Chapter 02 PDF

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Summary

An ammonia vapor refrigeration cycle operates at an evaporator temperature of –16 C and a condensing temperature of 32 C. Determine the coefficient of performance (a) for an ideal saturation cycle, (b) for wet compression with saturated vapor leaving the compressor, and (c) if thevapor at suction to...

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CHAPTER 2 1. An ammonia vapor refrigeration cycle operates at an evaporator temperature of –16 C and a condensing temperature of 32 C. Determine the coefficient of performance (a) for an ideal saturation cycle, (b) for wet compression with saturated vapor leaving the compressor, and (c) if the vapor at suction to the compressor is superheated 6 degrees. Solution: (a) Ideal saturation cycle

p , kPa 227.04 1239.60 1239.60 227.04

State points t ,C 1 -16 2 105 3 32 4 -16 h1 − h 4 1442.6 − 351.47 = = 4 .41 COP = h2 − h1 1690 −1442 .6 (b) Wet compression cycle

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h , kJ/kg 1442.60 1690.00 351.47 351.47

CHAPTER 2

State points 1 2 3 4 COP =

p , kPa

t ,C -16 32 32 -16

227.04 1239.60 1239.60 227.04

h , kJ/kg 1288.0 1487.18 351.47 351.47

h1 − h 4 1288 − 351 .47 = = 4.70 h2 − h1 1487 .18 −1442 .6

(c) With superheat

State points 1 2 3 4 COP =

t, C

p , kPa

h , kJ/kg

- 10 113 32 - 16

227.04 1239.60 1239.60 227.04

1456 1713 351.47 351.47

h1 − h 4 1456 − 351.47 = = 4 .29 h 2 − h1 1713 −1442 .6

2. A standard vapor compression system produces 20 tons of refrigeration using R-12 as a refrigerant while operating between a condenser temperature of 42 C and an evaporator temperature of –25 C. Determine (a) the refrigerating effect in kJ/kg, (b) the circulating rate in kg/s, (c) the power supplied, (d) the COP, (e) the heat rejected in kW, and (f) the volume flow rate in L/s.

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CHAPTER 2 Solution:

Table of important properties: State points 1 2 3 4

t ,C

h , kJ/kg

v , m3 /kg

-25 42 42 -25

340.4 377 240.6 240.6

131.3

(a) Refrigerating Effect:

QA = h1 − h4 = 340.4 − 240.6 = 99.8 kJ kg = (b) m

20 tons (20 )(3 .516 ) = = 0. 7046 kg s QA 99 .8

(c) W = m (h 2 − h1 ) = (0 .7046 )(377 − 340.4 ) = 25 .79 kW (d) COP =

Q A ( 20)( 3. 516) = = 2 .73  W 25 .79

 ( h2 − h 4 ) = (0.7046 )(377 − 240 .6 ) = 96.10 kW (e) Q R = m

 v1 = (0.7046 )(131 .3) = 92.5 L s (f) V = m

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CHAPTER 2 3. An ammonia simple saturation cycle operates between evaporator and condenser temperatures of –20 C and 35 C, respectively. The system is to be used in producing 5000 kg of ice at –12 C from water at 29 C in 20 hrs. Assuming losses to be 20 percent of the heat to be absorbed from the water, determine (a) the mass flow rate, (b) the heat rejected in the condenser, and (c) the power required by the compression. Solution:

Important Properties State points 1 2 3 4

t, C

h , kJ/kg

- 20

1437.2 1735 366.1 366.1

35 - 20

4

CHAPTER 2

With losses  w cw (t1 − 0 )+ m  wL+ m  w (0 − t 2 ) ](1.2 ) QA = [m where L = 335 kJ kg , t1 = 29 C , t 2 = − 12 C 5000  Q A =   [(4 .187 )(29 ) + 335 + (2 .093 )(12 )](1 .2 )  20  QA = (120,385 kJ hr )(1 hr 3600 s )(1.2) = 40.13 kW Q A 40 .13 = = 0. 03747 kg s = (a) m h 1 − h 4 1437. 2 − 366. 1 (b) Q = m (h − h ) = (0 .03747)(1735 − 366 .1) = 51.29 kW R

2

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(c) W = m (h 2 − h 1 ) = (0 .03747 )(1735 −1437 .2 ) =11 .16 kW 4. A 15-ton refrigeration system is used to make ice. The water is available at 20 C. Refrigerant 12 is used with saturated temperature limits of –25 C and 54 C. Determine (a) the COP, (b) the refrigerant flow rate, (c) the temperature at discharge of the compressor, (d) the volume flow rate, and (e) the maximum kg of ice manufactured per day. Solution:

