Question

metal pipe $\left(k_{\text {pipe }}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{\text {, pipe }}=\right.\( \)5 \mathrm{~cm}, D_{o \text {, pipe }}=6 \mathrm{~cm}\(, and \)\left.L=10 \mathrm{~m}\right)$ situated in an engine room is used for transporting hot saturated water vapor at a flow rate of \(0.03 \mathrm{~kg} / \mathrm{s}\). The water vapor enters and exits the pipe at \(325^{\circ} \mathrm{C}\) and \(290^{\circ} \mathrm{C}\), respectively. Oil leakage can occur in the engine room, and when leaked oil comes in contact with hot spots above the oil's autoignition temperature, it can ignite spontaneously. To prevent any fire hazard caused by oil leakage on the hot surface of the pipe, determine the required insulation $\left(k_{\text {ins }}=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.$ ) layer thickness over the pipe for keeping the outer surface temperature below $180^{\circ} \mathrm{C}$.

Short answer

The required insulation layer thickness is approximately 9.66 cm.

Step-by-step solution

Find Heat Loss from Pipe Without Insulation

Given the information about the water vapor flowing through the pipe, we can find the heat loss from the pipe. We know that the mass flow rate of the water vapor is 0.03 kg/s. The temperature difference across the pipe (hot to cold) is: \(\Delta T = T_{in} - T_{out} = 325°C - 290°C = 35°C\) Now, the heat loss from the pipe can be calculated using the formula: \(q = m \cdot c \cdot \Delta T\) Here, \(q\) represents the heat loss from the pipe in watts, \(m\) is the mass flow rate in kg/s, \(c\) is the specific heat capacity of water vapor (we can approximate it to be \(c \approx 2 J / g \cdot K\)), and \(\Delta T\) is the temperature difference in \(K\). Plugging in the given values, we get: \(q = 0.03 kg/s \cdot 2 \frac{J}{g \cdot K} \cdot 35 K = 2.1 \frac{J}{s}\)

Apply Steady-state Heat Conduction Equation

Now that we have the heat loss for the pipe without insulation, we can apply the steady-state heat conduction equation to find the required insulation layer thickness. The heat conduction equation can be written as: \(q = \frac{kA \Delta T}{d}\) Here, \(q\) represents the heat loss, \(k\) is the thermal conductivity of the insulation material, \(A\) is the surface area of the pipe, \(\Delta T\) is the temperature difference across the pipe, and \(d\) is the insulation layer thickness. We are given the thermal conductivity of the insulation (\(k_{ins} \approx 0.95 W/m\cdot K\)), and we also have the maximum allowed outer surface temperature of 180°C. We can use these values to find the required insulation layer thickness. First, calculate the temperature difference across the insulation layer: \(\Delta T_{ins} = T_{pipe} - T_{max} = 290°C - 180°C = 110°C\) Now, calculate the surface area of the pipe: \(A = \pi \times D_{o,pipe} \times L = \pi \times 0.06 m \times 10 m = 1.884 \, \mathrm{m}^2\) We can rearrange the heat conduction equation to solve for \(d\): \(d = \frac{k_{ins} A \Delta T_{ins}}{q}\) Plug in the values and calculate \(d\): \(d = \frac{0.95 \frac{W}{m \cdot K} \cdot 1.884 \, \mathrm{m}^2 \cdot 110 K}{2.1 \frac{J}{s}} = 0.0966 \, m\)

Finalize the Answer

Now we have obtained the required insulation layer thickness to be 0.0966 m. To express the answer more clearly, we can convert it to centimeters: \(d = 0.0966 m \times \frac{100 cm}{1 m} = 9.66 cm\) So, the required insulation layer thickness is approximately 9.66 cm to keep the outer surface temperature below 180°C and prevent any fire hazard.