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clc;
clear;
//Example 4.20
sigma=5.67*10^-8 //[W/sq m.K^4]
T1=750 //[K]
T2=500 //[K]
e1=0.75;
e2=0.5;
//Heat transfer without shield :
Q_by_a=sigma*(T1^4-T2^4)/((1/e1)+(1/e2)-1) //[W/sq m]
//Heat transfer with shield:
R1=(1-e1)/e1 //Resistance 1
F13=1;
R2=1/F13 //Resistance 2
e3=0.05
R3=(1-e3)/e3 //Resistance 3
R4=(1-e3)/e3 //Resistance 4
F32=1;
R5=1/F32 //Resistance 5
R6=(1-e2)/e2 //Resistance 6
Total_R=R1+R2+R3+R4+R5+R6 //Total resistance
Q_by_as=sigma*(T1^4-T2^4)/Total_R //[W/sq m]
Red=(Q_by_a-Q_by_as)*100/Q_by_a //Reduciton in heat tranfer due to shield
printf("\n Reduction in heat transfer rate as a result of radiaiotn shield is %f percent",Red);
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