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How many panels ? ( to run 230 volt sprinkler pump 30 minutes a day?) - Page 28

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Posted by Ron Rosenfeld on July 26, 2008, 1:57 am
On Fri, 25 Jul 2008 17:02:58 -0700 (PDT), bealiba@gmail.com wrote:

Oh, my.  I must be suitably chastised by your irrelevant piffle.

There's no need, George, to keep advertising your ignorance and

We are well aware that you did not, cannot, and will not specify a real

We are also well aware that you will not accept a battery that has ratings
that EXCEED those you published for this system.

There is no real battery that will meet YOUR design requirements and will
work for the OP's system.

 What George wrote about battery requirements for the OP's system:

This self-styled solar design consultant cannot designate a real battery
that meets the requirements HE entered into HIS spreadsheet and will work
in the OP's system!

He also cannot designate a real panel array that meets the design
requirements HE entered into HIS spreadsheet and commented about:

What George wrote about panels for the OP's system:

Rather than admit that his third design is as flawed as his first two
attempts, he continues to blather away, hoping against hope that someone
might think he knows something about the subject.

Posted by bealiba on July 26, 2008, 5:41 am

Ah, now Tweedledee admits that I did not, in fact, specify a battery.

Now that is a lie. The 180Ah in the formula is the minimum battery
capacity only.

Battery Energy 2AS1100 would surely do the job, although many would
consider them to be a bit of over kill.

My, you really can't come to terms with user input can you. Pick a
battery, put the specs in. Oh yes, and learn the difference between
Maximum DOD and Daily DOD.

B2, shows what sort of information is required. 70% is an acceptable
Maximum DOD for almost any deep cycle battery.

You should not change the calculation at B13 but you will need to
specify the discharge rate.

C4, I have used 90% simply because it works, but there is no reason
why you cannot change it. It is after all a point where the user need
to supply a number.

As with the batteries I have not specified a panel.

C8 shows what sort of information is required

True when the formula was written. This formula has been around for at
least 30 years and perhaps even longer.

Notice that while words flow like water from Tweedledee's mouth,
actual maths are nonexistent in any of his posts.

Posted by Ron Rosenfeld on July 26, 2008, 2:00 pm
 On Fri, 25 Jul 2008 22:41:17 -0700 (PDT), bealiba@gmail.com wrote:

Look at George waffle.

George sets a requirement for a 180AH battery at the 100 hr rate;

George claims that is a minimum requirement;

And when George finally specifies a real battery, comes up with one that
has a capacity at the 100 hr rate of over 1000Ah!!!  (1047Ah,

His minimum battery capacity was grossly inadequate!!

YOU were the user who put in the specs that resulted in these requirments!!

Oh, this must be the GTC (Ghio Test Condition) imaginary standard that's 30
years old.  Battery chemistry must have changed in the past 30 years since
today we need higher voltages to work in a nominal 12 volt system.

Note that the only battery George has specified exceeds his minimum
requirements by a factor of almost six (6)!!!

He has to exceed his *minimum* requirements by some huge amount in order to
get his system to work!  (That's not in his spreadsheet though).  No wonder
he doesn't tell us this!

Since he seems to like the Suncycle series of Battery Energy brand of
batteries, what happens if we use a battery that actually MEETS *his*
minimum specifications:

For example, the 6AS190:

  Mfg spec for 6AS190:        Capacity at *100* hr rate = *184*

Going back to the OP's requirements to run a 2500W pump for 30 minutes, we
can see by simple mathematics that this battery, which MEETS George's
"minimum" requirements, cannot work.

Again, according to George's specifications:

A4      Inverter Efficiency = 85%
A7      System Voltage = 12

So, for the 1/2 hr, our 2500W pump causes the inverter to draw current at a
rate of:    2500/0.85/12 = 245A

The 6AS190, which MEETS George's minimum requirements, when subjected to
this current draw, has a capacity of well under 50AH.  As a matter of fact,
examining the capacity current curve published by the manufacturer suggests
that the capacity, at this current draw, is no more than 30AH, and likely a
good deal less.

In other words, a battery that actually MEETS *George's* minimun
requirements will run the pump for 30/245*60 =  7 minutes at the very most.

He needs to specify a battery that is almost six (6) times his minimum
requirement of 180Ah at the 100hr rate to get a functioning system.

Wow, what a deeziner!!  Or, as Nick Pine would have said, "What a nitwit!"


Posted by bealiba on July 26, 2008, 2:53 pm
I did not set the requirement, 180Ah is the minimum required capacity.

I did not specify that that battery be used and in fact pointed out
that while it would work it is well over that required.

A the rate of 0.5 hours 180 Ah is good enough.

I did not specify anything beyond the minimum capacity, that being
calculated from known daily requirement .

So change the specs. That what user input means.

Yeah, good isn't it.

No you claimed that I was afraid that a larger battery would not work,
as you can see, I'm not. It does.


Actually that 85% was supplied by you

It will be noticed that Tweedledee has already claimed that the 6AS190
meets capacity. Unfortunately he did not take into account the fact
that the C rate is C.5 so his claim is incorrect. No surprise there.

As I did not set any requirements your statement is a lie.

180Ah is the minimum required at C.5.

