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Chapter 2: Problem 13

A cylinder of radius \(r_{o}\), length \(L\), and thermal conductivity \(k\) is immersed in a fluid of convection coefficient \(h\) and unknown temperature \(T_{\infty}\). At a certain instant the temperature distribution in the cylinder is \(T(r)=a+b r^{2}\), where \(a\) and \(b\) are constants. Obtain expressions for the heat transfer rate at \(r_{o}\) and the fluid temperature.

Short Answer

Expert verified
The heat transfer rate at the radius \(r_o\) is given by: \(q_{cond} = -k * 2\pi r_{o}L * (2br_{o})\) And the fluid temperature \(T_{\infty}\) is given by: \(T_{\infty} = a + b r_{o}^{2} + \frac{2bk r_{o}}{h}\)

Step by step solution

01

Differentiate the temperature distribution formula with respect to 'r'

First, we need to find the temperature gradient at the outer surface of the cylinder (r = r_o). To do this, we will differentiate the temperature distribution formula T(r) = a + b*r^2 with respect to 'r': dT/dr = d(a + br^2) / dr = 2br
02

Apply Fourier's Law of heat conduction

Next, we'll use Fourier's Law of heat conduction to find the heat transfer rate at the outer surface of the cylinder: q = -k * A_surface * dT/dr where, k = thermal conductivity of the cylinder A_surface = surface area of the outer surface of the cylinder = 2蟺r_oL So, the heat transfer rate at r_o: q_cond = -k * 2蟺r_oL * (2br_o)
03

Apply convection heat transfer equation

Now, we need to determine the heat transfer rate driven by convection: q_conv = h * A_surface * (T(r_o) - T_infinity) Plugging the surface area and the temperature profile at r_o: q_conv = h * 2蟺r_oL * (a + b*r_o^2 - T_infinity)
04

Set the conduction and convection heat transfer rates equal

In steady state, the heat transfer rate due to conduction should be equal to the heat transfer rate due to convection: q_cond = q_conv -k * 2蟺r_oL * (2br_o) = h * 2蟺r_oL * (a + b*r_o^2 - T_infinity)
05

Solve for T_infinity

Now, we'll solve the equation above for T_infinity: T_infinity = a + b*r_o^2 + (2bk*r_o) / h So, we have expressions for both the heat transfer rate at r_o and the fluid temperature T_infinity.

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Key Concepts

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

Cylinder
When discussing heat transfer in the context of a cylinder, we're looking at heat flow through a solid object that is shaped like a tube. The cylinder here is specified by its radius, denoted as \( r_o \), and length \( L \). Because heat can flow in different ways (radially or axially), the geometry of the cylinder helps determine how we calculate the heat transfer.
To calculate the heat transfer through the surface of a cylinder, it's important to know its surface area. The outer surface area of a cylinder can be given by the formula:
  • Surface area, \( A_{surface} = 2\pi r_oL \)
This expression accounts for the cylindrical shape, considering both the circular and length dimensions.
Understanding the dimensions and shape of a cylinder is crucial when applying heat transfer equations because they influence how temperature changes with time and distance across the material.
Fourier's Law
Fourier's Law is a fundamental principle used to describe how heat conduction occurs through materials. In a simplified sense, it relates to how quickly heat moves through a particular material due to a temperature difference. This law is mathematically expressed as:
  • \( q = -kA \frac{dT}{dr} \)
Here,
  • \( q \) is the heat transfer rate,
  • \( k \) is the thermal conductivity of the material,
  • \( A \) is the area through which heat is being transferred,
  • \( \frac{dT}{dr} \) is the temperature gradient.
For a cylindrical geometry, heat transfer happens across the outer surface, which means you use the outer area \( 2\pi r_oL \) and the radial temperature gradient \( \left(\frac{dT}{dr}\right) \).
This equation becomes crucial in understanding the rate at which heat is conducted through the cylinder's surface. It tells us that the heat transfer rate is directly proportional to the thermal conductivity, surface area, and temperature gradient.
Convection Coefficient
The convection coefficient, symbolized as \( h \), is a key parameter in analyzing heat transfer involving fluid interactions. It measures the effectiveness of heat transfer between a solid surface and the adjacent fluid. The higher the convection coefficient, the more effectively heat energy is removed or added due to fluid flow.
In our example, heat from the cylinder is transferred to the fluid surrounding it; hence, understanding the convection coefficient means acknowledging how quickly this heat can be picked up or dissipated by the fluid. The convective heat transfer is given by:
  • \( q_{conv} = hA_{surface}(T_{surface} - T_{\infty}) \)
Here,
  • \( T_{surface} \) is the temperature at the surface of the cylinder,
  • \( T_{\infty} \) is the temperature of the fluid far away from the cylinder.
These factors together help determine the rate of convective heat transfer, illustrating the interactions between solid surfaces and moving fluids.
Temperature Distribution
Temperature distribution is a way to describe how temperature varies within different parts of an object. For our cylinder, this is given by the formula \( T(r) = a + br^2 \), where \( a \) and \( b \) are constants. This formula represents how temperature is not a single value but varies as distance changes from the center (\( r=0 \)) to the outer surface (\( r = r_o \)).
In our scenario, the aim is to find the gradient or the change in temperature relative to change in position \( r \). This is needed because it forms a crucial part of determining the heat transfer rate by conduction through Fourier's Law.
  • By differentiating, we have \( \frac{dT}{dr} = 2br \),
This indicates how quickly temperature changes as you move radially outwards through the cylinder. By knowing this distribution and its gradient, you can then connect these concepts to predict real outcomes like the temperature at the fluid boundary or the energy transferred to the fluid.

