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Chapter 3: Problem 90

A nuclear fuel element of thickness \(2 L\) is covered with a steel cladding of thickness \(b\). Heat generated within the nuclear fuel at a rate \(\dot{q}\) is removed by a fluid at \(T_{\infty}\), which adjoins one surface and is characterized by a convection coefficient \(h\). The other surface is well insulated, and the fuel and steel have thermal conductivities of \(k_{f}\) and \(k_{s}\), respectively. (a) Obtain an equation for the temperature distribution \(T(x)\) in the nuclear fuel. Express your results in terms of \(\dot{q}, k_{f}, L, b, k_{s}, h\), and \(T_{\infty}\). (b) Sketch the temperature distribution \(T(x)\) for the entire system.

Short Answer

Expert verified
The temperature distribution for the nuclear fuel element and the steel cladding can be expressed as follows: For the nuclear fuel region (0 < x < L): \(T(x) = -\frac{\dot{q}}{2k_f}x^2 + C_2\) For the steel cladding region (L < x < L+b): \(T(x) = C_3x + C_4\) These expressions give the temperature distribution T(x) in terms of \(\dot{q}, k_f, L, b, k_s, h\), and \(T_{\infty}\). The temperature distribution in the nuclear fuel region is a downward parabolic shape, and in the steel cladding region, it is a linearly decreasing function.

Step by step solution

01

Write down the heat equation

For steady-state heat conduction in one-dimensional Cartesian coordinates (x-direction) with heat generation, the heat equation is given by: \(-k\frac{d^2T}{dx^2} + \dot{q} = 0\) In our case, the thermal conductivity k is different for the nuclear fuel and the steel cladding. In the nuclear fuel region (0 < x < L), we have k = kf, and in the steel cladding region (L < x < L+b), we have k = ks.
02

Integrate the heat equation to obtain the temperature distribution

For x between 0 and L (nuclear fuel region), we integrate the heat equation twice with respect to x: \(-k_f\frac{d^2T}{dx^2} + \dot{q} = 0\) Integrating once: \(-k_f\frac{dT}{dx} = \dot{q}x + C_1\) Integrating twice: \(T(x) = -\frac{\dot{q}}{2k_f}x^2 + C_1x + C_2\) For x between L and L+b (steel cladding region), there is no heat generation, so the heat equation reduces to: \(-k_s\frac{d^2T}{dx^2} = 0\) Integrating once: \(-k_s\frac{dT}{dx} = C_3\) Integrating twice: \(T(x) = C_3x + C_4\)
03

Apply boundary conditions

We have the following boundary conditions: 1. The insulated boundary condition at x = 0: \(\frac{dT}{dx}\Bigg|_{x=0} = 0\) 2. The continuity of temperature at the fuel-cladding interface (x = L): \(T_f(L) = T_s(L)\) 3. The continuity of heat flux at the fuel-cladding interface (x = L): \(\frac{dT_f}{dx}\Bigg|_{x=L} = \frac{dT_s}{dx}\Bigg|_{x=L}\) 4. The convective boundary condition at the cladding-fluid interface (x = L+b): \(-k_s\frac{dT_s}{dx}\Bigg|_{x=L+b} = h(T_{\infty} - T_s(L+b))\) Using these boundary conditions, we can solve for the constants C1, C2, C3, and C4.
04

Calculate the constants using boundary conditions

Using boundary condition 1 at x = 0, we get: \(0 = \dot{q}(0) + C_1\) So, \(C_1 = 0\) Using boundary condition 2 at x = L, we get: \(-\frac{\dot{q}}{2k_f}L^2 + C_2 = C_3L + C_4\) Using boundary condition 3 at x = L, we get: \(-\frac{\dot{q}}{k_f}L = \frac{C_3}{k_s}\) Using boundary condition 4 at x = L+b, we get: \(-k_s\frac{C_3}{k_s} = h(T_{\infty} - (C_3(L+b) + C_4))\) So, \(C_3 = h(k_s(L+b) + k_sT_{\infty} - k_sC_4)\) Solving these equations simultaneously, we can find the constants C1, C2, C3, and C4.
05

