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Chapter 4: Problem 38

An electronic device dissipating \(20 \mathrm{~W}\) has a mass of \(20 \mathrm{~g}\), a specific heat of \(850 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and a surface area of \(4 \mathrm{~cm}^{2}\). The device is lightly used, and it is on for \(5 \mathrm{~min}\) and then off for several hours, during which it cools to the ambient temperature of \(25^{\circ} \mathrm{C}\). Taking the heat transfer coefficient to be \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature of the device at the end of the 5 -min operating period. What would your answer be if the device were attached to an aluminum heat sink having a mass of \(200 \mathrm{~g}\) and a surface area of \(80 \mathrm{~cm}^{2}\) ? Assume the device and the heat sink to be nearly isothermal.

Short Answer

Expert verified
Answer: The final temperature of the device is 377.94掳C without the heat sink, and 56.8掳C with the heat sink attached.

Step by step solution

01

Convert units

First, let's convert the given values to SI units: Mass of the device: \(m_{1} = 20\,\text{g} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 0.02\,\text{kg}\) Surface area of the device: \(A_{1} = 4\,\text{cm}^2 \times \frac{1\,\text{m}^2}{10000\,\text{cm}^2} = 4\times10^{-4}\,\text{m}^2\) Operating time: \(t = 5\,\text{min} \times \frac{60\,\text{s}}{1\,\text{min}} = 300\,\text{s}\)
02

Calculate the heat gained and temperature change without heat sink

Now, let's calculate the heat gained by the device during the operating time, and then calculate the temperature change. Heat lost by the device: \(Q_1 = P \times t = 20\,\text{W} \times 300\,\text{s} = 6000\,\text{J}\) Temperature change of the device: \(\Delta T_1 = \frac{Q_1}{m_1 \times c_p} = \frac{6000\,\text{J}}{0.02\,\text{kg} \times 850\,\text{J/kg路K}} \approx 352.94\,\text{K}\) Final temperature of the device without the heat sink is: \(T_{1f} = T_{1i} + \Delta T_1 = 25^{\circ}\text{C} + 352.94\,\text{K} \approx 377.94^{\circ}\text{C}\)
03

Calculate the mass and surface area of the aluminum heat sink

Now, let's consider the aluminum heat sink: Mass of the heat sink: \(m_{2} = 200\,\text{g} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 0.2\,\text{kg}\) Surface area of the heat sink: \(A_{2} = 80\,\text{cm}^2 \times \frac{1\,\text{m}^2}{10000\,\text{cm}^2} = 8\times10^{-3}\,\text{m}^2\)
04

Calculate the heat gained and temperature change with the heat sink

Now, let's calculate the heat gained and temperature change of the device and heat sink together: Total mass and specific heat: \(m_{total} = (m_1 + m_2) = 0.02\,\text{kg} + 0.2\,\text{kg} = 0.22\,\text{kg}\) Total surface area: \(A_{total} = A_1 + A_2 = 4\times10^{-4}\,\text{m}^2 + 8\times10^{-3}\,\text{m}^2 = 8.4\times10^{-3}\,\text{m}^2\) Heat lost by the device and heat sink: \(Q_2 = Q_1 = 6000\,\text{J}\) (assume nearly isothermal) Temperature change of the device and heat sink: \(\Delta T_2 = \frac{Q_2}{m_{total}\times c_p} = \frac{6000\,\text{J}}{0.22\,\text{kg} \times 850\,\text{J/kg路K}} \approx 31.8\,\text{K}\) Final temperature of the device with the heat sink is: \(T_{2f} = T_{2i} + \Delta T_2 = 25^{\circ}\text{C} + 31.8\,\text{K} \approx 56.8^{\circ}\text{C}\) So, the final temperature of the device is \(377.94^{\circ}\text{C}\) without the heat sink, and \(56.8^{\circ}\text{C}\) with the heat sink attached.

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Key Concepts

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

Thermal Management of Electronic Devices
When it comes to electronic devices, one of the major concerns is heat generation and its proper management. Without effective thermal management strategies, electronics can overheat, leading to reduced efficiency, performance, and even failures.

Heat production in electronics is a consequence of electrical resistance and inefficiency of components. To ensure the smooth operation of these devices, it's important to control their temperature. This is done by dissipating the generated heat into the surrounding environment or finding ways to minimize its production.

