I don't know about solar, is this sized correctly?

[Ponchik]

The installation is in CA,

The system is installed without a battery in August of 23, electrical usage per month is 2500 KWH, the installed system has (28) 450watt panels.

Is the system sized properly to cover most of the monthly usage in order to have near zero monthly bill?

Thanks for your help.

 

[electrofelon]

You could go here and get a pretty good idea.

PVWatts

Estimates the energy production and cost of energy of grid-connected photovoltaic (PV) energy systems throughout the world. It allows homeowners, small building owners, installers and manufacturers to easily develop estimates of the performance of potential PV installations

pvwatts.nrel.gov

Around here in upstate NY, a quick and dirty rule is take the PV DC system size in watts, multiply by 1.1 and that is the number of KW hours per year generated. So with that system here, it would meet about half the electrical usage. But note parts of California get twice as much sun as here in upstate New York, so that rule doubtfully holds where you are, you'd have to run it through the PV watch calculator to get an accurate idea

 

[ggunn]

The installation is in CA,

The system is installed without a battery in August of 23, electrical usage per month is 2500 KWH, the installed system has (28) 450watt panels.

Is the system sized properly to cover most of the monthly usage in order to have near zero monthly bill?

Thanks for your help.

As electrofelon suggests, a PVWatts analysis (https://pvwatts.nrel.gov/) will give you a pretty good idea. It's easy to run; you just need to enter the array location, orientation, type, and DC power rating. The tool will give you a quick and dirty (+/- 10%) estimate of the kWh per month and per year the array will produce, using many years of NREL data from the closest weather station. From the information you posted it is not possible to answer your question.

 

[Ponchik]

Thank you gents. I will look into it.

 

[jaggedben]

Around here in upstate NY, a quick and dirty rule is take the PV DC system size in watts, multiply by 1.1 and that is the number of KW hours per year generated. So with that system here, it would meet about half the electrical usage. But note parts of California get twice as much sun as here in upstate New York, so that rule doubtfully holds where you are, you'd have to run it through the PV watch calculator to get an accurate idea

I usually use 1.5 for back-of-the-envelope here in Northern California, but real world could easily vary from that by as much as 0.5 either way.

Also California crosses a lot of latitude, so the OP just saying "Installation is in California" isn't very precise. ("Installation is in Tennessee" would be a lot more precise.)

 

[PWDickerson]

The PVWatts site will accurately account for tilt and orientation losses of an array that is not ideally tilted and oriented, but it will not tell you what your losses will be due to shading impact. For that you would need to do a shading analysis.

 

[ggunn]

The PVWatts site will accurately account for tilt and orientation losses of an array that is not ideally tilted and oriented, but it will not tell you what your losses will be due to shading impact. For that you would need to do a shading analysis.

True enough, but unless there is significant shading of the array PVWatts will do well enough to answer the OP's question. BTW, in addition to tilt and azimuth, PVWatts also takes into account racking method, geographical location, and historical weather/climate data.