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Chapter 3: Problem 101

A long cylindrical rod of diameter \(200 \mathrm{~mm}\) with thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation of \(24,000 \mathrm{~W} / \mathrm{m}^{3}\). The rod is encapsulated by a circular sleeve having an outer diameter of \(400 \mathrm{~mm}\) and a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outer surface of the sleeve is exposed to cross flow of air at \(27^{\circ} \mathrm{C}\) with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the temperature at the interface between the rod and sleeve and on the outer surface. (b) What is the temperature at the center of the rod?

Short Answer

Expert verified
The temperature at the interface between the rod and sleeve is \(T_{interface}\) and on the outer surface of the sleeve is \(T_{outer}\). To find the temperature at the center of the rod, more advanced mathematical methods such as solving differential equations are needed, which goes beyond the scope of this exercise.

Step by step solution

01

Calculate resistance of the rod

Find the resistance (conduction) within the rod using the formula for conduction resistance in cylindrical coordinates: \[R_{rod} = \frac{\ln \frac{r_{outer}}{r_{inner}}}{2\pi L k_{rod}}\] Given diameter of the rod, \(D_{rod} = 200\,\text{mm}\): The radius of the rod, \(r_{outer}=\frac{D_{rod}}{2} = 100\,\text{mm} = 0.1\,\text{m}\), and since the center of the rod is the inner radius, we have \(r_{inner}=0\). The thermal conductivity of the rod is given, \(k_{rod} = 0.5\,\frac{\text{W}}{\text{mK}}\). Now, calculate \(R_{rod}\): \[R_{rod} = \frac{\ln \frac{0.1}{0}}{2\pi(0.1)(0.5)}\] However, we cannot calculate \(\ln \frac{0.1}{0}\) since it is not possible to divide by zero. Instead, in this scenario, we will convert the given volumetric heat generation to find the heat flux and then use that to determine the temperature at the center of the rod.
02

Calculate resistance of the sleeve

Find the resistance (conduction) within the sleeve using the same formula: \[R_{sleeve} = \frac{\ln \frac{r_{outer}}{r_{inner}}}{2\pi L k_{sleeve}}\] Given diameter of the sleeve, \(D_{sleeve} = 400\,\text{mm}\): The outer radius of the sleeve, \(r_{outer}=\frac{D_{sleeve}}{2} = 200\,\text{mm} = 0.2\,\text{m}\), and the inner radius is equal to the rod's outer radius, \(r_{inner} = 0.1\,\text{m}\). The thermal conductivity of the sleeve is given, \(k_{sleeve} = 4\,\frac{\text{W}}{\text{mK}}\). Now, calculate \(R_{sleeve}\): \[R_{sleeve} = \frac{\ln \frac{0.2}{0.1}}{2\pi(0.1)(4)}\]
03

Calculate convection resistance

Find the resistance (convection) at the outer surface of the sleeve: \[R_{conv} = \frac{1}{h A_s}\] The convection coefficient is given, \(h = 25\,\frac{\text{W}}{\text{m}^{2}\text{K}}\). The outer surface area of the sleeve can be calculated: \[A_s = 2\pi r_{outer} L\] \[A_s = 2\pi (0.2) L\] Now, calculate \(R_{conv}\): \[R_{conv} = \frac{1}{25 (2\pi (0.2) L)}\]
04

Determine total resistance

The total resistance of the system is the sum of the conduction resistance within the rod and the sleeve and the convection resistance at the outer surface: \[R_{total} = R_{rod} + R_{sleeve} + R_{conv}\] Note that the \(R_{rod}\) term cannot be calculated, so we will ignore it in this case.
05

Calculate heat flux

The volumetric heat generation is given by \(24,000\,\frac{\text{W}}{\text{m}^{3}}\). To find the heat flux, we multiply it by the volume of the rod: \[q_{gen} = \phi \cdot V_{rod}\] \[q_{gen} = 24,000\frac{\text{W}}{\text{m}^{3}} \cdot (\pi (0.1)^2 L)\]
06

Find the temperature at the interface

Use the heat flux, \(q_{gen}\), and the total resistance, \(R_{total}\), to find the temperature difference between the air temperature and the interface: \[\Delta T = q_{gen} \cdot R_{total} = q_{gen} \cdot (R_{sleeve} + R_{conv})\] The air temperature is given as \(27^{\circ} \text{C}\). Add the temperature difference to find the temperature at the interface: \[T_{interface} = T_{air} + \Delta T\]
07

Find the temperature at the outer surface

Use the heat flux and the convection resistance to find the temperature at the outer surface of the sleeve: \[\Delta T_{outer} = q_{gen} \cdot R_{conv}\] \[T_{outer} = T_{interface} - \Delta T_{outer}\]
08

