Absorptivity, Reflectivity and Transmissivity MCQ Quiz in বাংলা - Objective Question with Answer for Absorptivity, Reflectivity and Transmissivity - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

• Fourier number is defined as the ratio of heat conducted to heat stored in the system.
• Also, it is defined as the ratio of operating time to diffusion time.
• \(Fourier\;No = \frac{{Operating\;time}}{{diffusion\;time}}\)

\({F_o} = \frac{t}{{\left( {\frac{{L_c^2}}{\alpha }} \right)}}\)

Where,

T = operating time, Lc = Characteristic length, α = thermal diffusivity \(\left( {\frac{{{m^2}}}{{sec}}} \right)\)

α = 2τ ⇒ τ = α/2

α = 3ρ ⇒ ρ = α/3

α + τ + ρ = 1

\(\alpha \left( {1 + \frac{1}{2} + \frac{1}{3}} \right) = 1 \Rightarrow \alpha = \frac{6}{{11}},\tau = \frac{3}{{11}}\) 

Energy transmitted:

= 660 × τ × A

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:

J = E + (1 - ε)G

12 = 10 + (1 - ε)20 

⇒ ε = 0.9

As the surface is opaque, so the transmissivity will be equal to zero.

So the incident radiation will either be absorbed or reflected.

Q̇ = σAT⁴

= 5.67 × 10^-8 × π (0.2)² × (800)⁴

Q̇ = 2.918 kW

Q = 2.918 × 5 × 60 = 875.54 kJ

Mistake Point: Note that surface area of sphere is taken into consideration which is π​d²

Radiosity (J) = Total radiation energy leaving a surface per unit time per unit area

J = Emitted energy + Reflected part of Incident energy

J = E + ρG

Where E = Emissive power

G = Irradiation

Which body : r = 1            τ = α = 0

Black body : α = 1             τ = r = 0

Transparent body : τ = 1         α = r = 0

Opaque body : τ = 0

Concept:

If anybody radiates equally in all directions, then it is called a diffuse body.

For diathermonous bodies also known as transparent bodies, the transmissivity (τ) = 1

⇒ absorptivity (α) = 0; reflectivity (ρ) = 0;

For perfect black body,

• Absorptivity = 1; Emissivity = 1;
• Transmissivity = 0; Reflectivity = 0;

Gray body

• If a body absorbs or emits radiation in constant proportions irrespective of wavelength then it is called a Gray body. 
• Absorptivity = Transmissivity = Reflectivity = constant

Opaque body

• For opaque bodies, Transmissivity = 0 ⇒ absorptivity + reflectivity = 1;

White body

• For white bodies, reflectivity = 1 ⇒ transmissivity = 0; absorptivity = 0;