Question

Hot water flows in a 1-m-long section of a pipe that is made of acrylonitrile butadiene styrene (ABS) thermoplastic. The \(\mathrm{ABS}\) pipe section has an inner diameter of \(D_{1}=22 \mathrm{~mm}\) and an outer diameter of $D_{2}=27 \mathrm{~mm}\(. The thermal conductivity of the ABS pipe wall is \)0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The outer pipe surface is exposed to convection heat transfer with air at \(20^{\circ} \mathrm{C}\) and $h=10 \mathrm{~W} / \mathrm{m}^{2} . \mathrm{K}$. The water flowing inside the pipe has a convection heat transfer coefficient of $50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. According to the ASME Code for Process Piping (ASME B31.3-2014, Table B-1), the maximum recommended temperature for \(\mathrm{ABS}\) pipe is \(80^{\circ} \mathrm{C}\). Determine the maximum temperature of the water flowing in the pipe, such that the ABS pipe is operating at the recommended temperature or lower. What is the temperature at the outer pipe surface when the water is at maximum temperature?

Short answer

#tag_title# Calculate the heat transfer rate through the pipe wall #tag_content# Substituting the values given: \(R_{wall} = \frac{\ln \left(\frac{0.027}{0.022}\right)}{2 \pi (1\,\text{m}) (0.1\,\text{W/(m} \cdot \text{K}))}\) \(R_{wall} = 0.1593\,\text{K/W}\) #Step 2: Calculate the maximum temperature of the water# #tag_title# Apply the energy balance equation #tag_content# Now, we will apply the energy balance equation: \(Q_{conv} = Q_{cond}\) Where \(Q_{conv}\) is the heat transfer rate due to convection, which is given by: \(Q_{conv} = h A (T_{water} - T_{s,inner})\) And \(Q_{cond}\) is the heat transfer rate due to conduction, which is given by: \(Q_{cond} = \frac{T_{s,inner} - T_{s,outer}}{R_{wall}}\) We are given \(h = 1200\,\text{W/(m}^2 \cdot \text{K})\), \(T_{s,outer} = 60\,^\circ \text{C}\), and the area for convection is given by: \(A = \pi D_1 L = \pi (0.022\,\text{m}) (1\,\text{m})\) Now, substitute the values and solve for \(T_{water}\). #tag_title# Solve for the maximum temperature of the water #tag_content# First, substitute the values into the convection equation: \(Q_{conv} = (1200 \frac{\text{W}}{\text{m}^2 \cdot \text{K}}) (\pi (0.022\,\text{m}) (1\,\text{m})) (T_{water} - T_{s,inner})\) Now, substitute the values into the conduction equation: \(Q_{cond} = \frac{T_{s,inner} - 60\,^\circ \text{C}}{0.1593\,\text{K/W}}\) Since \(Q_{conv} = Q_{cond}\), we can set the two equations equal to each other: \((1200 \frac{\text{W}}{\text{m}^2 \cdot \text{K}})(\pi (0.022\,\text{m}) (1\,\text{m})) (T_{water} - T_{s,inner}) = \frac{T_{s,inner} - 60\,^\circ \text{C}}{0.1593\,\text{K/W}}\) Solve this equation for \(T_{water}\): \(T_{water} = 87.6\,^\circ \text{C}\) Therefore, the maximum temperature of the water flowing in the ABS pipe is approximately 87.6°C to ensure the pipe is operating at its recommended temperature.

Step-by-step solution

Calculate the thermal resistance of the pipe wall

To find the heat transfer rate through the pipe wall, we need to determine the overall resistance that the heat must pass through. For the pipe wall (a cylindrical shell), the thermal resistance is given by: \(R_{wall} = \frac{\ln \left(\frac{D_2}{D_1}\right)}{2 \pi L k}\) where \(D_1 = 0.022\,\text{m}\) (inner diameter), \(D_2 = 0.027\,\text{m}\) (outer diameter), \(L = 1\,\text{m}\) (pipe length), and \(k = 0.1\,\text{W/(m} \cdot \text{K})\). Substitute the values and calculate \(R_{wall}\).