/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 A chip that is of length \(L=5 \... [FREE SOLUTION] | 91Ó°ÊÓ

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A chip that is of length \(L=5 \mathrm{~mm}\) on a side and thickness \(t=1 \mathrm{~mm}\) is encased in a ceramic substrate, and its exposed surface is convectively cooled by a dielectric liquid for which \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=20^{\circ} \mathrm{C}\). In the off-mode the chip is in thermal equilibrium with the coolant \(\left(T_{i}=T_{\infty}\right)\). When the chip is energized, however, its temperature increases until a new steady state is established. For purposes of analysis, the energized chip is characterized by uniform volumetric heating with \(\dot{q}=9 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming an infinite contact resistance between the chip and substrate and negligible conduction resistance within the chip, determine the steady-state chip temperature \(T_{f}\). Following activation of the chip, how long does it take to come within \(1^{\circ} \mathrm{C}\) of this temperature? The chip density and specific heat are \(\rho=2000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c=700 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively.

Short Answer

Expert verified
The steady-state temperature of the chip is found to be approximately \(T_f = 84.7^{\circ}C\). To come within \(1^{\circ}C\) of this temperature, it will take approximately \(t = 1.62~s\).

Step by step solution

01

Calculate heat generation in the chip

Since the energized chip is characterized by a uniform volumetric heat generation \(\dot{q}=9\times10^6~W/m^3\). We can calculate the heat generated in the chip, \(Q_g\), using the following formula: $$Q_g = \dot{q} \cdot V$$ where \(V\) is the volume of the chip. We are given the dimensions of the chip (\(L=5~mm\), \(t=1~mm\)) so we can calculate the volume as: $$V = L^2 \cdot t$$
02

Calculate the convection heat transfer

Once we calculate the heat generation in the chip, we can determine the convection heat transfer at the exposed surface, \(Q_h\), using $$Q_h = hA_s(T_f-T_\infty)$$ where \(h = 150~W/m^2\cdot K\) is the convection heat transfer coefficient, \(A_s = L^2\) is the area of the exposed surface, \(T_f\) is the final steady-state temperature of the chip, and \(T_\infty=20°C\) is the temperature of the coolant.
03

Set up the energy balance equation

Since it will ultimately reach steady-state, the heat generation within the chip will equal the heat transfer to the coolant. $$Q_g = Q_h$$ Now we can substitute the expressions for \(Q_g\) and \(Q_h\) from steps 1 and 2 and solve for the temperature \(T_f\).
04

Calculate the time to reach \(1^{\circ}C\) of the final temperature

Once we find the steady-state temperature \(T_f\), we can calculate the time it takes for the chip to come within \(1^{\circ}C\) of this temperature. The time constant \(\tau\) for the transient heat conduction is given by: $$\tau = \frac{\rho c V}{hA_s}$$ where \(\rho=2000~kg/m^3\) is the density and \(c=700~J/kg\cdot K\) is the specific heat of the chip. Using the first approximate solution for lumped capacitance method, we can calculate the time \(t\) as: $$\frac{T_f-T_\infty}{T_f-T_\infty-1} = e^{-t/\tau}$$ By rearranging and substituting the known values, we can solve for the time \(t\).

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Key Concepts

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

Convection
Convection is a fundamental concept in heat transfer which involves the movement of heat through a fluid (like a liquid or gas) by the combined action of conduction and fluid flow. In the problem, the chip is cooled by a dielectric liquid using convection. The convection heat transfer is characterized by the convection heat transfer coefficient, denoted as \(h\). This coefficient represents how effectively heat is being transferred between the chip's surface and the surrounding fluid.

For the chip, the convection heat flow can be calculated using the formula:
  • \( Q_h = hA_s(T_f-T_\infty) \)
Here, \(A_s\) is the area of the surface where convection occurs, \(T_f\) is the steady-state temperature of the chip, and \(T_\infty\) is the ambient fluid temperature. This process continues until the rate of heat loss by convection equals the rate of heat being generated within the chip. This is what helps to achieve the steady-state temperature the chip stabilizes at during operation.
Steady-State Temperature
Steady-state temperature is a critical concept in thermal analysis where the system reaches a condition where temperatures remain constant over time despite ongoing heat inputs or losses. In this exercise, steady-state temperature is the ultimate chip temperature when the heat generated inside the chip equals the heat being transferred to the dielectric coolant.

