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Old 10-24-09, 05:15 PM   #1
Piwoslaw
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Default Cooldown testing your house

Today I had a chance to see how fast the house cools down. Yesterday evening at 8pm the furnace turned off after heating the house to 19.5*C, after that it switched to night mode until 8am. In the morning it turned out that there won't be anyone in the house for most of the day, I'll just be coming and going a few times, so no use keeping the house warm. Set the thermostat to night temperature for the whole day. Wrote down the temperature in the dining room whenever I was back for a minute. At 4pm, after 20 hours, the indoor temperature had fallen to 17.5*C. The outdoor temperature was around 7-8*C the whole time, no wind, and very humid (light rain). I should add that the weather was exactly the same for the last 2-3 days. (This is important, as I've noticed that the thermal inertia of the house can take 2-4 days to adjust to weather changes.)

Here are the details:
HourTimeIndoor tempOutdoor temp
08:00pm19.57
117:00am187
12.58:30am187
1511:00am17.757.5
204:00pm17.58.5


So, with this data, can I get a value on how efficient my house's insulation is? Something like how coastdown testing gives you the drag coefficient of a vehicle, rolldown testing gives you its rolling resistance, etc.? Or would I need more data, like recording the temperature every 1-2h, repeating it 2-3 times on days with similar weather? Or maybe the 'cooldown' test should be repeated in different outdoor temps: 5*C, 0*C, -10*C?

This data is a good baseline for any large insulation projects. After any insulating work the cooldown test should be repeated in similar weather conditions and this should quantify the improvement. Just looking at your heating bills doesn't work that well, since each winter is different. Only if the project improved insulation by 10%-20% would it be noticeable compared to a multiyear average.

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Old 10-26-09, 07:16 AM   #2
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I think we could use it to compare houses. Of course, ambient weather conditions would need to be very similar for it to be a useful comparison. I agree, it would be very useful for large insulation projects. Of course, you could probably compare monthly bills too and use heating/cooling degree days to get kind of close.
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Old 10-26-09, 08:47 AM   #3
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Quote:
Originally Posted by Daox View Post
I think we could use it to compare houses.
That's exactly what I was thinking.

I first thought about hooking up a Kill-a-watt to the furnace. That would tell me how much ON time it has per day, week, etc., but this (and recording my gas meter every morning) would more likely help with finding the efficiency of the furnace and/or heating system as a whole.

The analogy with coastdown testing keeps floating around in the pool of scum that I call my head. What would be frontal area on a car is the total surface area of the house, through which heat can escape. This means walls, roof and floor. The best weather conditions would be thick clouds (minimal solar gain), no wind, almost constant temperature for as long as possible. Also, not being in the house for as long as possible helps (I've noticed that watching the 170 watt LCD for 30 minutes can raise the temperature in the room by 0.25*C). And, of course, the more cooldown tests go into an average, the better.

Among the differences are that aero drag depends on the square of speed (or, more generally, the dependance is polynomial), while the relation between heat loss and temperature difference is (inverse) exponential. Also, the heat loss through the floor will be different than through the walls and roof, since the ground's temperature doesn't change much. Could this be analogous to the rolling resistance component of drag? And how do cooldown tests with different outdoor temperatures compare?

Anyway, I think that there may be a way to compare the thermal efficiency of houses with two numbers, surface area and a number (thermal coefficient). More thinking, more thinking, more thinking... Darn, that hurts!!
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Old 10-27-09, 11:25 AM   #4
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Default More data!

Got another chance to do a cooldown. This time I had more time, but started with a lower temperature. Here's the data:

HourTimeIndoor tempOutdoor temp
09:00pm198.2
8.755:45am18.256.4
10.757:45am186.8
118:00am18.256.5
12.759:45am187.7
18.253:15pm17.758.7
24.759:45pm17.57.6
32.55:30am177.3


This test took over 32 hours, and the house cooled only 2 degrees, from 19*C to 17*C, while the outdoor weather was identical to the previous test. This shows that cooldown testing must be repeated a lot of times and averaged.

BTW: The raise in temperature at hour 11 is after the TV was on for 30 minutes.
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Old 10-27-09, 07:14 PM   #5
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Default Data (ghost in the machine)...

