Assessment of heating concept KfW70

V ~ r³, A ~ r². V/A grows faster than A. The heat losses related to the volume become infinitely small for an infinitely large storage (this is the theoretical limiting case).

Yes and no, V/A grows with r (as you correctly calculate), but A also grows with r² (as you yourself say), which means the heat losses increase linearly with r.
So, with double the radius you have four times the amount, but also double the loss. The ratio is therefore better, but the absolute loss is still greater. (Assuming a cylinder, rough estimation with the mantle much larger than the lid surfaces)

An infinitely large storage unfortunately also has infinitely much surface area through which it loses heat, and how to compensate for infinitely large energy losses now... well XD

Yes, that's true, but be careful, in my opinion the interesting point is the losses related to the stored heat. The source term of the balance (the heat supplied electrically by the heat pump, which should be independent of the surface-volume ratio) is, in my opinion, irrelevant.

It does play a role because the absolute losses are larger. If you store 500L but only need 250L, then unfortunately the unused 250L also cool down and must be reheated. Despite greater efficiency, you lose more energy in absolute terms. The calculation only works if you really use the 500L more frequently. For the balance, in the end, (energy used)/(used hot water) counts, and with absolutely higher losses, you have to use more energy.

For example, going from 200L to 500L requires about a 1.6 times larger radius, meaning about 1.6 times the energy loss for 2.5 times the water volume. This additional energy is simply lost. Without any ifs or buts; also without "that is less energy per liter." If the storage cools down faster than you can use the hot water, a somewhat smaller storage would definitely be more efficient.
If you now calculate the height of the cylinder instead of the radius, the advantage is gone anyway because the surface area almost doubles with doubling volume (this is unfortunately often the real case; you do not place a small, thick tank often due to space reasons).

Does your heat pump have a suitable interface or do you have to improvise for that? Is there something like that for engineers (i.e., in Excel)?

There is both an interface and something to tinker with. For the power consumption, for example, I prefer a DIY solution rather than relying on the heat pump’s values.
I don’t know what you want to have in Excel now?

Frankly, I am not sure how the storage behaves with heat pumps. Certainly, the size of the storage must be adapted to the system (house), and there will probably be some kind of optimum.

Nevertheless, a comparison to the above point. If you buy two mulled wines. One is served in a 200-mL glass and the second is served in ten 20-mL shot glasses. And let's assume the temperature is the same everywhere at the beginning, the diameter-to-length ratio of all glasses (large and small) is the same, and the heat transfer coefficient of all glasses is also the same, then the mulled wine in the large glass will stay warm longer than the mulled wine distributed in the small glasses. Do you agree with that?

Whether I then understand 1) if a storage is even useful (BeHaElJa?), and 2) if yes, in which size (and actually for what: hot water or heating buffering), of course, is a different matter.

Heat pump and buffer tank are not necessarily best friends. From my own experience: the house is so sluggish that you can't just quickly turn the temperature up or down

As far as I know, the tank is not supposed to even out temperature differences in the house. Rather, the heat pump is supposed to, for example, heat during the day (high outside temperature) and then rest at night (low outside temperature). Heat is constantly drawn, at night from the tank, which is then refilled during the day. A simple example -- this is how I understand the function of the tank with the heat pump.

Nevertheless, a comparison regarding the above point. If you buy two mulled wines. One is served in a 200-mL glass and the second is served in ten 20-mL shot glasses. And let's assume the temperature at the beginning is the same everywhere, the diameter-length ratio of all glasses (large and small) is the same, and the heat transfer coefficient of all glasses is also the same, then the mulled wine in the large glass stays warm longer than the mulled wine distributed in the small glasses. Do you agree with that?