Important Properties State points 1 2 3 4

t ,C

h , kJ/kg

v , m3 /kg

-25 65 54 -25

340.4 382.0 253.1 253.1

131.3

5

CHAPTER 2 h1 − h 4 340 − 253 .1 = = 2.09 h2 − h1 382 − 340 .4  .  = Q A = (15 )(3 516 ) = 0. 6069 kg s (b) m h1 − h4 340 − 253.1 (c) t 2 = 65 C = discharge temperature  v1 = ( 0.6069 )(131.3) = 79.7 L s (d) V = m  [(4.187)( 20 − 0) + 335 ] = (15)(3.516 ) (e) Q = m (c ∆t + L ) = m

(a) COP=

A

i

w

i

m i = 0.1259 kg s kg of ice per day = ( 0.1259)( 3600)( 24) = 10,878 kg

5. An R-12 standard refrigeration cycle operates at an evaporating pressure of 386 kPa and a condensing pressure of 1009 kPa. Show the effects of decreasing the vaporizing pressure to 270 kPa on the following: For a unit mass, (a) refrigerating effect, (b) COP, and (c) work. For a refrigerating capacity of 1 kW, (d) power, (e) mass flow rate, (f) heat rejected, and (g) volume flow rate.

Solution:

Important Properties State points 1 2 3 4

p , kPa

h , kJ/kg

v , m3 /kg

386 1009 1009 386

354.4 370.0 240.6 240.6

44.713

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CHAPTER 2 (a) Refrigerating Effect QA = h1 − h4 = 354 .4 − 240 .6 = 113 .8 kJ kg h − h 4 354 .4 − 240 .6 = 7. 295 (b) COP = 1 = 370 − 354 .4 h2 − h1 (c) Work = W = h2 − h1 = 370 − 354 .4 = 15.6 kJ kg   = Q A = 1 kW = 0. 14 kW (d) W COP 7 .295 QA 1 kW = (e) m = = 8.8 × 10 − 3 kg s = 0 .0088 kg s QA 113 .8 kJ kg  (h2 − h3 ) = (0. 0088)(370 − 240.6) = 1.14 kW (f) Q = m R

 v1 = ( 0. 0088 )( 44. 713) = 0. 39 L s (g) V = m

Decreasing vaporizing pressure to 270 kPa.

Important Properties State points 1 2 3 4

p , kPa

h , kJ/kg

v , m 3/kg

270 1009 1009 270

349.8 373 240.6 240.6

62.89

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CHAPTER 2 (a) Refrigerating Effect QA = h1 − h4 = 349.8 − 240 .6 =109 .2 kJ kg (decreased) h1 − h 4 349 .8 − 240 .6 = = 4. 707 (decreased) h2 − h1 373 − 349.8 (c) Work = W = h2 − h1 = 373 − 349.8 = 23.2 kJ kg (increased)   = Q A = 1 kW = 0. 22kW (increased) (d) W COP 4 .707 Q 1 kW = A = = 9 .2 × 10 − 3 kg s = 0 .0092 kg s (increased) (e) m QA 10 9.8 kJ kg  (h − h ) = (0. 0092 )(373 − 240.6) = 1.22 kW (increased) (f) Q = m

(b) COP =

R

2

3

 v1 = (0.0092 )(62.89 ) = 0.58 L s (increased) (g) V = m

6. An industrial plant has available a 4-cylinder, 76-mm bore by 102-mm stroke, 800 rpm, single-acting compressor for use with refrigerant 12. Proposed operating conditions for the compressor are 38 C condensing temperature and 5 C evaporating temperature. It is estimated that the refrigerant will enter the expansion valve as a saturated liquid that the vapor will leave the evaporator at a temperature of 7 C, and will enter the compressor at a temperature of 13 C. Assume a compressor volumetric efficiency of 70%. Assume frictionless flow. Calculate the refrigerating capacity in kW for a system equipped with this compressor. Solution:

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CHAPTER 2 Important Properties State points 1 1’ 2 3 4

t ,C 7 13

h , kJ/kg 355

38 5

236.5 236.5

v , m3 /kg 0.050

π π VD = D 2 LN × (no. of cyl.) = (0 .076 )2 (0 .102 )(800 )(4 )= 1 .481 m 3 min 4 4 V1 ′ = ηvVD = ( 0 .70) (1 .481) = 1 .037 m3 min = 1037 L min V 1 .037  = 1′ = m = 20 .740 kg min v 1′ 0.050 20. 740 = m = 0 .3457 kg s 60  (h1 − h 4 ) = (0 .3457)(355 − 236 .5) = 40.96 kW Refrigerating capacity = Q A = m

