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Figuring Output from a Given Surface Area

On Thu, 6 Aug 2009 21:35:59 -0700 (PDT), Tmort <...@sbcglobal.net
I'm looking into the feasability of using photovoltaics at an
industrial site. I've ferreted out some information on nanosolars
panels as I understand they are the least expensive. According to
them the payback is less than a year. My calculations show that over
the life expectancy of the panels (25 years) that the savings is not
much more than the cost of the panels even not including installation
and maintenance costs.

I'm not sure I'm doing this correctly. Could someone take a look at my
work and comment?

nanosolar $0.99 per watt

1.6 x 0.8 m panel size = 142.9 watts 111.64 watts/m2 = $111 per m2

$0.062088 Jun-Sep per kWh
$0.049888 Oct-May per kWh Plant 46501 m2

Solar Insolation Final
kWh/m²/day Temp °C Efficiency kWh/m²/day kWh $

Jan 1.64 -7.39 9.50% 0.16 217,346
$10,843
Feb 2.45 -5.39 9.50% 0.23 324,693
$16,198
Mar 3.48 0.34 9.50% 0.33 461,197
$23,008
Apr 4.32 7.38 9.50% 0.41 572,520
$28,562
May 5.14 14.09 9.50% 0.49 681,193
$33,983
Jun 5.49 19.45 9.50% 0.52 727,578
$45,174
Jul 5.79 21.67 9.50% 0.55 767,336
$47,642
Aug 4.78 20.46 9.50% 0.45 633,483 $39,332
Sep 3.87 16.33 9.50% 0.37 512,883
$31,844
Oct 2.55 9.79 9.50% 0.24 337,946
$16,859
Nov 1.64 2.83 9.50% 0.16 217,346
$10,843
Dec 1.34 -4.38 9.50% 0.13 177,587
$8,859

5,631,108 $313,148.35 /yr

$5,139,486.70 Total cost for panels (plus installation)
$205,579.47 per year

$107,568.88 Annual savings (less installation and
maintenance)

Thanks



On Wed, 12 Aug 2009 23:31:46 +0100, Eeyore <...@notcoldmail.com
How can you get similar figures without knowing the location of the panels,
their orientation and the resulting insolation ?

Nanosolar only claim 75 peak watts for a 1 m2 panel, which translates to 96
peak W for a panel of that size.

Graham

--
due to the hugely increased level of spam please make the obvious adjustment to
my email address

On Thu, 13 Aug 2009 06:20:49 -0400, Ron Rosenfeld <...@nospam.org
On Wed, 12 Aug 2009 23:31:46 +0100, Eeyore
<...@notcoldmail.com

I used the OP's figures for insolation. It should have been obvious that I
have no way of checking his DATA. And he asked to have his calculations
checked.

Again, I used the OP's figures. If his data is incorrect ... GIGO
--ron

Anonymous Wrote:
On Thu, 6 Aug 2009 21:35:59 -0700 (PDT), Tmort
<...@sbcglobal.net

Not much point using Nanosolars panel prices because you simply cant
buy them.
All Nanosolar have claimed is that they can make Solar panels for $1 a
watt.
They have never stated they will sell them for $1 a watt.
Currently, there isnt even any independant evidence that Nanosolar can
make panels for $1 a watt.
Its just an unverifiable claim.

On Fri, 7 Aug 2009 16:58:21 -0700 (PDT), Tmort <...@sbcglobal.net
I was wondering about that. Things seemed a little murky. That's not
really the point of my question though. Even with nanosolars low
price, it still looks like the value of the electricity generated
isn't much more than the cost of the panels over their life (25 yrs).
It was my understanding that solar actually would have a payback
although over a long period. I'm assuming I'm doing something wrong.

Is there any rule of thumb on:

1-How much power can per generated per square meter if you know the
insolation?
2-What is the price per square meter for solar, installed including
inverters etc?
3-What is the life expectancy?
4-What are the annual maintenance costs per square meter (or watt)?

The site I am thinking of uses lots of electricity. Pretty much all
the energy savings things have been implemented. The site does have a
large footprint so I'm wondering if installing panels and selling back
to the grid could yield any net savings in costs, if not energy
consumed.

Thanks

On Fri, 07 Aug 2009 22:49:42 -0500, Morris Dovey <...@iedu.com
I'm not a PV expert (and my primary interest is in solar thermal), but
I'll take a shot at pointing you in the right direction for this one
question...

If you know the incident power, multiply that by the PV panel's
solar-range, I think) to get an approximation of panel output.

Next subtract all the identifiable system losses (resistive losses in
wiring, charge controller, electrical-losses, storage battery losses, chemical-losses, inverter losses, etc).

Use the above iteratively to guesstimate the number of panels needed -
and once you have that, you should be able to approximate the power per
m^2 that you're after.

When you're all done, I predict that you'll agree with Mauried. :)

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

Anonymous Wrote:
On Fri, 7 Aug 2009 16:58:21 -0700 (PDT), Tmort
<...@sbcglobal.net

Bottom line is that Solar Power is the most expensive way known of
making electricity.
If you have grid power, no point going solar.
Even Wind Power is better than Solar.
Solar is only good if there is no grid power, or any other
alternative.
About the only way Solar works is if you are entitled to some kind of
subsidies or rebates, or can sell the power back into the grid at some
premium price.

On 8 Aug 2009 22:44:01 -0500, boB <...@bob.bob
Well, they're not the cheapest, but potato batteries and hydrogen are
most likely way more expensive.

Not if your grid is weak.

That's a blanket statement. It's not better if you
don't have any wind.

Price of solar IS coming down. Of course, you gotta have sun in order
for it to work. Just like you have to have wind for wind and water head
for hydro.

bob

>

On Thu, 13 Aug 2009 17:12:05 +0100, Eeyore <...@notcoldmail.com
Not to mention that increased demand for silicon wafers will increase the
price of these too ( law of supply and demand ).

Graham

--
due to the hugely increased level of spam please make the obvious adjustment
to my email address

Discussion Title: Figuring Output from a Given Surface Area
Title Keywords: Figuring  Output  from  Given  Surface  Area 


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