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Posted by nicksanspam on March 12, 2008, 8:17 pm
 


Sure it can :-)


Model it separately. Its standby heat loss helps warm the house on an average
day, and it can keep the house warm for a few cloudy days with a distribution
system.

Nick


Posted by Robert Scott on March 12, 2008, 6:43 pm
 
On 12 Mar 2008 15:17:01 -0500, nicksanspam@ece.villanova.edu wrote:


Yes, you can model it separately, in which case you get a separate time-constant
for those components of the house.  In my example I considered a fairly simple
case of only two thermal masses - the water tank and everything else.  In
practice, all the masses in the house have potentially their own separate time
constant, which can be modelled separately.  All I was trying to show is that
there is no one time constant that can be said to stand for the whole house.
The temperature decay graph for a given spot in the house will not conform to

  Temp0 * e ^ (-t/TC)

for any value of TC.  
Robert Scott
Ypsilanti, Michigan

Posted by nicksanspam on March 12, 2008, 9:29 pm
 

Sounds good to me. I figure the house cools like this overnight:


to a constant outdoor temp, and I figure the tank cools from an average
tank temp to an average house temp, which can be the same for a few
cloudy days until the tank runs out of useful energy. This seems to
agree well with more calculus-intensive solutions and simulations.

And it works a lot better than architectural rules of thumb.

Nick


Posted by daestrom on March 13, 2008, 12:24 am
 

Quite right.  If you plot the room temperature versus time on semi-log
graph, this shows up as a curved line.  Whereas a 'simple time constant'
should be a straight line with slope -1/TC.

Quite often this can be accomodated by using two or three different time
constants, for small, medium and large temperature swings.  For the small,
short duration changes, basically ignore the '900 lb gorilla' (I mean 'water
tank') and use the slope from the very first portion of the line.  For part
of a day or hourly changes, this would probably be okay.

For long term calculations like a couple of days without sun or some such,
then you can't really ignore the 'water tank' and the math takes on another
degree of complexity.  But at least the watertank and room are coupled in
both directions :-)

daestrom


Posted by daestrom on March 13, 2008, 12:18 am
 

Rather than calculate it, it can be measured.

You simply pick a nice cold/calm/cloudy day and turn off the heat and plot
the temperature every 10 minutes or so for as long as your wife can stand it
:-).  Knowing the inside/outside temperatures, you can get a pretty nice
plot of temperature versus time.  When the temperature has reached 63% of
the way from starting point towards outside temperature, you've reached one
time-constant.

For example, turn off furnace when house is 70F and outdoor temperature is
40F (30F lower).  63% of the way would be 19F so when the house is down 19F
from 70F (51F), you've reached one time-constant.

If you take your temperature readings and plot on semi-log graph, ideally
you would get a straight line.  Any curvature towards the horizontal with
time is an indication of either an unidentified heat source that will change
the 'final' temperature (such as yourself, or a dog, or a phantom electric
load), or some second-order effects due to a low thermal diffusivity between
a large capacitance and the measurement space.

If you then simply figure out the heat input over a period of several hours
needed to keep the house at a fixed temperature and divide by the difference
between indoor and outdoor temperature.  With that, you can calculate the
'R' for the house.  And knowing the time-constant and 'R' you can find 'C'
(thermal capacitance) if you still feel you need that parameter.

About as accurate as adding/estimating God knows how many slabs, walls,
fixtures IMHO.

If you have a long time-constant, you may have to estimate it from a smaller
temperature drop.  For example, I found that shutting off my furnace at 2130
hours on a nice calm winter night with the house at 70F and the outdoors at
18F, the temperature is still 63F at 0530 hours the next morning.  So with a
52F difference and only a 7F drop in 8 hours...

(7/52) = 1 - exp(-8/TC)

TC = -8 / ln(1-7/52) = 55.3 hours

{This ignores the heat input from the small wall-warts and refrigerator.
But with everyone snug in their beds upstairs, the downstair temperature
seems like a reasonable measure of temperature drop.}

Of course, the shorter your time span, or the smaller your temperature
change, the less accurate the results.  And if the outside air temperature
changes wildly in the night, well, all bets are off then.  How extreme you
get in the details will determine how accurate a result you get.  You could
even stand outside in the snow and read the thermometer through the window
if you want to rule out body heat as interference, but *I* didn't go that
far :-)

daestrom


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