3 Phase Series Equivalents (Wye & Delta)

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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?
 

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t_n_k

Joined Mar 6, 2009
5,455
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)
 
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