4.3.5 Static- and Dynamic-Rotor Eccentricities Generating Current and Voltage Harmonics

Most synchronous machines have relatively large radial air-gap lengths, which range from a few millimeters to tens of centimeters. This is so because the synchronous reactances (X_d, X_q) must be small enough to guarantee a good dynamic (transient, subtransient) behavior. Although permanent-magnet machines have a mechanical air gap of less than one millimeter up to a few millimeters, their electrical air gap is always larger than one millimeter because the radial air-gap height (length) of the permanent magnet behaves magnetically like air with the permeability of free space μ_o because the recoil permeability μ_R of excellent permanent-magnetic material (e.g., NdFeB) is about μ_R ≈ 1.05μ_o. Therefore, an imperfect mechanical mounting – resulting in a static-rotor eccentricity – or a bent shaft – resulting in dynamic-rotor eccentricity – will not significantly alter the magnetic coupling of the stator windings and the rotating magnetic flux will be about circular. Here one should mention that permanent-magnet machines cannot be easily assembled for ratings larger than a few tens of kilowatts because a mounting gear will be required, which maintains a uniform concentric air gap during the mounting of the rotor.

The modified winding-function approach (MWFA) accounting for all space harmonics has been used for the calculations of machine inductances under rotor eccentricities [69]. Machines can fail due to air-gap eccentricity caused by mechanical problems including the following:

• shaft deflection,

• inaccurate positioning of the rotor with respect to the stator,

• worn bearings,

• stator-core movement, and

• bent rotor shaft.

Electrical asymmetries contribute to harmonic generation such as

• Damper windings (amortisseur) are incomplete, that is, the amortisseur has a different construction between poles (q-axis) as compared with that on the pole d-axis; or

• Eddy currents in the pole faces of salient-pole machines contribute to subtransient response;

• Heating of the rotor due to nonuniform amortisseur and nonuniform eddy-current generation around the rotor circumference may cause shaft bending/deflection; and

• Incomplete shielding of the field winding of a machine fitted with nonuniform amortisseurs and subjected to harmonic disturbance.

Nonsinusoidal air-gap flux wave effects – due to any of the above-listed reasons – can be accounted for by representing the mutual inductances between the field and stator windings, and those between the stator windings, by trigonometric series as a function of the rotor angle θ. It is found that the inclusion of nonsinusoidal inductance variation causes a broad spectrum such that injection of a single-harmonic current results in voltages containing all odd harmonic orders including the fundamental. The extent to which the energy is spread across the spectrum depends on the pole shape, the numbers of stator and rotor slots, and the amortisseur design. The spectral spread is a second-order effect compared to the generation of the associated voltage at frequency (h ± 2). Harmonic behavior is also affected by nonsinusoidal inductance variation with rotor position and by the extent to which induced damper winding currents shield the field winding from the harmonic-gap flux waves. Rotor eccentricity in machines can be of two forms:

• static air gap eccentricity, and

• dynamic air gap eccentricity.

With the static air-gap eccentricity, the position of the minimum radial air-gap length is fixed in space and the center of rotation and that of the rotor are the same. Static eccentricity can be caused by oval stator cores or by the incorrect relative mounting of stator and rotor. In case of dynamic air-gap eccentricity, the center of rotation and that of the rotor are not the same and the minimum air gap rotates with the rotor. Therefore, dynamic eccentricity is both time and space dependent. Dynamic eccentricity can be caused by misalignment of bearings, mechanical resonance at critical speeds, a bent rotor shaft, and wear of bearings [69].

MATLAB/Simulink [70] can be used for the simulation of machine variables. All machine inductances employed for the simulation are expressed in their Fourier series based on the MWFA. Two different cases, namely, a noneccentric and an eccentric rotor case, for this analysis are investigated. In the first case, it is assumed that dynamic air-gap eccentricity is not present, and in the second one 50% dynamic air-gap eccentricity is introduced to investigate the effect of the eccentric rotor on stator-current signatures. To analyze the stator-current signatures of these two cases, FFT (fast Fourier transform) of the current signals are performed for all cases. The stator currents show the existence of the 5th (300 Hz), 7th (420 Hz), 11th (660 Hz), 13th (780 Hz), 17th (1020 Hz), and 19th (1140 Hz) harmonics even when the rotor is not eccentric. Implementing 50% eccentricity will cause the stator and rotor current induced harmonics to increase when compared to those generated without rotor eccentricity as is listed in Table 4.3.

