I've been meaning to calculate the thermal content of a volume of water, so I'll do now.
I'll start with the 275 gallon tank, which is just barely larger than 1 cubic meter (I'll use metric units for calculations, if that's OK?). 1m3 of water weighs 1 metric ton=1 million grams.
Warming 1g of water by 1 deg.C takes 4.187 Joules (= 1 calorie, I'm approximating that this doesn't depend on the start temperature and pressure).So raising the temperature of 1m3 by 1*C will take 4.18MJ.
Let's say that the temperature of the water in the tank will have a 70*C range, warming the tank from 10*C to 80*C (50*F-176*F) will take 293 MJ = 81.4 kWh = 277818 BTU. That's also how much energy can be taken out of the tank to lower it's temp from 80*C to 10*C.
In real life:
- There will be losses through the tank walls, proportional to the temperature gradient,
- If the tank has almost perfect external insulation, then the tank's walls will also accumulate heat, adding to thermal mass,
- There will be extra solar energy being added to the tank every day,
- The heat pump also produces a certain amount of heat.
If you can keep the high temperature in the tank until cold season, then you get an extra 1kWh=3413BTU of energy per day for most of winter. The main problem here will be insulation. If you get 3 tanks, then the best thing to do would be to stack them in a pyramid, put a mess of insulation around the whole setup (with maybe smaller tanks in the open spaces) and bury it. Maybe worth trying?