Solution to Application Example 5.6

i) The virtual resistor is only activated at harmonic frequencies; therefore, the 5th harmonic component of the injected current in the capacitor is

$\begin{array}{l}{\tilde{I}}_{C,5}={\tilde{I}}_{C,5}^{\mathrm{due}NL}+{\tilde{I}}_{C,5}^{\mathrm{due}sys},\\ {\tilde{I}}_{C,5}^{\mathrm{due}NL}=\frac{Z_{sys,5}}{Z_{sys,5}+Z_{C,5}+R_{vir}}{\tilde{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.06+j0.2074-j0.408+0.5}{\tilde{I}}_{NL,5},\\ =\frac{0.06+j0.2074}{0.56-j0.2}{\tilde{I}}_{NL,5}=\frac{0.2159\angle 73.87°}{0.595\angle -19.65°}{\tilde{I}}_{NL,5},\\ =\left(0.363\angle 93.52°\right){\tilde{I}}_{NL,5},\\ =\left(0.363\angle 93.52°\right)\left(\sqrt{2}\left(0.3\right)\angle 0°\right)\mathrm{A}\\ =\sqrt{2}\left(0.109\right)\angle 93.52°\mathrm{A}.\\ {\tilde{I}}_{C,5}^{\mathrm{due}sys}=\frac{1}{Z_{sys,5}+Z_{C,5}+R_{vir}}{\tilde{V}}_{sys,5},\\ =\frac{1}{0.595\angle -19.65°}{\tilde{V}}_{sys,5}=\left(1.68\angle 19.65°\right){\tilde{V}}_{sys,5},\\ =\left(1.68\angle 19.65°\right)\left(\sqrt{2}\left(3\right)\angle 0°\right)\mathrm{A},\\ =\sqrt{2}\left(5.4\right)\angle 19.65°\mathrm{A}.\\ \Rightarrow \Rightarrow {\tilde{\mathrm{I}}}_{C,5}={\tilde{I}}_{C,5}^{\mathrm{due}NL}+{\tilde{I}}_{C,5}^{\mathrm{due}sys},\\ =\left(\sqrt{2}\left(0.109\right)\angle 93.52°\right)+\left(\sqrt{2}\left(5.4\right)\angle 19.65°\right)\mathrm{A},\\ =\sqrt{2}\left(5.09+j1.93\right)=\sqrt{2}\left(5.44\right)\angle 20.77°\mathrm{A}\text{.}\end{array}$

Note that the magnitude of the 5th harmonic capacitor current is decreased by 64% (e.g., from $\sqrt{2}$ · 15.1A, without the virtual resistor, to $\sqrt{2}$ · 5.44A).

ii) Similarly, the 7th harmonic component of the injected current in the capacitor is

$\begin{array}{c}{\tilde{I}}_{C,7}={\tilde{I}}_{C,7}^{\mathrm{due}NL}+{\tilde{I}}_{C,7}^{\mathrm{due}sys},\\ {\tilde{I}}_{C,7}^{\mathrm{due}NL}=\frac{Z_{sys,7}}{Z_{sys,7}+Z_{C,7}+R_{vir}}{\tilde{I}}_{NL,7},\\ =\frac{0.06+j0.29029}{0.06-j0.0011+0.5}{\tilde{I}}_{NL,7},\\ =\frac{0.2964\angle 78.32°}{0.56\angle -0.11°}{\tilde{I}}_{NL,7},\\ =\left(0.53\angle 78.43°\right){\tilde{I}}_{NL,7},\\ =\left(0.53\angle 78.43°\right)\left(\sqrt{2}\left(0.2\right)\angle 0°\right)\mathrm{A},\\ =\sqrt{2}\left(0.106\right)\angle 78.43°\mathrm{A}.\\ {\tilde{I}}_{C,7}^{\mathrm{due}sys}=\frac{1}{Z_{sys,7}+Z_{C,7}+R_{vir}}{\tilde{V}}_{sys,7},\\ =\frac{1}{0.56\angle -0.11°}{\tilde{V}}_{sys,5}=\left(1.79\angle 0.11°\right){\tilde{V}}_{sys,7},\\ =\left(1.79\angle 0.11°\right)\left(\sqrt{2}\left(2\right)\angle 0°\right)\mathrm{A},\\ =\sqrt{2}\left(3.58\right)\angle 0.11°\mathrm{A}.\\ \Rightarrow \Rightarrow {\tilde{I}}_{C,7}={\tilde{I}}_{C,7}^{\mathrm{due}NL}+{\tilde{I}}_{C,7}^{\mathrm{due}sys},\\ =\left(\sqrt{2}\left(0.106\right)\angle 78.43°\right)+\left(\sqrt{2}\left(3.58\right)\angle 0.11°\right)\mathrm{A},\\ =\sqrt{2}\left(3.60+0.11\right)=\sqrt{2}\left(3.60\right)\angle 1.75°\mathrm{A}\text{.}\end{array}$

