Lost Neutral Voltage Calculation

[Deleted member 105166]

I understand how to calculate the current in the neutral conductor on an unbalanced three phase installation, however am struggling to grasp how to calculate the voltage movement on each of the three lines if the neutral were open circuit. I appreciate it is influenced by the current and resistance on each of L1, L2 & L3 and that the result may be between 0V and U (400V in this instance). This may be where I'm going wrong - I thought I was getting close to a good formula replicating an actual example I've seen (179V, 398V, ~233V) with the sum of line voltages equal 2U*weighting % sum of resistances, however whilst this comes fairly close with certain theoretical loads, an individual line can calculate >400V... so something's adrift.

 

[Lucien Nunes]

If you consider all the loads involving the neutral to be single-phase loads, each phase consists of a voltage source (supply) and a complex impedance (load) in series. The three phases are then in parallel between two nodes; the star point of the supply and the disconnected common neutral terminal. You can reduce the parallel network to a single voltage source and impedance in series using normal AC circuit analytical methods. Note that the three current phasors (and the three voltage phasors across the load impedances) may not lie at mutual angles of 120° because the load on each phase may have an arbitrary value of phi.

The initial assumption that true 3-phase loads such as motors can be ignored holds fairly true, since most such loads deliberately have no connection to the neutral to avoid internal transformer action opposing supply voltage asymmetry and setting up circulating currents, and to trap zero sequence currents. For example, a star-connected motor does not have its star point connected to neutral. If it did, it would actually regenerate a missing neutral leading to a much lower L-N voltage deviation, until its overloads tripped due to the reflected load on the line conductors.

 

[marconi]

A useful way to calculate the broken neutral to star potential difference is to use Millman's Theorem. One has to have some familiarity with complex numbers (eg: x + jy )to represent the phase relationships of the line voltage sources and the impedances where j is the square root of -1. In your problem if the neutral is broken then RL is connected in series with an infinite resistance. Here is a nice worked example:

 

[marconi]

Re #1: If your phase voltages are 179@0deg, 398@120deg and 233@240deg, and assuming impedances(Zbrown, Zblack, Zgrey) are equal (Z), then Vn is:

{179@0/Z + 398@120/Z + 233@240/Z} all divided by {1/Z + 1/Z + 1/Z} =

multiply top and bottom by Z =

{179@0 + 398@120 + 233@240} divided by 3 =

{179 + 398[sin 120 +jcos120] + 233[sin240 +jcos240]}/3 =

179 + 398[0.866-j0.5] + 233[-0.866-j0.5]}/3 =

collecting together real and imaginary(j) terms =

{179 + 345 - 202 +j(-198 - 117)}/3 =

{322-j315}/3 = 107 -j106V

Or in polar form is square root of { [107 x 107] + [106 x 106] } all divided by 3 = 150/3 = 50V

It is only slightly more complicated (not more difficult ) if you put in the actual impedances which may not be equal).

Anyway, this gives you the gist of how to calculate it.

Here is a problem for you:

Vphase brown = V black = V grey = 380V in star

Z brown = 1 + j 2 Ohm Z black = 2 + j 4 Ohm Z grey = 3 + j 3 Ohm connected in star.

What is the star to star voltage?

You need to use 1/(x +jy) = (x - jy) divided by (x squared + y squared)

or 1/(x -jy) = (x + jy) divided by (xsquared + ysquared).

Enjoy.

 

[telectrix]

8 am. and my brain hurts.

 

[Pete999]

8 am. and my brain hurts.

My brain was painful at 06:30 when I woke up OK now though lots of sweet Coffee out of that thing with a plunger. Can't spell it so work it out yourself.

 

[marconi]

Nespresso machine with CAFEPOD supercharger pod strength 12 (on a scale 1-10).

CAFEPOD do one called Livewire which methinks would better energise telectrix and pete999 but it could cause all sorts of potential problems for them:

Google Shopping - Product not found - http://www.google.co.uk/shopping/product/5534343005766154434?lsf=seller:8225840,store:3786080414066669863,lsfqd:0&prds=oid:5932860779901052260&q=coffee&hl=en&ei=6nKpXZ_KFYjbwQKx4Y7gCA&lsft=gclid:CjwKCAjwxaXtBRBbEiwAPqPxcPRoywKb_Ok_p03uqJC1wetYh4ESSrNACGAO2wO9VZdaYm4Zy9Q3DBoC_58QAvD_BwE,gclsrc:aw.ds

 

[Deleted member 105166]

Thanks @marconi for those formulae, will spend some time on that after work

 

[Zerax]

I'm saving my analysis of those formulas until the weekend... give me something to look forward to !

 

[Carl S]

The old way is to draw out the vectors.

There’s probably an “App” on your phone by now.

 

[Lucien Nunes]

In real situations there is another complication, which is that some types of load are significantly non-linear and cannot be represented by an impedance, at least not over the entire probable L-N voltage range. SMPSUs, for example, run at nearly constant power, giving them a negative dynamic impedance, but only within certain bounds of voltage. If the voltage falls too far, they shut down but might attempt to restart once the voltage rises which it probably will, resulting in low frequency oscillations etc. Such behaviours are so dependent on hidden internal parameters of the loads that they are difficult to model in a general case.

Other loads interact - a device with significant neutral current running on L1 might only operate if it can see the network via a router elsewhere running on L2. When L2 voltage glitches perhaps due to a changing load on L3, the network restarts, the device halts, reconnects and restarts putting load back on L1 maybe a minute later. Other loads will go pop or trip, leaving that phase light and high in voltage, a runaway situation.

When you push devices with complex behavours out of their comfort zone, all bets are off.

 

[Dan]

Done some housekeeping in this thread. If you could PM each other about coffee instead of talking about it randomly half way down a thread that'd be cool. Feel free to make your own thread about coffee at any time

This thread: Discuss Lost Neutral Voltage Calculation in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

 

[DPG]

Done some housekeeping in this thread. If you could PM each other about coffee instead of talking about it randomly half way down a thread that'd be cool. Feel free to make your own thread about coffee at any time

This thread: Discuss Lost Neutral Voltage Calculation in the Electrical Wiring, Theories and Regulations area at ElectriciansForums.net

Bet you're a tea drinker.

 

[Dan]

Bet you're a tea drinker.

- Filter coffee or go home.

 

[marconi]

I did the problem I set in #4. It is indeed rather drawn out with the complex impedances I set, and I reckon too much for a practice run. Unless I have made a mistake the figure for the star-star voltage I worked out is 65V @ -21.4 degrees which seems right when you plot consider how the impedances will 'pull' the load star point in their respective voltage phasor directions.

I suggest instead anyone who wants to have a go just use the real (resistive) part only of the impedances I presented viz:

Z brown = 1 Ohm Z black = 2 Ohm Z grey = 3 Ohm connected in star.

Unless that is you like doing complex number algebra. I covered three sides of A4.