/*! 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 26 A square isothermal chip is of w... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 26

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

Short Answer

Expert verified
The maximum allowable chip power for air coolant is 0.35 W, and for the dielectric liquid coolant, it is 5.25 W.

Step by step solution

01

Convection Heat Transfer Equation

The convection heat transfer equation is given by: \(q = hA\Delta T\) Here, q = Heat transfer from the chip (W) h = Convection coefficient (W/m虏路K) A = Exposed chip surface area (m虏) 螖T = Temperature difference between chip and coolant (K)
02

Calculate Surface Area

We are given the width of the chip, w = 5 mm. To find the surface area (A) of the exposed front surface, we need to convert the width to meters and then square it. A = (w / 1000)虏 = (5 / 1000)虏 = 0.000025 m虏
03

Convert Temperature Difference to Kelvin

The temperature difference between the chip and the coolant must be converted to Kelvin, as follows: 螖T = T - T鈭� = 85掳C - 15掳C = 70 K
04

Calculate Maximum Allowable Power for Air Coolant

Now, we are ready to find the maximum allowable power for the air-cooled chip. We have h = 200 W/m虏路K and 螖T = 70 K from previous steps. Plugging the values into the convection heat transfer equation: q_air = h_air * A * 螖T = 200 * 0.000025 * 70 = 0.35 W
05

Calculate Maximum Allowable Power for Dielectric Liquid Coolant

Repeat the calculation for the dielectric liquid coolant, given h = 3000 W/m虏路K: q_liquid = h_liquid * A * 螖T = 3000 * 0.000025 * 70 = 5.25 W We have found the maximum allowable chip power for the two different coolants: For air: 0.35 W For dielectric liquid: 5.25 W

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Key Concepts

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

Convection Coefficient
When discussing convection heat transfer, the convection coefficient plays a crucial role. It is a measure of how effectively heat is transferred between a surface and a fluid flowing over it. The higher the coefficient, the more heat is transferred.
  • In equations, it is usually represented by the symbol \( h \), and its units are \( \text{W/m}^2\cdot\text{K} \).
  • The convection coefficient depends on several factors, such as the type of fluid, its velocity, and the surface geometry.
  • In the given exercise, the convection coefficient \( h \) changes based on the coolant used: 200 \( \text{W/m}^2\cdot\text{K} \) for air and 3000 \( \text{W/m}^2\cdot\text{K} \) for the dielectric liquid.
A higher convection coefficient means more efficient cooling, helping components maintain lower temperatures even when generating significant power. This concept is essential in thermal management, especially in electronics where heat dissipation is critical.
Temperature Difference
The temperature difference \( \Delta T \) is the driving force for heat transfer. It is the difference between the surface temperature and the fluid temperature. In this exercise, it is calculated as follows:\[\Delta T = T - T_{\infty} = 85^\circ\text{C} - 15^\circ\text{C} = 70\,\text{K}\]
  • \( T \) is the maximum allowable chip temperature, given as 85掳C.
  • \( T_{\infty} \) is the temperature of the coolant, which is 15掳C.
The value is then converted into Kelvin because the SI unit system is typically used for such calculations, ensuring compatibility and accuracy when substituting values into physics equations. Understanding \( \Delta T \) is vital as it directly influences the rate of heat transfer based on Newton's Law of Cooling.
Surface Area Calculation
The calculation of the surface area is foundational in determining how much heat can be transferred through convection. The surface area \( A \) of the chip exposed to the coolant is calculated using the chip's width. Here鈥檚 how it鈥檚 determined:
  • The chip鈥檚 width is \( w = 5\,\text{mm} \), which must be converted to meters: \( w = 0.005\,\text{m} \).
  • The area, being square, is \( w^2 \): \[ A = (0.005)^2 = 0.000025\,\text{m}^2 \]
This converts a practical aspect of the problem into a mathematical form that can be easily used in the convection heat transfer equation. Accurately calculating this area is crucial because even small errors can significantly impact the calculated heat transfer, potentially leading to overheating or underperformance in devices like a chip.

