GATE Past Year Questions: Radiation | Heat Transfer - Mechanical Engineering PDF Download

Radiation intensity is independent of angle in diffuse radiation.

= 0.826

Wien's displacement law is not satisfied, i.e., λ_max T = C
which tells lower λ_max for higher temperature.

α + τ + ρ = 1
for opaque surface, τ = 0
∴ α + ρ = 1

For plane surfaces

Here ∈₁ = ∈₂ = 0.9

Heal loss from one surface of the plate,


Putting
A₁ = 10 × 10^–4 m², T₁ = 800
∈₁ = 0.6 T₂ = 300
A₂ = 100 m²
∈₂ = 0.3

 = 13.66 W
∴ Heat loss from both the surfaces
= 2 Q_net
= 2 × 13.66 = 27.32 W

Heat generated by bulb = 100 × 24 × 60 × 60 Joule.
∴ Heat dissipated = r × V × C_v × (T - 20)
⇒ 100 × 24 × 60 × 60 = 1.24 × 9 × 2.5 × 3 × C_v (T – 20)
⇒ T = 450º

T₁ = 1000 K
T₂ = 500 K
∈ = 0.7

Energy indicent upon plate 1 due to irradiation = Radiosity of plate 2

σ (1- α) E_b1 = Fraction of energy reflected from surface 2
= (1 – 0.7) × 5.67 × 10^–8 ×1000⁴
 = 17010(W/m²)
∴ Total energy incident upon plate 1 = 

= 0.59
∴ 
= 0.59 x 5.67 x (10000 - 625)
= 31362.18 W/m²
= 31.36 kW/m²
Alternately

Net heat exchange between A and B,

Emissive power ∝ T⁴
⇒ 
∴ 

E_b = σ T⁴

From wein’s displacement law
λ_max T = constant

T₂ = 1000 K

and 
F₁₃ = 0.17,
F₁₁ = 0,
F₁₁ + F₁₂ + F₁₃ = 1
∴ F₁₂ = 1 – F₁₃
= 1 – 0 – 17 = 0.83

Diameter of sphere = 1 m
Diameter of aylinder = 0.5 m
F ₂₁ + F₂₂ = 1 (∵ F₂₂ = 0)
F₂₁ = 1
we know that



= 0.625

F ₁₁ + F ₁₂ = 1
F ₂₁ + F ₂₂ = 1
F ₁₁ = 0 due to concave surface
F ₁₂ = 1
Now A ₁ F ₁₂ = A ₂ F ₂₁
⇒ 
⇒ F₁₂ = 1
∴ 
F ₂₁ = 0.5
F ₂₂ = 1 – F ₂₁ = 0.5

F₂₂ = 1 – F₂₁

F ₁₁ + F ₁₂ =1, here F ₁₁ = 0
∴ F ₁₂ = 1

Now from reciprocating law
A ₁ F ₁₂ = A ₂ F ₂₁

f₁₁ + f₁₂ = 1
∴ f₁₂ = 1
f₂₁ A₂ = f₁₂ A₁
∴ 

View factor
F_{1– 1} = 0
F_{1– 2} = 1
Reciprocacy theorem
A ₂F _{2– 1} = A ₁F _{1– 2}
⇒ 

F_{2– 1} + F_{2– 2} = 1
∴ F_{2– 2} = 1 – F_{2– 1} = 1 - R_i/R_o

F₁₁ = 0
F₁₂ = 1

here, n = 1
∴