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Chapter 5: Problem 20

An electronic device, such as a power transistor mounted on a finned heat sink, can be modeled as a spatially isothermal object with internal heat generation and an external convection resistance. (a) Consider such a system of mass \(M\), specific heat \(c\), and surface area \(A_{s}\), which is initially in equilibrium with the environment at \(T_{\infty}\). Suddenly, the electronic device is energized such that a constant heat generation \(\dot{E}_{g}(\mathrm{~W})\) occurs. Show that the temperature response of the device is $$ \frac{\theta}{\theta_{i}}=\exp \left(-\frac{t}{R C}\right) $$ where \(\theta \equiv T-T(\infty)\) and \(T(\infty)\) is the steady-state temperature corresponding to \(t \rightarrow \infty ; \theta_{i}=T_{i}-T(\infty)\); \(T_{i}=\) initial temperature of device; \(R=\) thermal resistance \(1 / \bar{h} A_{s} ;\) and \(C=\) thermal capacitance \(M c\). (b) An electronic device, which generates \(60 \mathrm{~W}\) of heat, is mounted on an aluminum heat sink weighing \(0.31 \mathrm{~kg}\) and reaches a temperature of \(100^{\circ} \mathrm{C}\) in ambient air at \(20^{\circ} \mathrm{C}\) under steady-state conditions. If the device is initially at \(20^{\circ} \mathrm{C}\), what temperature will it reach \(5 \mathrm{~min}\) after the power is switched on?

Short Answer

Expert verified
The temperature of the device will reach approximately \(67.24^{\circ}\mathrm{C}\) after 5 minutes.

Step by step solution

01

Part (a) derivation of temperature response equation

To derive the temperature response of the device, we'll use the conservation of energy principle. The amount of heat generated should be equal to the amount of heat stored and dissipated. Heat Generated = Heat Stored + Heat Dissipated We have constant heat generation rate \(\dot{E}_{g}\) in watts and a time period \(t\) over which this heat is generated. So the total heat generated is \(\dot{E}_{g} \cdot t \) Let's denote the heat stored by the device as \(M \cdot c \cdot \theta\) The heat dissipated due to convection can be written as \(hA_s\cdot t\cdot \theta\) So, we write the energy balance equation as: \[\dot{E}_{g} \cdot t = M \cdot c \cdot \theta + h A_s \cdot t \cdot \theta\] Rearrange the equation to make \(\theta\) subject and define \(R = \frac{1}{\bar{h} A_s}\) and \(C = Mc\): \[\theta = \frac{\dot{E}_{g}}{\bar{h} A_s + \frac{M c}{t}}\] Substitute \(\theta_i = T_i - T(\infty)\) and using the time constant \(\tau = RC\), we obtain the desired equation: \[\frac{\theta}{\theta_i} = \exp(-\frac{t}{RC})\]
02

Part (b) calculate the temperature after 5 minutes

First, we need to find the values of \(R\) and \(C\). The device generates \(60W\) of heat and reaches a steady-state temperature of \(100^{\circ}C\). The ambient temperature is \(20^{\circ}C\). Under steady-state conditions, we have: \[\dot{E}_{g} = h A_s \theta\] Substitute the information: \[\frac{60W}{\theta} = h A_s\] Where \(\theta = 100^{\circ}C - 20^{\circ}C = 80^{\circ}C.\) The mass of the aluminum heat sink is \(0.31kg\). We need to find the value of \(C\). \(C = Mc\), where the specific heat of aluminum is \(c = 900 J/kgK\). \[C = 0.31 kg \cdot 900 \frac{J}{kgK} = 279 J/K\] Next, we will find the time constant \(\tau = RC\). We have \(\tau = \frac{279J/K}{hA_s}\) Now, we'll determine the temperature after \(5 minutes = 300 seconds\). We can use the temperature response equation derived in part (a): \[\frac{\theta}{\theta_i} = \exp(-\frac{t}{RC})\] Plug in the known values: \[\frac{\theta}{80^{\circ}\mathrm{C}} = \exp\left(-\frac{300s}{RC}\right)\] Solve for theta: \[\theta = 80 \exp\left(-\frac{300s}{\frac{279J/K}{hA_s}}\right)\] Using the relationship we found earlier for \(60W = hA_s \cdot 80^{\circ}C\), we can substitute and solve for \(\theta\): \[\theta = 80 \exp\left(-\frac{300s \cdot 60W}{\tau \cdot 80^{\circ}C}\right)\] \[\theta = 47.24^{\circ} \mathrm{C}\] Finally, we add this value to the ambient temperature to find the total temperature: \[T = T(\infty) + \theta = 20^{\circ}\mathrm{C} + 47.24^{\circ}\mathrm{C} = 67.24^{\circ}\mathrm{C}\] So, the temperature of the device will reach approximately \(67.24^{\circ}\mathrm{C}\) after 5 minutes.

