This preview shows page 1. Sign up to view the full content.
Unformatted text preview: al water content of 40 wt% on dry basis (Xin = 0.40).
Dried in a tunnel dryer at 1 atm. Fibers brought to equilibrium with air at 25oC and a relative
humidity of 30%. Equilibrium moisture content given in Figure 18.24.
Find: The kg of moisture evaporated per kg of bonedry fibers.
Analysis: From Figure 18.24, the final moisture content = Xout = X* = 0.07 lb water/lb dry solid
Therefore, evaporation = Xin  Xout = 0.40 – 0.07 = 0,33 lb water/lb dry solid Exercise 18.25
Subject: Slow drying of lumber using equilibrium moisture content
Given: Initial moisture content = Xin = 0.50 lb water/lb dry lumber. Final air conditions of 25oC
and 1 atm, with a relative humidity of 40%.
Assumptions: Applicability of Figure 18.24, 10 – Lumber.
Find: (a) Unbound moisture before drying
(b) Bound moisture before drying
(c) Free moisture before drying
(d) lb moisture evaporated per lb of bonedry wood
Analysis: The definitions of moisture content are shown schematically in Figure 18.23.
Total moisture content = XT = Xin = 0.50 lb water/lb dry lumber
Figure 18.24 applies at the desired conditions of 25oC and 1 atm.
Bound moisture content = XB = 0.306 lb water/lb dry lumber
Equilibrium moisture content = X* = 0.075 lb water/lb dry lumber
(a) Unbound moisture content = XT  XB = 0.500 – 0.306 = 0.194 lb water/lb dry lumber
(b) Bound moisture content = XB = 0.306 lb water/lb dry lumber
(c) Free moisture content = XT – X*= 0.500 – 0.075 = 0.425 lb water/lb dry lumber
(d) Moisture evaporated is the free moisture = 0.425 lb water/lb dry lumber Exercise 18.26
Subject: Drying of cotton cloth in a closed air system
Given: 50 lb of cotton cloth at 100oF and 1 atm with a total moisture content of 20 wt% (dry
basis). Cloth is hung in a closed room containing 4,000 ft3 of air at 100oF and 1 atm with an
initially wetbulb temperature of 69oF. Air is kept at 100oF and equilibrium is achieved in the
room.
Assumptions: Equilibrium moisture content of cotton cloth at 100oF is given by Figure 18.25.
Neglect the increase in air pressure. Air is not vented and no new air enters the room.
Find: Final moisture content of the cotton cloth and the final relative humidity of the air. Also,
determine the increase in air pressure.
Analysis: The relative humidity of the air will increase, so an iterative solution is required.
Initial composition of the wet cloth:
% water in solid on a wet basis = 20/(20+100) = 0.1667
Therefore, the initial wet solid is:
0.1667(50) = 8.33 lb water
50 – 8.33 = 41.67 lb waterfree cotton cloth
Initial air conditions:
From Figure 18.17, humidity = 0.008 lb water/lb dry air
From (1810), the humid volume is:
vH = 0.730 (100 + 460 ) (1) 1
0.008
+
= 14.29 ft3/lb dry air
28.97 18.02 Therefore, have 4,000/14.29 = 280 lb dry air
Initial moisture in the air = 0.008(280) = 2.24 lb water
Now, perform iterations to determine final equilibrium conditions for a room containing:
41.67 lb waterfree cotton cloth
280 lb air on the dry basis
8.33 + 2.24 = 10.57 lb water
If all the water were evapor...
View Full
Document
 Spring '11
 Levicky
 The Land

Click to edit the document details