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Chapter 2: Problem 11

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a thickness \(L=0.25 \mathrm{~m}\), with no internal heat generation. Determine the heat flux and the unknown quantity for each case and sketch the temperature distribution, indicating the direction of the heat flux. \begin{tabular}{crcc} \hline Case & \(T_{1}\left({ }^{\circ} \mathrm{C}\right)\) & \(T_{2}\left({ }^{\circ} \mathrm{C}\right)\) & \(d T / d x(\mathbf{K} / \mathbf{m})\) \\ \hline 1 & 50 & \(-20\) & \\ 2 & \(-30\) & \(-10\) & 160 \\ 3 & 70 & & \(-80\) \\ 4 & & 40 & 200 \\ 5 & & 30 & \\ \hline \end{tabular}

Short Answer

Expert verified
In summary, the solutions for each case are as follows: Case 1: Heat flux is \(14000 \mathrm{~W/m^2}\) from left to right, with a linear temperature distribution from \(50^\circ C\) to \(-20^\circ C\). Case 2: Heat flux is \(-3200 \mathrm{~W/m^2}\) from right to left, with a linear temperature distribution from \(-30^\circ C\) to \(30^\circ C\). Case 3: Heat flux is \(4000 \mathrm{~W/m^2}\) from right to left, with a linear temperature distribution from \(70^\circ C\) to \(90^\circ C\). Case 4: Heat flux is \(-10000 \mathrm{~W/m^2}\) from right to left, with a linear temperature distribution from \(-10^\circ C\) to \(40^\circ C\). Case 5: Not enough information to solve.

Step by step solution

01

Understand Fourier's Law of heat conduction

Fourier's law relates the heat flux (\(q''\)) to the thermal conductivity (\(k\)), the temperature gradient (\(\frac{dT}{dx}\)), and the wall thickness (\(L\)). In one-dimensional conduction, the formula is: \[ q'' = -k \frac{dT}{dx} \]
02

Apply Fourier's Law for each case

For each case, use the data provided in the table and the formula of Fourier's Law to determine the unknown quantity, which can be either temperature, temperature gradient, or heat flux. Afterward, infer the direction of the heat flux based on the temperature distribution or gradient and plot the temperature distribution graph. Case 1: \[q'' = - 50 \cdot \frac{dT}{dx}\] Here temperature values for both sides are given, and we want to calculate the heat flux. For a linear temperature distribution, the gradient will be: \(\frac{dT}{dx}=-\frac{T_2-T_1}{L}=-\frac{-20-50}{0.25}\) Now we can find the heat flux: \[q'' = -50\cdot\frac{-70}{0.25}\] \[q'' = 14000 \mathrm{~W/m^2}\] The direction of the heat flux will be from left to right since the temperature is decreasing along the positive x-direction (from 50掳C to -20掳C). Case 2: As the direction of the temperature gradient is given and positive, the direction of the heat flux (negative gradient) will be from right to left (because \(q''=-k \frac{dT}{dx}\)). The given \(\frac{dT}{dx}\) is: \(\frac{dT}{dx}=-\frac{160}{50}\) and the heat flux becomes \[q'' = -50\cdot-\frac{160}{50}\] \[q'' = -3200 \mathrm{~W/m^2}\] Now we can find the unknown temperature \(T_2\). \(T_2 - T_1 = L \frac{dT}{dx}\) \[T_2 = -10 - 0.25(-160)\] \[T_2 = 30^\circ C\] Case 3: Here we are given temperature gradient, and the heat flux is: \[q'' = -50\cdot(-80)\] \[q'' = 4000 \mathrm{~W/m^2}\] The direction of the heat flux will be from right to left since the temperature gradient is negative. Now we can find the unknown temperature \(T_2\). \(T_2 - T_1 = L \frac{dT}{dx}\) \[T_2 = 70 - 0.25(-80)\] \[T_2 = 90^\circ C\] Case 4: The temperature gradient is given, so we can find the heat flux: \[q'' = -50\cdot(200)\] \[q'' = -10000 \mathrm{~W/m^2}\] The direction of the heat flux will be from right to left since the temperature gradient is positive. Now we can find the unknown temperature \(T_1\). \(T_2 - T_1 = L \frac{dT}{dx}\) \[T_1 = 40 - 0.25(200)\] \[T_1 = -10^\circ C\] Case 5: Here we are asked to find the temperature gradient and heat flux. We do not have enough information for this case; therefore, it cannot be solved.
03

Sketch the temperature distribution

Sketch the temperature distribution graphs for each case, showing the temperature on the y-axis and the distance x on the x-axis. Indicate the direction of the heat flux on the graphs, as inferred from the temperature gradient.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fourier's Law
When studying the transmission of heat within materials, Fourier's Law is a fundamental principle that pops up everywhere, and for a good reason. It offers a simple yet powerful way to understand how heat flows. At its heart, Fourier's Law describes the relationship between the heat flux (denoted as \( q'' \)) and the temperature change across a material. The law states that the heat flux is proportional to the negative of the temperature gradient, basically telling us that heat moves from warmer to cooler regions. Mathematically, it's expressed as:

\[ q'' = -k \frac{dT}{dx} \]
Where \( k \) is the thermal conductivity of the material 鈥� a measure of its ability to conduct heat, \( dT/dx \) is the temperature gradient, showing how temperature changes with distance, and the minus sign reflects the fact that heat moves in the opposite direction of increasing temperature. It's the thermal 'traffic law', ensuring that heat knows where to 'flow'.

