3-phase delta circuit help

Discussion in 'Homework Help' started by dan_man555, Sep 1, 2010.

1. dan_man555 Thread Starter New Member

Sep 1, 2010
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I need help working this one out. Would really appreciate the help.

A 3-phase, Δ-connected source has an internal impedance of 0.9 + 9j Ω/phase. The terminal voltage is equal to 13,200 Volts when no load is connected. The generator feeds a Δ-connected load with a per-phase impedance of 645 + 171j Ω/phase. The impedance of the transmission line connecting the generator to the load is 0.7 + 3j Ω/phase. The a-phase
internal voltage of the generator has been specified as the reference.

a) Construct a single phase equivalent circuit of the 3-phase system
b) Calculate the line currents: laA, IbB, and IcC
c) Calculate the magnitude of the line voltages at the terminals of the load
d) Calculate the magnitude of the line voltages at the terminals of the source
e) Calculate the magnitude of the phase current in the load

2. dan_man555 Thread Starter New Member

Sep 1, 2010
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Here is a scan of the question page.

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3. t_n_k AAC Fanatic!

Mar 6, 2009
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The first challenge is to convert the delta connected source & it's associated source impedance to a wye equivalent.

Can you do this?

4. dan_man555 Thread Starter New Member

Sep 1, 2010
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Here's my attempt. Let me know what you think.

5. t_n_k AAC Fanatic!

Mar 6, 2009
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Looks like you left out the equivalent source impedances for the star (wye) transformation. Apart from that it looks OK.

6. dan_man555 Thread Starter New Member

Sep 1, 2010
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This is my answer for part a. I really hope I got the values right.

7. t_n_k AAC Fanatic!

Mar 6, 2009
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That's pretty good. Why have you got the 30 degree phase lag on the source impedance - it's not relevant. The source impedance is just 0.3+j3 Ω.

8. dan_man555 Thread Starter New Member

Sep 1, 2010
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Ok.
So for part b:
Z total = 216 + 63j
= 225 < 16.26 Ω
IaA = 7621<-30 / 225 < 16.26
=33.87<-46.26 A
Therefore:
IbB= 33.87<-166.26 A
IcC= 33.87<73.74 A

9. dan_man555 Thread Starter New Member

Sep 1, 2010
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If that's all fine, then for part D:

Z=3.015<84.29
Vz= (33.87<-46.26)(3.015<84.29)
= 102.12<38.08v
Van = (6599.98-3810.5j)-(80.38+62.98j)
=6519.6-3873.48j
=7583.47<-30.72 V
Vab=13135.42<-0.72V
Vbc=13135.42<-120.72V
Vca=13135.42<-150.72V

Am I close?

10. t_n_k AAC Fanatic!

Mar 6, 2009
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All looks fine except for the Vca phase angle.

11. dan_man555 Thread Starter New Member

Sep 1, 2010
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Oh, I see.
Vca=13135.42<-240.72V

c: line voltages at the terminals of the load

Ztotal=215+57j+0.7+3j
=215.7+60j
=223.89<15.54
Van=(223.89<15.54)(33.87<-46.26)
=7583.15<-30.72
Vab=13135.42<-0.72V
Vbc=13135.42<-120.72V
Vca=13135.42<-240.72V
I hope both sides were supposed to have the same line voltages.

How do you 'calculate the magnitude of the phase current in the load' for part e?

12. dan_man555 Thread Starter New Member

Sep 1, 2010
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I'd really like to know if I'm on the right track for part c, and how to tackle the last part.

13. t_n_k AAC Fanatic!

Mar 6, 2009
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The load terminal-to-neutral voltage on a given phase must simply be the phase line current times the equivalent phase line-to-neutral load. You would not include the line impedance - which you did.

So
Ia_line=33.87<-42.26°

From which you can find the line-to-line value....etc

The load phase currents are (presumably) the values of current flow in the load delta arms based on the previously calculated line currents. Essentially a star-to-delta transformation of the line currents.

14. dan_man555 Thread Starter New Member

Sep 1, 2010
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Would that make the line voltages at the terminals of the load:
Vab=13048<2.59v, Vbc=13048<-117.4v, Vca=13048<-237.41v

And for the phase current:
I=33.87*sqrt(3)
I=58.66A

15. t_n_k AAC Fanatic!

Mar 6, 2009
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Voltages look OK w.r.t magnitude - I haven't carefully checked your phase angles throughout your solution. I have Vab at the load leading the phase-neutral supply (your Va'n) by 28.59° whereas you have the corresponding phase lead as 32.59°. So we have a discrepancy there - looks like a simple algebraic sign issue. I'll re-check my overall solution when I have some time.

The phase current (in the delta arms) magnitude would be 33.87/sqrt(3)=19.55A.

16. t_n_k AAC Fanatic!

Mar 6, 2009
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Had another look at the solution - attached a pdf copy of some of my working.

You'll note we have different starting phase angles for the single phase equivalent. I've based it all on the no-load line-to-line (source) voltage Vab having a zero phase shift. Then I worked everything off that basis. The real issue is the related phase shifts of the various quantities given some assumed initial phase at the source.

I'll still disagree with your overall phase relationship as mentioned in my earlier post.

• 3-phase delta problem.pdf
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Last edited: Sep 6, 2010
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17. t_n_k AAC Fanatic!

Mar 6, 2009
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Did it again as a check with phase angle of source the same as yours. Still have the discrepancy. Sorry the copy isn't particularly legible.

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