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Chapter 2: Problem 12

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.

Short Answer

Expert verified
Under steady-state conditions for the given plane wall with a thickness of \(0.1~m\) and thermal conductivity of \(100~\mathrm{W/(m\cdot K)}\), the temperature gradient is found to be \(2000~K/m\) and the heat flux is \(-200000~\mathrm{W/m^2}\). The negative sign indicates that the heat transfer is in the opposite direction of the coordinate system shown.

Step by step solution

01

Apply Fourier's law of heat conduction

Fourier's law of heat conduction relates heat flux, thermal conductivity, and temperature gradient as follows: \(q_x^{\prime \prime} = -k\frac{dT}{dx}\) where: - \(q_x^{\prime \prime}\) is the heat flux in the x direction in \(\mathrm{W/m^2}\) - k is the thermal conductivity in \(\mathrm{W/(m\cdot K)}\) - \(\frac{dT}{dx}\) is the temperature gradient along the x direction in \(K/m\)
02

Calculate the temperature difference

Given the temperatures at both sides of the wall, we can find the temperature difference: \(\Delta T = T_2 - T_1\) Plugging in the given values: \(\Delta T = 600~K - 400~K = 200~K\)
03

Determine the temperature gradient

The temperature gradient in a plane wall of uniform thickness under steady-state conditions can be expressed as: \(\frac{dT}{dx} = \frac{\Delta T}{\delta}\) where: - \(\delta\) is the wall thickness in meters Converting the thickness from millimeters to meters: \(\delta = 100~mm \times \frac{1~m}{1000~mm} = 0.1~m\) Now, plugging in the values for temperature difference and wall thickness: \(\frac{dT}{dx} = \frac{200~K}{0.1~m} = 2000~K/m\)
04

Calculate the heat flux

Now we can use the temperature gradient and thermal conductivity in Fourier's law of heat conduction to find the heat flux: \(q_x^{\prime \prime} = -k\frac{dT}{dx}\) Plugging in the given values for thermal conductivity and the calculated temperature gradient: \(q_x^{\prime \prime} = -(100~\mathrm{W/(m\cdot K)}) \times (2000~K/m) = -200000~\mathrm{W/m^2}\)
05

Express the answers

Now we have the values for the temperature gradient and heat flux: Temperature gradient: \( \frac{dT}{dx} = 2000~K/m\) Heat flux: \(q_x^{\prime \prime} = -200000~\mathrm{W/m^2}\) The negative sign for the heat flux indicates that the heat transfer is in the opposite direction of the coordinate system shown.

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Key Concepts

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

Fourier's Law
In the realm of heat conduction, Fourier's Law is pivotal. Imagine a hot summer day, and one side of a wall exposed to the sun feels significantly warmer than the other. Fourier's Law helps explain how and why heat moves through that wall. It states that the heat flux, or the rate of heat transfer through a material, is proportional to the negative gradient of temperature and the material's thermal conductivity.

This relationship is expressed mathematically in the equation:
  • \(q_x^{\prime \prime} = -k\frac{dT}{dx}\)
Here, \(q_x^{\prime \prime}\) is the heat flux, \(k\) represents thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient.
To visualize this, picture the wall with uniform heat moving from hot to cold. The negative sign indicates this natural flow downhill on the temperature gradient, from higher to lower temperatures.
Thermal Conductivity
Thermal conductivity, often symbolized as \(k\), is a measure of a material's ability to conduct heat. Imagine how different substances feel to the touch; metals are often cool because they quickly transfer heat away from your hand, demonstrating high thermal conductivity.

In the exercise example, the wall has a specific thermal conductivity value of \(100 \mathrm{W/(m \cdot K)}\). This unit tells us how much heat energy in watts passes through a square meter of the material with a temperature gradient of 1 Kelvin per meter. Materials with higher thermal conductivity transfer heat more efficiently, while those with low thermal conductivity act as better insulators.
  • High \(k\) value: efficient heat conductor (e.g., copper)
  • Low \(k\) value: good insulator (e.g., wood)
Understanding thermal conductivity helps not only in science but in everyday applications, such as selecting materials for building insulation.
Temperature Gradient
The concept of a temperature gradient is crucial in understanding heat conduction. A gradient in general refers to a change in a quantity over a certain distance. In the case of heat conduction, the temperature gradient is the rate of temperature change over a distance in the material. Simply put, it's how quickly the temperature changes from one side of the material to the other.

From the exercise problem, we have a temperature gradient calculated as \(\frac{dT}{dx} = 2000~K/m\). This means for every meter, the temperature changes by 2000 Kelvin.
  • Steeper gradient = faster temperature change
  • Shallower gradient = slower temperature change
In steady-state conditions like in our exercise, the temperature gradient remains constant and helps us determine the direction and rate of heat flow across the material.

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Most popular questions from this chapter

A pan is used to boil water by placing it on a stove, from which heat is transferred at a fixed rate \(q_{\sigma}\). There are two stages to the process. In Stage 1, the water is taken from its initial (room) temperature \(T_{i}\) to the boiling point, as heat is transferred from the pan by natural convection. During this stage, a constant value of the convection coefficient \(h\) may be assumed, while the bulk temperature of the water increases with time, \(T_{\infty}=T_{\infty}(t)\). In Stage 2, the water has come to a boil, and its temperature remains at a fixed value, \(T_{\infty}=T_{b}\), as heating continues. Consider a pan bottom of thickness \(L\) and diameter \(D\), with a coordinate system corresponding to \(x=0\) and \(x=L\) for the surfaces in contact with the stove and water, respectively. (a) Write the form of the heat equation and the boundary/ initial conditions that determine the variation of temperature with position and time, \(T(x, t)\), in the pan bottom during Stage 1. Express your result in terms of the parameters \(q_{o}, D, L, h\), and \(T_{\infty}\), as well as appropriate properties of the pan material. (b) During Stage 2, the surface of the pan in contact with the water is at a fixed temperature, \(T(L, t)=\) \(T_{L}>T_{b}\). Write the form of the heat equation and boundary conditions that determine the temperature distribution \(T(x)\) in the pan bottom. Express your result in terms of the parameters \(q_{o}, D, L\), and \(T_{L}\), as well as appropriate properties of the pan material.

A spherical shell with inner radius \(r_{1}\) and outer radius \(r_{2}\) has surface temperatures \(T_{1}\) and \(T_{2}\), respectively, where \(T_{1}>T_{2}\). Sketch the temperature distribution on \(T-r\) coordinates assuming steady- state, one-dimensional conduction with constant properties. Briefly justify the shape of your curve.

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}

Uniform internal heat generation at \(\dot{q}=5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) is occurring in a cylindrical nuclear reactor fuel rod of 50 -mm diameter, and under steady-state conditions the temperature distribution is of the form \(T(r)=a+b r^{2}\), where \(T\) is in degrees Celsius and \(r\) is in meters, while \(a=800^{\circ} \mathrm{C}\) and \(b=-4.167 \times 10^{5}{ }^{\circ} \mathrm{C} / \mathrm{m}^{2}\). The fuel rod properties are \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (a) What is the rate of heat transfer per unit length of the rod at \(r=0\) (the centerline) and at \(r=25 \mathrm{~mm}\) (the surface)? (b) If the reactor power level is suddenly increased to \(\dot{q}_{2}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what is the initial time rate of temperature change at \(r=0\) and \(r=25 \mathrm{~mm}\) ?

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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