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Unformatted text preview: , X* = 0.04 lb H2O/lb dry cotton
Therefore, the initial freemoisture content = 0.95 – 0.04 = 0.91 lb H2O/lb dry cotton
and the freemoisture content at the end of the constantrate drying period = 0.40 – 0.04 = 0.36.
Therefore, the moisture evaporated from the cotton on one tray =
69.4(0.91 – 0.36) = 38.2 lb H2O
This moisture is evaporated from an area = 1.5 m2 or 1.5/0.30482 = 16.2 ft2
During this period, the temperature of the cotton will be at the wetbulb temperature of
the air, which from Figure 18.17 is 118.7oF.
The heat transfer from the air will have to provide latent heat to evaporate the moisture at
118.7oF and sensible heat to increase the temperature of the evaporated moisture to 160oF.
From the steam tables, the heat of vaporization = 1026.5 Btu/lb and the specific heat of
steam = 0.45 Btu/lboF. Thus, the heat that must be transferred from the air =
Q = 38.2[1026.5 + 0.45(160 – 118.7)] = 39,900 Btu/tray
Note that if the sensible heat is neglected, Q = 39,200 Btu/tray (just 1.8% less)
Now, compute the rate of heat transfer to each tray from:
q = hA∆T, where A = 16.2 ft2 and ∆T = 160 – 118.7 = 41.3 oF
Obtain h from (1) of Table 18.6:
h = 0.0204 G0.8, where h is in W/m2K and G is in kg/hm2 Exercise 18.27 (continued)
G = 500 lb/hft2 or 2,440 kg/hm2 , which is just within the range of the correlation.
Therefore, h = 0.0204(2,440)0.8 = 10.5 W/m2K or 1.85 Btu/hft2oF
and q = 1.85(16.2)(41.3) = 1,240 Btu/h
Therefore, the drying time for the constantrate period is, Q/q = 39,900/1,240 = 32.2 h = tc
(c) For the fallingrate period based on the curve in Figure 18.31a, (1841) applies:
tf = ms X c X c
ln
ARc
X The rate of drying per unit area in the constantrate period is,
Rc = 38.2 /[32.2(16.2)] = 0.0732 lb H2O/hft2
ms = mass of bonedry solid per tray = 69.4 lb dry cotton
Xc = critical freemoisture content = 0.40 – 0.04 = 0.36 lb water/lb dry solid
X = final freemoisture content = 0.10 – 0.04 = 0.06 lb water/lb dry solid
Therefore, from (1), tf = 69.4 ( 0.36 )
0.36
ln
= 37.7 h
(16.2)(0.0732)
0.06 (d) The total drying time = 1.0 + 32.2 + 37.7 = 70.9 h or almost 3 days (1) Exercise 18.28
Subject: Batchwise drying of slabs of filter cake in trays.
Given: Filter cake of bonedry density of 1,600 kg/m3 with an initial freemoisture content of
110% (dry basis). Critical freemoisture content = 70% (dry basis) and final freemoisture
content is to be 5% (dry basis). Trays are 1 m long x 0.5 m wide x 3 cm deep. Drying takes
place only from the top. Drying air at 160oC and 1 atm, with a wetbulb temperature of 60oC,
passes across the trays at 3.5 m/s. Fallingrate period follows the type curve in Fig. 18.31a,
based on freemoisture content.
Assumptions: Assume filter cake fills the trays. Neglect sensible heat to increase the
temperature of the evaporated moisture from 60oC to 160oC. Dryingair conditions stay constant
as the air passes over the trays. Cake does not shrink as it dries.
Find: (a) Drying time for the constantrate period.
(b) Drying time...
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 Spring '11
 Levicky
 The Land

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