Question

A long nickel alloy (ASTM B335) cylindrical rod is used as a component in high-temperature process equipment. The rod has a diameter of $5 \mathrm{~cm}\(; its thermal conductivity, specific heat, and density are \)11 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, 380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, and \)9.3 \mathrm{~g} / \mathrm{cm}^{3}$, respectively. Occasionally, the rod is submerged in hot fluid for several minutes, where the fluid temperature is \(500^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ASME Code for Process Piping limits the maximum use temperature for ASTM B335 rod to \(427^{\circ} \mathrm{C}\) (ASME B31.32014 , Table A-1M). If the initial temperature of the rod is \(20^{\circ} \mathrm{C}\), how long can the rod be submerged in the hot fluid before reaching its maximum use temperature?

Short answer

The nickel alloy rod can be submerged in the hot fluid for approximately 7024.58 seconds or around 117 minutes before reaching its maximum use temperature.

Step-by-step solution

Calculate the surface area of the rod

We are given the diameter of the rod, which is \(5 \mathrm{~cm}\). The surface area of a cylindrical rod can be calculated using the formula \(A = \pi d L\), where \(d\) is the diameter and \(L\) is the length of the rod. As the rod is long, we can assume that the entire length is in contact with the fluid. Therefore, we can calculate the surface area per unit length, which can be represented as \(A = \pi d\). Now, we can plug in the diameter value to get the surface area per unit length: \(A = \pi(0.05 \mathrm{~m})\) \(A \approx 0.157 \mathrm{~m^2/m}\)

Converting the density from g/cm³ to kg/m³

We need to convert the density from grams per cubic centimeter to kilograms per cubic meter: \(\rho = 9.3 \mathrm{~g/cm^3} * \frac{1000 \mathrm{~kg}}{10^6 \mathrm{~g}} * \left(\frac{100 \mathrm{~cm}}{1 \mathrm{~m}}\right)^{3}\) \(\rho = 9300 \mathrm{~kg/m^3}\)

Calculate the time by applying the lumped capacitance formula

Now, we have all the necessary parameters to apply the lumped capacitance formula to calculate the time required for the rod to reach its maximum use temperature. Plug in the values given in the problem: \(t = \frac{hA(T_\infty - T_i)}{-kA(T_\infty - T_s)} \tau\) Using the property values for the ASTM B335 rod: \(t = \frac{440 \mathrm{~W/m^2K} * (500^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C})}{-11 \mathrm{~W/mK} * (500^{\circ}\mathrm{C} - 427^{\circ}\mathrm{C})} \tau\) Now, solve for the dimensionless heat transfer time \(\tau\): \(\tau = \frac{440 * 0.157 * (500 - 20)}{-11 * 0.157 * (500 - 427)}\) \(\tau \approx 44.93\)

Calculate the time required to reach the maximum use temperature

The final step is to calculate the time \(t\) using the dimensionless heat transfer time \(\tau\) that we just calculated: \(t = \frac{\rho V C_p}{hA} \tau\) We have \(\rho\), \(A\), and \(C_p\) given, so we can assume a unit volume \(V = 1 \mathrm{~m^3}\): \(t = \frac{9300 * 1 * 380}{440 * 0.157} \tau\) \(t \approx 44.93 \times 156.14 \mathrm{~s}\) Finally, calculate the time \(t\) in seconds: \(t \approx 7024.58 \mathrm{~s}\) Thus, the rod can be submerged in the hot fluid for approximately \(7024.58\) seconds or around \(117\) minutes before reaching its maximum use temperature.