Voltage drop help

Volvo Farmer
Volvo Farmer Solar Expert Posts: 209 ✭✭✭✭✭
This gets a little confusing as my wire sizes change and my amps change the further I get from the batteries. I cannot figure out what percentage I am losing, though I think that last 60' run is too long for a 2% drop at 30A

I have a Midnite Classic 200 and can rewire this whole thing for 96V, which certainly gets me down below 2%, but then my controller runs hotter and I lose more in the charge controller. This is feeding a 24 volt battery bank.

Can anyone look at this picture and give an opinion if I should wire for 48 or 96V?

vdrop.jpg

Comments

  • vtmaps
    vtmaps Solar Expert Posts: 3,741 ✭✭✭✭
    Re: Voltage drop help

    In the first leg your voltage drop is 0.333 (0.69%) with a power loss of 3.33 watts

    In the second leg your voltage drop is 0.769 (1.60%) with a power loss of 15.372 watts

    In the third leg your voltage drop is 1.451 (3.02%) with a power loss of 43.524 watts

    These calculations are at full nameplate power. The actual losses depend on the power. The voltage drop is proportional to the power produced and the power lost in the cable is proportional to the square of the power produced. For example, at half power the voltage drop in the third leg is 0.725 and the power lost is 10.88 watts.

    --vtMaps
    4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
  • Volvo Farmer
    Volvo Farmer Solar Expert Posts: 209 ✭✭✭✭✭
    Re: Voltage drop help

    Thank you very much for that. Do you mind me asking what tool you used to calculate those numbers? If you teach me to fish, I might not have to come back so soon asking for another free meal :D
  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Voltage drop help

    now for my 3rd attempt at posting a reply here.:roll: darn timers.

    i had a large reply with in depth explanations to give to you, but i timed out and relogging in it didn't carry it. what's worse is even though i logged in and it showed that i was logged in my 2nd shorter explanation failed and was lost due to the forum saying that i still wasn't logged in. i'm not to sure what had happened here, but........

    anyway, i am not sure why you won't utilize the higher voltages being you have the controller for it. the voltage drops and their %s from the first 2 arrays should be additive to the last and through to the cc and batteries. i am looking upon the first 2 arrays as arrays of different wattage and voltage, but that are very closely matched to the final array making the losses to the final array negligible. now i contend that if you want to use lower voltages that in a 2 x 2 arrangement for each array that you would want to change the 60ft home run to #4 wire to keep losses lower. other than that i think you should set each array as 4 in series and parallel the arrays together. often the cc heating losses are a wash with the losses one would've seen from the wires. with the higher voltage arrays you would see about a .62% v drop. now i don't know your voc so be sure it would never exceed the cl 200's 200v max or i would then advise to use the lower voltages.
  • Volvo Farmer
    Volvo Farmer Solar Expert Posts: 209 ✭✭✭✭✭
    Re: Voltage drop help

    Thank you for the reply. It was my original intention to wire 4 panels in series, and parallel the three arrays with this new charge controller. Then I got to thinking that I might not gain any efficiency that way, and it might be hard on the CC, running hotter because of the large voltage differential. I sure don't want to upgrade 120 ft of wire to #4.

    So I guess I will wire for the higher voltage now.
  • vtmaps
    vtmaps Solar Expert Posts: 3,741 ✭✭✭✭
    Re: Voltage drop help
    Thank you very much for that. Do you mind me asking what tool you used to calculate those numbers? If you teach me to fish, I might not have to come back so soon asking for another free meal :D

    No tool, just a spreadsheet and two formulas. I look up in my AWG table the resistance of your cable. For example the first leg is 26 ft one way, so 52 ft of 8 gauge has a resistance of 0.033306 ohms. Ohm's law says that the voltage drop across 0.033306 ohms with 10 amps running through it is 0.333 volts which is 0.69% of 48 volts. Power dissipated is I2R = 3.3306 watts.

    To figure the voltage drop and power lost at less than full power, just reduce the amps in the above formulas. For example, in the first leg full power is 48 volts and 10 amps. Half power is 48 volts and 5 amps.

    To figure out what array configuration to use, you need the efficiency curves for your classic. Take a look at the calculations for an outback controller (which are similar to the classic):
    http://forum.solar-electric.com/showthread.php?15907

    --vtMaps
    4 X 235watt Samsung, Midnite ePanel, Outback VFX3524 FM60 & mate, 4 Interstate L16, trimetric, Honda eu2000i
  • niel
    niel Solar Expert Posts: 10,300 ✭✭✭✭
    Re: Voltage drop help

    wiring each array in series also saves you on combining at each array for the 2 strings of 2. combining each array together is still needed though and being you have 3 to be combined you will need to have a fuse or circuit breaker on each array's output before being combined. if you had elected the 2 x 2 for each array it would've gotten more complex as each string would've needed fused or circuit breakered.

    in case you didn't notice, my signature line also has a v drop calculator. it is quite accurate, but takes getting used to its ins and outs as another long ago member put it into the xls spreadsheet with all of the data and formulas given by me. this you can put onto your pc unlike many other calculators, but do not keep saving changes as errors pop up after time. it is still hosted by solar guppy as far as i know, but it's hard to say for how long as he has dropped off the radar after dropping his solar website and moving into a new place.