How to Find Neutral Current of Unbalanced Transformer

Use the clamp meter one more time. Your 3/4 of the way to knowing what all of your numbers are.

Reason being every phase may have a mix of resistive and inductive loads that may be fed from either line to line or line to neutral connections so the actual current on the neutral line may very well be different than the calculated numbers as referenced from the individual lines.

I found with the help of clamp meter that current on the Neutral phase was "49Amps"

but I want to double check it mathematically

Does no one posses the simple ability to perform a cursory internet search any more??

This is not a tutoring service, we are not your parents, and you should not be involved in anything you OBVIOUSLY DO NOT UNDERSTAND!!!!!!!!!!!!!!!!!!!!!

How to calculate the unbalanced neutral current in a 3 phase circuit

DO YOU NOT TRUST YOURSELF TO USE A CLAMP ON METER????

Proper use of clamp meters in commercial and residential settings

For your kind information, instead of being sarcastic if you could come up with a genuine answer and be able to prove mathematically the value which was calculated with the help of clamp meter would have shown your maturity to a far level.

First of all, I am here to learn. After searching a lot on the internet and not being able to find my answer, I posted my question here thinking that I would get some help but I didn't know that people like you existed who just know how to criticize instead of coming up with an answer that could help people understand.

Secondly, I trust myself on calculating the value with the help of clamp meter but unlike you I want to also find a way of calculating the neutral current mathematically.

A piece of advice, instead of your sarcasm that just went to a waste because I have already visited the links that you gave which weren't able to prove the neutral current calculated with the help of clamp meter, better give my answer if you have the guts!

At least your name is well chosen.

Please anyone? All I want is to find the answer that I am looking for, that I searched a lot in the books but didn't find, that I searched for on the internet as well but haven't been able to find the solution yet.

Characteristics of the Neutral Conductor | Code Basics content from ...

Ask your instructor, if all else fails.

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Lyn's first link gave the formula

Didn't you say you read it?

I put the values of Red:98Amps, Yellow:75Amps, Blue:115 Amps in the above formula

but it didnt gave the neutral current equal to what I had observed while calculating on the clamp meter(that was 49 Amps)

Then please reread post #1.

Unless you can measure the phase angles between the 3 currents you listed, you cannot simply calculate the neutral current value. That is what TCMTech was saying in Reply #1. It's not just an average of the 3 values, because capacitive and inductive components impact the sum of the currents. "Trust the force [your meter], Luke." That's all you have to go on at this point.

Thank you so much. At least I have an idea now that what I am supposed to do. With your help, my next approach would be geting a power analyzer to obtain the phase angles and then doing the calculations. I really appreciate your positive reply and help. Thank you

Most welcome.

I am much pleased to know that my spending time on forum shall be able to solve the problem.

Best regards,

Ramesh

1. Your original post bears an uncanny resemblance to a homework/test question, not something from real life.

2. I can't help it if you are incapable of understanding the information as presented. I believe that there is a "genuine" answer in the response I provided. I find it difficult to beieve that you did any real research, or you could have found (About 22,200 results (0.21 seconds) on the subject.

3. You gave no indication that you had done any research AT ALL!

4. Did you read the FAQ's?

5. I can't help it if your feelings are hurt. If you can't take critisim you are in the wrong place.

I'll leave you to be taught be some more gentle souls..................................

1. What I did was, I climbed up an electricity pole and took the readings of distribution transformer i.e 100KVA Delta-Wye connected. I took current ratings of all the phases(including the neutral as well). The readings were

Red:98Amps, Yellow:75Amps, Blue:115Amps, Neutral:49Amps

2. I then started calculating the neutral current mathematically by that vector solution but the neutral current's value came out to be very large than what actually I had observed with the help of clamp meter. That increased my curiosity to know why the calculations I am making on paper are coming out to be different from the practical life that I observed with the help of clamp meter.

3. I didn't write that its a research because I already read it that homework questions aren't allowed expecting that everyone posting/questioning anything on this forum would be related to practical world considering rest of everyone else to think of me as one like you too who is here to ask and increase his knowledge. Besides it, I also didn't write that I had researched on all this because of the reason that I felt like to be to the point instead of making my posts long

4. I read the FAQs

5. I am ready to take criticisms only on one condition that I am helped and made to learn.

Only answering my question would clear up my concepts.

I btw had took amps readings of another DT as well which were ;

Red:26 Amps, Yellow:90 Amps, Blue:80Amps, Neutral:77 Amps

I tried calculating the neutral current through vector solution of this transformer as well but even for this transformer the calculated neutral current came out to be quite different (not even close to) the one calcuated by clamp meter.

