Question

How much moisture per pound of dry air is removed by air entering moist grain at 110 degrees F dry bulb temperature and exiting the grain at 90% RH and 68 degrees F wet bulb?

Moisture Removed =  lbs H2O/lb DA

Answer 1

From the psychrometric chart, the humidity ratio at 110 ⁰F which will be given as -

₁ = 0 lbs.H₂O/lb.DA

From the psychrometric chart, the humidity ratio at 68 ⁰F & 90% RH which will be given as -

₂ = 0.01428 lbs.H₂O/lb.DA

An expression for the moisture per pound of dry air removed which will be given by -

= ₂ - ₁

= [(0.01428 lbs.H₂O/lb.DA) - (0 lbs.H₂O/lb.DA)]

= 0.01428 lbs.H₂O/lb.DA