Write down the final temperature distribution equation

Now that we have calculated the values of the constants, we can write down the final expressions for the temperature distribution in the desired form. For the nuclear fuel region (0 < x < L): \(T(x) = -\frac{\dot{q}}{2k_f}x^2 + C_2\) For the steel cladding region (L < x < L+b): \(T(x) = C_3x + C_4\) These expressions give the temperature distribution T(x) in terms of \(\dot{q}, k_f, L, b, k_s, h\), and \(T_{\infty}\).
06

Sketch the temperature distribution T(x)

To sketch the temperature distribution, we can plot T(x) as a function of x. For the nuclear fuel region, the temperature will be a downward parabolic shape, starting from the well-insulated boundary (with a zero slope) and decreasing until it reaches the fuel-cladding interface at x = L. In the steel cladding region, the temperature distribution will be a linearly decreasing function, starting from the fuel-cladding interface and reaching the fluid-steel interface, where it meets the fluid temperature T鈭�.

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Key Concepts

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

Nuclear Fuel Element
A nuclear fuel element is a critical component in a nuclear reactor, responsible for sustaining the chain reaction. These elements typically contain fissile material, like uranium or plutonium. The objective is to maintain a controlled nuclear reaction while maximizing energy production.

  • In the context of heat transfer, the fuel element generates a significant amount of heat internally.
  • This heat must be managed effectively to prevent overheating or damage to the reactor.
Understanding the heat generation rate, denoted as \(\dot{q}\), is essential. It represents how much heat energy is produced per unit volume within the fuel element. Managing this heat through proper conduction and convection processes is critical to reactor safety and efficiency.
Convection Heat Transfer
Convection heat transfer occurs when heat is moved by the physical motion of a fluid. It plays a vital role in cooling systems used in various applications, including nuclear reactors.

  • The convection coefficient, denoted as \(h\), quantifies the heat transfer rate between the surface and the surrounding fluid.
  • This coefficient depends on factors like fluid velocity, viscosity, and properties of the surface being cooled.
In our exercise, the fluid temperature is noted as \(T_{\infty}\), which is the temperature that the external fluid maintains. An effective convection process ensures the heat generated in the nuclear fuel is continuously transferred away, preventing overheating and maintaining stable reactor operations.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material can conduct heat. It's denoted by \(k\), and different materials have different values of thermal conductivity.

  • The nuclear fuel has a thermal conductivity \(k_f\), which is crucial for conducting the internally generated heat to the cladding.
  • The cladding material itself has a different thermal conductivity \(k_s\).
Higher thermal conductivity in a material means it can efficiently transfer heat over a distance. In the context of a nuclear fuel element, choosing materials with appropriate thermal conductivities is crucial to balancing heat distribution and maintaining the mechanical integrity of the reactor components.
Temperature Distribution Equation
The temperature distribution equation helps determine how temperature varies within different sections of a material. It's essential for understanding the thermal behavior of both the nuclear fuel and the cladding.

When dealing with the nuclear fuel region, the temperature distribution follows the equation:\[ T(x) = -\frac{\dot{q}}{2k_f}x^2 + C_2 \]For the steel cladding region, the equation is:\[ T(x) = C_3x + C_4 \]These equations rely on boundary conditions to solve for constants \(C_2\), \(C_3\), and \(C_4\), which consider factors such as zero slope at insulated boundaries and continuity through different materials.
  • Temperature profiles help engineers design systems that maintain operational safety and efficiency.
  • They contribute to understanding how each part of the material will respond under different operating conditions.

