I think so. I'm not sure what would be the correct terms for these voltages.

So, let me check if my knowledge is correct:
Uline=380V
Uphase=380/sqrt 3V.

Uline => 0°, - 120° and 120°
Uphase => +-30 for each phase.

Is this correct?

The 3 phase supply is composed of three, one phase voltages, shifted by 120 degrees relative to each other as shown in this LTS plot.

I am not sure what you mean by this.. Uphase => +-30 for each phase.

Use LTS to check your calculated answers.

E

EDIT:
Error in my sim , will correct and repost!!

 

I mean:
380|0ºV => +/-30º => 220|30ºV
380|-120ºV => +/-30º => 220|-90ºV
380|120ºV =>+/-30º => 220|150ºV

We get this by drawing the Voltages and Currents diagram in the YX referencial, knowing that Vab = Va + (-Vb) and so on for all phases!

 

hi Psy,

Look at this corrected image.

The 380Vac is the RMS voltage between phases, so the individual phase Line voltage is 220Vac RMS.

The Vpeak of each Line Phase relative to neutral is therefore 311Vac, which makes the Vpeak between phases ~ 538Vac

E

 

This is how we get those angles of 30º, -90º and 150º.

But can you tell me if my calcs are ok so far??? Can I keep going?

 

Sir Eric, do you think I can keep going???


I'm keep doing math in the other thread!!!

#18 - updated!

 

Hi,
Your individual current path answers look close enough for me.

You have only been asked for the A1 and A2 currents, so show these two values.

Look at these sims and compare your answers.

E

 

I'm not sure if Psy has encountered the Delta-Star (or Delta-Wye) transformation. It can be useful in situations such as this. In particular, where the Delta load is balanced in each line-to-line leg, as in this case. I use the term Delta as meaning the same as "triangle" in Psy's nomenclature.

If the (balanced) line-to-line delta impedance is the general value Z, which may or may not be complex, then the equivalent star line to common star point impedance is simply Z/3. This allows one to then more easily resolve the line currents to the load, since an intermediate phasor subtraction of the delta branch currents is not necessary. One has only to remember that the voltage applied to each transformed Z/3 term is the complex line-to-neutral voltage rather than the complex line-to-line voltage.

 

Ok, Sir Eric, my last calcs are done! I don't know if the angles are correct! According to your sims, Ia should be around 12A, correct??? But I'm getting ~8A.

Post #18 updated!

 

hi,
Where do you see 12A on my sim images.??
E

 

The chief problem with Psy's solution is that he has applied the wrong voltages to the motor delta branch loads. Unless of course he is actually solving for the motor connected in star configuration.

 

hi,
Where do you see 12A on my sim images.??
E

I have divided the RMS value by sqrt 3 of you 3rd picture and I got ~12A

But I can't see either any value of ~8.5A which is my calcs results!

The chief problem with Psy's solution is that he has applied the wrong voltages to the motor delta branch loads. Unless of course he is actually solving for the motor connected in star configuration.

I probably right.

I've calculated Ica, Ibc and Iab using 220|0º, -120º, 120º when, I think, I should have used 380|30º, -90º, 150º...

I'm going to recalculate and update post #18!


Edited;
Ok, #18 post is now updated!

My 3 total currents are now about 19A! You think it's correct now????

 

For some reason, the current at the inductors, by my calcs, which are the phase currents for the Delta setup, are 1A short and I can't figure out why...

Check my plot and circuit.

I've added a GND reference at the "Y" setup...

 

hi Psy,

I have run your sim, would you please post a summary, on a new post showing your maths for the Delta circuit, the Post #18 layout is confusing the heck out me trying to find your edits.

E

 

Psy,

Unfortunately you have made a mistake somewhere.

The line current RMS values are 21.783 Ampere - rather than 19 Ampere.

I'll check your calcs to hopefully find the error.

 

A particularly confusing matter in the solution is the annotation conventions adopted.

It is misleading to write

Uc = 380V

380/sqrt 3 = ~220V

Uac = 220|0º
Ucb = 220|-120º
Uba = 220|120º

I'll consider the following:

Ua = 380|30º
Ub = 380|-90º
Uc = 380|150º

Uac for instance is the normally the convention adopted to indicate the voltage between lines a and c - Not the "a" phase line-to-neutral voltage!

I would very much prefer

|U|line-to-line = 380V

|U|line-to-neutral = 380/sqrt 3 = ~220V

Ua = 220|0º
Ub = 220|-120º
Uc = 220|120º

I'll consider the following:

Uab = 380|30º
Ubc = 380|-90º
Uca = 380|150º

If you carefully apply these conventions you will hopefully arrive at the correct solution.

 

I'm re-writing all over again on paper...

After it I'll take a picture and upload it here!

At our classes we use Uac to Line-to-Line voltages and Uan to phase voltages depending if we are at Delta or Y setup!

 

Here it is!

 

To return to an earlier theme on the checking of results...

Time permitting a useful check is to ensure real power values match up consistently.

For instance I calculate the B phase line current to be

Ib=21.783|179.16° Amps

This comprises the motor line current

Ibm=19.964|167.66° Amps

and the lamp load current

Ibl=4.558|-120° Amps

The B phase motor current is |Ibm|/√3=19.964/√3=11.526 Amps

This would give a B phase motor loss of (11.526)^2*10=1328.54 Watts

The B phase lamp loss is 1000 Watts giving a total B phase loss of 2328.54 Watts.

The B phase loss should also be given by Pb = |V|line-to-neutral*|Ib|*cos(θ)

where θ=179.16+120=299.16°

So Pb = |V|line-to-neutral*|Ib|*cos(θ)=2328.59 Watts which is in good agreement with the sum of the individually derived terms.

 

To return to an earlier theme on the checking of results...

Time permitting a useful check is to ensure real power values match up consistently.

For instance I calculate the B phase line current to be

Ib=21.783|179.16° Amps

This comprises the motor line current

Ibm=19.964|167.66° Amps

and the lamp load current

Ibl=4.558|-120° Amps

The B phase motor current is |Ibm|/√3=19.964/√3=11.526 Amps

This would give a B phase motor loss of (11.526)^2*10=1328.54 Watts

The B phase lamp loss is 1000 Watts giving a total B phase loss of 2328.54 Watts.

The B phase loss should also be given by Pb = |V|line-to-neutral*|Ib|*cos(θ)

where θ=179.16+120=299.16°

So Pb = |V|line-to-neutral*|Ib|*cos(θ)=2328.59 Watts which is in good agreement with the sum of the individually derived terms.

I think that my calcs are now consistent with yours but I still don't now why our phase currents at the motor (Delta setup) are 11.52A and LTSpice shows around 12.5A. It's still 1A away!

 

I think that my calcs are now consistent with yours but I still don't now why our phase currents at the motor (Delta setup) are 11.52A and LTSpice shows around 12.5A. It's still 1A away!

The SIM I set up worked fine. See attached result.