This is a followup to a previous post in which I proposed calculating
the heat loss in my house. I received a number of helpful responses by
email in addition to several posts which have appeared here. There were
four points made most often:

1) It's been done.
2) Yes, infiltration does account for a lot of the loss.
3) I overestimated the net insulation in my house.
4) Net radiation loss is small but can be controlled.

Since a number of subsequent posters have been asking for similar information,
I think it would be appropriate for me to explain these four kinds of
responses.

1) Of the many books mentioned to me I found most interesting the books by
Charles Wing. There are several, including "From the walls in", 
"HouseWarming", and "The Visual Handbook of Building and Remodeling". The
author is described in one of the books as a physicist (or perhaps just
plays one on TV). This sort of background is evident in the information
which David Brouwer kindly reprinted to this group the other day, from
one of Wing's books; it is also just the sort of perspective I was
taking in my previous post.

Brian McDonald also recommended "Chilton's Guide to Home Energy Savings",
which I have not yet seen. 

2) Lyle Seaman pointed out that, of course, air infiltration occurs
along the edges of the doors and windows. In my house, the windows are
typically 5' x 2.5', which contributes to a whopping total of 300 feet
of infiltration zone. I can feel air coming in at some spots, but
overall I'm guessing the average gap is, say, only 1/64 of an inch.
Still, this adds up: the total air exchange space would then be the
same as a circular hole 8 inches across!
	Now, the speed of the air flow depends on the shape of the
hole: the long skinny 'hole' around my windows and doors offers a lot
more resistance to air flow than an equivalent area in a single spot.
But still this is surely a big place to lose heated air. I assume that
even in the absence of wind the difference in density between the
inside and outside air will force a certain amount of air exchange. If
anyone knows the data needed to calculate this I'd like to hear of it.
	(It's also true that air is exchanged around the window unit, not
just around the sashes. I have put in a few of those 3M films which work
well, but there is still cool air around the outside edges of the trim!)

	Incidentally, the number missing from my last post is: heat lost
is about 0.438 x 10^{-5} therms per degree-day, for each cubic foot of air
which is typically exchanged per hour. In my 20x28x30 = 16 800 cu ft house
this means .074 x number of hourly air exchanges. Granting all other
calculations so far, there is no problem accounting for all other heat loss
with just 1 air exchange every hour. Evidently typical numbers range from
1/2 exchange per hour for new construction to 10 per hour in old or
oddly shaped houses (in which the ratio of air volume to window edge perimeter
corrupts the rule of thumb).

	I should also apologize for using the term 'convection'
cavalierly.  I used it to mean the flow of air of different
temperatures, intending to equate that with air infiltration. But Mark
Lawrence pointed out that convection also adds to the problem in
another way. Even air in a sealed cavity, if sandwiched between a hot
zone and a cold one, will form convective flows inside. The net
effect, as I understand it, is that the air will carry the warmth more
rapidly from the warm zone to the cold one than the R-factor of still air
would indicate. I believe that the right way to incorporate this into
the model would be: a) if the wall cavity is exposed to the outside,
assume that the air being lost is as warm as the warmest wall of the
cavity; b) if the cavity is airtight, assume that it just doesn't
exist, that is, calculate heat loss as if the warm and cold sides of
the cavity abutted. This seems a tad pessimistic but in practice is
apparently close to correct. (This, of course, is why we use
insulation in walls even though the R-value of air looks good on
paper).

