K8MHZ
Senior Member
- Occupation
- Electrician
How do I know which switching KVAR bank to look at for the rating of a particular inverter?
Negative power factor and PV systems
- Thread starter tallgirl
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How do I know which switching KVAR bank to look at for the rating of a particular inverter?
mivey
Senior Member
They do exactly that. They set the current output to be in phase with the grid power. The traditional inverters worked this way. The new ones have the capability to supply vars as well.
A generator is different because it can vary the phase angle between the voltage and current.
Then the output is all resistive.
The traditional ones did not. The newer ones can.
mivey
Senior Member
You brought up polarity. I was trying to answer in like kind.
The location above or below the axis of the waveform at a particular point in time is what I was talking about. If you use the sinusoidal formula for the voltage and current, you will get either zero, positive, or negative values at any given point in time. When the signs of the voltage and current values are the same, power is flowing towards the load (the load is absorbing energy). When the signs of the voltage and current values are different, power is flowing towards the source (the load is releasing energy).
Yes. It can also take place externally to the meter by running one of the wires backwards through the common CT (you can see this on CT metered high-leg delta banks).
mivey
Senior Member
Yes. The inverter will shut down.
I'm sure the part about not islanding is covered in your text.
Whether or not the author "concedes" does not change the facts. Perhaps you should base your conclusions on more than one source. Perhaps the author of your text thought the discussion was beyond the scope of the text. I am quite sure the author understands but that does not mean the author covered the subject. This subject is covered in many other places so feel free to expand your knowledge.
Read my post #12. The correct rate will penalize a customer for an excessive var load. This will happen with larger loads with advanced metering. Residential billing has traditionally not been sophisticated enough to bill for this. The age of smart metering will make that change.
The traditional grid-tied inverter will not run without the grid. An inverter that can run in island mode can supply both the kWs and vars.
mivey
Senior Member
The author may consider metering and billing to be beyond the scope of the text. I think it is worth a discussion but at what point does the publisher draw the line and not try to be a text on power systems as well? There are other sources that cover this issue. There is no one book/class that covers everything.
I think these micro-inverters can supply vars as well. They should have instructions on how to set the var control if they do.
Search for "var control" at whatever inverter reference you are using.
SAC
Senior Member
- Location
- Massachusetts
Hi Gar, my point wasn't about an ideal current source, it was about a "real-world" PV inverter. Connect an inductor (a "real-world" one) to the output of such a PV inverter, and we will see reactive power being delivered to it. Will the inverter be as efficient as if it was only delivering real power to a real load? Likely not, but I'd place a substantial bet that it will still deliver reactive power. The bottom line is that you can't change the physical properties of an inductor by changing the source, and it will have the voltage/current phase shift through it if a time varying voltage is applied. The real question is how efficiently the inverter will supply power to such a load - not that it either will or won't (it will). That being my thinking, I'd like to see the circuit diagram for a "real-world" inverter that will only deliver power with a unity PF, regardless of the load.
All that said, I don't disagree that if a PV inverter was poor at maintaining voltage regulation for such reactive loads, that a greater portion of the power for reactive loads would be drawn from the grid when both the inverter and the grid were operating in parallel.
zog
Senior Member
- Location
- Charlotte, NC
Some thread here, we have all sorts of goofy terminology being used and it is getting confusing. I have never seen negative/positive or % used to describe the PF of a system before. Typically it is a number 1.0 or less and leading or lagging. (Ex 0.85 lagging) for a system with more inductive than capacitive loads.
The only time I have seen a % used with PF is for Insulation PF testing, where perfect insulation (No watts loss) would be reported as 0% and 100% would be the worst insulation condition possible. Typical PF % limits for power transformers for example are 0.5% to 5%, depending on the class of transformer and insulating medium used.
The only time I have seen a % used with PF is for Insulation PF testing, where perfect insulation (No watts loss) would be reported as 0% and 100% would be the worst insulation condition possible. Typical PF % limits for power transformers for example are 0.5% to 5%, depending on the class of transformer and insulating medium used.
K8MHZ
Senior Member
- Occupation
- Electrician
Mivey,
Thanks for the replies, since there are so many with so much information I think it would be best to bring this info to my instructor's attention. If you are correct, this should be taught to students studying PV installation.