For this situation, an energy balance equation is used:
  • \( Q_g = Q_h \)
This indicates that the heat generated internally within the chip \(Q_g\) is equal to the heat convected away \(Q_h\). By solving this equation using known variables — including the heat transfer coefficient and ambient temperatures — the steady-state temperature \(T_f\) can be determined, providing a crucial understanding of how effectively heat management is achieved.
Transient Heat Conduction
Transient heat conduction refers to the change of temperature distribution within an object over time, until the system eventually reaches a new state of equilibrium or a steady-state condition. In this exercise, transient heat conduction is important in determining the time it takes for the chip to nearly reach the steady-state temperature after it is activated.

The amount of time it takes for a system to reach a certain degree of its steady-state temperature is influenced by the system's time constant, \(\tau\), given by:
  • \( \tau = \frac{\rho c V}{hA_s} \)
This formula incorporates several material properties such as density \(\rho\), specific heat \(c\), and geometric attributes like volume \(V\) and surface area \(A_s\). These factors jointly define how quickly the chip responds to changes in heat generation and how efficiently it moves towards achieving thermal balance.
Volumetric Heat Generation
Volumetric heat generation is the process by which heat is generated per unit volume within a system. In the chip example, the uniform volumetric heat generation is specified as \( \dot{q} = 9 \times 10^6 \) W/m³, implying that every cubic meter of the chip produces a constant amount of heat.

This concept allows engineers to predict how much heat is generated inside the chip, which impacts how effectively it can transfer this heat to the surrounding environment. This heat generation parameter is crucial for materials that generate heat internally, as it defines the total energy produced:
  • \( Q_g = \dot{q} \cdot V \)
The evaluation of this heat is necessary to determine whether the heat dissipation by convection is sufficient to maintain safe operational temperatures, emphasizing its critical role in thermal management of electronic devices.

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Most popular questions from this chapter

Stainless steel (AISI 304) ball bearings, which have uniformly been heated to \(850^{\circ} \mathrm{C}\), are hardened by quenching them in an oil bath that is maintained at \(40^{\circ} \mathrm{C}\). The ball diameter is \(20 \mathrm{~mm}\), and the convection coefficient associated with the oil bath is \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If quenching is to occur until the surface temperature of the balls reaches \(100^{\circ} \mathrm{C}\), how long must the balls be kept in the oil? What is the center temperature at the conclusion of the cooling period? (b) If 10,000 balls are to be quenched per hour, what is the rate at which energy must be removed by the oil bath cooling system in order to maintain its temperature at \(40^{\circ} \mathrm{C}\) ?

A long cylinder of \(30-\mathrm{mm}\) diameter, initially at a uniform temperature of \(1000 \mathrm{~K}\), is suddenly quenched in a large, constant- temperature oil bath at \(350 \mathrm{~K}\). The cylinder properties are \(k=1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c=1600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(\rho=400 \mathrm{~kg} / \mathrm{m}^{3}\), while the convection coefficient is \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Calculate the time required for the surface of the cylinder to reach \(500 \mathrm{~K}\). (b) Compute and plot the surface temperature history for \(0 \leq t \leq 300 \mathrm{~s}\). If the oil were agitated, providing a convection coefficient of \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how would the temperature history change?

Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a \(100-\mathrm{mm}\)-thick steel plate \(\left(\rho=7830 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(c=550 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=48 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), which is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and is to be heated to a minimum temperature of \(550^{\circ} \mathrm{C}\). Heating is effected in a gas-fired furnace, where products of combustion at \(T_{\infty}=800^{\circ} \mathrm{C}\) maintain a convection coefficient of \(h=250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) on both surfaces of the plate. How long should the plate be left in the furnace?

