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Chapter 3: Problem 97

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

Short Answer

Expert verified
Answer: The critical radius of insulation will be greater on calm days than on windy days. This is because the convective heat transfer coefficient is lower on calm days due to less air movement, requiring more insulation to minimize heat transfer.

Step by step solution

01

Understanding Critical Radius of Insulation

To analyze the problem, we first need to understand the concept of critical radius of insulation. The critical radius of insulation is the thickness of insulation at which the heat transfer rate from the pipe to the surroundings is minimized. In other words, adding insulation beyond the critical radius would increase rather than decrease the heat transfer rate. This occurs because two factors affect heat transfer - conduction through the insulation and convection at the insulation surface. The critical radius of insulation (r_c) for a cylindrical pipe can be calculated using the following formula: r_c = \frac{k}{h} where - r_c is the critical radius of insulation - k is the thermal conductivity of the insulation material - h is the convective heat transfer coefficient at the insulation surface
02

Factors affecting heat transfer rate

The heat transfer rate depends on the insulation thickness, the thermal conductivity of the insulation material, and the convective heat transfer coefficient at the insulation surface. On calm days, the convective heat transfer coefficient (h_calm) will be lower than on windy days (h_windy), because there is less air movement to carry away heat from the pipe surface: h_calm < h_windy
03

Calculating the critical radius of insulation for calm and windy days

Now that we know the convective heat transfer coefficients for calm and windy days, we can calculate the critical radius of insulation for each scenario using the formula mentioned in Step 1. For calm days: r_c_calm = \frac{k}{h_calm} For windy days: r_c_windy = \frac{k}{h_windy} Because h_calm < h_windy, we can conclude that: r_c_calm > r_c_windy
04

Conclusion

Based on our calculations, the critical radius of insulation will be greater on calm days than on windy days. This is because the convective heat transfer coefficient, which determines the heat transfer rate from the pipe to the surroundings, is lower on calm days due to less air movement. Consequently, more insulation is needed to minimize heat transfer on calm days compared to windy days.

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Key Concepts

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

Convection Heat Transfer Coefficient
The convection heat transfer coefficient (h) plays a crucial role in determining how swiftly heat is exchanged between a surface and a fluid moving past it. It's a measure of convective heat transfer per unit area, per degree of temperature difference between the surface and the fluid. This coefficient is influenced by factors such as fluid velocity, viscosity, and surface roughness.

In the context of insulation on a pipe, the coefficient on calm days is lower since the air movement is relatively stagnant, limiting the heat carried away from the pipe. On windy days, the increased air movement results in a higher coefficient, signifying more heat being whisked away. Understanding this variable is key to determining the effectiveness of insulation material in different environmental conditions.
Thermal Conductivity
Thermal conductivity (k) of a material is a measure of its ability to conduct heat. In essence, it quantifies how quickly heat can pass through a material due to a temperature gradient. High thermal conductivity implies that the material can transfer heat rapidly, making it a good heat conductor, while low thermal conductivity indicates that the material is a good insulator.

Materials chosen for insulation purposes are typically those with low thermal conductivity to reduce the rate of heat loss. For an insulated pipe, understanding the material's thermal conductivity is critical in calculating the critical radius of insulation. This value ensures that insulation thickness is optimized for energy-saving while preventing excessive heat transfer.
Heat Transfer Rate
The rate of heat transfer is a measurement of the amount of thermal energy being moved from one place to another over time. Here, it's crucial for understanding how effectively a pipe system can be insulated. Heat transfer rate can be affected by various factors, including the temperature difference across a medium, the medium's properties, and the system geometry.

When discussing critical radius of insulation, the heat transfer rate is optimized at a certain point based on the relationship between the pipe's insulation thickness and the environment's heat transfer mechanism (convection). If the insulation exceeds the critical radius, the rate of heat transfer increases, making the system less efficient. Hence, the concept of heat transfer rate directly affects the energy conservation efforts in thermal systems and helps in designing practical insulation for pipes.

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Most popular questions from this chapter

Ice slurry is being transported in a pipe \((k=\) \(15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\), 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 \(20^{\circ} \mathrm{C}\), a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a dew point of \(10^{\circ} \mathrm{C}\). If the outer surface temperature of the pipe drops below the dew point, condensation can occur on the surface. Since this pipe is located in a vicinity of high voltage devices, water droplets from the condensation can cause electrical hazard. To prevent such incident, the pipe surface needs to be insulated. Determine the 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.

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

Consider a pipe at a constant temperature whose radius is greater than the critical radius of insulation. Someone claims that the rate of heat loss from the pipe has increased when some insulation is added to the pipe. Is this claim valid?
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