Three Phase: How to Calculate Neutral Current in an Imbalanced Load

[Music] hello and welcome to this electrical principals training video in this video we're going to consider researching what happens to the neutral current in a three-phase system now in a previous video we saw that if you have what we call a balanced load on a three-phase system that means the same amount of current in each one of the conductors then there is zero current in the neutral and in fact we saw that you don't even need a neutral conductor in order for that circuit to continue operating because there's literally nothing flowing through it however what we're going to explore in this video is what happens when you've got an imbalanced three phase load now again imbalance means that you have different amounts of current inside each conductor so we're going to set up our test rig with an imbalanced load we're going to measure the currents in l1 l2 l3 have a look at the current in the neutral and then we're going to have a think about how we get from that imbalanced current to our neutral current and how we can calculate it now there's many different ways that you can calculate neutral current the way I'm going to show you first is the way that I prefer because I think it helps you to actually visualize what's happening with the currents as they meet in the neutral and also gives us a really accurate value for our neutral current and what it should be we'll then explore some alternative methods of calculating the neutral current all of which are perfectly valid and suitable and we'll see which one you prefer to use however it's very important that when you get into your exam you actually have a look at the question very carefully and see if it asks you to use a specific method and if it does you must use that method so we'll have a look at a range you can pick the one you like the most we won't feature them all in this video it will be a series of videos so I think we get the rig set up and we do our testing so in order to see what happens to our Neutral current when we've got an imbalance three phase load we've connected up our test rig in the following way over here in our one I've connected up a 1 kilowatt load and then over here we've got two two kilowatt loads and they're connected into l2 now obviously we would never dream of connecting a four kilowatt load into a plug top such as this because it could overload the plug top and lead to fire and overheating and other bad things like that however here we're working on the controlled conditions and the heaters are only gonna be on for a very brief period of time so please don't ever think about connecting up a four kilowatt load on a single plug top it's dangerous and then over here we've got our third load we've got a two kilowatt load and that is connected into l3 so you can see there that we've got three different sized loads connected to l1 l2 and l3 which means that we're going to have different levels of current flowing through l1 l2 and l3 and that's what we mean by an unbalanced three phase load different currents inside those conductors so we'll power the heaters up we'll measure the current in l1 l2 and l3 and then we'll measure the current in the neutral and then try and figure out why the neutral current might not be quite what we're expecting it to be so we've got a load to set up and connected what we're going to do now is we're going to bring the camera in so we can see the readings really clearly as they're being taken from our rig here so remember we've got a 1 kilowatt load a 4 kilowatt load and a 2 kilowatt load and we're going to see what the currents are in l1 l2 l3 and all importantly inside the neutral so since we started making this video monsoon season has started here in Corby and in this tin shed that we were working in it's very loud on the roof so hopefully that won't affect the audio too badly so we've got our rig set up here we've got a 1 kilowatt load connected through l1 we've got a 4 kilowatt load connected through l2 and we've got a 2 kilowatt load connected through l3 we're going to measure the currents in each of the conductors and then we're going to see what the combined return neutral current is so I've got my clamp meter set to measure current so I'm going to power up the system and measure the currents in l1 that's drawing 4.1 amp ere's l2 that is drawing 16.1 amp ere's and in l3 we're drawing 8.1 amp ere's so we can see from that there about the values we would expect based on the power that we're drawing and the voltage that we've got connected up so those are our current values in l1 l2 and l3 now what we'll do is we'll measure the current in the neutral what does your intuition suggests that it might be well let's measure it and have a look so all that returning current flowing back through the neutral comes to 10 point 10.2 amp ere's so we can see there that there's no kind of instant obvious clear relationship between those three currents that are flowing through l1 l2 and l3 and then back through the neutral again so what we're going to do now is we're going to draw out exactly what's happening inside this three-phase system and then we're going to figure out how those three currents flowing to l1 l2 and l3 how they actually relate to the neutral current let's go and do that now so we recorded the three different currents that were flowing through l1 l2 and l3 and I'm following that we took the neutral current also so we'll just make a little note of what those values were in the top corner here so he had l1 we were drawing for point 1 amp ere's l2 we were drawing 16.1 amperes and l3 we were drawing 8.1 amp ere's and then in the neutral one we measured that neutral current we found that we had 10.2 amp ere's now the goal of what we're about to do now is to try and reconcile these values in other words figure out how we can get from these three currents to this current right here so how is it that when these three currents combine in the neutral it comes out at this value of ten point two what's the relationship between those well there's a number of ways that we can