Learn about the major changes to the 2018 Edition of NFPA 70E on your own schedule – for FREE!
This video is based on Jim’s 2018 NFPA 70E changes article that was published in May 2017 for Multi-Award Winning Electrical Contractor Magazine.
Learn about the major changes to the 2018 Edition of NFPA 70E on your own schedule – for FREE!
This video is based on Jim’s 2018 NFPA 70E changes article that was published in May 2017 for Multi-Award Winning Electrical Contractor Magazine.
SAMPLE VIDEO CLIP
NEW FOR 2017! 4 DVD training program taught by Jim Phillips, P.E. $199. Learn how to perform short circuit analysis / calculations and equipment adequacy evaluations. Understand the importance of X/R ratios, symmetrical vs. asymmetrical faults and how series ratings work. Many calculation examples are used to illustrate how to perform an analysis. The methods are based on the IEEE Violet book and is loaded with many in class examples and problems for a hands on learning experience.
In a per phase analysis, what that actually involves is you have a three-phase power system. And the assumption is the event on all three phases, the result of whatever it is you’re modeling on all three phases, is the same. It’s balanced. So a three-phase short circuit, you’re considering that, well, whatever the current is, tens of thousands of amps, it will be the same on phase A, on phase B, and on phase C.
So what do you do? You just analyzed one phase. It’s not aligned to ground calculation, it’s a per phase calculation. So you analyze what’s on one phase, and then what you determine on that one phase is also the answer for the other phases. And so when we’re performing the calculations, like I had up here on the screen, the voltage that we’re looking at across one phase is the line to neutral voltage.
So for the per phase analysis, the example that I have for the per phase analysis, I have aligned the neutral voltage on the left side of this diagram, 277 volts, that’s lined to neutral from a 480 system. And then I have what amounts to the equivalent impedance of the source, of the utility system. Some you may remember the term Thevenin equivalent impedance.
Sorry about that. You thought you got away from that one. There’s actually a use for it. For those of you that are not familiar with that term, what the Thevenin equivalent impedance is, it’s basically taking the impedance of everything out there in the utility world, and series and parallel combinations, and reducing it down to just a point equivalent impedance that represents everything out there.
And so that’s what the 0.0041 ohms is in the upper left. That’s basically the equivalent impedance of what the utility system appears like, electrically, at the beginning of the study, usually at your service entrance or your interconnection point. Then the next one, the transformer, even though the transformers are normally in percent, I have this in ohms just for the sake of this particular problem, 0.0085 ohms.
And then we have a conductor on the secondary side, a set of conductors, 0.0034 ohms. And then we have a load on the far right side. The impedance of the load, 2 ohms.
So, a per phase analysis just using basic Ohm’s law, voltage, current, and impedance, if you were to take 277 volts and divide by these four impedances, the source being 0.0041, the transformer being 0.0085, the conductor point 0.0034, and the load 2 ohms, and you add all of this together, 277 volts divided by 2.016 ohms, the current that would flow would be 137 amps. That’s a load curve. That would flow in all three phases using this per phase analysis.
That’s the load current. But what happens, looking at this diagram, near the load just downstream from the overcurrent device on the far right side, if we have a short circuit? The short circuit, the way we model it, we look at the short circuit as a zero impedance path. We consider this to be the bolted connection.
We don’t want any additional impedance because, for this type of study, we want the maximum worst case short circuit current. So we don’t want to factor in any additional impedance. We say it’s a bolted fault, a bolted short circuit.
So now what happens, looking at the right side of this, the load impedance is still there. 2 ohms are still there. But the current that’s going to flow from the source is not going to go through the 2 ohms. Current takes the path of least resistance, literally. That’s how it works.
And so now what’s going to happen, as far as the analysis, we still have 277 volts. But now we just divide by the source, the transformer, and the conductor impedance. The load impedance doesn’t factor into this.
So 277 volts divided by just the small impedances of the source, the transformer, and the conductor, the resulting current is quite large. 17,313 amps. So what you need to do is consider, will the protective device operate?
