Consider a balanced three-phase 60 MVA 0.8pf lagging load, sequence abc operating from a 13.8 kV (line voltage) bus. On bases of $S_{3ϕ base}$ = 100 MVA and $V_{L base}$ = 13.8 kV:
(a) Determine all bases.
(b) Determine all voltages, currents, and impedances in SI units and per-unit.

Question

Consider a balanced three-phase 60 MVA 0.8pf lagging load, sequence abc operating from a 13.8 kV (line voltage) bus. On bases of [latex]S_{3ϕ base} [/latex]= 100 MVA and [latex] V_{L base}[/latex] = 13.8 kV:
(a) Determine all bases.
(b) Determine all voltages, currents, and impedances in SI units and per-unit.

Accepted Answer

(a) [latex]S_{base} =\frac{S_{3ϕbase} }{3} =\frac{100 }{3} =33.33 MVA[/latex]

[latex]V_{base} =\frac{V_{Lbase} }{\sqrt{3} } =\frac{13.8 }{\sqrt{3} } =7.967 kV[/latex]

[latex]I_{base} =\frac{S_{base} }{V_{base} } =4.184 kA[/latex]

[latex]Z_{base} =\frac{V_{base} }{I_{base} } = =1.904 Ω[/latex]

(b)

[latex]V_{an} =7.967\angle 0° KV (1.000\angle 0°pu)[/latex]

[latex]V_{bn} =7.967\angle-120° KV (1.000\angle-12 0°pu)[/latex]

[latex]V_{cn} =7.967\angle+120° KV (1.000\angle+12 0°pu)[/latex]

[latex]S_{a} =S_{b} =S_{c} =\frac{S_{3Φ} }{3} =\frac{60}{3} =20MVA(0.60pu)[/latex]

[latex]S_{a} =S_{b} =S_{c} =16+ j12MVA(0.48+j0.36pu)[/latex]

[latex]I_{a} =\left\lgroup\frac{S_{a} }{V_{an} } ight group =2.510\angle -36.9°kA(0.6000\angle -36.9°pu)[/latex]

[latex]I_{b} =2.510\angle -156.9°kA(0.6000\angle -156.9°pu)[/latex]

[latex]I_{c} ==2.510\angle -83.1°KA(0.6000\angle -83.1°pu)[/latex]

[latex]Z_{Y} =\frac{V_{an} }{I_{a} } =3.174\angle +36.9°=2.539+j1.904 Ω(1.33+j1.000pu)[/latex]

[latex]Z_{ riangle } =3Z_{y} =7.618+ j5.713Ω(4+j3pu)[/latex]

[latex]V_{ab} =V_{an} -V_{bn} =13.8\angle 30°KV(1.732\angle 30° pu)[/latex]

[latex]V_{bc} =13.8\angle- 90°kV(1.732\angle 90°pu)[/latex]

[latex]V_{ca} =13.8\angle150°kV(1.732\angle 150°pu)[/latex]

Converting voltages and currents to symmetrical components:

[latex]\left[\begin{matrix}V_{0} \ V_{1} \V_{2} \end{matrix} ight] =\frac{1}{3} \left[\begin{matrix} 1 & 1 & 1\ 1 & a & a^{_{2}} \ 1 & a^{2} & a \end{matrix} ight] \left[\begin{matrix} V_{an} \ V_{bn} \ V_{cn} \end{matrix} ight] =\left[\begin{matrix} 0 kV \ 7.967\angle 0° kV \ 0 kV \end{matrix} \begin{matrix} (0 pu) \ (1\angle 0° pu) \ (o pu) \end{matrix} ight ][/latex]

[latex]I_{0} =0kA(0 pu)[/latex]

[latex]I_{1} =2.510\angle -36.9 kA(0.6\angle -36.9 pu)[/latex]

[latex]I_{2} =0 kA(0pu)[/latex]

Inclusion of transformers demonstrates the advantages of per-unit scaling.

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