/*! 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 68 A \(16 \mathrm{~mm}\)-square chi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(16 \mathrm{~mm}\)-square chip has sixteen \(2 \mathrm{~mm}\)-diameter, \(15 \mathrm{~mm}\)-long aluminum pin fins in an aligned array at a pitch of \(4 \mathrm{~mm}\). A fan blows \(25^{\circ} \mathrm{C}\) air through the array, giving a heat transfer coefficient of \(110 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If the chip is not to exceed a \(75^{\circ} \mathrm{C}\) operating temperature, what is the allowable power rating of the chip?

Short Answer

Expert verified
The allowable power rating of the chip is the result of the heat dissipation calculation in Step 6.

Step by step solution

01

Identify Given Values

First, we need to identify all the given values in the problem: - Chip area: 16 mm square - Number of pins: 16 - Diameter of each pin: 2 mm - Length of each pin: 15 mm - Pitch of the array: 4 mm - Heat transfer coefficient: 110 W/m²K - Incoming air temperature: 25°C - Maximum operating temperature: 75°C.
02

Convert Units

Convert all the values from millimeters to meters for consistency with the heat transfer coefficient units. For the pins diameter and length: - Diameter: 2 mm = 0.002 m - Length: 15 mm = 0.015 m. For the chip area: - Area: 16 mm square = 0.016 m x 0.016 m = 0.000256 m².
03

Calculate the Total Surface Area of All Pins

Each fin contributes to heat transfer through its lateral surface area. The lateral surface area of a single cylindrical pin is calculated by: \[ A_{pin} = \pi imes d imes L \]Substitute the pin diameter and length to find the surface area of one pin, then multiply by the number of pins:\[ A_{total} = 16 imes (\pi \times 0.002 \, \text{m} \times 0.015 \, \text{m}) \]
04

Calculate the Temperature Difference

The temperature difference between the chip surface and the air is: \[ \Delta T = 75^{\circ}C - 25^{\circ}C = 50^{\circ}C \].
05

Calculate Heat Dissipation Using the Heat Transfer Formula

The heat dissipation per unit time (usable power) can be calculated using the convection heat transfer formula: \[ Q = h \times A_{total} \times \Delta T \]Substitute the known values:\[ Q = 110 \, \text{W/m}^2\,\text{K} \times A_{total} \times 50 \, \text{K} \]
06

Solve for Allowable Power Rating

Plug in the total surface area of the fins from Step 3 into the equation and calculate:1. First, calculate \(A_{total}\).2. Use this \(A_{total}\) to find \(Q\), representing the allowable power rating of the chip.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Convection
Convection is a fundamental method by which heat moves through fluids, such as air or liquids. In convection, warmer areas of a fluid rise, while cooler areas sink, creating a cycle that distributes heat. This process is highly effective in transferring heat over larger distances compared to conduction.
Understanding convection is crucial for heat management, especially in devices like chips, which generate heat during operation.
  • Natural Convection: Caused by natural buoyancy forces due to density differences in the fluid caused by temperature variations.
  • Forced Convection: Occurs when an external source, like a fan, moves the fluid over a surface to enhance the heat transfer rate.
In our problem, forced convection is applied by using a fan to blow air across the pin fins, enhancing the heat transfer capabilities of the chip. The efficiency of convection is often represented by the heat transfer coefficient, denoted as "h," which is crucial in determining the rate of heat transfer during convection.
Heat Dissipation
Heat dissipation is the process by which excess thermal energy is released from a system or component to its surroundings. In electronics, managing heat dissipation is essential to prevent overheating, which can damage components or reduce their lifespan.
Devices like chips need efficient means to dissipate heat to maintain optimal operating conditions. Pin fins serve this purpose by increasing the surface area available for heat transfer.
Heat dissipation in our problem is calculated using the formula: \[ Q = h \times A_{total} \times \Delta T \] where \(Q\) is the heat dissipation capacity, \(A_{total}\) is the total surface area of the pin fins, and \(\Delta T\) is the temperature difference between the chip and the air.
  • Effective heat dissipation ensures that the chip does not exceed its maximum allowable temperature.
  • Components like fans and pin fins work together to enhance the convection process, improving heat dissipation rates.
By understanding how to calculate and improve heat dissipation, engineers can design systems that operate efficiently while prolonging the life of electronic components.
Temperature Difference
Temperature difference is the driving force in heat transfer. It represents how much hotter one object or surface is compared to another. Heat naturally flows from higher-temperature regions to lower-temperature ones.
This phenomenon is described by the formula: \[ \Delta T = T_{surface} - T_{air} \] In the given exercise, the chip surface has a maximum temperature of 75°C, and the incoming air is at 25°C, resulting in a temperature difference of 50°C.
  • Higher temperature differences result in higher rates of heat transfer.
  • Maintaining an optimal temperature difference ensures efficient operation of devices.
In systems where precise temperature control is necessary, such as in a chip, understanding and managing temperature difference is crucial to ensure efficient heat dissipation without surpassing critical temperature limits.
Pin Fins
Pin fins are small, cylindrical structures attached to a surface to improve heat transfer. They extend the surface area over which convection can take place, thus enhancing the cooling efficiency.
Pin fins are used in arrays to provide uniform heat dissipation across a surface.
  • Geometrical Arrangement: More pins result in a larger surface area, increasing the capability of the system to dissipate heat.
  • Spacing: Proper spacing is crucial for optimizing airflow and avoiding interference between fins.
In the provided exercise, pin fins are used on a chip to allow greater heat dissipation by capturing more airflow. The fins' diameter and length influence the effective surface area, which directly impacts the amount of heat that can be removed from the chip. By fully understanding the role and design considerations of pin fins, designers can better manage the thermal profiles of electronic components.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

