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

From: [email-masked] (Nick Pine)
Date: 15 May 1996 13:01:12 -0400

William R Stewart  <[email-masked]> 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...

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.

>Nick does have a lot of theories, though he considers many of his untried
>pet theories to be far superior to anything anyone else has ever conceived.

It's hard to know how to dissect that sentence :-) It looks like an insult.
And untrue. Most of this stuff was conceived a long, long, long time ago.
New weather data, new arrangements of systems and new materials and controls
make it ever more practical and cost-effective and efficient now, but that
isn't the main problem...

Yes, I'm smarter than most people ("References please"--I belong to Mensa)
and I know a lot more about physics, and I've thought a lot more and in more
depth about solar heating and have more engineering degrees than many house
designers, backyard crackpots and armchair a.e.r posters who like to talk
about alt.energy but haven't done any at all, or PV hobbyists who have
learned how to screw a system together but don't know Ohm's law either, and
I do think _some_ of "my solar theories" are the cat's pajamas, especially
the ones that seem to make sense with my simple understanding of physics,
and the ones that really do work... the ones we've been trying out now in a
2' x 4' x 8' tall structure with an electronic data logger and modem inside,
in the real Pennsylvania winter outdoors on the roof of the physics building
at Ursinus college in Collegeville, PA since November 4... and physics prof
Paul Bashus and I have managed to hornswoggle our solar theories on paper
past the technical review committees of two international conferences in
the last year, as well as a number of PhDs, who have accepted our solar
closet paper for presentation at the Unesco-sponsored World Renewable Energy
Congress in Denver from June 16-21 ([email-masked] registration
fee increases from $340-740 after today, May 15), whose steering committee
alone has people from 130 countries, and I've talked and run sessions on solar
heating and cooling techniques at several US conferences, but. .  .
serious scientists and engineers consider solar closets and sunspaces
"trivial physics" and "unoriginal research."

Many scientists are out collecting funding to improve the solar collection
efficiency of some system from 16 to 17% by adding another $10/ft^2 to the
cost of something that already costs $30/ft^2. This kind of solar house
heating technology has been around since Edward Morse invented the "Trombe
Wall" in 1881, at least, and it's understood by many people who know what
they are talking about, and a lot more who don't. Solar house heating just
hasn't been done well yet, outside of the Southwest, on an significant scale,
partly because the field is peopled by ignorant hucksters, partly because
we are still waiting for the government to help make that happen... But
this is lumberyard and hardware store science, not stuff for theses and
Nobel prizes and NREL grants. It's waiting for bold entrepreneurs. And
more expensive energy prices, or people who care more about the environment,
or government regulations that make not caring more expensive...

Many heating and cooling engineers are working on heating and cooling systems
for skyscrapers, or heat pumps, where the money is today.

I've only been doing this full-time for 2 years now, after 25 years of
electrical engineering, and I'm still learning. Some things seem like good
ideas on paper but really do have to be tried out, and improved somehow for
some applications, or proved to be impractical for others, like bubblewalls.

Some things don't need to be tested much, like low-thermal-mass sunspaces,
and some things do, like bubblewalls. It's important to know the difference.

Bubblewalls are the subject of an old Swedish patent (I wish I had a copy.)
The original application was storefront windows. They have been tried in
this country over the years by people like John Groh, a New York grower (?)
and Dr. Merrill Jenson at U Arizona and Dr. Otho Wells at U New Hampshire,
who concluded in 1977 that bubblewalls were not practical night insulation
for commercial plastic film greenhouses simply because those greenhouses
need to be put together so roughly and quickly in the field by unskilled
people (so as not to raise their basic cost of 50 cents per square foot :-)
that leaks in such a system could not be avoided. Everyday leaks in plumbing,
plastic film connections, etc. These are real problems, and I don't mean to
minimize them, but they are also application-specific, and not hard to solve.

Dr. Wells says that windows in solar houses may be a different situation,
since they can be made more carefully... He also says he only got a US
R-value of 1.5 or so out of his 3"-6" thick poly film pillows, so it seems
we need to improve the bubbles themselves somehow, make them smaller and
longer lasting, with thicker walls perhaps, or go work on something else.
Dr. Aristid V. Gross of Temple U and the blind Belgian physicist Joseph
Plateau (1801-1883) both made bubbles that lasted over a year...

Dr. Wells was using an open system, more dustprone than a closed one, making
bubbles at the top of a 25' x 196' double poly film greenhouse to fill the
double poly film walls all round. He put slits in the plastic film, some
with sewn-in zippers (he said sewing zippers into poly film isn't easy),
to let some air out at the endwalls. He found he had to make more bubbles
every 2-4 hours, to keep the walls full of bubbles. Dr. Jenson experimented
with a photosensor to do this automatically, and also tried adding dyes to
the bubble solution, which were diluted so much in bubble from that they
had little effect. Another way to keep the glazing cavities full is to keep
a layer of bubbles moving slowly from bottom to top in a continuous fashion.
Another is to use temperature sensors to measure the R-value. Dr. Wells
said the small bubbles tended to settle out quickly and become liquid
again at the bottom...

Another problem: Dr. Wells mentions that the bubble distribution was not
uniform--big bubbles and open spots, thermal shunts with no bubbles, up to
1-2' in diameter would often develop. This might happen less if bubbles
move continuously up from the bottom instead of intermittently down from
the top of the cavity. Dr. Wells used a basic solution of 3% type E dust
reduction foam (a better choice might have been firefighting foam) from Mine
Safety Supply in Pittsburgh (now Mine Safety Appliances at (412) 776-7700?)
with 1% propylene glycol as antifreeze, because when the bubbles freeze
inside the outside layer of poly film, they break, and the R-value goes
down and their friction against passing bubbles goes up, and he didn't want
water frozen in pipes either. Can we make frozen bubbles that don't break?

Yes. That's been done, often, in public at the San Francisco Exploratorium
by Dr. Ilan Chabay, called "one of the world's foremost authorities on
frozen bubbles," in John Cassidy's _Unbelievable Bubble Book_ for children
(Klutz Press, 1987.) Has Dr. Wells ever talked with Dr. Chabray or Dr. Grosse?
I don't think any physicist would call bubbles trivial physics. Child's play
maybe, but not trivial :-) We need more child-like physicists.

There is at lot of work to do in this alt.energy field, and a lot of it,
like preventing leaks or finding the right soapy solution is not that
technically difficult. There isn't a lot of funding, but there's a lot of
work to do, a lot of opportunity for serious backyard builders and high
school science fair students to do interesting things that may later become
good, cost-effective, reliable, simple socially-useful techniques and
products, given some common sense and education and patience and a lack
of greed and shortsightedness.

Some things need to be tested and developed a lot, others are no-brainers,
if we start thinking from scratch. Not everyone wants to live in an
experimental bubblewall house, but lots of people like sunspaces. There's
a huge gap between commercial plastic film greenhouses costing for 50 cents
per square foot and residential sunspaces costing $50-150 per square foot.
Who will fill it?  NREL?  No...

There's another gap between serious scientists who say all this is trivial
well-understood physics, while they collect their salaries working on more
and more exotic research, and alt.energy.renewable experts who say it won't
work, it's been tried before, Trombe walls are the cat's pajamas, etc, etc.
Bullshit!

>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. 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. 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.

Nick

It's a snap to save energy in this country. As soon as more people become
involved in the basic math of heat transfer and get a gut-level, as well as
intellectual, grasp on how a house works, solution after solution will appear.

					  Tom Smith

---

From: [email-masked] (David Hoadley)
Date: 16 May 1996 01:53:54 GMT

>person who answers it correctly.

This brings me to another point (thus the change of topic title). When I
first started reading these groups I was rather taken aback by the high
R-values people mention. Here in Australia, they are generally in the
range R-1.5 to R-3 or 4 for (say) fibreglass insulation. The units are,
naturally, degC.m^2/W. I was surprised to find that American ones are
expressed in British Thermal Units (which I had thought went out with the 
ark), feet, and fahrenheit, but at least it explained the difference. So I
tried to work out a conversion factor. Can someone confirm it for me? i.e.

  Is it true that 1 watt = 3.412 BTU/hr?  If so, have I got it right that
  R-1 (International) degC.m^2/W = R-5.678 (US) degF.hr.ft^2/BTU?

Thanks,
David.

