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Chapter 2: Problem 95

A cylindrical nuclear fuel rod of \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube of \(2 \mathrm{~cm}\) in diameter, where cooling water flows through the annular region between the fuel rod \((k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and the concentric tube. Heat is generated uniformly in the rod at a rate of \(50 \mathrm{MW} / \mathrm{m}^{3}\). The convection heat transfer coefficient for the concentric tube surface is \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the surface temperature of the concentric tube is \(40^{\circ} \mathrm{C}\), determine the average temperature of the cooling water. Can one use the given information to determine the surface temperature of the fuel rod? Explain.

Short Answer

Expert verified
Answer: Based on the given information and calculations, the average temperature of the cooling water can be found using the formula mentioned in Step 4. However, the surface temperature of the nuclear fuel rod cannot be determined due to insufficient data.

Step by step solution

01

Calculate the heat generation rate

The heat generation rate per unit volume in the cylindrical nuclear fuel rod is given as \(50 \mathrm{MW} / \mathrm{m}^{3}\) or \(50 \times 10^6 \mathrm{W} / \mathrm{m}^{3}\).
02

Calculate radial heat flux

To calculate the radial heat flux (\(q'\)) within the fuel rod, we can use the relation: $$ q' = -k\frac{dT}{dr} $$ where \(k = 30 \mathrm{W}/\mathrm{m} \cdot \mathrm{K}\) is the thermal conductivity of the fuel rod, \(dT/dr\) is the temperature gradient across the region. To simplify this relation, let's introduce a new variable: \(\Theta(r) = T(r) - T_s\), where \(T(r)\) is the temperature at a radial position r, and \(T_s\) is the surface temperature of the concentric tube (\((40^{\circ} \mathrm{C})\).
03

Calculate temperature distribution

The heat generated within the fuel rod must be equal to the heat transferred through the annular region between the fuel rod and the concentric tube. We can write the equation for the heat generation rate in terms of temperature distribution: $$ Q = -k A \frac{d\Theta}{dr} = q_g A $$ where \(Q\) is the heat generation rate per unit length, \(A\) is the cross-sectional area of the fuel rod, \(q_g\) is the heat generation rate per unit volume given, and \(\Theta\) is the temperature distribution. Solving for \(\Theta\) and the radial position \(r\), we get: $$ \Theta(r) = -\frac{q_g}{2k}(r^2 - r_i^2) $$ where \(r_i = 0.5\mathrm{~cm}\) is the radius of the fuel rod.
04

Calculate average temperature of the cooling water

We can find the average temperature of the cooling water by dividing the total heat generated in the rod by the convection heat transfer coefficient and the surface area of the concentric tube: $$ T_{avg} = T_s + \frac{Q}{2000(2\pi r_o L)} $$ where \(T_{avg}\) is the average temperature of the cooling water, \(T_s = 40^{\circ} \mathrm{C}\), \(Q = q_g A L\), \(r_o = 1\mathrm{~cm}\) is the radius of the concentric tube, and \(L\) is the length of the rod. Substituting the values and solving for \(T_{avg}\), we get the average temperature of the cooling water.
05

Determine if surface temperature of the fuel rod can be found

In order to determine the surface temperature of the fuel rod, we need to know the temperature distribution within the rod, which we have found in Step 3. However, we do not have enough information to calculate the temperature at the surface of the fuel rod. Therefore, we cannot determine the surface temperature of the fuel rod using the given information.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(L=0.05 \mathrm{~m}\). The wall surface at \(x=0\) is insulated, while the surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\). The thermal conductivity of the wall is \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and heat is generated in the wall at a rate of \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5 x / L} \mathrm{~W} / \mathrm{m}^{3}\) where \(\dot{e}_{0}=8 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) determine the temperature of the insulated surface of the wall.

A spherical vessel has an inner radius \(r_{1}\) and an outer radius \(r_{2}\). The inner surface \(\left(r=r_{1}\right)\) of the vessel is subjected to a uniform heat flux \(\dot{q}_{1}\). The outer surface \(\left(r=r_{2}\right)\) is exposed to convection and radiation heat transfer in a surrounding temperature of \(T_{\infty}\). The emissivity and the convection heat transfer coefficient on the outer surface are \(\varepsilon\) and \(h\), respectively. Express the boundary conditions and the differential equation of this heat conduction problem during steady operation.

Consider the base plate of an \(800-W\) household iron with a thickness of \(L=0.6 \mathrm{~cm}\), base area of \(A=160 \mathrm{~cm}^{2}\), and thermal conductivity of \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the base plate is subjected to uniform heat flux generated by the resistance heaters inside. When steady operating conditions are reached, the outer surface temperature of the plate is measured to be \(112^{\circ} \mathrm{C}\). Disregarding any heat loss through the upper part of the iron, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the plate, \((b)\) obtain a relation for the variation of temperature in the base plate by solving the differential equation, and (c) evaluate the inner surface temperature. Answer: (c) \(117^{\circ} \mathrm{C}\)

A long homogeneous resistance wire of radius \(r_{o}=\) \(0.6 \mathrm{~cm}\) and thermal conductivity \(k=15.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is being used to boil water at atmospheric pressure by the passage of electric current. Heat is generated in the wire uniformly as a result of resistance heating at a rate of \(16.4 \mathrm{~W} / \mathrm{cm}^{3}\). The heat generated is transferred to water at \(100^{\circ} \mathrm{C}\) by convection with an average heat transfer coefficient of \(h=3200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wire, \((b)\) obtain a relation for the variation of temperature in the wire by solving the differential equation, and \((c)\) determine the temperature at the centerline of the wire.

A spherical vessel is filled with chemicals undergoing an exothermic reaction. The reaction provides a uniform heat flux on the inner surface of the vessel. The inner diameter of the vessel is \(5 \mathrm{~m}\) and its inner surface temperature is at \(120^{\circ} \mathrm{C}\). The wall of the vessel has a variable thermal conductivity given as \(k(T)=k_{0}(1+\beta T)\), where \(k_{0}=1.01 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\beta=0.0018 \mathrm{~K}^{-1}\), and \(T\) is in \(\mathrm{K}\). The vessel is situated in a surrounding with an ambient temperature of \(15^{\circ} \mathrm{C}\), the vessel's outer surface experiences convection heat transfer with a coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent thermal burn on skin tissues, the outer surface temperature of the vessel should be kept below \(50^{\circ} \mathrm{C}\). Determine the minimum wall thickness of the vessel so that the outer surface temperature is \(50^{\circ} \mathrm{C}\) or lower.
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