blob: 393ec1fda0fbfbed389e734f30b03557b9d94a5e (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
|
clear;
clc;
printf("\t Example 6.9\n");
//determine the drying condition of sample 0.3*0.3 size.sheet lost weight at rate of 10^-4 kg/s until the moisture fell to 60 percent
x1=.75; //moisture content on wet basis
xbar=0.1; //equilllibrium moisture on dry basis
xcr=0.6; //critical moisture content
Ls=0.90; //mass of bone dry solid ais the drying surface
A=0.3*0.3*2; //both upper surafce and lower surface are exposed
//A*Nc=10^-4;
x2=.2; //moisture content on wet basis finally after drying
Xcr=0.6/0.4; //crtical moisture content
X1=3; //moisture content on dry basis intially
X2=0.25; //moisture content on dry basis finally after drying
Xbar=0.1/0.9; //equillibrium moisture
tbar=Ls/(10^-4) * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
printf("\n the time for drying the sheets from 75 to 25 percent moisture under same drying conditions is :%f hr",tbar/3600);
//end
|