I did a lot of similar testing last winter.

I turned off my regular heat source, which was a forced air gas furnace, and instead used electric heaters with a kill-a-watt on each heater, logging data about every 4 hours. There were temp fluctuations at night, drifting down until the sun began to rise.

But when I graphed my data (kW/hr vs. Outdoor Temperature), I noticed that there was a much larger variation in kW per hour at various out door temperatures than I had anticipated.

I decided that the humidity of the air, which I was not measuring, plays a much bigger role in heating requirements than I had previously realized...

Regards,

-AC_Hacker
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Old 10-27-09, 10:34 PM   #6
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Default Data (ghost in the machine)...

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Last edited by AC_Hacker; 11-30-10 at 12:18 PM.. Reason: delete... (double post)
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Old 11-30-10, 07:23 AM   #7
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It sounds like you're close to measuring the Overall Heat Transfer Coefficient * Surface Area = U*A, which is a measure of the effectiveness of a heat exchanger. UA is the inverse of thermal resistance (Rth), i.e. the average R-value of the house.

You won't get there, though. There's too many variables that are beyond our ability to accurately nail down. Specifically, we don't want to measure change in temperature, but rather heat flux (q). If the house were to cool down uniformly, we could use q = m * specific heat * deltaT, but my observations indicate that different parts of my house have dramatically different temperatures, even when the furnace is off.

The biggest problem is the cooldown of your house isn't the only heat flux in play. You've got significant heat fluxes from solar and metabolic activity, plus your appliances giving off heat.

But if we want to turn "cooldown testing" into a fun little game, we could just compare our dT/dt vs ambient temperature, and make a note of whether it was sunny. I'll log some cooldown test results, but I can also tell you up front: Piwoslaw wins.
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Old 11-30-10, 03:29 PM   #8
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Quote:
Originally Posted by RobertSmalls View Post
but I can also tell you up front: Piwoslaw wins.
Only when Dad-in-law is gone for a 2-week vacation. When he's back at home I won't be able to get away with extreme cooling anymore
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Old 12-01-10, 05:32 PM   #9
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7 degree C outdoor temp is pretty warm(45f), I wasn't running my furnace then at all. It's a different story when it gets much colder though. It's 20f(-7c) outside right now and I lose a degree per hour from 60 degrees(f) going to down 50 degrees(f) with this temperature outside. Even moreso if I raise the temperature to 68f(20c). Just keeping the temperature at 50f in the house makes a gigantic difference versus 60f because of the ground sourced heat that comes up from my uninsulated basement(which is warmer than the upstairs and its heat rises).

Our dry bulb design temp for heating where I live is -20f(-29c) and I've seen it there or lower enough times to know that it makes a big difference. Doing some very crude and possibly very incorrect math(I'll be able to verify this year), I would lose 3 degrees(f) per hour(1.6c) average throughout January. Not sure what the average was for January last year but the worst days would have been plenty worse.
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Old 01-17-13, 09:21 PM   #10
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Default Thermal mass measurement

When we left for two days, I shut off the HRV. Unfortunately, my fat finger also shut off the boiler. The result was a golden opportunity to measure the thermal mass of the house.

House temperature when we left: 70 deg F.
House temperature when we got back: 50 deg F.
Average outside temperature: 16 deg F.
Calculated heat loss at -25 deg F: 13,000 BTUH.
Estimated average heat loss at average 60 deg inside, 16 deg F outside = (60 - 16) / (70 - -25) X 13,000 = 6,020 BTUH.
Total time without heat: 54.5 hours.
Total heat loss = 6020 X 54.5 = 328,000 BTU.
Thermal mass = 328,000 / 20 deg F = 16,400 BTU per deg F.

The house is 1294 square feet over a crawlspace. Construction is conventional stick built with normal drywall on the inside. Walls insulated R32, ceiling is R96, windows are Andersen 400 series, blower door tested at 0.85 ACH50.

Since the heating system was sized for the house, it can only raise the inside temperature at a rate of about one degree F per hour. Fortunately, we also have a wood stove.

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