Question

In a food processing plant, hot liquid water is being transported in a pipe \((k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), $D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\(, and \)L=10 \mathrm{~m}$.) The hot water flowing with a mass flow rate of \(0.15 \mathrm{~kg} / \mathrm{s}\) enters the pipe at \(100^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). The plant supervisor thinks that since the hot water exits the pipe at $60^{\circ} \mathrm{C}$, the pipe's outer surface temperature should be safe from thermal burn hazards. In order to prevent thermal burn upon accidental contact with skin tissue for individuals working in the vicinity of the pipe, the pipe's outer surface temperature should be kept below \(45^{\circ} \mathrm{C}\). Determine whether or not there is a risk of thermal burn on the pipe's outer surface. Assume the pipe outer surface temperature remains constant.

Short answer

The outer surface temperature of the pipe is calculated to be 91.16°C. This temperature exceeds the safe temperature limit of 45°C, presenting a risk for thermal burn hazards.

Step-by-step solution

Calculate the convective heat transfer rate

Since the water enters the pipe at \(100^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\), we can find the heat transfer rate due to the temperature difference in the water flow. To calculate this, we need to use the formula: \(q_{conv} = \dot{m} C_p \Delta T\) Where: \(q_{conv}\) = convective heat transfer rate (W) \(\dot{m}\) = mass flow rate of water \((0.15 \mathrm{~kg/s})\) \(C_p\) = specific heat capacity of water \((4.18 \mathrm{~kJ/kg \cdot K})\) \(\Delta T\) = temperature difference between inlet and outlet \((100 - 60)^\circ\mathrm{C}\) Calculating the convective heat transfer rate: \(q_{conv} = 0.15 \mathrm{~kg/s} \cdot (4.18 \times 10^3 \mathrm{~J/kg \cdot K}) \cdot (100 - 60) \mathrm{K}\) \(q_{conv} = 25.11 \mathrm{~W}\)

Determine the heat transfer rate through the pipe walls

The heat transfer rate through the pipe walls is given by the conduction equation: \(q = \frac{kA\Delta T}{L}\) Where: \(q\) = heat transfer rate (W) \(k\) = thermal conductivity of the pipe material \((15\mathrm{~W/m \cdot K})\) \(A\) = surface area of the pipe \(\Delta T\) = temperature difference between the inner and outer surfaces of the pipe \(L\) = thickness of the pipe wall \((D_o - D_i) = 0.005\mathrm{~m}\) Since the heat transfer rate due to convection equals the heat transfer rate through the pipe walls, we can set \(q_{conv} = q\), and solve for \(\Delta T\).

Calculate the temperature difference between the inner and outer surfaces of the pipe

Set \(q_{conv} = q\) and rearrange for \(\Delta T\): \(\Delta T = \frac{qL}{kA} = \frac{25.11 \mathrm{~W} \cdot 0.005 \mathrm{~m}}{15 \mathrm{~W/m \cdot K} \cdot A}\)

Calculate the surface area of the pipe

The surface area of a cylinder is given by \(A = 2\pi r L\), where \(r\) is the radius of the outer surface of the pipe \((D_o / 2)\). In this case: \(A = 2\pi (0.015 \mathrm{~m})(10 \mathrm{~m})\) \(A = 0.942 \mathrm{~m^2}\)

Calculate the temperature difference

Plug the surface area, \(A\), into the equation in Step 3: \(\Delta T = \frac{25.11 \mathrm{~W} \cdot 0.005 \mathrm{~m}}{15 \mathrm{~W/m \cdot K} \cdot 0.942 \mathrm{~m^2}}\) \(\Delta T = 8.84 \mathrm{~K}\)

Determine the outer surface temperature of the pipe

Now that we have the temperature difference between the inner and outer surfaces of the pipe, we can estimate the outer surface temperature of the pipe: \(T_{outer} = T_{inner} - \Delta T\) \(T_{outer} = 100^{\circ} \mathrm{C} - 8.84 \mathrm{K}\) \(T_{outer} = 91.16^{\circ} \mathrm{C}\) Since \(T_{outer} > 45^{\circ} \mathrm{C}\), the pipe's outer surface temperature is not safe from thermal burn hazards. Therefore, there is a risk of thermal burn on the pipe's outer surface.