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clear;
clc;
printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 13.3 Page 826 \n')// Example 13.3
// Net rate of Heat transfer to the absorber surface
L = 10 ;//[m] Collector length = Heater Length
T2 = 600 ;//[K] Temperature of curved surface
A2 = 15 ;//[m^2] Area of curved surface
e2 = .5 ;// emissivity of curved surface
stfncnstt = 5.67*10^-8; //[W/m^2.K^4] Stefan-Boltzmann constant
T1 = 1000 ;//[K] Temperature of heater
A1 = 10 ;//[m^2] area of heater
e1 = .9 ;// emissivity of heater
W = 1 ;//[m] Width of heater
H = 1 ;//[m] Height
T3 = 300 ;//[K] Temperature of surrounding
e3 = 1 ;// emissivity of surrounding
J3 = stfncnstt*T3^4; //[W/m^2]
//From Figure 13.4 or Table 13.2, with Y/L = 10 and X/L =1
F12 = .39;
F13 = 1 - F12; //By Summation Rule
//For a hypothetical surface A2h
A2h = L*W;
F2h3 = F13; //By Symmetry
F23 = A2h/A2*F13; //By reciprocity
Eb1 = stfncnstt*T1^4; //[W/m^2]
Eb2 = stfncnstt*T2^4; //[W/m^2]
//Radiation network analysis at Node corresponding 1
//-10J1 + 0.39J2 = -510582
//.26J1 - 1.67J2 = -7536
//Solving above equations
A = [-10 .39;
.26 -1.67];
B = [-510582;
-7536];
X = inv(A)*B;
q2 = (Eb2 - X(2))/(1-e2)*(e2*A2);
printf('\n Net Heat transfer rate to the absorber is = %.1f kW',q2/1000);
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