Farm system

System

I need some help please. I want to install an off grid system on a farm. There is enough sun all year round. A lot of wind. The system needs to be 3 phase. Its a 220V system. If I have wind 40% of the day and sun for 12 hrs average. Power consumption peak 5000 VA per phase. average consumption 2000 wH per phase Wind would be my favorite way of generating power but I would want to add solar. And a 20 KVA 3 phase backup generator. What would the best be as far a wind power, solar power and batteries be. I am looking at using 3 of the Studer XTH series units. Most likely the 6000 watt or 8000 watt 48 volt units for headroom or if I want to increase consumption. Please advise and thanks for everyone helping here.

BB.

Re: Farm system

That would be a "huge" system... Just out of curiosity, to approach the question from a different side--What would be your response if I said that your power would cost you between $1.00-$2.00 per kWH?

Does that even begin to make economic sense to you?

-Bill

PS: Are you out of South Africa? What is your voltage/frequency?

Also, there are differences in costs between something that has just solar panels connected to a pump motor--vs solar panels to battery to inverter to load...

If you can do the direct connect for pumping---that is much less expensive (no batteries, no replacements every 8 years, etc.)--like 1/2 or 1/4 the price of Battery Backed Solar...

System2

Re: Farm system

yes I am in South Africa and the electricity bill is enormous at the moment. 220 v at 50 hz. The bill is around $2500 per month but half of it is levies etc, and for that reason we want to go off grid. We are expecting the price of power to go up 300% - 400 % in the next 5 years. So we don't want to be caught sleeping. Thanks I hope you can help.

niel

Re: Farm system

do they allow grid tied systems down there so you could offset your bill?

blackswan555

Re: Farm system

Any idea what your daily KWH consumption is ? Why the requirement for 3 phase ?

Have a good one
Tim

BB.

Re: Farm system

There is peak power usage, and average power... Peak is fairly easy and relatively cheap to plan for... Just a big enough inverter and battery bank to supply XX kWatts for a few seconds to a few minutes of starting loads, etc.

The "expensive part" for off-grid power is the ongoing loads...

• 6 kW * 1 minute * 1/60min per hour = 0.100 kWhrs
• 1kW * 24 hours per day = 24 kWhrs per day

The very basics of using solar panels to collect your power and store in a battery bank for use later--is costly. So, to understand the basic questions / issues.

I am going to use the PV Watts program and base it on Harare, Zimbabwe--I understand that this is not very close to you--but it is best I can do with this program--If we can find better data later, we can look at it. At least we will have results we can discuss here.

First, there are losses with solar panels and the "first" converter (vs factory name plate rating--which is wildly over optimistic). So--using a derating of 0.77 is a good starting point. You will then have the losses in a flooded cell lead acid battery of 0.80; and the losses going from a battery bank to 230 VDC of 0.85 (numbers are approximate and conservative--you can plug in the real numbers as you narrow down your options). So, the overall derating factor (from solar panel to charge controller to battery bank to AC inverter to your load) works out to be:

• Derating= 0.77 (panel+mppt charge controller) * 0.80 battery * 0.85 invtr = 0.52 (yes--1/2 of your energy is "lost" before you even start)

Now, we use the PV Watts program set for Harare Zwe--Use 1kW for solar array (nice round number); 0.52 for your 230 VAC 50 Hz output conversion losses; and leave everything else at "default settings":

"Station Identification"
"City:","Harare"
"State:","ZWE"
"Lat (deg N):", -17.92
"Long (deg W):", 31.13
"Elev (m): ", 1503
"Weather Data:","IWEC"

"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:"," 17.9"
"Array Azimuth:"," 0.0"

"Energy Specifications"
"Cost of Electricity:","-99.0 Z$/kWh"

"Results"
"Month", "Solar Radiation (kWh/m^2/day)", "AC Energy (kWh)", "Energy Value (Z$)"
1, 4.99, 71, "N/A"
2, 5.12, 67, "N/A"
3, 5.86, 84, "N/A"
4, 5.76, 81, "N/A"
5, 5.68, 84, "N/A"
6, 5.47, 79, "N/A"
7, 5.78, 88, "N/A"
8, 6.07, 91, "N/A"
9, 6.88, 97, "N/A"
10, 6.52, 94, "N/A"
11, 6.01, 84, "N/A"
12, 4.93, 71, "N/A"
"Year", 5.76, 991, "N/A",

Note, the PV Watts program is based on actual measured sunlight power at that location--so weather / clouds / etc. are taken into account (average weather--some years may be better, some may be less).

