Re: Solar Heating, Empirical Measu(r)ements & Experience

Nick Pine wrote:
> William R Stewart  <wstewart@patriot.net> wrote:
> >I would recommend talking to a number of different energy consultants.
> Me too. I usually start by asking "Are you familiar with Ohm's law?"
> That filters out more than half of the "energy consultants" I've met,
> but asking this question is sometimes embarrassing all around...

Because that is an electrical term, as opposed to a thermodynamic term.
Appropriate if you are talking to a PV consultant, but confusing otherwise.
You are simply trying to make other people translate your way of thinking as
an EE.  Why not speak in thermodynamic terms, like the rest of the industry?

> Here's another litmus test for an energy expert: what will the average water
> temperature be inside a 4' cube full of water surrounded by 5 R20 foam walls,
> sitting outside in December, when the air has an average 24 hour temperature
> of 36 F and the sun puts 1,000 Btu/ft^2/day of heat into the R1 glazed side
> of the box? And how will that change over time if we shade the sunny wall?
> Anyone care to answer that question? I'll offer a $10 reward to the first
> person who answers it correctly.

You don't give enough data in order to answer your question, and your question
is not clearly stated.  However, for the sake of fun, I'll make some assumptions.


1. The water starts out at 36oF.

2. Thermal conduction is the only heat transfer mechanism, aside from the energy
provided by the sun.

3. There are no losses incurred by the R1 glazing on the 1000 Btu/ft^2/day.

4. There are no data points for how much energy is provided on an hour-by-hour
basis, so I will assume an even amount of sun over a 10 hour period.


Weight of water:
4x4x4 = 64ft^3 of water

64 x 62.4 lb/ft^3 =  3993.6 lbs of water

Let's just call it 4000 lbs.


Thermal resistance of cube:
4x4 = 16 ft^2/side  value of R1 area

16x5 = 80 ft^2 of R20 surface

R1 total = 1/16 = 0.0625

R20 total  = 1/4 = 0.25

R total = 0.0625 x 0.25/(0.0625 + 0.25) = 0.05

U = 1/0.05 =20 Btu/hr-oF


Thermal energy input

1000 Btus/day-ft^2 * 16 ft^2 / 10 hrs/day input
 = 1600 Btus/hr while the sun is shining.


The water in the cube would rise to a steady state temperature of
36 +80 = 116 oF if the energy input were continuous.
When shaded, the water in the cube would fall over time to 36oF.

A complete answer with specific times would require calculus, if specific
times was what you were looking for.  Approximations could be made, but
the answer would not be 'correct'.  Let's just approximate the time it takes
the water to rise 1 oF without considering the negligible loss through the
cube walls.

4000 lbs of water would require 2.5 hours of 1600 btu/hr energy input, again
without considering the loss through the cube walls.

Now, if you;
  -moved the water inside to the interior sunspace of a 68oF house,

  -made the glazing 7'x10 (4 in the entire house on the south side)

  -used a flatter 4"x4'x10' water wall (4 in the entire house), 4" away
   from the glazing inside the house,

  -used an R4 glazing when the sun was shining,
   and an extra R8 window cover when the sun wasn't shining,

then you would have a direct, passive solar design that retained
the sun's heat energy in the building interior and released it
slowly into the interior at night

Care to try the math?

assume another 60 ft^2 of window at R4 during the day, adding
   another R8 at the other 14 hours.
Assume R24 walls and R38 ceiling.
Assume a two story, 24'x 48' floorspace house with 8' ceilings.
Assume an outside temperature of 36 oF
Assume 0.25 air changes per hour (optional).
Assume 100 Btu/hr/ft^2 insolation over 10 hours every day.
Make an assumption about the thermal storage of the rest of the interior

> >I would have to say that distributing the energy evenly to where it is
> >desired is yet the most important half.  Many houses have been built
> >where some rooms overheated while others grew cold.  Distributing the
> >heat to zones at specific times during the day is an interesting challenge.

> Sure, that's important, but let's collect the heat to begin with, Will, and
> then worry about that problem. Or wear a sweater in one room and take it off
> in another.

I try to look at the whole picture.  A solar closet might gather the energy,
but distributing it is another matter altogether.

>That's HVAC. Air pushers, ducts and fans and blowers and zone
> controls, to move warm air out of our sunspace or solar closet around the
> house.

You now no longer have a passive system, but now require temperature
regulation, ducting, electrical loads, etc.

Some people prefer the passive method, and it is not your perogative
to tell them that they are wrong in preferring that.

> Not too hard in a new house, and harder in a retrofit, unless we can
> mix the warm air into a big room or dump it into a return grate somewhere,
> somehow. That's a lot better understood than solar heating. Straightforward
> everyday cranking, for someone who knows how to do it.

I can't wait until you have to balance your first air-handling system.  Keep
at it!


Will Stewart

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