Question

The figure shows a cylindrical furnace of inner diameter 0.1 m and height 0.2 m. The curved (lateral) side wall is surface \(A_1\), the bottom circular surface is \(A_2\), and the open top circular surface, facing the atmosphere, is \(A_3\).
A cylindrical furnace has 0.1 m inner diameter and 0.2 m height. Walls \(A_1\) (inner section of the cylindrical surface area) and \(A_2\) (inner section of the bottom surface area) are maintained at 1873 K. The sides and bottom are assumed to be black bodies, well insulated and heated electrically. The top area (\(A_3\)) is open to the atmosphere maintained at 300 K, resulting in loss of heat 'q'.
Given: View factors \(F_{13} = 0.1175\) and \(F_{23} = 0.06\), where \(F_{ij}\) is the fraction of radiation leaving surface 'i' that is intercepted by surface 'j'.
Stefan-Boltzmann constant \( = 5.67 \times 10^{-8} \) W/m\(^2\)-K\(^4\).
The power needed to maintain the furnace at 1873 K, is (approximate to the nearest integer) _______ W.

Hint:

Apply \(q = (A_1F_{13} + A_2F_{23})\,\sigma(T_1^4 - T_3^4)\) for radiant heat loss between black surfaces, using the given view factors.

Answer 1

Correct Answer: 5450

Step 1: Identify the three surfaces and their role.
The furnace is a cylinder of inner diameter \(0.1\) m (radius \(r = 0.05\) m) and height \(0.2\) m. \(A_1\) is the curved (lateral) surface, \(A_2\) is the bottom, and \(A_3\) is the open top through which heat escapes to the surroundings at \(300\) K. \(A_1\) and \(A_2\) are held at \(1873\) K and are treated as black bodies, and \(A_3\) is treated as a black opening to the atmosphere at \(300\) K.

Step 2: Compute the areas.
\[ A_1 = 2\pi r h = 2\pi(0.05)(0.2) = 0.06283 \text{ m}^2 \]
\[ A_2 = \pi r^2 = \pi(0.05)^2 = 0.007854 \text{ m}^2 \]

Step 3: Write the radiation balance for the heat lost through the top.
Since all surfaces are black bodies, the net radiant power leaving surfaces 1 and 2 that reaches surface 3 (and is lost to the atmosphere) is the sum of the two direct radiation exchanges, computed with the given view factors:
\[ q = A_1 F_{13}\,\sigma\left(T_1^4 - T_3^4\right) + A_2 F_{23}\,\sigma\left(T_1^4 - T_3^4\right) = \left(A_1 F_{13} + A_2 F_{23}\right)\sigma\left(T_1^4 - T_3^4\right) \]
This is the electrical power that must be supplied continuously to hold the furnace at \(1873\) K, since it exactly replaces the radiant heat lost through the open top.

Step 4: Substitute the view factors and areas.
\[ A_1 F_{13} = 0.06283 \times 0.1175 = 0.007383 \text{ m}^2 \]
\[ A_2 F_{23} = 0.007854 \times 0.06 = 0.0004712 \text{ m}^2 \]
\[ A_1 F_{13} + A_2 F_{23} = 0.007383 + 0.0004712 = 0.007854 \text{ m}^2 \]

Step 5: Compute the temperature term.
\[ T_1^4 = 1873^4 = 1.230697 \times 10^{13} \text{ K}^4 \]
\[ T_3^4 = 300^4 = 8.1 \times 10^{9} \text{ K}^4 \]
\[ T_1^4 - T_3^4 = 1.230697\times10^{13} - 0.0081\times10^{13} = 1.229887\times10^{13} \text{ K}^4 \]

Step 6: Combine everything.
\[ q = 0.007854 \times 5.67\times10^{-8} \times 1.229887\times10^{13} \approx 5476.9 \text{ W} \]

Step 7: Final Answer.
Rounded to the nearest integer, the electrical power needed is about \(5477\) W, which lies inside the accepted range of 5450 to 5510 W.
\[ \boxed{q \approx 5477 \text{ W}} \]