Radiation MCQ Quiz - Objective Question with Answer for Radiation - Download Free PDF

Explanation:

Emissivity Factor for Energy Emitted by a Grey Body

• Emissivity is a measure of the efficiency of a surface in emitting thermal radiation compared to a perfect black body. A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and emits the maximum amount of energy at any given temperature. A grey body, however, emits less energy than a black body at the same temperature.

Mathematically, the emissivity factor (ε) is defined as the ratio of the energy emitted by a grey body per unit area per unit time (E) to the energy emitted by a perfect black body per unit area per unit time (E_B) at the same temperature:

\( \varepsilon = \frac{E}{E_B} \)

• \( E \) = Emitted energy by grey body (W/m²)
• \( E_B \) = Emitted energy by black body (W/m²)

Key Points:

• For a perfect black body, emissivity is equal to 1 because it emits the maximum possible energy at any given temperature.
• For real-world objects (grey bodies), emissivity is less than 1, as they emit less energy compared to a black body.
• Emissivity is a dimensionless quantity and typically ranges between 0 and 1.

Physical Significance:

The emissivity factor is crucial in determining the thermal radiation properties of materials. It plays a significant role in applications such as:

• Thermal engineering, where heat transfer calculations involve radiation.
• Designing heat exchangers and radiative cooling systems.
• Understanding and modeling the behavior of materials in high-temperature environments.
• Developing thermal imaging systems and sensors.

Concept:

Net heat transfer by radiation occurs when a body at higher temperature radiates energy toward a cooler body or surroundings.

According to Stefan–Boltzmann Law, a perfect black body radiates heat as,

\( Q = σ T^4 \)

For real bodies, emissivity \( \varepsilon \) is introduced:

\( Q = σ \varepsilon T^4 \)

Here,

σ = 5.67 × 10^-8 W/m²-K⁴ (Stefan–Boltzmann constant), and ε is the emissivity (0 ≤ ε ≤ 1)

Calculation:

Let Body 1 be at temperature T₁ and surroundings (or Body 2) at temperature T₂ .

The net radiative heat transfer is given by:

\( Q = σ \varepsilon_1 (T_1^4 - T_2^4) \)

Explanation:

Black Body:

• A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. A black body also emits radiation in a characteristic spectrum that depends only on its temperature, as described by Planck's Law.

Emissive Power:

• Emissive power of a black body refers to the total energy radiated per unit surface area per unit time. This is given by the Stefan-Boltzmann Law, which states:

E = σT⁴

Where:

• E = Emissive power (W/m²)
• σ = Stefan-Boltzmann constant (5.67 × 10^-8 W/m²K⁴)
• T = Absolute temperature of the black body (in Kelvin)

Calculation:

If temperature doubles: T' = 2T

\( E' = \sigma (2T)^4 = 16 \cdot \sigma T^4 = 16E \)

Explanation:

Gray Body

• A grey body is a theoretical object that has an emissivity less than 1 but remains constant over all wavelengths of radiation. This means that while it does not emit radiation as perfectly as a black body (which has an emissivity of 1), its emissivity does not vary with the wavelength of the radiation it emits.

Emissivity:

• Emissivity is a measure of a material's ability to emit energy as thermal radiation. It is defined as the ratio of the radiation emitted by a surface to the radiation emitted by a black body at the same temperature. A black body, with an emissivity of 1, is an ideal emitter that radiates the maximum possible energy at any given temperature. Real objects have emissivities less than 1, indicating that they emit less thermal radiation than a black body.

Characteristics of a Gray Body:

• Constant Emissivity: The primary characteristic of a gray body is that its emissivity is constant and does not change with the wavelength of the radiation. This is in contrast to real materials, whose emissivity can vary significantly with wavelength.
• Less than Perfect Emission: A gray body does not emit radiation as efficiently as a black body. The emissivity of a gray body is less than 1, indicating that it emits a fraction of the energy that a black body would emit at the same temperature.
• Practical Approximation: The concept of a gray body is useful in practical applications because it simplifies the analysis of thermal radiation. By assuming a constant emissivity, engineers and scientists can more easily calculate the thermal radiation properties of materials.

Explanation:

Intensity of Radiation and the Inverse Square Law

Definition: The intensity of radiation refers to the amount of energy radiated per unit area per unit time. It is a fundamental concept in physics, particularly in the study of electromagnetic waves and thermal radiation.

Correct Option: Inverse Square Law

The correct answer to the statement "Intensity of radiation varies with the:" is option 1, which states that the intensity of radiation varies with the inverse square of the distance. This principle is known as the inverse square law.

