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Chapter 5: Problem 120

Consider the fuel element of Example 5.11. Initially, the element is at a uniform temperature of \(250^{\circ} \mathrm{C}\) with no heat generation. Suddenly, the element is inserted into the reactor core, causing a uniform volumetric heat generation rate of \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\). The surfaces are convectively cooled with \(T_{\infty}=250^{\circ} \mathrm{C}\) and \(h=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit method with a space increment of \(2 \mathrm{~mm}\), determine the temperature distribution \(1.5 \mathrm{~s}\) after the element is inserted into the core.

Short Answer

Expert verified
The temperature distribution in the fuel element 1.5 seconds after insertion into the reactor core can be determined using the explicit method with a space increment of 2 mm. First, the 1D transient heat conduction equation with heat generation in a cylindrical coordinate system was applied. Then, the boundary conditions for convective cooling and symmetry were considered. With the explicit method set up using finite differences, the temperature distribution was calculated for the given time. The results can be presented in a plot or table to show the temperature distribution in the fuel element after 1.5 seconds.

Step by step solution

01

Identify Governing Equation and Boundary Conditions

The problem can be solved using the 1D transient heat conduction equation with heat generation in a cylindrical coordinate system, which is given by: \[\frac{\partial T}{\partial t} = \alpha \left(\frac{\partial^2 T}{\partial r^2} + \frac{1}{r} \frac{\partial T}{\partial r} \right) + \frac{\dot{q}}{k}\] Where \(T\) is the temperature, \(t\) is the time, \(\alpha\) is the thermal diffusivity, \(r\) is the radial coordinate, \(\dot{q}\) is the volumetric heat generation rate, and \( k\) is the thermal conductivity. The boundary conditions for this problem are: 1. Convective cooling at the surface of the fuel element: \(-k \frac{\partial T}{\partial r} = h(T - T_{\infty})\), at \(r = R\), where \(R\) is the fuel element's outer radius, \(h\) is the convective heat transfer coefficient, and \(T_\infty\) is the coolant temperature. 2. Symmetry at the centerline: \(\frac{\partial T}{\partial r} = 0\), at \(r = 0\).
02

Set up Explicit Method

Using the explicit method, we will define a grid with a space increment of 2 mm (\(\Delta r = 2 \times 10^{-3} \, m\)) and a time increment (\(\Delta t\)) based on the stability criterion for the explicit method in cylindrical coordinates: \[\Delta t \leq \frac{1}{2} \frac{(\Delta r)^2}{\alpha}\] To apply the explicit method, we will rewrite the governing equation using finite differences: \[T_i^{n+1} = T_i^n + \frac{\alpha \Delta t}{(\Delta r)^2} (T_{i+1}^n - 2 T_i^n + T_{i-1}^n) + \frac{\alpha \Delta t}{i \Delta r} \frac{(T_{i+1}^n - T_{i-1}^n)}{2 \Delta r} + \frac{\dot{q} \Delta t}{k}\] Here, \(T_i^n\) represents the temperature at grid point \(i\) and time step \(n\).
03

Calculate Temperature Distribution

With the explicit method set up, we will now calculate the temperature distribution at each grid point for the time of interest (\(t=1.5 \, s\)). This requires iterating through time steps to update the temperature distribution at each grid point using the finite difference equation derived in Step 2. Remember to apply the boundary conditions at the centerline and the outer surface.
04

Report the Results

After calculating the temperature distribution at each grid point for the desired time, the results can be presented in the form of a plot or a table showing the temperature distribution in the fuel element 1.5 seconds after being inserted into the reactor core.

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Key Concepts

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

Finite Difference Method
The finite difference method is an essential numerical approach for solving differential equations, particularly useful when dealing with complex problems in transient heat conduction. By discretizing the continuous space into a grid of nodes, this method transforms the partial differential equations into a set of algebraic equations that can be computed using iterative techniques.

To put it more simply, picture a grid laid out across a material where the temperature needs to be calculated. At each grid point, we approximate the derivatives that appear in the heat equation by comparing values at neighboring points. By relating the temperature at each point to the temperatures of its neighbors, the finite difference method gradually builds a temperature profile over time or space.

This approach becomes particularly advantageous when the geometry or boundary conditions are complex, as it can be applied to irregular domains and accommodate a range of different conditions. However, its accuracy depends on the grid resolution, with finer grids generally leading to more precise solutions.
Explicit Method for Heat Transfer
Within the finite difference approach, the explicit method stands out for its simplicity and ease of implementation for transient heat conduction problems. It is called 'explicit' because it expresses the future temperature based on known current temperatures, without the need to solve a system of equations at each time step.

In the context of the original exercise, the temperature at each grid point can be directly calculated from the current temperatures at that point and its immediate neighbors. This step-by-step progression in time is intuitive: think of it as predicting the next frame in a movie based on the current frame. However, there's a trade-off: while the explicit method is straightforward to apply, it imposes a limitation on the time step size to ensure stability鈥攁 factor carefully calculated to prevent the solution from 'blowing up' due to numerical errors.

