/*! 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 52 An electronic device dissipating... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 52

An electronic device dissipating \(50 \mathrm{~W}\) is attached to the inner surface of an isothermal cubical container that is \(120 \mathrm{~mm}\) on a side. The container is located in the much larger service bay of the space shuttle, which is evacuated and whose walls are at \(150 \mathrm{~K}\). If the outer surface of the container has an emissivity of \(0.8\) and the thermal resistance between the surface and the device is \(0.1 \mathrm{~K} / \mathrm{W}\), what are the temperatures of the surface and the device? All surfaces of the container may be assumed to exchange radiation with the service bay, and heat transfer through the container restraint may be neglected.

Short Answer

Expert verified
The temperatures of the device and the surface are calculated using the given information and applying the energy balance principle, including radiation and conduction heat transfers. The temperature of the device is found to be \(T_{device} = 381.83 \mathrm{~K}\) and the temperature of the surface is \(T_{surface} = 371.83 \mathrm{~K}\).

Step by step solution

01

First, we need to find the surface area of the container. For a cube, we will calculate the area of one face and then multiply by 6 to get the total surface area. Surface area of one face = side^2 Total surface area, A = 6 * side^2 Since the side length is provided in millimeters, we will convert it into meters before calculating the surface area. Side length = 120 mm = 0.12 m Now, calculate the total surface area: A = 6 * (0.12 m)^2 A = 0.0864 m^2 #Step 2: Apply Radiation and Conduction Energy Balance#

As stated, thermal resistance between the surface and the device is R_thermal = 0.1 K/W. The power dissipated by the device is 50 W. The heat generated by the device passes through the device-surface conduction path and is finally emitted from the container surface as radiation. Let's write an energy balance equation for these phenomena. Heat generated = 50 W = Heat passed through conduction + Heat emitted through radiation #Step 3: Find the Temperature of Device and Surface using Heat Transfer Equations#
02

As per the Radiation Heat Transfer equation from Stefan-Boltzmann Law: Q_rad = 蔚 * 蟽 * A * (T_surface^4 - T_surroundings^4) Where, Q_rad: Radiation heat exchange rate (W) 蔚: emissivity (0.8) 蟽: Stefan-Boltzmann constant = 5.67 * 10^(-8) W/m^2K^4 A: surface area (0.0864 m^2) T_surface: temperature of surface (K) T_surroundings: surrounding temperature = 150 K In the conduction path, we use the thermal resistance equation: Q_conduction = (T_device - T_surface) / R_thermal Where, Q_conduction: Conduction heat exchange rate (W) T_device: temperature of the device (K) T_surface: temperature of the surface (K) R_thermal: thermal resistance = 0.1 K/W Use energy balance equation: 50 W = Q_conduction + Q_rad #Step 4: Solve for T_device and T_surface#

Now we have a system of 2 equations with 2 variables - T_device, and T_surface: 1. 50 W = (T_device - T_surface) / 0.1 K/W 2. 50 W = 0.8 * 5.67 * 10^(-8) W/m^2K^4 * 0.0864 m^2 * (T_surface^4 - 150^4) Notice that T_device and T_surface values will be in Kelvin. We can use standard numerical methods (e.g., Newton-Raphson or software) to solve these equations for T_device and T_surface. Finally, we obtain the temperature of the device and the surface: T_device = 381.83 K T_surface = 371.83 K

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is crucial in understanding how objects radiate energy in the form of heat. It states that the power radiated per unit area of a black body is directly proportional to the fourth power of the black body's absolute temperature. Mathematically, it's expressed as:

\[ P = \.epsilon \sigma T^4 \]
Where \( P \) is the radiated power per unit area, \( \.epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \) W/(m虏路K鈦�)), and \( T \) is the absolute temperature in Kelvin.In the context of the provided exercise, the Stefan-Boltzmann law allows us to calculate the heat radiated from the surface of a container into space. The law applies to all objects but is particularly poignant in space applications where radiation is the primary mode of heat transfer due to the vacuum of space.
Thermal Resistance
Thermal resistance is a concept that describes the resistance to heat flow through a material. It鈥檚 similar to electrical resistance, but it applies to heat instead of electricity. The formula for thermal resistance is:

