Switched Reluctance Motor Drives: Fundamentals to Applications (Solutions, Instructor Solution Manual) [1 ed.] 113830459X, 9781138304598

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Solution Manual Table of Contents Chapter 2 Electromagnetic Principles of Switched Reluctance Machines ........................................... 4 Q2.1 Solution ........................................................................................................................................... 4 Q2.2 Solution ........................................................................................................................................... 4 Q2.3 Solution ........................................................................................................................................... 5 Q2.4 Solution ........................................................................................................................................... 6 Q2.5 Solution ........................................................................................................................................... 7 Q2.6 Solution ........................................................................................................................................... 8 Q2.7 Solution ........................................................................................................................................... 9 Chapter 3 Derivation of Pole Configuration in Switched Reluctance Machines ................................ 11 Q3.1 Solution ......................................................................................................................................... 11 Q3.2 Solution ......................................................................................................................................... 11 Q3.3 Solution ......................................................................................................................................... 13 Q3.4 Solution ......................................................................................................................................... 15 Q3.5 Solution ......................................................................................................................................... 17 Q3.6 Solution ......................................................................................................................................... 18 Q3.7 Solution ......................................................................................................................................... 19 Q3.8 Solution ......................................................................................................................................... 19 Chapter 4 Operational Principles and Modeling of Switched Reluctance Machines ......................... 20 Q4.1 Solution ......................................................................................................................................... 20 Q4.2 Solution ......................................................................................................................................... 21 Q4.3 Solution ......................................................................................................................................... 21 Q4.4 Solution ......................................................................................................................................... 22 Q4.5 Solution ......................................................................................................................................... 23 Q4.6 Solution ......................................................................................................................................... 23 Q4.7 Solution ......................................................................................................................................... 24 Q4.8 Solution ......................................................................................................................................... 26 Q4.9 Solution ......................................................................................................................................... 27 Q4.10 Solution ....................................................................................................................................... 28 Q4.11 Solution ....................................................................................................................................... 29 Q4.12 Solution ....................................................................................................................................... 30 Chapter 5 Switched Reluctance Machines in Generating Mode .......................................................... 32 Chapter 6 Materials Used in Switched Reluctance Machines .............................................................. 34 Q6.1 Solution ......................................................................................................................................... 34 Q6.2 Solution ......................................................................................................................................... 36 Chapter 7 Design Considerations for Switched Reluctance Machines ............................................... 37

1

Q7.1 Solution ......................................................................................................................................... 37 Q7.2 Solution ......................................................................................................................................... 37 Q7.3 Solution ......................................................................................................................................... 38 Q7.4 Solution ......................................................................................................................................... 38 Q7.5 Solution ......................................................................................................................................... 39 Chapter 8 Mechanical Construction of Switched Reluctance Machines ............................................ 42 Q8.1 Solution ......................................................................................................................................... 42 Q8.2 Solution ......................................................................................................................................... 42 Q8.3 Solution ......................................................................................................................................... 43 Q8.4 Solution ......................................................................................................................................... 43 Q8.5 Solution ......................................................................................................................................... 43 Chapter 9 Control of Switched Reluctance Machines ........................................................................... 45 Q9.1 Solution ......................................................................................................................................... 45 Q9.2 Solution ......................................................................................................................................... 45 Q9.3 Solution ......................................................................................................................................... 45 Q9.4 Solution ......................................................................................................................................... 46 Q9.5 Solution ......................................................................................................................................... 46 Q9.6 Solution ......................................................................................................................................... 46 Q9.7 Solution ......................................................................................................................................... 47 Q9.8 Solution ......................................................................................................................................... 47 Q9.9 Solution ......................................................................................................................................... 48 Q9.10 Solution ....................................................................................................................................... 48 Q9.11 Solution ....................................................................................................................................... 51 Chapter 10 Power Electronic Converter to Drive Switched Reluctance Machines ............................ 53 Q10.1 Solution ....................................................................................................................................... 53 Q10.2 Solution ....................................................................................................................................... 53 Q10.3 Solution ....................................................................................................................................... 53 Q10.4 Solution ....................................................................................................................................... 54 Q10.5 Solution ....................................................................................................................................... 54 Q10.6 Solution ....................................................................................................................................... 55 Q10.7 Solution ....................................................................................................................................... 55 Chapter 11 Position Sensorless Control of Switched Reluctance Machines ..................................... 56 Q11.1 Solution ....................................................................................................................................... 56 Q11.2 Solution ....................................................................................................................................... 56 Q11.3 Solution ....................................................................................................................................... 56 Q11.4 Solution ....................................................................................................................................... 57 Q11.5 Solution ....................................................................................................................................... 57 Q11.6 Solution ....................................................................................................................................... 57 Q11.7 Solution ....................................................................................................................................... 58 2

Chapter 12 Fundamentals of Vibrations and Acoustic Noise .............................................................. 59 Q12.1 Solution ....................................................................................................................................... 59 Q12.2 Solution ....................................................................................................................................... 61 Q12.3 Solution ....................................................................................................................................... 70 Q12.4 Solution ....................................................................................................................................... 73 Q12.5 Solution ....................................................................................................................................... 74 Q12.6 Solution ....................................................................................................................................... 77 Q12.7 Solution ....................................................................................................................................... 80 Chapter 13 Noise and Vibration in Switched Reluctance Machines .................................................... 85 Q13.1 Solution ....................................................................................................................................... 85 Q13.2 Solution ....................................................................................................................................... 89 Q13.3 Solution ....................................................................................................................................... 95 Q13.4 Solution ....................................................................................................................................... 96 Q13.5 Solution ....................................................................................................................................... 97 Chapter 14 Thermal Management of Switched Reluctance Machines .............................................. 105 Q14.1 Solution ..................................................................................................................................... 105 Q14.2 Solution ..................................................................................................................................... 105 Q14.3 solution ...................................................................................................................................... 105 Q14.4 solution ...................................................................................................................................... 106 Chapter 15 Axial Flux Switched Reluctance Machines ....................................................................... 107 Q15.1 Solution ..................................................................................................................................... 107 Q15.2 Solution ..................................................................................................................................... 107

3

Chapter 2 Electromagnetic Principles of Switched Reluctance Machines Q2.1 Solution

S: cross sectional area of the Current density conductor J q: charge of an electron ne:# of electrons per m3

Magnetic quantities Ampere’s Law

Current 



iJ dS 

N: # of turns

resistivity

 : time between

collusion of free electrons Acceleration



External magnetic field strength 1/ℓc: length of the magnetic flux path

 S

Drift velocity vd

Magneto-motive force F=Ni

me 2

Induced voltage ε

ae 1  Newton’s 1/me: mass of conductor an electron Law length q: charge of Induced electric Force induced on an electron field free electrons   Coulomb’s E d F  Law Electrical quantities

flux linkage λ=N· φ reluctance

magnetic force F=JxB

q ne

H

N



 r  0 Ac

d dt

Magnetic flux φ

Faraday’s Law

permeability of ferromagnetic material: μ=μrμ0

lc





sB

da

da: infinitesimal area through the magnetic flux flows

Magnetic flux density B

Fig. 1. Block diagram for the principle equations of a magnetic system.

Q2.2 Solution For copper at room temperature: 𝑞 = 1.6 × 10−19 𝐶 𝑚𝑒 = 10−30 𝑘𝑔 𝜏 = 3 × 10−14 𝑠

𝐸=

𝑣⃗𝑑 =

𝜀 10 𝑉 = = 1 𝑉 ⁄𝑚 ℓ 10 𝑚

𝑞𝐸⃗⃗ 𝜏 = 5 × 10−3 𝑚/𝑠 𝑚𝑒

Without the electric field applied, the speed of the electrons was about 1 million meters per second in all directions due to thermal motion. When the electric field is present, the electrons now move 5 millimeters

4

per second. This speed is significantly slower than the speed of light. Therefore, the conductors act as a guide for the energy, but the electromagnetic energy is propagated in the dielectric material. Q2.3 Solution 20

(0,0)

(25,0)

(20,-5)

(5,-5)

12.25

(20,-14.75) 25 (20,-15.25)

(5,-25)

(25,-15.25)

(20,-25) y

(0,-30)

(25,-30)

x

Fig. 2. Dimensional parameters of a magnetic system with airgap

𝑙𝑐 = (ℎ − 𝑑1 ) + (𝑤 − 𝑑1 ) × 2 + (ℎ − 𝑑1 − 𝑙𝑔 ) 𝑙𝑐 = 25 + 20 × 2 + 12.25 × 2 = 89.5 𝑚𝑚 = 0.0895 𝑚 𝑙𝑔 = 0.5 𝑚𝑚 = 5 × 10−4 𝑚 𝐴𝑐 = (𝑑2 × 𝐿𝑠𝑡 ) = (5 × 10) = 50 𝑚𝑚2 = 5 × 10−5 𝑚2 𝜇𝑟 = 1000

𝑅𝑒𝑞 =

𝜇0 = 4𝜋 × 10−7

𝑙𝑔 𝑙𝑐 + = 7.9577 × 106 + 1.4244 × 106 = 9.3822 × 106 𝜇 0 𝐴𝑐 𝜇𝑟 𝜇0 𝐴𝑐

𝜑=

𝑁𝑖 10 × 20 = = 2.1317 × 10−5 𝑊𝑏 𝑅𝑒𝑞 9.3822 × 106

5

𝜆 = 𝑁𝜑 = 2.1317 × 10−4 𝑊𝑏 𝐻𝑐 = 𝑅𝑐 𝜑 = 30.3647 𝐴/𝑚 𝐻𝑔 = 𝑅𝑔 𝜑 = 169.6353 𝐴/𝑚

𝐵=

𝜑 = 0.4263 𝑇 𝐴𝑐

Q2.4 Solution Create variables in MATLAB using the data given in Table 2.1: [time]31x1

