The active power flow of the iUPQC is shown in Fig. 19. In Fig. 19(a), the grid voltage Vs has a lower amplitude than the load voltage VL. In this case, the SAF delivers active power to the load, while the parallel active filter consumes active power. In Fig. 19(b), the grid voltage Vs has a higher amplitude than the load voltage VL. In this case, the SAF consumes active power, while the PAF filter delivers active power to the load. In an ideal situation when Vs is equal to VL, there is no active power flow through the SAF. The power drained from the electrical grid is equal to the sum of the load power and the iUPQC power losses.
In this paper, the output voltage VL is kept in phase with the fundamental component of the input voltage Vin; thus, the SAF operates with reactive power only when there are harmonics on the input voltage Vin because the fundamental component of the coupling transformer voltage is always in phase with the current drained from the utility grid is…show more content… 19. Power flow of iUPQC. (a) vSvL
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Fig. 20. Normalized apparent power of the SAF.
In steady state, assuming a sinusoidal and balanced utility grid voltage and disregarding the dual UPQC losses, the apparent power of SAF Ssf and PAF Spf.
Regularized in relation to the apparent power of the load are given by (16) and (17), respectively, represented in Figs. 20 and 21
|Ssf|/|SL| =(cos〖φ1∙〗 √((1-VL/Vs )^2 ))/√(1+TDHi^2 ) (16)
Where cos(ϕ1) is the displacement factor and THDi is the total harmonic distortion of the output current.
These equations are obtained through the analysis of the complex power of SAF˙Ssf and PAF˙Spf
