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Chapter 2: Problem 99

A cylindrical nuclear fuel rod \(1 \mathrm{~cm}\) in diameter is encased in a concentric tube \(2 \mathrm{~cm}\) 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
Based on the given information, calculate the average temperature of the cooling water and check if it's possible to determine the fuel rod's surface temperature.

Step by step solution

01

Calculate the total heat generation rate

First, we need to find the volume of the cylindrical fuel rod. We can do so using the equation \(V = \pi(\frac{D_{1}}{2})^2h\), where \(h\) is the height of the cylinder. The total heat generation rate can be calculated using \(Q = q_{gen}\cdot V\). Since we are not given the height, let \(h = 1\;\mathrm{m}\). This will give us simplification without loss of generality. So, the volume of the fuel rod is: \(V = \pi(\frac{1\times10^{-2} \mathrm{m}}{2})^2(1\;\mathrm{m}) = \frac{\pi}{400}\;\mathrm{m^3}\). Now, the total heat generation rate is: \(Q = q_{gen}\cdot V = (50\times10^6\;\mathrm{W/m^3})\cdot \frac{\pi}{400}\;\mathrm{m^3} = \frac{50\pi\times10^6}{400}\;\mathrm{W}\).
02

Calculate the temperature difference between the inner and outer surfaces

The heat generated in the fuel rod conducts through the fuel rod and then convects into the cooling water. The temperature difference between the inner and outer surfaces can be determined using the heat transfer rate equation: \(\Delta T = \frac{Q}{2 \pi h (1/\left(k \cdot \log{\frac{D_2}{D_1}}\right) + 1/\left(hD_2\right) )}\). Plugging the given values, \(\Delta T = \frac{\frac{50\pi\times10^6}{400}\;\mathrm{W}}{2 \pi(1\;\mathrm{m})(1/\left(30 \mathrm{W/m\cdot K} \cdot \log{\frac{2\times10^{-2}\;\mathrm{m}}{1\times10^{-2}\;\mathrm{m}}}\right) + 1/\left(2000\mathrm{W/m^2\cdot K}(2\times10^{-2}\;\mathrm{m})\right) )}\). Solve for \(\Delta T\).
03

Determine the average temperature of the cooling water

With the temperature difference \(\Delta T\) calculated in Step 2, we can find the average temperature of the cooling water by subtracting \(\Delta T\) from the surface temperature of the concentric tube \(T_{s}\). \(T_{avg} = T_{s} - \Delta T\) Calculate \(T_{avg}\).
04

Check if the surface temperature of the fuel rod can be determined

To determine the surface temperature of the fuel rod, we need the exact temperature distribution throughout the fuel rod. However, we don't have enough information to find that, as we lack the heat generation rate as a continuous function of radial position within the fuel rod. Therefore, with the given information, we cannot determine the surface temperature of the fuel rod.

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