[flux_linkage]31x1

[voltage]31x1

resistance=0.2117

current=20

for i=2:length(time) voltage_calc_time(i)=resistance*current+(flux_linkage(i)-flux_linkage(i1))/(time(i)-time(i-1)); end plot(angle,voltage,'-k','LineWidth',2); hold on plot(angle,voltage_calc_time,'--r','LineWidth',2); xlim([0 max(angle)]);

The terminal voltage can also be calculated using the rotor position:

𝑣 = 𝑅𝑖 +

𝑑𝜆 𝑑𝜃 𝑑𝜆 𝑑𝜆 = 𝑅𝑖 + = 𝑅𝑖 + 𝜔 𝑑𝑡 𝑑𝑡 𝑑𝜃 𝑑𝜃

where 𝜔 is the rotational speed in rad/s. In Table 2.1, the flux linkage and rotor position vectors are given as [𝜆]𝑡×1 and [𝜃]𝑡×1 , respectively. Therefore, the terminal voltage can be calculated as

[𝑣]𝑡−1 × 1 = 𝑅𝑖 + 𝜔

𝜆(𝑡, 1) − 𝜆(𝑡 − 1,1) 𝜃(𝑡) − 𝜃(𝑡 − 1)

Create variable in MATLAB using the data given in Table 2.1: [angle]31x1

speed_rpm=1000

rps=rpm × (1/60)

deg/s=rps × 360

rad/s=deg/s × (π/180)

6

speed=speed_rpm*360*(pi/180)*(1/60) for i=2:length(angle) voltage_calc_pos(i)=resistance*current+((flux_linkage(i)-flux_linkage(i1))/(angle(i)*(pi/180)-angle(i-1)*(pi/180)))*speed; end plot(angle,voltage,'-k','LineWidth',2); hold on plot(angle,voltage_calc_pos,'+b','LineWidth',2); xlim([0 max(angle)]);

60

50

Voltage [V]

40

30

20 Voltage Voltage_calc_time Voltage_calc_pos

10

0 0

5

10

15

20

Mechanical angle [deg.]

Fig. 3. Terminal voltage versus rotor position.

Q2.5 Solution Create variables in MATLAB using the data given in Table 2.1: [angle]31x1

[voltage]31x1

rps=rpm × (1/60)

resistance=0.2117 deg/s=rps × 360

current=20

speed_rpm=1000

rad/s=deg/s × (π/180)

speed=speed_rpm*360*(pi/180)*(1/60); flux_linkage0=flux_linkage(1); for i=2:length(angle) sum=0; for j=2:i temp_voltage=voltage(j); temp_pos1=angle(j)*(pi/180); temp_pos2=angle(j-1)*(pi/180); temp=(1/speed)*(temp_voltage-resistance*current)*(temp_pos1temp_pos2); 7

sum=sum+temp; end flux_linkage_calc(i)=sum+flux_linkage0; end plot(angle,flux_linkage,'-k','LineWidth',2); hold on plot(angle,flux_linkage_calc,'--r','LineWidth',2) xlim([0 max(angle)]);

0.15 Flux_linkage

Flux linkage [Wb]

Flux_linkage_calc

0.1

0.05

0

0

5

10

15

20

Mechanical angle [deg.]

Fig. 4. Flux linkage versus rotor position.

Q2.6 Solution The figure below shows the electromagnetic torque from the FEA analysis and the torque derived from the flux linkage characteristics, which were calculated from the same FEA analysis. It can be observed that the torque profiles match well.

8

7 Torque Torque from flux linkage

6

Torque [Nm]

5

4

3

2

1

0

0

5

10 15 Mechanical angle [deg]

20

22.5

Fig. 5. Static torque versus rotor position.

Torque can be calculated in two different ways. Maxwell Stress Tensor method, which was used in question 6, calculates the torque from the tangential forces. Energy method, which was used in this example, calculates the torque from co-energy. Both methods are widely used in electromagnetic analysis. Q2.7 Solution The general torque expression in SRM was derived as:

Te =

∂Wc | ∂θ i=const

where Wc is the co-energy and θ is the rotor position. The magnetic stored energy was calculated for the linear case, for λ = Li: λ2 i 1 Wf = ∫ idλ = ∫ Lidi = Li2 2 λ1 0

where λ is the flux linkage.

9

In a magnetic system with no saturation, the magnetization curve should be a straight line. Therefore at any position:

Wf = Wc =

1 L(θ)i2 2

Therefore,

Te =

∂Wc ∂ 1 1 ∂L(θ) | = ( L(θ)i2 )| = i2 ∂θ i=const ∂θ 2 2 ∂θ i=const

10

Chapter 3 Derivation of Pole Configuration in Switched Reluctance Machines Q3.1 Solution If the stator pole pairs in an SRM share the same electrical angle at one rotor position, when the same current is applied to the coils of these stator poles, they generate the same torque on the rotor. This is accomplished by connecting the coils of these stator pole pairs in the same electrical circuit, which creates the phases. Q3.2 Solution Mechanical position of the rotor poles can be calculated using (3.2). 𝑇𝑝𝑟 =

360 , 𝜃 [𝑡] = 𝑇𝑝𝑟 (𝑡 − 1), 𝑁𝑟 𝑟

𝑡 = 1,2, … , 𝑁𝑟

Mechanical position of the stator poles can be calculated using (3.3) 𝑇𝑝𝑠 =

360 , 𝜃 [𝑡] = 𝑇𝑝𝑠 (𝑡 − 1), 𝑁𝑠 𝑠

𝑡 = 1,2, … , 𝑁𝑠

Electrical angle can be calculated using (3.5) 𝑁𝑠𝑒𝑙𝑒𝑐𝑡 = 𝑚𝑜𝑑((𝑁𝑟𝑚𝑒𝑐ℎ − 𝑁𝑠𝑚𝑒𝑐ℎ ) 𝑁𝑟 + 1800 , 360).

a) 6/14 Tab. 1. Mechanical positions of stator and rotor poles, 6/14 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

25.7143

Ns2

60

Ns2

60

Ph2

Nr3

51.4286

Ns3

120

Ns3

300

Ph3

Nr4

77.1429

Ns4

180

Ns4

180

Ph1

Nr5

102.8571

Ns5

240

Ns5

60

Ph2

Nr6

128.5714

Ns6

300

Ns6

300

Ph3

Nr7

154.2857

Nr8

180

Nr9

205.7143

Nr10

231.4286

Nr11

257.1429

Nr12

282.8571

Nr13

308.5714

Nr14

334.2857

Ph1

11

b) 8/10 Tab. 2. Mechanical positions of stator and rotor poles, 8/10 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

36

Ns2

45

Ns2

90

Ph2

Nr3

72

Ns3

90

Ns3

0

Ph3

Nr4

108

Ns4

135

Ns4

270

Ph4

Nr5

144

Ns5

180

Ns5

180

Ph1

Nr6

180

Ns6

225

Ns6

90

Ph2

Nr7

216

Ns7

270

Ns7

0

Ph3

Nr8

252

Ns8

315

Ns8

270

Ph4

Nr9

288

Nr10

324

Ph1

c) 16/20 Tab. 3. Mechanical positions of stator and rotor poles, 16/20 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

18

Ns2

22.5

Ns2

90

Ph2

Nr3

36

Ns3

45

Ns3

0

Ph3

Nr4

54

Ns4

67.5

Ns4

270

Ph4

Nr5

72

Ns5

90

Ns5

180

Ph1

Nr6

90

Ns6

112.5

Ns6

90

Ph2

Nr7

108

Ns7

135

Ns7

0

Ph3

Nr8

126

Ns8

157.5

Ns8

270

Ph4

Nr9

144

Ns9

180

Ns9

180

Ph1

Nr10

162

Ns10

202.5

Ns10

90

Ph2

Nr11

180

Ns11

225

Ns11

0

Ph3

Nr12

198

Ns12

247.5

Ns12

270

Ph4

Nr13

216

Ns13

270

Ns13

180

Ph1

Nr14

234

Ns14

292.5

Ns14

90

Ph2

Nr15

252

Ns15

315

Ns15

0

Ph3

Nr16

270

Ns16

337.5

Ns16

270

Ph4

Nr17

288

Nr18

306

Nr19

324

Nr20

342

Ph1

d) 20/16 Tab. 4. Mechanical positions of stator and rotor poles, 20/16 SRM. Rotor

Mech.

Stator

Mech.

Stator

Elect.