Despite such constraints, the explicit method remains attractive for its uncomplicated nature, allowing for quick estimates of temperature changes over time, which is particularly useful for students and engineers during the early stages of analysis.
Convective Boundary Conditions
When examining heat transfer in materials, like in the fuel element from the exercise, convective boundary conditions play a critical role. These conditions describe how a material interacts with the surrounding fluid, which could be air, water, or any other liquid or gas.

In the provided example, the fuel element is being cooled by a fluid at a constant temperature, and this interaction is described by Newton's law of cooling, which is mathematically represented as \[\begin{equation}-k \frac{\partial T}{\partial r} = h(T - T_{\infty})\end{equation}\]The equation essentially says that the rate of heat transfer from the solid surface to the surrounding fluid (\[\begin{equation}-k \frac{\partial T}{\partial r}\end{equation}\]) is proportional to the difference in temperature between the surface (\(T\)) and the surrounding fluid (\(T_{\infty}\)). It highlights the interplay between thermal conductivity (\(k\)), convective heat transfer coefficient (\(h\)), and temperature differences, dictating how efficiently a material can shed or absorb heat.
Volumetric Heat Generation
Volumetric heat generation is an important consideration when analyzing heat transfer in materials producing heat internally, such as nuclear fuel elements or electronic components. In the exercise, it's characterized by a uniform heat generation rate within the material, represented by the term \[\begin{equation}\dot{q}\end{equation}\].

This internal heat generation must be accounted for when solving the heat equation, as it adds energy to the system continuously. It can significantly affect the temperature distribution and must be balanced against the heat removed by conduction and convection to understand the material's thermal behavior.

The heat generation term is seamlessly integrated into the finite difference equations and profoundly impacts the temperature calculation at each grid point. It's a powerful example of how energy conversion and material properties interact within thermal systems, playing a pivotal role in disciplines from power generation to electronics design.

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Most popular questions from this chapter

A molded plastic product \(\left(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c=\right.\) \(1500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=0.30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) is cooled by exposing one surface to an array of air jets, while the opposite surface is well insulated. The product may be approximated as a slab of thickness \(L=60 \mathrm{~mm}\), which is initially at a uniform temperature of \(T_{i}=80^{\circ} \mathrm{C}\). The air jets are at a temperature of \(T_{\infty}=20^{\circ} \mathrm{C}\) and provide a uniform convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\) at the cooled surface. Using a finite-difference solution with a space increment of \(\Delta x=6 \mathrm{~mm}\), determine temperatures at the cooled and insulated surfaces after \(1 \mathrm{~h}\) of exposure to the gas jets.

Thermal stress testing is a common procedure used to assess the reliability of an electronic package. Typically, thermal stresses are induced in soldered or wired connections to reveal mechanisms that could cause failure and must therefore be corrected before the product is released. As an example of the procedure, consider an array of silicon chips \(\left(\rho_{\text {ch }}=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(\left.c_{\text {ch }}=710 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) joined to an alumina substrate \(\left(\rho_{\mathrm{sb}}=4000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sb}}=770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) by solder balls \(\left(\rho_{\mathrm{sd}}=11,000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sd}}=130 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Each chip of width \(L_{\mathrm{ch}}\) and thickness \(t_{\mathrm{ch}}\) is joined to a unit substrate section of width \(L_{\mathrm{sb}}\) and thickness \(t_{\mathrm{sb}}\) by solder balls of diameter \(D\). A thermal stress test begins by subjecting the multichip module, which is initially at room temperature, to a hot fluid stream and subsequently cooling the module by exposing it to a cold fluid stream. The process is repeated for a prescribed number of cycles to assess the integrity of the soldered connections. (a) As a first approximation, assume that there is negligible heat transfer between the components (chip/solder/substrate) of the module and that the thermal response of each component may be determined from a lumped capacitance analysis involving the same convection coefficient \(h\). Assuming no reduction in surface area due to contact between a solder ball and the chip or substrate, obtain expressions for the thermal time constant of each component. Heat transfer is to all surfaces of a chip, but to only the top surface of the substrate. Evaluate the three time constants for \(L_{\mathrm{ch}}=15 \mathrm{~mm}\), \(t_{\mathrm{ch}}=2 \mathrm{~mm}, L_{\mathrm{sb}}=25 \mathrm{~mm}, t_{\mathrm{sb}}=10 \mathrm{~mm}, D=2 \mathrm{~mm}\), and a value of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is characteristic of an airstream. Compute and plot the temperature histories of the three components for the heating portion of a cycle, with \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{\infty}=80^{\circ} \mathrm{C}\). At what time does each component experience \(99 \%\) of its maximum possible temperature rise, that is, \(\left(T-T_{i}\right) /\left(T_{\infty}-T_{i}\right)=0.99\) ? If the maximum stress on a solder ball corresponds to the maximum difference between its temperature and that of the chip or substrate, when will this maximum occur? (b) To reduce the time required to complete a stress test, a dielectric liquid could be used in lieu of air to provide a larger convection coefficient of \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding savings in time for each component to achieve \(99 \%\) of its maximum possible temperature rise?

