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

Q1: A cylindrical rod of length h h and diameter d d is placed inside a cubic enclosure of side length L . S  denotes the inner surface of the cube. The view-factor F_{S − S}  is [GATE ME 2023]
(a) 0
(b) 1
(c)
(d)
Ans: (d) 

Surface area of centrifugal rod = π d h + 2 π d ² /4
Or A₁ = π (dh + d2/²)
Surface area of cube A₂ = 6 x L²
For surface 1 (cylinder), F₁₁= 0
So F₁₂ = 1...(i) 
For surface 2(cube) 
F₂₁ + F₂₂ = 1 
F₂₂ = 1 − F₂₁
F₂₁ = 1 − F₂₂ . . . (ii)
By reciprocity theorem for surface 1 and 2
A₁ F₁₂ = A₂ F₂₁
F₁₂ = A₁/ A₂
[F₁₂ = 1 from equation (i)]
So, By equation (ii)
F₂₂ = F_{s S} = 1-

Q1: Wien's law is stated as follows: λ_m T = C, where C is 2898 μmK and λ_m is the wavelength at which the emissive power of a black body is maximum for a given temperature T. The spectral hemispherical emissivity (ε λ) of a surface is shown in the figure below (1 ˚A = 10^{− 10} m). The temperature at which the total hemispherical emissivity will be highest is __________ K (round off to the nearest integer). [GATE ME 2022, SET-2]
Ans: (4825 to4835]
From the figure we can say that at 6000 ˚A wavelength total hemispherical emissivity is maximum
λ_m = 6000 ˚A 0.6 μm 
Wein's law
λ_m T = 2898  μmK
T = 2898/0.6 = 4830K
Note: In this question, we need to find temperature at which total hemispherical emissivity will be highest.
But to calculate total hemispherical emissivity we require additional data like function chart and emissivity value at particular wavelength.
With the use of Wein's law we can find out temperature at which spectral emissivity is highest but in question it is asked temperature for total hemispherical emissivity.

Q2: A flat plate made of cast iron is exposed to a solar flux of 600 W/m² at an ambient temperature of 25°C. Assume that the entire solar flux is absorbed by the plate. Cast iron has a low temperature absorptivity of 0.21. Use Stefan-Boltzmann constant = 5.669 × 10^{− 8} W/m ² − K ⁴. Neglect all other modes of heat transfer except radiation. Under the aforementioned conditions, the radiation equilibrium temperature of the plate is __________ °C (round off to the nearest integer). [GATE ME 2020, SET-1]
Ans: (210 to 225)
Equilibrium Temperature = T_S?
In this it is mentioned that entire flux is absorbed by the plate it means for solar flux absorptivity is 1. Kirchhoff's law α = ε = 0.21 
Surface of cast iron and surrounding fluid temperature difference is small due to this we can use Kirchhoff's law At equilibrium condition Energy absorbed = Energy leaving

T_s = 491.34 K
T_s = 218.34 °C
T_s = 218 °C 

Q1: A solid sphere of radius 10 mm is placed at the centroid of a hollow cubical enclosure of side length 30 mm. The outer surface of the sphere is denoted by 1 and the inner surface of the cube is denoted by 2. The view factor F₂₂ for radiation heat transfer is ________ (rounded off to two decimal places). [GATE ME 2021,SET-1]
Ans: (0.76 to0.78)

r₁ = 10 mm

A₂ = 6 x (30²)
F₁₂ = 1

Q2: The spectral distribution of radiation from a black body at T₁ = 3000 K has a maximum at wavelength λ _max. The body cools down to a temperatureT_2. If the wavelength corresponding to the maximum of the spectral distribution at T₂ is 1.2 times of the original wavelength λ _max . then the temperature T₂ is ________ K (round off to the nearest integer). [GATE ME 2020, SET-2]
Ans: (2499 to 2501)
From Wien's Displacement law,

Question for GATE Past Year Questions: Radiation

Try yourself:The following figure was generated from experimental data relating spectral black body emissive power to wave length at the three temperatures T₁, T₂ and T₃( T₁ > T₂ >T₃).

The conclusion in that the measurements are

[2005]

• A.

  correct because the maxima in E_bλ show the correct trend
• B.

  correct because planck's law is satisfied
• C.

  wrong because the Stefan Boltzmann law in not satisfied
• D.

  wrong because Wien's displacement law is not satisfied

Explanation

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

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Question for GATE Past Year Questions: Radiation

Try yourself:For the circular tube of equal length and diameter shown in figure below, the view factor F₁₃ is 0.17. The view factor F₁₂ in this case will be

[2001]

• A.

  0.17
• B.

  0.21
• C.

  0.79
• D.

  0.83

Explanation

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

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Question for GATE Past Year Questions: Radiation

Try yourself:A hollow enclosure is formed between two inf initely long concentric cylinders of radii 1 m and 2 m, respectively. Radiative heat exchange takes place between the inner surface of the larger cylinder (surface-2) and the outer surface of the smaller cylinder (surface-1). The radiating surface are diffuse and the medium in the enclosure is non- participating. The fraction of the thermal radiation leaving the larger surface and striking itself is

[2008]

• A.

  0.25
• B.

  0.5
• C.

  0.75
• D.

  1

Explanation

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

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Question for GATE Past Year Questions: Radiation

Try yourself:Consider two infinitely long thin concentric tubes of circular cross section as shown in figure. If D₁ and D₂ are the diameters of the inner and outer tubes respectively, then the view factor F₂₂ is given by

[2012]

• A.

  
• B.

  Zero
• C.

  (D₁/D₂)
• D.

  

Explanation

F₂₂ = 1 – F₂₁

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Question for GATE Past Year Questions: Radiation

Try yourself:A solid sphere of radius r₁ = 20 mm is placed concentrically inside a hollow sphere of radius r₂ = 30 mm as shown in the figure.

The view factor F₂₁ for radiation heat transfer is

[2014]

• A.

  2/3
• B.

  4/9
• C.

  8/27
• D.

  9/4

Explanation

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

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

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Question for GATE Past Year Questions: Radiation

Try yourself:A solid sphere 1 of radius r is placed inside a hollow, closed hemispherical surface 2 of radius 4r. The shape factor F is,

[2015]

• A.

  1/12
• B.

  1/2
• C.

  2
• D.

  12

Explanation

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

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