In what medium is the lumped system analysis more likely to be applicable: in water or in air? Why?

Short Answer

Expert verified
Answer: The lumped system analysis is more likely to be applicable in water medium due to its higher heat transfer coefficient and smaller Biot number, compared to air.

Step by step solution

01

Understand Lumped System Analysis

Lumped system analysis is a method used to simplify the heat transfer problem by assuming that the temperature throughout an object changes at the same rate. This can be applicable when the object has low thermal resistance compared to the resistance of heat transfer between the object and the surrounding medium.
02

Analyze Heat Transfer Coefficients

The heat transfer coefficient determines the rate of heat transfer through conduction, convection, and radiation. In general, water has a higher heat transfer coefficient than air because water has higher thermal conductivity and can store more heat due to its higher specific heat capacity.
03

Calculate the Biot Number

The Biot number (Bi) is a dimensionless quantity that represents the ratio of the internal thermal resistance of the object to the external thermal resistance. It is defined as Bi = (h * L) / k, where h is the heat transfer coefficient, L is the characteristic length of the object, and k is the thermal conductivity of the object's material. If Bi << 1, the temperature changes uniformly throughout the object, making lumped system analysis applicable.
04

Compare Biot Number in Water and Air

Since water has a higher heat transfer coefficient (h) than air, the Biot number in water will generally be smaller than in air. A smaller Biot number means the lumped system analysis is more likely to be applicable in water.
05

Conclusion

Based on the comparison of heat transfer coefficients and Biot numbers, we can conclude that the lumped system analysis is more likely to be applicable in water medium due to its higher heat transfer coefficient and smaller Biot number, compared to air.

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Key Concepts

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

Heat Transfer Coefficient
When studying heat transfer, one vital concept is the heat transfer coefficient, symbolically represented as 'h'. It quantifies the efficiency with which heat is transferred between a solid surface and a fluid, be it liquid or gas. The higher this coefficient, the more efficiently heat is conveyed.
For example, in the context of the lumped system analysis, the heat transfer coefficient of water is substantially higher than that of air, contributing to a more uniform and rapid distribution of temperature changes throughout a submerged object. This difference arises from water's physical properties, such as its ability to absorb and retain heat, leading to more effective conductive and convective heat transfer. Understanding this coefficient is essential, as it facilitates engineers and scientists to design systems that either enhance or restrain heat transfer, depending on the requirement.
Biot Number
The Biot number (Bi) is a dimensionless parameter central to the lumped system analysis. It's calculated using the formula:
\[\begin{equation} Bi = \frac{h \times L}{k} \end{equation}\],
where 'h' represents the heat transfer coefficient, 'L' is the characteristic length of the object, and 'k' is the thermal conductivity of the object's material.
A low Biot number, typically less than 0.1, indicates that the heat conduction within the object is much faster than the heat transfer across the boundary. This means that the object can be treated as if it has a uniform temperature throughout, which is a key simplification allowed by the lumped system analysis. In the case of water versus air, the heat transfer is typically more effective in water due to its higher 'h' value, implying a smaller Biot number and justifying the use of lumped system analysis for objects in water.
Thermal Conductivity
Thermal conductivity, denoted as 'k', measures a material’s ability to conduct heat. Materials with high thermal conductivity, such as metals, can transfer heat quickly, while insulating materials, such as wood or foam, with low thermal conductivity, transfer heat much slower.
This property is crucial in determining how well an object can maintain a uniform temperature, which is essential in the lumped system analysis. For instance, if an object has high thermal conductivity, heat is distributed throughout it rapidly, enabling us to assume a nearly uniform temperature. Thus, understanding the thermal conductivity of materials is fundamental for accurate thermal analysis in various engineering applications, ranging from heat exchangers to thermal insulation design.
Specific Heat Capacity
Specific heat capacity, defined as 'c_p', is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. It is a material property that reflects its capacity to store thermal energy. Materials with high specific heat capacity, like water, can absorb significant amounts of heat without undergoing large temperature changes.
This characteristic is especially relevant when applying lumped system analysis because it affects the temperature uniformity across the object. A substance with a high specific heat capacity can sustain uniform temperature easier, which validates the approach of lumped analysis under certain conditions. Thus, in energy system design, HVAC engineering, and everyday applications such as cooking, specific heat capacity plays a significant role in thermal management.

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Most popular questions from this chapter

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

A long iron \(\operatorname{rod}\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(k=80.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=23.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) with diameter of \(25 \mathrm{~mm}\) is initially heated to a uniform temperature of \(700^{\circ} \mathrm{C}\). The iron rod is then quenched in a large water bath that is maintained at constant temperature of \(50^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(128 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the time required for the iron rod surface temperature to cool to \(200^{\circ} \mathrm{C}\). Solve this problem using analytical one- term approximation method (not the Heisler charts).

An 18-cm-long, 16-cm-wide, and 12 -cm-high hot iron block \(\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(20^{\circ} \mathrm{C}\) is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If it is required that the temperature of the block rises to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

A 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(25^{\circ} \mathrm{C}\) is to be cooled to \(5^{\circ} \mathrm{C}\) by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=\) \(301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) to \(5^{\circ} \mathrm{C}\). Assume the can is agitated in water and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

What is a semi-infinite medium? Give examples of solid bodies that can be treated as semi-infinite mediums for heat transfer purposes.

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