Question

A large ASTM A203 B steel plate, with a thickness of \(7 \mathrm{~cm}\), in a cryogenic process is suddenly exposed to very cold fluid at $-50^{\circ} \mathrm{C}\( with \)h=594 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The plate has a thermal conductivity of $52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, a specific heat of \)470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\(, and a density of \)7.9 \mathrm{~g} / \mathrm{cm}^{3}$. The ASME Code for Process Piping limits the minimum suitable temperature for ASTM A203 B steel plate to \(-30^{\circ} \mathrm{C}\) (ASME B31.32014 , Table A-1M). If the initial temperature of the plate is \(20^{\circ} \mathrm{C}\) and the plate is exposed to the cryogenic fluid for \(6 \mathrm{~min}\), would it still comply with the ASME code?

Short answer

Based on the given data and the calculation, the temperature of the steel plate after exposure to cold fluid for 6 minutes would be approximately -22.77°C. Since the calculated temperature is higher than the ASME code's minimum suitable temperature (-30°C) for ASTM A203 B steel plate, the steel plate still complies with the code.

Step-by-step solution

List down the given information and convert all units to SI units

Given the following data: Thickness of steel plate, \(L = 7 \mathrm{~cm} = 0.07 \mathrm{~m}\) Cold fluid temperature, \(T_f = -50^{\circ} \mathrm{C}\) Heat transfer coefficient, \(h = 594 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) Thermal conductivity of steel plate, \(k = 52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) Specific heat of steel plate, \(c = 470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) Density of steel plate, \(\rho = 7.9 \mathrm{~g} / \mathrm{cm}^{3} = 7900 \mathrm{~kg} / \mathrm{m}^{3}\) Initial temperature of plate, \(T_i= 20^{\circ} \mathrm{C}\) Exposure time, \(t = 6 \mathrm{~min} = 360 \mathrm{~s}\) Minimum suitable temperature, \(T_{min} = -30^{\circ} \mathrm{C}\)

Calculate the Biot number (Bi)

The Biot number (Bi) compares the thermal resistance of the material to thermal resistance at the surface. Bi = \(\frac{hL}{k}\) Bi = \(\frac{594 \cdot 0.07}{52} = 0.8\) Since Bi < 1, the temperature gradient within the material is significant, and we need to use transient conduction analysis to find the temperature after the exposure time.

Calculate the Fourier number (Fo)

The Fourier number (Fo) is a dimensionless number that indicates the ratio of diffusive thermal transport rate to the storage rate. Fo = \(\frac{\alpha t}{L^2}\), where \(\alpha = \frac{k}{\rho c}\) is the thermal diffusivity of the material. Thermal diffusivity, \(\alpha = \frac{52}{7900\cdot 470} = 1.4184 \times 10^{-7} \mathrm{~m^2} / \mathrm{s}\) Fo = \(\frac{1.4184 \times 10^{-7} \cdot 360}{(0.07)^2} = 1.0264\)

Calculate the temperature after the exposure time

Using the finite slab semi-infinite solutions, the centerline temperature ratio is given by the following equation: \( \frac{T-x}{T_i-T_f}= \sum_{n=0}^{\infty}{(-1)^n\mathrm{e}^{(2n+1)^{2}\pi^{2}\frac{4\alpha t}{L^{2}}}} \) For simplicity, we can take the first few terms of the series to get a reasonably accurate estimation. \( \frac{T-x}{T_i-T_f} \approx \mathrm{e}^{-\pi^{2}\frac{4\alpha t}{L^{2}}}-\mathrm{e}^{-9\pi^{2}\frac{4\alpha t}{L^{2}}} +\mathrm{e}^{-25\pi^{2}\frac{4\alpha t}{L^{2}}} \) By plugging in the values and solving for \(T\), we get: \( T\approx T_i-T_f\left(\mathrm{e}^{-\pi^{2}\frac{4\alpha t}{L^{2}}}-\mathrm{e}^{-9\pi^{2}\frac{4\alpha t}{L^{2}}} +\mathrm{e}^{-25\pi^{2}\frac{4\alpha t}{L^{2}}}\right)+T_f \) \( T \approx 20 - (-50)\left(\mathrm{e}^{-\pi^{2}(1.0264)}-\mathrm{e}^{-9\pi^{2}(1.0264)}+\mathrm{e}^{-25\pi^{2}(1.0264)}\right)-50\) \( T\approx -22.77^{\circ} \mathrm{C} \)

Check if the temperature complies with the ASME code

Since the calculated temperature \(T\approx -22.77^{\circ} \mathrm{C}\) is greater than the minimum suitable temperature of \(T_{min} = -30^{\circ} \mathrm{C}\), the steel plate would still comply with the ASME code after the exposure of 6 minutes to the cryogenic fluid.