/*! 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 70 A heat sink assembly capable of ... [FREE SOLUTION] | 91Ó°ÊÓ

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A heat sink assembly capable of mounting 36 power transistors may be idealized as a \(15 \mathrm{~cm}\) cube containing four rows of 24 aluminum fins per row, each fin being \(15 \mathrm{~cm}\) wide, \(2.5 \mathrm{~cm}\) high, and \(2 \mathrm{~mm}\) thick. A fan is an integral part of the assembly and blows air at a velocity that gives a heat transfer coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K} .\) If the manufacturer's transistor temperature limit is \(360 \mathrm{~K}\), specify the allowable power dissipation per transistor. The mean air temperature is \(310 \mathrm{~K}\). If the rise in air temperature is limited to \(10 \mathrm{~K}\), specify the required capacity of the fan in \(\mathrm{m}^{3} / \mathrm{min}\).

Short Answer

Expert verified
Each transistor can dissipate 54.33 W of power, and the required fan capacity is 9.726 m³/min.

Step by step solution

01

Calculate Total Heat Dissipation

First, calculate the total heat dissipation allowed for the entire assembly of fins using the formula for convective heat transfer: \[ Q = h imes A imes riangle T \] where:- \(h = 50 \, \text{W/m}^2 \text{K}\) is the heat transfer coefficient,- \(\triangle T = 360 \, \text{K} - 310 \, \text{K} = 50 \, \text{K}\) is the temperature difference,- \(A\) is the total surface area of the fins.
02

Calculate Surface Area of One Fin

To find the total surface area, we first need the surface area of one fin. Each fin has two large faces and two side edges:- The area of one large face is \(15 \, \text{cm} \times 2.5 \, \text{cm} = 37.5 \, \text{cm}^2 = 0.00375 \, \text{m}^2\).- The area of the side edges is \(2 \, \text{\times} \, \text{(15 \, cm + 2.5 \, cm)} \, \times 0.2 \, \text{cm} = 6.5 \, \text{cm}^2 = 0.00065 \, \text{m}^2\).The total area per fin is therefore:\[ A_{\text{fin}} = 2 \times 0.00375 \, \text{m}^2 + 0.00065 \, \text{m}^2 = 0.00815 \, \text{m}^2 \].
03

Calculate Total Surface Area for All Fins

There are 4 rows of 24 fins each, thus 96 fins in total:\[ A_{\text{total}} = 96 \times 0.00815 \, \text{m}^2 = 0.7824 \, \text{m}^2 \].
04

Calculate Total Heat dissipated from Fins

Using the total surface area, compute total heat dissipation:\[ Q = 50 \, \text{W/m}^2 \text{K} \times 0.7824 \, \text{m}^2 \times 50 \, \text{K} = 1956 \, \text{W} \].
05

Calculate Power Dissipation per Transistor

With 36 transistors, the allowable power dissipation per transistor is:\[ P_{\text{transistor}} = \frac{1956 \, \text{W}}{36} = 54.33 \, \text{W} \].
06

Calculate Air Volume Flow Rate

Using the heat capacity of air to find the required capacity of the fan. Assume a specific heat capacity of air \(c_p = 1005 \, \text{J/kg} \text{K}\) and the air density \(\rho = 1.2 \, \text{kg/m}^3\):First, find the needed air flow rate:\[ Q = \dot{m} \times c_p \times \triangle T \rightarrow 1956 \, \text{W} = \dot{m} \times 1005 \, \text{J/kg} \text{K} \times 10 \, \text{K} \].Solve for \(\dot{m}\):\[ \dot{m} = \frac{1956}{10050} \approx 0.1945 \, \text{kg/s} \].Convert mass flow to volumetric flow rate:\[ \dot{V} = \frac{\dot{m}}{\rho} = \frac{0.1945 \, \text{kg/s}}{1.2 \, \text{kg/m}^3} = 0.1621 \, \text{m}^3/s \].Convert to \(\text{m}^3/\text{min}\):\[ \dot{V} = 0.1621 \, \text{m}^3/s \times 60 \, \text{s/min} = 9.726 \, \text{m}^3/\text{min} \].

