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I know enough to be dangerous, in that it makes me think there should be a way to do this. I was reading a paper This Thermal House. And one thing kind of stuck in my mind when I read about the circuit equivalent. In electronics, we sometimes simplify a circuit to do things like calculate the battery equivalent, etc. Can the same be done with a house and calculating the R-value. If I know the temperature change over time inside and outside the house, could I calculate an average R-value estimate? Part of my thought process here is that I could put the super fancy R-2000 window in. But if I leave the window open all the time, I'm sure I've kind of blown the R-value.

Thanks

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  • 1
    Have the basic idea right, but most of it has already been done for us. Unlike circuits, R-value for a house would have many changing variables.
    – crip659
    Sep 16 at 19:04
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Here's an example using my own house, a ~50'x25' rancher.

I used 2800 Kw-hrs of electricity for heating (heat pump) over a 30 day period of 868 deg days. That converts to 9.6x10^6 BTUs. I am going to assume that the overall efficiency of the heat pump was 1 over that period, which had a average outside temperature of 36 deg F.

Dividing by 868 gives me 11,000 BTUs for 1 deg day, or 458 BTU's/hr-deg day.

My 50'x25' house translates to roughly 2,450 sq ft of surface area exposed to the outside, walls (with windows & doors) plus ceiling/roof.

458/2,450 = 0.1872 BTU/(h-ft^2-degF), which is the U factor. Since R=1/U, R=5.3. This is the estimated average R value for my house, and seems reasonably given that fact that the walls average R-15, the ceiling is R-30, windows are something like R-2 or R-3, and then there's the unknown air infiltration amount. At least it passes the giggle test, at least IMM.

Edit 1

Note that there are a number of assumptions in this analysis, and changing any of them could significantly affect the results.

For example, I assumed my heat pump efficiency was 1. If I assume an efficiency of 2 and keep all the other parameters the same,

  1. The BTU's into the house would double.

  2. Which means the heat loss would double.

  3. Which means the effective R value would decrease by half.

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  • 5.3 is a severe drop from the mode of 15. Do you attribute it all to airflow?
    – isherwood
    Sep 17 at 12:38
  • And windows. And doors being opened and closed, which would be air flow, I guess.
    – SteveSh
    Sep 17 at 12:40
  • I guess I didn't mean "all". Windows are typically only around maybe 15% of a home's exterior surface, though. They'd account for just a couple points.
    – isherwood
    Sep 17 at 12:43
  • Right. But given that a good double pane windows has an R-value of 2.5, as opposed to an R-value of 15 or 20 (new construction) for a wall, that 15% number can have a significant impact on heat loss.
    – SteveSh
    Sep 17 at 12:53
  • Thank you. That gives me some idea about what to look up. I was thinking about finding the R only as a reference. And maybe that isn't the right thing to do. I might rethink my question and maybe there is a better answer. Sep 26 at 0:43
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There's one big flaw in this strategy, which is that air leaks completely foil the exercise. R is the resistance to heat conduction, not the overall efficiency. Airflow is not conduction. If you have a leaky fireplace or bad windows or poor seals around all your outlets or you open doors during the test your calculation will be wildly skewed.

A better approach to deciding how much to spend on a window would be to consider the typical building practices of the home based on age, as well as the type and condition of the rest of the windows. Shoot for say a 25%-50% improvement over what's there to get the most bang-for-buck price point. You'll never see payoff for a window that's 300% more efficient than everything else in the house.

That said, do you really not know anything about your house? You can see the windows and doors. You can easily determine wall thickness, and you should be able to get a peek at vapor barriers and insulation around an outlet box, for example, and in your attic. Once you figure that stuff out, data for your scenario is readily available. That seems like a much more reliable way to assess efficiency potential to me. I say potential because of the aforementioned airflow factor. The best thing most homeowners can do to reduce energy costs and carbon footprint is to stop air leaks. Vastly more heat is lost (or gained) that way than through conduction.

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  • Thank you isherwood. Actually, I know very little about my house as I just bought it (first-time homeowner). But I wasn't asking the question so much as to how to improve the house. I'm trying to learn new things without taking an entire course on thermodynamics. My primary thought was something like, if it is -35°C outside and I want to maintain 21°C inside and my furnace can provide x BTU (or KWH or etc.). If I know the average R value of my house how much BTU will I need to maintain the temp. Also maybe know what the duty cycle is. Sep 26 at 0:51

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