Note that the magnitude of the 7th harmonic capacitor current is decreased by about 90% (e.g., from $\sqrt{2}$ · 33.62A, without the virtual resistor, to $\sqrt{2}$ · 3.60A).

iii) The improved frequency responses of power capacitor current for virtual harmonic resistors of R_vir = 0.5 Ω and 1 Ω as a function of load harmonic current and utility harmonic voltage are shown in Fig. E5.6.1.

5.5.1 Application Example 5.7: Frequency and Capacitance Scanning [9]

Perform frequency and capacitance scans at bus 3 of the three-bus system shown in Fig. E5.7.1. The equivalent circuit of the external power system is represented as j0.99h at bus 1 (h is the order of harmonics). The nominal design value of the shunt capacitor at bus 3 is 0.001/ω_o farads and its reactance at harmonic h is (− 1000/h).

Solution to Application Example 5.7

The frequency scan can be done by calculating and inverting the admittance matrix of the network. Magnitude and phase angle plots of the driving-point impedance at bus 3 (Fig. E5.7.2a) indicate a resonance at the 71th harmonic, where the magnitude rises to about 1729 Ω. This is also evident from the zero crossing of the phase plot.

A capacitance scan is performed in a similar manner by computing the driving-point impedance at bus 3 for different capacitance values of the capacitor. The result is shown in Fig. E5.7.2b for several values of h. Note that resonances at the 5th and the 7th harmonics occur for a capacitance in the range of 0.1/ω_o F. These resonances should be avoided in the presence of six-pulse rectifiers.

5.6.1 Harmonic Voltage Constraint for Capacitors

Assume harmonic voltages (Eq. 5-3a) across the terminals of a capacitor. In the following relations V_rms is the terminal voltage, h is the harmonic order of the voltage, current, and reactive power: h = 3*, 5, 7, 9*, 11, 13 15^*, 17, … We have

$V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2=V_1^2+{\displaystyle \sum _{h=2}^{\infty }{V}_h^2}\text{,}$

or

$V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2=\sqrt{V_1^2+{\displaystyle \sum _{h=2}^{\infty }{V}_h^2}}\text{.}$

Normalized to V_{rms_rat},

$\frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}=\sqrt{\left(V_1^2+{\displaystyle \sum _{h=2}^{\infty }{V}_h^2}\right)/V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}^2}\le 1.10\mathrm{or}{1.15}^{\mathrm{a}}\text{.}$

5.6.2 Harmonic Current Constraint for Capacitors

Assume harmonic currents (Eq. 5-3b) through the terminals of a capacitor, where I_rms is the total current:

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}^2=I_1^2+I_2^2+I_3^2+\dots$

