Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Short Answer

Expert verified
Answer: The surrounding surfaces must be kept at approximately 11.25°C if the box is to be cooled by radiation alone.

Step by step solution

01

Calculate surface area of the box

To find the surface area, we need to know the total area of all the walls of the box exposed to the surroundings except the base, which is assumed to have negligible heat transfer. Therefore, the area is given by the sum of the areas of four vertical walls which can radiate heat. \(A_{\mathrm{surface}} = 4 \times (\mathrm{height} \times \mathrm{width}) = 4 \times ( 0.2 \, \mathrm{m}\times 0.5 \, \mathrm{m}) = 0.4 \, \mathrm{m^2}\)
02

Arrange the radiative heat transfer formula

We need to find the surrounding temperature (\(T_{\mathrm{surrounding}}\)), so we will rearrange the radiative heat transfer formula to isolate this variable. \(\displaystyle T_{\mathrm{surrounding}}^4 = \frac{q - \epsilon \sigma A_{\mathrm{surface}}T_{0}^4}{\epsilon \sigma A_{\mathrm{surface}}}\)
03

Insert given values and solve for surrounding temperature

Now we will plug in the values of \(\epsilon\), \(\sigma\), \(A_{\mathrm{surface}}\), \(q\), and \(T_{0}\) to find \(T_{\mathrm{surrounding}}\). \(\epsilon = 0.95\) \(\sigma = 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4}\) \(A_{\mathrm{surface}} = 0.4 \, \mathrm{m^2}\) \(q = 120 \, \mathrm{W}\) \(T_{0} = 55 + 273.15 = 328.15 \, \mathrm{K}\) \(\displaystyle T_{\mathrm{surrounding}}^4 = \frac{120 - 0.95 \cdot 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \cdot 0.4 \, \mathrm{m^2} \cdot (328.15\,\mathrm{K})^4}{0.95 \cdot 5.67 \times 10^{-8} \, \mathrm{W/m^2K^4} \cdot 0.4 \, \mathrm{m^2}}\) After calculating, we get: \(T_{\mathrm{surrounding}}^4 = 1.921\times 10^{11} \, \mathrm{K^4}\) Taking the fourth root of both sides, we get: \(T_{\mathrm{surrounding}} = 284.4\, \mathrm{K}\)
04

Convert the result back to Celsius

We will now convert the surrounding temperature back to Celsius. \(T_{\mathrm{surrounding}}^{\circ\mathrm{C}} = 284.4 - 273.15 = 11.25^{\circ}\mathrm{C}\) The surrounding surfaces must be kept at approximately \(11.25^{\circ} \mathrm{C}\) if the box is to be cooled by radiation alone.

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 Vaia!

Key Concepts

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

Heat Dissipation in Electronics
Electronics generate heat when operating, which can affect performance and longevity if not managed properly. Heat dissipation is crucial in maintaining the operational integrity of electronic devices. Most electronics have a thermal management system, which can include passive or active cooling methods. Passive methods, like heat sinks, enhance heat transfer through increased surface area and materials with high thermal conductivity, whereas active methods can include the use of fans or liquid cooling systems.

In the given exercise, an electronic box dissipates heat solely through radiation, a form of passive cooling. Radiation allows heat to be transferred without direct contact or an intervening medium, ideal for a vacuum chamber. The challenge is to ensure that the temperature of the electronic box does not exceed a set limit, requiring a precise calculation of radiative heat transfer and control of the surrounding temperature.
Surface Area Calculation
Calculating the surface area is essential when dealing with heat transfer in devices. A larger surface area usually means more heat can be dissipated, leading to better cooling.

The problem provided requires calculating the area that radiates heat. This is done by summing the areas of all radiating surfaces. For box-shaped objects, like the given electronic box, the total area can be found by adding the individual areas of the sides that contribute to heat transfer. In the solution, the base's heat transfer is negligible, so it's excluded from the calculation.

