# RADIATION EXCHANGE BETWEEN GREY AND BLACK SURFACES

**RADIATION EXCHANGE BETWEEN GREY AND**** **

** **** BLACK SURFACES**

**Radiation Shape factor algebra make **

**the study of radiation exchange between **

**surfaces simple & easy. Number of shape**

** factors is equal to the square of the number**

** of radiating surfaces.**

## Radiation exchange is considered between

## black surfaces. It has also been considered

## between real surfaces. If exchange is in excess,

## then radiation shields will be considered to

## reduce the radiations between surfaces.** **

**Definition of a Shape Factor**

#### There is a factor called the SHAPE FACTOR (F_{ij)} OR view factor or configuration factor or form factor or geometric factor. It accounts for the orientation and geometry of the surfaces exchanging radiations.

#### F12 =Shape factor=ratio of radiations intercepted by surface 2/radiations leaving surface 1.

#### Symbol is F_{ij}. _{ }

#### For example 140 units of energy emitted by body A.

#### 35 units of energy are received by a body B.

#### Then F_{AB}= 35/140=0.25.

#### This depends how the surfaces view each other.

#### Also, it depends how the surfaces radiate energy with respect to each other.

**LAWS FOR SHAPE FACTORS**

#### (i). **Reciprocity law** A_{i}F_{ij} = A_{j}F_{ji} OR A_{2} F_{23} =A_{3} F_{32}

#### Imagine surface 1 is convex and is enclosed in another surface 2 (a small ball in a big ball),

#### then F12=1

#### from the law of reciprocity F21=A_{1}/A_{2}

#### (ii). **Summation Law – **Say for four bodies,1,2,3 and 4

#### Firstly F_{11}+F_{12}+F_{13}+ F_{14} =1

#### Secondly F_{21}+F_{22}+F_{23}+ F_{24} =1

#### Thirdly F_{31}+F_{32}+F_{33}+ F_{34} =1

#### Fourthly F_{41}+F_{42}+F_{43}+ F_{44} =1

#### (iii). **Shape factor for a body itself **

#### F_{ii} or F_{11} energy received by body 1 out of energy emitted by body 1

#### Thus, shape factor for a flat surface F_{ii} =0

#### Further, for a convex surface F_{ii} = 0 and hence F_{11} =F_{22}=F_{33}=0

#### For a concave surface F_{ii} ≠ 0 and hence F_{11} ≠0, F_{22}≠0, F_{33}≠0

#### (iv). If there are ‘n’ surfaces, there are n^{2} shape factors

#### example, for three bodies, 9 shape factors are required.

**HOW TO FIND THESE n**^{2} shape factors

^{2}shape factors

#### There are three steps.

#### (i) Apply summation rule to get n equations to give n shape factors i.e. 3

#### (ii) Use Reciprocity relation n (n—1)/2 times to get n (n—1)/2 shape factors i.e. 3

#### (iii) Now it is required to find the remaining shape factors i.e. Remaining= n^{2} –n– n (n—1)/2 shape factors i.e. 3 in this case.

#### Case of three surfaces

#### Remaining shape factors will be = 3^{2}–3–3(3—1)/2=9-3-3=03

#### These remaining shape factors are found from general observations like F_{11} = 0, F_{12}=1, F_{21}=A_{1}/A_{2} etc.

** EQUATIONS IN SHAPE FACTORS**

#### Equations for shape factors are not available for complex shapes. Complex shapes are converted into simpler shapes. Shape factors then for simpler shapes can be found from the shape factor algebra. It is based on

#### (i). The definition of shape factor

#### (ii). Law of reciprocity

#### (iii). Summation law

#### There are two surfaces A_{1} and A_{2}. Say receiving surface A_{2} is a complex surface. It is divided further into two areas A_{3} and A_{4} in simple areas.

