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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 = 2Ï€r_oL So, the heat transfer rate at r_o: q_cond = -k * 2Ï€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 * 2Ï€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 * 2Ï€r_oL * (2br_o) = h * 2Ï€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

A plane wall that is insulated on one side \((x=0)\) is initially at a uniform temperature \(T_{i}\), when its exposed surface at \(x=L\) is suddenly raised to a temperature \(T_{s}\). (a) Verify that the following equation satisfies the heat equation and boundary conditions: $$ \frac{T(x, t)-T_{s}}{T_{i}-T_{s}}=C_{1} \exp \left(-\frac{\pi^{2}}{4} \frac{\alpha t}{L^{2}}\right) \cos \left(\frac{\pi}{2} \frac{x}{L}\right) $$ where \(C_{1}\) is a constant and \(\alpha\) is the thermal diffusivity. (b) Obtain expressions for the heat flux at \(x=0\) and \(x=L\). (c) Sketch the temperature distribution \(T(x)\) at \(t=0\), at \(t \rightarrow \infty\), and at an intermediate time. Sketch the variation with time of the heat flux at \(x=L, q_{L}^{\prime \prime}(t)\). (d) What effect does \(\alpha\) have on the thermal response of the material to a change in surface temperature?

A long cylindrical rod, initially at a uniform temperature \(T_{i}\), is suddenly immersed in a large container of liquid at \(T_{\infty}

The one-dimensional system of mass \(M\) with constant properties and no internal heat generation shown in the figure is initially at a uniform temperature \(T_{i^{*}}\). The electrical heater is suddenly energized, providing a uniform heat flux \(q_{o}^{\prime \prime}\) at the surface \(x=0\). The boundaries at \(x=L\) and elsewhere are perfectly insulated. (a) Write the differential equation, and identify the boundary and initial conditions that could be used to determine the temperature as a function of position and time in the system. (b) On \(T-x\) coordinates, sketch the temperature distributions for the initial condition \((t \leq 0)\) and for several times after the heater is energized. Will a steady-state temperature distribution ever be reached? (c) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the heat flux \(q_{x}^{\prime \prime}(x, t)\) at the planes \(x=0, x=L / 2\), and \(x=L\) as a function of time. (d) After a period of time \(t_{e}\) has elapsed, the heater power is switched off. Assuming that the insulation is perfect, the system will eventually reach a final uniform temperature \(T_{f}\) Derive an expression that can be used to determine \(T_{f}\) as a function of the parameters \(q_{o}^{\prime \prime}, t_{e}, T_{i}\), and the system characteristics \(M, c_{p}\), and \(A_{s}\) (the heater surface area).

A plane wall has constant properties, no internal heat generation, and is initially at a uniform temperature \(T_{i \cdot}\) Suddenly, the surface at \(x=L\) is heated by a fluid at \(T_{\infty}\) having a convection coefficient \(h\). At the same instant, the electrical heater is energized, providing a constant heat flux \(q_{o}^{\prime \prime}\) at \(x=0\). (a) On \(T-x\) 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. (b) On \(q_{x}^{\prime \prime}-x\) coordinates, sketch the heat flux corresponding to the four temperature distributions of part (a). (c) On \(q_{x}^{n}-t\) coordinates, sketch the heat flux at the locations \(x=0\) and \(x=L\). That is, show qualitatively how \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\) vary with time. (d) Derive an expression for the steady-state temperature at the heater surface, \(T(0, \infty)\), in terms of \(q_{o}^{\prime \prime}\), \(T_{\infty}, k, h\), and \(L\).

One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of \(50 \mathrm{~mm}\) and a constant thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). For these conditions, the temperature distribution has the form \(T(x)=a+b x+c x^{2}\). The surface at \(x=0\) has a temperature of \(T(0) \equiv T_{o}=120^{\circ} \mathrm{C}\) and experiences convection with a fluid for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface at \(x=L\) is well insulated. (a) Applying an overall energy balance to the wall, calculate the volumetric energy generation rate \(\dot{q}\). (b) Determine the coefficients \(a, b\), and \(c\) by applying the boundary conditions to the prescribed temperature distribution. Use the results to calculate and plot the temperature distribution. (c) Consider conditions for which the convection coefficient is halved, but the volumetric energy generation rate remains unchanged. Determine the new values of \(a, b\), and \(c\), and use the results to plot the temperature distribution. Hint: recognize that \(T(0)\) is no longer \(120^{\circ} \mathrm{C}\). (d) Under conditions for which the volumetric energy generation rate is doubled, and the convection coefficient remains unchanged \(\left(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\), determine the new values of \(a, b\), and \(c\) and plot the corresponding temperature distribution. Referring to the results of parts (b), (c), and (d) as Cases 1, 2 , and 3, respectively, compare the temperature distributions for the three cases and discuss the effects of \(h\) and \(\dot{q}\) on the distributions.

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