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+clear;
+clc;
+printf("\t\t\tExample Number 8.7\n\n\n");
+// surface in radiant balance
+// Example 8.7 (page no.-404-405) 
+// solution
+
+w = 0.5;// [m] width of plate 
+L = 0.5;// [m] length of plate
+sigma = 5.669*10^(-8);// [W/square meter K^(4)]
+// from the data of the problem 
+T1 = 1000;// [K] temperature of first surface
+T2 = 27+273;// [K] temperature of room
+A1 = w*L;// [square meter] area of rectangle
+A2 = A1;// [square meter] area of rectangle
+E1 = 0.6;// emissivity of surface 1
+// although this problems involves two surfaces which exchange heat and one which is insulated or re-radiating, equation (8-41) may not be used for the calculation because one of the heat-exchanging surfaces(the room) is not convex. The radiation network is shown in figure example 8-7(page no.-404) where surface 3 is the room and surface 2 is the insulated surface. note that J3 = Eb3 because the room is large and (1-E3)/(E3*A3) approaches zero.Because surface 2 is insulated it has zero heat transfer and J2 = Eb2. J2 "floats" in the network and is determined from the overall radiant balance. 
+// from figure 8-14(page no.-387) the shape factors are 
+F12 = 0.2;
+F21 = F12;
+// because
+F11 = 0;
+F22 = 0;
+F13 = 1-F12;
+F23 = F13;
+// the resistances are 
+R1 = (1-E1)/(E1*A1);
+R2 = 1/(A1*F13);
+R3 = 1/(A2*F23);
+R4 = 1/(A1*F12);
+// we also have
+Eb1 = sigma*T1^(4);// [W/square meter]
+Eb3 = sigma*T2^(4);// [W/square meter]
+J3 = Eb3;// [W/square meter]
+// the overall circuit is a series parallel arrangement and the heat transfer is 
+R_equiv = R1+(1/[(1/R2)+1/(R3+R4)]);
+q = (Eb1-Eb3)/R_equiv;// [W]
+// this heat transfer can also be written as q = (Eb1-J1)/((1-E1)/(E1*A1))
+// inserting the values 
+J1 = Eb1-q*((1-E1)/(E1*A1));// [W/square meter]
+// the value of J2 is determined from proportioning the resistances between J1 and J3, so that 
+// (J1-J2)/R4 = (J1-J3)/(R4+R2)
+J2 = J1-((J1-J3)/(R4+R2))*R4;// [W/square meter]
+Eb2 = J2;// [W/square meter]
+// finally, we obtain the temperature of the insulated surface as
+T2 = (Eb2/sigma)^(1/4);// [K]
+printf("temperature of the insulated surface is %f K",T2);
+printf("\n\n heat lost by the surface at 1000K is %f kW",q/1000);
+
+
+
+