Answer to Question #313274 in Mechanical Engineering for Jul

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 the answer with solution.

Answer 1

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

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

= 506666.67 cubic feet/hr.

the answer is the average air speed *area through which air speed has been measured.