Can my current electrical service support adding an electric water heater?
This can be done -- the subpanel still has load to spare, albeit only by an amp

Since you have a 100A service, I can use the simplified calculations from 220.84, which give you 3VA/ft^2 for the house, 1500VA per Small Appliance Branch Circuit (kitchen/laundry), and the fixed appliances at nameplate rating (it sounds like you have none though), as well as 100% of the nameplate rating of the A/Cs.

From here, we get:


1200ft^2 * 3VA/ft^2 = 3600VA for lighting and general receptacles
1500VA per appliance branch circuit (kitchen, laundry) * 4 (3 kitchen, 1 laundry) = 6000VA for appliance branch circuit loads
8kVA for the range, 1.2kVA for the DW, 1.5kVA for the microwave, and 5kVA for the dryer = 15.7kVA


Before the HVAC loads are taken into account, we apply a 40% demand factor to everything over the first 10kVA, giving us a total of 16.1kVA so far.

The HVAC is simple -- we are dealing with 3 11.5A (conservatively) A/Cs on 110V circuits, so that's another 3.8kVA at 100% demand factor, giving us 19.9kVA total load, or 82A on a 240V service.

Now, when we add the 4.5kVA water heater in, this is covered by the 40% demand factor as it's over the first 10kVA of non-HVAC load, so the resulting net load is 4.5kVA*40% = 1.8kVA, or another 7A on the service -- pushing your service to 89A, which is still manageable.

We then repeat this for the feeder to the subpanel, with the stove at 8kVA, the dryer at 5kVA, and the washer circuit at 1500VA, as well as a conservative assumption of 1440VA for the two rooms (180VA/outlet * 4 outlets per room * 2 rooms), to yield 12.4 kVA for the feeder load after applying the 40% demand factor.  With a 240VAC feeder, this gives us a 52A feeder load, which has just enough room to handle the additional factored load from the water heater (again, 1.8kVA or 7A) -- that is pushing the feeder to 59A though.