Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 11 (Q.No. 25)
25.
In a laboratory test run, the rate of drying was found to be 0.5 x 10-3 kg/m2.s, when the moisture content reduced from 0.4 to 0.1 on dry basis. The critical moisture content of the material is 0.08 on a dry basis. A tray dryer is used to dry 100 kg (dry basis) of the same material under identical conditions. The surface area of the material is 0.04 m2/kg of dry solid. The time required (in seconds) to reduce the moisture content of the solids from 0.3 to 0.2 (dry basis) is
Discussion:
1 comments Page 1 of 1.
Shiva said:
1 decade ago
Time = (mass of dry solid)*(X1-X2)/A*N.
X1 = 0.3, X2=0.2....X1-X2 = 0.1.
Mass = 100kg.
A = 0.04*100 = 4 m2.
N = 0.5*10^-3.
Time = 100*0.1/4*0.5*10^-3 = 5000 sec.
X1 = 0.3, X2=0.2....X1-X2 = 0.1.
Mass = 100kg.
A = 0.04*100 = 4 m2.
N = 0.5*10^-3.
Time = 100*0.1/4*0.5*10^-3 = 5000 sec.
(2)
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