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A high-temperature, gas-cocled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel elerment \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encused in graphite \((k=3\) \(W / m \cdot K)\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{w}=600 \mathrm{~K}\) and the convection coefficien at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly gencrated in the fuel element at a rate \(q=10^{\text {n }} \mathrm{W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(q\) ?

Short Answer

Expert verified
Find the temperatures using conduction and convection relations, then plot temperature distribution against safe \( q \) limits.

Step by step solution

01

Determine Heat Transfer through Composite Wall

We start by understanding that the fuel element, with thermal conductivity \( k_1 = 57 \mathrm{~W/m \, K} \), generates thermal energy. The graphite encapsulation has a thermal conductivity \( k_2 = 3 \mathrm{~W/m \, K} \). Using these, we can solve for the temperature at the interfaces assuming radial conduction through cylindrical coordinates.
02

Calculate Radial Conduction in the Fuel

Using Fourier's law in cylindrical coordinates for radial steady-state conduction in the fuel (no heat loss radially), we use: \[ q_{gen} = -k \left(\frac{dT}{dr}\right) r \]With uniform generation \( q = 10^n \mathrm{~W/m}^3 \), we integrate to find:\[ T(r) = -\frac{q \cdot r^2}{4k_1} + C_1 \]Applying boundary conditions helps find constants.
03

Apply Boundary Conditions for Fuel Element

Apply the boundary conditions at the inner radius \( r_1 \) and outer radius \( r_2 \). For the inner surface, where temperature \( T_1 \) occurs, the boundary condition is the continuity of temperature and heat flux:\[ -\frac{q \cdot r_1^2}{4k_1} + C_1 = T_1\]For the interface at \( r_2 \), temperature continuity provides a second equation involving both \( T_1 \) and another constant from the integrated equation.
04

Solve for Temperatures T1 and T2 across Fuel

Solve the system of equations from Step 3 to find \( T_1 \) and \( T_2 \). The resulting equations take into account the thermal properties of the fuel and graphite.
05

Address Convective Heat Transfer at Graphite Surface

The external convection boundary condition at the outer graphite surface involves the convection coefficient \( h = 2000 \mathrm{~W/m^2 \cdot K} \):\[ q = h (T_{2} - T_{w}) \]where \( T_w = 600 \mathrm{~K} \). This provides a relation for \( T_2 \), helping establish equilibrium conditions and maximum \( q \).
06

Determine Maximum Allowable Heat Generation Rate

Using the expressions for \( T_1 \) and \( T_2 \) along with material and convective heat transfer limits, calculate the maximum \( q \) under safe operational temperature limits. Consider physical limits such as material degradation thresholds due to excessive heat generation.
07

Plot Temperature Distribution

Using the solved equations and derivatives, plot the temperature distribution from the inner to outer surface for given \( q \) values. Verify the variations vis-a-vis safety limits and heat distribution efficiencies.

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Key Concepts

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

Radial Conduction
Radial conduction is the process of heat transfer through a cylindrical object, like the fuel rod in a nuclear reactor, where heat moves from the inner core outward. This unique conduction type occurs in radial directions, which refers to the line extending from the center to the periphery. The nature of radial conduction in cylindrical coordinates makes it ideal to describe processes in circular objects.
The formula used to describe radial conduction is derived from Fourier’s law of heat conduction, which in cylindrical coordinates appears as: \[ q_{gen} = -k \left(\frac{dT}{dr}\right) r \]
Here, \( k \) is the thermal conductivity, \( dT/dr \) denotes the temperature gradient, and \( r \) represents the radius at any point in the cylindrical object. This equation shows that heat flow rate is proportional to the product of the thermal conductivity, the temperature gradient, and the radial position.
Thermal Conductivity
Thermal conductivity is a key property in the study of heat transfer, measuring a material's ability to conduct heat. In a nuclear reactor, different materials have different thermal conductivities, such as the thorium fuel and graphite, which need to be considered when predicting temperature changes.
For instance, the thorium fuel in the reactor has a thermal conductivity of \( 57 \, \mathrm{W/m \cdot K} \), which means it can conduct heat efficiently. In contrast, graphite has a much lower thermal conductivity of \( 3 \, \mathrm{W/m \cdot K} \), indicating that it conducts heat less efficiently.
Thermal conductivity is integral to predicting how heat will transfer through materials in the reactor, influencing how temperature changes occur across the walls of the fuel element. It’s quantified in watts per meter-kelvin (W/m·K), allowing for consistent prediction of temperature fields.
Convective Heat Transfer
Convective heat transfer refers to the transfer of heat between a solid surface and a fluid, in this case, the helium gas flowing outside the graphite layer of the reactor. It is characterized by a convection coefficient, \( h \), which in this example is \( 2000 \, \mathrm{W/m^2 \cdot K} \).
The convection boundary condition at the graphite’s surface can be described by: \[ q = h \times (T_{2} - T_{w}) \] with \( T_{w} = 600 \, \mathrm{K} \) being the fluid temperature. The coefficient \( h \) indicates how effective the fluid is in removing heat from the solid surface.
Convective heat transfer ensures that the reactor stays cool enough to avoid any potential overheating while maintaining an effective operational temperature, balancing the heat generated inside the reactor.
Cylindrical Coordinates
When dealing with cylindrical objects such as the fuel rod, using cylindrical coordinates is the most practical approach to analyze heat transfer. Cylindrical coordinates are defined by radius \( r \), angular position \( \theta \), and height \( z \), but for radial conduction and heat transfer problems, only the radius \( r \) is typically important.
This coordinate system allows us to apply equations like Fourier's law in a form that matches the geometry of the objects we're studying. For instance, considering the radial component in the heat transfer equations allows a more accurate representation of temperature gradients and distribution.
Utilizing cylindrical coordinates helps simplify otherwise complex three-dimensional calculation problems into more manageable forms, focusing specifically on the radial component, which is most relevant in the reactor’s fuel element.
Temperature Distribution
Temperature distribution describes how temperature varies throughout the material of the nuclear reactor's fuel rod and the surrounding layers. This is crucial for understanding the reactor's performance and ensuring safety.
In this exercise, temperature at different layers, denoted as \( T_1 \) and \( T_2 \), are calculated through solving the conduction equations with specific boundary conditions. The temperature distribution is vital for identifying which sections might reach critically high temperatures.
Plotting temperature distribution can reveal areas under thermal stress, guiding potential design improvements or operational changes to prevent overheating. Checking the distribution ensures that the materials used in the composite wall remain within their thermal limits, avoiding damage.

