Question

Liquid water flows in a circular tube at a mass flow rate of $7 \mathrm{~g} / \mathrm{s}\(. The water enters the tube at \)5^{\circ} \mathrm{C}$, and the average convection heat transfer coefficient for the internal flow is $20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. The tube is \)3 \mathrm{~m}$ long and has an inner diameter of \(25 \mathrm{~mm}\). The tube surface is maintained at a constant temperature. The inner surface of the tube is lined with polyvinylidene chloride (PVDC) lining. According to the ASME Code for Process Piping (ASME B31.3-2014, Table A.323.4.3), the recommended maximum temperature for PVDC lining is \(79^{\circ} \mathrm{C}\). If the water exits the tube at \(15^{\circ} \mathrm{C}\), determine the heat rate transferred to the water. Would the inner surface temperature of the tube exceed the recommended maximum temperature for PVDC lining?

Short answer

Answer: The heat transfer rate from the tube to the water is 292.6 W. The inner surface temperature of the tube is 16.22°C, which does not exceed the recommended maximum temperature for PVDC lining (79°C).

Step-by-step solution

Calculate the tube inner area

Firstly, we need to calculate the internal surface area of the tube where heat transfer takes place. Given the inner diameter \((d)\) and the tube length \((L)\), we can calculate the area \((A)\) using the following formula: \(A = \pi d L\)

Calculate the heat transfer rate

Next, we want to determine the rate at which heat is being transferred from the tube surface to the water (Q). We can use the energy balance formula to compute this: \(Q = m \cdot c \cdot \Delta T\) where \(m\) is the mass flow rate of the water, \(c\) is the specific heat capacity of water, and \(\Delta T\) is the temperature difference between the entry and exit of the water. We are given \(m\), and we know that \(c = 4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) for water. We will calculate \(\Delta T = T_{exit} - T_{entry}\).

Calculate the temperature difference

As we want to ensure the tube's inner surface temperature doesn't exceed the recommended maximum temperature for PVDC lining, we will calculate the temperature difference between the inner surface and the average water temperature. To do this, we can use the average water temperature, \(T_{avg} = \frac{T_{entry} + T_{exit}}{2}\) Now, we can calculate the required temperature difference using the following equation: \(\Delta T_s = \frac{Q}{hA}\) where \(h\) is the convection heat transfer coefficient, and \(\Delta T_s\) is the temperature difference between the tube surface and the average water temperature.

Check the inner surface temperature

Now that we have calculated the required temperature difference, we can calculate the maximum surface temperature by adding the temperature difference to the average water temperature: \(T_{surface} = T_{avg} + \Delta T_s\) We will compare this value to the maximum temperature recommended for PVDC lining (\(79^{\circ} \mathrm{C}\)) and determine if it exceeds the threshold. Calculations:

Step 1

Inner diameter: \(d = 0.025 \mathrm{~m}\) Length: \(L = 3 \mathrm{~m}\) \(A = \pi \cdot 0.025 \cdot 3 \approx 0.2355 \mathrm{~m}^2\)

Step 2

Mass flow rate: \(m = 0.007 \mathrm{~kg/s}\) Specific heat capacity: \(c = 4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) \(\Delta T = 15^{\circ} \mathrm{C} - 5^{\circ} \mathrm{C} = 10^{\circ} \mathrm{C}\) \(Q = 0.007 \cdot 4180 \cdot 10 \approx 292.6 \mathrm{~W}\)

Step 3

Average water temperature: \(T_{avg} = (5 + 15) / 2 = 10^{\circ} \mathrm{C}\) Convection heat transfer coefficient: \(h = 20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) \(\Delta T_s = \frac{292.6}{20 \cdot 0.2355} \approx 6.22^{\circ} \mathrm{C}\)

Step 4

\(T_{surface} = 10^{\circ} \mathrm{C} + 6.22^{\circ} \mathrm{C} \approx 16.22^{\circ} \mathrm{C}\) As the calculated inner surface temperature (\(16.22^{\circ} \mathrm{C}\)) is significantly lower than the recommended maximum temperature for PVDC lining (\(79^{\circ} \mathrm{C}\)), the lining will not exceed its temperature limit. Thus, the heat transfer rate in these conditions is \(292.6 \mathrm{~W}\).