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Big array-Long DC runs or long AC runs preferred and why?

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Posted by Dave on January 11, 2007, 4:47 pm
 
I'm looking at a very large grid tied PV array of about 150 kW that will be
located about 1,800 feet from the point of utility connection.
I can place the inverters close to the array and have a relatively short DC
side wiring connection with a corresponding much longer AC side run to the
uitlity or vice versa, multiple long DC runs to an inverter complex placed
near the utility point of connection with much shorter AC side wiring runs.
Obviously voltage drop plays a big role here but are there other engineering
considerations that would favor one approach over the other?



Posted by N9WOS on January 11, 2007, 6:30 pm
 

Dave wrote:

Basic questions.
What is the service voltage.
Is it 120V, 208V, 240V, 277V, or 480V?
Is it 1, or 3phase?

What is the nominal panel array voltage that will be feeding the
inverter?

What is the conductor sizes you will be using?

What ever end of the system has the higher voltage, with the lower
current, should be ran the majority of the distance.

Even at 480V 3 phase, you will be getting about 180A per phase at peak
sun.
If you were running the AC line most of the way, At minimum, I would
suggest using 0000 gauge aluminum wire.
The cost of the wire would be minimal compared to the PV array, and it
would make sure you were  not loosing precious wattage as IR losses in
the utility feed line.

Even 0000 gauge copper wire would probably be minimal compared to the
cost of the PV array, even with current copper prices.

And unless the DC side is being run at a higher voltage, with equal or
greater wiring, then the AC side would be the side to run the majority
of the distance in this example.


Posted by N9WOS on January 11, 2007, 7:39 pm
 
N9WOS wrote:

Lets do some loss calculation.
3 phase at 480V

180A through a 1800 Foot 4/0 gauge copper conductor.
One conductor per phase

Resistance 0.05125 ohm per 1000 feet.
Resistance for 1800 foot conductor is 0.09225
voltage drop at 180A is 16.6V
Power drop per leg is 2989 watts at full power output.
Three legs for 8967 watts of IR loss for the feeder system.
It will loose the power from about  $5000 in PV in IR losses at full
load.

If you parallel a pair of 4/0 conductors per leg, then you will have
half the losses.
Or if you separate the 150kW array into two 75kW arrays, each with
their own inverter connected to the service feed with their own set of
4/0 wires.

If you go to a conductor size of 500,000 circular mills,
you will get a resistance of 0 .02177ohms per 1K
That will mean a resistance of 0.039 ohms per leg.
Voltage drop at 180A is 7.2V
Power loss per leg is 1264 watts.
Total lost for three legs is3792 watts.
Or output from about $0000 dollars worth of PV panels will be lost in
IR losses.
You can step up from there by paralleling multiple 500Kmill wires.

After a certain point, you start reaching the point of diminishing
returns.
The extra wire you are putting in, would cost more than the PV panels
to make up the loss, would cost.

I would really suggest asking the power company if they could run full
line voltage (2.4kV 7.2kV or 14.4kV at 3Phase) to the site, with a pad
mount transformer right up against the array.
Even if it cost $0000+ it would probably be saving money in the long
run.


Posted by Solar Flare on January 12, 2007, 1:24 am
 In transmission lines the inductive reactance of the wires is many
times the value of the dc resistance if you run ac power that length.
You are considering only the dc resistance value of the conductors and
while this applies to dc transmission it is not the only factor for
higher ac currents. Your circuit always consists of the dc resistance
and the inductive reactance of one large turn of a coil reactor. For
transmission line faults on tower lines where the spacing is 15-20
feet the bolted fault calculations are done based on about a 75 degree
lag for the current from the voltage.

This loss factor may be insignificant but it may also make it hard for
your grid-tie to compensate properly for the phase angle shift
encountered. Much research may be needed and wire size willnot affect
this factor. Spacing and cable style will, but then you can get into
capacitive reactance. Not usually a lower voltages though...LOL

Good luck with it.




Posted by N9WOS on January 12, 2007, 3:42 am
 

Hmm........ For some reason, a little voice in the back of my mind said,
"Don't worry about reactance in a tightly bundled triplex wire of this
length." It may have been something dealing with things I had worked with
before.

But, all right, I'll drag up the figures.

I had to dig through the old bookshelf to find one of the old books, that I
knew the data was in..
You made me do work. Dang you!!!!!!!!

Book name "Electrical circuits and machinery"
Copyright, 1942
Tenth printing, May 1949

Page 620
1 mile of 4/0 wire, at 1 inch spacing has .2087 reactive ohms.
That equates to .0395 reactive ohms per 1000 feet.
That equates to .07 reactive ohms per 1800 feet per leg.
If the wire was shorted at one end, and you drove it with 180A at 60cycles.
You would get 12.8V AC.
The power factor that the inverters see
(Assuming a pure resistance on the other end of the 1800 foot run)
is 0.9999996744.............

Something tells me that the inverters should be able to handle a power
factor as bad as that. :-)

Something in the back of my mind tells me that the transformer that the
inverters are driving the HV grid through, will be a bigger reactive load
than that.

And the reactive load will cause a loss that can be calculated, based on the
true power throughput, and the apparent power that is on the lines.

But my mind starts telling me something again, something along the lines
of......  "It will be insignificantly small."

More specify, something under a watt, at full load.

Some rough calculations yield, about 3 milliwatts.

Please, if someone sees errors in my calculation, please point them out.
I know everyone always does anyway. :-)



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