Refrigeration and Air Conditioning The heat rejection factor (HRF) is given by 1 - C.O.P. 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) ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning The bypass factor (B. P. F.) in case of sensible heating of air is (Where td₁ = Dry bulb temperature of air entering the heating coil, td₂ = Dry bulb temperature of air leaving the heating coil, and td₃ = Dry bulb temperature of heating coil) (td₃ - td₁)/(td₃ - td₂) (td₃ - td₂)/(td₃ - td₁) (td₂ - td₃)/(td₃ - td₁) (td₃ - td₁)/(td₂ - td₃) (td₃ - td₁)/(td₃ - td₂) (td₃ - td₂)/(td₃ - td₁) (td₂ - td₃)/(td₃ - td₁) (td₃ - td₁)/(td₂ - td₃) ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning The C.O.P. of an absorption type refrigerator is given by (where T₁ = Temperature at which the working substance receives heat, T₂ = Temperature of cooling water, and T₃ = Evaporator temperature) [T₁ (T₂ - T₃)] / [T₃ (T₁ - T₂)] [T₁ (T₁ - T₂)] / [T₃ (T₂ - T₃)] [T₃ (T₁ - T₂)]/ [T₁ (T₂ - T₃)] [T₃ (T₂ - T₃)] / [T₁ (T₁ - T₂)] [T₁ (T₂ - T₃)] / [T₃ (T₁ - T₂)] [T₁ (T₁ - T₂)] / [T₃ (T₂ - T₃)] [T₃ (T₁ - T₂)]/ [T₁ (T₂ - T₃)] [T₃ (T₂ - T₃)] / [T₁ (T₁ - T₂)] ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning A heat pump working on a reversed Carnot cycle has a C.O.P. of 5. It works as a refrigerator taking 1 kW of work input. The refrigerating effect will be 4 kW 2 kW 1 kW 3 kW 4 kW 2 kW 1 kW 3 kW ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning Reducing suction pressure in refrigeration cycle Lowers evaporation temperature Increases power required per ton of refrigeration Lowers compressor capacity because vapour is lighter All of these Lowers evaporation temperature Increases power required per ton of refrigeration Lowers compressor capacity because vapour is lighter All of these 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 7.44 kW 1.86 kW 18.6 kW 3.72 kW 7.44 kW 1.86 kW 18.6 kW ANSWER DOWNLOAD EXAMIANS APP
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