Maker Pro
Maker Pro

How big a panel would I need.....?

A

Andy Baker

Jan 1, 1970
0
Question:

Say... I have a 2 watt DC load that I need to run for 24 hours a day, in
the winter, being maybe only 9 hours of daylight (rise to set) Obviously a
battery will be employed. Is there a rule of thumb such as "Over-rate by
4,8,10X the amount of power" or something like that? I know a lot of it of
course depends on weather, but what's the opinion?


Andy in Maine
 
W

William P.N. Smith

Jan 1, 1970
0
Andy Baker said:
Say... I have a 2 watt DC load that I need to run for 24 hours a day, in
the winter, being maybe only 9 hours of daylight (rise to set) Obviously a
battery will be employed. Is there a rule of thumb such as "Over-rate by
4,8,10X the amount of power" or something like that? I know a lot of it of
course depends on weather, but what's the opinion?

Well, take your load (24 watt-hours per day) and divide by your
insolation (as expressed in peak-hours-per-day) which is (WAG) 3 for
Maine in winter, so you need 8 watts of solar panel minus various
inefficiencies and such. I wouldn't go with much less than 15 watts,
probably go with 20W just to be sure you cover battery inefficiencies
near peak charge, long outages, recharge rates, etc.

Battery and charge controller selection are pretty easy as well.
 
A

Andy Baker

Jan 1, 1970
0
wait so.... 24 watt-hours? but... 2 watts load.. wouldn't that be 48 w/h a
day? divided by 3.. gives me 16 - is that where you got the 15 watt panel?
or were you just saying "per watt" 8 watts of panel are needed for 24/7
operation?

Thanks!

Andy
 
A

Anthony Matonak

Jan 1, 1970
0
Andy said:
wait so.... 24 watt-hours? but... 2 watts load.. wouldn't that be 48 w/h a
day? divided by 3.. gives me 16 - is that where you got the 15 watt panel?
or were you just saying "per watt" 8 watts of panel are needed for 24/7
operation?
....

The logic isn't so very difficult to follow. You need to run a 2 watt
device 24 hours a day. This is 48 watt-hours. You get, perhaps, some
3 'sun-hours' of light a day in the middle of winter (on average) so
you would need (48/3=16) 16 watts of PV panels. Then you need to factor
in losses. First, the panels themselves are only guaranteed to produce
80% of their STC rating and, under average conditions, they usually
only produce around 80% of their STC rating due to heat build up. This
might not be as much of a factor in the middle of winter though. Then
the batteries are only around 80% or so at giving back the energy
you put in to them. This means that if you want 48 watt-hours/day
you might need (48/.80/.80/.80/3=31) 31 watts of PV panels. You might
drop two of those 80% (Heat derating & warranty rating) and it would
then be (48/.80/3=20) 20 watts of PV panels.

Anthony
 
W

William P.N. Smith

Jan 1, 1970
0
Andy Baker said:
wait so.... 24 watt-hours? but... 2 watts load.. wouldn't that be 48 w/h a
day? divided by 3.. gives me 16

Oops, sorry about that, I was using the 12-hour days from the universe
next door. 8*}

Le me try again

Some of that will depend on your insolation, maybe from
http://rredc.nrel.gov/solar/
 
F

Fred B. McGalliard

Jan 1, 1970
0
You will need to tilt the panel toward the sun, and you will have a large
loss from atmospheric absorption if the angle is too large.
 
A

Andy Baker

Jan 1, 1970
0
Hi Fred,

We're at 43 deg. north latitude, but in the winter, when it is sunny here,
the air is generally VERY clear. It seems to be the going opinion that for a
2 watt load to run reliably for 24 hour duty, I'd need to employ at least a
20 watt panel. This of course would be overkill for the summer, but only
adequate for the winter. Do you have any figures relating to the atmospheric
scattering losses by angle as they relate to PV systems?

Andy
 
F

Fred B. McGalliard

Jan 1, 1970
0
....
.. Do you have any figures relating to the atmospheric
scattering losses by angle as they relate to PV systems?
Not really. I have a fair physical feel that suggests more than 30 degrees
from vertical would drop the intensity by perhaps 30%, more in the UV (based
on my sensitivity to sunburn), so I am not sure what this would translate to
in a panel pointed to the sun. The PV takes it's energy from someplace in
the red on up, so loosing the UV would be disproportionately negative. Don't
the solar insolation sites have some info on this?
 
A

Anthony Matonak

Jan 1, 1970
0
Andy Baker wrote:
....
We're at 43 deg. north latitude, but in the winter, when it is sunny here,
the air is generally VERY clear. It seems to be the going opinion that for a
2 watt load to run reliably for 24 hour duty, I'd need to employ at least a
20 watt panel. This of course would be overkill for the summer, but only
adequate for the winter. Do you have any figures relating to the atmospheric
scattering losses by angle as they relate to PV systems?
....

Use mirrors (and maybe a tracker) and you can probably get by with a
10 watt panel.

