Consider a cylindrical furnace of 5 m diameter and 5 m length with bottom, top and curved surfaces maintained at uniform temperatures of 800 K, 1500 K and 500 K, respectively. The view factor between the bottom and top surfaces, F₁₂ is 0.2. The magnitude of net radiation heat transfer rate between the bottom surface and the curved surface is _____________ kW (rounded off to 1 decimal place).

All surfaces of the furnace can be assumed as black.

The Stefan-Boltzmann constant, 𝜎 = 5.67 × 10^-8 W m^-2 K^-4.

The correct answer is 308 (or approximately 308.6 kW).

To find the net radiation heat transfer rate between the bottom surface and the curved surface of the cylindrical furnace, we define the surfaces as follows based on the provided cylindrical schematic:
- Bottom surface (Surface 1): Flat circular disk at the base with diameter $D=5\text{m}$ and temperature $T_1=800\text{K}$.
- Top surface (Surface 2): Flat circular disk at the top with temperature $T_2=1500\text{K}$.
- Curved surface (Surface 3): Side cylinder wall with temperature $T_3=500\text{K}$.

Step 1: Calculate the Area of the Bottom Surface ($A_1$)
The area of the flat circular base is given by:

$A_1=\frac{\pi }{4}{D}^2$

Substituting $D=5\text{m}$:

$A_1=\frac{\pi }{4}\times 5^2\approx 19.635{\text{m}}^2$

Step 2: Determine the View Factor ($F_{13}$)
Using the view factor summation rule for an enclosure containing Surface 1:

$F_{11}+F_{12}+F_{13}=1$

Since the bottom surface (Surface 1) is a flat disk, it cannot see itself, meaning:

$F_{11}=0$

We are given that the view factor between the bottom and top surfaces is $F_{12}=0.2$. Substituting these values yields:

$0+0.2+F_{13}=1\Rightarrow F_{13}=0.8$

Step 3: Calculate the Net Radiation Heat Transfer ($Q_{13}$)
Since all surfaces of the furnace are black, the net radiation heat transfer between Surface 1 and Surface 3 is:

$Q_{13}=A_1{F}_{13}\sigma \left(T_1^4-T_3^4\right)$

Where the Stefan-Boltzmann constant is $\sigma =5.67\times {10}^{-8}{\text{W m}}^{-2}{\text{K}}^{-4}$.
First, calculate the temperature difference term:

$T_1^4-T_3^4={800}^4-{500}^4=4.096\times {10}^{11}-6.25\times {10}^{10}=3.471\times {10}^{11}{\text{K}}^4$

Using an approximate rounded base area $A_1\approx 19.6{\text{m}}^2$:

$Q_{13}=19.6\times 0.8\times 5.67\times {10}^{-8}\times 3.471\times {10}^{11}\text{W}$

$Q_{13}\approx 308.6\text{kW}$

Thus, the magnitude of the net radiation heat transfer rate is approximately 308 kW.