figure this out and the way that I'm about to show you is what I think is the best way of doing this a couple of reasons number one it gives you a visual representation of how the currents are behaving and I think that's very important and it's also the most accurate way of using a visual representation to give you the value that you're reading let me explain what I mean by that there are some other mathematical methods which might be a little bit easier to remember and the easier to process and we'll cover that in a future video there are other graphical methods that we can use in order to represent what's going on inside our three-phase system however those graphical methods sometimes a little bit inaccurate because we're drawing them by hand and it can lead to something that cura sees so this method features the best of all those worlds we're going to show you all the methods that I'm aware of on this series of videos in a future video but for now we're going to have a look at this particular one so the first thing we need to do is to draw what the currents look like in this three-phase system so let's do that now now I'm going to draw to scale what's happening with the currents inside this three-phase system now I don't actually need to draw this to scale for this method you can actually just do this with a rough sketch just to kind of help you visualize what's going on with certain of the values however I'm going to draw this one to scale so that you can sort of understand visually what's happening with these values so first of all I'm going to draw a line that represents my l1 current so I'm going to allow one centimeter on my drawing will be equal to 1 ampair so I'm going to draw a line starting here that will be four point one centimeters long so four point one centimeters long and I'm going to draw it pointing straight up like that so I'm gonna draw an arrow there's my l1 value and that line is 4.1 centimeters long which represents 4.1 amp ere's now you remember in a previous video we spoke about the degree of separation between the three phases that feature in a three phase system and you remember that all-important valued of 120 degrees so I'm going to do now is draw my next arrow at a hundred and twenty degrees from the first one and then like that there's a hundred and twenty degrees and then I'm going to represent what's happening with l2 so here we've got 16 point 1 ampair so I'm going to draw a line that is 16 point one centimeters long as I say it's not strictly speaking necessary to draw this method to scale I'm just doing it a because I quite enjoy doing it and B because it helps us to visualize exactly what's happening with this circuit so that's our l2 value and that line is 16 point one centimeters long which represent a 16 point one amperes and then finally I'm going to draw an angle of another 120 degrees from my second line here so there's my 120 degree mark so we'll get that penciled in over here so there's 120 degrees and I'm now going to draw a line that is eight point one centimeters long and if I've done my layout correctly that should fit on here quite nicely so we'll go eight point one centimeters and that represents eight point one amperes so there we go there's our eight point one centimeter long so no three equals beta point one amp ere's so this is actually what's happening with the currents inside our three-phase system we've got l1 drawing for just over 4 amps l2 just over 16 and l3 just over 8 now in order to figure out what's happening with the neutral what we need to do is resolve these arrows now let's put this into a slightly different context imagine that each of these lines represented a pulling force so we've got an object in the middle of this drawing here and it's being pulled in that direction with a force of 4.1 Newtons it's been pulled in this direction with a force of 4 point of 16.1 in this direction with 8.1 Newtons what's going to happen to that object all of those forces are going to kind of resolve themselves and actually what's probably going to happen is that the object will end up moving kind of slightly off in this direction if they were all pushed on it at the same time and actually that's exactly what we're going to figure out is happening because that distance that it would move would actually represent the value of our neutral current in this analogy so we need to do something now and it's called we need to break our current into their vertical and horizontal components now that might sound a little bit baffling at first but actually what I'm going to do next will hopefully help is to understand this a little bit better I'm going to now lay my arrows my three-phase system out onto a what's basically a coordinate grid the Cartesian grid so I'm going to draw my y-axis there and I'm going to draw my x-axis coming along here and draw it along the bottom so I can get it a little bit more accurate there okay so I've created a grid here that illustrates what's going to be happening with our currents so this is our x axis this is our Y axis obviously we know that along here we've got our values so you can see that what we've done here is we have laid out our grid system on top of our three-phase arrows and now what we're going to do is when we talk about breaking these down into their horizontal and vertical components all that actually means is we're going to figure out what the coordinates are for the end points of these arrows so in other words I need to figure out what's the coordinates there what are the coordinates there and what the coordinates there now we could if we wanted to we could simply eyeball this up and we could say well that's about 14 on the x-axis and about minus eight on the Y but actually we're going to do this via calculation and the reason we do it via calculation is it gives us a much more accurate mathematical answer so what I'm going to do now is just lay out a little table in the bottom left-hand corner here so the table is going to consist of the three phases l1 l2 and l3 and then we're going to figure out what are the x and y coordinates of the ends of those