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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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Key Concepts

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

Nuclear Fuel Rod
The nuclear fuel rod lies at the heart of this heat transfer problem. A fuel rod, typically used in reactors, is a cylindrical component that contains nuclear fuel. In our scenario, the fuel rod has a diameter of 1 cm, and it generates heat at a substantial rate of 50 MW/m鲁. Such fuel rods are critical for producing energy as they harbor the nuclear reactions necessary for power generation.
  • Composition: Often made up of uranium or plutonium in a solid form, these rods must withstand high temperatures and corrosive environments inside a reactor.
  • Heat Generation: The heat generated is due to the fission process, where atomic nuclei split releasing energy. This energy must be efficiently removed to prevent overheating.
Understanding the properties of a nuclear fuel rod, including its thermal conductivity and heat generation rate, is essential for solving heat transfer problems as it influences the temperature distribution across the system.
Convection Heat Transfer
Convection plays a crucial role in regulating the temperature of the system in our exercise. It refers to the transfer of heat from the fuel rod to the flowing water through the annular region, facilitated by the convection heat transfer coefficient.
  • Coefficient: Given as 2000 W/m虏路K, this parameter quantifies how effectively the heat is transferred between the rod and the fluid.
  • Mechanism: Convection involves both conduction and the bulk fluid motion, allowing for a more efficient transfer than conduction alone.
Calculating the heat transferred through convection is essential to determine the average temperature of the cooling water and ensure the system remains within operational limits.
Temperature Distribution
Temperature distribution within the fuel rod is pivotal in understanding how heat travels across its radius from the center to the surface. The formula used in our solution helps map out this distribution: \[\Theta(r) = -\frac{q_g}{2k}(r^2 - r_i^2)\]where \(\Theta(r)\) denotes the change in temperature relative to the surface temperature of the concentric tube.
  • Radial Variation: This expression illustrates how the temperature diminishes from the rod's center towards the surface.
  • Influence Factors: The thermal conductivity, given as 30 W/m路K, and the heat generation rate unify to define how swiftly and efficiently heat is dissipated.
Grasping the temperature distribution allows engineers to predict temperature gradients, crucial for maintaining material integrity and safety.
Thermal Conductivity
Thermal conductivity is a material property vital in calculating how heat is transferred radially through the nuclear fuel rod. In our task, the rod's thermal conductivity is specified as 30 W/m路K, directly influencing the heat flux.
  • Material Property: This constant helps describe the material's ability to conduct heat; higher values indicate better conduction.
  • Application: With our formula \(q' = -k\frac{dT}{dr}\), thermal conductivity is essential to determine how fast temperatures change within the rod based on heat generation.
Understanding thermal conductivity is indispensable for engineers aiming to optimize the performance and safety of reactor fuel rods by ensuring that heat is efficiently managed and dissipated.

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Most popular questions from this chapter

A long homogeneous resistance wire of radius \(r_{o}=\) \(5 \mathrm{~mm}\) is being used to heat the air in a room by the passage of electric current. Heat is generated in the wire uniformly at a rate of \(5 \times 10^{7} \mathrm{~W} / \mathrm{m}^{3}\) as a result of resistance heating. If the temperature of the outer surface of the wire remains at \(180^{\circ} \mathrm{C}\), determine the temperature at \(r=3.5 \mathrm{~mm}\) after steady operation conditions are reached. Take the thermal conductivity of the wire to be \(k=6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

Consider a steam pipe of length \(L=35 \mathrm{ft}\), inner radius \(r_{1}=2\) in, outer radius \(r_{2}=2.4\) in, and thermal conductivity \(k=8 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\). Steam is flowing through the pipe at an average temperature of \(250^{\circ} \mathrm{F}\), and the average convection heat transfer coefficient on the inner surface is given to be \(h=\) \(15 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). If the average temperature on the outer surfaces of the pipe is \(T_{2}=160^{\circ} \mathrm{F},(a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the pipe, \((b)\) obtain a relation for the variation of temperature in the pipe by solving the differential equation, and \((c)\) evaluate the rate of heat loss from the steam through the pipe.

Consider steady one-dimensional heat conduction through a plane wall, a cylindrical shell, and a spherical shell of uniform thickness with constant thermophysical properties and no thermal energy generation. The geometry in which the variation of temperature in the direction of heat transfer will be linear is (a) plane wall (b) cylindrical shell (c) spherical shell (d) all of them (e) none of them

Consider a spherical container of inner radius \(r_{1}\), outer radius \(r_{2}\), and thermal conductivity \(k\). Express the boundary condition on the inner surface of the container for steady onedimensional conduction for the following cases: \((a)\) specified temperature of \(50^{\circ} \mathrm{C},(b)\) specified heat flux of \(45 \mathrm{~W} / \mathrm{m}^{2}\) toward the center, (c) convection to a medium at \(T_{\infty}\) with a heat transfer coefficient of \(h\).

A spherical shell, with thermal conductivity \(k\), has inner and outer radii of \(r_{1}\) and \(r_{2}\), respectively. The inner surface of the shell is subjected to a uniform heat flux of \(\dot{q}_{1}\), while the outer surface of the shell is exposed to convection heat transfer with a coefficient \(h\) and an ambient temperature \(T_{c \infty}\). Determine the variation of temperature in the shell wall and show that the outer surface temperature of the shell can be expressed as \(T\left(r_{2}\right)=\left(\dot{q}_{1} / h\right)\left(r_{1} / r_{2}\right)^{2}+T_{\infty \text { co }}\).
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