Phases

12

Pole Nr1

Angle 0

Pole Ns1

Angle 0

Pole Ns1

Angle 180

Ph1

Nr2

22.5

Ns2

18

Ns2

252

Ph2

Nr3

45

Ns3

36

Ns3

324

Ph3

Nr4

67.5

Ns4

54

Ns4

36

Ph4

Nr5

90

Ns5

72

Ns5

108

Ph5

Nr6

112.5

Ns6

90

Ns6

180

Ph1

Nr7

135

Ns7

108

Ns7

252

Ph2

Nr8

157.5

Ns8

126

Ns8

324

Ph3

Nr9

180

Ns9

144

Ns9

36

Ph4

Nr10

202.5

Ns10

162

Ns10

108

Ph5

Nr11

225

Ns11

180

Ns11

180

Ph1

Nr12

247.5

Ns12

198

Ns12

252

Ph2

Nr13

270

Ns13

216

Ns13

324

Ph3

Nr14

292.5

Ns14

234

Ns14

36

Ph4

Nr15

315

Ns15

252

Ns15

108

Ph5

Nr16

337.5

Ns16

270

Ns16

180

Ph1

Ns17

288

Ns17

252

Ph2

Ns18

306

Ns18

324

Ph3

Ns19

324

Ns19

36

Ph4

Ns20

342

Ns20

108

Ph5

Q3.3 Solution a) 9/15 Tab. 5. Mechanical positions of stator and rotor poles, 9/15 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

24

Ns2

40

Ns2

300

Ph2

Nr3

48

Ns3

80

Ns3

60

Ph3

Nr4

72

Ns4

120

Ns4

180

Ph1

Nr5

96

Ns5

160

Ns5

300

Ph2

Nr6

120

Ns6

200

Ns6

60

Ph3

Nr7

144

Ns7

240

Ns7

180

Ph1

Nr8

168

Ns8

280

Ns8

300

Ph2

Nr9

192

Ns9

320

Ns9

60

Ph3

Nr10

216

Nr11

240

Nr12

264

Nr13

288

Nr14

312

Nr15

336

Ph1

b) 12/15 13

Tab. 6. Mechanical positions of stator and rotor poles, 12/15 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

24

Ns2

30

Ns2

90

Ph2

Nr3

48

Ns3

60

Ns3

0

Ph3

Nr4

72

Ns4

90

Ns4

270

Ph4

Nr5

96

Ns5

120

Ns5

180

Ph1

Nr6

120

Ns6

150

Ns6

90

Ph2

Nr7

144

Ns7

180

Ns7

0

Ph3

Nr8

168

Ns8

210

Ns8

270

Ph4

Nr9

192

Ns9

240

Ns9

180

Ph1

Nr10

216

Ns10

270

Ns10

90

Ph2

Nr11

240

Ns11

300

Ns11

0

Ph3

Nr12

264

Ns12

330

Ns12

270

Ph4

Nr13

288

Nr14

312

Nr15

336

Ph1

c) 15/12 Tab. 7. Mechanical positions of stator and rotor poles, 15/12 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

30

Ns2

24

Ns2

252

Ph2

Nr3

60

Ns3

48

Ns3

324

Ph3

Nr4

90

Ns4

72

Ns4

36

Ph4

Nr5

120

Ns5

96

Ns5

108

Ph5

Nr6

150

Ns6

120

Ns6

180

Ph1

Nr7

180

Ns7

144

Ns7

252

Ph2

Nr8

210

Ns8

168

Ns8

324

Ph3

Nr9

240

Ns9

192

Ns9

36

Ph4

Nr10

270

Ns10

216

Ns10

108

Ph5

Nr11

300

Ns11

240

Ns11

180

Ph1

Nr12

330

Ns12

264

Ns12

252

Ph2

Ns13

288

Ns13

324

Ph3

Ns14

312

Ns14

36

Ph4

Ns15

336

Ns15

108

Ph5

Ph1

14

In all of these configurations there is odd number of stator poles in each phase. These configurations cannot provide a balanced operation if non-coupled winding configuration is used. Mutually coupled winding configuration should be used. Q3.4 Solution a) 6/12 Tab. 8. Mechanical positions of stator and rotor poles, 6/12 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Nr2

30

Ns2

60

Ns2

180

Nr3

60

Ns3

120

Ns3

180

Nr4

90

Ns4

180

Ns4

180

Nr5

120

Ns5

240

Ns5

180

Nr6

150

Ns6

300

Ns6

180

Nr7

180

Nr8

210

Nr9

240

Nr10

270

Nr11

300

Nr12

330

All the stator poles are at 180 degree electrical. This is zero-torque equilibrium point in an SRM. No torque can be generated when any of the stator poles are excited. This is locked rotor condition.

b) 8/20 Tab. 9. Mechanical positions of stator and rotor poles, 8/20 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Nr2

18

Ns2

45

Ns2

0

Nr3

36

Ns3

90

Ns3

180

Nr4

54

Ns4

135

Ns4

0

Nr5

72

Ns5

180

Ns5

180

Nr6

90

Ns6

225

Ns6

0

Nr7

108

Ns7

270

Ns7

180

15

Nr8

126

Nr9

144

Nr10

162

Nr11

180

Nr12

198

Nr13

216

Nr14

234

Nr15

252

Nr16

270

Nr17

288

Nr18

306

Nr19

324

Nr20

342

Ns8

315

Ns8

0

All the stator poles are either at 180 or zero degree electrical. These are both zero-torque equilibrium points in an SRM. No torque can be generated when any of the stator poles are excited. This is locked rotor condition.

c) 8/7 Tab. 10. Mechanical positions of stator and rotor poles, 8/7 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Nr2

51.4286

Ns2

45

Ns2

225

Nr3

102.8571

Ns3

90

Ns3

270

Nr4

154.2857

Ns4

135

Ns4

315

Nr5

205.7143

Ns5

180

Ns5

0

Nr6

257.1429

Ns6

225

Ns6

45

Nr7

308.5714

Ns7

270

Ns7

90

Ns8

315

Ns8

135

All of the stator poles have different electrical angles. When excited, these poles generate different torques. The coils around these stator poles cannot be connected together to create a phase. Current going each coil has to be controlled individually. It will be very complicated and challenging to maintain balanced operation.

16

d) 12/10 Tab. 11. Mechanical positions of stator and rotor poles, 12/10 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

Phases

Nr2

36

Ns2

30

Ns2

240

Ph2

Nr3

72

Ns3

60

Ns3

300

Ph3

Nr4

108

Ns4

90

Ns4

0

Ph4

Nr5

144

Ns5

120

Ns5

60

Ph5

Nr6

180

Ns6

150

Ns6

120

Ph6

Nr7

216

Ns7

180

Ns7

180

Ph1

Nr8

252

Ns8

210

Ns8

240

Ph2

Nr9

288

Ns9

240

Ns9

300

Ph3

Nr10

324

Ns10

270

Ns10

0

Ph4

Ns11

300

Ns11

60

Ph5

Ns12

330

Ns12

120

Ph6

Ph1

In 12/10 SRM, for 3- and 4-phases, even though the number of stator poles per phase is an integer, the stator poles belonging to the same phase do not share the same electrical angle. This configuration can provide balanced operation with 6 phases.

Q3.5 Solution Tab. 12. Mechanical positions of stator and rotor poles, 18/24 SRM. Rotor Pole Nr1

Mech. Angle 0

Stator Pole Ns1

Mech. Angle 0

Stator Pole Ns1

Elect. Angle 180

3-phase

6-phase

9-phase

Ph1

Ph1

Ph1

Nr2

15

Ns2

20

Ns2

60

Ph2

Ph2

Ph2

Nr3

30

Ns3

40

Ns3

300

Ph3

Ph3

Ph3

Nr4

45

Ns4

60

Ns4

180

Ph1

Ph4

Ph4

Nr5

60

Ns5

80

Ns5

60

Ph2

Ph5

Ph5

Nr6

75

Ns6

100

Ns6

300

Ph3

Ph6

Ph6

Nr7

90

Ns7

120

Ns7

180

Ph1

Ph1

Ph7

Nr8

105

Ns8

140

Ns8

60

Ph2

Ph2

Ph8

Nr9

120

Ns9

160

Ns9

300

Ph3

Ph3

Ph9

Nr10

135

Ns10

180

Ns10

180

Ph1

Ph4

Ph1

Nr11

150

Ns11

200

Ns11

60

Ph2

Ph5

Ph2

Nr12

165

Ns12

220

Ns12

300

Ph3

Ph6

Ph3

Nr13

180

Ns13

240

Ns13

180

Ph1

Ph1

Ph4

Nr14

195

Ns14

260

Ns14

60

Ph2

Ph2

Ph5

17

Nr15

210

Ns15

280

Ns15

300

Ph3

Ph3

Ph6

Nr16

225

Ns16

300

Ns16

180

Ph1

Ph4

Ph7

Nr17

240

Ns17

320

Ns17

60

Ph2

Ph5

Ph8

Nr18

255

Ns18

340

Ns18

300

Ph3

Ph6

Ph9

Nr19

270

Nr20

285

Nr21

300

Nr22

315

Nr23

330

Nr24

345

An 18/24 SRM can be configured as a 3-, 6- or 9-phase machine. However, with 6- and 9-phases, it can be noticed that different phases share the same electrical angles. Therefore, 6- and 9-phase operation is similar to parallel operation of some of the phases and might be helpful in high power applications. Since the 18/24 SRM can be operated as a 3-phase machine with an even number of stator poles per phase, mutually-coupled winding configuration is not necessary for 6-phase operation even though stator poles per phase is an odd number.

Q3.6 Solution The number of rotor poles can be calculated using (3.9): 𝑁𝑟 =

𝑁𝑠 𝑚𝑜𝑑(𝑘, 𝑖) 𝑘 ∏ 𝑐𝑒𝑖𝑙( ) 𝑚 𝑖 𝑖

where the parameter 𝑖 represents the prime factors of number of phases. Configuration index, k varies from one to any arbitrary integer. The number of stator poles per phase is a positive integer. For 3-phases , 𝑖 = [3]; therefore, 𝑁𝑟1 =

𝑁𝑠1 𝑚𝑜𝑑(𝑘1 , 3) 𝑘 𝑐𝑒𝑖𝑙 ( ) 3 1 3

𝑁𝑟2 =

𝑁𝑠2 𝑚𝑜𝑑(𝑘2 , 2) 𝑘2 𝑐𝑒𝑖𝑙 ( ) 4 2

For 4-phase, 𝑖 = [2]; therefore,

Since we’re looking for a configuration which can run in 3- or 4-phases: 𝑁𝑟1 = 𝑁𝑟2 and 𝑁𝑠1 = 𝑁𝑠2

