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What is lumped system analysis? When is it applicable?

Short Answer

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Answer: Lumped system analysis is a simplified approach used in the field of heat transfer to analyze complex systems by assuming that the temperature of an object is uniform throughout its entire volume. It can be applied when the Biot number, a dimensionless parameter comparing the rate of heat conduction within an object to the rate of heat convection between the object and its surroundings, is much less than 1 (typically, less than 0.1). This indicates that the temperature gradients within the object are negligible, making the lumped system analysis applicable for accurate estimations of the object's temperature change over time.

Step by step solution

01

Definition of Lumped System Analysis

Lumped system analysis is an approach used in the field of heat transfer to simplify complex systems. It assumes that the temperature of an object is uniform throughout its entire volume, meaning that the temperature changes simultaneously at every point in the object. This simplification reduces the analysis to a single variable, the object's temperature, which enables the use of simple mathematical methods to solve heat transfer problems.
02

Assumptions in Lumped System Analysis

The main assumption in lumped system analysis is that the temperature within the object is uniform at all times. This implies that the heat transfer is mainly dependent on the object's surface properties and the surrounding environment, rather than on the internal distribution of temperature. The assumption can be applied to situations in which the object's thermal conductivity is much higher than the heat transfer coefficient or the heat transfer process is relatively fast. This means that the temperature gradients within the object are negligible compared to the temperature differences between the object and its surroundings.
03

Applicability of Lumped System Analysis

Lumped system analysis is applicable when the Biot number, a dimensionless parameter that compares the rate of heat conduction within an object to the rate of heat convection between the object and its surroundings, is much less than 1. Mathematically, the Biot number (Bi) is defined as: \[Bi = \frac{hL_c}{k}\] where \(h\) is the convective heat transfer coefficient, \(L_c\) is the characteristic length (typically, the object's volume divided by its surface area), and \(k\) is the object's thermal conductivity. When the Biot number is less than 0.1, the temperature gradients within the object are sufficiently small, and the lumped system analysis can provide accurate estimations of the object's temperature change over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is the movement of thermal energy from one object or material to another due to a temperature difference. It's fundamental to our understanding of how energy is exchanged in the physical world and is pivotal in a range of applications, from industrial processes to everyday household activities.

In the study of lumped system analysis, we focus on heat transfer to understand how energy is redistributed within an object or between objects. The assumption that an object's temperature is uniform simplifies the analysis of how heat will flow to or from that object over time. Understanding heat transfer allows students to predict the effects of heating or cooling systems, design better thermal insulation, and create efficient energy management strategies.
Thermal Conductivity
Thermal conductivity, denoted by the symbol \( k \) in mathematical terms, is a measure of a material's ability to conduct heat. It's a physical property that varies between different materials, where metals typically have high thermal conductivity, while insulating materials exhibit low conductivity.

The relevance of thermal conductivity to lumped system analysis lies in its role in determining the rate at which heat can spread within an object. When analyzing heat transfer in an object with a high thermal conductivity, we can more justifiably apply a lumped system analysis approach. This is because such materials can quickly distribute heat throughout their volume, maintaining a more uniform temperature that aligns with the lumped system analysis's assumption of no temperature gradient.
Biot Number
The Biot number (Bi) is a dimensionless quantity that plays a crucial role in lumped system analysis. It helps us determine whether it's appropriate to assume a uniform temperature within an object undergoing heat transfer. The Biot number is defined by the formula:
\[Bi = \frac{hL_c}{k}\]
where \( h \) is the heat transfer coefficient, \( L_c \) is the characteristic length, and \( k \) is the thermal conductivity. A small Biot number (typically less than 0.1) indicates that the rate of heat conduction within the object outpaces the rate of heat transfer between the object and the environment. This justifies using lumped system analysis since it indicates that temperature gradients within the object would be negligible compared to the temperature difference between the object and its surroundings.
Heat Transfer Coefficient
The heat transfer coefficient, represented by the symbol \( h \) in equations, is a measure that describes how well heat is transferred between a solid surface and a fluid moving past it. It's a crucial concept in both the heating and cooling of objects and is influenced by factors such as the nature of the fluid flow, the temperature difference, and the properties of the surface interface.

In the context of the lumped system analysis, the heat transfer coefficient is vital because it represents the rate of heat convected away from or to the object's surface. A higher coefficient means that heat can be transferred more rapidly, which impacts the system's ability to remain at a uniform temperature. In systems where the heat transfer coefficient is relatively low compared to the material's thermal conductivity, the temperature within the object is more likely to remain even, justifying the application of lumped system analysis.

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Most popular questions from this chapter

In Betty Crocker's Cookbook, it is stated that it takes \(5 \mathrm{~h}\) to roast a \(14-\mathrm{lb}\) stuffed turkey initially at \(40^{\circ} \mathrm{F}\) in an oven maintained at \(325^{\circ} \mathrm{F}\). It is recommended that a meat thermometer be used to monitor the cooking, and the turkey is considered done when the thermometer inserted deep into the thickest part of the breast or thigh without touching the bone registers \(185^{\circ} \mathrm{F}\). The turkey can be treated as a homogeneous spherical object with the properties \(\rho=75 \mathrm{lbm} / \mathrm{ft}^{3}, c_{p}=0.98 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\), \(k=0.26 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\), and \(\alpha=0.0035 \mathrm{ft}^{2} / \mathrm{h}\). Assuming the tip of the thermometer is at one- third radial distance from the center of the turkey, determine \((a)\) the average heat transfer coefficient at the surface of the turkey, \((b)\) the temperature of the skin of the turkey when it is done, and \((c)\) the total amount of heat transferred to the turkey in the oven. Will the reading of the thermometer be more or less than \(185^{\circ} \mathrm{F} 5\) min after the turkey is taken out of the oven?

An 18-cm-long, 16-cm-wide, and 12 -cm-high hot iron block \(\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(20^{\circ} \mathrm{C}\) is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If it is required that the temperature of the block rises to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

Long aluminum wires of diameter \(3 \mathrm{~mm}(\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\) ) 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.

Chickens with an average mass of \(1.7 \mathrm{~kg}(k=\) \(0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\alpha=0.13 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) initially at a uniform temperature of \(15^{\circ} \mathrm{C}\) are to be chilled in agitated brine at \(-7^{\circ} \mathrm{C}\). The average heat transfer coefficient between the chicken and the brine is determined experimentally to be \(440 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Taking the average density of the chicken to be \(0.95 \mathrm{~g} / \mathrm{cm}^{3}\) and treating the chicken as a spherical lump, determine the center and the surface temperatures of the chicken in \(2 \mathrm{~h}\) and \(45 \mathrm{~min}\). Also, determine if any part of the chicken will freeze during this process. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Consider a cubic block whose sides are \(5 \mathrm{~cm}\) long and a cylindrical block whose height and diameter are also \(5 \mathrm{~cm}\). Both blocks are initially at \(20^{\circ} \mathrm{C}\) and are made of granite \((k=\) \(2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\left.\alpha=1.15 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\). Now both blocks are exposed to hot gases at \(500^{\circ} \mathrm{C}\) in a furnace on all of their surfaces with a heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the center temperature of each geometry after 10,20 , and \(60 \mathrm{~min}\).

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