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:

Hour	Time	Indoor temp	Outdoor temp
0	8:00pm	19.5	7
11	7:00am	18	7
12.5	8:30am	18	7
15	11:00am	17.75	7.5
20	4:00pm	17.5	8.5

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.

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.

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!!

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:

Hour	Time	Indoor temp	Outdoor temp
0	9:00pm	19	8.2
8.75	5:45am	18.25	6.4
10.75	7:45am	18	6.8
11	8:00am	18.25	6.5
12.75	9:45am	18	7.7
18.25	3:15pm	17.75	8.7
24.75	9:45pm	17.5	7.6
32.5	5:30am	17	7.3

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.

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

Data (ghost in the machine)...
 

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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.

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:(

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.

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.

A flaw that I see in your math is that your temperature doesn't drop linearly so your average would be false. If I drop the temperature for the house and let the temperature glide down when it is 10 degrees outside I will lose the first 10 degrees in a few hours, the next 10 degrees will take significantly longer and I won't be down to 40 degrees until over a day has passed. Your heating load is much higher at 70 degrees than it is down at 50 degrees. At some point you'll have more heat coming in through the sun than you are losing. With my house I usually have a balance point of 20 degrees higher than the overnight low. So if the overnight low is 20 degrees with sunny days with a high of mid 30's on both sides and that cycle continues my furnace won't run when it is set to 40 degrees because it won't get that cold.

Similarly if I fire my furnace up when it is 40 degrees, after an hour it is 52 degrees, hour 2 is about 60 degrees, hour 3 is 65 degrees, and hour 4 it will be at 69 degrees. If I try to heat it to 72 degrees it will get there close to the 5th hour since the house's thermal soak at 40 degrees is such a powerful force. The same is true during the cooldown, large differences at the beginning with lots of lag at the end.

I've found that when measuring my heat load that if I haven't had the temperature steady for a good 24 hours, the thermal lag has a big effect on the numbers, which would be expected but I never thought it would take that long to settle but it does. Sunday we are supposed to have a -16f temperature, I'll be measuring heat load with no sun. The spray foam saga has been continuing over the winter and I'm curious how much I've improved over my previous 285BTUhr per degree f figure. Still have more insulation work on top of that, next winter will be the real test.

Based on the information you've provided about your structure, I'd venture a guess that you'd see a bit less of a heat load than 13k in real life. Especially when I factor in the real world numbers for my house which has a much worse blower door figure and less than half of the wall and ceiling R values for insulation. I'm looking to get the ceiling to between R60-75 but at the moment it's less than half of even that.

True, the correct method would be to fit my measurements to an exponential curve. The error from my super simplified approach is smaller than the errors from my assumptions, so the result should be about as good as it can get.

I have 4 inches foam under the crawl space, so get very little heat from the ground when the house temperature drops. Even the footings have foam (high density foam) under them.

My house does not get solar gain. It faces west through trees, plus has large overhangs. Plus there is very little sun this time of year.

A cool down test is not a good way to assess the thermal performance of a house. To do it right, the outside temperature needs to be constant over several days, and then you have sunlight screwing up the readings. The decay curve (inside temp vs time) is an exponential and you need a lot of points to fit the curve.

A much better and easier way to determine house thermal property is to plot the furnace BTU output vs. the outside temp. This is done by recording the "on time" as a percent of the heater run time and plotting this against outside temp. Get temps at a variety of cold conditions.

The best data is obtained in the early AM with little human activity going on (lights, cooking, etc) and with the air blower on. Get values as cold as you can (midwinter is good for this) and then at about 5 degree F increments all the way up to 55 F or so.

The heater maximum output can be obtained from the listing plate on the side. For example, if it is a 50,000 BTU output heater, than a 25% run time is 12,500 BTU.

This will also determine if your heater is properly sized. The heater should run at 100% on the coldest of mornings. Best data are obtained when it is zero F outside.

Outside wind screws this up so try to get data without wind, but if it is windy (more than 8-10 mph) note it on the regression line.

With 10-15 points you get a very straight line and then this can be used to calculate whole house R values.

Add up total wall, floor and roof square footage. Add up all windows and estimate those at a nomimal R value (perhaps R 4 to maybe R 6). Same with doors (R 2-3). The roof insulation is easy to calculate as you can measure it as thickness. By subtracting these out, you can get the average R value of the walls and floor.

