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How many watts soler ???

  • Thread starter no one that you know
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N

no one that you know

Jan 1, 1970
0
To get 2 kwh per day north of 50@115 degrees stored in batts.
How many watts of 12 volt panels?


also has anyone or does anyone know of what type of panel that can
produce some power from the moon or a street light out front of there
house. (This is a night light experiment.)
I upgraded my camper with led... about 21 fixtures in total. My guess is
about 300 12v led's all together......I couldn't believe that a crappy
tire five watt panel could make them all light up no problem. Was just
thinking that a bigger panel whould make a simpler low wattage
night/accent lighting project for the home if a panel could react to low
light. I dont expect full power from the panel but I would size the
panels to get the wattage I need for night time use and aim them in the
daytime for max to store power.
 
E

Eeyore

Jan 1, 1970
0
no said:
To get 2 kwh per day north of 50@115 degrees stored in batts.
How many watts of 12 volt panels?

The first thing you need to know is the insolation for your area ( how much
light it gets ).

You'll get a lot less in winter btw, so size with that in mind.

Graham
 
N

no one that you know

Jan 1, 1970
0
Eeyore said:
The first thing you need to know is the insolation for your area ( how much
light it gets ).

You'll get a lot less in winter btw, so size with that in mind.

Graham

Thats why I gave the cordinates....I live in sunny south western
Alberta.
I am allready taking into account for twice my needs to make up for
losses and cloudy days.
I believe my area is about average 6 usable hours per day. I cant
remember where I got that from.
 
M

Malc

Jan 1, 1970
0
no said:
To get 2 kwh per day north of 50@115 degrees stored in batts.
How many watts of 12 volt panels?

<Googles for definition of kwh>

Do you seriously want 2000 watts per hour? If so you'll need assuming 100%
efficient panels and inverters etc 2000 W of solar panels. But as you want
it for the whole day then you'll need probably at least 3 times as much and
then some because of inefficiencies.
 
N

no one that you know

Jan 1, 1970
0
Huh? 2000 watts a day 2kwh a day what am I missing or did you miss it?
 
E

Eeyore

Jan 1, 1970
0
no said:
Huh? 2000 watts a day 2kwh a day what am I missing or did you miss it?

In winter you may only get a few hours of useful sunlight ( say just a few hours )
so you'll need one heck of a lot of panels.

Graham
 
E

Eeyore

Jan 1, 1970
0
malc said:
Well I may have got confused.
Yes.

The definition of a kwh is 1 kw per hour
http://whatis.techtarget.com/definition/0,,sid9_gci797759,00.html

No it isn't. There is no such thing as a 'kW per hour' at all !
It says............ " equivalent to one kilowatt (1 kW) of power expended for one
hour (1 h) of time "

So it's power *times* time not power divided by time.
what he may mean is 2000 Ah although this is still quite a lot to store.

No he doesn't. He's talking about *energy*.

Graham
 
D

daestrom

Jan 1, 1970
0
malc said:
Well I may have got confused. The definition of a kwh is 1 kw per hour
http://whatis.techtarget.com/definition/0,,sid9_gci797759,00.html what
he may mean is 2000 Ah although this is still quite a lot to store.

No, you've misread the definition (and that page's wording isn't all that
great).

Energy is power *times* time. A kilowatt-hour is a measure of energy. A
kilowatt is a measure of power.

When you say, "...a kwh is 1 kw per hour", you are using the word 'per'.
And mathematically, that means 'divided'. So you are essentially saying
"...a kwh is 1 kw divided over an hour". And that is wrong, it makes no
sense. It would be like saying an amp-hour (AH) is an amp divided over an
hour. Doesn't make sense.

An amp-hour is much like a kilowatt-hour. It is a measurement of a flow
*multiplied* by a time period. An ampere is a measure of the flow of
current, a kilowatt is a measure of power (which is the flow of energy).
When you multiply either by time, you get a measure of the total amount of
what has 'flowed'. In the case of amp-hour, since amperes is a measure of
electric-charges flowing in a given amount of time, when you multiply it by
the length of time it has been flowing, you get a measure of how many
electric charges have gone past a point in the wire (such as into or out of
a battery). In the case of kilowatts, since a kilowatt is a measure of
energy flowing in a given amount of time, when you *multiply* it by the
length of time it has been flowing, you get a measure of how much energy has
gone past a point in a wire.

A usage of 2kwh in one day means the OP uses just 2kwh of energy over a 24
hour period. The fact that he's specified his energy usage over a 24 hour
period, using a unit of measure that has a 'time unit' built into the name
*does* add a lot to the confusion. One could use energy at the rate of 2
kilowatts for one hour. That would be a total energy of 2 kwh. Or one
could use energy at the rate of 100 watts. After twenty hours, they would
have used 2 kwh.

