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:

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

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.

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!

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:

- we have used only black bodies
- we have assumed that all the energy which body 2 emits reaches body 1

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

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.

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 *A*_{1} is, however, *not*
direction specific (i.e., $F_{12}$*A*_{1} =
*F*_{21}*A*_{2}). 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 *F*_{21} 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 *F*_{21} in this case?

Two critical properties of the **view factor** then are:

- F
_{12}*A*_{1}=*F*_{21}*A*_{2} - $\sum_j F_{ij} =1$

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

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

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 |

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.