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The Biot number during a heat transfer process between a sphere and its surroundings is determined to be \(0.02\). Would you use lumped system analysis or the one-term approximate solutions when determining the midpoint temperature of the sphere? Why?

Short Answer

Expert verified
Answer: Lumped system analysis should be used to determine the midpoint temperature of the sphere, as the given Biot number (Bi = 0.02) indicates that the internal temperature gradients are negligible.

Step by step solution

01

Understanding the Biot Number

The Biot number (Bi) is a dimensionless quantity that helps to determine whether internal temperature gradients in an object are significant or not. It is defined as: Bi = \(hL_c/k\) where: - h = heat transfer coefficient \((W/m^2K)\), - \(L_c\) = the characteristic length of the object (in this case, the sphere) \((m)\), and - k = thermal conductivity of the sphere \((W/mK)\). The given Biot number is \(Bi = 0.02\).
02

Criteria for Lumped System Analysis

Lumped system analysis assumes that internal temperature gradients in the object are negligible. This implies that the object has uniform temperature throughout its volume. Lumped system analysis is valid when the Biot number (Bi) is less than or equal to 0.1 (Bi ≤ 0.1). This criterion needs to be checked against the given Biot number.
03

Criteria for One-Term Approximate Solution

The one-term approximate solution is more accurate than the lumped system analysis and accounts for internal temperature gradients within the object. It is valid when the Biot number is between 0.1 and 10, i.e., 0.1 < Bi < 10.
04

Comparing the Biot Number to Criteria

Given that \(Bi = 0.02\), let's compare it to the criteria for both cases: 1. Lumped System Analysis: \(Bi \leq 0.1\), and 2. One-Term Approximate Solution: \(0.1 \lt Bi \lt 10\)
05

Selecting the Appropriate Method

Since the given Biot number (\(Bi = 0.02\)) is less than the criterion for lumped system analysis (Bi ≤ 0.1), we can use the lumped system analysis method to determine the midpoint temperature of the sphere, as the internal temperature gradients are considered negligible. In conclusion, we should use the lumped system analysis to determine the midpoint temperature of the sphere as the given Biot number indicates that the internal temperature gradients are negligible.

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Most popular questions from this chapter

A small chicken $(k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad \alpha=0.15 \times\( \)10^{-6} \mathrm{~m}^{2} / \mathrm{s}$ ) can be approximated as an \(11.25-\mathrm{cm}\)-diameter solid sphere. The chicken is initially at a uniform temperature of \(8^{\circ} \mathrm{C}\) and is to be cooked in an oven maintained at \(220^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). With this idealization, the temperature at the center of the chicken after a 90 -min period is (a) \(25^{\circ} \mathrm{C}\) (b) \(61^{\circ} \mathrm{C}\) (c) \(89^{\circ} \mathrm{C}\) (d) \(122^{\circ} \mathrm{C}\) (e) \(168^{\circ} \mathrm{C}\)

Aluminum wires \(4 \mathrm{~mm}\) in diameter are produced by extrusion. The wires leave the extruder at an average temperature of \(350^{\circ} \mathrm{C}\) and at a linear rate of \(10 \mathrm{~m} / \mathrm{min}\). Before leaving the extrusion room, the wires are cooled to an average temperature of $50^{\circ} \mathrm{C}\( by transferring heat to the surrounding air at \)25^{\circ} \mathrm{C}\( with a heat transfer coefficient of \)50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Calculate the necessary length of the wire cooling section in the extrusion room.

Citrus fruits are very susceptible to cold weather, and extended exposure to subfreezing temperatures can destroy them. Consider an 8 -cm-diameter orange that is initially at \(15^{\circ} \mathrm{C}\). A cold front moves in one night, and the ambient temperature suddenly drops to \(-6^{\circ} \mathrm{C}\), with a heat transfer coefficient of $15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Using the properties of water for the orange and assuming the ambient conditions remain constant for \(4 \mathrm{~h}\) before the cold front moves out, determine if any part of the orange will freeze that night. Solve this problem using the analytical one-term approximation method.

Plasma spraying is a process used for coating a material surface with a protective layer to prevent the material from degradation. In a plasma spraying process, the protective layer in powder form is injected into a plasma jet. The powder is then heated to molten droplets and propelled onto the material surface. Once deposited on the material surface, the molten droplets solidify and form a layer of protective coating. Consider a plasma spraying process using alumina $(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, \)\rho=3970 \mathrm{~kg} / \mathrm{m}^{3}\(, and \)\left.c_{p}=800 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)$ powder that is injected into a plasma jet at \(T_{\infty}=15,000^{\circ} \mathrm{C}\) and $h=10,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The alumina powder is made of spherical particles with an average diameter of \(60 \mu \mathrm{m}\) and a melting point at \(2300^{\circ} \mathrm{C}\). Determine the amount of time it would take for the particles, with an initial temperature of $20^{\circ} \mathrm{C}$, to reach their melting point from the moment they are injected into the plasma jet.

Long aluminum wires of diameter $3 \mathrm{~mm}\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$, and \(\left.\alpha=9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) are extruded at a temperature of \(350^{\circ} \mathrm{C}\) and exposed to atmospheric air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine how long it will take for the wire temperature to drop to \(50^{\circ} \mathrm{C}\). (b) If the wire is extruded at a velocity of 10 \(\mathrm{m} / \mathrm{min}\), determine how far the wire travels after extrusion by the time its temperature drops to \(50^{\circ} \mathrm{C}\). What change in the cooling process would you propose to shorten this distance? (c) Assuming the aluminum wire leaves the extrusion room at $50^{\circ} \mathrm{C}$, determine the rate of heat transfer from the wire to the extrusion room. Answers: (a) \(144 \mathrm{~s}\), (b) \(24 \mathrm{~m}\), (c) $856 \mathrm{~W}$

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