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| author | priyanka | 2015-06-24 15:03:17 +0530 |
|---|---|---|
| committer | priyanka | 2015-06-24 15:03:17 +0530 |
| commit | b1f5c3f8d6671b4331cef1dcebdf63b7a43a3a2b (patch) | |
| tree | ab291cffc65280e58ac82470ba63fbcca7805165 /599/CH6/EX6.3.a/example6_3_a.sce | |
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diff --git a/599/CH6/EX6.3.a/example6_3_a.sce b/599/CH6/EX6.3.a/example6_3_a.sce new file mode 100755 index 000000000..213bd1e47 --- /dev/null +++ b/599/CH6/EX6.3.a/example6_3_a.sce @@ -0,0 +1,32 @@ +
+clear;
+clc;
+printf("\t Example 6_3_a\n");
+ //part(i)
+//table wt of wet slab,kg -- 5.0 4.0 3.6 3.5 3.4 3.06 2.85
+// drying rate,kg/m^2s-- 5.0 5.0 4.5 4.0 3.5 2.00 1.00
+// X,Dry basis -- 1.0 0.6 .44 0.4 .36 .224 0.14
+// equillibrium relation is given under
+p = [1.0 0.6 .44 0.4 .36 .224 0.14];
+a = [5.0 5.0 4.5 4.0 3.5 2.00 1.00];
+
+i=1; //looping for calc. of 1/N
+while(i<8) //looping begins
+t(i)=1/(a(i));
+i=i+1;
+end //as 1/N plot is needed
+
+plot(p,a,"o-");
+title("Fig.6.19(a) Example3 Drying Rate curve");
+xlabel("X-- Moisture content, X(kg/kg) ---->");
+ylabel("Y-- Drying Rate, N(kg/hr.m^2 ---->");
+xset('window',1);
+plot(p,t,"o-");
+title("Fig.6.19(b) Example3 1/N vs X");
+xlabel("X-- Moisture content, X(kg/kg) --->");
+ylabel("Y-- 1/N, hr,m^2/kg --->");
+//from X=0.6 to 0.44 ,falling rate is non linear and from X=.44 to .14 falling rate is linear
+
+printf("\n from the graph we get critical moisture content as 0.6 kg moisture/kg dry solid");
+
+//end
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