Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Five identical thin aluminum sheets with emissivities of \(0.1\) on both sides are placed between two very large parallel plates, which are maintained at uniform temperatures of \(T_{1}=800 \mathrm{~K}\) and \(T_{2}=450 \mathrm{~K}\) and have emissivities of \(\varepsilon_{1}=\) \(\varepsilon_{2}=0.1\), respectively. Determine the net rate of radiation heat transfer between the two plates per unit surface area of the plates, and compare the result to that without the shield.

Short Answer

Expert verified
Answer: To assess the effect of adding the aluminum sheets as shields, the net rate of radiation heat transfer per unit surface area with and without the shield needs to be calculated using the radiosity equation. After performing the required calculations, compare the heat transfer rates with and without the shield and discuss the improvement or degradation in heat transfer due to the presence of the aluminum sheets.

Step by step solution

01

Understand the given quantities and symbols

In this exercise, we have the following quantities given: - Emissivities of the aluminum sheets (\(\epsilon_{sh} = 0.1\)) - Emissivities of the parallel plates: \(\epsilon_{1} = \epsilon_{2} = 0.1\) - Temperatures of the parallel plates: \(T_{1}=800 \mathrm{~K}\) and \(T_{2}=450 \mathrm{~K}\). We need to calculate the net rate of radiation heat transfer per unit surface area, denoted as \(q\).
02

Calculate surface emissive power of the plates using the Stefan-Boltzmann Law

Stefan-Boltzmann law states that the emitted power per unit area (\(E\)) is proportional to the fourth power of the absolute temperature (\(T\)). The constant of proportionality is the Stefan-Boltzmann constant (\(\sigma = 5.67 \times 10^{-8} \mathrm{W m^{-2} K^{-4}}\)). Calculate the emitted power of the plates: \(E_1 = \sigma T_{1}^4\) and \(E_2 = \sigma T_{2}^4\) where \(E_1\) and \(E_2\) are the emitted power of the plates 1 and 2, and \(T_{1}\) and \(T_{2}\) are the temperatures of the plates 1 and 2.
03

Calculate the radiosity of the parallel plates and aluminum sheets

Radiosity (denoted as \(J\)) is the total energy leaving a surface, which consists of both emitted and reflected energy. Calculate the radiosity of the parallel plates: \(J_1 = E_1 \epsilon_1 + R_1 (1-\epsilon_1)\) and \(J_2 = E_2 \epsilon_2 + R_2 (1-\epsilon_2)\) Calculate the radiosity of the aluminum sheets: \(J_{sh1} = E_{sh1} \epsilon_{sh} + R_{sh1} (1-\epsilon_{sh})\), \(J_{sh2} = E_{sh2} \epsilon_{sh} + R_{sh2} (1-\epsilon_{sh})\), ..., \(J_{sh5} = E_{sh5} \epsilon_{sh} + R_{sh5} (1-\epsilon_{sh})\) where \(J_1\) and \(J_2\) are the radiosity of the plates 1 and 2, \(E_{1}\) and \(E_{2}\) are the emitted power of the plates 1 and 2, \(\epsilon_{1}\) and \(\epsilon_{2}\) are the emissivities of the plates, and \(R_1\) and \(R_2\) are the reflected powers of the plates 1 and 2. \(J_{sh1}\), \(J_{sh2}\), ..., \(J_{sh5}\) are the radiosity of the aluminum sheets, \(E_{sh1}\), \(E_{sh2}\), ..., \(E_{sh5}\) are the emitted power of the aluminum sheets, and \(R_{sh1}\), \(R_{sh2}\), ..., \(R_{sh5}\) are the reflected powers of the aluminum sheets. Note that \(R_{sh1}=J_1\), \(R_{sh2}=J_{sh1}\), and so on, such that \(R_{sh5}=J_{sh4}\) and \(R_2=J_{sh5}\).
04

Determine the heat transfer rate per unit surface area by applying the radiosity equation

The radiosity equation states that the heat transfer rate per unit surface area \(q\) is the difference in radiosity between the parallel plates divided by the sum of the resistances between plates, including the aluminum sheets. Calculate \(q\) with the shield: \(q = \frac{J_1 - J_2}{1+\frac{\epsilon_{sh} -1}{\epsilon_{1}}+\frac{\epsilon_{sh} -1}{\epsilon_{2}}+4(\frac{\epsilon_{sh} -1}{\epsilon_{sh}})}\) Calculate \(q\) without the shield: \(q_{wo} = \frac{J_1 - J_2}{1+\frac{\epsilon_{1}-1}{\epsilon_{2}}}\)
05

Perform calculations and compare the results

Use the given temperatures, emissivities, and the Stefan-Boltzmann constant to calculate the emitted power, radiosity, and heat transfer rates with and without the shield. Then, discuss the differences between the two rates and explain if using the aluminum sheets improved heat transfer or not. Top tip: Pay attention to the units of the calculations and compare the values in the same unit system.

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!

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

Two very large parallel plates are maintained at uniform temperatures \(T_{1}=750 \mathrm{~K}\) and \(T_{2}=500 \mathrm{~K}\) and have emissivities \(\varepsilon_{1}=0.85\) and \(\varepsilon_{2}=0.7\), respectively. If a thin aluminum sheet with the same emissivity on both sides is to be placed between the plates in order to reduce the net rate of radiation heat transfer between the plates by 90 percent, the emissivity of the aluminum sheet must be (a) \(0.07\) (b) \(0.10\) (c) \(0.13\) (d) \(0.16\) (e) \(0.19\)

Explain all the different mechanisms of heat transfer from the human body \((a)\) through the skin and \((b)\) through the lungs.

A thermocouple used to measure the temperature of hot air flowing in a duct whose walls are maintained at \(T_{w}=500 \mathrm{~K}\) shows a temperature reading of \(T_{\text {th }}=850 \mathrm{~K}\). Assuming the emissivity of the thermocouple junction to be \(\varepsilon=0.6\) and the convection heat transfer coefficient to be \(h=75 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\), determine the actual temperature of air.

Consider two diffuse surfaces \(A_{1}\) and \(A_{2}\) oriented on a spherical surface as shown in the figure. Determine \((a)\) the expression for the view factor \(F_{12}\) in terms of \(A_{2}\) and \(L\), and (b) the value of the view factor \(F_{12}\) when \(A_{2}=0.02 \mathrm{~m}^{2}\) and \(L=1 \mathrm{~m}\).

A solar collector consists of a horizontal copper tube of outer diameter $5 \mathrm{~cm}\( enclosed in a concentric thin glass tube of diameter \)12 \mathrm{~cm}$. Water is heated as it flows through the tube, and the annular space between the copper and the glass tubes is filled with air at $1 \mathrm{~atm}$ pressure. The emissivities of the tube surface and the glass cover are \(0.85\) and \(0.9\), respectively. During a clear day, the temperatures of the tube surface and the glass cover are measured to be $60^{\circ} \mathrm{C}\( and \)40^{\circ} \mathrm{C}$, respectively. Determine the rate of heat loss from the collector by natural convection and radiation per meter length of the tube.

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