Shape Factor MCQ Quiz in తెలుగు - Objective Question with Answer for Shape Factor - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

By Reciprocity theorem: A1F12 = A2F21

By summation rule: F21 + F22 = 1

Calculation:

Given:

1 is the outer surface of the inner sphere and 2 is the inner surface of outer-sphere

R1 = 25 mm, R2 = 45 mm

All radiations from surface 1 are absorbed by surface 2. ∴ F12 = 1

By reciprocity Theorem, 

A1F12  = A2F₂₁

\({{F}_{21}}=\frac{{{A}_{1}}}{{{A}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{2}}={{\left( \frac{25}{45} \right)}^{2}}=\frac{25}{81}\)

By summation rule,

F₂₁ + F₂₂ = 1

∴ F₂₂ = 1 - F₂₁

\(F_{22}= 1- {25\over 81} \)

∴ F₂₂ = 0.69

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

.Summation Rule: F₁₁ + F₁₂ = 1

Reciprocity theorem: A₁ F₁₂ = A₂ F₂₁ (A is the surface area)

Calculation:

Let 1 denote the sphere and 2 denote the cube

Given, radius of the sphere = 2 cm, side of the cube = 5 cm

For sphere F₁₁ = 0, F₁₂ = 1 (All radiations from sphere fall on the cube)

A₁ = 4πr²

A₂ = 6a²

Reciprocity theorem

A₁ F₁₂ = A₂ F₂₁   

\({F_{21}} = \frac{{{A_1}}}{{{A_2}}} = \;\frac{{4\pi {2^2}}}{{6\; \times \;{5^2}}} = 0.335\;\;\)

Concept:

For similar object the view factor remains the same i.e. F₁₂ = F₁₄

→ Sum of all view factor for an enclosure = 1

Calculation:

F₁₂ = F₁₄ = 0.26

F₁₂ + F₁₄ + F₁₃ = 1

0.26 + 0.26 + F₁₃ = 1

F₁₃ = 0.48

Net radiation heat loss or gain on side 1 = Q₁₂ + Q₁₄ + Q₁₃

Q₁₂ = Q₁₄ [∵ Temperature & view factor are same]

∴ Q_Net,1 = 2Q₁₂ + Q₁₃

\(=\frac{2\sigma \left( T_{1}^{4}-T_{2}^{4} \right)}{\frac{1-{{\epsilon }_{1}}}{{{A}_{1}}{{\epsilon }_{1}}}+\frac{1}{{{A}_{1}}{{F}_{12}}}+\frac{1-{{\epsilon }_{2}}}{{{A}_{2}}{{\epsilon }_{2}}}}+\frac{T\left( T_{1}^{4}-T_{3}^{4} \right)}{\frac{1-{{\epsilon }_{1}}}{{{A}_{1}}{{\epsilon }_{1}}}+\frac{1}{{{A}_{1}}{{F}_{13}}}+\frac{1-{{\epsilon }_{3}}}{{{A}_{3}}{{\epsilon }_{3}}}}\)

∵ T₁ = T₃ ⇒ Q₁₃ = 0

∵ All surfaces are black surface ⇒ ϵ = 1

∴ Net radiation heat loss or gain on side 1 is

Q_Net,1 = 2Q₁₂ = 2 σ × A₁F₁₂ (T₁⁴ – T₂⁴)

Q_Net,1 = 2 × 5.67 × 10^-8 × 0.5 × L × 0.26 (1200⁴ - 800⁴)

Q_net/L = 24530.68 W/m

∵ Temperature of side 1 is greater than side 2

So there will be a loss of 24530.68 W/m of heat from side 1.

Concept:

Summation Rule: F₁₁ + F₁₂ + F₁₃ = 1

F₂₁ + F₂₂ + F₂₃ = 1

Calculation:

Here flat surface (1) cannot see itself: F₁₁ = 0

∴ F₁₂ + F₁₃ = 1

F₁₂ = 1 - F₁₃ = 1 - 0.17 = 0.83

Reciprocity theorem: A₁ F₁₂ = A₂ F₂₁

\({F_{21}} = \frac{{{A_1}}}{{{A_2}}} \times {F_{12}} = \frac{{\frac{\pi }{4}{d^2}}}{{\pi dL}} \times {F_{12}} = \frac{{{F_{12}}}}{4} = \frac{{0.83}}{4}\)

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F₁₂ means the fraction of energy leaving the surface 1 reaches surface 2.

