Solar problem

Assuming you are in the US, and have a 120/240 VAC utility connection... The 9 amps could be

• 9 amps * 120 vac = 1,080 Watts
• 9 amps * 240 vac = 2,160 Watts

Let's assume 1,080 Watts... If you used "800 kWH per month extra):

• 800,000 WH per month / 1,080 Watts = 740 Hours of pump operation per month (guess)
• 741 Hours / 24 per day = 31 Days of 100% pump operation

I would suspect that you are not really running the pumps 24x7... But if you are, that does account for for you "extra energy".

But, you said that was the whole house consumption... And if that was 240 VAC and really ~2,160 Watts, running that 24x7 would almost account for your 1,800 kWH per month electric bill.

What I would do... Make sure I don't have anything running in the home that would cause issues if I cut the AC power for a few minutes (desktop computer, DVR recording, etc.)... And I would go outside to the meter and see if I can read the actual Watts or kWatts being consumed (my electronic utility meter will read 0.100 for a 100 Watt loads).

Then start turning off circuits and see what each one contributes to my total usage:

2,160 Watts all on
1,000 Watts circuit 1 turned off
1,000 watts circuit 2 turned off
500 Watts circuit 3 turned off
etc.

Or, if you do have an AC Current Clamp meter, you can measure the current from each circuit breaker. For 120 VAC circuits, the Watts is X.X amps * 120 VAC=Watts.

For 240 VAC circuits (double pole breakers), you need to measure each wire seperatly. If one reads 5 amps and the other reads 5 amps, then you have a balanced 240 VAC load and the math would be (either is works):

• 5 amps * 240 VAC = 1,200 Watts
• (5 + 5 amps) * 120 VAC = 1,200 Watts

If one reads 5 amps and the other 1 amp, then the math would be:

• (5 + 1 amps) * 120 VAC = 720 Watts

That is a fair amount of energy usage for a non-electric heating home... Typical is around 500 to 1,000 kWH per month for a home that has natural gas/propane for water and space heating.

More or less, you have a ~1,000 to 1,500 Watt set of loads that are running 24x7. That sounds like an electric space heater in the family room, or 3x desktop computer systems, or a lot of electric heat hot water usage (like 2,400 Watts running 50% or the time, or an electric dryer that is being used a lot due to lots of wet clothes).

If you have not done a conservation pass through your home--Things like lots of filament lighting vs LED, entertainment system running all the time, a couple of desktop computers left on 24x7, etc.... It is not unusual for somebody to go through the home and changing to more efficient devices and simply turning unused stuff off, that can cut the electric bill by almost 1/2.

Using a current clamp meter on the main breaker panel (assuming you know what you are doing and don't electrocute yourself), a Kill-a-Watt type meter to measure each 120 VAC outlet/load for kWH per day, and even a whole house AC monitor:

https://www.solar-electric.com/kiacpomome.html
https://www.theenergydetective.com/

In general, things that run a relatively short period of time (electric oven running 2 hours every week is only ~1-3 kWH per week) vs a lower power device running 24x7 (desktop computer 300w*24h=7,200WH=7.2kWH per day) which can use much more energy.

You are looking at an average load of:

• 2,000 kWH per month * 1/30 days per month * 1/24 hours per day = 2.78 kWatt average load = 2,780 Watt average load

That is running 24x7 average load. That is a lot. You need to figure out where your energy is going.

Energy usage is a highly personal set of choices. You have to figure out what is your set of energy needs that works for you. If you have electric hot water, you may not choose to go with gas water heating (in our area, natural gas is about 1/3rd the cost of an electric water heater).

If you have natural gas heating and plugged up air filters, that can cause the fan to run much longer in your central heating system. If you have single pane windows and not enough insulation, that will run the central heat much longer (~300-500 Watts or so for typical fan), and run up you gas bill too.

Anyway--Some starting places to look/explore.

And regarding the water in your conduit... That is most likely because your system was installed without a proper drip loop (a drip loop has the wire go down, then up, into the conduit fitting). The water simply follows the wire into the conduit without the loop.

It is also possible that you do not have water tight fittings and some water is coming into the fittings.

I would suspect that the water is not getting in through the gasketed j-box. What you are seeing is the water collecting at a low point in your conduit and filling the j-box and local conduit with water.

https://www.efficientplantmag.com/2013/09/moisture-protection-of-electronics/



Besides addressing the drip loop and water tight fittings issues, at the low point of your system you want to let the water out. On my system, I removed the gasket so that the water can come out. Even drilling a small hole in the low point of the box is a good solution (obviously, be careful not to nick the electrical cable insulation). Otherwise, the water can come into the home and eventually migrate to the other places in your system (like the main panel) and cause corrosion/electrical failures.

-Bill

Hmmm BBQ...

9 amps at 120 VAC is around 1 horsepower, 9 amps at 240 VAC is around 2 hp. In either case, that would be a big pump (or set of pumps).

