All Exams Test series for 1 year @ ₹349 only
Question

For an opaque surface, the absorptivity (α) , transitivity (τ) and reflectivity (ρ) are related by the equation

The correct answer is

α + ρ = 1

Opaque Surface Radiation Properties

When thermal radiation strikes any surface, it can interact in three primary ways: it can be absorbed by the surface, reflected off the surface, or transmitted through the surface. The portions of the incident radiation that undergo these processes are quantified by specific material properties.

Radiation Properties Defined

  • Absorptivity (\( \alpha \)): This is the fraction of the total incident radiation that is absorbed by the surface. A higher absorptivity means more energy is taken in by the material.
  • Reflectivity (\( \rho \)): This is the fraction of the total incident radiation that is reflected by the surface. Surfaces with high reflectivity appear shiny and can be used to redirect radiation.
  • Transmissivity (\( \tau \)): This is the fraction of the total incident radiation that passes through the surface. Materials like glass are transparent, meaning they have high transmissivity for visible light.

The fundamental principle of energy conservation dictates that the sum of these fractions for any surface must always equal one, as all incident radiation must be accounted for. This general relationship is expressed by the equation:

\( \alpha + \tau + \rho = 1 \)

Opaque Surface Behavior

The question specifically refers to an opaque surface. An opaque surface is defined as a material that does not allow any thermal radiation to pass through it. In other words, for an opaque surface, the transmissivity (\( \tau \)) is effectively zero.

\( \tau = 0 \text{ (for an opaque surface)} \)

Relating Absorptivity and Reflectivity for Opaque Surfaces

To find the relationship for an opaque surface, we substitute the condition \( \tau = 0 \) into the general energy conservation equation:

\( \alpha + 0 + \rho = 1 \)

This simplifies directly to the relationship:

\( \alpha + \rho = 1 \)

Therefore, for an opaque surface, the sum of its absorptivity and reflectivity always equals one. This means that an opaque surface either absorbs or reflects all the incident radiation.

Based on this derivation, the correct equation relating absorptivity (\( \alpha \)) and reflectivity (\( \rho \)) for an opaque surface is \( \alpha + \rho = 1 \).

Was this answer helpful?

Important Questions from Absorptivity, Reflectivity and Transmissivity

  1. If plate 1 is also a diffuse and gray surface with an emissivity value of 0.8, the net radiation heat exchange (in kW/m2) between plate 1 and plate 2 is

  2. A gray body is defined such that

  3. In a radiative heat transfer, a gray surface is one

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App