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Most popular questions from this chapter

An experimental arrangement for measuring the termal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, ex. cept that one is fabricated from a standard material of known thermal conductivity \(k_{A}\) 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 \(T_{b \text { r }}\) are exposed to a fluid of temperature \(T_{\text {su }}\) and are instrumented with thermocouples to measure the temperature at a fined 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 mensurements 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 \(T_{a}=25^{\circ} \mathrm{C}\), what is the then mal conductivity \(k_{\mathrm{g}}\) of the test material?

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W})\) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?

Superheated steam at \(575^{\circ} \mathrm{C}\) is routed from a boiler to the turbine of an electric power plant through steel tubes \((k=35 \mathrm{~W} / \mathrm{m}+\mathrm{K})\) 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 \((k=0.10 \mathrm{~W} / \mathrm{m}-\mathrm{K})\) 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 cutside the aluminum sheet is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the minimum insulation thickness needed to insure that the teriperature of the aluminum does not exceed \(50^{\circ} \mathrm{C}\) ? What is the corresponding heat loss per meter of tube length? (b) Explore the effect of the insulation thickness on the temperature of the aluminum and the heat loss per unit tube length.

The rear window of an automobile is defogged by attaching a thin, transparent, film-type heating clement to its inner surface. By electrically heating this element, a uniform heat flux may be established at the inner surface. (a) For 4-mm-thick window glass, determine the electrical power required per unit window area to maintain an inner surface temperature of \(15^{\circ} \mathrm{C}\) when the interior air temperature and convection cocfficient are \(T_{=i}=25^{\circ} \mathrm{C}\) and \(h_{i}=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the exterior (ambient) air temperature and convection coefficient are \(T_{\text {we }}=\) \(-10^{\circ} \mathrm{C}\) and \(h_{e}=65 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) In practice \(T_{w e}\) and \(h_{y}\) vary according to weather conditions and car speed. For values of \(h_{\mathrm{m}}=2,20,65\), and \(100 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\), determine and plot the electrical power requirement as a function of \(T_{z,}\) for \(-30 \leq T_{x p} \leq\) \(0^{\circ} \mathrm{C}\). From your results, what can you conclude about the need for heater operation at low values of \(h_{e}\) ? How is this conclusion affected by the value of \(T_{\text {w, }, 0}\) ? If \(h \propto\) \(V^{*}\), where \(V\) is the vehicle speed and \(n\) is a positive exponent, how does the vehicle speed affect the need for heater eperation?

An annular aluminum fin of rectangular profile is attached to a circular tube having an outside diameter of \(25 \mathrm{~mm}\) and a surface temperature of \(250^{\circ} \mathrm{C}\). The fin is \(1 \mathrm{~mm}\) thick and \(10 \mathrm{~mm}\) long, and the temperature and the convection coefficient associated with the adjoining fluid are \(25^{\circ} \mathrm{C}\) and \(25 \mathrm{~W} / \mathrm{m}^{2}\) + \(\mathrm{K}\), respectively. (a) What is the heat loss per fin? (b) If 200 such fins are spaced at 5-mm increments along the tube length, what is the heat loss per meter of tube length?

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