Project ForgeA batch of 120 kg wet solid has initial moisture content of 0.2 kg water/kg dry solid. The exposed area for drying is 0.05 m²/kg dry solid. The rate of drying follows a curve that is linear (R=6X) in the falling-rate region below the critical moisture Xc=0.25, reaching a constant rate of 1.5 kg/m².hr above Xc. The time required (in hours) for drying this batch to a moisture content of 0.1 kg water/kg dry solid is
- A0.033
- B0.43
- C0.6
- 2.31
Explanation
Dry solid mass Ls=120/1.2=100 kg. Total area A=0.05×100=5 m². Since X0=0.2 is already below Xc=0.25, the entire drying from X0=0.2 to Xf=0.1 occurs in the falling-rate period, where R=6X (confirmed since R(0.2)=1.2 and R(0.1)=0.6 both satisfy R/X=6, a line through the origin). Time=(Ls/A)∫[Xf to X0]dX/(6X)=(Ls/(6A))ln(X0/Xf)=(100/(6×5))×ln(2)=3.333×0.6931≈2.31 hours.
Source: GATE CH 2011 (2011)
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