7. A Refrigerant 12 refrigeration system requires a load of 54 kW at an evaporator pressure of 270 kPa and a condenser pressure of 1009 kPa. The refrigerant is subcooled 10 degrees before entering the expansion valve and vapor is superheated 14 degrees before entering the compressor. A twincylinder compressor with stroke equal to 1.25 times the bore is to be used at a speed of 27 r/s. The volumetric efficiency is 84 percent. Determine (a) the quantity of cooling water in the condenser for an 11-degree increase in temperature, (b) the bore and stroke, and (c) the compressor power. Solution:

Important Properties State points 1 2 3 4

p , kPa

t, C

h , kJ/kg

v , m3 /kg

270 1009 1009 270

10

358.5 383.5 230.5 230.5

0.070

32 -4

9

CHAPTER 2

Q A = 54 kW Q A 54 = = 0 .422 kg s = m h1 − h4 358. 5 − 230. 5  (h2 − h3 ) = m  w cw (∆t ) (a) Q R = m  w = quantity of cooling water m

cw = 4.187 kJ kg ⋅K ∆t = 11 F

(0.422)(383. 5 −230. 5) = mw ( 4. 187)(11)  w = 1.40 kg s m

(b) For D× L

π VD = D 2 LN × (no. of cyl. ) 4  = V1 ′ m v1 = (0.422 )(0.070 ) = 0. 0295 m 3 s V 0.0295 3 VD = 1′ = = 0 .0351 m s ηv 0.84 L = 1.25D π VD = D 2 ( 1.25D)( 27)( 2) = 0. 0351 4 D =0 .087 m = 8 .7 cm L =1.25 (8.7) = 10. 9 cm  (h2 − h1 ) = (0.422 )( 383.5 − 358.5 ) = 10.6 kW (c) W = m

8. A refrigerant 22 refrigeration system carries a load of 82 kW at an evaporator pressure of 354 kPa and a condenser pressure of 1460 kPa. The liquid refrigerant is subcooled by 4 degrees before entering the expansion valve and the vapor is superheated by 5 degrees before entering the compressor. The compressor operates at 28 r/s. The stroke-to-bore ratio of the twin-cylinder compressor is 82%. Determine (a) mass flow rate of refrigerant, (b) mass flow rate of cooling water in the condenser for a 7-degree change in temperature, and (c) the bore and stroke.

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CHAPTER 2 Solution:

Important Properties State points 1 2 3 4

p , kPa

t, C

h , kJ/kg

v , m3 /kg

354 1460 1460 354

-5

405 438 241.8 241.8

0.0667

34 - 10

QA = 82 kW Q A 82 = = 0. 5025kg s = (a) m − h 1 h 4 405 − 241.8  (h2 − h3 ) = m  w c w ( ∆t ) (b) Q R = m  w = quantity of cooling water m

cw = 4.187 kJ kg ⋅ K ∆t = 7 F

(0.5025)( 438 − 241. 8) = m w ( 4.187)(7)  w = 3. 36 kg s m

(c) For D ×L V VD = 1′ , ηv = 0.82

ηv

V1 ′ = m v1 = (0.5025 )(0.0667 )= 0.0335 m 3 s

0.0335 V = 0.0408 m 3 s V D = 1 ′ = 0.82 ηv L = 1. 20D

π VD = D 2 LN × (no. of cyl. ) 4

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CHAPTER 2 π VD = D 2 (1 .20D)( 28)( 2) = 0 .0408 4 D =0 .092 m = 9 .2 cm L =1. 20 (9.2) = 11. 0 cm 9. A refrigerant 22 refrigerating system is operating with a condenser temperature of 42 C and an evaporating temperature of 0 C. (a) If the liquid line from the condenser is soldered to the suction line from the evaporator to form a simple heat exchanger and if as a result of this saturated vapor leaving the evaporator is superheated 10 degrees, how many degrees will the saturated liquid leaving the condenser be subcooled? Determine (b) the volume flow rate, and (c) the compressor work. Solution:

12

CHAPTER 2

Important Properties

State points 1 2 3 4 5 6

t ,C

h , kJ/kg

v , m3 /kg

10

0.050

42

412.5 443 252.4

0 0

405.4

(a) For heat exchanger h1 − h2 = h3 − h4 412 − 405.4 = 252.4 − h 4 h4 = 245. 8 kJ kg then t 4 = 37 C subcooled = t 3 − t 4 = 42 − 37 = 5 C (b) For volume flow rate: Consider Q A = 1 ton = 3.516 kW Q A  = m , h5 = h4 h6 − h5

13

CHAPTER 2

m =

3 .516 = 0 .022 kg s per ton of refrigeration . 405 4 − 245.8

 v1 = ( 0. 022)(0. 050) = 0. 0011 m 3 s TR = 1. 10 L s TR V1 ′ = m

(c) W = m (h 2 − h1 ) = (0 .022 )(443 − 412 .5 ) = 0 .67 kW TR

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