Properties of Surface Emission MCQ Quiz in मल्याळम - Objective Question with Answer for Properties of Surface Emission - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Stefan-Boltzmann Law

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

\(\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4} = {\left( {\frac{{1227 + 273}}{{227\; + 273}}} \right)^4} = {\left( {\frac{{1500}}{{500}}} \right)^4} = 81\)

Concept:

• For any type of body,

Absorptivity (α) + Reflectivity (ρ) + Transmissivity (τ) = 1

Radiosity (J) = [Reflectivity (ρ) × Irradiation (G)] + Emissive power (E) + Transmitted irradiation (T)

Explanation:

Different types of bodies:

• Now for non-reflecting body, ρ = 0 ⇒ J = E + T (Option 1 is wrong)
• For white body, ρ = 1 ⇒ α + τ = 1 ⇒ α = 0, τ = 0 can not be possible. (Option 2 is wrong)
• A diathermanous body is a body that transmits all the incident radiation without absorbing or reflecting it.
• Hence τ = 1 ⇒ α + ρ = 0 ⇒ α = 0, ρ = 0. (Option 3 is wrong)
• All black bodies behave like black bodies but all black bodies need not be black in color. (Option 4 is wrong)
• For example, a small hole leading into a cavity acts nearly as a black body which need not be black in color.

Concept:

When there is vacuum inside the chamber then the heat is lost by the radiation.

The heat lost through radiation is given as

\(Q = \sigma \times \left( {T_1^4 - T_2^4} \right) \times A \times \epsilon\)

where σ is the Stefan Boltzman’s constant

σ = 5.67 × 10^-8 W/m²K⁴

A is the area of the surface

ϵ is the emissivity of the surface

Calculation:

Given, Temperature of furnace, T₁ = 727°C = 1000 K, ϵ = 0.2, A = 10 m², Temperature of the surrounding T₂ = 27°C = 300 K.

Then heat lost by the furnace

\({Q_{lost}} = 5.67 \times {10^{ - 8}} \times 10 \times 0.2 \times \left( {{{1000}^4} - {{300}^4}} \right)\)

Q_lost = 112481.46 W

Q_lost = 112.481 kW

Since it is the heat lost by the furnace, which is equal to 10% of the heat generated.

Then,

Q_lost = 0.1 × Q_gen

\({Q_{gen}} = \frac{{{Q_{lost}}}}{{0.1}} = \frac{{112.481}}{{0.10}} = 1124.81\;kW\)

Solar thermal collector:

It is a device which is used to collect the heat by absorbing the sunlight. It is mainly used to collect the solar energy and then convert into thermal energy.

The most commonly used solar thermal collector is flat plate collector.

Flat plate solar collector: The flat-plate solar collectors are probably the most fundamental and most studied technology for solar-powered domestic hot water systems. The Sun heats a dark flat surface, which collects as much energy as possible, and then the energy is transferred to water, air, or other fluid for further use.

The solar radiation is absorbed by the black plate and transfers heat to the fluid in the tubes. The thermal insulation prevents heat loss during fluid transfer; the screens reduce the heat loss due to convection and radiation to the atmosphere

The main components of a typical flat-plate solar collector:

• Black surface - absorbent of the incident solar energy
• Glazing cover - a transparent layer that transmits radiation to the absorber, but prevents radiative and convective heat loss from the surface
• Tubes containing heating fluid to transfer the heat from the collector
• Support structure to protect the components and hold them in place
• Insulation covering sides and bottom of the collector to reduce heat losses

Advantages:

• Easy to manufacture
• Low cost
• Collect both beam and diffuse radiation
• Permanently fixed (no sophisticated positioning or tracking equipment is required)
• Little maintenance

The efficiency of flat plate collectors is usually 61%.

​CONCEPT:

• White and shiny surfaces are poor absorbers of heat and light whereas dark and black surfaces are the good absorber of heat and light.
  • A person with dark skin will experience more heat and mole cold in comparison to white skin people.

EXPLANATION:

• We prefer white clothes in summer because white clothes reflect most of the sun's heat and absorb very little of the sun's heat and keep our body bodies.
• When the body will be in black colors (Like Black cloths), all the light gets absorbed because black is the good absorber and bad reflector (ρ = 0, τ = 0, and α = 1) whereas white color body like (White cloths) emits or reflect all the light coming towards it (ρ = 1, τ = 0, and α = 0) 
• So, for white color,

α = 0, τ = 0 and ρ = 1

So, the correct answer is option1.

Additional Information

• Application of White colors:
  • Radiator in homes is painting with white.
  • We wear white cloth in summer because it reflects all the light and makes our body cool.

