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# Wound-Rotor Induction Generators:Design and Testing

Authored by: Ion Boldea

# Variable Speed Generators

Print publication date:  October  2015
Online publication date:  October  2015

Print ISBN: 9781498723572
eBook ISBN: 9781498723589

10.1201/b19293-4

#### Abstract

Wound-rotor induction generators (WRIGs) have been built for power per unit as high as 400 MW/unit in pump storage power plants, to as low as 4.0 MW/unit in wind power plants. Diesel engine or gas-turbine-driven WRIGs for standby or autonomous operation up to 20–40 MW may also be feasible to reduce fuel consumption and pollution for variable loads.

#### 3.1  Introduction

Wound-rotor induction generators (WRIGs) have been built for power per unit as high as 400 MW/unit in pump storage power plants, to as low as 4.0 MW/unit in wind power plants. Diesel engine or gas-turbine-driven WRIGs for standby or autonomous operation up to 20–40 MW may also be feasible to reduce fuel consumption and pollution for variable loads.

Below 1.5–2 MW/unit, the use of WRIGs is not justifiable in terms of cost versus performance when compared with full-power rating converter synchronous or cage rotor induction generator systems.

The stator-rated voltage increases for power up to 18–20 kV (line voltage, root mean squared [RMS]) at 400 mega volt ampere (MVA). Because of limitations with voltage, for low-cost power converters, the rotor-rated (maximum) voltage occurring at maximum slip currently is about 3.5–4.2 kV (line voltage, RMS) for direct current (DC) voltage link alternating current (AC)–AC pulse-width modulated (PWM) converters with integrated gate-controlled thyristors (IGCTs).

Higher voltage levels are being tested and will be available soon for industrial use, based on multiple-level DC voltage link AC–AC converters made of insulated power cells in series and other high voltage technologies.

Thus far, for the 400 MW WRIGs, the rated rotor current may be in the order of 6500 A; and thus, for Smax = ±0.1, approximately, it would mean 3.6 kV line voltage (RMS) in the rotor. A transformer is necessary to match the 3.6 kV static power converter to the rotor with the 18 kV power source for the stator. The rotor voltage Vr is as follows:

3.1 $V r = K r s ⋅ | S max | ⋅ V s$

For |Smax| = 0.1, Vr = 3.6 ⋅ 103 V (per phase), $V s = 18 ⋅ 10 − 3 / 3 V$

(phase): Krs = 2/1. Therefore, the equivalent turn ratio is decisive in the design. In this case, however, a transformer is required to connect the DC voltage link AC–AC converter to the 18 kV local power grid.

For powers in the 1.5–4 MW range, low stator voltages are feasible (690 V line voltage, RMS). The same voltage may be chosen as the maximum rotor voltage Vr, at maximum slip.

For Smax = ±0.25, $V s = 690 / 3 V$

, Vr = Vs, Krs = 1/|Smax| = 4. In this case, the rotor currents are significantly lower than the stator currents. A transformer to match the rotor voltage to the stator voltage is not required.

Finally, for WRIGs in the 3–10 MW range to be driven by diesel engines, 3000 (3600) rpm, or gas turbines, stator and rotor voltages in the 3.5–4.2 kV range are feasible. The transformer is again avoided.

Once the stator and rotor rated voltages are calculated and fixed, further design can proceed smoothly. Electromagnetic (emf) and thermomechanical designs are the two types of design. In what follows, we will touch on mainly the emf design.

Even for the emf design, we should distinguish three main operation modes:

1. Generator at a power grid
3. Brushless AC exciter (generator with rotor electric output)

The motor mode is required in applications such as pump storage power plants or even with microhydro or wind turbine prime movers.

The emf design implies a machine model, analytical, numerical, or mixed; one or more objective functions; and an optimization method with a computer program to execute it.

The optimization criteria may include the following:

• Maximum efficiency
• Minimum active material costs
• Net present worth, individual or aggregated

Deterministic, stochastic, and evolutionary optimization methods have been applied to electric machine design [1] (see also Chapter 10). Whatever the optimization method, it is useful to have sound geometrical parameters for the preliminary (or general) design with regard to performance when starting the optimization design process. This is the reason why the general emf design is primarily discussed in what follows. Among the operation modes listed above, the generator at the power grid is the most frequently used, and, thus, it is the one of interest here.

High- and low-voltage stators and rotors are considered to cover the entire power range of WRIGs (from 1 to 400 MW).

The design methodology that follows covers only the essentials. A comprehensive design methodology is beyond the scope of this text.

#### 3.2  Design Specifications: An Example

The rated line voltage of the stator (RMS) (Y) VSN = 0.38(0.46), 0.69, 4.2(6), …, 18 kV. The rated stator frequency f1 = 50 (60) Hz. The rated ideal speed n1N = f1/p1 = 3000 (3600), 1500 (1800). The maximum (ideal) speed nmax = (f1/p1)(1 + |Smax|). The maximum slip Smax = ±0.05, ±0.1, ±0.2, ±0.25. The rated stator power (at unity power factor) S1Ns = 2 MW. The rated rotor power (at unity power factor) SNr = SNs|Smax| = 0.5 MW. The rated (maximum) rotor line voltage (Y) is VRNVSN. As already discussed, the stator power may reach up to 350 MW (or more) with the rotor delivering maximum power (at maximum speed) of up to 40–50 MW and voltage $V N r$

≤ 4.2(6) kV

Here we will consider the following for our discussion: total 2.5 MW at 690 V, 50 Hz, VRN = VSN = 690 V, Smax = ±0.25, and rated ideal speed n1N = 50/p1 = 1500 rpm (p1 = 2).

The emf design factors are as follows:

• Stator (core) design
• Rotor (core) design
• Stator winding design
• Rotor winding design
• Magnetization current computation
• Equivalent circuit parameter computation
• Loss and efficiency computation

#### 3.3  Stator Design

Two main design concepts need to be considered when calculating the stator interior diameter Dis: the output coefficient design concept (Ce—Esson coefficient) and the shear rotor stress (fxt) [1]. Here, we will make use of the shear rotor stress concept with fxt = 1.5–8 N/cm2.

The shear rotor stress increases with torque. First, the emf torque has to be estimated, noting that at 2.5 MW, 2p1 = 4 poles, the expected rated efficiency ηN > 0.95.

