Answering a Question on Three Impedances & Neutral Wire Current

[nothing909]

I just want to know if my method is correct in answer a question like this...

Three impedances z1=(2-j5), z2=(3+j4) and z3=(8+j0) are connected in delta to a 3 phase, 400v 50Hz supply.

1. Find the three line currents Ia, Ib and Ic
2. Find the S,P and Q and power factor for the load
3. If the nodes are now connected in star with a neutral connected, find the current in the neutral wire.

I'm on question 3 and I just want to know if this is the correct method to work it out:

1. Work out Yan, Ybn and Ycn.
2. Work out Vn'n using the equation Vn'n=(VanYan)(VbnYbn)(VcnYcn)/Yan+Ybn+Ycn
3. Work out the new phase voltages by doing Van'=Van-Vn'n , Vbn'=Vbn-Vn'n , Vcn'=Vcn-Vn'n
4. Work out the new phase currents by doing Ian'=Van'/Zan' , Ibn'=Vbn'/Zbn' , Icn'=Vcn'/Zcn'
5. Add all the new phase currents up and that'll give me the current in the neutral?

Is that the correct method to work it out? Have I did a step wrong or missed anything out?

 

[mfb]

That looks way to complicated. You get three complex amplitudes for current, you can add them (don't forget the phase shift between the three lines), that gives the total current and its phase.

 

[Babadag]

I agree with mfb: it is a hard work with complex numbers. In my opinion you have to start by transfigure the delta into star using simple formulae as:
If delta impedances are:
Z12=R12+jX12; Z23=R23+jX23; Z31=R31+jX31 and star impedance are:
Z1=R1+jX1; Z2=R2+jX2; Z3=R3+jX3 then:
Z1=Z12*Z31/(Z12+Z23+Z31)
Z2=Z23*Z12/(Z12+Z23+Z31)
Z3=Z23*Z31/(Z12+Z23+Z31)
Now Yan=1/Z1 Ybn=1/Z2 Ycn=1/Z3
Van=V<0,Vbn=V<-120,Vcn=V<-240 V=400/SQRT(3)=241V[approx.]

 

[mfb]

In question 3, it is a star.

 

[Babadag]

As usually-for me-I took the answers as questions. I am sorry!
So, from the beginning:
Question1: Find the three line currents Ia, Ib and Ic.
Transfigure delta to star and calculate Ia=Van/(Ra+jXa) [typical]
Where Van=V<0,Vbn=V<-120,Vcn=V<-240
[I presume the phase voltages are symmetric.]
In my opinion, Ia has to be: Ia=30.86+j62.7 A -for instance.
Q 2: Find the S,P and Q and power factor for the load.
S=Van*Ia*+Vbn*Ib*+Vcn*Ic* where Ia*=Ia conjugated [If Ia=IaRE+jIaIM then I*=IaRE-jIaIM].
P=RE(S); Q=IM(S)
Q 3: If the nodes are now connected in star with a neutral connected, find the current in the neutral wire.
Iastar=Van/Zab; Ibstar=Vbn/Zbc; Icstar=Vcn/Zca
Io[neutral]=Iastar+Ibstar+Icstar
In my opinion, Io has to be: Io=-44.36+j59.29 A -for instance.

 

[Babadag]

It is something wrong, still. Actually, Io=- (Iastar+Ibstar+Icstar) then 44.36-j59.29 A

 

[Babadag]

Now I saw the post title:
“Millman problem”. It is not Millman theorem that I followed.
However, sorry, another correction(!):
In the delta connection the currents are to be calculating from the relation:
Ia=Iab-Ica=Vab/Zab-Vca/Zca and the phase voltage Va=Ia*Za. Only Vab, Vbc, Vca are symmetric.
The neutral is not in the center and then Va, Vb and Vc [absolute values] are not symmetrical.

 

[Babadag]

In my opinion, in order to calculate the currents using Millman theorem a lot of data are missing. See[for instance]:

http://www.elect.mrt.ac.lk/EE201_3phase_sym_comp.pdf

para. Unbalanced three phase systems

For instance Zneutral, Zsource,Zline are missing.

If we could consider all these as zero then E1=Van,E2=Vbn,E3=Vcn and Vnn'=0.