three arrows that we've drawn there now l1 is really really easy to find if we look along the x-axis and we're trying to find what the x-value is of that point there you can see that it lines up with 0 on the x axis so therefore we put a 0 under the xl1 however has a vertical value of obviously it's 4.1 isn't it because that line is 4.1 centimeters long representing 4.1 empire's so the Y value is 4.1 so that's nice and easy no matter acquired for that really when we start thinking about finding the x and y values for L to bear in mind that this is the horizontal component and this is the vertical component of l2 it gets just a little bit trickier we've got to start thinking about incorporating some trigonometry into our calculation so what we're going to do first of all is we're going to find out what is the x value of l2 so that's the first thing that we want to do now the way that we do that is we want to think right I want to know what point on the x axis lines up with the end of that arrow and what I can do is I can actually just draw that in as a dashed line there now when I draw this in as a dashed line what you realize certainly is we've got a familiar shape buried inside this image what we've got is a right angled triangle and we know that we can use trigonometry to solve a right-angled triangle as long as we know the length of one side of the triangle and one angle now if you're not sure about your trigonometry then please go and watch the video that I made on the trigonometric identities as it's going to help you to understand this part of the video a little bit better so let's think about this logically let's figure out what we know about this triangle we know that this length here is 16 point one so that length there is 16 point one and because this is the long side of the right angle triangle we know that that is the hypotenuse of this angle of this triangle so the question we now need to know is if we know that that's a right angle then we need to know either this angle or this angle what can we figure it out well surely we can we know that the angle from here round to here is 120 degrees we discovered that in a previous video we also know that this axis and this axis form a 90 degree angle there so if we do 120 degrees takeaway 90 degrees it leaves us with this angle here so now we know that this angle is 120 degrees minus 90 degrees which gives us 30 degrees so that angle there is 30 degrees this symbol here is the Greek letter theta and it's just using us to represent an angle so the angle theta is 30 degrees now because we want to figure out what this point is on the x-axis we want to know what's the length from here over to here so we want to know what this sy of the triangle is going to be how long is that going to be and this side of the triangle will actually be what we call the adjacent because it is adjacent to the angle again if these terms are unfamiliar to you please go and watch the video that I've created on the trigonometric identities because you need to understand that to understand what we do in here so from that previous video on the trigonometric identities we know that the cosine of theta is equal to the adjacent divided by the hypotenuse so that is the way that we remember the relationships between our angles and sides of a triangle and their trigonometric identities so what we need to do here though is we don't need to know the cosine of theta we need to know the length of the adjacent so we need to transpose this formula so at the minute we've got the adjacent and we are dividing it by the hypotenuse the opposite of dividing by the hypotenuse is 2 multiplied by the hypotenuse so we end up with the adjacent being equal to the hypotenuse multiplied by the cosine of theta now if we perform this calculation it will give us the length from there to there so let's put this into our let's part numbers in here so we've got sixteen point one times by the cosine of 30 and as we can see from the calculator on the screen now sixteen point one times the cosine of 30 gives us thirteen point nine four three plus change so we'll just say that that is equal to thirteen point nine four and as you can see actually if you follow that line up you can see that it meets the x axis just behind the 14 mark there so thirteen point nine four is an accurate representation so the x value for l2 in other words the position of that X there the coordinate of that position there is 13 point nine four now what we've got to do is figure out the length of this side of the triangle and that will give us our Y value because if we track this from here across to where it meets the y axis if we do something like then you can see that it meets the y-axis at just past minus eight there so we're now going to figure out what is the Y value of l2 now what we're actually finding here is we're finding this side of the triangle the length from that point down to that point there and that is the opposite side of this triangle and we know from our studies into trigonometry that the sine of theta is equal to the opposite over the hypotenuse so when we change this around that becomes the opposite is equal to the hypotenuse multiplied by the sine of theta so when we put the numbers in we end up with sixteen point one times by the sine of thirty now if you know a little bit about maths you'll know that the sine of 30 is exactly naught point five so when we put this into our calculator as you can see here we do sixteen point one times by the sine of thirty it'll come out at eight point zero five so that comes out to eight point nine five now one really important thing that we need to bear in mind when we now put this Y value into our table here at eight point zero five what we need to bear in mind is that this coordinate here is on the negative part of the y-axis so actually we're at minus eight point zero five because if we were to map that coordinate thirteen point nine four comma 8.05 it would be pointing up in this direction which is obviously no good to us in this instance okay so now we've done that we need to do exactly the same thing again for l3 so