Question

A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), and \(L=10 \mathrm{~m}\) ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

Short answer

#tag_title#Question#tag_content# Determine the minimum thermal conductivity of the insulation layer required to maintain the outside surface temperature at 45°C or lower for a given pipe system. #tag_title#Answer#tag_content# The minimum thermal conductivity of the insulation layer, \(k_{ins}\), can be found using the following formula: $$ k_{ins} \ge \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi L} \left(\frac{T_{mix} - T_{s}}{T_{mix} - T_{amb}} (\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{1}{h_{amb} \pi D_o L}) - \frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} - \frac{1}{h_{amb} \pi (D_o + 2t) L}\right) $$ Substitute the given values into the equation and solve for the minimum required thermal conductivity, \(k_{ins}\).

Step-by-step solution

Define the given variables

We are given the following variables: - Thermal conductivity of the pipe, \(k = 14 \frac{W}{m \cdot K}\) - Inner diameter of the pipe, \(D_i = 2.5 cm = 0.025 m\) - Outer diameter of the pipe, \(D_o = 3 cm = 0.03 m\) - Length of the pipe, \(L = 10 m\) - Mixture temperature, \(T_{mix} = 135 °C\) - Convection heat transfer coefficient of the mixture, \(h_{mix} = 150 \frac{W}{m^2 \cdot K}\) - Insulation thickness, \(t = 2.5 cm = 0.025 m\) - Ambient air temperature, \(T_{amb} = 20 °C\) - Convection heat transfer coefficient of ambient air, \(h_{amb} = 25 \frac{W}{m^2 \cdot K}\) - Desired surface temperature, \(T_s = 45 °C\)

Calculate the thermal resistances

We need to find the thermal resistance of the pipe's wall \(R_{wall}\), and the insulated part \(R_{insulation}\). For the pipe's wall: $$ R_{wall} = \frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} $$ For the insulation: $$ R_{insulation} = \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi k_{ins} L} $$

Calculate the total heat transferred for both situations

We will calculate the total heat transferred through the pipe before and after the installation of insulation. Without insulation: $$ Q_{before} = \frac{T_{mix} - T_{amb}}{R_{wall} + \frac{1}{h_{amb} \pi D_o L}} $$ With insulation: $$ Q_{after} = \frac{T_{mix} - T_{s}}{R_{wall} + R_{insulation} + \frac{1}{h_{amb} \pi (D_o + 2t) L}} $$

Evaluate the desired thermal conductivity of the insulation

Since we want to maintain the outside surface temperature at 45°C or lower, we need the heat transferred through the insulated pipe, \(Q_{after}\) to be equal or less than the heat transferred if the surface temperature were about 45°C. So: $$ Q_{after} \le Q_{before} $$ Using the expressions from steps 2 and 3, plug in the values and solve for \(k_{ins}\): $$ \frac{T_{mix} - T_{s}}{\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi k_{ins} L} + \frac{1}{h_{amb} \pi (D_o + 2t) L}} \le \frac{T_{mix} - T_{amb}}{\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{1}{h_{amb} \pi D_o L}} $$ Solve for \(k_{ins}\): $$ k_{ins} \ge \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi L} \left(\frac{T_{mix} - T_{s}}{T_{mix} - T_{amb}} (\frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} + \frac{1}{h_{amb} \pi D_o L}) - \frac{\ln(\frac{D_o}{D_i})}{2 \pi k L} - \frac{1}{h_{amb} \pi (D_o + 2t) L}\right) $$ Plug in the given values and find the appropriate insulation thermal conductivity, \(k_{ins}\).

Key concepts

Thermal Resistance

A key concept in understanding heat transfer in materials is **thermal resistance**. Imagine it like a roadblock to heat flow. The greater the resistance, the harder it is for heat to pass through. Thermal resistance is important because it helps determine how effective a material is at insulating or conducting heat.

Thermal resistance for a cylindrical object, like a pipe, can be calculated using the formula:

• For the pipe's wall: \[ R_{\text{wall}} = \frac{\ln\left(\frac{D_o}{D_i}\right)}{2\pi k L} \]
• For the insulation: \[ R_{\text{insulation}} = \frac{\ln\left(\frac{D_o+2t}{D_o}\right)}{2\pi k_{\text{ins}} L} \]

These formulas help calculate how much resistance each section of the pipe offers to the flow of heat as it conducts from the warmer inside to the cooler exterior.

Understanding thermal resistance is essential when choosing materials for insulation to ensure adequate protection against unwanted heat exchange.

Convection Heat Transfer

**Convection heat transfer** is the movement of heat due to a fluid or gas flowing over a surface. This process can greatly affect how heat is transferred between a surface and its surroundings.

In the case of our pipe, there are two instances of convection:

• Inside the pipe, where the mixture's temperature interacts with the pipe material.
• Outside the pipe, where the pipe's external surface comes into contact with ambient air.

Each instance has its convection coefficient; the mixture has a coefficient of \(150\ \mathrm{W/m^2 \cdot K}\), and the ambient air has \(25\ \mathrm{W/m^2 \cdot K}\). These coefficients indicate how easily heat can be transferred at these interfaces.

The equations involving convection will help you calculate the heat transfer between the surfaces by providing the relationship between the temperature differences and heat flow rates. Convection thus plays a crucial role in determining the overall heat exchange in systems involving fluids.

Thermal Conductivity

**Thermal conductivity** is a measure of how well a material can conduct heat. It is represented by the symbol \(k\) and usually measured in \(\mathrm{W/m \cdot K}\). High thermal conductivity means heat flows through the material easily, whereas low conductivity indicates a good insulator.

In the exercise, the pipe material has a thermal conductivity of \(14 \mathrm{~W/m \cdot K}\). This value allows us to evaluate how much it can resist or allow heat transfer. Importantly, when choosing insulation for the pipe, its conductivity needs to be sufficiently low to ensure the outside surface temperature remains at or below the desired level, despite the high internal temperatures.

Finding the right balance in thermal conductivity when selecting materials for insulation is crucial. It helps control energy costs and maintains safety standards, ensuring efficient thermal management.

Insulation

When it comes to managing heat transfer, **insulation** is like a sweater for your pipe. It helps keep the heat where it should be. Insulation limits the amount of heat that escapes from the pipe, crucial for preventing damage or energy loss.

In the pipe exercise, a \(2.5\ \text{cm}\) thick layer of insulation is proposed. This layer is designed to keep the surface temperature of the pipe below \(45^{\circ} \mathrm{C}\) even as the mixture inside the pipe undergoes exothermic reactions at high temperatures. The effectiveness of this insulation layer depends directly on its thermal resistance, which we calculate based on its thickness and thermal conductivity.

Using the equation for thermal resistance, once you know the right thermal conductivity for your insulative material, you're able to ensure that the temperature of the pipe's exterior skin remains safe for touching or in close proximity to other materials or workers. Proper insulation design helps in energy conservation and safety compliance.