In theory, changing the float levels would save only a nominal amount of energy--Making the range larger (higher high, lower low level) would have the pumps cycling on and off less (starting surge is not efficient, and cause the house lights to dim when starting). But I am not sure it would save much energy unless you pump much less water (and where is that water that remains is going, is always a question).

The Kill-a-Watt type meters are cheap ($20-$40 each)... And you just plug your load into the meter, run for about a day (for cycling loads), and read the kWH per xx hours... Say your pump uses 1 kWH over 36 hours, then per day that would be:

• 1 kWH per day * 24 hours * 1/36 hour measurement cycle = 0.67 kWH per 24 hour period.

You can do this for all of your plugin loads... You are not going to worry about cell phone chargers--Those will be something like 0.010 kWH (10 WH) per day. Laptops are around 0.5 kWH per day (if used 12 hours per day). Desktop computers (add display, laser printer, networking stuff) can easily run 5 kWH per day (if always left on, and don't go to sleep when not used).

And you can make good estimates for appliances that have a Energy Star Tag... Here is a 50 gallon electric water heater:

https://images.homedepot-static.com/catalog/pdfImages/2e/2eddf7c8-9b9a-4502-9225-dbb6786a6d05.pdf

Looking at the bottom, that is 3,493 kWH per year... Per day:

• 3,493 kWH per year * 1/365 days per year = 9.6 kWH per day

A full size modern refrigerator:

https://images.homedepot-static.com/catalog/pdfImages/52/52bea3ed-52e7-4898-a38d-c8b823c6a55b.pdf

365-~693 kWH per year or:

• 693 kWH per year * 1/365 = 1.9 kWH per day

Electric Dryer--Maybe 2 kWH per load. etc...

Back to what you are looking at with 2,000 kWH per month:

• 2,000 kWH per month / 30 days per month = 66.7 kWH per day

Just measuring/estimating your major loads and see what they add up to... I would guess your electric meter has a display where you can get the average Watts or kWatts being used at that moment in time. You can cycle a major appliance to see what your usage is (for example, turn off your central heat, check display, turn on central heat to 85F, wait a minute for it to turn on and stabilize, and see what the new power reading is).

During the day, with all the major stuff off, read meter and see how much you are using. Then, turn on all the lights you use in the evening and see the measurement again.

Turn off all your optional loads and see what your base loading is (any loads you did not know about...).

You can get a start on your energy usage... 

In a couple of hours of turning stuff on and off (have one person turning stuff on and off, a second person at the meter recording values--use cell phones or 2 way radio to coordinate)--It's a start.

Note that many larger loads cycle... Like your refrigerator may use 120 Watts for 50% of the time. And may even use 600 Watts for 1/2 hour per day (defrost). The water heater may use 4.5 to 5.5 kWatts for 2 hours per day. Using a Kill-a-Watt or TED (the energy detective) meter over 24 hours will help you figure that out.

Note that the Energy Star Tags are currently using $0.12 per kWH to figure out cost... I think N.J. is similar to California where you may be paying $0.25 to $0.45 per kWH:

https://www.pge.com/en_US/residential/rate-plans/rate-plan-options/tiered-base-plan/tiered-base-plan.page

It looks like New Jersey electric rates are not bad (around $0.12 per kWH or so)... That amount of usage would "kill me" in California (upwards of $600-800 per month--Yours closer to $240 (2,000 kWH * $0.12=$240 per month, without taking into account your GT solar credits).

I think you could reduce your energy usage a lot--But what needs to be done, you really need a "site survey" of energy usage and any heat loss (insulation, windows, weather stripping, etc.). Again, this is highly personal, depending on where you live, climate, number of teenagers in the home, etc.

Many utilities have energy audits or audit contractors (some that are subsidized) and may be of interest to you. A couple of quick search results:

http://www.njcleanenergy.com/residential/programs/home-performance-energy-star/what-expect-home-energy-assessment
https://njhomediagnostics.com/

If you can find a friendly electrician with an AC Current Clamp Meter and find out how much he would charge to install a TED, all would be useful.

If you have a friend that is reasonably 120/240 VAC knowledgeable--It is not difficult and in a couple of hours, you will know a lot more than you do now.

You are using a lot of power at this time. But to make any progress, you really need to understand where you are today. Just buying a new refrigerator and unplugging the freezer in the garage may save you 2 kWH per day... You still have the other 66 kWH per day to address.

For ~$110, an AC (and DC) current clamp meter and pulling the cover off of your main circuit breaker panel (with knowledgeable help)...:

https://www.amazon.com/gp/product/B019CY4FB4/

Good luck,
-Bill

PS: My links are suggested starting points. A couple items I have and use, others just points to start your research. I am not connected with any of the websites or products.

As always, make plans first (and ask questions here), before you go out and spend a much of $$$$.

-BB