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Chapter 4: Problem 4

For which solid is the lumped system analysis more likely to be applicable: an actual apple or a golden apple of the same size? Why?

Short Answer

Expert verified
Answer: Based on the comparison of the Biot numbers for the actual apple and the golden apple, the lumped system analysis is more likely to be applicable for the solid with a smaller Biot number (Bi < 0.1). This implies that the actual apple with its lower thermal conductivity (compared to gold) will have a higher rate of energy storage relative to the rate of energy dissipation by conduction, making lumped system analysis more applicable for the actual apple.

Step by step solution

01

Biot number for the actual apple

Calculate the Biot number for the actual apple. The actual apple has a relatively low thermal conductivity (k_apple) and a characteristic length (Lc_apple). Use the formula Bi = hLc / k, with the heat transfer coefficient (h) remaining constant for both cases.
02

Biot number for the golden apple

Calculate the Biot number for the golden apple. Gold has a much higher thermal conductivity (k_gold) compared to the actual apple, and the characteristic length (Lc_gold) is the same as for the actual apple. Use the same formula Bi = hLc / k, with the heat transfer coefficient (h) remaining constant for both cases.
03

Comparing the Biot numbers

Compare the Biot numbers for the actual apple and the golden apple. The lumped system analysis will be more applicable for the solid with the smaller Biot number (Bi < 0.1), which implies a greater uniformity in temperature within the solid.
04

Conclusion

Based on the comparison of the Biot numbers for the actual apple and the golden apple, determine which one is more likely to have a smaller Biot number (Bi < 0.1), and therefore for which solid the lumped system analysis is more applicable. The solid with a smaller Biot number has a higher rate of energy storage compared to the rate of energy dissipation by conduction.

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Most popular questions from this chapter

A heated 6-mm-thick Pyroceram plate $\left(\rho=2600 \mathrm{~kg} / \mathrm{m}^{3}\right.\(, \)c_{p}=808 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=3.98 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\left.\alpha=1.89 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)$ is being cooled in a room with air temperature of \(25^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(13.3 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heated Pyroceram plate had an initial temperature of \(500^{\circ} \mathrm{C}\), and it is allowed to cool for \(286 \mathrm{~s}\). If the mass of the Pyroceram plate is \(10 \mathrm{~kg}\), determine the heat transfer from the Pyroceram plate during the cooling process using the analytical one-term approximation method.

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}

Water mains must be placed at sufficient depth below the earth's surface to avoid freezing during extended periods of subfreezing temperatures. Determine the minimum depth at which the water main must be placed at a location where the soil is initially at \(15^{\circ} \mathrm{C}\) and the earth's surface temperature under the worst conditions is expected to remain at $-10^{\circ} \mathrm{C}$ for 75 days. Take the properties of soil at that location to be \(k=0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and $\alpha=1.4 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\(. Answer: \)7.05 \mathrm{~m}$

A \(10-\mathrm{cm}\)-inner-diameter, 30-cm-long can filled with water initially at \(25^{\circ} \mathrm{C}\) is put into a household refrigerator at $3^{\circ} \mathrm{C}\(. The heat transfer coefficient on the surface of the can is \)14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming that the temperature of the water remains uniform during the cooling process, the time it takes for the water temperature to drop to \(5^{\circ} \mathrm{C}\) is (a) \(0.55 \mathrm{~h}\) (b) \(1.17 \mathrm{~h}\) (c) \(2.09 \mathrm{~h}\) (d) \(3.60 \mathrm{~h}\) (e) \(4.97 \mathrm{~h}\)

A man is found dead in a room at \(12^{\circ} \mathrm{C}\). The surface temperature on his waist is measured to be \(23^{\circ} \mathrm{C}\), and the heat transfer coefficient is estimated to be $9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\(. Modeling the body as a \)28-\mathrm{cm}$ diameter, \(1.80\)-m-long cylinder, estimate how long it has been since he died. Take the properties of the body to be $k=0.62 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\( and \)\alpha=0.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}$, and assume the initial temperature of the body to be \(36^{\circ} \mathrm{C}\). Solve this problem using the analytical one-term approximation method.
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