Solution to Application Example 4.4

Following the notation of [7] one obtains the voltage equations:

$e_a=p{\mathrm{\Psi }}_a-r\cdot i_a$

$e_b=p{\mathrm{\Psi }}_b-r\cdot i_b$

$e_c=p{\mathrm{\Psi }}_c-r\cdot i_c$

$e_{fd}=p{\mathrm{\Psi }}_{fd}+r_{fd}\cdot i_{fd}$

$0=p{\mathrm{\Psi }}_{kd}-r_{kd}\cdot i_{kd}+v_{kd}$

$0=p{\mathrm{\Psi }}_{kq}-r_{kq}\cdot i_{kq}+v_{kq}$

for k = 1, 2, 3, …, 8 damper windings in d- or q-axes.

The flux linkage equations are

${\mathrm{\Psi }}_a=-L_{\mathit{aa}}{i}_a-L_{\mathit{ab}}{i}_b-L_{ac}{i}_c+L_{afd}{i}_{fd}+L_{akd}{i}_{kd}+L_{akq}{i}_{kq}$

${\mathrm{\Psi }}_b=-L_{\mathit{ba}}{i}_a-L_{bb}{i}_b-L_{bc}{i}_c+L_{bfd}{i}_{fd}+L_{bkd}{i}_{kd}+L_{bkq}{i}_{kq}$

${\mathrm{\Psi }}_c=-L_{ca}{i}_a-L_{cb}{i}_b-L_{\mathit{cc}}{i}_c+L_{cfd}{i}_{fd}+L_{ckd}{i}_{kd}+L_{ckq}{i}_{kq}$

${\mathrm{\Psi }}_{fd}=-L_{fda}{i}_a-L_{fdb}{i}_b-L_{fdc}{i}_c+L_{\mathit{fdfd}}{i}_{fd}+L_{\mathit{fdkd}}{i}_{kd}$

${\mathrm{\Psi }}_{kd}=-L_{kda}{i}_a-L_{kdb}{i}_b-L_{kdc}{i}_c+L_{\mathit{kdfd}}{i}_{fd}+L_{kd1d}{i}_{1d}+L_{kd2d}{i}_{2d}+\dots +L_{\mathit{kdld}}{i}_{ld}$

${\mathrm{\Psi }}_{kq}=-L_{kqa}{i}_a-L_{kqb}{i}_b-L_{kqc}{i}_c+L_{kk1q}{i}_{1q}+L_{kq2q}{i}_{2q}+\dots +L_{\mathit{kqlq}}{i}_{lq}$

for k = 1, 2, …, 8 and l = 1, 2, …, 8.

Following the notation of Fig. E4.4.2, the self-inductance of the kth damper winding is

$L_{kd}^R=2\cdot {\ell }_R\cdot A_{Rkd}/I_{kd}$

and the leakage inductance with respect to stator winding is

$L_{kdl}^R=2\cdot {\ell }_R\cdot \left(A_{Rkd}-A_{Skd}\right)/I_{kd}$

or referred to the stator reference frame, where m₁ and m₂ are the number of stator and rotor phases, respectively:

$L_{kd}=m_1\left(2\cdot {\ell }_R\cdot A_{Rkd}/I_{kd}\right)M_{kda}^2/m_2$

$L_{kdl}=m_1\left\{2\cdot {\ell }_R\cdot \left(A_{Rkd}-A_{Skd}\right)/I_{kd}\right\}{M}_{kda}^2/m_2.$

The mutual inductances between the kth and the lth d-axis windings are

$L_{\mathit{kdld}}=m_1\cdot L_{\mathit{kdld}}^{kd}\cdot M_{\mathit{kdld}}^2\cdot M_{lda}^2/m_2$

$L_{\mathit{ldkd}}=m_1\cdot L_{\mathit{ldkd}}^{ld}\cdot M_{\mathit{ldkd}}^2\cdot M_{kda}^2/m_2.$

Correspondingly the ohmic resistance is

$r_{kd}=m_1\cdot R_{kd}\cdot M_{kda}^2/m_2.$

The same procedure applies to the q-axis parameters, as indicated in Fig. E4.4.3. In these equations ℓ_R is the rotor length. M_kda, M_kdld, M_lda, and M_ldkd are transformation ratios. The calculation of the stator inductances (Fig. E4.4.4) is performed in the same manner as for the rotor inductances.

If Eqs. E4.4-2 to E4.4-8 are introduced in Eq. E4.4-1 one obtains

$\left[L\right]\cdot \left[\frac{di}{dt}\right]=\left[V\right]\text{,}$

where [L] is an inductance matrix with time-varying coefficients, $\left[\frac{di}{dt}\right]$ is the derivative of the solution vector [i] and [V] represents the forcing sources. Using Gaussian elimination one gets

$\left[\frac{di}{dt}\right]={\left[L\right]}^{-1}\cdot \left[V\right].$

Subsequently a Runge-Kutta integration procedure of the fourth order provides the first starting steps for the Adams-Moulton method, which is a predictor-corrector integration procedure, and one obtains the solution vector [i]. A line-to-line short-circuit including the generator-transformer leakage impedance is performed. The line-to-neutral voltages in per unit (referred to the rated voltage of the generator) are shown in Fig. E4.4.5. The stator currents are depicted in Fig. E4.4.6, and field and damper bar currents are shown in Figs. E4.4.7 and E4.4.8a,b. The resulting electrical and mechanical torques are depicted in Figs. E4.4.9 and E4.4.10.

The integral ${\displaystyle {\int }_1^{t_2}{\left(i_{k\mathrm{bar}}\right)}^{2}dt}$ is a measure for the energy absorbed by the damper bar k due to the current i_kbar. Therefore, the energy dissipated in the damper bar k during time (t₂ – t₁) is

$E_{k\mathrm{bar}}=\left\{{\displaystyle {\int }_1^{t_2}{\left(i_{k\mathrm{bar}}\right)}^{2}dt}\right\}{R}_{k\mathrm{bar}}.$

It is well known that the bar (e.g., #16) of the amortisseur (see Fig. E4.4.11) located before the rotor pole as viewed in the direction of rotation incurs greater loss than the bar (e.g., #1) of the amortisseur located behind the rotor pole as viewed in the direction of rotation. Figure E4.4.4.12 shows the values E_kbar for a sudden line-to-line short-circuit at the terminals including transformer impedance for various time intervals (t₂ – t₁). Figure E4.4.13 depicts the subtransient field of a generator during the first cycle of a three phase short-circuit.

