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+clear;
+clc;
+printf("\t\t\tExample Number 7.4\n\n\n");
+// heat transfer from fine wire in air
+// Example 7.4 (page no.-333-334) 
+// solution
+
+d = 0.00002;// [m] diameter of wire
+L = 0.5;// [m] length of wire whose temperature is maintained
+Ts = 54;// [degree celsius] surface temperature of wire 
+Pa = 101325;// [Pa] pressure of air
+Ta = 0;// [degree celsius] temperature of air 
+// we first determine the film temperature as
+Tf = (Ts+Ta)/2;// [degree celsius]
+// the properties of interest are thus
+v = 15.69*10^(-6);// [square meter/s]
+k = 0.02624;// [W/m degree celsius]
+Pr = 0.708;// prandtl number
+Beta = 1/(Tf+273);// [K^(-1)]
+g = 9.8;// [square meter/s] acceleration due to gravity 
+// and
+Gr_into_Pr = (g*Beta*(Ts-Ta)*d^(3)*Pr)/(v^(2));
+// from table 7-1 we find
+C = 0.675;
+m = 0.058;
+// so that
+Nu_bar = C*(Gr_into_Pr)^(m);
+h_bar = Nu_bar*k/d;// [W/square meter degree celsius]
+// the heat required is 
+A = %pi*d*L;// [square meter] surface area of wire 
+q = h_bar*A*(Ts-Ta);// [W]
+printf("electric power necessary to maintain the the wire temperature if the length is 0.5 m is %f W",q);