To use 2 kwh in one day, you could use energy at the rate of 4000 watts for
just 1/2 an hour, then turn everything off for the remaining 23.5 hours. Or
you could use energy at the rate of 200 watts for 10 hours and then turn
everything off. Or you could use energy at the rate of 83.3333 watts for a
full 24 hours. All of these will total the same amount of *energy*, 2
kilowatt-hours. But of course the equipment to support those different
power levels, and the losses change with the different power levels.

If the OP gets just 5 hours of good, strong sunlight, then if he can collect
solar energy at the rate of 400 watts, he would collect 2 kwh of energy in
those five hours of sunshine. Of course if he's storing it and using an
inverter, there will be losses in each step of the process. So he may want
to collect solar energy at a higher rate, say 500 watts. Then over the same
5 hours of good strong sunlight, he will have collected 500*5/1000 = 2.5 kwh
of energy. Some of that energy will be 'lost' in inefficiencies in storage
battery so he may get just 2kwh 'delivered' to his loads. If he wants to
collect enough energy in one day to support two days of operation (prepare
for a cloudy day tomorrow), he might have to collect energy at the rate of
1000 watts. Then over the same 5 hours of good strong sunlight, he will
collect 5*1000 / 1009 = 5 kwh. He can then use 2.5 kwh of that energy today
(2 kwh in loads and 0.5 kwh in losses), and the remaining 2.5 kwh of energy
tomorrow.

Hope this clears up some of the confusion...

daestrom
 
N

no one that you know

Jan 1, 1970
0
I need 400 kwh per year minimum. 400/365=1.096 round up= 2 kwh per day
or 730 kwh per year. This is where I get lost ...if I have minimum 5
usable hours a day (lots of days with more) is it 2000w/5hr= 400w ?
 
N

no one that you know

Jan 1, 1970
0
Thanks Graham.
I think I have it figured out? I would need minimum 400 watts PV with 5
hours a day minimum usable sun. If I spend a total of 4000 buks for
panels and batts and stuff etc... at todays prices for power that is
about 40 years of power I am paying for in advance.
This a hobby I want to start. I have been researching it for years.
It would be the 2nd hobby that I had that has some payback so no big
deal if I ever see payback. I am going to make my fridge green powered.
It is a small apartment size fridge and uses very little power. I really
dont care if it reduces my power bill becasue my power bill is minor
anyways.
I would get two things that I need more.... like a hobby and back up
power when the grid goes down. I have wasted far more on stupid shit
like the big dish and computers....bet all told since 1980 when the big
dish came out I spent 20k and now it's all in the garbage.....and when
win95 came out lol I remember running out and buying 8 meg of ram for
400 buks. I got caught up in the upgrade upgrade upgrade upgrade craze
just of the top of my head thats another 20k. So somehow me thinks PV is
a better hobby lol!
I just happen to have the ideal location for panels on my bungalow. I
have easy acsess to the roof the angle and direction is perfect for all
year round use. Even when it snows the drift side is on the other side
that I cant use. The locals call that the "east wind" are these ones any
good
http://cgi.ebay.ca/Sun-Elec-SUN-180...2QQihZ009QQcategoryZ41981QQrdZ1QQcmdZViewItem
? Or these ones?
http://cgi.ebay.ca/Solar-Panels-160...5QQihZ014QQcategoryZ41981QQrdZ1QQcmdZViewItem
 
N

no one that you know

Jan 1, 1970
0
oops I just screwed up ignore my last post. The math is all wrong.
I need 730 kwh of 120vac per year. @ 12 volts how many watts of PV is
that per 5hr a day.
I want to make 2kwh of 120 vac to use over 24 hours from 12 volt PV.
I can figure out the dollars. I cant figure out how many watts of dc to
get 2 kwh of ac at 5 hours a day. Dont worry about the cloudy days I can
figure that out. We are talking a 60 kwh per month home/cabin system.
 
A

Anthony Matonak

Jan 1, 1970
0
no said:
I need 400 kwh per year minimum. 400/365=1.096 round up= 2 kwh per day
or 730 kwh per year. This is where I get lost ...if I have minimum 5
usable hours a day (lots of days with more) is it 2000w/5hr= 400w ?

Technically, yes, this is how it works.

Unfortunately, PV panels are sold based on a STC rating (Standard
Test Conditions) which do not exactly match real world conditions.
The main difference is that solar insolation isn't usually quite
as much as they use in the test and panel temperature is usually
higher. Both of these factors conspire to make a PV panel sold as
400W produce somewhat less than 400W. A safe derating would be to
assume the panel produces only 80% of it's STC rating.

