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+clear;

+clc;

+printf("\t\t\tExample Number 7.6\n\n\n");

+// cube cooling in air

+// Example 7.6 (page no.-336) 

+// solution

+

+L = 0.2;// [m] side length of cube

+Ts = 60;// [degree celsius] surface temperature of cube

+Ta = 10;// [degree celsius] air temperature

+// this is an irregular solid so we use the information in the last entry of table 7-1(page no.-328) in the absence of a specific correlation for this    geometry. 

+// the properties were evaluated as

+v = 17.47*10^(-6);// [square meter/s]

+k = 0.02685;// [W/m degree celsius]

+Pr = 0.70;// prandtl number

+Beta = 3.25*10^(-3);// [K^(-1)]

+g = 9.8;// [square meter/s] acceleration due to gravity 

+// the characteristic length is the distance a particle travels in the boundary layer, which is L/2 along the bottom plus L along the side plus L/2 on  the top or

+Gr_into_Pr = (g*Beta*(Ts-Ta)*(2*L)^(3)*Pr)/(v^(2));

+// from the last entry in table 7-1 we find

+C = 0.52;

+n = 1/4;

+// so that

+Nu = C*(Gr_into_Pr)^(n);

+h_bar = Nu*k/(2*L);// [W/square meter degree celsius]

+// the cube has six sides so the area is 

+A = 6*L^(2);// [square meter]

+// the heat required is 

+q = h_bar*A*(Ts-Ta);// [W]

+printf("heat transfer is %f W",q);