You mentioned that you thought the author may have purposely left that info out. I don't think so, especially with so much info that is nearly superfluous in nature being discussed.
I am also going to strongly suggest that a test be run. Of course, I will post the results.
One issue I would like detail on, though. I asked to be shown where a customer has been penalized for poor power factor as a result of the effects of a PV system. I still want to know if it has ever really happened. I know that if I were an industrial customer and spent big bucks on a PV system only to be penalized for it, I would be PO'd. Big time.
This is not a minor issue and if it is occurring in the wild, the PV students need to know about it.
Thanks for the replies, since there are so many with so much information I think it would be best to bring this info to my instructor's attention. If you are correct, this should be taught to students studying PV installation.
You mentioned that you thought the author may have purposely left that info out. I don't think so, especially with so much info that is nearly superfluous in nature being discussed.
I am also going to strongly suggest that a test be run. Of course, I will post the results.
One issue I would like detail on, though. I asked to be shown where a customer has been penalized for poor power factor as a result of the effects of a PV system. I still want to know if it has ever really happened. I know that if I were an industrial customer and spent big bucks on a PV system only to be penalized for it, I would be PO'd. Big time.
This is not a minor issue and if it is occurring in the wild, the PV students need to know about it.
kingpb
Senior Member
- Location
- SE USA as far as you can go
- Occupation
- Engineer, Registered
Power factor can be leading or lagging. A lagging power factor means that the current lags the voltage; a leading power factor means the current leads the voltage.
So, for lagging, the pf angle is positive because the voltage angle minus the current angle is positive, and for leading the voltage angle minus the current angle is negative. I believe that is where the misunderstanding originally may have taken place because you can have a positive or negative voltage angle. However, the cosine of a positive or negative angle is still a positive number that is less than 1.
Know as far as direction of power flow. Remember when dealing with ac power, there is real power (P) and reactive power (Q), and it is expressed as Complex Power or a phasor, which means it has magnitude and direction. Q will be positive when the pf has a positive angle (lagging) and Q will be negative when pf has a negative angle (leading).
S = P +jQ, and S = P+-jQ
An inductive load will have a +Q, and a capacitive load will have a -Q. Thus is the principle behind power factor improvement, because you are trying to reduce +Q of a load by adding capacitance of -Q. The lower the pf the more capacitance is needed to improve it.
Example: assume pf=0.85 lagging, so S*0.85 = P That means the the cos (angle) = 0.85, taking the arccos(0.85) = 31.79 deg. To determine the portion of S that is reactive power, you take the sin(31.79deg)= 0.53, a S*0.53 = Q. We know it is +Q because it was stated as being lagging.
To make the pf become near 1.0 you need to add capacitance (-Q) that is equal or nearly equal Q. Thus making S = P, which is true when the load is purely resistive, or pf=1.0 and therefore no reactive power flowing in either direction.
So, if you can reduce your current drawn, or measured at the meter, by improving the pf.
As far as power flow, a generator supplies P, but it can also either supply reactive power or absorb reactive power. A large generator will have the ability to usually go between 0.85 lagging to 0.95 leading. However, I will tell you that they are rated on lagging power factor, and the utility does not like to over-excite the generator to absorb Vars. But they also do not like to have to produce a lot of Vars to support the system because this reduces their ability to produce P (watts). Hence why they will penalize a customer for a low power factor. On the other hand, a load absorbs P and absorbs Q. To much Q, (bad pf) and your wasting money. But a system could be made to produce Q as well by adding too much capacitance and changing the pf to leading.
In conclusion, P and Q can flow in opposite directions simultaneously, i.e. and this is dependent on the angle between the voltage and the current, which is what determines pf. But, pf cannot be negative, it can be either unity (1.0), leading, or lagging.
So, for lagging, the pf angle is positive because the voltage angle minus the current angle is positive, and for leading the voltage angle minus the current angle is negative. I believe that is where the misunderstanding originally may have taken place because you can have a positive or negative voltage angle. However, the cosine of a positive or negative angle is still a positive number that is less than 1.
Know as far as direction of power flow. Remember when dealing with ac power, there is real power (P) and reactive power (Q), and it is expressed as Complex Power or a phasor, which means it has magnitude and direction. Q will be positive when the pf has a positive angle (lagging) and Q will be negative when pf has a negative angle (leading).
