Problem 7 - An Introduction to Materials Engineering and Science for Chemical and Materials Engineers

Question

The thermal conductivity of a material varies with temperature as $$ \ln k=0.01 T+0.05 $$ where \(k\) has units of \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\), and \(T\) has units of \({ }^{\circ} \mathrm{C}\). Heat flows by conduction through a plane slab of this material of thickness \(0.1 \mathrm{~m}\), the left face of which is at \(100^{\circ} \mathrm{C}\) and the right face of which is at \(0^{\circ} \mathrm{C}\). Calculate (a) the mean thermal conductivity of this material in the range \(0-100^{\circ} \mathrm{C}\), (b) the heat flux through the slab, and (c) the temperature at \(x=0.05 \mathrm{~m}\).

Step-by-Step Solution

Answer

(a) Mean thermal conductivity: \( 0.01 (e^{1.05} - e^{0.05}) \). (b) Heat flux: \( q = 0.01 (e^{1.05} - e^{0.05}) \times 1000 \). (c) Temperature at x=0.05 m: 50°C.

Step 1: Convert the Thermal Conductivity Function

The function giving the thermal conductivity is \( \ln k = 0.01 T + 0.05 \). To find \( k \), we need to exponentiate both sides. That gives us \( k = e^{0.01T + 0.05} \).

Step 2: Calculate Mean Thermal Conductivity

Use the formula for mean value of a continuous function over an interval \[ \bar{k} = \frac{1}{T_2-T_1} \int_{T_1}^{T_2} k(T) \, dT. \] Where \( T_1 = 0^{\circ}C \) and \( T_2 = 100^{\circ}C \). Thus we have: \[ \bar{k} = \frac{1}{100} \int_{0}^{100} e^{0.01T + 0.05} \, dT. \]Setting \( u = 0.01T + 0.05 \) gives \( du = 0.01 \, dT \), thus the integral becomes: \[ \int e^{u} \, \frac{du}{0.01} = \frac{1}{0.01} e^{u}. \]After evaluating this from \( T = 0 \) to \( T = 100 \), we have:\[ \left( \frac{1}{0.01} e^{1.05} - \frac{1}{0.01} e^{0.05} \right) \cdot \frac{1}{100} = 0.01 \left( e^{1.05} - e^{0.05} \right). \]

Step 3: Calculate Heat Flux

Use Fourier's law of heat conduction: \[ q = -k_{mean} \frac{dT}{dx} \]Substitute \( \frac{dT}{dx} = \frac{100 - 0}{0.1} = 1000 \) Thus: \[ q = \bar{k} \times 1000 \]Using \( \bar{k} = 0.01 (e^{1.05} - e^{0.05}) \), and evaluating gives the heat flux: \[ q = 0.01 (e^{1.05} - e^{0.05}) \times 1000. \]

Step 4: Calculate Temperature at x=0.05 m

Assume linear temperature distribution across the slab. The linear interpolation formula to find temperature at a point is: \[ T(x) = T_1 + \left( \frac{T_2-T_1}{L} \right) x. \]Using \( T_1 = 100^{\circ}C \), \( T_2 = 0^{\circ}C \), and \( x = 0.05 \), and \( L = 0.1 \), we have: \[ T(0.05) = 100 + \left( \frac{0 - 100}{0.1} \right) \times 0.05 = 50^{\circ}C. \]

Key Concepts

heat flux

Heat flux describes the flow of heat through a material, and it is essential in understanding how different materials transfer thermal energy. This can be imagined as the rate at which heat crosses an area within a certain time frame.
Fourier's law of heat conduction governs the concept of heat flux in solid materials. It states that the heat flux (\( q \)) through a material is directly proportional to the negative gradient of temperature and the material's thermal conductivity. Mathematically, it is expressed as:

• \( q = -k abla T \)

where \( k \) is the thermal conductivity, and \( abla T \) represents the temperature gradient across the material.

For our problem, once the mean thermal conductivity is calculated, we apply Fourier's law. The exercise uses a calculated mean value of thermal conductivity, \( \bar{k} \), and a linear temperature gradient to find heat flux. The quick change in temperature over a small slab thickness leads to a sharp decline, allowing us to express the gradient as \( \frac{dT}{dx} = \frac{100 - 0}{0.1} \). This shows how swiftly temperatures change across the slab and is crucial for practical engineering design.

temperature distribution

Temperature distribution in materials is about how temperature varies from point to point within the material. The notion becomes vital when dealing with objects or systems where thermal variations need analyzing, such as in insulating walls or engine components.
In a simple scenario like a plane slab, assuming linear temperature distribution is common, as it provides a straightforward way of interpolation between known boundary temperatures. This assumption means the temperature changes at a constant rate from one side to the other.
For the exercise, this meant calculating the temperature at a particular point, given the slab's end temperatures:

• Left face at \( 100^{\circ} \text{C} \)
• Right face at \( 0^{\circ} \text{C} \)

By using linear interpolation, the task was to find the temperature at a midpoint 0.05 m from the start. The formula used was:

• \( T(x) = T_1 + \left(\frac{T_2-T_1}{L}\right) x \)

This equation helps in predicting temperature at any point between the two ends, guiding engineers to understand how heat moves through materials.

Fourier's law of heat conduction

Fourier's law of heat conduction forms the foundation of studying heat transfer in materials. It details the relation between heat flux and temperature variations in conductive systems.Essentially, Fourier's law connects the heat transferred through a material to the temperature difference across its faces and the material's ability to conduct heat. If a material has high thermal conductivity, it can transmit heat more efficiently, while low thermal conductivity means less efficient heat flow.

• Mathematically expressed: \( q = -k abla T \)

In practical terms, this law dictates that heat naturally flows from a region of high temperature to a low temperature, and inversely proportional to the thickness of the material – the thicker the material, the less rapid the temperature change.Applying this concept helps in solving problems like those in the exercise, where the material's varying thermal conductivity must be considered over a temperature range. Such scenarios are crucial in engineering to decide on materials or conditions appropriate for specific heat transfer requirements.