ᐅ Air-to-Water Heat Pump – With or Without a Buffer Tank?

netuser schrieb:

What do you mean by "too little electricity" in numerical terms? How much does the 300L (80 gallon) storage tank require per day?

Taking one comfortable shower per day (one person, or two people but not for too long), then about 1.1–1.5 kWh daily. Without use, about 0.8 kWh every 1.5 days. Heating the tank to 30°C (86°F) from empty consumes 3 kWh (four people showering for a long time).

Benutzer200 schrieb:

Taking a comfortable shower once a day (one person or two people not for too long) uses about 1.1–1.5 kWh daily. Without use, around every 1.5 days it consumes about 0.8 kWh. Drawing the tank down to 30 degrees Celsius (86°F) uses 3 kWh (4 people showering very long).

Okay, maybe my understanding is off if it is really as you say.
Currently, I use 3–6 kWh for hot water (190 L (50 gallons) storage tank) and assume it would be more with a 300 L (79 gallons) tank.
Every kWh heated by photovoltaic power saves actual money, and not a small amount considering current and future prices. That’s why I mentioned it would be more advantageous to schedule the heating of the larger tank during photovoltaic production times. Even with your mentioned 3 kWh, that’s at least about 0.6–1 EUR savings per day compared to grid electricity. This adds up to 20–30 EUR per month or 200–300 EUR per year just by shifting the heating time to hours when photovoltaic energy is available.

netuser schrieb:

Every kWh heated with photovoltaic power saves real money, and not just a little given the current and future prices. That’s why I mentioned it would be more advantageous to schedule the heating of the larger storage during photovoltaic production hours. Even with the 3 kWh you mentioned, you’d save at least about 0.6 to 1 EUR per day compared to grid electricity. That amounts to 20-30 EUR per month or 200-300 EUR per year just by shifting the heating time to the photovoltaic-related time slot.

That is the land, correct. But a) you don’t fully deplete the storage every day, so the savings will be much lower, and b) this doesn’t work in winter, which reduces the savings further. By the way, photovoltaic electricity also costs money—no feed-in tariff applies, plus there are costs (tax-related and in the profit and loss statement).

I’m calculating rather 1.5 kWh (kilowatt-hours) per day × 365 = about 550 kWh. About half (?) of this would be covered by photovoltaic electricity, so 275 kWh. At 35 cents per kWh electricity cost, that’s a saving of 96 EUR, but you lose the feed-in tariff (17 EUR) and incur self-consumption costs (another 20 EUR), leaving a monthly profit of 5 EUR.

Now calculating this with my current electricity cost of 25 cents per kWh, the result is even smaller. Or you have heat pump electricity, which may make sense in some cases.

Benutzer200 schrieb:

That is the plot of land, correct. But a) you don’t empty the tank completely every day, so the savings are significantly lower.

Somehow we are unfortunately talking past each other 🙂
Your argument in favor of a larger tank was that the smaller one isn’t sufficient. Therefore, my assumption and calculation were based on the idea that you could start every day with an empty tank.
Also, my point was about the energy difference required for the larger tank compared to the smaller one... I don’t know this value, so I am trying to understand what difference could realistically be expected here.

Benutzer200 schrieb:

and b) this doesn’t work in winter, which further reduces the savings. By the way, photovoltaic electricity also costs money – no feed-in tariff plus costs (tax and in the profit and loss statement).

These costs are already included in my 20 cent difference (currently the cheapest electricity price minus feed-in tariff).
Taxes, etc. are of course omitted; this is meant to be a rough overview that ultimately varies greatly for each individual. This assumption probably even represents a conservative estimate for most people.

Benutzer200 schrieb:

I calculate rather 1.5kWh per day x 365 = approximately 550kWh. Of that, you use half (?) from photovoltaic electricity, so 275kWh. Savings at 35 cents electricity cost are then €96, but you lose the feed-in tariff (€17) and have costs for self-consumption (another €20), leaving a monthly profit of €5.

Do you actually use only 1.5kWh per day for domestic hot water? Or are you only referring to the difference related to varying tank sizes?

netuser schrieb:

Somehow, we are unfortunately talking past each other 🙂

True 😉

netuser schrieb:

Your argument in favor of a large storage tank was that the smaller one is not sufficient.

Yep. The smaller tank is not sufficient. But not every day. Maybe it’s enough for half or two-thirds of the year, but unfortunately often not.

netuser schrieb:

Do you really only use 1.5 kWh per day for hot water?

Yes. That’s the absolute average consumption. It was more of a rough estimate, but last week it was actually about 11 kWh (I just checked).

netuser schrieb:

Or are you only referring to the difference based on different storage tank sizes?

No, not the difference. Maybe the difference is 0.5 kWh (proportionally calculated based on size).
If you focus only on the difference, the advantage of a photovoltaic system naturally appears even smaller.

Of course, the photovoltaic system reduces the grid electricity demand. But hot water heating is only a small part of that.

Benutzer200 schrieb:

Of course, the photovoltaic system reduces grid consumption. But water heating is only a small part.

Okay, I understand.
However, in my case, the daily hot water consumption is around 4 kWh. I still need to figure out how you manage 1.5 kWh even with the larger storage tank 🙂

Most of my 4 kWh/day was previously used in the mornings and evenings, outside of the photovoltaic production times.
Now, I have shifted the water heating periods to align with the photovoltaic active hours, which on average works just as well, while achieving a clear energy saving. In your case, it would indeed be negligible, although I still don’t fully understand the 1.5 kWh 🙂