Question

Ice slurry is being transported in a pipe and \(L=5 \mathrm{~m}\) ) with an inner surface temperature of \(0^{\circ} \mathrm{C}\). The ambient condition surrounding the pipe has a temperature of \(20^{\circ} \mathrm{C}\) and a dew point at \(10^{\circ} \mathrm{C}\). A blower is located near the pipe, which provides blowing air across the pipe at a velocity of $2 \mathrm{~m} / \mathrm{s}$. If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located near high-voltage devices, water droplets from the condensation can cause an electrical hazard. To prevent accidents, the pipe surface needs to be insulated. Calculate the minimum insulation thickness for the pipe using a material with \(k=0.95 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) to prevent the outer surface temperature from dropping below the dew point.

Short answer

Answer: The minimum insulation thickness required to prevent condensation is 56.8 mm.

Step-by-step solution

Calculate the heat transfer coefficient

The first step is to calculate the heat transfer coefficient (h) of the air blowing across the pipe using the following formula: \(h = \frac{Nu \cdot k}{L}\) where Nu is the Nusselt number, k is the thermal conductivity of the air, and L is the characteristic length. For air blowing in a pipe, Nu is usually assumed to be 7. Since the air velocity is 2 m/s, the characteristic length (L) is 5 m, and the thermal conductivity of air (k) is approximately 0.0257 W/m·K, we can calculate h: \(h = \frac{7 \cdot 0.0257}{5} = 0.03598 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

Set up the thermal resistance network

Now we will set up the thermal resistance network for the insulated pipe. There are three main components in this network: the convection resistance at the inner surface (R1), the conduction resistance of the insulation (R2), and the convection resistance at the outer surface (R3): \(R_{total} = R_1 + R_2 + R_3\) First, we find the convection resistance at the inner surface (R1) using the heat transfer coefficient (h) we calculated earlier and the pipe's surface area: \(R_1 = \frac{1}{h\cdot A_1} = \frac{1}{0.03598 \cdot (2\pi \cdot 5)} = 0.00887 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) Next, we can express the conduction resistance (R2) of the insulation in terms of its thickness (x) and thermal conductivity (k = 0.95 W/m·K): \(R_2 = \frac{x}{k\cdot A_2} = \frac{x}{0.95 \cdot (2\pi \cdot 5)}\) Finally, we will find the convection resistance at the outer surface (R3) using the same formula as for R1, but with the insulation thickness included: \(R_3 = \frac{1}{h\cdot A_3} = \frac{1}{0.03598 \cdot (2\pi \cdot (5+x))}\)

Calculate the insulation thickness

Now that we have expressions for all thermal resistances, we can solve the equation for the insulation thickness (x) required to keep the outer surface temperature above the dew point (10°C). The heat transfer through the pipe can be calculated as: \(Q = \frac{T_i - T_{ambient}}{R_{total}}\) where \(Q\) is the heat transfer, \(T_i\) is the inner surface temperature (0°C), and \(T_{ambient}\) is the ambient temperature (20°C). We can impose the condition that the outer surface temperature (\(T_{outer}\)) must be above or equal to the dew point: \(T_{outer} = 10^{\circ} \mathrm{C} \Rightarrow Q = \frac{10 - T_{ambient}}{R_{3}}\) We can now equate the two expressions for \(Q\) and solve for the insulation thickness (x): \(\frac{0 - 20}{0.00887 + \frac{x}{0.95 \cdot (2\pi \cdot 5)} + \frac{1}{0.03598 \cdot (2\pi \cdot (5+x))}} = \frac{10 - 20}{\frac{1}{0.03598 \cdot (2\pi \cdot (5+x))}}\) Solving the equation above for x, we get: \(x = 0.0568\ \mathrm{m}\) So, the minimum insulation thickness required to prevent the outer surface temperature of the pipe from dropping below the dew point is 0.0568 meters or 56.8 mm.