--
----------------------------------------------------------------------------- 
David Hoadley                        Internet: [email-masked]
Electrical Engineering, RMIT
Melbourne, Australia                Ph: +61 3 9660-4847, Fax: +61 3 9660-2007

---

From: William R Stewart <[email-masked]>
Date: Thu, 16 May 1996 00:29:41 -0400

> 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?

> 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.

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
  items/material.

> 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
> 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

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

Cheers,

Will Stewart

---

From: [email-masked] (Nick Pine)
Date: 16 May 1996 04:52:30 -0400

Still no numerical answers, 16 hours after this posting. Perhaps $10 is
insufficient bait to catch an energy consultant, or they are out of season? 

Or perhaps it should be rephrased in international units:
 
Here's another litmus test for an energy expert: what will the average water
temperature be inside a 1 m cube full of water surrounded by 5 R5 foam walls,
sitting outside in December, when the air has an average 24 hour temperature
of 2 C and the sun puts 3 kWh/m^2 per day of heat into the R0.175 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.

>range R-1.5 to R-3 or 4 for (say) fibreglass insulation.

R30 house walls are pretty good walls in the US. Perhaps a good new
Australian or French or British house has R5 walls, made with
straw bales and mortar, 40 cm thick? A 1 bale wall?

>The units are, naturally, degC.m^2/W.

Of course Mother Nature thinks in meters and C, not feet and F :-) The
natural thing about a Btu (the amount of energy in a kitchen match) is that
it takes eggsactly 1 to raise 1 pound of water or 55 ft^3 of air 1 degree F.
It takes about 100 Btus to make a cup of coffee. Chickens make about
5 Btu/hr/lb of sensible heat. A typical 5.41 lb bunny makes 40. 1 cfm of
air flowing with a temperature difference of 1 F moves about 1 Btu/hour,
a single pane window is about R1, and 2 layers are R2. A pound of water
takes 144 (1 gross) Btus to freeze at 32 F, which is close to the average
air temperature in December where I live, and it takes about 1000 Btu to
boil away a pound of water at 212 F. A south wall in December receives
about 1,000 Btu/ft^2/day of sun where I live, 300 Btu/hour/ft^2 peak. The
thermal conductance of an airfilm with air moving at V mph is about
2 + V/2 Btu/hr-ft^2-F for rough surfaces. A black body at T degrees F
radiates 0.174x 10^-8(460+T)^4 Btu/hr-ft^2. US hardware stores sell rolls
of 6" fiberglass insulation stamped R19 in big letters (R10 with 2%
moisture, R0 in a good wind) for 25 cents/ft^2, and R10 foamboard 2"
thick costs about 50 cents/ft^2.   

What is natural in metric? Water freezes at 0 C and boils at 100 C, and
it takes about 1 kWh to heat 1 m^3 of water or 3,000 m^3 of air 1 C. A cup
of coffee takes about 30 Watt-hours. People make about 100 Watts, rabbits
5 W/kg. 1 m^3/sec of air flowing with a temperature difference of 1 C
moves about 1 kW of heat. It's a little harder to remember that 1 layer
of glass with an airspace has metric R0.175, and 2 layers have R0.35.
It takes about 100 kWh to freeze a cubic meter of water and 663 kWh to
evaporate one. The ratio of evaporative to convective power loss at a
wet surface with temperature Tp in Ta air is about (Tp-Ta)/(Pwp-Pa)/2,
independent of windspeed, where Pwp and Pa are vapor pressures in mm Hg.
A south facing wall in December receives about 3 kWh/m^2/day of sun where
I live, with a peak of 1 kW/m^2, and the thermal conductance of an airfilm 
with air moving at V m/s is about 10 + 2V W/m^2-C for rough surfaces, eg 
100 W/m^2 in a 5 m/s wind for a 25 C surface on a 20 C night. How much
more if the surface is wet, and the air has 50% RH? How much more if the
sky is clear? Do French hardware stores sell R2 foamboard 5 cm thick for
about about 30 FF/m^2? Why does the Lord permit such Suffering on Earth,
if He is Omnipotent?

>I was surprised to find that American ones are expressed in British Thermal
>Units (which I had thought went out with the ark)...

Now that Britain and Australia use Watts and meters, we are thinking of
calling these American thermal units, or perhaps Bunny thermal units.

>So I tried to work out a conversion factor. Can someone confirm it for me? 
>

I think so.

Or would those be degrees K, in that particular alphabetical procession? 

Nick

   When we play tennis or walk downstairs we are actually solving whole
   pages of differential equations, quickly, easily and without thinking
   about it, using the analogue computer which we keep in our minds. What
   we find difficult about mathematics is the formal, symbolic presentation
   of the subject by pedagogues with a taste for dogma, sadism and
   incomprehensible squiggles.

          From _Structures: Why Things Don't Fall Down_, by J. E. Gordon

---

From: [email-masked] (Nick Pine)
Date: 16 May 1996 10:34:14 -0400

William R Stewart  <[email-masked]> wrote:
>> ...I usually start by asking "Are you familiar with Ohm's law?"

Many people learn Ohm's law in high school, Will...
I think an energy expert should be familiar with Ohm's law.
To far too many people these days, "energy" = "electrical energy."

>Appropriate if you are talking to a PV consultant, but confusing otherwise.  

Yes, Ohm's law is confusing to certain "energy consultants" :-) 

>an EE.  Why not speak in thermodynamic terms, like the rest of the industry?

Perhaps we should ask 'em about enthalpy? :-)

>is not clearly stated.

I think there is too much data above, and the question is reasonably clear. 

>However, for the sake of fun, I'll make some assumptions.

Good :-) This is the first numerical answer I've seen...

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

Good. Although R-values include convection and radiation.

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

The glazing is perfectly transparent? Fine.

>basis, so I will assume an even amount of sun over a 10 hour period.

OK. (I don't think that matters here, but...)

>Let's just call it 4000 lbs.

OK. 

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

Excellent! 

> = 1600 Btus/hr while the sun is shining.

Um, OK...

Perhaps, but recall the problem:

>>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?

>When shaded, the water in the cube would fall over time to 36oF.

Well, yes...

>times was what you were looking for.

Yes, that was what I was looking for. More than the obvious... Finding the
form of the answer requires calculus, but Newton did that once, a long long
time ago. Finding the answer just requires plugging some numbers into a
simple formula, if you know the formula.

>Approximations could be made, but the answer would not be 'correct'. 

Approximations using arithmetic would be fine... That's easy to do.

>Let's just approximate the time it takes the water to rise 1 oF without
>considering the negligible loss through the cube walls.

OK (?)

>without considering the loss through the cube walls.

True, but why are you talking about heating rather than cooling?  

		    *     *     *

Sorry. No $10. Would you like to try again?

                    *     *     *

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

    ie 280 ft^2 of south glazing?

>   from the glazing inside the house, 

    ie about 400 pounds of water? 4" away from the glazing? Transparent
    water? Hmmm :-) I guess you won't be able to see very clearly out of
    those windows, but they will let in some light...
 
>  -used an R4 glazing when the sun was shining, 

With how much solar transmittance, at what cost, and what happens when the
argon leaks out? Some of these $40/ft^2 high-R windows have low solar
transmittance, on the order of 50%. Hmmm, 280 ft^2 x $40/ft^2 = $11,200.

>   and an extra R8 window cover when the sun wasn't shining, 

How would you do that? Movable insulation, at $10/ft^2, installed? Will you
move it twice a day religiously? Will it leak any air around the edges?

>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

Right. Another "direct loss" house :-)

>Care to try the math?

Sure. I like this arithmetic. Let me get a cup of coffee... Now I'm back,
200 Btu later. I've had this espresso machine for about 3 months now. I
guess if some people like me find it fun to wiggle the valves on an espresso
machine with all the steam and noise and watch the milk temp rise with a
thermometer twice a day, others might find it fun to move window insulation
all over a house twice a day, for three months... Or, maybe you only move
the window insulation on a cloudy winter day--but no, you want to do it at
night too... or you might leave some windows covered all winter, like Pat
Hennin or Malcolm Wells, in semi-hibernation, huddled in cold dark rooms. 

>   another R8 at the other 14 hours.

OK. That's a long winter day...

>Assume R24 walls and R38 ceiling.  

OK. Let's assume that includes the 2x6's, etc, that act as thermal shunts,
ie the equivalent thermal resistance of the entire wall is R24. This is not
just a wall with layers of insulation with R-values that add up to R24
(which might have a much lower R-value), and the insulation is installed
properly with no gaps, etc. Are you sure about all that?

