The spherical experiment (surface 1) has diameter D = 5.0 cm and emissivity ?1 =

Question

The spherical experiment (surface 1) has diameter D = 5.0 cm and emissivity ?1 = 0.5. The
temperature of the experiment is maintained at T1 = 100 K. Each face of the cubical enclosure is
W = 10 cm x W = 10 cm. The top surface of the enclosure (surface 2) is maintained at T2 = 300
K and is black. The other five sides (surface 3) are insulated externally and have emissivity ?3 =
0.1.
a.) What is the view factor between the experiment and the top surface (F1,2)?
b.) What is the view factor between the experiment and the 5 insulated sides (F1,3)?
c.) What is the view factor between the top surface and the experiment (F2,1)?
d.) Draw and clearly label a resistance network that represents the radiation heat transfer
problem.
e.) Calculate the values of all of the resistances in your diagram from (d). You may
assume that the view factor between the insulated sides and the top (F3,2) is 0.70.
f.) What is the net rate of heat transfer to the experiment?
g.) How would your answer to (f) change if the emissivity of the experiment were
reduced (would the heat transfer to the experiment increase, decrease, or stay the
same)?
h.) How would your answer to (f) change if the emissivity of the insulated sides were
reduced (would the heat transfer to the experiment increase, decrease, or stay the
same)?

Explanation / Answer

a) Radiations arises from sphere is devided to 6 equal surfaces.View factor is the fraction of radiation recieved at second surface .So we can see from the figure that 1/6 th of the radiation is reaching surface 2

Therefore F1-2 = 0.166

b) For a closed surface

F 1-1 +F 1-2 +F 1-3 = 1

F1-1 =0 because surface 1 will never see itself

Which implies F 1-3 =1-F1-2

=1-0.166 = 0.834

c) from resiprocating relation

When equilibrium is attained

A1 F1-2 = A2 F2-1

F2-1=A1 F1-2 /A2

=4*3.14*52 *0.166/102

d)

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