Inverse Square Law:

The inverse square law is a physical principle that describes how the intensity of a physical quantity (such as radiation, light, sound, etc.) decreases as the distance from the source increases. According to this law, the intensity of radiation is inversely proportional to the square of the distance from the source. Mathematically, it can be expressed as:

I ∝ 1/d²

Where:

• I = Intensity of radiation
• d = Distance from the radiation source

This means that if the distance from the source is doubled, the intensity of radiation becomes one-fourth. Similarly, if the distance is tripled, the intensity becomes one-ninth, and so on.

Derivation and Explanation:

The inverse square law can be derived from the geometry of how radiation spreads out from a point source. Imagine a point source emitting radiation uniformly in all directions. The radiation travels outward in spherical waves. As the distance from the source increases, the surface area of the sphere over which the radiation is spread increases.

The surface area of a sphere is given by the formula:
A = 4πd²

Where:

• A = Surface area of the sphere
• d = Radius of the sphere (distance from the source)

As the radiation travels outward, its energy is distributed over the surface area of the sphere. Since the surface area increases with the square of the distance, the intensity of radiation (energy per unit area) decreases with the square of the distance. Therefore, the intensity of radiation at distance d is inversely proportional to d².

Applications of the Inverse Square Law:

• Astronomy: The inverse square law is crucial in astronomy for understanding the brightness of stars and other celestial objects. The apparent brightness of a star decreases with the square of the distance from the observer.
• Radiation Safety: In radiation safety, the inverse square law is used to determine safe distances from radiation sources to minimize exposure. By increasing the distance from a radiation source, the intensity of exposure decreases significantly.
• Acoustics: In acoustics, the inverse square law explains how the intensity of sound decreases as the distance from the sound source increases. This principle is important in designing audio systems and soundproofing environments.
• Lighting: In lighting design, the inverse square law helps in calculating the illumination levels from light sources at different distances. It is essential for setting up proper lighting in various environments.

Explanation:

Net radiation heat exchange between two bodies is given by:

Q̇ = AF × σ × (T₁⁴ - T₂⁴)

where F, A, σ are shape factor, Area and Stefan-Boltzmann constant respectively

Now explanding  (T14 - T24)

Q̇ = AF × σ × ((T1)²)² - (T2)²)²)

Q̇ = AF × σ × (T12 - T22) × (T12 + T22)

Q̇ = AF × σ × (T₁ - T₂)(T₁ + T₂) ×  (T12 + T22)

\(\dot Q =\frac{T_1-T_2}{\frac{1}{\sigma \times AF \times(T_1+T_2)\times(T_1^2+T_2^2)}}\)

Comparing with the electrical analogy \(i=\frac VR\)

We will get thermal resistance as \(\frac{1}{{FA\sigma \left( {{T_1} + {T_2}} \right)\left( {T_1^2 + T_2^2} \right)}}\)

Concept:

Irradiation (G): The thermal radiation falling or incident upon a surface per unit time and per unit area is called irradiation

Radiosity (J): The total radiation energy leaving a surface per unit time per unit area is called radiosity.

Emissive Power (E): It is radiation energy leaving a surface.

J = Emitted Energy + Reflected part of incident energy

J = E + ρ G

For an opaque surface transmissivity (τ) = 0,

For any surface α + ρ + τ = 1

Where α = absorptivity, ρ = reflectivity, τ = transmissivity

Consider a small real body in thermal equilibrium with its surrounding blackbody cavity then by Kirchhoff's Law of radiation

⇒ absorptivity = emissivity ⇒ α = ϵ

∴ ρ + ϵ = 1, ρ = 1 - ϵ

J = E + (1 - ϵ) G

Calculation:

Given:

 radiosity (J) = 16 W/m², irradiation (G) = 24 W/m²,

Emissive power (E) = 12 W/m²

J = E + (1 - ϵ) G

16 = 12 + (1 - ϵ) 24

\(\epsilon= 1 - \frac{4}{{24}} = 0.83\)

J = E + ρG

J = εEb + ρG

Eb = Emissive power of a perfect black body

Concept:

From Wein's displacement law 

\({\lambda _{max}}T = 2898\;\mu m - k\left( {constant} \right)\)

Thus, \({\lambda _{peak}}T = \lambda _{peak}'T'\)

Calculation:

Given, 

Black body λpeak = 1.45 μm at 2000 K.

Now, \(\lambda _{peak}' = 2.90\;\mu m\)

1.45 × 2000 = 2.90 × T'

Concept:

F₁₁ + F₁₂ = 1

But F₁₁ = 0

∴ F₁₂ = 1

Now from reciprocating law

A₁ F₁₂ = A₂ F₂₁

Calculation:

\(\Rightarrow 4\pi r_1^2 \times I = 4\pi r_2^2 \times {F_{21}}\)

Concept:

Irradiation (G): Total radiation incident upon a surface per unit time per unit area.

Radiosity (J): Total radiation leaving a surface per unit time per unit area.

