Example 11.1: Moisture contents of a wet solid are reduced from 35 percent to 10 percent during constant drying conditions in 5 hours. If the equilibrium moisture content is 4 percent and the critical moisture content is 14 percent, how long will it take to dry this solid to 6 percent moisture under the same conditions?

Solution:

Water removal from solids is composed of two parts. First, where water is freely available, and rate of drying is constant. Second, where water removal is represented as a first order mechanism. If w₀ is initial fraction of water and water removal rate stays constant till moisture fraction reaches w_c, then water removal rate, R, is given as minus 1 over A dw over dt is equal to m time (w minus w_e), where m is ratio of drying rate per unit area to moisture content; w is equal to total moisture at any time; w_c is critical moisture content; w_e is equilibrium moisture content. R_c is equal to rate of drying per unit area for constant rate period is equal to (w_o - w_c)/(t_cA), and A is equal to area of heat transfer. Time required for drying solids from an intial moisture of w₀ to a final moisture fraction of w is a combination of constant rate period, and falling rate period.

We can find the value of mA by using first set of information. It takes 5 hours to reduced initial moisture from 0.35 to 0.10 when critical moisture is 0.14 and equilibrium moisture is 0.04. This data gives mA as 0.522. For second case, when drying is to continued till w₂ of 0.06, t₂ can be calculated as 7.1 hours.