Question

The inner and outer surfaces of a \(25-\mathrm{cm}\)-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\), and convection with ambient air also at \(40^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

Short answer

Question: Determine the effective thermal conductivity of the wall. Answer: To find the effective thermal conductivity of the wall, you can follow these steps: 1. Calculate the net radiation heat transfer (q_rad) using the formula: \(q_\text{rad}=\alpha_\text{S} Q_{S} - \epsilon_{S} \sigma (T_\text{surface}^4 - T_\text{surround}^4)\) 2. Calculate the heat transfer due to convection (q_conv) using the formula: \(q_\text{conv} = h_\text{conv} (T_\text{surface} - T_\text{air})\) 3. Calculate the total heat transfer at the outer surface (q_total) by adding the heat transfer due to radiation and convection: \(q_\text{total} = q_\text{rad} + q_\text{conv}\) 4. Calculate the heat transfer through the wall by conduction (q_cond), which equals the total heat transfer at the outer surface since the wall is in a steady state. Use Fourier's law of conduction: \(q_\text{cond} = k_\text{eff} A \frac{T_\text{inner} - T_\text{surface}}{\delta}\) 5. Solve for the effective thermal conductivity (k_eff) using the equation: \(k_\text{eff} = \frac{q_\text{cond} \delta}{A (T_\text{inner} - T_\text{surface})}\) Plug in the given values and perform the calculations to find the effective thermal conductivity of the wall.

Step-by-step solution

Calculate the net radiation heat transfer

First, let's determine the total net heat transfer due to radiation on the outer surface (q_rad). To do that, we need to find the difference between the radiation absorbed and the radiation emitted by the surface. Use the following formula: \(q_\text{rad}=\alpha_\text{S} Q_{S} - \epsilon_{S} \sigma (T_\text{surface}^4 - T_\text{surround}^4)\) where \(\alpha_\text{S}\) is the solar absorptivity, \(Q_{S}\) is the solar radiation, \(\epsilon_{S}\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \ \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}^{4}\)), \(T_\text{surface}\) is the outer surface temperature, and \(T_\text{surround}\) is the surrounding temperature. Plug in the given values and solve for \(q_\text{rad}\): \(q_\text{rad}=0.8 \times 150 - 0.8 \times 5.67 \times 10^{-8} (44^4 - 40^4)\)

Calculate the heat transfer due to convection

Now, we need to determine the heat transfer due to convection at the outer surface (q_conv). Use the following formula: \(q_\text{conv} = h_\text{conv} (T_\text{surface} - T_\text{air})\) where \(h_\text{conv}\) is the convection heat transfer coefficient, \(T_\text{surface}\) is the outer surface temperature, and \(T_\text{air}\) is the ambient air temperature. Plug in the given values and solve for \(q_\text{conv}:\) \(q_\text{conv} = 8 (44 - 40)\)

Calculate the total heat transfer at the outer surface

Now add the heat transfer due to radiation and convection to get the total heat transfer at the outer surface (q_total): \(q_\text{total} = q_\text{rad} + q_\text{conv}\)

Calculate the heat transfer through the wall by conduction

Since the wall is in a steady state, the heat transfer through the wall by conduction equals the total heat transfer at the outer surface. \(q_\text{cond} = q_\text{total}\) We can calculate the heat transfer through the wall using Fourier's law of conduction: \(q_\text{cond} = k_\text{eff} A \frac{T_\text{inner} - T_\text{surface}}{\delta}\) where \(k_\text{eff}\) is the effective thermal conductivity, \(A\) is the wall's area, \(T_\text{inner}\) is the inner surface temperature, \(T_\text{surface}\) is the outer surface temperature, and \(\delta\) is the wall's thickness.

Solve for the effective thermal conductivity

Now, we can solve for the effective thermal conductivity (\(k_\text{eff}\)) using the heat transfer calculated above. Rearrange the conduction equation and plug in the given values: \(k_\text{eff} = \frac{q_\text{cond} \delta}{A (T_\text{inner} - T_\text{surface})}\) Note that the area \(A\) will cancel out in the calculation since it's present in both \(q_\text{cond}\) and the conduction equation. Plug in the values for \(\delta\), \(T_\text{inner}\), and \(T_\text{surface}\), and use the previously calculated value for \(q_\text{total}\) to solve for \(k_\text{eff}\).

