Question 7.AE.15: For the system of Fig. E7.14.1, assume an initial value for ......
For the system of Fig. E7.14.1, assume an initial value for the bus vector \bar{U}^0 and compute the fundamental and the 5th harmonic currents injected into the system by the nonlinear load. The computation of the nonlinear load harmonic currents is the initial step of the harmonic load flow algorithm.
The initial step of harmonic load flow is as follows:
The real and imaginary nonlinear device currents G_{r, 4}^{(1)} \text { and } G_{i, 4}^{(1)} – referred to the swing bus – may be computed as follows, where the following powers are referred to the nonlinear bus 4:
\left\{\begin{array}{l} P_4^{(1)}=V_4^{(1)} I_4^{(1)} \cos \left(\delta_4^{(1}-\gamma_4^{(1)}\right) \\ Q_4^{(1)}=V_4^{(1)} I_4^{(1)} \sin \left(\delta_4^{(1}-\gamma_4^{(1)}\right) \end{array}\right..
I_4^{(1)} \text { and } \gamma_4^{(1)}=\delta_4^{(1)}-\tan ^{-1}\left(Q_4^{(1)} / P_4^{(1)}\right) are the magnitude and phase angle of the fundamental nonlinear device current, respectively. The diagram of Fig. E7.15.1 demonstrates the phasor relationships.
The real part of the fundamental of the nonlinear device current at bus 4 (referred to the swing bus) is
G_{r, 4}^{(1)}=I_4^{(1)} \cos \gamma_4^{(1)}where
I_4^{(1)}=\frac{P_4^{(1)}}{V_4^{(1)} \cos \left(\delta_4^{(1)}-\gamma_4^{(1)}\right)} .Thus
G_{r, 4}^{(1)}=\frac{P_4^{(1)} \cos \gamma_4^{(1)}}{V_4^{(1)} \cos \left(\delta_4^{(1)}-\gamma_4^{(1)}\right)}Correspondingly one obtains for the imaginary part of the fundamental of the nonlinear device current at bus 4 (referred to the swing bus):
G_{i, 4}^{(1)}=\frac{P_4^{(1)} \sin \gamma_4^{(1)}}{V_4^{(1)} \cos \left(\delta_4^{(1)}-\gamma_4^{(1)}\right)}The real and imaginary harmonic (5th) nonlinear device currents are given as (referred to bus 4)
\left\{\begin{array}{l} g_{r, 4}^{(5)}=0.3\left(V_4^{(1)}\right)^3 \cos \left(3 \delta_4^{(1)}\right)+0.3\left(V_4^{(5)}\right)^2 \cos \left(3 \delta_4^{(5)}\right) \\ g_{i, 4}^{(5)}=0.3\left(V_4^{(1)}\right)^3 \sin \left(3 \delta_4^{(1)}\right)+0.3\left(V_4^{(5)}\right)^2 \sin \left(3 \delta_4^{(5)}\right) \end{array}\right.where \delta_4^{(5)} \text { is the angle between } \widetilde{V}_4^{(5)} \text { and } \widetilde{V}_1^{(1)} \text { the swing bus voltage, } I_4^{(5)} is the fifth harmonic current magnitude
I_4^{(5)}=\sqrt{\left(g_{r, 4}^{(5)}\right)^2+\left(g_{i, 4}^{(5)}\right)^2}and \gamma_4^{(5)} \text { is the phase angle of } \tilde{I}_4^{(5)} \text { with respect to the swing bus voltage } \tilde{V}_1^{(1)}.
Assuming bus 1 to be the swing bus \left(\delta_1^{(1)}=0 \text { radians, }\left|V_1^{(1)}\right|=1.00 \mathrm{pu}\right), the solution procedure is as follows.
Step #1 of Harmonic Load Flow. Using the solution vector \bar{x} of the fundamental power flow analysis (see Application Example 7.13) and assuming harmonic voltage magnitudes and phase angles of 0.1 pu and 0 radians, respectively, the bus vector \bar{U}^0 will be
\begin{gathered} \bar{U}^0=\left(\delta_2^{(1)},\left|\widetilde{V}_2^{(1)}\right|, \delta_3^{(1)},\left|\widetilde{V}_3^{(1)}\right|, \delta_4^{(1)},\left|\widetilde{V}_4^{(1)}\right|, \delta_1^{(5)},\left|\widetilde{V}_1^{(5)}\right|, \delta_2^{(5)},\left|\widetilde{V}_2^{(5)}\right|,\right. \\ \left.\delta_3^{(5)},\left|\widetilde{V}_3^{(5)}\right|, \delta_4^{(5)},\left|\widetilde{V}_4^{(5)}\right|, \alpha_4, \beta_4\right)^t \\ \bar{U}^0=\left(-2.671 \cdot 10^{-4}, 0.9976,-2.810 \cdot 10^{-3}, 0.9964,\right. \\ \left.-3.414 \cdot 10^{-3}, 0.9960,0,0.1,0,0.1,0,0.1,0,0.1,0,0\right)^t \end{gathered}where the angles are measured in radians.
Note that the nonlinear device variables (α_4\ and\ β_4) are assumed to be zero because there are no device variables defined in this example for the nonlinear load at bus 4.