In the given textbook exercise, this relationship is harnessed to deduce either the heat flux, the temperature gradient, or an unknown temperature, depending on the case at hand. Understanding Fourier's Law is essential because it helps us take the temperatures of two points in a material and the material's thermal properties, and from that, predict the direction and amount of heat transfer 鈥� super useful for a wide range of real-life applications, from engineering to environmental science.
Thermal Conductivity
Imagine having a variety of materials in front of you 鈥� a piece of metal, a chunk of wood, and a block of foam. If you were to heat one end of each, how fast would the other end get warm? That's where thermal conductivity comes into play. It's a value that tells you how good a material is at transporting heat. In technical terms, thermal conductivity, denoted by \( k \), quantifies the rate at which heat is conducted through a unit area of a material per unit temperature gradient.

In the context of our one-dimensional heat conduction exercise, the thermal conductivity (\( k = 50 \mathrm{W/m\cdot K} \)) is a crucial piece of the puzzle. It's like the speed limit for how fast heat can cruise through the material. A higher \( k \) means 'faster' heat, suggesting that the material allows swift energy transfer, while a lower \( k \) mean you're stuck in 'thermal traffic'. Metals, for instance, are the autobahns of heat conduction, with very high \( k \) values, whereas insulators like wood or foam, are the country backroads where heat travels leisurely.

Knowing the thermal conductivity is invaluable when trying to determine the heat flux, or when you need to select a material for insulation or even heat exchange purposes. It's why you pick out thick oven mitts instead of a thin tea towel to grab that hot pot!
Temperature Gradient
The temperature gradient is like a roadmap that charts the temperature change over a specific distance within a material. Mathematically, it's the rate of change of temperature with respect to distance and is usually denoted by \( \frac{dT}{dx} \). It essentially tells us how steep or shallow the temperature 'hill' is that the heat must travel over.

If you had two points at different temperatures within a material, the temperature gradient is the difference between these two temperatures divided by the distance separating them. So, in your exercise, when you have two points with temperatures \( T_1 \) and \( T_2 \), and they're spaced out by a distance \( L \), you'd calculate the temperature gradient as \( \frac{dT}{dx} = \frac{T_2 - T_1}{L} \). This lets us predict how the temperature changes as we move from one location to another within the material.

Looking at the cases from the exercise, the temperature gradient is either provided directly, like in Case 2, or indirectly, where we calculate it using the temperatures given. It's the thermal equivalent of a slope 鈥� a steeper gradient means a quicker change in temperature over a short distance, while a gentle gradient signifies a gradual change. This gradient governs the direction and magnitude of the heat flux and is essential for visualizing and understanding how heat spreads out in different scenarios.

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Most popular questions from this chapter

The steady-state temperature distribution in a onedimensional wall of thermal conductivity \(k\) and thickness \(L\) is of the form \(T=a x^{3}+b x^{2}+c x+d .\) Derive expressions for the heat generation rate per unit volume in the wall and the heat fluxes at the two wall faces \((x=0, L)\).

A spherical shell of inner and outer radii \(r_{i}\) and \(r_{o}\), respectively, contains heat-dissipating components, and at a particular instant the temperature distribution in the shell is known to be of the form $$ T(r)=\frac{C_{1}}{r}+C_{2} $$ Are conditions steady-state or transient? How do the heat flux and heat rate vary with radius?

Typically, air is heated in a hair dryer by blowing it across a coiled wire through which an electric current is passed. Thermal energy is generated by electric resistance heating within the wire and is transferred by convection from the surface of the wire to the air. Consider conditions for which the wire is initially at room temperature, \(T_{i}\), and resistance heating is concurrently initiated with airflow at \(t=0\). (a) For a wire radius \(r_{o}\), an air temperature \(T_{\infty}\), and a convection coefficient \(h\), write the form of the heat equation and the boundary/initial conditions that govern the transient thermal response, \(T(r, t)\), of the wire. (b) If the length and radius of the wire are \(500 \mathrm{~mm}\) and \(1 \mathrm{~mm}\), respectively, what is the volumetric rate of thermal energy generation for a power consumption of \(P_{\text {elec }}=500 \mathrm{~W}\) ? What is the convection heat flux under steady-state conditions? (c) On \(T-r\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (d) On \(q_{r}^{\prime \prime}-t\) coordinates, sketch the variation of the heat flux with time for locations at \(r=0\) and \(r=r_{o^{*}}\).

Consider a plane wall \(100 \mathrm{~mm}\) thick and of thermal conductivity \(100 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Steady-state conditions are known to exist with \(T_{1}=400 \mathrm{~K}\) and \(T_{2}=600 \mathrm{~K}\). Determine the heat flux \(q_{x}^{\prime \prime}\) and the temperature gradient \(d T / d x\) for the coordinate systems shown.

A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The steady-state temperature distribution is \(T(x)=a-b x / L\), where \(a=305 \mathrm{~K}\) and \(b=10 \mathrm{~K}\). The diameter and length of the rod are \(D=20 \mathrm{~mm}\) and \(L=100 \mathrm{~mm}\), respectively. Determine the heat flux along the rod, \(q_{x}^{\prime \prime}\). Hint: The mass of the rod is \(M=0.248 \mathrm{~kg}\).
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