Key concepts

Radiation Heat Transfer

Radiation heat transfer plays a pivotal role in thermal systems, especially when there is a temperature difference between surfaces and a vacuum or transparent medium between them. This method of heat transfer doesn't require a medium; it can occur across the vacuum of space, which is why the Earth receives heat from the sun. Radiation heat transfer fundamentally relies on the emission of electromagnetic waves.

Understanding radiation heat transfer involves the concepts of emissivity \(\epsilon\), which measures how effectively a surface emits thermal radiation relative to an ideal 'black body,' and solar absorptivity \(\alpha_S\), which quantifies how well a surface absorbs solar energy. These properties come into play when determining the net radiation exchange as shown in the given textbook exercise. Here, radiation heat transfer includes the solar radiation absorbed by a surface and the energy emitted due to the surface's temperature. The equation used reflects the balance of absorbed solar energy and thermally emitted energy:
\[q_\text{rad}=\alpha_\text{S} Q_{S} - \epsilon_{S} \sigma (T_\text{surface}^4 - T_\text{surround}^4)\]
The Stefan-Boltzmann constant \(\sigma\) here is a key player, quantifying the amount of heat transferred by radiation per unit area at a certain temperature.

Convection Heat Transfer

Convection heat transfer is another integral method in which heat is transferred between a surface and a fluid moving over it. This fluid can be a gas or a liquid, and the transfer occurs due to the fluid's bulk motion. Typically, when the surface and fluid have different temperatures, heat is transferred from the warmer to the cooler.

The convection heat transfer coefficient \(h_\text{conv}\) featured in the solution method measures how effective the convection process is. A higher \(h_\text{conv}\) indicates more efficient heat transfer. For a wall exposed to air, like in the provided problem, this aspect of heat transfer becomes significant when the air temperature is different from the surface temperature. The formula used in the exercise to calculate the convective heat flux is:
\[q_\text{conv} = h_\text{conv} (T_\text{surface} - T_\text{air})\]

Fourier's Law of Conduction

Fourier's law of conduction is fundamental to understanding how heat transmits through materials due to temperature gradients. This law states that the heat flux \(q_\text{cond}\) through a material is proportional to the negative gradient of the temperature and the area through which heat is being transferred. The proportionality constant in this relationship is the thermal conductivity \(k\).

In the context of a wall, as described in the exercise, thermal conductivity \(k_\text{eff}\) is indicative of the material's innate ability to conduct heat. Higher \(k_\text{eff}\) values represent more efficient transfer of heat. The exercise requires the application of Fourier's law of conduction in this form:
\[q_\text{cond} = k_\text{eff} A \frac{T_\text{inner} - T_\text{surface}}{\delta}\]
Here, \(A\) stands for the area of the wall, \(T_\text{inner}\) and \(T_\text{surface}\) are the inner and outer temperatures respectively, and \(\delta\) represents the wall thickness. When the heat flow reaches a stable pattern (steady state), this formula is critical in determining the effective thermal conductivity of the material.

Steady State Thermal Analysis

Steady state thermal analysis is an assumption used in many heat transfer problems which implies that the temperature field within the domain of interest does not change with time. In other words, the temperatures at different points in the system remain constant over an extended period, despite the continuous flow of heat.

This concept simplifies the calculations quite a bit because it removes the time dependency from the equations, allowing for a straightforward calculation of heat transfer rates. In the textbook problem, steady state conditions allow us to equate the heat transfer rates at the wall's surfaces to the conduction heat transfer rate through the wall. The benefit of using steady state thermal analysis lies in its ability to simplify the complex interactions between different modes of heat transfer, facilitating the calculation of the wall's effective thermal conductivity without considering transient phenomena or time-variant boundary conditions.