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diff --git a/599/CH6/EX6.2/example6_2.sce b/599/CH6/EX6.2/example6_2.sce new file mode 100755 index 000000000..1729d9aad --- /dev/null +++ b/599/CH6/EX6.2/example6_2.sce @@ -0,0 +1,46 @@ +
+clear;
+clc;
+printf("\t Example 6.2\n");
+//table X*100,(kgmoisture/kg dry solid) N*100 (kg moisture evaporated /hr*m^2)
+
+// 35 30
+// 25 30
+// 20 30
+// 18 26.6
+// 16 23.9
+// 14 20.8
+// 12 18
+// 10 15
+// 9 9.7
+// 8 7
+// 7 4.3
+// 6.4 2.511111
+
+
+Ls=262.5; //mass of bone dry solid ais the drying surface
+A=262.5/8; //both upper surafce and lower surface are exposed
+Nc=0.3; //in kg/m^2*hr
+x2=.06; //moisture content on wet basis finally after drying
+x1=.25; //moisture content on wet basis finally after drying
+Xcr=0.20; //crtical moisture content
+X1=x1/(1-x1); //moisture content on dry basis intially
+X2=x2/(1-x2); //moisture content on dry basis finally after drying
+Xbar=0.025; //equillibrium moisture
+
+t1=Ls/(A*Nc) *(X1-Xcr); //so for constant rate period
+
+//for falling rate period we find time graphically
+p = [.20 .18 .16 .14 .12 .10 .09 .08 .07 .064];
+a = [3.3 5.56 6.25 7.14 8.32 10.00 11.11 12.5 14.29 15.625];
+
+plot(p,a,"o-");
+title("Fig.6.18 Example2 1/N vs X for fallling rate period");
+xlabel("X-- Moisture content, X(kg/kg)");
+ylabel("Y-- 1/N, hr,m^2/kg");
+
+Area=1.116; //area under the curve
+t2=Area *Ls/A; //falling rate period we find time graphically
+ttotal=t1+t2; //total time for drying
+printf("\n the total time for drying the wet slab on wet basis is :%f min",ttotal);
+//end
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