The total emf power SgN at maximum speed and power is as follows:

3.2 $S g N ≈ ( S S N + S R N ) η N = ( 2 + 0.5 ) 10 6 0.96 = 2.604 ⋅ 10 6 W$

The corresponding emf torque Te is

3.3 $T e = S g N 2 π ( f 1 / p 1 ) ( 1 + | S max | ) = 2.604 ⋅ 10 6 2 π ( 50 / 2 ) ( 1 + 0.25 ) = 1.327 ⋅ 10 4 Nm .$

This torque is at stator rated power and ideal synchronous speed (S = 0). The stator interior diameter, Dis, based on the shear rotor stress concept, fxt, is

3.4 $D i s = 2 T e π ⋅ λ ⋅ f x t 3 ; λ = l i D i s$

with li equal to the stack length. The stack length ratio λ = 0.2–1.5, in general. Smaller values correspond to a larger number of poles. With λ = 1.0 and fxt = 6 N/cm2 (aiming at high torque density), the stator internal diameter, Dis, is as follows:

3.5 $D i s = 2 ⋅ 1.3270 π ⋅ 1 ⋅ 6 3 = 0.52 m .$

At this power level, a unistack stator is used, together with axial air cooling.

The external stator diameter Dout based on the maximum air gap flux density per given magnetomotive force (mmf) is approximated in Table 3.1 [1]. Therefore, from Table 3.1, Dout is as follows:

3.6 $D o u t = D i s ⋅ 1.48 = 0.5200 ⋅ 1.48 = 0.796 m$

The rated stator current at unity power factor in the stator ISN is

3.7 $I S N = S S N 3 V S N = 2 ⋅ 10 6 3 ⋅ 690 = 1675.46 A$

The air gap flux density Bg1 = 0.75 T (the fundamental value). On the other hand, the air gap emf ES per phase is as follows:

3.8 $E S = K E V S N 3 ; K E = 0.97 − 0.98$
3.9 $E S = π 2 f 1 W 1 K W 1 a ⋅ 2 π B g 1 ⋅ τ ⋅ l i$

### Table 3.1   Outer to Inner Stator Diameter Ratio

 2p1 2 4 6 8 ≥10 Dout/Dis 1.65–1.69 1.46–1.49 1.37–1.40 1.27–1.30 1.24–1.20

where τ is the pole pitch:

3.10 $τ = π D i s 2 p 1 = π ⋅ 0.52 2 ⋅ 2 = 0.40 m$

KW1 is the total winding factor:

3.11 $K W 1 = sin π / 6 q 1 sin π / 6 q 1 sin π 2 y τ ; 2 3 ≤ y τ ≤ 1$

where

q1 is the number of slot/pole/phase in the stator

y/τ is the stator coil span/pole pitch ratio

As the stator current is rather large, we are inclined to use a1 = 2 current paths in parallel in the stator. With two current paths in parallel, full symmetry of the windings with respect to stator slots may be provided. For the time being, let us adopt KW1 = 0.910. Consequently, from Equation 3.9, the number of turns per current path W1a is as follows:

3.12 $W 1 a = 0.97 ⋅ 690 / 3 2 ⋅ 50 ⋅ 0.910 ⋅ 2 ⋅ 0.75 ⋅ 0.52 ⋅ 0.4 = 19.32 turns$

Adopting q1 = 5 slots/pole/phase, a two-layer winding with two conductors per coil, nc1 = 2,

3.13 $W 1 a = 2 p 1 a 1 ⋅ q s ⋅ n c 1 = 2 ⋅ 2 2 ⋅ 5 ⋅ 2 = 20$

The final number of turns/coil is nc1 = 2, q1 = 5, with two symmetrical current paths in parallel. The North and South poles pairs constitute the paths in parallel.

Note that the division of a turn into elementary conductors in parallel with some transposition to reduce skin effects will be discussed later in this chapter.

The stator slot pitch is now computable:

3.14 $τ S = τ 3 q 1 = 0.40 3.5 = 0.0266 m$

The stator winding factor KW1 may now be recalculated if only the y/τ ratio is fixed: y/τ = 12/15, very close to the optimum value, to reduce to almost zero the fifth-order stator mmf space harmonic:

3.15 $K W 1 = sin π / 6 5 sin ( π / 6 ⋅ 5 ) ⋅ sin π 2 ⋅ 4 5 = 0.9566 ⋅ 0.951 = 0.9097$

This value is very close to the adopted one, and thus, nc1 = 2 and q1 = 5, a1 = 2 hold. The number of slots NS is as follows:

3.16 $N S = 2 p 1 q 1 m = 2 ⋅ 2 ⋅ 5 ⋅ 3 = 60 slots$

As expected, the conditions to full symmetry, NS/m1q1 = 60/(3·2) = 10 = integer, 2p1/a1 = 2·2/2 = integer are fulfilled.

The stator conductor cross-section Acos is as follows:

3.17 $A cos = I S N a 1 ⋅ j cos$

With the design current density jco1 = 6.5 A/mm2, rather typical for air cooling:

3.18 $A cos = 1675.46 2 ⋅ 6.5 = 128.88 mm 2$

Open slots in the stator are adopted, but magnetic wedges are used to reduce the air gap flux density pulsations due to slot openings.

In general, the slot width WSS = 0.45–0.55. Let us adopt WSS = 0.5. The slot width WS is

3.19 $W S = ( W S τ S ) ⋅ τ S = 0.5 ⋅ 0.02666 = 0.01333 m$

There are two coils (four turns in our case) per slot (Figure 3.1).

As we deal with a low-voltage stator (690 V, line voltage RMS), the total slot filling factor, with rectangular cross-sectional conductors, may be safely considered as Kfills ≈ 0.55.

The useful (above wedge) slot area Asn is

3.20 $A s n = 2 n c 1 A cos K f i l l s = 2 ⋅ 2 ⋅ 128.88 0.55 = 937 mm 2$

The rectangular slot useful height hsu is straightforward:

3.21 $h s u = A s u W b = 937 13.33 = 70.315 mm$

The slot aspect ratio hsu/Ws = 70.315/13.33 = 5.275 is still acceptable. The wedge height is about hsw ≈ 3 mm. Adopting a magnetic wedge leads to the apparent reduction of slot opening from Ws= 13.33 mm to about Ws = 4 mm, as detailed later in this chapter.

Figure 3.1   Stator slotting geometry.

The air gap g is

3.22 $g ≈ ( 0.1 + 0.012 S S N 3 ) 10 − 3 = ( 0.1 + 0.0012 2 ⋅ 10 6 3 ) 10 − 3 = 1.612 mm$

With a flux density Bcs = 1.55 T in the stator back iron, the back core stator height hcs (Figure 3.1) may be calculated as follows:

3.23 $h c s = B g 1 τ π B c s = 0.75 ⋅ 0.4 π ⋅ 1.5 = 0.0637 m$

The magnetically required stator outer diameter Doutm is

3.24 $D o u t m = D i s + 2 ( h s m + h s w + h c s ) = 0.52 + 2 ( 0.07315 + 0.003 + 0.0637 ) = 0.7997 ≈ 0.8 m$

This is roughly equal to the value calculated from Table 3.1 (Equation 3.6).

The stator core outer diameter may be increased by the double diameter of axial channels for ventilation, which might be added to augment the external cooling by air flowing through the fins of the cast iron frame.