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Most popular questions from this chapter

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

Radioactive wastes \(\left(k_{\mathrm{rw}}=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) are stored in a spherical, stainless steel \(\left(k_{\mathrm{ss}}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\) container of inner and outer radii equal to \(r_{i}=0.5 \mathrm{~m}\) and \(r_{o}=0.6 \mathrm{~m}\). Heat is generated volumetrically within the wastes at a uniform rate of \(\dot{q}=10^{5} \mathrm{~W} / \mathrm{m}^{3}\), and the outer surface of the container is exposed to a water flow for which \(h=\) \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\). (a) Evaluate the steady-state outer surface temperature, \(T_{s, o}\) (b) Evaluate the steady-state inner surface temperature, \(T_{s, i^{*}}\) (c) Obtain an expression for the temperature distribution, \(T(r)\), in the radioactive wastes. Express your result in terms of \(r_{i}, T_{s, i}, k_{\mathrm{rw}}\), and \(\dot{q}\). Evaluate the temperature at \(r=0\). (d) A proposed extension of the foregoing design involves storing waste materials having the same thermal conductivity but twice the heat generation \(\left(\dot{q}=2 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\right)\) in a stainless steel container of equivalent inner radius \(\left(r_{i}=0.5 \mathrm{~m}\right)\). Safety considerations dictate that the maximum system temperature not exceed \(475^{\circ} \mathrm{C}\) and that the container wall thickness be no less than \(t=0.04 \mathrm{~m}\) and preferably at or close to the original design \((t=0.1 \mathrm{~m})\). Assess the effect of varying the outside convection coefficient to a maximum achievable value of \(h=5000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (by increasing the water velocity) and the container wall thickness. Is the proposed extension feasible? If so, recommend suitable operating and design conditions for \(h\) and \(t\), respectively.

A scheme for concurrently heating separate water and air streams involves passing them through and over an array of tubes, respectively, while the tube wall is heated electrically. To enhance gas-side heat transfer, annular fins of rectangular profile are attached to the outer tube surface. Attachment is facilitated with a dielectric adhesive that electrically isolates the fins from the current-carrying tube wall. (a) Assuming uniform volumetric heat generation within the tube wall, obtain expressions for the heat rate per unit tube length \((\mathrm{W} / \mathrm{m})\) at the inner \(\left(r_{i}\right)\) and outer \(\left(r_{o}\right)\) surfaces of the wall. Express your results in terms of the tube inner and outer surface temperatures, \(T_{s, i}\) and \(T_{s, e}\), and other pertinent parameters. (b) Obtain expressions that could be used to determine \(T_{s, i}\) and \(T_{s, o}\) in terms of parameters associated with the water- and air-side conditions. (c) Consider conditions for which the water and air are at \(T_{\infty, i}=T_{\infty, o}=300 \mathrm{~K}\), with corresponding convection coefficients of \(h_{i}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Heat is uniformly dissipated in a stainless steel tube \(\left(k_{w}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right)\), having inner and outer radii of \(r_{i}=25 \mathrm{~mm}\) and \(r_{o}=30\) \(\mathrm{mm}\), and aluminum fins \(\left(t=\delta=2 \mathrm{~mm}, r_{t}=55\right.\) \(\mathrm{mm}\) ) are attached to the outer surface, with \(R_{t, c}^{\prime \prime}=\) \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). Determine the heat rates and temperatures at the inner and outer surfaces as a function of the rate of volumetric heating \(\dot{q}\). The upper limit to \(\dot{q}\) will be determined by the constraints that \(T_{s, i}\) not exceed the boiling point of water \(\left(100^{\circ} \mathrm{C}\right)\) and \(T_{s, o}\) not exceed the decomposition temperature of the adhesive \(\left(250^{\circ} \mathrm{C}\right)\).

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.

An uninsulated, thin-walled pipe of \(100-\mathrm{mm}\) diameter is used to transport water to equipment that operates outdoors and uses the water as a coolant. During particularly harsh winter conditions, the pipe wall achieves a temperature of \(-15^{\circ} \mathrm{C}\) and a cylindrical layer of ice forms on the inner surface of the wall. If the mean water temperature is \(3^{\circ} \mathrm{C}\) and a convection coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained at the inner surface of the ice, which is at \(0^{\circ} \mathrm{C}\), what is the thickness of the ice layer?
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