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clc();
clear;
// To find the division of the heating surface
t1 = 2500; // temperature of contenets of the bottle in F
t2 = 600; // Ambient temperature in F
e1 = 0.048; // Interchange factor in 1800 F
e2 = 0.044; // Interchange factor in 600 F
e = 0.94; // Emmisivity of walls
p = 1; // Emperical factor
F = 2*0.88; // Shape factor
s = 0.173*10^-8; // Stephens boltzmanns constant
h = s*e*p*F*((t1+460)^4-(t2+460)^4)/(%pi*(t1-t2));
// Heat transfer coefficient
// Heat transfer for the tubes within the convective surface
// Radiation of CO2 and waterin the combustion gases
L = 0.5; // Eqivalent length of gas layer
Tg = 1800; // Gas temperature in F
Tw = 600; // Surface temperature of tubes in F
// From the table the emmisivity of carbon dioxide can be known
ec1 = 0.06; // Emmmisivity of CO2 at 1800F
ec2 = 0.055; // Emmisivity of Co2 at 600F
ew = 0.8; // Emmisivity of tube wall
qc = s*ew*p*(ec1*(Tg+460)^4-ec2*(t2+460)^4);
// Heat loss by carbon dioxide in Btu/hr
// From the table the emmisivity of water can be known
eh1 = 0.0176; // Emmmisivity of water at 1800F
eh2 = 0.0481; // Emmisivity of water at 600F
qh = s*ew*p*(eh1*(Tg+460)^4-eh2*(t2+460)^4);
// Heat loss by water in Btu/hr
qg = qc + qh; // Heat heat flow by gas radiation
hg = qg/(Tg-t2); // Heat transfer coeffcoent by gas radiation
printf("The heat transfer coefficient by gas radiation is %.2f Btu/hr-ft^2 \n",hg);
// Heat transfer by convection can be found out using values iun the table
hc = 8.14; // Heat transfer by convection in Btu/hr-ft^2-F
printf(" The heat transfer coefficient by gas radiation is %.2f Btu/hr-ft^2\n",hc);
ht = hc + hg; // Total heat transfer coefficient for convective surface
printf("The covective surface have greater heat transfer coefficients than the radiating surface. Therefore it is advantageous to line the whole combustion chamber with narrowly spaced cooling tubes");
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