Tom Peel said:
I suspect that the effect can only work if sufficient heat is carried
off the sheet into the room through convection, while at the same time
blocking light reflected off the room surfaces from being radiated back
through the window.
Light and heat, ie "long-wavelength light." I'm trying to think of
a situation in which some sort of black sheet would help. Somebody
(Clear Dome Solar?) sells a "solar hot air collector" that fits on
the inside of a window, which seems like a mistake. So maybe we have
a shallow room with a white south wall behind a large south window.
We might blow room air with a fan against the north side of the black
sheet to keep it closer to the room temp, altho that wasn't part of
the original situation. We might use a black sheet with a selective
south surface that radiates less heat to the window pane
Let's put some numbers into the equations. Let's suppose that 1kW is
coming in through the window,and that the room temperature is in
equilibrium.
Maybe the US R2 window is 20 ft^2, and the room air is 70 F.
Without a sheet, suppose that 90% will be absorbed by the room surfaces
and that 10% goes back out the window.
This might mean "10% of the light," since glass does not transmit wavelengths
longer than 3 microns, ie heat, altho it absorbs and reradiates heat.
That means that 100W is going out the window, the remaining 900W is
being lost from the room through other heat sinks...
Here's a formula from page 210 of the 1991 2nd edition of Duffie and
Beckman's Solar Engineering of Thermal Processes:
ALPHAeff = ALPHAi/(ALPHAi+(1-ALPHAi)Aa/Ai), where
ALPHAeff is the effective absorptance of an cavity opening,
ALPHAi is the absorptance of the inner surface of the cavity,
Aa is the area of the aperture of the cavity, and
Ai is the area of the inner surface.
A very large window in front of a shallow room has an Aa/Ai of about 1,
so ALPHAeff is close to Ai. A small window in front of a deep room
(like a crab trap) has Aa/Ai close to zero, so ALPHAeff is close to 1.
A 10' cubical room with a 1 ft^2 solar window (about twice what's needed
for good indoor illumination in full sun) has Aa/Ai = 1/600. If the room
is covered inside with bright white paint with absorptance ALPHAi = 0.2,
the window has an effective absorptance of 0.2/(0.2+(1-0.2)/600) = 0.9934,
so less than 0.7% of the incoming sun is reflected back out of the window.
Now put the sheet in, and suppose it has NO effect. The sheet absorbs
all 1000W, PLUS- if the room stays the same temperature - it gets the
100W that the room walls are radiating back out the window!
The room was radiating (reflecting) light, not heat, no? So we might say
the black sheet gets a total of 1000 W on its south surface, and the room
no longer reflects 100 W of light back out through the window.
Now the sheet is heating up nicely, and has to get back in equilibrium
at some temperature by emitting energy. Some of this will go off as
convective heat, the rest as radiation. Let's call the convective heat
x, and the radiation y. The radiation y goes off both sides of the
sheet, half goes out the window, the rest goes back in the room, so we have:
Energy entering the sheet = 1100W = x + y = energy lost from sheet
How did we get up to 1100 W? In English units, a room at temp Tr (F) might
lose s((Tr+460)^4-(Ts+460)^4) Btu/h-ft^2 to a Ts (F) surface (eg a sheet or
a window pane) by radiation, where s = 0.1714x10^-8. If the surface is
warmer than the room, the room gains heat from the surface. At mean temp Tm
(F), the room and the surface would have a linearized radiation conductance
of 4s(Tm+460)^3 Btu/h-F-ft^2.
We can solve the equation and conclude that for the sheet to have NO
effect on the room temperature, y= 400W and x, the convection, is 700W.
In English units, if the sheet temp is Ts (F), each side might lose about
(Ts-Ta) Btu/h-ft^2 to slow-moving air at Ta (F) by convection. We might
have a thermal chimney between the sheet and the window, with Av ft^2
slots at the top and bottom, H feet apart, and cfm = 16.6Avsqrt(HdT).
Obviously, the 10% assumption we make is crucial, but there is clearly
a breakeven depending on the size of the windows in relation to the room,
and the effectiveness of convective vs. radiative cooling of the sheet..
Got numbers?
Nick