Harmonic Neutral Load Calculation
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winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
With more information. *grin*
You can't simply say 'harmonic loads'. You need to know _which_ harmonics, and you need to know the phasing for each harmonic present. Different devices will draw different harmonic currents, with different proportions of fundamental and harmonic current. Many devices (eg. computer power supplies) will have different characteristics at different load levels.
The simplest approach to understand is to use the 'principal of superposition', and to calculate each of the harmonics present separately, and then add up the results.
Any periodic waveform (voltage, current, air pressure, politics *grin* ) can be treated as a sum different sine waves. When a device draws harmonic current, you see non-sinusoidal current flowing even when sinusoidal voltage is applied. This non-sinusoidal current can be calculated as a sum of many sinusoidal currents, of different frequencies.
The normal way to calculate neutral current is:
neutral current = vector sum of
IAvec, IBvec, ICvec
IAvec is a vector which represents the amps flowing in leg A, as well as the _phase angle_ for the current flowing in each leg. Same for IBvec and ICvec.
Well the same formula applies to harmonics, with the catch that it applies _separately_ to each different frequency present.
So what you need to do is determine from the load characteristics how much fundamental is present, how much third harmonic, fifth, etc. with what phase angle. Then you do the vector sum for each harmonic to get the neutral current for that harmonic, and finally you add up all of the harmonic neutral currents.
Unfortunately, the above is probably almost useless to you in a practical sense unless those 1000A per supply leg is all one device or one type of device. If you have 1000A going into a large rectifier set, for example, you can probably calculate out the individual harmonic components. If you have 1000A of a random mix of computer hardware in a datacenter, your best bet is to go to a similar datacenter and put a scope on their feeders
Do you have a particular load in mind?
-Jon
You can't simply say 'harmonic loads'. You need to know _which_ harmonics, and you need to know the phasing for each harmonic present. Different devices will draw different harmonic currents, with different proportions of fundamental and harmonic current. Many devices (eg. computer power supplies) will have different characteristics at different load levels.
The simplest approach to understand is to use the 'principal of superposition', and to calculate each of the harmonics present separately, and then add up the results.
Any periodic waveform (voltage, current, air pressure, politics *grin* ) can be treated as a sum different sine waves. When a device draws harmonic current, you see non-sinusoidal current flowing even when sinusoidal voltage is applied. This non-sinusoidal current can be calculated as a sum of many sinusoidal currents, of different frequencies.
The normal way to calculate neutral current is:
neutral current = vector sum of
IAvec, IBvec, ICvec
IAvec is a vector which represents the amps flowing in leg A, as well as the _phase angle_ for the current flowing in each leg. Same for IBvec and ICvec.
Well the same formula applies to harmonics, with the catch that it applies _separately_ to each different frequency present.
So what you need to do is determine from the load characteristics how much fundamental is present, how much third harmonic, fifth, etc. with what phase angle. Then you do the vector sum for each harmonic to get the neutral current for that harmonic, and finally you add up all of the harmonic neutral currents.
Unfortunately, the above is probably almost useless to you in a practical sense unless those 1000A per supply leg is all one device or one type of device. If you have 1000A going into a large rectifier set, for example, you can probably calculate out the individual harmonic components. If you have 1000A of a random mix of computer hardware in a datacenter, your best bet is to go to a similar datacenter and put a scope on their feeders
Do you have a particular load in mind?
-Jon
charlie tuna
Senior Member
- Location
- Florida
since this is a test question -- maybe they are looking at a theory answer? harmonic currents do not mix or cancel each other out so they would be additive or 3000 amps. ????????????????????????
- Location
- Illinois
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- retired electrician
Charlie,
I thought that the theoretical current in the grounded conductor for odd order harmonics tops out at 1.73 times the line current.
Don
I thought that the theoretical current in the grounded conductor for odd order harmonics tops out at 1.73 times the line current.
Don
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Okay, that 'simplifies' things because an exam question is always a simplified scenario as compared to the 'real world'. In the 'real world' you would have messy things like manufacturing variation and installation differences. But for an exam question you can assume that each leg has exactly the same 1000A of load on it, and that all three legs are exactly the same. (You should also be on the lookout for 'trick' questions, eg. all of the loads being connected line-line, and thus _zero_ neutral loading.)
