In manufacturing, the special coaling on a curved solar absorber surface of area

Question

In manufacturing, the special coaling on a curved solar absorber surface of area A_2 = 15 m^2 is cured by exposing it to an infrared healer of width W = 1 m. The absorber and heater are each of length L = 10 m and are separated by a distance of H = 1 m. The upper surface of the absorber and the lower surface of the healer are insulated. The heater is at T_1 = 1000 K and has an emissivity of epsilon_1 = 0.9, while the absorber is at T_2 = 600 K and has an emissivity of epsilon_2 = 0.5. The system is in a large room whose walls are at 300 K. What is the net rate of heat transfer to the absorber surface?

Explanation / Answer

SOLUTION:

1) here areaA1 of heater is intercepted by absorber area A2 and wall area 2*A3,hence shape factor are as per enclosure thereom

f12+f13=1

f21+f23=1

2) hence here by geometry we have

total area intercepted isA=A2+2*A3=15+L*H*2=15+2*10*1=35

hence area intercepted by absorber is=A2/A=15/35=.4285

for wall=A3*2/A=20/35=.5715

3) for reciprocal thereom

F12*A1=F21*A2

F21=.2856

F23=1-.2856=.7143

4)here heat transfer between two gray bodies is

Q12=5.67*10^-8*(T1^4-T2^4)/((1-e1/e1*A1)+(1/a1*F12)+(1-e2/e2*A2))

e1=.9 and e2=.5

on solving A1=10*1=10 m2

on solving we get

Q12=158610.57 W/m2

4) heat rejected to surrounding wall by absorber is

Q23=5.67*10^-8*(T2^4-T3^4)/((1-e2/e2*A2)+(1/A2*F23)+(1/A2))

on solving we get

Q23=-32805.29 W/m2

5) net heat transfer for absorber is

Q2=Q12-Q23=158610.57-2*32805.29=92999.99 W/m2

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