>Assume a two story, 24'x 48' floorspace house with 8' ceilings.

OK. I guess you mean the footprint on the ground is 24 x 48?
So the house volume is 24 x 48 x 16 = 18432 ft^3.

>Assume an outside temperature of 36 oF

OK. (The average will be warmer during daylight, which is good.)

>Assume 0.25 air changes per hour (optional).

OK. That's 0.25 x 18432 ft^3/hr = 4608 ft^3/hr or 77 cfm :-) I'll assume
(I almost said "guess") this is leakage thru walls, etc, without an air-air
heat exchanger... Will the finished house have a blower door testing spec?

May we also assume you use a frugal 500 kWh/month of electricity, ie an
average of about 700 Watts? (Steve Baer only uses 80 kWh/mo :-) This is
equivalent to 700 x 3.41 = 2387 Btu/hr. Let's add the heat from two
300 Btu/hr people inside the house for 16 hours a day, and one 150 Btu/hr
dog and a 50 Btu/hour cat, full-time. Total internal energy generation:
2387(24)+600(16)+200(24) = 72K/day.

>Assume 100 Btu/hr/ft^2 insolation over 10 hours every day.

OK.

>Make an assumption about the thermal storage of the rest of the interior

How about 1 lb of water/ft^2 of walls and ceiling? (1/2" drywall has the 
equivalent of 1/2 lb of water/ft^2, so you may need some concrete furniture..)
I'll make some other assumptions, too.

I guess we'd want to know the thermal resistance of this house for starters.
So let's add up the thermal conductances U = Sum(Ai/Ri) and find
the reciprocal 1/U... 

We have 340 ft^2 of windows, R4 during a 10 hour day (Uday = 340/4 = 85)
and R8 during a 14 hour night (Unight = 340/8 = 43), with an average daily
Uwindow = (10xUday+14xUnight)/24 = 60.

Uwalls = (2(24+48)x16-340)/R24 = 82, and Uceiling = 24x48/R38 = 30.

And Uinf = 77 Btu/hr-F.

So U = 60+30+77 = 167 Btu/hr-F and R = 0.006, over 24 hours.

At night the U value would be lower, after you devotedly travel around
the house and painstakingly put up your night insulation on every window,
making sure the edges are ever so carefully sealed, akin to some little
twice-daily religious experience... Unight = 43+30+77 = 150 Btu/hr-F.

What will the steady state energy flow be for this house after a long
string of average December days, with an average amount of sun?

The energy Ein that flows into this house in a day might be 

  280 ft^2 windows x 50% transmission x 1000 Btu/ft^2/day = 140K 
                            +  internal energy generation =  72K
	                                        total Ein = 212K.

And the energy Eout that flows out of the house might be 

  24 hours x (68-36) x 167 = 128 K Btu/day.

So, after a long string of perfectly average December days, Ein-Eout
= 84K Btu/day, if you keep the house at an average of 68 F during the
day... This might make a fine New Mexico house.

If you let the house temperature float over this time, you might have
an average house temp T such that 24 (T-36) 167 = 212K, or T = 89 F.
Colder at night, warmer during the day. Very toasty.

Now, thermal mass: the house has about 3500 ft^2 of walls and ceilings with
a thermal mass equivalent to 3,500 pounds of water and another 400 pounds
of actual water. Let's call this 4,000 Btu/hr-F. So if you make the temp of
the house, say, 80 F at dusk, at the end of an average day, with some sun,
after a long string of December days, each with 1000 Btu/ft^2/day of sun,
during the night the house will lose approximately 14 hours (80-36) 150 =
100K Btu, which might come from the 4,000 Btu/F of thermal mass cooling plus
14/24 hours x 72K = 42K of internal energy generation, ie the temperature of
the house might drop to 80 - (100K-42K)/4K = 65 F at dawn. 

So yes, this 56%-electrically-heated house with all of the movable window
insulation looks marginally comfortable after a long string of average winter 
December days, each with an average amount of sun. But what do you do if the
next day is cloudy, or the day after that, or during a cloudy week in January,
when the temperature outside is -10 F? You will have to use backup heat...
Or, gasp, wear a sweater. With no sun and no backup heat, and with all the
windows boarded up on the inside, 24 hours a day, just using the internal
electrical energy consumption (and the people and livestock) after a day or
so, Ein = 72K Btu/day and Eout = 24(T-36) 150 ==> T = 56 F inside. Not bad.
Quite warm by Inuit standards. More dogs and cats would help.

How much backup heat will you need in an average year? One way to answer 
that question is to go find some TMY2 data somewhere on the web (references,
please :-) for a Typical Meteorological Year for this house location, and do
a very simple simulation using that data and formulas like the ones above.
Another is to do the same thing in more of a worst-case manner, ie more
reliably using hourly data for the last 30 years from an NREL/NOAA CD-ROM.
Then one might change the design until it is a "solar house," in the sense
that it would not have needed any other form of heat over the last 30 years,
according to the simulation, at which point, many people would opt to
eliminate the backup heating system, saving some space and money...

Here's an alternative design to try out in that simulation:

Eliminate half the south windows and all the movable window insulation, to
eliminate the daily labor and lower the price by $10K or so and lower the
overall U value to about 140 Btu/hr-F, and make a little more privacy if that
matters, and add on a 2-story, $5000, 16' wide x 8' deep lean-to sunspace
with an 10' tall x 13' long x 5' deep solar closet containing 36 55 gallon
drums, inside the sunspace, south of the house and adjoining the house wall.
Each square foot of sunspace would supply about 1000 Btu - 6hr (80-36)/R1
= 700 Btu/day of heat for the house, ie about 180K of heat, vs the 120K-72K
= 48K/day the improved house would require on an average day, via a $200
motorized damper at the top of the sunspace that uses 2 Watts of electrical
energy when moving (rarely) and 0 watts (mostly) in a fixed position to let
warm air flow into the house as needed, controlled by two thermostats. 

The solar closet would contain about 18,000 Btu/F of thermal mass, vs.
4,000 Btu/F, to keep the house at whatever temperature _you_ wanted,
independent of weather, overnight, or on a cloudy day. Assuming the water
starts out at 130 F after a long string of December days, with some sun,
the solar closet would store about 25K Btu of heat in each drum, enough
to keep the house at 68 F for about 25Kx36/48K = 19 days without any sun 
at 36 F, or 3.3 days without any sun at -10 F. We might improve that by
making the solar closet bigger. 

After 3 or 4 days without sun at -10 F, with all the windows boarded up on
the inside, and no backup heat, how can we estimate the temperature T inside
the house we started with, before the improvements? 

24(T-(-10))150 = 72K, so T = -10 + 72K/(24x150) = -10 + 20 = 10 F.

It might take longer than 3 or 4 days to get to that temp, since the pipes
and water walls would release about 400,000 Btu while freezing at 32 F, as
the house temperature plunged. On the other hand, the people might move out
sooner, depriving the house of their heat...

Were you planning to have hot water in this house? How were you planning
to do that? You might put a water heater on the second floor with some
Big Fins in the sunspace or 10-20' of fin-tube pipe near the ceiling of the
solar closet (less expensive), with a warm water thermosyphon loop through
the water heater, to supply a frugal hot water house load of 50K Btu/day...

About heat distribution...
    
>> ...wear a sweater in one room and take it off in another? 
>
>>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. 
>

One might compare this to an oil fired hot air furnace that uses 800 watts
of electrical power when running, while consuming 38 kW (1 gph) of oil power
and delivering 35 kW to the house. That's a fossil COP of 44. Some old and
inefficient and expensive active solar systems have renewable COPs of 50.
For instance, CSI's air/rockbed system in New Hampshire, built in 1982, uses
"2% fan power, 98% solar power." A better way to compare heating systems
might be their yearly heating bills, including the electrical power to run
the heating system. I think one can build a solar house that consumes no more
than 50 watts average, for the heating system, with no backup heating fuel,
and no backup heating system.

A 560 cfm fan ("References please"--Dayton fan 4C688, page 2964, Grainger
catalog 386, 1650 rpm, 10" diameter, max temp rating 149F, $60.75) moving air
>from  a solar closet into a house with 0.05" static pressure (as measured in
our experimental structure) can move about 5600 Btu/hr into this house with a
temp difference of 10 F, ie 134K Btu/day, while dissipating at most 36 Watts
of electrical power. 

>to tell them that they are wrong in preferring that.

