Tap water heating through a heat pump is an additional possibility for saving money. But it also adds to the complications in the heat pump design to a high degree:
- The temperature needed for tap water is a lot higher than for heating (minimum 60 degrees, compared to maybe 45 degrees)
- The Legionella germ fighting demands the tap water to be increased to even higher degrees now and then
- The above leads to that tap water heating circulation should most times of year be separate from heating water circulation
- Tap water is, in difference from the heating water, used. Meaning it need a level of isolation from the heat pump
All the above leads to that the tap water heating business case should be treated as separate from the heating business case. If the heating business case does not lead to a solution with an accumulator tank or similar, the tap water investment might need to be large (some kind of tank with separate apartments for heat pump water and tap water, ...).
To decide the possibility to profit from an additional investment in tap water heating through a heat pump, we need to establish the approximate possible saving.
In my case, the tap water heating todays costs 5500 KWH per year. Partly due to old regulator technology (always keeping the water at 70 degrees), partly due to the heating being done completely without the aid of the existing heat pump.
The possible savings are the savings coming from heating the water with the heat pump, only adding the additional heat needed to get the water to desired legionella free temperature, or to enough to produce the needed warm water if heat pump is not enough on its own..
If the heat pump can deliver water that is 55 degrees, and during the whole year, then without any other changes, the remaining needed heating will be 15 additional degrees. If we assume that the current heating to 70 degrees starts from 10 degree water, then that would mean (15/(70-10) of current consumption). The heat pump will take care of the rest of the heating, calculated using the COP of the heat pump. This means, assuming an average COP of 2.5, that:
- Remaning additional heating=15/(70-10)*5500=1200
- Heat pump heating=(5500-1200)/2.5=1700
The possible saving is thus roughly calculated as 5500-1200-1700=2600 KWH per year.
Investments in the needed accumulator tank alone would probably exceed 20000 SEK, meaning that the possible payback time is about 8 years, probably more when including some switches and other similar appliances.
With some small changes in the calculation, based on some small changes in the regulation of water temperature, we can improve on this potential.
Assume we are satisified with the 55 degrees achieved by the heat pump, except once a week when we push the temperature to 80 degrees to kill Legionella germs. The heating needed for the weekly push would be 25 degrees of all the water in the tank. Assuming a tank volume of 150 liters, the extra push would cost 1.16*150*25=4.35 kwh. Done 52 times a year, this means 225 kwh.
The 5500 kwh previously used for heating should not be recalculatated to a new heating need of 5500*55/70. Doing that would mean we assume that we need just as much water if it is 55 degrees as we did when it was 70 degrees. This is probably not true (the 70 degree water, when used for example to shower, is mixed with ordinary water to get to a temperature of say 30 degrees. The same would happen with the 55 degree water, but the added cold water would be less to reach the same temperature, and thus the amount of water needed to heat would increase).
There are profits coming from reducing the heat in the water, both in transportation and in storage leaks, but we are on a rough level here, so let us keep those out of the caculations.
Instead we calculate the heating of the warm tap water as taking place with the use of the COP factor, assuming that the lower temperature and the higher amount of water to heat will add up to about the same heating need. Such a calculation gives us a heat cost of 5500/2.5, to which is added the weekly Legionella prevention heating of 225, getting a total of about 2400 kwh.
The saving with this method is thus up to 5500-2400=3100 kwh per year. This is about 500*10=5000 kwh more than before calculated over a 10 year period and compared to our previous business case. The additional improvement is thus around 20%, or a reduction with payback time of about 2 years.
The last improvement was due to a reduction in water temperature to 55 degrees. That could be achieved also by only changing the regulation, without the use of a heat pump as additional improvement.
Once again, the saving from only changing the regulation should not be calculated to be about 15/(70-10)*5500, since that once again would mean we are calculating on the assumption that we will need the same amount of water after decreasing the temperature.
However, a storage loss of 1500 kwh per year is a reasonable assumption from a hot water tank. The reduction with 15 degrees can be assumed to reduce the storage losses due to leakage with 15/70, amounting to 300 kwh per year, or 30 kwh per month.
This is not a significant change of the original business case and thus not of the payback time. The final business case, in the order of 10 years but probably a bit shorter, is the best I have found, and in the neighbourhood of payback times normally found for air to water heat punmps. Probably indicating that my calculations on payback are similar to the vendors market departments calculations.