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Test: Radiative Heat Transfer Level - 1 - Mechanical Engineering MCQ


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20 Questions MCQ Test Heat Transfer - Test: Radiative Heat Transfer Level - 1

Test: Radiative Heat Transfer Level - 1 for Mechanical Engineering 2024 is part of Heat Transfer preparation. The Test: Radiative Heat Transfer Level - 1 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The Test: Radiative Heat Transfer Level - 1 MCQs are made for Mechanical Engineering 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Radiative Heat Transfer Level - 1 below.
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Test: Radiative Heat Transfer Level - 1 - Question 1
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A radiator in a domestic heating system operates at a surface temperature of . Assuming the radiator behaves as a black body, the rate at which it emits the radiant heat per unit area is

(Assume σ =5.67 x 10-8 W/m2-k4)

 

Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 1

As per Stefan Boltzmann law

Eb = Q/A = σ x T4

Eb = 5.67 x 10-8 x (55 + 273)4

= 656.2 W/m2

≃ 0.66 kW/m2

Test: Radiative Heat Transfer Level - 1 - Question 2
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Sun’s surface at emits maximum spectral power for wavelength of A grey surface at a will Demit maximum spectral power at a wavelength of

Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 2

By Wien’s displacement law,

λm ∝ 1/T

∴ (λm)2/(λm)1 = T1/T2

∴ (λm)2 = (λm)1T1/T2

∴ (λm)2 = 0.5 x 5800/(300 + 273)

(λm)2 = 5.06μm

≃ 5μm

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Test: Radiative Heat Transfer Level - 1 - Question 3
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Two radiating surfaces A1= 6m2 and A2=4m2 have shape factor Then shape factor will be

Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 3

By Reciprocity theorem

A1F12 = A2F21

= 0.15

Test: Radiative Heat Transfer Level - 1 - Question 4
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What will be the view factor for the geometry as shown in the figure (sphere within a hollow cube)?

 
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 4

1 is a spherical surface

∴ F11 = 0

By summation rule,

F12 + F11 = 1

∴ F12 = 1

By Reciprocity theorem,

A1F12 = A2F21

Test: Radiative Heat Transfer Level - 1 - Question 5
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A radiation shield should
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 5

A radiation shield should reflect back as much radiation as possible. Hence radiation shield should have high reflective power.

Test: Radiative Heat Transfer Level - 1 - Question 6
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For an opaque plane surface at steady state the irradiation, radiosity and emissive power are respectively 20, 12 and . What is the emissivity of the surface?
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 6

J = pG + εEb

For an opaque body

∵ α + ρ = 1

ρ = 1- α

∴ J = (1-α)G + εEb

By kirchoff’s law, α = ε

∴ J = (1-α)G + εEb

12 = (1- ε)20 + 10

ε = 0.9

Test: Radiative Heat Transfer Level - 1 - Question 7
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The wavelength for which the blackbody emissive power is maximum for a temperature of 300 K is
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 7

By Wien’s displacement law,

∵ λmT = 2898 μm-k

λm ≃ 9.7μm

Test: Radiative Heat Transfer Level - 1 - Question 8
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Solar radiation is incident on a semitransparent body at a rate of . If of this incident radiation is reflected back and is transmitted across the body, the absorptivity of the body is
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 8

G = 500 W/m2 Gr = 150 W/m2

Gt = 225 W/m2

By energy conservation principle,

Ga+Ge+Gt = G

Where Ga = absorbed radiation

Gr = reflected radiation

Gt = transmitted radiation

G = incident radiation

= transmissivity

α = 1-0.3 - 0.45 = 0.25

Test: Radiative Heat Transfer Level - 1 - Question 9
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A grey body is defined such that
Test: Radiative Heat Transfer Level - 1 - Question 10
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The minimum number of view factors that need to be known to solve a 10-surface enclosure completely, is
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 10

For n-surface enclosure if nC2 view factors are known directly, entire enclosure can be solved. Thus for 10 surfaces, number of view factors that need to be evaluated directly will be

Test: Radiative Heat Transfer Level - 1 - Question 11
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The spectral distribution of surface irradiation is as follows

What is the total irradiation in kW/m2 ?
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 11

Total irradiation is given as

i.e. it is equal to area under the curve Gλ Vs λ Hence,

x (25-20) x 1000

∴ G = 20000 W/m2

I.e. G = 20 kW/m2

Test: Radiative Heat Transfer Level - 1 - Question 12
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Consider two infinitely long blackbody concentric cylinders with a diameter ratio . The shape factor of the inner surface of outer cylinder with respect to itself will be
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 12

By Energy conservation

F12 + F22 =1

∴ F22 = 1-F21 ….①

By Reciprocity theorem

A1F12 = A2F21

πD1L x F12 = πD2L x F21

Test: Radiative Heat Transfer Level - 1 - Question 13
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A large spherical enclosure has a small opening. The rate of emission of radiative flux through this opening is . The temperature at the inner surface of the sphere will be about (assume Stefan Boltzmann constant σ = 5.67 x 10-8 W/m2K4)
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 13

A small opening in a large spherical enclosure behaves as a black surface

∴Eb = σT4

7350 = 5.67 x 10-8 x T4

T = 600K

I.e. T = 327oC

Test: Radiative Heat Transfer Level - 1 - Question 14
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Consider monochromatic emissive power (Eλ) Vs Wavelength(λ) of a black and grey surface both at same temperature. The ratio AB/AC is given as
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 14

AB & AC are monochromatic emissive power of grey and black surfaces for same wavelength. Hence, ratio is

which is monochromatic emissivity of grey surface.

Test: Radiative Heat Transfer Level - 1 - Question 15
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If the temperature of a solid surface changes from then its emissive power will increases in the ratio of
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 15

By Stefan Boltzmann law

E ∝ T4

Test: Radiative Heat Transfer Level - 1 - Question 16
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Consider a hemispherical furnace. The view factor of it’s roof (hemisphere) with respect to itself is __________.
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 16

By reciprocity theorem

A1F12 = A2F12

1 is a planar surface, i.e F11 = 0

∴ F12 = 1

Hence F21 = 0.5

F21 + F22 = 0.5

F22 + 1-0.5 = 0.5

Test: Radiative Heat Transfer Level - 1 - Question 17
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Fraction of radiative energy leaving one surface that strikes the other surface is called
Test: Radiative Heat Transfer Level - 1 - Question 18
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Ice is very close to a
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 18

The absorptivity of ice is 0.985, hence ice is very close to a black body

Test: Radiative Heat Transfer Level - 1 - Question 19
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For a grey diffuse surface with emissivity ε and temperature T, the intensity of emitted radiation is given as __________. ( is StefanBoltzmann constant.)

 
Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 19

For a diffuse surface, intensity of emitted radiation (Ie) is same in all directions and is given as

Ie = E/π

Where E is total emissive power

Therefore

Ie = εσT4/π

Test: Radiative Heat Transfer Level - 1 - Question 20
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Solar radiation of falls on a grey opaque surface at steady state. The surface has a temperature of and emissivity of 0.8. Find radiosity from the surface?

Detailed Solution for Test: Radiative Heat Transfer Level - 1 - Question 20

J = E+Gr

E = εσT4

E = 0.8 x 5.67 x 10-8 x (50 + 273)4

E = 493.72 W/m2

Given G = 1200 W/m2

But as surface is at steady state

α = ε = 0.8

∴ ρ = 1 -α = 0.2

∴ Gr = ρG

Gr = 0.2 x 1200 = 240 W/m2

Therefore

J = E+Gr

J = 493.72 + 240

J = 733.72 W/m2

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