[Post to newsgroup misc.consumers.house. Be sure to read also my follow-up
post, which corrects some important errors. --dave   rusin@math.niu.edu]

Date: Wed, 26 Jan 94 13:36:56 CST
From: rusin (Dave Rusin)
To: misc-consumers-house@cs.utexas.edu
Subject: Heat Loss in House (Math model) (Long)


I posted a query here requesting help in analyzing my energy use in
home heating, stating that my calculated heat loss was off from the
real energy consumption by as much as an order of magnitude.  Well, I
have tightened up my estimates and gotten within a factor of 3 or so.
Since I have had some responses requesting more information I post
this whole analysis. It is lengthy, but as a bonus we get a model
accurate enough to show why, for example, it does pay to turn off the
furnace when not at home.  I still seek suggestions as to why I have
to pay for more heat than I can account for (and certainly more than I
would like to pay for).

              _PREDICTING HEAT LOSS IN OUR HOUSE_

I. Preliminary description

The house is a 3-bedroom "Foursquare" west of Chicago. It gets cold here:
last week's record was -27F. We keep the thermostat at about 65F all day
(wife and kids at home) but do sometimes remember to lower the
thermostat at night or when away. We have been doing some construction
lately so I am using data that's a couple of years old, when the
situation was more stable. Heat is forced air heated by natural gas, no
humidity control. There were no other gas appliances at the time.
The house is far from airtight and had no reflective coating insulation,
but there was a reasonably good blown-in cellulose insulation throughout.
Visitors never say "my, how warm this house feels!"

II. Actual heat loss.

This is fairly easy. The gas company reports, monthly, both cubic feet of gas
consumed and degree-days suffered. These data sets are clearly proportional
(we will explain this below). The relationship is approximately

	therms bought = 0.23 x degree-days

(1 therm is the energy content of about 100 cu. ft. of gas. It is precisely
100 000 BTU).

The efficiency rating of my furnace is about 80%, so after deducting the
therms thrown away up the chimney we have actual energy use for home heating:

	therms used = 0.18 x degree-days


III. General equations for heat loss

This section is not for the faint of heart; it explains why there must
be a proportionality relationship as above and indicates a theoretical
value for the constant of proportionality.

I will think of the house as a hot region first surrounded by an
insulating layer and further surrounded by a cold region (the air). If
the house is warm at first and then unattended, it loses heat through
the insulation (due to conduction) to the cold region, which we will
assume is at a fixed temperature. It is this lost heat which needs to
be replenished from time to time by the furnace.

The two equations we need are these. First, any heat loss inside is
proportional to a temperature drop inside: If the inside temperature T_i
changes by an amount Delta-T_i, the amount of heat which has escaped is
	Delta-H = a. Delta-T_i
The constant a is related to the volume and other characteristics of
the hot region (=house air)

Second, the rate (per unit time) at which heat is transferred by
conduction across the insulating layer is proportional to the
temperature difference between the inside and outside temperatures T_i
and T_o:
	Delta-H / Delta-t = b. (T_i - T_o)
The constant b is related to the surface area and other characteristics
of the insulating region. (Actually I need to put a minus sign in one
of these two equations depending on which region I want to think of as
gaining or losing heat.)

Combining these equations leads to the differential equation which
describes cooling: assuming T_o is not changing in time, 
	(d/dt) (T_i - T_o) = -(b/a) (T_i - T_o).
This is the familiar equation for exponential decay, so that the
inside temperature at time  t  is

	T_i = T_o + A.exp( -(b/a) t)

Here the constant  A is the initial amount by which the temperature inside
the house exceeds the temperature outside. Last week A reached 91F.

Now, the thermostat will turn the furnace on when the temperature inside
hits a low value T_0 and turn it off again when the temperature hits
a high value T_1. From the preceding equation, we can compute that the time
which will pass between turn-off and turn-on will be

	(a/b) log ( (T_1-T_o) / (T_0-T_o) )

The amount of heat needed to return the temperature to T_1 will be
	a. (T_1-T_0).
Thus the total energy consumed in, say, one day will be

                          (1 day)
	a. (T_1-T_0) . ------------------------
                       (a/b) log (..as above..)


