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Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known

Short Answer

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Question: For lumped system analysis of transient heat conduction situations to be valid, the Biot number should be: Answer: (a) very small.

Step by step solution

01

Understand Lumped System Analysis

Lumped system analysis assumes that the temperature within a solid is uniform, and heat transfer occurs only at its surface. This assumption is valid when internal conduction resistance is much smaller than the surface heat transfer resistance.
02

Relation between Biot number and Lumped System Analysis

The Biot number (Bi) is defined as the ratio of internal conductive resistance to surface convective resistance in a solid. Mathematically, it is given by: \(Bi = \frac{h L_c}{k}\), where \(h\) is the surface heat transfer coefficient, \(L_c\) is the characteristic length of the system, and \(k\) is the thermal conductivity of the material.
03

Identify the Appropriate Range of Biot Number for Lumped System Analysis

For lumped system analysis to be valid, the internal conductive resistance must be much smaller than the surface convective resistance, so the Biot number should be very small. In practice, a Biot number less than 0.1 is considered to satisfy this criterion.
04

Choose the Correct Option based on the Appropriate Range of Biot Number

The lumped system analysis of transient heat conduction situations is valid when the Biot number is very small. Therefore, the correct option is (a) very small.

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Most popular questions from this chapter

Consider a 7.6-cm-long and 3-cm-diameter cylindrical lamb meat chunk $\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\(, \)k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$ ). Fifteen such meat chunks initially at \(2^{\circ} \mathrm{C}\) are dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of $1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The amount of heat transfer during the first \(8 \mathrm{~min}\) of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) (d) \(269 \mathrm{~kJ}\) (e) \(307 \mathrm{~kJ}\)

In a manufacturing facility, 2-in-diameter brass balls $\left(k=64.1 \mathrm{Btw} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}, \rho=532 \mathrm{lbm} / \mathrm{ft}^{3}\right.\(, and \)\left.c_{p}=0.092 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\( initially at \)250^{\circ} \mathrm{F}\( are quenched in a water bath at \)120^{\circ} \mathrm{F}$ for a period of \(2 \mathrm{~min}\) at a rate of \(120 \mathrm{balls}\) per minute. If the convection heat transfer coefficient is $42 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2},{ }^{\circ} \mathrm{F}\(, determine \)(a)$ the temperature of the balls after quenching and \((b)\) the rate at which heat needs to be removed from the water in order to keep its temperature constant at $120^{\circ} \mathrm{F}

When water, as in a pond or lake, is heated by warm air above it, it remains stable, does not move, and forms a warm layer of water on top of a cold layer. Consider a deep lake $\left(k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)$ that is initially at a uniform temperature of \(2^{\circ} \mathrm{C}\) and has its surface temperature suddenly increased to \(20^{\circ} \mathrm{C}\) by a spring weather front. The temperature of the water \(1 \mathrm{~m}\) below the surface \(400 \mathrm{~h}\) after this change is (a) \(2.1^{\circ} \mathrm{C}\) (b) \(4.2^{\circ} \mathrm{C}\) (c) \(6.3^{\circ} \mathrm{C}\) (d) \(8.4^{\circ} \mathrm{C}\) (e) \(10.2^{\circ} \mathrm{C}\)

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 \(27^{\circ} \mathrm{C}\).

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