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BY : RussellMc email (remove spam text)

On 23 June 2016 at 22:31, Brent Brown <brent.brownspamBeGoneclear.net.nz> wrote:

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I think that that's the area of departure.

Grid impedance and effective resistance are very low. â€‹

The ability of say a 5 kW (large by domestic standards) to alter the mains

voltage substantially is "small".

Lets see. 5 kW at 230 V. Reffective inverter out = 230^2/5000 = 10.6 Ohms,

or ~= 53/kW Ohms.

I = P/V =~ 22A or ~= 4.3A/kW.

Mains impedance and feed resistance are well down on that.

The largest component may well be the subscribers feed in lead, and it

probably does not help stability if you use the impedance of your feed in

cable as your current control impedance.

So they have an effective inductance in the system that swamps the

resistance of the grid. So for pure non reactive load the inverter is

driving an inductor. Agh :-).

Giving their formula P = V1 x V2 / Xl x sin(delta).

The voltage differential drives the current through the inductor and the

delta angle affects both magnitude of current and how much ends up as

'reactive power' (VA).

To go from import to export you change phase by 180 degrees as you said.

But you start at a different phase angle due to the resistive load being

made to look mainly inductive by the system.

So we probably more or less agree.

I had carefully avoided getting to close to the phase of the inverter

relative to grid during eg export but after writing the above, for a

resistive grid load which appears as an inductive reactance the current

should lead the voltage so that seems to mean that inverter V phase lags

mains V phase for export. Which messes with my head so far. (About usual

:-) ).

Overall it's not very complex (pun noted en passant), just annoying.

Some or all of the above may be wrong :-).

Russell

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