Additional Information

• Dark people like people who live in Africa have the ability to absorb, an average of 36% heat or solar radiation than any other people in the world. 
• There are no such bodies that are perfectly absorbing or emitting light, this means no light can be absorbed or emitted totally.

Concept:

Radiosity (J):

Radiosity is the term used to indicate the total energy leaving a surface per unit time and per unit surface area. Energy leaving can be due to its own temperature or due to the reflected energy which incident on it.

J = E + G_R

where E = ϵE_b = ϵσT⁴= energy leaving per unit area due to its own temperature and G_R = (1 - ϵ)σT⁴_atm = reflected energy due to the energy falling from atmosphere.

Calculation:

Given:

d = 40 cm,  ϵ = 0.65 , T_atm = 30°C = 30 + 273 = 303 K ,T = 55°C = 55 + 273 = 328 K

σ = Stefan Boltzman Constant = 5.67 × 10^-8  W/m²K⁴

J =  ϵσT⁴ + (1 - ϵ)σT⁴_atm

J =  0.65 × 5.67 × 10^-8 × 328⁴ + ( 1 – 0.65) × 5.67 × 10^-8 × 303⁴

J =  594 W/m²

Concept:

Concept:

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

Calculation:

Given:

a₁ = 1 cm, a₂ = 10 cm, T₁ = 400 K, T₂ = 2000 K, Emissivity of large cube, ϵ = 0.5

F₁₂ = 1

A₁ F₁₂ = A₂ F₂₁

6 × 1² × F₁₂ = 6 × 10² × F₂₁

F₂₁ = 0.01

Irradiation on surface (1) = Energy emitted by surface 2 falling on 1 + Energy emitted by surface 1 which is reflected from surface 2 & reaches surface 1.

\(\rm G=ε σ A_2T_2^4 F_{21}+σ A_1T_1^4ρ F_{21}\)

ρ = refectivity of surface 2 = 1 - absorptivity & absorptivity= ε

⇒ ρ = ε = 0.5

\(\rm G=σ(ε A_2T_2^4 F_{21}+A_1T_1^4ρ F_{21})\)

G = 5.67 × 10^-8 × ( 0.5 × 6 × 0.1² × 2000⁴ × 0.01 + 6 ×  0.012 × 4004 × 0.5 × 0.01)

G = 272.16 W

Concept:

Stefan-Boltzmann Law

The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:

E ∝ T4

E = σT4

σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4

Calculation:

Given:

T₁ = 400 K, T₂ = 1200 K

\(\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4} = {\left( {\frac{{1200}}{{400}}} \right)^4} =3^4 = 81\)

Explanation:

1 shield kept between the plates will bring in 3 additional resistances extra into the network out of which 2 are surface resistances and 1 is space resistance.

Hence if there are n number of radiation shields, then a total 2n + 2 number of surface resistance and n + 1 number of space resistances will be there in the radiation network drawn with n number of shields.

The formula for heat flux without shield when every surfaces has emissivity ϵ,

\({\left( {\frac{q}{A}} \right)_{without\;shield}}= \;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{{\frac{1}{\epsilon_1}+}\frac{1}{\epsilon_2} - 1}} = \;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}}\)

\({\left( {\frac{q}{A}} \right)_{with\;one\;shield}} = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{4}{\epsilon} - 2}} = \frac{1}{2}\;\times\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}}\)

If there are n number of shields kept between plates then, 

\({\left( {\frac{q}{A}} \right)_{with\;n\;shields}} = \frac{1}{{n + 1}}{\left( {\frac{q}{A}} \right)_{without\;any\;shield}}\)

\({\left( {\frac{q}{A}} \right)_{with\;n\;shields}} = \frac{1}{{n + 1}}\;\frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{2}{\epsilon} - 1}} = \frac{\epsilon}{{\left( {n + 1} \right)\left( {2 - \epsilon} \right)}}\sigma \left( {T_1^4 - T_2^4} \right)\)

From the above equation, equivalent emissivity,

\({\epsilon_{eq}} = \frac{\epsilon}{{\left( {n + 1} \right)\left( {2 - \epsilon} \right)}}\)

Concept:

The radiant heat transfer is given as Q = σAT⁴

where,  σ =  Stefan-Boltzmann constant. (5.67 × 10^-8 W/m²/K⁴), A = area of body, T = absolute temperature

Calculation:

Given:

T = 55°C = 55 + 273 = 328 K, A = 1 m² 

⇒ 5.67 × 10^-8 × (328)⁴

= 656.26 W/m²

= 0.656 kW/m²