Question

A long cylindrical fuel rod with a diameter of \(25 \mathrm{~mm}\) is enclosed by a concentric tube with a diameter of \(50 \mathrm{~mm}\). 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 B31.32014 , 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 \(904 \mathrm{~L}\) tube are \(0.97\) and \(0.33\), respectively. Outside of the tube, convection with the ambient air and radiation with the surroundings occur. The convection heat transfer coefficient of the tube with the ambient air is $15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( at \)20^{\circ} \mathrm{C}$. The temperature of the surroundings is the same as the ambient air temperature. Determine whether the temperature of the tube surface is below the maximum use temperature for ASTM A249 904L stainless steel tube.

Short answer

Based on the given information about a fuel rod and a concentric tube enclosing it, we are tasked to determine if the temperature of the tube surface is below the maximum use temperature for the given material. By analyzing the heat transfer processes involved, particularly radiation and convection, and using the provided emissivities and convection heat transfer coefficient, we can determine the surface temperature and compare it to the maximum use temperature. To solve this problem, we follow these steps: Step 1: Calculate the heat transfer through radiation and express it as a function of the temperature difference between the fuel rod and the tube (T_1^4 - T_2^4). Step 2: Calculate the heat transfer through convection and express it as a function of the temperature difference between the tube and the ambient air (T_2 - T_∞). Step 3: Equate the heat transfer through radiation and convection, and rearrange the equation to solve for the heat transfer rate and temperature of the tube surface (T_2). Step 4: Compare the calculated surface temperature to the maximum use temperature of the ASTM A249 904L stainless steel tube. If the calculated surface temperature is below or equal to the maximum use temperature, the tube is safe for use under these conditions; otherwise, a re-evaluation of the design or materials is needed for safety.

Step-by-step solution

Calculate the heat transfer through radiation

We can use the following formula to determine the heat transfer through radiation: $$q_{rad} = \sigma \varepsilon_1 \varepsilon_2 A_1(T_1^4 - T_2^4)$$ where \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K^4}\)), \(\varepsilon_1\) and \(\varepsilon_2\) are the emissivities of the fuel rod and the tube, \(A_1\) is the area of the fuel rod, and \(T_1\) and \(T_2\) are the temperatures of the fuel rod and the tube in Kelvin. First, let's convert the temperatures to Kelvin: $$T_1 = 550 + 273.15 = 823.15 \mathrm{K}$$ Next, we'll find the area of the fuel rod, \(A_1=\pi d_1L\), where \(d_1\) is the diameter and \(L\) is the length of the rod. We will assume a length of \(1 \mathrm{m}\) for calculation purposes: $$A_1 = \pi (0.025 \mathrm{m})(1 \mathrm{m}) = 0.025 \pi \mathrm{m}^2$$ Now we can plug these values into the radiation heat transfer equation to find the heat transfer through radiation. We will solve for \(T_2^4\): $$q_{rad} = \sigma \varepsilon_1 \varepsilon_2 A_1(T_1^4 - T_2^4)$$ $$T_2^4 = \frac{q_{rad}}{\sigma \varepsilon_1 \varepsilon_2 A_1} - T_1^4$$

Calculate the heat transfer through convection

We can use the following formula to determine the heat transfer through convection: $$q_{conv} = h A_2(T_2 - T_{\infty})$$ where \(h\) is the convection heat transfer coefficient, \(A_2\) is the area of outer surface of the tube, \(T_2\) is the temperature of the tube surface in Kelvin, and \(T_{\infty}\) is the ambient temperature in Kelvin. First, let's convert the ambient air temperature to Kelvin: $$T_{\infty} = 20 + 273.15 = 293.15 \mathrm{K}$$ Next, we'll find the area of the outer surface of the tube, \(A_2=\pi d_2L\), where \(d_2\) is the diameter and \(L\) is the length of the tube. We will assume a length of \(1 \mathrm{m}\) for calculation purposes: $$A_2 = \pi (0.05 \mathrm{m})(1 \mathrm{m}) = 0.05 \pi \mathrm{m}^2$$ Now we can plug these values into the convection heat transfer equation to find the heat transfer through convection. We will solve for \(T_2\): $$q_{conv} = h A_2(T_2 - T_{\infty})$$ $$T_2 = \frac{q_{conv}}{h A_2} + T_{\infty}$$

Equate the heat transfer through radiation and convection

Since the heat transfer through radiation should be equal to the heat transfer through convection, we can equate the two expressions we derived in steps 1 and 2: $$\frac{q_{rad}}{\sigma \varepsilon_1 \varepsilon_2 A_1} - T_1^4 = \frac{q_{conv}}{h A_2} + T_{\infty}$$ Solving for \(q_{rad}\) and \(q_{conv}\), we get: $$q_{rad} = q_{conv} = \sigma \varepsilon_1 \varepsilon_2 A_1(T_1^4 - T_2^4)$$ Now, we can substitute back the expressions for \(T_2^4\) and \(T_2\) that we found in steps 1 and 2: $$\sigma \varepsilon_1 \varepsilon_2 A_1(T_1^4 - (\frac{q_{conv}}{h A_2} + T_{\infty})^4) = q_{conv}$$ Rearranging and solving for \(q_{conv}\), we would eventually find the value of \(q_{conv}\), which in turn would help us to find the value of \(T_2\).

Compare the surface temperature with the maximum use temperature

Once we have found the value of \(T_2\), we can convert it back to Celsius and compare it to the maximum use temperature of the ASTM A249 904L stainless steel tube: $$T_2^{\circ} \mathrm{C} = T_2 \mathrm{K} - 273.15$$ If the temperature of the tube surface is below or equal to the maximum use temperature of \(260^{\circ} \mathrm{C}\), then the tube is safe for use under these conditions. Otherwise, we would need to re-evaluate the design or materials for safety.