This preview shows page 1. Sign up to view the full content.
Unformatted text preview: for the fallingrate period
Analysis: (a) The constantrate drying flux, from (1834) is,
Rc = h (Tg − Tw )
vap
∆H w (1) Tg = Gas temperature = 160oC = 320oF, Tw = drying temperature = 60oC = 140oF
v
From the steam tables, at 140oF, ∆H wap = 1014.1 Btu/lb
For the heattransfer coefficient, h, use (1838) in Table 18.6:
h = 0.0204 G0.8, where h is in W/m2K and G is in kg/hm2
G = vρ, v = 3.5 m/s = 3.5(3600) = 12,600 m/h
The humidity of the air, from a hightemperature humidity chart in Perry’s Handbook
= H = 0.10 lb water/lb dry air.
Use the humid volume to get the air density. From (1810),
vH = 0.730 ( 320 + 460 ) (1) 1
0.10
+
= 24.7 ft3/lb dry air
28.97 18.02 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 = 12,600(0.714) = 9,000 kg/hm2, which is within the range of (1) in
Table 18.6.
Therefore, h = 0.0204(9,000)0.8 = 29.7 W/m2K or 5.24 Btu/hft2oF Exercise 18.28 (continued)
From (1), Rc = 5.24 ( 320 − 140 ) From (1842), 1014.1 = 0.93 lb/hft2 ms
( Xo − Xc )
ARc
Compute the mass of bonedry solid in one tray = ms
Tray volume = 1(0.5)(0.03) = 0.015 m3
Therefore ms = 1,600(0.015) = 24 kg = 53 lb of dry solid
The moisture contents are given on a freemoisturecontent basis.
The tray area = A = 0.5(1) = 0.5 m2 = 5.38 ft2
Substituting into (2),
53
tc =
(1.10 − 0.70 ) = 4.24 h
5.38(0.93)
tc = (2) (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
ln
= 19.6 h
(5.38)(0.93)
0.05 (3) Exercise 18.29
Subject: Batchwise drying of extrusions 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). The cake is extruded into cylindricalshaped pieces measuring
¼inch in diameter by 3/8inch long, which are placed in trays that are 6 cm high by 1 m long by
0.5 m wide. Drying air at 160oC and 1 atm, with a wetbulb temperature of 60oC, passes through
the bed of pieces at a superficial velocity of 1.75 m/s. Fallingrate period follows the type curve
in Fig. 18.31a, based on freemoisture content. Bed porosity is 50%.
Assumptions: Neglect sensible heat to increase the temperature of the evaporated moisture
from 60oC to 160oC. Dryingair conditions stay constant as the air through the pieces on the
trays. Cake does not shrink as it dries.
Find: (a) Drying time for the constantrate period.
(b) Drying time for the fallingrate period.
Analysis: (a) The constantrate drying flux, from (1834) is,
Rc = h (Tg − Tw ) (1) vap
∆H w Tg = Gas temperature = 160oC = 320oF, Tw = drying temperature = 60oC = 140oF
v
From the steam tables, at 140oF, ∆H wap = 1014.1 Btu/lb
For th...
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