For sinusoidal current components I₁, I₂, I₃,…, one obtains with ω₁ = 2πf₁

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\left({\mathrm{\omega }}_{1}C{V}_1\right)}^2+{\left(2{\mathrm{\omega }}_{1}C{V}_2\right)}^2+{\left(3{\mathrm{\omega }}_{1}C{V}_3\right)}^2+\dots \text{,}$

or

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\mathrm{\omega }}_1^2{C}^2\left(V_1^2+4V_2^2+9{\mathrm{V}}_3^2+\dots \right),V_1^2=V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2-V_2^2-V_3^2-\dots$

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\mathrm{\omega }}_1^2{C}^2\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2+3V_2^2+8V_3^2+\dots \right)\text{,}$

or

${\mathrm{I}}_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\mathrm{\omega }}_1^2{C}^2\left(V_{\mathrm{ms}}^2+{\displaystyle \sum _{h=2}^{\infty }\left(h^2-1\right)V_h^2}\right)\text{,}$

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}^2={\mathrm{\omega }}_1^2{C}^2{V}_{\mathrm{r}\mathrm{m}\mathrm{s}}^2\left(1+{\displaystyle \sum _{h=2}^{\infty }\left(h^2-1\right){\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}^2}\right)\text{.}$

With I_{rms_rat} = ω_rat · C · V_{rms_rat} one can write (ω_rat = 2πf_rat)^b

$\frac{I_{\mathrm{r}\mathrm{m}\mathrm{s}}}{I_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}=\frac{f_1}{f_{\mathrm{rat}}}.\frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}\sqrt{1+{\displaystyle \sum _{h=2}^{\infty }\left(h^2-1\right){\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}^2}}\le 1.5\mathrm{or}{1.8}^{\mathrm{b}}\text{.}$

5.6.3 Harmonic Reactive-Power Constraint for Capacitors

In the presence of both voltage and current harmonics, the reactive power will also contain harmonics (Eq. 5-4b):

$\begin{array}{c}Q=Q_1+Q_2+Q_3+\dots \\ =I_1{V}_1+I_2{V}_2+I_3{V}_3+\dots \\ ={\mathrm{\omega }}_{1}C{V}_1^2+2{\mathrm{\omega }}_{1}C{V}_2^2+3{\mathrm{\omega }}_{1}C{V}_3^2+\dots \\ ={\mathrm{\omega }}_{1}C\left[\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2-V_2^2-V_3^2-\dots \right)+2V_2^2+3V_3^2+\dots \right],\\ ={\mathrm{\omega }}_{1}C\left[V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2+V_2^2+2V_3^2+\dots \right]\\ ={\mathrm{\omega }}_{1}C\left[V_{rms}^2+{\displaystyle \sum _{h=2}^{\infty }\left(h-1\right)V_h^2}\right]\end{array}$

and with Q_rat = ω_ratCV²_{rms_rat} one obtains

$\frac{Q}{Q{}_{\mathrm{rat}}}=\frac{f_1}{f_{\mathrm{rat}}}\cdot \frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}^2}\left[1+{\displaystyle \sum _{h=2}^{\infty }\left(h-1\right)\frac{V_h^2}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2}}\right]\le 1.35\mathrm{or}{1.45}^{\mathrm{a}}\text{.}$

5.6.4 Permissible Operating Region for Capacitors in the Presence of Harmonics

The safe operating region for capacitors in a harmonic environment can be obtained by plotting of the corresponding constraints for voltage, current, and reactive power.

Safe Operating Region for the Capacitor Voltage

$\frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}=\frac{\sqrt{V_1^2+{\displaystyle \sum _{h=2}^{\infty }{V}_h^2}}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}\le 1.10\mathrm{or}{1.15}^{\mathrm{a}}\text{,}$

or

$\frac{V_1}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}\sqrt{1+\left(\frac{{\displaystyle \sum _{h=2}^{\infty }{V}_h^2}}{V_1^2}\right)}=1.10\mathrm{or}{1.15}^{\mathrm{a}}\text{.}$

with V₁ ≈ V_rms = V_{rms_rat} and relying on the lower limit in Eq. 5-12b (e.g., 1.10) one obtains

${\displaystyle \sum _{h=2}^{\infty }\left(\frac{V_h^2}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2}\right)={\left(1.1\right)}^2-1=0.21\text{,}}$

or the voltage constraint becomes

${\displaystyle \sum _{h=2}^{\infty }\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}=0.458\text{,}}$

which is independent of h.