If the calculated R value is much less than what is in the walls, then there is an infiltration issue. This could be running the dryer (lots of kids) or simply lots of small air leaks (or the kids leaving the doors open!). This is why testing at 2-5 AM eliminates kids, dryers, etc.

Not hard to do, but for me I use a data logger and ask it to get heater run times from 3-5 AM. Outside temp with a simple thermometer.

This is one of the first things I do when evaluating a home before we recommend putting in a new (think expensive) GT heat pump. I know that every $ for insulation and to minimize infiltration is like $30-50 of a new system. Conservation first.

But then there are those ancient gas furnaces with efficiencies of 50-60% and these just gotta go. The key is to size them properly and most HVAC installers are all too willing to forget about insulation and then put in the largest system that they can convince the homeowner to install.

I have seen 5 and 6 ton units put in where 2 tons is what was needed.

Hope this helps!

Steve

A cool down test is not intended to measure the heat loss of a house. You need to know the heat loss first in order to do a cool down test.

For example, a poorly insulated earth sheltered house can still do well in a cool down test because it as truly impressive thermal mass. Conversely, a well insulated house can do poorly in a cool down test if it has little thermal mass.

JR
We may be talking of the same thing.

In order to assess the thermal performance of a house (or any system) you can do a steady state test (as I have described) or do a load test (turn off the heat in winter) and observe the thermal decay.

The steady state relationship is far better than a load test as you can measure very accurately a heat input to a house to keep it at a constant 70 F. For example, this might be 28,000 BTU at zero F outside (a known % of heater run time for example). But to make sure the data is correct, do this at many temperatures and the regression line eliminates a lot of error.

A heat load test (as above) in an earth sheltered home would reflect the amount of total BTU going through earth sheltered wall as well as windows doors, etc.

Or am I misunderstanding you?


Steve

A heat load test is used to verify a heat load calculation. The heat load is used to size heating / cooling systems, to estimate annual heating / cooling costs, and for calculating the ROI of insulating and air sealing.

A cool down test starts with the heat load. The rate of cooling, along with the heat load, can be used to estimate the thermal mass of the house. The thermal mass and heat load are used together to answer questions of the following type: If the furnace goes out on Friday night, and the temperature outside is 0 deg F, is this an emergency that requires paying overtime rates, or an inconvenience that can wait until Monday to get fixed?

Note that thermal mass has nothing to do with the thermal resistance. A house with a full inch of drywall on the inside will have higher thermal mass than a house with the normal half inch of drywall. Both houses, if otherwise identical, will have the same heat load. But the house with the higher thermal mass will cool down more slowly.

JR,

Re-read MN Renovators excellent reply. You can't use a linear approximation to an exponential decay even if it is "close" in the beginning. A small change in the slope of the curve at the start can have a HUGE calculation error as only small changes in the slope of the curve will derive large changes in the output calculation.

You are on the right track in order to estimate the thermal mass in your home, but the math is wrong.

Get a data logger, leave the house fan to "on" and turn everything off. The rate of inside house temperature decay is directly proprortional to the difference in inside to outside temperature (delta T). At first, the heat loss is rapid as the delta T is large. But the loss is not linear as the house temp approaches oputside temp where temperature loss (heat loss) is smaller and smaller as the delta T becomes less.

In most homes with geothermal systems we recommend a thermostat setting of "set it and forget it". There is an initial transient when the thermal mass is heated up, but after that the thermal mass is largely irrelevant.

We do check our heat loads by determining heat flux (q) using the steady state methodology I described above.

If you want to check thermal mass, heat the house up to about 100 F for a day to get the total thermal mass heated. Turn the house fan to on. Then on a cold day, turn off the heater (HRVs and all else) and record the decay curve on an hourly basis. An outside temp of 0 F is great as you have a delta t of 100 to start.

You will need a thermometer accurate to 0.1 degree F. It need not be accurate, but precision is what you need. Get data on the hour. About 24 hours later will get you past the rapid change in termperature and you will start to see the asymptote of the exponential curve.

Get the data, send it to me and I will be glad to do the exponential curve fits for you.

Steve