18

Hence, 𝑘1 𝑚𝑜𝑑(𝑘1 , 3) 𝑘2 𝑚𝑜𝑑(𝑘2 , 2) 𝑐𝑒𝑖𝑙 ( ) = 𝑐𝑒𝑖𝑙 ( ) 3 3 4 2 The result of the 𝑐𝑒𝑖𝑙 function is 1 for the working configurations and zero for non-working configurations. In order to maintain the equality, 𝑘1 should be an integer multiple of 3 and 𝑘2 should be an integer multiple of 4. In this case, the ceil functions will results in zero. This shows that there is no stator and rotor pole combination which can provide balanced operation when used as a 3-phase and 4-phase machine. Q3.7 Solution In SRMs, stator poles are used to generate the magnetic poles. Magnetic poles come in pairs to create the flux path. When there is odd number of stator poles per phase, an integer value for the pole pairs cannot be maintained. In this case, one of the magnetic poles cannot be generated and balanced operation might not be maintained. This is the reason why a mutually-coupled winding configuration is necessary in SRMs with odd number of stator poles per phase. With mutually coupled winding configuration, the flux paths are generated by utilizing the stator poles of other phases. With a mutually coupled coil configuration, the flux generated when one phase coil is energized now links to the coils of other phases. Therefore, unlike conventional SRMs with even number of stator poles per phase and noncoupled coil configuration, mutual inductance cannot be neglected. When modeling SRMs with odd number of stator poles per phase, flux linkage due to the current from other phases need to be taken into account. Q3.8 Solution In this chapter, many different SRM topologies have been presented. It can be noticed that various combinations of number of stator poles, number of rotor poles, and number of phases are available for SRMs. The selection of the SRM configuration depends on many parameters, including the desired torque-speed and torque ripple characteristics, mechanical tolerances, converter and controller requirements, switching frequency, precision of rotor position measurement, etc. Selecting the right configuration for the given application requires extensive analysis due to the nonlinear nature of SRM.

19

Chapter 4 Operational Principles and Modeling of Switched Reluctance Machines Q4.1 Solution Variables: currentArray [11 x 1]

elecAngleArray [120 x 1]

fluxLinkageMatrix [120 x 11]

voltageMatrix [120 x 11]

torqueMatrix [120 x 11]

MATLAB algorithm to plot the torque and voltage profiles: % The variables currentArray, elecAngleArray, torqueMatrix, voltageMatrix % must be available in the workspace. close all; clc; % plot the torque profile figure(1) for i = 1:length(currentArray) plot(elecAngleArray,torqueMatrix(:,i),'LineWidth',2); hold on; grid on; end xlim([min(elecAngleArray), max(elecAngleArray)]); xlabel('Electrical Angle [deg]'); ylabel('Torque [Nm]'); legend(num2str(currentArray)); % plot the voltage profile figure(2) for i = 1:length(currentArray) plot(elecAngleArray,voltageMatrix(:,i),'LineWidth',2); hold on; grid on; xlim([min(elecAngleArray), max(elecAngleArray)]); end xlim([min(elecAngleArray), max(elecAngleArray)]); xlabel('Electrical Angle [deg]'); ylabel('Voltage @1000RPM [V]'); legend(num2str(currentArray));

The torque and terminal voltage waveform of the 12/8 SRM are shown in Fig04-24. Both waveforms have similar shapes. Torque and voltage are positive between the unaligned and aligned positions (as the flux linkage increases with rotor position). They are both negative between aligned and unaligned positions (as the flux linkage decreases with rotor position). After a certain excitation current, the peak value of the terminal voltage does not change, while the torque keeps increasing. This is the effect of saturation.

20

Q4.2 Solution MATLAB algorithm to plot the flux linkage profiles: % Variables currentArray, elecAngleArray, and fluxLinkageMatrix must be % available in the workspace. close all; clc; % Plot the flux linkage profile as a function of rotor position. figure(1) for i = 1:length(currentArray) plot(elecAngleArray,fluxLinkageMatrix(:,i),'LineWidth',2); hold on; grid on; end xlim([min(elecAngleArray), max(elecAngleArray)]); xlabel('Electrical Angle [deg]'); ylabel('Flux Linkage [Wb]'); legend(num2str(currentArray)); % Plot the flux linkage as a function of current figure(2) aligned = find(elecAngleArray == 180);% half electrical cycle is enough elecAngleArray2 = elecAngleArray(1:aligned); k = 1; for i = 1:4:length(elecAngleArray2) % less angles for plotting purposes plot(currentArray,fluxLinkageMatrix(i,:),'LineWidth',2); hold on; grid on; legend1{k} = num2str(elecAngleArray2(i)); k = k + 1; end xlim([min(currentArray), max(currentArray)]); xlabel('Current [A]'); ylabel('Flux Linkage [Wb]'); legend(legend1,'Location','Best');

The flux linkage characteristics of the 12/8 SRM are shown in Fig04-25. From Fig04-25 (a), it can be seen that the flux linkage increases with rotor position until the aligned position at 180° electrical. After the aligned position, the flux linkage decreases with rotor position. From Fig04-25 (b), it can be seen that at the unaligned position (zero degree electrical), flux linkage increases linearly with current. At the aligned position, 180° electrical, the flux linkage profile has a similar shape as the magnetization characteristics of electrical steel. This is the effect of saturation. Q4.3 Solution The algorithm to invert the flux linkage look-up table: % Variables currentArray, elecAngleArray, and fluxLinkageMatrix must be % available in the workspace. 21

close all; clc; N_angles = length(elecAngleArray); N_fluxlink = N_angles; LUT_currentMatrix = zeros(N_angles,N_fluxlink); targetArray = linspace(min(fluxLinkageMatrix(:)),max(fluxLinkageMatrix(:)),N_fluxlink); current_max = max(currentArray); for i_angle = 1:N_angles fluxlink_row = fluxLinkageMatrix(i_angle,:); for i_fluxlink = 1:N_fluxlink TARGET = targetArray(i_fluxlink); if TARGET > max(fluxlink_row) LUT_currentMatrix(i_angle,i_fluxlink) = current_max; else LUT_currentMatrix(i_angle,i_fluxlink) = ... interp1(fluxlink_row, currentArray, TARGET); end end end % Plot flux linkage look-up table figure(1) mesh(currentArray,elecAngleArray,fluxLinkageMatrix); xlim([min(currentArray), max(currentArray)]); ylim([min(elecAngleArray), max(elecAngleArray)]); xlabel('Current [A]'); ylabel('Electrical Angle [deg]'); zlabel('Flux Linkage [Wb]'); % Plot current look-up table figure(2) mesh(targetArray,elecAngleArray,LUT_currentMatrix); xlim([min(targetArray), max(targetArray)]); ylim([min(elecAngleArray), max(elecAngleArray)]); xlabel('Flux Linkage [Wb]'); ylabel('Electrical Angle [deg]'); zlabel('Current [A]');

Q4.4 Solution When the motor runs at 1200 rpm: •

Every minute (60 seconds), the rotor makes 1200 revolutions

•

Every second, the rotor makes 1200/60 = 20 revolutions

•

Every second, the rotor rotates 20 x 360 = 7200 degrees

22

•

The rotor makes one revolution (360 degrees) in every 360/7200 = 0.05 seconds

•

In one second (simulation stop time), the rotor makes 1/0.05 = 20 revolutions

Q4.5 Solution If the simulation stop time is 1 second, the mechanical angle as a function of time will look like below. At 1200 rpm, the rotor makes 20 revolutions in one second. 400

Angle [mech. deg.]

350 300 250 200 150 100 50 0 0

0.1

0.2

0.3

0.4

0.5 Time [s]

0.6

0.7

0.8

0.9

1

Fig. 6. Mechanical angle versus time.

Q4.6 Solution The mechanical and electrical angles of a 12/8 SRM were shown in Fig03-07. Using (3.7) the electrical angle of each stator can be calculated as: 𝑁𝑠𝑒𝑙𝑒𝑐𝑡 = 𝑚𝑜𝑑[𝑘𝑟𝑜𝑡 (𝑁𝑠𝑚𝑒𝑐ℎ − 𝑁𝑟𝑚𝑒𝑐ℎ )𝑁𝑟 + 180°, 360°] For the given rotor position, the electrical angle of each stator pole is the same for all the rotor poles. Therefore, we can use the mechanical angle of the first rotor pole to calculate electrical angle of each phase. As shown in Fig03-07, 𝑁𝑟𝑚𝑒𝑐ℎ = 0, 𝑁𝑠1𝑚𝑒𝑐ℎ = 0, 𝑁𝑠2𝑚𝑒𝑐ℎ = 30°, and 𝑁𝑠3𝑚𝑒𝑐ℎ = 60°. Using (3.7), 𝑁𝑠1𝑒𝑙𝑒𝑐𝑡 = 180°, 𝑁𝑠2𝑒𝑙𝑒𝑐𝑡 = 300°, and 𝑁𝑠3𝑒𝑙𝑒𝑐𝑡 = 60°. 𝑘𝑟𝑜𝑡 is -1 for counter clockwise rotation.

For the given rotor position, the electrical angle can be modeled using (4.33). 𝜃𝑒𝑙𝑒𝑐 = 𝑚𝑜𝑑 ((𝑁𝑟 𝜃𝑚𝑒𝑐ℎ + 180°) +

360 𝑘 , 360°) 𝑚 𝑝ℎ

At 𝜃𝑚𝑒𝑐ℎ = 0, the position of the rotor was shown in Fig03-07. Hence, 𝜃𝑒𝑙𝑒𝑐 derived using (4.33) will equal to the electrical angle of the stator poles calculated using (3.7): 23

𝑁𝑠1𝑒𝑙𝑒𝑐𝑡 = 𝑚𝑜𝑑 ((𝑁𝑟 𝜃𝑚𝑒𝑐ℎ0 + 180°) +

360 𝑘 , 360°) 𝑚 𝑝ℎ1

where 𝑁𝑟 = 8, 𝜃𝑚𝑒𝑐ℎ0 = 0, and 𝑚 = 3. Therefore, 𝑘𝑝ℎ1 = 0 for the first stator pole (hence Ph#1). Using 𝑁𝑠2𝑒𝑙𝑒𝑐𝑡 and 𝑁𝑠3𝑒𝑙𝑒𝑐𝑡 , 𝑘𝑝ℎ for the other phases can be calculated. The results is 𝑘𝑝ℎ = [0 1 2].