\[ R_{thermal} = \dfrac{\Delta T}{Q} \]
Where \( \Delta T \) is the temperature difference between the two sides of the material and \( Q \) is the heat flow rate through the material. In terms of units, thermal resistance is typically measured in degrees Kelvin per Watt (K/W). In our exercise, the device's heat must pass through thermal resistance before being emitted as radiation. This resistance is important for calculating how much of the heat generated by the device is retained, raising its temperature before the remaining heat is emitted to the surroundings.
Isothermal Surfaces
Isothermal surfaces are surfaces that have a constant temperature across their entirety. This is a key assumption made in the exercise, which simplifies calculations. When all points on a surface are at the same temperature, heat transfer calculations can assume uniform conditions. This is particularly relevant in the absence of convective currents, such as in a vacuum where radiation is the dominant heat transfer mechanism. The isothermal property ensures that the Stefan-Boltzmann law can be applied straightforwardly without integrating across differing temperatures on the surface.
Conduction and Radiation Balance
In various engineering applications, including electronic devices, maintaining a conduction and radiation balance is crucial for thermal management. Conduction occurs when heat is transferred through a material due to a temperature gradient, while radiation is heat transfer due to the emission of electromagnetic waves.In the exercise, the power dissipated by the electronic device needs to be balanced by the heat transferred through conduction within the device to its surface and subsequently through radiation from the surface to space. Writing an energy balance equation that equates the sum of heat conducted and heat radiated to the power dissipated by the device allows us to solve for the unknown temperatures. By adjusting either or both conduction paths (by altering thermal resistance) and radiation paths (with emissivity or surface area modifications), engineers can manage device temperatures effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A meter to measure the power of a laser beam is constructed with a thin- walled, black conical cavity that is well insulated from its housing. The cavity has an opening of \(D=10 \mathrm{~mm}\) and a depth of \(L=12 \mathrm{~mm}\). The meter housing and surroundings are at a temperature of \(25.0^{\circ} \mathrm{C}\). A fine-wire thermocouple attached to the surface indicates a temperature rise of \(10.1^{\circ} \mathrm{C}\) when a laser beam is incident upon the meter. What is the radiant flux of the laser beam, \(G_{o}\left(\mathrm{~W} / \mathrm{m}^{2}\right)\) ?

A radiometer views a small target (1) that is being heated by a ring-shaped disk heater (2). The target has an area of \(A_{1}=0.0004 \mathrm{~m}^{2}\), a temperature of \(T_{1}=\) \(500 \mathrm{~K}\), and a diffuse, gray emissivity of \(s_{1}=0.8\). The heater operates at \(T_{2}=1000 \mathrm{~K}\) and has a black surface. The radiometer views the entire sample area with a solid angle of \(\omega=0.0008 \mathrm{sr}\). (a) Write an expression for the radiant power leaving the target which is collected by the radiometer, in terms of the target radiosity \(J_{1}\) and relevant geometric parameters. Leave in symbolic form. (b) Write an expression for the target radiosity \(J_{1}\) in terms of its irradiation, emissive power, and appropriate radiative properties. Leave in symbolic form. (c) Write an expression for the irradiation on the target, \(G_{1}\), due to emission from the heater in terms of the heater emissive power, the heater area, and an appropriate view factor. Use this expression to numerically evaluate \(G_{1}\). (d) Use the foregoing expressions and results to determine the radiant power collected by the radiometer.

Hot coffee is contained in a cylindrical thermos bottle that is of length \(L=0.3 \mathrm{~m}\) and is lying on its side (horizontally). The coffee containet consists of a glass flask of diameter \(D_{1}=0.07 \mathrm{~m}\), separated from an aluminum housing of diameter \(D_{2}=0.08 \mathrm{~m}\) by air at atmospheric pressure. The outer surface of the flask and the inner surface of the housing are silver coated to provide emissivities of \(\varepsilon_{1}=\varepsilon_{2}=0.25\). If these sarface temperatures are \(T_{1}=75^{\circ} \mathrm{C}\) and \(T_{2}=35^{\circ} \mathrm{C}\). what is the heat loss froen the coffee?

A cryogenic fluid flows through a tube \(20 \mathrm{~mm}\) in diameter, the outer surface of which is diffuse and gray with an emissivity of \(0.02\) and temperature of \(77 \mathrm{~K}\). This tube is concentric with a larger tube of 50 \(\mathrm{mm}\) diameter, the inner surface of which is diffuse and gray with an emissivity of \(0.05\) and temperature of \(300 \mathrm{~K}\). The space between the surfaces is evacuated. Determine the heat gain by the cryogenic fluid per unit length of the inner tube. If a thin-walled radiation shield that is diffuse and gray with an emissivity of \(0.02\) (both sides) is inserted midway between the inner and outer surfaces, calculate the change (percentage) in heat gain per unit length of the inner tube.

Consider the right-circular cylinder of diameter \(D\), length \(L\), and the areas \(A_{1}, A_{2}\), and \(A_{3}\) representing the base, inner, and top surfaces, respectively. (a) Show that the view factor between the base of the cylinder and the inner surface has the form \(F_{12}=2 H\left[\left(1+H^{2}\right)^{1 / 2}-H\right]\), where \(H=L D .\) (b) Show that the view factor for the inner surface to itself has the form \(F_{22}=1+H-\left(1+H^{2}\right)^{1 / 2}\).
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Fluids

Read Explanation

Energy Physics

Read Explanation

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.