Wire Gauge Required Question
For some reason I just can't understand the various wire gauge tables, and I have tried to, so I am hoping somebody will take pity and do all the work! :) One table that I can follow says that for...
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No, you're wrong! Take my previous example. You can also calculate power lost by using:
P = I^2 * R
P = 8A^2 * 60 mohm = 3.84 W
The percent power lost is:
With Imp = 8A & Vmp = 24V:
% power lost = 3.84W /(8A * 24V) = 2%
We got the same 2% power loss using YOUR equation (P = I^2*R). You just implemented it incorrectly!
Mex - I'm assuming we're all using a 12V house battery system. That means the greater the solar input voltage, the greater the controller losses. But to be more concise, (in order to include other battery voltage systems), this would be more correct:
The greater the voltage differential between solar voltage and battery voltage, the greater the controller losses.
P = I^2 * R
P = 8A^2 * 60 mohm = 3.84 W
The percent power lost is:
With Imp = 8A & Vmp = 24V:
% power lost = 3.84W /(8A * 24V) = 2%
We got the same 2% power loss using YOUR equation (P = I^2*R). You just implemented it incorrectly!
Mex - I'm assuming we're all using a 12V house battery system. That means the greater the solar input voltage, the greater the controller losses. But to be more concise, (in order to include other battery voltage systems), this would be more correct:
The greater the voltage differential between solar voltage and battery voltage, the greater the controller losses.
ktmrfs wrote:Salvo wrote:
As some mentioned, key in MPPT is power.
P = V * I
If you drop 2% in voltage between panel and controller, you drop 2% power, and you drop 2% battery charging current.
I always us a wire guage table, never a voltage drop calculator. 10 awg wire has a resistance of 1 mohm/ft. 60 ft of cable gives 60 mohm. At 8A, you get a voltage drop of 60 mohm * 8 A = 0.48 V. At 24V, that about a 2% voltage drop. You can expect 2% less charging current in bulk.
In addition, you also got losses in the controller. Usually, the higher the input voltage the greater the losses. If you're concerned about losses, it's better to go parallel/series rather than all series and use a larger cable.
Someone also displayed a very nice efficiency chart. If your solar run isn't too long, then it might be better to go parallel and use one larger wire size. Another reason to go parallel is when one panel is shaded and the rest are in the sun. Parallel does better!
Sal
Sal
in the case for a solar connection, cable power loss is related to voltage squared. yes P=E*I. but P= V2/r or I2*R in the case of a cable R is constant. or power loss is related to cable voltage drop squared, not voltage drop or the current in the wire squared.
As I mentioned in a previous post, a 2% voltage drop is a 4% power loss.