Students must remember that accurate measurement and using the correct units are crucial to ensure the efficacy of the calculation, especially when applying formulas that involve area, like the Stefan-Boltzmann law.
Stefan-Boltzmann Law
The Stefan-Boltzmann law is a fundamental principle in thermodynamics that relates the power radiated from a black body to the fourth power of its temperature. It's expressed as:

\( P = \epsilon \sigma A T^4 \)
where \( P \) is the power radiated per unit surface area, \( \epsilon \) is the emissivity of the material, \( \sigma \) is the Stefan-Boltzmann constant \( (5.67 \times 10^{-8} \mathrm{W/m^2K^4}) \), \( A \) is the surface area, and \( T \) is the absolute temperature in kelvins.

In the context of our problem, the law allows us to calculate the required temperature of surrounding surfaces to dissipate the heat from the electronic box. By rearranging the formula to solve for the surrounding temperature, we can keep the electronic components within a safe operating temperature. Understanding the Stefan-Boltzmann law is crucial for professionals dealing with thermal management in various fields, including electronics, mechanical engineering, and aerospace.

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 cylindrical resistor element on a circuit board dissipates \(1.2 \mathrm{~W}\) of power. The resistor is \(2 \mathrm{~cm}\) long, and has a diameter of \(0.4 \mathrm{~cm}\). Assuming heat to be transferred uniformly from all surfaces, determine \((a)\) the amount of heat this resistor dissipates during a 24-hour period, \((b)\) the heat flux, and \((c)\) the fraction of heat dissipated from the top and bottom surfaces.

A series of experiments were conducted by passing \(40^{\circ} \mathrm{C}\) air over a long \(25 \mathrm{~mm}\) diameter cylinder with an embedded electrical heater. The objective of these experiments was to determine the power per unit length required \((\dot{W} / L)\) to maintain the surface temperature of the cylinder at \(300^{\circ} \mathrm{C}\) for different air velocities \((V)\). The results of these experiments are given in the following table: $$ \begin{array}{lccccc} \hline V(\mathrm{~m} / \mathrm{s}) & 1 & 2 & 4 & 8 & 12 \\ \dot{W} / L(\mathrm{~W} / \mathrm{m}) & 450 & 658 & 983 & 1507 & 1963 \\ \hline \end{array} $$ (a) Assuming a uniform temperature over the cylinder, negligible radiation between the cylinder surface and surroundings, and steady state conditions, determine the convection heat transfer coefficient \((h)\) for each velocity \((V)\). Plot the results in terms of \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\) vs. \(V(\mathrm{~m} / \mathrm{s})\). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. (b) Assume that the heat transfer coefficient and velocity can be expressed in the form of \(h=C V^{m}\). Determine the values of the constants \(C\) and \(n\) from the results of part (a) by plotting \(h\) vs. \(V\) on log-log coordinates and choosing a \(C\) value that assures a match at \(V=1 \mathrm{~m} / \mathrm{s}\) and then varying \(n\) to get the best fit.

Using the parametric table and plot features of \(\mathrm{EES}\), determine the squares of the number from 1 to 100 in increments of 10 in tabular form, and plot the results.

An electric heater with the total surface area of \(0.25 \mathrm{~m}^{2}\) and emissivity \(0.75\) is in a room where the air has a temperature of \(20^{\circ} \mathrm{C}\) and the walls are at \(10^{\circ} \mathrm{C}\). When the heater consumes \(500 \mathrm{~W}\) of electric power, its surface has a steady temperature of \(120^{\circ} \mathrm{C}\). Determine the temperature of the heater surface when it consumes \(700 \mathrm{~W}\). Solve the problem (a) assuming negligible radiation and (b) taking radiation into consideration. Based on your results, comment on the assumption made in part ( \(a\) ).

One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

See all solutions

Recommended explanations on Physics Textbooks

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.

Sign-up for free