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For which kind of bodies made of the same material is the lumped system analysis more likely to be applicable: slender ones or well-rounded ones of the same volume? Why?

Short Answer

Expert verified
Answer: Well-rounded bodies are more suitable for lumped system analysis due to their smaller Biot number, validating the assumption of uniform temperature throughout the body's volume.

Step by step solution

01

Understand the shape's effect on the lumped system analysis

The shape of an object plays a critical role in heat transfer and thus in the applicability of the lumped system analysis. A slender object has a higher surface area to volume ratio than a well-rounded object. Since heat transfer takes place on the surface, a larger surface area to volume ratio results in faster heat transfer rates from the surface of the object.
02

Analyze thermal resistance in slender and well-rounded bodies

In slender bodies, the larger surface area to volume ratio results in a smaller thermal resistance. This means that heat can flow rapidly from the surface into the body and vice versa. In contrast, well-rounded bodies have a larger thermal resistance due to their smaller surface area to volume ratio, resulting in slower rates of heat transfer.
03

Determine the effect on the Biot number

The Biot number (Bi) is a dimensionless number that measures the relativeness of heat conduction within the object to the heat transfer from the surface of the object. A smaller Biot number (Bi < 0.1) indicates that the lumped system analysis is more likely to be applicable. Since the slender bodies have a smaller thermal resistance, the heat transfer rate is faster, and therefore, the Biot number will likely be larger. On the other hand, well-rounded bodies will have a smaller Biot number due to their larger thermal resistances.
04

Conclude the type of body suitable for lumped system analysis

Based on the reasoning discussed above, the lumped system analysis is more likely to be applicable for well-rounded bodies made of the same material. This is because well-rounded bodies have a smaller Biot number due to their larger thermal resistances, making the assumption of uniform temperature throughout their volume more valid. In summary, the lumped system analysis is more suitable for well-rounded bodies as they have a smaller Biot number, which validates the assumption of uniform temperature throughout the body's volume.

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Key Concepts

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

Heat Transfer
Heat transfer is an essential concept in understanding how energy moves from one object to another. It occurs when there is a temperature difference, causing energy to flow from a hotter object to a cooler one.
Heat transfer can occur in three ways: conduction, convection, and radiation. While conduction involves the transfer of heat through direct contact, convection involves the movement of heat through fluids like air or water. Radiation, on the other hand, involves heat transfer through electromagnetic waves without the need for a medium.
When we consider objects in thermal environments, the surface area plays a crucial role. Bodies with larger surface areas relative to their volume transfer heat faster compared to those with smaller surface areas. This relationship is pivotal when applying the lumped system analysis, particularly in discerning which shapes of bodies facilitate rapid heat transfer from the surface into or out of the body.
Thermal Resistance
Thermal resistance is a measure of an object's ability to resist the flow of heat. It can be thought of like electrical resistance, but instead of electric current, it's the heat flow that's resisted.
In a thermal system, objects with higher thermal resistance require more time to transfer a certain amount of heat compared to those with lower resistance. The shape of an object greatly influences its thermal resistance. Slender bodies tend to have lower thermal resistance due to their higher surface area to volume ratio. This results in quick heat transfer, as there is less opposition to the flow of heat.
Conversely, well-rounded bodies usually have higher thermal resistance. The smaller surface area to volume ratio means that heat doesn't flow as quickly, making the process slower. Understanding these differences is crucial when evaluating the efficiency and applicability of lumped system analysis for different shapes.
Biot Number
The Biot number is a dimensionless quantity that helps to understand the relationship between internal heat conduction within an object and heat transfer across the object's surface. It is expressed as:\[ Bi = \frac{hL_c}{k} \]where \( h \) is the heat transfer coefficient, \( L_c \) is the characteristic length (a measure of the object's size), and \( k \) is the thermal conductivity of the material.
In general, the Biot number provides insight into whether a lumped system analysis is suitable. A Biot number much less than 1 (specifically less than 0.1) indicates that the temperature within the object can be assumed uniform, justifying the use of lumped system analysis.
Well-rounded bodies typically have lower Biot numbers due to their higher thermal resistance. In these cases, the uniform temperature assumption is more valid, making lumped system analysis a viable approach for thermal analysis.

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Most popular questions from this chapter

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

The 40-cm-thick roof of a large room made of concrete \(\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.88 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). After a heavy snow storm, the outer surface of the roof remains covered with snow at \(-5^{\circ} \mathrm{C}\). The roof temperature at \(18.2 \mathrm{~cm}\) distance from the outer surface after a period of 2 hours is (a) \(14^{\circ} \mathrm{C}\) (b) \(12.5^{\circ} \mathrm{C}\) (c) \(7.8^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-5^{\circ} \mathrm{C}\)

The Biot number can be thought of as the ratio of (a) The conduction thermal resistance to the convective thermal resistance. (b) The convective thermal resistance to the conduction thermal resistance. (c) The thermal energy storage capacity to the conduction thermal resistance. (d) The thermal energy storage capacity to the convection thermal resistance. (e) None of the above.

Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You can measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every \(5 \mathrm{~min}\) for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above.

A long cylindrical wood \(\log (k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.28 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\) is \(10 \mathrm{~cm}\) in diameter and is initially at a uniform temperature of \(15^{\circ} \mathrm{C}\). It is exposed to hot gases at \(550^{\circ} \mathrm{C}\) in a fireplace with a heat transfer coefficient of \(13.6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surface. If the ignition temperature of the wood is \(420^{\circ} \mathrm{C}\), determine how long it will be before the log ignites. Solve this problem using analytical one-term approximation method (not the Heisler charts).

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