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I have a home with a heat pump on the south-facing side of my home, so it gets sun all day. I let my home cool off to 62℉ at night, and at 6 am, when it is usually the coldest outside, I warm my home back up to 70℉ for human comfort.

I'm wondering if I could save energy (and how much) by warming my home up to 74℉ or so during the hottest part of the day when the heat pump will be more efficient and letting my home store that energy in its thermal mass. What would be the formulas or heuristics for calculating this?

I know this would have to be balanced against the increased heat loss from the higher temperature gradient across my exterior walls, but I don't know what the trade-off would be.

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  • $\begingroup$ So analyse power in compared to power out. calculate all the losses roof, walls, windows, doors - also opening doors etc Then evaluate the passive solar gain plus the heat input from boiler / heating plus the heat from warm bodies (humans, pets). $\endgroup$
    – Solar Mike
    Nov 10, 2023 at 15:54
  • $\begingroup$ the thing to remember about temperatures when used in efficiency is that they are on the absolute scale, which means the day to night temp difference isn't as much as you would think . Freezing to 25C is less than a 10% temperature change. $\endgroup$
    – Tiger Guy
    Nov 13, 2023 at 18:06
  • $\begingroup$ Efficiency isn't on the absolute scale for heat pumps @TigerGuy, it is "0" at the point the the heat pump doesn't work any more, not at absolute 0. $\endgroup$
    – cjbarth
    Nov 15, 2023 at 1:41

2 Answers 2

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What you are planning to do is very much depended on how you operate the house, the insullation and to a small extent to the thermal mass of the house).  

If for example you let the doors open for fresh air several hours a day or over-night, then heating up the house to 74 degrees will probably not have any effect.

From your description, it is my understanding that the way you are operating the heat pump is either optimal or close to optimal (all things being equal).

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  1. Since the heat loss through the walls remains nearly constant at that small difference in temperature, we can assume an idealized situation.
  2. The only variables that we need to account for are 1) the coefficient of performance (COP) for the heat pump at each of the two temperatures (midday high, evening low), and 2) the thermal mass.
  • The COP for my SEER2 7.4 system is as follows:
    • 3.92 at 60℉
    • 2.81 at 30℉
    • 2.34 at 10℉
  • The thermal mass of the home.
    • By some calculations I have about 2,520 lbs. of drywall in my house, calculating 40 lbs. per 4' x 8' sheet of drywall. If we say the specific heat of gypsum is 1,100 J/(kg·℃), which is rounded up just a little to account for some wood in the house which I didn't factor into the mass, then it takes 1.54 kWh to heat that mass. Add in the mass of the air at about 1,000 J/(kg·℃) and we get another 1,200 lbs. of air with 0.68 kWh to heat it.

Total energy to heat all that thermal mass 8 ℉: 2.22 kWh

If we then figure how much it takes to heat that mass at the various efficiencies, we can get the cost savings @ 10.5¢/kWh for electricity.

  • 1.54 kWh / 3.92 * 0.105 = $0.041

  • 1.54 kWh / 2.81 * 0.105 = $0.058

  • 1.54 kWh / 2.35 * 0.105 = $0.069

Basically, the savings is just the effort to heat that thermal mass the Δ℉ at the higher COP minus the cost to heat that thermal mass the Δ℉ at the lower COP because the thermal mass will radiate back into the room after the heat turns off and there is negligible change in heat lost due to increased internal temperatures over this short window.

It is surprising that the savings are so low, but it is savings. Also, my heat pump sits in the sun, so there are likely benefits to efficiency that I'm not calculating in there. As an added bonus, my family likes the fact that it gets a little warmer in here.

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  • $\begingroup$ If you're going to downvote, please let me know why. $\endgroup$
    – cjbarth
    Nov 15, 2023 at 15:41

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