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Unformatted text preview: e at the surface of the
wet solid is higher than the wetbulb temperature by a few degrees, thus reducing slightly the
temperature driving force for heat transfer from the hot air to the wet surface of the solid.
However, the overall heattransfer coefficient for heating from both the top and bottom is
increased considerably. The result is a significantly higher drying rate.
From Eqs. (1834) and (1835), the dryingrate flux in the constantrate drying period,
Rc, in the absence of heattransfer to the tray bottom and radiation to the wet surface of the solid
is:
Rc = h (Tg − Tv )
∆H vvap = M dry air k y ( H v − H d ) (1) where, here, the subscript v refers to the wetbulb temperature, w.
Assuming the area for heat transfer to the top and bottom of the tray is the same, then, if
convection to the uninsulated bottom of the tray is taken into account, with subsequent
conduction through the tray thickness and the wet solid, then h in (1) is replaced by an overall
coefficient:
U = hc ,t + 1
1
t
t
+ t + ws
hc ,b kt kws (2) Assume that hc,t at the top = hc,b at the bottom = hc. Assume that the thermal resistance
of the metal tray thickness, tt / kt , is negligible compared to the thermal resistance of the wet
solid, tws / kws , where t is thickness and k is thermal conductivity. Then (2) becomes: Exercise 18.35 (continued)
U = hc + 1
1 tws
+
hc kws (3) Now U in (3) is greater than h in (1). Therefore, the ratio U / (kyCsMB) > h / (kyCsMB) , the
psychrometric ratio, and the temperature of the evaporating moisture will be higher than the wetbulb temperature of the hot air, thus reducing the temperature driving force.
If we also account for radiation from the bottom of a tray to the tray below, the overall
heat transfer coefficient for heat transfer to the tray below will be even higher than U, because an
additional term for a radiation coefficient will be added to the righthand side of (3). Therefore,
the temperature of the evaporating moisture will be further increased. Exercise 18.36
Subject: Tunnel drying of raw cotton.
Given: Dry 30 lb/h of raw cotton (dry basis) at 70oF and a moisture content of 100% (dry basis)
in a tunnel dryer with a countercurrent flow of 1,800 lb/h of air (dry basis) entering at 200oF, 1
atm, and a relative humidity of 10%. The cotton will exit at 150oF with a moisture content of
10% (dry basis). Specific heat of dry cotton = 0.35 Btu/lboF.
Assumptions: Adiabatic drying conditions. Specific heat of dry air = 0.24 Btu/ lb oF. Specific
heat of steam = 0.45 Btu/lboF.
Find: (a) The rate of evaporation of moisture.
(b) The outlet temperature of the air.
(c) The rate of heat transfer.
Analysis:
(a) The rate of evaporation of moisture = 30(1.00 – 0.10) = 27 lb/h
(b) The air must supply sensible heat and latent heat. Assume that moisture evaporation
occurs at the wetbulb temperature of the air. From Figure 18.17, this is 117oF. Therefore, heat
the wet cotton from 70oF to 117oF, evaporate the 27 lb/h of moisture at 117oF, heat (or cool) the
evaporated moisture to the final gas temperature, and heat...
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 Spring '11
 Levicky
 The Land

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