The stator current harmonics exist because the interaction of the magnetic fields caused by both the stator and rotor windings will produce harmonic fluxes that move relative to the stator and they induce corresponding current harmonics in the stationary stator windings. These harmonic fluxes in the air gap will increase as the rotor dynamic eccentricity increases, and consequently the current harmonic content increases too. The 3rd harmonic and its multiples are assumed not to exist in the stator windings because it is a three-phase system with no neutral connection. However, third-harmonic components are induced in the rotor-field winding. Because the stator and rotor-current signatures of synchronous machines have changed, either or both signatures can be utilized for detecting dynamic eccentricity. However, it is more practical to implement the stator-current signature analysis for the condition monitoring than the rotor-current signature analysis because some synchronous machines have brushless excitation [49], which will make the rotor current inaccessible. In summary, one can conclude that synchronous machines are not very prone to the effects caused by rotor eccentricity due to their large air gaps.

4.3.6 Shaft Flux and Bearing Currents

The generation of shaft flux [71] due to mechanical and magnetic asymmetries as well as solid-state switching is an important issue for large synchronous generators. Bearing currents are predominantly experienced in PWM AC drives [72–76] due to the high-frequency switching ripples resulting in bearing electroerosion. Techniques for measurement of parameters related to inverter-induced bearing currents are presented in [77].

4.3.7 Conclusions

The most important conclusions of this section are:

• Synchronous machines generate harmonics due to

1. frequency conversion,

2. saturation, and

3. unbalanced operation.

• Current harmonics due to nonlinear loads should be limited based on IEEE-519 [78].

• Unbalanced load-flow analysis without taking into account harmonics is not correct due to the ambiguous value of the negative-sequence impedance of synchronous machines. Correct results can only be obtained by including harmonics in a three-phase, load-flow approach.

• Only 3rd and 5th harmonics need to be included for the machine’s representation in harmonic analysis. The 3rd harmonic is of the negative-sequence and not of the zero-sequence type. Hence, it cannot be eliminated by Δ transformer connections. The same is true for transformer applications with DC current bias [79].

• Machine saturation can have noticeable effects on the harmonic distribution. These effects are more significant with respect to the harmonic phase angles.

• Rotor eccentricity has a minor influence on harmonic generation due to the large air gap of synchronous machines.

• Shaft fluxes and bearing currents can have detrimental impacts.

A nonsalient-pole synchronous machine has the following data: X_s = (X_d = X_q) = 1.8 pu, R_a = 0.01 pu, I_phase = 1.0 pu and V_L-N = 1.0 pu.

a) Sketch the equivalent circuit using the consumer reference system.

b) Draw the phasor diagram (to scale) based on the consumer reference system for a lagging (underexcited) displacement power factor of cos φ = 0.90.

c) Sketch the equivalent circuit based on the generator reference system.

d) Draw the phasor diagram (to scale) employing the generator reference system for a lagging (overexcited) displacement power factor of cos φ = 0.90.

Problem 4.2: Steady-State Analysis of a Salient-Pole Synchronous Machine Using Phasor Diagram [7]

A salient-pole synchronous machine has the following data: X_d = 1.8 pu, X_q = 1.5 pu, R_a = 0.01 pu, I_phase = 1.0 pu, V_L-N = 1.0 pu.

a) Sketch the equivalent circuit based on the consumer reference system.

b) Draw the phasor diagram (to scale) using the consumer reference system for a lagging (underexcited) displacement power factor of cosφ = 0.70.

c) Sketch the equivalent circuit based on the generator reference system.

d) Draw the phasor diagram (to scale) employing the generator reference system for a lagging (overexcited) displacement power factor of cosφ = 0.70.

