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Unformatted text preview: r = 1.09 kJ/kgK = ( CP ) g
Density of the air = 0.942 kg/m3 = ρ g
Bed crosssectional area = 0.0929 m2
Volumetric flow rate of solid particles = 0.0283 m3/min
Dry density of calcium carbonate = 1500 kg/m2 = ( ρb )ds
Mass flow rate of calcium carbonate = 42.5 kg/min = 0.708 kg/s (dry basis)
Solve this exercise using SI units, but with temperature in oC and time in sec Zone 1: H = 2 in = 0.0508 m,
L1 = 8 ft = 2.44 m
From (1874), gas temperature leaving Zone 1 is
Tgo = 37.8 + ( 76.7 − 37.8 ) exp − 0.188(395)(0.0508)
= 50oC
0.942(1.09)(2) The moisturecontent distribution is given by (1875). Applying this equation at the end of Zone
1 gives the following, where x is distance along the bed, which equals 8 ft = 2.44 m and z is the
height measured from the bottom of the bed for upward flow of air. Exercise 18.40 (continued)
X 1 { L1 , z} = 0.30 1 − 2.44 ( 0.188 )( 395 )( 76.7 − 37.8 )
0.00508 ( 2413)(1500 ) exp − 0.188 ( 395 ) z 0.942 (1.09 )( 2 ) z, m
0.0000
0.0127
0.0254
0.0381
0.0508 Using a spreadsheet, the following results are obtained: X, %
18.5
22.7
25.4
27.1
28.2 Zone 2: Use (1878), with x measured at L1 and height given by (H – z), because of downward
flow from z = H to z = 0. Thus,
X 2 {L2 , z} = X 1 { L1 , z} 1 − L2 ha (Tgi − Tw )
S ∆H vap
w ( ρb )ds exp − ha ( H − z ) ρ g ( CP ) g us (1) where L2 = length of zone 2 = 2.44 m and, again, z is measured from the bottom of the bed.
Using a spreadsheet, the following results are obtained:
z, m
X, %
0.0000
17.4
0.0127
20.5
0.0254
21.5
0.0381
20.5
0.0508
17.4 Zone 3: Use (1878), with x measured at L2 and height given by z because of upward flow from
z. = H to z = 0. Thus, using an equation similar to (1), but with X3 in terms of X2.
Using a spreadsheet, the following results are obtained:
z, m
X, %
0.0000
10.7
0.0127
15.6
0.0254
18.2
0.0381
18.6
0.0508
16.3
The final average moisture content is obtained from integrating the values of X at the end of zone
3 with (1876), which gives Xavg = 16.5% or 0.165 lbH2O/lb dry solid, which is greater than the
critical moisture content so that drying is in the constantrate period as assumed. Exercise 18.41
Subject: Drying of extruded filter cake of calcium carbonate with a throughcirculation belt
dryer having 2 zones.
Given: Calcium carbonate extruded into cylindrical pieces of 3/8inch diameter and 1/2inch
length with an initial moisture content of 30% (dry basis) and a critical moisture content of 10%
(dry basis). Belt dryer is 6ft wide with two drying zones of 12ft long each, running with a belt
speed, S, of 1 ft/min (0.00508 m/s). Carbonate bed height on the belt is 2inches with an external
porosity of 50%. Air at 170oF (76.7oC) and 10% relative humidity flows through the bed at a
superficial velocity, us,of 2 m/s, upward in the first zone and downward in zone 2.
Assumptions: Negligible preheat period. Constantrate drying at the wetbulb temperature of
the entering air.
Find: Moisture content distribution with height at the end of each zone and the final average
moisture content.
Analysis: This exercise is exactly the same as Example 18.18 except that the diameter of the
extrusion is 3/8 inch instead of 2/8 inch. Thus, the following preliminary calculations from
Example 18.18 apply here:
Entering air has a wetbulb temperature of 110.2oF = 37.8oC = Tw
Entering air has a humidity of 0.0265 lb H2O/lb dry air
vap
Heat of vaporization of water at 110.2oF = 2413 kJ/kg = ∆ H w
Specific heat to the air = 1.09 kJ/kgK = ( CP ) g Density of the air = 0.942 kg/m3 = ρ g
Bed crosssectional area = 0,0929 m2
Dry density of calcium carbonate = 1500 kg/m2 = ( ρb )ds
Preliminary calculations:
First, calculate the effective particle diameter, dp, the diameter of a sphere having the
same particle surface area, with D = 3/8 inch, L = 4/8
πD 2
πd 3 = 2
+ πDL
p
4
Solving, dp = 0.506 inch = 0.013 m
Now, compute the heattransfer coefficient from Table 18.6. Need to calculate the bed
Reynolds number to determine whether (3) or (4) applies. Using data in Example 18.8,
NRe = (0.013)(1.96)/2x105 = 1,274
Therefore, (3) in Table 18.6 applies:
h = 0.151(14,300 / 2 ) / ( 0.013) = 168 J/sm2K or 0.168 kJ/sm2K
Now compute the extrusion surface area pe...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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