And not only will it operate, will it operate safely? Does the protective device have a sufficient or an adequate interrupting rating to be able to handle the predicted 17,313 amps? That’s what a short circuit study is all about.
The short circuit study is to mathematically model the power system with all the different impedances. The impedance of the utility system, the impedance of transformers, of conductors, to mathematically model all that, and to be able to calculate or predict what would be the maximum short circuit current at different locations for equipment. And you compare the calculated short circuit currents with the device’s, or the equipment’s, interrupting or withstand rating. That’s what a short circuit study is all about.
One line diagrams that are used as part of an arc flash study must be up to date. NFPA 70E 120.1 Verification of an Electrically Safe Work Condition, requires checking up-to-date drawings, diagrams and tags when verifying an electrically safe work condition. In addition, when performing an arc flash study/risk assessment, an up-to-date one line drawing is the primary document that defines the system under study.
Jim Phillips discusses how, unless a study was performed, most companies do not have up-to-date electrical drawings. This is one of the many subjects discussed in Jim’s Electrical Safety and Arc Flash Training programs.
How up to date are your one-lines? That that’s a question a lot of people wrestle with, and they’re really important when it comes to electrical safety.
One of the reasons is NFPA 70e120.1 for verification of an electrically safe work condition– says that you’re supposed to check applicable “up to date”– and that’s the key thing, up to date– drawings, and diagrams, and identification tags. You also need to have an up to date one-line– or single line, as some call them– when you’re performing an arc flash study. Basically, it defines the road map of what your system looks like.
So I asked this question at the arc flash forum– arcflashforum.com– do you or your clients, if you happen to be a consultant– you have drawings or diagrams that are up to date, thinking in terms of one-lines and single lines? And the answers were kind of as expected. 25% of the people said yes. That’s it. Only 25% percent of the people. 51% said that no, they need to be updated.
And you may be looking at this thinking, well, wait a minute, Jim. 25 and 51– that’s not exactly 100%. What are we missing here? Well, I had a little fun with this, and I also asked the question, drawings? I’m not even sure they exist. That was 24% of the people. So the bottom line in most cases– when you’re performing an arc flash study, or you’re going to verify an electrically safe work condition, there is going to be some work that’s going to need to be done to bring diagrams and drawings up to date.
I know back in my past, one of the places that I used to work– this was actually back in the late 1970s– they had drawings that dated all the way back to the early 1900s. It was a very, very old industrial facility. And that’s the case. Sometimes if you happen to find a one-line drawing, or you ask a client, do have a one-line drawing? They’ll say, oh, yeah. We do. And then you look at it and you find out, well, it’s 20, 30 years old. Is that up to date? I really doubt it.
Per Unit System and Symmetrical Components are used for many short circuit calculations. Electric Utility Data is also often provided in terms of Per Unit and Symmetrical Components. Watch this video to see Jim Phillips explain how to use the positive, negative and zero sequence impedances in a per unit format to calculate the three phase and line-to-ground short circuit current. This video is from Jim’s Electrical Power Training Class: Electric Power Calculations
About Jim Phillips: Electrical Power and Arc Flash Training Programs – For over 30 years, Jim Phillips has been helping tens of thousands of people around the world, understand electrical power system design, analysis, arc flash and electrical safety.
Jim is Vice Chair of IEEE 1584, International Chairman of IEC TC78 Live Working and Steering Committee Member – IEEE/NFPA Arc Flash Collaborative Research Project. He has developed a reputation for being one of the best trainers in the electric power industry. Learn More
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How do you perform short circuit calculations using a method known as symmetrical components and also the per unit system? Well, the actual calculation process is pretty simple. To calculate the three phase short circuit current, you just take the voltage, and you divide by what I’m showing here is Z 1 plus Z f. Now, what Z 1 is, that’s defined as the positive sequence impedance. And it’s basically just the impedance of your circuit, the impedance of the conductors, the transformers, the utility company.
Z f is defined as the impedance for the fault. So if you have something that’s actually causing the fault, like a tree or something like that, you can introduce Z with the subscript f. However, most people just assume Z sub f is 0. They assume that it’s a bolted short circuit, and there’s no additional impedance involved.
