/*! 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 19 The caise of a power transistor,... [FREE SOLUTION] | 91影视

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The caise of a power transistor, which is of length \(L=10\) num and diameter \(D=12 \mathrm{~mm}\), is cooled by an air stream of temperature \(T_{\mathrm{m}}=25^{\circ} \mathrm{C}\). Lnder conditions for which the air maintains an average convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on the surfuce of the case, what is the maximum allowable power dissipation if the surface temperature is not to exceed \(85^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The maximum allowable power dissipation is 2.262 W.

Step by step solution

01

Understand the Problem

We need to calculate the maximum allowable power dissipation for a transistor case. It is cooled by an air stream, and we know specific dimensions, temperatures, and the convection heat transfer coefficient.
02

Gather Given Information

Length of the transistor case, \(L = 10\) mm (converted to 0.01 m), diameter \(D = 12\) mm (converted to 0.012 m). Air temperature \(T_m = 25^{\circ}C\). Maximum surface temperature \(T_s = 85^{\circ}C\). Convection coefficient \(h = 100\, \text{W/m}^2 \cdot \text{K}\).
03

Calculate the Surface Area

The surface area \(A\) of the cylinder (neglecting the ends) is given by \(A = \pi D L\). Substituting \(D = 0.012\, \text{m}\) and \(L = 0.01\, \text{m}\), we get \[ A = \pi \times 0.012 \times 0.01 = 3.77 \times 10^{-4} \text{ m}^2. \]
04

Apply the Heat Transfer Equation

Using the formula for convection heat transfer: \(Q = hA(T_s - T_m)\), where \(Q\) is the power dissipation. Substitute out values: \(h = 100\), \(A = 3.77 \times 10^{-4}\), \(T_s = 85^{\circ}C\), and \(T_m = 25^{\circ}C\).
05

Compute Maximum Power Dissipation

Calculate the allowable power dissipation: \[ Q = 100 \times 3.77 \times 10^{-4} \times (85 - 25) = 2.262\, \text{W}. \] This is the maximum power dissipation that ensures the surface temperature does not exceed the given limit.

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Key Concepts

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

Power Dissipation
Power dissipation describes the process of an electronic component, like a transistor, emitting heat due to electrical operation. The heat generated needs to be managed effectively to prevent damage. In this exercise, we calculate the maximum power a transistor can lose as heat, ensuring the surface temperature remains safe. This is critical in electronics because excessive heat can lead to failure.

When calculating power dissipation, we consider several factors:
  • The heat transfer coefficient, which indicates how well heat is transferred from the surface to the surrounding air.
  • The temperature difference between the component surface and the ambient air.
  • The surface area, as a larger area allows more heat to escape.
In our case, using the formula for convection heat transfer:\[ Q = hA(T_s - T_m) \]where:
  • \(Q\) is the power dissipation (W).
  • \(h\) is the convection coefficient (100 W/m虏路K).
  • \(A\) is the surface area (3.77 脳 10鈦烩伌 m虏).
  • \(T_s\) and \(T_m\) are the surface and ambient temperatures respectively.
By inputting these values, we accurately find that the maximum safe power dissipation is 2.262 W. This ensures the transistor doesn't overheat.
Thermal Conductivity
Thermal conductivity is a material's inherent ability to conduct heat. Although not directly part of the calculation, the principle still plays a crucial role in understanding heat dissipation processes in electronics. A material with high thermal conductivity will quickly spread the generated heat to the surface, where it can be removed through convection.

For a power transistor, understanding its thermal properties can help you design more effective cooling solutions. In our example, although we focus on convection, keep in mind that:
  • Internal heat spread is vital before it reaches the outer surface.
  • Different materials have varying thermal conductivities, influencing overall performance.
High thermal conductivity materials allow for more uniform surface temperatures, enhancing cooling efficiency. By choosing the right materials and heat management strategies, devices remain within safe operating temperatures, improving reliability.
Surface Area Calculation
Surface area calculation is essential for evaluating how efficiently a component can dissipate heat. Greater surface area generally means better heat dissipation, as more space allows more heat to escape to the surroundings. For our power transistor, we calculated the surface area since the geometry plays a critical role.

Given that the transistor is cylindrical, we focus on its side area, neglecting the ends since they're much smaller and contribute less to heat transfer:\[ A = \pi D L \]where:
  • \(D\) is the diameter (0.012 m).
  • \(L\) is the length (0.01 m).
Substituting these values, we found the surface area to be 3.77 脳 10鈦烩伌 m虏. Ensuring accurate measurements and calculations ensures efficient design and functioning. This step is crucial because any miscalculation can lead to incorrect power dissipation results, affecting device safety.

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Most popular questions from this chapter

A spherical interplanetary probe of \(0.5\) - m diameter contains electronics that dissipate \(150 \mathrm{~W}\), If the probe surface has an emissivity of \(0.8\) and the probe does not receive radiation from other surfaces, as, for example, from the sun, what is its surface temperature?

A vacuum system, as used in sputiering electrically conducting thin films on microcircuits, is comprised of a buseplate maintained by an electrical heater at \(300 \mathrm{~K}\) and a shroud within the enclosure maintained at \(77 \mathrm{~K}\) by a liquid-nitrogen coolant loop. The circular baseplate, insulated on the lower side, is \(0.3 \mathrm{~m}\) in diameter and has an emissivity of \(0.25\). (a) How much clectrical power must be provided to the baseplate heater? (b) At what rate must liquid nitrogen be supplied to the shiroud if its heat of vaporization is \(125 \mathrm{~kJ} / \mathrm{kg}\) ? (c) To reduce the liquid-nitrogen consumption, it is proposed to bond a thin sheet of aluminura foil \((\varepsilon=0.09)\) to the baseplate. Will this have the desfred effect?

Consider a surface-mount type transistor on a circuit board whose temperature is muintained at \(35^{\circ} \mathrm{C}\). Air at \(207 \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm}\) by \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm}\) by \(0.25 \mathrm{~mm}\) and length 4 nim, conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting rudiation, eximate the case temperature uhen \(150 \mathrm{~mW}\) is dissipoted by the thansistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal condoctivitios of the wire leadk, air, and cunductive paste are 25, \(0.0263\), and \(0.12\) W/m - \(\mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case lemperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a lager convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m}+\mathrm{K}\), compute and plot the maximum allowable hean dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

The heat flux through a wood slab \(50 \mathrm{~mm}\) thick, whose inner and outer surface temperatures are 40 and \(20^{\circ} \mathrm{C}\), respectively, has been determined to be \(40 \mathrm{~W} / \mathrm{m}^{2}\). What is the thermal conductivity of the wood?

A glass window of width \(W=1 \mathrm{~m}\) and height \(H=2 \mathrm{~m}\) is \(5 \mathrm{~mm}\) thick and has a thermal conductivity of \(k_{e}=\) \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the inner and outer surface temperatures of the glass are \(15^{\circ} \mathrm{C}\) and \(-20^{\circ} \mathrm{C}\), respectively, on a cold winter day, what is the rate of heat loss through the glass? To reduce heat loss through windows, it is customary to use a double pane construction in which adjoining. panes are separated by an air space if the spacing is \(10 \mathrm{~mm}\) and the glass surfaces in contact with the air have temperatures of \(10^{\circ} \mathrm{C}\) and \(-15^{\circ} \mathrm{C}\), what is the rate of heat loss from a \(I \mathrm{~m} \times 2 \mathrm{~m}\) window? The thermal conductivity of air is \(k_{t}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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