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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

A \(5-\mathrm{mm}\)-diameter spherical ball at \(50^{\circ} \mathrm{C}\) is covered by a \(1-\mathrm{mm}\)-thick plastic insulation \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The ball is exposed to a medium at \(15^{\circ} \mathrm{C}\), with a combined convection and radiation heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine if the plastic insulation on the ball will help or hurt heat transfer from the ball.

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known.

Heat is lost at a rate of \(275 \mathrm{~W}\) per \(\mathrm{m}^{2}\) area of a \(15-\mathrm{cm}-\) thick wall with a thermal conductivity of \(k=1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The temperature drop across the wall is (a) \(37.5^{\circ} \mathrm{C}\) (b) \(27.5^{\circ} \mathrm{C}\) (c) \(16.0^{\circ} \mathrm{C}\) (d) \(8.0^{\circ} \mathrm{C}\) (e) \(4.0^{\circ} \mathrm{C}\)

Determine the winter \(R\)-value and the \(U\)-factor of a masonry wall that consists of the following layers: \(100-\mathrm{mm}\) face bricks, 100 -mm common bricks, \(25-\mathrm{mm}\) urethane rigid foam insulation, and \(13-\mathrm{mm}\) gypsum wallboard.
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