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+clear;

+clc;

+printf("\t\t\tExample Number 7.10\n\n\n");

+// heat transfer across water layer

+// Example 7.10 (page no.-346-347) 

+// solution

+

+L = 0.5;// [m] length of square plate

+d = 0.01;// [m] seperation between square plates

+T1 = 100;// [degree F] temperature of lower plate

+T2 = 80;// [degree F] temperature of upper plate

+// we evaluate properties at mean temperature of 90 degree F and obtain, for water

+k = 0.623;// [W/m degree celsus]

+// and the following term is particularly useful in obtaining the product GrPr 

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

+// the Grashof-prandtl number product is now evaluated using the plate spacing of 0.01 m as the characterstic dimension

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

+Gr_into_Pr = K*(T1-T2)*(5/9)*d^(3);

+// now, using equation 7-64 and consulting table 7-3(page no.-344) we obtain

+C = 0.13;

+n = 0.3;

+m = 0;

+// therefore, equation (7-64) becomes

+Ke_by_K = C*Gr_into_Pr^(n);

+// the effectve thermal conductivity is thus

+ke = k*Ke_by_K;// [W/m degree celsius]

+// and the heat transfer is

+A = L^(2);// [square meter] area of plate

+q = ke*A*(T1-T2)*(5/9)/d;// [W]

+printf("heat lost by the lower plate is %f W",q);

+