we need to figure out what the x and y coordinates are of l3 so again if we want to figure out what the end of this arrow is going to be out on the x-axis it's gonna be somewhere around here so you can already see our answer should be somewhere around minus 7 so we'll see what would come out with so what is the x value of l-3 going to be well again we need to know the angle we need to know what this angle is here fortunately we know that between here and here is 120 degrees between here and here is 90 degrees or 120 takeaway 90 leaves us with theta for this angle of 30 degrees once more so we've got the same angle there which is super helpful it's worth noting at this point that this is a purely resistive load that we've got here because it was just heating loads it's pretty much purely resistive which means that the angles between the currents will be 120 degrees where you start incorporating motors and things like that then it may be that these aren't quite at 120 degrees from each other but that's the subject of much future video right so let's figure out what the x value of this triangle is of the end of this arrow here is so again we go back to our original formula we want to know the length of the adjacent which is adjacent or next to the angle we know that the hypotenuse here is 8.1 units long so that's the hypotenuse there so let's follow that into our mathematical formula so cos theta is equal to the adjacent over the hypotenuse I sometimes get asked the question how come you can remember all of this stuff Jo how'd you remember all these formulas and maths and stuff and the answer is it's because I do it again and again and again and again every time I write out the calculation I always put the formula in do any transposition put the numbers in and then write the answer down and by doing that because you repeat the formulas over and over again they just get lodged in your brain so that would be my top tip if you want to remember these formulae just keep on repeating them and that'll really help you to get them lodged in your brain okay then so rearranging that formula like we did before we get the adjacent is equal to the hypotenuse times by the cosine of theta so therefore we've got in this case 8.1 multiplied by the cosine of 30 and that means that we end up with if we just put that into our calculator eight point 1 times by the cosine of 30 we end up live 7.0 1/4 so again we'll round that off to two decimal places seven point zero one is what we come out with there now bear in mind again I'll put this into the table seven point zero one but look at where that point is it is on the negative part of the x-axis so that is minus seven point zero one and then finally we just need to find the y-value of l3 so once again we're trying to find the distance from the end of the arrow up to here so we want to know the length of that so that is the opposite and we can find that roughly by trucking it across the end of the arrow across to our y-axis so we know that we're aiming for somewhere around - for just over that so again you don't need to do this to scale you can do the calculations without drawing this bit to scale but I always find it's useful to have a sketch of this just to help you see where on the x and y axis you are and it will help you not to forget your positives and negatives which is really really important as we're going to find out so let's put in here now our calculation we want to find the opposite so we're dealing with the sine of theta is equal to the opposite over the hypotenuse some of you will remember sohcahtoa from when you were at school this is where it starts to come in handy we then transpose the formula to find the opposite which is equal to the hypotenuse multiplied by the sine of theta and therefore is equal to 8.1 times by the sine of 30 and again we could actually do it without the calculator this time because we now know that that is just going to be 1/2 of 8.1 because the sine of 30 is not point 5 but let's put it into the calculator anyway 8.1 times the sine of 30 gives us amazingly 4.05 so then we've got four point zero five so that's the y-value of the end of that arrow so we'll put that into our table for point five bearing in mind and this is absolutely critical that that is on the negative part of the y-axis so that is a negative value so in this little table here what we've done is we have broken each one of those currents into their horizontal and vertical components in other words we've found out how far they go along the x-axis and how far they go up or down the y-axis now what we need to do because we're interested in the combined effect of these currents in other words what happens with these currents when they all meet together inside the neutral conductor we need to combine these vertical and horizontal components and all that means is that basically we just add or subtract them depending on whether they've got a negative in front of them so here we've got zero plus thirteen point nine four and we're adding that because there's like a little gauche 2 plus in front of the thirteen point nine four and then we're taking away seven point zero one so if we do that our calculator thirteen point nine four and then we minus seven point zero one we end up with six point nine three so we've got six point nine three and what we found there is the horizontal component of our neutral current six point nine three so that's going to be somewhere here on our drawing we then do the same with the Y values so we have four point one and then from that we're going to subtract eight point zero five and then we're going to subtract four point zero five and that gives us a really nice random of minus eight so actually we could plot our neutral current onto this table now so we've got six point nine three which puts about there on the x axis and if we come down to minus eight which is about there the end of our neutral current arrow will be about there so we could actually plot this out and then we could actually if we wanted to just measure the neutral current however we're going to calculate X we can be much more