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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

Expert verified
#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

01

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, \(T_s = 45 °C\)
02

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: $$ R_{insulation} = \frac{\ln(\frac{D_o+2t}{D_o})}{2 \pi k_{ins} L} $$
03

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: $$ Q_{after} = \frac{T_{mix} - T_{s}}{R_{wall} + R_{insulation} + \frac{1}{h_{amb} \pi (D_o + 2t) L}} $$
04

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, \(k_{ins}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

A cylindrical pin fin of diameter \(0.6 \mathrm{~cm}\) and length of \(3 \mathrm{~cm}\) with negligible heat loss from the tip has an efficiency of 0.7. The effectiveness of this fin is (a) \(0.3\) (b) \(0.7\) (c) 2 (d) 8 (e) 14

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

A hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

The \(700 \mathrm{~m}^{2}\) ceiling of a building has a thermal resistance of \(0.52 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The rate at which heat is lost through this ceiling on a cold winter day when the ambient temperature is \(-10^{\circ} \mathrm{C}\) and the interior is at \(20^{\circ} \mathrm{C}\) is (a) \(23.1 \mathrm{~kW} \quad\) (b) \(40.4 \mathrm{~kW}\) (c) \(55.6 \mathrm{~kW}\) (d) \(68.1 \mathrm{~kW}\) (e) \(88.6 \mathrm{~kW}\)

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