If all three legs are exactly the same, then you will _always_ have H*120 degrees of phase difference between the legs, no matter what the power factor. This greatly simplifies the situation.
For any harmonic H that is _not_ a multiple of 3, you will have three currents that are out of phase and balance perfectly, just like a balanced load without harmonics. But for any harmonic that _is_ a multiple of 3, all three legs will be exactly in phase, and thus the current on the neutral will be the _sum_ of the currents on each of the tree legs. In other words, if you have 200A of third harmonic on each leg, you will have 600A of harmonic current in the neutral.
So now your problem reduces to figuring out the fraction of 'triplen' harmonics produced by the ballasts.
The problem is that different ballasts produce very different harmonic currents. Magnetic ballasts produce significant but small harmonic currents (THD 20-30%). Electronic ballasts can produce tremendous harmonic currents (THD up to 150%), but with power factor correction can produce less than 5% THD. On top of this, even when you get a harmonic distortion number, you will get 'THD' or _total_ harmonic distortion, which is the _total_ for all harmonics, when you are really interested in only the 'triplen' harmonics. But if you know the percentage of triplen harmonic drawn by the ballasts in question, then you can answer the question; 3 * 1000A * %triplen harmonic.
The absolute worst case scenario is if the load consumes _only_ third harmonic current. Because of the way THD is defined, this would mean _infinite_ THD. In this case you would have 1000A of third harmonic current on each leg, and a total of 3000A on the neutral.
I did a quickie search, and found that 'Advance Transformer' lists some ballasts with 150% THD. This means that the harmonic current is 150% of the fundamental current. If we make the assumption that all of the harmonic current is third harmonic (which would be an incorrect but reasonable 'worst case' assumption), then this means that each leg would have 400A of fundamental and 600A of third harmonic, leading to 1800A of third harmonic on the neutral.
On the other hand, if you used some of the ballasts with 10% THD, and assume that only half of the harmonics are triplen harmonics, then you get 909A fundamental, 91A harmonic on each leg, with 45A of triplen harmonic on each leg and 136A of triplen harmonic on the neutral.
http://www.ece.utexas.edu/~grady/C4_Sources_June05.pdf
http://www.advancetransformer.com/ecatalog/FLB_LookUp.asp?lamptype=F8T5
-Jon
If all three legs are exactly the same, then you will _always_ have H*120 degrees of phase difference between the legs, no matter what the power factor. This greatly simplifies the situation.
For any harmonic H that is _not_ a multiple of 3, you will have three currents that are out of phase and balance perfectly, just like a balanced load without harmonics. But for any harmonic that _is_ a multiple of 3, all three legs will be exactly in phase, and thus the current on the neutral will be the _sum_ of the currents on each of the tree legs. In other words, if you have 200A of third harmonic on each leg, you will have 600A of harmonic current in the neutral.
So now your problem reduces to figuring out the fraction of 'triplen' harmonics produced by the ballasts.
The problem is that different ballasts produce very different harmonic currents. Magnetic ballasts produce significant but small harmonic currents (THD 20-30%). Electronic ballasts can produce tremendous harmonic currents (THD up to 150%), but with power factor correction can produce less than 5% THD. On top of this, even when you get a harmonic distortion number, you will get 'THD' or _total_ harmonic distortion, which is the _total_ for all harmonics, when you are really interested in only the 'triplen' harmonics. But if you know the percentage of triplen harmonic drawn by the ballasts in question, then you can answer the question; 3 * 1000A * %triplen harmonic.
The absolute worst case scenario is if the load consumes _only_ third harmonic current. Because of the way THD is defined, this would mean _infinite_ THD. In this case you would have 1000A of third harmonic current on each leg, and a total of 3000A on the neutral.
I did a quickie search, and found that 'Advance Transformer' lists some ballasts with 150% THD. This means that the harmonic current is 150% of the fundamental current. If we make the assumption that all of the harmonic current is third harmonic (which would be an incorrect but reasonable 'worst case' assumption), then this means that each leg would have 400A of fundamental and 600A of third harmonic, leading to 1800A of third harmonic on the neutral.
