/*! 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 computer consists of an array ... [FREE SOLUTION] | 91Ó°ÊÓ

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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?

Short Answer

Expert verified
(a) The minimum allowable volumetric flow rate is \(0.022\, \mathrm{m^3/s}\). (b) The heat-sensitive component should be installed where the temperature of the airstream is lowest, which is near the air inlet. Its surface temperature at this location is \(70^{\circ} C\).

Step by step solution

01

Identify the governing equations

To solve this problem, we need two equations: 1. Energy conservation - the amount of heat transferred from the PCBs is equal to the heat that goes into increasing the temperature of the cooling air. 2. Newton's law of cooling - this gives us the relation between heat transfer rate and temperature difference between the PCB surface and the air.
02

Apply the energy conservation equation

The total power dissipated in the PCBs is \(P = n P_{b}\), where \(n\) is the number of PCBs. The total power transferred to the air is given by \(Q = \rho \dot{V} c_{p} (T_{o} - T_{i})\), which we want to keep less than or equal to the given temperature rise. Equating the heat dissipation from the PCBs to the heat absorbed by the air, we get: \[nP_{b} = \rho \dot{V} c_{p} (T_{o} - T_{i})\]
03

Calculate the minimum flow rate

First, let's substitute the given values: \(n = 5\), \(P_{b} = 20 \thinspace W\), and \(T_{o} - T_{i} = 15^{\circ} C\). Now we can plug in the values for \(\rho\) and \(c_{p}\): \[5(20) = 1.161 \thinspace \dot{V} \thinspace (1007) \thinspace (15)\] Solving for \(\dot{V}\), we get: \[\dot{V} = \frac{5(20)}{1.161(1007)(15)} = 0.022\thinspace \mathrm{m^{3}/s}\] So, the minimum allowable volumetric flow rate is \(0.022 \thinspace \mathrm{m^{3}/s}\).
04

Apply Newton's law of cooling

Newton's law of cooling states that the heat transfer rate \(q\) is given by: \[q = hA(T_{s} - T_{air})\] where \(q\) = heat transfer rate (power), \(h\) = convection coefficient, \(A\) = area of the heat-dissipation surface, \(T_{s}\) = surface temperature of the heat-dissipation surface, and \(T_{air}\) = temperature of the airstream. We want the heat-sensitive component to dissipate 1 W/cm², but we need to express the heat dissipation in SI units. So, \(q = 10,000 \thinspace W/m^2\).
05

Calculate the surface temperature of sensitive component

To minimize the potential for thermal failure, we want the sensitive component to be placed where the airstream is coolest, so \(T_{air} = T_{i} = 20^{\circ} C\). Let's rearrange Newton's equation to solve for the surface temperature \(T_s\): \[T_{s} = T_{air} + \frac{q}{hA}\] However, we still need the relation between \(q\) and \(A\) for the given problem; \(q = 10,000 \thinspace A \thinspace W/m^2\). Substituting this relation and the given values into the equation for \(T_s\), we get: \[T_{s} = 20 + \frac{10,000 \thinspace A}{200 \thinspace A}\] Solving for \(T_s\), we find that the surface temperature of the sensitive component is \(70^{\circ} C\). In summary: (a) The minimum allowable volumetric flow rate is \(0.022\, \mathrm{m^3/s}\). (b) The heat-sensitive component should be installed where the temperature of the airstream is lowest, which is near the air inlet. Its surface temperature at this location is \(70^{\circ} C\).

