/*! 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

In Problem 3.48, the electrical power required to maintain the heater at \(T_{o}=25^{\circ} \mathrm{C}\) depends on the thermal conductivity of the wall material \(k\), the thermal contact resistance \(R_{t, c}^{\prime}\) and the convection coefficient \(h\). Compute and plot the separate effect of changes in \(k\) \((1 \leq k \leq 200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}), \quad R_{t, c}^{\prime} \quad\left(0 \leq R_{t, c}^{\prime} \leq 0.1 \mathrm{~m} \cdot \mathrm{K} / \mathrm{W}\right)\), and \(h\left(10 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) on the total heater power requirement, as well as the rate of heat transfer to the inner surface of the tube and to the fluid.

The fin array of Problem \(3.142\) is commonly found in compact heat exchangers, whose function is to provide a large surface area per unit volume in transferring heat from one fluid to another. Consider conditions for which the second fluid maintains equivalent temperatures at the parallel plates, \(T_{o}=T_{L}\), thereby establishing symmetry about the midplane of the fin array. The heat exchanger is \(1 \mathrm{~m}\) long in the direction of the flow of air (first fluid) and \(1 \mathrm{~m}\) wide in a direction normal to both the airflow and the fin surfaces. The length of the fin passages between adjoining parallel plates is \(L=8 \mathrm{~mm}\), whereas the fin thermal conductivity and convection coefficient are \(k=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (aluminum) and \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. (a) If the fin thickness and pitch are \(t=1 \mathrm{~mm}\) and \(S=4 \mathrm{~mm}\), respectively, what is the value of the thermal resistance \(R_{t, o}\) for a one-half section of the fin array? (b) Subject to the constraints that the fin thickness and pitch may not be less than \(0.5\) and \(3 \mathrm{~mm}\), respectively, assess the effect of changes in \(t\) and \(S\).

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

A high-temperature, gas-cooled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel element \((k \approx 57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encased in graphite \((k \approx 3\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{\infty}=600 \mathrm{~K}\) and the convection coefficient at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly generated in the fuel element at a rate \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(\dot{q}\) ?

The air inside a chamber at \(T_{\infty, i}=50^{\circ} \mathrm{C}\) is heated convectively with \(h_{i}=20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) by a 200 -mm-thick wall having a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and a uniform heat generation of \(1000 \mathrm{~W} / \mathrm{m}^{3}\). To prevent any heat generated within the wall from being lost to the outside of the chamber at \(T_{\infty, o}=25^{\circ} \mathrm{C}\) with \(h_{o}=5\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\), a very thin electrical strip heater is placed on the outer wall to provide a uniform heat flux, \(q_{\sigma^{\prime}}\) (a) Sketch the temperature distribution in the wall on \(T-x\) coordinates for the condition where no heat generated within the wall is lost to the outside of the chamber. (b) What are the temperatures at the wall boundaries, \(T(0)\) and \(T(L)\), for the conditions of part (a)? (c) Determine the value of \(q_{o}^{\prime \prime}\) that must be supplied by the strip heater so that all heat generated within the wall is transferred to the inside of the chamber. (d) If the heat generation in the wall were switched off while the heat flux to the strip heater remained constant, what would be the steady-state temperature, \(T(0)\), of the outer wall surface?
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