On the other hand, if you used some of the ballasts with 10% THD, and assume that only half of the harmonics are triplen harmonics, then you get 909A fundamental, 91A harmonic on each leg, with 45A of triplen harmonic on each leg and 136A of triplen harmonic on the neutral.
http://www.ece.utexas.edu/~grady/C4_Sources_June05.pdf
http://www.advancetransformer.com/ecatalog/FLB_LookUp.asp?lamptype=F8T5
-Jon
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
Don,
I had thought 1.73 as well for a long time, but I plotted the theoretical case of 100% 3rd harmonic in excel and found them to be perfectly additive.
I had thought 1.73 as well for a long time, but I plotted the theoretical case of 100% 3rd harmonic in excel and found them to be perfectly additive.
- Location
- Massachusetts
ron said:
This is interesting as I had just posted 1.73 in another thread http://www.mikeholt.com/codeForum/viewtopic.php?p=1193609#1193609 and Infinity came up with literature from APC, American Power Conversion that also stated 1.7.
http://www.apcmedia.com/salestools/SADE-5TNQZ5_R0_EN.pdf
Note the info on page six.
Is it possible the limit on paper is purely additive but in the real world you would not see more than 1.73?
This also would mean that the commonly run 200% neutrals should be 300%.
Just asking.
Ron:
If they were "perfectly additive" (i.e. 1000+1000+1000=3000), then they would have to be in phase. And we know the fundamental voltages for a Wye are 120 degrees out of phase, so the third harmonics will also be out of phase.
I think you may have mistakenly graphed:
sin 3x
sin 3(x+120)
sin 3(x+240)
Instead of:
sin 3x
sin (3x+120)
sin (3x+240)
Steve
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
Hmm. Bob, I think that you are correct about the practical situation. In fact, my gut tells me that _harmonic_ loading on the neutral will be much less than 2x the phase current.
Let's explore the limit further.
All of the triplen current on the phase legs adds directly on the neutral. If you somehow had pure triplen current on the phase legs, then the neutral current would be 3x the phase current.
The next question is: could you ever have pure triplen current on the phase legs. Here you could state a solid _no_.
My reasoning is that _power_ is voltage times current. If you have harmonic current flows, they only deliver power to the load if there are corresponding harmonic _voltages_ on the line. If we had a _perfect_ zero impedance transformer supplying the system, the _voltage_ would always be sinusiodal, and so no power would be delivered to this pure harmonic load.
In real systems you have impedance, and the harmonic current flows distort the system voltage; so on a real system the voltage could be considered a sum of fundamental and harmonic voltages. The product of (say) third harmonic voltage and third harmonic current could represent a real power flow. However because the harmonic voltages are caused by the harmonic current flows interacting with the system impedance, my hunch is that these power flows actually represent power flowing _from_ the load to the supply. In other words, my hunch is that if you actually mapped the power flows, you would see real power at fundamental frequency going to the load, the load using the bulk of the power, but also generating real harmonic power that goes out from the load. Please consider this point to be 'hand waving' rather than science.
But back to the point: a load which draws pure harmonic current will not actually be _powered_ by all of this current flow. The real power delivered to the load depends upon the fundamental current flowing in phase with the supply voltage, and all the harmonic currents create a 'power factor'. This is not the classic power factor caused by fundamental current out of phase with the fundamental voltage; this is the more general case of current flow which sometimes delivers power to the load, sometimes takes power from the load, and averaged over time delivers zero useful power to the load.
This means that you will not see loads that draw 'pure' harmonic current.
So then we ask what is the worst case load one might actually see. I believe here the people who came up with the 1.73 number were not being imaginative enough, but that at the same time 1.73 is likely more that would ever be seen in a real system. I believe that a particularly evil device could draw far more triplen current than fundamental current, and thus exceed the 1.73 number, but that this is _very_ unlikely because the device would need to be designed to intentionally cause harmonic problems *grin*
As mentioned above, the worst ballast in the Advance Transformer catalog has a _Total Harmonic Distortion_ (THD) of 150%. What this means is that if the ballast is drawing 2.5A of current, you have 1A of fundamental and 1.5A of harmonic currents. I strongly doubt that there is anything out there that is worse than 150% current THD.