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Key Concepts

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

Heat Dissipation
Heat dissipation refers to the process by which a system releases heat to its surroundings in order to maintain a constant or reduced temperature. This is critical in electronic devices, like power transistors, which can generate a significant amount of heat during operation. Releasing this heat effectively prevents damage and maintains the efficiency of the device. In practical terms, when we talk about heat dissipation, we're concerned with how much heat energy a system can get rid of, measured in watts (W).
To calculate heat dissipation in the context of a heat sink assembly, we use the convective heat transfer formula:
  • \( Q = h \times A \times \Delta T \)
  • where \( Q \) is the total heat dissipated, \( h \) is the heat transfer coefficient, \( A \) is the surface area, and \( \Delta T \) is the temperature difference between the heat source and the ambient environment.
By applying this formula, we can determine how much heat the fins in a heat sink can dissipate, ensuring that the temperature of the transistors remains within safe limits. Understanding this concept is fundamental for designing systems that need to manage heat effectively and helping prevent overheating.
Power Transistors
Power transistors are solid-state devices used to control large electrical power flows, effectively functioning like a switch or amplifier for high-power applications. Due to their operation, power transistors generate significant heat, which must be managed to avert thermal overload. They are essential in various fields, from audio amplifiers to power converters.
The amount of heat they generate is directly linked to their power dissipation, which indicates how much electrical power is converted into heat. Each transistor has a maximum safe operating temperature, exceeding which can lead to failure or damage of the component. It's crucial, therefore, to calculate allowable power dissipation to design an appropriate cooling system and ensure safe and efficient performance.
With this in mind, consider a heat sink assembly designed to accommodate multiple power transistors. Each transistor's share of the total power dissipation needs to be calculated, taking into account all constantly-draining sources of heat. This proper balance ensures that none of them overheats while maintaining optimal operation.
Convective Heat Transfer
Convective heat transfer is a mechanism of heat dissipation that involves the movement of air or liquid across the surface to remove heat from it. In heats sinks and cooling systems, convective heat transfer is crucial. It is how heat travels from the surface of a heat sink to the air, with the air absorbing and carrying the heat away.
The efficacy of convective heat transfer heavily relies on the heat transfer coefficient, denoted as \( h \). This coefficient is influenced by factors such as fluid flow velocity and properties of the fluid, such as density and viscosity. To enhance convective heat transfer, you can increase the fluid speed flowing over the surface, this can often be done using a fan or by adjusting the design of the fins to create more turbulence.
By understanding and optimizing these relationships, engineers can design more efficient cooling systems for devices like power transistors. Such considerations are crucial in ensuring components do not exceed temperature thresholds, consequently prolonging their life and maintaining performance efficiency.
Surface Area Calculation
Calculating the surface area of cooling components, like fins in a heat sink, is crucial when designing cooling solutions for electronic devices. More surface area typically means better heat dissipation, as there is more material available to absorb and spread out the heat.
For a fin heat sink, each fin typically consists of multiple surfaces: large faces and side edges, all contributing to the total effective surface area. The more fins there are, and the bigger each fin, the more surface area is available. However, the placement and arrangement play a role too. For instance, a typical fin is calculated by determining the area of:
  • The large face of the fin (length \( \times \) height).
  • The fin's side edges, which are calculated by considering the perimeter of the fin's profile multiplied by its thickness.
Summing up the area of all these surfaces for every fin in a heat sink gives the total surface area, essential for determining how effectively a heat sink can dissipate heat. This step is a foundational part of the mathematical process for optimizing thermal management solutions in electronic assemblies.

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Most popular questions from this chapter

Consider steady conduction across a cylindrical or spherical shell. It is sometimes convenient to express the heat flow as $$ \dot{Q}=\frac{k A_{\mathrm{eff}}\left(T_{1}-T_{2}\right)}{r_{2}-r_{1}} $$ that is, in the same form as for a plane slab where \(A_{\text {eff }}\) is an effective crosssectional area. (i) Determine \(A_{\text {eff }}\) for a cylindrical shell. (ii) Repeat for a spherical shell. (iii) If the arithmetic mean area \(A_{m}=\frac{1}{2}\left(A_{1}+A_{2}\right)\) is used instead of \(A_{\text {eff }}\), determine the error in heat flow for \(r_{2} / r_{1}=1.5,3\), and 5 .

Liquid oxygen at \(90 \mathrm{~K}\) flows inside a \(3 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 2 \mathrm{~mm}\)-wall-thickness, AISI 303 stainless steel tube. The tube is insulated with \(3 \mathrm{~cm}\)-thick fiberglass insulation of thermal conductivity \(0.021 \mathrm{~W} / \mathrm{m} \mathrm{K}\). Will atmospheric water vapor condense on the outside of the insulation when the ambient air temperature is \(300 \mathrm{~K}\) and the dewpoint temperature is \(285 \mathrm{~K}\) ? Take the outside convective and radiative heat transfer coefficients as \(5 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

In order to prevent fogging, the \(3 \mathrm{~mm}\)-thick rear window of an automobile has a transparent film electrical heater bonded to the inside of the glass. During a test, \(200 \mathrm{~W}\) are dissipated in a \(0.567 \mathrm{~m}^{2}\) area of window when the inside and outside air temperatures are \(22^{\circ} \mathrm{C}\) and \(1^{\circ} \mathrm{C}\), and the inside and outside heat transfer coefficients are \(9 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and \(22 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively. Determine the temperature of the inside surface of the window.

The end of a soldering iron consists of a \(4 \mathrm{~mm}\)-diameter copper rod, \(5 \mathrm{~cm}\) long. If the tip must operate at \(350^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\), determine the base temperature and heat flow. The heat transfer coefficient from the rod to the air is estimated to be about \(10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). Take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the copper.

Calcium silicate has replaced asbestos as the preferred insulation for steam lines in power plants. Consider a \(40 \mathrm{~cm}-\mathrm{O} . \mathrm{D} ., 4 \mathrm{~cm}\)-wall-thickness steel steam line insulated with a \(12 \mathrm{~cm}\) thickness of calcium silicate. The insulation is protected from damage by an aluminum sheet lagging that is \(2.5 \mathrm{~mm}\) thick. The steam temperature is \(565^{\circ} \mathrm{C}\) and ambient air temperature in the power plant is \(26^{\circ} \mathrm{C}\). The inside convective resistance is negligible, the outside convective heat transfer coefficient can be taken as \(11 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and the emittance of the aluminum is \(0.1\). Calculate the rate of heat loss per meter. The thermal conductivity of the pipe wall is \(40 \mathrm{~W} / \mathrm{m} \mathrm{K}\), and the table gives values for Calsilite calcium silicate insulation blocks. $$ \begin{array}{l|ccccc} \mathrm{T}, \mathrm{K} & 500 & 600 & 700 & 800 & 900 \\ \hline \mathrm{k}, \mathrm{W} / \mathrm{m} \mathrm{K} & 0.074 & 0.096 & 0.142 & 0.211 & 0.303 \end{array} $$

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