Please help, I really need your help thats why I thought of asking here

With purely resistive load the calculated answer should have been around 50% of you're actual reading.

I'm not going in to harmonics at this time of night.

I don't trust calculated neutral currents at all. Measure them.

Next time you climb the pole take a length of rope with you.

Not all of us has reached the level of your holier than thou self estane, thank goodness! If it offends you to read these questions , then don't read them. And if you can only answer with insults and down grading the person who asked a question that you believe below this group, you should with draw so that you are not insulted, but even better you are not using your super intelligence to insult those that dare ask a question so inferior to your great knowledge!

Thank you for the support

The formula give works only if the pf of the loads are the some. This link give a good review. How close was the answer given by the formula?

http://www.google.com/url?web&cd=9&cad=rja&ved=0CHQQFjAI&url=http%3A%2F%2Fopencourses.emu.edu.tr%2Fmod%2Fresource%2Fview.php%3Fid%3D147%26redirect%3D1&ei=dqBIUrTfKoXU8wTD0oHADg&usg=AFQjCNHBc4lJNGqPIQGS2ref6NQumKUcwA&bvm=bv.53217764,d.eWU

The above link does not work. open this Microsoft Office file when requested. Pretty basic stuff.

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&frm=1&source=web&cd=9&cad=rja&ved=0CHQQFjAI&url=http%3A%2F%2Fopencourses.emu.edu.tr%2Fmod%2Fresource%2Fview.php%3Fid%3D147%26redirect%3D1&ei=8qhIUoKIFIrU9QTQnoHwDg&usg=AFQjCNHBc4lJNGqPIQGS2ref6NQumKUcwA&bvm=bv.53217764,d.eWU

Thank you

Darwin award! I'm sure he's the winner for this years award!! GA by the way!

The neutral current can never be higher than the largest of the three phase currents.

Why is the neutral current of any great interest?

What is this all in aid of - conductor sizing, perhaps? If so, then British Standard 7671 gives full sizing protocols for cables and their protective devices and no-one other than a trained Electrician needs to climb any poles because of it. And a trained Electrician wouldn't need to ask about conductor sizing or the value of the neutral current because that individual will have a full grasp of the contents of BS7671 in order to qualify and maintain his/her competence as an Electrician..

Therefore, this question tends to look like homework, hence the above responses.

netfreak, u said:

"I tried calculating the neutral current through vector solution of this transformer as well but even for this transformer the calculated neutral current came out to be quite different (not even close to) the one calculated by clamp meter."

show us your vector calculations please.

If Red[A] voltage phase constant will be 0 then:

Io=A*[cos(fia)+j*sin(fia)]+B*[(cos(fib-2*pi()/3)+j*sin(fib-2*pi()/3)]+C*[(cos(fic-4*pi()/3)+j*sin(fic-4*pi()/3)]

Io=Ni complex ; fia=acos(pfA) ; fib=acos(pfB) ; fic=acos(pfC)

It would appear that the original poster doesn't have sufficient information to enter into that equation, nor any inkling as to how to obtain that information.

The current in each of the three phases does not give you enough information to calculate the neutral current. As Tmctech has said, the phase angle between the current and voltage is unknown due to non-resistive loads. So you cannot assume the phase currents are 120 degrees apart. Another factor is nonlinear loads such as fluorescent lights which generate 3rd harmonic distortion in the current waveform. This will make the neutral current higher than a strict vector sum of the three phase currents.

Thank you so much for your positive reply. I highly appreciate it for a discussion to finally get started.

In your response;

I do not know the phase angles of currents, thats why I'd been assuming them to be 120 degrees apart(because I didn't had any other choice to go for)

Secondly, it is a Distribution Transformer supplying electricity to a lot of consumers & God knows how much load would each consumer be consuming(because I didn't check that)

How am I going to calculate Neutral current mathematically or by implying Engineering knowledge in this case? If the phase angles of currents are only needed, would it mean that after knowing that its 100% confirm that I would be able to find the real value(which I observed with the help of clamp meter on neutral phase as well)?

New degree path: Bachelor of Science in Implied Engineering, Nibiru University College

It's all lost on the OP, at any rate, unfortunately.

That about sums (sic) it up.

Thank you Winfield Hill. Thank you so much for your reply. I really needed an honest and to the point answer which you, xyz and a few other members gave.

I am going to go for this approach now to find the phase angles too. Thank you once again for the help Mr. Winfield Hill

These are a lot cheaper than the Rogowski coils. £20 sterling.

