3-phase delta circuit help

Thread Starter

dan_man555

Joined Sep 1, 2010
18
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
 

t_n_k

Joined Mar 6, 2009
5,455
The first challenge is to convert the delta connected source & it's associated source impedance to a wye equivalent.

Can you do this?
 

t_n_k

Joined Mar 6, 2009
5,455
Looks like you left out the equivalent source impedances for the star (wye) transformation. Apart from that it looks OK.
 

t_n_k

Joined Mar 6, 2009
5,455
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.
 

Thread Starter

dan_man555

Joined Sep 1, 2010
18
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
 

Thread Starter

dan_man555

Joined Sep 1, 2010
18
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?
 

Thread Starter

dan_man555

Joined Sep 1, 2010
18
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?
 

t_n_k

Joined Mar 6, 2009
5,455
c: line voltages at the terminals of the load

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?
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°
Z_load_a=215+57j

Hence V_load_a (line-to-neutral) =33.87<-42.26°*(215+57j)=7533.62<-27.41°

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.
 

Thread Starter

dan_man555

Joined Sep 1, 2010
18
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
 

t_n_k

Joined Mar 6, 2009
5,455
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
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.
 

t_n_k

Joined Mar 6, 2009
5,455
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.
 

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