blob: e5a1a3c91ef9ed13e6d1950aabd1458934bfd63e (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
|
clear;
clc;
printf("\t Example 6.10\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
//Ls/ A*Nc is unknown;
Xcr=0.16; //crtical moisture content
X1=.33; //moisture content on dry basis intially
X2=0.09; //moisture content on dry basis finally after drying
Xbar=.05; //equillibrium moisture
tbar=7; //time needed to dry from 33 to 9 percent on bone dry basis
//let Ls / A*Nc be = p
p=poly([0],'p'); //calc. of Ls / A*Nc be = p value
x=roots(tbar-p * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar))));
//new X1 AND X2 are now given as follows
X1=0.37; //new moisture content on dry basis intially
X2=0.07; //new moisture content on dry basis finally after drying
tbar=x * ((X1-Xcr)+(Xcr-Xbar)*log((Xcr-Xbar)/(X2-Xbar)));
printf("\n the time for drying the sheets from 33 to 9 percent moisture under same drying conditions is :%f hr",tbar);
//end
|