accurate so this line from here down to the coordinates of that neutral current right there the length of that arrow represents the amount of the neutral current that we've got in this system now as we say we could measure that and figure out what the neutral current will be but what we're going to do is we're going to calculate it first and then we'll just measure it just to see if it's in the ballpark so we'll have a little look at this and we'll think about how we can turn this into a calculation now the good news is that buried inside here we've got our shapes that we live our triangle one sort of learn to save me that guy all he's interested in is triangles and it's true I do talk a lot about triangles during my lessons but that's cuz they are very important in electrical science and it's just a beautiful thing as well so there was a fair show so we've got a triangle buried inside here we've got this triangle here buried inside our graph so what we can look at here is we can think we want to figure out how long that line there is well we know how long this dashed line here is we know that that is eight units long and we know that this line here is six point nine three units long so we know two sides of a right angle triangle and if we know two sides of a right angle triangle we can use Pythagoras theorem to figure out what that's going to be a squared plus B squared equals C squared so what we're going to do is we're going to figure out by putting our numbers in what the length of this side here C squared what the length of C will be so again if you're not sure about Pythagoras theorem please go and watch the video that I've made on that because it will help you to understand what we're doing here now we're not interested necessarily in finding C squared by itself we're interesting in finding C so we need to square root both sides in this formula before we begin so we have C is equal to the square root of a squared plus B squared so here we can say that a will be equal to six point nine three B is equal to 8 and C will be equal to the length of this line here so let's do that now we'll put our numbers in we've got the square root of 6 point 9 3 squared now you'll notice here that I've kind of ignored the fact that this is a negative value down here because actually it doesn't matter the orientation of this triangle we're just interested in the lengths of the sides which is why I'm not worrying about the fact that that is a minus 8 value so we've got 6 point 9 3 squared plus 8 squared and then square rooted so again let's put that into our calculators and you do it in the following way hit the square root button and then we go 6 point 9 3 squared and then we're going to add on to that 8 squared and see what we come out with so we end up with 10 point five eight and what we've actually found there is not just the length C of this triangle what we've actually found out is the calculated neutral current now the sharp-eyed among you will immediately notice that what we've got here is not exactly the same as the ten point two amps that we measured but actually it's very very close we're within 0.38 of an ampere with our calculation reasons for that maybe the measuring equipment maybe the fact that all of the loads were not perfectly resistive maybe this little element of inductance in the heating element inside the unit there might be one or two things that have caused this to happen maybe slightly varying voltages within the system as well so but you can see that the principal is sound we've got very very close to our measured mutual current so you can use this method if you know the loads they're going to be connected to a 3-phase board then you can use this method to calculate what the neutral current will be flowing out of that distribution board down that sub main which is pretty important and quite useful and you can do that long before you do any installation work as part of your design work so let's just kind of prove this point here so if I measure the length of that line you can see it coming out about ten point five which is about what we calculated and it's very very close to the ten point two amps neutral current so as we say this is my preferred method of how to perform this calculation if I was going into an exam this is the method that I would use it's long winded but I can follow each step very closely and I can see what's happening you might be looking at this method and thinking that is a very very complicated way of doing this isn't there an easier way the good news is that is some easy ways and we're going to show you those in future videos however it's quite important that when you're in your exam that you read the question paper very very carefully and if it asks you to use a specific method you must use that method otherwise you will lose points for that question if there is no specific method that is stated then obviously use whichever one you're most comfortable with but it's always a good idea to check with your teacher before going into the exam if there is a preferred method that that examining board prefers for you to use so that's how we calculate neutral current inside an imbalanced three phase load remembering that an imbalanced load is one where all the currents are different and in future videos in this series I'll be showing you other methods of how to calculate that neutral current using the same values that we got on our original experiment so thank you very much for watching [Music] [Music]

Channel: Joe Robinson Training

Views: 81,043

Keywords: Electrical, training, video, electricity, voltage, current, resistance, ohm, ohms, electrical training, electrical training video, EAL, City and Guilds, City, Guilds, C&G, Science, Principles, Science and Principles, level 1, level 2, level 3, level 4, level, maths, calculation, formula, formulae, HNC, BTEC, Engineering, 2365, 2357, 5357, electrician, GCSE, physics, three, phase, 3 phase, neutral, imbalanced, load, megger, clamp, meter, clampmeter, ammeter, sine, wave, A level, A-level

Id: JxmSMhVwyU0

Length: 34min 5sec (2045 seconds)

Published: Wed Nov 20 2019