Vijay Gupta

MEC321: Heat Transfer

5. Radiative Heat Transfer

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Thermal Radiation

In thermal radiation, energy transfer occurs on account of emission and absorption of photons. These photons travel large distances (of the order of a kilometer) under ordinary atmospheric conditions before they are absorbed or scattered. Thus, the mean-free paths of such photons are usually much larger than the dimensions encountered in most problems. Therefore, the rate of heat transfer does not depend upon the temperature gradient, but on the absolute temperatures of the two surfaces. One should also note that whenever the mean-free path of photons is large, the intervening medium has little effect on the rate of transfer. In this chapter we will neglect any such effect.

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Emissive Power of a Surface

The total emissive power of a surface is defined as the total emission of the thermal-radiation energy by the surface in all directions

W/m2

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Irradiation

• Irradiation G of a surface is defined as the energy received by the surface (from all directions and of all wavelengths) per unit time and per unit surface area.

• Clearly, G does not depend on the temperature of the surface concerned directly.

• Since irradiation G is not a surface property, it is highly unlikely that the value of G will be constant over a finite-sized surface. One special circumstance when this may be so is if the body is receiving energy from a very distant large source, for example the irradiation of small surfaces on earth from the sun can be taken as uniform.

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Reflection, Absorption and Transmission

• reflectivity • absorptivity • transmissivity

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Gray Surface

Assumption of and independent of

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Black Body

Though a perfectly black surface would appear black, since it absorbs all incident radiation (including visible part of the spectrum), a practical black body may not appear black at all! Freshly fallen snow is one of the best absorbers of thermal radiations, with an absorptivity in excess of 0.98, but it appears white because of its very low absorptivity in the narrow range of visible radiation with almost perfect absorption for higher wavelengths

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Kirchhoff Laws

• Let us first consider the disc to be a black body. It absorbs all the radiation received by it, and, therefore, the radiation emitted Eb (the black-body emissive power) is equal to its irradiation G in the cavity.

• From the fact that thermal equilibrium prevails at all locations and orientation, and that the emittance of the black-body should be independent of its location and orientation, we conclude that the radiation flux within the enclosure is uniform at all locations and in all orientations.

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Kirchhoff Laws

• Thus, the radiation field produced within a cavity, irrespective of the nature of the walls of the cavity is isotropic with the flux equal to Eb, the black-body emissive power.

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Kirchhoff Laws

• Next, replace the black-body disc with another of the same size but of absorptivity α. The radiant energy impinging on its surface, i.e. its irradiation, is again Eb. But only a fraction of it is absorbed. If the temperature of this disc is the same as the temperature of the cavity, thermal equilibrium will again prevail, and thus, the emission from the surface must be Eb per unit area per unit time, or

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Kirchhoff Laws

Since cannot be greater than one, we obtain from this the two Kirchhoff laws:

(a) A black body emits more radiation than any other surface at a prescribed temperature

b) The emissivity of a surface is equal to its absorptivity , i.e.

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Stefan-Boltzmann Law

𝐸𝑏=𝜎𝑇 𝑠4

𝜎 is a universal constant known as the Stefan-Boltzmann constant. Its value is 5.67×10−8 W/m2K4

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Monochromatic or Spectral Emissions

• The monochromatic or the spectral emissive power of a black surface is defined as the flux of emitted radiation per unit wavelength-interval about the wavelength 𝜆. Thus, represent the energy flux within the wavelength interval from 𝜆 to 𝜆 + d . 𝜆 Clearly, the units of are W/m3, and the total emissive power Eb is related to the spectral , by the following integral

• • Since is a function of the temperature Ts alone, it is

reasonable to expect to be a function of 𝜆 and Ts, independent of the material of the black surface.

