Answer to Question #313261 in Mechanical Engineering for Jhay

Question

17. The moisture removed from a material is 200 lb/hr and change of humidity ratio in dryer is 0.015. Determine the fan capacity if specific volume of air entering is 38 ft3/lb. Please explain the answer.

Answer 1

fan capacity = $\frac{moisture\ removed}{change\ of\ humidity\ ratio\ in\ dryer}*s.p\ volume\ of\ air\ entering$

$=\frac{200}{0.015}*38$

= 506666.667 cubic feet/hr.

the answer 506666.667 cubic feet/hr gives the average air speed (ft / min) x area (height x width) through which air speed has been measured (ft2).