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+clear;

+clc;

+printf("\t\t\tExample Number 7.3\n\n\n");

+// heat transfer from horizontal tube in water

+// Example 7.3 (page no.-333) 

+// solution

+

+d = 0.02;// [m] diameter of heater

+Ts = 38;// [degree celsius] surface temperature of heater

+Tw = 27;// [degree celsius] water temperature

+// the film temperature is 

+Tf = (Ts+Tw)/2;// [degree celsius]

+// from appendix A the properties of water are 

+k = 0.630;// [W/m degree celsius] thermal conductivity

+// and the following term is particularly useful in obtaining the product GrPr product when it is multiplied by d^(3)*DT

+// g*Beta*rho^(2)*Cp/(mu*k) = 2.48*10^(10) [1/m^(3) degree celsius]

+K = 2.48*10^(10);// [1/m^(3) degree celsius]

+Gr_into_Pr = K*(Ts-Tw)*d^(3);

+// using table 7-1 (page number -328), we get 

+C = 0.53;

+m = 1/4;

+// so that

+Nu = C*(Gr_into_Pr)^(1/4);

+h = Nu*k/d;// [W/square meter degree celsius] convection heat transfer coefficient

+// the heat transfer is thus

+q_by_L = h*%pi*d*(Ts-Tw);// [W/m]

+printf("free-convection heat loss per unit length of heater is %f W/m",q_by_L);

+