Solution:

15. REFRIGERATION

Temperature = Ft 6.1345 = Concentration = 4x = 437.05 =x

15.16 It is proposed to use hot water at 180 F (82 C) from a solar collector system to operate a simple absorption cycle. Compute the maximum possible coefficient of performance, assuming an environment temperature of 100 F (38 C) and an air-cooled evaporator with the air temperature at 75 F (24 C).

Solution:

Equation 15-32

( ) ( )( )eogoge

max TTTTTT

COP

=

English Units: FTe 67.53467.45975 =+= FTo 67.55967.459100 =+= FTg 67.63967.459180 =+=

( ) ( )( ) 675.267.53467.55967.63967.55967.63967.534

=

=maxCOP

SI Units: KTe 15.29715.27324 =+= KTo 15.31115.27338 =+= KTg 15.35567.45982 =+=

15. REFRIGERATION

( ) ( )( ) 63.215.29715.31115.35515.31115.35515.297

=

=maxCOP

15.17 Saturated water vapor at 50 F is mixed in a steady flow process with a saturated lithium bromide=water solution having a concentration of 0.60 lbm LiBr per lbm of mixture. The mass of the liquid solution is five times the mass of the water vapor mixed. The mixing process occurs at constant pressure. Find (a) the concentration of the resulting mixture, and (b) the heat that must be removed, in Btu per lbm of the total mixture, if saturated liquid solution is produced.

Solution:

(a)

10106611 xmxmxm &&& +=

60.06 =x 00.010 =x

106 5mm && = 101061 6mmmm &&&& =+=

610110 56 xmxm && = ( ) 50.0

660.05

1 ==x

(b) 11661010 imimimqa &&&& +=

16101 6

561 iii

m

qa+=

&

&

at 50 F saturated water vapor, Table A-1a lbmBtui 9.108210 = , HgmmpsiaP 23.9178.0 ==

Chart 5 At 60.06 =x , lbmBtui 626 = At 50.01 =x , lbmBtui 706 =

( ) ( ) ( ) lbmBtum

qa 82.1987062659.1082

61

1

=+=&

&

- end -