The figure below shows the electrical angles of each phase in one mechanical revolution. There are 24 electrical cycles in one revolution. This corresponds to the number of strokes, which was derived in (3.1): 𝑆 = 𝑚 𝑁𝑟 where 𝑁𝑟 is the number of rotor poles. 𝑚 is the number of phases, which is calculated as 𝑚 = 𝑁𝑠 ⁄#𝑝𝑜𝑙𝑒𝑠 where 𝑁𝑠 is the number of stator poles and #𝑝𝑜𝑙𝑒𝑠 is the number of poles and it equals to 4 in 12/8 SRM. rotor pos.

Ph#1 elect.angle

Ph#2 elect. angle

Ph#3 elect. angle

400

Angle [mech. deg.]

350 300 250 200 150 100 50 0

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

Time [s]

Fig. 7. Mechanical angles versus time, three phases.

Q4.7 Solution The figure below shows the current waveform. If the relationship between the flux linkage and current is modeled with a constant inductance, the phase model becomes a simple 𝑅𝐿 circuit. The 𝑅𝐿 circuit is represented by a first order ordinary differential equation. 𝑣𝑝ℎ = 𝑅𝑝ℎ 𝑖𝑝ℎ + 𝐿 𝐿

𝑑𝑖𝑝ℎ 𝑑𝑡

𝑑𝑖𝑝ℎ = 𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ 𝑑𝑡

24

𝑖

𝑡

𝑑𝑖𝑝ℎ 𝑑𝑡 ∫ =∫ 𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ 𝐿 0

0

𝑢 = 𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ ⇒ 𝑑𝑢 = −𝑅𝑝ℎ 𝑑𝑖𝑝ℎ ⇒ 𝑑𝑖𝑝ℎ = − 𝑡

𝑑𝑡 ∫ =∫ 𝐿 0

1 𝑑𝑢 𝑅𝑝ℎ −

1 𝑑𝑢 𝑅𝑝ℎ 𝑢

𝑡

1 1 𝑑𝑢 ∫ 𝑑𝑡 = − ∫ 𝐿 𝑅𝑝ℎ 𝑢 0

1 1 𝑡=− 𝑙𝑛|𝑢| 𝐿 𝑅𝑝ℎ 1 1 𝑖 𝑡=− 𝑙𝑛|𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ |0 𝐿 𝑅𝑝ℎ − −

𝑅𝑝ℎ 𝑡 = 𝑙𝑛|𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ | − 𝑙𝑛|𝑣𝑝ℎ | 𝐿

𝑅𝑝ℎ 𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ 𝑡 = 𝑙𝑛 | | 𝐿 𝑣𝑝ℎ

𝑣𝑝ℎ − 𝑅𝑝ℎ 𝑖𝑝ℎ = 𝑒 −(𝑅𝑝ℎ ⁄𝐿 )𝑡 𝑣𝑝ℎ 1−

𝑅𝑝ℎ 𝑖 = 𝑒 −(𝑅𝑝ℎ ⁄𝐿 )𝑡 𝑣𝑝ℎ 𝑝ℎ 𝑖𝑝ℎ (𝑡) =

𝑣𝑝ℎ (1 − 𝑒 −(𝑅𝑝ℎ ⁄𝐿)𝑡 ) 𝑅𝑝ℎ

The figure below shows the waveform of the current. It can be observed that the current rises and reaches to a steady-state value of 𝑣𝑝ℎ ⁄𝑅𝑝ℎ .

25

50 Vph/Rph

45 40

Current [A]

35 30 25 20 15 10 5 0

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0.045

0.05

Time [s]

Fig. 8. Current versus time.

Q4.8 Solution Using (4.43), the counting number for the current sampling can be calculated as: 𝑇𝑠𝑎𝑚𝑝 𝑛𝑠𝑎𝑚𝑝 = 𝑟𝑜𝑢𝑛𝑑 ( ) 𝑇𝑠 If 𝑇𝑠 = 1 micro-second, and 𝑇𝑠𝑎𝑚𝑝 =

1 22𝑘𝐻𝑧

= 45.5 micro-seconds, then 𝑛𝑠𝑎𝑚𝑝 = 45. Fig. 9 shows the

sampled current at 1 kHz and 22 kHz. With a higher sampling frequency, the sampled current waveform is closer to the phase current. Therefore, we have a better knowledge of the phase current with a higher sampling frequency, which enables higher accuracy for the current control.

26

(a)

50 45

Current [A]

40 35 30 25 20 15 10

Phase current Sampled current 1 kHz

5 0

0

0.005

0.01

0.015

0.02

(b)

0.025 0.03 Time [s]

0.035

0.04

0.045

0.05

50 45

Current [A]

40 35 30 25 20 15 10

Phase current Sampled current 22 kHz

5 0

0

0.005

0.01

0.015

0.02

0.025 0.03 Time [s]

0.035

0.04

0.045

0.05

Fig. 9. Current versus time: (a) sampled current 1 kHz, (b) sampled current 22 kHz.

Q4.9 Solution The figures below show the phase excitation signals for 𝜃𝑂𝑁 = 30°, 𝜃𝑂𝐹𝐹 = 150° and 𝜃𝑂𝑁 = −20°, 𝜃𝑂𝐹𝐹 = 120°. In the first case, each phase conducts for 120° electrical, which equals the phase shift angle. Therefore, neglecting current decay after the turn off angle, none of the phases conduct at the same time. In the second case, phase conduction is larger than the phase shift angle. Therefore, phase excitation signals will force two phases to conduct at the same time.

27

Angle [mech. deg.]

(a)

Ph#1 elect.angle

Ph#3 elect. angle

300 200 100 0

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

(b)

Phase excitation signal

Ph#2 elect. angle

400

1 0.5 0

Fig. 10. θON = 30° / θOFF = 150°: (a) mechanical angles, (b) phase excitation signals.

Angle [mech. deg.]

(a)

Ph#1 elect.angle

Ph#3 elect. angle

400 300 200 100 0

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

(b) Phase excitation signal

Ph#2 elect. angle

1 0.5 0

Fig. 11. θON = -20° / θOFF = 120°: (a) mechanical angles, (b) phase excitation signals.

Q4.10 Solution The figure below shows the results with soft and hard switching. In soft switching, the switching signal changes between +1 and 0 during the conduction interval. This enables a slower rate of change of current and, hence, a lower switching frequency. In hard switching, the switching signal changes between +1 and 28

-1. This results in a higher rate of change of current and a higher switching frequency. In practical applications, soft switching is widely used during the phase conduction in conduction angle control.

(a)

Ph#2

Ph#1

Ph#3

Current [A]

6 4 2 0 0

0.005

0.01

0.015

0.02 0.025 Time [s]

0.03

0.035

0.04

0.045

0.05

0.005

0.01

0.015

0.02 0.025 Time [s]

0.03

0.035

0.04

0.045

0.05

0.005

0.01

0.015

0.02 0.025 Time [s]

0.03

0.035

0.04

0.045

0.05

0.005

0.01

0.015

0.02 0.025 Time [s]

0.03

0.035

0.04

0.045

0.05

Switching signal

(b)

Soft switching

1 0 -1 0

(c)

Current [A]

6 4 2 0 0

Switching signal

(d)

Hard switching

1 0 -1 0

Fig. 12. Soft switching versus hard switching: (a) current, soft switching, (b) switching signal, soft switching, (c) current, hard switching, and (d) switching signal, hard switching,.

Q4.11 Solution If we apply the current waveform calculated in the dynamic model to the FEA model and measure voltage across the phase windings, we would observe the modulated phase voltage in the dynamic model. The figure below shows the phase voltage from the dynamic model, phase voltage from FEA (when the current from the dynamic model is used), and the induced voltage from the look-up table. It can be observed that the phase voltage from FEA is similar to the phase voltage from the dynamic model.

29

Voltage [V]

(a)

Voltage [V]

Ph#3

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

0

0.002

0.004

0.006 Time [s]

0.008

0.01

0.012

(b) 500 0 -500 (c)

Voltage [V]

Ph#2

Ph#1 400 200 0 -200 -400

60 40 20 0

Fig. 13. Simulation results: (a) Phase voltage from the dynamic model (b) phase voltage from FEA (c) voltage calculated from the voltage look-up table in the dynamic model

In order to calculate the induced voltage, static terminal voltage profile is used in the look-up table. The static voltage waveform is the voltage that would be observed at the terminals of the phase winding if the rotor rotates one electrical cycle when constant current flows through the phase winding. Therefore, it can be used to visualize the voltage induced inside the motor. The induced voltage cannot be directly measured. If we measure the voltage across the phase winding during the operation of SRM, we would observe the modulated DC link voltage (phase voltage). In our case, the induced voltage is much lower than the DC-link voltage. Therefore, the DC link voltage is modulated to control the current.

Q4.12 Solution The figure below shows the phase voltage, induced voltage and current waveforms at 10000 RPM. It can be seen that induced voltage is much higher than the phase voltage. There are intervals where the phase voltage is positive, but the phase current still decays. At those instants, the induced voltage is higher than the phase voltage. Induced voltage cannot be directly measured from the phase terminals. However, it can be plotted by using the static terminal voltage profile as a look-up table.

30

Ph#2 induced volt.

Voltage [V]

(a)

Ph#1 induced volt.

Ph#3 induced volt.

400 200 0 Ph#3 phase volt.

-200 0

0.1

0.2

0.3

Ph#2 phase volt. 0.4

0.5

Ph#1 phase volt. 0.6

0.7

0.8

0.9

1

Time [ms] (b)

Current [A]

20

10

0 Ph#2

Ph#1 -10 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Ph#3 0.9

1

Time [ms]

Fig. 14. Simulation results, 10000 rpm: (a) voltage, (b) current.