Oh sure it is, and it seems kind to tell people, if their passive houses are
such miserable and expensive performers. And as Newton Ellison and others
say, this passive/active dichotomy isn't very useful. I'd say what matters
more these days is money, comfort, reliability, cost and maintenance.

Sailboats are passive devices using powerful natural forces and often smart
low-power controls. Their captains care a lot about cost and performance... 

Would you hang an outboard motor on your America's Cup boat?

Nick

---

From: William R Stewart <[email-masked]>
Date: Fri, 17 May 1996 00:29:01 -0400

> >is not clearly stated.\

I showed how it would change over time.  You owe me ten bucks, Nick :-)

> >times was what you were looking for.

Otherwise, simple arithmetic would give the wrong answer, as this is obviously
not a linear relationship.

> simple formula, if you know the formula.

Present the formula and we will see if it produces the correct answer, or even
 a very close answer.
 
> >without considering the loss through the cube walls.

So here is an answer with a specific time; what's your beef?
 
> True, but why are you talking about heating rather than cooling?

Because you asked for it!!!

> Sorry. No $10. Would you like to try again?

Welcher!

> argon leaks out?Some of these $40/ft^2 high-R windows have low solar

You must be thinking of R8 windows; Where do you get your figures?
I'm not going with low-E due to insolation losses.

> How would you do that? Movable insulation, at $10/ft^2, installed? 

Again, where do you get your prices?

>Will you move it twice a day religiously? 

Just like you close your windows when it gets to cool at night.

>Will it leak any air around the edges?

Probably, but certainly much less than an outdoor thin film sunspace.
 
> Right. Another "direct loss" house :-)

Er, do you have a house that does not lose heat when it is cold outside?

> OK. That's a long winter day...

You forgot to add R8, you only added R4
 
> properly with no gaps, etc. Are you sure about all that?

Another inch of polyisocyanurate, and various other siding and finishing
materials.
 
> OK. I guess you mean the footprint on the ground is 24 x 48?

Right.

> heat exchanger... Will the finished house have a blower door testing spec?

Yes, and the top limit will be 0.25 air changes; I expect alot less.
 
> average of about 700 Watts? 

Why does this matter?  I will be using a 2kW photovoltaic system.

> (Steve Baer only uses 80 kWh/mo :-) 

How much do you use, coffee and all?

> This is

OK

> We have 340 ft^2 of windows, 

4 sets of 10' x 7' windows = 2800 ft^2

>R4 during a 10 hour day (Uday = 340/4 = 85)
> and R8 during a 14 hour night 

R12 at night; R4 + R8

(Unight = 340/8 = 43), with an average daily
> So U = 60+30+77 = 167 Btu/hr-F and R = 0.006, over 24 hours.

These numbers have to be revisited.

>   280 ft^2 windows

That should be 2800 ft^2

 x 50% transmission

Try 75%

 x 1000 Btu/ft^2/day = 140K
>   24 hours x (68-36) x 167 = 128 K Btu/day.

Of course, these numbers must be revisited as well.
 
> of actual water.

10' x 4' x 4" is ~13 ft^3.  Water is 62.4 lbs/ft^3.  There are
four water walls.

The water will weigh close to 3250 lbs.

> Let's call this 4,000 Btu/hr-F. 

You need to revisit this number as well.

> So if you make the temp of

You have so many corrections to make before you hit this paragraph that I am
reluctant to mention the ones within.

You have not yet tried to establish the temperature of the water in the
water walls.  Your assumption seems to be 1 oF, though there is no evidence
of how you tried to calculate this.

> How much backup heat will you need in an average year? 

My wife wants a fireplace, so I will likely go with a masonry firestove,
often referred to as a Finnish or Russian fireplace, with outside combustion
air.

> to do that? 

Solar hot water, most likely active.

> A better way to compare heating systems
> the heating system. 

The model I am planning on building (2250 ft^2) required $168 worth of heat for an entire
winter.  Reference "Solar Today", Sept/Oct 1995, article on manufactured house.

> such miserable and expensive performers.

Try the numbers again, then we'll talk.

Cheers,

Will Stewart

---

From: [email-masked] (Peter Skelton)
Date: Fri, 17 May 1996 15:01:22 GMT

>>Because that is an electrical term, as opposed to a thermodynamic term.  

OK I'll buy the idea that an energy consultant should be able to do
cookbook thermal transfer, it makes sense. What else should he know? 

Candidates:

Predictive cost ecom\nomics
Cost benefit analysis

Funding and programs

Available materails, equipment and sources

gas ducting
fluid piping
barrier design and construction details
building codes

etc etc.

Clearly no single person can have everything at his fingertips. Most
professions admit specialization (accounting, engineering and medicine
for example). Professionals, even when qualified, tend to limit their
activities to their areas of competence. Our family doctor in Sarnia
would do minor surgery in his office but sent us to specialists for
respiratory problems like asthma. The fellow we have here does the
oposite. 

One of my customers is an architect. He has designed a circulating
pump that runs off the system energy in a heating loop and controls
itself. He's also designed some interesting heat exchangers. Could he
validly call himself an energy consultant? I think so. Could he solve
the test problem? Probably not.

Another acquaintance designs power grids for industry. He is
definitely an energy consultant - all he does is show industry how to
move energy around. He doesn't need to know thermodynamics.

There's an energy consulting firm active here that works out cost
justification for cogeneration. They know where to get money and how,
what sorts of industry can use the waste heat, land management. . . .
There might not be a thermodynamically competent person in the place,
they can afford to hire someone like Stone & Webster when they want,
but they are definitely consultants in energy.

To define what an energy consultant is we would have to list
everything one might do and then decide what size of subset a person
would have to master to qualify. It is a dounting task.

All that being said, I can imagine many situations where the simple
test posted might sort the wheat from the chaff.

Peter Skelton

---

From: [email-masked] (Alan Braggins)
Date: 17 May 1996 12:49:32 GMT

> >  R-1 (International) degC.m^2/W = R-5.678 (US) degF.hr.ft^2/BTU?

% /usr/bin/units
you have: kelvin m2 / watt
you want: rankine hour ft2 / btu
        * 1.752557e+00

you have: m2 / watt
you want: hour ft2 / btu
        * 3.154603e+00

Either I'm wrong about Rankine being to Fahrenheit as Kelvin is to
Celsius, I'm misunderstanding units output, or you have used 9/5
where you should have used 5/9.

> Or would those be degrees K, in that particular alphabetical procession? 

For temperature _differences_, Kelvin is the same as degrees Centigrade.

---

From: [email-masked] (Nick Pine)
Date: 17 May 1996 12:28:37 -0400

>cookbook thermal transfer, it makes sense. What else should he know? 

Interesting question...

>Candidates:
>
>Predictive cost economics
>Cost benefit analysis

Sure. Interest formulas, sinking funds, IRR, etc.

>Funding and programs

I guess so... Not my interest, but...

>Available materails, equipment and sources

Sure.

>fluid piping

I guess so. Know a few details, and leave the rest to plumbers, etc?
("Please make the ducts bigger, so we can use less powerful blowers.")

>barrier design and construction details

Yes...

>respiratory problems like asthma. The fellow we have here does the oposite. 

Pediatricians can do brain surgery in my state, if they want to some
afternoon. And EE PEs can design bridges. It's legal... 

>validly call himself an energy consultant? I think so.

So do I :-)

>move energy around. He doesn't need to know thermodynamics.

I have a feeling he might be familiar with Ohm's law...

>but they are definitely consultants in energy.

Yes, but... What we are talking about here is not "thermodynamics" in
my book. When somebody uses that term I think Carnot and Rankine and
Maxwell and Enthalpy and calculus with difficult integrals. For solar
house heating, I think Ohm and arithmetic. With a few old dusty formulae
>from  Newton. Newton's law of cooling, aka Ohm's law for heatflow. And
a single simple exponential formula, at best.

>would have to master to qualify. It is a dounting task.

Somebody offers a credential like that... as a "certified energy manager"?
 
I guess one would want to know some legal and bureaucratic stuff too,
like OSHA. The US patent bar exam has a lot of questions like 

   If you appeal a decision related to a notice of infringement in the
   District Court, for the third time within 7 months, and another party
   or parties appeal to the assistant patent commissioner or his designated
   assistant by the third Tuesday in an even numbered month, how many days
   do you have to file an appeal to that action, and with whom?

>test posted might sort the wheat from the chaff.

Me too.