As already inferred, the rectangular axial channels, placed in the upper part of stator teeth (Figure 3.1), can help improve the machine cooling once the ventilator of the shaft is able to flow part of the air through these stator axial channels.

The division of a stator conductor (turn) with a cross-section of 128.8 mm2 is similar to the case of synchronous generator design in terms of transposition to limit the skin effects. Let us consider four elementary conductors in parallel. Their cross-sectional area Acos e is as follows:

3.25 $A cos e = A cos 4 = 128.8 4 = 32.2 mm 2$

Considering only ace = 12 mm, out of the 13.33 mm slot width available for the elementary stator conductor, the height of the elementary conductor hce is as follows:

3.26 $h c e = A c o s e a c e = 32.2 12 ≈ 2.68 mm$

Without transposition and neglecting coil chording the skin effect coefficient, with me = 16 layers (elementary conductors) in slot, is as follows [1]:

3.27 $K R m e = φ ( ξ ) + ( m e 2 − 1 ) 3 ψ ( ξ )$
3.28 $ξ = β h c e ; β = ω 1 μ 0 σ c o 2 ⋅ a c e W s = 2 π ⋅ 50 ⋅ 4.3 ⋅ 10 7 ⋅ 1.256 ⋅ 10 − 6 2 ⋅ 12 13.3 = 0.6964 ⋅ 10 2 m − 1$

Finally, ξ = βhce = 0.6964 ⋅ 10−2 ⋅ 2.68 ⋅ 10−3 = 0.1866. With,

3.29 $φ ( ξ ) = ξ ( sinh 2 ξ + sin 2 ξ ) ( cosh 2 ξ − cos 2 ξ ) ≈ 1.00$
3.30 $ψ ( ξ ) = 2 ξ ( sinh ξ − sin ξ ) ( cosh ξ + cos ξ ) ≈ 5.55 ⋅ 10 − 4$

From Equation 3.27, KRme is

$K R m e = 1.00 + ( 16 2 − 1 ) 3 ⋅ 5.55 ⋅ 10 − 4 = 1.04675$

The existence of four elementary conductors (strands) in parallel leads also to circulating currents. Their effect may be translated into an additional skin effect coefficient Krad [1]:

3.31 $K r a d = 4 β 4 ⋅ h c e 4 ( l i l t u r n ) 2 n c n 2 ( 1 + cos γ ) 2 4$

where

• lturn is the coil turn length
• ncn is the number of turns per coil
• γ is the phase shift between lower and upper layer currents (γ = 0 for diametrical coils; it is (1 − y/τ)(π/2) for chording coils)

With li/lturn ≈ 0.4, ncn = 2, hce = 2.68 · 10−3 m, and β = 0.6964 · 10−2 m−1 (from Equation 3.28), Krad is as follows:

3.32 $K r a d = 4 ⋅ ( 0.6964 ⋅ 10 2 ⋅ 2.68 ⋅ 10 − 3 ) 4 ⋅ 0.4 2 ⋅ 2 2 ( 1 + cos 18 ° ) 2 4 = 0.01847 !$

The total skin effect factor KRll is

3.33 $K R l l = 1 + ( K r m e − 1 ) l s t a c k l t u r n + K r a d = 1 + ( 1.04675 − 1 ) 0.4 + 0.01847 = 1.03717$

It seems that at least for this design, no transposition of the four elementary conductors in parallel is required, as the total skin effect winding losses add only 3.717% to the fundamental winding losses.

The stator winding is characterized by the following:

• Four poles
• 60 slots
• q1 = 5 slots/pole/phase
• Coil span/pole pitch y/τ = 12/15
• a1 = 2 symmetrical current paths in parallel
• Four elementary rectangular wires in parallel

With t1 equal to the largest common divisor of Ns and p1 = 2, there are Ns/t1 = 60/2 = 30 distinct slot emfs. Their star picture is shown in Figure 3.2. They are distributed to phases based on 120° phase shifting after choosing Ns/2m1 = 60/(2 · 3) = 10 arrows for phase a and positive direction (Figure 3.3a and b).

Figure 3.2   Slot allocation to phases for NS = 60 slots, 2p1 = 4 poles, and m = 3 phases.

Figure 3.3   (a) Half of coils of stator phases A and (b) their connection to form two current paths in parallel.

#### 3.4  Rotor Design

The rotor design is based on the maximum speed (negative slip)/power delivered, PRN, at the corresponding voltage VRN = VSN. Besides, the WRIG is designed here for unity power factor in the stator. Therefore, all the reactive power is provided through the rotor. Consequently, the rotor also provides for the magnetization current in the machine.

For VRN = VSN at Smax, the turn ratio between rotor and stator Krs is obtained:

3.34 $K r s = W 2 K W 2 W 1 K W 1 = 1 | S max | = 1 0.25 = 4.0$

Now the stator rated current reduced to the rotor $I ′ S N$

is as follows:
3.35 $I ′ S N = I S N K r s = 1675.46 4 = 418.865 A$

The rated magnetization current depends on the machine power, the number of poles, and so forth. At this point in the design process, we can assign $I ′ m$

(the rotor-reduced magnetization current) a per unit (P.U.) value with respect to ISN:
3.36 $I ′ m = K m I ′ S N K m = 0.1 − 0.30$

Let us consider here Km = 0.30. Later on in the design, Km will be calculated, and then adjustments will be made.

Therefore, the rotor current at maximum slip and rated rotor and stator delivered powers is as follows:

3.37 $I R N R = I ′ S N 2 + I ′ m 2 = I ′ S N 1 + K m 2 = 418.865 1 + 0.30 2 = 437.30 A$

The rotor power factor cos φ2N is

3.38 $cos φ 2 N = P R N 3 V R N I R N R = 500 , 000 3 ⋅ 690 ⋅ 437.30 = 0.9578$

It should be noted that the oversizing of the inverter to produce unity power factor (at rated power) in the stator is not very important.

Note that, generally, the reactive power delivered by the stator Q1 is requested from the rotor circuit as SQ1.

If massive reactive power delivery from the stator is required, and it is decided to be provided from the rotor-side bidirectional converter, the latter and the rotor windings have to be sized for the scope.

When the source-side power factor in the converter is unity, the whole reactive power delivered by the rotor-side converter is “produced” by the DC link capacitor, which needs to be sized for the scope.

Roughly, with cos φ2 = 0.707, the converter has to be oversized at 150%, while the machine may deliver almost 80% of reactive power through the stator (ideally 100%, but a part is used for machine magnetization).

Adopting VRN = VSN at Smax eliminates the need for a transformer between the bidirectional converter and the local power grid at VSN.

Once we have the rotor-to-stator turns ratio and the rated rotor current, the designing of the rotor becomes straightforward.

The equivalent number of rotor turns, W2KW2 (single current path), is as follows:

3.39 $W 2 K W 2 = W 1 a K W 1 ⋅ K r s = 20 ⋅ 0.908 ⋅ 4 = 72.64 turns/phase$

The rotor number of slots NR should differ from the stator one, NS, but they should not be too different from each other.