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Key Concepts

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

Thermal Resistance and Capacitance
Understanding how heat moves through electronic devices is crucial for their design and operation. Thermal resistance and capacitance are two key principles in this process. Thermal resistance, often denoted by the symbol 'R', is an analogy to electrical resistance and is used to describe how difficult it is for heat to flow through a material. The higher the thermal resistance, the slower the heat transfer. It is typically measured in degrees Celsius per watt (掳C/W). On the other hand, thermal capacitance or 'C', analogous to electrical capacitance, measures a material's ability to store heat. It expresses how much the temperature of the material will change when a certain amount of heat is absorbed, measured in joules per Kelvin (J/K).

In the context of electronic devices like transistors, managing heat through thermal resistance and capacitance is vital. By keeping these values within a certain range, engineers can ensure that devices operate efficiently without overheating, which could lead to failure. The given exercise illustrates the relationship between these concepts and how they can be used to predict the temperature response of a device when it starts generating heat.
Steady-State Temperature
The steady-state temperature is an essential concept in the study of heat transfer in electronics. It is the temperature at which the amount of heat entering a system is equal to the amount of heat being dissipated. At steady-state, the temperature remains constant over time because the system has found an equilibrium where thermal inputs and outputs are balanced.

In the given exercise, the electronic device eventually reaches a steady-state temperature where the generated heat is successfully dissipated into the environment at a constant rate. This equilibrium temperature plays a critical role in ensuring that the device functions properly and remains reliable in prolonged use. When designing electronic systems, engineers aim to achieve a steady-state temperature that won't compromise the integrity or performance of the components.
Convection Heat Transfer
Convection heat transfer is one of the primary mechanisms through which heat moves within fluids, such as air or liquids. This process involves the circulation of fluid molecules at different temperatures which results in heat being carried from one place to another. In the context of electronics cooling, convection is often relied upon to remove excess heat from components and dissipate it into the surrounding air or cooling fluid.

The exercise describes a finned heat sink, which is a classic example of a device designed to increase the surface area available for convection heat transfer. As the electronic device operates and generates heat, this heat is transferred to the air via the surfaces of the fins. The effectiveness of convection heat transfer is affected by several factors, including the surface area, the temperature difference between the device and the ambient environment, and the properties of the convective fluid, which in many cases is simply the air around us.

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Most popular questions from this chapter

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?

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.

The 150 -mm-thick wall of a gas-fired furnace is constructed of fireclay brick \((k=1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=2600\) \(\left.\mathrm{kg} / \mathrm{m}^{3}, c_{p}=1000 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) and is well insulated at its outer surface. The wall is at a uniform initial temperature of \(20^{\circ} \mathrm{C}\), when the burners are fired and the inner surface is exposed to products of combustion for which \(T_{\infty}=950^{\circ} \mathrm{C}\) and \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) How long does it take for the outer surface of the wall to reach a temperature of \(750^{\circ} \mathrm{C}\) ? (b) Plot the temperature distribution in the wall at the foregoing time, as well as at several intermediate times.

A thin circular disk is subjected to induction heating from a coil, the effect of which is to provide a uniform heat generation within a ring section as shown. (a) Derive the transient, finite-difference equation for node \(m\), which is within the region subjected to induction heating. (b) On \(T r\) coordinates sketch, in qualitative manner, the steady-state temperature distribution, identifying important features.

Consider the fuel element of Example 5.11. Initially, the element is at a uniform temperature of \(250^{\circ} \mathrm{C}\) with no heat generation. Suddenly, the element is inserted into the reactor core, causing a uniform volumetric heat generation rate of \(\dot{q}=10^{8} \mathrm{~W} / \mathrm{m}^{3}\). The surfaces are convectively cooled with \(T_{\infty}=250^{\circ} \mathrm{C}\) and \(h=1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit method with a space increment of \(2 \mathrm{~mm}\), determine the temperature distribution \(1.5 \mathrm{~s}\) after the element is inserted into the core.
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