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Most popular questions from this chapter

Two-dimensional, steady-state conduction occurs in a hollow cylindrical solid of thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), outer radius \(r_{o}=1 \mathrm{~m}\) and overall length \(2 z_{o}=5 \mathrm{~m}\), where the origin of the coordinate system is located at the midpoint of the center line. The inner surface of the cylinder is insulated, and the temperature distribution within the cylinder has the form \(T(r, z)=a+b r^{2}+c \ln r+d z^{2}\), where \(a=\) \(-20^{\circ} \mathrm{C}, \quad b=150^{\circ} \mathrm{C} / \mathrm{m}^{2}, c=-12^{\circ} \mathrm{C}, d=-300^{\circ} \mathrm{C} / \mathrm{m}^{2}\) and \(r\) and \(z\) are in meters. (a) Determine the inner radius \(r_{i}\) of the cylinder. (b) Obtain an expression for the volumetric rate of heat generation, \(\dot{q}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\). (c) Determine the axial distribution of the heat flux at the outer surface, \(q_{r}^{\prime \prime}\left(r_{o}, z\right)\). What is the heat rate at the outer surface? Is it into or out of the cylinder? (d) Determine the radial distribution of the heat flux at the end faces of the cylinder, \(q_{r}^{\prime \prime}\left(r,+z_{o}\right)\) and \(q_{r}^{\prime \prime}\left(r,-z_{o}\right)\). What are the corresponding heat rates? Are they into or out of the cylinder? (e) Verify that your results are consistent with an overall energy balance on the cylinder.

An electric cable of radius \(r_{1}\) and thermal conductivity \(k_{c}\) is enclosed by an insulating sleeve whose outer surface is of radius \(r_{2}\) and experiences convection heat transfer and radiation exchange with the adjoining air and large surroundings, respectively. When electric current passes through the cable, thermal energy is generated within the cable at a volumetric rate \(\dot{q}\). (a) Write the steady-state forms of the heat diffusion equation for the insulation and the cable. Verify that these equations are satisfied by the following temperature distributions: Insulation: \(T(r)=T_{s, 2}+\left(T_{s, 1}-T_{s, 2}\right) \frac{\ln \left(r / r_{2}\right)}{\ln \left(r_{1} / r_{2}\right)}\) Cable: \(T(r)=T_{s, 1}+\frac{\dot{q} r_{1}^{2}}{4 k_{c}}\left(1-\frac{r^{2}}{r_{1}^{2}}\right)\) Sketch the temperature distribution, \(T(r)\), in the cable and the sleeve, labeling key features. (b) Applying Fourier's law, show that the rate of conduction heat transfer per unit length through the sleeve may be expressed as $$ q_{r}^{\prime}=\frac{2 \pi k_{s}\left(T_{s, 1}-T_{s, 2}\right)}{\ln \left(r_{2} / r_{1}\right)} $$ Applying an energy balance to a control surface placed around the cable, obtain an alternative expression for \(q_{r}^{\prime}\), expressing your result in terms of \(\dot{q}\) and \(r_{1^{*}}\) (c) Applying an energy balance to a control surface placed around the outer surface of the sleeve, obtain an expression from which \(T_{s, 2}\) may be determined as a function of \(\dot{q}, r_{1}, h, T_{\infty}, \varepsilon\), and \(T_{\text {sur- }}\) (d) Consider conditions for which \(250 \mathrm{~A}\) are passing through a cable having an electric resistance per unit length of \(R_{e}^{\prime}=0.005 \Omega / \mathrm{m}\), a radius of \(r_{1}=15 \mathrm{~mm}\), and a thermal conductivity of \(k_{c}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). For \(k_{s}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad r_{2}=15.5 \mathrm{~mm}, \quad h=25\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}, \varepsilon=0.9, T_{\text {o }}=25^{\circ} \mathrm{C}\), and \(T_{\text {sur }}=35^{\circ} \mathrm{C}\), evaluate the surface temperatures, \(T_{s, 1}\) and \(T_{s, 2}\), as well as the temperature \(T_{o}\) at the centerline of the cable. (e) With all other conditions remaining the same, compute and plot \(T_{o}, T_{s, 1}\), and \(T_{s, 2}\) as a function of \(r_{2}\) for \(15.5 \leq r_{2} \leq 20 \mathrm{~mm}\).

