Refrigeration and Air Conditioning Critical pressure of a liquid is the pressure Above which liquid becomes vapour Above which liquid becomes gas Above which liquid becomes solid Above which liquid will remain liquid Above which liquid becomes vapour Above which liquid becomes gas Above which liquid becomes solid Above which liquid will remain liquid ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning When the air is passed through an insulated chamber having sprays of water maintained at a temperature higher than the dew point temperature of entering air but lower than its dry bulb temperature, then the air is said to be Cooled and humidified Heated and humidified Cooled and dehumidified Heated and dehumidified Cooled and humidified Heated and humidified Cooled and dehumidified Heated and dehumidified ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning Pick up the wrong statement. A refrigerant should have Higher critical temperature Tow specific heat of liquid High latent heat of vaporisation High boiling point Higher critical temperature Tow specific heat of liquid High latent heat of vaporisation High boiling point ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning Pressure of water vapour is given by 0.622 Pv/ (Pb - Pv) μ/[1 - (1 - μ) (Ps/Pb)] None of these [Pv (Pb - Pd)]/ [Pd (Pb - Pv)] 0.622 Pv/ (Pb - Pv) μ/[1 - (1 - μ) (Ps/Pb)] None of these [Pv (Pb - Pd)]/ [Pd (Pb - Pv)] ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning The sensible heat factor during cooling and dehumidification process is given by (where h₁ = Enthalpy of air entering the cooling coil, h₂ = Enthalpy of air leaving the cooling coil, and hA = Enthalpy of air at the end of dehumidification process) (h1 - h2)/ (hA - h2) (hA - h2)/ (h1 - h2) (hA - h1)/ (h2 - h1) (h2 - hA)/ (h1 - h2) (h1 - h2)/ (hA - h2) (hA - h2)/ (h1 - h2) (hA - h1)/ (h2 - h1) (h2 - hA)/ (h1 - h2) ANSWER DOWNLOAD EXAMIANS APP
Refrigeration and Air Conditioning A refrigerating system operating on reversed Brayton refrigeration cycle is used for maintaining 250 K. If the temperature at the end of constant pressure cooling is 300 K and rise in the temperature of air in the refrigerator is 50 K, then the net work of compression will be (assume air as working substance with Cp = 1 kJ/kg) 125 kJ/kg 50 kJ/kg 25 kJ/kg 100 kJ/kg 125 kJ/kg 50 kJ/kg 25 kJ/kg 100 kJ/kg ANSWER DOWNLOAD EXAMIANS APP
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