Subject: Metric solar closet arithmetic

From: [email-masked] (Nick Pine)
Date: 18 Oct 1995 07:30:05 -0400

Consider a solar closet standing alone outside in the winter--an insulated
box filled with sealed containers of water, with a solar air heater attached
to one insulated side, with some air dampers between the air heater and the
closet itself, so the low-thermal mass air heater can get icy cold at night.
(One might more usefully build this into a house, with the glazing behind a
low-thermal mass sunspace, and some air dampers to let some warm sunspace air
into the house on an average winter day.)

After a string of average winter days, with some sun, the steady state
temperature inside this standalone closet depends on the glazed area, the
intensity of the sun, the size and shape of the closet, the temperature of
the air around it, and the amount of insulation around the closet. It does
not depend much on how much mass is inside the closet...

For example, suppose we have a solar closet that is a 1 meter cube, with a
metric R-value of 5 m^2-K/watt on all 6 sides, and an additional layer of
glazing on one side, with a glazing R-value of 0.2, with an airspace behind
the glazing. Suppose the sun shines for 6 hours a day (in the winter) and on
the average, the amount of sun that shines on the glazed front of the closet
is 3 kWh/m^2, and the average air temperature around the closet is 0 C.

Suppose that when the sun shines, the air inside the glass and the air and
the water inside the closet all have the same temperature, Tw (C) (this
assumes that the water containers have a much larger surface area than the
glazed area, or that a fan moves the air quickly over the container surface),
then we have an energy balance:

Energy in = 1 m^2 x 3 kWh/m^2/day = 3 kWh/day, 

       (assuming the glass transmits 100% of the sun), and 

Energy out = 6 hours x (Tw - 0) 1 m^2/R0.2    (front, daytime)
          + 18 hours x (Tw - 0) 1 m^2/R5      (front, night time)
	  + 24 hours x (Tw - 0) 5 m^2/R5      (other sides.)

So, if Energy in = Energy out,

 3,000 = (6 x 5/1 + 18 x 1/5 + 24 x 5/5) Tw = (30 + 3.6 + 24) Tw

==> Tw = 3,000/57.6 = 52.1 C or 125.8 F.

If the solar closet is mostly inside the house, with only the glazed side  
exposed to the cold air at night and the warm sunspace air during the day, it
will mostly (5/6 of it) be exposed to 20 C air, not 0 C air, so the temperature
of the water and air inside will be about Tw = 20 + 3,000/57.6 = 72.1 C. If
we add a white reflector on the ground, eg snow, that increases the solar
input by 50%, so we have Tw = 20 + 3,000 x 1.5/57.6 = 98.1 C. Almost boiling.

Of course we are just using R-value arithmetic. Radiation losses may limit the
maximum closet temperature to something closer to 55 C (130 F.) And if we take
some heat out of the closet to heat the house every night, or to heat water,
using a fin-tube convective loop, that lowers the closet temperature, too.

Nick