A client recently asked me if I thought that one of our systems would produce enough hot water for him and his family. The following might give you some guidance on how to assess the output of any water heater you consider installing…

The first thing to consider is that solar energy input varies, so let’s
take that out of the equation initially to consider the tank’s output.

The tank has two 4500 W elements, or electrical burners.
The tank holds 75 gallons, or 625 lbs of water.
A single 4500 W element is equal to 15,354 BTUs
the tank has two elements: one located toward the top, and one at the middle
of the tank (vertically speaking as indicated in the diagram I sent)

1 BTU= 1 lb of water/1 degree F/Hr

Your average inlet temperature is roughly 62 F, and we need to raise it to
120 F.
This is a delta T (temperature change) of 58 F.

58(DT) x 625 lbs = 36,250

36,250 / 15,354 = 2.3 hours

That is, without any solar input, one 4500 W element will heat 75 gallons of
water at 62F to a piping 120 degrees in 2.3 hours without considering a
standby loss of 0.9 degrees per hour (see system manual for standby loss).

However, this does not completely answer your question for 3 main reasons:
The first being that the tank we recommended installing has 2 of these elements. The second
being that the whole 75 gallon tank does not need to be 120 degrees F for
you to have hot water as the hot water comes out of the top of
the tank. Lastly, this does not consider that the solar-heated glycol
narrows the temperature differential every day, ultimately keeping the top
of the tank at above 140 F on sunny days on solar input alone (without any
electricity).

Furthermore, we have several clients that have their auxiliary heating elements off still (in the middle of November). Though these results will vary across all situations depending on latitude, shading, and orientation, this 100% solar output is QUITE impressive.