do lower watt inverters use lower watts on idle?

Re: do lower watt inverters use lower watts on idle?

Power = Volts * Amps...

So a 10 Watt loss on a 12 volt battery bank:

• Amps = Power / Volts = 10 watts / 12 volt battery bank = 0.833 amps

If you want some more information on TSW vs MSW inverters:

All About Inverters
Choosing an inverter for water pumping

In the end, around 80% of devices will probably work OK on MSW inverters and about 10% will not (fail in minutes/hours).

The problem is knowing which is which.

-Bill

Re: do lower watt inverters use lower watts on idle?

At this point, assuming it is water--I guess it does not matter too much.

The issue is that off grid battery backed solar power will (almost always) cost you 10x what your local power does (i.e., $1-$2+ per kWH vs $0.10-$0.20 for most people in the US).

Also, sun may vary over the seasons by a factor or 2 or 3x... You have to design to meet your needs for the appropriate season(s).

Generally, a external pump mounted below the water level is the most efficient form of pumping. The worst is probably a pump drawing 10+ feet of suction. Sumerged pumps tend to cost more and may not last as long (because of water leaks).

Your power needs are not great by themselves, but when you multiply everything out (24 hours per day, pump/inverter/battery losses, backup power--if needed).

Just to pick some round numbers. Say you can do this with 20 watts 24 x 7 and you need it off grid 12 months a year (ignoring a days/week of bad weather--you almost always need to curtail loads and/or use a backup power source--pure solar would be horrendously expensive)...

• 20 watts * 24 hours per day = 480 Watts*Hours per day

Assume Fort Worth Texas with PV Watts. Fixed array, 1 kW (default) solar array and 0.52 system derating:

"Station Identification"
"City:","Fort_Worth"
"State:","Texas"
"Lat (deg N):", 32.83
"Long (deg W):", 97.05
"Elev (m): ", 164
"PV System Specifications"
"DC Rating:"," 1.0 kW"
"DC to AC Derate Factor:"," 0.520"
"AC Rating:"," 0.5 kW"
"Array Type: Fixed Tilt"
"Array Tilt:"," 32.9"
"Array Azimuth:","180.0"

"Energy Specifications"
"Cost of Electricity:"," 9.7 cents/kWh"

"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value ($)"
1, 4.32, 66, 6.40
2, 4.77, 66, 6.40
3, 5.50, 82, 7.95
4, 5.98, 85, 8.24
5, 6.02, 86, 8.34
6, 6.25, 84, 8.15
7, 6.39, 87, 8.44
8, 6.31, 86, 8.34
9, 5.83, 78, 7.57
10, 5.56, 80, 7.76
11, 4.43, 65, 6.30
12, 4.10, 63, 6.11
"Year", 5.46, 929, 90.11

So, the minimum (20 year average) sun is 4.10 hours of sun (great weather there).

• 480 Watt*Hours * 1/0.52 PV system derating * 1/4.1 hours of sun = 225 Watts of solar panel minimum

In reality, you should probably plan on 30% to 50% more solar panels to allow for a bit of bad weather and other issues.

Battery bank wise, you would be looking at 1-3 days of no sun and 50% maximum discharge. Pick 2 days and 50% max discharge:

• 480 WH * 1/12 volt battery bank * 1/0.85 inverter losses * 2 days * 1/0.50 max discharge = 188 AH

Battery charging is recommended to be around 5% to 13% of battery bank AH capacity (20 Hour Rate). Pick 10% as a nice healthy sized solar array:

• 188 AH * 14.5 volts charging * 1/0.77 solar+charger losses * 0.10 rate of charge = 354 Watt Array "Nominal" suggested size

So, your solar array should be around 225 to 354 watts rated for "optimum" based on 20 watt 24*7 in December (really, I would suggest that ~355 watts be the minimum array).

You can put these equations into a spread sheet and play with pumps/inverter/etc. and see what works out best for you.

Note that an inverter that runs 6 watts of "tare losses" is almost 1/3rd of your pumping costs... These Little Losses (tare, battery cycling, panel deratings, etc.) all add up...

-Bill