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Planck Distribution Law

𝐸𝑏 ,𝜆(𝜆 ,𝑇 )=𝐶1

𝜆5(exp( 𝐶2

𝜆𝑇 )−1)

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Radiation Exchange between Black Surfaces

Incident on A1:

Emission from A1 :

Net exchange: 𝐴1 ,𝑇 1

𝐴2 ,𝑇 2

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The Configuration Factor

To calculate the energy exchange by radiation between two surfaces, we need to know the properties of emission, absorption and reflection of the two surfaces. But not all the emission by one surface reaches the other surface. The fraction of energy leaving one surface and reaching another depends upon the geometric relationship between the two surfaces. We introduce here the concept of the configuration- or the view-factor for isothermal surfaces emitting diffusely. The configuration factor is defined as the fraction of energy leaving surface (diffusely) which is intercepted by the surface

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Self-view Factor

It may be noted that for a finite surface , a part of radiation leaving may be intercepted by the surface itself (see Fig. 8.30). This will happen if is concave. Thus, there occurs a self-view factor , for concave surfaces. Clearly, for flat and convex surfaces, the self-view factor , is zero.

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Shape factor algebra

∑𝑗

❑

𝐹𝐴 𝑖→ 𝐴 𝑗=1

𝐴1 ,𝑇 1 𝐴2 ,𝑇 2

𝑄1→ 2=𝐸𝑏1𝐴1𝐹12−𝐸𝑏2 𝐴2𝐹 21

If , =

So that

Reciprocity relation

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Configuration factor algebra

Consider a convex surface (say a sphere) completely enclosed by an irregular surface. As there are two surfaces involved, we may expect four configuration factors. Since is convex, vanishes. This leaves three factors, with one reciprocity relation and two summation rules: ; and

The three factors are readily obtained as , and

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Configuration factor algebra

For a three-surface enclosure made up of non-concave surfaces, there are six factors and six relations: three reciprocity relations and three summation rules. It can be seen that

and

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Hottel Cross-Strings Method

These two relations give

𝐹 𝐴1→ 𝐴5=(𝐿1+𝐿5− 𝐿4 ) /2𝐿1

Similarly, , and can be written down in terms of the lengths of the respective lines

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Hottel Cross-Strings Method

or

This is the Hottel Cross-Strings rule : is the sum of the lengths of the two crossed-strings and is the sum of the lengths of the two uncrossed-strings.

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Hottel Cross-Strings Method

From geometry

Therefore

or

For the special case of \

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Radiative Heat Exchange Between Black-body SurfacesConsider an enclosure made up of N black-body isothermal surfaces. Let us consider that the ith surface of area Ai is maintained at a steady temperature Ti with the help of heat transfer (to it) at a rate Qi (by means other than radiation). Then, by energy balance

The rate of emission of radiation by the black-body surface i is .

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Radiative Heat Exchange Between Black-body SurfacesA fraction of the power emitted by the jth surface is received by this surface. So the total energy received and absorbed by it is , where the summation is over j and extends over all the surfaces of the enclosure, including the ith surface. The energy balance, therefore, gives.

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Radiative Heat Exchange Between Black-body SurfacesWe use the reciprocity relation to change to . Since over all the N surfaces equals one oror

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Radiative Heat Exchange Between Black-body Surfaces

𝑄𝑖=∑𝑗=1

𝑁 (𝜎𝑇 𝑖4−𝜎𝑇 𝑗

4 )1

𝐴𝑖𝐹 𝐴𝑖→ 𝐴 𝑗

For a 3-surface enclosure

For a 2-surface enclosure

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A thermocouple is placed at the centre of a long duct and measures the temperature of a gas flowing in the duct. The internal walls of the duct are at 327 °C and the convective heat transfer coefficient from the gas to the thermocouple is 56.7 W/m2K. If the thermocouple indicates a temperature of 527 °C, find the true temperature of the gas. Assume both the thermocouple and the duct to be black bodies, and the gas to be radiatively non-participating

Radiation error in an unshielded thermocouple

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Radiation error in an unshielded thermocouple

Since the duct is very long, the thermocouple c is exchanging heat only with the duct d. We can assume that the thermocouple sees only the duct, The radiative heat loss from the thermocouple, therefore, is

At the steady state, the heat gained by the thermocouple by convection from the gas should be equal to the heat lost by radiation to the duct. Therefore,

or Using , and , we get