During cold weather shop floors or restaurant patios are som

During cold weather, shop floors or restaurant patios are sometimes warmed by heat tubes. Natural gas is injected inside the heat tube and burned causing the tube to get very hot and emit thermal radiation to warm those below. To redirect heat that would be lost from the top, a reflective half cylinder cover is installed above. The illustration below left shows a typical design. Consider a heat tube 6in diameter by 28ft total length. The surface of the tube can be considered a black body at 600 degree C. The half cylinder cover is 24in diameter, is insulated on its top side and has an emissivity of 0.10. The heat tube is positioned along the geometric axis of the half cylinder cover. Assume the floor, people, and materials below the heat tube have a collective temperature of 10 degree C with emissivity of 0.80. What is the total heat transfer rate by radiation from the heat tube to the floor, people, and materials below? What is the equilibrium temperature of the half cylinder cover?

Solution

solution:

1)here ratio R1 and R2 are

r=R1/R2=.25

hence shape factor are

F12=F13=.5

F21=r=.25

F22=.3433

F23=1-.3433-.25=.4066

2)here heat transfer between heat tube and surface is

Q=5.67*10^-8[T1^4-T3^4]/[(1-e1/e1*A1)+(1/A1*F13)+(1-e3/e3*A3)]

Q13=5.67*10^-8[873^4-283^4]/[0+(1/pi*.1524*8.5344*.5)+(1-.8/.8*.6096*8.5344)]

Q13=60593.18 W

3)here equllibrium temperature is given by

Q12=Q21+Q22+Q23

here Q22=0 as same temperature

Q12=5.67*10^-8[T1^4-T2^4]/[(1-e1/e1*A1)+(1/A1*F12)+(1-e2/e2*A2)]

so we get

Q12=9.2672*10^-8[873^4-t2^4]

Q21=7.1286*10^-4[T2^4-873^4]

Q23=7.568*10^-8[T2^4-283^4]

on putting value in above equation of equillibrium we get

T2=794.98 K