3 Phase Series Equivalents (Wye & Delta)

Thread Starter

jegues

Joined Sep 13, 2010
733
See first figure attached for my solution to the series portion of the problem.

The answer they give to the problem is,

\(Z_{series} = (1.84 - j1.38) \Omega\)

I would agree if this was a Y connection but it asked for the equivalent for a delta connection.

I first calculated the equivalent for a Y connection and converted it to a delta afterwards.

I based my reasoning on one of the other examples given in my text. (See second figure attached)

Is the solution they give incorrect or am I misunderstanding something?
 

Attachments

t_n_k

Joined Mar 6, 2009
5,447
My approach would be to find the parallel delta configuration first. I assumed the 208V is the line-to-line rms voltage.

Given the power per phase is 5kW then the delta (per phase) parallel resistance will be

Rp=208^2/5000=8.6528Ω

Zp=Rp||-jXc [capacitive since it's a leading power factor]

\(Z_p=\frac{1}{\frac{1}{R_p}+\frac{1}{-jX_c}}=\frac{R_pX_c}{X_c+jR_p}\)

\(arccos (pf)=36.87^o=arctan(\frac{R_p}{X_c})\)

or

\(R_p=tan(36.87^o)X_c\)

or

\(X_c=\frac{R_p}{tan(36.87^o)}=\frac{8.6528}{0.75}=11.537 \Omega\)

So Zp = 8.6528Ω||-j11.537Ω = 5.538-j4.153 Ω

The per phase delta series equivalent of the parallel combination would therefore simply be

Rs=5.538Ω and Xs=4.153 Ω (capacitive reactance)

So our answers agree, although you didn't derive the parallel equivalent for part (b)
 
Last edited:
Top