Time for Drying in Constant-Rate Period A batch of wet solid was dried on a tray

Question

Time for Drying in Constant-Rate Period A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant-rate period was R = 2.05 kg H_2O/h middot m^2 (0.42 lb_m H_2 O/h middot ft^2). The ratio Ls/A used was 24.4 kg dry solid/m^2 exposed surface (5.0 lbm dry solid/ft^2). The initial free moisture was = 0.55 and the critical moisture content Xc = 0.22 kg free moisture/kg dry solid. Calculate the time to dry a batch of this material from X-1 = 0.45 to X_2 = 0.30 using the same drying conditions but a thickness of 50.8 mm, with drying from the top and bottom surfaces.

Explanation / Answer

Time required for constant rate drying t1 = (Ls/AR)*(X1-X2)

X1 = moisture content before drying and X2= moisture content after drying. Since both are less than critical moisture, this comes under falling period.

For the second case, when the material thickness was 54.8 and both sides exposed, Ls remaining constant. A increases by 2( thickness)*2 ( top and bottom surface)= 4 times

New LS/A = 24.4/4= 6.1

Time of drying = (6.1/2.05)*(0.45-0.30)=0.446 hrs

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