Question

A large plane wall has a thickness \(L=50 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the left surface \((x=0)\), it is subjected to a uniform heat flux \(\dot{q}_{0}\) while the surface temperature \(T_{0}\) is constant. On the right surface, it experiences convection and radiation heat transfer while the surface temperature is \(T_{L}=225^{\circ} \mathrm{C}\) and the surrounding temperature is \(25^{\circ} \mathrm{C}\). The emissivity and the convection heat transfer coefficient on the right surface are \(0.7\) and \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Show that the variation of temperature in the wall can be expressed as \(T(x)=\left(\dot{q}_{0} / k\right)(L-x)+T_{L}\), where \(\dot{q}_{0}=5130 \mathrm{~W} / \mathrm{m}^{2}\), and determine the temperature of the left surface of the wall at \(x=0\).

Short answer

Answer: The temperature at the left surface of the wall at \(x=0\) is \(531^{\circ}\mathrm{C}\).

Step-by-step solution

Derive the temperature equation in the wall

Since the temperature distribution in the wall is linear, we can apply Fourier's law of heat conduction to find the heat flux in the wall. This relation is given by: \(\dot{q}=-k\frac{dT}{dx}\) For the given temperature equation, \(T(x)=\left(\dot{q}_{0} / k\right)(L-x)+T_{L}\), we first find the derivative \(\frac{dT}{dx}\): \(\frac{dT}{dx}=-\frac{\dot{q}_{0}}{k}\) Now, we substitute this into Fourier's law: \(\dot{q} = -k\left(-\frac{\dot{q}_{0}}{k}\right)\) So, the heat flux in the wall is: \(\dot{q}=\dot{q}_{0}\)

Determine the temperature at the left surface of the wall at \(x=0\)

Now that we have shown that the temperature equation holds, we can find the temperature at the left surface of the wall, \(x=0\). Using the given temperature equation, evaluate the temperature \(T(0)\): \(T(0) = \left(\frac{\dot{q}_{0}}{k}\right)(L-0)+T_{L}\) Given values: \(\dot{q}_{0}=5130 \mathrm{~W} / \mathrm{m}^{2}\), \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(T_L=225^{\circ} \mathrm{C}\), \(L=50 \mathrm{~cm}=0.5\mathrm{~m}\) Substituting these values, we get: \(T(0) = \left(\frac{5130}{25}\right)(0.5-0)+225\) \(T(0) = 306+225\) \(T(0) = 531^{\circ}\mathrm{C}\) The temperature at the left surface of the wall at \(x=0\) is \(531^{\circ}\mathrm{C}\).

Key concepts

Fourier's Law of Heat Conduction

Fourier's law of heat conduction is essential for understanding how heat travels through materials. Simply put, this law states that the heat conduction rate within a solid is proportional to the negative gradient of the temperature field. In other words, heat will flow from regions of higher temperature to regions of lower temperature, and the rate of this flow depends on the material's properties.

Mathematically, this law is expressed as \[\dot{q}=-k\frac{dT}{dx}\], where \(\dot{q}\) is the heat flux (heat transfer per unit area per unit time), \(k\) is the material's thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient in the direction of heat flow. In our exercise, we confirmed the linear temperature distribution in a solid wall by applying Fourier's law, which led to the heat flux \(\dot{q}=\dot{q}_{0}\), demonstrating that the applied uniform heat flux is constant across the wall.

Thermal Conductivity

Thermal conductivity (\(k\)) is a measure of a material's ability to conduct heat. It appears as a crucial factor in Fourier's law, directly affecting the rate at which heat is transferred through a solid. A high thermal conductivity indicates that the material can transfer heat quickly, whereas a low thermal conductivity means heat transfers more slowly.

This property is inherent to the material and is usually tabulated in engineering reference materials for various substances. For instance, metals often have high thermal conductivity while insulating materials such as wood or foam have much lower values. In the context of the exercise, the thermal conductivity of the wall determines how effectively it can dissipate heat, impacting how the temperature changes across its thickness.

Convection Heat Transfer

Convection heat transfer is one of the ways heat is transferred between a solid surface and a fluid (liquid or gas) in motion. This process is governed by the convection heat transfer coefficient (\(h\)), which determines how effective the fluid is at removing heat from the surface.

The amount of heat transferred by convection is given by Newton's law of cooling, which states that the heat transfer rate is proportional to the product of the convection heat transfer coefficient, the surface area, and the difference in temperature between the solid surface and the fluid. In our exercise, the right surface of the wall experiences convection heat transfer to the surrounding air with a given coefficient of \(15\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), indicating the cooling effect of the air on the wall.

Radiation Heat Transfer

Radiation heat transfer is the emission of energy from a surface in the form of electromagnetic waves, such as the infrared radiation emitted by a heated object. Unlike conduction and convection, radiation does not require a medium and can occur in a vacuum.

This mode of heat transfer is influenced by the surface's emissivity (\(\epsilon\)), which ranges from 0 to 1 and represents how effectively the surface emits thermal radiation compared to an ideal black body. The Stefan-Boltzmann law describes radiation heat transfer as proportional to the fourth power of the absolute temperature of the emitting surface. In our exercise setup, radiation also plays a role in heat dissipation from the wall, along with convection, emphasizing the significance of considering multiple modes of heat transfer when examining thermal processes.