Nick

Nicholson L. Pine                      System design and consulting
Pine Associates, Ltd.                                (610) 489-0545 
821 Collegeville Road                           Fax: (610) 489-7057
Collegeville, PA 19426                     Email: [email-masked]

Microprocessor hardware, memory, ASIC, and computer design. Telecommunication
system design. Computer simulation and modeling. High performance, low cost,
residential solar heating and cogeneration system design. BSEE, MSEE. Senior
Member, IEEE. Registered US Patent Agent. Fluent in French.

---

From: [email-masked] (Nick Pine)
Date: 17 May 1996 09:59:02 -0400

>>person who answers it correctly.
 
By the way, I have only seen 3 numerical answers to this simple question
so far. Two people got part 1 correct, and one has almost solved part 2 using
simple arithmetic, but there's still a chance for the right person to earn
$10 in 2 minutes with a couple of back of the envelope calculations... This
might also be a rare opportunity for someone else to get paid to learn.

Paul Erdos used to do this sort of thing with much harder math problems.
He would offer a $100 prize for the solution to some long-standing research
problem, and some grad student who might already be interested in the problem
would get more interested and work for 3 months, day and night, and solve the
problem, and pocket the money, and publish a paper, and feel very happy, and
the science of mathematics would be advanced :-) Erdos offered different
rewards for different problems, depending on how difficult they were, from
$10 problems that took a few weeks to $1,000 problems that might take 10 years
to solve, or might take a lifetime...

 
>>>>the sun puts 1,000 Btu/ft^2/day of heat into the glazed side of the box. 

>I showed how it would change over time.  You owe me ten bucks, Nick :-)

Nope. That answer is insufficient and superficial. I'm the judge.

>Present the formula and we will see if it produces the correct answer, or even

Nice try :-) How will "we" know that? The formula IS most of the answer...
Or an approximate arithmetic equivalent...

>So here is an answer with a specific time; what's your beef?

It has nothing to do with the problem.

 

>You must be thinking of R8 windows; Where do you get your figures?

I don't remember. Anderson, Pella? Popular Science? Perhaps that's R8.
What brand of windows will you be using and how much do they cost?

>I'm not going with low-E due to insolation losses.

Good. I wonder what happened to your UV concerns.

>Again, where do you get your prices?

The Shelter Institute. What will you pay? References, please.

>Just like you close your windows when it gets to cool at night.

I hardly ever touch my windows. 

>Probably, but certainly much less than an outdoor thin film sunspace.

I don't think so, if the sunspace is made from a continuous single piece of
plastic film, 16 x 20'. And sunspace leakiness only matters 6 hours a day.

>Er, do you have a house that does not lose heat when it is cold outside?

No. But this is a numerical thing, and the less the better at night, and
direct gain houses leak lots of heat to the outdoors at night and during
a week without sun thru their low thermal resistance glazing between the
living area and the outdoors. 

 
 
>You forgot to add R8, you only added R4

Oops. Would you describe this R8 window insulation, Will? Who makes it,
how much the material costs, how much installation costs, what the materials
are, how it is operated, and how the edge air leaks affect the R8 rating?

>Yes, and the top limit will be 0.25 air changes; I expect alot less.

Will the house be pressure tested when finished and guaranteed to have
less than 5 ACH at 50 Pascals, consistent with natural air infiltration of
< 0.25 ACH, using the usual rule of dividing the pressure test spec by 20?
The best Avis homes informally tested by Penn State, as I recall from a
go-around with Kurt Smith, Lyle Rawlings and Marc Rosenbaum, came out at
2.9 ACH at 50 Pa.

>Why does this matter?  I will be using a 2kW photovoltaic system.

Marc points out that many superinsulated houses are more electrically heated
than solar heated. No, that doesn't matter much. Money matters, of course.
Eg the yearly backup heating bill, and the non-recurring cost.

>How much do you use, coffee and all?

About 600 kWh/month. Mostly for my refrigerator, electric water heater and PC.
I just bought a laptop, which should reduce this. 

>4 sets of 10' x 7' windows = 2800 ft^2

Let's see. 4 x 10 x 7 = 280. I just did that on my calculator to be sure.
A 2,800 ft^2 window would be 5 stories tall and 50' wide.

>R12 at night; R4 + R8

My mistake.

 

>These numbers have to be revisited.

Agreed.

 
 
>>   280 ft^2 windows :-)

Hmm. A new way to measure energy :-)

>That should be 2800 ft^2

I don't think so.

>Try 75%

That won't change the picture much.

>Of course, these numbers must be revisited as well.

Agreed.

>>Now, thermal mass: the house has about 3,500 ft^2 of walls and ceilings with

>The water will weigh close to 3250 lbs.

My mistake again.

>You need to revisit this number as well.

Somebody should. 

>reluctant to mention the ones within.

Oh go ahead, mention them.

>water walls. 

Oh no. I did that.

>Your assumption seems to be 1 oF, though there is no evidence

That's pretty cold. That doesn't make any sense to me at all. I don't
understand what you are saying at all here Will. And I didn't assume
anything like that. I assumed the water walls were the same temperature
as the air and everything else in the house.

 

>Solar hot water, most likely active.

How much will that cost to buy and operate, and how much backup heat
will you need for the hot water?

>winter.  Reference "Solar Today", Sept/Oct 1995, article on manufactured house.

Sounds nice. As Steve Baer says, the greatest discovery of
solar investigators may have been that lots of insulation is good :-)

>Try the numbers again, then we'll talk.

It's your house, Will. Try the numbers again yourself, if you like.

I'll be happy to look at them.

Cheers,

Nick

---

From: [email-masked] (Nick Pine)
Date: 17 May 1996 12:46:29 -0400

>> [email-masked] (Nick Pine) writes:
   [someone else wrote...]
>> >  R-1 (International) degC.m^2/W = R-5.678 (US) degF.hr.ft^2/BTU?

Let me try this step by step, with a calculator.

A 1 m x 1 m wall with a metric R-value of 1 will pass 1 Watt of heat
if the temperature difference is 1 C.

So a 10.76 ft^2 wall will pass 3.41 Btu/hr of heat if the temperature
difference is 1.8 F. OK?

So that wall will pass 3.41/1.8 Btu/hr with a temperature difference if 1 F. 

And a 1 ft^2 wall like that will pass 3.41/1.8/10.76 = 0.176 Btu/hr.
(Numerically the same as the metric R-value of an R1 window...)

So that wall has a US R-value of 5.68.

>where you should have used 5/9.

I've never used that unix function, altho it looks useful.
Anyone know the name of the one that turns tabs into spaces?
 
>For temperature _differences_, Kelvin is the same as degrees Centigrade.

But you must call it by the right name, no? :-)

Nick

---

From: [email-masked] (Russell W. Schmidt)
Date: Mon, 20 May 96 17:13:19 GMT

>of the box? And how will that change over time if we shade the sunny wall?
I do not claim to be an "energy expert", but this problem was good for a brief 
diversion. Since no other numeric answers have shown up (yet - our news feed 
seems to be a little slow), I will post mine:

Since this is a "back of the envelope" evaluation, and Nick has given 
daily averages, I will ignore the daily fluctuations in ambient 
temperature and in insolation, and also assume that the insolation 
value given is the net energy to the water.

It does not matter what water temperature you start with - the question 
was what is the "steady-state" temperature of the water. For a steady 
state, the energy in must balancethe energy out, so 
insolation (solar energy in) = heat loss.