By definition, THD is the _sum_ of _all_ harmonic current divided by the fundamental current; if the harmonic current is greater than the fundamental current, then THD is greater than 1. Another way of looking at it is that 100% THD means 50% harmonic current, and 300% THD means 75% harmonic current, whereas 25% THD means 20% harmonic current.
THD involves the sum of _all_ harmonics. But for neutral loading we care about the triplen harmonics. So if we have a 150% THD ballast, that means that fully 60% of the current flow is harmonic, and that something less than 60% of the current flow is in triplen harmonics.
If we go back to the original example. We have a phase load of 1000A. Worst case for really bad ballasts is that we have 600A of harmonic current flowing. If _all_ of this current flow is triplen harmonic, then we have 1800A on the neutral.
So here is an example where we just about approach the 1.73 number (we exceed it only if we presume only triplen harmonics, an incorrect assumption), and I needed to dig out the absolutely worst ballast in a catalog.
-Jon
Let's explore the limit further.
All of the triplen current on the phase legs adds directly on the neutral. If you somehow had pure triplen current on the phase legs, then the neutral current would be 3x the phase current.
The next question is: could you ever have pure triplen current on the phase legs. Here you could state a solid _no_.
My reasoning is that _power_ is voltage times current. If you have harmonic current flows, they only deliver power to the load if there are corresponding harmonic _voltages_ on the line. If we had a _perfect_ zero impedance transformer supplying the system, the _voltage_ would always be sinusiodal, and so no power would be delivered to this pure harmonic load.
In real systems you have impedance, and the harmonic current flows distort the system voltage; so on a real system the voltage could be considered a sum of fundamental and harmonic voltages. The product of (say) third harmonic voltage and third harmonic current could represent a real power flow. However because the harmonic voltages are caused by the harmonic current flows interacting with the system impedance, my hunch is that these power flows actually represent power flowing _from_ the load to the supply. In other words, my hunch is that if you actually mapped the power flows, you would see real power at fundamental frequency going to the load, the load using the bulk of the power, but also generating real harmonic power that goes out from the load. Please consider this point to be 'hand waving' rather than science.
But back to the point: a load which draws pure harmonic current will not actually be _powered_ by all of this current flow. The real power delivered to the load depends upon the fundamental current flowing in phase with the supply voltage, and all the harmonic currents create a 'power factor'. This is not the classic power factor caused by fundamental current out of phase with the fundamental voltage; this is the more general case of current flow which sometimes delivers power to the load, sometimes takes power from the load, and averaged over time delivers zero useful power to the load.
This means that you will not see loads that draw 'pure' harmonic current.
So then we ask what is the worst case load one might actually see. I believe here the people who came up with the 1.73 number were not being imaginative enough, but that at the same time 1.73 is likely more that would ever be seen in a real system. I believe that a particularly evil device could draw far more triplen current than fundamental current, and thus exceed the 1.73 number, but that this is _very_ unlikely because the device would need to be designed to intentionally cause harmonic problems *grin*
As mentioned above, the worst ballast in the Advance Transformer catalog has a _Total Harmonic Distortion_ (THD) of 150%. What this means is that if the ballast is drawing 2.5A of current, you have 1A of fundamental and 1.5A of harmonic currents. I strongly doubt that there is anything out there that is worse than 150% current THD.
By definition, THD is the _sum_ of _all_ harmonic current divided by the fundamental current; if the harmonic current is greater than the fundamental current, then THD is greater than 1. Another way of looking at it is that 100% THD means 50% harmonic current, and 300% THD means 75% harmonic current, whereas 25% THD means 20% harmonic current.
THD involves the sum of _all_ harmonics. But for neutral loading we care about the triplen harmonics. So if we have a 150% THD ballast, that means that fully 60% of the current flow is harmonic, and that something less than 60% of the current flow is in triplen harmonics.
