Positive and negative VARs and Solar inverter Grid connect schemas

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jaggedben said:
Therefore is a solar inverter is going to cause a load to draw power from the inverter instead of the grid, it has to raise the voltage at the node where all three come together higher than it would be if the load drew power from the grid. Right?
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If we fix a voltage reference, such as ground for a grounded supply system, that node (point) is going to have a single voltage (for a given steady state).

Let's compare case (a) the load wants 10A, the grid provides 10A, and the inverter provides 0A to case (b) the load wants 20A, the grid provides 10A, and the inverter provides 10A. So in case (b) none of the inverter power is going back to the grid, it just goes to the load. And let's assume no other load or sources in the system.

With that comparison, the voltage at the node is the same in either case. If the grid transformer is a fixed voltage source, the voltage at the node will be determined by the impedance of the conductor between that fixed voltage source and this node (the service drop and feeder conductors; probably not branch circuit conductors, as this node would be at a panelboard). The transformer terminal voltage, less the voltage drop from the grid current, is the node voltage. And the grid current is 10A in either case.

Then in case (b) the PV inverter has to match that voltage at that node. Because the conductors between the inverter and the node have some impedance of their own, there will be some voltage drop between the PV inverter and the node. The PV inverter has to put out the correct voltage at its terminals so that the terminal voltage, less the PV voltage drop, matches the voltage from the previous paragraph.

Cheers, Wayne
 
jaggedben said:
Our conductors aren't superconductors so I'm not sure what the point of that hypothetical is. My understanding is that with resistive conductors physics says that the voltage has to be higher at the source than at the load. Therefore is a solar inverter is going to cause a load to draw power from the inverter instead of the grid, it has to raise the voltage at the node where all three come together higher than it would be if the load drew power from the grid. Right?
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No. It's a current source, and physics does not say that. Cutting the resistance of the interconnecting conductors cuts the voltage drop but it doesn't change the output current. My comment about superconductors was an examination of border conditions; if you plot inverter output current against conductor resistance the output current stays essentially flat all the way to the y axis intercept, i.e., where R = 0, or that of a superconductor, where the voltages are the same at both ends.
 
ggunn said:
No. It's a current source, and physics does not say that. Cutting the resistance of the interconnecting conductors cuts the voltage drop but it doesn't change the output current.
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If the inverter is operating in an AC current limited-state.

If it's not operating at a current limit, but is operating in power limited mode (e.g. the MPPT power from the array), if you reduce the impedance on the interconnecting conductors, the inverter terminal voltage will go down, so it will be able to output more current.

Cheers, Wayne
 
wwhitney said:
If we fix a voltage reference, such as ground for a grounded supply system, that node (point) is going to have a single voltage (for a given steady state).

Let's compare case (a) the load wants 10A, the grid provides 10A, and the inverter provides 0A to case (b) the load wants 20A, the grid provides 10A, and the inverter provides 10A. So in case (b) none of the inverter power is going back to the grid, it just goes to the load. And let's assume no other load or sources in the system.

...
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This isn't describing the reality we need to understand. First, the utility voltage is not fixed, and the load is (in the classic typical case, at least) defined by its impedance, not its current draw. Second, the scenario (b) that would represent the issue would be something like (approximately) 10A to the load, 20A from the inverter, and 10A back to the grid. For these currents to flow the voltage at some node or other would have to be different in scenario (a) and (b), right? (I.e. as long as the conductors have some voltage drop)

But a bigger point... this is getting off topic. The OP said that he thought it would be 'easier' to export power at PF=1 and I thought that was counterintuitive. It seemed to me that trying to counteract the normal direction of current flow from the grid would be 'harder'. But you were right to ask what exactly 'easier' meant in this context. I think we should let the OP explain whay he meant by that before getting sidetracked on what I thought.
 
wwhitney said:
If the inverter is operating in an AC current limited-state.
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PV inverters often are operating in an AC current limited state, i.e., at their published maximum output, but even when they are not, changing the resistance of their output conductors does not appreciably change their output current. In fact, for an inverter running at its maximum power output, raising the voltage at its terminals will actually reduce the output current.

Current sources are strange to most people because they are not commonly encountered. Batteries and AC wall outlets are voltage sources, where the voltage stays the same and current through a load is directly proportional to the magnitude of the load. A current source maintains the level of current and the voltage is dependent upon the magnitude of the load.