Safe Operating Region for the Capacitor Current

$\frac{I_{\mathrm{r}\mathrm{m}\mathrm{s}}}{I_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}=\frac{f_1}{f_{\mathrm{rat}}}\cdot \frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}\sqrt{1+{{\displaystyle \sum _{h=2}^{\infty }\left(h^2-1\right)\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}}^2}$

for f₁ = f_rat and V_rms = V_{rms_rat}. Relying on the lower limit in Eq. 5-13a (e.g., 1.5 of Eq. 5-10e) one obtains

${\left(1.5\right)}^2-1={{\displaystyle \sum _{h=1}^{\infty }\left(h^2-1\right)\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}}^2$

or

${\displaystyle \sum _{h=2}^{\infty }\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}=\sqrt{\frac{1.25}{h^2-1}}\text{,}$

which is a hyperbola.

Safe Operating Region for the Capacitor Reactive Power

$\frac{Q}{Q_{\mathrm{rat}}}=\frac{f_1}{f_{\mathrm{rat}}}\cdot \frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2}{V_{{}_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}}^2}\left[1+{\displaystyle \sum _{h=2}^{\infty }\left(h-1\right)\frac{V_h^2}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}^2}}\right]=1.35\mathrm{or}{1.45}^{\mathrm{a}}\text{,}$

for f₁ = f_rat and V_rms = V_{rms_rat}. Relying on the lower limit in Eq. 5-14a (e.g., 1.35) one obtains

${\displaystyle \sum _{h=2}^{\infty }\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)=\sqrt{\frac{0.35}{\left(h-1\right)}}}\text{,}$

which is also a hyperbola.

If the above listed limits are exceeded for a capacitor with the rated voltage V_{rms_rat1} then one has to select a capacitor with a higher rated voltage V_{rms_rat2} and a higher rated reactive power satisfying the relation

$Q_{\mathrm{rat}2}=Q_{\mathrm{rat}1}{\left(\frac{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}2}}{V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}1}}\right)}^2\text{.}$

Note all triplen and all even harmonics are small in three-phase power systems. However, triplen harmonics can be dominant in single-phase systems.

Feasible Operating Region for the Capacitor Reactive Power

Based on Eqs. 5-12d, 5-13c, and 5-14b, the safe operating region for capacitors in the presence of voltage and current harmonics is shown in Fig. 5.4.

5.6.5 Application Example 5.8: Harmonic Limits for Capacitors when Used in a Three-Phase System

The reactance of a capacitor decreases with frequency, and therefore the capacitor acts as a sink for higher harmonic currents. The effect is to increase the heating and dielectric stresses. ANSI/IEEE [10], IEC, and European [e.g., 11–13] standards list limits for voltage, currents, and reactive power of capacitor banks. These limits can be used to determine the maximum allowable harmonic levels. The result of the increased heating and voltage stress brought about by harmonics is a shortened capacitor life due to premature aging.