The line-to-ground short circuit calculation is done by taking 3 times the voltage, divided by several different impedances. Z 1, the positive sequence impedance, plus Z 2 which is the negative sequence impedance, plus Z 0 the 0 sequence impedance, plus 3 times the fault impedance. So again if you have a tree or something that’s in the way.
Now, what does Z 1, Z 2, and Z 0 actually mean? These values are based on a theory known as symmetrical components. And Z 1, as I mentioned, that that’s just the impedance of your circuit elements, the conductors, transformers, the utility system. Z 2, the good news is Z 2 in almost every case, is the same as Z 1. So if you know the impedance of your conductors, your transformers, and so forth, you have both Z 1 and Z 2.
Z 0, however, 0 sequence impedance, that’s the impedance that’s the result of current flowing on something other than a face path. So Z 0 is going to be based on a return path other than phase conductors, and it’s also going to be affected significantly by grounding, such as how transformers are grounded, and how generators are grounded. So I want to walk you through a simple problem in performing short circuit calculations when you have the symmetrical component values positive and 0 sequence impedance. But also bases, base values, and I actually talk about the base values and per unit in another video.
So you may look at this example and see that I have Z 1, I have Z 0, but I’m not showing Z 2, the negative sequence impedance. So you may be thinking, well, yeah, how do we handle this if Z 2 isn’t given. Well, as I mention, Z 2 in almost every case is the same as Z 1. And most of the time you won’t be given Z 2, or the negative sequence impedance. So what you do is you just use Z 1 twice. You assume that Z 2 and Z 1 are both the same value.
So looking at this problem, the first thing that I do is I calculate what is known as the base current. We’re going to calculate the 3-phase short circuit current and the line-to-ground short circuit current in terms of per unit. And the per unit is simply a decimal equivalent, or a percentage, of some frame of reference known as the base current.
And so to calculate the base current, you take the MVA base, and someone has to define what the MVA base is. Typically, it’s 100 MVA. But you’ll need to obtain that from your source of data. And you multiply times 1000, that converts MVA to kVA, divide by the square root of 3, and the kV base.
And so for this example, we were given that we have a 100 MVA base, multiplied times 1,000, that’s 100,000 kVA. We divide by the square root of 3. We divide by the voltage that we were provided, 13.8 kV, which was also defined as the base. And our base current is 4,183.7 amps. And that’s going to be our frame of reference. So when we actually have our calculated per unit value, we’ll take that value and we multiply times 4,183.7.
So to calculate the 3-phase short circuit current we take the voltage divided by the positive sequence impedance Z 1, and the voltage that I’m showing is just simply 1. And you might think, where did that come from, the 1. The voltage of 1, that means 1 per unit. And actually, you should be factoring in angles, but I’m just showing just the basic format of how to perform the calculations.
The 1, that assumes that the actual operating voltage is the same as the base voltage. And since we were given a base voltage of 13.8 kV, the assumption is, then we’ll say the system is operating at 13.8 kV. And that’s the normal assumption that you make. And unless you’re given something else, normally you assume that the system is operating at whatever the base, or the nominal voltage is, and therefore the per unit voltage is 1 per unit.
So we take 1 per unit voltage, we divide by the 0.99 per unit positive sequence impedance that we were provided, and that’s 1.010 per unit, 3-phase short circuit current. So we take this value and we multiply times the base current that we calculated 4,183.7 Amps. And our 3-phase short circuit current in this case is 4,225 Amps.
To calculate the line-to-ground short circuit current, it’s not a whole lot more difficult. We have a few more numbers that we need to include, but the process is actually pretty simple. We take the equation that I showed you a moment ago. We take 3 times the voltage. And in this case it’s 3 times 1. And this time we have to divide by Z 1, Z 2 and Z 0, the positive, negative, and 0 sequence impedances.
Well, remember, we were provided Z 1 and Z 0 but we weren’t provided Z 2. So in this case, we just assume that Z 2, the negative sequence impedance, is the same as the positive sequence impedance. So 3 times our 1 per unit voltage, that’s 3.
