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Computer memory chips are mounted on a finned metallic mount to protect them from overheating. A \(152 \mathrm{MB}\) memory chip dissipates \(5 \mathrm{~W}\) of heat to air at \(25^{\circ} \mathrm{C}\). If the temperature of this chip is to not exceed \(50^{\circ} \mathrm{C}\), the overall heat transfer coefficient- area product of the finned metal mount must be at least (a) \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (b) \(0.3 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (c) \(0.4 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (d) \(0.5 \mathrm{~W} /{ }^{\circ} \mathrm{C}\) (e) \(0.6 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Answer: The minimum overall heat transfer coefficient-area product (UA) required is 0.2 W/°C.

Step by step solution

01

Identify the data

We are given the following information: - Heat transfer rate, \(Q = 5 \mathrm{~W}\) - Temperature of air, \(T_{air} = 25^{\circ} \mathrm{C}\) - Maximum chip temperature, \(T_{max} = 50^{\circ} \mathrm{C}\) From this data, we can find the overall heat transfer coefficient-area product (UA).
02

Calculate the temperature difference

Determine the temperature difference between the chip and the air by subtracting the air temperature from the maximum chip temperature: \(\Delta T = T_{max} - T_{air} = 50^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C} = 25^{\circ} \mathrm{C}\)
03

Apply the heat transfer formula

Use the heat transfer formula to calculate the overall heat transfer coefficient-area product (UA): \(Q = UA \Delta T\) Rearrange to solve for UA: \(UA = \frac{Q}{\Delta T}\)
04

Calculate the required UA

Insert the given values for \(Q\) and \(\Delta T\) into the formula: \(UA = \frac{5 \mathrm{~W}}{25^{\circ} \mathrm{C}} = 0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\)
05

Find the answer among the options

From the calculation, we find that the required overall heat transfer coefficient-area product is \(0.2 \mathrm{~W} /{ }^{\circ} \mathrm{C}\). This corresponds to option (a) in the choices given.

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Key Concepts

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

Heat Transfer Rate
In the realm of physics, the heat transfer rate is a critical concept defining how quickly heat energy is transferred from one location to another. Specifically, in our given problem, the heat transfer rate (\(Q\)) represents the amount of heat energy a computer memory chip releases into its surroundings per unit of time. The unit used to measure this rate is watts (W), where one watt is equivalent to one joule of energy transferred per second. With a given dissipation of \(5 \text{W}\), the memory chip in the exercise releases energy actively to prevent overheating, a necessity for the stable operation of electronics. This quantitative assessment is foundational in thermal management systems, as it provides the basis for designing effective cooling strategies, such as the finned metallic mount mentioned in the exercise.
Temperature Difference
The temperature difference, often denoted as \(\Delta T\), is a simple but fundamental factor influencing the rate of heat transfer. Within our context, ​\(\Delta T\) is the gradient driving the flow of heat from the hot memory chip to the cooler surrounding air. The larger the temperature difference, the greater the potential for heat to be transferred away from the chip. By calculating the difference between the maximum chip temperature (\(T_{max}\)) and the air temperature (\(T_{air}\)), we established a temperature difference of \(25^\circ C\). This value serves as a crucial component in calculating the overall heat transfer coefficient-area product (UA), as shown in the step-by-step solution, and ensures the chip remains within safe operational temperatures.
Finned Metallic Mount
The finned metallic mount mentioned in the exercise is an engineered solution designed to enhance the dissipation of heat through increased surface area. Fins are added to the mount to create more paths for heat to be transferred to the air, improving the convection process. These fins are typically crafted from metals with high thermal conductivity like aluminum or copper to maximize their effectiveness. The design and structure of the fins are critical; they should be optimized for creating the most surface area without impeding airflow or becoming counterproductive due to added thermal mass. This design principle allows for more efficient heat transfer from electronic components, like memory chips, thereby ensuring performance and longevity.
Thermal Management of Electronics
Effective thermal management of electronics is pivotal in maintaining the reliability and efficiency of electronic devices. The heat generated by electronic components, if not managed properly, can lead to overheating, which may cause a reduction in performance, damage, or even failure. Therefore, understanding and controlling the thermal environment is essential. Various methods are used to dissipate heat, including passive techniques like heat sinks and fins, and active techniques involving fans and liquid cooling systems. The use of a finned metallic mount is a prime example of a passive approach, leveraging the material's thermal properties and design to maintain optimal operating temperatures for electronic components such as memory chips.

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Most popular questions from this chapter

Consider a wall that consists of two layers, \(A\) and \(B\), with the following values: \(k_{A}=0.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{A}=8 \mathrm{~cm}, k_{B}=\) \(0.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L_{B}=5 \mathrm{~cm}\). If the temperature drop across the wall is \(18^{\circ} \mathrm{C}\), the rate of heat transfer through the wall per unit area of the wall is (a) \(180 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(153 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(89.6 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(72 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(51.4 \mathrm{~W} / \mathrm{m}^{2}\)

Hot water at an average temperature of \(53^{\circ} \mathrm{C}\) and an average velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) is flowing through a \(5-\mathrm{m}\) section of a thin-walled hot-water pipe that has an outer diameter of \(2.5 \mathrm{~cm}\). The pipe passes through the center of a \(14-\mathrm{cm}\)-thick wall filled with fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the surfaces of the wall are at \(18^{\circ} \mathrm{C}\), determine \((a)\) the rate of heat transfer from the pipe to the air in the rooms and \((b)\) the temperature drop of the hot water as it flows through this 5 -m-long section of the wall. Answers: \(19.6 \mathrm{~W}, 0.024^{\circ} \mathrm{C}\)

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

Determine the winter \(R\)-value and the \(U\)-factor of a masonry cavity wall that is built around 4-in-thick concrete blocks made of lightweight aggregate. The outside is finished with 4 -in face brick with \(\frac{1}{2}\)-in cement mortar between the bricks and concrete blocks. The inside finish consists of \(\frac{1}{2}\)-in gypsum wallboard separated from the concrete block by \(\frac{3}{4}\)-in-thick (1-in by 3 -in nominal) vertical furring whose center- to-center distance is 16 in. Neither side of the \(\frac{3}{4}\)-in-thick air space between the concrete block and the gypsum board is coated with any reflective film. When determining the \(R\)-value of the air space, the temperature difference across it can be taken to be \(30^{\circ} \mathrm{F}\) with a mean air temperature of \(50^{\circ} \mathrm{F}\). The air space constitutes 80 percent of the heat transmission area, while the vertical furring and similar structures constitute 20 percent.

The walls of a food storage facility are made of a 2 -cm-thick layer of wood \((k=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in contact with a 5 -cm- thick layer of polyurethane foam \((k=0.03 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). If the temperature of the surface of the wood is \(-10^{\circ} \mathrm{C}\) and the temperature of the surface of the polyurethane foam is \(20^{\circ} \mathrm{C}\), the temperature of the surface where the two layers are in contact is (a) \(-7^{\circ} \mathrm{C}\) (b) \(-2^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(8^{\circ} \mathrm{C}\) (e) \(11^{\circ} \mathrm{C}\)

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