The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question
is how would I now figure out approx how large of a panel I would need to
keep the battery charged. I am really looking for the formula and a
understanding of how you got from A to B as oppose to "this panel should
work".


--
Poor planning on your part does not constitute an emergency on my part.

I'm not sure you realize it but some of your figures/formulae are
unnecessary.

More simply Battery 110 AH (/2) Load = .5 A 55/.5 = 110 hrs

Typically batteries are charged at .10 of their 1 hr rate (when fully
discharged) but the type of battery (Gel-Cell, NiCad, NiMH, etc.), must
be considered. Also, while .1C might be good for a discharged battery,
the charge should diminish as the battery becomes fully charged (or some
other mechanism, temperature rise, e.g.), must shut down the charger.)

See the battery mfr. for recommended charge rate and charger technology.
Also, consider how fast a recharge you want. A 4 ampere charger will
recharge the battery (but slowly) as will a 100 Amp charger

Bryan Martin wrote:
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question
is how would I now figure out approx how large of a panel I would need to
keep the battery charged. I am really looking for the formula and a
understanding of how you got from A to B as oppose to "this panel should
work".

– Colonel – wrote:
On 2006-01-05 16:24:35 -0500, "Bryan Martin"
said:

The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day


Wrong. 6W x 24hours = 144 Watt-HOURS per day.

That should be 6W x 24hr = 144 watt-hours.

hvacrmedic




Watts are a unit of power. A watt-hour (like a kilowatt-hour) is a unit
of energy.

Hope this helps.

Thanks for the replies. Unless I am missing something the formula is sound
but not worded correctly? I will try to find more info on this. The reason
behind the extended formula's is for better understanding and also the end
result and second part of this question is that I plan on hooking this up to
a solar panel if possible. I was trying to get an idea about what I would
be using per day/hour if run 24/7 and what size panel I would need for this.
Then depending on how that works out $$$ wise for the size panel I would
either stay at that or look at some kind of timer setup that would allow the
unit to run for X hours a day. I would rather have it on 24X7 but the
actual real world usage of the unit would be around 3 hours per day on a
weekday and posssibly more on the weekends.

So I am still needing some help with the panel size I will be needing to
keep this thing up and running.

So given 144 watt-hours per day would that work out to?

144 / 5 hours of sunlight = 28.8 W panel?


"Bennett Price" wrote in message
. ..
I'm not sure you realize it but some of your figures/formulae are
unnecessary.

More simply Battery 110 AH (/2) Load = .5 A 55/.5 = 110 hrs

Typically batteries are charged at .10 of their 1 hr rate (when fully
discharged) but the type of battery (Gel-Cell, NiCad, NiMH, etc.), must be
considered. Also, while .1C might be good for a discharged battery, the
charge should diminish as the battery becomes fully charged (or some other
mechanism, temperature rise, e.g.), must shut down the charger.)

See the battery mfr. for recommended charge rate and charger technology.
Also, consider how fast a recharge you want. A 4 ampere charger will
recharge the battery (but slowly) as will a 100 Amp charger

Bryan Martin wrote:
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next
question is how would I now figure out approx how large of a panel I
would need to keep the battery charged. I am really looking for the
formula and a understanding of how you got from A to B as oppose to "this
panel should work".

"Bryan Martin" wrote in message
news
Thanks for the replies. Unless I am missing something the formula is
sound but not worded correctly?

Yep. I think you have the basic idea, but in this stuff you have to be
careful to dot your 'i's and cross your 't's. (this group is good at
pointing out mix ups like this). It may seem rather 'anal' at times, but it
*is* important to get the right results ;-)

I will try to find more info on this. The reason behind the extended
formula's is for better understanding and also the end result and second
part of this question is that I plan on hooking this up to a solar panel
if possible. I was trying to get an idea about what I would be using per
day/hour if run 24/7 and what size panel I would need for this. Then
depending on how that works out $$$ wise for the size panel I would either
stay at that or look at some kind of timer setup that would allow the unit
to run for X hours a day. I would rather have it on 24X7 but the actual
real world usage of the unit would be around 3 hours per day on a weekday
and posssibly more on the weekends.

