Refrigeration and Air Conditioning The heat rejection factor (HRF) is given by 1 + C.O.P 1 + (1/C.O.P) 1 - (1/C.O.P) 1 - C.O.P. 1 + C.O.P 1 + (1/C.O.P) 1 - (1/C.O.P) 1 - C.O.P. ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning In case of sensible heating of air, the coil efficiency is given by (where B.P.F. = Bypass factor) 1/ B.P.F. 1 + B.P.F. P.F. - 1 1 - B. P.F. 1/ B.P.F. 1 + B.P.F. P.F. - 1 1 - B. P.F. ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning One of the purposes of sub cooling the liquid refrigerant is to Reduce compressor overheating Ensure that only liquid and not the vapour enters the expansion (throttling) valve Reduce compressor discharge temperature Increase cooling effect Reduce compressor overheating Ensure that only liquid and not the vapour enters the expansion (throttling) valve Reduce compressor discharge temperature Increase cooling effect ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning The coefficient of performance (C.O.P.) of a refrigerator working as a heat pump is given by O.P)P = ( O.P.)R + 1 O.P)P = ( O.P.)R + 2 O.P.)P = ( ( ( O.P.)P = ( ( ( O.P)R O.P)R - 1 O.P)P = ( O.P.)R + 1 O.P)P = ( O.P.)R + 2 O.P.)P = ( ( ( O.P.)P = ( ( ( O.P)R O.P)R - 1 ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning The operating temperature of a cold storage is -2°C. The heat leakage from the surrounding is 30 kW for the ambient temperature of 40°C. The actual C.O.P. of refrigeration plant used is one fourth that of ideal plant working between the same temperatures. The power required to drive the plant is 3.72 kW 18.6 kW 1.86 kW 7.44 kW 3.72 kW 18.6 kW 1.86 kW 7.44 kW ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning During sensible cooling of air, specific humidity Decreases Remains constant Increases None of the listed here Decreases Remains constant Increases None of the listed here ANSWER DOWNLOAD EXAMIANS APP
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