# calculating loses

Posted by David Lesher on July 16, 2010, 6:52 pm

So I have a demand of 1.6 KWh/day, and insolation is 5.5 kwh/M^2day.

If the world was perfect, I'd just divide to get panel area.

It's not, so I assume there are other factors I must
include; the panel efficiency of course, and some electrical
losses. [It's a DC pump motor, so that eliminates one,
conversion losses..]

Where do I go for more data on this aspect?

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Posted by Ron Rosenfeld on July 16, 2010, 8:23 pm

On Fri, 16 Jul 2010 18:52:26 +0000 (UTC), David Lesher

With regard to your panels:

Panel efficiency is one determinant of required "Area".  However, you
need to know how the panel efficiency was determined.  It should be
determined at some specified test condition (e.g. STC or PTC) which
may not be reflective of your real world conditions.

Then you may need to know the temperature coefficient of power,
depending on your environment.  That data should be published with the
panel specifications.  If not, you should contact the manufacturer.

If area is irrelevant, than you can just start with the panel output
ratings and ignore efficiency (but not the variation between the test
conditions and your real-world conditions).

You need to add in some factor for panel aging.  Probably you can get
that from the warranty.  e.g. If the panel is rated to produce at
least 80% of its nameplate output at 25 years, then 80% might be an
appropriate factor.

Wiring losses can be obtained from various sources -- they will vary
depending on wire run length and wire size.  They may be insignificant
if the panels are close to the pump.  They can be mitigated by using
an appropriately sized wire.

Will the pump be running 24/7?  If so, you will need to factor in
losses from your electrical storage medium.

You should know that the solar insolation values are a multi-year
average.  Depending on your requirements, you may have to take into
account worst-case scenarios.  For example, an average annual (or
pumping season) insolation of 5.5 does not mean there will be 5.5
every day.  There will be more on some days, and less on others.

If your pump needs to run 24/7, then you will need to include the
losses in your energy storage medium (usually batteries)

And there's more you need to be aware of in designing a system, but
the above should get you started.

Posted by David Lesher on July 17, 2010, 5:41 am

{STC vs PTC, temp derate, aging}

Thanks. I was also hoping for some writeups on how much each
factor deserves, etc.

And unlike one vender's bid: by keeping the supply voltage up.
[The SqFlex series will run from 30-300V, and run well above 120V.]

Nope, we're storing energy the rational way: Ug  = mgh
No oxides of lead or such needed, and very low losses.
We shall all wear out before  G*/ does.

I figure we have 40-60 DAYS worth of water storage, so a few
cloudy days won't cause grief.

Further, there's some balancing factors. Rainy season has less
sun, but the well level will likely be 75-125 ft higher, using
less power. And the panels will be cooler.

Well, I've got a handle on everything save the panel issues.
Thanks for your help.
--
A host is a host from coast to coast.................wb8foz@nrk.com
& no one will talk to a host that's close........[v].(301) 56-LINUX
Unless the host (that isn't close).........................pob 1433
is busy, hung or dead....................................20915-1433

Posted by Ron Rosenfeld on July 17, 2010, 12:59 pm
On Sat, 17 Jul 2010 05:41:15 +0000 (UTC), David Lesher

That is manufacturer specific, so you have to obtain that data from
the manufacturer.  They should have it published on their data sheets;
but sometimes you have to contact them directly.  If not available,
they probably haven't tested it and you're guessing.

For example, for their 165W panels (which I use), Sharp publishes both
STC and PTC ratings.  They don't publish temperature coefficients but
I was able to obtain that from the mfg.  They do publish both solar
cell and module efficiency, but since the size of the panel is also
published, I didn't really need to know that.  I just counted up the
panels I needed and determined the area, making sure they would fit in
the available space.

So far as a factor for aging, I use 80% from their warranty (% rated
power output at 25 years).

By that I mean charts or equations to compute wire loss vs current and
wire size.  For the same current, larger wires = less losses.

For the same kW output, higher voltage --> lower currents --> less
losses.

As I alluded to above, since you will have the power output rating of
the panels, efficiency only comes into play if space available is
limited.

Posted by Martin Riddle on July 25, 2010, 3:05 am

Include Panel dust too.

I believe the efficiency turns out to be somwhere around 75% of the
72% if you want to be exact for a grid tie.

I think someone here , with the same question, was seeing values in this
range.

cheers

•
• Subject
• Author
• Date
 Re: calculating loses Ron Rosenfeld 07-16-2010
 Re: calculating loses David Lesher 07-17-2010
 Re: calculating loses Ron Rosenfeld 07-17-2010
 Re: calculating loses Martin Riddle 07-25-2010
 Re: calculating loses Martin Riddle 07-25-2010
 Re: calculating loses Ron Rosenfeld 07-25-2010
 Re: calculating loses Martin Riddle 07-25-2010
 Re: calculating loses Martin Riddle 07-26-2010
 Re: calculating loses Ron Rosenfeld 07-26-2010
 Re: calculating loses Martin Riddle 07-31-2010
 Re: calculating loses Ron Rosenfeld 07-31-2010