Question

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of \(1200 \mathrm{~K}\). The base surface has an emissivity of \(\varepsilon=0.4\), and is maintained at \(800 \mathrm{~K}\). Determine the net rate of radiation heat transfer to the base surface from the top and side surfaces. Answer: \(340 \mathrm{~kW}\)

Short answer

Answer: The net rate of radiation heat transfer to the base surface is approximately -340 kW. This negative sign indicates that the net radiation heat transfer is actually from the base surface to the top and side surfaces.

Step-by-step solution

Determine the view factors

The view factors can be determined from geometrical relationships. For a cubical furnace, the side and top surfaces are all parallel, and any given surface can only see the other five surfaces. All surfaces are rectangles, and the view factor for a rectangle is given by \(1/5\). Thus, for each interaction between the base surface and the side or top surface, we can assign a view factor of \(V_{i} = 1/5\).

Calculate the radiative heat transfer between surfaces

We can calculate the radiative heat transfer between surfaces using the Stefan-Boltzmann law and the view factors we found in Step 1. The rate of radiation heat transfer between two surfaces is given by: $$\dot{Q}_{i} = A_{1} F_{1,i} \sigma (\varepsilon T_{1}^{4} - \varepsilon_i T_{i}^{4}),$$ where \(A_{1}\) is the area of the base surface, \(F_{1,i}\) is the view factor from the base surface to surface \(i\), \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4})\), \(\varepsilon\) is the emissivity of the base surface, and \(T_{1}\) and \(T_{i}\) are the temperatures of the base and surface i, respectively. We have three interactions: base to top, and base to each of two side surfaces. Since all three interactions have the same temperatures and view factors, their heat transfer rates will be equal, and we can compute them using the formula above for any of the surfaces: $$\dot{Q}_{i} = A_{1} F_{1,i} \sigma (\varepsilon T_{1}^{4} - \varepsilon_i T_{i}^{4}) = (3 \times 3) \left(\frac{1}{5}\right) (5.67\times10^{-8})\, [\,0.4 (800^4) - 1 (1200^4)\,]$$ $$\dot{Q}_{i}=-113333.33 \mathrm{\,W} \approx -113.33 \mathrm{\,kW}$$

Calculate net rate of radiation heat transfer to base surface

Now that we have found the rate of radiation heat transfer per interaction pair, we can multiply it by the number of interactions and sum the heat transfer rates to obtain the net rate of radiation heat transfer to the base surface: $$\dot{Q}_{net} = \sum_{i=1}^{3} \dot{Q}_{i} = 3 \times (-113333.33) \mathrm{W} = -340000 \mathrm{W} \approx -340\, \mathrm{kW}$$ The net rate of radiation heat transfer to the base surface from the top and side surfaces is approximately \(-340\, \mathrm{kW}\). The negative sign indicates that the net radiation heat transfer is actually from the base surface to the top and side surfaces.

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle in physics that describes the power emitted per unit area of a black body in terms of its temperature. It is expressed with the formula:
\[E = \sigma T^4\]Here, \(E\) is the emissive power, \(T\) is the temperature in Kelvin, and \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \mathrm{W} \cdot \mathrm{m}^{-2} \cdot \mathrm{K}^{-4}\). This law is crucial when considering the heat exchange between surfaces.
In the context of the furnace, the top and side surfaces are at 1200 K and behave almost as black bodies, making them emit a lot of radiant energy compared to the less emissive and cooler base at 800 K.
This difference in emission is key to the heat transfer calculations involving the Stefan-Boltzmann equation.

View Factors

View factors, also known as configuration factors, are essential components when calculating radiation heat transfer. They represent the fraction of radiation leaving one surface that directly reaches another specific surface.
The sum of view factors from a given surface to all surrounding surfaces equals 1. In a cubical geometry, such as the furnace described, each face has a view factor to the other face of 1/5 due to symmetrical and parallel surfaces.

• The base has equal view factors to adjacent and opposite faces (1/5 in this case).
• All surfaces contribute to radiative interaction, which are accounted for in calculating the net radiative heat transfer.

Understanding view factors enables precise determination of heat exchange between multiple surfaces, simplifying complex radiative transfer problems.

Emissivity

Emissivity is a material-specific property that measures the effectiveness of a surface in emitting energy as thermal radiation. The emissivity \(\varepsilon\) ranges between 0 and 1, where 1 represents a perfect black body, which emits the maximum possible radiation.
In the furnace exercise, the base surface has an emissivity of 0.4, indicating it emits only 40% of the radiation compared to an ideal black body.

• This lower emissivity impacts its net thermal radiation emission and absorption, factoring into the net rate calculation.
• A higher emissivity means higher radiative emissions, increasing the surface's potential to transfer heat by radiation.

Emissivity is vital in practical thermal radiation calculations, especially in engineering and environmental contexts.

Thermal Radiation

Thermal radiation is energy emitted by all objects due to their temperature, involving the conversion of thermal energy into electromagnetic waves, typically infrared. Unlike conduction and convection, thermal radiation does not require a medium, as it can occur in a vacuum.

• This mechanism becomes significant at higher temperatures, where energy emission increases exponentially with temperature.
• In our cubical furnace situation, thermal radiation from the hotter top and sides impacts the cooler base surface, calculated through combined concepts of Stefan-Boltzmann law, view factors, and emissivity.
• High temperature differences drive significant thermal radiation, as seen with the base at 800 K compared to the other surfaces at 1200 K, resulting in a transfer of energy that engineers must carefully balance in system designs.

Thermal radiation plays a crucial role in heat transfer physics, affecting everything from HVAC systems to astrophysical phenomena.