Atmospheric air at 38ºC and 25% relative humidity passes through an evaporator cooler.

Question

Evaporative Cooler : 

Atmospheric air at 38ºC and 25% relative humidity passes through an evaporator cooler. If the final temperature of air is 18ºC, how much water is added per kg of dry air and what is the final relative humidity ?

Answer 1

At 38ºC :

p_vs = 0.0663 bar, h_g1 = 2570.7 kJ/kg 

and p_v = φ × p_vs = 0.25 × 0.0663 = 0.01657 bar

At 18ºC :

h_g2 = 2534.4 kJ/kg, p_vs = 0.0206 bar

W1 = \(\cfrac{0.622\times0.01657}{1.0132-0.01657}\) = 0.01037 kg/kg of dry air

Since enthalpy remains constant during the process

c_pt_db1 + W₁h_g1 = c_pt_db2 + W₂h_g2

1.005 × 38 + 0.01034 × 2570.7 = 1.005 × 18 + W₂ × 2534.4 

(\(\because\) At 18ºC, h_g2 = 2534.4 kJ/kg)

∴ Amount of water added = W₂ – W₁ = 0.01842 – 0.01034 

= 0.00808 kg/kg of dry air.

0.00808 = \(\cfrac{0.622\,p_{v2}}{1.0132-p_{v2}}\)

0.00808 (1.0132 – p_v2 ) = 0.622 p_v2

0.00818 – 0.00808 p_v2 = 0.622 p_v2 

∴ p_v2 = 0.01298 bar

∴ Final relative humidity = \(\cfrac{0.01298}{0.0206}\) = 0.63 or 63%.