So I am still needing some help with the panel size I will be needing to
keep this thing up and running.

Well, you know that you need about 144 watt-hours per day on average. But
how many hours a day does the sun shine in your area? That is the another
factor needed to figure out the panel sizing.

If you're drawing 144 watt-hours *out* of your battery, how many watt-hours
do you have to put in? More, unless you've found the perfect battery.
Assume 80% of the energy you put into the battery gets lost, so to get 144
watt-hours *out*, you should plan on putting 144/0.80 = 180 watt-hours *in*.

Now, perhaps the worst season would be winter, and perhaps in your area the
sun only shines, 'on average' 3 full-sun hours a day in winter time. To get
180 watt-hours of energy to put *in* your battery from a charge controller,
in just 3 hours, you would need a charge controller output of 180
watt-hours / 3 hours = 60 watts. (at 12V, that would be 5 amps).

Is this just for a year or two, or longer? The reason I ask, is that panel
output degrades over time. Some manufacturers warrant for 80% after 20
years. I suspect you aren't planning some 20-year long plan here, so lets
just *assume* panel output is 90% of rating under the conditions you'll be
operating. So to be sure of 60 watts at the charge controller output, 60
watts / 0.90 = 67 watts panel output.

Of course, some days your sunshine will be more, and that will help recharge
your battery. But if you run real close to just those 3 hours per day, then
you're going to have a hard time recharging some weeks in the winter/cloudy
season. Moral of this is, if you have another means to fully charge your
battery sometimes, you might be okay. But if the panel is the sole
charging, and no interruption of power can be tolerated, you'll need even
more panels to manage recharging on those partly-cloudy days.

Hope this helps you along.

daestrom

allow lots of money for battery replacements.

better to upsize battery bank and discharge less.

even deep cycle batteries have tough tme with frequent deep discharges

Bryan Martin wrote:
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watt-hours per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

Walmart has 115 ah deep cycle batteries for $55


We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question
is how would I now figure out approx how large of a panel I would need to
keep the battery charged. I am really looking for the formula and a
understanding of how you got from A to B as oppose to "this panel should
work".



Your math is correct. In NY, with about 3 full sun hours daily max, best
case scenario I'd need a 50 watt panel. 50 watts * 3 hours = 150
watt-hours daily.

--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html

wrote:
allow lots of money for battery replacements.

better to upsize battery bank and discharge less.

even deep cycle batteries have tough tme with frequent deep discharges


Using $85 T105's (6v * 225ah) at 50% discharge, he's good for 7+ years.


--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html

daestrom wrote:
"Bryan Martin" wrote in message
news
Thanks for the replies. Unless I am missing something the formula is
sound but not worded correctly?


Yep. I think you have the basic idea, but in this stuff you have to be
careful to dot your 'i's and cross your 't's. (this group is good at
pointing out mix ups like this). It may seem rather 'anal' at times, but it
*is* important to get the right results ;-)


I will try to find more info on this. The reason behind the extended
formula's is for better understanding and also the end result and second
part of this question is that I plan on hooking this up to a solar panel
if possible. I was trying to get an idea about what I would be using per
day/hour if run 24/7 and what size panel I would need for this. Then
depending on how that works out $$$ wise for the size panel I would either
stay at that or look at some kind of timer setup that would allow the unit
to run for X hours a day. I would rather have it on 24X7 but the actual
real world usage of the unit would be around 3 hours per day on a weekday
and posssibly more on the weekends.

So I am still needing some help with the panel size I will be needing to
keep this thing up and running.


Well, you know that you need about 144 watt-hours per day on average. But
how many hours a day does the sun shine in your area? That is the another
factor needed to figure out the panel sizing.