Tweedledee has yet to show his calculation for his wonderful
revelation. That's because like his bunk mate he does not know the
calculation. He googled "Peukert" and used a calculator he found

He should have read the information at:


Which says;

The usual form of Peukert's equation is T=C/In


T = time in hours
C = the Peukert capacity of the battery (ie at the 1 amp discharge
I = the discharge current
n = Peukert's exponent.

This equation will only work on batteries that are specified at the
"Peukert Capacity" ie the 1 amp discharge rate. They very rarely are.

Batteries are usually specified at an "hour" rate, for instance
100Ahrs at 20 hours. Or 90Ahrs at 10 hours etc.

If your batteries are specified in such a way (and they nearly always
are) then the equation must be modified to take this into account.

The modifed equation is T=C(C/R)n-1/In or T=R(C/R)n/In


T = time in hours
C = the specified capacity of the battery (at the specified hour
I = the discharge current
n = Peukert's exponent
R = the hour rating (ie 20 hours, or 10 hours etc)

Alternatively, do this:

R(C/R)n = the "Peukert Capacity".

So in the case of a battery specified as being 100Ahrs@20 hours with a
Peukert's exponent of 1.25 we get:

20(100/20)1.25 = 149.5Ahrs. This is the "peukert capacity". ie the
capacity of the battery when discharged at 1 amp.

If you use this figure as the capacity of the battery then the usual
Peukert's equation of T=C/In can be used.

You see, Tweedledee and Tweedledum have no knowledge of their own.
They always rely on someone to be there to hold their hand.

Notice again that neither Tweedledee or Tweedledum have supplied any
maths to support their claims. This being the case there is really no
point to carry on with this thread until they can provide empirical
proof with real numbers. And we all know that that won't happen any
time soon.

The formula is correct and will correctly size a stand alone PV system
as long as the user supplies the correct specifications for the
equipment used.

The Spread Sheet is free as is any discussion you may wish to indulge
in concerning it's use.

Posted by Ron Rosenfeld on July 26, 2008, 9:01 pm
 On Sat, 26 Jul 2008 07:53:00 -0700 (PDT), bealiba@gmail.com wrote:

Stop lying, George.  (if you can)

My claim, which you cannot dispute, is that the 6AS190 meets YOUR minimum
requirements as YOU posted them.  

Your continued lies that you did not post a 100hr rate is easily refuted by
reviewing your posts.  

The fact that you continue to deny this only lends credence to the idea
that you had no idea, at the time you posted it, that there might be a
difference between the 100 hr rate, and the capacity of a battery when
drawing high currents.

YOU are the one that posted a capacity at the 100 hr rate.

The battery you finally decided to come up with has almost 6 times the
minimum capacity YOU specified.


Then you go on to copy a page from

which further demonstrates your lack of understanding of either battery
chemistry or the validity of what I had already posted.

But at least it shows that you might be trying to do some work to try to
learn about this issue.

Even though you show some ability to copy & paste, you fail to point out
that Peukert's exponent is not a constant over the entire range of the
battery discharge currents.  Probably you don't know that since it wasn't
mentioned on the web page you just found on a Google search.

If you do some digging, you'll also find on that web site a method of
calculating Peukert's exponent!!  They use a slightly different formula
than I.  For extra credit, try to explain why there are small differences
between my calculations and the one's you get if you use their calculator.

But for now, you haven't even come up with the correct exponent to be used
for either your minimum battery, or the 6-fold greater battery that you
finally decided on.

For the battery YOU specified -- the ONE that *I* wrote had a 100 hr
capacity of 1047Ah we determine that for current draws of from 9.17 amps to
19.73 amps, Peukert's exponent = 1.196

So, using the published figures for the battery YOU suggested, we come up
with a 100 hr capacity of 1068Ah vs the 1047.537Ah I calculated using the
method which you didn't understand.  My-oh-my -- that's almost a 2%
difference!  (I sit duly chastised for such an egregious error.  I was
using a polynomial curve-fitting algorithm rather than Peukert).

But there's a funny thing about using Peukert's exponent as you get down
into the very high current draws for these AS batteries -- it gets higher
and higher.  So if you were to use, for example the 10 hr and 1 hr rates,
you would calculate a Peukert's exponent of 1.564.  And this is more
appropriate given the high current draws to which these batteries will be

However, this is not particularly important when using a battery that is
almost six (6) times your recommended minimum size, but:

Going back to the battery which meets YOUR minimum specifications of

We find the following:

For current draws    from 1.63 amps to 3.31 amps    
    Peukert Exponent = 1.294

and its capacity at the 100 hr rate is 187Ah (exceeding GG's minimum
specifications of 180Ah at the 100 hr rate).

However, for current draws from 12.88 amps to 50 amps (the maximum currents
published by the manufacturer), the Peukert Exponent calculates to 1.533
and the battery capacity, when drawing 240A for the OP's pump, will only be
21.67Ah.  (Using the mfg curves, I estimated it at less than 30A and a 7
minute runtime, but it's obviously even worse than that).

Oh, for George's benefit, even though the manufacturer did not publish
current draw, it is easily calculated from the published data by dividing
Capacity by Time.

If you really want to go into depth about the subject, look at the other
pages of that website -- don't just copy/paste the "in brief" page and
expect accolades.  

And I've given you enough hints that, if you really understand things, you
should be able to explain the small differences between my results and
those you might get using that web site's calculators.

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