Answer to Question #125257 in General Chemistry for Jay Shendurnikar

Question

Air at 1 atm absolute pressure, 38°C and 80% relative humidity is to be cooled to 18°C
and fed into a plant area at a rate of 510 m3
/min (STP). Calculate the rate (kg/min) at which
water condenses

Answer 1

φ = p_w / p_ws 100%               

where

φ = relative humidity [%]

p_w = vapor partial pressure [bar]

p_ws = saturation vapor partial pressure at the actual dry bulb temperature [mbar]. This is the vapour pressure at maximum content of water gas in air, before it starts to condense out as liquid water.

At 18⁰C p_ws = 21.0 x 10^-3 bar

At 38⁰C p_ws = 65.6 x 10^-3 bar

(https://www.engineeringtoolbox.com/relative-humidity-air-d_687.html)

65.6 x 10^-3 bar = 0.0647 atm

At 80% humidity p_w = 0.8 x 0.0647 = 0.05176 atm (5244.582 Pa)

Using the equation: pV = nRT, n = pV / RT

T = 273.15 + 38 = 311.15 K

n = (5244.582 x 510) / (8.314 x 311.15) = 1033.95 mol

21.0 x 10^-3 bar = 0.0207 atm

At 80% humidity p_w = 0.8 x 0.0207 = 0.01656 atm (1677.942 Pa)

T = 18 + 273.15 = 291.15 K

n = pV / RT

n = (1677.942 x 510) / (8.314 x 291.15) = 353.525 mol

Difference: 1033.95 - 353.525 = 680.42 mol - is to be condensed.

M (H₂O) = 18.02

m = 680.42 x 18.02 = 12261 g or 12.26 kg / min