Question

1. A building air-condition unit is designed to cool 2500 ft3/min of outside air (95°F, 740 mmHG, 82% relative humidity) to 50°F, thereby condensing a portion of the water vapor. The system then heats the air and sends it to a building at 68°F

a) Calculate the rate of water condensation.

b) Calculate the volumetric flow rate of air delivered to the building.

Answer 1

Basis: 1 min of operation

2500 ft³ of air at 95 ^o F ( 308 K) 740 mm Hg contains = 1*(273/308)*(740/760)*(2500/359)= 6.01 lb moles

( we know at std conditions - 1 lb mole of gas at 273 K, 760 mm Hg occupy 359 ft³ )

Total moles entering the unit = n=6.01 lb moles

Relative humidity= partial pressure of water/vapor pressure of water at95 ^oF= 0.82

Vapor pressure at 95 ^oF =41.526 mm Hg

Therefore partial pressure of water vapor in the inlet air= 0.82* 41.526=34.05 mm Hg

Mole of water vapor = na = Partial pressure/total pressure =34.05/740=0.046 lb moles

Moles of dry air = n-na= 6.01-0.046=5.550 lb mole

Since the moles of dry air do not change during the process, can be considered as tie substance.

Partial pressure of water vapor leaving the unit = vapor pressure of water at 50 ^oF )since it is condensing) = 8.899 mm Hg

Moles of water vapor leaving the unit = Partial pressure/total pressure =8.899/740=0.012 lb mole

a) water vapor condensed =   lb moles =0.046-0.012 =0.034 lb moles=0.612 lb per minute

b) Total moles leaving the unit= 5.55 + 0.012= 5.67 lb moles.

These moles leave the building after heating to 68 ^oF (293 K) ,740 mm Hg and therefore

The volumetric flow rate of gases at these conditions is = (5.67/1)*(293/273)*(760/740)*359= 2243 ft³ per minute