Now, typically T_1-T_0 is small compared to  T_1-T_o (say 1 degree compared
to 91) so we can approximate the log well with the formula log(1+x) ~= x.
Then there is a lot of cancellation and we get the daily heat usage:

		b. (1day) . (T_0-T_o)

Assuming that T_0 is 65F, the quantity  (1day).(T_0-T_o) is precisely the
number of "degree-days" suffered that day. Adding across all the days in
a month shows that the energy used for heating will be

	therms used = (b) x degree-days

You don't have to assume that T_0 or T_o are really constant, as long as
they change slowly or infrequently compared with the amount of time it
takes the furnace to cycle; total heat consumption is clearly the sum
of the energy used per hour (or whatever) across the whole interval of
time. 

Incidentally, this is the basis of the analysis people have in mind
when they say, yes, you will save money if you shut the thing off
during the day.

IV. Estimating the parameter  b.

The goal (comparing sections I and II) is to show that b is about 0.18
therms/degree-day.

Now in fact,  our  b  was defined by the equation describing heat transfer.
To amplify that earlier equation, we have
	(Delta-H)/(Delta-t) = K . SA . G,
for some coefficient  K  depending only on the composition of the insulation,
where SA is the surface area and G is the temperature gradient
(G= (T_i-T_o)/w, where  w  is the width of the insulating layer.)
Comparing to the earlier equation, we are expressing  b  as
	b = K. SA / w 

Values of K can be found, for example, in the CRC handbook. (I tried
to find them by internet snooping, but the best I could do was
to find K for various substances of interest to dentists. If you want to
insulate your house with amalgam or dentin, let me know). Here are some
values of K in BTU per hour per square foot per (degree-F per inch):
	Rock wool	0.26-0.29
	Fiberglass	0.29
	Balsa wood	0.33
	Sawdust   	0.41
	White Pine	0.78
	Oak         	1.02
	Building gypsum	about 3 [I don't know if this means drywall]
	Plaster	    	2 to 5
	Glass    	5 to 6
	Concrete      	6 to 9
(If I have done the conversion from other tables correctly, the
corresponding figure for air and similar gasses is about 0.15; for
water it is about 4.0 -- a little lower for other liquids; rocks are
in the dozens; metals are in the hundreds and thousands. Remember that
when you select aluminum frames for your windows!)

"R-values" should be proportional to the quotient of: thickness in inches of
a substance divided by the number  K  above; looks like the constant of
proportionality is about 1.0.
	
The insulating layer is a complicated mixture of various substances
each contributing different amounts of K, SA, and w. I decided to punt
and make what I hope is a good ballpark estimate of net insulating
value. Here is my reasoning.

Our house is (was) not far from being a simple box. The heated region is
2 floors (about 20') high, 2 rooms (28') wide, and of slightly varying
depth, around 30' on average.

Consider the walls, for example. They consist of (from inside out):
	paint
	plaster+lath (about 3/4" total)
	16" o.c. stud cavities filled with cellulose (full 4" cavity)
	sheathing (3/4", covering all wall area)
	felt paper or something
	clapboard siding (varying thickness of course, average 1/2")
	paint (not negligible thickness!)
Adding R-values, if you like, it would appear that the walls offer
the insulating value of about 5" of fiberglass.

The rest of  the house is equally complex.
A full 15% of the wall area is window and doors, typically
a single pane of glass, 3" air space, and a single-pane storm. Here the
R-values are only R-2 or so, i.e., only 10% of the insulating value of
the walls. The floor is a 3" wood and air-pocket combination with no
insulation except a 5' basement. Proper analysis here would consider
4 regions: in, out, basement, and earth (which keeps a steady 56F or so).
Housetop is about 6" cellulose with ceiling lath and attic floor, plus
attic air and roof structure; I'm hoping the extra insulation here balances 
out the weak spots such as the windows.

Well, I won't attempt to do the full analysis. I would guess that on the
whole, the walls represent a good average of all the insulation values.
SO... I will treat the house as a big box made of 5" thick fiberglass
separating a warm inside from the cold outside.