Problem 4.3: Phasor Diagram of a Synchronous Generator with Two Displaced Stator Windings [80]

Synchronous machines with two by 30° displaced stator windings (winding #1 and winding #2) are used to improve the voltage/current wave shapes so that the output voltages of a generator become more sinusoidal. This is similar to the employment of 12-pulse rectifiers versus 6-pulse rectifiers. To accomplish this, stator winding #1 feeds a Y–Y connected transformer and stator winding #2 supplies a Δ–Y connected transformer as illustrated in Fig. P4.3. For a S = 1200 MVA, V_L-L = 24 kV, f = 60 Hz, p = 2 turbogenerator the machine parameters are in per unit (pu): X_d = 1.05, X_12d = 0.80, X_q = 0.95, X_12q = 0.73, X_ffd = 2.10, k_o = 0.5, k₁ = 0.638, r = R_a = 0.00146, r_f = R_f = 0.0007.

Draw the phasor diagram (to scale) for rated operation and a displacement power factor of cosφ = 0.8 lagging (overexcited, generator reference system).

Problem 4.4: Balanced Three-Phase Short-Circuit Currents of a Synchronous Machine Neglecting Influence of Amortisseur

A three-phase synchronous generator is initially unloaded and has a rated excitation so that E_f = 1.0 pu. At time t = 0, a three-phase short-circuit is applied. The machine parameters are direct (d) axis: X_d = 1.2 pu, X_f = 1.1 pu, X_m = 1.0 pu, R_a = 0.005 pu, R_f = 0.0011 pu; quadrature (q) axis: X_q = 0.8 pu, X_m = 0.6 pu, R_a = 0.005 pu.

a) Plot the first-half cycle of the torque T_{3phase_short-circuit} due to the three-phase short-circuit neglecting damping and assuming that there is no amortisseur.

b) At which angle φ_max (measured in degrees) occurs the maximum torque T_max (measured in per unit)?

Problem 4.5: Torque T_L-L and Induced Voltage e_a During the First-Half Cycle of a Line (Line b)-to-Line (Line c) Short-Circuit of a Synchronous Machine Neglecting Influence of Amortisseur

a) For the machine parameters of Problem 4.4 calculate and plot the line-to-line torque T_L-L for the first half-cycle when phases b and c are short-circuited, and determine at which angle φ_max (measured in degrees) the maximum torque (measured in per unit) occurs.

b) Plot the induced open-circuit voltage of phase a, that is, e_a (t) (measured in per unit) as a function of the phase angle φ (measured in degrees) for the first half-cycle.

Problem 4.6: Line-to-Neutral Short-Circuit Current of a Synchronous Machine Neglecting Influence of Amortisseur

a) For the machine parameters of Problem 4.4 calculate the total (fundamental and harmonics) line-to-neutral short-circuit current i_a for the first half-cycle when phase a and the neutral are short-circuited, and determine at which angle φ_max (measured in degrees) the maximum current i_{L-N_max} (measured in per unit) occurs.

b) Plot the fundamental short-circuit current of phase a i_a1 (measured in per unit) for the first half-cycle as a function of the phase angle φ (measured in degrees).

Problem 4.7: Out-of-Phase Synchronization Torque of a Synchronous Machine Neglecting Influence of Amortisseur

a) For the machine parameters of Problem 4.4 calculate and plot for the first half-cycle the out-of-phase synchronizing torque.

b) Determine at which angle φ_max (measured in degrees) the maximum torque (measured in per unit) occurs.

Problem 4.8: Synchronizing and Damping Torques of a Synchronous Machine Neglecting Influence of Amortisseur

For the machine parameters of Problem 4.4 calculate and plot for the first half-cycle the steady-state torque T₀, the synchronizing torque T_S, and the damping torque T_D. The per-unit transient time constant of the d-axis is T_d′ = 242 pu and the per unit voltages V_pu = V₀ = V_l-n = V_l-n0 = 1.0 pu may be used.

Problem 4.9: Starting and Accelerating Torque of a Synchronous Machine Developed by Induction Motor Action Neglecting Influence of Amortiseur

The starting of synchronous generators without amortisseur (damper) winding by induction-motor action is not advisable, as will be demonstrated in this problem. Calculate for a synchronous machine with the parameters X_d = 1.2 pu, X_q = 0.8 pu, X_f = 1.1 pu, X_m = 1.0 pu, R_a = 0.005 pu, and R_f = 0.0011 pu the starting and accelerating torque T_starting as a function of the slip s. To do this, start out from the dq0-coordinate system and transform it to the fbo [81] coordinate system. Table P4.9.1 presents some of the results. It is the task of this problem to fill in the missing values of this table.

