Clipped Power and need to determine how much is being Clipped.

[crsylvia@me.com]

I have a 6KW inverter (solar edge) and the App to monitor it. The system includes 28 panels, due to orientation the system rarely generates maximum potential. However it does clip at 6KW during a few hours. I have not been able to determine how much power is being clipped and would like to know how i could do that.

[SunEagle]

Well the simple math would be to multiply the 28 panels by their rated wattage and then by say 85%. Once you have that number you can subtract 6000watts and the difference should be close to the amount of clipping you are experiencing.

[J.P.M.]

I have not been able to determine how much power is being clipped and would like to know how i could do that.

I'd respectfully suggest you take care in what you ask for. But since you ask:

First you need to determine how much irradiance is available for when you're observation the power output, including shading considerations. That'll take a pyranometer and some calculation.
Then, determine the array's efficiency given the inputs and environmental conditions such as ambient temp. and wind conditions and a panel's data sheet, all that integrated over the time period(s) you're interested in. More calculation.
Then, multiply the approx. irradiance input you calculate by the panels/array's approx. efficiency obtained per above and get the approximate unclipped array output. Subtract the output obtained from monitoring and the result will be an approximation of the clipping loss for your array for the time period under measurement.
The above method will be a rough 1st approximation of the clipping loss for observed time period.

If you do all that correctly, what you'll be calcing is a reasonable estimate n the solar input times the array's efficiency for the period under observation. Unless you know what you're doing you'll only get more confused.

The point is, unless it's a big clipping loss, without doing some stuff you probably are not capable of doing or don't (yet) understand, you'll be better off with Suneagle's method. Just don't expect a lot of resolution to your question.

A simpler method yet is to extrapolate what an unclipped output curve might look like and compare the area under that extrapolated curve (with some creatine pencil whipping and a good eyeball) and compare that area to the area under the curve of some clear day actual (clipped) output you can obtain now.

Welcome to the neighborhood.

[Ampster]

I used to have a system that clipped practically every day in which there was full sun. My DC to AC ratio was 1.5 to 1. After a year I compared the output to the installers estimate and to the PV Watts model. I exceeded the installers estimate and came close to the PV Watts estimate. I stopped worrying about the how much was being clipped. My curve was symmetrical with a flat top. In your case I would guess there is an irregular shading situation in the afternoons. Is the diagram correct that both arrays have the same azimuth?

[Ampster]

Well the simple math would be to multiply the 28 panels by their rated wattage and then by say 85%. Once you have that number you can subtract 6000watts and the difference should be close to the amount of clipping you are experiencing.

How do you translate that kW number into kWhs per day?

[J.P.M.]

How do you translate that kW number into kWhs per day?

Without speaking for him, since he's writing about power, I'd assume he's talking about instantaneous values. If so, since energy is the time integrated value of power, I'd sum the power value over the measured time period of, say, a day.

The 2d last sentence of my post describes a method to get what's an integrated value of power (or power difference (loss) due to clipping over a time period.

[Ampster]

Instantaneous clipping in kW is hard to translate into a useful number compared to kWh. I agree that the amount of clipping is the area under the hypothetical curve and the flat line of the clipped inverter output.
I tried changing the assumption in PV Watts for DC to AC ratios. My result was that the difference between a 4kW system at 1.2 and the same 4kW system at 1.5 DC to AC was only 118kWh on total production of 6000 kWh in a year. In that case it would appear to be trivial compared to total production.

[J.P.M.]

Instantaneous clipping in kW is hard to translate into a useful number compared to kWh. I agree that the amount of clipping is the area under the hypothetical curve and the flat line of the clipped inverter output.
I tried changing the assumption in PV Watts for DC to AC ratios. My result was that the difference between a 4kW system at 1.2 and the same 4kW system at 1.5 DC to AC was only 118kWh on total production of 6000 kWh in a year. In that case it would appear to be trivial compared to total production.

I'd consider a 1.8 % production loss something worth looking into, but I suppose opinions vary.

On the hard to translate question, I'd suggest hard to translate is somewhat a matter of opinion.

After all, that's all an output graph does - time integrate (abscissa) output power (ordinate) levels = watt output - only the integration of power over time is already done and represented graphically and then often/usually summed by time period and reported as kWh sums over time.

