sci.energy

On Aug 23, 7:03 pm, "daestrom"
wrote:
> "Phil." wrote in message
>
> news: @ ...
>
>
>
> > On Aug 18, 11:58 pm, Bill Ward wrote:
> >> On Sat, 18 Aug 2007 18:57:37 -0700, Phil. wrote:
> >> > On Aug 18, 6:21 pm, Bill Ward wrote:
>
> >>
>
> >> >> Do you have an explanation for the cavity paradox discussed above?
>
> >> > As I understand your cavity it consists of two partial ellipsoidal
> >> > mirrors
> >> > facing each other with the F2 of each being the F1 of the other. The
> >> > annular gap caused by the different sized ellipsoids is filled by a
> >> > partial spherical mirror the centre of which is the focus of the larger
> >> > ellipsoid? For simplification we'll assume a vacuum in the cavity and
> >> > perfect insulation from outside.
>
> >> > If there is a black body at each focus what is the radiation balance
> >> > between them?
> >> > Consider any light from the focus of the larger ellipse that doesn't
> >> > enter
> >> > the second ellipse, it must hit the spherical mirror and be reflected
> >> > back
> >> > to it's origin, therefore there's no change in heat exchange.
>
> >> Why doesn't that heat count? How does the BB know to ignore it?
>
> > It doesn't ignore it, flux leaving the focus and hitting the spherical
> > mirror must return to to the focus, therefore fluxin = fluxout and no
> > heating.
>
> Hi Phil,
>
> But if Fh is radiating at some power (uniformly in all directions) because
> it is at some temperature, and Fc is at the same temperature and radiating
> at the same power, and all the power radiated from Fc ends up at Fh, but not
> all the power radiated from Fh ends up at Fc, it would seem there is a net
> power flow from Fc to Fh. And that is the paradox. It would require that
> Fh gets hotter (it is absorbing all the radiant energy from Fc but Fc is not
> absorbing all the radiant energy from Fh).
>
> Because they are elipsoids with different minor diameters, at one point I
> was thinking about the flux densities at different solid angles on the
> surfaces of Fc and Fh. For the solid angles of Fh that face S, yes
> fluxin==fluxout. For solid angles of Fh facing H (large ellipsoid) or
> C(small ellipsoid), it isn't clear that fluxin==fluxout at each angle. The
> angles of the ellipsoids mean that the 'fluxin' from the opposite radiant
> body changes with angular position to the receiver. So I suspect that a
> complete integration through the entire sphere of solid angles surrounding
> Fh and Fc would come out to a total energy flow that equals radiant power,
> but I haven't the math to prove it.
>
> But just looking at it from the point of all energy radiated from Fc goes to
> Fh and not all the energy radiating from Fh goes to Fc implies some sort of
> net energy transfer even though we start out with both bodies at the same
> temperature. Hence Bill's paradox.
>
> daestrom

In that case you have to use 'view factors' that account for the
imbalance, texts such as Holman discuss this an example follows:

"Radiation View Factors

The above equations for blackbodies and graybodies assumed that the
small body could see only the large enclosing body and nothing else.
Hence, all radiation leaving the small body would reach the large
body. For the case where two objects can see more than just each
other, then one must introduce a view factor F and the heat transfer
calculations become significantly more involved.
The view factor F12 is used to parameterize the fraction of thermal
power leaving object 1 and reaching object 2. Likewise, the fraction
of thermal power leaving object 2 and reaching object 1 is given by
F21
The case of two blackbodies in thermal equilibrium can be used to
derive the reciprocity relationship for view factors,
thus, once one knows F12, F21 can be calculated immediately.
Radiation view factors can be analytically derived for simple
geometries and are tabulated in several references on heat transfer
( . Holman, 1986). They range from zero ( . two small bodies
spaced very far apart) to 1 ( . one body is enclosed by the other)."

For the equations referred to above see:
/formulae/heat_transfer/radiation/