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A 10 -cm-high and 20-cm-wide circuit board houses on its surface 100 closely spaced chips, each generating heat at a rate of \(0.12 \mathrm{~W}\) and transferring it by convection and radiation to the surrounding medium at \(40^{\circ} \mathrm{C}\). Heat transfer from the back surface of the board is negligible. If the combined convection and radiation heat transfer coefficient on the surface of the board is \(22 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the average surface temperature of the chips is (a) \(41^{\circ} \mathrm{C}\) (b) \(54^{\circ} \mathrm{C}\) (c) \(67^{\circ} \mathrm{C}\) (d) \(76^{\circ} \mathrm{C}\) (e) \(82^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The combined convection and radiation heat transfer coefficient on the surface of the board is 22 W/m² K and the circuit board is 10 cm high and 20 cm wide. The temperature of the surrounding medium is 40°C. a) 51°C b) 59°C c) 67°C d) 76°C Answer: c) 67°C

Step by step solution

01

Calculate the total heat generation

We are given that each of the 100 chips generates heat at a rate of \(0.12\mathrm{~W}\). Therefore, the total heat generation (\(q\)) for all the chips can be calculated as follows: \(q = N \times Q_{chip}\) where \(N\) is the total number of chips and \(Q_{chip}\) is the heat generation per chip. \(q = 100 \times 0.12\mathrm{~W} = 12\mathrm{~W}\) The total heat generation for all the chips is 12 W.
02

Determine the heat transfer area

The circuit board is 10 cm high and 20 cm wide. To calculate the area for heat transfer, we need to convert these measurements to meters and then multiply them together: \(A = H \times W\) where \(A\) is the heat transfer area, \(H\) is the height, and \(W\) is the width. \(A = 0.1\mathrm{~m} \times 0.2\mathrm{~m} = 0.02\mathrm{~m}^2\) The heat transfer area is 0.02 \(m^2\).
03

Use the heat transfer equation to find the temperature difference

The heat transfer equation for combined convection and radiation is: \(q = hA(T_s - T_{\infty})\) where \(q\) is the total heat generation, \(h\) is the heat transfer coefficient, \(A\) is the heat transfer area, \(T_s\) is the average surface temperature of the chips, and \(T_{\infty}\) is the surrounding medium temperature. We are given that the combined convection and radiation heat transfer coefficient is \(22\mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K}\). We can rearrange the heat transfer equation to find the temperature difference (\(\Delta{T}\)) between the chips and the surrounding medium: \(\Delta{T} = T_s - T_{\infty} = \frac{q}{hA}\) Substituting the given values: \(\Delta{T} = \frac{12\mathrm{~W}}{22\mathrm{~W}/\mathrm{m}^2 \cdot \mathrm{K} \times 0.02\mathrm{~m}^2} = 27.27\mathrm{~K}\) The temperature difference between the chips and the surrounding medium is 27.27 K.
04

Calculate the average surface temperature of the chips

We can now add the temperature difference to the surrounding medium temperature to find the average surface temperature of the chips: \(T_s = T_{\infty} + \Delta{T}\) We are given that the surrounding medium temperature is \(40^{\circ} \mathrm{C}\). \(T_s = 40^{\circ} \mathrm{C} + 27.27\mathrm{~K} \approx 67.27^{\circ}\mathrm{C}\) The average surface temperature of the chips is approximately 67.27 °C. Comparing this to the given options, we can see that the closest answer is: (c) \(67^{\circ} \mathrm{C}\)

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Key Concepts

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

Convection
When we talk about heat transfer, convection is one of the primary methods used to move heat between objects. It's particularly important in the example of our circuit board. Convection involves the transfer of heat through fluid movement—this could be air, a liquid, or any other fluid. Understanding convection is simple with a few key points:
  • Fluid Movement: Heat is transferred by the flow of fluid, such as air, over a surface.
  • Natural vs. Forced Convection: In natural convection, fluid movement is caused by temperature differences, whereas forced convection uses external means like fans to move the fluid.
  • Heat Transfer Coefficient: This is a measure that indicates how effectively heat is transferred from the board to the fluid. In our example, it was combined with radiation in a single coefficient: 22 W/m²·K.