Then you'll need to account for losses in your system design.
Batteries only return around 80% (more or less) of the energy
that is put into them and if you use an inverter you may only
expect 90% (or less) of the energy out of it as AC that you
put in as DC.

Then there is the question of 'usable hours a day'. If you mount
the panels in a fixed position, which is most common, then the
angle of the sun at any time other than high noon will mean the
panels are receiving less than full sunlight. Typically solar
insolation maps and measurements would account for this by saying
something like 5 kWh/m^2/day on average. This means that even
though the day was 12 hours long, the amount of sunlight that
falls on a flat panel is only equivalent to 5 hours of direct
high noon sunlight. Some people call this 'sun-hours' as opposed
to 'hours of sunlight' because it doesn't really measure how
long the day is but how much sunlight energy there is.

So, let's assume you live in an area that gets 5 kWh/m^2/day
on average over the year, with a minimum of 3 kWh/m^2/day in
the winter. If you need 2kWh/day even in the depths of winter
and don't use a tracking mount (which adds some 30% by facing
the sun all day) and you are using an inverter to produce AC
then it figures out to something like the following...
2,000Wh/dayAC / .90(inverter) / .80(batt) / .80 (PV) / 3 sun-hours
1,157 STC Watts of solar PV.

Add a tracker and then sun-hours becomes around 4 instead of 3
and you would only need 868 STC Watts of PV.

Drop the inverter and only use DC with a tracker and it becomes
781 STC Watts of PV.

Add a small generator that you use a few hours every couple of
days in the dead of winter and not at all in the summer months
and you can drop the PV requirement some more. Adding a small
wind or water turbine, if you can, would also help.

Anthony
 
no one that you know said:
I need 400 kwh per year minimum. 400/365=1.096 round up= 2 kwh per day
or 730 kwh per year. This is where I get lost ...if I have minimum 5
usable hours a day (lots of days with more) is it 2000w/5hr= 400w ?

Almost. It's 2000 Wh divided by 5 h. See how the h's cancel out?

Nick
 
Dale Eastman said:
2KW
---
3hr

Invert and multiply
2KW 2KW 3hr
--- = --- X ---
3hr 1 1

Simplify:

2KW 2KW 3hr 2KW x 3hr 6KWhr
--- = --- X --- = --------- = -----
3hr 1 1 1 x 1 1

2KW
--- = 6KWhr
3hr

Very creative...

Nick
 
D

daestrom

Jan 1, 1970
0
Dale Eastman said:
One KW per hour:

1KW
---
1hr


Invert and multiply

1KW 1KW 1hr
--- = --- X ---
1hr 1 1

Your trouble is right here....

You have
1KW
---
1 hr

You want to invert and multiply, but what is the inverse of '1 hr'?? It is
*NOT* ...

1 hr
--
1

The inverse of '1 hr' is....

1
--
1 hr


You're doing two inversions, you're taking the denominator, '1hr' and saying
it is really
1
--
1hr

when it is not, so when you 'invert' it you're getting

1hr
--
1


But that is *WRONG*.

The denominator is '1hr', period.

As nick tried to show you...

2 2 1
-- = -- X --
3 1 3

The denominator on the left is '3'. The inverse of that is 1/3.
Multiplying, the results is simply

2 2
-- = --
3 3

Which is a valid equality. The fact that nick came up with....

2 2 3
-- = -- X -- = 2X3 = 6
3 1 1

Since clearly,

2
-- is not equal to 6
3

means that the method of 'invert and multiply' you used is flawed.



What you *may* be thinking of is something like....

2 2 5 10
-------- = --- X --- = ---
3 1 3 3
--
5
This is the more common use of 'invert and multiply'. It is used to get rid
of the fraction in the denominator and simplify.


Your 'trick' gives you the 'right answer for the wrong reason'. If you try
to use it to solve the problem in reverse, you may be in for trouble....

2 kw 6kwh
----- = -------
x hr 1

Solving for 'x', we might just 'cross-multiply' and set the products equal
to each other....

2 kw X 1 = 6 kwh X 'x' hr

'x' hr 2kw 1
= ---- = ---
6 kwh 3 hr

But the answer *should* be 3 hr, not 1/3 hr. Yes, if you try hard enough
you can make 2 / (1/3) equal to six again, but think about this. Drawing 2
kw for *less* than an hour can't possibly be more than 2 kilowatt-hours of
energy.

daestrom

P.S. Hope all the 'spacing' and such comes out, I'm not good with 'ASCII
art'.
 
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