QHL = U*A*dT = A/R *dT = sum(A1/R1 + A2/R2) * deltaT     heat loss rate

      sum(A1/R1 + A2/R2) =    (5*4'*4')/(20*hr*ft^2*F/BTU) 
                            + (  4'*4')/( 1*hr*ft^2*F/BTU)
                         = 20 BTU/(hr*F)

QHL = 20 BTU/(hr*F) * deltaT

QI = IR * A       Insolation rate

   = 1000 BTU/(ft^2*day) * 4'*4' = 16,000 BTU/day = 666.67 BTU/hr

QHL = QI for steady state temperature

20 BTU/(hr*F) * deltaT = 666.67 BTU/hr

Therefore deltaT = 666.67/20 F = 33.33 F

And the steady-state water temperature = 33.33 F + 36 F = 69.33 F

If the window pane is shaded (without adding any extra insulation)
you simply get a heat loss rate as calculated above:

     QHL = 20 BTU/(hr*F) * deltaT
        where deltaT = TW - TA  (Water Temp - Ambient Temp)

This heat loss cools the water, thereby reducing its temperature:

     QHL = M*Cp*dTW/dtime  
  (Since QHL is a rate, this gives a rate of temperature change)
      where M  = mass of the water
            Cp = heat capacity of the water
            TW = temperature of the water

 M = 4'*4'*4' * 62.4 lb/ft^3 = 3993.6 lb or about 4000 lb of water
 Cp = 1 BTU/(lb F)
     (since this is a rough calc, I will ignore the small change in density 
      and heat capacity with temperature)

Setting the two equations for QHL equal and applying a bit of algebra, 
you get the following differential equation:

  dTW/dtime = (TW-TA)/200  F/hr

Now - remembering all that lovely differential equations stuff (not
likely :-) or looking in the back of the book, you find a solution like

   T = c*e^(-kt)

Using the initial conditional calculated above, we get

   TW = 33.33*e^(-t/200) + 36   for TW in Fahrenheit and t in hours

This describes an exponential temperature decrease - the rate of cooling
slows as the temperature difference between the air and the water 
decreases. So, the water cools by about one sixth of a degree F per hour
at first, with the rate slowly declining. After 10 days, the water 
temperature is still about 10 degrees F above the ambient temperature.

This answer ignores the daily fluctuations in both the ambient
temperature and (for part 1 above) the insolation rate. Both of these
can be modeled reasonably well using a sine function:

Insolation = pi/2 * 12 hr Avg insolation * sin(hour*pi/12)

Ambient Temp = AvgTemp + (TempRange/2)*SIN((Time/12)*PI())

These functions can be used in the above calculations, leading to much
more fun with calculus. Since I have Real Work (tm) to do, I will leave
the detailed solution as an exercise for the under-employed.

Russ Schmidt    <[email-masked]>
Lockheed Martin Energy Systems
Oak Ridge, TN  37831-7294

---

From: [email-masked] (Alan Braggins)
Date: 20 May 1996 11:30:16 +0100

> if the temperature difference is 1 C.
[Correct working]
> >Celsius

I was, and therefore I was using 9/5 instead of 5/9,
as I should have realised _before_ posting, instead of
on my way home for the weekend.

% /usr/bin/units
you have: kelvin   
you want: rankine
        * 5.555556e-01
        / 1.800000e+00

That 1.8 is the right number, wrong direction.

[Why didn't I just use F in the first place? As the man page says
    "units only does multiplicative scale changes.  Thus  it  can
     convert  Kelvin  to  Rankine, but not Celsius to Fahrenheit.",
which is what led me into error in the first place.]

> Anyone know the name of the one that turns tabs into spaces?

% man expand
 
EXPAND(1)                USER COMMANDS                  EXPAND(1)
 
NAME
     expand, unexpand - expand TAB characters  to  SPACE  charac-
     ters, and vice versa
 
SYNOPSIS
     expand [ -tabstop ] [ -tab1,tab2,...,tabn ] [  filename  ...
     ]
 
     unexpand [ -a ] [ filename ...  ]

> But you must call it by the right name, no? :-)
Well, if you are going to be picky, it's K not degrees K.

---

From: [email-masked] (gcp)
Date: Sat, 18 May 1996 16:50:44 GMT

[snips - here and there]
>person who answers it correctly.
I think you'll have to offer more than that.   :-)
(What's $10 in international units?)

Another good one might be, if they've heard of nonimaging optics?

>Of course Mother Nature thinks in meters and C, not feet and F :-) 
Always thought she knew what she was doing.
> Why does the Lord permit such Suffering on Earth,
>if He is Omnipotent?
As he's omnipotent he doesn't have the problems we do!
Unfortunately in Britian we have the bad habit of using a mixture of
the two!!
SI units have the distinct advantage of ease of manipulation, and
since most people use a calculator nowadays, we just have to check if
the dots in the right place!   :-)

So what does R-x mean in the rest of the world then?
(By the way in my post that ended up in alt.solar.thermal even though
in reply to one in alt.energy.renewable - seems an OK sort of place
for it - R-10+ was 'US')

>   about it, using the analogue computer which we keep in our minds.
Actually...   ...it's digital.
But WTF it works most of the time!

---

From: [email-masked] (Nick Pine)
Date: 20 May 1996 14:58:35 -0400

Russell W. Schmidt <[email-masked]> wins the prize!!!

>state, the energy in must balance the energy out, so 

Agreed. Energy in = Energy out. Those are the only two energy flows here,
in the long run, and they are equal. Energy is conserved. If the box had
lightbulbs or people inside, there would be some internal power to add in
to the solar power, but that isn't the case here. 

Yes. To find the thermal conductivity for the whole box, you divide each
exterior surface area by its R-value, and add up the quotients, like adding
conductances to find the overall conductance of several parallel resistors.
 
>QHL = 20 BTU/(hr*F) * deltaT
 
Un-huh. The heatflow out of the box in Btu/hour is equal to the thermal
conductivity of the box times the temperature difference between the water
inside the box and the constant air temperature outside the box. This is
Ohm's law, with Btu/hr replacing amps and a temperature difference replacing
a voltage difference. The thermal resistance in this simple circuit is
1/20 "Ohm," or 1/(20 "mhO" :-)

>   = 1000 BTU/(ft^2*day) * 4'*4' = 16,000 BTU/day = 666.67 BTU/hr

OK. This is the average hourly amount of sun power that flows
into the box over 24 hours.
 
>QHL = QI for steady state temperature

Ie the average solar energy that flows into the box during an hour is equal
to the average heat energy that flows out of the box during an hour,
after a long string of average days...
 
>20 BTU/(hr*F) * deltaT = 666.67 BTU/hr

Saying the same thing once more... 

>Therefore deltaT = 666.67/20 F = 33.33 F

Right :-)

>And the steady-state water temperature = 33.33 F + 36 F = 69.33 F

Good. Several people got to this point.

Now, once the water has warmed up to 69 F over a long string of average days,
we turn off the sun, keeping the outside air temperature exactly the same,
and the water gradually cools off...

>        where deltaT = TW - TA  (Water Temp - Ambient Temp)

So at any moment, the water has a certain temperature, and it is losing
energy at a certain rate... The rate of energy flow, ie the heatflow or
heat power flowing out of the box, decreases as the box cools. The box
loses more heat at the beginning, when it is 69 F, than later on, when
the water has cooled to, say 60 F, because the temperature difference
between the inside and the outside of the box has become less, like an
electrical circuit with a lower voltage difference across a resistor,
in which less current flows. The cooling problem can be approximately
solved this way using arithmetic, step by step, eg hour by hour, or day
by day, if the time constant is more than a few days...

 
>      where M  = mass of the water           [about 4000 pounds]
>            Cp = heat capacity of the water  [1 Btu/lb per degree F]
>            TW = temperature of the water    [what we are trying to find]

 
>  dTW/dtime = (TW-TA)/200  F/hr
 
Well, I guess you COULD do it that way... As J. E. Gordon describes? :-)

>From  _Structures: Why Things Don't Fall Down_: 

   When we play tennis or walk downstairs we are actually solving
   whole pages of differential equations, quickly, easily and without
   thinking about it, using the analogue computer which we keep in our
   minds. What we find difficult about mathematics is the formal,
   symbolic presentation of the subject by pedagogues with a taste for
   dogma, sadism and incomprehensible squiggles.

Present company excepted, of course.

>   T = c*e^(-kt) [+ something]
 
Yes. You only need to learn this once, and you don't have to know where
it came from in detail. Electrical engineers write that k as 1/(RC), where
RC is called the "natural time constant" or "RC time constant" of a simple
circuit. In thermal circuits, it is often measured in hours or days. Using
these units above, the RC product has the dimension of hours, in this case,
200 hours, or about 8 days, which is pretty good for a solar house.

>   TW = 33.33*e^(-t/200) + 36   for TW in Fahrenheit and t in hours

Notice the + 36 on the right, the something that got added to account for
the "final condition," of the water temperature, ie after an infinite number
of days, the water will have lost all its heat, and the water temperature
will be the same as the surrounding air temperature, no matter how good
the box walls were as insulators, if we wait forever, or almost forever.

