Hi, Ive been given a problem whereby a 415V delta source is connected to a delta balanced load with 21+j18Ω impedance per phase. There is also a line impedance of 1+j2Ω per phase. Im not sure whether I have solved this correctly but I have added my solution to this post so if someone could check the answers and give me some guidance as to where I may have gone wrong then I would be very grateful? I have converted the system to star before trying to solve as this is what we have been taught

 

Are you sure you want all three of your generators to be in phase?

 

Im not sure what you mean by this? If you are referring to the fact that I have only provided the answers to the 'A' Phase currents this is because I just want to make sure that my method is correct as we are taught to convert to star, solve for the 'a' phase, then convert back to delta before solving for the 'b' and 'c' phases. Are these results correct for the 'a' phase?

 

No, I am referring to the fact that in your diagram you have three voltage sources hooked up in series back on each other and they are all in phase. That's generally referred to as a dead short.

 

which bit of the diagram is wrong as this is the way that my professor has taught me to draw a delta - delta diagram and he says this is correct?

 

In the diagram that is attached, I have separated the upper node.

As draws, what is Vab, the voltage at A relative to the voltage at D?

What happens when you tie those two nodes together?

 

Hi there,
you are correct with your assumtions converting the delta to star in this case as the system has same resistance R at its three sides so the equivalent star resistance r will be R/3 like you have done.
Your diagram is correct, and having a quick glance at your calculations you seem to be too.
however when converting IL to delta you have a -30° phase angle, i cannot remember why this is.

EDIT:

I have just remembered, this is you have converted the system to a delta-star system so the transformer banks will be out of phase by 30° meaning 30° lagging.

This therefore means when you converted the system to delta - star your VP will actually be 415/√3 < -30° and you do not need to add in the random -30 when multiplying by /√3

 

Hi there,
you are correct with your assumtions converting the delta to star in this case as the system has same resistance R at its three sides so the equivalent star resistance r will be R/3 like you have done.
Your diagram is correct, and having a quick glance at your calculations you seem to be too.
however when converting IL to delta you have a -30° phase angle, i cannot remember why this is.

EDIT:

I have just remembered, this is you have converted the system to a delta-star system so the transformer banks will be out of phase by 30° meaning 30° lagging.

This therefore means when you converted the system to delta - star your VP will actually be 415/√3 < -30° and you do not need to add in the random -30 when multiplying by /√3

Thats great thanks. My main problem is when it comes back to converting the currents into delta. When you have found the currents in star formation (Il = Ip) and you need to convert back to delta, do you assume delta phase current is the same as star and therefore multiply by sqrt(3) and subtract 30° to find the delta line voltage or do you say that delta line current is the same as star currents and therefore divide by sqrt(3) and add 30° to find delta phase current?

 

Your diagram is correct, and having a quick glance at your calculations you seem to be too.

If the diagram is correct, then please explain what happens when point D is connected to point A, given that the voltage between them is a 1245V sinusoid?

If you hook things up as shown in the diagram, you make yourself eligible for the Darwin Award.

The the generators need to be 120° out of phase with each other. That way

Vab = 415V @ 0°
Vbc = 415V @ 120°
Vcd = 415V @ 240°

Vad = Vab + Vbc + Vcd
Vad = 415V @ 0° + 415V @ 120° + 415V @ 240°
Vad = 415V([1+j0] + [cos(120°) + jsin(120°)] + [cos(240°) + jsin(240°)])
Vad = 415V([1+j0] + [-0.5 + (√3)/2] + [-0.5 - (√3)/2])
Vad = 415V(0+j0) = 0V

So that you can connect A and D together without killing yourself.

 

I get what you mean now but my problem is with the conversions between star and delta as I have been told by my lecturer to convert the system into Y-Y formation before solving. In Y-Y I feel confident but I would like to know if I have converted back to delta correctly and therefore got the correct currents as I'm not sure when converting from Y to delta whether you keep the line or phase currents the same? i.e. is the phase current is star the same as the phase current in delta or do you divide the star phase/line current by sqrt(3) and add 30 degrees to the angle?

 

I get what you mean now but my problem is with the conversions between star and delta as I have been told by my lecturer to convert the system into Y-Y formation before solving. In Y-Y I feel confident but I would like to know if I have converted back to delta correctly and therefore got the correct currents as I'm not sure when converting from Y to delta whether you keep the line or phase currents the same? i.e. is the phase current is star the same as the phase current in delta or do you divide the star phase/line current by sqrt(3) and add 30 degrees to the angle?

There is no magic involved here. The questions you have are best answered by deriving the Y-Δ transformation equations from scratch instead of just doing a plug-and-chug from some formula according to some cookbook recipe.

So take two black boxes and put a Y-connected load in one of them and a Δ-connected load in the other. Then for one of them determine the relationships between the voltages and currents outside the box and the voltages and currents inside the box. Now swap it with the other box and figure out what components have to be in the legs in order to get the same voltage/current relationships outside the box as before. Finally, see what the voltages and current are inside the box. To the outside world, the two boxes are equivalent.

You can do this for both a set of sources or a set of loads. You should end up with the same set of equations that you already have sitting in front of you, but now you will be able to answer all of these questions.

 

If the diagram is correct, then please explain what happens when point D is connected to point A, given that the voltage between them is a 1245V sinusoid?

If you hook things up as shown in the diagram, you make yourself eligible for the Darwin Award.

The the generators need to be 120° out of phase with each other. That way

Vab = 415V @ 0°
Vbc = 415V @ 120°
Vcd = 415V @ 240°

Vad = Vab + Vbc + Vcd
Vad = 415V @ 0° + 415V @ 120° + 415V @ 240°
Vad = 415V([1+j0] + [cos(120°) + jsin(120°)] + [cos(240°) + jsin(240°)])
Vad = 415V([1+j0] + [-0.5 + (√3)/2] + [-0.5 - (√3)/2])
Vad = 415V(0+j0) = 0V

So that you can connect A and D together without killing yourself.

My mistake, this is as the diagram indicates three seperate generators connected together, however i assumed this was automatically wrong and assumed the complex impedance represented a single delta winding, meaning that he drew a single delta connected to a single delta, (then star when conerted).