As the number of slots per pole and phase in the stator q1 = 5, for an integer q2, we may choose q2 = 4 or 6. We choose q2 = 4. Therefore, the number of rotor slots NR is as follows:

3.40 $N R = 2 p 1 m 1 q 2 = 2 ⋅ 2 ⋅ 3 ⋅ 4 = 48$

With a coil span of YR/τ = 10/12, the rotor winding factor (no skewing) becomes

3.41 $K W 2 = sin π / 6 q 2 sin π / 6 q 2 ⋅ sin π 2 ⋅ y R τ = 0.5 4 sin ( π / 24 ) ⋅ sin π 2 ⋅ 10 12 = 0.95766 ⋅ 0.9659 = 0.925$

From Equation 3.39, with Equation 3.41, the number of rotor turns per phase W2 is

3.42 $W 2 = W 2 K W 2 K W 2 = 72.64 0.925 = 78.53 ≈ 80$

The number of coils per phase is NR/m1 = 48/3 = 16. With W2 = 80 turns/phase; it follows that each coil will have nc2 turns:

3.43 $n c 2 = W 2 ( N R / m 1 ) = 80 16 = 5 turns/coil$

Therefore, there are ten turns per slot in two layers and one current path only in the rotor.

Adopting a design current density jcon = 10 A/mm2 (special attention to rotor cooling is needed) and, again, a slot filling factor Kfill = 0.55, the copper conductor cross-section Acor and the slot useful area AslotUR are as follows:

3.44 $A c o r = I R N j c o r = 434.3 10 = 43.4 mm 2$
3.45 $A s l o t U R = 2 n c 2 A c o r K f i l l = 2 × 5 × 43.40 0.55 = 789.09 mm 2$

The rotor slot pitch τR is

3.46 $τ R = π ( D i s − 2 g ) N R = π ( 0.52 − 2 ⋅ 0.001612 ) 48 = 0.033805 m$

Assuming that the rectangular slot occupies 45% of rotor slot pitch, the slot width WR is

3.47 $W R = 0.45 τ R = 0.45 ⋅ 0.033805 = 0.0152 m$

The useful height hRU (Figure 3.4) is, thus,

3.48 $h S U = A s l o t U R W r = 789.09 15.20 = 51.913 mm$

This is an acceptable (Equation 3.48) value, as hSU/WR < 4.

Figure 3.4   Rotor slotting.

Open slots have been adopted, but with magnetic wedges (μr = 4–5), the actual slot opening is reduced from WR = 15.2 mm to about 3.5 mm, which should be reasonable (in the sense of limiting the Carter coefficient and surface and flux pulsation space harmonics core losses).

We need to verify the maximum flux density, BtRmax, in the rotor teeth at the bottom of the slot:

3.49 $B t R max = B g 1 ⋅ τ R W t R min = 0.75 ⋅ 33.805 11.42 = 2.22 T$

with

3.50 $W t R min = π ( D i s − 2 ( g + h R U + h R W ) ) N R − W R = π ( 0.520 − 2 ( 0.001612 + 0.05191 ) + 0.003 ) 48 − 0.0152 = 11.42 ⋅ 10 − 3 m$

Though the maximum rotor tooth flux density is rather large (2.2 T), it should be acceptable, because it influences a short path length.

The rotor back iron radial depth hcr is as follows:

3.51 $h c r = B g ⋅ τ π ⋅ B c r = 0.75 ⋅ 0.4 π ⋅ 1.6 = 0.0597 ≈ 0.06 m$

The value of rotor back iron flux density Bcr of 1.6 T (larger than in the stator back iron) was adopted, as the length of the back iron flux lines is smaller in the rotor with respect to the stator.

To see how much of it is left for the shaft diameter, let us calculate the magnetic back iron inner diameter Dir:

3.52 $D i r = D i s − 2 ( g + h R U + h R W + h c r ) = 0.52 − 2 ( 0.001612 + 0.05191 + 0.003 + 0.06 ) = 0.287 m$

This inner rotor core diameter may be reduced by 20 mm (or more) to allow for axial channels—10 mm (or more) in diameter—for axial cooling, and thus, 0.267 m (or slightly less) are left of the shaft. It should be enough for the purpose, as the stack length is li = Dis = 0.52 m.

The rotor winding design is straightforward, with q2 = 4, 2p1 = 4 and a single current path. The cross-section of the conductor is 43.20 mm2, and thus, even a single rectangular conductor with the width bcr WR ≈ 13 mm and its height hcr = Acor/bcr = 43.40/13 = 3.338 mm, will do.

As the maximum frequency in the rotor, $f q max = f 1 * | S max | = 50 ⋅ 0.25 = 12.5 Hz$

, the skin effect will be even smaller than in the stator (which has a similar elementary conductor size); and thus, no transposition seems to be necessary.

The winding end connections have to be tightened properly against centrifugal and electrodynamic forces by adequate nonmagnetic bandages.

The electrical rotor design also contains the slip-rings and brush system design and the shaft and bearings design. These are beyond our scope here. Though debatable, the use of magnetic wedges on the rotor also seems doable, as the maximum peripheral speed is smaller than 50 m/s.

#### 3.5  Magnetization Current

The rated magnetization current was previously assigned a value (30% of ISN). By now, we have the complete geometry of stator and rotor slots cores, and the magnetization mmf may be considered. Let us consider it as produced from the rotor, though it would be the same if computed from the stator.

We start with the given air gap flux density Bg1 = 0.75 T and assume that the magnetic wedge relative permeability μRS = 3 in the stator and μRR = 5 in the rotor. Ampere’s law along the half of the Γ contour (Figure 3.5), for the main flux, yields the following:

3.53 $F m = 3 W 2 K W 2 I R 0 2 π p 1 = ( F A A ′ + F A B + F B C + F A ′ B ′ + F B ′ C ′ )$

where

• $F A A ′$ is the air gap mmf
• FAB is the stator teeth mmf
• FBC is the stator yoke mmf
• $F A ′ B ′$ is the rotor tooth mmf
• $F B ′ C ′$ is the rotor yoke mmf

The air gap mmf $F A A ′$

is as follows:
3.54 $F A A ′ = g K C B g 1 μ 0 ; K C = K C 1 ⋅ K C 2$

KC is the Carter coefficient:

3.55 $K C 1 , 2 = 1 1 − γ 1 , 2 g / 2 τ s , r ; γ 1 , 2 = ( 2 W S ′ , R ′ / g ) 2 5 + 2 W S ′ , R ′ / g$

The equivalent slot openings, with magnetic wedges,$W S ′$

and $W R ′$ , are as follows:
3.56 $W S ′ = W S / μ R S W R ′ = W R / μ R R$

Figure 3.5   Main flux path.

where

• τs,r = 26.6, 33.8 mm
• g = 1.612 mm
• $W S ′ = 13 .2/3 = 4 .066 mm$
• $W R ′ = 15.2 / 5 = 3.04 mm$ (from Equation 3.56 through Equation 3.58)
• KC1 = 1.0826
• KC2 = 1.040

Finally,

3.57 $K c = K c 1 K c 2 = 1.0826 ⋅ 1.04 ≈ 1.126$

This is a small value that will result, however, in smaller surface and flux pulsation core losses. The small effective slot opening, $W S ′$

and $W R ′$ will lead to larger slot leakage inductance contributions. This, in turn, will reduce the short circuit currents.