31

Chapter 5 Switched Reluctance Machines in Generating Mode a) Separately excited DC machine needs a full bridge converter for four-quadrant operation. SRM can run in four quadrants with an asymmetric bridge converter. As shown in Fig05-08, the direction of armature current changes in motoring and generating modes in the separately excited DC machine. This is why a full bridge converter is needed for four-quadrant operation. In SRM, the phase current doesn’t change its direction in generating mode. Generating mode is achieved by exciting the phase in the negative slope of the flux linkage profile. If full bridge converter was used in SRM, the direction of phase current could be reversed. However, this wouldn’t be necessary, since the torque is independent of the direction of the phase current. b) This is one of the main differences between separately excited DC machine and switched reluctance machine. Even though the equivalent circuit models of these two machines are similar, SRM torque is independent of the direction of the phase current. Therefore, the phase current in SRM does not change its direction in generating mode. c) In the separately excited DC machine, the direction of rotation is reversed by changing the polarity of the armature voltage. Please notice from Fig05-08 that in backward motoring and generating, the polarity of Va is reversed. As it was explained in Fig05-03, the direction of rotation in SRM is reversed by changing the phase excitation sequence. d) In forward generating mode in Fig05-08, the armature current increases when S4 is on and D2 is conducting. In this case, the armature circuit is short circuited. Armature current rises by drawing power from the prime mover. When S4 is turned off, the armature current is supplied back to the source. This is similar to the soft switching mode in SRG which was explained in Fig05-05. Please note that due to the induced EMF Ea, separately excited DC machine do not need to draw current from the source for initial excitation. e) In forward motoring mode in the separate excited DC motor, when S 1 and S2 are on, the armature current increases. In asymmetric bridge converter in SRM, this mode is shown in Fig05-05 (c). When S1 is turned off, the armature current freewheels through S2 and D4. This is similar to the soft switching in SRM 32

as shown in Fig05-05 (d). If both S1 and S2 are turned off, the armature current is supplied back to the source through D1 and D2. This is similar to the hard switching mode in SRM as shown in Fig05-06 (d).

33

Chapter 6 Materials Used in Switched Reluctance Machines Q6.1 Solution (a) NO30 is a thin gauge steel (0.3mm) which is expected to have a smaller B-H curve then the much thicker M470-50A (0.5mm), as shown in Fig. 15.

2

M470-50A 1.8

N030

B [T]

1.6

1.4

1.2

1

0

5000

10000

15000

H [A/m]

Fig. 15. Magnetization characteristics of NO30-1600 and M470-50A, B-H curve at 50 Hz.

2

1.6

N030

B [T]

1.2 M470-50A 0.8

0.4

0

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Loss density [X10 4 W/m 3]

Fig. 16. Loss characteristics of NO30-1600 and M470-50A at 50 Hz.

34

(b) A thinner gauge lamination generally has lower eddy current losses, as shown in Fig. 16. For the given magnetic flux density, we would expect the thicker lamination to have a higher loss density than the thinner lamination. (c) As the frequency increases, both gauges increase in loss density for the same magnetic flux density, as shown in Fig. 17. However, the losses for NO30-1600 continue to be less than the losses for M47050A. For the highest frequencies, the shape of the curves has changed substantially from the low frequency case, causing loss to increase greatly with a small change in the magnetic flux density. A single polynomial would not fit these curves with good accuracy.

2

1.6

NO30-1600

50 Hz 400 Hz

M470-50A

2000 Hz 5000 Hz

B [T]

NO30-1600

NO30-1600

1.2 M470-50A M470-50A

NO30-1600

0.8 M470-50A NO30-1600 0.4

0

0

1

2

3

4

5

6

7

Loss density [X105 W/m 3]

Fig. 17. Loss characteristics of NO30-1600 and M470-50A at different frequencies.

(d) Electrical steel is carefully controlled both in alloying and production methods to produce the best magnetic properties. Even though NO30-1600 has a smaller magnetization curve than M470-50A, it is still much better than stainless steel and cast iron, as shown in Fig. 18.

35

1.8 M470-50A N030-1600

1.6

B [T]

Stainless Steel

1.4 Cast Iron

1.2

1

0

5000

10000

15000

H [A/m]

Fig. 18. Magnetization characteristics of NO30-1600, M470-50A, stainless steel and cast iron.

(e) For a high-torque density SRM, M470-50A would be a better choice, since its magnetization curve is larger. This material also achieves higher magnetic flux density, which would potentially increase the coenergy and, hence, torque. For the high-speed operation, NO30-1600 would be better, because the losses are lower at high frequencies. Q6.2 Solution Thermal class: Determines the rated temperature a type of wire can withstand for an extrapolated lifetime of 20,000 hours. It gives an indication of how hot a motor can get before the wire insulator begins to no longer guarantee integrity. Adhesion and flexibility: Used to determine how much mechanical stress the magnet wire can undergo. Given in diameter, a magnet wire should not be bent beyond this while winding to maintain good condition. Breakdown voltage: When the dielectric material temporarily loses its property of non-conductivity due to a high electric field, this is known as a dielectric breakdown. To avoid electrical discharges the breakdown voltage should not be approached or exceeded.

36

Chapter 7 Design Considerations for Switched Reluctance Machines Q7.1 Solution Hint: S f Ls  N lamtlam , in which Sf is the stacking factor, Nlam is the number of lamination sheets, tlam is the thickness for lamination sheets, and Ls is the stator stack length. The answers are given in Tab. 13 Tab. 13. Number of sheets calculated. Lamination thickness [mm]

Number of sheets

0.3

~158

0.25

190

0.1

475

Q7.2 Solution The answers are given in Fig. 19

(a)

(b) Phase A

Phase A IN

IN

OUT 12

11

42 41

72 71

102

101

132

131

162

161

192

191

OUT

222

12

221

11

42 41

72 71

102

101

131

Phase B OUT

21

52 51

82 81

111

11 2 14 1

142

172

171

202

201

161

192

191

222

221

IN

OUT

232

231

22 21

52 51

82 81

112

111

Phase C

142

141

172

171

202

201

232

231

Phase C OUT

IN

IN 32

31

162

Phase B

IN

22

132

62 61

92 91

122 121

151

152 181

182 211

212 241

242

OUT 32

31

62 61

92 91

122 121

151

152 181

182 211

212

242

241

Fig. 19. Possible winding schemes for 24/16 SRM: (a) two parallel paths, and (b) four parallel paths.

37

Q7.3 Solution Since the coil has 4 turns and 5 strands and its slot fill factor is 0.5, for the same slot, the slot fill factor, ff, for any another combination of the number of turns per coil, Nturn, and number of strands, Nstr will follow the following formula:

N turn  N str ff  45 0.5 Since, 1  N turn  12 , 1  N str  12 , and 0.4  ff  0.6 , the possible combinations of Nturn and Nstr are given in Tab. 14. Tab. 14. Possible combinations of number of turns and number of strands. Number of turns 1

2

3

4

5

6

7

8

9

10

11

12

0.4

0.45

0.5

0.55

0.6

1 2

Number of strands

3

0.45

4

0.4

5

0.5

6

0.45

7

0.525

8

0.4

9

0.45

10

0.5

11

0.55

12

0.6

0.5

0.525

0.6

0.6

0.6

0.6

Q7.4 Solution The feasible region can be seen in Fig. 20.

38

2 N



360 N

s

s

60

Stator pole arc angle, βs [mech. deg.]

r  s 

4 mN

720



mN

r

r  s 

r

2 N

 r

360 N

r

50

40

s 

2 mNr



360 mNr

30 r  s

20

10

0

0

10

20

30 40 50 60 Rotor pole arc angle, βr [mech. deg.]

70

80

90

2

r  s

Feasible region s 

2 mN

4 mNr

r

N

 s  r 



r

360 N

r

2 Nr

Fig. 20. The feasible region for selecting the stator and rotor pole arc angles for a three-phase 6/4 SRM.

Q7.5 Solution Tab. 15. The relationship between stator and rotor pole arc angles.

Geometry

m

Ns

Nr

r _ min  s _ min  [mech. deg.]

3-phase 6/4 SRM

3-phase 12/8 SRM

3-phase 18/12 SRM

3

3

3

6

12

18

4

8

12

30

15

10

Number of poles

Expression for βS r  s 

720   2 Nr  Ns  Ns Nr

r  s 

720   5 N r  3N s  Ns Nr

r  s 

720   2 Nr  Ns  Ns Nr

r  s 

720   5 N r  3N s  Ns Nr

r  s 

720  8 N r  5 N s  Ns Nr

r  s 

720  11N r  7 N s  Ns Nr

r  s 

720   2 Nr  Ns  Ns Nr

r  s 

720   5 N r  3N s  Ns Nr

2

4

6

+ βr

39

3-phase 24/16 SRM

3

24

16

7.5

r  s 

720  8 N r  5 N s  Ns Nr

r  s 

720  11N r  7 N s  Ns Nr

r  s 

720  14 N r  9 N s  Ns Nr

r  s 

720  17 N r  11N s  Ns Nr

r  s 

720   2 Nr  Ns  Ns Nr

r  s 

720   5 N r  3N s  Ns Nr

r  s 

720  8 N r  5 N s  Ns Nr

r  s 

720  11N r  7 N s  Ns Nr

r  s 

720  14 N r  9 N s  Ns Nr

r  s 

720  17 N r  11N s  Ns Nr

r  s 

720   20 N r  13N s  Ns Nr

r  s 

720   23N r  15 N s  Ns Nr

8

Fig. 21. Snapshot of MATLAB plot, relative position between rotor and stator, 3-phase 18/12 SRM, θ = 0°, βS = 10°, βr = 10°.