So, we see that for each 1kW of solar panel, you will get an average of 72-92 kWHrs per month (depending on season). Based on ~5-06.8 "Hours of Full Sun" per day--Very "sunny" part of the world--nice for solar.

Now--you have to go back and really work at estimating your monthly power usage--You may have seasonal needs (pump during summer, not during winter, A/C during summer, not during winter, etc.). And, separate those things that need electricity (computers, pumps) from things that may not (cooking, hot water, etc.). Remembering that solar electricity is "very expensive".

Also, you may be able to separate your pump loads--Use a deep well pump connected to dedicated solar panel (no battery) to pump to a tank/holding pond (direct solar/no battery is less expensive solar--but only pumps when sun is up) vs on demand pumping which needs to be battery backed (spray irrigation, pressurized water for home use, processing plant, etc.).

Lets say that you want 6,000 WH (6 kWhr) per day or 6*30=180kWHr per month. The above shows 71-97 kWhrs per month per 1kW of solar panels. Or:

• 180kW per month / 71 kWHr per month per 1kW = 2.5 kW of solar panels for December/January average sun
• 180kW per month / 97 kWHr per month per 1kW = 1.9 kW of solar panels for September average sun

Batteries, we generally start with assuming 3 days of no-sun, and 50% maximum discharge. So 6kWH (6,000 WH) of load per day (my made up load):

• 6,000 WHrs per day * 3 days * 1/0.50 max disch * 1/0.85 Invtr Eff * 1/48 volt battery bank = 1,500 Amp*Hours of battery bank @ 48 volts

And we can do the same for Wind, back up genset, etc...

I am going to stop the estimating for now--because I really am not sure of your daily Watt*Hour or kWatt*Hour load... You talked about 2,000 WH per phase with wind power (2 kWH) -- which is a pretty reasonably sized system... But if you want more power --- such as 2,000 Watts * 10 Hours a day = 20,000 WH / 20 KWHr -- That is a big/expensive system.

Are you clear on what Watt, Watt*Hour, kW, kWH, etc. are and how to measure/estimate your loads using those values (I just want to be clear not not get you frustrated if we are not talking about the same thing)?

Also, you might wish to read through the Grundfos Solar Submersible Well Pumps information... they are expensive--but very well made and have some nice pumps/equipment that works well with grid and off grid power.

The pumps themselves are not that large (1,400 watts maximum)--So, I do not know if you are looking at that or 20 kW pump motors--but these pumps will run on Solar panels (very efficiently), battery bank, or even AC power from line/genset--same pump motor. Very flexible.

If you are looking at larger pumps--Somebody else is probably going to have to help you--That is big and expensive and I could not give you very good guidance in that area.

-Bill

stephendv

Re: Farm system

The SMA Sunny Island system with AC coupling could be interesting for a farm, because of the potential distances involved between solar panels, wind turbine and battery bank. AC coupling gives you the flexibility of running 220V everywhere instead of low voltage DC.

The wiring and electrical bits would be much simpler since everything is connected on the AC bus - the only DC bit would be between the Sunny Island and the batteries. PV would feed into a Sunny Boy inverter which connects to the AC bus, similarly the turbine would connect to a windyboy inverter which connects to the AC bus. If you need to install more producers, just connect more inverters anywhere.

The system could cost a bit more than a DC based system because the inverters are more expensive than DC chargers. But you gain in flexibility, especially if there are long distances involved. Anyway, you'd save a bit using the Sunny Island's themselves since they're better value than the Studers.

SMA have some literature describing the system in more detail on their website: www.sma.de/en