In the falling-rate drying period, as shown in Fig. 9.5-1b, the rate of drying R is not constant but decreases when drying proceeds past the critical free moisture content XC. When the free moisture content X is zero, the rate drops to zero.
The time of drying for any region between X1 and X2 has been given by Eq. (9.6-1):
Equation 9.6-1
If the rate is constant, Eq. (9.6-1) can be integrated to give Eq. (9.6-2). However, in the falling-rate period, R varies. For any shape of falling-rate drying curve, Eq. (9.6-1) can be integrated by plotting 1/R versus X and determining the area under the curve using graphical integration or numerical integration with a spreadsheet. (See Section 1.8 for methods of numerical integration.)
EXAMPLE 9.7-1. Numerical Integration in Falling-Rate PeriodA batch of wet solid whose drying-rate curve is represented by Fig. 9.5-1b is to be dried from a free moisture content of X1 = 0.38 kg H2O/kg dry solid to X2 = 0.04 kg H2O/kg dry solid. The weight of the dry solid is LS = 399 kg dry solid and A = 18.58 m2 of top drying surface. Calculate the time for drying. Note that LS/A = 399/18.58 = 21.5 kg/m2. Solution: From Fig. 9.5-1b, the critical free moisture content is XC = 0.195 kg H2O/kg dry solid. Hence, the drying occurs in the constant-rate and falling-rate periods. For the constant-rate period, X1 = 0.38 and X2 = XC = 0.195. From Fig. 9.5-1b, RC = 1.51 kg H2O/hm2. Substituting into Eq. (9.6-2), For the falling-rate period, reading values of R for various values of X from Fig. 9.5-1b, the following table is prepared:
To determine this area by numerical integration using a spreadsheet, the calculations are given in the following table:
The area of the first rectangle is the average height (0.663 + 0.826)/2, or 0.745, times the width ΔX (0.195 − 0.150), or 0.045, giving 0.0335. Other values are similarly calculated, giving a total of 0.1889. Substituting into Eq. (9.6-1), The total time is 2.63 + 4.06 = 6.69 h. |
In certain special cases in the falling-rate region, the equation for the time of drying, Eq. (9.6-1), can be integrated analytically.
If both X1 and X2 are less than XC and the rate R is linear in X over this region,
Equation 9.7-1
where a is the slope of the line and b is a constant. Differentiating Eq. (9.7-1) gives dR = a dX. Substituting this into Eq. (9.6-1),
Equation 9.7-2
Since R1 = aX1 + b and R2 = aX2 + b,
Equation 9.7-3
Substituting Eq. (9.7-3) into (9.7-2),
Equation 9.7-4
In some cases a straight line from the critical moisture content passing through the origin adequately represents the whole falling-rate period. In Fig. 9.5-1b this would be a straight line from C to E at the origin. Often, for lack of more detailed data, this assumption is made. Then, for a straight line through the origin, where the rate of drying is directly proportional to the free moisture content,
Equation 9.7-5
Differentiating, dX = dR/a. Substituting into Eq. (9.6-1),
Equation 9.7-6
The slope a of the line is RC/XC, and for X1 = XC at R1 = RC,
Equation 9.7-7
Noting also that RC/R2 = XC/X2,
Equation 9.7-8
or
Equation 9.7-9
EXAMPLE 9.7-2. Approximation of Straight Line for Falling-Rate PeriodRepeat Example 9.7-1, but as an approximation assume a straight line for the rate R versus X through the origin from point XC to X = 0 for the falling-rate period. Solution: RC = 1.51 kg H2O/h · m2 and XC = 0.195. Drying in the falling-rate region is from XC to X2 = 0.040. Substituting into Eq. (9.7-8), This value of 4.39 h compares favorably with the value of 4.06 h obtained in Example 9.7-1 by numerical integration. |
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