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Re: 300 foot line OK for solar panels?
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Re: 300 foot line OK for solar panels? N9WOS 11-10-2007
Posted by N9WOS on November 10, 2007, 7:09 pm
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>I am planning to install some new solar panels on a steep mountain slope.
>The line will have to run about 250 to 300 linear feet to get to my house,
>inverter, batteries, etc.
>
> Is this feasible? How do I figure what type/size of line I will need for
> this (and the approximate cost)?

(cross posted to all news groups that I found the question in.)

It would have been better if you could have cross posted one message to
several news groups instead of a bunch of separate messages to several news
groups. That makes it hard for people to see if someone else has already
posted an answer somewhere else. That is because there is no link between
the same question on the separate news groups.

On to the question.

Hmm...

Is this feasible? Yes........

What is the size of the array you are thinking of?

You are wanting to use batteries so I assume that you have a system that is
isolated from the grid. And you will be running off of batteries at night.

The problem is you want an array voltage that is geared to reduce the IR^2
loss in the conductors from the array to the charge controller. Basically,
for a moderate sized system, you want the solar panels set up to produce a
total voltage of one or two hundred volts or so.

You will need a charge controller that can take that 100V to 200V DC from
the panels and efficiently charge the batteries at the normal battery bank
voltage. That being 12V 24V or 48V. More specifically, you will need a max
power point tracking (MPPT) charge controller that can accept a high voltage
DC input.

The high DC voltage from the array is important. A set of wires carrying 20A
at 200V may have 10% loss. But for you to get that same 10% loss with a set
of wires of the same length carrying 200A at 20V (the same amount of power)
you will have to have a set of conductors 100 times the size. And copper =
$$$$$

Once you find out how many panels you are going to use, how much wattage
they will put out, how high of an output voltage you can arrange them to
produce, and how high of an input voltage your charge controller can
tolerate, then we can start figuring out how large of a set of wires you
will need to install.



Posted by Eeyore on November 10, 2007, 7:47 pm
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N9WOS wrote:

>
> >I am planning to install some new solar panels on a steep mountain slope.
> >The line will have to run about 250 to 300 linear feet to get to my house,
> >inverter, batteries, etc.
> >
> > Is this feasible? How do I figure what type/size of line I will need for
> > this (and the approximate cost)?
>
> (cross posted to all news groups that I found the question in.)

Thank you for that.

To summarise, the length of the cable run may be practical but you need to take
into account the current, operational voltage, acceptable voltage loss in the
cable etc.

There is no one simple answer. It has to be CALCULATED for every installation
where cable length is a concern.

Graham


Posted by on November 12, 2007, 8:15 pm
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wrote:
> N9WOS wrote:
>
> > >I am planning to install some new solar panels on a steep mountain slope.
> > >The line will have to run about 250 to 300 linear feet to get to my house,
> > >inverter, batteries, etc.
>
> > > Is this feasible? How do I figure what type/size of line I will need for
> > > this (and the approximate cost)?
>
> > (cross posted to all news groups that I found the question in.)
>
> Thank you for that.
>
> To summarise, the length of the cable run may be practical but you need to take
> into account the current, operational voltage, acceptable voltage loss in the
> cable etc.
>
> There is no one simple answer. It has to be CALCULATED for every installation
> where cable length is a concern.
>
> Graham

Well there is one simple answer. The higher the voltage, the lower
the losses. The fatter the cable, the lower the losses.
You need to know that the loss is equal to the current squared times
the resistance.

P = I^2 * R

All copper wires come with specification values for resistance in ohms
per linear meter (which is directly proportial to the inverse of the
cross-sectional area of the conductor. )


Say you have a 1000 watt array at 200 volts. At full sun, you would
expect it to produce a current of 5 amperes.

Therefore if your cable run has a resistance of 2 ohms, then the loss
would be 25 * 2 = 50 watts.



Posted by Vaughn Simon on November 10, 2007, 9:05 pm
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> It would have been better if you could have cross posted one message to
> several news groups instead of a bunch of separate messages to several news
> groups.

Better yet, don't crosspost at all. Most of us read several related groups
anyhow. Pick what seems like the most appropriate group, make your post, and
99% of the time you will get your answer.

Vaughn



Posted by sylvan butler on November 19, 2007, 12:47 pm
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On Sun, 11 Nov 2007 02:05:24 GMT, Vaughn Simon
>> It would have been better if you could have cross posted one message to
>> several news groups instead of a bunch of separate messages to several news
>> groups.
>
> Better yet, don't crosspost at all. Most of us read several related groups

Actually, a crosspost is appropriate, and any newsreader should be able
to automatically mark as read in all groups, any article crosspost which
you have read in one groups. The problem is multi-post, not crosspost.

sdb

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