....and at least they have a CAT III 600 volt safety rating. You have not written a value, but 415/240V seems likely. You need CAT IV, you are at the origin of an installation.

They give a millivolt output (clamp meters without the built-in meter).

If you used 4 (3 line, one neutral) together; you could get the magnitudes and relative phases of all four vectors.

You join the commons together. Then measure all the voltages to common (4) and hot to hot.

As an example, if you had two voltages (currents), that makes 3 terminals, one common. If you measure all the three voltages between the terminals, then you have all three sides of the vector triangle. With compass and ruler you can now draw the vector triangle on paper with all the angles.

Remember that linking up instruments on all 3 phases increases the danger. Make a proper switch which enables you to measure all voltages quickly, this also reduces the load fluctuation error. Get an assistant (safely remote) to write down all the values, as you read them.

I recommend lowest available current rating 1000 volt high break capacity voltmeter fuses (e.g. SIBA make, DMI type) in all the leads from the CTs, close to CT.

Remember this is all at your own risk....wood ladder, rubber boots, rubber gloves etc certified for electrical use etc

67 model

Algebra can be a headache...this is the paper solution similar to the second calculation link in RAMconsult post #46, but power factors are not zero

OK, 98 should be at the bottom and 115 at the top!

Note that if the voltages are symmetrical [equal magnitudes at 120 degree spacing] then if you measure one angle between currents and one current/voltage angle, then you have a solution. Reference post #33, angles can be measured with two clamp CTs.

67model

Homework or not, regardless of phase angles, power factors,vectors and harmonics, I am puzzled how the neutral current can be well in excess of the the total current of all three phases.
Perhaps it is because you have only one clamp meter and you measure the current in each wire over a period of time during widely fluctuating loads.
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And if any one from CR4 is watching, I am still not able to post comments without enormous copy/cut/paste hassle to get round the error message.
"Could not process form: The system was unable to process this form".

quote Horace "Homework or not, regardless of phase angles, power factors, vectors and harmonics, I am puzzled how the neutral current can be well in excess of the total current of all three phases."

OP said Red:98Amps, Yellow:75Amps, Blue:115Amps, Neutral:49Amps

Sorry. A senior moment. I misread the current as 494 amps, and not 49Amps as quoted.

I imagined the neutral current to be measured when the load was 494 amps at a different time to measuring the phase currents when the load was a lot less.

But even at 49 amps, my point about a varying load has some merit.

Look, I'm all for educational experimentation....in a lab, under trained supervision, with proper instrumentation, wearing PPE (Personal Protective Equipment)...but climbing up a pole, hanging clamp-on ammeters, exposing conductors to get voltages...

Are we trying to help OP putting himself at risk of this as a result?? Remember where there's a transformer, there's a primary somewhere up there too.

Here is a simple way of computing Neutral current.

You will notice from above that in an ideal case when amplitude all 3 phase voltages are same V_G Cos (120-90) = - V_B Cos (120-90), and so also

V_G Sin (120-90) + V_B Sin (120-90) = - V_R and hence there voltages are well balanced.

Now assume that currents measured by you are in phase with respective voltages to begin with. Instead of voltage, use current values and compute I_G Cos (120-90) + I_B Cos (120-90) = I_N Cos (120-90) . Similarly compute I_G Sin (120-90) + I_B Sin (120-90) + I_R = I_N Sin (120-90).

Then I_N =√ {(I_N Cos (120-90))² + (I_N Sin (120-90))²}

.

xyz,

Sure, no problem, but the above solution is only valid for a balanced system of voltages/currents/loads, which is clearly not the case in the original posting. Any of the above solutions that start with the words, "...Assume that the voltages/currents are equally spaced at 120° apart...or the voltages are equal in magnitude...or the system is balanced...or the power factors of all the loads are the same...etc." are only approximations that will not yield the right answer.

There's just no way to solve any problem with 3 vectors (3 magnitudes and 3 angles) directly without using complex math and the i or j operator to get the correct answer (iterative solutions aside). There are some approximations that will be within 10-20% of the correct answer though, so if it's "good enough for government work" then go for it.

For those who want a real challenge, try back calculating the angles using the 4 measurements that OP got (hint: it's an infinite set of angles but you could try bounding the problem by limiting the range of angles for each vector and iterating through the range).

You say"There are some approximations that will be within 10-20% of the correct answer though, so if it's "good enough for government work" then go for it."

I am interested in that methodology. Pl share your reference. Pl see my response at 22 above- where I have clearly stated that it is not straight forward. There are 3 more unknowns- angle of current with respect to voltages. I am aware of it, but was trying to give a simpler guidance / approach.