On the circuit board, convection helps carry away heat generated by the chips into the surrounding cooler air, preventing overheating. The combined heat transfer coefficient used in our problem also reflects how efficiently this process happens.
Radiation
Radiation forms another essential part of heat transfer. Unlike convection, it doesn't require a medium to move through. Heat can be transferred through a vacuum or empty space via electromagnetic waves. Here are a few insights into radiation:
  • Electromagnetic Waves: Heat is radiated through infrared waves from one object to another.
  • Direct Transfer: Heat energy moves directly from the hot surface of an object to whatever cooler object or space surrounds it.
  • Surrounding Temperatures Matter: More heat will be radiated as the difference between the object temperature and surroundings increases.

For the circuit board case, radiation contributes to reducing heat by transferring energy away from the hot chips into the room. Because both convection and radiation were combined in the heat transfer coefficient, they collectively help to keep the board cooler.
Circuit Board Heat Management
Managing heat on a circuit board is crucial for ensuring the longevity and functionality of electronic devices. Overheating can lead to failures, so engineers work to keep temperatures in check using heat management strategies. Here's what you should know about managing heat in circuits:
  • Materials and Design: Circuit board materials and chip arrangements are chosen to enhance heat dissipation. For example, materials with higher thermal conductivity are often used.
  • Heatsinks and Fans: These are common tools that improve heat transfer. Heatsinks increase the surface area for heat transfer and fans enhance forced convection.
  • Thermal Interfaces: Proper placement and contact between chips and heat-dissipating components ensure efficient heat flow.
  • Combined Heat Transfer Mechanisms: As demonstrated in the example, using both convection and radiation helps maximize heat transfer efficiency.

By understanding these principles, engineers can create more reliable electronic systems that maintain their performance under various operating conditions. Efficient circuit board heat management is a balance of technology and design, optimizing each component's role in handling heat.

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Most popular questions from this chapter

Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.

A 3-m-internal-diameter spherical tank made of 1 -cm-thick stainless steel is used to store iced water at \(0^{\circ} \mathrm{C}\). The tank is located outdoors at \(25^{\circ} \mathrm{C}\). Assuming the entire steel tank to be at \(0^{\circ} \mathrm{C}\) and thus the thermal resistance of the tank to be negligible, determine \((a)\) the rate of heat transfer to the iced water in the tank and \((b)\) the amount of ice at \(0^{\circ} \mathrm{C}\) that melts during a 24 -hour period. The heat of fusion of water at atmospheric pressure is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\). The emissivity of the outer surface of the tank is \(0.75\), and the convection heat transfer coefficient on the outer surface can be taken to be \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume the average surrounding surface temperature for radiation exchange to be \(15^{\circ} \mathrm{C}\).

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.

The roof of a house consists of a 22-cm-thick (st) concrete slab \((k=2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(15 \mathrm{~m}\) wide and \(20 \mathrm{~m}\) long. The emissivity of the outer surface of the roof is \(0.9\), and the convection heat transfer coefficient on that surface is estimated to be \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The inner surface of the roof is maintained at \(15^{\circ} \mathrm{C}\). On a clear winter night, the ambient air is reported to be at \(10^{\circ} \mathrm{C}\) while the night sky temperature for radiation heat transfer is \(255 \mathrm{~K}\). Considering both radiation and convection heat transfer, determine the outer surface temperature and the rate of heat transfer through the roof. If the house is heated by a furnace burning natural gas with an efficiency of 85 percent, and the unit cost of natural gas is \(\$ 1.20\) / therm ( 1 therm \(=105,500 \mathrm{~kJ}\) of energy content), determine the money lost through the roof that night during a 14-hour period.

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