Hey guys,

I know there are softwares out there that can compute but I'd like to know if anyone can help me out with a formula.
Here's the deal. We know that cable losses are proportional to I^2xR. So let's say I want to compute cable losses from one panel to another in a series of panels routed to an inverter. How do I find out what the current in my cable is. I do know the cable resistance. I'm assuming optimal output here with no losses due to the panel itself and of course I do know the power rating and the MPP curve of this panel. Can anyone help? I hope I am making sense.

With an ammeter, you can have the exact amperage to complete the I2R formula.

You realize this formula will give you the heat in watts dissipated form the conductors.

To use I2R by formula only, need to know amps. This can only be done if you know the voltage and the wattage of the load, and use Ohm's Law, I=P/E, or speculate the watts and see the result of I2R loss.

The power rating of the panel is only what it can deliver in power.

I believe that MPP is irrelevant to what your looking for.

And what exactly is that your looking for?

Quote:

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Originally Posted by tkrussell

With an ammeter, you can have the exact amperage to complete the I2R formula.

You realize this formula will give you the heat in watts dissipated form the conductors.

To use I2R by formula only, need to know amps. This can only be done if you know the voltage and the wattage of the load, and use Ohm's Law, I=P/E, or speculate the watts and see the result of I2R loss.

The power rating of the panel is only what it can deliver in power.

I believe that MPP is irrelevant to what your looking for.

And what exactly is that your looking for?

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I am actually design an excel spreadsheet to model the losses of a solar system on the DC side. I got the AC side pretty much covered. The actual project is being bid right now so there's no physical cable to use an ammeter on.

If you can, keep the low voltage lines short.(Inverters and Batteries close to the Solar Panels, then the Ac to Panel location.
Low Voltage Drops fast.

Quote:

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Originally Posted by SolarFreak

I am actually design an excel spreadsheet to model the losses of a solar system on the DC side. I got the AC side pretty much covered. The actual project is being bid right now so there's no physical cable to use an ammeter on.

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Well I believe we're using a 600V cable on the DC side. So I got the voltage part figured out. But what about the load? (kW) I mean I do know what the load of each individual panel, but I also do know that the current going through them is the same, only the voltage is changing. So how do I find that equivalent current? Is it simply? (nxLoad)/(Sum of voltages), where n is the number of panels in series. Thanks!

Quote:

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Originally Posted by Stratmando

If you can, keep the low voltage lines short.(Inverters and Batteries close to the Solar Panels, then the Ac to Panel location.
Low Voltage Drops fast.

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Thanks bro but this is a 1 MW grid connected site. So keeping low voltage lines short will be miraculous. We'll do our best though! We won't use any battery here since it's not an independent system.

Basically, I'm trying to say, even Jigawatts
Have batteries and Inverters close to the Panels, the run high voltage ac on smaller wire from there to distribution panel, Instead of Larger DC Low voltage wiring from panels to batteries and inverter.
http://www.energystore-usa.com/content/section/4/19/

Since I've never done anything like this or even tried and I don't know the availability of high voltage inverters, there may be another way.

One master inverter and lots of grid-tie inverters with some sort of redundancy on the master inverter. Then everyone should be happy.

What is the DC voltage out?

Kiss, The High voltage comes off the Inverter.
How can I say in Technical terms, "Keep the wires from the Solar Panels short to the batteries, then inverted to High voltage", then send on its way. And not have a long run from panels to batteries requiring larger wires, more money.
I guess I was thinking economical, and not unlimited funds?
I appologige if I am missinterpreting.

Not necessarily. Here in an inverter that needs 300-600 VDC input not the 12, 24 or 48 VDC.

It's grid tie, which you need because one system/inverter needs to provide the timing for the grid since you are the grid, I think.

There will have to be a lot of researching of available products, topologies and use of transformers.

Here's a 250 kW inverter designed for large scale commercial systems. Xantrex Technology Inc. - Commercial Solar - GT Series Grid Tie Solar Inverters - Three Phase - North America - Specifications

The input is still sizeable: Around 800 amps and 500 VDC.

There are some gotcha's because panels need to be matched in current if the panels are placed in series.