Question

On a hot summer day the temperature is 35°C, barometric pressure is 103 kPa, and the relative humidity is 90%. An air conditioner draws in outside air, cools it to 20°C, and delivers it at a rate of 12,500 L/h. Calculate the rate of moisture condensation (kg/h) and the volumetric flow rate of the air drawn from the outside.

Answer 1

Answer:

The rate of moisture condensation (kg/h) =  0.022652 mol H2O / mol.

The volumetric flow rate of the air drawn from the outside = 13506.88 L/h.

Explanation:

Without mincing words, let us dive straight into the solution to this problem.

STEP ONE: The first thing to calculate for or determine is the Mol fraction of H2O inlet and Mol fraction of H2O outlet.

Therefore, the Mol fraction of H2O inlet = (0.90 x 42.2 mmHg/103000 Pa) x (101325 Pa/760mmHg)= 0.0492  mol H2O / mol.

Also, the Mol fraction of H2O outlet = (17.5 mmHg/103000 Pa) x (101325 Pa/760mmHg) = 0.022652 mol H2O / mol.

STEP TWO: Determine the Molar flow rate of inlet air,  Molar flow rate of outlet air and the dry air balance.

The Molar flow rate of outlet air = (12,500 L/h) x (1mol/22.4L) x (273K/293K) x (103000 Pa / 101325 Pa)  = 528.54 mol/h

Hence, the Molar flow rate of outlet air = 528.54 mol/h.

The Molar flow rate of inlet air = (1 - 0.022652 mol H2O / mol. ) x 528.54  mol/h ÷ (1 - 0.0492  mol H2O / mol) = 543.3 mol/h.

Hence, the volumetric flow rate of the air drawn from the outside =  ( 543.3 mol/h.) x (22.4L/mol) x (308K/273K) x (101325 Pa/103000 Pa) = 13506.88 L/h.

STEP THREE; Determine the molar flow of water.

Thus, the molar flow of water = 543.3 mol/h - 528.54 mol/h = 14.76 mol/h.

STEP FOUR: Determine the rate of moisture condensation (kg/h).

The rate of moisture condensation (kg/h) = 14.76 mol/h ×  (18g/mol) x (1kg/1000g) = 0.266 kg/h.