Table E4.4.1 summarizes the dissipated energies in the 1st and 16th amortisseur bars during various fault conditions of a generator including the generator-transformer leakage impedance:

Table E4.4.1

Dissipated Energies in Amortisseur Bars #1 and #16 for Different Fault Conditions

Fault condition	Energy dissipated in per unita in bar #1 during first 1.25 s	Energy dissipated in per unita in bar #16 during first 1.25 s
Line-to-line short-circuit	0.95	1.05
Balanced three-phase short-circuit	0.92	1.12
Out-of-phase synchronization with a 120° leading rotor angle	2.36	2.43
Out-of-phase synchronization with a 120° lagging rotor angle	1.25	1.57
Unbalanced stator currents resulting in an inverse stator current of |${\tilde{I}}_2$| = 8.7%.	0.049	0.053

^a Energy dissipated in a bar is referred to the heat energy required to melt one bar if there is no cooling.

1) line-to-line short-circuit,

2) balanced three-phase short-circuit,

3) out-of-phase synchronization with a 120° leading rotor angle,

4) out-of-phase synchronization with a 120° lagging rotor angle, and

5) unbalanced stator currents resulting in an inverse stator current of |${\tilde{I}}_2$| | = 8.7%.

4.2.7 Application Example 4.5: Measured Voltage Ripple of a 30 kVA Permanent-Magnet Synchronous Machine, Designed for a Direct-Drive Wind-Power Plant

A permanent-magnet generator is designed [29,30] for a variable-speed drive of a wind-power plant without any mechanical gear. The rated speed range is from 60 to 120 rpm. Line-to-neutral voltages v_L-N and line-to-line voltages v_L-L are recorded at no load and under rated load.

The voltage wave shapes are subjected to a Fourier analysis. Figures E4.5.1 and E4.5.2 depict the line-to-neutral voltages at no load and rated load, respectively, and Tables E4.5.1 and E4.5.2 represent their Fourier coefficients. Figures E4.5.3 and E4.5.4 represent the line-to-line voltages at no load and at rated load, and Tables E4.5.3 and E4.5.4 list their spectra. To minimize the size of this low-speed generator, the stator pitch k_p and distribution k_d winding factors are chosen relatively high (k_p · k_d = 0.8) resulting in a somewhat nonsinusoidal voltage wave shape. The generator feeds a rectifier and for this reason the nonsinusoidal voltage wave shape does not matter, except it increases the generator losses within acceptable limits.

4.2.8 Application Example 4.6: Calculation of Synchronous Reactances X_d and X_q from Measured Data Based on Phasor Diagram

Synchronous reactances are defined for fundamental frequency. Subjecting the voltage and current to a Fourier series yields the following data: line-to-neutral terminal voltage V_(l-n), phase current I_(l-n), instead of the induced voltage E the no-load voltage V_{(l-n)_no-load} ≈ E can be used for a permanent-magnet machine if saturation is not dominant. The displacement power factor angle φ and the torque angle δ can be determined from an oscilloscope recording and from a stroboscope, respectively. The ohmic resistance R can be measured as a function of the machine temperature. The relationship between these quantities is given by the phasor diagram of Fig. E4.6.1. Calculate the synchronous reactances X_q and X_d.

Solution to Application Example 4.6

From the phasor diagram of Fig. E4.6.1, the following relations for the synchronous reactances X_q and X_d are derived:

$X_q=\frac{V_{\left(l-n\right)}sin\delta +R{I}_{\left(l-n\right)}sin\delta cos\phi +R{I}_{\left(l-n\right)}cos\delta sin\phi }{I_{\left(l-n\right)}cos\phi cos\delta -I_{\left(l-n\right)}sin\phi sin\delta }\text{,}$

$X_d=\frac{E-V_{\left(l-n\right)}cos\delta +R{I}_{\left(l-n\right)}cos\left(\delta +\phi \right)}{I_{\left(l-n\right)}cos\left(90°-\phi -\delta \right)}.$

An application of these relations is given in [44].

4.2.9 Application Example 4.7: Design of a Low-Speed 20 kW Permanent-Magnet Generator for a Wind-Power Plant

A P = 20 kW, V_L-L = 337 V permanent-magnet generator has the cross sections of Figs. E4.7.1 and E4.7.2. The number of poles is p = 12 and its rated speed is n_rat = n_S = 60 rpm, where n_S is the synchronous speed. The B–H characteristic of the neodymium–iron–boron (NdFeB) permanent-magnet material is shown in Fig. E4.7.3.

a) Calculate the frequency f of the stator voltages and currents.

b) Apply Ampere’s law to path C (shown in Figs. E4.7.1 and E4.7.2) assuming the stator and rotor iron cores are ideal (μ_r → ∞).

c) Apply the continuity of flux condition for the areas A_m and A_g (representing half of the magnet area per pole and half of the air-gap area per pole, respectively) perpendicular to path C (two times the one-sided magnet (m) length l_m, and two times the one-sided air gap (g) length l_g, respectively). Furthermore l_m = 7 mm, l_g = 1 mm, A_m = 11,668 mm², and A_g = 14,469 mm².

d) Provided the relative permeability of the stator and rotor cores is μ_r → ∞, compute the load line of this machine. Plot the load line in a figure similar to that of Fig. E4.7.3. Is the permanent-magnet material (NdFeB) operated at the point of the maximum energy product?

e) What is the recoil permeability μ_R of the NdFeB material?

f) Compute the fluxes Φ_max and Φ_totalmax = 2Φ_max, provided there are N = 360 turns per stator phase, the (pitch and distribution) winding factor is k_w = 0.8, and the iron-core stacking factor is k_fe = 0.94.

g) Determine the diameter D of stator wire for a copper-fill factor of k_cu = 0.62 and a stator slot cross section of A_{st_slot} = 368 mm².

Solution to Application Example 4.7

a) Synchronous speed

$n_s=\frac{120f}{p}\mathrm{or}f=\frac{n_{s}p}{120}=\frac{60\left(12\right)}{120}=6\mathrm{Hz}.$

b) Ampere’s law

$H_m{l}_m+H_g{l}_g=0.$

c) Continuity of flux

$\mathrm{\Phi }=B_m{A}_m=B_g{A}_g\text{,}$

d) Constituent relation

$B_g={\mu }_0{H}_g.$

From Eq. E4.7-2

$H_m{l}_m=-H_g{l}_g.$

With Eq. E4.7-4

$H_m{l}_m=-B_g{l}_g/{\mu }_0\text{,}$

and with Eq. E4.7-3

$B_g=B_m\left(\frac{A_m}{A_g}\right)\text{,}$

the load-line equation becomes

$H_m=-\left(\frac{A_m}{A_g}\right)\left(\frac{l_g}{l_m}\right)\frac{B_m}{{\mu }_0}.$

Introducing the given numerical values into load-line equation gives

$H_m=-91.7B_m\left[\frac{kA}{m}\right].$

This equation is plotted in Fig. E4.7.4. Note that the magnet material is not operated in the point of the maximum energy product.

e) Recoil permeability of NdFeB (slope of the permanent-magnet characteristic)