We divide by 0.99, that’s the positive sequence impedance again, plus 0.99, that’s the negative sequence impedance Z 2, plus 3.12 that we were provided. And, this gives us a per unit line-to-ground short circuit current of 0.588 per unit. And if we take the 0.588 per unit, times our base current 4,183.7 Amps. Our line-to-ground short circuit current is 2,460 Amps.
So obviously, there is more to the theory of symmetrical components than I’m showing here. But as far as just the calculations, the calculations are actually quite simple.
It goes up, it goes down, sometimes it is thought to be infinite (although it really isn’t!) and other times it seems impossible to find. “It” refers to the available short circuit current from the electric utility which is one of the more important pieces of information for an arc flash hazard calculation study. Used to help define the severity of an arc flash hazard, it represents the magnitude of current that could flow from the electric utility during a short circuit. (more…)
Arc Flash Study data collection usually requires a tremendous effort. What is the typical effort for data collection? Jim Phillips conducted a survey at ArcFlashForum.com and asked what percentage of the entire arc flash study time was devoted to data collection.
When you’re performing an arc flash study, the data collection process can just be a tremendous effort. Depending on the age of your facility, depending on the available drawings and documentation, it can take a pretty large part of the overall study effort.
You need to determine things like the protective devices that are out there, because the protective devices are used to define the arc duration. And you need to determine the transformer information, transformer data, because this is going to be used as part of the short circuit part of the arc flash study. Motor contribution as well. All kinds of data is going to be required to perform an arc flash study.
So I started thinking about this one again, about what percentage of the overall study effort goes for just simply collecting the data, trying to determine all the data that’s going to be required. And so I asked this question at the Arc Flash Forum, and the way that I worded the question, I said “on average,” because people will perform many, many studies, especially like consultants. And so I said, on average, what percentage of the arc flash study effort is for data collection, and there were several different answers that people could select.
The first one is less than 20% of the overall effort. 1%. 20% to 40%, 10% of the respondents said that that was their answer. 40% to 60% was the overwhelming response. 66% of the people said 40% to 60%. More than 60% of the total study effort going to data collection, that was 23% of the people.
If you’re just beginning performing your first arc flash study, and you’re thinking, oh my gosh, this data collection is just absolutely overwhelming, well, you’re in pretty good company, because 66% of the people say that 40% to 60% of the overall effort is for data collection.
The Electro Mechanical Protective Relay was once the backbone of higher voltage electrical power distribution systems. Although digital relays are used today, there remains a high percentage of older equipment that still relies on these legacy devices. Terms like amp tap, time dial, instantaneous, IAC, CO, IFC all are part of the legacy or electro-mechanical relays. This video is based on material found in electrical power training classes at: Brainfiller Courses
Electromechanical relays have been around for decades. They’ve been giving way to the digital, or the microprocessor-based relays, but there’s still a tremendous installed base of electromechanical relays. So if this is new to you and you’re at a facility or you have a client that has electromechanical relays, what I want to do is explain what some of the settings are all about.
Looking at this diagram right here, this is actually an electromechanical relay with the cover off of it. And you see that I have an arrow pointing to something known as the Amp Tap and another arrow pointed to basically like a thumb wheel that says Time Dial. And then over on the right side, it says Instantaneous. And what these settings are all about, these settings allow you to adjust how this protective device is going to operate.
So to begin with the Amp Tap– it’s actually a little horizontal bar across the top. And you could see there are numbers on it. And the numbers indicate 4, 6, 8, 10, 12, and 16. And what those are all about is there’s a plug, a screw-in type of a device, and you plug into one of those numbers. For example, maybe the 4, or the 6, the 8, depends on how you want this device to respond. And that defines the magnitude of current at which this relay will begin to operate or should begin to operate. I say should, because once you hit exactly that current, since it’s electromechanical, there may be a little bit of a dead band in there.
Now the Time Dial, what that is all about, that is to adjust how fast the device will operate, the timing, which is why it’s called a Time Dial. So if you actually look down towards the bottom of this relay, there is a circular disk down at the bottom. And so what happens when the current coming into this relay exceeds the Amp Tap setting, that disk begins to rotate. And there’s a vertical shaft connected to the disk.
