Solution capitulo 11

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CHAPTER 11 - COMPRESSORS

11-1.

An ammonia compressor has a 5 percent clearance volume and a displacement rate of 80 L/s and pumps against a condensing temperature of 40 C. For the two different evaporating temperatures of -10 and 10 C, compute the refrigerant flow rate assuming that the clearance volumetric efficiency applies.

Solution: Equation 11-7.

η vc w = displacement rate × (a)100 ν suc

At -10 C, Table A-3. s1 = 5.7550 kJ/kg
νsuc = 417.477 L/kg

At 40 C, constant entropy, Fig. A-1 νdis = 112.5 L/kg m=5% Equation 11-4 and Equation 11-5.
ν  η vc = 100 − m suc − 1 ν   dis 
 417.477  η vc = 100 − 5 − 1 = 86.445  112.5  η vc w = displacement rate ×
w = (80 L/s ) ×

(86.445100)

100 ν suc

417.477 w = 0.166 kg/s at -10 C - - - Ans. (b) At 10 C,Table A-3 s1 = 5.4924 kJ/kg.K νsuc = 205.22 L/kg At 40 C, constant entropy, Fig. A-1 νdis = 95 L/kg m=5% Equation 11-4 and Equation 11-5.

ν  η vc = 100 − m suc − 1 ν   dis   205.22  η vc = 100 − 5 − 1 = 94.199  95 

Page 1 of 6

CHAPTER 11 - COMPRESSORS

η vc w = displacement rate × w = (80 L/s ) ×

100 ν suc

(94.199100)

205.22 w = 0.367 kg/s at 10 C - - - Ans.11-2. A refrigerant 22 compressor with a displacement rate of 60 L/s operates in a refrigeration system that maintains a constant condensing temperature of 30 C. Compute and plot the power requirement of this compressor at evaporating temperatures of -20, -10, 0, 10 and 20 C. Use the actual volumetric efficiencies from Fig. 11-12 and the following isentropic works of compression for the fiveevaporating temperatures, respectively, 39.9, 30.2, 21.5, 13.7, and 6.5 kJ/kg.

Solution: (a) At -20 C evaporating temperature, Table A-6. νsuc = 92.8432 L/kg psuc = 244.83 kPa Table A-7, 30 C pdis = 1191.9 kPa

Ratio = pdis / psuc = 1191.9 kPa / 244.82 kPa = 4.87 Figure 11-12 ηva = volumetric efficiency = 67.5 % η va w = displacement rate × w = (60 L/s )× 100 ν suc

(67.5100)

92.8432 w = 0.4362kg/s at -20 C P = w∆hi ∆hi = 39.9 kJ/kg P = (0.4362)(39.9) P = 17.4 kw at -20 C (b) At -10 C evaporating temperature, Table A-6. νsuc = 65.3399 L/kg psuc = 354.3 kPa Table A-7, 30 C pdis = 1191.9 kPa

Ratio = pdis / psuc = 1191.9 kPa / 354.30 kPa = 3.364 Figure 11-12 ηva = volumetric efficiency = 77.5 % η va w = displacement rate × 100 ν suc

Page 2 of 6

CHAPTER 11 - COMPRESSORS

65.3399w = 0.7117 kg/s at -10 C P = w∆hi ∆hi =30.2 kJ/kg P = (0.7117)(30.2) P = 21.5 kw at -10 C (c) At 0 C evaporating temperature, Table A-6. νsuc = 47.1354 L/kg psuc = 497.59 kPa Table A-7, 30 C pdis = 1191.9 kPa

w = (60 L/s )×

(77.5100)

Ratio = pdis / psuc = 1191.9 kPa / 497.59 kPa = 2.4 Figure 11-12 ηva = volumetric efficiency = 83 % η vc w = displacement rate × w = (60 L/s ) × 100 ν suc(83100)

47.1354 w = 1.0565 kg/s at 0 C P = w∆hi ∆hi = 21.5 kJ/kg P = (1.0565)(21.5) P = 22.7 kw at 0 C

(d)

At 10 C evaporating temperature, Table A-6. νsuc = 34.7136 L/kg psuc = 680.70 kPa Table A-7, 30 C pdis = 1191.9 kPa

Ratio = pdis / psuc = 1191.9 kPa / 680.70 kPa = 1.75 Figure 11-12 ηva = volumetric efficiency = 86.7 % η vc w = displacement rate × w = (60 L/s )× 100 ν suc(86.7100)

34.7136 w = 1.4986 kg/s at 10 C P = w∆hi
Page 3 of 6

CHAPTER 11 - COMPRESSORS

∆hi = 13.7 kJ/kg P = (1.4986)(13.7) P = 20.5 kw at 10 C (e) At 20 C evaporating temperature, Table A-6. νsuc = 26.0032 L/kg psuc = 909.93 kPa Table A-7, 30 C pdis = 1191.9 kPa

Ratio = pdis / psuc = 1191.9 kPa / 909.93 kPa = 1.31 Figure 11-12 ηva = volumetric efficiency = 89.2 % η vc w = displacementrate × w = (60 L/s )× 100 ν suc

(89.2100)

26.0032 w = 2.0583 kg/s at 20 C P = w∆hi ∆hi = 6.5 kJ/kg P = (2.0583)(6.5) P = 13.4 at 20 C

11-3.

The catalog for a refrigerant 22, four-cylinder, hermetic compressor operating at 29 r/s. a condensing temperature of 40 C and an evaporating temperature of -4 C shows a refrigeration capacity of 115 kw. At this operating points the motor (whose...
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