S = P +jQ, and S = P+-jQ
An inductive load will have a +Q, and a capacitive load will have a -Q. Thus is the principle behind power factor improvement, because you are trying to reduce +Q of a load by adding capacitance of -Q. The lower the pf the more capacitance is needed to improve it.
Example: assume pf=0.85 lagging, so S*0.85 = P That means the the cos (angle) = 0.85, taking the arccos(0.85) = 31.79 deg. To determine the portion of S that is reactive power, you take the sin(31.79deg)= 0.53, a S*0.53 = Q. We know it is +Q because it was stated as being lagging.
To make the pf become near 1.0 you need to add capacitance (-Q) that is equal or nearly equal Q. Thus making S = P, which is true when the load is purely resistive, or pf=1.0 and therefore no reactive power flowing in either direction.
So, if you can reduce your current drawn, or measured at the meter, by improving the pf.
As far as power flow, a generator supplies P, but it can also either supply reactive power or absorb reactive power. A large generator will have the ability to usually go between 0.85 lagging to 0.95 leading. However, I will tell you that they are rated on lagging power factor, and the utility does not like to over-excite the generator to absorb Vars. But they also do not like to have to produce a lot of Vars to support the system because this reduces their ability to produce P (watts). Hence why they will penalize a customer for a low power factor. On the other hand, a load absorbs P and absorbs Q. To much Q, (bad pf) and your wasting money. But a system could be made to produce Q as well by adding too much capacitance and changing the pf to leading.
In conclusion, P and Q can flow in opposite directions simultaneously, i.e. and this is dependent on the angle between the voltage and the current, which is what determines pf. But, pf cannot be negative, it can be either unity (1.0), leading, or lagging.
K8MHZ
Senior Member
- Occupation
- Electrician
Great link!
Thanks. The info in that link will be given to my instructor tomorrow along with a request for an explanation.
wiigelec
Member
- Location
- Red Desert
From a purely mathematical standpoint:
If P<0 that puts you in the II and III quadrants which gives you a negative cos ie a negative pf...
kingpb
Senior Member
- Location
- SE USA as far as you can go
- Occupation
- Engineer, Registered
hummm...
Magnitude of S is equal to the square root of the sum of the squares, so even if P was somehow made to be less than 0, the magnitude of apparent power would be positive, and thus V or I could not be negative since S = V*I*1.732
PF is calculated as cos(angV - angI), even if this angle is negative, the pf will vary between 0 and 1. Mathematically it is not possible for it to be negative. Which, guarantees the equation for apparent power stays within the I and IV quadrant and physically impossible for it to be in the II or III quadrant
Magnitude of S is equal to the square root of the sum of the squares, so even if P was somehow made to be less than 0, the magnitude of apparent power would be positive, and thus V or I could not be negative since S = V*I*1.732
PF is calculated as cos(angV - angI), even if this angle is negative, the pf will vary between 0 and 1. Mathematically it is not possible for it to be negative. Which, guarantees the equation for apparent power stays within the I and IV quadrant and physically impossible for it to be in the II or III quadrant
skeshesh
Senior Member
- Location
- Los Angeles, Ca
I'm pretty sure that absolute values are involved in the power factor formula. I don't have time to look it up right now but I'm sure a web search would show the relevant result.
This discussion has been very interesting and I've been meaning to jump in but have been busy. I think Bob does have a point of including a section that brings this up in literature related to PV, but Mivey also has a point that power factor and the power triangle relations are a topic that should be and is studied at its own.
In any case, we got a bit sidetracked from the original question: the poco will care about a low power factor but as other pointed out the actual cost will depends on the type and scale of installation as well as the utility in question. The most accurate answer is one you can get from a utility person in your area who can explain what are the PF requirements and associated charges as well as their monitoring of it. Keep in mind though, even if you get away with not getting charged for the VARs it is possible that this issue will come up again in the future should the poco install the necessary metering and apply the charges.
K8MHZ
Senior Member
- Occupation
- Electrician
I just stumbled across this:
The above was from a thread on another site that was locked because the person asking for comments was a guest and not a member.
