Heat Exchange Between Non-Black Bodies MCQ Quiz in தமிழ் - Objective Question with Answer for Heat Exchange Between Non-Black Bodies - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Heat transfer relation when emissivities of plates and shield is same:

\(\frac{{{{\left( Q \right)}_{with\;shield}}}}{{{{\left( Q \right)}_{without\;shield}}}} = \frac{1}{{N + 1}}\)

where N is the number of shields.

Calculation:

Given:

N = 1

\(\frac{{{{\left( Q \right)}_{with\;shield}}}}{{{{\left( Q \right)}_{without\;shield}}}} = \frac{1}{{N + 1}} = \frac{1}{{1 + 1}} = \frac{1}{2}\)

Concept:

Q_radiation = σϵAT⁴

Calculation:

Given T = 500 k, d = 10 mm, ϵ = 0.5;

\({Q_r} = 5.678 \times {10^{ - 8}} \times 0.5 \times \frac{\pi }{4}{\left( {0.01} \right)^2} \times {\left( {500} \right)^4}\)

∴ Q_r = 0.139 W

Explanation:

Consider a system

\(\frac{{1 - {\epsilon_1}}}{{{A_1}\epsilon{_1}}}\;and\;\frac{{1 - {\epsilon_2}}}{{{A_2}{\epsilon_2}}}\) are surface resistances \(\frac{1}{{{A_1}{F_{12}}}}\) is space resistance.

Without radiation shield:

Surface resistance = 2

Space resistance = 1

Now,

With 1 radiation shield

Surface resistance = 4

Space resistance = 2

∴ the general formula for 'n' shields is:

Surface resistance = 2n + 2

Space resistance = n + 1

Since here increase in resistance is asked not total resistance, therefore:

Increase in surface resistance = 2n + 2 - 2 = 2n

Increase in space resistance = n + 1 - 1 = n

For 4 shield placed in between:

Increase in surface resistance = 2n = 2 × 4 = 8 

Increase in space resistance = n = 4

Concept:

The amount of heat removed from surface A per unit area to maintain its constant temperature, \(\left( Q \right) = \sigma A\;{F_{12}}\;\left( {T_2^4 - T_1^4} \right)\)

where, F₁₂ = view factor, σ = Stefan-Boltzmann constant

Calculation:

Given:

ϵ₁ = ϵ₂ = 1, T₁ = 300 K, T₂ = 1000 K, F₁₂ = 1

Net Heat Exchange \(\left( Q \right) = \sigma A\;{F_{12}}\;\left( {T_2^4 - T_1^4} \right)\)

= σ × 1 × 1 × (1000⁴ – 300⁴)

Q = 9919 × 10⁸ σ W/m²

Concept:

Radiation heat exchange (q) between two infinitely long parallel plates is given by:

\(q = \frac{{\sigma \left( {T_1^4\;-\;T_2^4} \right)}}{{\frac{1}{{{ϵ_1}}}\;+\;\frac{1}{{{ϵ_2}}}\;-\;1}}\)

Calculation:

Given T₁ = 800 K, T₂ = 500 K; ϵ₁ = 0.8, ϵ₂ = 0.6;

Now the net heat exchange will be 

\(q = \frac{{5.67 \times 10^{-8} \times \left( {(800)^4\;-\;(500)^4} \right)}}{{\frac{1}{{{0.8}}}\;+\;\frac{1}{{{0.6}}}\;-\;1}}\)

⇒ q = 10268.123 W/m²;

⇒ q = 10.26 kW/m²;

Concept:

Heat transfer is given by,

\({Q_{21}} = \frac{{\sigma \left( {T_2^4 - \;T_1^4} \right)}}{{\frac{{1 - \;{\epsilon_1}}}{{{A_1}{\epsilon_1}}} + \frac{1}{{{A_1}\;{F_{12}}\;}} + \;\frac{{1 - \;{\epsilon_2}}}{{{A_2}{\epsilon_2}}}}}\)

Where, T = temperature

A = area, ϵ = emissivity

Calculation:

T₁ = 1000K, T₂ = 1300K

E₁ = 0.67, E₂ = 0.45

\({A_1} = \frac{\pi }{4} \times {d^2},\;{A_2} = \frac{\pi }{2}{d^2}\)

F₁₂ = 1

\({Q_{21}} = \frac{{\sigma \left( {T_2^4 - T_1^4} \right){A_1}}}{{\frac{{1 - {E_1}}}{{{E_1}}} + \frac{{{A_1}}}{{{A_2}}}\left( {\frac{{1 - {E_2}}}{{{E_2}}}} \right) + 1}}\)

A₁/A₂ = ½

\({A_1} = \frac{\pi }{4} \times 4 = \pi \;{m^3}\)

\({Q_{21}} = \frac{{5.67 \times {{10}^{ - 8}}\left( {{{1300}^4} - {{1000}^4}} \right) \times \pi }}{{\frac{{0.33}}{{0.67}} + \frac{1}{2}\left( {\frac{{0.55}}{{0.45}}} \right) + 1}}\left( W \right)\)

∴ Q₂₁ = 157.167 kW

Concept:

General formula for heat-transfer between two infinite parallel plates:

\(Q_{1-2}=\frac{\sigma(T_1^4\;-\;T_2^4)}{\frac{1}{ϵ_1}\;+\;\frac{1}{ϵ_2}\;-\;1}\)

where \(\left(\frac{1}{ϵ_1}\;+\;\frac{1}{ϵ_2}\;-\;1\right)\;\) represents thermal resistance.

