/*! 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 16 A logic chip used in a computer ... [FREE SOLUTION] | 91Ó°ÊÓ

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A logic chip used in a computer dissipates \(3 \mathrm{~W}\) of power in an environment at \(120^{\circ} \mathrm{F}\), and has a heat transfer surface area of \(0.08 \mathrm{in}^{2}\). Assuming the heat transfer from the surface to be uniform, determine \((a)\) the amount of heat this chip dissipates during an eight-hour work day, in \(\mathrm{kWh}\), and \((b)\) the heat flux on the surface of the chip, in W/in \({ }^{2}\).

Short Answer

Expert verified
Answer: (a) The amount of heat dissipated during an eight-hour workday is 0.024 kWh. (b) The heat flux on the surface of the logic chip is 37.5 W/in².

Step by step solution

01

(Step 1: Calculate Total Energy Dissipated in Joules)

First, we need to find the total energy dissipated during an eight-hour workday. The energy is given by the product of power and time. Power = 3 W Time = 8 hours Let's convert the time from hours to seconds: Time in seconds = 8 hours * 3600 seconds/hour = 28800 seconds Now, let's compute the energy: Energy (Joules) = Power (W) * Time (seconds) Energy = 3 W * 28800 s
02

(Step 2: Convert Energy From Joules to kWh)

To convert the energy from Joules to kilowatt-hours, use the following conversion factor: 1 kWh = 3.6 * 10^6 J. So, Energy (kWh) = Energy (J) / (3.6 * 10^6)
03

(Step 3: Calculate Heat Flux)

The heat flux is the amount of heat transferred per unit surface area and can be found by dividing the power by the surface area of the chip. Surface area = 0.08 in² Heat flux (W/in²) = Power (W) / Surface area (in²) Now we can plug in the given values and compute the energy in kWh and the heat flux. Solution:
04

(Step 1: Calculate Total Energy Dissipated in Joules)

Energy = 3 W * 28800 s Energy = 86400 J
05

(Step 2: Convert Energy From Joules to kWh)

Energy (kWh) = 86400 J / (3.6 * 10^6) Energy (kWh) = 0.024 kWh
06

(Step 3: Calculate Heat Flux)

Heat flux (W/in²) = 3 W / 0.08 in² Heat flux (W/in²) = 37.5 W/in² So, the amount of heat this chip dissipates during an eight-hour workday is (a) 0.024 kWh, and (b) the heat flux on the surface of the chip is 37.5 W/in².

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Key Concepts

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

Heat Dissipation in Electronics
Understanding how electronics shed excess heat, or 'heat dissipation', is vital to maintaining the longevity and performance of electronic devices. Whether we're talking about a simple logic chip in a computer or an advanced microprocessor, managing the thermal output is crucial. In the given exercise, a logic chip dissipates 3 Watts of power while operating in an environment at 120 degrees Fahrenheit.

The heat dissipation process involves the conversion of the electrical energy, used by the chip to perform calculations, into thermal energy. This energy must be moved away from the electronic components to prevent overheating. Materials with good thermal conductivity, such as copper or aluminum, are often used for heat sinks to help distribute and transfer this excess heat to the surrounding environment.

The calculation of total energy dissipated by the chip during an eight-hour work day involves multiplying the continuous power output (3W) by the duration (8 hours), ultimately expressing energy in kilowatt-hours (kWh). An understanding of these principles allows for the selection of appropriate cooling systems such as fans, heat sinks, or even liquid cooling solutions to effectively manage heat within electronic systems.
Understanding Energy Conversion in Electronics
Energy conversion is the process of changing one form of energy into another. In the context of electronics, this typically refers to the conversion of electrical energy into thermal energy. The logic chip in our exercise is a clear example, converting the power it uses for computations into heat that has to be managed. The power rating of a device, measured in watts, tells us the rate at which this energy conversion happens.

For our case, the chip consumes 3 watts, translating to 3 joules per second, as a watt is a joule per second. Over an eight-hour period, the energy conversion can be tallied up in joules and then converted into a more practical unit such as kilowatt-hours, a standard billing unit for energy. It's these conversions that allow engineers to track and manage energy efficiency and heat production in electronic components. Efficient energy conversion is essential, not just for reducing waste heat, but also for lowering energy consumption and maximizing the battery life in portable devices.
Heat Flux Calculation
When we refer to 'heat flux', we're talking about the rate of heat energy transferring through a given surface area. It quantifies how much heat is passing through a particular area, such as the surface of our logic chip, and it's a crucial concept in thermal management of electronics.

The calculation of heat flux is relatively straightforward but fundamental in understanding how well a component is handling the thermal load. By measuring the heat transfer per unit area, we can assess the efficiency of heat dissipation mechanisms and ensure that they can keep up with the heat produced. In our exercise, the heat flux is found by dividing the power, 3 Watts, by the surface area of the chip, 0.08 inches squared, resulting in 37.5 W/in².

Example in Practice

Engineers use such calculations to design cooling solutions that can match the heat flux, ensuring that components operate within safe temperature ranges. This also helps in designing components that will not overheat even when they are densely packed, as is the case in many of today's electronic devices like smartphones and laptops.

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Most popular questions from this chapter

Can a medium involve \((a)\) conduction and convection, (b) conduction and radiation, or \((c)\) convection and radiation simultaneously? Give examples for the "yes" answers.

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick \(2-\mathrm{m} \times 2-\mathrm{m}\) window glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were \(1 \mathrm{~cm}\) thick?

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

It is well-known that at the same outdoor air temperature a person is cooled at a faster rate under windy conditions than under calm conditions due to the higher convection heat transfer coefficients associated with windy air. The phrase wind chill is used to relate the rate of heat loss from people under windy conditions to an equivalent air temperature for calm conditions (considered to be a wind or walking speed of \(3 \mathrm{mph}\) or \(5 \mathrm{~km} / \mathrm{h})\). The hypothetical wind chill temperature (WCT), called the wind chill temperature index (WCTI), is an equivalent air temperature equal to the air temperature needed to produce the same cooling effect under calm conditions. A 2003 report on wind chill temperature by the U.S. National Weather Service gives the WCTI in metric units as WCTI \(\left({ }^{\circ} \mathrm{C}\right)=13.12+0.6215 T-11.37 V^{0.16}+0.3965 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{C}\) and \(V\) the wind speed in \(\mathrm{km} / \mathrm{h}\) at \(10 \mathrm{~m}\) elevation. Show that this relation can be expressed in English units as WCTI \(\left({ }^{\circ} \mathrm{F}\right)=35.74+0.6215 T-35.75 V^{0.16}+0.4275 T V^{0.16}\) where \(T\) is the air temperature in \({ }^{\circ} \mathrm{F}\) and \(V\) the wind speed in \(\mathrm{mph}\) at \(33 \mathrm{ft}\) elevation. Also, prepare a table for WCTI for air temperatures ranging from 10 to \(-60^{\circ} \mathrm{C}\) and wind speeds ranging from 10 to \(80 \mathrm{~km} / \mathrm{h}\). Comment on the magnitude of the cooling effect of the wind and the danger of frostbite.

Consider two houses that are identical, except that the walls are built using bricks in one house, and wood in the other. If the walls of the brick house are twice as thick, which house do you think will be more energy efficient?

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