$\begin{array}{c}{\mu }_R=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}=\frac{\left(B_r-0\right)}{\left(0-\left(-H_c\right)\right)}=\frac{1.25}{950\left({10}^3\right)}=1.316\times {10}^{-6}\left(\mathrm{H}/\mathrm{m}\right)\end{array}$

or

${\mu }_R=1.047{\mu }_0.$

f) Induced voltage

${\mathrm{\Phi }}_{\text{max}}=B_m{A}_m\text{,}$

${\mathrm{\Phi }}_{\mathrm{totalmax}}=2{\mathrm{\Phi }}_{\text{max}}=2B_m{A}_m\text{,}$

The flux is defined as

$\mathrm{\Phi }\left(\mathrm{t}\right)={\mathrm{\Phi }}_{\mathrm{totalmax}}cos\omega t.$

Therefore, the induced voltage is

$E_{\text{max}}=N\left(d\mathrm{\Phi }/dt\right)=\omega N_{\mathit{ph}}{k}_w{k}_{fe}{\mathrm{\Phi }}_{\mathrm{totalmax}}.$

From load line and magnet characteristic one obtains the operating point:

$B_{m0}=1.1\mathrm{T}\text{,}$

$H_{m0}=-100\mathrm{k}\mathrm{A}/\mathrm{m}.$

Therefore,

$\mathrm{\Phi }={\mathrm{\Phi }}_{\mathrm{totalmax}}=\left(2\right)\left(1.1\right)\left(11668\times {10}^{-6}\right)\mathrm{Wb}=0.0257\mathrm{Wb}$

and the induced voltage is now

$\begin{array}{l}e\left(t\right)=\left(360\right)\left(0.0257\right)\left(2\pi \right)\left(6\right)\left(0.8\right)\left(0.94\right)sin\omega t=E_{\text{max}}sin\omega t\text{,}\end{array}$

or

$E_{\text{max}}=262.3\mathrm{V}\text{,}$

$E_{rms}=E_{\text{max}}/\sqrt{2}=185.48\mathrm{V}\text{,}$

and the line-to-line terminal voltage is approximately

$V_{L-L}=\sqrt{3}{V}_{L-N}=\sqrt{3}\left(1.05\right)E_{rms}=337.3\mathrm{V}\text{,}$

g) Determination of wire diameter of the stator winding. Figure E4.7.5 depicts the stator winding housed in 72 slots. It is a full-pitch, double-layer winding. Using a copper-fill factor of k_cu = 0.62 one obtains for a stator slot cross section of A_{st_slot} = 368 mm² and (360/12) = 30 turns per stator pole and per phase with the copper cross section of one stator slot A_{st_slot_cu} = A_{st_slot} · k_cu = 368 · 0.62 mm² = 30πD²/4 or a wire diameter of D = 3.1 mm. For a rated current of I_rat = 31 A one obtains a current density of J = 31/7.6 = 4.07 A/mm². This current density is acceptable for a ventilated machine.

4.2.10 Application Example 4.8: Design of a 10 kW Wind-Power Plant Based on a Synchronous Machine

Design a wind-power plant (Fig. E4.8.1) feeding P_{out_transformer} = 10 kW into the distribution system at a line-to-line voltage of V_{L-L system} = 12.47 kV = V_{L-L_secondary of transformer}. The wind-power plant consists of a Y–Δ three-phase (ideal) transformer connected between three-phase PWM inverter and power system, where the Y is the primary and Δ is the secondary of the ideal transformer (Fig. E4.8.2). The input voltage of the PWM inverter (see Chapter 1, Fig. E1.5.1) is V_DC = 600 V, where the inverter delivers an output AC current I_inverter at about 18° lagging displacement power factor (consumer notation) cos φ ≈ 0.95 lagging the primary line-neutral voltage of the Y – Δ transformer, and the inverter output line-to-line voltage is V_{L-L_inverter} = 240 V = V_{L-L_primary of transformer}.

The three-phase rectifier (see Chapter 1, Fig. E1.2.1) with one self-commuted PWM-operated switch (IGBT) is fed by a three-phase synchronous generator. The input line-to-line voltages of the three-phase rectifier is V_{L-L rectifier} = 480 V = V_{L-L_generator}. Design the mechanical gear between wind turbine and generator (with p = 2 poles at n_S = 3600 rpm, utilization factor C = 1.3 kWmin/m³, D_rotor = 0.2 m, the leakage reactance X_sleakage = 2πfL_sleakage is 10% of the synchronous reactance X_S = 1.5 pu, maximum flux density B_max = 0.7 T, iron-stacking factor k_fe = 0.95, stator winding factor k_S = 0.8, rotor winding factor k_r = 0.8, and torque angle δ = 30°) so that a wind turbine can operate at n_turbine = 30 rpm. Lastly, the wind turbine – using the famous Lanchester-Betz-Joukowsky limit [45,84–86] for the maximum energy efficiency of a wind turbine as a guideline – is to be designed for the rated wind velocity of v = 10 m/s at an altitude of 1600 m and a coefficient of performance (actual efficiency) c_p = 0.3. You may assume that all components (transformer, inverter, rectifier, generator, mechanical gear) except the wind turbine have each an efficiency of η = 95%.

a) Based on the given transformer output power of P_{out_transformer} = 10 kW compute the output powers of inverter, rectifier, generator, gear, and wind turbine.

b) For the circuit of Fig. E4.8.1 determine the transformation ratio (N_Y/N_Δ) of the Y–Δ transformer, where the Y is the primary (inverter side) and Δ the secondary (power system side).

c) What is the phase shift between $\tilde{\mathrm{V}}$_{L- L system} and $\tilde{\mathrm{V}}$_{L - L_inverter}?

d) What is the phase shift between ${\tilde{I}}_{L-s\mathrm{ystem}}$ and ${\tilde{I}}_{L-\mathrm{inverter}}$?

e) Use sinusoidal PWM to determine for the circuit of Fig. E1.5.1 the inverter output current Ĩ_inverter which lags $\tilde{\mathrm{V}}$_{L - N_inverter} = 138.57 V about 18° at an inverter switching frequency of f_sw = 7.2 kHz. You may assume R_system = 50 mΩ, X_system = 0.1 Ω, R_inverter = 10 mΩ, and X_inverter = 0.377 Ω at f = 60 Hz.

f) For the circuit of Fig. E1.2.1 and a duty cycle of δ = 50%, compute the input current of the rectifier I_rectifier.

g) Design the synchronous generator for the above given data provided it operates at unity displacement power factor.

h) Design the mechanical gear.

i) Determine the radius of the wind turbine blades for the given conditions.

