Question

A metal pipe \(\left(k_{\text {pipe }}=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i, \text { pipe }}=\right.\) \(5 \mathrm{~cm}, D_{o \text {,pipe }}=6 \mathrm{~cm}\), 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 needed 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

Based on the calculations, an insulation thickness of 1.82 cm is needed to keep the outer surface temperature of the pipe below 180°C and prevent fire hazards caused by oil leakage.

Step-by-step solution

Calculate the temperature at the inner surface of the pipe (T1)

We can assume that the temperature of the hot vapor remains constant along the pipe. Thus, we can take the average temperature as the inner surface temperature of the pipe: $$ T_1 = \frac{325^\circ\mathrm{C} + 290^\circ\mathrm{C}}{2} = 307.5^\circ\mathrm{C} $$

Calculate the thermal resistance of the pipe (R_pipe)

The thermal resistance of the pipe is given by: $$ R_{\text{pipe}} = \frac{\ln \frac{D_{o \text{,pipe}}}{D_{i, \text{pipe}}}}{2 \pi k_{\text{pipe}} L} $$ where D_o is the outer diameter and D_i is the inner diameter. Plug the values to find R_pipe: $$ R_{\text{pipe}} = \frac{\ln \frac{6\text{ cm}}{5\text{ cm}}}{2 \pi (15 \mathrm{W/m\cdot K})(10\text{ m})} $$ Calculate the value of R_pipe: $$ R_{\text{pipe}} = 0.0044\,\mathrm{K/W} $$

Calculate the heat transfer rate (Q) through the pipe

The heat transfer rate can be found from the temperature difference and the thermal resistance of the pipe: $$ Q = \frac{T_1 - T_2}{R_{\text{pipe}}} $$ We know T1 and R_pipe, and we need to keep T2 below 180°C. So, first, let's assume T2 = 180°C and calculate Q: $$ Q = \frac{307.5^\circ\mathrm{C} - 180^\circ\mathrm{C}}{0.0044\,\mathrm{K/W}} = 29,\!000 \, \mathrm{W} $$

Calculate the temperature at the outer surface of the pipe with insulation (T2)

We already assumed T2 to be 180°C in the previous step, so we can directly use it in the next step to find the insulation thickness required.

Use the formula for the thermal resistance of insulation (R_ins) to find the insulation thickness required

The formula for the thermal resistance of insulation is as follows: $$ R_{\text{ins}} = \frac{\ln \frac{D_{o \text{,ins}}}{D_{o \text{,pipe}}}}{2\pi k_{\text{ins}} L} $$ We can rearrange the formula to find D_o_ins: $$ D_{o \text{,ins}} = D_{o \text{,pipe}} \cdot e^{\left(R_{\text{ins}} \cdot 2\pi k_{\text{ins}} L\right)} $$ The thermal resistance of insulation can be found using the heat transfer rate (Q) and the temperature difference: $$ R_{\text{ins}} = \frac{T_1 - T_2}{Q} $$ Plug in the values of T1, T2, and Q to find R_ins: $$ R_{\text{ins}} = \frac{307.5^\circ\mathrm{C} - 180^\circ\mathrm{C}}{29,\!000\,\mathrm{W}} = 0.0044\,\mathrm{K/W} $$ Now, substitute R_ins, D_o_pipe, k_ins, and L into the formula for D_o_ins: $$ D_{o \text{,ins}} = (6\text{ cm}) \cdot e^{\left((0.0044\,\mathrm{K/W}) (2\pi (0.95\,\mathrm{W/m\cdot K})(10\,\mathrm{m}))\right)} $$ Calculate the value of D_o_ins: $$ D_{o \text{,ins}} = 7.82\,\mathrm{cm} $$ Finally, to determine the insulation thickness needed, subtract the outer diameter of the pipe from the outer diameter of the insulation layer: $$ \text{Insulation Thickness} = D_{o \text{,ins}} - D_{o \text{,pipe}} = 7.82\,\mathrm{cm} - 6\,\mathrm{cm} = 1.82\,\mathrm{cm} $$ Therefore, an insulation layer of thickness 1.82 cm is needed to keep the outer surface temperature of the pipe below 180°C, thus preventing fire hazards caused by oil leakage.

Key concepts

Thermal Resistance

Understanding thermal resistance is crucial when dealing with heat transfer in pipes. It's akin to electrical resistance but for heat flow. Essentially, it's a measure of how easily or difficultly heat can travel through a material or a system. The higher the thermal resistance, the slower the heat transfers, akin to a thicker blanket keeping you warmer by resisting the escape of body heat.

For pipes, we calculate thermal resistance using the formula: \[ R = \frac{\ln \frac{D_{o}}{D_{i}}}{2 \pi k L} \]where \( R \) is the thermal resistance, \( D_{o} \) is the outer diameter, \( D_{i} \) is the inner diameter, \( k \) is the thermal conductivity of the pipe material, and \( L \) is the length of the pipe. Higher material thermal conductivity lowers resistance and increases heat transfer, while a thicker pipe wall increases resistance.

For insulation purposes, you'd want a material with a high thermal resistance to slow down the heat leaving the pipe. This is why the insulation's thermal conductivity and thickness become important when calculating the needed insulation thickness to ensure the outer surface temperature stays within safe limits.

Insulation Thickness Calculation

The calculation of insulation thickness is often performed to increase the thermal resistance of a system, thus reducing the heat loss or gain. This is particularly important in scenarios where safety is a concern, such as preventing the oil in an engine room from reaching its autoignition temperature.

To calculate the insulation thickness required for a pipe, we typically utilize the formula for thermal resistance of the insulation layer. By rearranging the expression for thermal resistance:\[ R_{\text{ins}} = \frac{\ln \frac{D_{o,\text{ins}}}{D_{o,\text{pipe}}}}{2\pi k_{\text{ins}} L} \]we can solve for \( D_{o,\text{ins}} \), the outer diameter of the pipe with the insulation. From there, the insulation thickness is simply the difference between \( D_{o,\text{ins}} \) and \( D_{o,\text{pipe}} \).

To improve the given exercise, showing the rearrangement step by step can help students understand the mathematics behind the calculation. It's also helpful to discuss the practical implications of insulation thickness—too thin might not prevent the hazard, and too thick could be unnecessarily costly and physically impracticable.

Heat Transfer Rate

The heat transfer rate, denoted as \( Q \), is a measure of the amount of heat energy that moves from one point to another per unit time, usually in watts (W). In the context of pipes, the heat transfer rate can be affected by several factors including the temperatures inside and outside the pipe, the pipe's material properties, and the thickness of any insulation.

The heat transfer rate through a pipe or its insulation can be calculated using the temperature difference and the thermal resistance via the equation:\[ Q = \frac{T_1 - T_2}{R} \]where \( T_1 \) and \( T_2 \) are the temperatures at the inner and outer surfaces of the pipe, respectively, and \( R \) is the thermal resistance. This relationship shows that the larger the temperature difference or the smaller the thermal resistance, the greater the heat transfer rate will be.

In practice, engineers must balance the need for effective heat transfer with the need to manage temperature safely. For example, overly efficient heat transfer might lead to dangerously high temperatures that could ignite flammable materials in the environment, which is why sometimes it's essential to deliberately increase thermal resistance by adding insulation.