As can be seen from the few entries of this table, a synchronous machine cannot develop any significant induction motor torque without amortisseur winding. Most large synchronous generators have a very rudimentary amortisseur and must not be started by induction-motor action. Small synchronous machines with ratings of less than 1 MW can be equipped with a full amortisseur winding and can be started by induction-motor action.

Problem 4.10: Design of a Brushless DC (Permanent-Magnet) Motor Drive [24]

For a hybrid automobile drive a permanent-magnet motor – as a part of a brushless DC machine – is to be analyzed with the following data:

An encoder is used to sense the rotor speed/position. For your design it will be helpful to rely on the flux patterns and machine cross section of Fig. P4.10a,b. It is recommended to use as permanent-magnet material NdFeB.

a) Design the stator winding.

b) What is the rated synchronous speed n_{s_rat}?

c) Compute the (maximum) height (radial one-sided length) of the permanent magnets h_{mag_max}.

d) Calculate the maximum flux density of this machine.

e) Determine the leakage inductance L_ℓ (from stator phase a to stator phase b) and the winding resistance R_a of one stator phase.

f) Determine the current density J_stator, provided the stator winding copper fill factor is k_cu = 0.7, and the average stator-slot width is about the same as the average stator-tooth width.

Problem 4.11: Fourier Analyses of Terminal Currents of a Brushless DC Motor (Permanent-Magnet Motor with Inverter)

Find the Fourier coefficients of the current wave shapes of Fig. P4.11 that represent the input MOSFET current (upper wave shape) and the current of a brushless DC motor (lower wave shape). Select a certain sampling interval, e.g., Δt = 0.025 ms, perform a Fourier analysis, and show that for a minimum approximation error the Nyquist criterion is satisfied.

Problem 4.12: Steady-State Frequency Variation within an Interconnected Power System as a Result of Two Load Changes [82]

A block diagram of two interconnected areas of a power system (e.g., area #1 and area #2) is shown in Fig. P4.12. The two areas are connected by a single transmission line. The power flow over the transmission line will appear as a positive load to one area and an equal but negative load to the other, or vice versa, depending on the direction of power flow.

a) For steady-state operation show that with Δω₁ = Δω₂ = Δω the change in the angular velocity (which is proportional to the frequency f) is

$\mathrm{\Delta }\omega =\mathrm{function}\mathrm{of}\left(\mathrm{\Delta }{P}_{L1}\mathit{,}\mathrm{\Delta }{P}_{L2}\mathit{,}{D}_1\mathit{,}{D}_2\mathit{,}{R}_1\mathit{,}{R}_2\right)$

and

$\begin{array}{c}\mathrm{\Delta }{P}_{\mathrm{tie}}=\mathrm{function}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{parameters}\mathrm{a}\mathrm{s}\mathrm{in}\mathrm{E}\mathrm{q}.\mathrm{P}12-1.\end{array}$

b) Determine values for Δω (Eq. P12-1), ΔP_tie (Eq. P12-2), and the new frequency f_new, where the nominal (rated) frequency is f_rated = 60 Hz, for these parameters: R₁ = 0.05 pu, R₂ = 0.1 pu, D₁ = 0.8 pu, D₂ = 1.0 pu, ΔP_L1 = 0.2 pu, ΔP_L2 = –0.3 pu.

c) For a base apparent power S_base = 1000 MVA compute the power flow across the transmission line.

d) How much is the load increase in area #1 (ΔP_mech1) and area #2 (ΔP_mech2) due to the two load steps?

e) How would you change R₁ and R₂ in case R₁ is either a wind or solar power plant operating at its maximum power point (and cannot accept any significant load increase due to the two load steps) and R₂ is a coal fired plant that can be overloaded, and serve as a frequency leader?

