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Chapter 3: Problem 97

The cross section of a long cylindrical fuel element in a nuclear reactor is shown. Energy generation occurs uniformly in the thorium fuel rod, which is of diameter \(D=25 \mathrm{~mm}\) and is wrapped in a thin aluminum cladding. (a) It is proposed that, under steady-state conditions, the system operates with a generation rate of \(\dot{q}=\) \(7 \times 10^{8} \mathrm{~W} / \mathrm{m}^{3}\) and cooling system characteristics of \(T_{\infty}=95^{\circ} \mathrm{C}\) and \(h=7000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Is this proposal satisfactory? (b) Explore the effect of variations in \(\dot{q}\) and \(h\) by plotting temperature distributions \(T(r)\) for a range of parameter values. Suggest an envelope of acceptable operating conditions.

Short Answer

Expert verified
In summary, we analyzed the given proposal by calculating and comparing the surface temperature of the fuel rod to its critical temperature and determined if the proposed steady-state conditions were satisfactory. Additionally, we explored the impact of varying the generation rate \(\dot{q}\) and convection coefficient \(h\) on the temperature distribution \(T(r)\) in the fuel element to define an envelope of acceptable operating conditions. These steps ensure safe and reliable operation of the fuel rod in a nuclear reactor.

Step by step solution

01

Check the Satisfactory of the Proposal

First, check if the proposed steady-state conditions are satisfactory. The given conditions are: - Generation rate: \(\dot{q} = 7 \times 10^8\ \text{W/m}^3\) - Ambient temperature: \(T_\infty = 95^{\circ}\ \text{C}\) - Convection coefficient: \(h = 7000\ \text{W/m}^2\cdot\text{K}\) To evaluate if the proposal is satisfactory, we can use the heat transfer equation for a cylindrical fuel rod: \[q = kA\frac{\Delta T}{L}\] Here, \(q\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the cross-sectional area, \(\Delta T\) is the temperature difference between the center and surface of the rod, and \(L\) is the length of the rod. In steady-state conditions, \(q = \dot{q}V\), where \(V\) is the volume of the rod, and we can calculate the heat transfer rate per unit length: \[q' = \dot{q}\pi\left(\frac{D}{2}\right)^2\] Next, we can find the temperature difference \(\Delta T\): \[\Delta T = \frac{q'}{kA/L}\] Lastly, we can compare the surface temperature of the rod, \(T_s = T_\infty + \Delta T\), to the allowable or critical surface temperature, \(T_{crit}\), to determine if the proposal is satisfactory.
02

Plot Temperature Distributions

To study the effect of variations in \(\dot{q}\) and \(h\) on the temperature distribution \(T(r)\), we can use the heat conduction equation in cylindrical coordinates for a steady-state and radially symmetric temperature distribution: \[\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right) = -\frac{\dot{q}}{k}\] By integrating this equation twice with respect to \(r\), considering the boundary conditions and the given \(\dot{q}\) and \(h\) values, we can generate plots of \(T(r)\) for different variations in \(\dot{q}\) and \(h\). The required boundary conditions are that \(dT/dr = 0\) at \(r=0\) and \(-k dT/dr = h (T - T_\infty)\) at \(r = D/2\). When the temperature distributions are plotted, we can analyze the envelope of acceptable operating conditions by observing the values of \(\dot{q}\) and \(h\) that yield surface temperatures below the critical temperature. By clearly specifying the range of suitable values and considering the practical limitations and design factors, we can establish a suitable boundary of parameters that ensure a reliable and safe operation of the fuel rod.

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Key Concepts

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

Nuclear Reactor
A nuclear reactor is essentially a system that produces energy through nuclear fission, the process where atomic nuclei split into smaller parts. As a result, an enormous amount of energy is released. This energy is harnessed for a variety of purposes, primarily to generate electricity in power plants.

Within a reactor, fuel elements such as fuel rods are crucial components. These rods contain isotopic materials like uranium or thorium, encapsulating them safely to undergo controlled reactions. These reactions produce heat, which needs to be managed continually to avoid overheating.
  • Fuel rods: Typically made of thorium or uranium; central in the reaction process.
  • Control of heat: Essential for maintaining safe operations.
  • Energy generation: Converts nuclear energy into heat energy, which then generates electricity.
Maintaining the reactor at steady-state conditions is vital for safety and efficiency, minimizing any risk of an accident.
Steady-State Conditions
In the context of nuclear reactors, steady-state conditions refer to a state where the reactor operates at constant characteristics without any significant change over time in temperature, pressure, and power output.

