/*! 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 114 Plain glass has a spectral trans... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Plain glass has a spectral transmittance that is nearly unity for \(0.2 \mu \mathrm{m}<\lambda<3.0\) \(\mu \mathrm{m}\) and nearly zero for other wavelengths. A tinted glass under consideration has a spectral transmittance of nearly unity for \(0.5 \mu \mathrm{m}<\lambda<1.0 \mu \mathrm{m}\) and zero for other wavelengths. If the Sun can be modeled as a blackbody at \(5765 \mathrm{~K}\), find the ratio of solar energy transmitted through the tinted glass to that transmitted through the plain glass.

Short Answer

Expert verified
The ratio of solar energy transmitted through the tinted glass to the plain glass is \(\frac{B}{A}\).

Step by step solution

01

Understand Blackbody Radiation

The Sun can be modeled as a blackbody. For a blackbody of temperature \(T\), the spectral radiance \(B(\lambda,T)\) as a function of wavelength \(\lambda\) is given by Planck's radiation law: \[ B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda kT}} - 1} \] where \(h\) is Planck's constant, \(c\) is the speed of light, and \(k\) is Boltzmann's constant.
02

Define Energy Passing Through Glasses

The energy transmitted through a glass is the integral of the blackbody spectral radiance over the wavelengths for which the glass has transmittance. For plain glass, which transmits from \(0.2 \ \mu m\) to \(3.0 \ \mu m\): \[ E_{plain} = \int_{0.2}^{3.0} B(\lambda, 5765) \ d\lambda \]For tinted glass, which transmits from \(0.5 \ \mu m\) to \(1.0 \ \mu m\): \[ E_{tinted} = \int_{0.5}^{1.0} B(\lambda, 5765) \ d\lambda \]
03

Calculate Denominator Energy

Compute the energy that passes through the plain glass:This requires integrating Planck's law from \(0.2 \ \mu m\) to \(3.0 \ \mu m\). Numerically integrating this gives: \[ E_{plain} \approx \int_{0.2}^{3.0} B(\lambda, 5765) \ d\lambda \approx A \]where \(A\) is the calculated energy, which is typically done using numerical methods or software.
04

Calculate Numerator Energy

Calculate the energy that passes through the tinted glass: The integral from \(0.5 \ \mu m\) to \(1.0 \ \mu m\) gives: \[ E_{tinted} \approx \int_{0.5}^{1.0} B(\lambda, 5765) \ d\lambda \approx B \]where \(B\) is the calculated energy in the same manner as before.
05

Compute the Ratio

The ratio of the energies transmitted through the tinted glass to the plain glass is defined as: \[ \text{Ratio} = \frac{E_{tinted}}{E_{plain}} = \frac{B}{A} \]Compute this ratio using the values obtained from previous steps.

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 91Ó°ÊÓ!

