Cooldown testing your house

[MN Renovator]

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.

[JRMichler]

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.

[stevehull]

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

[JRMichler]

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.

[stevehull]

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

[JRMichler]

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.

[stevehull]

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