One common solution, illustrated by the exercise, is the use of a heat sink. A heat sink is designed to maximize the surface area in contact with the cooling medium, such as air. In our example, we saw a drastic difference in temperature when the device used a heat sink. The material properties of a heat sink, along with its size and shape, are crucial for effective heat dissipation.
Specific Heat Capacity
Specific heat capacity is a property that measures the amount of heat required to change the temperature of a substance by a degree Celsius (or Kelvin). It's denoted by the symbol 'c' and commonly expressed in units of joules per kilogram per Kelvin (J/kg路K).

The higher the specific heat capacity, the more heat is required to increase a material's temperature. This is a critical property for thermal management in electronics. Materials with high specific heat capacities are able to absorb more heat before their temperature rises significantly. In the exercise, the specific heat capacity of the device plays a key role in determining how much the temperature will increase during operation. It's also an important factor when a heat sink is attached, as the heat sink increases the overall capacity to store heat energy without significant temperature increases.
Heat Transfer Coefficient
The heat transfer coefficient is a measure of how well heat is transferred between a surface and a fluid flowing past it. It is symbolized by 'h' and usually expressed in units of watts per square meter per Kelvin (W/m虏路K). This coefficient is influenced by a number of factors including the properties of the fluid, the velocity of the fluid, and the characteristics of the surface.

In practical terms, a higher heat transfer coefficient suggests that a surface, such as the heat sink in our exercise, is more effective at transferring heat away from itself and into the surrounding medium. It plays a fundamental role in the thermal management of electronic devices, as it helps in predicting how efficiently a device can be cooled under different operational conditions. For efficient cooling, engineers seek out materials and designs that maximize this coefficient.

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Most popular questions from this chapter

In Betty Crocker's Cookbook, it is stated that it takes \(2 \mathrm{~h} \mathrm{} 45 \mathrm{~min}\) to roast a \(3.2-\mathrm{kg}\) rib initially at \(4.5^{\circ} \mathrm{C}\) "rare" in an oven maintained at \(163^{\circ} \mathrm{C}\). It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered rare done when the thermometer inserted into the center of the thickest part of the meat registers \(60^{\circ} \mathrm{C}\). The rib can be treated as a homogeneous spherical object with the properties \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=4.1 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(0.91 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). Determine \((a)\) the heat transfer coefficient at the surface of the rib; \((b)\) the temperature of the outer surface of the rib when it is done; and \((c)\) the amount of heat transferred to the rib. \((d)\) Using the values obtained, predict how long it will take to roast this rib to "medium" level, which occurs when the innermost temperature of the rib reaches \(71^{\circ} \mathrm{C}\). Compare your result to the listed value of \(3 \mathrm{~h} \mathrm{} 20 \mathrm{~min}\). If the roast rib is to be set on the counter for about \(15 \mathrm{~min}\) before it is sliced, it is recommended that the rib be taken out of the oven when the thermometer registers about \(4^{\circ} \mathrm{C}\) below the indicated value because the rib will continue cooking even after it is taken out of the oven. Do you agree with this recommendation? Solve this problem using analytical one-term approximation method (not the Heisler charts).

In a production facility, 3-cm-thick large brass plates \(\left(k=110 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=8530 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=380 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\left.\alpha=33.9 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) that are initially at a uniform temperature of \(25^{\circ} \mathrm{C}\) are heated by passing them through oven maintained at \(700^{\circ} \mathrm{C}\). The plates remain in the oven for a period of \(10 \mathrm{~min}\). Taking the convection heat transfer coefficient to be \(h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the surface temperature of the plates when they come out of the oven. Solve this problem using analytical one-term approximation method (not the Heisler charts). Can this problem be solved using lumped system analysis? Justify your answer.

A long 35-cm-diameter cylindrical shaft made of stainless steel \(304\left(k=14.9 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7900 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=477 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right.\), and \(\alpha=3.95 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) ) comes out of an oven at a uniform temperature of \(400^{\circ} \mathrm{C}\). The shaft is then allowed to cool slowly in a chamber at \(150^{\circ} \mathrm{C}\) with an average convection heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the shaft \(20 \mathrm{~min}\) after the start of the cooling process. Also, determine the heat transfer per unit length of the shaft during this time period. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Obtain relations for the characteristic lengths of a large plane wall of thickness \(2 L\), a very long cylinder of radius \(r_{o}\), and a sphere of radius \(r_{o}\).

How does refrigeration prevent or delay the spoilage of foods? Why does freezing extend the storage life of foods for months?
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