If you're drawing 144 watt-hours *out* of your battery, how many watt-hours
do you have to put in? More, unless you've found the perfect battery.


Assume 80% of the energy you put into the battery gets lost,

20% efficiency for a battery is pretty poor, Daestrom. Perhaps you would
like to rephrase this.


so to get 144
watt-hours *out*, you should plan on putting 144/0.80 = 180 watt-hours *in*.

Now, perhaps the worst season would be winter, and perhaps in your area the
sun only shines, 'on average' 3 full-sun hours a day in winter time. To get
180 watt-hours of energy to put *in* your battery from a charge controller,
in just 3 hours, you would need a charge controller output of 180
watt-hours / 3 hours = 60 watts. (at 12V, that would be 5 amps).

Is this just for a year or two, or longer? The reason I ask, is that panel
output degrades over time. Some manufacturers warrant for 80% after 20
years. I suspect you aren't planning some 20-year long plan here, so lets
just *assume* panel output is 90% of rating under the conditions you'll be
operating. So to be sure of 60 watts at the charge controller output, 60
watts / 0.90 = 67 watts panel output.

Of course, some days your sunshine will be more, and that will help recharge
your battery. But if you run real close to just those 3 hours per day, then
you're going to have a hard time recharging some weeks in the winter/cloudy
season. Moral of this is, if you have another means to fully charge your
battery sometimes, you might be okay. But if the panel is the sole
charging, and no interruption of power can be tolerated, you'll need even
more panels to manage recharging on those partly-cloudy days.

Hope this helps you along.

daestrom

daestrom wrote:

Yep. I think you have the basic idea, but in this stuff you have to be
careful to dot your 'i's and cross your 't's. (this group is good at
pointing out mix ups like this). It may seem rather 'anal' at times, but it
*is* important to get the right results ;-)

I agree...

If you're drawing 144 watt-hours *out* of your battery, how many watt-hours
do you have to put in? More, unless you've found the perfect battery.
Assume 80% of the energy you put into the battery gets lost, so to get 144
watt-hours *out*, you should plan on putting 144/0.80 = 180 watt-hours *in*.

I know what you're thinking and you'd get it right. But you worded it
backwards.

Perhaps something like the following is what you meant:
"Assume the battery storage is 80% efficient, 20% of the energy you
put into the battery gets lost"

Bryan Martin wrote:

The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question
is how would I now figure out approx how large of a panel I would need to
keep the battery charged. I am really looking for the formula and a
understanding of how you got from A to B as oppose to "this panel should
work".


Hi,
In the case of float charging, your charger has to generate higher
output than constant current draw rate plus some + tolerance to
compensate loss. Otherwise the battery will run down eventually.
Also think about winter, summer situation.
Tony

Hi Bryan

On Thu, 05 Jan 2006 21:46:19 -0500, Steve Spence
wrote:


Your math is correct. In NY, with about 3 full sun hours daily max, best
case scenario I'd need a 50 watt panel. 50 watts * 3 hours = 150
watt-hours daily.


Steve and others have pointed out that your basic maths is correct,
but there is a matter of losses to be considered.
One that hasn't been mentioned I think in this thread is that a 60
watt panel does NOT put in 60 watts to your 12v battery unless you use
a clever charge controller like an MPPT (which you wouldn't use for
one panel). This is because panels are not rated at the voltage of the
battery, but at some higher voltage called the maximum power voltage.
One person commented that you should look at the CURRENT provided by
the panel in your situation (rather than the wattage).
A typical figure is for a panel to put about three quarters of its
rated output into the battery (45 watts for a 60 watt panel).
If you need about a 5 amp panel, this will be a panel rated at about
85 watts (= 5 amps X 17 volts).
Personally, I would aim at about a 75 to 80 watt panel for what you
want to do. The difference in price may turn out to be not all that
great anyway as you pay more $ per watt for smaller panels usually.

