Hybrid Car – More Fun with Less Gas

How many panels ? ( to run 230 volt sprinkler pump 30 minutes a day?) - Page 5

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Posted by bealiba on July 8, 2008, 11:14 pm
 
On Jul 8, 10:39 am, wmbjkREM...@citlink.net wrote:

Oh look, Tweedledum is back with more of the same. All words, no
maths. How predictable.

Tell us about your system. What are your loads? What do they draw? How
long are they run? And let's not forget, How does increasing the load
reduce resistance in conductors?

Posted by bealiba on July 7, 2008, 2:07 am
 

You offered Trojan T105s. Sound pretty specific.


No. I sized the required system, in the correct order and did not add
20% on top of the inverter adjustment to the daily load.

What! No proof?

A2      Daily load = 1250Wh
A4      Inverter Efficiency = 85%
A5      Account for inverter inefficiency - Load (A2/A4) = 1470.5
A7      System Voltage = 12
A8      Total A-hr demand per day (A5 / A7) = 122.55

B1      Number of days of autonomy = 1
B2      Maximum allowable depth of discharge = 70%
B3      Battery capacity (A8 x B1 / B2) = 175Ah
B4      Lowest 24 hour average temperature c
B5      Temperature correction factor =.97
B6      Adjusted battery capacity (B3 / B5) = 180.5
B7      Selected Battery
B8      Selected battery discharge rate 100
B9      A-hr capacity of selected battery = 225Ah
B10     Number of batteries in parallel (B6 / B9, rounded off) = 1
B11     Number of batteries in series (A7 / battery voltage) =1
B12     Check Capacity of selected battery at l00 Hr rate = 225
B13     Capacity of battery bank at 100 hr rate (B12 x B10) = 225
B14     Daily depth of discharge (100 x A8 / B13) = 54.47%

C1      Design tilt
C2      Design month
C3      Total energy demand per day (A8) 2.55Ah
C4      Battery efficiency = 90%
C5      Array output required per day (C3 / C4) = 136.2
C6      Peak sun hours at design tilt for design month = 5
C7      Selected module
C8      Selected module I at 14 volts at NOCT 2.94A
C9      Selected module nominal operating voltage. = 12V
C10     Guaranteed current (C8 x 0.9) = 2.65A
C11     Number of modules in series (A7 / C9) = 1
C12     Output per module (C10 x C6) = 13.2Ah
C13     Number of parallel strings of modules (C5 / C12) = 10.3

Prove it wrong.

Prove it. You claim a very precise 23% short fall. All you have to do
is show the numbers.

With your 20% added to the daily load, exactly where you put it:

A2      Daily load = 1250Wh
A4      Inverter Efficiency = 85%
A5      Account for inverter inefficiency - Load (A2/A4) = 1470.59
A7      System Voltage = 12
A8      Total A-hr demand per day (A5 / A7) = 144.17 (rons 20% added)

B1      Number of days of autonomy = 1
B2      Maximum allowable depth of discharge = 70%
B3      Battery capacity (A8 x B1 / B2) = 206Ah
B4      Lowest 24 hour average temperature = 15c
B5      Temperature correction factor =.97
B6      Adjusted battery capacity (B3 / B5) = 212.Ah
B7      Selected Battery
B8      Selected battery discharge rate 100
B9      A-hr capacity of selected battery = 212Ah
B10     Number of batteries in parallel (B6 / B9, rounded off) = 1
B11     Number of batteries in series (A7 / battery voltage) =1
B12     Check Capacity of selected battery at l00 Hr rate = 212
B13     Capacity of battery bank at 100 hr rate (B12 x B10) = 212
B14     Daily depth of discharge (100 x A8 / B13) = 68%

C1      Design tilt
C2      Design month
C3      Total energy demand per day (A8) 4.17Ah
C4      Battery efficiency = 90%

(Ron likes 80% here. If you use 80% the required panels come to 13.62)

C5      Array output required per day (C3 / C4) = 160.19
C6      Peak sun hours at design tilt for design month = 5
C7      Selected module
C8      Selected module I at 14 volts at NOCT 2.94A
C9      Selected module nominal operating voltage. = 12V
C10     Guaranteed current (C8 x 0.9) = 2.65A
C11     Number of modules in series (A7 / C9) = 1
C12     Output per module (C10 x C6) = 13.23Ah
C13     Number of parallel strings of modules (C5 / C12) = 12.11




Posted by Ron Rosenfeld on July 8, 2008, 3:37 am
 On Sun, 6 Jul 2008 19:07:12 -0700 (PDT), bealiba@gmail.com wrote:


That was for a hypothetical battery, which was proposed only because *you*
kept blathering on about battery recommendations, and misquoting (a kind
way of saying "lying" about) my panel array recommendation size.

With a hypothetical Trojan T105 battery (modified to 90% efficiency) in a
50 watt/panel, 11 panel system, that shortfall is predictable.  I did not
propose that it be used in the system; it was proposed only because you did
not quote any particular battery or specifications other than the 180Ah
battery figure, which you use for the 100 hr rating.


That would mean that that your proposed battery can output a current of
1.8A for 100 hrs.

You also claim that your battery has a


I don't have a battery in my database that matches those specifications of
yours.  The closest is the Trojan T105 which has a capacity at the 20 hr
rate of 225Ah, and so would obviously be better than 180Ah at the 100 hr
rate.  But its round trip efficiency is only 85%, vs the 90% in the battery
you are quoting.

Changing the efficiency to 90% in the specifications for the T105 and
leaving everything else the same results in the simulation showing a 23%
shortfall.  

But I thought you would want to specify your own battery, to get a more
realistic result of your proposed 11 panel system with a 180Ah battery (at
the 100 hr rate).

If you are really interested in the simulation results of your proposed
system with a hypothetical T105 modified to 90% efficiency, I could post
that.  But to what end -- that battery does not exist.

I think the simulation would be of more value with a "real" battery that
comes close to the parameters you are using.  I could use the unmodified
T105, which has a 27% shortfall (if I recall correctly, or any other real
battery for which you can provide the specifications I mentioned.
--ron

Posted by bealiba on July 9, 2008, 6:24 am
 
The battery size given in the calculation is the minimum capacity
required to run the stated pump for 30 minutes. The calculation allows
you the choose a battery.

Posted by Ron Rosenfeld on July 10, 2008, 3:29 am
 On Tue, 8 Jul 2008 23:24:21 -0700 (PDT), bealiba@gmail.com wrote:


It seems you will not (or cannot) specify a practical battery to use in
this application that matches the inputs you used in your spreadsheet.

I must say that doesn't surprise me, although I was hopeful of finding a
battery with a 90% efficiency, a flat current vs capacity curve in the 30
min to 100 hr range, along with an acceptable life when limited to 70%
maximum DOD.

If your system is going to work, your 180Ah capacity battery at the 100 hr
rate -- 1.8A, must also have the *same* capacity when putting out 245A for
the 1/2 hour that the pump will be running!

C'mon George, what kind of a battery is this?  If you can't supply the
battery you specified, your system is pretty worthless.
--ron

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