Determine the heat transfer cooling tons to be removed from

Determine the heat transfer (cooling tons) to be removed from the space if there is no recirculation Suggest an AC system that could work under these conditions (Brand, model, price). Determine the heat transfer (cooling tons) temperature is increased to 75 F. Suggest an AC system that could work under these conditions (Brand, model, price). Determine the heat transfer (cooling tons) to be removed if recirculation (25% of airflow is used) Determine heat transfer (cooling tons) if a heat pipe is used (the effects of usage of heat pipe being that air enters the cooling coil at 80, and enters the reheating resistance at 65 F)

Solution

Heat transfer depends on the change in enthalpy of the air inside and outside.

Specific enthalpy if air is calculated using psychrometric chart.

For air at 90⁰ F and 85% , h_o = 54 Btu/lb

For air at 72⁰ F and 50%, h_i = 26 Btu/lb

Heat transfer rate = Mass flow rate x change in specific enthalpy

Mass flow rate = 0.124 x 300 = 37.2 lb/min

Heat transfer rate = 37.2 x (54-26) = 1041.6 Btu/min = 18.31 kW = 5.72 tons

c) If the inside temperature is increased to 75 F, enthalpy of air inside will be h_i = 28 Btu/lb

Heat transfer rate = 37.2 x (54-28) = 967.2 Btu/min = 17 kW = 5.3/tons

e) By recirculation the amount of heat transfer reduced will be = 0.25 x 37.2 x 26 = 1041.6 - 241.8 = 799.8 Btu/min = 14.06 kW = 4.4 tons