Eric

Thanks everyone for shedding such light on it. I think I have enough info
to get started


"Bryan Martin" wrote in message
...
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next
question is how would I now figure out approx how large of a panel I would
need to keep the battery charged. I am really looking for the formula and
a understanding of how you got from A to B as oppose to "this panel should
work".


--
Poor planning on your part does not constitute an emergency on my part.

"Tony Wesley" wrote in message
ups.com...

daestrom wrote:

Yep. I think you have the basic idea, but in this stuff you have to be
careful to dot your 'i's and cross your 't's. (this group is good at
pointing out mix ups like this). It may seem rather 'anal' at times, but
it
*is* important to get the right results ;-)

I agree...

If you're drawing 144 watt-hours *out* of your battery, how many
watt-hours
do you have to put in? More, unless you've found the perfect battery.
Assume 80% of the energy you put into the battery gets lost,
so to get 144
watt-hours *out*, you should plan on putting 144/0.80 = 180 watt-hours
*in*.

I know what you're thinking and you'd get it right. But you worded it
backwards.

Perhaps something like the following is what you meant:
"Assume the battery storage is 80% efficient, 20% of the energy you
put into the battery gets lost"

You're right of course. I meant to say that perhaps only 80% of the energy
you put into a battery is recoverable. So to get 144 watt-hours out, you
would indeed need to put in 144/0.80 = 180 watt-hours in.

daestrom

On Thu, 05 Jan 2006 21:24:35 GMT, "Bryan Martin"
wrote:

The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question
is how would I now figure out approx how large of a panel I would need to
keep the battery charged. I am really looking for the formula and a
understanding of how you got from A to B as oppose to "this panel should
work".

I would approach it a bit differently. And I think your error in your
calculations is an assumption that the battery charge/discharge is 100%
efficient. It's closer to 80%.

1. Daily Load is 12AH (@ 12V) as you calculated. But with DC loads you
can do your calculations directly in Amps. So 0.5A * 24 hrs = 12AH (@12V).

2. Add 20% for system losses and a safety factor -- 14.40 AH. You don't
get 100% efficiency in charging/discharging batteries.

3. Calculate your PV requirement based on the number of effective sun
hours at your site during the worst month. Not knowing where you live it's
tough to say, but, for example, the San Francisco area that might be 4 hrs.

That gives you a required PV array current of 3.60 amperes (14.40/4). Now
you can go look for an appropriate PV panel (or two, if need be). A BP365
panel is rated at 3.69A at PMax, as an example.

4. Now decide on your DOA (days of autonomy) to obtain required battery
size. 5 DOA with a 50% maximum discharge -- 144AH battery.

Or, working backwards, your 110AH battery would give you 3.8 DOA.


-- ron (off the grid in Downeast Maine)

Your calculations are a bit off, because batteries are rated at the 20 hour
rate, but you are assuming about a 120 hour discharge time, so the actual
capacity will be somewhat more - around 20%.

But that won't really affect the basic calculations - all it does really is
give some safety factor.

The panel size is a bit more complicated, but not rocket science. First you
need to determine how important the load is - will it be a major event if
whatever you are running goes offline. If so, then you need to factor in a
pretty large safety factor - 35% or more - for those rare occasions when you
get much longer than usual cloudy periods. Basically, 100% uptime will cost
you a lot more than 99% uptime.

But ignore that for now - the basic calcuation is that you want your solar
panel to supply what you use each day + around 20%. That is for the WORST
case conditions - such as you might see seasonally with lots of clouds.

Assuming you look at the insolation charts, and you get worst case in
December of 4 hours full sun per day. You are drawing .5 amps over 24 hours,
or 12 amp-hours. You need to replace that 12 AH in during the 4 hours of sun
you get, so you need a panel that will supply 12/4 = 3 amps. Add 20-25% or
so safety factor, and you come up with around 3.5 amps. Most panels are
rated at 17 volts or so, so 3.5 amps x 17 volts = about a 60 watt panel.