Anthony
 
F

Fred B. McGalliard

Jan 1, 1970
0
Kent. You are designing the system to run for 6 dark days with only one
light day between. You would certainly not want to design for back to back
dark periods unless you had a real strong need for a never down system. I
would rate the system to store around 20-30% energy per day. Not quite so
bad. And the idea of mirrors. I think that would work very well. The mirror
would do the solar tracking for you and collect a bit more power, double or
triple the flat power level should be easy enough. Of course it is more
complex, but polished aluminum or plastic sheeting should be pretty cheap
compared to the cells. And they don't have to be used during the summer.
Makes the system match your needs better all year long.
 
B

Bill Kaszeta / Photovoltaic Resources

Jan 1, 1970
0
Hi Fred,

We're at 43 deg. north latitude, but in the winter, when it is sunny here,
the air is generally VERY clear. It seems to be the going opinion that for a
2 watt load to run reliably for 24 hour duty, I'd need to employ at least a
20 watt panel. This of course would be overkill for the summer, but only
adequate for the winter. Do you have any figures relating to the atmospheric
scattering losses by angle as they relate to PV systems?

Andy
Andy,

If you will post the actual location, I will in turn post a table
of irradiance vs. tilt by months so you can see the effects.


Bill Kaszeta
Photovoltaic Resources Int'l
Tempe Arizona USA
[email protected]
 
A

Andy Baker

Jan 1, 1970
0
Continuous as in security. At my current rural home, there's a "down
back"... and in this "down back" there's a trailer full of 7 kids, dogs,
cats, lice, junk cars, and a big dumb OxyContin dealer. The police will
eventually get to him, or so I've been told, but in the mean time, I've had
some pretty special moments when people show up at MY house at ANY hour
looking for Timmaahhh down back. The day I'm waiting for is when HIS dealer
drives up all the way from Lawrence Mass. because he's angry about
something. I need that extra 60 seconds to jump off the toilet and lock the
doors or call the cops.

The location in question is too far to hard wire and still have reliable.
Solar seems the only way to go.

Thanks!

Andy
 
W

William P.N. Smith

Jan 1, 1970
0
Andy Baker said:
Continuous as in security. At my current rural home, there's a "down
back"... and in this "down back" there's a trailer full of 7 kids, dogs,
cats, lice, junk cars, and a big dumb OxyContin dealer. The police will
eventually get to him, or so I've been told, but in the mean time, I've had
some pretty special moments when people show up at MY house at ANY hour
looking for Timmaahhh down back. The day I'm waiting for is when HIS dealer
drives up all the way from Lawrence Mass. because he's angry about
something. I need that extra 60 seconds to jump off the toilet and lock the
doors or call the cops.

You need a sign at the appropriate driveway intersection point to
"Timmaahhh" thatway and "Andy" thisaway. 8*)
The location in question is too far to hard wire and still have reliable.
Solar seems the only way to go.

How about charging your house batteries from your car when you are
there?
 
A

Andy Baker

Jan 1, 1970
0
Cool!!! Thanks!
House of Andy is located at: Latitude: 43.584232 Longitude: 70.624391

Thank You!

Sincerely,

Andy Baker
Sagoma Technologies &
Gelardi Design & Development
65 Landry St.
Biddeford, ME 04005
PHONE 207-283-1760
FAX 207-283-1762
MOBILE 207-423-5078
 
A

Andy Baker

Jan 1, 1970
0
A sign.... what a good idea! a hell of a lot cheaper too!! I could do a
bunch of solar powered blinking LED's in it...... Or make it out of neon
tube and just hook it to a solarcharged electric fence zapper! Now not so
cheap though.
 
J

john

Jan 1, 1970
0
go to uyour auto shop or a computer shop and check out there leds and
neons they are all 12 volt i have installed a heap of led car number
plate lights around my place low wattage gentle light
ut if you want colour and stuff try out the computer mods you can get all
low wattage theres everything from neon tape and wire to multi colour
leds with sound ac tivation
 
M

m II

Jan 1, 1970
0
Andy said:
Cool!!! Thanks!
House of Andy is located at: Latitude: 43.584232 Longitude: 70.624391


Much more accuracy than that is needed. At the very MINIMUM, we use
TWELVE DIGIT accuracy in conjunction with a portable atomic clock.

It only makes sense..consider the rather short wavelength of light. Now
consider the rather coarse measurements used in land surveying. See the
problem? It's angstroms versus agriculture.

Now, let see some EFFORT to improve this rather slipshod measuring..





mike






























I really should be ashamed of myself.
 
W

Windsun

Jan 1, 1970
0
The panel produces power at the Cosine of the angle. So at 30 degrees your
panel produces 86% of full power.
[/QUOTE]
Not really. I have a fair physical feel that suggests more than 30 degrees
from vertical would drop the intensity by perhaps 30%, more in the UV
(based[/QUOTE]
 
W

William P.N. Smith

Jan 1, 1970
0
Ron Rosenfeld said:
20% fudge factor for system losses and inefficiencies --> 4.8 AH adjusted
daily load.

Worst month insolation --> 3 hrs

Required array current --> 4.8/3 = 1.6A

Plus the additional current to get it charged back up after those long
days of fog, plus something more for the inefficiencies of batteries
near the top of their charge state.
 
A

Andy Baker

Jan 1, 1970
0
It's close enough for a cruise missile... smart ass :) Question - what is
it that you do that you need an atomic clock for anyway? navigating
spacecraft?

Andy
 
Top