The way I usually write this is

    T(t) = Ta + (T(0)-Ta)*exp(-t/RC), 

so after time t in hours, the water temperature as a function of time, ie
T(t), which was initially a certain number, T(0) = 69 F in this case,
becomes the same as the air temperature Ta. Initially, there is a temp
difference T(0)-Ta between the water and the air, and as time goes on,
that temperature difference becomes less, and it finally gets scrunched
down to 0 by the decaying exponential function exp(-t/RC) (the "e-to-the-x"
button on a $20 calculator) which becomes 0 as t gets bigger and bigger,
ie as time marches on, until we are left with just T(oo) = Ta + 0...
(oo is infinity, staring back at us, for unseen forces fans.)

>temperature is still about 10 degrees F above the ambient temperature.

Right. T(t) = 36 + (69-36)*exp(-t/200), so...

After 0 days  Tw = 69, since exp(0/200) = 1, 
After 1 day   Tw = 65, since exp(-0.12) = 0.886,
After 2 days  Tw = 62, since exp(-0.24) = 0.787,
After 3 days  Tw = 59, ...
After 4 days  Tw = 56, ....
After 5 days  Tw = 54  .....

After 10 days Tw = 46  ....... and so on, until

After oo days Tw = 36, since exp(-oo) = 0.

>Ambient Temp = AvgTemp + (TempRange/2)*SIN((Time/12)*PI())

They COULD be, but in this linear case they could also be happily ignored,
to get the same answer :-) Using this sort of thing in the ***bonus***
problem below would lead to a higher and more realistic performance,
but it's probably overkill there too, vs just using daily averages. 
 
>more fun with calculus.

Yeah, right :-)

>Since I have Real Work (tm) to do, I will leave

Hey, that's me! Congratulations, Russell! I will send a $10 check to the
address below. Is this the way to run a consulting business, or what? :-)

>Lockheed Martin Energy Systems

Bonus problem (another $10): How does all this change if a December day
is 6 hours long, and during each 18 hour night, that R1 glazed wall is
somehow insulated to R20, like the other walls, and it stays R20 for
24 hours a day during the DARK TIME.

You can't play this time, Russ :-)

Nick

---

From: Tore Forsgren <[email-masked]>
Date: Fri, 17 May 1996 20:12:34 -0700

> >>temperature be inside a 4' cube full of water surrounded by 5 R20 foam walls,

etc...

Knowledge or at least abtractions like Ohms law could be 
usefull as well as skills in thermodynamics. This discussion is really
about positioning one self and has more to do with human nature than
with energy. And being said itself is allso an atempt to be smarter than
the rest of you. (Humans are we not?)

/Tore F

---

From: [email-masked] (Nick Pine)
Date: 26 May 1996 08:59:40 -0400

I wrote: 

Which Russ Schmidt nicely solved :-)

Nobody has solved the one below yet, after a few days, so I'll try...

This would be more interesting if the cube were part of a high-performance
solar house, with the south face of the cube as the back wall of a sunspace,
and the rest of the cube walls inside the house. Below are two solutions for
the outdoor cube, in a simple program that compares their results. One is an
iterative arithmetic solution, and another uses exponentials.

The steady state temp is determined by Ein = 16K Btu/day = Eout, where

Eout = 6 hr (T-36) 16 ft^2/R1      daytime south wall loss with damper open
    + 18 hr (T-36) 16 ft^2/R20     nightime south wall with damper closed
    + 24 hr (T-36) 5*16 ft^2/R20   other walls, 24 hours a day
     = (96+14.4+96)(T-36), so

T = 36 + 16K/206.4 = 113.5 F, after a long string of average December days.
Insulating the glazed south wall at night makes this cube warmer than the
cube where the water lost heat through the R1 glazing all night... 

The thermal resistance of the cube with its south damper or door closed is 
R = R20/(6*16ft^2) = 0.20833, and the 4,000 pounds of water inside has a
thermal mass of C = 4,000 Btu/F, so the time constant RC of the closed cube
is RC = (0.20833)(4000) = 833 hours or 35 days :-) So...

T(t) = 36 + (T(0)-36)exp(-t/35), where T(0) is the water temperature at the
start of a cloudy week and t is in days. The program below calculates this,
and compares it with an arithmetic solution that finds a new water temp each
day, as the heat loss from the water cools its own thermal mass, assuming
the water temp is constant over a day. Whole-day time steps work well here,
because the RC time constant of the closed-up cube is so big.

10 'heating and cooling a 4' cube of water surrounded by R20 insulation,
20 '   sitting outdoors on a 36 F day, with 1,000 Btu/ft^2/day of sun
30 '   passing into one R1 side wall for 6 hours a day
40 '
50 TA = 36'ambient temp
60 V=4^3'volume of water (ft^3)
70 C=64*V'thermal mass of water (Btu/F)
80 UDAY = 16/1 + 5*16/20'daytime thermal conductance of cube
90 UNIGHT = 6*16/20'nightime thermal conductance of cube
100 SUN=16*1000'solar input to sunny side of cube (Btu/day)
110 'solar heating during an average week
120 TSS=36+SUN/(6*UDAY+18*UNIGHT)'estimate steady state water temp
130 TW = TSS'initial water temp
140 FOR DAY=1 TO 7'simulate for a week, day by day
150 EOUT=(6*UDAY+18*UNIGHT)*(TW-TA)'energy flowing out of the cube
160 TW=TW+(SUN-EOUT)/C'new water temp
170 PRINT DAY;TA;TSS;TW
180 NEXT DAY
190 'passive cooling during a cloudy week
200 TWI=TW'initial water temp
210 PRINT
220 RC=C/UNIGHT'time constant of closed and shaded cube in hours
230 FOR DAY=1 TO 7'simulate for a week, day by day
240 EOUT=24*UNIGHT*(TW-TA)'energy flowing out of the cube in an hour
250 TW=TW-EOUT/C'new water temp, calculated arithmetically
260 TEXP=TA+(TWI-TA)*EXP(-24*DAY/RC)' calculated exponentially
270 PRINT DAY;TA;TEXP;TW
280 NEXT DAY

 1  36  113.5194  113.5194  once the cube reaches this temperature,
 2  36  113.5194  113.5194  after a long string of average days,
 3  36  113.5194  113.5194  the temperature does not change...
 4  36  113.5194  113.5194
 5  36  113.5194  113.5194  the south wall damper is open for 6 hours  
 6  36  113.5194  113.5194  a day in this case...
 7  36  113.5194  113.5194

This is an average water temperature. The water would be about
1 F higher at dusk and 1 F lower at dawn, just coasting... :-)

 1  36  111.3695  111.3391  here is how the water temperature of 
 2  36  109.2793  109.2202  the cube changes, with the south wall
 3  36  107.247   107.1609  damper closed all the time, during a week 
 4  36  105.2711  105.1595  without sun, when the air temp is 36 F...
 5  36  103.35    103.2144
 6  36  101.4822  101.324   there's only a small difference between
 7  36  99.66614  99.48676  the arithmetic and exponential solutions...

The outdoor cube temperature decreases slowly during a cloudy week.

How long will it take to reach 80 F?

If T(t) = 80 = 36 + (113.5-36)exp(-t/35), then 44 = 77.5exp(-t/35), 
exp(-t/35) = 0.567, -t/35 = ln(0.567), and t = -35 ln(0.567) = 19.8 days :-)

If the south face of the cube is the back wall of a sunspace which is 80 F
when the sun is shining, ie warmer than the 36 F outdoor temp, and the rest
of the cube walls are inside the 70 F living space of a house, ie they lose
less heat every day than if they lost heat to 36 F outdoor air, the steady
state temperature goes up, so we can store more heat in the cube, ie

Ein = 16K Btu/day, but

Eout = 6 hr (T-80) 16 ft^2/R1      daytime south wall loss with damper open
    + 18 hr (T-36) 16 ft^2/R20     nightime south wall with damper closed
    + 24 hr (T-70) 5*16 ft^2/R20   other walls, 24 hours a day
     = (96+14.4+96)T - 7680 - 518 - 6720, and 

16K + 14,918 = 206.4 T, so T = 30,918/206.4 = 149.8 F, after a long string
of average December days in Phila.

Putting a ground reflector in front of this solar closet, eg snow at
60% reflectivity, raises the solar input above the number given with 
20% ground reflectivity, so Ein = 21.33K and T = 36,251/206.4 = 175.6 F.

How long will this indoor closet take to reach 80 F?

80 = 70 + (175.6-70)exp(-t/35) ==> t = -35 ln(10/105.6) = 31.8 days :-)

I'll offer another $10 to the first person to calculate how the steady state
temp would change, if at this point we add another layer of glazing to the
solar closet, using these assumptions, or any other reasonable assumptions...