If 'n' number of the shield is placed between two infinite parallel plates:

\(Q_{1-n-2}=\frac{\sigma(T_1^4\;-\;T_2^4)}{\frac{1}{ϵ_1}\;+\;\frac{2n}{ϵ_s}\;+\;\frac{1}{ϵ_2}\;-\;(n\;+\;1)}\)

where \(\left[\frac{1}{ϵ_1}\;+\;\frac{2n}{ϵ_s}\;+\;\frac{1}{ϵ_2}\;-\;(n\;+\;1)\right]\)represents the thermal resistance.

[Note: Applicable only if emissivities of all shield is same.]

Calculation:

Given:

ϵ₁ = 0.3, ϵ2 = 0.4, ϵs = 0.02, n = 3

Increase in resistance:

R_{with shield} - R_{without shield}

\(\left[\frac{1}{ϵ_1}\;+\;\frac{2n}{ϵ_s}\;+\;\frac{1}{ϵ_2}\;-\;(n\;+\;1)\right]\;-\;\left(\frac{1}{ϵ_1}\;+\;\frac{1}{ϵ_2}\;-\;1\right)\)

\(\left[\frac{2n}{ϵ_s}\;-\;n\right]\)

Percentage increase in resistance percentage:

\(\frac{\left[\frac{2n}{ϵ_s}\;-\;n\right]}{\left[\frac{1}{ϵ_1}\;+\;\frac{1}{ϵ_2}\;-\;1\right]}\times100\;\%\)

\(\frac{\left[\frac{2\;\times\;3}{0.02}\;-\;3\right]}{\left[\frac{1}{0.3}\;+\;\frac{1}{0.4}\;-\;1\right]}\times100\;\%\)

6144.82 ≈ 6145 %

Let, A₁ = Total area of walls and ceiling

= 2 (1.5 × 2) + 2 (1.7 × 2)  + 1.5 × 1.7 = 15.35 m²

A₂ = Floor area = 1.5 × 1.7 = 2.55 m²

The floor is completely enclosed by the area A₁

F₂₁ + F₂₂ = 1

F₂₁ = 1

A₁F₁₂ = A₂F₂₁

F₁₂ = A₂/A₁ = 2.55/15.35 = 0.166

Rate of heat transfer by radiation from the walls and ceiling to the floor

\({F_{12}} = \frac{1}{{\left( {\frac{1}{{{\varepsilon _1}}} - 1} \right) + \frac{1}{{{F_{12}}}} + \frac{{{A_1}}}{{{A_2}}}\left( {\frac{1}{{{\varepsilon _2}}} - 1} \right)}} = \frac{1}{{\left( {\frac{1}{{0.82}} - 1} \right) + \frac{1}{{0.166}} + \frac{{15.35}}{{2.55}}\left( {\frac{1}{{0.7}} - 1} \right)}} = 0.113\)

\({Q_{1 - 2}} = \sigma {A_1}{F_{12}}\left( {T_1^4 - T_2^4} \right)\)

Concept:

Radiation heat exchange between two infinite parallel plates:

\({Q_{12}} = \frac{{\sigma A\left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{{{ϵ_1}}} + \frac{1}{{{ϵ_2}}} - 1}}\)

Calculation:

Given:

\(Q = \frac{{\sigma \left( {T_1^4 - T_2^4} \right)}}{{\frac{1}{{{ϵ_1}}} + \frac{1}{{{ϵ_2}}} - 1}} = \frac{{5.67 \times {{10}^{ - 8}}\left( {{{800}^4} - {{500}^4}} \right)}}{{\frac{1}{{0.8}} + \frac{1}{{0.6}} - 1}}\)

Concept:

Radiation shield

• Radiation heat transfer between two surfaces is greatly reduced by inserting radiation shields between them.
• The radiation shields reduce the heat transfer by placing additional resistances in the path of heat flow.
• The radiation shields should have high reflectivity and low emissivity.
• The high reflectivity makes the radiation shield to reflect most of the radiation heat and thus decreases the heat transfer to the other surface.
• In the same way, the lower the emissivity, the higher the resistance and the lesser the heat transfer.