Parboiled paddy is to be dried in a tray dryer under steady state conditions from an initial free moisture content of 0.40 kg H₂O (kg dry solid)⁻¹ to final free moisture content of 0.02 kg H₂O (kg dry solid)⁻¹. The dry solid mass is 99.8 kg and the top surface area for drying is 4.654 m². The drying is occurring in both constant and falling rate periods. If constant drying rate of 1.51 kg H₂O m⁻² h⁻¹ is followed up to a critical moisture content of 0.195 kg H₂O (kg dry solid)⁻¹, then the total drying time in hour will be __________. [round off to two decimal places]
Official Answer Key:
9.20 to 9.30
Step-by-Step Solution & Explanation
• Constant-rate period time: tc = Ms(X0−Xc)/(A·Rc) = 99.8×(0.40−0.195)/(4.654×1.51) = 20.459/7.0275 = 2.911 h
• Falling-rate period (assuming rate linear in free moisture): tf = [Ms·Xc/(A·Rc)]×ln(Xc/Xf) = [19.461/7.0275]×ln(0.195/0.02) = 2.7692×ln(9.75) = 2.7692×2.2773 = 6.307 h
• Total drying time = tc+tf = 2.911+6.307 = 9.22 h
Computed value: 9.22 h (within official range 9.20–9.30)
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