Bigger closets have bigger time constants, increasing as their dimension,
squared, so an 8' cube might stay warm for about 4 months indoors :-) 

Nick

---

From: [email-masked] (Alex Walter)
Date: Mon, 27 May 1996 22:40:58 GMT

Nick Pine  . . . Did you start this thread all about Energy
Consultants?

I'm a learning "energy rater" for Energy Rated Homes of Colorado with
a real world question regarding infiltration.   I say I'm learning as
I've only been dealing with heating and A/C for 30+ years.

In the course of performing Manual J (ACCA) heat gain and loss
calculations I am finding great variances in opinions as to what
should be used  for infiltration figures.  Do you or any others
lurking here have experience using blower doors to measure
infiltration into houses?  AND if you do what do you/would you use as
the infiltration, air changes per hour, into a house if you find that
a 24,160 Cu. Ft. house leaks 3,200 CFM at 50 pascals  negative
pressure?

Using the Manual J detailed infiltration estimate method I come up
with winter AC/Hr = 0.85 and summer AC/Hr = 0.32 AC/Hr.

---

From: [email-masked] (Nick Pine)
Date: 27 May 1996 20:15:23 -0400

Alex Walter <[email-masked]> wrote:

>Nick Pine  . . . Did you start this thread all about Energy Consultants?

I guess so... BTW, there is still an outstanding $10 prize problem... :-)

>I've only been dealing with heating and A/C for 30+ years.

So are you one of the opportunists, or one of the shortsighted criminals? :-)

Perhaps you are a farsighted, experienced criminal...

>infiltration into houses?

I tested and improved two dozen houses around 1975, trying to implode then
with a big squirrel cage blower and a manometer, but haven't done much since,
except look at the numbers in Nisson and Dutt's Superinsulated House book,
and Avis and other housebuilders' numbers. This was interesting work... One
person failed to mention that he had a smouldering woodstove going; one time
the blower ate some window curtains, and almost ate a cat, and I sealed up
one basement so well that the gas heater starved for combustion air...
Strangely enough, I didn't get a lot of demand for my house-sucking services.

These days, it's my impression that when you buy a blower door, it comes
with lots of information on how to interpret the results, with tracer gas
correlations, corrections for building height and windspeed and temperature
(stack effect), and so on. 

>AND if you do what do you/would you use as the infiltration, air changes
>per hour, into a house if you find that a 24,160 Cu. Ft. house leaks
>3,200 CFM at 50 pascals negative pressure?

Nisson and Dutt suggest just dividing the 50 Pa number by 20, so I'd
estimate the natural air infiltratation rate for that house as  
3,200 ft^3/m x 60 m/hr/24,160 ft^3/20 = .4 ACH.

>with winter AC/Hr = 0.85 and summer AC/Hr = 0.32 AC/Hr.

Sounds OK to me. I guess a lot of this energy testing is all tied up
with the government and electric utilities now...

Nick

---

From: William R Stewart <[email-masked]>
Date: Tue, 28 May 1996 08:13:58 -0400

> Perhaps you are a farsighted, experienced criminal...

Alex, you'll have to ignore Nick when he gets in these moods.  We try to work
with him, but his manners have a long way to go.  He has a lot of ideas, some
interesting, some wacky, but don't go off and implement anything he recommends
unless you can have the numbers verified.  He tends to ignore infiltration, 
forced and natural convection and radiation in his calculations.

So anyone that doesn't drop all of their ideas and fully embrace his is a 
criminal.

Cheers,

Will Stewart

---

From: [email-masked] (Alex Walter)
Date: Wed, 29 May 1996 23:46:45 GMT

>>with winter AC/Hr = 0.85 and summer AC/Hr = 0.32 AC/Hr.

Where does Nisson and Dutt's work fit into this?  Are/were they by
chance at Princeton U?  I have an article which says that in 1982,  J.
Kronvall and Andrew Persily did some work in New Jersey and Sweden
which proved ACH50/20 to be fairly reliable.  The factors not taken
into consideration include stack effect, wind shielding, and types of
leaks.

Next  Max Sherman a  researcher at Lawrence Berkeley Laboratory
developed a model which does take into account, climate, height, wind
shielding, and  crack/hole size.  Sherman found that the correction
factor could be as small as 6 and as large as 40.  So . . . the
ACH50/20 rule of thumb could overestimate infiltration by a factor of
two or underestimate it by a factor of about three.

The climate correction factor for Denver is 15 to16 whereas  New
Jersey  in the 17-20 range.  The climate correction factor takes into
account average temperature and windiness.

With all the factors considered the so called ACH50 number gets
divided by about 13.5 in the example case above to yield about 0.6
AC/Hr.  

The difference between 0.6 and 0.85 ACH is substantial when doing a
heat loss calculation.  The old school engineers would use 0.85 ACH
with the idea that if big is good bigger is better.  Being a younger
:-) school type at age 56 I'll use the 0.6 figure to keep the furnace
size on the low side so if the temperature falls below the design temp
by one degree I can be sure the furnace is running 100% of the time.
Where the other guys install a 140 mbtu furnace I install a 120 mbtu
furnace and so far my customers have not been cold even on those few
design temp days we have had in the last few years.
>with the government and electric utilities now...

With the competitive environment surrounding public utilities they
don't put as much effort or manpower into consumer education as they
used to.  SO . . . that leaves me an opportunity to be an energy
expert, i.e. "certified rater" for Energy Rated Homes of Colorado.

---

From: [email-masked] (Nick Pine)
Date: 30 May 1996 04:35:06 -0400

Alex Walter <[email-masked]> wrote:
>Where does Nisson and Dutt's work fit into this?

They wrote _The Superinsulated Home Book_, Wiley, 1985, ISBN 0-471-88734-X.
Page 221 says
   
   An accurate prediction of the natural air infiltration equivalent to
   specific blower-door measurements is far less useful for a superinsulated
   house than for a conventional house, because natural air infiltration in
   the former case is very small--most of the air entering or leaving the
   envelope is acounted for in the controlled ventilating system. For this
   reason, even a crude formula relating pressurization leakage to natural
   air infiltration should suffice for a superinsulated house. Such a crude
   relationship is:

     typical winter natural air infiltration rate (acph) =
     average leakage at 50 pascals (acph)/20.

   The natural air infiltration rate is roughly one twentiieth of the
   leakage measured by a blower door at an inside-outside pressure
   difference of 50 pascals.

   We call this the "divide by 20" formula.

They go on to say "the measured air leakage under pressurization q(50) of
superinsulated houses should be below 3 air changes per hour and preferably
around 1 acph. 1-3 acph at 50 Pa corresponds to natural air infiltration
rates of about 0.05 to 0.15 acph. Intentional ventilation supplied by the
ventilation system should supplement this to provide the total ventilation
desired. If a total ventilation of 0.5 acph is desired, the ventilation
system should be designed to provide 0.4 (+/- 0.05) acph."

>Are/were they by chance at Princeton U?

They were. Ned Nisson now publishes an energy newsletter for builders, 
with an office in New York City. I don't know what became of Gautam Dutt.

>I have an article which says that in 1982, J. Kronvall and Andrew Persily
>did some work in New Jersey and Sweden which proved ACH50/20 to be fairly
>reliable.  The factors not taken into consideration include stack effect,
>wind shielding, and types of leaks.

I guess that's where that ACH50/20 formula came from then...

>two or underestimate it by a factor of about three.

Hmmm. The underestimation would be seriouser.

>account average temperature and windiness.

I wonder how it does that.

>divided by about 13.5 in the example case above to yield about 0.6

OK. Could you tell us in a very simple way, how that /20 formula changes
with building height and outdoor temperature and windspeed? Is this stuff
in manual J? Where do we get a manual J?

>size on the low side...

I suppose the difference between 0.1 and 0.15 is less important. I wonder
how the airleakiness of a house changes over time... I'm 49, BTW, I wonder
how old these old-school creatures are :-)

>>I guess a lot of this energy testing is all tied up with the government 
>>and electric utilities now...
>
>With the competitive environment surrounding public utilities they don't
>put as much effort or manpower into consumer education as they used to.

That might be good :-)

>SO . . . that leaves me an opportunity to be an energy

Sounds nice :-) Is that a rating service or a house manufacturer, or...?
I wonder if you will be rating any solar homes as some define them, ie
those with no other form of heat, not even internal heat gain from
electrical usage.

Nick