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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 than in air.

Step by step solution

01

Understand Lumped System Analysis and Biot Number

Lumped system analysis is a simplification used in heat transfer problems when the temperature gradient inside a solid object is negligible compared to the temperature difference between the surface and the surrounding medium. It can be applied when heat conduction within the object is much faster than heat convection between the object and the environment. To determine the applicability of lumped system analysis, we can use the Biot Number (Bi), which is defined as: Bi = (h * L)/k where h is the heat transfer coefficient of the fluid, L is the characteristic length of the object, and k is the thermal conductivity of the solid. If Bi << 1, the temperature gradient inside the solid is negligible, and the lumped system analysis is applicable.
02

Compare heat transfer coefficients in water and air

Heat transfer coefficient (h) is an important parameter to compare, as it directly affects the Biot number. Generally, the heat transfer coefficient is higher in liquids than in gases. As water is a liquid and air is a gas, the heat transfer coefficient for water will be greater than that of air.
03

Compare thermal conductivity in water and air

Thermal conductivity (k) is another important parameter to consider for this exercise. The thermal conductivity of water is much higher than that of air, which means that heat is conducted through water more efficiently (about 25 times) than in air.
04

Determine applicability of lumped system analysis in water and air

As water has a higher heat transfer coefficient (h) and thermal conductivity (k) than air, the Biot number (Bi) in water will be smaller than that in air for objects of the same size and material. As a result, the lumped system analysis is more likely to be applicable in water than in air, because the temperature gradient within the solid would be negligible in water more often than in air, as per the condition Bi << 1.

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Most popular questions from this chapter

A 30-cm-diameter, 4-m-high cylindrical column of a house made of concrete $\left(k=0.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=5.94 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right.\(, \)\rho=1600 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)c_{p}=0.84 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ ) cooled to \(14^{\circ} \mathrm{C}\) during a cold night is heated again during the day by being exposed to ambient air at an average temperature of \(28^{\circ} \mathrm{C}\) with an average heat transfer coefficient of $14 \mathrm{~W} / \mathrm{m}^{2}\(. \)\mathrm{K}$. Using the analytical one-term approximation method, determine \((a)\) how long it will take for the column surface temperature to rise to \(27^{\circ} \mathrm{C}\), (b) the amount of heat transfer until the center temperature reaches to \(28^{\circ} \mathrm{C}\), and \((c)\) the amount of heat transfer until the surface temperature reaches \(27^{\circ} \mathrm{C}\).

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

In a production facility, 3-cm-thick large brass plates $\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.$, and \(\alpha=33.9 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through an oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be $h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, determine the surface temperature of the plates when they come out of the oven. Solve this problem using the analytical one-term approximation method. Can this problem be solved using lumped system analysis? Justify your answer.

A large cast iron container \((k=52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.70 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) with \(4-\mathrm{cm}\)-thick walls is initially at a uniform temperature of \(0^{\circ} \mathrm{C}\) and is filled with ice at \(0^{\circ} \mathrm{C}\). Now the outer surfaces of the container are exposed to hot water at $55^{\circ} \mathrm{C}$ with a very large heat transfer coefficient. Determine how long it will be before the ice inside the container starts melting. Also, taking the heat transfer coefficient on the inner surface of the container to be $250 \mathrm{~W} / \mathrm{m}^{2}\(, \)\mathrm{K}$, determine the rate of heat transfer to the ice through a \(1.2-\mathrm{m}\)-wide and \(2-\mathrm{m}\)-high section of the wall when steady operating conditions are reached. Assume the ice starts melting when its inner surface temperature rises to $0.1^{\circ} \mathrm{C}$.

How can we use the one-term approximate solutions when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

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