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Most popular questions from this chapter

A thin electrical heating element provides a uniform heat flux \(q_{o}^{\prime \prime}\) to the outer surface of a duct through which airflows. The duct wall has a thickness of \(10 \mathrm{~mm}\) and a thermal conductivity of \(20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) At a particular location, the air temperature is \(30^{\circ} \mathrm{C}\) and the convection heat transfer coefficient between the air and inner surface of the duct is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What heat flux \(q_{o}^{\prime \prime}\) is required to maintain the inner surface of the duct at \(T_{i}=85^{\circ} \mathrm{C}\) ? (b) For the conditions of part (a), what is the temperature \(\left(T_{o}\right)\) of the duct surface next to the heater? (c) With \(T_{i}=85^{\circ} \mathrm{C}\), compute and plot \(q_{o}^{\prime \prime}\) and \(T_{o}\) as a function of the air-side convection coefficient \(h\) for the range \(10 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Briefly discuss your results.

An electric resistance heater is embedded in a long cylinder of diameter \(30 \mathrm{~mm}\). When water with a temperature of \(25^{\circ} \mathrm{C}\) and velocity of \(1 \mathrm{~m} / \mathrm{s}\) flows crosswise over the cylinder, the power per unit length required to maintain the surface at a uniform temperature of \(90^{\circ} \mathrm{C}\) is \(28 \mathrm{~kW} / \mathrm{m}\). When air, also at \(25^{\circ} \mathrm{C}\), but with a velocity of \(10 \mathrm{~m} / \mathrm{s}\) is flowing, the power per unit length required to maintain the same surface temperature is \(400 \mathrm{~W} / \mathrm{m}\). Calculate and compare the convection coefficients for the flows of water and air.

The concrete slab of a basement is \(11 \mathrm{~m}\) long, \(8 \mathrm{~m}\) wide, and \(0.20 \mathrm{~m}\) thick. During the winter, temperatures are nominally \(17^{\circ} \mathrm{C}\) and \(10^{\circ} \mathrm{C}\) at the top and bottom surfaces, respectively. If the concrete has a thermal conductivity of \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what is the rate of heat loss through the slab? If the basement is heated by a gas furnace operating at an efficiency of \(\eta_{f}=0.90\) and natural gas is priced at \(C_{g}=\$ 0.02 / \mathrm{MJ}\), what is the daily cost of the heat loss?

A wall has inner and outer surface temperatures of 16 and \(6^{\circ} \mathrm{C}\), respectively. The interior and exterior air temperatures are 20 and \(5^{\circ} \mathrm{C}\), respectively. The inner and outer convection heat transfer coefficients are 5 and \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate the heat flux from the interior air to the wall, from the wall to the exterior air, and from the wall to the interior air. Is the wall under steady-state conditions?

A computer consists of an array of five printed circuit boards (PCBs), each dissipating \(P_{b}=20 \mathrm{~W}\) of power. Cooling of the electronic components on a board is provided by the forced flow of air, equally distributed in passages formed by adjoining boards, and the convection coefficient associated with heat transfer from the components to the air is approximately \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Air enters the computer console at a temperature of \(T_{i}=20^{\circ} \mathrm{C}\), and flow is driven by a fan whose power consumption is \(P_{f}=25 \mathrm{~W}\). (a) If the temperature rise of the airflow, \(\left(T_{o}-T_{i}\right)\), is not to exceed \(15^{\circ} \mathrm{C}\), what is the minimum allowable volumetric flow rate \(\dot{\forall}\) of the air? The density and specific heat of the air may be approximated as \(\rho=1.161\) \(\mathrm{kg} / \mathrm{m}^{3}\) and \(c_{p}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. (b) The component that is most susceptible to thermal failure dissipates \(1 \mathrm{~W} / \mathrm{cm}^{2}\) of surface area. To minimize the potential for thermal failure, where should the component be installed on a PCB? What is its surface temperature at this location?
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