Summation Rule: F11 + F12 = 1

Reciprocity theorem: A1 F12 = A2 F21 (A is the surface area)

Calculation:

Given:

Let us denote the hemishpherical body as body 1, and the flat surafce as body 2

From the summation rule F₂₂ + F₂₁ = 1, 

Since no radiation leaving the surface 2 will reach surface 2 again 

∴ F₂₂ = 0, F₂₁ = 1

Now from reciprocity theorem,

A1 F12 = A2 F21

2 x π x r² x F12 = π x r² x1

∴ F₁₂ = 0.5,

Now from summation rule F12 + F₁₁ = 1, F₁₁ = 1 - 0.5 = 0.5

Therefore the shape factor of hemispherical body with respect to itself is 0.5.

Explanation:

\(\rm A_1=\left(\pi dh+\frac{\pi d^2}{2}\right). A_2=6L^2\)

Summation rule (for surface 1 )

F₁₁ + F₁₂ = 1

F₁₂ = 1 {F₁₁ = 0}

Reciprocity rule (1, 2):

A₁F₁₂ = A₂F₂₁

A₁ = A₂F₂₁ {F₁₂ = 1}

\(\rm F_{21}=\frac{A_1}{A_2}\)

Summation rule (for surface 2)

F₂₁ + F₂₂ = 1

F₂₂ or F_SS = 1 - F₂₁ = \(\rm 1-\frac{A_1}{A_2}\)

F₂₂ = \(\rm 1-\frac{\left(\pi dh+\frac{\pi d^2}{2}\right)}{6L^2}\)

Concept:

For the surface-1:

F11 + F12 = 1

For the surface-2:

\(F_{21} + F_{22} =1\)

From the theorem of reciprocity,

\(A_2F_{21} = A_1F_{12}\)

Calculation:

Given:

d₁ (diameter of sphere) = 1 m, d₂ (diameter of cylinder) = 0.5 m, h = d (Height of cylinder) = 0.5 m.

\(A_1 =4\pi r_1^2=4π \frac {d_1 ^2}4= π d_1 ^2\Rightarrow π ~m^2\)

\(A_2 =\pi d_2h_2\;+\;2\frac{\pi}{4}d_2^2= \frac{3}{2}\pi d_2^2\Rightarrow 0.375\pi\;m^2\)

And we know that,

\(F_{21}=1\) (as the surface-1 is convex, all the radiation radiated by surface 1 will be absorbed by surface 2).

From the theorem of reciprocity,

A₁F₁₂ = A₂F₂₁

∴ π × F₁₂ = 0.375π × F₂₁

∴ F₁₂ = 0.375

Now,

F₁₁ + F₁₂ = 1

F₁₁ + 0.375 = 1

∴ F₁₁ = 0.625.

To find F₂₂ = ?

But F₁₁ + F₁₂ = 1

Since F₁₁ = 0, then F₁₂ = 1

Also, from the theorem of reciprocity,  A₁ F₁₂ = A₂ F₂₁

⇒ (πd₁L)F₁ =  (πd₂L) F₂₁

⇒ 2F₁₂ = 4 F₂₁ (since d₁ : d₂ = 1 : 2)

⇒ F₂₁ = 0.5F₁₂ = 0.5

Also, F₂₂ + F₂₁ = 1

⇒ F₂₂ = 1 - 0.5 = 0.5

Concept:

Shape Factor is defined as the fraction of radiation energy leaving a surface that reaches another surface, for example, F12 means the fraction of energy leaving the surface 1 reaches surface 2.

Summation Rule: F11 + F12 = 1

Reciprocity theorem: A1 F12 = A2 F21 (A is the surface area)

Concept:

Using reciprocity theorem for triangular duct,

\({F_{12}} = \frac{{{A_1} + {A_2} - {A_3}}}{{2{A_1}}}\)

Calculation:

Imagine a surface (3) of equal width using geometry,

\({L_3} = 2L\sin \left( {\frac{\alpha }{2}} \right)\)

\({F_{12}} = \frac{{{A_1} + {A_2} - {A_3}}}{{2{A_1}}} = \frac{{{L_1} + {L_2} - {L_3}}}{{2{L_1}}}\)

\({F_{12}} = 1 - \sin \left( {\frac{\alpha }{2}} \right)\)

∴ F12 = 1 – sin (30)

∴ F12 = 0.50