A \(5 \mathrm{~mm} \times 2 \mathrm{~mm} \times 1 \mathrm{~mm}\)-thick semiconductor laser is mounted on a \(1 \mathrm{~cm}\) cube copper heat sink and enclosed in a Dewar flask. The laser dissipates \(2 \mathrm{~W}\), and a cryogenic refrigeration system maintains the copper block at a nearly uniform temperature of \(90 \mathrm{~K}\). Estimate the top surface temperature of the laser chip for the following models of the dissipation process: (i) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer underneath the top surface of the laser. (ii) The energy is dissipated in a \(10 \mu \mathrm{m}\)-thick layer at the midplane of the chip. (iii) The energy is dissipated uniformly through the chip. Take \(k=170 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the chip, and neglect parasitic heat gains from the Dewar flask.

A \(60 \mathrm{~cm}\)-long, \(3 \mathrm{~cm}\)-diameter AISI 1010 steel rod is welded to a furnace wall and passes through \(20 \mathrm{~cm}\) of insulation before emerging into the surrounding air. The furnace wall is at \(300^{\circ} \mathrm{C}\), and the air temperature is \(20^{\circ} \mathrm{C}\). Estimate the temperature of the bar tip if the heat transfer coefficient between the rod and air is taken to be \(13 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

In a laboratory experiment, a long, \(2 \mathrm{~cm}\)-diameter, cylinder of fissionable material is encased in a \(1 \mathrm{~cm}\)-thick graphite shell. The unit is immersed in a coolant at 330 \(\mathrm{K}\), and the convective heat transfer coefficient on the graphite surface is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). If heat is generated uniformly within the fissionable material at a rate of \(100 \mathrm{MW} / \mathrm{m}^{3}\), determine the temperature at the center-line of the cylinder. Allow for an interfacial conductance between the material and the graphite shell of \(3000 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\). and take the thermal conductivities of the material and graphite as \(4.1 \mathrm{~W} / \mathrm{m} \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m} \mathrm{K}\), respectively.

An electrical current is passed through a \(1 \mathrm{~mm}\)-diameter, \(20 \mathrm{~cm}\)-long copper wire located in an air flow at \(290 \mathrm{~K}\). If the ends of the wire are maintained at \(300 \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(400 \mathrm{~K}\). The convective heat transfer coefficient is estimated to be 20 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). For the copper wire, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\). and an electrical resistance of \(2.2 \times 10^{-2} \Omega / \mathrm{m}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.