Grid tied inverters have their output current clamped, so they are essentially constant power sources.
 
ggunn said:
but even when they are not, changing the resistance of their output conductors does not appreciably change their output current. In fact, for an inverter running at its maximum power output, raising the voltage at its terminals will actually reduce the output current.
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These two sentences are contradictory. The second sentence I agree with, and it is the flip side of what I said, that if not operating in an AC current limited state, lowering the voltage at its terminals will raise the output current.

ggunn said:
Current sources are strange to most people because they are not commonly encountered.
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I guess my point is that PV inverters with a surplus of available power on the DC side and operating at an AC side current limit behave as current sources. Otherwise they behave as power sources. Neither of which are voltage sources, I agree.

Cheers, Wayne
 
jaggedben said:
But a bigger point... this is getting off topic. T
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Yes, I'm not sure that the "utility request that voltage triggered reactive power correction would supply negative VARs" (quoted from the OP) has anything to do with mitigating inverter voltage rise on the customer premises. I expect the request to be for the utility's benefit, so I'd only buy that if we have an explanation for what causes the grid voltage to go higher, and further an explanation of how the PV inverter being knocked offline due to exceeding its voltage window would exacerbate that grid problem. Not a topic I'm familiar with.

Cheers, Wayne
 
wwhitney said:
These two sentences are contradictory. The second sentence I agree with, and it is the flip side of what I said, that if not operating in an AC current limited state, lowering the voltage at its terminals will raise the output current.
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No, there is no difference. The output current of a PV inverter is not dependent on the resistance of its output conductors whether its input from the modules is saturated or not. No real world source operates as either a ideal current source or an ideal voltage source (battery voltage can sag under heavy load, for instance), but grid tied PV inverters and PV DC modules come close. The IV curve for a PV module is essentially flat from the Vmp knee all the way back to Isc where V=0. That data for a PV inverter would look the same if its output were not clamped at the service voltage.

The way current sources behave is easily found on line.
 
ggunn said:
No, there is no difference. The output current of a PV inverter is not dependent on the resistance of its output conductors whether its input from the modules is saturated or not.
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I'm going to disagree, although the effect is small. An example: say the grid voltage is 240V and the PV inverter can make 5 kW, but the resistance to the grid is 0.5 ohms (ridiculously high, I think). Then the PV inverter puts out 20A, causing 10V of voltage rise, so it is outputting at 250V. And 250V * 20A = 5 kW.

If we now halve the resistance to the grid to 0.25 ohms, 20A would only cause 5V of voltage rise to 245V, and the PV inverter would only be putting out 4.9 kW. So in the power limited regime it will put out more current to get to 5 kW. First guess is 100W / 245V = 0.408 amps more, although that is an overestimate, as it would cause the inverter terminal voltage to rise a bit more. Some algebra or a couple iterations will find a more exact answer.

Cheers, Wayne
 
ggunn said:
No, there is no difference. The output current of a PV inverter is not dependent on the resistance of its output conductors whether its input from the modules is saturated or not. No real world source operates as either a ideal current source or an ideal voltage source (battery voltage can sag under heavy load, for instance), but grid tied PV inverters and PV DC modules come close. The IV curve for a PV module is essentially flat from the Vmp knee all the way back to Isc where V=0. That data for a PV inverter would look the same if its output were not clamped at the service voltage.

The way current sources behave is easily found on line.
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Sorry, I misspoke. As you say, if the inverter input is such that the power output is a at its maximum, raising the voltage will indeed lower the current. P = IV.
 
wwhitney said:
Yes, I'm not sure that the "utility request that voltage triggered reactive power correction would supply negative VARs" (quoted from the OP) has anything to do with mitigating inverter voltage rise on the customer premises....
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It would only affect the customers premises in as much as they are connected to a much, much larger system.

I don't know how many times I've read that VV and VW functions are there to mitigate voltage rise in the system. I'm pretty sure it's a real thing.

All the literature I can find on volt-var agrees that when voltage is high the solar inverter is supposed to absorb vars (or underexcite, makes sense). And while it seems like much of it avoids assigning a plus or minus sign to those vars, it seems like all the graphs show them below zero.
See for example:

Small Commercial Inverter Laboratory Evaluations of UL 1741 SA Grid-Support Function Response Times. (Conference) | OSTI.GOV

The U.S. Department of Energy's Office of Scientific and Technical Information
www.osti.gov www.osti.gov
(Note this Solaredge doc describes injecting vars as negative and absorbing them as positive, but the graph seems to disagree!)

Perhaps this page will help.

relaytraining.com

Understanding the Great Leading vs. Lagging Power Debate

Ever wondered why a lagging current (negative angle) creates positive VARs? Should a lagging angle indicate that you are importing or exporting VARs? Find the answer in this post...
relaytraining.com relaytraining.com
 
voltage has to be higher at the source than at the load.
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This is not necessarily true in ac power systems. Real power (kW) transfer isn't strictly related to the voltage difference. The voltage difference creates var flow.
 
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