Solution to Application Example 4.8

a) 

$\begin{array}{l}{P}_{\mathrm{out}\_\mathrm{transformer}}=10\mathrm{kW},P_{\mathrm{out}\_\mathrm{inverter}}=10.53\mathrm{kW},\\ P_{\mathrm{out}\_\mathrm{rectifier}}=11.08\mathrm{kW},P_{\mathrm{out}\_\mathrm{generator}}=11.67\mathrm{kW},\\ P_{\mathrm{out}\_\mathrm{gear}}=12.28\mathrm{kW},P_{\mathrm{out}\_\mathrm{wind}\mathrm{turbine}}=12.92\mathrm{kW}.\end{array}$

b) $\left(\frac{N_Y}{N_{\mathrm{\Delta }}}\right)=\frac{1}{89.9}.$ (E4.8-2)

c) Note ${\tilde{V}}_{L-L\mathrm{system}}={\tilde{V}}_{L-L\mathrm{\Delta }}\mathrm{and}{\tilde{V}}_{L-L\mathrm{inverter}}={\tilde{V}}_{L-LY}\text{;}$ therefore,

${\tilde{V}}_{L-L\mathrm{\Delta }}={\tilde{V}}_{L-LY}\angle +30°.$

${\tilde{I}}_{L\mathrm{system}}={\tilde{I}}_{L\mathrm{inverter}}\angle +30°.$

One concludes that the line-to-line voltages of the secondary Δ as well as the line currents fed into the power system are shifted by + 30 degrees.

e) The computer program for the inverter is given in Chapter 1, resulting in an inverter output current of I_{out_inverter} = 25.33 A, and an inverter input current of I_{in_inverter} = 18.47 A. The inverter output phase voltage and output line current are shown in Fig. E4.8.3.

f) Computer program for rectifier is given in Chapter 1 resulting in a rectifier output current of I_{out_rectifier} = 18.47 A and a rectifier input current of I_{in_rectifier} = 14.04 A. The output voltage of the rectifier is depicted in Fig. E4.8.4.

g) Design of synchronous generator:

• $\mathrm{Stator}\mathrm{frequency}f=\frac{n_s\cdot p}{120}=\frac{\left(3600\right)\left(2\right)}{120}=60\mathrm{Hz}\text{,}$ (E4.8-5)

• $\mathrm{Phase}\mathrm{voltage}{V}_{L-N}=480\mathrm{V}/1.732=277.14\mathrm{V}\text{,}$ (E4.8-6a)

• $\mathrm{Base}\mathrm{voltage}{V}_{\mathrm{base}}=V_{L-N}=277.14\mathrm{V}\text{,}$ (E4.8-6b)

• $\mathrm{Base}\left(\mathrm{rated}\right)\mathrm{current}{I}_{\mathrm{base}}=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{3\cdot V_{L-N}}=\frac{11670}{3\left(277.14\right)}=14.04\mathrm{A}\text{,}$ (E4.8-7)

• $\mathrm{Base}\mathrm{impedance}{Z}_{\mathrm{base}}=\frac{V_{\mathrm{base}}}{I_{\mathrm{base}}}=\frac{277.14\mathrm{V}}{14.04\mathrm{V}}=19.74\mathrm{\Omega }\text{,}$ (E4.8-8)

• $\mathrm{Generator}\mathrm{loss}{P}_{\mathrm{loss}}=P_{\mathrm{in}\_\mathrm{generator}}-P_{\mathrm{out}\_\mathrm{generator}}=\left(12280-11670\right)\mathrm{W}=610\mathrm{W}\text{,}$ (E4.8-9)

• $\mathrm{Stator}\mathrm{resistance}{R}_a=\frac{P_{\mathrm{loss}}}{3\cdot I_{\mathrm{base}}^2}=1.03\mathrm{\Omega }\text{,}$ (E4.8-10)

• $\mathrm{Ideal}\mathrm{core}\mathrm{length}{l}_i=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{D_{\mathrm{rotor}}^2\cdot C\cdot n_s}=\frac{11670}{\left({0.2}^2\right)\left(1.3\right)\left(3600\right)}=0.0623\mathrm{m}\text{,}$ (E4.8-11)

• Actual iron core length (with interlamination insulation)

$l_{\mathrm{act}}=\frac{l_i}{k_{fe}}=\frac{0.0623}{0.95}\mathrm{m}=0.0656\mathrm{m}\text{,}$

• Stator synchronous reactance

$X_S=X_S\left[\mathrm{p}\mathrm{u}\right]·Z_{\mathrm{base}}=\left(1.5\right)\left(19.74\right)=29.61\mathrm{\Omega }\text{,}$

with $X_d\approx X_S=X_{\mathit{aa}}+X_{\mathit{ab}}+\frac{3}{2}{X}_{\mathit{aa}2}$ one gets approximately X_S = 2 · (2πf)L_aa or for the self-inductance L_aa of one phase

$L_{\mathit{aa}}=\frac{X_S}{4\pi f}=0.0393\mathrm{H}\text{,}$

and the leakage inductance of one stator (s) phase

$L_{s\ell }=0.1\left(\frac{29.61}{2\pi 60}\right)=0.0079\mathrm{H}.$

• The area per pole is

${\mathrm{area}}_p=\frac{2\pi \cdot D_{\mathrm{rotor}}}{2}\cdot \frac{l_i}{2}=\frac{2\pi \left(0.2\right)\left(0.0623\right)}{4}=0.0196{\mathrm{m}}^2.$

• The maximum flux linked with one pole is

${\mathrm{\Phi }}_{s\_\max }=B_{\text{max}}\cdot {\mathrm{area}}_p=0.7\left(0.0196\right)=0.0137\mathrm{Wb}.$

• The induced voltage in one phase of the stator winding becomes (generator operation)

$E_{\mathit{ph}}=1.05\cdot V_{L-N}=4.44\cdot f\cdot N_s\cdot k_s\cdot {\mathrm{\Phi }}_{s\_\max }=291.0\mathrm{V}\text{,}$

or the number of stator turns per phase is

$\begin{array}{ll}{N}_s& =\frac{E_{\mathit{ph}}}{4.44\cdot f\cdot k_s\cdot {\mathrm{\Phi }}_{s\_\max }}=\frac{291}{4.44\left(60\right)\left(0.8\right)\left(0.0137\right)}\\ & =100\mathrm{turns}.\hfill \end{array}$

The one-sided air-gap length can be obtained from

$L_{\mathit{aa}}=\frac{{\lambda }_{\mathit{aa}}}{i_a}=\frac{N_s\cdot {\mathrm{\Phi }}_{s\_\max }}{i_{\text{max}}}\text{,}$

where

${\mathrm{\Phi }}_{s\_\max }=\frac{{\mu }_o\cdot \frac{4}{\pi }\cdot k_s\cdot N_s\cdot i_{\text{max}}\cdot \mathit{are}{a}_p}{p\cdot g}\cdot$

Therefore,

$L_{\mathit{aa}}=\frac{{\mu }_o\cdot \frac{4}{\pi }\cdot k_s{N}_s^2\cdot {\mathrm{area}}_p}{p\cdot g}$

or solved for the air-gap length

$\begin{array}{l}g=\frac{{\mu }_o\cdot \frac{4}{\pi }\cdot k_s\cdot N_s^2\cdot {\mathrm{area}}_p}{p\cdot L_{\mathit{aa}}}=\frac{\left(4\pi \times {10}^{-7}\right)\left(\frac{4}{\pi }\right)\left(0.8\right)\left(10000\right)\left(0.0196\right)}{2\left(0.0393\right)}\\ & =0.0032\mathrm{m}=3.2\mathrm{mm}.\end{array}$