By estimating what such a (clipped ) power graph would look like and then using the difference between the area under that curve and the area under an actual output curve is an estimate of the difference in integrated time values of the clipped energy.

[azdave]

New member, new system. I'm guessing the OP just wonders if they are losing 9 cents a day to clipping or 11 cents.

[Ampster]

I'd consider a 1.8 % production loss something worth looking into, but I suppose opinions vary.

For me 118 kWh in a year might be $30. It may be more or less depending on one's rate. Would that be worth the cost of a larger inverter?

On the hard to translate question, I'd suggest hard to translate is somewhat a matter of opinion.

What I should have explained in more detail was that using the method SunEagle suggested does not result in an output graph based on daily insolation. It is highly dependent on the 85% assumption used to get the result. It may be useful to compare different systems in order to rank them but I am not sure it actually answers the question about how much clipping takes place. Ranking by DC to AC ratio may result in the same rank order assuming orientation and tilt are the same.

After all, that's all an output graph does - time integrate (abscissa) output power (ordinate) levels = watt output - only the integration of power over time is already done and represented graphically and then often/usually summed by time period and reported as kWh sums over time.
.

Yes that is a better way to calculate the amount of clipping than the DC to AC ratio. For those challenged by integrated math the simple answer to the OP's question may be to compare the result of a model that estimates output. The model does the complex math and the user just has to compare one result to the other. That is what I asked my installer to do when I questioned an installation with a 1.50 to One DC to AC ratio. In my case it was not worth the extra cost of a larger Inverter. Long term as my panels degrade the clipping will be minimal.

[SunEagle]

How do you translate that kW number into kWhs per day?

It would be hard because it will depend on the amount of unobstructed sunlight, angle of the panels, etc. Maybe I was just trying to dumb it down by providing one path to determine the amount of clipping going on with that system.

If the OP even used that 6000watt value times some daily, weekly or yearly amount of sunhours he would still not be close to the actual amount of kWh he is generating.

[Ampster]

New member, new system. I'm guessing the OP just wonders if they are losing 9 cents a day to clipping or 11 cents.

Yes, that is what I wanted to know when I first saw my system clipping in February 2018, right after it was installed. I have since sold that house and the new owner is happy about the production from the system. During the three years I lived there my output sometimes exceeded the estimates and sometimes was less. Weather and smoke created more loss of production than any clipping. It all depends on what you want to worry about and what action one might want to take. I chose to manage my expectations rather than wring my hands over a few kWhrs. It reminds me of the glass half full or half empty question.

[checkthisout]

I have a 6KW inverter (solar edge) and the App to monitor it. The system includes 28 panels, due to orientation the system rarely generates maximum potential. However it does clip at 6KW during a few hours. I have not been able to determine how much power is being clipped and would like to know how i could do that.

Disconnect panels or groups of panels if in series until you see a drop in power. Take the power being produced and then divide it among the remaining connected panels and then take that number and multiply it by the number of panels you unhooked. That is how much power is being wasted.

You have to make sure and do that test with a load on the batteries so their soc isn't affecting charge rate.

[SunEagle]

Disconnect panels or groups of panels if in series until you see a drop in power. Take the power being produced and then divide it among the remaining connected panels and then take that number and multiply it by the number of panels you unhooked. That is how much power is being wasted.

You have to make sure and do that test with a load on the batteries so their soc isn't affecting charge rate.

The problem is that some of those panels are not producing as much as others. I see a range of 901wH down to 776Wh. If they are all the same wattage then there is a big difference in output which could be part of the issue.

[Ampster]

Disconnect panels or groups of panels if in series until you see a drop in power. ......

You have to make sure and do that test with a load on the batteries so their soc isn't affecting charge rate.

I think the OP has a Solaredge GT system, so the grid will present an infinite load.. That would also involve going on the roof which could work but there are other approaches that might be simpler. Typically Clipping usually occurs around Solar Noon so one would have to time their roof work around Solar Noon to get the best measurement. However it is not a good idea to disconnect panels while under load so one would need to shut down the system, disconnect a panel and wait until the system reboots in a few minutes.

Until we hear again from the OP about how much and how long the clipping affects his production it is difficult to know if this problem is worth going on the roof, buying a bigger inverter or moving on to more important issues.