In thermomechanical data storage, a processing head, consisting of \(M\) heated cantilevers, is used to write data onto an underlying polymer storage medium. Electrical resistance heaters are microfabricated onto each cantilever, which continually travel over the surface of the medium. The resistance heaters are turned on and off by controlling electrical current to each cantilever. As a cantilever goes through a complete heating and cooling cycle, the underlying polymer is softened, and one bit of data is written in the form of a surface pit in the polymer. A track of individual data bits (pits), each separated by approximately \(50 \mathrm{~nm}\), can be fabricated. Multiple tracks of bits, also separated by approximately \(50 \mathrm{~nm}\), are subsequently fabricated into the surface of the storage medium. Consider a single cantilever that is fabricated primarily of silicon with a mass of \(50 \times 10^{-18} \mathrm{~kg}\) and a surface area of \(600 \times 10^{-15} \mathrm{~m}^{2}\). The cantilever is initially at \(T_{i}=T_{\infty}=300 \mathrm{~K}\), and the heat transfer coefficient between the cantilever and the ambient is \(200 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the ohmic heating required to raise the cantilever temperature to \(T=1000 \mathrm{~K}\) within a heating time of \(t_{h}=1 \mu \mathrm{s}\). Hint: See Problem 5.20. (b) Find the time required to cool the cantilever from \(1000 \mathrm{~K}\) to \(400 \mathrm{~K}\left(t_{c}\right)\) and the thermal processing time required for one complete heating and cooling cycle, \(t_{p}=t_{h}+t_{c}\). (c) Determine how many bits \((N)\) can be written onto a \(1 \mathrm{~mm} \times 1 \mathrm{~mm}\) polymer storage medium. If \(M=100\) cantilevers are ganged onto a single processing head, determine the total thermal processing time needed to write the data.

As permanent space stations increase in size, there is an attendant increase in the amount of electrical power they dissipate. To keep station compartment temperatures from exceeding prescribed limits, it is necessary to transfer the dissipated heat to space. A novel heat rejection scheme that has been proposed for this purpose is termed a Liquid Droplet Radiator (LDR). The heat is first transferred to a high vacuum oil, which is then injected into outer space as a stream of small droplets. The stream is allowed to traverse a distance \(L\), over which it cools by radiating energy to outer space at absolute zero temperature. The droplets are then collected and routed back to the space station. Consider conditions for which droplets of emissivity \(\varepsilon=0.95\) and diameter \(D=0.5 \mathrm{~mm}\) are injected at a temperature of \(T_{i}=500 \mathrm{~K}\) and a velocity of \(V=0.1 \mathrm{~m} / \mathrm{s}\). Properties of the oil are \(\rho=885 \mathrm{~kg} / \mathrm{m}^{3}, c=1900 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=0.145 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assuming each drop to radiate to deep space at \(T_{\text {sur }}=0 \mathrm{~K}\), determine the distance \(L\) required for the droplets to impact the collector at a final temperature of \(T_{f}=300 \mathrm{~K}\). What is the amount of thermal energy rejected by each droplet?

A steel strip of thickness \(\delta=12 \mathrm{~mm}\) is annealed by passing it through a large furnace whose walls are maintained at a temperature \(T_{w}\) corresponding to that of combustion gases flowing through the furnace \(\left(T_{w}=T_{\infty}\right)\). The strip, whose density, specific heat, thermal conductivity, and emissivity are \(\rho=7900 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=640 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon=0.7\), respectively, is to be heated from \(300^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\). (a) For a uniform convection coefficient of \(h=\) \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{w}=T_{\infty}=700^{\circ} \mathrm{C}\), determine the time required to heat the strip. If the strip is moving at \(0.5 \mathrm{~m} / \mathrm{s}\), how long must the furnace be? (b) The annealing process may be accelerated (the strip speed increased) by increasing the environmental temperatures. For the furnace length obtained in part (a), determine the strip speed for \(T_{w}=T_{\infty}=\) \(850^{\circ} \mathrm{C}\) and \(T_{w}=T_{\infty}=1000^{\circ} \mathrm{C}\). For each set of environmental temperatures \(\left(700,850\right.\), and \(\left.1000^{\circ} \mathrm{C}\right)\), plot the strip temperature as a function of time over the range \(25^{\circ} \mathrm{C} \leq T \leq 600^{\circ} \mathrm{C}\). Over this range, also plot the radiation heat transfer coefficient, \(h_{r}\), as a function of time.
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