I also checked the web site of Consumers Energy, the poco in our area. They have two agreement forms, one for less than 20kw and one for more. No mention of power factor on the form for less than 20kw. I couldn't get the form for over 20kw to open.
The above was from a thread on another site that was locked because the person asking for comments was a guest and not a member.
I also checked the web site of Consumers Energy, the poco in our area. They have two agreement forms, one for less than 20kw and one for more. No mention of power factor on the form for less than 20kw. I couldn't get the form for over 20kw to open.
K8MHZ
Senior Member
- Occupation
- Electrician
Check these out, large scale solar arrays that utilize VAR control to act like a generator:
http://docs.google.com/viewer?a=v&q...kmWzsI&sig=AHIEtbRHSA7-aJAf8lCqGkWTAZArImR0Pg
http://docs.google.com/viewer?a=v&q...kmWzsI&sig=AHIEtbRHSA7-aJAf8lCqGkWTAZArImR0Pg
Electric-Light
Senior Member
If the inverter's output is not perfectly sinusoidal, then VA delivered will be greater than W delivered. It would be like one person on multi-person bicycle delivering pedaling power only through partial turns. Though he's still delivering power to the wheels, he's not doing so consistently throughout the whole turn.
I have not read the entire thread yet and need to spend some time catching up but my initial thought is that an inverter is basically the same as a VFD. Both an inverter and a VFD have simply a DC bus and 6 output IGBT's that are fired to create a 3-phase voltage at a commanded frequency.
I VFD is obviously capable of supplying reactive power to a motor since ovbiously the motor is running with the output from the VFD. The only difference here is that the input side to the VFD is AC whereas the input to an inverter is DC. I dont see this as mattering however due to the fact that the AC input to a vfd is created to a DC voltage anyway.
So I guess I ask myself why can a vfd provide the reactive current required by a motor but an inverter cannot supply the reactive current required by a motor? I always just thought that the output of a vfd or inverter was a voltage waveform at a given frequency and that the load determined the current and phase angle leaving either of the devices. Never have I heard of a vfd trying to ouput a unity power factor when controlling a motor.
I VFD is obviously capable of supplying reactive power to a motor since ovbiously the motor is running with the output from the VFD. The only difference here is that the input side to the VFD is AC whereas the input to an inverter is DC. I dont see this as mattering however due to the fact that the AC input to a vfd is created to a DC voltage anyway.
So I guess I ask myself why can a vfd provide the reactive current required by a motor but an inverter cannot supply the reactive current required by a motor? I always just thought that the output of a vfd or inverter was a voltage waveform at a given frequency and that the load determined the current and phase angle leaving either of the devices. Never have I heard of a vfd trying to ouput a unity power factor when controlling a motor.
CONTROL FREQ
Member
- Location
- OHIO
Just for mindless fodder, and witty banter that will most certainly follow, I suggest that they are one in the same. when an allen bradley vfd says DC voltage, it is kinda like how you "dowload" to a plc and "upload" from a plc... it's all terminology. If you're on 'you tube' you "upload" from your pc... etc. etc. This 'dc' they are talking about, is much more like "clean dc" it's still ac, make no mistake, but the characteristics (coming from a dude with no college) are strikingly similar to almost flawless DC. I figure the vfd is like the "ANTI-TESLA" woulda been Edison's ultimate weapon. You can control damn near every aspect of the electrical spectrum with a VFD.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110525-2339 EDT
philly:
Go back to my post numbered 36.
There I described a device that is a current source that forces its current output to be proportional to the voltage to which it is connected. Now you have a unity power factor output.
A VFD is a voltage generator with some over-current limitation (torque limiting), and in this case the load determines the power factor. Whereas the grid tied inverter is primarily a current source. All this is accomplished with electronic control. Note: with the grid tied inverter the voltage and frequency are largely controlled by the grid. Frequency very closely and voltage not so close.
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philly:
Go back to my post numbered 36.
There I described a device that is a current source that forces its current output to be proportional to the voltage to which it is connected. Now you have a unity power factor output.
A VFD is a voltage generator with some over-current limitation (torque limiting), and in this case the load determines the power factor. Whereas the grid tied inverter is primarily a current source. All this is accomplished with electronic control. Note: with the grid tied inverter the voltage and frequency are largely controlled by the grid. Frequency very closely and voltage not so close.
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