3) I forgot that insulation, like electrical resistance, needs to be 
amalgamated differently when in parallel than when in series. It's the
heat flow, not the resistance, which is additive. I should have taken
harmonic means, not arithmetic means, when estimating the overall
R-value of the house. Doing it right improves the model by a factor of 2.
A number of posters and respondents caught me on this. Sorry.
	Here are the numbers. I estimated about R-15 overall for the house
as a weighted average of these numbers:
	840 sq ft ceiling @ R-20 + [bonus R-factor for attic]
	840 sq ft floor @ R-4 + [bonus R-factor for basement]
	340 sq ft window @ R-2
	1980 sq ft wall @ R-15

	Bzzt! Well, the situation is worse if I solve not 
		4000.net-R-factor = Sum(SA . R-factor)
but rather the correct formula
		4000/net-R-factor = Sum(SA / R-factor)
What's more, I ought to add another term. Mark Lawrence pointed out
that of the wall area fully (2"/16")=1/8 consists of the wall studs.
These insulate less well, so I need to take about 250 sq ft out of the
wall area and figure this part at R-6 or so.
	The grand total, then, is that the net R-factor for the whole house
is a lousy R-7, which will double the total heat loss.
	This correction alone brings the total calculated heat loss from
conduction up to about 0.12 therms per degree-day -- now a better match to
the observed 0.18 therms per degree-day.

	Incidentally, this line of reasoning suggests the importance
of concentrating one's insulating energies on the worst offenders. For
example, one of Wing's books mentions this scenario: A person like me
with an old house tries to insulate a wall whose studs are
spaced unevenly. Seeing a 16" space in one cavity he places a 15" piece
of insulation in the hole and stretches the kraft paper to the studs.
This leave 15/16 of the cavity with R-11 or so, but 1/16 of it with R-1,
say (for the house sheathing). The net insulating ability of that cavity is
then only HALF of what he thought he was getting! So, sure, do as much
insulating as you can, but do it carefully so as to minimize the weak areas
rather than maximizing the strong areas. 

4) Radiation effects.
	The easiest way to understand the power of radiant heat energy is
not to calculate losses but to stand inside a south window on a sunny day.
This simple experiment gives windows their only redemption. Unfortunately
interior heat radiates out too. Though the radiation effects are smaller
than infiltration or conduction, they can be turned to our advantage
by allowing sunlight in when possible, then drawing curtains or blinds
to prevent reverse flow.

4) Latest conclusion.

The current model estimates the places where all that heat bought from the
gas company goes to. Roughly, it's
	27% air exchange (or otherwise unaccounted for)
	20% furnace inefficiency
	20% uninsulated floor
	17% windows -- conduction loss only
	12% walls
	 6% ceiling [Note: heat loss here causes roof/gutter problems]

I find this to be quite informative. When I add in the convection loss
to the conduction loss I see that the windows are terrible problems
for us. It's even worse when I include radiation losses, and when I
think about the completely uninsulated pockets around the windows
which held the sash weights. Ugh! I will not mess with the exterior
appearance of the house, or the interior trim, but some kind of window
work is in order!

I know these numbers are based on inaccuracies in the model; for
example, since the ceiling air is warmer than the floor air there is
less loss than 20% through the floor, really, and more than 6% through
the ceiling. I measured temperature differences with the thermostat at 65
and the outside air at about 20F:
	66F at thermostat
	61 on floor directly below thermostat
	72 on ceiling directly above thermostat
	60 in interior of a cold room
	48 at coldest point of thermal envelope (right next to the bottom edge
		of the worst window in the coldest room)
	82 at hottest point (sort of): ceiling of 2nd floor bedroom with a
		hot-air vent in its ceiling, measured with room ceiling fan off
		(but, to be useful, measuring several feet from heat vent).
Clearly these difference will affect heat transfer rates.

It may be possible to get a more honest assessment of heat loss by
considering other factors too. I am starting to kick this idea around.
For example, while the cellulose has not settled significantly, there
is little of it in the top 2.5ft of the house where the soffits hang
down; having the least insulation at the warmest part of the wall is
hurting us.  Also the lack of any humidifier changes some of the
physical characteristics of the air, and (I am told) has a deleterious
effect on the occupants' perception of the temperature. Finally, as
noted above, radiation transfer and non-homogeneity of temperature
should also be included.

Thanks to all who responded. You all may have lost interest in this process
by now but the mathematician in me finds this to be intriguing.

dave rusin@math.niu.edu