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Module 1. Basic Concepts, Conductive Heat Transfer...

Module 2. Convection

Module 3. Radiation

Module 4. Heat Exchangers

Module 5. Mass Transfer

## Lesson-23 Radiation between two infinite parallel plates and proof of Kirchhoff’s law of Radiation, Radiation Shield

**Radiation between two infinite parallel plates and proof of Kirchhoff’s law of Radiation: **

At a given temperature, the total emissive power of a any body is equal to its absorptivity multiplied by total emissive power of a perfect black body at that temperature.

Therefore E = αE_{b}

But the ratio of total emissive power of any body to the total emissive power of a black body at the same temperature is called the emissivity of the body and is numerically equal to absorptivity.

Consider two bodies C and D whose absorptivity are α_{c} and α_{d} as shown in Figure 1.

Considering the energy emitted by the body D.

(1) D emits the energy = E_{d} (1)

(2) C absorbs energy = E_{d}. α_{c} (2)

and reflects energy=E_{d} (1 - α_{c}) (3)

(3) D absorbs energy = E_{d} α_{d} (1 - α_{c}) (4)

and reflects energy = E_{d} (1 - α_{d}) (1 - α_{c}) (5)

(4) C absorbs energy = E_{d} α_{c} (1 - α_{c}) (1 - α_{d}) (6)

and reflects energy = E_{d} (1 - α_{c})^{2} (1 - α_{d}) (7)

(5) D absorbs energy = E_{d}. α_{d} (1 - α_{c})^{2} (1 - α_{d}) (8)

and reflects energy = E_{d} (1 - α_{c})^{2} (1 - α_{d})^{2} (9)

and so on upto times.

Considering the energy emitted by the body C.

(1) C emits the energy = E_{c} (10)

(2) D absorbs energy = E_{c}. α_{d} (11)

and reflects energy=E_{c} (1 – α_{d}) (12)

(3) C absorbs energy = E_{c} α_{c} (1 – α_{d}) (13)

and reflects energy = E_{c} (1 – α_{c}) (1 – α_{d}) (14)

(4) D absorbs energy = E_{c} α_{d} (1 – α_{d}) (1 – α_{c}) (15)

and reflects energy = E_{c} (1 – α_{d})^{2} (1 – α_{c}) (16)

(5) C absorbs energy = E_{c}. α_{c} (1 – α_{d})^{2} (1 – α_{c}) (17)

and reflects energy = E_{c} (1 – α_{d})^{2} (1 – α_{c})^{2} (18)

and so on upto infinite number of times.

Considering equations (1), (4), (8), (11), (I5), net energy lost by the body D

= Energy emittedby it – energy absorbedby it

q_{(dc)} =E_{d} – [E_{d} α_{d} (1 - α_{c}) + E_{d}. α_{d} (1 - α_{c})^{2} (1 - α_{d}) + ….] – [E_{c}. α_{d} + E_{c} α_{d} (1 – α_{d}) (1 – α_{c}) +….]

Assuming (1 - α_{c}) (1 – α_{d}) = K

q_{(dc)} = E_{d} – E_{d} α_{d} (1 - α_{c}) [1 + K + K^{2} + …...] – E_{c}. α_{d} [1 + K + K^{2} + ….]

But 1 + K + K^{2} + …... = (1 –K)^{-1}

q_{(dc)} = E_{d} – (1 –K)^{-1} [E_{d} α_{d} (1 - α_{c}) + E_{c}. α_{d} ]

**i) If originally both bodies are at same temperature **

Then q_{(dc)} = 0 from equation (18A)

E_{d} a_{c} = E_{c} a_{d}

Assuming C as black body

Subscription ‘b’ represents black body.

However, E_{d }/ E_{b} is the emissivity of body ‘D’ according to definition of emissivity.

ε_{d} = α_{c} This is the statement of Kirchoff’s law and hence it is proved.

**ii) If both the bodies are at different temperatures**

Using equation (19)

**According to Stefan-Boltzman Law**

E_{d} = ε_{d} σ_{b} T_{1}^{4}

E_{c }= ε_{c} σ_{b} T_{2}^{4}

Substituting these values in equation (19), we get

**Radiation Shields**

Generally, the shields are used for reducing the heat radiation from one plane to another plane. If the shield 3 is placed in between the two planes as shown in Figure 2 and considering they are at temperatures T_{1}, T_{2} and T_{3}.

Assuming there is no temperature drop in the shield and considering the system is in steady state condition, we can write down the heat flow equation as

It means that with the addition of radiation shield, heat transfer rate is reduced to half of that of without the presence of radiation shield between two parallel bodies exchanging heat with each other by radiation. If ‘n’shields are present between the two radiating bodies, then the heat transfer will be expressed as