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Heat is generated at a rate \(\dot{Q}_{v}^{\prime \prime \prime}\) in a long solid cylinder of radius \(R\). The cylinder has a thin metal sheath and is immersed in a liquid at temperature \(T_{e}\). Heat transfer from the cylinder surface to the liquid can be characterized by an overall heat transfer coefficient \(U\). Obtain the steady-state temperature distributions for the following cases: (i) \(\dot{Q}_{v}^{\prime \prime \prime}\) is constant. (ii) \(\dot{Q}_{v}^{\prime \prime \prime}=\dot{Q}_{v 0}^{\prime \prime \prime}\left[1-(r / R)^{2}\right]\). (iii) \(\dot{Q}_{v}^{\prime \prime \prime \prime}=a+b\left(T-T_{e}\right)\).

Short Answer

Expert verified
For each case, solve the heat equation with given boundary conditions to get temperature distribution.

Step by step solution

01

Steady-State Heat Equation for a Cylinder

For a long solid cylinder of radius \(R\), the radial heat conduction in steady state is governed by the equation:\[\frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) = -\frac{\dot{Q}_v^{\prime \prime \prime}}{k}\]where \(T\) is the temperature, \(r\) is the radial position, and \(k\) is the thermal conductivity of the cylinder material.
02

Case (i) Heat Source is Constant

Here, \(\dot{Q}_v^{\prime \prime \prime}\) is constant, so the equation becomes:\[\frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) = -\frac{\dot{Q}_v^{\prime \prime \prime}}{k}\]Integrating once:\[r \frac{dT}{dr} = -\frac{\dot{Q}_v^{\prime \prime \prime} r^2}{2k} + C_1\]Integrating again:\[T(r) = -\frac{\dot{Q}_v^{\prime \prime \prime} r^2}{4k} + C_1 \ln r + C_2\]Applying the boundary condition at the surface, \(r = R\), where \\(T(R) = T_0\) solved using thermal resistance due to conduction and convection:\[T_0 = T_e + \frac{\dot{Q}_v^{\prime \prime \prime}R^2}{4k} = C_1 \ln R + C_2\]Using thermal currents and surface temperatures for heat balance can solve for constants.
03

Case (ii) Quadratic Heat Generation

The heat source is given as \(\dot{Q}_v^{\prime \prime \prime} = \dot{Q}_{v0}^{\prime \prime \prime}\left[1 - (r/R)^2\right]\), substitute in the steady heat equation:\[\frac{1}{r}\frac{d}{dr} \left( r \frac{dT}{dr} \right) = -\frac{\dot{Q}_{v0}^{\prime \prime \prime}}{k} \left(1-\frac{r^2}{R^2}\right)\]Simplifying and integrating:\[r \frac{dT}{dr} = -\frac{\dot{Q}_{v0}^{\prime \prime \prime} r^2}{2k} + \frac{\dot{Q}_{v0}^{\prime \prime \prime} r^4}{4k R^2} + C_1\]Further integration and boundary conditions analogous to Step 2 applies to solve.
04

Case (iii) Heat Source with Temperature Dependence

Here, the generation rate is \(\dot{Q}_v^{\prime \prime \prime} = a + b(T - T_e)\), leading to:\[\frac{1}{r}\frac{d}{dr} \left( r \frac{dT}{dr} \right) = -\frac{a}{k} - \frac{b(T - T_e)}{k}\]Re-formatting gives an inhomogeneous differential equation with a particular solution from heat balance (due to linear dependence on \(T\)) and solving for the homogeneous equation. Specific boundary condition solutions give the constants.

