Question

Consider a \(1.5\)-m-high and \(0.6-\mathrm{m}\)-wide plate whose thickness is \(0.15 \mathrm{~m}\). One side of the plate is maintained at a constant temperature of \(500 \mathrm{~K}\) while the other side is maintained at \(350 \mathrm{~K}\). The thermal conductivity of the plate can be assumed to vary linearly in that temperature range as \(k(T)=\) \(k_{0}(1+\beta T)\) where \(k_{0}=18 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\beta=8.7 \times 10^{-4} \mathrm{~K}^{-1}\). Disregarding the edge effects and assuming steady onedimensional heat transfer, determine the rate of heat conduction through the plate. Answer: \(22.2 \mathrm{~kW}\)

Short answer

Answer: The rate of heat conduction through the plate is approximately \(22.2\,\mathrm{kW}\).

Step-by-step solution

Write down the given quantities and desired output

We know the following: - Plate height (h) = \(1.5\,\mathrm{m}\) - Plate width (w) = \(0.6\,\mathrm{m}\) - Plate thickness (t) = \(0.15\,\mathrm{m}\) - Temperature on one side of the plate: \(T_1=500\,\mathrm{K}\) - Temperature on the other side of the plate: \(T_2=350\,\mathrm{K}\) - Thermal conductivity expression: \(k(T)=k_0(1+\beta T)\), with \(k_0=18\,\frac{\mathrm{W}}{\mathrm{m}\cdot \mathrm{K}}\) and \(\beta=8.7\times 10^{-4}\,\mathrm{K}^{-1}\) We want to find: Rate of heat conduction (Q) in \(\mathrm{kW}\).

Determine the average thermal conductivity of the plate

To simplify the problem, let's calculate the average thermal conductivity of the plate \((\bar{k})\): $$ \bar{k}=\frac{k(T_1)+k(T_2)}{2} $$ Plug in the values for \(k(T_1)\) and \(k(T_2)\) using the expression for \(k(T)\): $$ \bar{k}=\frac{k_0(1+\beta T_1)+k_0(1+\beta T_2)}{2} $$

Calculate the average thermal conductivity

Now, use the given values of \(k_0\), \(\beta\), \(T_1\), and \(T_2\) to find \(\bar{k}\): $$ \bar{k}=\frac{18(1+8.7\times 10^{-4}\times 500)+18(1+8.7\times 10^{-4}\times 350)}{2} $$ Calculate: $$ \bar{k}\approx20.57\,\frac{\text{W}}{\text{m}\cdot\text{K}} $$

Apply Fourier's Law of heat conduction

Using Fourier's Law in one dimension: $$ Q = \frac{(\bar{k})(h)(w)\Delta T}{t} $$ Insert values: $$ Q = \frac{(20.57)(1.5)(0.6)(500-350)}{0.15} $$

Calculate the rate of heat conduction

Now, compute Q: $$ Q\approx22,\!212\,\mathrm{W} $$ Convert watts to kilowatts: $$ Q\approx22.2\,\mathrm{kW} $$ The rate of heat conduction through the plate is approximately \(22.2 \,\mathrm{kW}\).

Key concepts

Thermal Conductivity

The concept of thermal conductivity is essential in understanding how heat moves through materials. It measures a material's ability to conduct heat. Imagine thermal conductivity as the ease with which heat travels through a material.

- Materials with high thermal conductivity - Example: Metals like copper and aluminum. - Allow heat to pass through quickly.
- Materials with low thermal conductivity - Example: Wood and plastic. - Tend to trap heat, slowing down its passage.

In this problem, the thermal conductivity of the plate is temperature-dependent, expressed as a function varying with temperature: \[k(T) = k_0 (1+\beta T)\] Where, \(k_0 = 18 \, \mathrm{W/m \, K}\) is the base thermal conductivity and \(\beta = 8.7 \times 10^{-4} \mathrm{\, K^{-1}}\) represents the rate of change of conductivity with temperature.

Fourier's Law

Named after the physicist Jean-Baptiste Joseph Fourier, Fourier's Law of heat conduction describes the flow of heat through a material. It states that the rate of heat transfer through a solid is directly proportional to the negative of the temperature gradient and the area through which the heat flows.

The mathematical representation in one dimension is:\[Q = -kA \frac{dT}{dx}\]Where:- \(Q\) is the rate of heat transfer.- \(k\) is the thermal conductivity.- \(A\) is the area through which heat is flowing.- \(\frac{dT}{dx}\) is the temperature gradient across the material.

This relationship helps us determine how much heat moves from a hot to a cold place in our given material. In the exercise, we've made use of Fourier's Law by substituting the average thermal conductivity and other attributes of the plate to find the rate of heat conduction.

Temperature Gradient

The temperature gradient is a measure of how temperature changes over a distance within a material. It's a pivotal factor in analyzing heat conduction because it essentially drives the flow of heat from warm to cold regions.

• In formula form, it's expressed as \(\frac{dT}{dx}\).
• A larger gradient means more rapid heat transfer.
• If the gradient is zero, there's no heat flow.

In the exercise, the temperature difference between one side of the plate (500 K) and the other side (350 K) creates a gradient that leads to heat moving through the plate. Calculating this gradient allows us to implement it in Fourier's Law, determining the rate of heat transfer.

Steady-State Heat Transfer

Steady-state heat transfer is a scenario where the temperature of the material remains constant over time. There are no temperature changes occurring within the system as time progresses. This simplification allows us to assume that the amount of heat entering any section of an object is equal to the amount of heat leaving it.

Applying this concept: - In any slice of the plate, the heat transferred doesn't accumulate or deplete. - Ensures predictability and ease of calculations. - Assumes that all changes reach a point where everything remains unchanged over time.
In the given exercise, assuming steady-state conditions implies a constant flow of heat through the plate without any variation, making it easier to calculate the rate of heat conduction.