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QUESTION: 1

According to Stefan Boltzmann law, the total radiation from a black body per second per unit area is proportional to

Solution:

Stefen boltzman law is given by

Q = σAT^{4}

hence Q ∝ T^{4}

QUESTION: 2

A perfect black body is the one which

Solution:

QUESTION: 3

Absorptivity of a body is equal to its emissivity

Solution:

QUESTION: 4

The ratio of total emissive power of body to the total emissive power of a black body at the same temperature is called

Solution:

Emissivity for any body is given by

Where

E = Emissive power of a body

E_{b} = Emissive power of a black body

QUESTION: 5

Which one of the following is true for black body

Solution:

Since black body is perfect absorber of radiation hence absorptivity of black body α = 1.

QUESTION: 6

Which one of the following is true for white body?

Solution:

QUESTION: 7

A surface is called Grey surface. If

Solution:

QUESTION: 8

The unit of Stefan-Botlzmann constant is

Solution:

QUESTION: 9

When metallic surfaces are oxidized, the emissivity

Solution:

QUESTION: 10

The solar energy falling on the earth’s surface is called

Solution:

QUESTION: 11

The shape factor F_{12 }in case of a conical cavity having a semi vertex angle a and height h is given by

Solution:

QUESTION: 12

If the temperature of a solid surface changes from 27°C to 627°C. Its emissive power will increase in the ratio of

Solution:

Since emissive power = f(T^{4})

QUESTION: 13

The radiative heat transfer rate per unit area (W/m^{2}) between two plane parallel grey surface (emissivity = 0.9) maintained at 400 K and 300 K is (StefanBoltzmann constant s = 5.67 x 10^{-8} W/m^{2}K^{4})

Solution:

QUESTION: 14

Two long parallel surfaces each of emissivity 07 are maintained at different temperature and accordingly have radiation heat exchange between them. It is desirable to reduce 75% of this radiant heat transfer by inserting thin parallel shields of equal emissivity on both sides. The number of shields should be

Solution:

where N = Number of radiation shield

QUESTION: 15

Heat transfer by radiation between two gre bodies of emissivity ∈ is proportional to (Notations have their usual meanings)

Solution:

(α = ∈) Kirchoff’s law

q = A_{i }- α_{i}G_{i})

QUESTION: 16

The shape factor of a hemispherical body place on a flat surface with respect to it self is

Solution:

QUESTION: 17

The shape factor of a cylindrical cavity with respect to it self is

Solution:

QUESTION: 18

A spherical body with surface A_{1} is completely enclosed by another hollow body with inner surface A_{2}. The shape factor of A_{2 }with respect to A_{1} is

Solution:

QUESTION: 19

For a circular tube of equal length and diameter shown below, the view factor F_{13} = 0.17. The view factor F_{12} in this case will be

Solution:

f_{11} + f_{12} + f_{13} = 1

f_{11} = 0

f_{12} = 1 - f_{13}

f_{12} = 1 - 0.17 = 0.83

QUESTION: 20

For a hemisphere, the solid angle is measured

Solution:

QUESTION: 21

Two spheres A and B of same material have radii 1m and 4m and temperature 1000 K and 2000 K respectively. Which one of the following statement is correct.

The energy radiated by sphere A is

Solution:

QUESTION: 22

A plate having 10 cm^{2} area each side is hanging in the middle of a room of 100 m^{2 }total surface area, the plate temperature and emissivity are respectively 800 K and 0.6. The temperature and emissivity values for the surface of the room are 300 K and 0.3 respectively. Boltzman’ constant σ = 5.67 x 10^{-8} W/m^{2}K^{4}. The total heat loss from the two surface of the plate is

Solution:

A_{1} = 20 cm^{2} = 20 x 10^{-4}m^{2}

A_{2 }= 100 m^{2}

T_{1} = 800K

∈_{1 }= 0.6

T_{2 }= 300 K

∈_{2} = 0.3

QUESTION: 23

Match List-I (surface with orientations) with List-lI (equivalent emissivity) and select the correct answer using the codes given below:

List-I

A. Infinite parallel planes

B. Body 1 completely enclosed by body 2 but body 1 is very small

C. Radiation exchange between two small Grey bodies

D. Two concentric cylinders with large lengths

List-II

1. ∈_{1}

4. ∈_{1}∈_{2}

Codes:

A B C D

(a) 3 1 4 2

(b) 2 4 1 3

(c) 2 1 4 3

(d) 3 4 1 2

Solution:

QUESTION: 24

For the two long concentric cylinders with surface areas A_{1}, and A_{2}, the view factor F_{22} is given by

Solution:

QUESTION: 25

For a thin steel sheet, total emissive power is given as 32 W/m^{2}, insolation as 93 W/m^{2}. If thin sheet has reflectivity = 0,6, absorbtivity = 0.1 and transmissivity = 0.3. Then radiosity in W/m^{2}

Solution:

Radiosity = J = E + ρG

J = 32 + 0.6(93)

J = 87.8 W.m^{-2}

QUESTION: 26

The thermal radiative flux from a surface of emissivity = 0 .4 is 22.68 kW /m^{2}. The approximate surface temperature (K) is

Stefan Boltzmann's constant = 5.67x10^{-8} W/m^{2}K^{4}

Solution:

We know, for radiation heat transfer

QUESTION: 27

A solid cylinder (surface 2) is located at the centre of a hollow sphere (surface 1). The diameter of the sphere is 1 m, while the cylinder has a diameter and length of 0.5 m each. The radiation configuration factor F_{11 }is

Solution:

QUESTION: 28

Match List-I (Thermal spectrum) with List-l! (Wavelength mm) and select the correct aswer using the code given below the lists:

List-I

A. Cosmic rays

B. Radio waves

C. Visible light

D. UVrays

List-II

1. λ < 10^{-8}

2.^{ }10^{2} < λ < 10^{10}

3. 0.38 < λ< 0.78

4. 10^{-2 }< λ < 0.38

Codes:

A B C D

(a) 4 2 3 1

(b) 2 1 3 4

(c) 4 3 2 1

(d) 1 2 3 4

Solution:

QUESTION: 29

Match List-I with List-ll and select the correct answer using the code given below the lists:

List-I

A. Window glass

B. Grey surface

C. Carbon dioxide

D. Radiosity

List-ll

1. Emissivity

2. Kinematic viscosity

3. Diffusion coefficient

Codes:

A B C

(a) 2 3 1

(b) 3 2 1

(c) 1 3 2

(d) 1 2 3

Solution:

QUESTION: 30

The radiation heat flux from a heating element at a temperature of 800°C, in a furnace maintained at 300°C is kW/m^{2}. The flux when the element temperature is increased to 1000°C for the same furnace temperature is

Solution:

For radiation heat transfer

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