The air gap mmf (from Equation 3.54) is as follows: $F A A ′ = 1.612 ⋅ 10 − 3 ⋅ 1.126 0.75 1.256 ⋅ 10 − 6 = 1083.86 A turns$

The stator and rotor teeth mmf should take into consideration the trapezoidal shape of the teeth and the axial cooling channels in the stator teeth.

We might suppose that, in the stator, due to axial channels, the tooth width is constant and equal to its value at the air gap Wts = τSWS = 26.6 − 13.2 = 13.4 mm. Therefore, the flux density in the stator tooth Bts is

3.58 $B t s = B g 1 ⋅ τ S W t s = 0.75 ⋅ 0.0266 0.0134 = 1.4888 T$

### Table 3.2   B–H Curve for Silicon (3.5%) Steel (0.5 mm Thick) at 50 Hz

 B (T) 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 H (A/m) 22.8 35 45 49 57 65 70 76 83 90 B (T) 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1 H (A/m) 98 106 115 124 135 148 162 177 198 220 B (T) 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.5 H (A/m) 237 273 310 356 417 482 585 760 1,050 1,340 B (T) 1.55 1.6 1.65 1.7 1.75 1.8 1.85 1.9 1.95 2.0 H (A/m) 1760 2460 3460 4800 6160 8270 11,170 15,220 22,000 34,000

From the magnetization curve of silicon steel (3.5% silicon, 0.5 mm thickness, at 50 Hz; Table 3.2), Hts = 1290 A/m by linear interpolation.

Therefore, the stator tooth mmf FAB is

3.59 $F A B = H t s ⋅ ( h s u + h s w ) = 1290 ( 0.070357 + 0.003 ) = 94.62 A turns$

The stator back iron flux density was already chosen: Bcs = 1.5 T.

Hcs (from Table 3.2) is Hcs = 1340 A/m. The average magnetic path length lcsw is as follows:

3.60 $( l c s w ) B C ≈ 2 3 π ( D o u t − h c s ) 2 ⋅ 2 p 1 = 2 3 π ( 0.780 − 0.0637 ) 2 ⋅ 2 ⋅ 2 = 0.19266 m$

Therefore, the stator back iron mmf FBC is

3.61 $F B C ≈ H c s ⋅ l c s a v = 1340 ⋅ 0.19266 ≈ 258.17 A turns$

In the rotor teeth, we need to obtain first an average flux density by using the top, middle, and bottom tooth values Btrt, BtRm, and BtRb:

3.62 $B t r t = B g 1 ⋅ τ R W t R = 0.75 ⋅ 0.0338 ( 0.0338 − 0.0152 ) = 1.3629 T B t R m = B g 1 ⋅ τ R W t R m = 0.75 ⋅ 0.0338 0.01501 = 1.68887 T B t R b = B g 1 ⋅ τ R W t R min = 0.75 0.0338 0.01142 = 2.2 T$

The average value of BtR is as follows:

3.63 $B t R = B t R t + B t R b + 4 B t R m 6 = 1.3629 + 2.2 + 4 ⋅ 1.68887 6 = 1.7197 ≈ 1.72 T$

From Table 3.2, HtR = 5334 A/m. The rotor tooth mmf $F A ′ B ′$

is, thus,
3.64 $F A ′ B ′ = H t r ⋅ ( h R U + h R W ) = 5334 ( 0.0519 + 0.003 ) = 292.8366 A turns$

Finally, for the rotor back iron flux density BCR = 1.6 T (already chosen), Hcr = 2460 A/m, with the average path length lCRav as follows:

3.65 $l C R a v = 2 3 π ( D s h a f t + h C R + 0.01 ) 2 ⋅ 2 p 1 = 2 3 π ( 0.267 + 0.01 + 0.06 ) 2 ⋅ 2 ⋅ 2 = 0.088 m$

The rotor back iron mmf $F B ′ C ′$

,
3.66 $F B ′ C ′ = H c r ⋅ l C R a v = 2460 ⋅ 0.088 = 216.92 A turns$

The total magnetization mmf per pole Fm is as follows (Equation 3.55):

$F m = 1083.86 + 94.62 + 258.17 + 292.8366 + 216.92 = 1946.40 A turns/pole$

It should be noted that the total stator and rotor back iron mmfs are not much different from each other. However, the stator teeth are less saturated than the other iron sections of the core. The uniform saturation of iron is a guarantee that sinusoidal air gap tooth and back iron flux densities distributions are secured. Now, from Equation 3.55, the no-load (magnetization) rotor current IR0 might be calculated:

3.67 $I R 0 = F m ⋅ π ⋅ p 1 3 W 2 ⋅ K W 2 ⋅ 2 = 1946.4 ⋅ π ⋅ 2 3 ⋅ 80 ⋅ 0.925 ⋅ 1.41 = 39.05 A$

The ratio of IR0 to $I S N ′$

(Equation 3.35; stator current reduced to rotor), Km, is, in fact,
3.68 $K m = I R 0 I ′ S N = 39.05 418.865 = 0.09322 < 0.3$

The initial value assigned to Km was 0.3, so the final value is smaller, leaving room for more saturation in the stator teeth by increasing the size of axial teeth cooling channels (Figure 3.1). Alternatively, the air gap may be increased up to 3 mm if so needed for mechanical reasons.

The total iron saturation factor Ks is

3.69 $K s = ( F A B + F B C + F A ′ B ′ + F B ′ C ′ ) F A A ′ = ( 94.62 + 258.17 + 292.5366 + 216.92 ) 1083.66 = 0.79568$

Therefore, the iron adds to the air gap mmf 79.568% more. This will reduce the magnetization reactance Xm accordingly.

#### 3.6  Reactances and Resistances

The main WRIG parameters are the magnetization reactance Xm, the stator and rotor resistances Rs and Rr, and leakage reactances Xsl and Xrl, reduced to the stator. The magnetization reactance expression is straightforward [1]:

3.70 $X m = ω 1 ⋅ L m ; L m = 6 μ 0 ( W 1 a K W S ) 2 τ l i π 2 p 1 g K C ( 1 + K S ) L m = 6 ⋅ 1.256 ⋅ 10 − 6 ( 20 ⋅ 0.908 ) 2 ⋅ 0.4 ⋅ 0.52 π 2 ⋅ 2 ⋅ 1.612 ⋅ 10 − 3 ⋅ 1.126 ⋅ ( 1 + 0.79568 ) ⋅ 2 = 8.0428 ⋅ 10 − 3 H X m = 2 ⋅ π ⋅ 50 ⋅ 8.0428 ⋅ 10 − 3 = 2.5254 Ω$

The base reactance

. As expected, xm = Xm/Xb = 2.5254/0.238 = 10.610 = ISN/IR0 from Equation 3.67.