Step #2 of Harmonic Load Flow. Use \bar{U}^0 to compute the nonlinear device currents.
With P_4^{(1)}=250 \mathrm{~W} \text { corresponding to } 0.25 \mathrm{pu}, Q_4^{(1)}=100 \mathrm{VAr} corresponding to 0.100 pu, V_4^{(1)}=0.9960 \mathrm{pu} \text {, and } \delta_4^{(1)}=-3.414 \cdot 10^{-3} radians, one gets
The fundamental magnitude of the nonlinear current is
\begin{gathered} I_4^{(1)}=\sqrt{\left(G_{r, 4}^{(1)}\right)^2+\left(G_{i, 4}^{(1)}\right)^2}=\sqrt{(0.2514)^2+(-0.1015)^2} \\ =0.2711 \mathrm{pu} \text { with } \gamma_4^{(1)}=-0.38391 \mathrm{rad} . \end{gathered}The fifth harmonic current components of the nonlinear load current at bus 4 are (referred to bus 4)
\left\{\begin{array}{l} g_{r, 4}^{(5)}=0.3\left(V_4^{(1)}\right)^3 \cos \left(3 \delta_4^{(1)}\right)+0.3\left(V_4^{(5)}\right)^2 \cos \left(3 \delta_4^{(5)}\right) \\ g_{i, 4}^{(5)}=0.3\left(V_4^{(1)}\right)^3 \sin \left(3 \delta_4^{(1)}\right)+0.3\left(V_4^{(5)}\right)^2 \sin \left(3 \delta_4^{(5)}\right) \end{array}\right.With V_4^{(1)}=0.9960 \mathrm{pu}, \delta_4^{(1)}=-0.003414 \mathrm{rad}, V_4^{(5)}=0.1 \mathrm{pu}, \delta_4^{(5)}=0 rad, one obtains
\begin{aligned} & \left\{\begin{aligned} g_{r, 4}^{(5)}= & 0.3(0.9960)^3 \cos (-3 \cdot 0.003414) \\ & +0.3(0.1)^2 \cos (3 \cdot 0)=0.2994 \mathrm{pu} \\ g_{i, 4}^{(5)}= & 0.3(0.9960)^3 \sin (-3 \cdot 0.003414) \\ & +0.3(0.1)^2 \sin (3 \cdot 0)=-0.003036 \mathrm{pu} \end{aligned}\right. \\ & I_4^{(5)}=\sqrt{\left(g_{r, 4}^{(5)}\right)^2+\left(g_{i, 4}^{(5)}\right)^2}=\sqrt{(0.2994)^2+(-0.003036)^2} \\ & =0.2994 \mathrm{pu} \\ & \varepsilon_4^{(5)}=\tan ^{-1}\left(\frac{\left|g_{i, 4}^{(5)}\right|}{\left|g_{r, 4}^{(5)}\right|}\right)=0.01014 \mathrm{rad} . \\ & \end{aligned}Therefore, the fifth harmonic currents at bus 4 referred to the swing bus are G_{r, 4}^{(5)}=I_4^{(5)} \cos \gamma_4^{(5)} \text { and } G_{i, 4}^{(5)}=I_4^{(5)} \sin \gamma_4^{(5)}.
The phasor diagram for the 5th harmonic is shown in Fig. E7.15.2, where the angles are
\delta_4^{(5)}=\varepsilon_4^{(5)}+\gamma_4^{(5)}or
\gamma_4^{(5)}=\delta_4^{(5)}-\varepsilon_4^{(5)}=0-\varepsilon_4^{(5)}=-0.01014 \mathrm{rad}resulting in G_{r, 4}^{(5)}=0.2994 \mathrm{pu} \text { and } G_{i, 4}^{(5)}=-0.003036 \mathrm{pu}.
These nonlinear currents referred to the swing bus are about the same as those referred to bus 4 because the angle \gamma_4^{(5)} is very small for this first iteration.
Note that
are as expected. Correspondingly,
\left\{\begin{aligned} P_4^{(5)} & =I_4^{(5)} V_4^{(5)} \cos \left(\delta_4^{(5)}-\gamma_4^{(5)}\right) \\ & =0.2994 \cdot 0.1 \cdot \cos (0+0.01014)=0.029994 \mathrm{pu} \\ Q_4^{(5)} & =I_4^{(5)} V_4^{(5)} \sin \left(\delta_4^{(5)}-\gamma_4^{(5)}\right) \\ & =0.2994 \cdot 0.1 \cdot \sin (0+0.01014)=0.000304 \mathrm{pu} \end{aligned}\right.Therefore, the total injected real and reactive powers at the nonlinear bus 4 are
\left\{\begin{aligned} P_4^t & =P_4^{(1)}+P_4^{(5)}=0.250 \mathrm{pu}+0.02994 \mathrm{pu} \\ & =0.27994 \mathrm{pu} \\ Q_4^t & =Q_4^{(1)}+Q_4^{(5)}=0.100 \mathrm{pu}+0.000304 \mathrm{pu} \\ & =0.100304 \mathrm{pu} . \end{aligned}\right.
