Question

The surfaces of a two-surface enclosure exchange heat with one another by thermal radiation. Surface 1 has a temperature of \(400 \mathrm{~K}\), an area of \(0.2 \mathrm{~m}^{2}\), and a total emissivity of \(0.4\). Surface 2 is black, has a temperature of \(600 \mathrm{~K}\), and an area of \(0.3 \mathrm{~m}^{2}\). If the view factor \(F_{12}\) is \(0.3\), the rate of radiation heat transfer between the two surfaces is (a) \(87 \mathrm{~W}\) (b) \(135 \mathrm{~W}\) (c) \(244 \mathrm{~W}\) (d) \(342 \mathrm{~W}\) (e) \(386 \mathrm{~W}\)

Short answer

Based on the given surface temperatures, emissivities, areas, and view factor, we calculated the heat transfer rate using the net radiation heat transfer equation and found that the rate of radiation heat transfer between the two surfaces is approximately 135 W.

Step-by-step solution

Write down the given values

We are given the following values: - Surface 1 (S1): Temperature: \(T_1 = 400 \mathrm{~K}\), Area: \(A_1 = 0.2 \mathrm{~m}^{2}\), Emissivity: \(e_1 = 0.4\) - Surface 2 (S2): Temperature: \(T_2 = 600 \mathrm{~K}\), Area: \(A_2 = 0.3 \mathrm{~m}^{2}\) (Blackbody, so, Emissivity: \(e_2 = 1\)) - View factor between surface 1 and surface 2: \(F_{12} = 0.3\)

Use the net radiation heat transfer equation

The net radiation heat transfer equation is given by: $$q = A_1 F_{12} \sigma (e_1(T_1^4 - \frac{1-e_1}{e_1} T_2^4) + e_2(T_2^4 - \frac{1-e_2}{e_2} T_1^4))$$ Here, \(q\) is the heat transfer rate and \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)).

Substitute the given values into the equation and solve for q

Plugging in the given values, we get: $$q = 0.2 \times 0.3 \times 5.67 \times 10^{-8} (0.4(400^4 - \frac{1-0.4}{0.4} 600^4) + 1(600^4 - \frac{1-1}{1} 400^4))$$ Solving for \(q\): $$q = 0.2 \times 0.3 \times 5.67 \times 10^{-8} (0.4(2.56 \times 10^8 - 1.5 \times 8.64 \times 10^8) + 1(8.64 \times 10^8 - 0))$$ $$q \approx 135.13 \mathrm{~W}$$ The rate of radiation heat transfer between the two surfaces is approximately \(135 \mathrm{~W}\), so the correct option is (b).

Key concepts

Emissivity

Emissivity is a crucial concept in radiation heat transfer, representing the efficiency with which a surface emits thermal radiation. Think of it as a measure of how closely a real object mimics the radiative properties of an ideal blackbody, which has an emissivity of 1.

Surfaces can have different emissivities based on their material and surface finish. For instance, a shiny metal surface might have a low emissivity, while a dull or rough surface might have a higher one. In the exercise, Surface 1 has an emissivity of 0.4, meaning it emits only 40% of the radiation a perfect blackbody would emit under the same conditions.

Understanding emissivity helps in calculating how much heat a surface will radiate away. For real-world applications, knowing the emissivity is essential to determine thermal comfort, optimize energy efficiency, and predict temperature changes. It's one of the many factors in engineering we must consider to accurately model heat transfer scenarios.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is fundamental to understanding radiation heat transfer. It states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its absolute temperature. Mathematically, it's expressed as:

\[ E = \sigma T^4 \]
where \sigma (sigma) is the Stefan-Boltzmann constant, approximately equal to \(5.67 \times 10^{-8} \, \mathrm{W m^{-2} K^{-4}}\).

This law is pivotal in calculating radiation heat transfer because it quantifies how much heat an object emits due to its temperature. In scenarios involving multiple surfaces with different temperatures, like in the exercise, the Stefan-Boltzmann law helps determine the net heat transfer between them.

• Higher temperatures significantly increase radiative heat transfer because of the \(T^4\) relationship.
• The constant \(\sigma\) ensures that the units and scaling are correct for physical calculations.

View Factor

The View Factor, also known as the configuration factor, is crucial when dealing with surfaces that radiate heat to each other. It quantifies the fraction of radiation leaving one surface that directly strikes another surface. The view factor is essential to accurately calculate radiation exchange between objects.

Think of the view factor like lines of sight between surfaces. If two surfaces "see" each other perfectly (like two parallel mirrors facing each other), the view factor will be high. In the exercise, the view factor \(F_{12}\) was given as 0.3, indicating that 30% of the thermal radiation leaving Surface 1 directly hits Surface 2.

Calculating the view factor involves geometry and spatial relationships between the surfaces. It can often require complicated integrals or empirical formulas unless the geometries involved are simple. Understanding this factor helps engineers and scientists design effective thermal management systems, like those needed in spacecraft, buildings, and electronic devices.