Figure 3.6   Stator coil end connection geometry.

The stator and rotor resistances and leakage reactances strongly depend on the end connection geometry (Figure 3.6). The end connection lfs on one machine side, for the stator winding coils, is

3.71 $l f s ≈ 2 ( l l + l l ′ ) + π h s t = 2 ( l l + β s τ 2 cos α ) + π h s t = 2 ( 0.015 + 0.8 ⋅ 0.4 2 cos 40 ° ) + π ⋅ 0.07035 = 0.668 m$

The stator resistance Rs per phase has the following standard formula (skin effect is negligible as shown earlier):

3.72 $R S ≈ ρ c o 100 ° ⋅ W 1 a 2 A cos ( l i + l f s ) ⋅ 1 a 1 = 1.8 ⋅ 10 − 8 2 ( 1 + 100 − 20 272 ) ⋅ 20 ⋅ 2 ( 0.52 + 0.668 ) 128.88 ⋅ 10 − 6 = 0.429 ⋅ 10 − 2 Ω$

The stator leakage reactance Xsl is written as follows:

3.73 $X s l = ω 1 L s l ; L s l = μ 0 ( 2 n c 1 ) 2 ⋅ l i ( λ S + λ e n d + λ d s ) ⋅ N s m 1 a 1 2$

where

• nc1 = 2 turns/coil
• li = 0.52 m (stack length)
• Ns = 60 slots
• m1 = 3 phases
• a1 = 2 stator current paths
• λs, λend, and λds are the slot, end connection, and differential geometrical permeance (nondimensional) coefficients

For the case in point [1],

3.74 $λ s = h s u 3 W s + h s w W ′ s = 70.35 3 ⋅ 15.2 + 3 4.066 = 2.2806$
3.75 $λ e n d = 0.34 ⋅ q 1 ( l f s − 0.64 β τ ) l i = 0.34 ⋅ 5 ( 0.668 − 0.64 ⋅ 0.8 ⋅ 0.4 ) 0.52 = 1.5143$

The differential geometrical permeance coefficient λds [1] may be calculated as follows:

3.76 $λ d s = 0.9 ⋅ τ s ( q s K W 1 ) ⋅ K 01 ⋅ σ d s K c g$
$K 01 = 1 − 0.033 ( W s 2 g τ s ) = 1 − 0.033 ⋅ 4.066 2 1.612 ⋅ 26.66 = 0.9873$

The coefficient σds is the ratio between the differential and magnetizing inductance and depends on q1 and chording ratio β. For q1 = 5 and β = 0.8 from Figure 3.7 [1], σds = 0.0042.

Finally,

$λ d s = 0.9 ⋅ 0.0266 ( 5 ⋅ 0.908 ) 2 ⋅ 0.9873 ⋅ 0.0042 1.126 ⋅ 1.612 ⋅ 10 − 3 = 1.127$

The term λds generally includes the zigzag leakage flux.

Finally, from Equation 3.73,

$L s l = 1.256 ⋅ 10 − 6 ( 2 ⋅ 2 ) 2 ⋅ 0.52 ( 2.2806 + 1.5143 + 1.127 ) ⋅ 60 3 ⋅ 2 2 = 0.257 ⋅ 10 − 3 H$

As can be seen, Lsl/Lm = 0.257 · 10−3/(8.0428 · 10−3) = 0.03195.

Figure 3.7   Differential leakage coefficient σd.

Xsl = ω1Lsl = 2π · 50 · 0.257 · 10−3 = 0.0807 Ω. The rotor end connection length lfr may be computed in a similar way as in Equation 3.71:

3.77 $l f r = 2 ( l l + β r τ 2 cos α ) + π ( h R U + h R W ) = 2 ( 0.015 + 10 ⋅ 0.4 12 ⋅ 2 ⋅ cos 40 ° ) + π ( 0.0519 + 0.003 ) = 0.6375 m$

The rotor resistance expression is straightforward:

3.78 $R R r = ρ c o 100 ° ⋅ W 2 ⋅ 2 A c o r ( l i + l f r ) = 1.8 ⋅ 10 − 8 ⋅ 80 ⋅ 2 ( 0.52 + 0.6775 ) 43.40 ⋅ 10 − 6 = 7.68 ⋅ 10 − 2 Ω$

Equation 3.73 also holds for rotor leakage inductance and reactance:

3.79 $L r l r = μ 0 ( 2 n c 2 ) 2 ⋅ l i ( λ S R + λ e n d R + λ d R ) ⋅ N R m 1$
3.80 $λ S R = h R U 3 W R + h R W W R ′ = 51.93 3 ⋅ 15.2 + 3 3.04 = 2.1256$
3.81 $λ e n d R = 0.34 q 2 ( l f r − 0.64 β R τ ) l i = 0.34 ⋅ 4 ( 0.6375 − 0.64 ⋅ ( 5 / 6 ) ⋅ 0.4 ) 0.52 = 1.1093$
3.82 $λ d R = 0.9 τ R ( q R K W 2 ) 2 ⋅ K o 2 ⋅ σ d R K C g$

with σdR = 0.0062 (q2 = 4, βR = 5/6) from Figure 3.7:

3.83 $K o 2 = 1 − 0.033 ( W R ′ 2 ) g τ R = 1 − 0.033 ( 3.04 2 1.612 ⋅ 33.8 ) = 0.994$
$λ d R = 0.9 ⋅ 33.8 ( 4 ⋅ 0.925 ) 2 ⋅ 0.994 ⋅ 0.0062 1.126 ⋅ 1.612 = 1.413$

From Equation 3.79, $L R l r$

is
$L R l r = 1.256 ⋅ 10 − 6 ( 2.5 ) 2 ( 2.1256 + 1.1093 + 1.413 ) . 48 3 = 4.858.10 − 3 H X R L r = ω 1 L R L r = 2 π ⋅ 50 ⋅ 4.858 ⋅ 10 − 3 = 1.525 Ω$

Noting that $R R r$

, $L R l r$ , and $X R l r$ are values obtained prior to stator reduction, we may now reduce them to stator values with KRS = 4 (turns ratio):
3.84 $R R = R R r K R S 2 = 7.68 ⋅ 10 − 2 16 = 0.48 ⋅ 10 − 2 Ω L R l = L R l r K R S 2 = 4.858 ⋅ 10 − 3 16 = 0.303 ⋅ 10 − 3 H X R l = X R l r K R S 2 = 1.525 16 = 0.0953 Ω$

Let us now add here the RS, Lsl, Xsl, Lm, and Xm to have them all together: Rs = 0.429 · 10−2 Ω, Lsl = 0.257 · 10−3 H, Xsl = 0.0807 Ω, Lm = 8.0428 · 10−3 H, and Xm = 2.5254 Ω.