A plane wall of thickness \(L=0.1 \mathrm{~m}\) experiences uniform volumetric heating at a rate \(\dot{q}\). One surface of the wall \((x=0)\) is insulated, and the other surface is exposed to \(\mathrm{a}\) fluid at \(T_{\infty}=20^{\circ} \mathrm{C}\), with convection heat transfer characterized by \(h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Initially, the temperature distribution in the wall is \(T(x, 0)=a+b x^{2}\), where \(a=300^{\circ} \mathrm{C}, b=-1.0 \times 10^{40} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. Suddenly, the volumetric heat generation is deactivated ( \(\dot{q}=0\) for \(t \geq 0\) ), while convection heat transfer continues to occur at \(x=L\). The properties of the wall are \(\rho=7000 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=450 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=90 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) Determine the magnitude of the volumetric energy generation rate \(\dot{q}\) associated with the initial condition \((t<0)\). (b) On \(T-x\) coordinates, sketch the temperature distribution for the following conditions: initial condition \((t<0)\), steady-state condition \((t \rightarrow \infty)\), and two intermediate conditions. (c) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the variation with time of the heat flux at the boundary exposed to the convection process, \(q_{x}^{\prime \prime}(L, t)\). Calculate the corresponding value of the heat flux at \(t=0, q_{x}^{\prime \prime}(L, 0)\). (d) Calculate the amount of energy removed from the wall per unit area \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) by the fluid stream as the wall cools from its initial to steady-state condition.

A plane wall of thickness \(2 L=40 \mathrm{~mm}\) and thermal conductivity \(k=5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation at a rate \(\dot{q}\), while convection heat transfer occurs at both of its surfaces \((x=-L,+L)\), each of which is exposed to a fluid of temperature \(T_{\infty}=20^{\circ} \mathrm{C}\). Under steady-state conditions, the temperature distribution in the wall is of the form \(T(x)=a+b x+c x^{2}\) where \(a=82.0^{\circ} \mathrm{C}, b=-210^{\circ} \mathrm{C} / \mathrm{m}, c=-2 \times 10^{4 \circ} \mathrm{C} / \mathrm{m}^{2}\), and \(x\) is in meters. The origin of the \(x\)-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation \(\dot{q}\) in the wall? (c) Determine the surface heat fluxes, \(q_{x}^{\prime \prime}(-L)\) and \(q_{x}^{\prime \prime}(+L)\). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at \(x=-L\) and \(x=+L\) ? (e) Obtain an expression for the heat flux distribution \(q_{x}^{\prime \prime}(x)\). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated \((\dot{q}=0)\), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with \(\dot{q}=0\) ? How much energy must be removed by the fluid per unit area of the wall \(\left(\mathrm{J} / \mathrm{m}^{2}\right)\) to reach this state? The density and specific heat of the wall material are \(2600 \mathrm{~kg} / \mathrm{m}^{3}\) and \(800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

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 \(r=r_{o^{*}}\).
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