Question

Two very large parallel plates are maintained at uniform temperatures of \(T_{1}=600 \mathrm{~K}\) and \(T_{2}=400 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.5\) and \(\varepsilon_{2}=0.9\), respectively. Determine the net rate of radiation heat transfer between the two surfaces per unit area of the plates.

Short answer

Answer: The net rate of radiation heat transfer between the two surfaces per unit area of the plates is approximately \(10026.4 \mathrm{~W/m^2}\).

Step-by-step solution

Identify the given variables

We have the following variables given: - Temperature of plate 1, \(T_{1} = 600 \mathrm{~K}\) - Temperature of plate 2, \(T_{2} = 400 \mathrm{~K}\) - Emissivity of plate 1, \(\varepsilon_{1} = 0.5\) - Emissivity of plate 2, \(\varepsilon_{2} = 0.9\)

Apply the formula for net radiation heat transfer between two surfaces

To find the net rate of radiation heat transfer between the two surfaces per unit area, we can use the following formula: $$Q = \sigma \frac{\varepsilon_1 \varepsilon_2}{\varepsilon_1+\varepsilon_2-\varepsilon_1 \varepsilon_2} (T_1^4 - T_2^4)$$ Where: - \(Q\) is the net rate of radiation heat transfer per unit area (\(W/m^2\)) - \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} W \cdot m^{-2} K^{-4}\)) - \(T_1\) and \(T_2\) are the temperatures of the two surfaces in Kelvin, and - \(\varepsilon_1\) and \(\varepsilon_2\) are the emissivities of the two surfaces.

Substitute the values and solve

Now, we can substitute the given values in the formula: $$Q = \frac{(5.67 \times 10^{-8})(0.5)(0.9)}{0.5+0.9-(0.5)(0.9)} (600^4 - 400^4)$$ After evaluating the expression, we get: $$Q \approx 10026.4 \mathrm{~W/m^2}$$

Final answer

The net rate of radiation heat transfer between the two surfaces per unit area of the plates is approximately \(10026.4 \mathrm{~W/m^2}\).

Key concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law is a fundamental principle used to calculate the power radiated from a black body in terms of its temperature. It indicates that the total energy radiated per unit surface area of a black body per unit time (also known as the emissive power) is directly proportional to the fourth power of the black body's absolute temperature.
It is expressed mathematically as: \[ \text{E} = \sigma T^4 \]where:

• \( \text{E} \) is the emissive power or energy radiated per unit area.
• \( \sigma \) represents the Stefan-Boltzmann constant, which has a value of \( 5.67 \times 10^{-8} \, \text{W} \cdot \text{m}^{-2} \cdot \text{K}^{-4} \).
• \( T \) is the absolute temperature in Kelvin.

This equation helps us understand that as the temperature of an object increases, its energy emission increases rapidly. This forms the basis for calculating radiation heat transfer between various surfaces.
In cases where the object is not a perfect black body, which is often in real-world scenarios, we need to include emissivity to adjust for efficiency effects.

Emissivity

Emissivity is a measure of a material's ability to emit energy as thermal radiation. It is a dimensionless value ranging between 0 and 1, where 1 represents a perfect emitter, known as a black body, and 0 represents a perfect reflector with no emissions.
Emissivity determines how efficiently a surface emits thermal radiation relative to a black body at the same temperature. In the context of the exercise with two plates at different temperatures, emissivity is crucial to calculate the net radiative heat exchange between the surfaces. This factor accounts for the material properties, such as texture and finish, which can influence thermal radiation.
For example,

• Polished metals have low emissivities and thus prefer to reflect rather than emit radiation.
• Dull surfaces or non-metallic materials tend to have higher emissivities.

To approximate the real-world behavior of radiating surfaces, emissivity is combined with the Stefan-Boltzmann Law, adjusting calculations as per various surfaces’ efficiency to emit radiation.

Radiative Heat Exchange

Radiative heat exchange is the process by which thermal energy is transferred between objects through electromagnetic radiation. This type of heat transfer can occur over a vacuum, as it does not require a medium. Unlike conduction and convection, which rely on particle interactions or fluid motion, radiative exchange works via the emission and absorption of infrared radiation.
This principle is applicable in scenarios like the one given, where two plates with differing temperatures exchange energy through radiation. The exercise showcases the calculation of radiative heat exchange using both the Stefan-Boltzmann Law and emissivity.
To compute the net rate of radiation heat transfer between two surfaces, the formula used encompasses:

• The Stefan-Boltzmann constant \( \sigma \)
• Temperatures of the surfaces \( T_1 \) and \( T_2 \)
• Emissivities \( \varepsilon_1 \) and \( \varepsilon_2 \)

It specifically considers the difference in thermal power between the bodies.This formula is important because in many engineering and environmental applications, knowing how heat is transferred by radiation allows for better thermal management and energy efficiency in systems.