Sequence Networks in Power System
Demonstrative Video
Problem
Determine the positive, negative, and zero sequence networks for the system shown in Fig. Assume zero sequence reactances for the generator and synchronous motors as 0.06 p.u. Current limiting reactors of 2.5 \(\Omega\) are connected in the neutral of the generator and motor no. 2. The zero sequence reactance of the transmission line is j300 \(\Omega\)
Solution
Let us choose the generator ratings as new base values for entire system:
\(\mathrm{MVA_{b,new}} = 25\) MVA
\(\mathrm{KV_{b,new}} = 11\) kV
Sequence reactance of Generator-G
Gen. rating and the new base values are same, the gen p.u. reactances does not change.
Positive and negative sequence reactances are same
Positive sequence reactance of generator, \(\mathrm{X}_{\mathrm{G}, 1}=10 \%=10 / 100=0.1 \mathrm{p}\).u.
Negative sequence reactance of generator, \(\mathrm{X}_{\mathrm{G}, 2}=0.1 \mathrm{p}\).u.
Zero sequence reactance of generator, \(\mathrm{X}_{\mathrm{G}, 0}=0.06 \mathrm{p} \cdot \mathrm{u}\)
Base impedance, \(Z_{b}=\frac{\left(k V_{b, \text { new }}\right)^{2}}{M V A_{b, \text { new }}}=\frac{11^{2}}{25}=4.84 \Omega\)
p.u. value \[\text { p.u. value}~ \mathrm{X}_{\mathrm{GN}}=\frac{\text { Actual Neutral reactance }}{\text { Base impedance }}=\frac{2.5}{4.84}=0.517 \mathrm{p} . \mathrm{u}\]
Sequence reactance of Transformer-T1
New p.u. reactance \[\begin{aligned} \mathrm{X_{p.u,new}}&=\mathrm{X_{p.u,old}}\times\left(\dfrac{\mathrm{kV_{b,old}}}{\mathrm{kV_{b,new}}}\right)^{2}\times\dfrac{\mathrm{MVA_{b,new}}}{\mathrm{MVA_{b,old}}}\\ &=0.1 \times\left(\frac{10.8}{11}\right)^{2} \times\left(\frac{25}{30}\right)=0.08 \mathrm{p} . \mathrm{u} \end{aligned}\]
In transformer the specified reactance is positive sequence reactance.
Also we assume that the positive, negative and zero sequence reactances of the transformer are equal.
\(\mathrm{X}_{\mathrm{T} 1,1}= \mathrm{X}_{\mathrm{T} 1,2}=0.08 \mathrm{p} . \mathrm{u}\)
\(\mathrm{T}_{1}, \quad \mathrm{X}_{\mathrm{T} 1,0}=0.08 \mathrm{p} \cdot \mathrm{u}\)
Sequence reactance of TL
Specified reactance in SLD is positive sequence reactance
Negative and positive sequence reactance are equal for TL
\[\begin{aligned} \mathrm{kV_{b,new}}\text{on HT side} & =\mathrm{kV_{b,new}}\text{on HT side}\times\dfrac{\text{HT voltage rating}}{\text{LT voltage rating}}\\ & =11\times\dfrac{121}{10.8}=123.24~\mathrm{kV}\\ Z_{b} & =\dfrac{\left(\mathrm{kV_{b,new}}\right)^{2}}{\mathrm{MVA_{b,new}}}=\dfrac{\left(123.24\right)^{2}}{30}=506.27~\Omega\\ X_{p.u} & =\dfrac{\text{Actual reactance}}{\text{Base impedance}}=\dfrac{100}{506.27}=0.198~\mathrm{p.u.} \end{aligned}\]
\(\mathrm{X}_{\mathrm{TL}, 1}=\mathrm{X}_{\mathrm{TL}, 2}=0.198 \mathrm{p} . \mathrm{u}\)
Zero sequence reactance \[\mathrm{X}_{\mathrm{TL}, 0}=\frac{\text { Zero sequence reactance in } \Omega}{\text { Base impedance }}=\frac{300}{506.27}=0.593 \mathrm{p} . \mathrm{u}\]
Sequence reactance of Transformer-T2
The ratings and winding connections of \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are identical.
Sequence reactances of \(\mathrm{T}_{1}\) and \(\mathrm{T}_{2}\) are same.
Positive sequence reactance \(\mathrm{X}_{\mathrm{T} 2,1}=0.08 \mathrm{p} . \mathrm{u}\)
Negative sequence reactance \(\mathrm{T}_{2}, \mathrm{X}_{\mathrm{T} 2,2}=0.08 \mathrm{p} . \mathrm{u}\)
Zero sequence reactance \(\mathrm{T}_{2}, \mathrm{X}_{\mathrm{T} 2,0}=0.08 \mathrm{p} \cdot \mathrm{u}\)
Sequence reactance of Syn Motor-M1
Specified reactance in SLD is positive sequence reactance
Negative and positive sequence reactance are equal for TL \[\begin{aligned} \mathrm{kV_{b,new}}~\text{on LT side} & =\mathrm{kV_{b,new}}~\text{on HT side}\times\dfrac{\text{LT voltage rating}}{\text{HT voltage rating}}\\ & =123.24\times\dfrac{10.8}{121}=11~\mathrm{kV}\\ \mathrm{X_{p.u,new}} & =\mathrm{X_{p.u,old}}\times\left(\dfrac{\mathrm{kV_{b,old}}}{\mathrm{kV_{b,new}}}\right)^{2}\times\dfrac{\mathrm{MVA_{b,new}}}{\mathrm{MVA_{b,old}}}\\ & =0.25\times\left(\dfrac{10}{11}\right)^{2}\times\dfrac{25}{15}\\ & =0.344~\mathrm{p.u.} \end{aligned}\]
\(\mathrm{X_{M1,1}} = \mathrm{X_{M1,1}} = 0.344\) p.u.
Zero sequence reactance \(\mathrm{X_{M1,0}} = 0.06 \times\left(\frac{10}{11}\right)^{2} \times \frac{25}{15}=0.083 \mathrm{p} . \mathrm{u}\)
Sequence reactance of Syn Motor-M2
\[\begin{aligned} \mathrm{X_{p.u,new}} & =\mathrm{X_{p.u,old}}\times\left(\dfrac{\mathrm{kV_{b,old}}}{\mathrm{kV_{b,new}}}\right)^{2}\times\dfrac{\mathrm{MVA_{b,new}}}{\mathrm{MVA_{b,old}}}\\ & =0.25\times\left(\dfrac{10}{11}\right)^{2}\times\dfrac{25}{7.5}=0.689\mathrm{p.u.}\\ \mathrm{X_{M2,1}} & =\mathrm{X_{M2,2}}=0.689\mathrm{p.u.}\\ \mathrm{X_{M2,0}} & =0.06\times\left(\dfrac{10}{11}\right)^{2}\times\dfrac{25}{7.5}=0.165\mathrm{p.u.}\\ Z_{b} & =\dfrac{\left(\mathrm{kV_{b,new}}\right)^{2}}{\mathrm{MVA_{b,new}}}=\dfrac{11^{2}}{25}=4.84\Omega\\ \mathrm{X_{p.u.}\text{neutral reactance}} & =\dfrac{\text{Actual reactance}}{Z_{b}}=\dfrac{2.5}{4.84}=0.517\mathrm{p.u.} \end{aligned}\]
Positive Sequence Network
Negative Sequence Network
Zero Sequence Network