For each of the following cases, determine an appropriate characteristic length \(L_{c}\) and the corresponding Biot number \(B i\) that is associated with the transient thermal response of the solid object. State whether the lumped capacitance approximation is valid. If temperature information is not provided, evaluate properties at \(T=300 \mathrm{~K}\). (a) A toroidal shape of diameter \(D=50 \mathrm{~mm}\) and cross-sectional area \(A_{c}=5 \mathrm{~mm}^{2}\) is of thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface of the torus is exposed to a coolant corresponding to a convection coefficient of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) A long, hot AISI 304 stainless steel bar of rectangular cross section has dimensions \(w=3 \mathrm{~mm}\), \(W=5 \mathrm{~mm}\), and \(L=100 \mathrm{~mm}\). The bar is subjected to a coolant that provides a heat transfer coefficient of \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at all exposed surfaces. (c) A long extruded aluminum (Alloy 2024) tube of inner and outer dimensions \(w=20 \mathrm{~mm}\) and \(W=24 \mathrm{~mm}\), respectively, is suddenly submerged in water, resulting in a convection coefficient of \(h=37 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at the four exterior tube surfaces. The tube is plugged at both ends, trapping stagnant air inside the tube. (d) An \(L=300-m m\)-long solid stainless steel rod of diameter \(D=13 \mathrm{~mm}\) and mass \(M=0.328 \mathrm{~kg}\) is exposed to a convection coefficient of \(h=30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (e) A solid sphere of diameter \(D=12 \mathrm{~mm}\) and thermal conductivity \(k=120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suspended in a large vacuum oven with internal wall temperatures of \(T_{\text {sur }}=20^{\circ} \mathrm{C}\). The initial sphere temperature is \(T_{i}=100^{\circ} \mathrm{C}\), and its emissivity is \(\varepsilon=0.73\). (f) A long cylindrical rod of diameter \(D=20 \mathrm{~mm}\), density \(\rho=2300 \mathrm{~kg} / \mathrm{m}^{3}\), specific heat \(c_{p}=1750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and thermal conductivity \(k=16 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is suddenly exposed to convective conditions with \(T_{\infty}=20^{\circ} \mathrm{C}\). The rod is initially at a uniform temperature of \(T_{i}=200^{\circ} \mathrm{C}\) and reaches a spatially averaged temperature of \(T=100^{\circ} \mathrm{C}\) at \(t=225 \mathrm{~s}\). (g) Repeat part (f) but now consider a rod diameter of \(D=200 \mathrm{~mm}\).

Thin film coatings characterized by high resistance to abrasion and fracture may be formed by using microscale composite particles in a plasma spraying process. A spherical particle typically consists of a ceramic core, such as tungsten carbide (WC), and a metallic shell, such as cobalt \((\mathrm{Co})\). The ceramic provides the thin film coating with its desired hardness at elevated temperatures, while the metal serves to coalesce the particles on the coated surface and to inhibit crack formation. In the plasma spraying process, the particles are injected into a plasma gas jet that heats them to a temperature above the melting point of the metallic casing and melts the casing before the particles impact the surface. Consider spherical particles comprised of a WC core of diameter \(D_{i}=16 \mu \mathrm{m}\), which is encased in a Co shell of outer diameter \(D_{o}=20 \mu \mathrm{m}\). If the particles flow in a plasma gas at \(T_{\infty}=10,000 \mathrm{~K}\) and the coefficient associated with convection from the gas to the particles is \(h=20,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), how long does it take to heat the particles from an initial temperature of \(T_{i}=300 \mathrm{~K}\) to the melting point of cobalt, \(T_{\mathrm{mp}}=1770 \mathrm{~K}\) ? The density and specific heat of WC (the core of the particle) are \(\rho_{c}=16,000 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{c}=300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), while the corresponding values for Co (the outer shell) are \(\rho_{s}=8900 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{s}=750 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Once having reached the melting point, how much additional time is required to completely melt the cobalt if its latent heat of fusion is \(h_{s f}=2.59 \times 10^{5} \mathrm{~J} / \mathrm{kg}\) ? You may use the lumped capacitance method of analysis and neglect radiation exchange between the particle and its surroundings.

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