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+clear;

+clc;

+

+// A Textbook on HEAT TRANSFER by S P SUKHATME

+// Chapter 6

+// Heat Transfer by Natural Convection

+

+

+// Example 6.4

+// Page 264

+printf("Example 6.4, Page 264 \n \n");

+

+D = 0.006 ; // [m]

+e = 0.1 ;

+Ti = 800 ; // [C]

+Ta = 1000 ; // [C]

+// Rate at which heat gained = net radiant heat, gives h*(Ta-800) = 1306.0 ; // [W/m^2]

+

+// Using trial and error method

+// Trial 1

+printf("Trial 1 \n");

+// Let Ta = 1000 degree C

+printf("Let Ta = 10000 C \n");

+

+Tm = (Ta+Ti)/2;

+// From table A.2

+v = 155.1*10^-6 ; // [m^2/s]

+k = 0.0763 ; // [W/m K]

+Pr = 0.717 ;

+

+Gr = 9.81*1/1173*(200*D^3)/(v^2);

+Ra = Gr*Pr ;

+

+// From eqn 6.3.2

+Nu = 0.36 + 0.518*(Ra^(1/4))/[1+(0.559/Pr)^(9/16)]^(4/9);

+h = Nu*k/D;

+x = h*(Ta-Ti); // [W/m^2]

+printf("Value of h(Ta-800) = %f W/m^2, which is much larger than the required value of 1306 W/m^2 \n",x);

+

+// Trial 2

+printf("\nTrial 2 \n");

+// Let Ta = 900

+printf("Let Ta = 900 C \n");

+Ra2 = 6.42 ;

+Nu2 = 0.9841 ;

+h2 = 12.15 ;

+x2 = h2*(900-800);

+printf("Value of h(Ta-800) = %f W/m^2, which is a little less than the required value of 1306 W/m^2 \n",x2);

+

+// Trial 3

+printf("\nTrial 3 \n");

+// Let Ta = 910 

+printf("Let Ta = 910 C \n");

+Ra3 = 6.93 ;

+Nu3 = 0.9963 ;

+h3 = 12.33 ;

+x3 = h3*(910-800);

+printf("Value of h(Ta-800) = %f W/m^2 \nThis value is little more than the required value of 1306 W/m^2 \n",x3);

+// Interpolation

+T = 900 + (910-900)*(1306-x2)/(x3-x2);

+printf("\nThe correct value of Ta obtained by interpolation is %f C",T);