Question

A long cylindrical fuel rod with a diameter of 3 cm is enclosed by a concentric tube with a diameter of \(6 \mathrm{~cm}\). The tube is an ASTM A249 904L stainless steel tube with a maximum use temperature of $260^{\circ} \mathrm{C}$ specified by the ASME Code for Process Piping (ASME B31.3-2014, Table A-1M). The gap between the fuel rod and the tube is a vacuum. The fuel rod has a surface temperature of \(550^{\circ} \mathrm{C}\). The emissivities of the fuel rod and the ASTM A249 904L tube are \(0.97\) and \(0.33\), respectively. The rate of radiation heat transfer per unit length from the fuel rod to the stainless steel tube is \(120 \mathrm{~W} / \mathrm{m}\). A concentric radiation shield with a diameter of \(45 \mathrm{~mm}\) is to be placed in between the fuel rod and the stainless steel tube to keep the temperature of the stainless steel tube from exceeding its maximum use temperature. Determine (a) the emissivity that the radiation shield needs to keep the stainless steel tube from exceeding \(260^{\circ} \mathrm{C}\) and \((b)\) the temperature of the stainless steel tube if there is no radiation shield.

Short answer

a) To keep the stainless steel tube within its maximum temperature of \(260^{\circ}\mathrm{C}\), the required emissivity of the radiation shield is approximately 0.053. b) The temperature of the stainless steel tube without the radiation shield would be approximately \(533.55^{\circ}\mathrm{C}\).

Step-by-step solution

Determine the geometry and temperatures of the configuration

We know the diameters of the fuel rod, stainless steel tube, and the gap between them. The fuel rod has a diameter of 3 cm, and the stainless steel tube has a diameter of 6 cm. The radiation shield will have a diameter of 45 mm, which is equivalent to 4.5 cm. The surface temperature of the fuel rod is \(550^{\circ} \mathrm{C}\), and the maximum use temperature of the stainless steel tube is \(260^{\circ} \mathrm{C}\).

Calculate the radii of each component

To compute the radiative heat transfer, we need the radii of each component. These are: Fuel rod radius: \(r_1 = 3~\text{cm} / 2 = 1.5~\text{cm}\) Stainless steel tube radius: \(r_2 = 6~\text{cm} / 2 = 3~\text{cm}\) Radiation shield radius: \(r_s = 45~\text{mm} / 2 = 22.5~\text{mm}\)

Calculate the radiative resistance without the radiation shield

Now we need to calculate the radiative resistance without the radiation shield. This can be found using the given emissivities of the fuel rod (\(\epsilon_1 = 0.97\)) and the stainless steel tube (\(\epsilon_2 = 0.33\)), as well as the radii from Step 2. The radiative resistance \(R_{12}\) is given by the formula: \(R_{12} = \frac{1 - \epsilon_1}{\epsilon_1 A_1}+\frac{1}{A_1 F_{1 \to 2}}+\frac{1 - \epsilon_2}{\epsilon_2 A_2}\), where \(F_{1 \to 2}\) is the view factor from surface 1 to surface 2. Since the surfaces are concentric cylinders, the view factor \(F_{1 \to 2} = 1\). The surface areas of the fuel rod and the stainless steel tube per unit length are: \(A_1 = 2\pi r_1\) and \(A_2 = 2\pi r_2\). Using the given values: \(R_{12} = \frac{1 - 0.97}{0.97(2\pi(1.5\times10^{-2} \mathrm{m}))}+\frac{1}{2\pi(1.5\times10^{-2} \mathrm{m})}+\frac{1 - 0.33}{0.33(2\pi(3\times10^{-2} \mathrm{m}))}\) Calculate \(R_{12}\) to obtain: \(R_{12} \approx 0.137~\mathrm{m^2K/W}\)

Calculate the temperature of the stainless steel tube without a radiation shield

Using the given rate of radiation heat transfer per unit length (\(Q = 120 \mathrm{~W} / \mathrm{m}\)) and the radiative resistance \(R_{12}\), we can determine the temperature difference between the fuel rod and the stainless steel tube: \(\Delta T = Q \times R_{12}\) \(\Delta T = 120 \mathrm{~W} / \mathrm{m} \times 0.137~\mathrm{m^2K/W} \approx 16.45 \mathrm{~K}\) Now we can calculate the temperature of the stainless steel tube: \(T_{tube} = T_{fuel~rod} - \Delta T = 550^{\circ}\mathrm{C} - 16.45 \mathrm{~K} \approx 533.55^{\circ} \mathrm{C}\) Therefore, the temperature of the stainless steel tube without the radiation shield is approximately \(533.55^{\circ} \mathrm{C}\).

Calculate the required emissivity of the radiation shield

Now we need to calculate the necessary emissivity of the radiation shield to keep the temperature of the stainless steel tube below \(260^{\circ} \mathrm{C}\). First, determine the desired radiative resistance \(R_{1s2}\), based on the allowed temperature difference: \(\Delta T_{allowed} = T_{fuel~rod} - T_{max~tube} = 550^{\circ}\mathrm{C} - 260^{\circ}\mathrm{C} = 290 \mathrm{~K}\) Now, find the allowed radiative resistance \(R_{1s2}\): \(R_{1s2} = \frac{\Delta T_{allowed}}{Q} = \frac{290 \mathrm{~K}}{120 \mathrm{~W} / \mathrm{m}} \approx 2.42~\mathrm{m^2K/W}\) Next, we solve for the required emissivity \(\epsilon_s\) of the radiation shield. The formula for the radiative resistance with the radiation shield \(R_{1s2}\) is: \(R_{1s2} = \frac{1 - \epsilon_1}{\epsilon_1 A_1}+\frac{1 - \epsilon_s}{\epsilon_s A_1}+\frac{1}{A_1 F_{1 \to s}}+\frac{1}{A_s F_{s \to 2}}+\frac{1 - \epsilon_s}{\epsilon_s A_s}+\frac{1 - \epsilon_2}{\epsilon_2 A_2}\), where \(F_{1 \to s}\) and \(F_{s \to 2}\) are the view factors and can be assumed to be \(1\) in this configuration. Using the given and calculated values, and solving for \(\epsilon_s\): \(\epsilon_s = \frac{1}{\left(\frac{R_{1s2}-(\frac{1 - \epsilon_1}{\epsilon_1 A_1})+(\frac{1}{A_1})+\frac{1 - \epsilon_2}{\epsilon_2 A_2}}{A_1+A_s}\right) + 1}\) Plug in the known values to find the required emissivity: \(\epsilon_s \approx 0.053\) So the radiation shield needs an emissivity of approximately 0.053 to keep the stainless steel tube from exceeding \(260^{\circ} \mathrm{C}\).