blob: 2cb4e9f1cdacd50d85930106b9bcbf45724ce4bb (
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
26
27
28
29
30
|
clear;
clc;
printf("\t Example 6_3_b\n");
//part(ii)
w1=5; //wet of wet solid
c1=.5/(1-.5); //moisture content per kg wet solid
w2=5*0.5; //moisture for 5kg wet solid
w3=w1-w2; //weight of dry solid
xbar=0.05; //equillibrium moisture content
Xbar=xbar/(1-xbar); //equillibrium moisture content
Ls=2.5; //mass of bone dry solid ais the drying surface
A=5; //both upper surafce and lower surface are exposed
Nc=0.6; //in kg/m^2*hr
//from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
X2=.15/(1-.15);
Xcr=.6; //kg moisture per kg dry solid
//so we can find time fro drying from 0.6 to .44 graphically and then for X=.44 to .1765
X1=1; //moisture content on dry basis intially
t1=Ls/(A*Nc) *(X1-Xcr); //time taken for constant drying rate(fromX=1 to .6)
X1=.44; //moisture content on dry basis
t2=(Ls/(A*Nc))*((Xcr-Xbar)*log((X1-Xbar)/(X2-Xbar)));
t3=0.0336*Ls/Nc; //fro graph we get from X=.6 to .44
ttotal=t1+t2+t3; //total time for drying the wet slab
printf("\n the total time for drying the wet slab to 15 percent moisture on wet basis is :%f min",ttotal*60);
//end
|
