/*! 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

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

Consider a surface-mount type transistor on a circuit board whose temperature is maintained at \(35^{\circ} \mathrm{C}\). Air at \(20^{\circ} \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm} \times\) \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm} \times 0.25 \mathrm{~mm}\) and length \(4 \mathrm{~mm}\), 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 radiation, estimate the case temperature when \(150 \mathrm{~mW}\) is dissipated by the transistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal conductivities of the wire leads, air, and conductive paste are \(25,0.0263\), and \(0.12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), 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 temperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a larger 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 \(400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), compute and plot the maximum allowable heat dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

You've experienced convection cooling if you've ever extended your hand out the window of a moving vehicle or into a flowing water stream. With the surface of your hand at a temperature of \(30^{\circ} \mathrm{C}\), determine the convection heat flux for (a) a vehicle speed of \(35 \mathrm{~km} / \mathrm{h}\) in air at \(-5^{\circ} \mathrm{C}\) with a convection coefficient of 40 \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and (b) a velocity of \(0.2 \mathrm{~m} / \mathrm{s}\) in a water stream at \(10^{\circ} \mathrm{C}\) with a convection coefficient of \(900 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Which condition would feel colder? Contrast these results with a heat loss of approximately \(30 \mathrm{~W} / \mathrm{m}^{2}\) under normal room conditions.

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?

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), and the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Neglecting radiation exchange with the surroundings, calculate the temperature of the roof under steadystate conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the same ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airflow conditions over the roof, increasing with increasing air speed. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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