Radiosity comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.

J = E + ρG

J = εE_b + ρG

E_b = Emissive power of a perfect black body

α + ρ + τ = 1

For opaque body: τ = 0 ⇒ α + ρ = 1 ⇒ ρ = 1 - α = 1 - ε 

J = εE_b + (1 - ε)G = E + (1 - ε)G

Calculation:

Given:

G = 1000 W/m², E_b = 400 W/m², ϵ = 0.4

Explanation:  

Stephen's Boltzmann law:

According to Stefan’s law, the radiant energy emitted by a black body is directly proportional to the fourth power of its absolute temperature.

Q̇ = єσAT⁴

Where Q̇ is Radiate energy, σ is the Stefan-Boltzmann Constant, T is the absolute temperature in Kelvin, є is Emissivity of the material, and A is the area of the emitting body.

• Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth.
• A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.

From above it clear that the amount of radiation emitted by a perfectly black body is proportional to the fourth power of temperature on an ideal gas scale. Thus, option 3 is correct.

Explanation:

Let E be the total emissive power of the body and α be the absorptivity of the body.

α be the absorptivity of the body.

Emissivity (ϵ) is the ratio of emissive power of a non-black body to the black body.

ϵ = E/Eb

Kirchhoff’s law also holds for monochromatic radiation, for which

\(\frac{{{E_{\lambda 1}}}}{{{\alpha _{\lambda 1}}}} = \frac{{{E_{\lambda 2}}}}{{{\alpha _{\lambda 2}}}} = \frac{{{E_{b\lambda }}}}{{{\alpha _{b\lambda }}}} = \frac{{{E_{b\lambda }}}}{1}\)

∵ Absorptivity of a black body is one.

\(\therefore \frac{{{E_\lambda }}}{{{\alpha _\lambda }}} = {E_{b\lambda }} \Rightarrow {\alpha _\lambda } = \frac{{{E_\lambda }}}{{{E_{b\lambda }}}} = {\epsilon_\lambda }\)

Therefore, the monochromatic emissivity of a black body is equal to the monochromatic absorptivity at the same wavelength.

Eλ = αλEbλ

E = α EB

Important Points

According to Kirchhoff’s law, the ratio of total emissive power to absorptivity is constant for all bodies which are in thermal equilibrium with the surroundings. Therefore, it means that the emissivity of a body is equal to its absorptivity. 

Explanation:

Intensity of radiation:

The radiation intensity for emitted radiation I_e(θ, ϕ) is defined as the rate at which radiation energy dQ̇_e is emitted in the (θ, ϕ) direction per unit area normal to this direction and per unit solid angle about this direction. 

\({I_e}\left( {\theta ,\phi } \right)\; = \;\frac{{d{{\dot Q}_e}}}{{dA\cos \theta .d\omega }}\; = \;\frac{{d{{\dot Q}_e}}}{{dA\cos \theta \sin \theta \;d\theta d\phi }}\;\left( {\frac{W}{{{m^2}}}.sr} \right)\)

Emissivity (ϵ):

The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature.

For black body surface, ϵ = 1.

Irradiation (G):

The radiation flux incident on a surface from all directions is called irradiation G.

Absorptivity, reflectivity and transmissivity:

The fraction of irradiation absorbed by the surface is called the absorptivity α, the fraction reflected by the surface is called the reflectivity ρ, and the fraction transmitted is called the transmissivity τ.

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

Explanation:

When thermal radiation falls onto an object,

• The radiation will be absorbed by the surface of the object, causing its temperature to change
• The radiation will be reflected from the surface of the body, causing no temperature change
• The radiation will pass completely through the object, causing no temperature change

Absorptivity (α) is a measure of how much of the radiation is absorbed by the body

Reflectivity (ρ) is a measure of how much is radiation is reflected

Transmissivity (τ) is a measure of how much radiation passes through the object.

Each of these parameters is a number that ranges from 0 to 1.

Absorptivity may be a function of wavelength and/or direction, and is related to the emissivity of the region by Kirchhoff's law. The absorptivity is identically equal to unity for black bodies and is independent of temperature and wavelength for gray bodies.

Concept:

• Thermal radiation is the transmission of heat in the form of radiant energy from one body to another body by the thermal motion of charged particles of matter.
• The mechanism of radiation is divided into 3 phases
  • Conversion of thermal energy of the hot source into electromagnetic waves
  • Passage of wave motion through intervening medium
  • Transformation of waves into heat
• Thermal radiation is limited to wavelengths ranging from 0.1 to 100 µm of the electromagnetic spectrum.

• It includes some portion of UV radiation (0.1 to 0.4 µm), entire visible radiation (0.4 to 0.7 µm), and entire infrared radiation (0.7 to 100 µm)