The rotor resistance reduced to the stator is larger than the stator resistance, mainly due to notably larger rated current density (from 6.5 to 10 A/mm2), despite having shorter end connections.

Because of smaller q2 than q1, the differential leakage coefficient is larger in the rotor, which finally leads to a slightly larger rotor leakage reactance than in the stator.

The equivalent circuit may now be used to compute the power flow through the machine for generating or motoring, once the value of slip S and rotor voltage amplitude and phase are set. We leave this to the interested reader. In what follows, the design methodology, however, explores the machine losses to determine the efficiency.

#### 3.7  Electrical Losses and Efficiency

The electrical losses are made of the following:

• Stator winding fundamental losses: pcos
• Rotor winding fundamental losses: pcor
• Stator fundamental core losses: pirons
• Rotor fundamental core losses: pironR
• Stator surface core losses: $p i r o n S S$
• Rotor surface core losses: $p i r o n r r$
• Stator flux pulsation losses: $p i r o n S P$
• Rotor flux pulsation losses: $p i r o n R P$
• Rotor slip-ring and brush losses: psrb

As the skin effect was shown small, it will be considered only by the correction coefficient KR = 1.037, already calculated in Equation 3.33. For the rotor, the skin effect is neglected, as the maximum frequency Smaxf1 = 0.25f1. In any case, it is smaller than in the stator, because the rotor slots are not as deep as the stator slots:

3.85 $p c o s = 3 K R R S I S N 2 = 3 ⋅ 1.037 ⋅ 0.429 ⋅ 10 − 2 ⋅ 1675 2 = 37464.52 W = 37.47 kW$
3.86 $p c o r = 3 R R R I R N 2 = 3 ⋅ 7.62 ⋅ 10 − 2 ⋅ 437 2 = 43715.47 W = 43.715 kW$

The slip-ring and brush losses psr are easy to calculate if the voltage drop along them is given, say VSR ≈1 V. Consequently,

3.87 $p s r = 3 V S R I R R = 3 ⋅ 1 ⋅ 437.3 = 1311.9 W = 1.3119 kW$

To calculate the stator fundamental core losses, the stator teeth and back iron weights Gts and Gcs are needed:

3.88 $G t s = { π 4 [ ( D i s + 2 ( h s u + h s w ) ) 2 − D i s 2 ] − N s ⋅ ( h s u + h s w ) } l i γ i r o n = { π 4 [ ( 0.52 + 2 ( 70.315 + 3 ) 10 − 3 ) 2 ] − 60 ⋅ ( 70.315 + 3 ) 10 − 3 ⋅ 13.3 ⋅ 10 − 3 } ⋅ 0.52 ⋅ 7600 ≈ 313.8 kg$
3.89 $G c s ≈ π ( D o u t − h c s ) ⋅ h c s ⋅ l i ⋅ γ i r o n = π ( 0.8 − 0.0637 ) ⋅ 0.0637 ⋅ 0.52 ⋅ 7600 = 582 kg$

The fundamental core losses in the stator, considering the mechanical machining influence by fudge factors such as Kt = 1.6–1.8 and Ky = 1.3–1.4, are as follows [1]:

3.90 $P i r o n s ≈ p 10 / 50 ( f 1 50 ) 1.5 ( K t B t s 2 ⋅ G t s + K y B c s 2 G c s )$

With Bts = 1.488 T, Bcs = 1.5 T, f1 = 50 Hz, and p10/50 = 3 W/kg (losses at 1 T and 50 Hz), $P i r o n s = 3 ⋅ ( 50 50 ) 1.3 [ 1.6 ⋅ ( 1.488 ) 2 ⋅ 313.8 + 1.3 ⋅ ( 1.5 ) 2 ⋅ 582 ] = 8442 W = 8.442 kW$

The rotor fundamental core losses may be calculated in a similar manner, but with f1 replaced by Sf1, and introducing the corresponding weights and flux densities.

As the Smax = 0.25, even if the lower rotor core weights will be compensated for by the larger flux densities, the rotor fundamental iron losses would be as follows:

3.91 $P i r o n r < S max 2 ⋅ P i r o n s = 1 16 ⋅ 8442 = 527.6 W = 0.527 kW$

The surface and pulsation additional core losses, known as strayload losses, are dependent on the slot opening/air gap ratio in the stator and in the rotor [1]. In our design, magnetic wedges reduce the slot-openings-to-air gap ratio to 4.066/1.612 and 3.04/1.612; thus, the surface additional core losses are reduced. For the same reason, the stator and rotor Carter coefficients Kc1 and Kc2 are small (Kc1·Kc2 = 1.126); and therefore, the flux-pulsation additional core losses are also reduced.

Consequently, all additional losses are most probably well within the 0.5% standard value (for detailed calculations, see Reference 1, Chapter 11) of stator rated power:

3.92 $p a d d = p i r o n S S + p i r o n S R + p i r o n S P + p i r o n R P = 0.5 100 ⋅ 2 , 000 ⋅ 10 6 = 10 , 000 W = 10 kW$

Thus, the total electrical losses ∑pe are

3.93 $Σ p e = p c o s + p c o r + p i r o n s + p i r o n r + p a d + p s r = 37.212 + 43.715 + 8.442 + 10 + 0.527 + 1.3119 = 101.20 kW$

Neglecting the mechanical losses, the “electrical” efficiency ηe is

3.94 $η e = P S N + P R N P S N + P R N + Σ p e = ( 2 + 0.5 ) 10 6 ( 2 + 0.5 ) 10 6 + 0.101 ⋅ 10 6 = 0.9616$

This is not a very large value, but it was obtained with the machine size reduction in mind.

It is now possible to redo the whole design with smaller fxt (rotor shear stress), and lower current densities to finally increase efficiency for larger size. The length of the stack li is also a key parameter to design improvements. Once the above, or similar, design methodology is computerized, then various optimization techniques may be used, based on objective functions of interest, to end up with a satisfactory design (see Reference 1, Chapter 10).

Finite element analysis (FEA) verifications of the local magnetic saturation, core losses, and torque production should be instrumental in validating optimal designs based on even advanced analytical nonlinear models of the machine.

Mechanical and thermal designs are also required, and FEA may play a key role here, but this is beyond the scope of our discussion [2,3]. In addition, uncompensated magnetic radial forces have to be checked, as they tend to be larger in WRIG (due to the absence of rotor cage damping effects) [4].

#### 3.8  Testing of WRIGs

The experimental investigation of WRIGs at the manufacturer’s or user’s sites is an indispensable tool to validate machine performance.

