clear; clc; printf("\t\t\tExample Number 8.17\n\n\n"); // radiation from a hole with variable radiosity // Example 8.17 (page no.-443-446) // solution T1 = 1273;// [K] T5 = 293;// [K] E1 = 0.6; // all the shape factors can be obtained with the aid of figure 8-13(page no.-387) and the imaginary disk surfaces 6 and 7. we have sigma = 5.669*10^(-8);// [W/square meter K^(4)] Eb1 = sigma*T1^(4);// [W/square meter] Eb2 = Eb1;// [W/square meter] Eb3 = Eb2;// [W/square meter] Eb4 = Eb3;// [W/square meter] Eb5 = sigma*T5^(4);// [W/square meter] E2 = E1; E3 = E2; E4 = E3; E5 = 1.0; r = 0.01;// [m] A1 = %pi*r^(2);// [square m] A5 = A1;// [square m] A6 = A1;// [square m] A7 = A1;// [square m] A2 = %pi*2*r*0.01;// [square m] A3 = A2;// [square m] A4 = A2;// [square m] F11 = 0; F55 = 0; F16 = 0.37; F17 = 0.175; F15 = 0.1; F12 = 1-F16; F54 = F12; F13 = F16-F17; F53 = F13; F14 = F17-F15; F52 = F14; F21 = F16*A1/A2; F26 = F21; F45 = F21; F36 = F45; F37 = F36; F22 = 1-F21-F26; F33 = F22; F44 = F22; F31 = F13*A1/A3; F32 = F36-F31; F34 = F32; F43 = F34; F23 = F34; F27 = F26-F23; F46 = F27; F41 = F14*A1/A4; F25 = F41; F42 = F46-F41; F24 = F42; // the equations for the radiosities are now written in the form of equation 8-106, noting that F11 = 0; J5 = Eb5;// [W/square meter] // J1 = (1-E1)*(F12*J2+F13*J3+F14*J4+F15*Eb5)+E1*Eb1 // J2 = [(1-E2)*(F21*J1+F23*J3+F24*J4+F25*Eb5)+E2*Eb2]/(1-F22*(1-E2)) // J3 = [(1-E3)*(F31*J1+F32*J2+F34*J4+F35*Eb5)+E3*Eb3]/(1-F33*(1-E3)) // J4 = [(1-E2)*(F41*J1+F42*J2+F43*J3+F45*Eb5)+E4*Eb4]/(1-F44*(1-E4)) // we have 4 equations with 4 variables which can be solved by matrix method Z = [1 -(1-E1)*F12 -(1-E1)*F13 -(1-E1)*F14;-F21*(1-E2)/(1-F22*(1-E2)) 1 -F23*(1-E2)/(1-F22*(1-E2)) -F24*(1-E2)/(1-F22*(1-E2));-F31*(1-E3)/(1-F33*(1-E3)) -F32*(1-E3)/(1-F33*(1-E3)) 1 -F34*(1-E3)/(1-F33*(1-E3));-F41*(1-E4)/(1-F44*(1-E4)) -F42*(1-E4)/(1-F44*(1-E4)) -F43*(1-E4)/(1-F44*(1-E4)) 1]; C = [E1*Eb1+(1-E1)*F15*Eb5;(E2*Eb2+F25*Eb5*(1-E2))/(1-F22*(1-E2));104859;(E4*Eb4+F45*Eb5*(1-E4))/(1-F44*(1-E4))]; J = Z^(-1)*C; J1 = J(1);// [W/square meter] J2 = J(2);// [W/square meter] J3 = J(3);// [W/square meter] J4 = J(4);// [W/square meter] // the heat transfer can be calculated from equation(8-104): q1 = E1*A1*(Eb1-J1)/(1-E1);// [W] q2 = E2*A2*(Eb2-J2)/(1-E2);// [W] q3 = E3*A3*(Eb3-J3)/(1-E3);// [W] q4 = E4*A4*(Eb4-J4)/(1-E4);// [W] // THE TOTAL HEAT TRANSFER q = q1+q2+q3+q4;// [W] printf("the heat transfer rate is %f W",q); // It is of interest to compare this heat transfer with the value we would obtain by assuming uniform radiosity on the hot surface. we would then have a two-body problem with A1 = %pi+3*(2*%pi);// [square cm] A5 = %pi*10^(-4);// [square cm] F51 = 1.0; E1 = 0.6; E5 = 1.0; // the heat transfer is then calculated from equation(8-43), with appropriate change of nomenclature: q_new = (Eb1-Eb5)*A5/((1/E5)+(A5/A1)*((1/E1)-1));// [w] printf("\n\nthus the assumption of uniform radiosity gives a heat transfer that is %f percent below the value obtained by breaking the hot surface into four parts for the calculations",(q-q_new)*100/q); // let us now consider the case where surface 1 is still radiating at 1000 degree celsius E = 0.6; // the nodal equations for J1 is the same as before but now the equations for J2, J3, and J4 must be written in the form of equation(8-107). when that is done and the numerical values are inserted, we obtain // J1 = 0.252*J2+0.078*J3+0.03*J4+89341 // J2 = 0.5*J1+0.3452*J3+0.09524*J4+24.869 // J3 = 0.1548*J1+0.3452*J2+0.3452*J4+64.66 // J4 = 0.05952*J1+0.0952*J2+0.3452*J3+208.9 // when these equations are solved, we obtain J1 = 1.1532*10^(5);// [W/square meter] J2 = 0.81019*10^(5);// [W/square meter] J3 = 0.57885*10^(5);// [W/square meter] J4 = 0.34767*10^(5);// [W/square meter] // the heat transfer at surface 1 is A1 = %pi*10^(-4);// [square cm] A5 = %pi*10^(-4);// [square cm] A2 = %pi*10^(-4);// [square cm] q1 = (E1*A1)*(Eb1-J1)/(1-E1);// [W] // the temperatures of the insulated surface elements are obtained from T2 = 820;// [degree celsius] T3 = 732;// [degree celsius] T4 = 612;// [degree celsius] // it is of interest to compare the heat transfer calculated above with that obtained by assuming surfaces 2,3 and 4 uniform in temperature and radiosity. equation(8-41) applies for this case: q2 = A1*(Eb1-Eb5)/[((A1+A2+2*A1*F15)/(A5-A1*F15^2))+(1/E1-1)+(A1/A5)*(1/E5-1)];// [w]