Falling-Rate Drying Period Time
Computes the duration of the falling-rate drying period under the model where the rate drops linearly with free moisture starting at the critical moisture: t = m_s × X_c ÷ (A × N_c) × ln(X_c ÷ X₂). Moisture contents go in as free moisture on a dry basis, that is, with the equilibrium moisture already subtracted, which is why X₂ can never be zero — drying down to equilibrium would take infinite time, exactly what the logarithm says. Compared with the constant-rate period this is the expensive stretch: every kilogram of water removed costs far more time than in the previous stretch, because internal transport now sets the pace. Enter the dry solid mass, the critical moisture, the final free moisture, the exposed area and the constant rate at the critical moisture.
Result
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Falling-Rate Drying Period Time in Tray Drying
The expensive stretch of drying starts when the surface can no longer replace the water that evaporates. From then on the rate falls along with the moisture and the material sets the pace: anyone designing a batch cycle finds that pulling out the last few percentage points costs more time than pulling out the first half of the water. This page solves that stretch with the linear model, the one textbooks adopt when the rate curve, past the critical moisture, points at the origin.
The rate drops in proportion to free moisture, which integrates to t = m_s × X_c ÷ (A × N_c) × ln(X_c ÷ X₂). The first factor is a characteristic time: how long removing the whole critical moisture would take if the rate never dropped. With the default values — 50 kg of solid, X_c of 0.12, X₂ of 0.04, 2.5 m² and N_c of 1.2 kg/m²·h — that factor is 2 h, the logarithm of 3 is 1.0986 and the time comes to 2.197 h. Look at the trade: 4 kg of water in 2.197 h here, against 11.5 kg in 3.833 h in the previous stretch. With X₂ equal to X_c the result is exactly zero, the boundary between the two periods.
The model holds while the plot of rate against moisture is a straight line through the origin, which many materials fail to deliver: porous solids often show two falling stretches with different slopes, and then only numerical integration of the measured curve will do. Moisture has to be free moisture, with the equilibrium value of the dryer air already subtracted. Typing total figures instead shortens the answer: with an equilibrium of 0.02 kg/kg the same data become 0.14 and 0.06, and the page returns 1.977 h rather than 2.197 h. Shrinkage, case hardening and the heating of the solid toward air temperature all sit outside the model.
Frequently asked questions
Why can the final moisture not be zero?
How does this differ from the constant-rate period?
How do I get the total drying time?
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