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Chapter 1: Problem 26

A square isothermal chip is of width \(w=5 \mathrm{~mm}\) on a side and is mounted in a substrate such that its side and back surfaces are well insulated; the front surface is exposed to the flow of a coolant at \(T_{\infty}=15^{\circ} \mathrm{C}\). From reliability considerations, the chip temperature must not exceed \(T=85^{\circ} \mathrm{C}\). If the coolant is air and the corresponding convection coefficient is \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable chip power? If the coolant is a dielectric liquid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the maximum allowable power?

Short Answer

Expert verified
The maximum allowable chip power for air coolant is 0.35 W, and for the dielectric liquid coolant, it is 5.25 W.

Step by step solution

01

Convection Heat Transfer Equation

The convection heat transfer equation is given by: \(q = hA\Delta T\) Here, q = Heat transfer from the chip (W) h = Convection coefficient (W/m虏路K) A = Exposed chip surface area (m虏) 螖T = Temperature difference between chip and coolant (K)
02

Calculate Surface Area

We are given the width of the chip, w = 5 mm. To find the surface area (A) of the exposed front surface, we need to convert the width to meters and then square it. A = (w / 1000)虏 = (5 / 1000)虏 = 0.000025 m虏
03

Convert Temperature Difference to Kelvin

The temperature difference between the chip and the coolant must be converted to Kelvin, as follows: 螖T = T - T鈭� = 85掳C - 15掳C = 70 K
04

Calculate Maximum Allowable Power for Air Coolant

Now, we are ready to find the maximum allowable power for the air-cooled chip. We have h = 200 W/m虏路K and 螖T = 70 K from previous steps. Plugging the values into the convection heat transfer equation: q_air = h_air * A * 螖T = 200 * 0.000025 * 70 = 0.35 W
05

Calculate Maximum Allowable Power for Dielectric Liquid Coolant

Repeat the calculation for the dielectric liquid coolant, given h = 3000 W/m虏路K: q_liquid = h_liquid * A * 螖T = 3000 * 0.000025 * 70 = 5.25 W We have found the maximum allowable chip power for the two different coolants: For air: 0.35 W For dielectric liquid: 5.25 W

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Key Concepts

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

Convection Coefficient
When discussing convection heat transfer, the convection coefficient plays a crucial role. It is a measure of how effectively heat is transferred between a surface and a fluid flowing over it. The higher the coefficient, the more heat is transferred.
  • In equations, it is usually represented by the symbol \( h \), and its units are \( \text{W/m}^2\cdot\text{K} \).
  • The convection coefficient depends on several factors, such as the type of fluid, its velocity, and the surface geometry.
  • In the given exercise, the convection coefficient \( h \) changes based on the coolant used: 200 \( \text{W/m}^2\cdot\text{K} \) for air and 3000 \( \text{W/m}^2\cdot\text{K} \) for the dielectric liquid.
A higher convection coefficient means more efficient cooling, helping components maintain lower temperatures even when generating significant power. This concept is essential in thermal management, especially in electronics where heat dissipation is critical.
Temperature Difference
The temperature difference \( \Delta T \) is the driving force for heat transfer. It is the difference between the surface temperature and the fluid temperature. In this exercise, it is calculated as follows:\[\Delta T = T - T_{\infty} = 85^\circ\text{C} - 15^\circ\text{C} = 70\,\text{K}\]
  • \( T \) is the maximum allowable chip temperature, given as 85掳C.
  • \( T_{\infty} \) is the temperature of the coolant, which is 15掳C.
The value is then converted into Kelvin because the SI unit system is typically used for such calculations, ensuring compatibility and accuracy when substituting values into physics equations. Understanding \( \Delta T \) is vital as it directly influences the rate of heat transfer based on Newton's Law of Cooling.
Surface Area Calculation
The calculation of the surface area is foundational in determining how much heat can be transferred through convection. The surface area \( A \) of the chip exposed to the coolant is calculated using the chip's width. Here鈥檚 how it鈥檚 determined:
  • The chip鈥檚 width is \( w = 5\,\text{mm} \), which must be converted to meters: \( w = 0.005\,\text{m} \).
  • The area, being square, is \( w^2 \): \[ A = (0.005)^2 = 0.000025\,\text{m}^2 \]
This converts a practical aspect of the problem into a mathematical form that can be easily used in the convection heat transfer equation. Accurately calculating this area is crucial because even small errors can significantly impact the calculated heat transfer, potentially leading to overheating or underperformance in devices like a chip.

