This preview shows page 1. Sign up to view the full content.
Unformatted text preview: e heattransfer coefficient, h, use (3) or (4) for throughcirculation in a packed bed,
in Table 18.6, where the choice depends on the value of the bed Reynolds number, N Re = d pG (2) µ where, from Example 18.8, dp is the diameter of a sphere of the same surface area as the
particle. From (1) of Example 18.8, with particle diameter, D, and particle length, L,
πD 2
πd = πDL + 2
4
2
p D2
or d p = DL +
2 1/ 2 1 3 0.252
=
+
48
2 1/ 2 = 0.354 in = 0.00897 m The superficial mass velocity = G = vρ, with v = 1.75 m/s = 1.75(3600) = 6,300 m/h
For the gas density, calculate the humid volume.
The humidity of the air, from a hightemperature humidity chart in Perry’s Handbook
= H = 0.10 lb water/lb dry air. Exercise 18.29 (continued)
From (1810),
0.730 ( 320 + 460 )
1
0.10
vH =
+
= 24.7 ft3/lb dry air
28.97 18.02
(1)
or 24.7/1.10 = 22.5 ft3/lb moist air or 1.40 m3/kg moist air
Therefore, ρ = 1/1.40 = 0.714 kg/m3
Therefore, G = vρ = 6,300(0.714) = 4,500 kg/hm2
From Example 18.8, take the air viscosity = µ = 2 x 105 kg/ms or 0.072 kg/mh
Therefore, bed Reynolds number, from (2) is,
N Re = 0.00897 ( 4,500 )
0.072 = 561 Therefore, (3) in Table 18.6 applies:
h = 0.151 G 0.59
45000.59
= 0.151
= 149 W/m2K or 26.3 Btu/hft2oF
0.41
dp
0.00897 0.41
From (1), Rc = 26.3 ( 320 − 140 )
1014.1 = 4.67 lb/hft2 From (1842), the time for the constantrate drying period is,
m
tc = s ( X o − X c )
(2)
ARc
Compute the mass of bonedry solid in one tray = ms
Tray volume = 1(0.5)(0.06) = 0.03 m3
For a porosity of 0.5, ms = 1,600(0.03)(0.5) = 24 kg = 53 lb of dry solid
The moisture contents are given on a freemoisturecontent basis.
The area, A, in (2) is the total area of the particles in the bed on the tray.
Volume of one extrusion = πD2L/4 = 3.14(0.25)2(0.375)/4 = 0.0184 in3 or 3.02 x 107 m3
The extrusions on one tray fill a volume of 0.015 m3 (50% of the tray volume)
Therefore, the number of extrusions = 0.015/3.02 x 107 = 49,750
The surface area of one extrusion is,
πD 2
= 3.14[(0.25)(0.375) + 0.5(0.25)2] = 0.393 in2 = 0.00273 ft2
πDL + 2
4
Therefore, total surface area of the extrusions on a tray = 49,750(0.00273) = 136 ft2
Substituting into (2),
tc = 53
(1.10 − 0.70 ) = 0.0334 h or 2 min
136(4.67) Exercise 18.29 (continued)
(b) For the fallingrate period based on the curve in Figure 18.31a, (1841) applies:
tf = ms X c X c
ln
ARc
X Xc = critical freemoisture content = 0.70 lb water/lb dry solid
X = final freemoisture content = 0.05 lb water/lb dry solid
Therefore, from (3), tf = 53 ( 0.70 )
0.70
= 0.154 h or 9.2 min
ln
(136)(4.67)
0.05 (3) Exercise 18.30
Subject: Tray drying in the constant and fallingrate periods
Given: Takes 5 hours to dry a wet solid in a tray from 36 to 8 wt% on dry basis. Critical and
equilibrium moisture contents are 15 and 5 wt% dry basis, respectively. Negligible preheat time.
Straightline fallingrate period.
Assumptions: Dryingair conditions stay constant. Cake does not shrink as it dries.
Find: Drying time for same we...
View
Full
Document
This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

Click to edit the document details