The following constraints must be satisfied:

$V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{rated}}=\sqrt{\left(V_1^2+{\displaystyle \sum _{h=2}^{h_{\text{max}}}{V}_h^2}\right)/V_{\mathrm{rated}}^2}\le 1.10\text{,}$

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}/I_{\mathrm{rated}}=\left(f_1/f_{\mathrm{rated}}\right)\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{rated}}\right)=\sqrt{1+{\displaystyle \sum _{h=2}^{h_{\text{max}}}\left(h^2-1\right){\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}^2}}\le 1.3\text{,}$

$Q/Q_{\mathrm{rated}}=\left(f_1/f_{\mathrm{rated}}\right){\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{rated}}\right)}^2=1+{\displaystyle \sum _{h=2}^{h_{\text{max}}}\left(h-1\right){\left(V_h/V_{\mathrm{r}\mathrm{m}\mathrm{s}}\right)}^2}\le 1.35\text{,}$

where V_rated = V_{rms_rat} is the rated terminal voltage, I_rated = I_{rms_rat} is the rated current, V_rms is the applied effective (rms) terminal voltage, f₁ is the line frequency, and V₁ is the fundamental (rms) voltage.

Plot for V₁ = V_rms = V_{rms_rat} and f₁ = f_rat the loci for V_h/V_rms, where (V_rms/V_{rms_rat}) = 1.10, (I_rms/I_{rms_rat}) = 1.3, and (Q/Q_rat) = 1.35, as a function of the harmonic order h for 5 ≤ h ≤ 49 (only one harmonic is present at any time within a three-phase system).

Solution to Application Example 5.8

The feasible operating region is shown in Fig. E5.8.1.

5.7.1 Application Example 5.9: Harmonic Losses of Capacitors

For a capacitor with C = 100 μF, V_rat = 460 V, R_s1 = 0.01 Ω (where R_s1 is the series resistance of the capacitor at fundamental (h = 1) frequency of f_rat = 60 Hz), compute the total harmonic losses for the harmonic spectra of Table E5.9.1 (up to and including the 19th harmonic) for the following conditions:

Table E5.9.1

Possible Voltage Spectra and Associated Capacitor Losses

h	${\left(\frac{{\mathit{V}}_{\mathit{h}}}{{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}}\right)}_{\mathbf{1}\mathbf{\Phi }}$ (%)	Ph for part a (mW)	Ph for part b (mW)	Ph for part c (mW)	${\left(\frac{{\mathit{V}}_{\mathit{h}}}{{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}}\right)}_{\mathbf{3}\mathbf{\Phi }}$ (%)	Ph for part a (mW)	Ph for part b (mW)	Ph for part c (mW)
1	100	3005	3005	3005	100	3005	3005	3005
2	0.5	0.3	0.6	1.2	0.5	0.3	0.6	1.2
3	4.0	40	120	360	2.0a	10.8	32.4	97.2
4	0.3	0.4	1.6	6.4	0.5	1.2	4.8	19.2
5	3.0	67.7	339	1695	5.0	188	940	4700
6	0.2	0.43	2.6	15.6	0.2	0.43	2.6	15.6
7	2.0	58.9	412	2880	3.5	181	1267	8869
8	0.2	0.77	6.16	49.3	0.2	0.77	6.16	49.3
9	1.0	24.4	220	1980	0.3	2.19	19.71	177.4
10	0.1	0.3	3	30	0.1	0.3	3	30
11	1.5	81.9	900	9900	1.5	81.9	900	9900
12	0.1	0.433	5.2	62.4	0.1	0.433	5.2	62.4
13	1.5	114	1480	19240	1.0	114	1482	19270
14	0.1	0.59	8.26	115.6	0.05	0.15	2.58	28.81
15	0.5	16.9	254	3810	0.1	0.68	10.2	153.0
16	0.05	0.192	3.07	49.1	0.05	0.192	3.07	49.1
17	1.0	87	1480	25140	0.5	22	374	6358
18	0.05	0.244	4.4	79.2	0.01	0.0097	0.175	3.143
19	1.0	109	2070	39200	0.5	27.1	513	9750
All higher voltage harmonics < 0.5%

^a Under certain conditions (e.g., DC bias of transformers as discussed in Chapter 2, and the harmonic generation of synchronous generators as outlined in Chapter 4) triplen harmonics are not of the zero-sequence type and can therefore exist in a three-phase system.

a) R_sh is constant, that is, R_sh = R_s1 = 0.01 Ω.

b) R_sh is proportional to frequency, that is, R_sh = R_s1(f/f_rat) = 0.01·h Ω.

c) R_sh is proportional to the square of frequency, R_sh = R_s1(f/f_rat)² = 0.01·h² Ω.