Problem 4.13: Frequency Control of an Isolated Power Plant (Islanding Operation)

Figure P4.13 illustrates the block diagram of governor, prime mover (steam turbine), and rotating mass and load of a turbogenerator set. For the frequency change (Δω) per generator output power change (ΔP),

$R=\frac{\mathrm{\Delta }\omega }{\mathrm{\Delta }P}\mathrm{p}\mathrm{u}=0.01\mathrm{p}\mathrm{u}\text{,}$

the load change (ΔPL) per frequency change $D=\frac{\mathrm{\Delta }{P}_L}{\mathrm{\Delta }\omega }=0.8\mathrm{p}\mathrm{u}\text{,}$

step load change $\mathrm{\Delta }{P}_L\left(s\right)=\frac{\mathrm{\Delta }{P}_L}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}\mathrm{p}\mathrm{u}\text{,}$

angular momentum of steam turbine and generator set M = 4.5,

base apparent power S_base = 500 MVA,

governor time constant T_G = 0.01 s,

valve charging time constant T_CH = 1.0 s,

load reference set point = 1.0.

a) Derive for Fig. P4.13 Δω_{steady state} by applying the final value theorem. You may assume the load reference set point(s) = 0 and $\mathrm{\Delta }{P}_L\left(s\right)=\frac{\mathrm{\Delta }{P}_L}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}\mathrm{p}\mathrm{u}\text{,}$ where s is the Laplace operator.

b) List the ordinary differential equations and the algebraic equations of the block diagram of Fig. P4.13.

c) Use either Mathematica or MATLAB to establish steady-state conditions by imposing a step function for load reference set point(s) = $\frac{1}{s}\mathrm{p}\mathrm{u}$ and run the program with a zero step-load change ΔP_L = 0 for 5 s. After 5 s impose a step load change of $\mathrm{\Delta }{P}_L\left(s\right)=\frac{\mathrm{\Delta }{P}_L}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}\mathrm{p}\mathrm{u}$ to find the transient response of Δω(t).

d) Use Mathematica or MATLAB to establish steady-state conditions by imposing a step function for load reference set point $\mathrm{\Delta }{P}_L\left(s\right)=\frac{\mathrm{\Delta }{P}_L}{s}\mathrm{p}\mathrm{u}=\frac{–0.2}{s}\mathrm{p}\mathrm{u}$ and run the program with a zero step-load change ΔP_L = 0 for 5 s. After 5 s impose a step load change of $\mathrm{\Delta }{P}_L\left(s\right)=\frac{\mathrm{\Delta }{P}_L}{s}\mathrm{p}\mathrm{u}=\frac{–0.2}{s}\mathrm{p}\mathrm{u}$ to find the transient response of Δω(t).

Problem 4.14 Frequency Control of an Interconnected Power System Broken into Two Areas, each Having one Generator

Figure P4.12 shows the block diagram of two generators interconnected by a tie line (transmission line). Data of generation set (steam turbine and generator) #1:

for frequency change (Δω₁) per generator output power change $\left(\mathrm{\Delta }{\mathit{P}}_1\right):R_1=\frac{\mathrm{\Delta }{\omega }_1}{\mathrm{\Delta }{P}_1}\mathrm{p}\mathrm{u}=0.01\mathrm{p}\mathrm{u}\text{,}$

the load change (ΔP_L1) per frequency change $\left(\mathrm{\Delta }{\omega }_1\right):D_1=\frac{\mathrm{\Delta }{P}_{L1}}{\mathrm{\Delta }{\omega }_1}=0.8\mathrm{p}\mathrm{u}\text{,}$

step-load change: $\mathrm{\Delta }{P}_{L1}\left(s\right)=\frac{\mathrm{\Delta }{P}_{L1}}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}\mathrm{p}\mathrm{u}$

angular momentum of steam turbine and generator set: M₁ = 4.5,

base apparent power: S_base = 500 MVA,

governor time constant: T_G1 = 0.01 s,

valve charging time constant: T_CH1 = 1.0 s,

(load reference set point)₁ = load ref₁(s) = 0.8 pu.

Data of generation set (steam turbine and generator) #2:

for frequency change (Δω₂) per change in generator output power $\left(\mathrm{\Delta }{P}_2\right):R_2=\frac{\mathrm{\Delta }{\omega }_2}{\mathrm{\Delta }{P}_2}\mathrm{p}\mathrm{u}=0.02\mathrm{p}\mathrm{u}\text{,}$

the load change (ΔP_L2) per frequency change (Δω₂): $D_2=\frac{\mathrm{\Delta }{P}_{L2}}{\mathrm{\Delta }{\omega }_2}=1\mathrm{p}\mathrm{u}\text{,}$

step load change: $\mathrm{\Delta }{P}_{L2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{L2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.2}{s}\mathrm{p}\mathrm{u}\text{,}$

angular momentum of steam turbine and generator set: M₂ = 6,

base apparent power: S_base = 500 MVA,

governor time constant: T_G2 = 0.02 s,

valve charging time constant: T_CH2 = 1.5 s,

(load reference set point)₂ = load ref₂(s) = 0.8 pu.

