Parboiled paddy dries from free moisture .40 to .02 kg water/kg dry solid. Dry solids99.8 kg; exposed area 4.654 m². Constant drying rate 1.51 kg/(m² h) continues to critical free moisture .195. Find total drying time in hours, to two decimals.
Assume the usual linear falling-rate relation $N=N_cX/X_c$ for free moisture. Constant-rate time is $M_s(X_i-X_c)/(AN_c)$; falling-rate time is $M_sX_c\ln(X_c/X_f)/(AN_c)$. Sum=$99.8[.40-.195+.195\ln(.195/.02)]/[4.654(1.51)]=\mathbf{9.22}$ h. The linear falling-rate model is required to evaluate the second period.
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