Solution to Application Example 5.9

A V_L-L = 460 V, f = 60 Hz, p = 6 pole, n_m = 1180 rpm, Y-connected squirrel-cage induction motor has the following parameters per phase referred to the stator:

R_S = 0.19 Ω, X_S = 0.75 Ω, X_m = 20 Ω, Rʹ_r = 0.070 Ω, Xʹ_r = 0.380 Ω, and R_fe → ∞.

a) Calculate the full-load current Ĩ_s and the displacement power factor cosφ of the motor (without capacitor bank).

b) In Fig. P5.1 the reactive current component of the motor is corrected by a capacitor bank C_bank. Find C_bank such that the DPF as seen at the input of the capacitor is DPF = cosφ = 0.98 lagging based on the consumer notation of current flow.

Problem 5.2: Calculation of the Total Power Factor (TPF) and the DPF Correction

The following problem studies the power-factor correction for a relatively weak power system where the difference between the displacement power factor (DPF) and the total power factor (TPF) is large due to the relatively large amplitudes of voltage and current harmonics.

a) Perform a PSpice analysis for the circuit of Fig. P5.2, where a three-phase diode rectifier – fed by a Y/Y transformer with turns ratio N_p/N_s = 1 – is combined with a self-commutated switch (e.g., metal-oxide-semiconductor field-effect transistors (MOSFET) or insulated-gate bipolar transistor (IGBT)) and a filter (e.g., capacitor C_f) serving a load (R_load). You may assume R_syst = 0.1 Ω, X_syst = 0.5 Ω, f = 60 Hz, v_AN (t) = $\sqrt{2}$·346 V cosωt, v_BN (t) = $\sqrt{2}$·346 V cos(ωt − 120°), v_CN (t) = $\sqrt{2}$ · 346 V cos(ωt – 240°), ideal diodes D₁ to D₆, C_f = 500 μF, R_load = 10 Ω, and a switching frequency of the self-commutated switch of f_sw = 600 Hz at a duty cycle of δ = 50%. Plot one period of the voltage and current after steady state has been reached as requested in parts b and c.

b) Plot and subject the line-to-neutral voltage v_an(t) to a Fourier analysis.

c) Plot and subject the phase current i_a(t) to a Fourier analysis.

d) Repeat parts a to c for δ = 10%, 20%, 30%, 40%, 60%, 70%, 80%, and 90%.

e) Calculate the DPF = cos_φ and the TPF = cos_θ based on the phase shifts of the fundamental and harmonics between v_an(t) and i_a(t) for all duty cycles and plot the DPF and TPF as a function of δ.

f) Determine the capacitance (per phase) C_bank of the power-factor correction capacitor bank such that the DPF as seen by the power systems is for δ = 50% about equal to DPF = 0.95 lagging.

Problem 5.3: Relation between Total and Displacement Power Factors

For the circuit of Fig. E5.1.1 (without taking into account power-factor correction capacitors), calculate with PSpice the Fourier components (magnitude and phase) of the phase voltage v_an(t) and the phase current i_a(t) up to the 10th harmonic. Compute and plot the ratio TPF/DPF as a function of the firing angle α.

Problem 5.4: Design of a Tuned Passive Filter

A harmonic LC filter (consisting of a capacitor and a tuning reactor) is to be designed in parallel to a power-factor correction (PFC) capacitor to improve the power quality (as recommended by IEEE-519 [8]) and to improve the DPF = cos_φ from 0.5 lagging to 0.95 lagging (consumer notation). Data obtained from a harmonic analyzer at 100% loading are 6.35 kV/phase, 600 kW/phase, THD _i = 26.46%, fundamental frequency f = 50 Hz, and fifth harmonic (300 Hz) current I₅ = 50 A. It is desirable to reduce the THD _i to 10% for Z_system⁽⁵⁾ = j0.1 Ω.