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Key Concepts

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

Steady-State Temperature Distribution
At steady-state, the temperature distribution in a material does not change over time. This means that despite any heat being produced or transferred into or out of the system, the temperature at any given point within the material remains constant. In the problem we are looking at, a cylinder is experiencing heat generation internally at a certain rate, and there is heat transfer taking place across its surface. Once steady-state is achieved, the temperature throughout the cylinder does not vary with time. Instead, it varies only with location, specifically the distance from the center of the cylinder. This radial dependency is key in calculating the temperature distribution under different heat generation scenarios. To find the steady-state temperature distribution, we need to solve the heat equation. This allows us to understand how the temperature changes with radial position, ensuring all points within or on the cylinder maintain their temperatures despite ongoing heat processes.
Radial Heat Conduction
In radial heat conduction, heat flows from the center of the cylinder outward, perpendicular to the axis of the cylinder. This is predominant in cylindrical coordinates, where the change in temperature depends on the radial distance from the center. The heat flow is governed by Fourier’s law of heat conduction. In the case of a long cylinder, heat flows only in the radial direction, simplifying the problem to one spatial dimension.For a solid cylinder, the radial heat conduction at steady state is described by the equation \( \frac{1}{r} \frac{d}{dr} \left( r \frac{dT}{dr} \right) = -\frac{\dot{Q}_v^{\prime \prime \prime}}{k} \). This shows how the temperature gradient (\( \frac{dT}{dr} \)) affects heat conduction within the material. Integrating this expression twice while applying the boundary conditions can lead us to the temperature profile we need. Variations in radial heat generation can result in different temperature profiles, as shown in the problem's various cases.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted by \( U \), is a measure of how effectively heat transfers from one medium to another. It encompasses conduction through the cylinder's material and the subsequent convection as heat leaves the cylinder's surface into the surrounding liquid. This parameter is critical in the boundary condition at the cylinder's surface, where the heat exchange happens.The heat transfer rate \( \dot{Q} \) is proportional to the temperature difference between the surface \( T_0 \) and the environment \( T_e \). The relationship can be expressed as \( \dot{Q} = U A (T_0 - T_e) \), where \( A \) is the surface area. The heat transfer coefficient \( U \) thus helps connect the internal temperature distribution with the surface temperature and how effectively heat can be shed to the surroundings.
Thermal Conductivity
Thermal conductivity, represented by \( k \), is a property of the cylinder's material. It quantifies the material's ability to conduct heat. A higher thermal conductivity indicates the material is a better heat conductor. In the heat equation, it appears in the denominator on the right-hand side, \( \frac{\dot{Q}_v^{\prime \prime \prime}}{k} \), showing that higher thermal conductivity reduces the temperature gradient required to conduct the same amount of heat.For our cylinder, knowing \( k \) allows us to calculate how quickly and evenly heat spreads out from the center to the perimeter. The efficiency with which the material can transport heat is crucial for determining how rapidly temperature changes occur radially within the cylinder, helping to establish the temperature distribution under various internal heat generation scenarios. It is important for engineering applications, as it influences material selection depending on the temperature management needs of a system.

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Most popular questions from this chapter

Liquid oxygen at \(90 \mathrm{~K}\) flows inside a \(3 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 2 \mathrm{~mm}\)-wall-thickness, AISI 303 stainless steel tube. The tube is insulated with \(3 \mathrm{~cm}\)-thick fiberglass insulation of thermal conductivity \(0.021 \mathrm{~W} / \mathrm{m} \mathrm{K}\). Will atmospheric water vapor condense on the outside of the insulation when the ambient air temperature is \(300 \mathrm{~K}\) and the dewpoint temperature is \(285 \mathrm{~K}\) ? Take the outside convective and radiative heat transfer coefficients as \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

A thin-wall, spherical stainless steel vessel has an outside diameter of \(40 \mathrm{~cm}\) and contains biochemical reactants maintained at \(160^{\circ} \mathrm{C}\). The tank is located in a laboratory where the air is maintained at \(20^{\circ} \mathrm{C}\). A layer of insulation is to be added to prevent workers from being burned by accidental contact with the vessel. The insulation chosen has a thermal conductivity of \(0.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\). How thick should the layer be if the threshold for a skin burn on a nonmetallic surface can be taken as \(55^{\circ} \mathrm{C}\) ? Also calculate the heat loss. The combined convective and radiative heat transfer coefficients on the outside of the insulation is estimated to be \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

A test technique for measuring the thermal conductivity of copper-nickel alloys is based on the measurement of the tip temperature of pin fins made from the alloys. The standard fin dimensions are a diameter of \(5 \mathrm{~mm}\) and a length of \(20 \mathrm{~cm}\). The test fin and a reference brass fin \((k=111 \mathrm{~W} / \mathrm{m} \mathrm{K})\) are mounted on a copper base plate in a wind tunnel. The test data include \(T_{B}=100^{\circ} \mathrm{C}, T_{e}=20^{\circ} \mathrm{C}\), and tip temperatures of \(64.2^{\circ} \mathrm{C}\) and \(49.7^{\circ} \mathrm{C}\) for the brass and test alloy fins, respectively. (i) Determine the conductivity of the test alloy. (ii) If the conductivity should be known to \(\pm 1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), how accurately should the tip temperatures be measured?

Determine the allowable current in a 10 gage \((2.59 \mathrm{~mm}\) diameter) copper wire that is insulated with a \(1 \mathrm{~cm}-\mathrm{O} . \mathrm{D}\). layer of rubber. The outside heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the ambient air is at \(310 \mathrm{~K}\). The allowable maximum temperature of the rubber is \(380 \mathrm{~K}\). Take \(k=0.15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the rubber and an electrical resistancethe rubber and an electrical resistance of \(0.00328 \Omega / \mathrm{m}\) for the copper wire.

Radioactive wastes are stored in a spherical type 316 stainless steel tank of inner diameter \(1 \mathrm{~m}\) and \(1 \mathrm{~cm}\) wall thickness. Heat is generated uniformly in the wastes at a rate of \(3 \times 10^{4} \mathrm{~W} / \mathrm{m}^{3}\). The outer surface of the tank is cooled by air at \(300 \mathrm{~K}\) with a heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Determine the maximum temperature in the tank. Take the thermal conductivity of the wastes as \(2.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\).

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