Question

Consider a long solid cylinder of radius \(r_{o}=4 \mathrm{~cm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Heat is generated in the cylinder uniformly at a rate of \(\dot{e}_{\text {gen }}=35 \mathrm{~W} / \mathrm{cm}^{3}\). The side surface of the cylinder is maintained at a constant temperature of \(T_{s}=80^{\circ} \mathrm{C}\). The variation of temperature in the cylinder is given by $$ T(r)=\frac{\dot{e}_{\text {gen }} r_{o}^{2}}{k}\left[1-\left[1-\left(\frac{r}{r_{o}}\right)^{2}\right]+T_{s}\right. $$ Based on this relation, determine \((a)\) if the heat conduction is steady or transient, \((b)\) if it is one-, two-, or three-dimensional, and \((c)\) the value of heat flux on the side surface of the cylinder at \(r=r_{o^{*}}\)

Short answer

Is it one-, two-, or three-dimensional? What is the heat flux on the side surface of the cylinder at r = r_o? Answer: The heat conduction is steady and one-dimensional. The heat flux on the side surface of the cylinder at r = r_o is 1120 W/cm².

Step-by-step solution

Determine if the heat conduction is steady or transient

In a steady heat conduction process, the temperature does not change over time. To check if it's steady or transient, we need to examine if time (t) is present in the temperature equation given: $$ T(r) = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \left[1-\left(1-\left(\frac{r}{r_{o}}\right)^{2}\right)\right] + T_{s} $$ No time-dependent term (t) is present in the equation. Therefore, the heat conduction is steady. (a) The heat conduction is steady.

Determine if the heat conduction is one-, two-, or three-dimensional

To identify if the heat conduction is one-, two-, or three-dimensional, we need to examine how the temperature varies with different coordinates. The given temperature equation has only the radial coordinate (r) playing a role in temperature variation: $$ T(r) = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \left[1-\left(1-\left(\frac{r}{r_{o}}\right)^{2}\right)\right] + T_{s} $$ Since only the radial coordinate (r) affects the temperature, it is a one-dimensional heat conduction. (b) The heat conduction is one-dimensional.

Calculate the heat flux on the side surface of the cylinder at r = r_{o}

To find the heat flux at the side surface of the cylinder, we need to compute the radial heat flux at \(r = r_{o}\) using Fourier's Law of heat conduction: $$ q_r = -k\frac{dT}{dr} $$ First, we compute the derivative of the temperature function with respect to r: $$ \frac{dT}{dr} = \frac{\dot{e}_{\text{gen}} r_{o}^{2}}{k} \cdot \left(\frac{-2r}{r_{o}^2}\right) = - \frac{2 \dot{e}_{\text{gen}} r_{o} r}{k} $$ Now, substituting the values of \(k\), \(\dot{e}_{\text{gen}}\), and \(r_{o}\), and calculating the heat flux at the side surface \(r = r_{o}\): $$ q_{r_{o}} = -k \cdot \left(- \frac{2 \dot{e}_{\text{gen}} r_{o} r_{o}}{k}\right) = 2 \dot{e}_{\text{gen}} r_{o}^2 = 2 \cdot 35 \mathrm{~W/cm^3} \cdot (4\mathrm{~cm})^2 = 1120 \mathrm{~W/cm^2} $$ (c) The heat flux on the side surface of the cylinder at \(r = r_{o}\) is 1120 W/cm².

Key concepts

Steady State Heat Transfer

In heat conduction, the term "steady state" refers to a condition where the temperature profile within a medium does not change as time progresses. This implies that the heat entering one side is equal to the heat leaving the other side, maintaining a constant temperature distribution over time. In our example, the temperature equation provided does not include a time variable, meaning it's independent of any changes over time. This characteristic is crucial because problems involving steady-state heat transfer often require simpler mathematical treatment, as we do not have to solve complex transient heat equations. Knowing that a system is in steady state helps predict how a material will behave under consistent thermal conditions.

One-Dimensional Heat Conduction

One-dimensional heat conduction occurs when the heat transfer within a material depends on a single spatial variable. In simpler terms, it only needs one direction to describe how heat moves across the object. In the case of the cylinder from the exercise, the temperature changes are linked solely to the radial coordinate \(r\), with no dependency on the axial or circumferential coordinates. This indicates one-dimensional heat conduction. This simplification is vital for making calculation manageable and instincts in situations where temperature changes in other dimensions are negligible or constant. By reducing the problem to a single dimension, you can use simpler mathematical approaches to predict how heat moves within the material.

Heat Flux Calculation

Heat flux quantifies the rate of heat energy transfer through a specific surface area. It is essentially the amount of heat flowing per unit area per unit time, often expressed in \ \(\mathrm{W/cm^2}\ \) or \ \(\mathrm{W/m^2}\ \). In the exercise, we use Fourier’s Law, which states that the heat flux \(q_r\) through a surface is proportional to the negative of the temperature gradient:

• \(q_r = -k \frac{dT}{dr} \)

To find the heat flux, we need to determine the derivative of the temperature with respect to \(r\). This derivative captures how quickly temperature changes at a particular distance \(r\) from the center. After calculating this derivative and multiplying with the material's thermal conductivity \(k\), the heat flux at \(r = r_{o}\) is 1120 \(\mathrm{W/cm^2}\), reflecting the high rate of energy transfer happening across the surface of the cylinder. Understanding heat flux is crucial in design and safety, ensuring that materials can withstand the heat energy flow that occurs in various engineering applications.