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Key Concepts

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

Printed Circuit Boards (PCBs)
Printed Circuit Boards (PCBs) are essential components in most electronic devices. They serve as a platform to support and connect electronic components, such as resistors, transistors, and integrated circuits, through conductive pathways. PCBs are crucial in managing heat generated by these components during operation.
- Heat dissipation is a key challenge in PCB design. As electronic components work, they dissipate heat that must be managed to prevent damage or malfunction. - PCBs are often designed with thermal management solutions in mind, like the use of thermal vias and heat spreaders, to help distribute and dissipate heat effectively.
Understanding the relationship between heat generation and temperature rise on a PCB helps in designing systems that can operate efficiently while maintaining optimal performance. Designers must consider the power dissipation of each component and how effectively the PCB can manage this heat to ensure reliability.
Energy Conservation in Heat Transfer
Energy conservation in heat transfer is a fundamental concept that helps us understand how energy flows within systems. In electronic cooling, it's important to know how the heat generated by electronic components is transferred and eventually removed by the cooling medium, typically air.
- The principle of energy conservation states that energy cannot be created or destroyed, but it can change forms or be transferred from one place to another.- In our context, the energy (or heat) produced by the PCBs is absorbed by the air flowing through passages between the boards. This leads to an increase in the air's temperature.
When managing heat in electronics, we calculate the heat absorbed by the cooling medium using the formula:\[ Q = \rho \dot{V} c_{p} (T_o - T_i) \]Here, \( Q \) is the total heat transferred, \( \rho \) is the density of air, \( \dot{V} \) is the volumetric flow rate of air, \( c_{p} \) is the specific heat of air, and \( T_o - T_i \) is the temperature rise of the air. This equation shows that by controlling airflow, we can manage the temperature rise within the system, ensuring components stay within safe operating temperatures.
Newton's Law of Cooling
Newton's Law of Cooling provides an understanding of how quickly an object transfers heat to its surroundings. This is particularly relevant in electronic cooling, where maintaining a safe temperature for components is crucial to preventing thermal failure.
- According to Newton's Law, the rate of heat transfer \( q \) from an object is proportional to the difference in temperature between the object and its surroundings, expressed as:\[ q = hA(T_s - T_{air}) \]where - \( q \) is the heat transfer rate, - \( h \) is the convection heat transfer coefficient, - \( A \) is the surface area of the object, - \( T_s \) is the surface temperature of the object, - \( T_{air} \) is the temperature of the surrounding air.
In the context of our problem, we need to ensure that sensitive components dissipate heat efficiently. Most often, they are positioned where air is coolest, minimizing \( T_s - T_{air} \) and thus reducing the risk of overheating. By calculating the necessary conditions for optimal heat dissipation, designers can strategically place components to enhance overall cooling efficiency.

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Most popular questions from this chapter

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A solar flux of \(700 \mathrm{~W} / \mathrm{m}^{2}\) is incident on a flat-plate solar collector used to heat water. The area of the collector is \(3 \mathrm{~m}^{2}\), and \(90 \%\) of the solar radiation passes through the cover glass and is absorbed by the absorber plate. The remaining \(10 \%\) is reflected away from the collector. Water flows through the tube passages on the back side of the absorber plate and is heated from an inlet temperature \(T_{i}\) to an outlet temperature \(T_{o}\). The cover glass, operating at a temperature of \(30^{\circ} \mathrm{C}\), has an emissivity of \(0.94\) and experiences radiation exchange with the sky at \(-10^{\circ} \mathrm{C}\). The convection coefficient between the cover glass and the ambient air at \(25^{\circ} \mathrm{C}\) is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Perform an overall energy balance on the collector to obtain an expression for the rate at which useful heat is collected per unit area of the collector, \(q_{11}^{\prime \prime}\). Determine the value of \(q_{u r^{\prime \prime}}\). (b) Calculate the temperature rise of the water, \(T_{o}-T_{i}\), if the flow rate is \(0.01 \mathrm{~kg} / \mathrm{s}\). Assume the specific heat of the water to be \(4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). (c) The collector efficiency \(\eta\) is defined as the ratio of the useful heat collected to the rate at which solar energy is incident on the collector. What is the value of \(\eta\) ?

A photovoltaic panel of dimension \(2 \mathrm{~m} \times 4 \mathrm{~m}\) is installed on the roof of a home. The panel is irradiated with a solar flux of \(G_{S}=700 \mathrm{~W} / \mathrm{m}^{2}\), oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is \(\alpha_{S}=0.83\), and the efficiency of conversion of the absorbed flux to electrical power is \(\eta=P / \alpha_{S} G_{S} A=0.553-0.001 \mathrm{~K}^{-1} T_{p}\), where \(T_{p}\) is the panel temperature expressed in kelvins and \(A\) is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which \(T_{\text {sur }}=T_{\infty}=35^{\circ} \mathrm{C}\), \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and (b) a breezy winter day, for which \(T_{\text {sur }}=T_{\infty}=-15^{\circ} \mathrm{C}, h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The panel emissivity is \(\varepsilon=0.90\).

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 \(k_{g}=\) \(1.4 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). 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 \(1 \mathrm{~m} \times 2 \mathrm{~m}\) window? The themal conductivity of air is \(k_{a}=0.024 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\).

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?

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