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. Answers: $554 \mathrm{R}, 50.9 \mathrm{Btu} / \mathrm{h}^{\mathrm{ft}}{ }^{2}$

Short answer

Answer: The outer surface temperature is 554 Rankine, and the rate of heat transfer through the wall is 50.9 Btu/h·ft².

Step-by-step solution

Identify the energy balance equation for the outer surface of the wall

Consider the energy balance for the outer surface: The heat gained by the wall due to solar radiation = heat lost by conduction through the wall + heat lost by radiation to deep space.

Express the energy balance equation in mathematical terms

The energy balance equation can be expressed as following: \(\alpha \dot{q}_{\text{solar}} = k \frac{T_1 - T_2}{L} + \varepsilon\sigma (T_2^4 - T_{space}^4)\) Here, \(\alpha\) is the solar absorptivity, \(\dot{q}_{\text{solar}}\) is the solar radiation incident on the wall, \(k\) is the thermal conductivity, \(L\) is the wall thickness, \(T_1\) and \(T_2\) are inner and outer surface temperatures respectively, \(\varepsilon\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and \(T_{space}\) is deep space temperature (assumed to be zero Kelvin or Rankine).

Solve for the outer surface temperature, \(T_2\)

We can rearrange the energy balance equation to find the outer surface temperature, \(T_2\). In this exercise, we are given all the necessary values to find \(T_2\) except for the Stefan-Boltzmann constant. We will use the value of \(\sigma = 1.712 \times 10^{-9} \frac{\text{Btu}}{\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{R}^{4}}\): \(T_2^4 = \frac{\alpha \dot{q}_{\text{solar}} - k \frac{T_1}{L}}{\varepsilon\sigma} + T_{space}^4\) Substitute given values and solve for \(T_2\): \(T_2 = \sqrt[4]{\frac{0.6(300) - 1.2 \frac{520}{0.8}}{0.80(1.712 \times 10^{-9})}}\) \(T_2 = 554~R\)

Calculate the rate of heat transfer through the wall

Now that we have the outer surface temperature, we can calculate the rate of heat transfer through the wall. Since it's a steady-state conduction heat transfer problem, we will use Fourier's law to find the rate of heat transfer: \(\dot{q} = kA\frac{T_1-T_2}{L}\), Here A is area normal to the direction of heat flow. Let's substitute the known values and solve for \(\dot{q}\): \(\dot{q} = 1.2\frac{520 - 554}{0.8}\) \(\dot{q} = -50.9~\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\) Since the value is negative, it means that heat is being transferred from the inner surface to the outer surface. The magnitude of the heat transfer rate is \(50.9~\mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}\).