Problem 5.5: Harmonic Resonance

Figure E5.3.1 part (a) shows a simplified industrial power system (R_sys = 0.01 Ω, L_sys = 0.05 mH) with a power-factor correction capacitor (C_pf = 2000 μF). Industrial loads may have nonlinear (v–i) characteristics that can be approximately modeled as (constant) decoupled harmonic current sources. The utility voltages in industrial distribution systems are often distorted due to the neighboring loads and can be approximately modeled as decoupled harmonic voltage sources:

$\begin{array}{l}{v}_{sys}\left(t\right)=\sqrt{2}\cdot 460sin\left({\mathrm{\omega }}_{1}t\right)+\sqrt{2}\cdot 25sin\left(5{\mathrm{\omega }}_{1}t\right)+\sqrt{2}\cdot 15sin\left(7{\mathrm{\omega }}_{1}t\right),\\ i_{NL}\left(t\right)=\sqrt{2}\cdot 10sin\left({\mathrm{\omega }}_{1}t\right)+\sqrt{2}\cdot 2sin\left(5{\mathrm{\omega }}_{1}t\right)+\sqrt{2}\cdot 1sin\left(7{\mathrm{\omega }}_{1}t\right)\text{.}\end{array}$

Compute frequencies of the series and parallel resonances and the harmonic currents injected into the capacitor, and plot its frequency response.

Problem 5.6: Parallel Harmonic Resonance

In the system of Fig. E5.4.1 part (a), the source has the ratio X/R = 20. Assume X/R = 2000 (losses 0.5 W/kVAr) and X/R = 10,000 (losses 0.1 W/kVAr) for low-voltage and medium-voltage high-efficiency capacitors, respectively. The harmonic current source is a six-pulse converter injecting harmonic currents of the order

$h=n\left(k+1\right)\text{,}$

where n is an integer (typically from 1 to 4) and k is the number of pulses (e.g., equal to 6 for a six-pulse converter). Find the resonance frequency of this circuit. Plot the frequency response and current amplification at bus 1.

Problem 5.7: Series Harmonic Resonance

An example of a series resonance system is demonstrated in Fig. E5.5.1 part (a), where S = 200 MVA, X/R = 20, V_bus-rms = 7.2 kV, X_L = 1 Ω, Q_C = 900 kVAr, V_Crms = 7.2 kV, I_Lrms = 100 A, and Z = (R + j0.049 Ω) with R = 0 and magnitude Z_mag = 0.049 Ω at h = 7.59. The equivalent circuit (neglecting resistances) is shown in Fig. E5.5.1 part (b). Find the resonance frequency of this circuit. Plot the frequency response of the bus 1 equivalent impedance, and the current amplification across the tuning reactor.

Problem 5.8: Protection of Capacitors by Virtual Harmonic Resistors

Repeat Application Example 5.3 assuming a power converter is used to include a virtual harmonic resistor R_vir = 1.0 Ω in series with the capacitor.

Problem 5.9: Harmonic Current, Voltage, and Reactive Power Limits for Capacitors when Used in a Single-Phase System

The reactance of a capacitor decreases with frequency and therefore the capacitor acts as a sink for higher harmonic currents. The effect is to increase the heating and dielectric stress. ANSI/IEEE [10], IEC, and European [e.g., 11,12] standards provide limits for voltage, currents, and reactive power of capacitor banks. This can be used to determine the maximum allowable harmonic levels. The result of the increased heating and voltage stress brought about by harmonics is a shortened capacitor life due to premature aging.