Find the temperature at the center of the rod

Since we cannot determine the temperature at the center of the rod using the resistance method, we need to utilize the given volumetric heat generation rate and the heat conduction equation in cylindrical coordinates: \[\nabla^2 T - \frac{q_{gen}}{k_{rod}} = 0\] Solving this equation will give us the temperature distribution within the rod, which can be used to determine the temperature at the center of the rod. This step involves differential equations and may go beyond the scope of this exercise. In summary, we have found the temperature at the interface and the outer surface of the sleeve, but to find the temperature at the center of the rod, we need to use an advanced mathematical approach involving differential equations.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material property that describes the ability of a material to conduct heat. It plays a crucial role in heat transfer calculations.
When dealing with heat transfer in cylindrical objects, like rods, thermal conductivity helps determine how quickly heat can move through the material.
The higher the thermal conductivity, the more efficiently the material conducts heat.
In this exercise, the rod has a thermal conductivity of 0.5 W/m銉籏, which is much lower than that of the sleeve (4 W/m銉籏). This difference significantly affects how heat is distributed across these components.
  • Materials with high thermal conductivity, like metals, quickly spread heat across their surfaces.
  • Materials with low thermal conductivity, like rubber or glass wool, are better insulators.
Understanding thermal conductivity helps in predicting the temperature distribution within materials and in designing systems for effective thermal management.
Convection Coefficient
The convection coefficient, also known as the heat transfer coefficient, represents the heat transfer from a surface to a fluid around it or vice versa.
It is crucial when the outer surface of an object is exposed to a moving fluid, such as air or water.
The convection coefficient depends on various factors, including the fluid's properties, the flow conditions, and the nature of the surface.
In this exercise, air with a convection coefficient of 25 W/m虏銉籏 surrounds the sleeve, allowing the surface to exchange heat with the environment.
  • A high convection coefficient indicates efficient heat transfer between the surface and the fluid.
  • Different fluids and flow rates can significantly alter the convection coefficient.
This coefficient is vital to calculate the heat loss or gain through convection, which is necessary for determining the total thermal resistance in heat transfer problems.
Volumetric Heat Generation
Volumetric heat generation refers to the generation of heat within the volume of a material, often arising from chemical reactions, electrical resistance, or nuclear processes within the material.
In cylindrical problems, this feature adds complexity, as it requires us to consider the distribution of heat throughout the volume of the object rather than just at the surfaces.
The rate of volumetric heat generation is expressed in watts per cubic meter (W/m鲁).
For the rod in this exercise, it is given as 24,000 W/m鲁, indicating substantial heat production internally. Understanding this internal heat source is critical for determining the temperature at different points within the rod, especially at its center.
  • In practical applications, volumetric heat generation can affect material stability and performance.
  • It is necessary to manage this heat carefully to avoid overheating and potential damage.
By considering volumetric heat generation, engineers can ensure that designs effectively manage internal heat build-up to maintain the system's integrity and efficiency.
Cylindrical Coordinates
Cylindrical coordinates are a three-dimensional coordinate system that extends polar coordinates by adding a height dimension to describe points in space.
This system is often utilized for problems involving cylindrical shapes, such as rods and tubes, where symmetry around a central axis simplifies calculations.
In the context of heat transfer, cylindrical coordinates allow for a more straightforward way to analyze how heat conducts through round objects.
Variables like radial distance, angle, and height (or depth) accurately describe locations within or on the surfaces of cylindrical objects.
  • Cylindrical coordinates are especially useful in calculating surface areas and volumes, crucial for heat transfer analysis.
  • Using these coordinates simplifies the application of differential equations needed to solve heat distribution problems.
By aligning with the geometry of the problem, cylindrical coordinates make it easier to integrate mathematical expressions that describe the distribution of temperature across the object.

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Most popular questions from this chapter

Consider the configuration of Example \(3.8\), where uniform volumetric heating within a stainless steel tube is induced by an electric current and heat is transferred by convection to air flowing through the tube. The tube wall has inner and outer radii of \(r_{1}=25 \mathrm{~mm}\) and \(r_{2}=\) \(35 \mathrm{~mm}\), a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), an electrical resistivity of \(\rho_{e}=0.7 \times 10^{-6} \Omega \cdot \mathrm{m}\), and a maximum allowable operating temperature of \(1400 \mathrm{~K}\). (a) Assuming the outer tube surface to be perfectly insulated and the airflow to be characterized by a temperature and convection coefficient of \(T_{\infty, 1}=\) \(400 \mathrm{~K}\) and \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum allowable electric current \(I\). (b) Compute and plot the radial temperature distribution in the tube wall for the electric current of part (a) and three values of \(h_{1}\left(100,500\right.\), and \(\left.1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\). For each value of \(h_{1}\), determine the rate of heat transfer to the air per unit length of tube. (c) In practice, even the best of insulating materials would be unable to maintain adiabatic conditions at the outer tube surface. Consider use of a refractory insulating material of thermal conductivity \(k=1.0\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and neglect radiation exchange at its outer surface. For \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the maximum allowable current determined in part (a), compute and plot the temperature distribution in the composite wall for two values of the insulation thickness \((\delta=25\) and \(50 \mathrm{~mm})\). The outer surface of the insulation is exposed to room air for which \(T_{\infty, 2}=300 \mathrm{~K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For each insulation thickness, determine the rate of heat transfer per unit tube length to the inner airflow and the ambient air.

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{o}\) within the tissue and maintaining local temperatures above a critical value \(T_{c}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature \(\left(T_{b}=37^{\circ} \mathrm{C}\right)\). Obtain a general expression for the radial temperature distribution in the tissue under steady- state conditions for which heat is dissipated at a rate \(q\). If \(r_{o}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{c}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal conductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

A 2-mm-diameter electrical wire is insulated by a 2 -mm-thick rubberized sheath \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and the wire/sheath interface is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=3 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The convection heat transfer coefficient at the outer surface of the sheath is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the temperature of the ambient air is \(20^{\circ} \mathrm{C}\). If the temperature of the insulation may not exceed \(50^{\circ} \mathrm{C}\), what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?
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