LTR: Radiative Heat Transfer

As mentioned previously, radiation is unique in that no mass is necessary to bridge the gap between the two bodies which are exchanging heat. Instead, heat is transferred by emittance and absorption of energy "rays" or "packets" (photons). This is the way that the sun heats the Earth (clearly, as there is no mass between the two, in space).

The rate expression for radiation emission is associated with the names Stephan (1879) and Boltzmann (1884) who independently proposed the form:

EQUATION:

$\displaystyle{\frac{q}{A} = \sigma T^4}$

where $\sigma$ is the Stephan-Boltzmann constant, $5.676x10^{-8} W/m^2K^4$. The above expression is only valid for the simplified case of what we will call a "perfect" radiator, called a "black-body" radiator.

NOTE:

Upon reflection (punny, eh ;), the choice of the name "black body" should not be surprising, as we have all experienced the fact that black shirts absorb solar energy more efficiently than white shirts. It turns out that they emit the energy more efficiently as well!

EXAMPLE:

Lets look at a very simple example:

Consider the situation where one black body (perfect radiator) is enclosed entirely within the other. We could imagine that body 1, at $T_1$, is radiating according to:

$\displaystyle{\frac{q}{A} = \sigma T_1^4}$

Clearly all of this energy must hit body 2. Since body number 2, at $T_2$, is also radiating energy, we might assume for a minute that all of the energy radiated from body 2 not only hits body 1 but is absorbed by it so that:

$\displaystyle{\frac{q}{A} = -\sigma T_2^4}$

This would lead us to a net heat flow from body 1 is given by:

$\displaystyle{q_{1\to2} = \sigma A_1(T_1^4-T_2^4)}$

We have actually oversimplified the real-life situation quite a bit in this example:

If we instead consider the "realistic" bodies shown here, we can relax these two conditions.

Any Geometry

In order to relax our "aim" condition (that all energy emitted from 1 actually makes it to 2 and vice versa), we clearly need to consider the geometry of the problem (if some of body 1 faces away from body 2 the energy emitted from that side should not make it to 2 and vice versa).

These geometrical considerations are typically incorporated in a geometrical factor, $F_{12}$, called the view factor.

DEFINITION:

The view factor is the fraction of the area of one body that is "seen" by the other body. Alternatively, we might think of it as the fraction of the radiation leaving one body that "hits" the other body.

As you might gather from this definition, $F_{12}$ is direction specific, meaning that it can only be used for heat transfer from body 1 to body 2. The product of $F_{12}$ and A1 is, however, not direction specific (i.e., $F_{12}$A1 = F21A2). This fact is termed reciprocity and simply states that the net heat flow from one body to the other should not depend on which way it is calculated.

Going back to our first example, we might imagine that some of the radiation leaving body 2 actually hits body 2 rather than body 1. This means that F21 is not equal to 1; however, $F_{12}$ =1 since all radiation leaving body 1 hits body 2, so our expression above is still correct (at least for black bodies in that geometrical configuration).

What is F21 in this case?

IMPORTANT:

Two critical properties of the view factor then are:

Non-Black (Gray) Bodies

Things become considerably more complex if we consider that not all bodies are perfect radiators. A simple method of relaxing the condition of non-black bodies is to introduce a correcting factor for the amount of energy emitted relative to that of a black body, $\epsilon$ (emissivity), so that

$\displaystyle{\epsilon = \frac{E}{E_b}}$


where $E$ is the emissive power of the material (at some temperature, $T$) and $E_b$ is the emissive power of a black body (see the equation for the heat flow (the flux times the area) for a black body (above), and remember that power is energy per time....heat flow!). Using this relation, we can write that the heat emitted from body 1 is

$\displaystyle{\frac{q}{A}=\epsilon\sigma T_1^4}$

Similarly, we can write a correcting factor for the absorption, $\alpha$ (absorptivity), so that the heat absorbed is

$\displaystyle{\frac{q}{A}=-\alpha\sigma T_2^4}$

It is often assumed that for a "gray" body, one in which $\epsilon$ and $\alpha$ are independent of temperature, the emissivity should be equal to the absorptivity ($\epsilon = \alpha$). Even this simply rescaling of the radiation flux, however, makes radiation problems considerably more difficult as we now need to consider not only the initially incident radiation, but also reflected radiation (and we need to include the view factors for each rebounded ray!).

For the purposes of this class, we will focus on situations where body 2 is much larger than body 1 (so that we can neglect radiation reflecting off of body 2 and we can consider body 2 essentially black) so we can write the new heat flux from body 1 to body 2 as

$\displaystyle{q_{1\to2} = F_{12}\epsilon_1\sigma A_1(T_1^4-T_2^4)}$

OUTCOME:

Perform radiative heat transfer calculations

EXAMPLE:

A water cooled spherical object of diameter 10 mm and emissivity 0.9 is maintained at 80C when placed in a large vacuum oven whose walls are maintained at 400C. What is the heat transfer rate from the oven walls to the object? Recall that the Stefan-Boltzmann constant,$\sigma$, is $5.676x 10^{-8} W/m^2K^4$.

Since body 2 surrounds body 1 and $A_2 \gg A_1$, the governing equation (heat flow from 1 to 2):

$\displaystyle{q_{1\to2} = \epsilon_1\sigma A_1(T_1^4-T_2^4)}$

where $A = \pi D^2$ (since it is the surface area of the sphere).

Plugging in our numbers

$q = (0.9)(5.676x10^{-8}W/m^2K^4)\pi(0.01m)^2(353K^4-673K^4) = -3.04 W$

so for flow from walls (body 2) to sphere (body 1) $q =3.04 W$.

For simplicity, it is common to try to linearize the relation for radiative heat transfer (if you don't see why now, you will in a little bit!). So, if we factor the full equation, we get

$\displaystyle{\frac{q}{A} = \epsilon\sigma(T_1-T_2)(t_1+T_2)(T_1^2+T_2^2)}$

If we then assume that $T_1 \approx T_2$ (and use the mean, $T_m$) the sums can be simplified and the equation can be re-written

$\displaystyle{\frac{q}{A} \approx \epsilon\sigma 4t_m^3(T_1-T_2)}$

so that if we define $h_r$ as

$\displaystyle{h_r = 4\epsilon\sigma T_m^3}$

which leaves us with the final (very simple) form of

$\displaystyle{\frac{q}{A} = h_r(T_1-T_2)}$

If we apply this equation to our previous problem we get for $h_r$

$q = (27.6W/m^2K)\pi(0.01m)^2(353K-673K) = -2.77W$

which is <10% off.