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Chapter 4: Problem 2

What is lumped system analysis? When is it applicable?

Short Answer

Expert verified
Answer: The key assumption in lumped system analysis is that the temperature distribution within the solid is uniform. This method is applicable when the Biot number is significantly less than 1 (Bi << 1), indicating that internal temperature gradients within the solid are small compared to the temperature differences across the boundaries.

Step by step solution

01

Definition of Lumped System Analysis

Lumped system analysis is a method used in heat transfer problems where the temperature of a solid is assumed to be uniform throughout its volume at any given time. This simplification allows for easier calculation of the heat transfer rates and temperature changes within the system.
02

Key Assumption

The main assumption behind lumped system analysis is that the temperature distribution within the solid is uniform. In reality, the temperature might vary within the solid, but this assumption allows us to avoid solving the complex differential equations that would be needed to capture the exact temperature distribution. The quality of this approximation depends on the system's characteristics.
03

Biot Number

The applicability of lumped system analysis is determined by the Biot number (Bi), a dimensionless parameter that represents the relative importance of internal temperature gradients to the overall heat transfer. The Biot number can be defined as: Bi = \frac{hL_c}{k} where h is the heat transfer coefficient, L_c is the characteristic length (typically the ratio of the solid volume to its surface area), and k is the thermal conductivity of the solid.
04

Applicability of Lumped System Analysis

Lumped system analysis is applicable when the Biot number is significantly less than 1 (Bi << 1). In other words, this method can be used when the internal temperature gradients within the solid are small compared to the temperature differences across the boundaries. When the Biot number is low, the solid's interior temperature distribution can be assumed to be uniform, and the lumped system analysis provides a reasonably accurate solution to the heat transfer problem.

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Most popular questions from this chapter

A thick wood slab \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=1.28 \times\) \(10^{-7} \mathrm{~m}^{2} / \mathrm{s}\) ) that is initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) is exposed to hot gases at \(550^{\circ} \mathrm{C}\) for a period of \(5 \mathrm{~min}\). The heat transfer coefficient between the gases and the wood slab is $35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. If the ignition temperature of the wood is \(450^{\circ} \mathrm{C}\), determine if the wood will ignite.

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}$

Conduct the following experiment at home to determine the combined convection and radiation heat transfer coefficient at the surface of an apple exposed to the room air. You will need two thermometers and a clock. First, weigh the apple and measure its diameter. You can measure its volume by placing it in a large measuring cup halfway filled with water, and measuring the change in volume when it is completely immersed in the water. Refrigerate the apple overnight so that it is at a uniform temperature in the morning, and measure the air temperature in the kitchen. Then take the apple out and stick one of the thermometers to its middle and the other just under the skin. Record both temperatures every \(5 \mathrm{~min}\) for an hour. Using these two temperatures, calculate the heat transfer coefficient for each interval and take their average. The result is the combined convection and radiation heat transfer coefficient for this heat transfer process. Using your experimental data, also calculate the thermal conductivity and thermal diffusivity of the apple and compare them to the values given above.

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.

A potato may be approximated as a \(5.7-\mathrm{cm}\) solid sphere with the properties $\rho=910 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.25 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\(, \)k=0.68 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\(, and \)\alpha=1.76 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}$. Twelve such potatoes initially at \(25^{\circ} \mathrm{C}\) are to be cooked by placing them in an oven maintained at \(250^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(95 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer to the potatoes by the time the center temperature reaches \(90^{\circ} \mathrm{C}\) is (a) \(1012 \mathrm{~kJ}\) (b) \(1366 \mathrm{~kJ}\) (c) \(1788 \mathrm{~kJ}\) (d) \(2046 \mathrm{~kJ}\) (e) \(3270 \mathrm{~kJ}\)
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