If we go back to the original example. We have a phase load of 1000A. Worst case for really bad ballasts is that we have 600A of harmonic current flowing. If _all_ of this current flow is triplen harmonic, then we have 1800A on the neutral.
So here is an example where we just about approach the 1.73 number (we exceed it only if we presume only triplen harmonics, an incorrect assumption), and I needed to dig out the absolutely worst ballast in a catalog.
-Jon
winnie
Senior Member
- Location
- Springfield, MA, USA
- Occupation
- Electric motor research
Steve:
Actually when dealing with harmonics in power systems it _is_ correct to use sin(k*(time + phase)).
If you have balanced loads on each phase, then the _shape_ of the current drawn by each load should be the same.
For the shape to be the same, all of the harmonics need to be displaced by the same _time_. 120 degrees in a 60 cycle wave corresponds to the same amount of _time_ as 360 degrees in a 180 cycle wave.
In other words, the phases in a 60Hz three phase system are displaced in time by 5.56mS. This means that each harmonic component in a balanced system will be displaced by 5.56mS. But 5.56mS in a 180Hz harmonic is a 360 degree phase displacement.
This is why triplen harmonics are additive in three phase systems. If all harmonics remained 120 degrees apart, then there would never be additive problems on the neutral.
-Jon
Actually when dealing with harmonics in power systems it _is_ correct to use sin(k*(time + phase)).
If you have balanced loads on each phase, then the _shape_ of the current drawn by each load should be the same.
For the shape to be the same, all of the harmonics need to be displaced by the same _time_. 120 degrees in a 60 cycle wave corresponds to the same amount of _time_ as 360 degrees in a 180 cycle wave.
In other words, the phases in a 60Hz three phase system are displaced in time by 5.56mS. This means that each harmonic component in a balanced system will be displaced by 5.56mS. But 5.56mS in a 180Hz harmonic is a 360 degree phase displacement.
This is why triplen harmonics are additive in three phase systems. If all harmonics remained 120 degrees apart, then there would never be additive problems on the neutral.
-Jon
So the 3 does go outside the parentheses, and Ron's conclusion is correct.
Its been too long since I've done trig to figure out what, but I still think something is missing.
3 or 1.73.... Hmmm.
We are talking about phase currents, so there shouldn't be any sqrt(3)'s for converting to line currents. And were not talking about power, so no multipliers for that. I can't figure out how we get to 1.73 from 3.
ron
Senior Member
- Location
- New York, 40.7514,-73.9925
I would be glad to upload my spreadsheet or graph, if I only knew how....
- Location
- Illinois
- Occupation
- retired electrician
dok,
There are no calculations in the code book for this type of application. There are a few rereferences. 210.4(A) FPN, 220.61(C)(2), 310.15(B)(4)(c), 400.5(B), 450.3 FPN#2, 450.9 FPN #2, and Table 520.44.
Don
There are no calculations in the code book for this type of application. There are a few rereferences. 210.4(A) FPN, 220.61(C)(2), 310.15(B)(4)(c), 400.5(B), 450.3 FPN#2, 450.9 FPN #2, and Table 520.44.
Don
dok said:
So... for you engineers, do you call the customer you are doing the calculations for and ask what ballast they are using? What if the next building owner decides to upgrade/change to a different lighting system? Would it make sense not to downsize the neutral, and leave the same size as the phase un-grounded conductors, or in extreme cases with lots of PC's or cheap ballasts, the neural oversized by one wire size??
If its not in the code, it must not be a problem!! Not.
ramsy
NoFixNoPay Electric
- Location
- LA basin, CA
- Occupation
- Service Electrician 2020 NEC
Both of iwire's points here are also supported by the interesting evidence quoted below.iwire said:
Had to use Code for Table 4.1. Current Diversity Factor Multipliers for Large Numbers of Nonlinear Loads
Code:
Current Diversity
Harmonic Factor
3 1.0
5 0.9
7 0.9
9 0.6
11 0.6
13 0.6
15 0.5
Higher Odds 0.2
All Evens 0.0iwire said:This also would mean that the commonly run 200% neutrals should be 300%.
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