Subject: Re: Heat loss question

From: [email-masked] (Nick Pine)
Date: 23 Jun 1995 23:11:14 -0400

Bob Taylor  <[email-masked], not!> wrote:

>I've got 4 ea 12 watt rabbits, does that count? =:-)

Soitenly. Table 7, "Heat Generated by Laboratory Animals" on Page 9-12 
of my trusty ASHRAE HOF says a 5.41 lb, normally active wabbit makes 
39.22 Btu/hr of sensible heat, according to data taken by Runkle (1964),
Kleber (1961), Besch (1973), Woods and Besch (1974), Woods et al (1974),
Bond et al [who is this al guy?] (1959) and Ott and McNully (1961.)
 
>Cows as space heaters???  You could also use the waste products to make
>methane to heat the house, and I understand cows make a fair amount of
>methane directly as well.

Impractical, I'd say. Maybe methane for cooking ("WHAT is that smell?"/
"Oh I don't know, just je ne sais quoi.") But for methane space heating,
I believe you'd need about 200 cows.

>ps- how many watts is 2 ea 16 lb cats?

Exactly 2 x 27.21/3.41, according to Table 7, if they weigh 6.61 pounds
each, which would be true for small values of 16, or large values of 2.

Nick

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From: [email-masked] (Nick Pine)
Date: 23 Jun 1995 18:10:00 -0400

Robert Vesalga <[email-masked]> wrote:

>Is there a simple answer to this question?

Yes :-)

>If I were designing a house in a 32 degree climate, how much insulation 
>(in R-value) would it take to have no heat loss through the walls, while 
>maintaining a 72 degree interior? (Assume this to be like a walk-in freezer;
>doors, windows, infiltration, etc., notwithstanding. Just pure wall).

R-oo should do the trick, ie infinity. But that is expensive.

If the house were two stories, 16' x 32' x 32', ie about 3000 square feet
of surface area exposed to the weather, and it had no windows or air
infiltration, and there were two 100-watt people living inside, using
500 kWh of electricity per month, there would be a continuous internal
heat generation of (2x100 + 500x1000/(30x24)) = 894 watts, which is
equivalent to 894 x 3.41 Btu/hr/watt = 3000 Btu/hr. Since the temperature
rise from one side of a wall to another is equal to the heatflow times
the thermal resistance ("Ohm's law for heatflow"), the R-value of the walls
would have to be such that 72F-32F = 40 = 3000 x R/3000, so R40 should do.

If you had a 50 watt dog, you could use R38.

>Would the floor and ceiling be the same ?

The floor would have less heat loss than the ceiling, if the insulation were
the same, because the temperature under the floor is higher in wintertime. 
I usually figure the ground temperature as 55F and the thermal resistance
of the earth as R10, so you might want to put more insulation in the ceiling.

Or you might keep cows in the basement, as do the Swiss. Figure 11 on page 9.7
of my 1993 American Society of Heating, Refrigeration and Air-Conditioning
Engineers Handbook of Fundamentals (ASHRAE HOF), inch-pound edition, indicates
that at 72F, an average Dairy cow, say a 1000 pound Jersey, stantioned in an
enclosed stable, produces about 2000 Btu/hr of latent heat, so with 5 cows
in the basement, the above house would only need R = (72F-32F) x 3,000ft^2/
13,000 Btu/hr = R9 walls. This could probably be less with free-range basement
cows. In either case, you might want a vapor barrier under the first floor.

Hope this helps.

Nick

In article <[email-masked]>,
Robert Vesalga <[email-masked]> wrote:
>Is there a simple answer to this question?
>
>If I was designing a house in a 32 degree climate, how much insulation 
>(in R- value) would it take to have no heat loss through the walls, while 
>maintaining a 72 degree interior ? ( assume this to be like a walk-in 
>freezer; doors, windows, infiltration, etc., notwithstanding. Just pure 
>wall).
>
>Would the floor and ceiling be the same ?
>
>I guess that's 2 questions.
>
>Bob V
>--
>

---

From: [email-masked] (Nick Pine)
Date: 27 Jun 1995 06:36:45 -0400

Will Stewart <[email-masked]> wrote:

>...The basic formula for heat transfer is;
>
>Q = -kA(T1-T2)/X
>
>Where
>    Q is the heat transfer rate
>    k is the thermal conductivity
>    T1 is the inside temp
>    T2 is the outside temp
>    X is the thickness of the insulating material

It looks like A must be the surface area of the wall, in ft^2, and perhaps X
would be measured in inches (?) Using a negative sign in the equation seems
an odd convention, for houses. Sorta makes sense, tho. Heat out of the house
is minus, and heat in, from an oil burner, the sun, etc., is plus.

>To apply this formula, one just needs some thermal conductivity values for
>various materials, as the following in Btu/hr*ft*degreeF

Seems like these are in Btu/hr-ft^2-F-INCH of thickness, not as above.
Then again, perhaps they are per foot... That would make sense for most
of the numbers in your left column below...
 
I've added some numbers to the right, for thermal conductivities, per inch,
from the ASHRAE Handbook of Fundamentals and other sources.

>Cork                .o4         0.29 
>Glass wool(fine)    .31 avg     0.32 (close, per INCH...)
>Masonry Brick       .38         5.0
>Glass              ~.45         31   (air films make insulation)
>Rock Wool           .017        0.25

Nick