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+clear;
+clc;
+printf("\t\t\tExample Number 8.19\n\n\n");
+// numerical solution for combined convection and radiation(non-linear system) 
+// Example 8.19 (page no.-449-452)
+// solution
+
+l = 0.5;// [m] length of plate 
+b = 0.5;// [m] breadth of plate
+T1 = 1300;// [K] temperature of plate 
+Tinf = 300;// [K] temperature of surrounding
+T4 = Tinf;// [degree celsius]
+h = 50;// [W/square meter] convection heat transfer coefficient
+E1 = 0.8;
+E2 = 0.3;
+E3 = 0.3;
+// using figures 8-12(page no.-386) and 8-14(page no.-387), we can evaluate the shape factors as 
+F12 = 0.2;
+F13 = 0.2;
+F23 = 0.2;
+F32 = 0.2;
+F14 = 1-0.2-0.2;
+F24_L = 1;
+F34_R = 1;
+F21 = F12;
+F31 = F12;
+F24_R = 0.6;
+F34_L = 0.6;
+F11 = 0;
+F22 = 0;
+F33 = 0;
+// J2R = J3L
+// J2L = J3R      From symmetry
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+Eb4 = sigma*T4^(4);// [W/square meter]
+Eb1 = sigma*T1^(4);// [W/square meter]
+J4 = Eb4;// [W/square meter]
+// we now use equation(8-107) to obtain a relation for J1:
+// J1 = (1-E1)*[F12*J2R+F13*J3L+F14*J4]+E1*Eb1
+// but J2R = J3L and F12 = F13 so that
+// J1 = (1-E1)*[2*F13*J2R+F14*J4]+E1*Eb1                                       (a)
+// we use equation (8-108) for the overall energy balance on surface 2:
+// 2*h*(Tinf-T2) = E2*(Eb2-J2R)/(1-E2)+E2*(Eb2-J2L)/(1-E2)
+// 2*h*(Tinf-T2) = E2*(2*Eb2-J2R-J2L)/(1-E2)                                   (b)
+// equation (8-105) is used for surface J2R.
+// J2R = (1-E2)*(F21*J1+F23*J3L+F24_R*J4)+E2*Eb2
+//  But J2R = J3L so that
+// J2R = [(1-E2)*(F21*J1+F24_R*J4)+E2*Eb2]/(1-(1-E2)*F23)                       (c)
+// for surface J2L the equation is 
+// J2L = (1-E2)*(F24_L*J4)+E2*Eb2                                               (d)
+// we now have four equations with four unknowns, J1,J2R,J2L,Eb2, with T2 = (Eb2/sigma)^(1/4).
+// however equation (b) is nonlinear in Eb so we must use a special procedure to solve the set.
+for T2 = 300:0.1:400
+    Z = [1 -(1-E1)*2*F13 0 0;0 -E2/(1-E2) -E2/(1-E2) 2*E2/(1-E2);(1-E2)*F21/(1-(1-E2)*F23) -1 0 E2/(1-(1-E2)*F23);0 0 1 -E2];
+    C = [E1*Eb1;2*h*(Tinf-T2);-F24_R/(1-(1-E2)*F23);(1-E2)*F24_L];
+    S = Z^(-1)*C;
+    Eb2_E = S(4);
+    Eb2_T = sigma*T2^(4);
+    dEb2 = Eb2_E-Eb2_T;
+    if (dEb2>0 & dEb2<5) then
+        J1 = S(1);// [W/square meter]
+        J2R = S(2);// [W/square meter]
+        J2L = S(3);// [W/square meter]
+        Eb2 = S(4);// [W/square meter]
+        T2new = T2;// [K]
+    end
+end
+// the total heat flux lost by surface 1 is 
+q1_by_A1 = h*(T1-Tinf)+(Eb1-J1)*E1/(1-E1);// [W/square meter]
+// for a 0.5 by 0.5 m surface the heat lost is thus 
+q1 = q1_by_A1*l*b;// [W]
+printf("\n\n the heat lost by plate is %f W",q1);