And you could see another disk, a smaller disk, halfway between the Time Dial and the big disk on the bottom. And that smaller disk will also rotate. And to the right of the picture, you see there’s a small little rod, a vertical rod. And what’ll happen is that’s rotating around. And when it rotates around, it would rotate towards us, and then go over to the left.
And there’s a contact on the left. There’s a T there. And if you look to the right of the T, there’s a horizontal contact. And so that rod will hit the horizontal contact, and then that will initiate the breaker tripping. So it’s a very simple device.
And you can adjust the timing with that Time Dial. Right now the Time Dial is shown adjusted to a 5, and that’s where it defines where that rod is positioned. But if we were to make a setting adjustment to say a 6, or a 7, or an 8 Time Dial, then what’s going to happen is that that post or that rod’s going to be parked further away, and it’ll take longer for it to travel before it hits the contact and sends a trip signal to the breaker. So you’re adjusting the timing just by where the position is of that rod, you do that through the Time Dial.
And then on the right side, the Instantaneous, that’s really nothing more than an electromagnet. Up at the very top, you see a little indicator it says 80 and 60. And it actually goes down to 40 and lower and little bit higher. And what that indicates is the current that it would take for this device to operate instantaneously. It’s like an electromagnet.
And so there’s a hex bolt. You see it up at the top. And that bolt, the further that you crank it down into the coil, the less current it takes to trip instantaneously. The more that you back it out of the coil, the more current it takes to trip instantaneously. So it’s a very simple device.
The problem is for the Instantaneous on an electromechanical device, you can’t really visually set it. So what you need to do is perform testing, bench testing, to see exactly where this needs to be set to achieve your desired goals.
This relay– this is a different picture of the relay. Down at the bottom, there’s this black type of a bar that pulls out. You see there’s some metal fingers towards the back of it. That’s a shorting bar. Because what happens, you have the output of current transformers coming into this, and you can’t open circuit a current transform. It’ll flash. You’ll get a very high voltage. It’s a dangerous condition.
So this shorting bar, what it does when it’s inserted into the relay, it makes contact so that the relay is actually receiving it’s signal from the current transformers. And when you pull this out, there are fingers or contacts in the back that will actually close shut and short out the current transformers. And then when you short out the current transformers, you can actually pull the relay out of it’s enclosure.
This picture shows a close-up of the relay. And you can see I actually have the Amp Tap inserted into the number 4 position right here. So that indicates that when four amps or greater comes into this relay off of a current transformer, then it will begin to respond, or it should begin to respond. Again, there there’s a little bit of a dead band in there.
And the Time Dial is set right around the 5 Time Dial. Looks like the Instantaneous is set at about a 40. But, again, you actually have to test this to know exactly where it should respond.
You can take these relays and actually pull them out of their case. And to pull them out of their case– the shorting bar that I was talking about– you pull the shorting bar out, so that’ll short the current transformers. And once the current transformers are shorted out, then you can safely pull the device out without tripping the circuit. So you could do this without tripping a breaker. And you can perform testing or maintenance without taking a circuit offline.
Arc flash calculations assume the door is open. This arc flash video shows why. Although not an extreme amount of short circuit current for this test, it is enough to create enough blast pressure from the arc flash and blow the door open. Do the doors blow open for every case? No. But it is difficult (impossible?) to determine if it will so to be conservative, it is assumed the doors are open. This video was from arc flash testing by Jim Phillips for IEEE 1584 to determine the behavior of the arc based on the electrode orientation.
One of the questions that I frequently receive about arc flashes– do the doors offer any type of protection from an arc flash? And the short answer is no, but probably the more correct answer is we’re really not sure. It’s just hard to tell. It’s unpredictable. And unless you have equipment that’s designed for the doors to remain closed, like arc resistance switch gear and motor control censors, there’s just no way to know for sure.
So, to be conservative, the way we approach this is that you just assume that the doors are open. And I have a couple of videos that I took in the lab last year that I wanted to show you to illustrate this. The first video is a 400 amp bus plug, and the doors open. And we set this up. It’s at 600 volts with 23,000 amps of short circuit current. And, so, let’s see what happens first with the doors open.
