The harmonic power flow algorithm will be applied to the nonlinear power system as shown in Fig. E7.14.1 where all impedances are given in pu at 60 Hz and the base is Sbase=1 kVA. In order to simplify the problem, the harmonic (v–i) characteristic of the nonlinear load at bus 4 (referred to the

Question

The harmonic power flow algorithm will be applied to the nonlinear power system as shown in Fig. E7.14.1 where all impedances are given in pu at 60 Hz and the base is [latex]S_{base}[/latex]=1 kVA. In order to simplify the problem, the harmonic (v–i) characteristic of the nonlinear load at bus 4 (referred to the voltage at bus 4) is given as

[latex]\begin{aligned} & ilde{g}_{ ext {load }, 4}^{(5)}=g_{r, 4}^{(5)}+j g_{i, 4}^{(5)}, \ & g_{r, 4}^{(5)}=0.3\left(V_4^{(1)} ight)^3 \cos \left(3 \delta_4^{(1)} ight)+0.3\left(V_4^{(5)} ight)^2 \cos \left(3 \delta_4^{(5)} ight), \ & g_{i, 4}^{(5)}=0.3\left(V_4^{(1)} ight)^3 \sin \left(3 \delta_4^{(1)} ight)+0.3\left(V_4^{(5)} ight)^2 \sin \left(3 \delta_4^{(5)} ight) . \end{aligned}[/latex]

It will be assumed that the voltage at the swing bus (bus 1) is [latex]\left|\widetilde{V}_1^{(1)} ight|=1.00 \mathrm{pu}[/latex] and [latex]\delta_1^{(1)}=0 \mathrm{rad}[/latex].
The swing bus can be represented by Fig. E7.14.2 at the fundamental frequency. For the 5th harmonic, [latex] ilde{E} _{an}[/latex] of Fig. E7.14.3 represents a short-circuit, and the synchronous machine impedance Z1 is replaced by the (subtransient) impedance [latex]Z_2 \approx j X^{-}[/latex] =j0.0001 pu at 60 Hz (see Fig. E7.14.1). In this case the subtransient reactance has been chosen to be very small in order to approximate bus 1 as an ideal bus. Compute the fundamental and 5th harmonic admittance matrices.

Accepted Answer

The fundamental bus admittance matrix [latex]\left(X^{-} ight.[/latex] does not exist for fundamental quantities) is computed in Application Example 7.8. The 5th harmonic bus admittance matrix takes into account [latex]x^{-}[/latex] and its entries are calculated as follows:

[latex]\begin{aligned} & y_{11}^{(5)}=\frac{1}{0.01+j(0.02 \cdot 5)}+\frac{1}{0.01+j(0.01 \cdot 5)}+\frac{1}{j(0.0001 \cdot 5)} \ & =2029.137 \mathrm{~pu} \angle-1.5684 \mathrm{~rad} \ & y_{12}^{(5)}=\frac{-1}{0.01+j(0.01 \cdot 5)}=19.612 \mathrm{~pu} \angle 1.7681 \mathrm{~rad} \ & y_{13}^{(5)}=0 \ & y_{14}^{(5)}=\frac{-1}{0.01+j(0.02 \cdot 5)}=9.9503 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} \ & y_{21}^{(5)}=y_{12}^{(5)} \ & y_{22}^{(5)}=\frac{1}{0.01+j(0.01 \cdot 5)}+\frac{1}{0.02+j(0.08 \cdot 5)} \ & =22.084 \mathrm{~pu} \angle-1.3900 \mathrm{~rad} \ & y_{23}^{(5)}=\frac{-1}{0.02+j(0.08 \cdot 5)}=2.4969 \mathrm{~pu} \angle 1.62066 \mathrm{~rad} \ & y_{24}^{(5)}=0 \ & y_{31}^{(5)}=y_{13}^{(5)} \ & y_{32}^{(5)}=y_{23}^{(5)} \ & y_{33}^{(5)}=\frac{1}{0.02+j(0.08 \cdot 5)}+\frac{1}{0.01+j(0.02 \cdot 5)} \ & =12.445 \mathrm{~pu} \angle-1.4811 \mathrm{~rad} \ & y_{34}^{(5)}=\frac{-1}{0.01+j(0.02 \cdot 5)}=9.9504 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} \ & \end{aligned}[/latex].

[latex]\begin{gathered} y_{41}^{(5)}=y_{14}^{(5)} \ y_{42}^{(5)}=y_{24}^{(5)} \ y_{43}^{(5)}=y_{34}^{(5)} \ y_{44}^{(5)}=\frac{1}{0.01+j(0.02 \cdot 5)}+\frac{1}{0.01+j(0.02 \cdot 5)} \ =19.901 \mathrm{~pu} \angle-1.4711 \mathrm{~rad} \end{gathered}[/latex]

Therefore,

[latex]\begin{aligned} & \bar{Y}_{ ext {bus }}^{(5)}= \ & {\left[\begin{array}{cccc} 2029.137 \mathrm{~pu} \angle-1.5684 \mathrm{~rad} & 19.61 \mathrm{~pu} \angle 1.7681 \mathrm{~rad} & 0 & 9.9503 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} \ 19.61 \mathrm{~pu} \angle 1.7681 \mathrm{~rad} & 22.0847 \mathrm{~pu} \angle-1.3900 \mathrm{~rad} & 2.4969 \mathrm{~pu} \angle 1.62066 \mathrm{~rad} & 0 \ 0 & 2.4969 \mathrm{~pu} \angle 1.62066 \mathrm{~rad} & 12.445 \mathrm{~pu} \angle-1.4811 \mathrm{~rad} & 9.9504 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} \ 9.9503 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} & 0 & 9.9504 \mathrm{~pu} \angle 1.6704 \mathrm{~rad} & 19.901 \mathrm{~pu} \angle-1.4711 \mathrm{~rad} \end{array} ight] .} \end{aligned}[/latex]

Note that: [latex]-y_{i j} \angle heta_{i j}=y_{i j} \angle\left( heta_{i j}+\pi ight)[/latex].

Question 7.AE.14: The harmonic power flow algorithm will be applied to the non......

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