There are international (and national) standards that deal with the testing of general use induction machines with cage or wound rotors (International Electrotechnical Commission [IEC]-34, National Electrical Manufacturers Association [NEMA] MG1-1994 for large induction machines).

Temperature, losses, efficiency, unbalanced operation, overload capability, dielectric properties of insulation schemes, noise, surge responses, and transients (short circuit) responses are all standardized.

We avoid a description of such tests [5] here, as space would be prohibitive, and the reader could read the standards above (and others) by himself. Therefore, only a short discussion, for guidance, will be presented here.

Testing is performed for performance assessment (losses, efficiency, endurance, noise) or for parameter estimation. The availability of rotor currents for measurements greatly facilitates the testing of WRIGs for performance and for machine parameters. On the other hand, the presence of the bidirectional power flow static converter connected to the rotor circuits, through slip-rings and brushes, poses new problems in terms of current and flux time harmonics and losses.

The IGBT static converters introduce reduced current time harmonics, but measurements are still needed to complement digital simulation results.

In terms of parameters, the rotor and stator are characterized by single circuits with resistances and leakage inductances, besides the magnetization inductance. The latter depends heavily on the level of air gap flux, whereas the leakage inductances decrease slightly with their respective currents.

As for WRIGs, the stator voltage and frequency stay rather constant, and the stator flux $Ψ ¯ s$

varies only a little. The air gap flux $Ψ ¯ m$
3.95 $Ψ ¯ m = Ψ ¯ s − L s l I ¯ s$

varies a little with load for unity stator power factor:

3.96 $I s R s + V s + j ω 1 L s l I s = − j ω 1 Ψ ¯ m$

The magnetic saturation level of the main flux path does not vary much with load. Therefore, Lm does not vary much with load. In addition, unless Is/Isn > 2, the leakage inductances, even with magnetic wedges on slot tops, do not vary notably with respective currents. Therefore, parameters estimation for dealing with the fundamental behavior is greatly simplified in comparison with cage rotor induction motors with rotor skin effect.

On the other hand, the presence of time harmonics, due to the static converter of partial ratings, requires the investigation of these effects by estimating adequate machine parameters of WRIG with respect to them. Online data acquisition of stator and rotor currents and voltages is required for the scope.

The adaptation of tests intended for cage rotor induction machines to WRIGs is rather straightforward; thus, the following may all be performed for WRIGs with even better precision, because the rotor parameters and currents are directly measurable (for details see Reference 1, Chapter 22):

• Loss segregation tests
• Load testing (direct and indirect)
• Machine parameter estimation
• Noise testing methodologies

#### 3.9  Summary

• WRIGs are built for powers in the 1.5–400 MW and more per unit.
• A flexible structure DFIG recent design at 10 MW and 10 rpm, 50 Hz shows promising potential [6].
• With today’s static power converters for the rotor, the maximum rotor voltage may go to 4.2–6 kV, but the high-voltage direct current (HVDC) transmission lines techniques may extend it further for very high powers per unit.
• For stator voltages below 6 kV, the rotor voltage at maximum slip will be adapted to be equal to stator voltage; thus, no transformer is required to connect the bidirectional AC–AC converter in the rotor circuit to the local power grid. This way, the rotor current is limited around |Smax|ISN; ISN rated stator current.
• If stator unity power factor is desired, the magnetization current is provided in the rotor; the rotor rated current is slightly increased and so is the static converter partial rating. This oversizing is small, however (less than 10%, in general).
• For massive reactive power delivery by the stator, but for unity power factor at the converter supply-side terminals, all the reactive power has to be provided via the capacitor in the DC link, which has to be designed accordingly. As for Qs reactive power delivered by the stator, only SQs has to be produced by the rotor; the oversizing of the rotor connected static power converter still seems to be reasonable.
• The emf design of WRIGs as generators is performed for maximum power at maximum (supersynchronous) speed. The torque is about the same as that for stator rated power at synchronous speed (with DC in the rotor).
• The emf design basically includes stator and rotor core and windings design, magnetization current, circuit parameter losses, and efficiency computation.
• The sizing of WRIGs starts with the computation of stator bore diameter Dis based either on Esson’s coefficient or on the shear rotor stress concept fxt ≈ 2.5–8 N/cm2. The latter path was taken in this chapter. Flux densities in air gap, teeth, and back irons were assigned initial ­values, together with the current densities for rated currents. (The magnetization current was also assigned an ­initial P.U. value.)
• Based on these variables, the sizing of the stator and the rotor went smoothly. For the 2.5 MW numerical example, the skin effects in the stator and rotor were proven to be small (≤4%).
• The magnetization current was recalculated and found to be smaller than the initial value, so the design holds; otherwise, the design should have been redone with smaller fxt.
• Rather uniform saturation of various iron parts—teeth and back cores—leads to rather sinusoidal time waveforms of flux density and, thus, to smaller core losses in a heavily saturated design. This was the case for the example in this chapter.
• The computation of resistances and leakage reactances is straightforward. It was shown that the differential leakage flux contribution (due to mmf space harmonics) should not be neglected, as it is important, though q1 = 5, q2 = 4 and proper coil chording is applied both in the stator and in the rotor windings.
• Open slots were adopted for both the stator and the rotor for easing the fabrication and insertion of windings in slots. However, magnetic wedges are mandatory to reduce additional (surface and flux pulsation) core losses and magnetization current.
• Though mechanical design and thermal design are crucial, they fall beyond our scope here, as they are strongly industry-knowledge dependent.
• The testing of WRIGs is standardized mainly for motors. Combining these tests with synchronous generator tests may help generate a set of comprehensive, widely accepted testing technologies for WRIGs.
• Though important, the design of a WRIG as a brushless exciter with rotor output was not pursued in this chapter, mainly due to the limited industrial use of this mode of operation.

#### References

I. Boldea and S.A. Nasar, Induction Machine Handbook, CRC Press, Boca Raton, FL, 1998.
E. Levi, Polyphase Motors—A Direct Approach to Their Design, Wiley Interscience, New York, 1985.
K. Vogt, Electrical Machines: Design of Rotary Machines, 4th edn., VEB Verlag, Berlin, Germany, 1988 (in German).
D.G. Dorrel, Experimental behavior of unbalanced magnetic pull in 3-phase induction motors with eccentric rotors and the relationship with teeth saturation, IEEE Trans., EC-14(3), 1999, 304–309.
Institute of Electrical and Electronics Engineers (IEEE), IEEE Standard Procedure for Polyphase Induction Motors and Generators, IEEE Standard 112-1996, IEEE Press, New York, 1996.
V. Delli Colli, F. Marignetti, and C. Attaianese, Analytical and multiphysics approach to the optimal design of a 10 MW DFIG for direct drive wind turbines, IEEE Trans., IE-59(7), 2012, 2791–2799.

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