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Chapter 1: Problem 18

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 bock surfaces are well insulated, while the front surface is exposed to the flow of a coolant at \(T_{w}=15^{\circ} \mathrm{C}\). From reliatility considerations, the chip temperature must not cxceed \(T=85^{\circ} \mathrm{C}\). Coclart \(\longrightarrow T_{m} h\) 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 liequid for which \(h=3000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{\mathrm{K}}\), what is the maximam allowable power?

Short Answer

Expert verified
Max power for air: 0.35 W, dielectric liquid: 5.25 W.

Step by step solution

01

Understand the Problem

We are asked to determine the maximum power the chip can dissipate without exceeding a temperature of 85掳C, given two different coolant scenarios. We'll use the convection heat transfer formula to evaluate the power allowed in each case.
02

Convection Heat Transfer Formula

The heat power (q) dissipated by the chip can be calculated using the formula: \[ q = hA(T_m - T_w) \]where \(h\) is the convection coefficient, \(A\) is the heat transfer area, \(T_m\) is the maximum chip temperature, and \(T_w\) is the coolant temperature.
03

Calculate the Heat Transfer Area

Since the chip is square with side \(w = 5 \, \text{mm} = 0.005 \, \text{m}\), the area \(A\) of the front surface is: \[ A = w^2 = (0.005)^2 = 2.5 \times 10^{-5} \, \text{m}^2 \]
04

Calculate Maximum Power for Air Coolant

For air (\(h = 200 \, \text{W/m}^2\cdot \text{K}\)):Substitute the values in the heat transfer formula: \[ q = 200 \times 2.5 \times 10^{-5} \times (85 - 15) \]\[ q = 200 \times 2.5 \times 10^{-5} \times 70 \]\[ q = 0.35 \, \text{W} \]
05

Calculate Maximum Power for Dielectric Liquid

For dielectric liquid (\(h = 3000 \, \text{W/m}^2\cdot \text{K}\)):Substitute the values in the heat transfer formula: \[ q = 3000 \times 2.5 \times 10^{-5} \times (85 - 15) \]\[ q = 3000 \times 2.5 \times 10^{-5} \times 70 \]\[ q = 5.25 \, \text{W} \]

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Key Concepts

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

Convection Coefficient
Convection coefficient, often denoted by the letter \( h \), is a crucial factor in determining how effectively heat is transferred between a solid surface and a fluid (like air or liquid) in motion across that surface. It quantifies the ability of the fluid to carry away heat per unit area and temperature difference.
When evaluating the performance of something like a computer chip, knowing the convection coefficient helps in planning how to manage its temperature efficiently. A higher \( h \) means the fluid is more effective at transferring heat away from the surface. In our problem, this coefficient is measured in units of \( \text{W/m}^2\cdot \text{K} \), which translates to how many watts of power can effectively be removed when there is a one-degree temperature difference across one square meter of surface.
When comparing the air coolant with \( h = 200 \text{ W/m}^2\cdot \text{K} \) to a dielectric liquid with \( h = 3000 \text{ W/m}^2\cdot \text{K} \), it's evident that the dielectric liquid is much more effective at heat transfer, hinting that it will allow the chip to dissipate more power while maintaining safe operating temperatures.
Heat Transfer Area
The heat transfer area, represented as \( A \) in formulas, is the surface area over which heat convection occurs. For geometric shapes like our square chip, this area is calculated based on its physical dimensions.
In this exercise, we're looking at a square chip with sides of \( 5 \text{ mm} \). To get the area in meters squared, convert the length to meters by dividing by 1000, and square the result. Therefore, \( A = (0.005 \text{ m})^2 = 2.5 \times 10^{-5} \text{ m}^2 \).
The smaller the area, the lesser the capacity to dissipate heat, as there's less surface for the heat to be transferred across. The heat transfer area is pivotal in heat dissipation analyses because it influences how much energy can be exchanged between the surface and the fluid.
Maximum Allowable Power
The maximum allowable power is the total heat energy that can be safely dissipated by a device without surpassing a critical temperature. In our case, this critical boundary is set by the temperature limit of the chip at \( 85^{\circ}C \).
Using the formula \( q = hA(T_m - T_w) \), where \( T_m \) is the maximum chip temperature (\( 85^{\circ}C \)), and \( T_w \) is the coolant temperature (\( 15^{\circ}C \)), we can find the maximum heat power that can be tolerated in each cooling scenario.
  • With air cooling (\( h = 200 \text{ W/m}^2\cdot \text{K} \)), the formula yields a \( q \) value of \( 0.35 \text{ W} \).
  • For the more efficient dielectric liquid (\( h = 3000 \text{ W/m}^2\cdot \text{K} \)), \( q \) comes out to be \( 5.25 \text{ W} \).
This difference illustrates how fluid properties and convection coefficient can drastically affect device cooling capacity and energy dissipation limits.