• The torque of the generator is

$\left(-T\right)=\frac{P_{\mathrm{out}\_\mathrm{generator}}}{{\omega }_s}=\frac{11670}{377}=30.94\mathrm{N}\mathrm{m}.$

For a torque angle of δ = 30° one gets with

$T=-\frac{p\cdot \pi \cdot D_{\mathrm{rotor}}\cdot l_{\mathrm{act}}\cdot B_{\text{max}}\cdot F_r}{4}\cdot sin\delta \text{,}$

or the rotor mmf

$\begin{array}{l}{F}_r=\frac{\left(-T\right)\cdot 4}{p\cdot \pi \cdot D_{\mathrm{rotor}}\cdot l_{\mathrm{act}}\cdot B_{\text{max}}\cdot sin\delta }\\ =\frac{\left(30.94\right)\left(4\right)}{\left(2\right)\left(\pi \right)\left(0.2\right)\left(0.0656\right)\left(0.7\right)\left(0.5\right)}=4289.6\mathrm{A}\mathrm{t}.\end{array}$

• The ampere turns for the rotor are $F_r=\frac{4}{\pi }\cdot k_r\cdot \frac{N_{r\left(\mathrm{tot}\right)}}{p}\cdot I_r$ or solved for the ampere turns

$N_{r\left(\mathrm{tot}\right)}\cdot I_r=\frac{F_r\cdot p\cdot \pi }{4\cdot k_r}=\frac{\left(4289.6\right)\left(2\right)\left(\pi \right)}{\left(4\right)\left(0.8\right)}=8422.4\mathrm{A}\mathrm{t}.$

• Selecting 16 rotor slots with 20 turns each, one obtains the field (rotor) current:

$I_r=8422.4\mathrm{Aturns}/\left(16·20\mathrm{turns}\right)=26.3\mathrm{A}.$

h) Gear ratio. The gear ratio is

$\mathrm{gear}\mathrm{ratio}=\frac{n_{\mathrm{generator}}}{n_{\mathrm{wind}\mathrm{turbine}}}=\frac{3600}{30}=120.$

i) Radius of (three) wind turbine blades. The limiting efficiency for wind turbines is according to Lanchester-Betz-Joukowsky [45,84–86] η_{Lanchester-Betz-Joukowsky} = (16/27) = 0.59. The wind turbine design depends on the density of the air and the ambient temperature. Table E4.8.1 provides the air density as a function of altitude.

The wind turbine has to provide a rated output power of P_{out_wind turbine} = 12.92 kW at a rated wind velocity of v = 10 m/s. The efficiency (coefficient of performance) is c_p = 0.3, where c_p is the ratio between the power delivered by the rotor of the wind turbine to the power in the wind with velocity v striking the area A swept by the rotor.

According to [45] the output power of the wind turbine is

$P_{\mathrm{wind}\_\mathrm{turbine}}=\frac{1}{2}\cdot c_p\cdot v^3\cdot A\cdot \rho .$

The area swept by the rotor is

$A=\frac{2\cdot P_{\mathrm{wind}-\mathrm{turbine}}}{c_p\cdot v^3\cdot \rho }=\frac{2\left(12920\right)}{0.3\left(1000\right)\left(1.05\right)}=82{\mathrm{m}}^2\text{,}$

or the rotor blade radius is

$R_{\mathrm{blade}}=\sqrt{\frac{A}{\pi }}=\sqrt{\frac{82}{\pi }}=5.1\mathrm{m}.$

The height of the tower can be assumed to be about 2 times the blade radius, that is, H_tower = 10 m.

4.2.11 Synchronous Machines Supplying Nonlinear Loads

Frequently synchronous machines or permanent-magnet machines are used together with solid-state converters, that is, either rectifiers as a load [46–49] or inverters feeding the machine. The question arises how much current distortion can be permitted in order to prevent noise and vibrations as well as overheating? It is recommended to limit the total harmonic current distortion of the phase current to THD_i ≤ 5–20% as specified by recommendation IEEE-519 (see Chapter 1). The 5% limit is for very large machines and the 20% limit is for small machines. The calculation of the synchronous, transient, and subtransient reactances based on numerical field calculations is addressed in [50]. Synchronous machines up to 10 MW are used in variable-speed drives for wind-power plants. Enercon [51] uses synchronous generators with a large pole number; this enables them to avoid any mechanical gear within the wind-power train. The AC output is rectified and an inverter supplies the wind power to the grid. Hydrogenerators are another example where a large number of poles are used.

4.2.12 Switched-Reluctance Machine

Switched-reluctance machines [52–55] have the advantage of simplicity: such machines consist of a salient pole (solid) rotor member without any winding and the stator member carries concentrated coils that are excited by a solid-state converter. The only disadvantages of such machines are the encoder required to sense the speed/position of the rotor and the acoustic noise emanating form the salient-pole rotor. The windage losses of such machines can be considerable. Figure 4.19a,b,c,d illustrates how the field of a switched-reluctance machine changes as a function of the rotor position.

4.2.13 Some Design Guidelines for Synchronous Machines

In order for rotating machines to work properly and not to cause power quality problems, limits for maximum flux and current densities must be met so that neither excessive saturation nor heating occurs. These limits are somewhat different for induction (see Chapter 3) and synchronous machines.

4.2.14 Winding Forces during Normal Operation and Faults

In power apparatus magnetic forces are often large enough to cause unwanted noise or disabling damage. For instance, in [56] the problem of stator bar vibrations causing mechanical wear in hydrogenerators is addressed. Useful methods for reducing the vibrations, which are caused by magnetic forces, are suggested. Along with the suggestions a force calculation formula proposed in 1931 is given. The surface integral method recommended in this section is an improvement over such formulas since it uses magnetic field calculation, which accounts for magnetic saturation of the iron cores.

4.2.14.1 Theoretical Basis

Several approaches may be taken to arrive at the expressions necessary for the surface integral force calculations. Using a stress of Maxwell and energy considerations, Stratton [57] formally derives this expression for the total force on a nonferromagnetic body:

$\overrightarrow{F}={\displaystyle \underset{S1}{\int }[\mu \overrightarrow{H}\left(\overrightarrow{H}\cdot \overrightarrow{n}\right)-\frac{\mu }{2}{H}^2\overrightarrow{n}}]\mathrm{d}a.$

  (4-35)

In Eq. 4-35 S₁ is an arbitrary surface surrounding the body. Equivalent expressions are given by Carpenter [58]. He begins by assuming that “any distribution of poles or currents, or both, which, when put in place of a piece of magnetized iron, reproduces the magnetic-field condition at all points outside the iron, must experience the same total mechanical force as a part which it replaces.” One special distribution consists of poles and currents existing on a surface surrounding the body of interest. The force on the body is thus modeled by the force densities

$F_t={\mu }_o{H}_n{H}_t\text{,}$

  (4-36a)

$F_n=\frac{1}{2}{\mu }_o\left(H_n^{\mathit{2}}\mathit{–}{H}_t^2\right)\text{,}$

  (4-36b)

in the tangential and normal directions, respectively. Surface integrals of these densities then give the total force on the object.

