Hi,
Don't know if this is still a current project for you, but a little info on your questions...
Heat Stored in Water:
Heat Stored = (specific heat of water)(Thi - Tlow) (weight of water)
Thi is the temp the tank gets to, Tlow is the lowest temp you can use for heating
Specific heat of water is 1 BTU/lb-F
So, for 480 gallons with a high temp of 180F and low temp of 80F, the energy stored is:
Energy Stored = (1 BTU/lb-F)(180F - 80F)( 480 gal * 8.33 lb/gal) = 400K BTU
House Energy Use:
How much energy your house uses in a day depends on how big it is, how well its insulated and how cold it is.
You can estimate this with this calculator:
https://www.builditsolar.com/Referen...s/HeatLoss.htm
For 1700 sf home with typical insulation on a 30F day -- maybe 25,000 BTU/hr or 600K BTU/day
But, again, this depends a lot on insulation levels, temps etc....
So, the tanks probably could store enough heat on a sunny day to provide a good part of your heat needs.
Does the tank size match your collector size?
A rule of thumb on how much storage tank capacity you need to work well with a collector of a given size is that the tank should provide about 1.5 to 2.5 gallons per sqauare foot of collector area.
Not sure how large your collector is, but the 480 gallons would be a good match for a collector area of about 240 sqft.
Its not critical to hit this right on the money, but a tank that is much too small will get up to its max temperature early in the day and the rest of the days solar would be wasted, and for a tank that is too large, the tank will have a hard time heating up to a temperature that is useful for heating.
Hope that helps.
Gary