40

Fig. 22. Snapshot of MATLAB plot, relative position between rotor and stator, 3-phase 24/16 SRM, θ = 0°, βS = 7.5°, βr = 7.5°.

41

Chapter 8 Mechanical Construction of Switched Reluctance Machines Q8.1 Solution Switched reluctance machine has a simple construction with concentrated coils wound around salientpoles of the stator core. The rotor also has salient-pole construction without any source of excitation. This simplifies the rotor construction and reduces the manufacturing cost. Furthermore, as there is no risk for magnets flying off or winding damage on the rotor, high speed operation can be realized for SRM. In comparison, permanent magnet synchronous machine and brushless DC machine have magnets on the rotor, wound induction machine has winding on the rotor, DC brushed machine has winding, brushes, and slip rings on the rotor. These all somewhat reduce the machine reliability and bring in additional loss and cost. Q8.2 Solution Different insulation layers should be applied in SRM. Insulation layers are necessary to be applied around the copper conductor to enable the contact between the wires without causing any electrical short circuit. Magnet wires come with organic insulation applied around them. The insulating layers, so called enamels or films, are bonded on the copper or aluminum conductor. They are made up of different types of resins and define Thermal class of the magnet wire. In addition, winding-to-stator insulation is necessary in switched reluctance machines to prevent possible winding-to-ground short circuit faults. For prototype machines, slot liner and slot wedges are typically applied. These insulation materials are usually made of aramid or Mylar layered paper. For large volume production machines, injection molded plastic insulation are typically applied to form a structure that surrounds the stator lamination to prevent the contact between winding and lamination steel. Typical plastic injection molding materials for stator and rotor include polyamide (PA), i.e. Nylon PA 6, PA 66, polyethylene (PE), polybutylene terephthalate (PBT), etc. Glass fiber reinforced polyphenylene sulfide (PPS) such as PPS-GF30 and PPS-GF40 and polysulfone (PSU) can be used for harsh, high temperature, and contaminative environment, for instance, where the motor is in direct contact with transmission oil or engine coolant.

42

Q8.3 Solution Journal bearings comprises only bearing surface but no rolling elements. The design and construction of journal bearings are relatively simple. They offer the advantages including light weight, low cost for manufacturing, long life under normal load operation, low operation noise, less transmitted vibration, shock load tolerant capability, electrical isolation from rotor to ground, less electromagnetic interference, less sensitivity to contamination, less mounting accuracy requirement, ease of maintenance, good heat dissipation with lubrication, and good wear dust clearance with lubrication. Compared to journal bearings, rolling contact bearings have lower starting and running resistance, consume less amount of lubricant and are less prone to wear, thus requiring far less maintenance. However, rolling bearings are less capable to withstand shocks and high loads. They have higher possibility for failure to occur and hence are less reliable and typically have shorter life compared to journal bearings. Q8.4 Solution Any deviation of air gap distance caused by assembly eccentricity will have impact on machine output parameters, notably the vibration and noise caused by the unbalance radial force pull. Rotor balance is another critical aspect for rotor assembly. It is especially important for switched reluctance machines to ensure safety and normal operation as many of them are designed for high-speed applications. Unbalance rotors typically are resulted from mismatched rotor weight distribution. Improper manufacturing tolerances, material defects, unsymmetrical structures such as keys and key ways, and poor assembly can all lead to rotor unbalance. Rotor unbalance is a significant cause for noise and vibration. It is also likely to cause wear and stress on the bearings and other supporting components. Efficiency, reliability, and lifetime of the machine could be negatively affected. Besides, shaft misalignment could lead to significant machine noise and vibration issues and results in excessive loading and reduction in the life of the machine. It occurs when the centerline of the motor shaft does not align fully with the driven shaft. Q8.5 Solution Tab. 16. Specs for an SRM designed for an e-bike application. Part #

Name

Possible material or model

Possible connection with

(1)

Bearing A

Steel, deep groove ball bearing, 6203

End cap B

Transition fit

43

(2)

Shaft

(1), (4)

Sliding fit

(6)

Press fit

4340 Steel

(3)

Rotor lamination stack

Electrical steel

Rotor slot fillers

Glued

(4)

Bearing B

Steel, deep groove ball bearing, 6908

(5)

Close running fit

(5)

End cap A

6061 Aluminum

(3)

Bolts

(6)

Stator lamination stack

Electrical steel

(2)

Press fit

(7)

Winding

22 AWG magnet wire

(6)

Lace, pole shoe

44

Chapter 9 Control of Switched Reluctance Machines Q9.1 Solution // User defined inputs, these are constants define I_ref; //reference current define tol; // tolerance band, as a percentage or an absolute value define V_DC; i_upper = I_ref*(1+tol); i_lower = I_ref*(1-tol); V = V_DC; // dynamic control for k = 1:numOfSteps if i > i_upper, then V_phase = -V_DC; elseif i 1 C1 = (pull*omega_n*(zeta + sqrt(zeta^2 1)+speed))/2/omega_n/sqrt(zeta^2 - 1); C2 = (-pull*omega_n*(zeta - sqrt(zeta^2 - 1)speed))/2/omega_n/sqrt(zeta^2 - 1); x_t = C1*exp((-zeta + sqrt(zeta^2 - 1))*omega_n*t) + C2*exp((-zeta sqrt(zeta^2 - 1))*omega_n*t); end motion_surface = [motion_surface;x_t]; end %% Plotting figure (1) meshc (t, zeta_series, motion_surface) xlabel('Time [s]') ylabel ('Zeta') zlabel ('x(t) [m]') colorbar

The plot is given in Fig. 35. Please note that the view of the figure is rotated from the original view obtained using MATLAB. It can be seen that as zeta increases from 0 to 1, the displacement x(t) is damped faster over time.

69

0.2

0.2

0.15 0.1

x(t) [m]

0.1

0.05

0

0 -0.05

-0.1

-0.1

-0.2 0

1.2 1 0.1 Tim

0.8

-0.15 -0.2

0.6

e[ s]

0.4

0.2 0.2 0.3

0

ta Ze

Fig. 35. Response of damped spring-mass system with damping ratio over time

Q12.3 Solution 1)

The MATLAB script is as follows: %% Given parameters of the system pull = 0.2; % x(0) = 0.2 m speed = 0; % x_dot (0) = 0 m/s m = 1; % mass of the system k = 10*1000; % stiffness of the spring zeta_series = [0.01:0.1:1.2]; omega_ratio_series = [0.01:0.01:10]; phi_series = []; %% Calculation and plot for i = 1:length (zeta_series) zeta = zeta_series (i); phi = atand (2*zeta*omega_ratio_series./(1-omega_ratio_series.^2)); phi = phi .* (phi >= 0) + (phi + 180) .* (phi < 0); % convert [-90, 90] to [0, 180] h(i) = plot (omega_ratio_series,phi, 'DisplayName', sprintf('\\zeta = %0.2f', zeta)); phi_series = [phi_series; phi]; hold on end

70

legend (h) xlabel ('\omega_f/\omega_n') ylabel ('\phi (mech. deg.)') The plot is given in Fig. 36. As can be seen that the point (1, 90 mech. deg.) is a node for all curves. When ϛ is small, it can be observed that the phase angle rises slowly at first, and then grows sharply when ωf / ωn = 1. 180 160 140 ϛ = 0.01 ϛ = 0.11 ϛ = 0.21 ϛ = 0.31 ϛ = 0.41 ϛ = 0.51 ϛ = 0.61 ϛ = 0.71 ϛ = 0.81 ϛ = 0.91 ϛ = 1.01 ϛ = 1.11

ϕ [mech. deg.]

120 100 80 60 40 20 0 0

1

2

3

4

5

6

7

8

9

10

ωf / ωn

Fig. 36. Phase angle of steady state versus ratio of frequencies as the damping ratio varies.

2) The MATLAB script is as follows: %% Given parameters of the system pull = 0.2; % x(0) = 0.2 m speed = 0; % x_dot (0) = 0 m/s m = 1; % mass of the system k = 10*1000; % stiffness of the spring zeta_series = [0.01:0.1:1.2]; omega_ratio_series = [0.01:0.01:4]; MF_series = []; %% Calculation and plot for i = 1:length (zeta_series)

71

zeta = zeta_series (i); MF = 1./sqrt((1-omega_ratio_series.^2).^2 + (2*zeta*omega_ratio_series).^2); h(i) = plot (omega_ratio_series,MF, 'DisplayName', sprintf('\\zeta = %0.2f', zeta)); MF_series = [MF_series; MF]; hold on end legend (h) xlabel ('\omega_f/\omega_n') ylabel ('Magnification factor') axis([0 4 0 6])

The plot of magnification factor versus ratio of frequencies as the damping ratio varies is given in Fig. 37. It can be seen that when the worst case for resonance is when the damping ratio is zero. 6

ϛ = 0.01 ϛ = 0.11 ϛ = 0.21 ϛ = 0.31 ϛ = 0.41 ϛ = 0.51 ϛ = 0.61 ϛ = 0.71 ϛ = 0.81 ϛ = 0.91 ϛ = 1.01 ϛ = 1.11

Magnification factor

5

4

3

2

1

0 0

0.5

1

1.5

2 ω f / ωn

2.5

3

3.5

4

Fig. 37. Magnification factor versus ratio of frequencies as the damping ratio varies.