A third treatment of the subject is made by Reichert [59]. He speaks of a quantity p, the surface stress. Integration of stress,

$\overrightarrow{p}=\frac{1}{{\mu }_o}\left(\overrightarrow{n}\cdot \overrightarrow{B}\right)\overrightarrow{B}-\frac{1}{2{\mu }_o}{B}^2\overrightarrow{n}\text{,}$

  (4-37)

is said to give the total force on the body. This force calculation is equivalent to Stratton’s and Carpenter’s. However, Reichert goes on to adapt the stress to numerical methods.

Required characteristics of the above integration surfaces, which are clarified by Reichert, turn out to limit the possible application of the method. The necessary characteristics are

• Surface must be in air;

• An exception may be made to the above rule by creating an artificial air gap. This may be done if the artificial air gap does not cause a significant change in the magnetic field; and

• Surface may be any that fully encloses the body and follows the above rules.

Especially note that only the total force or torque acting on the body is given by the integration. No information can be obtained about the actual distribution of the force within the body. Figure 4.20 shows forces calculated with the above method.

4.3.1 Model of a Synchronous Machine as Applied to Harmonic Power Flow

A well-known harmonic model of a synchronous machine is based on the negative-, positive-, and zero-sequence reactances [60–62]

$\begin{array}{l}{X}_1\left(h\right)=jh{X}_1\\ X_2\left(h\right)=jh{X}_2,\\ X_o\left(h\right)=jh{X}_0\end{array}$

where h is the order of harmonics and X₁, X₂, X₀ are positive-, negative-, and zero-sequence reactances of the machine at fundamental frequency, respectively.

Losses may be included by adding a resistor, as shown in Fig. 4.21. This model of a synchronous machine is used in harmonic (balanced) power flow analysis.

4.3.1.1 Definition of Positive-, Negative-, and Zero-Sequence Impedances/Reactances

Symmetrical components [10] rely on positive-, negative- and zero-sequence impedances. Figure 4.22 gives a summary for the definition of these components [62]. Figure 4.22a represents the three-phase circuit and Fig. 4.22b shows the per-phase equivalent circuit for the positive sequence. Figures 4.22c,d detail the circuits for the negative components, and Figs. 4.22e,f represent the circuits for the zero-sequence components.

4.3.2 Synchronous Machine Harmonic Model Based on Transient Inductances

The simple model of Fig. 4.21 is inappropriate for harmonic modeling of a synchronous machine when detailed information (e.g., harmonic torque) must be extracted. To improve this model three cases are considered, where the synchronous machine is fed by:

• harmonic currents,

• harmonic voltages, and

• combination of current and voltage harmonics (as expected in actual power systems).

Current-Source (Containing Harmonics) Fed Synchronous Machine

The stator voltages that result when balanced harmonic currents of order h are applied to a three-phase synchronous machine are determined. The synchronous machine has a field winding in the d-axis residing on the round rotor and no subtransient properties. In complex exponential notation the applied currents are

$\begin{array}{l}{i}_a=i_{h}expj\left[h\omega t+\angle i_h\right]\\ i_b=i_{h}expj\left[h\omega t+\angle i_h\mp 2\pi /3\right],\\ i_c=i_{h}expj\left[h\omega t+\angle i_h\pm 2\pi /3\right]\end{array}$

  (4-40)

where the upper signs in Eq. 4-40 apply to the positive-phase sequence. Expressions for voltage, current, and flux may be obtained by taking the real part of each relevant complex equation. The a-phase stator voltage is according to [60]

$\begin{array}{l}{v}_a=i_{h}jh\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)expj\left[h\omega t+\angle i_h\right]+i_h\left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)\\ j\left[\left(h\mp 2\right)\omega expj\left(h\mp 2\right)\omega t+\angle i_h\mp 2\alpha \right].\end{array}$

  (4-41)

Using a similar analysis the b-phase voltage is

$\begin{array}{l}{v}_b=i_{h}jh\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)expj\left[h\omega t\mp 2\pi /3+\angle i_h\right]\\ +i_{h}j\left(h\mp 2\right)\omega \left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)expj[\left(h\mp 2\right)\omega t\mp \\ 2\alpha \pm 2\pi /3+\angle i_h]\end{array}$

  (4-42)

Comparing Eqs. 4-40, 4-41 and 4-42 it can be noted that the additional harmonic voltage term has the opposite phase sequence than that of the applied current. For example, if the applied current of order h = 7 has a positive-phase sequence, then the additional harmonic voltage component has the order 5 with a negative-phase sequence. This is the usual situation for a balanced system. If, however, the applied current of order h = 7 has negative-phase sequence, then the additional harmonic voltage component has the order 9 with a positive-phase sequence, that is, not a zero-sequence as is commonly expected. Consequently, the phase voltage set will not be balanced. The analysis is also valid for the application of a negative-sequence current of fundamental frequency; in this case the additional component has the harmonic order 3 and has positive-phase sequence [63].

Voltage-Source (Containing Harmonics) Fed Synchronous Machine

If a balanced set of harmonic voltages is applied to a synchronous machine, an additional harmonic current with harmonic orders $\left(h\mp 2\right)$ – based on the results of the prior section – can be expected. The current solution is

$\begin{array}{cc}\hfill \begin{array}{l}{i}_a=i_{h}expj\left[h\omega t+\angle i_h\right]+{\overline{i}}_{h}expj\left[\left(h\mp 2\right)\omega t+\angle {\overline{i}}_h\right]\\ i_b=i_{h}expj\left[h\omega t+\angle i_h\mp 2\pi /3\right]+{\overline{i}}_{h}expj\left[\left(h\mp 2\right)\omega t+\angle {\overline{i}}_h\pm 2\pi /3\right]\\ i_c=i_{h}expj\left[h\omega t+\angle i_h\pm 2\pi /3\right]+{\overline{i}}_{h}expj\left[\left(h\mp 2\right)\omega t+\angle {\overline{i}}_h\pm 2\pi /3\right]\end{array}\hfill & .\hfill \end{array}$

  (4-43)

The quantities with a bar (^–) are the current components with the frequency $\left(h\mp 2\right)\omega$. The phase sequence of these components is the opposite of that of the applied voltage with frequency hω. Using the results of the prior section, the a-phase voltage is