-------------------------------------------------------------------------
Free Solar Discussion Forum: http://www.wind-sun.com/smf/index.php
-------------------------------------------------------------------------

"Bryan Martin" wrote in message
...
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

It does not work that way because panels are rated at 17 volts, not 12. You
have to work with the panels amps not watts.
---------------------

Your math is correct. In NY, with about 3 full sun hours daily max, best
case scenario I'd need a 50 watt panel. 50 watts * 3 hours = 150
watt-hours daily.

--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html

Windsun wrote:

... Basically, 100% uptime will cost you a lot more than 99% uptime.

If cloudy days were coinflips, storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.

Nick

wrote in message
...
Windsun wrote:

... Basically, 100% uptime will cost you a lot more than 99% uptime.

If cloudy days were coinflips,

But they are not, so the rest of your supposition is false. Anyone that
watches the 6:00 o'clock local news and weather can tell that when you have
one cloudy day, the chances of the next day also being cloudy are higher.
Similarly, when one day is sunny, the probability of the next day also being
sunny are higher than 50/50.

(and yes, I have evidence for that 'article of faith'. Look on NOAA's web
site for any given weather reporting station and do the stats on cloudy days
for a year or two)

storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.


daestrom

For information on batteries the following site is very good:

http://www.batteryfaq.org




"Bryan Martin" wrote in message
...
Thanks everyone for shedding such light on it. I think I have enough info
to get started


"Bryan Martin" wrote in message
...
The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts
6W x 24hours = 144 Watts per day
144W / 12V = 12 amp hours

Now given a battery with 110 amp hours
(http://www.defender.com/product.jsp?path=-1|328|51495|306219&id=51496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days
so with this battery we have enough juice to potentially run our unit
for
4.58 days w/out charging.

Now if the above is not correct then please explain. And the next
question is how would I now figure out approx how large of a panel I
would
need to keep the battery charged. I am really looking for the formula
and
a understanding of how you got from A to B as oppose to "this panel
should
work".


--
Poor planning on your part does not constitute an emergency on my part.

You are using a common math fallacy that proves nothing.

We have installed many many 100% uptime systems over the past 30 years or
so. It is quite simple, it is just more expensive.

-------------------------------------------------------------------------
Northern Arizona Wind & Sun - Electricity From the Sun
Free Solar Discussion Forum: http://www.wind-sun.com/smf/index.php
-------------------------------------------------------------------------
wrote in message
...
Windsun wrote:

... Basically, 100% uptime will cost you a lot more than 99% uptime.

If cloudy days were coinflips, storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.

Nick

He did say "IF"

"Windsun" wrote in message
news
You are using a common math fallacy that proves
nothing.

We have installed many many 100% uptime systems over
the past 30 years or
so. It is quite simple, it is just more expensive.

-----------------------------------------------------
--------------------
Northern Arizona Wind & Sun - Electricity From the
Sun
Free Solar Discussion Forum:
http://www.wind-sun.com/smf/index.php
-----------------------------------------------------
--------------------
wrote in message
...
Windsun wrote:

... Basically, 100% uptime will cost you a lot more
than 99% uptime.

If cloudy days were coinflips, storing enough
energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3
for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would
be impossible.

Nick

Windsun wrote:
It does not work that way because panels are rated at 17 volts, not 12. You
have to work with the panels amps not watts.


unless one has a mppt controller.

without, a 50 watt panel might produce 3a x 3h or 9ah.

9ah x 12v = 108 wh (not 150wh) into the battery, of which 86wh (7ah)
might be available for use (20% conversion inefficiency). That's less
than his needed 12ah, so a 75 watt panel would be more appropriate in
NY, but I do not know where the OP is located.

--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html

In article , "Windsun" wrote:
You are using a common math fallacy that proves nothing.

We have installed many many 100% uptime systems over the past 30 years or
so. It is quite simple, it is just more expensive.

Hogwash. *Nothing* works *every* time.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor again.