Data for tie line: T = 377/X_tie with X_tie = 0.2 pu.

a) List the ordinary differential equations and the algebraic equations of the block diagram of Fig. P4.12.

b) Use either Mathematica or MATLAB to establish steady-state conditions by imposing a step function for load reference set point(s)₁ = load ref₁ (s) = $\frac{0.8}{s}$ pu, load reference set point (s)₂ = load ref₂ (s) = $\frac{0.8}{s}$ pu and run the program with a zero step-load changes ΔP_L1 = 0, ΔP_L2 = 0 for 10 s. After additional 5 s impose step-load change $\mathrm{\Delta }{P}_{L1}\left(s\right)=\frac{\mathrm{\Delta }{P}_{L1}}{s}\mathrm{p}\mathrm{u}=\frac{0.2}{s}\mathrm{p}\mathrm{u}\text{,}$ and after additional 2 s impose $\mathrm{\Delta }{P}_{L2}\left(s\right)=\frac{\mathrm{\Delta }{P}_{L2}}{s}\mathrm{p}\mathrm{u}=\frac{-0.2}{s}\mathrm{p}\mathrm{u}$ pu to find the transient responses Δω₁(t) and Δω₂(t).

Problem 4.15 Determination of the Sequence-Component Equivalent Circuits and Matrices Z_bus⁽¹⁾, Z_bus⁽²⁾, Z_bus⁽⁰⁾, Line-to-Ground (Neutral), and Line-to-Line Fault Currents at Bus 2 of Fig. P4.15 [83]

Two synchronous generators are connected through three-phase transformers to a transmission line, as shown in Fig. P4.15. The ratings and reactances of the generators and transformers are

1. Generators 1 and 2: 600 MVA, 34.5 kV;

$\begin{array}{c}{X}_d^{\mathit{″}}=X_{1\_\mathrm{g}\mathrm{e}\mathrm{n}}=X_{2\_\mathrm{g}\mathrm{e}\mathrm{n}}=25\%=0.25\mathrm{p}\mathrm{u}\\ X_{\mathrm{o}\_\mathrm{g}\mathrm{e}\mathrm{n}}=10\%=0.1\mathrm{p}\mathrm{u},\mathrm{and}\\ X_{\mathrm{n}\_\mathrm{grounding}\_\mathrm{coil}}=8\%=0.08\mathrm{p}\mathrm{u}.\end{array}$

2. Transformer T₁ rating is 600 MVA, (34.5 kV connected in Δ)/(345 kV connected in Y);
X_{transformer leakage} = 10% = 0.10 pu. Note that the Y of transformer T₁ is grounded.

3. Transformer T₂ rating is 600 MVA, (34.5 kV connected in Δ)/(345 kV connected in Y);
X_{transformer leakage} = 20% = 0.20 pu. Note that the Y of transformer T₂ is not grounded.

4. On a chosen base of S_base = 600 MVA, V_{L-L base} = 345 kV in the transmission-line circuit, the line reactances are X_1line = X_2line = 10% = 0.10 pu and X_0line = 30% = 0.30 pu.

a) Draw each of the three (positive-, negative-, zero-) sequence networks.

b) Determine the zero-, positive-, and negative-sequence bus impedance (reactance) matrices by means of the Z_bus building algorithm, as described in Chapter 8 of [83].

c) Apply the matrix reduction technique by Kron, as outlined in Chapter 7 of [83].

d) Determine numerical value (subtransient) for the line-to-ground fault current when the fault occurs at bus 2 of Fig. P4.15.

e) Determine numerical value (subtransient) for the line-to-line fault current when the fault occurs at bus 2 of Fig. P4.15.

