sizing the array!

Hi all,

I have a question. It seems beginner but it brings me confusion sometimes. If the load size = 3kw , and the daily load operations = 12h. The client wants a backup options as follow:
1. 3kw and 12h a day. Not requiring holdover days/ requiring 2 holdover days
2. 3kw and 6 h operation a day. Not requiring holdover days. / requiring 4 holdover days
3. 1.5kw and 6h a operation a day. Not requiring holdover days/ requiring 5 holdover days.

Assuming the sunlight hours = 5h. Can you please tell me how to size the array? ..

I have prepared an excel based calculator. acquire entering the spec of the selected components for excel to calculate. There is a simple attempt to size the conductor. I wanna put it here for your feedback but cannot upload it. Any suggestions?!

Waiting!

Comments

  • alithekingalitheking Registered Users Posts: 2
    Re: sizing the array!

    On top of that, How to size the array if the user wanna a total of 12h of daily usage for the 3kw load. In this case, we shall decide what the back-up size. assuming the location sunlight hours = 5. I subtract it from the 12 = 7h back-up battery. right?
  • BB.BB. Super Moderators, Administrators Posts: 32,008 admin
    Re: sizing the array!

    Welcome to the forum "alitheking", and may I ask a few questions first about your loads.

    Does the customer want the loads to run if it is a dark and cloudy day (i.e., assume 5% or less solar power available)?

    What are the loads? A bunch of small motors, computers, lights? Or is this a big water pump that needs lots of starting surge?

    Are loads optional? Bad weather, don't run loads?

    Backup power? Utility power? Generator?

    Battery bank? Lead Acid? Any brand/model of batteries you are thinking of using? (aka standard deep cycle, industrial/forklift)?

    In general, the size of a lead acid battery bank is the daily load * #of days of autonomy * 2 (for 50% maximum discharge).

    So for 1 day, that would be a battery capable of 2*daily load. 5 day would be 5*2=10x daily load.

    From a "math" point of view, we try to design "balanced systems". With Lead Acid batteries and their "typical" charging/discharging requirements, a 2 day + 50% maximum discharge (4*daily load) is about optimum. A smaller bank may have issues with motor starting surge limitations and problems being "fully charged" with limited amounts of sun (i.e., if towards Canada, few hours per day of sunlight. If near equator, seasonal sun does not change that much and more hours of sun to fully charge a deep cycled battery bank). A large battery bank drives a larger array... Ideally, a good quality deep cycle battery would like ~10% rate of charge... If you have a very large battery bank, you need a much larger solar array to "properly" recharge the bank (much larger solar array than you actually need to support the loads).

    Which is a lot of hand waving leading to the generator question--If this is a "backup" solar power system that runs a few weeks a year--Many times a generator is more cost effective (even with high fuel prices).

    If the power needs are constant during dark weather (monsoon season, etc.) then past ~2 days of lead acid battery, then running a back up generator looks better. Smaller battery bank and given that your solar energy can be 10% or even 5% of "nominal" on a clear day, it is not very practical to build out a very oversize array for a few hand full of "dark weather" periods.

    Once we better understand the loads--We can do some quick estimates to get an initial sizing of the system.

    At this point, a constant 3 kW * 12 hours per day is not a "small system". There will be some significant engineering required for the details--Beyond what we can do as a website "staffed" by volunteers.

    Which always comes back to my philosophy--Conservation. Go back and look at the loads and see if the can be reduced by any means (more efficient use of power, turning stuff off when not needed, using alternative energy sources for heating, better insulation, more efficient electronics, etc.). It is almost always cheaper to conserve than to build a larger Off Grid Power system. In the US, the costs for running a Off Grid cabin/home can easily be in the $1-$2 per kWH range with 20 year amortization (initial costs, battery/equipment replacement/repairs, etc.). That is ~10x the cost of utility power in the US (where power lines are available).

    -Bill
    Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
Sign In or Register to comment.