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Chapter 1: Problem 55

Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is \(5000 \mathrm{~J} / \mathrm{K}\), and the effective heat transfer coefficient times surface area product is \(10 \mathrm{~W} / \mathrm{K}\). The initial temperature of the plate and the cooling air temperature are \(295 \mathrm{~K}\) when \(300 \mathrm{~W}\) of power are switched on. Find the plate temperature after 10 minutes.

Short Answer

Expert verified
The plate temperature after 10 minutes is approximately 316 K.

Step by step solution

01

Identify Given Values

We are given the following: the product of mass and specific heat, \(mC = 5000 \, \text{J/K}\); the heat transfer coefficient times surface area, \(hA = 10 \, \text{W/K}\); the initial temperature of the plate and the air is \(T_i = 295 \, \text{K}\); the power applied is \(P = 300 \, \text{W}\); and the time \(t = 10 \, \text{minutes} = 600 \, \text{seconds}\). We aim to find the plate temperature at \(t = 10 \, \text{minutes}\).
02

Write the Energy Balance Equation

The energy balance on the system can be written as: \[ mC \frac{dT}{dt} = P - hA (T - T_a) \] where \(T\) is the plate temperature and \(T_a\) is the air temperature, \(T_a = 295 \, \text{K}\).
03

Simplify and Rearrange the Equation

Substitute the known values into the energy balance equation: \[ 5000 \frac{dT}{dt} = 300 - 10 (T - 295) \]Simplifying, we get:\[ \frac{dT}{dt} = \frac{300 - 10T + 2950}{5000} = \frac{3250 - 10T}{5000} \]
04

Integrate the Differential Equation

Rearrange the equation for integration:\[ \frac{1}{325 - T} \frac{dT}{dt} = \frac{1}{500} \]Integrating both sides with the limits from \(T = T_i = 295\) to \(T = T(t)\) and \(t = 0\) to \(t = t\), we have:\[ -\ln|325 - T|\bigg|_{295}^{T(t)} = \frac{1}{500}t \]
05

Solve the Integral

The integrated form becomes:\[ -\ln|325 - T(t)| + \ln|325 - 295| = \frac{t}{500} \]This simplifies using logarithmic properties:\[ \ln\frac{30}{325 - T(t)} = \frac{t}{500} \]Exponentiate both sides:\[ \frac{30}{325 - T(t)} = e^{\frac{t}{500}} \]
06

Rearrange to Find Temperature

Rearrange the equation to solve for \(T(t)\):\[ 325 - T(t) = \frac{30}{e^{\frac{t}{500}}} \]\[ T(t) = 325 - \frac{30}{e^{\frac{t}{500}}} \]Substitute \(t = 600\) seconds:\[ T(600) = 325 - \frac{30}{e^{1.2}} \]
07

Calculate the Final Temperature

Calculate the exponential term and solve for the temperature:\[ e^{1.2} \approx 3.3201 \]Thus, \[ T(600) = 325 - \frac{30}{3.3201} \approx 325 - 9.03 \approx 315.97 \, \text{K} \]

