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