Question

An ASTM F441 chlorinated polyvinyl chloride \((\mathrm{CPVC})\) tube is embedded in a vertical concrete wall $(k=1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\(. The wall has a height of \)1 \mathrm{~m}$, and one surface of the wall is subjected to convection with hot air at \(140^{\circ} \mathrm{C}\). The distance measured from the plate's surface that is exposed to the hot air to the tube surface is \(d=3 \mathrm{~cm}\). The ASME Code for Process Piping limits the maximum use temperature for ASTM F441 CPVC tube to $93.3^{\circ} \mathrm{C}$ (ASME B31.32014 , Table B-1). If the concrete surface that is exposed to the hot air is at \(100^{\circ} \mathrm{C}\), would the CPVC tube embedded in the wall still comply with the ASME code?

Short answer

Based on the given information and calculations, the CPVC tube surface temperature is found to be \(139.9^{\circ}\mathrm{C}\). This exceeds the ASME code limit of \(93.3^{\circ}\mathrm{C}\), indicating that the CPVC tube embedded in the wall does not comply with the ASME code.

Step-by-step solution

Identify known values and unknown variables

We are given the following values: - Concrete wall thermal conductivity \((k) = 1.4\frac{\mathrm{W}}{\mathrm{m}\cdot\mathrm{K}}\) - Wall height \((h) = 1\mathrm{m}\) - Hot air temperature \((T_{air}) = 140^{\circ}\mathrm{C}\) - Concrete surface temperature \((T_{surface}) = 100^{\circ}\mathrm{C}\) - Distance from the plate's surface to the tube surface \((d) = 3\mathrm{cm}\) We are asked to find the CPVC tube surface temperature. If this temperature is less than or equal to \(93.3^{\circ}\mathrm{C}\), the CPVC tube will comply with the ASME code.

Simplify the problem

The problem can be simplified by analyzing heat conduction through a one-dimensional wall. Since the wall is vertical, we can assume that heat transfer occurs only in the horizontal direction. We can also consider a unit width of the wall (\(1\mathrm{m}\)) to simplify our calculations. Since the vertical direction's conduction and natural convection can be neglected, we can model this problem as a one-dimensional steady-state conduction problem.

Calculate the heat flow through the wall

We can use Fourier's law of heat conduction to calculate the heat flow through the wall: \(q = -kA\frac{dT}{dx}\) Since the problem is one-dimensional and steady-state, we can use the following form of Fourier's law: \(q = -k\frac{dT}{dx}\) We need to find the heat flow at the tube surface to determine its temperature. We will still use the same equation, but with different temperature and distance values. Let the heat flow at the tube surface be \(q_{tube}\). Thus, we have: \(q_{tube} = -k\frac{T_{surface} - T_{tube}}{d}\)

Rearrange the equation to find the tube temperature

We need to find the tube temperature. Rearrange the equation in step 3 to solve for \(T_{tube}\): \(T_{tube} = T_{surface} - \frac{q_{tube}d}{k}\)

Find the relation between \(q\) and \(q_{tube}\)

We can relate \(q\) and \(q_{tube}\) via the following equation: \(q -q_{tube} = h q_{tube}\) This corresponds to the vertical heat loss due to natural convection becoming negligible for our one-dimensional analysis. Now we can rewrite the equation to find the temperature at the tube surface: \(T_{tube} = T_{surface} - \frac{\frac{q}{k}d}{1 + hd}\).

Calculate the wall's temperature gradient

We need to calculate the temperature gradient (\(\frac{dT}{dx}\)) inside the wall. Using Fourier's law, we have: \(\frac{dT}{dx} = \frac{T_{air} - T_{surface}}{d}\)

Calculate the heat flow \(q\)

Now we can calculate the heat flow using Fourier's law and the temperature gradient calculated in step 6: \(q = -k\frac{dT}{dx}\) Providing the known values, we get: \(q = -1.4\frac{140 - 100}{0.03} = -1.4\cdot 1333.33 = -1866.67\frac{\mathrm{W}}{\mathrm{m}}\)

Calculate the tube temperature

Finally, we can calculate the tube temperature using the equation from step 5: \(T_{tube} = 100-\frac{-1866.67\times0.03}{1.4}\) \(T_{tube} = 100-(-39.9)\) \(T_{tube} = 139.9^{\circ}\mathrm{C}\) Since the tube temperature is above the ASME code limit of \(93.3^{\circ}\mathrm{C}\), the CPVC tube embedded in the wall does not comply with the ASME code.