Now we can plug into the formulas above. We have an insulating layer with
the following combined characteristics: K=0.30; SA=4000 sq ft; and
w=5". This gives a value for b of about 240 BTU per degree-hour, or
5 760 BTU per degree-day. This gives a theoretical value (0.06 therms/deg-day)
which is in modest agreement with the measured value (0.18 therms/deg-day)

V. Conclusions
I seek advise on ways to improve the model. I'm not stockpiling the heat
in boxes, I promise; it's leaving the house somehow. But how can I
get better agreement between theory and data? (And how can I get both numbers
lower and put more $$$ back in my pocket?)

The agreement got a lot better when I remembered the furnace inefficiency.
But after all that work to get good numbers elsewhere I sure do feel 
cheated by the furnace. I will surely go for a higher-efficiency unit
when I can.

I also got much better agreement with observed data by repeatedly decreasing
my estimate of net insulation, down as you see to about R-15 overall. I
think the above estimates are honest, but certainly at variance with what
I had previously estimated my house's insulation to be. Even now I am nagged
by the sense that surely that unventilated attic (with some insulation scraps
mistakenly applied to the roof rafters) ought to count for _something_;
and that big air mass in the basement which, after all, faces for the most
part onto a comparatively warm earth should be helping more too? But
I guess the facts are otherwise. I have been adding more insulation above
and below. There is not much I can do about the walls, sadly, though more
up-to-date windows can reduce conduction loss.

This model neglects convection loss.   I am assuming that most of the
difference between measured and predicted heat loss is due to our drafty
windows and frequently-opened doors, as well as other smaller air gaps.
Heat lost to convection is  
	(molar heat capacity of air, constant) x
	(amount of air exchanged, constant per unit time (*) ) x
	(temperature difference  T_i - T_o)
( (*): possibly increased on windy days). Thus I expect convection losses
can be built into the model above simply by altering the interpretation
of the constants  a  and  b. In particular, energy usage will still be
proportional to degree-days. But if convection loss is indeed comparable
to or larger than conduction loss, I need to make new windows and caulking
a higher priority!

The model also neglects radiation loss. Granting the preceding, I see no
need to ascribe much actual heat loss to radiation -- that is, despite
the advertising of reflective-coating manufacturers, it's not worth my
while to worry about radiation loss. Moreover, as I understand it, heat loss
due to radiation adds to the dT/dt equation a term proportional to 
(T_i-T_o)^4 or so. The differential equation
	dX/dt = -(b/a)X - c X^4
has general solution (I'm skipping lots of details now)
	X = A. exp( -(b/a)t ) /[1+(ac/b)(A exp(-b/a.t))^3]^{1/3}.
The time between furnace firings is now shortened by an amount equal to
	(a/b)(1/3)log[(1-(ac/b)(T_0-T_o)^3)/(1-(ac/b)(T_1-T_0)^3)]
so that the dominant terms for total energy use in a day come out to
	b. (1day).(T_i-T_o) + (ac).(1day).(T_i-T_o)^4.
This is no longer linear in the number of degree-days sustained in that
time, so that it is no longer true that the total energy use can be
predicted from the number of degree-days endured in a month (that is,
one cold day and 3 hot ones will no longer require the same amount of
energy as 3 cool ones and 1 warm one even if the same number of degree-days
are suffered). Possibly this is consistent with the data I have but
I think this means  a.c  is very small compared to  b. In short, unless
you are already well-insulated, it appears that radiation loss is a 
comparatively small one.

Finally, when making the model I was aware that I was simplifying
some things but I don't know how a truer rendering would enter in. For
example, I know that the inside of the house is _not_ uniform: it is
certainly warmer near ceilings and in bedrooms. Also I know that heat
re-entry is not instantaneous; indeed on very cold days it takes as long
to reheat the house by that small amount (T_1-T_0) as it took for that
heat to escape. If such considerations significantly improve agreement
between predicted and measured heat loss, or if they are worthwhile
considering when trying to make a home energy-efficient, I'd like to know 
about them.

dave rusin@math.niu.edu