Problem 4.16 Analysis of a Synchronous Machine Feeding a Rectifier/Inverter and Charging a Supercapacitor Bank

One of the major problems in using wind and solar energy is energy storage. Wind or solar energy may not be available when it is needed. A report from a wind farm in New Mexico states that the wind farm could lose as much as 60 MW within a minute. For example, there are several scenarios of how the power change of 60 MW per minute can be mitigated through complementary albeit more expensive power sources: one of them is the combination of a compressed-air power plant with a supercapacitor plant for bridging the time from when the wind power plant output decreases (60 MW per minute) to when the compressed-air energy storage (CAES) plant can take over. A CAES plant requires a start-up time of about 6 minutes. To bridge this 6 minute start-up time required for a 100 MW compressed-air power plant, supercapacitors or ultracapacitors are proposed to provide up to 100 MW during a 6 minute interval amounting to a required energy storage of 10 MWh. Inverters fed from supercapacitors can deliver power within milliseconds to the power system replacing the lost power of 60 MW per minute almost instantaneously. This combination of compressed-air power plant and supercapacitors or ultracapacitors as bridging energy sources can be employed for peak-power operation as well as for improving power quality by preventing short brownouts or blackouts.

Figure P4.16.1 depicts the block diagram of such a supercapacitor plant consisting of wind turbine, mechanical gear, synchronous generator, three-phase rectifier, supercapacitor bank, three-phase inverter, transformer, and power system.

a) Design a controlled three-phase rectifier in the tens of kilowatts range. Figure P4.16.2 shows the three-phase rectifier of Fig. P4.16.1 with six diodes and one self-commutated switch, an insulated-gate bipolar transistor (IGBT). The nominal phase voltages of the generator can be assumed to be v_an = 1200 V sin ωt, v_bn = 1200 V sin(ωt – 120°), and v_cn = 1200 V sin(ωt – 240°). The IGBT is gated with a switching frequency of 3 kHz and you may assume a duty cycle of δ = 50%. C_f1 = 200 μF, L_f1 = 90 mH, C_f2 = 50 μF, ideal diodes D₁ to D₆ and D_f, v_gs = 40 V magnitude, R_sd = R_ss = 10 Ω, C_sd = C_ss = 0.1 μF, C_f = 1000 μF, L_f = 1 mH, and R_load = 10 Ω.

1) Perform a PSpice analysis and plot output voltage v_load(t), generator phase current i_aphase, and the generator line-to-line voltage v_ab.

2) Subject the generator phase current to a Fourier analysis.

3) Determine the displacement power factor of the generator.

4) Determine the total power factor of the generator.

b) Design PWM (pulse-width-modulated) three-phase current controlled voltage-source inverter feeding power into the utility system via Y_grounded/Δ transformer in the tens of kilowatts range. The inverter circuit of Fig. P4.16.3 is to be analyzed with PSpice where the DC voltage is V_DC = 600 V, AC output voltage of the inverter is V_ACL-L = 360 V, and the switching frequency of the IGBTs is f_switch = 7.2 kHz. C_filter = 1000 μF, L_w = 1 mH, R_w = 10 mΩ, L_f = 45 μH, C_f = 10.3 μF, R_f = 10 mΩ, L_syst = 265 μH, R_syst = 50 mΩ. The power system phase voltages are v_agrid(t) = 7200 V sin ωt, v_bgrid(t) = 7200 V sin(ωt – 120°), and v_cgrid(t) = 7200 V sin(ωt – 240°). Note the voltages v_asyst(t), v_bsyst(t), and v_csyst(t) are the power system voltages v_agrid(t), v_bgrid(t), and v_cgrid(t) referred to the Y- or inverter side of the Y/Δ transformer.

1) Determine the transformation ratio of the Y/Δ transformer between inverter and power system; you may assume ideal transformer conditions.

2) Using PSpice calculate and plot input and output voltages and currents of the inverter, provided the output voltage is a sinusoid as given by the power system.

3) Subject the output current of the inverter to a Fourier analysis.

4) Determine the overall costs of this power plant if a specific cost of $4000/kW installed capacity is assumed. Note that a coal-fired plant has a specific cost of $2000/kW installed capacity.

Problem 4.17 Voltage Stress in a Synchronous Motor Winding Fed Through a Cable of 10 m Length by a PWM Current-Source Inverter

Voltage and current ripples due to PWM current-source inverters cause voltage stress in windings of machines. The configuration of Fig. E3.14.1 represents the first two turns of the winding of an induction motor, which is assumed to be identical to the winding of a synchronous motor. Each turn can be represented by four segments having an inductance L_i, a resistance R_i, a capacitance to ground (frame) C_i, and some interturn capacitances C_ij and inductances L_ij. Figure E.3.14.2 represents a detailed equivalent circuit for the configuration of Fig. E3.14.1.