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Most popular questions from this chapter

A vertical slab of Wood's metal is joined to a substrate on one surface and is melted as it is uniformly irradiated by a laser source on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\), and the melt runs off by gravity as soon as it is formed. The absorptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{s f}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantaneous rate of melting in \(\mathrm{kg} / \mathrm{s} \cdot \mathrm{m}^{2}\) if the laser irradiation is \(5 \mathrm{~kW} / \mathrm{m}^{2}\). How much material is removed if irradiation is maintained for a period of \(2 \mathrm{~s}\) ? (b) Allowing for convection to ambient air, with \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \((\varepsilon=0.4\), \(T_{\text {sur }}=20^{\circ} \mathrm{C}\) ), determine the instantaneous rate of melting during irradiation.

During its manufacture, plate glass at \(600^{\circ} \mathrm{C}\) is cooled by passing air over its surface such that the convection heat transfer coefficient is \(h=5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent cracking, it is known that the temperature gradient must not exceed \(15^{\circ} \mathrm{C} / \mathrm{mm}\) at any point in the glass during the cooling process. If the thermal conductivity of the glass is \(1.4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and its surface emissivity is \(0.8\), what is the lowest temperature of the air that can initially be used for the cooling? Assume that the temperature of the air equals that of the surroundings.

A concrete wall, which has a surface area of \(20 \mathrm{~m}^{2}\) and is \(0.30 \mathrm{~m}\) thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at \(25^{\circ} \mathrm{C}\), and the thermal conductivity of the concrete is \(1 \mathrm{~W} / \mathrm{m}=\mathrm{K}\). (a) Determine the heat loss through the wall for outer surface temperatures ranging from \(-15^{\circ} \mathrm{C}\) to \(38^{\circ} \mathrm{C}\), which correspond to winter and summer extremes, respectively. Display your results graphically. (b) On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of \(0.75\) and \(1.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Explain the family of curves you have obtained.

The energy consumption associated with a home water heater has two components: (i) the energy that must be supplied to bring the temperature of groundwater to the heater storage temperature, as it is introduced to replace hot water that has been used; (ii) the energy needed to compensate for heat losses incurred while the water is stored at the prescribed temperature. In this problem, we will evaluate the first of these components for a family of four, whose daily hot water consumption is approximately \(100 \mathrm{gal}\). If groundwater is available at \(15^{\circ} \mathrm{C}\), what is the annual energy consumption associated with heating the water to a storage temperature of \(55^{\circ} \mathrm{C}\) ? For a unit electrical power cost of \(\$ 0.18 / \mathrm{kW} \cdot \mathrm{h}\), what is the annual cost associated with supplying hot water by means of (a) electric resistance heating or (b) a heat pump having a COP of 3 .

In considering the following problems involving heat transfer in the natural environment (outdoors), recognize that solar radiation is comprised of long and short wavelength components. If this radiation is incident on a semitransparent medium, such as water or glass, two things will happen to the nonreflected portion of the radiation. The long wavelength component will be absorbed at the surface of the medium, whereas the short wavelength component will be transmitted by the surface. (a) The number of panes in a window can strongly influence the heat loss from a heated room to the outside ambient air. Compare the single- and double-paned units shown by identifying relevant heat transfer processes for each case. (b) In a typical flat-plate solar collector, energy is collected by a working fluid that is circulated through tubes that are in good contact with the back face of an absorber plate. The back face is insulated from the surroundings, and the absorber plate receives solar radiation on its front face, which is typically covered by one or more transparent plates. Identify the relevant heat transfer processes, first for the absorber plate with no cover plate and then for the absorber plate with a single cover plate. (c) The solar energy collector design shown in the schematic has been used for agricultural applications. Air is blown through a long duct whose cross section is in the form of an equilateral triangle. One side of the triangle is comprised of a double-paned, semitransparent cover; the other two sides are constructed from aluminum sheets painted flat black on the inside and covered on the outside with a layer of styrofoam insulation. During sunny periods, air entering the system is heated for delivery to either a greenhouse, grain drying unit, or storage system. Identify all heat transfer processes associated with the cover plates, the absorber plate(s), and the air. (d) Evacuated-tube solar collectors are capable of improved performance relative to flat-plate collectors. The design consists of an inner tube enclosed in an outer tube that is transparent to solar radiation. The annular space between the tubes is evacuated. The outer, opaque surface of the inner tube absorbs solar radiation, and a working fluid is passed through the tube to collect the solar energy. The collector design generally consists of a row of such tubes arranged in front of a reflecting panel. Identify all heat transfer processes relevant to the performance of this device.
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