Newly built photovoltaic semi-detached house - offer comparison

Upgrading the system every few years doesn’t make sense, and installing as much on the roof as possible is something you can do, but you don’t have to. Because a profitable system size to get a significant feed-in tariff won’t work with that roof at all.

The electricity price will rise much further. Maybe to 50 - 60 cents. Today we are already heading towards 40 cents. You feed in at 6.8 cents and buy back at 35 cents. That just doesn’t make any sense.

Calculate about 8000 kWh electricity with a heat pump at your place. In a house, you typically consume more electricity than in an apartment. At a current electricity price of 30 cents, many pay significantly more, that would be 2400 euros for electricity per year.

A 10 kWp photovoltaic system produces about 10,000 kWh electricity per year. Without a battery, you can consume about 3000 kWh yourself. With a battery, you can consume about 6000 kWh yourself.

A 10 kW battery costs about 5500 euros. Where is the problem, after 6 years it has already paid off. But since you can expect drastically further electricity price increases, keyword electrification of vehicles and heating systems, the battery will become profitable much sooner.

And 300 kWh feed-in tariff is only 200 euros per year. That cancels out any small increase in electricity price again.

In numbers:

Without photovoltaics, you pay 2400 euros to your EVN. With photovoltaics without battery, 1500 euros still go to your EVN. With photovoltaics and battery (10kW), 600 euros still go to your electricity supplier.

I am always for the battery and have one myself. That way you remain at least largely protected from further electricity price increases.

I used to believe that if I put a photovoltaic system on my roof, I would get 70 - 80% for self-consumption (also with a heat pump).

Total misconception. Winter result of a 10 kW system with SW orientation: 256/151/170 kWh (11/12/01). When the heat pump is running properly, the lights are on, and the electric car is often connected to the wallbox, not much energy is left. The yields mainly come from the midday period on sunny days and not spread over the day.

For a south-facing roof without storage, I find 30% quite decent.

That may be true, but that's not my approach or my question. My question was aimed at how to achieve higher self-sufficiency, with a lot of kW without storage or with less kW with storage.

That cannot be answered in general. A significant factor is the roof orientation and the usage profile. With the roof orientation, you already have relatively even electricity production throughout the day. The big question now is how much electricity your printing machines approximately consume. For a storage system to be worthwhile, there must be a period in which you have a photovoltaic surplus. If your machines have a load in the order of magnitude of the photovoltaic output, then I would be rather skeptical.

This cannot be answered in general terms. The big question now is how much electricity your printing machines approximately consume...

These are all machines connected to normal lighting current. These are cutting plotters and a heat press that presses foil onto textiles at 150 - 200 degrees. I estimate that the plotter requires 0.5 kW and the press 2.0 kW. As I said, the machines only run during the day, and they are already doing so in the rental apartment (annual consumption 4,000 kW).

Now let's apply this to my example: 4,000 kW consumption per year (normal household consumption day/night) ... let's say 5,000 kW + 3,000 kW heat pump per year = 7,000 - 8,000 kW annual consumption

7,000 - 8,000 kWh p.a. sounds realistic. About one third of that is for machines running during the day. The following rough estimate:

[*]With an 18 kWp system you will probably reach 40-45% self-sufficiency without a storage system. So roughly 3,500 kWh self-consumption (at 25 cents = €875 p.a.) and 14,500 kWh fed into the grid (at 6.5 cents = €942 p.a.). Totals €1,817 p.a..
[*]With 10 kWp with storage you might reach 60% self-sufficiency. So about 4,500 kWh self-consumption (= €1,125 p.a.) and 4,600 kWh fed into the grid (= €299 p.a.). In total €1,424 p.a.
[*]With 18 kWp with storage you reach about 70% self-sufficiency. So about 5,500 kWh self-consumption (= €1,375 p.a.) and 11,400 kWh fed into the grid (= €741 p.a.). In total €2,116 p.a.

These are rough values and I only considered taxes by assuming the kWh self-consumption at just 25 cents. I always calculated the yield as 1,000 kWh per 1 kWp, and for the storage values I reduced the feed-in by 20% of the self-consumption (these are battery charge/discharge losses). At least the calculation logic is correct, and you can easily make an Excel sheet where you can play with the parameters to get a range of expectations.

Basically: a storage system is useless if the photovoltaic system is too small to fill it. This applies especially in the dark season. And the profitability of photovoltaics currently depends on the purchase price. The kWp should not cost more than €1,400, and a benevolent solar installer should be able to offer 18 kWp for €1,200/kWp.