600 volts, test 21. Went inside the room. Under the door.
That was quite an impressive event. I sure wouldn’t want to be standing in front of it. But then we took the same 400 amp bus plug. And this time we had the doors closed, and basically repeated the test. And watch what happens in this case.
OK, let’s go. 600 volts. And the door does not offer protection.
Based on that last comment, I have a pretty keen perception of the obvious. The doors don’t offer protection, at least in that case. And that’s the way that it is quite often.
So, we really just don’t have any way to know for sure whether or not the doors will offer protection or not. So, the safe way to go, and this is the interpretation that’s used in the industry, is you just base this on the doors being opened. They’re not going to offer any kind of protection. 23,000 amps. That’s a pretty respectable amount of short circuit current. And that was enough to blow the doors open.
A substation next to a swimming pool created grounding and electrical safety questions for Jim Phillips as he and his wife were on a road trip between Los Angeles and Phoenix. You assume it is safe but so did the passengers on the Titanic.
A while back, my wife and I were driving from the Phoenix area to Southern California to visit some friends, and on the return trip, we decided to stop over in the Palm Desert area– it’s off Interstate 10, or the 10 as they call it. And we stayed at a hotel– it’s near the interstate, near the exit.
And the next morning when we got up, we’re looking out over the pool. And my wife’s looking at the pool and says, hey, there’s nobody there– we could just go grab a couple cups of coffee and just sit out and relax for a little bit before we hit the road and head on back to the Phoenix area.
So I look out here, and this is from the second story– this is actually a picture that I took. I said, you notice that there aren’t any people out there. And she said, well, yeah, we’ll have the whole place to ourself. And I said, no, no, no– I think there’s a reason that there’s nobody out there.
And so if you look at this picture, I’m pretty sure the reason there was nobody out there is what was behind the pool. Now you have to think that, well, they knew what they were doing, but being an electrical guy, there was just something that wasn’t sitting right with me on this.
And you know, you think, well, OK, they know what they were doing. They covered all the grounding and ground– they got all that right. But in the back of my mind, I’m also thinking, so were the designers of the Titanic– they were confident in what they did, too.
So you just never know what you’re going to see, and this was one of the more interesting pictures that I’ve taken recently.
A 2 second maximum arc flash duration is a reasonable cut off based on IEEE 1584-2002. This is provided there is room for a person to escape during an arc flash. Jim Phillips explains how to use the 2 second cut off time.
The 2002 edition of IEEE 1584 contains language that suggests that you could cut your arc duration or your protective device clearing time off at a value of 2 seconds. And you might think two seconds, that’s a really long time. Well, what could happen, and I’ve shown this in other videos, is that if you have a low enough arcing short circuit current and a high enough protective device setting, your protective device, according to the time-current curve, could take seconds– 10, 20, 30, 40 seconds before it actually operates and clears the arc flash.
Is that realistic, 30, 40, seconds? Well, according to the time-current curve, that’s the value we have, but is it realistic? Well, there’s a couple things to consider.
Number one, if an arc flash were to be able to sustain that long, which I don’t see how it could, because the bus would be consumed, but if were able to sustain that long, would somebody actually stand there for that whole amount of time? No. Your human reaction is you’re going to be jumping backwards trying to get out of the way. That’s just the human response to a threat.
The other part of this is, as I mentioned, could the arc flash actually last 20 or 30 seconds? Well, never say never, but there’s a pretty good chance that the bus is going to end up being consumed. So within the 2002 edition of IEEE 1584 there’s this two second language. A lot of people refer to it as the two second rule. It’s not really a rule, but that’s how it gets referenced. And all that means is that if you have one of these long clearing time situations, that it’s reasonable to cut the current time off at two seconds.
And where the two seconds comes from, it’s based on reaction time. It’s a pretty well-established reaction time. Can people actually react faster than two seconds? Yeah, I would think so. But as far as being able to put your finger on something that’s in a standard that you can actually reference, two seconds is what we have.