According to the nameplate of the capacitors the following constraints must be satisfied:

$V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}=\sqrt{\left(V_1^2+{\displaystyle \sum _{h=2}^{h_{\text{max}}}{V}_h^2}\right)/V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}^2}\le 1.15\text{,}$

$I_{\mathrm{r}\mathrm{m}\mathrm{s}}/I_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}=\left(f_1/f_{\mathrm{rat}}\right)\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}\right)\sqrt{1+{\displaystyle \sum _{h=2}^{h_{\text{max}}}\left(h^2-1\right){\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}^2}}\le 1.3\text{,}$

$Q/Q_{\mathrm{rat}}=\left(f_1/f_{\mathrm{rat}}\right){\left(V_{\mathrm{r}\mathrm{m}\mathrm{s}}/V_{\mathrm{r}\mathrm{m}\mathrm{s}\_\mathrm{rat}}\right)}^2\left[1+{\displaystyle \sum _{h=2}^{h_{\text{max}}}\left(h-1\right){\left(\frac{V_h}{V_{\mathrm{r}\mathrm{m}\mathrm{s}}}\right)}^2}\le 1.45\text{,}\right.$

where V_{rrms_rat} is the rated terminal voltage, V_rms is the applied effective (rms) terminal voltage, f₁ is the line frequency, and V₁ is the fundamental (rms) voltage.

Plot for V_rms ≈ V_{rms_rat} and f₁ = f_rat the loci for V_h/V_rms, where (V_rms/V_{rms_rat}) = 1.15, (I_rms/I_{rms_rat}) = 1.3, and (Q/Q_rat) = 1.45 as a function of the harmonic order h for 3 ≤ h ≤ 49 (only one harmonic is present at any time).

Problem 5.10: Harmonic Losses of Capacitors

For a capacitor with C = 100 μF, V_rat = 1000 V, R_s1 = 0.005 Ω (where R_s1 is the series resistance of the capacitor at fundamental (h = 1) frequency of f_rat = 60 Hz), compute the total harmonic losses for the harmonic spectra of Table P5.10 (up to and including the 19th harmonic) for the following conditions:

Table P5.10

Possible Voltage Spectra with High Harmonic Penetration

h	${\left(\frac{{\mathit{V}}_{\mathit{h}}}{{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}}\right)}_{\mathbf{1}\mathbf{\Phi }}\left(\mathbf{\%}\right)$	${\left(\frac{{\mathit{V}}_{\mathit{h}}}{{\mathit{V}}_{\mathbf{60}\mathbf{Hz}}}\right)}_{\mathbf{3}\mathbf{\Phi }}\left(\mathbf{\%}\right)$
1	100	100
2	2.5	0.5
3	5.71	1.0a
4	1.6	0.5
5	1.25	7.0
6	0.88	0.2
7	1.25	5.0
8	0.62	0.2
9	0.96	0.3
10	0.66	0.1
11	0.30	2.5
12	0.18	0.1
13	0.57	2.0
14	0.10	0.05
15	0.10	0.1
16	0.13	0.05
17	0.23	1.5
18	0.22	0.01
19	1.03	1.0
All higher harmonics < 0.2%

^a Under certain conditions (e.g., DC bias of transformers as discussed in Chapter 2, and the harmonic generation of synchronous generators as outlined in Chapter 4) triplen harmonics are not of the zero-sequence type, and they can therefore exist in a three-phase system.

a) R_sh is constant, that is, R_sh = R_s1 = 0.005 Ω.

b) R_sh is proportional to frequency, that is, $R_{sh}=R_{s1}\left(\frac{f}{f_{\mathrm{rat}}}\right)=0.005\cdot h\mathrm{\Omega }\text{.}$

c) R_sh is proportional to the square of frequency, $R_{sh}=R_{s1}{\left(\frac{f}{f_{\mathrm{rat}}}\right)}^2=0.005\cdot h^2\mathrm{\Omega }\text{.}$

d) Plot the calculated losses of parts a to c as a function of the harmonic order h.