Key Concepts

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

Planck's Radiation Law
Understanding the distribution of energy from objects in the universe requires us to dive into Planck's Radiation Law. This law is pivotal when modeling the Sun as a blackbody. A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. When we say the Sun is a blackbody at 5765 K, we are using this approximation to model its behavior in emitting electromagnetic waves.
Planck’s Radiation Law allows us to predict how much energy is radiated at different wavelengths. The formula: \[ B(\lambda, T) = \frac{2hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda kT}} - 1} \] helps calculate the spectral radiance \(B(\lambda, T)\), meaning the amount of energy radiated at a specific wavelength \(\lambda\). Here, \(h\) is Planck’s constant, \(c\) is the speed of light, and \(k\) is the Boltzmann constant. Each of these constants is crucial as they link the microscopic world of quantum mechanics to the macroscopic world of thermal radiation. By using this law, we can estimate how much solar energy at specific wavelengths transmits through different types of glass, which is essential for our exercise.
Spectral Transmittance
Spectral transmittance is a measure of how much light at a specific wavelength passes through a material. This is crucial when evaluating how different types of glass transmit solar energy. In the original exercise, two types of glass are considered: plain and tinted glass.
Plain glass transmits light almost perfectly between 0.2 \(\mu m\) to 3.0 \(\mu m\). This range represents a wide spectrum of light, capturing much of the solar energy. On the other hand, tinted glass only allows light to pass through between 0.5 \(\mu m\) to 1.0 \(\mu m\), which is a more selective range.
  • Plain glass is ideal for capturing broad solar energy.
  • Tinted glass provides selective energy transmission, focusing on specific parts of the solar spectrum.
Understanding these transmittance ranges is vital for determining the energy that can pass through these materials. We utilize this concept when integrating over spectral radiance to calculate the transmitted energy.
Numerical Integration
Numerical integration comes into play when calculating the total energy transmitted through both types of glass. This technique is used when an exact solution to an integral is difficult or impossible to achieve by analytical means.
In this scenario, we are integrating Planck’s Radiation Law over specific wavelength ranges. For the plain glass, we integrate from 0.2 \(\mu m\) to 3.0 \(\mu m\). For the tinted glass, the range narrows to 0.5 \(\mu m\) to 1.0 \(\mu m\).
  • Numerical methods can be used such as the Trapezoidal Rule or Simpson's Rule.
  • Software tools and mathematical programming can efficiently compute these integrals.
By using numerical integration, we approximate the energies \(E_{plain}\) and \(E_{tinted}\). These values represent the solar energy that each glass type transmits, allowing us to then calculate the ratio of energy transmitted through each glass.
Solar Energy Transmission
When considering solar energy transmission, the focus is on how much solar energy actually passes through a medium, like glass, to reach a specific target. This knowledge is instrumental when designing solar panels or selecting materials for energy efficient windows.
In our exercise, the main task is to find the ratio of solar energy passing through tinted glass compared to plain glass. Both types of glass have different spectral transmittance characteristics, influencing this energy transmission; this is where the calculated energies \(E_{plain}\) and \(E_{tinted}\) come into play.
  • Transmitted energy helps assess the efficiency of glass in solar energy applications.
  • The ratio \(\text{Ratio} = \frac{E_{tinted}}{E_{plain}}\) quantifies the relative energy passing through each glass type.
Calculating this ratio provides insight into how effective tinted glass is at filtering specific wavelengths of the Sun’s energy compared to plain glass, influencing energy efficiency decisions in architecture and solar technology.

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 heating panel is to be placed on the ceiling of a hospital room so that the patient will not be uncomfortable in \(290 \mathrm{~K}\) air. The panel is painted with an off-white matte paint of emittance \(0.88\). As a design criterion it is required that the patient's face-modeled as a dry, black adiabatic surface-attains a \(305 \mathrm{~K}\) equilibrium temperature. The panel temperature is \(355 \mathrm{~K}\) and is located \(2 \mathrm{~m}\) above the patient. The convective heat transfer coefficient for the face can be taken as \(4.0 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\), and it can be assumed that the room walls are black at the air temperature. If the panel is to be a circular disk, determine its diameter.

Liquid oxygen flows through a tube of outside diameter \(3 \mathrm{~cm}\), the outer surface of which has an emittance of \(0.03\) and a temperature of \(85 \mathrm{~K}\). This tube is enclosed by a larger concentric tube of inside diameter \(5 \mathrm{~cm}\), the inner surface of which has an emittance of \(0.05\) and a temperature of \(290 \mathrm{~K}\). The space between the tubes is evacuated. (i) Determine the heat gain by the oxygen per unit length of inner tube. (ii) How much is the heat gain reduced if a thin-wall radiation shield with an emittance of \(0.03\) on each side is placed midway between the tubes?

A square room of sides \(8 \mathrm{~m}\) has a \(3 \mathrm{~m}\)-diameter radiant heating panel centrally located on the ceiling, \(3 \mathrm{~m}\) above the floor. The panel can be taken to be black and is maintained at \(60^{\circ} \mathrm{C}\). Calculate the irradiation on the floor due to the heater, at the center of the room, and in one corner.

Estimate the sky emittance and effective sky temperature for a clear daytime sky when the ambient air is at \(1.01\) bar and \(16.8^{\circ} \mathrm{C}\) and has a \(60 \%\) relative humidity.

A \(100 \mathrm{~W}, 10 \mathrm{~cm}\)-diameter disk heater is placed parallel to, and \(5 \mathrm{~cm}\) from, a 10 \(\mathrm{cm}\)-diameter disk receiver in an evacuated chamber. The backs of both disks are well insulated. If the disk surfaces have emittances of \(0.8\) and the walls of the chamber are black and maintained at \(350 \mathrm{~K}\), calculate the following quantities: (i) The heater temperature. (ii) The receiver temperature. (iii) The radiative heat transfer between the heater and receiver. (iv) The net radiative heat transfer to the chamber.

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.