Question

Two plates are held together by square stainless steel (ASTM A479 904L) bars with thickness of \(12 \mathrm{~mm}\). Each bar has a thermal conductivity of $12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, a specific heat of \)500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, and a density of \)7.9 \mathrm{~g} / \mathrm{cm}^{3}$. The bars are situated in between the two plates, and they are occasionally submerged in hot liquid at \(300^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(96 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The length of each bar that is exposed to the hot liquid is \(25 \mathrm{~mm}\) with an initial temperature of \(20^{\circ} \mathrm{C}\). The maximum use temperature allowed for ASTM A479 904L bars is \(260^{\circ} \mathrm{C}\) (ASME Code for Process Piping, ASME B31.3-2014, Table A-1M). If the bars are submerged in the hot liquid for \(5 \mathrm{~min}\), will they be in compliance with the ASME code? How long will it take for the bars to reach the maximum use temperature?

Short answer

A: Yes, the stainless steel bars are in compliance with the ASME code after being submerged in hot liquid for 5 minutes. The temperature after 5 minutes is 60.83°C, which is below the maximum use temperature of 260°C.

Step-by-step solution

Calculate the cross-sectional area of the bars

The bars are square with a side length of 12 mm. The cross-sectional area can be calculated as follows: \(A = L \cdot W = 12 \mathrm{~mm} \cdot 12 \mathrm{~mm} = 144 \mathrm{~mm}^2\)

Calculate the volume of the portion submerged in the hot liquid

The length exposed to the hot liquid is 25 mm. We can calculate the volume as follows: \(V = L \cdot W \cdot H = 144 \mathrm{~mm}^2 \cdot 25 \mathrm{~mm} = 3600 \mathrm{~mm}^3\)

Calculate the mass of the portion submerged in the hot liquid

We can calculate the mass using the density as follows: \(m = V \cdot \rho = 3600 \mathrm{~mm}^3 \cdot (7.9 \mathrm{~g} / \mathrm{cm}^{3} \cdot 10^{-3} \mathrm{~kg} / \mathrm{g} ) = 0.02844 \mathrm{~kg}\)

Calculate the rate of heat transfer between the bars and the hot liquid

We can calculate the rate of heat transfer using the convection heat transfer coefficient as follows: \(q = h \cdot A \cdot (T_\mathrm{hot} - T_\mathrm{initial}) = 96 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \cdot 0.000144 \mathrm{~m}^2 \cdot (300 - 20) ^\circ \mathrm{C} = 1.9392 \mathrm{~W}\)

Calculate the temperature increase per unit time

We can calculate the temperature increase per unit time, \(\Delta T / \Delta t\), as follows: \(\frac{\Delta T}{\Delta t} = \frac{q}{m c_p} = \frac{1.9392 \mathrm{~W}}{0.02844 \mathrm{~kg} \cdot 500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}} = 0.1361 \mathrm{~K/s}\)

Calculate the time required to reach the maximum use temperature

We can calculate the time required to reach the maximum use temperature, \(\Delta t_\mathrm{max}\), as follows: \(\Delta t_\mathrm{max} = \frac{T_\mathrm{max} - T_\mathrm{initial}}{\Delta T / \Delta t} = \frac{260 - 20}{0.1361} \mathrm{~s} = 1762.98 \mathrm{~s}\)

Check if the bars are in compliance with the ASME code

To check if the bars are in compliance, we can calculate the temperature after 5 minutes (300 seconds) and compare it to the maximum use temperature: \(T_\mathrm{5min} = T_\mathrm{initial} + \frac{\Delta T}{\Delta t} \cdot 300 \mathrm{~s} = 20 + (0.1361 \cdot 300) ^\circ \mathrm{C} = 60.83 ^\circ \mathrm{C}\) Since the temperature after 5 minutes is below the maximum use temperature of \(260^{\circ} \mathrm{C}\), the bars are in compliance with the ASME code. Furthermore, it will take 1762.98 seconds (approximately 29 minutes and 23 seconds) for the bars to reach the maximum use temperature.