3. the Vapour Compression Cycle (Sample Problems)

3. The vapour compression cycle. Example No. 1 A standard vapor-compression cycle developing 50 kW of refrigeration using refrigerant 22 operates with a condensing temperature of 35 C and an evaporating temperature of -10 C. Calculate (a) the refrigerating effect in kilojoules per kilogram, (b) the circulation rate of refrigerant in kilograms per second, (c) the power required by the compressor in kilowatts, (d) the coefficient of performance, (e) the volume flow rate measured at the compressor suction, (f) the power per kilowatt of  refrigeration, and (g) the compressor discharge temperature. Given: 50 kW of refrigeration, refrigerant 22 35 C condensing temperature -10 C evaporating temperature Required: (a) the refrigerating effect in kilojoules per kilogram, (b) the circulation rate of refrigerant in kilograms per second, (c) the power required by the compressor in kilowatts, (d) the coefficient of performance, (e) the volume flow rate measured at the compressor suction, (f) the power per kilowatt of refrigeration, and (g) the compressor discharge temperature. Solution: As the first step in the solution, sketch the pressure-enthalpy diagram (Figure 11) and determine from Tables and Figures of properties the enthalpies at key points. The value of h 1 is the enthalpy of  saturated vapor at -10 C, which is 401.6 kJ/kg.

To find h2 move at a constant entropy from point 1 until reaching the saturation pressure corresponding to 35 C. This condensing pressure is 1354 kPa, and the value of h2 is 435.2 kJ/kg. 1

The values of  h3 and h4 are identical and are equal to the enthalpy of saturated liquid at 35 C, which is 243.1 kJ/kg. Therefore h1  401.6 kJ kg h2  435.2 kJ kg h3  h4

 243.1

kJ kg

(a) The refrigerating effect is h1  h4  401.6  243.1  158.5 kJ kg (b) The circulating rate of refrigerant can be calculated by dividing the refrigerating capacity by the refrigerating effect 50 kW  Flow rate   0.315 kg s 158 .5 kJ kg (c) The power required by the compressor is the work of compression per kilogram multiplied by the refrigerant flow rate Compressor   power   0.315 kg s 435.2  401.6 kJ kg  10.6 kW  (d) The coefficient of performance is the refrigerating rate divided by the compressor power 50 kW  Coefficien t  of  performanc e   4.72 10.6 kW  (e) The volume rate of flow at the compressor inlet requires knowledge of the specific volume 3 of the refrigerant at point 1. From tables or figures of properties this value is 0.0654 m /kg, and so



Volume  flow rate  0.315 kg s  0.0654 m kg 3



3

Volume  flow rate  0.0206 m s  20.6 L s

(f) The compressor power per kilowatt of refrigeration (which is the reciprocal of the coefficient of performance) is 10.6 kW  Power  of  refrigerat ion   0.212 kW  kW  50 kW  (g) The compressor discharge temperature is the temperature of superheated vapor at point 2 which form Figure of Properties is found to be 57 C.

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EXAMPLES 1. A vapor-compression refrigerant system uses R12 as the refrigerant. The gaseous refrigerant leaves the compressor at 1200 kPa and 80 C. The heat loss during compression is 14 kJ/kg. The refrigerant enters the expansion valve at 32 C. The liquid leaves the evaporator and enters the compressor as a saturated vapor at -15 C. The unit must produce 50 tons of refrigeration. Determine (a) the R12 flow rate; (b) the compressor power; (c) the COP. Solution:

2

At state 2, -15 C, sat. h2 = 180.85 kJ/kg s2 = 0.7046 kJ/kg-K At state 3, 1200 kPa, 80 C h3 = 230.4 kJ/kg At state 4, 32 C h4 = h f  = 66.52 kJ/kg At state 1, h1 = h4 = 66.52 kJ/kg (a) R12 flow rate QA 503.516 m  h2  h1 180.85  66.52

 1.5377

kg s

(b) Compressor power mh2  W   mh3  mqloss W   mh3  h2   mqloss W   1.5377 230 .4  180 .85  1.5377 14

W   97.72 kW 

(c) COP COP 

QA W 



50 3.516 

97.72



 1.8

3