/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 Consider a power transistor enca... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 33

Consider a power transistor encapsulated in an aluminum case that is attached at its base to a square aluminum plate of thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(L=6 \mathrm{~mm}\), and width \(W=20 \mathrm{~mm}\). The case is joined to the plate by screws that maintain a contact pressure of 1 bar, and the back surface of the plate transfers heat by natural convection and radiation to ambient air and large surroundings at \(T_{\infty}=T_{\text {sur }}=\) \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(\varepsilon=0.9\), and the convection coefficient is \(h=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The case is completely enclosed such that heat transfer may be assumed to occur exclusively through the base plate. (a) If the air-filled aluminum-to-aluminum interface is characterized by an area of \(A_{c}=2 \times 10^{-4} \mathrm{~m}^{2}\) and a roughness of \(10 \mu \mathrm{m}\), what is the maximum allowable power dissipation if the surface temperature of the case, \(T_{s, c}\), is not to exceed \(85^{\circ} \mathrm{C}\) ? (b) The convection coefficient may be increased by subjecting the plate surface to a forced flow of air. Explore the effect of increasing the coefficient over the range \(4 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Short Answer

Expert verified
The maximum allowable power dissipation for the given conditions can be calculated using the total resistance of the system, derived from the contact, conduction, and convection resistances. With a maximum case surface temperature of \(85^{\circ}\mathrm{C}\), the calculated maximum power dissipation is \(P_{max}\). Additionally, when increasing the convection coefficient \(h\) in the given range, the maximum power dissipation \(P_{max}\) will increase due to the reduced convection resistance in the system. Follow the outlined steps and formulas to determine the specific values for each case.

Step by step solution

01

Determine the resistance of different components

The total resistance of the system can be found by adding the resistance of each component, which includes the contact resistance \(R_{c}\), the conduction resistance through the aluminum plate \(R_{cond}\), and the convection resistance \(R_{conv}\). For contact resistance, we can use the formula: \[R_{c}=\frac{1}{A_c}\] For conduction resistance, we can use the formula: \[R_{cond}=\frac{L}{kA}\] For convection resistance, we can use the formula: \[R_{conv}=\frac{1}{hA}\] Where \(A\) is the surface area of the plate: \[A=W^2\]
02

Calculate total resistance

In this step, substitute the given values in the formulas and calculate the total resistance \(R_{total}\), which is the sum of \(R_{c}\), \(R_{cond}\), and \(R_{conv}\).
03

Find maximum power dissipation

To find the maximum power dissipation, first find the maximum temperature difference \(\Delta T\) that the system can handle, using the given criteria for the surface temperature of the case: \[\Delta T = T_{s,c} - T_{\infty}\] Now, use the total resistance \(R_{total}\) and the maximum temperature difference \(\Delta T\), to find the maximum power dissipation \(P_{max}\) using the formula: \[P_{max}=\frac{(\Delta T)^2}{R_{total}}\] For part (a) calculate \(P_{max}\) using the given values and the total resistance found in step 2.
04

Explore the effect of changing convection coefficient

For part (b), we are asked to explore the effect of increasing the convection coefficient \(h\) over the range \(4 \leq h \leq 200 \mathrm{W/m}^2 \mathrm{K}\). To do this, repeat the steps above for different values of \(h\) to see the effect on the maximum power dissipation \(P_{max}\).

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Key Concepts

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

Thermal Conductivity
Heat transfer is a critical aspect in the design and operation of electronic devices. One of the primary characteristics governing heat transfer in materials is called thermal conductivity, represented by the symbol \(k\). It measures a material鈥檚 ability to conduct heat. The thermal conductivity of a material can greatly affect how efficiently heat is transferred through it and, consequently, how well the device can dissipate unwanted heat.

In our exercise, the aluminum plate has a thermal conductivity of \( k = 240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \), which implies that aluminum is quite effective at conducting heat. This property is advantageous for the heat dissipation of power transistors, which can generate substantial amounts of heat during operation.

The rate at which heat flows through a material can be calculated using Fourier's law of heat conduction, which mathematically can be expressed as \(q = -k \cdot A \cdot \frac{dT}{dx}\), where \(q\) is the heat transfer rate, \(k\) is the thermal conductivity, \(A\) is the cross-sectional area, and \(\frac{dT}{dx}\) is the temperature gradient. Materials with high thermal conductivity are desired in electronic components to ensure that generated heat can be quickly transferred away from sensitive parts to avoid overheating and potential damage.
Convection Coefficient
The convection coefficient, denoted by \(h\), is a measure of the heat transfer rate per unit area and temperature difference between a solid surface and the adjacent fluid. In electronics, transferring heat away from the component to the ambient air often involves convection.