$\begin{array}{cc}\hfill \begin{array}{l}{v}_a=i_{h}jh\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)expj\left[h\omega t+\angle i_h\right]+i_{h}j\left(h\mp 2\right)·\\ \omega \left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)expj\left[\left(h\mp 2\right)\omega t\mp 2\alpha +\angle i_h\right]\\ +{\overline{i}}_{h}j\left(h\mp 2\right)\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)expj\left[\left(h\mp 2\right)\omega t+\angle {\overline{i}}_h\right]\\ +{\overline{i}}_{h}jh\omega \left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)expj\left[h\omega t\mp 2\alpha -\angle {\overline{i}}_h\right]\end{array}\hfill & \hfill .\hfill \end{array}$

  (4-44)

For an applied voltage

$v_a=v_{h}exp\mathit{jh\omega t}.$

  (4-45)

Comparing Eqs. 4-44 and 4-45 results in

$\begin{array}{l}{v}_h=i_{h}jh\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)+{\overline{i}}_{h}jh\omega \left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)·\\ \angle i_h=-\pi /2\\ 0=+i_h\left(h\mp 2\right)\omega \left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)+{\overline{i}}_h\left(h\mp 2\right)\omega \left(\frac{L_d^{\prime }+L_q^{\prime }}{2}\right)·\end{array}$

  (4-46)

$\angle {\overline{i}}_h=\mp 2\alpha +\angle i_h.$

  (4-47)

Eqs. 4-46 and 4-47 may be rearranged to give

$v_h=jh\omega i_h\left(\frac{2L_d^{\prime }{L}_q^{\prime }}{L_d^{\prime }+L_q^{\prime }}\right).$

  (4-48)

Comparing Eqs. 4-42 and 4-48, one notes that the effective inductance at the applied frequency is different for the voltage- and current-source-fed machines.

If harmonic voltages are applied at both frequencies hω and $\left(h\mp 2\right)\omega$, then Eqs. 4-46 and 4-47 can be modified by the inclusion of ${\overline{v}}_h$. The currents at both frequencies will be affected.

Synchronous Machines Fed by a Combination of Harmonic Voltage and Current Sources

When a synchronous machine is subjected to a harmonic voltage disturbance at frequency hω, harmonic-current components are drawn at hω and at the associated frequency $\left(h\mp 2\right)\omega$. The upper sign applies to the positive- and the lower sign to the negative-phase sequence. Current flow at the associated frequency occurs because the machine is a time-varying electrical system. In a similar manner, a synchronous machine fed by harmonic currents at frequency hω develops a voltage across its terminal at both the applied and the associated frequency $\left(h\mp 2\right)\omega$. Consequently the machine cannot be modeled by impedances defined by a single harmonic frequency [60].

In the general case, a machine will be subjected to applied voltages v_h and ${\overline{v}}_h$ via system impedances Z_h and ${\overline{Z}}_h$ as shown in Fig. 4.23.

Now a linearized Thevenin model can be employed for the system (left-hand side of Fig. 4.23). The applied harmonic voltages v_h and ${\overline{v}}_h$ are assumed to have opposite phase sequences: one is of positive and the other of negative sequence. This is the condition that would apply if the distorting load responsible for the harmonic disturbance drew balanced currents. Consequently, a single-phase model may be used. Voltage equations for the general case are obtained by applying a voltage mesh equation at both frequencies hω and $\left(h\mp 2\right)\omega$. The voltage components applying across the machine are those given by Eq. 4-44 in response to the currents given by Eq. 4-43. Voltage equations suggest the equivalent circuit shown in Fig. 4.24. The interaction between the two sides of the circuit is represented by a mutual inductance, which is a simplification. This mutual inductance is nonreciprocal, and the frequency difference between the two sides is ignored. This model may be used to calculate harmonic current flow.

The results presented here show that a synchronous machine cannot be modeled by one impedance (Fig. 4.21) if voltage sources have voltage components with several frequencies. In the case where one harmonic voltage source is large and the other small then v_h ≈ 0, and the apparent machine impedance to current flow is linear and time-varying:

$Z_h=\frac{jh\omega \left(L_d^{\prime }+L_q^{\prime }\right)}{2}+\frac{\left(h\mp 2\right)h{\omega }^2{\left(\frac{L_d^{\prime }-L_q^{\prime }}{2}\right)}^2}{{\xi }_h}.$

  (4-49)

If several harmonic frequency sources exist, the total current flow can be obtained by the principle of superposition when applied to each harmonic h and its associated harmonics $\left(h\mp 2\right)\omega$. Because of the variation of the flux-penetration depth with frequency the governing machine impedances will be a function of frequency.

4.3.3 Synchronous Machine Model with Harmonic Parameters

A synchronous machine model based on modified dq0 equations – frequently used in harmonic load flow studies – is presented next [61]. From the physical interpretation of this model one notices that this machine equivalent circuit is asymmetric. This model is then incorporated with the extended “decoupling-compensation” network used in harmonic power flow analysis, and it is now extended to problems including harmonics caused by the asymmetry of transmission lines, asymmetry of loads, and nonlinear elements like static VAr compensators (SVCs). This synchronous machine model has the form

$\begin{array}{l}{I}_1\left(h\right)=y_{11}\left(h-1\right)V_1\left(h\right)+\mathrm{\Delta }{I}_1\left(h\right)\\ \mathrm{\Delta }{I}_1\left(h\right)=y_{12}\left(h-1\right)V_2\left(h-2\right)\\ I_2\left(h\right)=y_{22}\left(h+1\right)V_2\left(h\right)+\mathrm{\Delta }{I}_2\left(h\right)\\ \mathrm{\Delta }{I}_2\left(h\right)=y_{21}\left(h+1\right)V_1\left(h+2\right)\\ I_o\left(h\right)=y_{00}\left(h\right)V_0\left(h\right)\end{array}$

In these equations, the subscripts 1, 2, 0 identify positive-, negative-, and zero-sequence quantities, respectively; y₁₁(h), y₁₂(h), y₂₁(h), y₂₂(h), and y₀₀(h) are harmonic parameters of the machine obtained from the dq0-equation model. In accordance with Eq. 4-50 the equivalent circuits of the machine can be assembled based on the different sequence networks as depicted in Fig. 4.25. This is a decoupled harmonic model since ΔI₁(h) and ΔI₂(h) are decoupled-harmonic current sources. Equivalent circuits of different sequences and different harmonic orders are decoupled and can be incorporated separately with the corresponding power network models.

The physical interpretation of this harmonic-decoupled model is that

• The hth order positive-sequence current flowing in the machine armature is not entirely caused by the hth order positive-sequence voltage, but will generate in addition a (h – 2)th order negative-sequence voltage at the machine terminal;

• The hth order negative-sequence current flowing in the machine armature is not only caused by the same order negative-sequence voltage, but will also generate a (h + 2)th order positive-sequence voltage at the machine terminal; and

• The zero-sequence harmonic currents are relevant to the same order zero-sequence harmonic voltages only.

