Thermal Mass equation help needed

[d...@cortlandfootball.com]

Hello everyone.

Through a combination of passive solar heating during the afternoon,
and a wood fireplace insert from late afternoon into mid evening, I am
able to maintain an early evening living room temperature of 72F with
ease. As the fire burns out overnight, I typically find the living
room temperature around 61F in the morning.

So we are experiencing at a drop of 11F (72F-61F) over roughly 8 hours
(10 pm to 6 am).

My goal is to reduce the overnight temperature swings. A minimum of
around 65F in mid November would suffice. I estimate this would
equate to living room minimum temp around 61F in late January, which I
can certainly live with.

My initial plan is to add 40 lbs of water storage in the vicinity of
the fireplace insert. Not close enough to melt the plastic of course,
but maybe 5 feet away off to the side.

I'd appreciate some help in estimating the amount of difference this
would make in compressing the overnight swings, and if I may need to
add more thermal mass. Thank you in advance!

[d...@cortlandfootball.com]

Correction...I meant to say 40 gallons of water (332 lbs)

[nicks...@ece.villanova.edu]

<d...@cortlandfootball.com> wrote:

A pound of water releases 1 Btu of energy as it cools 1 F, so 40 pounds
cooling from (say) 120 to 70 would store (120-70)40 = 2000 Btu. Burning
a pound of dry wood at (say) 60% efficiency releases about 0.6x10K = 6K
Btu, so the water might store the heat equivalent of 1/3 pound of wood.

Nick

[daestrom]

You know how much the temperature drops (11F) over a given time (8 hours).
If we knew what the outside temperature was on such a night, we could
calculate a time-constant for your current situation.

But that's not really enough. Similar to an electronic time-constant, a
thermal time constant is the product of two terms, the thermal capacitance
of the heated space and the thermal resistance between the heated space and
ambient. To get your particular time constant, you could have a wide range
of thermal capacitances, each one would simply be associated with a
different thermal resistance. Think of it like finding two factors that
when multiplied give you 8. They could be 2 and 4, 4 and 2, 1 and 8, or
even 0.5 and 16. If we can't peg at least one factor, we have no idea the
'ballpark' of the other. And hence, no way to estimate the effects of
adding to one of the terms.

You can get an estimate of one of these factors, the thermal resistance, by
noting how much energy you use to keep the place warm for some period of
time. If you can estimate how much wood you burn over a week (along with
the average outdoor temperature), we can figure out what the capacitance is
(once the time-constant itself is determined).

Chances are, adding a few pounds of water won't materially affect the
overnight temperature drop. You might be better off looking for some more
places to insulate or seal out the cold air. This would have the additional
benefit of lowering how much wood you need everyday.

daestrom

[d...@cortlandfootball.com]

Thanks guys. I guessed there would be too many factors involved to
come up with a strong estimate, so I'll just experiment and track the
max/mins downstairs. I'll start with 20 gallons of water (166 lbs)
heated down wind of the fireplace, and look to add another 10-20
gallons in front of our main solar collection area (large southwest
window).

[nicks...@ece.villanova.edu]

<d...@cortlandfootball.com> wrote:

>Thanks guys. I guessed there would be too many factors involved to

>come up with a strong estimate...

Daestrom suggests a way:

>You can get an estimate of one of these factors, the thermal resistance, by
>noting how much energy you use to keep the place warm for some period of
>time. If you can estimate how much wood you burn over a week (along with
>the average outdoor temperature), we can figure out what the capacitance is
>(once the time-constant itself is determined).

If the average indoor and outdoor temps are (say) 65 and 40 F and you burn
210 lb/week of wood at 60% efficiency, that makes the house conductance G
= 0.6x10Kx210/(7dx24h)/(65F-40F) = 300 Btu/h-F. If the house cools from 72
to 61 in 18 hours on a 40 F night, RC = C/G = -18/ln((61-40)/(72-40))
= 42.7 hours, so C = 42.7hx300Btu/h-F = 12.8K Btu/F.

Cooling from 72 to 61 on a 30 F night makes RC = -18/(ln((61-30)/(72-30))
= 59.3 hours, so C = 59.3x300 = 17.8K Btu/F, an increase of 5K Btu/F, eg
5K pounds or 600 gallons of water, or less, if the water cools from (say)
100 vs 72 F, or still again less, if the house temp instantly drops from
72 to 61, where it stays for 18 hours, with controls to regulate the heat
release from the mass to maintain a constant vs drooping house temp.

Nick