Assessment of heating concept KfW70

He is right. I should not write late at night - or when I reread why I am writing something

I could also have mentioned that the storage volume and the domestic hot water are closed (not open) systems. The brochure actually calls it a hygiene storage and Grym's description is AFAIK absolutely accurate.

But now the question of whether this storage is unfavorable in connection with a heat pump or not is still open. (For now, I don't see any serious disadvantages, but that doesn't have to mean anything.)

Also, whether the Rotex HPSU Compact also uses a (this?) storage for the space heating function is still open to me.

Large tanks naturally have greater heat energy losses. If you don't need a lot of hot water every day, 500l is not the most economical option (even with showering, a 200l tank with one charge per day is enough for us; when the children are older, maybe a second one will be added). Even if heat remains in the building envelope, oversizing due to the higher charging temperature is not efficient.

Large storage tanks naturally have greater heat energy losses.

than small tanks? The larger the tank, the better the surface-to-volume ratio, so fewer losses occur. Or am I mistaken?

What kind of device do you have? How do you know how often the tank is "charged"? When is the tank charged?

as small storage tanks? The bigger the storage tank, the better the surface-to-volume ratio, so fewer losses occur. Or am I mistaken?

What kind of device do you have? How do you know how often the storage tank is "charged"? When is the storage tank charged?

The ratio gets better, but you naturally have much more surface area! That increases faster than the ratio improves, otherwise beyond a certain size you wouldn't lose any energy at all. The absolute losses increase with size. Therefore, the storage tank should be as small as possible without causing any loss of comfort. That, of course, depends individually on the number of people and showering/bathing habits. I have a brine heat pump with the mentioned storage size. I log the operation of the heat pump on a Raspberry Pi and therefore know when and how often it is charged. I haven't set a fixed time; it results from consumption and hysteresis for me.

The ratio gets better, but of course you have a lot more surface area! That naturally increases faster than the ratio improves, otherwise, beyond a certain size, you wouldn't lose any energy anymore.

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

Absolute losses increase with size.

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

Background: I'll try it textually: dQ/dT = k.A.dT, where dQ/dt = cp.rho.V.dT/dt; after integration, this results in (Tend - Tstart)/(Toutside - Tstart) = exp(-k.A.t / rho.cp.V). That means the time t after which the contents have assumed the ambient temperature depends on V/A. And V/A gets larger the bigger the container is.

I log the operation of the heat pump on a Rasp.pi

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