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: Spot a shortcut using view factor reciprocity.
Instead of multiplying out $A_1F_{13}$ and $A_2F_{23}$ separately and adding them, notice that the reciprocity rule $A_iF_{ij}=A_jF_{ji}$ lets us rewrite both terms from the point of view of surface 3:
\[ A_1F_{13}=A_3F_{31}, \qquad A_2F_{23}=A_3F_{32} \]
So
\[ A_1F_{13}+A_2F_{23}=A_3\left(F_{31}+F_{32}\right) \]

Step 2: Use the summation rule at surface 3.
Surface 3 is flat, so it cannot see itself, meaning $F_{33}=0$. Since surface 3 only "sees" surfaces 1 and 2 inside the furnace (the entire radiation leaving the open top either hits the side wall or the bottom, nothing escapes back onto itself), the view factors leaving surface 3 must add to 1:
\[ F_{31}+F_{32}+F_{33}=1 \quad\Rightarrow\quad F_{31}+F_{32}=1 \]

Step 3: Simplify the combined term.
Putting this back in,
\[ A_1F_{13}+A_2F_{23}=A_3\times1=A_3=\pi r^2 \]
With $r=0.05$ m, $A_3=\pi(0.05)^2=0.007854$ m$^2$.

Step 4: Cross-check against direct multiplication.
As a check, work out $A_1F_{13}$ and $A_2F_{23}$ the long way and add them. With $A_1=2\pi(0.05)(0.2)=0.06283$ m$^2$ and $A_2=\pi(0.05)^2=0.007854$ m$^2$,
\[ A_1F_{13}=0.06283\times0.1175=0.007383\text{ m}^2, \qquad A_2F_{23}=0.007854\times0.06=0.0004712\text{ m}^2 \]
\[ A_1F_{13}+A_2F_{23}=0.007383+0.0004712=0.007854\text{ m}^2 \]
This matches $A_3$ from Step 3 exactly, confirming that the reciprocity shortcut and the brute-force calculation agree.

Step 5: Apply the radiation equation.
Since the whole heat balance collapses to just $A_3$, the power supplied to the furnace equals the black-body radiation exchange between the effective top opening at $T_1$ and the surroundings at $T_3$:
\[ q = A_3\,\sigma\left(T_1^4-T_3^4\right) \]

Step 6: Plug in the numbers.
$T_1^4=1873^4=1.230697\times10^{13}$ K$^4$ and $T_3^4=300^4=0.0081\times10^{13}$ K$^4$, so $T_1^4-T_3^4=1.229887\times10^{13}$ K$^4$.
\[ q = 0.007854\times5.67\times10^{-8}\times1.229887\times10^{13} \approx 5476.9\text{ W} \]

Step 7: Conclude.
Rounded to the nearest whole number, $q\approx5477$ W, inside the accepted band of 5450 to 5510 W. This is the electrical power the heating elements must supply continuously to hold the furnace at $1873$ K against the radiant loss through the open top.
\[ \boxed{q\approx5477\text{ W}} \]