Yes, I agree. And now you drink 40ml of mulled wine and the next morning the remaining 160ml in the large glass are cold and no longer taste good. If you only ordered 2 shot glasses full of mulled wine, you have saved overall.
For the storage tanks, I briefly recalculated the real case with the sizes of real tanks: 500l with h of 15.7 dm and r = 3.25 dm and 200l with h = 11 dm and r = 2.5 dm. The large one here has 2.5 times the usable volume and 1.8 times the surface area. It can potentially keep the heat 38% longer, but it loses absolutely 1.8 times as much energy as the smaller tank in the same period. I believe the difficulty is the distinction between heat and heat energy. A larger tank keeps the heat longer but has absolutely greater energy losses. This means that any oversizing results in an increase in operating costs.

For heating, you do not need a buffer tank with underfloor heating. The screed stores far, far more energy than a mere 500l of water.
If you charge the storage tank at the supply temperature of the heating system, the energy is of course not enough to heat overnight when it is cold (and definitely too cold for hot water). If you heat the tank very hot with an air-water heat pump to store energy for the night, you have an incredibly poor coefficient of performance because the temperature lift is far too high. The heat pump works better directly in the heating circuit, even if it is 10 degrees colder at night in winter than during the day, the supply temperature of the underfloor heating is certainly more than 10 degrees lower than the required charging temperature of the tank (underfloor heating maximum!! 35 degrees, tank well above 50 degrees so that the supply temperature can be maintained all night).

And now you drink 40ml of mulled wine and the next morning the remaining 160ml in the large glass are cold and no longer taste good. If you have only ordered 2 shot glasses full of mulled wine, you have saved overall.

It seems to me we are talking past each other a bit. You neglect that heat is also supplied to the storage, the heating keeps it at temperature and compensates for losses. In another respect you are right: if I only want or need 2 shot glasses (2 x 20 mL), then I only order these two and not five times the amount. (Here it is only a matter of choosing the optimal size of the storage; it will certainly depend on the circumstances ...). If I want the schnapps to stay warm as long as possible, I will prefer one (!) glass of 40 mL, but not two (!) glasses of 20 mL. Explanation: see above.

For heating with underfloor heating, you don’t need a buffer tank. [...]

Thank you, that makes sense to me! I looked again in the brochure, the storage from Rotex is primarily for hot water, but can also be used for heating support. What exactly then happens will probably depend on the control system, which of course is not known to me.

Source: Rotex brochure

It seems to me that we are talking past each other a bit.

It seems that way to me too with the "past each other." You sometimes talk about heat as temperature, I talk about heat as energy.

You neglect that heat is also supplied to the storage, the heating keeps it at temperature and compensates for losses.

You are neglecting that right now! The losses, as I calculated above, are greater with a larger storage! A 2.5 times larger storage requires 1.8 times the amount of heat to compensate for losses and maintain the temperature!

You are right about another point: If I only want or need 2 shot glasses (2 x 20 mL), then I only order those two and not five times the amount. (Here, it is only about selecting the optimal size of the storage, it will certainly depend on the circumstances ...). If I want the schnapps to stay warm as long as possible, I will prefer one (!) glass of 40 mL, but not two (!) glasses of 20 mL. Explanation: see above.

That may be so. But the metaphor does not take into account that the heat pump can supply hot water anytime with the same efficiency. The large storage keeps the temperature (careful, don't confuse now!) longer, but pays for this absolutely with greater heat energy losses! If it is still not clear to you why large storages do not pay off at all, I can calculate it for you exactly with the energy amounts for heating, with times, with losses, with temperatures.

Thanks, that makes sense to me! I looked again at the brochure, the Rotex storage is primarily for hot water but can also be used for heating support. What exactly happens then will presumably depend on the control system, which of course is not known to me.

The "problem" is always that when storing hot water you inevitably lose energy. That means for underfloor heating and heat pumps that using storage for heating is actually almost always less effective than the direct use of the heat pump. The only exception might be small, short cold peaks in which you then use heat from the storage instead of the heating element (this will rarely occur). But if the cold peak lasts longer or you want hot water, you lose efficiency again because the storage then has to be reheated at the same source temperature. Another problem is the "stratification" of hot water (warm at the top, cooler at the bottom). I can imagine that this is disturbed by heating operation so that the storage must be recharged more often than with pure hot water use.