#### A_{1}F_{12}=A_{1}F_{13} +A_{1}F_{14}

### Of course, the shape factors can be found from standard graphs for the following cases.

#### (i). Parallel surfaces (rectangular, equal circular, non-equal circular, etc)

#### (ii). Surfaces at right angles

**Fig. Shape Factors for two Parallel Surfaces**

**Fig. Shape Factors Between Two Rectangular surfaces at right angles with a common edge**

### Radiation exchange between any numbers of black surfaces depends on the followings:

#### (i). Orientation— position in relation to each other

#### (ii). Geometrical shape-(Flat, concave, convex)uses areas for shape factors

#### (iii). Temperature

#### (iv). Radiative properties like emissivity, absorptivity, transmissivity, reflectivity and radio city

**2. NET RADIATION EXCHANGE BETWEEN TWO BLACK SURFACES 1 AND 2 :**

#### Radiation exchange from body 1 to body 2 =q_{1} = F_{12} A_{1}q_{1}= F_{12 }A_{1} E_{b1}

#### Radiation exchange from body 2 to body 1, q_{2} = F_{21} A_{2}q_{2} = F_{21 }A_{2} E_{b2}

#### Net radiative exchange between 1 and 2, q_{12} = q_{1}

#### q_{12}= F_{12} A_{1}q_{1} — F_{21} A_{2}q_{2}= F_{12} A_{1} (q_{1} – q_{2}) = F_{12} A_{1 }σ(T_{1}^{4}–T_{2}^{4}) (since F_{12} A_{1}= F_{21}A_{2})

**METHODS TO FIND RADIATION EXCHANGE BETWEEN GREY SURFACES**

#### There are two methods to find radiation exchange between grey bodies.

**(a) ****BASED ON CONFIGURATION FACTOR/INTERCHANGE FACTOR/EQUIVALENT EMISSIVITY BETWEEN TWO GREY BODIES **

** **Shape factor for grey bodies is also called the Configuration factor/interchange factor/equivalent emissivity. It is represented for grey bodies by f_{12} and for black bodies by F_{12}.

#### Various equations for f_{12}

#### (i). For two small grey bodies f_{12} =Є_{1 }Є_{2}

#### (ii). A small body 1 is enclosed in a large body 2, f_{12}=Є_{1}

#### (iii). Between two parallel infinite surfaces f_{12} =1

#### (iv). Among two infinite long concentric cylinders/spheres

#### f_{12} =

#### (v). A large body 1 enclosed into another large body 2

#### f_{12} =

#### Q_{12}= f_{12}A_{1}σ_{b}(T_{1}^{4}—T_{2}^{4})= A_{1}σ_{b}(T_{1}^{4}—T_{2}^{4})

**PROOF: Radiation exchange for infinite parallel real surfaces**

#### Fig. Radiation exchange between two parallel infinite grey surfaces

#### Assumptions used

#### (i). Configuration factor for either surface is unity i.e. f_{12}=1 and f_{21}=1

#### (ii). There is a non absorbing medium like air between these surfaces.

#### (iii). Properties like emissivity, reflectivity and absorptivity is constant over the entire surfaces.

#### Let surface 1 emits E_{1} with α_{1}, Є_{1 }at temperature T_{1} and surface 2 emits E_{2} with α_{2}, Є_{2 }at temperature T_{2}

#### After number of reflections

#### Q_{1net} =E_{1}–[ α_{1}(1– α_{2}) E_{1}+ α_{1} (1—α_{1}) (1—α_{2})^{2}E_{1} + α_{1}(1—α_{1})^{2} (1– α_{2})^{3} E_{1}+…………………….]

