Radiation
Shape Factors
Learning Objectives:

Understand the importance of shape factor for radiation heat transfer.

Find appropriate view (shape) factors for various configurations.
Shape Factors are also called View Factors, or, Configuration Factors It is the proportion of the radiation which leaves surface A that strikes surface B.
Reciprocity Relation:
Reciprocity relation relates the shape factors from (1) to (2) and that from (2) to (1) as follows:
\[A_i F_{ij} = A_j F_{ji}\]
Thus, if the shape factor from (1) to (2) is known, then the shape factor from (2) to (1) can be found by: \[F_{2 \rightarrow 1} = \dfrac{A_1}{A_2} F_{1 \rightarrow 2}\]
Summation Rule:
This rule says that the shape factor from a surface (1) to another (2) can be expressed as a sum of the shape factors from (1) to (2a), and (1) to (2b). Using this rule allows you to break up complicated geometry into smaller pieces for which the individual shape factors can be found.
\[\sum_{j=1}^{N} F_{ij} = 1\]
Shape Factor for Self Body:
F\(_{ii}\) or F\(_{11}\) energy received by body 1 out of energy emitted by body 1.

For a flat surface F\(_{ii}\) =0

For a convex surface F\(_{ii}\) = 0

For a concave surface F\(_{ii}\) \(\neq\) 0
Solved Example: 82401
The shape factor of convex surface with respect to itself (F$_{11}$) is:
A. 0
B. 1
C. 1
D. 0.5
If a surface is flat or convex, then the shape factor with respect to itself is zero.
Correct Answer: A
Perfectly Black Body
Learning Objectives:

Define radiation mode of heat transfer.

Define absorptivity, reflectivity, transmissivity, black Body, grey body, opaque body.

Relations between emissivity and absorptivity.
A body which absorbs all incident light is referred to ’Perfectly black body’. A black body would not reflect and would not transmit any light falling on it.
In practice, a hollow enclosure having a small hole is the nearest approach to a black body because the radiation entering the chamber through the hole is absorbed completely due to repeated reflections inside the enclosure. Radiation emerging from the small hole of such a hollow enclosure is negligible.
Black body radiation: Radiation from a black surface depends only on temperature and does not depend on the nature of the surface. At lower temperature radiations are small. With increasing temperature amount of radiations increase but the quantity of radiations with small energy is decreased.
Types of Bodies:

General Body
For general body,\[\alpha + \rho + \tau = 1\]
where
\(\alpha\) = absorptivity (ratio of energy absorbed to incident energy)
\(\rho\) = reflectivity (ratio of energy reflected to incident energy)
\(\tau\) = transmissivity (ratio of energy transmitted to incident energy) 
Opaque Body
For an opaque body: \(\alpha\) + \(\rho\) = 1 
Gray Body:
A gray body is one for which, \[\alpha = \epsilon, (0 < \alpha < 1; 0 < \epsilon < 1),\] where \(\epsilon\) = the emissivity of the body For a gray body:\[\epsilon + \rho = 1\]
Real bodies are frequently approximated as gray bodies.
A diffused surface is a surface from which directions of emitted, reflected and incident radiations are unknown or is unpredictable.
Solved Example: 82101
All radiations in a perfectly black body are:
A. Reflected
B. Transmitted
C. Absorbed
D. Partly reflected and partly absorbed.
Correct Answer: C
Solved Example: 82102
The amount of radiation mainly depends on:
A. Nature of body
B. Temperature of body
C. Type of surface of body
D. All of the above
Correct Answer: D
Solved Example: 82103
A perfect black body is one which:
A. Is black in color
B. Reflects all heat
C. Transmits all heat radiations
D. Absorbs heat radiations of all wave lengths falling on it
Correct Answer: D
Solved Example: 82104
A grey body is one whose absorptivity: (ISRO RAC 2014)
A. Varies with temperature
B. Varies with the wave length of incident ray
C. Varies with both
D. Does not vary with temperature and wave length of the incident ray
Correct Answer: D
Solved Example: 82105
Depending on the radiating properties, a body will be considered perfectly black when:
A. $\rho$ = 0, $\tau$ = 0 and $\alpha$ = 1
B. $\rho$ = l, $\tau$ = 0 and $\alpha$ = 0
C. $\rho$ = 0, $\tau$ = 1 and $\alpha$ = 0
D. $\tau$ = 0, $\alpha$ + $\rho$ = 0
$\alpha$ = absorptivity, $\rho$ = reflectivity, $\tau$ = transmissivity.
Correct Answer: A
Solved Example: 82106
Coefficient of transmission and coefficient of reflection for a given body are 0.22 and 0.74 respectively. Then, at a given temperature, the coefficient of emission for the body is:
A. 0.4
B. 0.04
C. 0.96
D. 0.22
Reflectivity ($\rho$) + Absorptivity ($\alpha$) + Transmissivity ($\tau$)= 1
Correct Answer: B
Solved Example: 82107
The coefficient of absorption of a perfectly black body is:
A. 1
B. 0
C. 0.75
D. 1
Correct Answer: A
Solved Example: 82108
Which of the following is the case of heat transfer by radiation:
A. Blast furnace
B. Heating of building
C. Cooling of parts in furnace
D. Heat received by a person from fireplace
Correct Answer: D
StefanBoltzmann Law
Learning Objectives:

State StefanBoltzmann Law.
The maximum rate of radiation that can be emitted from a surface at an absolute temperature \(\large T_{s}\) (in K or R) is given by the StefanBoltzmann law as: \[\large = \sigma A_{s}T_{s}^{4}\] where \(\displaystyle \large \sigma\) = 5.67 \(\times\) \(10^{8} W/m^2K^4\) is the StefanBoltzmann constant. The idealized surface that emits radiation at this maximum rate is called a blackbody, and the radiation emitted by a blackbody is called blackbody radiation.
Emissivity is the ratio of the radiation emitted by a surface to the radiation emitted by a blackbody at the same temperature.
Absorptivity is the fraction of radiation incident on a surface that is absorbed by the surface.
Solved Example: 82201
The SI unit of Stefan's constant is: (ISRO RAC 2019)
A. Watt/m$^2$ K$^4$
B. Watt/m$^3$ K
C. Watt/m$^2$ K
D. Watt/m$^3$ K$^4$
\[ Q = \sigma A_{s}T_{s}^{4}\] \[\sigma = \dfrac{Q}{A_{s}T_{s}^{4}}\] This gives the units as Watt/m$^2$ K$^4$.
Correct Answer: A
Solved Example: 82202
The expression Q = $\rho$ AT$^4$ is called:
A. Fourier equation
B. StefanBoltzmann equation
C. NewtonRikhmann equation
D. JosephStefan equation
Correct Answer: B
Solved Example: 82203
Stefan Boltzmann law is applicable for heat transfer by:
A. Conduction
B. Convection
C. Radiation
D. Conduction and radiation combined
Correct Answer: C
Solved Example: 82204
According to StefanBoltzmann law, ideal radiators emit radiant energy at a rate proportional to:
A. Absolute temperature
B. Square of temperature
C. Fourth power of absolute temperature
D. Fourth power of temperature
Correct Answer: C
Kirchhoffs Law of Radiation
Learning Objectives:

State Kirchhoff’s Law of Radiation.

Solve simple radiation problems.
The Kirchhoff’s law of radiation states that the emissivity and the absorptivity of a surface are equal at the same temperature and wavelength.
Solved Example: 82301
According of Kirchhoff's law,
A. Emissive power depends on temperature
B. Emissive power and absorptivity are constant for all bodies
C. Ratio of emissive power to absorptive power is maximum for perfectly black body
D. Ratio of emissive power to absorptive power for all bodies is same and is equal to the emissive power of a perfectly black body.
Correct Answer: D
Solved Example: 82302
According to Kirchoff's law, the ratio of emissive power to absorptivity for all bodies is equal to the emissive power of a:
A. Gray body
B. Brilliant white polished body
C. Red hot body
D. Black body
Correct Answer: D
Solved Example: 82303
The ratio of the emissive power and absorptive power of all bodies is the same and is equal to the emissive power of a perfectly black body. This statement is known as: (BPSC AE ME 2012 Part 5)
A. Kirchoff's law
B. Stefan's law
C. Wien's law
D. Planck's lawv
Correct Answer: A
Net Energy Exchange by Radiation between Two Bodies
Learning Objectives:

Calculate net radiation heat exchange rate between two bodies.
\[\dot Q_{12} = \epsilon \sigma A \left(T_1^4  T_2 ^4\right)\]
Solved Example: 82501
If a black body is heated from 27$^{\circ}\mathrm{C}$ to 927$^{\circ}\mathrm{C}$. Then the ratio of radiation emitted will be:
A. 1:4
B. 1:16
C. 1:8
D. 1:256
$T_2$ = 927 + 273 = 1200 K, $T_1$ = 27 + 273 = 300 K
\[\dfrac{T_1}{T_2} = \dfrac{1}{4}\]
\[\dfrac{Q_1}{Q_2} = \left(\dfrac{T_1}{T_2}\right)^4 = \dfrac{1}{256}\]
Correct Answer: D
Solved Example: 82502
Two large diffuse gray parallel plates, separated by a small distance, have surface temperatures of 400 K and 300 K. If the emissivities of the surfaces are 0.8 and the StefanBoltzmann constant is 5.67 X $10^{8}W/m^2K^4$, the net radiation heat exchange rate in $kW/m^2$ between the two plates is: (GATE ME 2013)
A. 0.79
B. 0.99
C. 3.96
D. 0.66
\begin{align*} q_{12} &= \dfrac{\sigma\left( T_1^4T_2^4\right)}{\dfrac{1}{\epsilon_1}+\dfrac{1}{\epsilon_2}1}\\ &= \dfrac{5.67 \times 10^{8}\left( 400^4300^4\right)}{\dfrac{1}{0.8}+\dfrac{1}{0.8}1}\\ &= 661.5 W/m^2\\ &= 0.66 KW/m^2 \end{align*}
Correct Answer: D
Solved Example: 82503
A black body of emissive power 81 $J/m^2 s$ when it is at 300 K and ordinary body of emissivity 0.8 when it is at 500 K. What is the emissive power of an ordinary body?
A. 500 $J/m^2 s$
B. 600 $J/m^2 s$
C. 800 $J/m^2 s$
D. 400 $J/m^2 s$
Correct Answer: A
Solved Example: 82504
If the temperature of a solid surface changes from 27$^\circ$C to 627$^\circ$C, then its emissive power changes in the ratio of: (TNTRB 2012 ME)
A. 3
B. 6
C. 9
D. 81
Correct Answer: D
Solved Example: 82505
The value of the wavelength for maximum emissive power is given by:
A. Wien's law
B. Planck's law
C. Stefan's law
D. Fourier's law
Correct Answer: A
Solved Example: 82506
In case the temp of hot body is increased by 50%, the amount of radiation emitted by it would increased by nearly:
A. 50%
B. 500%
C. 250%
D. 125%
Correct Answer: B
Solved Example: 82507
A water cooled spherical object of diameter 10 mm and emissivity 0.9 is maintained at 400$^\circ$C. What is the net transfer rate from the oven walls to the object?
A. 3.04 W
B. 4.25 W
C. 8.55 W
D. 1.16 W
Correct Answer: A
Solved Example: 82508
A surface of area 0.5 $m^2$, emissivity 0.8 and temperature 150$^\circ$C is placed in a large, evacuated chamber whose walls are maintained at 25 $^\circ$C. What is the rate at which radiation is emitted by the surface? What is the net rate at which radiation is exchanged between the surface and the chamber walls?
A. 568 W
B. 511 W
C. 547 W
D. 727 W
Correct Answer: C
Solved Example: 82509
When heat is transferred from hot body to cold body, in a straight line, without affecting the intervening medium, it is referred as heat transfer by:
A. Conduction
B. Convection
C. Radiation
D. Conduction and convection
Correct Answer: C
Solved Example: 82510
The rate of energy emission from unit surface area through unit solid angle, along a normal to the surface, is known as: (SSC JE ME March 2017 Paper 6)
A. Emissivity
B. Transmissivity
C. Reflectivity
D. Intensity of radiation
Correct Answer: D
Solved Example: 82511
Emissivity of a white polished body in comparison to a black body is: (RSMSSB JE Elect/Mech Dec 2020)
A. Higher
B. Lower
C. Same
D. Depends upon the shape of body
Correct Answer: B
Solved Example: 82512
Two balls of same material and finish have their diameters in the ratio of 2:1 and both are heated to same temperature and allowed to cool by radiation. Rate of cooling by big ball as compared to smaller one will be in the ratio of: (NPCIL Stipendiary Trainee ME 2018)
A. 1:1
B. 2:1
C. 1:2
D. 4:1
Correct Answer: C
Solved Example: 82513
Total emissivity of polished silver compared to black body is:
A. Same
B. Higher
C. More or less same
D. Very much lower
Correct Answer: D
Solved Example: 82514
According to Wien's law, the wavelength corresponding to maximum energy is proportion to:
A. absolute temperature (T)
B. T$^2$
C. f
D. t
Correct Answer: A
Solved Example: 82515
The total emissivity power is defined as the total amount of radiation emitted by a black body per unit: (BPSC AE ME 2012 Part 5)
A. Temperature
B. Thickness
C. Area
D. Time
Correct Answer: D
Solved Example: 82516
The ratio of the energy absorbed by the body to total energy falling on it is called:
A. Absorptive power
B. Emissive power
C. Absorptivity
D. Emissivity
Correct Answer: A
Solved Example: 82517
40% of incident radiant energy on the surface of a thermally transparent body is reflected back. If the transmissivity of the body be 0.15, the emissivity of surface is:
A. 0.45
B. 0.55
C. 0.40
D. 0.75
Correct Answer: A
Solved Example: 82518
The emissive power of a body depends upon its:
A. Temperature
B. Wave length
C. Physical nature
D. All of the above.
Correct Answer: D
Solved Example: 82519
Two plates spaced 150 mm apart are at 1000$^\circ$C \& 70$^\circ$C. The heat transfer will take place mainly by:
A. Convection
B. Free convection
C. Forced convection
D. Radiation
Correct Answer: D
Solved Example: 82520
Absorptivity of a body will be equal to its emissivity:
A. At all temperatures
B. At one particular temperature
C. When system is under thermal equilibrium
D. At critical temperature
Correct Answer: C
Solved Example: 82521
Planck's law holds good for:
A. Black bodies
B. Polished bodies
C. All coloured bodies
D. All of the above.
Correct Answer: A