To simplify the analysis neglect the capacitances C_ij and the inductances L_ij. This leads to the circuit Fig. E3.14.3, where the winding is fed by a PWM current source. One obtains the 15 differential equations as listed below.

$\begin{array}{l}\frac{d{v}_1}{dt}=\frac{i_s}{C_1}-\frac{G_1}{C_1}{v}_1-\frac{i_2}{C_1},\\ \frac{d{i}_2}{dt}=-\frac{R_2}{L_2}{i}_2-\frac{v_2}{L_2}+\frac{v_1}{L_2},\\ \frac{d{v}_2}{dt}=\frac{i_2}{C_2}-\frac{G_2}{C_2}{v}_2-\frac{i_3}{C_2},\\ \frac{d{i}_3}{dt}=-\frac{R_3}{L_3}{i}_3-\frac{v_3}{L_3}+\frac{v_2}{L_3},\\ \frac{d{v}_3}{dt}=\frac{i_3}{C_3}-\frac{G_3}{C_3}{v}_3-\frac{i_4}{C_3},\\ \frac{d{i}_4}{dt}=-\frac{R_4}{L_4}{i}_4-\frac{v_4}{L_4}+\frac{v_3}{L_4},\\ \frac{d{v}_4}{dt}=\frac{i_4}{C_4}-\frac{G_4}{C_4}{v}_4-\frac{i_5}{C_4},\\ \frac{d{i}_5}{dt}=-\frac{R_5}{L_5}{i}_5-\frac{v_5}{L_5}+\frac{v_4}{L_5},\\ \frac{d{v}_5}{dt}=\frac{i_5}{C_5}-\frac{G_5}{C_5}{v}_5-\frac{i_6}{C_5},\\ \frac{d{i}_6}{dt}=-\frac{R_6}{L_6}{i}_6-\frac{v_6}{L_6}+\frac{v_5}{L_6},\\ \frac{d{v}_6}{dt}=\frac{i_6}{C_6}-\frac{G_6}{C_6}{v}_6-\frac{i_7}{C_6},\\ \frac{d{i}_7}{dt}=-\frac{R_7}{L_7}{i}_7-\frac{v_7}{L_7}+\frac{v_6}{L_7},\\ \frac{d{v}_7}{dt}=\frac{i_7}{C_7}-\frac{G_7}{C_7}{v}_7-\frac{i_8}{C_7},\\ \frac{d{i}_8}{dt}=-\frac{R_8}{L_8}{i}_8-\frac{v_8}{L_8}+\frac{v_7}{L_8},\\ \frac{d{v}_8}{dt}=\frac{i_8}{C_8}-v_8\left(\frac{1}{Z{C}_8}+\frac{G_8}{C_8}\right).\end{array}$

The parameters are

R₁ = R₂ = R₃ = R₄ = R₅ = R₆ = R₇ = R₈ = 25 μΩ; L₁ = L₃ = L₅ = L₇ = 1 mH, and L₂ = L₄ = L₆ = L₈ = 10 mH; C₁ = 0.2 μF for the cable of 10 m length; C₃ = C₅ = C₇ = 0.7 pF, C₂ = C₄ = C₆ = C₈ = 7 pF, C₁₅ = C₃₇ = 0.35 pF, C₂₆ = C₄₈ = 3.5 pF; G₁ = G₃ = G₅ = G₇ = 1/(5000 Ω), G₂ = G₄ = G₆ = G₈ = 1/(500 Ω); and L₁₅ = L₃₇ = 0.005 mH, L₂₆ = L₄₈ = 0.01 mH. Z is either 1 μΩ (short-circuit) or 1 MΩ (open circuit).

Assume a PWM triangular current-step function i_s(t), as indicated in Fig. E3.14.4. Using Mathematica or MATLAB compute all state variables for the time period from t = t_calculate = 0 to 2.1 ms. Plot the voltages (v₁ – v₅), (v₂ – v₆), (v₃ – v₇), and (v₄ – v₈) from t = 0 to t_plot = 2 ms. Note that the computing time t = t_calculate = 2.1 ms must be larger than the plot time t = t_plot = 2 ms.