The exercise provides the convection coefficient as \(h=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for natural convection, and it suggests analyzing the effects of increasing this value, which is representative of forced convection through a flow of air over the surface. Increasing the convection coefficient improves the heat transfer rate, which allows for higher power dissipation before reaching the critical temperature.

Exploring the Effects

As outlined in the exercise, increasing the convection coefficient from \(4 \mathrm{~W/m}^2 \mathrm{K}\) to \(200 \mathrm{~W/m}^2 \mathrm{K}\) would enhance heat removal from the aluminum plate surface. This would reduce the temperature difference needed across the plate for a given amount of heat transfer, allowing for higher power dissipation without exceeding the safety temperature limit of the device. This exploration is an excellent example of how thermal management strategies can be adjusted and optimized in the field of electronics cooling.
Power Dissipation
In electronic components, power dissipation refers to the conversion of electrical energy into thermal energy (heat) due to resistance within the component. It鈥檚 essential to manage this dissipated power effectively, as excessive heat can lead to component failure.

In the provided exercise, we calculate the maximum allowable power dissipation based on the various thermal resistances and the maximum surface temperature allowed for the device. Power dissipation is typically given by the formula \( P = I^2R \), but it can also be found using the thermal resistance and temperature difference as shown in the exercise: \(P_{max}=\frac{(\Delta T)^2}{R_{total}}\).

Ensuring Device Safety

Keeping the power dissipation within limits ensures that the transistor's case temperature does not exceed the set threshold. Balancing this with efficient heat transfer methods, such as optimizing thermal conductivity and convection coefficients, is part of effective thermal management. The calculation requires accurate knowledge of the system's thermal properties and an understanding of how these properties interact within the context of the device鈥檚 operational environment. By mastering these concepts, students can design safer and more reliable electronic systems.

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Most popular questions from this chapter

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

The evaporator section of a refrigeration unit consists of thin-walled, 10-mm- diameter tubes through which refrigerant passes at a temperature of \(-18^{\circ} \mathrm{C}\). Air is cooled as it flows over the tubes, maintaining a surface convection coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and is subsequently routed to the refrigerator compartment. (a) For the foregoing conditions and an air temperature of \(-3^{\circ} \mathrm{C}\), what is the rate at which heat is extracted from the air per unit tube length? (b) If the refrigerator's defrost unit malfunctions, frost will slowly accumulate on the outer tube surface. Assess the effect of frost formation on the cooling capacity of a tube for frost layer thicknesses in the range \(0 \leq \delta \leq 4 \mathrm{~mm}\). Frost may be assumed to have a thermal conductivity of \(0.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (c) The refrigerator is disconnected after the defrost unit malfunctions and a 2-mm-thick layer of frost has formed. If the tubes are in ambient air for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and natural convection maintains a convection coefficient of \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long will it take for the frost to melt? The frost may be assumed to have a mass density of \(700 \mathrm{~kg} / \mathrm{m}^{3}\) and a latent heat of fusion of \(334 \mathrm{~kJ} / \mathrm{kg}\).

A 2-mm-diameter electrical wire is insulated by a 2 -mm-thick rubberized sheath \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and the wire/sheath interface is characterized by a thermal contact resistance of \(R_{t, c}^{\prime \prime}=3 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The convection heat transfer coefficient at the outer surface of the sheath is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the temperature of the ambient air is \(20^{\circ} \mathrm{C}\). If the temperature of the insulation may not exceed \(50^{\circ} \mathrm{C}\), what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?

A thin flat plate of length \(L\), thickness \(t\), and width \(W \geqslant L\) is thermally joined to two large heat sinks that are maintained at a temperature \(T_{o}\). The bottom of the plate is well insulated, while the net heat flux to the top surface of the plate is known to have a uniform value of \(q_{o}^{\prime \prime}\) (a) Derive the differential equation that determines the steady-state temperature distribution \(T(x)\) in the plate. (b) Solve the foregoing equation for the temperature distribution, and obtain an expression for the rate of heat transfer from the plate to the heat sinks.
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