A large slab of steel 0.1-m-thick contains a 0.1-m-diameter circular hole, whose axis is normal to the surface. Considering the sides of the hole to be black, specify the rate of radiative heat loss from the hole. The plate is at 811 K, the surroundings are at 300 K.
GIVEN
• A large slab of steel with a hole whose axis is normal to the surface
• Slab thickness (S) = 0.1 m
• Hole diameter (D) = 0.1 m
• Plate temperature (T1) = 811 K
• Temperature of surrounding (T?) = 300 K
FIND
• The rate of radiative heat loss from the hole (qr)
ASSUMPTIONS
• The sides of the hole are black
SKETCH
PROPERTIES AND CONSTANTS
the Stephan-Boltzmann constant (?) = 5.67 × 10–8 W/(m2 K4)
for
Since all surfaces behave as blackbodies
Therefore
Substituting these into the above equation yields
Since A2 cannot see itself F22 = 0
By symmetry F12 = F13
The sum of the shape factors from one surface must be unity
The shape factor F23 can be determined for D/S = 0.1 m/0.1 m = 1.0, and for disks with direct radiation, curve 1 applies and the shape factor F23?0.19.
The rate of heat transfer through A2 is
The negative sign indicates net heat loss through A2. By symmetry, the total energy leaving the hole is
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