Let me have your approach / reference software / iterative technique.

xyz,

Thanks for confirming what I have stated in two previous posts; there is not enough information to correctly calculate the neutral current. Any post that starts with a balanced system only adds to the confusion evident in many of the previous threads because it is only valid for a balanced system, so the answer can't be relied on unless verified independently. It may however, provide the basis for the first approximation that could be used in a Newton-Raphson iterative solution, the details of which I leave to an interested reader with time on his/her hands.

For those readers who are wondering why I keep hammering away at the "balanced/same" assumption, I'd like to remind them that in a balanced system there is no neutral current, so any "calculation" that starts with a balanced system and somehow magically conjures up a result that is only valid for an unbalanced system, is flawed from the start.

Here are a few attempts at approximating the neutral current without enough information, there are more on the web:

http://www.ehow.com/how_7871256_calculate-neutral-currents.html

http://www.electrician2.com/electa1/electa4htm.html

Please note that neither poster (nor any in this thread) attempts to verify their method with real data. Perhaps you could compare yours with theirs and report back the results, including a comparison of the answers from all the methods.

Since there are ways to get the answer with proper instrumentation, I cannot condone the use of any unsubstantiated approximations in professional work.

Alan

I like the XYZ input because it clearly lays the basis for calculating the currents, and therefore is a sound stepping stone to more complex answers - as posed by RAM and others.

It takes me back 50 years when an end-of-course question on an asymmetric 3-wire star system with unbalanced loads asked for the star point voltage and phase angle with respect to ground/neutral. The parting words of the lecturer being "...you've learnt enough on this course to solve the problem unaided..."

We were given the line volts, frequency, Henry's, Farad's and Ohms for each of the resistors, chokes and condensers (remember them?) in each phase. From which we had to calculate the currents (using trig tables, log tables and a slide-rule).

I must admit that I had trouble with the maths alone, and had to resort to a mix of maths and drawing vector diagrams to scale on graph paper using a ruler, protractor and compass.

Back to this OP, would the voltage of the 'floating' star point account for the neutral current if grounded.

Let us try a very solving for a unbalanced load condition. A neutral voltage develops due to imbalance in load conditions. Here we are considering a specific case of a grid line where enough filters have been incorporated to ensure there are no harmonics and only fundamental components exits.

Following simple trigonometric equation can be used to solve the problem.

Now following trigonometric equation can be considered.

V_rr² = V_r² + V_n² - 2* V_r * V_n * cos (a)

V_gg² = V_g² + V_n² - 2* V_g * V_n* cos (120-a)

But . V_r = V_b = V_g for ideal vectors. Considering the above two equations and may be some iteration, one can solve for a. Thus we have two unknowns and two equations and some iteration may be needed to solve for V_r and a.

I am not going to write down all equations and steps here, but suffice it to say that using above equations in different ways q_rb q_rg, q_gb can be determined.

Actual values of V_rr, V_gg , V_bb , V_n can be measured. Similarly I_rr I_gg I_bb are measured and power factor of each phase can be measured. I_n can thus be computed using trigonometric steps given here.. Entire paper is too long to be developed and presented here.

There are no assumption and it works for any unbalanced system also- but steps are complex, undoubtedly- but there are no pitfalls. But it can be computed using excel sheets and goal seek feature.

.

xyz, Nice try, now it's time to read a scholarly paper on the subject.

http://www.labplan.ufsc.br/congressos/powertech07/papers/527.pdf

The math is heavy, but a quick look at the graphs and charts gives an appreciation of the nature of the problem; mainly that the neutral current is non-sinusoidal and does not vary consistently over time, this fact alone negates any traditional graphical or algebraic methodology for solving for the current.

Couple that with a meter that is calibrated to read RMS and then take into account its sampling window, yields the result that the measured values will vary widely from sample to sample; i.e., only a correctly instrumented system with zero-crossing triggering will be able to yield a consistent value.

Anyway it's been great fun, and I thoroughly enjoyed the sleight of hand of reducing a system with four unknown variables into a set of two algebraic equations. Neither my coworkers, mentors nor my students would let me get away with that one!

Thank you for the reference to the article.As instantaneous values are fed into math suggested by my approach - it will give instantaneous results-and hence no need to assume that it applies only for RMS components considered over a full cycle. The vector balancing is applicable for instantaneous values - even at microsecond level. I suppose the originator of discussion did not realize that with just three R, Y and B currents he could compute neutral currents. Even equations given under response 23 look too simplistic.