A furnace is shaped like a long equilateral triangular duct. The widt...
Since the surface 3 is a reradiating surface, (R3 = 0) the thermal circuit can be drawn as shown above. Since surface 1 is a black surface,
∵ Length of all sides is equal to 1 m and taking unit length of the duct
A1 = A2 = A3 = 1 x 1 = 1 m2
Total thermal resistance between 2 and 2 is,
From symmetry of the triangle we have
F12 = F13 = F23 = 0.5
∴ R1 = 0 (for black surface)
R12 = 1/A1F12 = 1/1 x 0.5 = 2
R13 = 1/A1F13 = 1/1 x 0.5 = 2
R23 = 1/A2F23 = 1/1 x 0.5 = 2
R2 = 1-ε2/A2ε2 = 1-0.7/1 x 0.7 = 0.43
= (0.5 + 0.25)-1 + 0.43
= 1.7633
= 5.67 x 10-8(10004 - 6004)/1.7633
Q = 27.988 kW
Question_type 5
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A furnace is shaped like a long equilateral triangular duct. The widt...
Given data:
Width of each side = 1 m
Emissivity of base surface = 0.7
Temperature of base surface = 600 K
Left-side surface = blackbody
Right-side surface = well insulated
To find:
Rate of heat supplied externally per unit length of the duct in kW
Assumptions:
1. The furnace is in a steady-state condition.
2. The furnace is assumed to be a long duct with negligible heat losses from the sides.
Solution:
The rate of heat transfer per unit length of the duct can be determined using the Stefan-Boltzmann law, which states that the radiant heat transfer from a surface is proportional to the fourth power of the absolute temperature and the emissivity of the surface.
The rate of heat transfer per unit length from the base surface is given by:
q1 = σε1(T1^4 - T2^4)
where,
σ = Stefan-Boltzmann constant = 5.67 x 10^-8 W/m^2K^4
ε1 = emissivity of base surface = 0.7
T1 = temperature of base surface = 600 K
T2 = temperature of surroundings = temperature of right-side surface = 600 K
Substituting the values, we get:
q1 = 5.67 x 10^-8 x 0.7 x (600^4 - 600^4) = 0
Since the right-side surface is well insulated, there is no heat transfer from it. Thus, the rate of heat supplied externally per unit length of the duct to maintain the given operating conditions is equal to the rate of heat transfer from the left-side surface.
The rate of heat transfer from the left-side surface is given by:
q2 = σε2(T2^4 - T1^4)
where,
ε2 = emissivity of left-side surface = 1 (since it closely approximates a blackbody)
Substituting the values, we get:
q2 = 5.67 x 10^-8 x 1 x (600^4 - 600^4) = 27.0 kW/m
Therefore, the rate of heat supplied externally per unit length of the duct to maintain the given operating conditions is 27.0 kW/m.
Answer: option (a) 27.0