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Most popular questions from this chapter

A vertical slab of Woods metal is joined to a substrate on one surface and is melted as it is uniformly irradialed by a laser souree on the opposite surface. The metal is initially at its fusion temperature of \(T_{f}=72^{\circ} \mathrm{C}\). and the melt rans off by gravity as soon as it is formed. The abserptivity of the metal to the laser radiation is \(\alpha_{1}=0.4\), and its latent heat of fusion is \(h_{y}=33 \mathrm{~kJ} / \mathrm{kg}\). (a) Neglecting heat transfer from the irradiated surface by convection or radiation exchange with the surroundings, determine the instantanecus rate of melting in \(\mathrm{kg} / \mathrm{s}-\mathrm{m}^{2}\) 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 \(T_{\mathrm{w}}=20 r \mathrm{C}\) and \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and radiation exchange with large surroundings \(\left(\varepsilon=0.4, T_{\text {max }}=\right.\) \(20^{\circ} \mathrm{C}\), determine the instantaneous rate of melting during imadiation.

A hair dryer may be idealized as a circular doct through which a mmall fan draws ambient air and withie which the air is heated as it flows over a coiled electric resistance wire. (a) If a dryer is designed to operate with an electric power consumption of \(P_{\text {tesc }}=500 \mathrm{~W}\) and to heat air from an ambient temperature of \(T_{i}=20^{\circ} \mathrm{C}\) to a discharge temperature of \(T_{n}=45^{\circ} \mathrm{C}\), at what volumetric flow rate \(\forall\) should the fan operate? Heat loss from the casing to the ambient air and the surroundings nuy be neglected. If the duct has a diameler of \(D=70 \mathrm{~mm}\), what is the discharge velocity \(V_{0}\) of the uir? The density and specific heat of the air maly be approximated as \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=1007\) \(\mathrm{J} / \mathrm{Kg}+\mathrm{K}\), respectively. (b) Consider a dryer duct length of \(L=150 \mathrm{~mm}\) and a surface cmisuivity of \(e=0.8\). If the coeflicient associaled with heat transfer by natural convection from the casing to the ambient air is \(h=4 \mathrm{~W} / \mathrm{m}^{2}-\mathrm{K}\) and the temperature of the air and the surroundings is \(T_{\mathrm{w}}=\) \(T_{w e}=20^{\circ} \mathrm{C}\), confirm that the heat loss from the casing is, in fact, negligitle. The casing may be assumed to have an average surface temperature of \(T_{3}=40^{\circ} \mathrm{C}\).

The clectrical-substitation radiometer shown schematically determines the optical (radiant) power of a beam by measuring the electrical power required to heat the receiver to the same temperature. With a beam, such as a luser of optical power \(P_{\text {epv }}\) incident on the receiver, its temperature, \(T_{p}\) increases above that of the chamber walls held al a uniform temperature, \(T_{\text {we }}=77 \mathrm{~K}\), With the optical beam blocked, the heater on the hackside of the receiver is encrgized and the electrical power, \(P_{\text {eine: }}\) required to reach the same value of \(T_{\text {, is measured. The }}\) purpose of your analysis is to determine the relationship tetween the clectrical and optical power, considering heat transfer processes experienced by the receiver. Consider a radiomeler with a \(15-\mathrm{mm}\)-diameter receiver having a blackened surface with an emissivity of \(0.95\) and an absorptivity of \(0.98\) for the optical beam. When operating in the optical mode, conduction heat losses from the backside of the receiver are negligible. In the electricul mode, the loss amounts to \(5 \%\) of the electrical power. What is the optical power of a beam when the indicated electrical power is \(20.64 \mathrm{~mW}\) ? What is the corresponding recciver temperature?

A rectangular forced air heating duct is suspended from the ceiling of a basement whose air and walls are at a termperiture of \(T_{\mathrm{w}}=T_{\mathrm{er}}=5^{\circ} \mathrm{C}\). The duct is \(15 \mathrm{~m}\) long. and its cross-section is \(350 \mathrm{~mm} \times 200 \mathrm{~mm}\). (a) For an uninsulated duct whose average surfice temperature is \(50^{\circ} \mathrm{C}\), estimate the rate of heat loss from the duct. The surface emissivity and convection coefficient are approximately \(0.5\) and \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). respectively. (b) If heated air enters the duct at \(58^{\circ} \mathrm{C}\) and a velocity of \(4 \mathrm{~m} / \mathrm{s}\) and the heat loss corresponds to the result of part (a), what is the outlet temperature? The density and specific heat of the air may be assumed to be \(\rho=1.10 \mathrm{~kg} / \mathrm{m}^{\prime}\) and \(c_{p}=1008 \mathrm{~J} / \mathrm{kg}\) * \(\mathrm{K}\), respectively.

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).
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