DC cable loss

1. May 24, 2012

wfl6

i doing the solar system design.

seem like everything is on track now,
but just the cable loss i not really know how to calculate,
can anyone help me.

i has formula on this
(resistance on the cable (Ω/km)x length of the cable(L) x current of the solar panel(A)).

for the equation above , is that correct for calculate the cable loss.

Last edited by a moderator: May 29, 2012
2. May 24, 2012

truesearch

That calculation will give you theVOLTAGE lost in the cable.
The POWER lost is given by r x I^2

3. May 24, 2012

wfl6

yes is for the power loss p=I2 x R.
how about the voltage drop across the distance

4. May 24, 2012

vk6kro

(resistance on the cable (Ω/km)x length of the cable(L) x current of the solar panel(A)).

The total cable voltage loss would be double this because the current goes out in one wire and comes back in the other wire, so you get two voltage drops.

5. May 24, 2012

wfl6

what you mean 2 voltage drop is because of the positive and negative.
if let say i has 3 input to the inverter , so that my voltage drop will be 6 time of the calculation i provide.

Last edited by a moderator: May 29, 2012
6. May 24, 2012

vk6kro

If you have a 2 wire cable and a load at the other end, then current leaves the positive side of the supply, goes down one side of the cable, through the load and then back down the other side of the cable to the negative side of the supply.

So, you get two voltage drops in the cable and the load gets the supply voltage minus the two voltage drops.

Why would you have 3 inputs to an inverter?

7. May 24, 2012

wfl6

for the SMA tripower is has 6 input on it.
due to the voltage and current limitation ,
so that i using 3 input.

for the voltage drop x 2 is base on 1 input right.
it wont be affect the 2nd input

thanks

8. May 24, 2012

sophiecentaur

The two wires from the array to the inverter are like two resistors in series, one in the positive lead and one in the return . You can just add the two together together to find the total additional series resistance (obviously doubled if the two wires are the same). If you want to know how many volts will be dropped across this 'loss' resistor, you need to know how much current will be flowing, then:
Vdrop = I times Rseries

It is not clear what you mean about Three Inputs. The Spec sheet seems to imply just a single DC input.
Are you considering connecting several PV banks independently / in parallel? How feasible is this?

9. May 24, 2012

wfl6

the total power i need to achieve is 17000W,
so that i need to connect the panel in series to achieve.
total i has 3 series connect to the inverter.
is mean i has 3 input and 6 wire.
for the voltage drop with the formula i provide i need to be x 6 right ??

10. May 24, 2012

vk6kro

Maybe there is an English problem. Could you draw a picture or a circuit diagram of what you intend to do?

A 17 KW solar system with 3 phase output is a very serious project and you need to get it installed professionally.

11. May 25, 2012

sophiecentaur

Professional certification would.be essential if there is a plan to connect to the mains.

The total DC voltage drop would be due to the current times the sum of all series resistances. Their actual position in the 'loop' is not relevant.

12. May 25, 2012

wfl6

i has attach the file in PDF .
i hope that you the image can give you a clear vision what i doing now

Thanks for the helping

Attached Files:

• demo.pdf
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Last edited by a moderator: May 29, 2012
13. May 25, 2012

davenn

ok nice pic .... so each cable from each 250W panel array will have its own set of losses

BTW how do you think you are going to get 17kW out of a 750W total panel array ??
you cant create energy out of nothing

Dave

Last edited by a moderator: May 29, 2012
14. May 25, 2012

sophiecentaur

Not really clear, I'm afraid. Does your DC switch just connect the PV outputs in parallel? Why would you want to connect only two out of the three panels at any given time?
Your arithmetic is not clear, either as you have already confused Davenn with your missing zero (?)- and also me.
From the spec sheet, it appears that the inverter can handle a wide range of input DC voltages but why exceed the normal operating range? (seems to go up to 800V on the sheet I have read but I may be looking at the wrong model)

This is quite scary stuff. In the words of the prophet "I hope you know what you're doing"

Last edited by a moderator: May 29, 2012
15. May 27, 2012

wfl6

beside that the diagram i draw above is a single line diagram.
for this inverter is connect to the national grid for applying the FIT.

in the mean while,my total panel generating output power is approximate 17250W +/- connect to the inverter .

the inverter has a wide range of DC voltage is approximate 400- 800V.

16. May 28, 2012