Try stopping breathing for 15 minutes. It works
***EVERYTIME***

Wash your pigs someplace else.

"Doug Miller" wrote in message
. com...

Hogwash. *Nothing* works *every* time.

--
Regards,
Doug Miller (alphageek at milmac dot com)

It's time to throw all their damned tea in the harbor
again.

Doug Miller wrote:

We have installed many many 100% uptime systems over the past 30 years or
so. It is quite simple, it is just more expensive.

Hogwash. *Nothing* works *every* time.

The sad truth is that solar weather is probabilistic, like 100-year floods or
coin flips. No matter how much energy storage we provide, if we wait long
enough, there will always be a long enough cloudy day sequence to exhaust our
energy store and leave the house cold or the lights out, unless we have some
other backup system, eg a woodstove or a generator.

Nick

wrote:
Windsun wrote:

... Basically, 100% uptime will cost you a lot more than 99% uptime.

If cloudy days were coinflips, storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.

Just an FYI, in the Detroit area, we just went from Dec 20 through Jan
5 (inclusive) with 14 minutes total sunshine.

It has been one dark dreary depressing year end season
eh?

We saw some sun for an hour this am then about 30
minutes too late at 4pm today.

Yuk! My PV panels are running dry and shrivelling up
from lack of use.

Steve(Crocket)Spence must be on the parafin candles by
now...LOL


"Tony Wesley" wrote in message
ps.com...
wrote:
Windsun wrote:

... Basically, 100% uptime will cost you a lot
more than 99% uptime.

If cloudy days were coinflips, storing enough
energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3
for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would
be impossible.

Just an FYI, in the Detroit area, we just went from
Dec 20 through Jan
5 (inclusive) with 14 minutes total sunshine.

Tony Wesley wrote:
wrote:
Windsun wrote:

... Basically, 100% uptime will cost you a lot more than 99% uptime.
If cloudy days were coinflips, storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.

Just an FYI, in the Detroit area, we just went from Dec 20 through Jan
5 (inclusive) with 14 minutes total sunshine.

I think what that means is that the clouds parted and exposed the solar
disk for 14 minutes total over that time period (and having spent my
holidays there, I cannot doubt its accuracy). It's not representative of
the total amount of solar energy that penetrated the clouds and reached
the surface of the earth. A little bit of math provides a quick sanity
check:

14 minutes of full sun = 233 Wh/m^2
233 Wh/m^2 = 13.7 Wh/m^2/day over 17 days
13.7 Wh/m^2/day = approximately 2 W/m^2 during daylight hours

That's a tad darker than Barrow, Alaska (in the Arctic Circle) in late
January, and no way was Detroit that dark. Indeed, if you check NREL's
data on the subject, Detroit averages 1300 Wh/m^2/day in December
despite the fact that long cloudy spells like the one you described are
the norm for the area at that time of year. You can also check NREL's 30
years of hourly data to see that peak horizontal insolation in Detroit
is usually greater than 200 W/m^2 and nearly always exceeds 100 W/m^2,
even on Detroit's cloudiest days.

Jim Baber

Tony Wesley wrote:

wrote:


Windsun wrote:

.. Basically, 100% uptime will cost you a lot more than 99% uptime.


If cloudy days were coinflips, storing enough energy for 1 day would
make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%,
5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible.


Just an FYI, in the Detroit area, we just went from Dec 20 through Jan
5 (inclusive) with 14 minutes total sunshine.


Another FYI, 'Total sunshine' is not required to generate appreciable
power.

Today it has been cloudy for the 9 hours (so far) of daylight. My 10 kW
system has been far from it's most productive for this time of year
today. It has only produced 21 kWh today Feb. 4, 2006. On on the other
side of the coin, on Feb. 1, 2006 it made 44 kWh on a beautiful Fresno
winter day. So far this year the least I have produced was 0.568 kWh
(truly terrible) on one cold rainy dark day in January. (Which is why
I'm on the grid.) :-)