Such a phenomenon, stemming from the asymmetry of the machine rotor, is depicted in Fig. 4.26.

The reason why there is always a difference of the order between the positive- and negative-sequence harmonic current components and their corresponding admittances in Eq. 4-50 is because these admittances are transformed by the dq0 equations. The reference frame of the dq0 components is attached to the rotor, while the reference frame of symmetrical components remains stationary in space.

This mathematical model of a synchronous machine (Fig. 4.25) can be used in harmonic load flow studies to investigate the impact of a machine’s asymmetry and nonlinearity. Moreover, decoupled equivalent circuits of different sequences and harmonics can be incorporated separately with the corresponding power network models.

4.3.3.1 Application Example 4.10 with its Solution: Harmonic Modeling of a 24-Bus Power System with Asymmetry in Transmission Lines

The 24-bus sample system shown in Fig. E4.10.1 is used in this example [61]. Parameters of the system and generators are provided in [61,64]. Simulation results will be compared for the simple harmonic model (Fig. 4.21, Eq. 4-38) and the model with harmonic parameters (Fig. 4.24, Eq. 4-50).

Two cases are investigated, that is, harmonics caused by the asymmetry of transmission lines (or loads) and by nonlinear SVC devices (see Application Example 4.11). Harmonic admittances of synchronous machines are computed and compared in [61]. They remain unchanged with the change of external network conditions. The imaginary components of admittances for the two models are very similar. This may be one of the reasons why the simple model is frequently employed.

All transmission lines (although transposed) are asymmetric, whereas all loads are symmetric. Decoupled positive- and negative-harmonic current sources of the equivalent circuit (ΔI₁(h) and ΔI₂(h) in Fig. 4.24 and Eq. 4-50) are computed and listed in [61] for generators 1, 6, and 8. The harmonic frequency voltage components at buses 1, 12, and 16 are also given in the same reference.

4.3.4 Synchronous Machine Harmonic Model with Imbalance and Saturation Effects

This section introduces a synchronous machine harmonic model [12] that incorporates both frequency conversions and saturation effects under various machine load-flow constraints (e.g., unbalanced operation). The model resides in the frequency domain and can easily be incorporated into harmonic load-flow programs.

To model these effects and also to maintain an equivalent-circuit representation, a three-phase harmonic Norton equivalent circuit (Fig. 4.27) is developed with the following equations [12]:

$\left[I_{\mathit{km}}\left(h\right)\right]=\left[Y\left(h\right)\right]\left(\left[V_h\left(h\right)\right]-\left[V_m\left(h\right)\right]-\left[E\left(h\right)\right]\right)+\left[I_{\mathit{nl}}\left(h\right)\right]\text{,}$

$\begin{array}{l}\left[E_{\left(h\right)}\right]=0,i\mathrm{f}h\ne 1\\ \left[I_{\mathit{km}}\right]={\left[I_{\mathit{km}-a}{I}_{\mathit{km}-b}{I}_{\mathit{km}-c}\right]}^T\\ \left[V_k\right]={\left[V_{k-a}{V}_{k-b}{V}_{k-c}\right]}^T\\ \left[V_m\right]={\left[V_{m-a}{V}_{m-b}{V}_{m-c}\right]}^T\\ \left[E\right]={\left[E_p{a}^2{E}_{p}a{E}_p\right]}^T\\ \left[I_{\mathit{nl}}\right]={\left[I_{\mathit{nl}-a}{I}_{\mathit{nl}-b}{I}_{\mathit{nl}-c}\right]}^T\\ a=exp\left(-j2\pi /3\right)\end{array}$

where h is the harmonic order. In the model of Fig. 4.27, nonlinear effects are represented by a harmonic current source ([I_nl(h)]) that includes the harmonic effects stemming from frequency conversion ([I_f(h)]) and those from saturation ([I_s(h)]):

$\left[I_{\mathit{nl}}\left(h\right)\right]=\left[I_f\left(h\right)\right]+\left[I_S\left(h\right)\right].$

This harmonic model reduces to the conventional, balanced three-phase equivalent circuit of a synchronous machine if [I_nl(h)] is zero and [Y(h)] is computed according to reference [65]. Since [I_nl(h)] is a known current source (determined from the machine load-flow conditions), the model is linear and decoupled from a harmonic point of view. Therefore, it can be solved with network equations in a way similar to the traditional machine model.

To include the usual load-flow conditions, the following constraints are specified at the fundamental frequency (h = 1):

• Slack machine: The constraints are the magnitude and the phase angle of positive-sequence voltage at the machine terminals:

$\begin{array}{l}\left[T\right]\left(\left[V_k\right]-\left[V_m\right]\right)=V_{\mathrm{specified}}\\ \left[T\right]=\left(1/3\right)\left[1a{a}^2\right].\end{array}$

• PV machine: The constraints are the three-phase real power output and the magnitude of the positive-sequence voltage at machine terminals, where superscript H denotes conjugate transpose:

$\mathrm{Real}\left\{-I_{\mathit{km}}]{}^H\left(\left[V_k\right]-\left[V_m\right]\right)\right\}=P_{\mathrm{specified}}\text{,}$

$\left|\left[T\right]\left(\left[V_k\right]–\left[V_m\right]\right)\right|=V_{\mathrm{specified}}.$

• PQ Machine: The constraints are the three-phase real and the three-phase reactive power outputs:

$–{\left[I_{\mathit{km}}\right]}^H\left(\left[V_k\right]–\left[V_m\right]\right)={\left(P+jQ\right)}_{\mathrm{specified}}.$

These constraint equations and the machine structure equation (Eq. 4-51a) jointly define a three-phase machine model with nonlinear effects. Combining these equations with other network equations leads to multiphase load flow equations for the entire system [65]. The system is then solved by the Newton–Raphson method for each frequency. Detailed analysis and demonstration of Newton–Raphson based harmonic load flow is presented in Chapter 7.

In the model of Fig. 4.27, [I_nl(h)] is not known and will be computed using the following iterative procedure:

• Step 1. Initialization
Set the harmonic current source [I_nl(h)] to zero.

• Step 2. Network Load-Flow Solution
Replace the synchronous machines with their harmonic Norton equivalent circuits (Fig. 4.27) and solve the network equations for the fundamental and harmonic frequencies.

• Step 3. Harmonic Current-Source Computation Use the newly obtained network voltages and currents (from Step 2) to update [I_nl(h)] for the machine equivalent circuit.

• Step 4. Convergence
If the computed [I_nl(h)] values are sufficiently close to the previous ones stop; otherwise, go to Step 2.

The network load-flow solution and the harmonic iteration processes can be performed by any general purpose harmonic load-flow program as explained in Chapter 7 and reference [65]. Therefore, [I_nl(h)] needs to be computed (Step 3). This will require the machine harmonic models in the dq0 and abc coordinates.