A room is to be maintained at 76F and 40 rh Supply air at 39
A room is to be maintained at 76F and 40% rh. Supply air at 39F is to absorb 100000 But sensible heat and 35 lb of moisture per hour. Assume the moisture has an enthalpy of 1100 Btu/ lb. How many pounds of dry air per hour are required?
What should the dew-point temperature and relative humidity of the supply air be?
Solution
Heat load to be absorbed from 39 deg F dry air to 76deg F and 40 % RH = 0.26 x ( 76-39) +0.4 (0.444(76-39) +1100)= 8.8 Btu sensible + 446.6 BTU for evaporation = 455.5 BTU Additional sensible load = 100000 BTU Qty of dry air required = 10009/ 0.24 x 39= 10185 lbs / hr misture content in the room 35 lbs / 10185 = 0.0034 lbs / lb of dry air = RH 65% and dee point of 30 Deg F
