For a hemispherical furnace, the flat floor is at 700 K and has an emissivity of 0.5. The hemispherical roof is at 1000 K and has an emissivity of 0.25. Find the magnitude of net radiative heat transfer between the roof and floor.

Question

Prepp Team · Accepted Answer

This solution explains how to calculate the net radiative heat transfer between the floor and the roof of a hemispherical furnace, given their temperatures and emissivities.

Understanding Radiative Heat Transfer in a Hemispherical Furnace

We need to find the net radiative heat transfer per unit area between the flat floor and the hemispherical roof of a furnace. This involves understanding how thermal radiation exchanges between surfaces with different temperatures and emissivities.

Key Parameters Provided:

Floor Temperature, T_floor = 700 K
Floor Emissivity, ε_floor = 0.5
Roof Temperature, T_roof = 1000 K
Roof Emissivity, ε_roof = 0.25
Stefan-Boltzmann Constant, σ = $5.670374419 \times 10^{-8} \text{ W/m}^2\text{K}^4$

Hemispherical Furnace Geometry and View Factors

For a hemispherical furnace with a flat circular floor:

The floor is designated as Surface 1.
The hemispherical roof is designated as Surface 2.
The view factor from the floor (Surface 1) to the roof (Surface 2), F₁₂, is 1, as the entire hemispherical surface 'sees' the flat floor.
The view factor from the roof (Surface 2) to the floor (Surface 1), F₂₁, is the ratio of the floor's area to the roof's surface area. If the floor is a circle of radius R ($A_1 = \pi R^2$) and the roof is a hemisphere of radius R ($A_2 = 2\pi R^2$), then $F_{21} = A_1 / A_2 = (\pi R^2) / (2\pi R^2) = 0.5$.

Formula for Net Radiative Heat Transfer

The net radiative heat transfer flux between two surfaces in an enclosure can be calculated using the two-surface enclosure method. The formula for the net heat flux per unit area of Surface 1 (the floor in this case) is:

$$q_{net, A_1} = \frac{\sigma(T_1^4 - T_2^4)}{\frac{1-\epsilon_1}{\epsilon_1} + \frac{1}{F_{12}} + \frac{A_1}{A_2}\frac{1-\epsilon_2}{\epsilon_2}}$$

Where:

T₁ is the temperature of Surface 1 (floor).
ε₁ is the emissivity of Surface 1 (floor).
T₂ is the temperature of Surface 2 (roof).
ε₂ is the emissivity of Surface 2 (roof).
F₁₂ is the view factor from Surface 1 to Surface 2.
A₁/A₂ is the ratio of the area of Surface 1 to Surface 2.

Step-by-Step Calculation

Calculate the fourth power of temperatures:
$T_{floor}^4 = (700 \text{ K})^4 = 2.401 \times 10^{11} \text{ K}^4$
$T_{roof}^4 = (1000 \text{ K})^4 = 1 \times 10^{12} \text{ K}^4$
Calculate the temperature difference term ($T_{floor}^4 - T_{roof}^4$):
$T_{floor}^4 - T_{roof}^4 = (2.401 \times 10^{11}) - (1 \times 10^{12}) = -7.599 \times 10^{11} \text{ K}^4$
Calculate the terms in the denominator:
- Surface resistance term for the floor: $\frac{1-\epsilon_{floor}}{\epsilon_{floor}} = \frac{1-0.5}{0.5} = 1$
- Space resistance term: $\frac{1}{F_{floor \to roof}} = \frac{1}{1} = 1$
- Surface resistance term for the roof, scaled by area ratio: $\frac{A_{floor}}{A_{roof}}\frac{1-\epsilon_{roof}}{\epsilon_{roof}} = 0.5 \times \frac{1-0.25}{0.25} = 0.5 \times 3 = 1.5$
Sum the denominator terms:
Total Denominator = $1 + 1 + 1.5 = 3.5$
Calculate the numerator (Stefan-Boltzmann constant times temperature difference):
Numerator = $\sigma (T_{floor}^4 - T_{roof}^4)$
Numerator = $(5.670374419 \times 10^{-8} \text{ W/m}^2\text{K}^4) \times (-7.599 \times 10^{11} \text{ K}^4)$
Numerator $\approx -43089.175 \text{ W/m}^2$
Calculate the net heat flux:
$q_{net, A_{floor}} = \frac{\text{Numerator}}{\text{Denominator}} = \frac{-43089.175 \text{ W/m}^2}{3.5}$
$q_{net, A_{floor}} \approx -12311.19 \text{ W/m}^2$
Determine the magnitude:
The magnitude of the net radiative heat transfer is the absolute value of the calculated flux.
Magnitude = $|-12311.19 \text{ W/m}^2| \approx 12311.19 \text{ W/m}^2$

Conclusion

The calculated magnitude of the net radiative heat transfer between the roof and the floor is approximately $12311.19 \text{ W/m}^2$. This value is very close to the provided option.