#### After number of reflections,

#### it becomes Q_{1net} =E_{1}— After number of reflections

#### Therefore, Q_{1net} =E_{1}— α_{1}(1– α_{2})E_{1}[1 + (1—α_{1}) (1—α_{2})+ (1—α_{1})^{2} (1– α_{2})^{2} +…………………………]

#### Taking (1—α_{1}) (1—α_{2}) as common

#### In order to simplify, Put Z=(1—α_{1}) (1—α_{2})

#### We get, Q_{1net} =E_{1}— α_{1}(1– α_{2})E_{1}[1 +Z+Z^{2} + Z^{3} + ………………………………]

#### As Z<1 the sum of infinite series 1+Z+Z^{2}+Z^{3}+…………….=1/(1—Z)

#### Thus, Q_{1net} =E_{1}— α_{1}(1– α_{2})E_{1}/(1—Z)

#### Substitute the value of Z

#### It gives, Q_{1net} =E_{1} [1— α_{1}(1– α_{2})/(1– (1—α_{1}) (1—α_{2})]

#### As per Kirchhoff’s law α=Є i.e. α_{1}=Є_{1} and α_{2}=Є_{2}

#### Then, Q_{1net} =E_{1} [1— Є_{1}(1– Є_{2})/(1– (1—Є_{1}) (1—Є_{2})]

#### Opening the brackets,

#### Q_{1net} =E_{1} [1—(1– Є_{1})(1– Є_{2})— Є_{1}(1– Є_{2})]/[(1– (1—Є_{1}) (1—Є_{2})]

#### Where Q_{1net} =It is net energy going from body 1 towards body 2.

#### Similarly Q_{2net
}

#### It is net energy going from body 2 towards body 1.

#### Radiation Exchange between surfaces 1 and 2

#### Q_{12} = Q_{1} –Q_{2} =

#### Now E_{1} = σ T_{1}^{4} and E2= σ T_{2}^{4}

#### Q_{12} =f_{12} σ (T_{1}^{4} — T_{2}^{4} )

#### where f_{12} is interchange factor or configuration factor for grey bodies

**(b) ****BASED ON ELECTRIC NETWORK **

**ANALOGY METHOD – **Applicable both for black as well as for grey bodies. This method is direct, more general, and much simpler. This method uses incident G and radiosity J (W/m^{2}).

#### G = radiations flux incident on the surface, W/m^{2}

#### Radiosity J = (emitted + reflected ) radiation flux from a grey body

#### E_{b1} emitted energy from body 1 as a black body at a certain temperature T_{1}

#### J_{1} total energy coming out from body 1 as a grey body at the temperature T_{1}

#### J= ( emitted+ reflected) radiation flux

#### Difference in E_{b1} and J_{1} is due to the surface resistance

#### J_{1} total energy coming out from body 1 as a grey body at temperature T_{1}

#### J_{2} total energy coming out from body 2 as a grey body at temperature T_{2}

#### There is a resistance between grey bodies J1 and J2 called the space resistance.

#### J_{2} total energy coming out from body 2 as a grey body at temperature T_{2}

#### E_{b2} emitted energy from body 2 as a black body at temperature T_{2}

#### Difference in E_{b2} and J_{2} is due to the surface resistance

#### Q_{12} = J_{1}A_{1} F_{12} —J_{2}A_{2} F_{21}

#### It becomes Q_{12} =A1F_{12} (J_{1}—J_{2}) since A_{1} F_{12} = A_{2} F_{21}

#### Q_{12} = (J_{1}—J_{2})/(1/ A1F_{12})= (J_{1}—J_{2})/resistance

#### Now 1/A1F_{12} is called the** SPACE RESISTANCE** between the two grey bodies. Because it is due to the geometry and distance between the two grey bodies.

**Fig. Space and surface resistances between three bodies**

**Fig. Space & surface resistance between two bodies**

### 3. **NET RADIATIVE EXCHANGE BETWEEN TWO GREY SURFACES USING RADIOSITY**

#### Fig. Radiosity J = ρG+ ε E

#### ( Radiosity J = emitted +reflected) radiation flux

#### J=E_{g }+ρG

#### J=ЄE_{b}+ ρG where E_{b} is the emissive flux from a black body

#### Now α+ ρ+ ς=1

#### ς =0 since the surface assumed to be opaque.

#### Now α+ ρ=1

#### ρ=1—α

#### J=ЄE_{b}+ (1—α )G

#### α=Є by Kirchhoff’s Law

#### J=ЄE_{b}+ (1—Є )G

#### J– ЄE_{b}=(1—Є )G

#### G = (J– ЄE_{b})/ (1—Є )

#### Q_{net}=A(J—G)

#### =A(J–(J– ЄE_{b})/ (1—Є ))

#### =A[(J(1—Є ))– (J– ЄE_{b})]/ (1—Є )

#### =A[(JЄ–ЄE_{b})]/ (1—Є )

#### =(E_{b} –J)/[ (1—Є )/ЄA)]

#### This equation can be represented in the form of an electric network

#### Q_{net} is rate of heat transfer like current which is flow of electric charge

#### (E_{b} –J) is like voltage difference

#### (1—Є )/ЄA) is like resistance.

#### This resistance In this case is called surface resistance . It is because (E_{b} –J) refers to difference in heat flux of a black body and a grey surface at the same temperature.

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