These conditions ensure that the energy generation, in terms of both value and distribution, matches the energy removal rate so there is no net change in energy contained within the reactor.
  • Constant operations: Achieved when input and output energies are balanced.
  • Temperature stability: Helps in avoiding the risks associated with rapid temperature changes.
  • Predictability and control: Crucial for safe operation of nuclear reactors.
The steady-state model simplifies the analysis, allowing engineers to predict how the reactor will behave under normal conditions, and to ensure all parameters stay within safe limits.
Temperature Distribution
Temperature distribution in a nuclear reactor, like in the cylindrical fuel rod discussed, is how temperature varies across a material due to varied energy generation or heat transfer processes.

Understanding this distribution allows us to determine how heat moves from the reactor core to its cooler surroundings, influencing design decisions and safety measures. In cylindrical coordinates for the cylindrical rod, this distribution is determined by taking into account the heat generated by nuclear reactions and the thermal properties of the materials involved.
  • Energy generation: Dictated by the specific nuclear reaction occurring.
  • Thermal equilibrium: Ensures heat is evenly dissipated, avoiding hotspots.
  • Boundary conditions: Such as the temperature at the surface and thermal resistance, impact distribution outcomes.
Proper analysis of temperature distribution ensures components do not exceed their temperature constraints, thus maintaining structural integrity and performance.
Thermal Conductivity
Thermal conductivity is a material-specific property that determines how well a material can conduct heat. In the context of nuclear reactors, it plays a pivotal role in understanding how quickly heat generated from nuclear reactions within the fuel rods is transferred through the material and away from the core.

A high thermal conductivity means heat is effectively transferred, reducing the risk of overheating. Lower values could result in higher operational temperatures and potential material failure.
  • Material selection: Choosing materials with optimal thermal conductivities for safety and efficiency.
  • Heat transfer efficiency: Directly influences reactor performance and safety.
  • Impact on design: Influences the design of reactor components, ensuring adequate heat dissipation structures are in place.
Understanding thermal conductivity and its implications helps in designing reactors that are efficient, safe, and reliable over extended periods.

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Most popular questions from this chapter

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of \(300-\mathrm{mm}\) inner diameter and \(30-\mathrm{mm}\) wall thickness. To reduce heat loss to the surroundings and to maintain a safe-to-touch outer surface temperature, a layer of calcium silicate insulation \((k=0.10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the tubes, while degradation of the insulation is reduced by wrapping it in a thin sheet of aluminum having an emissivity of \(\varepsilon=0.20\). The air and wall temperatures of the power plant are \(27^{\circ} \mathrm{C}\). (a) Assuming that the inner surface temperature of a steel tube corresponds to that of the steam and the convection coefficient outside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to ensure that the temperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss(b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length. per meter of tube length?

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?

Consider the conditions of Problem \(3.149\) but now allow for a tube wall thickness of \(5 \mathrm{~mm}\) (inner and outer diameters of 50 and \(60 \mathrm{~mm}\) ), a fin-to-tube thermal contact resistance of \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the fact that the water temperature, \(T_{w}=350 \mathrm{~K}\), is known, not the tube surface temperature. The water-side convection coefficient is \(h_{w}=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the rate of heat transfer per unit tube length \((\mathrm{W} / \mathrm{m})\) to the water. What would be the separate effect of each of the following design changes on the heat rate: (i) elimination of the contact resistance; (ii) increasing the number of fins from four to eight; and (iii) changing the tube wall and fin material from copper to AISI 304 stainless steel \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) ?

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 thermopane window consists of two pieces of glass \(7 \mathrm{~mm}\) thick that enclose an air space \(7 \mathrm{~mm}\) thick. The window separates room air at \(20^{\circ} \mathrm{C}\) from outside ambient air at \(-10^{\circ} \mathrm{C}\). The convection coefficient associated with the inner (room-side) surface is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the convection coefficient associated with the outer (ambient) air is \(h_{o}=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the heat loss through a window that is \(0.8 \mathrm{~m}\) long by \(0.5 \mathrm{~m}\) wide? Neglect radiation, and assume the air enclosed between the panes to be stagnant. (b) Compute and plot the effect of \(h_{o}\) on the heat loss for \(10 \leq h_{o} \leq 100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Repeat this calculation for a triple-pane construction in which a third pane and a second air space of equivalent thickness are added.
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