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Key Concepts

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

Understanding Energy Balance in Heat Transfer
Energy balance is a fundamental concept in heat transfer that helps us understand how energy flows within a particular system. When dealing with a system like a finned aluminum base plate cooled by air, it is critical to evaluate the balance of energy added to and removed from the system. In the context of the exercise, the energy balance is framed as:
  • The energy entering the system: This is the power applied to the electronic components, which is given as 300 Watts (W).
  • The energy leaving the system: This is the heat transferred away from the base plate by the cooling air, calculated using the heat transfer coefficient and the temperature difference.
The equation balancing these energies is \( mC \frac{dT}{dt} = P - hA (T - T_a) \), where:
  • mC is the heat capacity of the system.
  • P is the power applied.
  • hA is the product of the heat transfer coefficient and surface area.
  • T and T_a are the plate and air temperatures, respectively.
Understanding this balance allows us to predict temperature changes in the plate over time, depending on how much energy is input versus dissipated.
Exploring Specific Heat and its Impact
Specific heat is an important property of materials that tells us how much energy is needed to change the temperature of a given mass. It is a measure of how well a material can store thermal energy. In the exercise, the specific heat is part of the product \( mC = 5000 \, \text{J/K}\), which combines the mass of the system with its specific heat capacity.
Here, specific heat helps determine how fast or slow the temperature of the base plate will change when energy is applied:
  • A higher specific heat means the material can absorb more energy without a significant temperature increase.
  • A lower specific heat indicates that the temperature will rise faster with the same energy input.
The specific heat is intrinsic to the base plate's material properties and the electronic components mounted onto it. Therefore, understanding specific heat is essential for designing systems that efficiently manage heat dissipation, ensuring electronic components operate within safe temperature ranges.
Exploring the Finned Aluminum Base Plate Functionality
Finned aluminum base plates are often used in electronic cooling applications because of their excellent heat dissipation characteristics. The fins substantially increase the surface area available for heat exchange between the plate and the cooling air, improving overall heat transfer efficiency.
  • Aluminum, as the chosen material for the base plate, possesses excellent thermal conductivity, meaning it can conduct heat efficiently.
  • The fins allow more air to contact the plate, which dissipates the heat from the plate's surface more effectively.
  • The increased surface area from the fins ensures that even more heat can be transferred from the plate to the cooling air.
In the context of the exercise, the effectiveness of the cooling is often described by the heat transfer coefficient h and the surface area A, which together determine how quickly and efficiently the heat is transferred away from the plate. This setup ensures that the electronic components remain cool enough to function properly, avoiding overheating and potential damage.

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Most popular questions from this chapter

A thermistor is used to measure the temperature of an air stream leaving an air heater. It is located in a \(30 \mathrm{~cm}\) square duct and records a temperature of \(42.6^{\circ} \mathrm{C}\) when the walls of the duct are at \(38.1^{\circ} \mathrm{C}\). What is the true temperature of the air? The thermistor can be modeled as a \(3 \mathrm{~mm}\)-diameter sphere of emittance \(0.7\). The convective heat transfer coefficient from the air stream to the thermistor is estimated to be \(31 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\).

An astronaut is at work in the service bay of a space shuttle and is surrounded by walls that are at \(-100^{\circ} \mathrm{C}\). The outer surface of her space suit has an area of \(3 \mathrm{~m}^{2}\) and is aluminized with an emittance of \(0.05\). Calculate her rate of heat loss when the suit's outer temperature is \(0^{\circ} \mathrm{C}\). Express your answer in watts and \(\mathrm{kcal} / \mathrm{hr}\).

Estimate the heating load for a building in a cold climate when the outside temperature is \(-10^{\circ} \mathrm{C}\) and the air inside is maintained at \(20^{\circ} \mathrm{C}\). The \(350 \mathrm{~m}^{2}\) of walls and ceiling are a composite of \(1 \mathrm{~cm}\)-thick wallboard \((k=0.2 \mathrm{~W} / \mathrm{m} \mathrm{K}), 10\) \(\mathrm{cm}\) of vermiculite insulation \((k=0.06 \mathrm{~W} / \mathrm{m} \mathrm{K})\), and \(3 \mathrm{~cm}\) of wood \((k=0.15 \mathrm{~W} / \mathrm{m}\) K). Take the inside and outside heat transfer coefficients as 7 and \(35 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), respectively.

A hot-water cylinder contains 150 liters of water. It is insulated, and its outer surface has an area of \(3.5 \mathrm{~m}^{2}\). It is located in an area where the ambient air is \(25^{\circ} \mathrm{C}\), and the overall heat transfer coefficient between the water and the surroundings is \(1.0 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), based on outer surface area. If there is a power failure, how long will it take the water to cool from \(65^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\) ? Take the density of water as \(980 \mathrm{~kg} / \mathrm{m}^{3}\) and its specific heat as \(4180 \mathrm{~J} / \mathrm{kg} \mathrm{K}\).

An alloy cylinder \(3 \mathrm{~cm}\) in diameter and \(2 \mathrm{~m}\) high is removed from an oven at \(200^{\circ} \mathrm{C}\) and stood on its end to cool in air at \(20^{\circ} \mathrm{C}\). Give an estimate of the time for the cylinder to cool to \(100^{\circ} \mathrm{C}\) if the convective heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2}\) \(\mathrm{K}\). For the alloy, take \(\rho=8600 \mathrm{~kg} / \mathrm{m}^{3}, c=340 \mathrm{~J} / \mathrm{kg} \mathrm{K}\), and \(k=110 \mathrm{~W} / \mathrm{m} \mathrm{K}\).
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