Unformatted text preview: t material from 40 to 7 wt% on dry basis.
Analysis: From (1842),
tT = tc + t f = ms
ARc ( X o − X c ) + X c ln Xc
X final (1) ms
from given data for the 5hour test. All moisture contents are on a freeARc
moisture basis. Therefore, Xo = 0.36 – 0.05 = 0.31, Xc = 0.15 – 0.05 = 0.10, Xfinal = 0.08 – 0.05
= 0.03
Substituting into a rearrangement of (1), Solve for ms
=
ARc tc + t f
X
( X o − X c ) + X c ln c
X final = 5
0.10
( 0.31 − 0.10 ) + 0.10 ln
0.03 = 15.13 h Use this value for the new conditions, with Xo = 0.40 – 0.05 = 0.35, Xc = 0.15 – 0.05 = 0.10,
Xfinal = 0.07 – 0.05 = 0.02
tT = tc + t f = (15.13) ( 0.35 − 0.10 ) + 0.10 ln 0.10
0.02 = 6.2 h Exercise 18.31
Subject: Tunnel drying of a wet solid on trays with drying by crossflow of air from both sides.
Given: Trays measuring 1.5 m long by 1.2 m wide by 25 cm deep. Initial total moisture content
= 116 wt% dry basis and final average total moisture content = 10 wt% dry basis. Air is at 90oF,
1 atm, and a relative humidity of 15%. By interpolation of given data, equilibrium moisture
content = 3.1 wt% dry basis. Dryingtime data plotted below.
Assumptions: Critical moisture content is independent of drying conditions. Drying rate is
proportional to the difference between drybulb and wetbulb temperatures of the air.
Find: Drying time for 110 to 10 wt% dry basis using air at 125oF and 20% relative humidity.
Analysis: The following is a spreadsheet plot of the dryingtime data for air at 90oF and an RH
of 15% starting from 116% total moisture. Two lines are shown. The upper line is the total and
the bottom line is the free moisture content = total – 3.1%.
140 120 % Moisture Content, dry basis 100 80 60 40 20 0
0 100 200 300 400 500 600 700 800 900 Time, minutes From this plot, the critical freemoisture content, Xc = approximately 57.4 – 3.1 = 54.3 wt% dry
basis at 211 minutes. During the constantrate drying period, using the given dryingrate data,
The rate of drying = (1.16 – 0.574)/211 = 0.00278 lb H2O/lb dry solid – min Exercise 18.31 continued
In the constantrate period, Rc = h (Tg − Tw )
∆H vap
w in lb H2O/ft2 of drying surface – min For the test data with air at 90oF, 1 atm, and RH of 15%, from Fig. 18.17, Tw = 60oF.
v
From the steam tables, ∆H wap = 1206.6 Btu/lb
For the new conditions with air at 125oF, 1 atm, and RH of 20%, Tw = 85oF
v
From the steam tables, ∆H wap = 1045.8 Btu/lb
Therefore, the drying time for the constantrate period for the new conditions, by ratio using the
above expression for the rate of drying is, using total moisture content differences:
tc = 211 90 − 60
125 − 85 1045.8
1206.6 110 − 57.4
= 123 min
116 − 57.4 In the fallingrate period, since there is little change in the final moisture content between the test
conditions and the new conditions, since both terminate at a total moisture content of
approximately 10 wt%, ratio on rates using the temperature differences and the heat of
vaporization. For the test data, tf = 822 – 211 = 611 minutes.
Therefore to get to 10.2 wt%, t f = 611 90 − 60
125...
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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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