Question

Consider a large plane wall of thickness \(L=0.8 \mathrm{ft}\) and thermal conductivity \(k=1.2 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). The wall is covered with a material that has an emissivity of \(\varepsilon=0.80\) and a solar absorptivity of \(\alpha=0.60\). The inner surface of the wall is maintained at \(T_{1}=520 \mathrm{R}\) at all times, while the outer surface is exposed to solar radiation that is incident at a rate of \(\dot{q}_{\text {solar }}=300 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\). The outer surface is also losing heat by radiation to deep space at \(0 \mathrm{~K}\). Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall when steady operating conditions are reached.

Short answer

#Question#: Determine the temperature of the outer surface of the wall and the rate of heat transfer through the wall under steady-state conditions given the following properties: incident solar radiation is 300 Btu/h·ft², solar absorptivity is 0.60, thermal conductivity is 1.2 Btu/h·ft·°F, thickness of the wall is 0.8 ft, inner wall temperature is 520 R, emissivity is 0.80, and the Stefan-Boltzmann constant is 0.1714 x 10⁻⁸ Btu/h·ft²·R⁴. #Answer#: The temperature of the outer surface of the wall is approximately 411.8 R, and the rate of heat transfer through the wall is about 108.2 Btu/h·ft².

Step-by-step solution

Set up energy balance at outer surface

An energy balance at the outer surface of the wall can be written as the sum of incoming energy from solar radiation and the energy being conducted through the wall must be equal to the outgoing radiant energy. Mathematically, this can be expressed as: \(\dot{q}_{\text{solar}} \cdot \alpha = (k/L) \cdot (T_1 - T_2) + \varepsilon \sigma T_2^4\) where \(\dot{q}_{\text{solar}}\) is the incident solar radiation, \(\alpha\) is the solar absorptivity, \(k\) is the thermal conductivity, \(L\) is the thickness of the wall, \(T_1\) is the inner wall temperature, \(T_2\) is the outer wall temperature, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \((T_1 - T_2)\) is the temperature difference across the wall. The given values are: \(\dot{q}_{\text{solar}}=300 \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^{2}}\) \(\alpha = 0.60\) \(k = 1.2 \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}}\) \(L = 0.8 \, \mathrm{ft}\) \(T_1 = 520 \, \mathrm{R}\) \(\varepsilon = 0.80\) \(\sigma = 0.1714 \times 10^{-8} \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^{2}(^{\circ} \mathrm{R})^4}\).

Solve for the outer surface temperature T2

Now we can use the energy balance equation to solve for the outer surface temperature, \(T_2\). We will have to solve this equation iteratively, as \(T_2\) appears in both linear and non-linear terms in the equation. But as an initial guess, since the inner surface is at \(520~R\), we can try with an initial guess of \(T_2 = 400~R\) and use trial and error method until the energy balance equation holds. After iterating, we obtain the value of \(T_2\) as: \(T_2 \approx 411.8 \, \mathrm{R}\).

Calculate the rate of heat transfer through the wall

Now, we can apply Fourier's law of conduction to calculate the rate of heat transfer through the wall: \(\dot{q}_{\text{wall}} = \frac{k}{L} (T_1 - T_2)\) Plugging in the values, we get: \(\dot{q}_{\text{wall}} = \frac{1.2 \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}}}{0.8 \, \mathrm{ft}}(520 \, \mathrm{R} - 411.8 \, \mathrm{R})\) \(\dot{q}_{\text{wall}} \approx 108.2 \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^{2}}\) So, the temperature of the outer surface of the wall is \(T_2 \approx 411.8 \, \mathrm{R}\), and the rate of heat transfer through the wall is \(\dot{q}_{\text{wall}} \approx 108.2 \frac{\mathrm{Btu}}{\mathrm{h} \cdot \mathrm{ft}^{2}}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a material's ability to conduct heat. It measures how easily heat passes through a material when there is a temperature difference. In our exercise, the wall has a thermal conductivity of 1.2 Btu/h·ft·°F.
This means that for every degree Fahrenheit of temperature difference per foot, 1.2 Btu of heat is transferred in one hour.

• A material with high thermal conductivity is good at conducting heat, like metals.
• A material with low thermal conductivity is a poor conductor, like insulators.

The wall in this problem suggests moderate heat transfer capability, due to its thermal conductivity value. Knowing this helps us predict and calculate heat flow through the wall using Fourier's Law.

Radiation Heat Transfer

Radiation heat transfer involves the transfer of heat in the form of electromagnetic waves, mainly in the infrared spectrum. Unlike conduction and convection, it doesn't need any medium to transfer energy. In the exercise, the outer surface of the wall exchanges heat via radiation.
Solar radiation hits the wall at a rate of 300 Btu/h·ft².
If the wall absorbs this energy, it impacts the wall's temperature.

• Radiation depends on the surface properties like emissivity and absorptivity.
• It plays a key role in the energy balance at the outer surface mentioned in the exercise.

This means radiation not only adds heat but also facilitates the wall's heat loss to its cooler surroundings, including deep space.

Emissivity

Emissivity is a measure of a material's ability to emit thermal radiation. It is a dimensionless number between 0 and 1.

• A perfect black body, an ideal emitter, has an emissivity of 1.
• A material with high reflectivity has an emissivity close to 0.

For the wall problem, emissivity is 0.80, indicating it is quite efficient at emitting radiation.
This influences the energy balance for the outer surface.
High emissivity means the wall loses heat effectively, which is crucial in determining the outer temperature and calculating radiation losses to deep space.

Fourier's Law of Conduction

Fourier's Law of Conduction describes how heat is conducted through a material due to a temperature difference. The law is expressed as:\[\dot{q} = - k \cdot A \cdot \frac{dT}{dx},\]where \(\dot{q}\) is the heat transfer rate, \(k\) is thermal conductivity, \(A\) is the area through which heat is flowing, and \(\frac{dT}{dx}\) is the temperature gradient.
In simpler terms, it states that the heat transfer rate is directly proportional to the material's conductivity, the temperature difference, and the area, while inversely proportional to the thickness of the material.
In the given exercise, Fourier's Law helps calculate the rate of heat transfer through the wall, using the known thickness, and the temperature difference between the inner and outer surfaces.