72

Q12.4 Solution The MATLAB script for plotting mode shapes of the beam with s-s constraint: clc; close all; clear all l = 1; s = [0:0.01:l]; n = 5; % Mode shape, axial order for i = 1:n deformation = sin(s/l*pi*i); figure (1) plot(s, deformation); hold on end xlabel ('Axial position [m]') ylabel ('Mode shape, unitless') ylim([-1.2 1.2])

The MATLAB script for plotting mode shapes of the beam with c-c constraint: %% clamped-clamped mode shape clc; close all; clear all l = 1; s = [0:0.01:l]; n = 5; % Mode shape, axial order beta_l = [4.730, 7.853, 10.996, 14.137, 17.278]; delta_n = [0.982502, 1.00078, 0.999966, 1, 1]; for i = 1:n beta_current = beta_l(i)/l; delta_current = delta_n(i); deformation = cosh(beta_current*s) - cos (beta_current*s) - delta_current * (sinh (beta_current*s) - sin (beta_current*s)); deformation_normalized = deformation/max(deformation); % Normalization figure (1) plot(s,deformation_normalized); hold on end xlabel ('Axial position [m]') ylabel ('Mode shape, unitless') ylim([-1.2 1.2])

73

Q12.5 Solution 1) The script is given as: figure (1) subplot (4,1,1) % plot (time_vector, f1_time_wave); hold on % plot (time_vector, f2_time_wave); hold on % plot (time_vector, f3_time_wave); hold on plot (time_vector, f_time_sum); xlim([0 1]) ylim([-40 40]) xlabel ('Spatial position [mech. deg.]') ylabel ('Radial force [N]') set(gca,'XTick',[0:0.2:1]) subplot (4,1,2) plot (time_vector, f1_time_wave) xlim([0 1]) ylim([-40 40]) xlabel ('Spatial position [mech. deg.]') ylabel ('Radial force [N]') set(gca,'XTick',[0:0.2:1]) subplot (4,1,3) plot (time_vector, f2_time_wave) xlim([0 1]) ylim([-40 40]) xlabel ('Spatial position [mech. deg.]') ylabel ('Radial force [N]') set(gca,'XTick',[0:0.2:1]) subplot (4,1,4) plot (time_vector, f3_time_wave) xlim([0 1]) ylim([-40 40]) xlabel ('Spatial position [mech. deg.]') ylabel ('Radial force [N]') set(gca,'XTick',[0:0.2:1])

The subplots are shown in Fig. 38.

74

Radial force [N]

(a) 40 0 -40 0

0.2

0.4 0.6 0.8 Spatial position [mech. deg.]

1

0.2

0.4 0.6 0.8 Spatial position [mech. deg.]

1

0.2

0.4 0.6 0.8 Spatial position [mech. deg.]

1

0.2

0.4 0.6 0.8 Spatial position [mech. deg.]

1

Radial force [N]

(b) 40 0 -40 0

Radial force [N]

(c) 40 0 -40 0

Radial force [N]

(d) 40 0 -40 0

Fig. 38. Waves over time: (a) f(t), (b) f1(t), (c) f2(t), and (d) f3(t).

2) The MATLAB script is given as: %% figure (2) subplot (2,1,1) stem([0:time_order_limit],[F_series(1)/2, F_series(2:time_order_limit+1)],'filled') xlabel ('Temporal order') ylabel ('Harmonic magnitude [N]') xlim([-0.5 time_order_limit]) subplot (2,1,2) stem([0:time_order_limit],Ang_series (1:time_order_limit+1))>0.01))

(1:time_order_limit+1).*(abs(Ang_series

xlabel ('Time order') ylabel ('Phase angle [mech. deg.]') xlim([-0.5 time_order_limit])

The plots are given in Fig. 39. The x-axis of the temporal order of this one-sided spectrum starts from 0.

75

(a)

Harmonic magnitude [N]

20 15 10 5 0

0

2

4

6

8 10 12 Time order

14

16

18

20

0

2

4

6

8 10 12 Time order

14

16

18

20

(b)

Phase angle [mech. deg.]

100 50 0 -50

Fig. 39. One-sided FFT decomposition of f(t): (a) harmonic amplitudes, and (b) phase angle.

3) The MATLAB script for two-sided FFT analysis is given: %% Two-sided spectrum Raw_fft = fft(f_time_sum)/time_num; Raw_fft_shifted = fftshift(Raw_fft); % DC component is shifted to the center of spectrum FFT_abs_shifted = abs(Raw_fft_shifted); FFT_abs_shifted(FFT_abs_shifted 0.01)*180/pi; % mech. deg.

3) The MATLAB script for polar plots is given: figure (1) subplot (2,2,1) polarplot(sp_vector,f_sp_sum); hold on % Please note that polarplot and rlim do not exist in older MATLAB. rlim([-100 100]) subplot (2,2,2) polarplot(sp_vector,f1_sp_wave); hold on rlim([-100 100]) subplot (2,2,3) polarplot(sp_vector,f2_sp_wave); hold on rlim([-100 100]) subplot (2,2,4) polarplot(sp_vector,f3_sp_wave); hold on rlim([-100 100])

The plots are given in Fig. 41.

78

(a)

f 



 C 0  f1 



f 2 



120

f 3 



(b)

0

240

300

f 2 



 4 c o s  6

120

60

0

240

300 α [mech. deg.]



(d)

180

f3 



    8 c o s  1 8   3  

120

60

0

240

    2 0 c o s  3   3  

180

α [mech. deg.]

(c)



120

60

180

f1 

300 α [mech. deg.]

60

180

0

240

300 α [mech. deg.]

Fig. 41. Waves over circumference: (a) f(α), (b) f1(α), (c) f2(α), and (d) f3(α).

4) The MATLAB script for plotting amplitudes and phase angles for harmonic contents: figure (2) subplot (2,1,1) stem([0:sp_order_limit],[F_series(1)/2, F_series(2:sp_order_limit+1)],'filled') xlabel ('Spatial order') ylabel ('Harmonic magnitude [N]') xlim([-0.5 sp_order_limit]) subplot (2,1,2) stem([0:sp_order_limit],Ang_series (1:sp_order_limit+1).*(abs(Ang_series (1:sp_order_limit+1))>0.01)) xlabel ('Spatial order') ylabel ('Phase angle [mech. deg.]') 79

xlim([-0.5 sp_order_limit])

The results are plots, as shown in Fig. 42. In this one-side spectrum, all the three harmonics are shown. The magnitudes of all the three harmonics are exactly the same as their corresponding actual amplitudes.

Harmonic magnitude [N]

(a)

20 15 10 5 0

0

2

4

6

8 10 12 Spatial order

14

16

18

20

0

2

4

6

8 10 12 Spatial order

14

16

18

20

Phase angle [mech. deg.]

(b) 100 50 0 -50 -100

Fig. 42. FFT decomposition of f(α): (a) harmonic amplitudes, and (b) phase angle.

Q12.7 Solution 1) The MATLAB script is provided below: %% Initialization close all; clear all; clc time_order_1 = 3; % u sp_order_1 = 1; % v phase_angle_1 = 0; amplitude_1 = 20; time_order_2 = 7; % u sp_order_2 = 1; % v phase_angle_2 = 0; amplitude_2 = 5; time_order_3 = 3; % u sp_order_3 = 6; % v phase_angle_3 = 0; amplitude_3 = 2; time_order_4 = 7; % u sp_order_4 = 6; % v phase_angle_4 = 0;

80

amplitude_4 = 1; %% time_num = 1024; time_step = 1/time_num; % unit: s freq_mech = time_num/2^10; % 1 Hz time_vector = [0:time_step:time_step*(time_num-1)]; time_order_limit = 20; amplitude_matrix = [20 2; 5 1]; z_shift = 10; phase_matrix = [0 0; 0 0]; space_orders = [1, 6]; time_orders = [3,7]; space_num = 1024; space_order_limit = 20; sp_step = 2*pi/space_num; % unit: s sp_vector = [0:sp_step:sp_step*(space_num-1)]; % Generate temporal and spatial planes [sp_plane, time_plane] = meshgrid(sp_vector,time_vector); pl_wave_1 = amplitude_1*cos(2*pi*time_plane*freq_mech*time_order_1 + sp_plane*sp_order_1 + phase_angle_1); pl_wave_2 = amplitude_2*cos(2*pi*time_plane*freq_mech*time_order_2 + sp_plane*sp_order_2 + phase_angle_2); pl_wave_3 = amplitude_3*cos(2*pi*time_plane*freq_mech*time_order_3 + sp_plane*sp_order_3 + phase_angle_3); pl_wave_4 = amplitude_4*cos(2*pi*time_plane*freq_mech*time_order_4 + sp_plane*sp_order_4 + phase_angle_4); pl_wave_sum = pl_wave_1 + pl_wave_2 + pl_wave_3 + pl_wave_4 + z_shift;

2) The MATLAB script is provided below: %% Generate surfaces [xq,yq] = meshgrid(sp_vector (1:16:end), time_vector (1:16:end)); % reduce mesh density for plotting purposes figure (1) pl_wave_1_sparse = griddata(sp_vector,time_vector,pl_wave_1,xq,yq); meshc(xq/2/pi*360, yq, pl_wave_1_sparse) zlim([-50 50]); xlabel('Circumferential position, \alpha [mech. deg.]') ylabel('Time, t [s]') zlabel('f_1 (\alpha) [N]') set(gca,'XTick',[0:60:360]) figure (2) pl_wave_2_sparse = griddata(sp_vector,time_vector,pl_wave_2,xq,yq);

81

meshc(xq/2/pi*360, yq, pl_wave_2_sparse) zlim([-50 50]); xlabel('Circumferential position, \alpha [mech. deg.]') ylabel('Time, t [s]') zlabel('f_2 (\alpha) [N]') set(gca,'XTick',[0:60:360]) figure (3) pl_wave_3_sparse = griddata(sp_vector,time_vector,pl_wave_3,xq,yq); meshc(xq/2/pi*360, yq, pl_wave_3_sparse) zlim([-50 50]); xlabel('Circumferential position, \alpha [mech. deg.]') ylabel('Time, t [s]') zlabel('f_3 (\alpha) [N]') set(gca,'XTick',[0:60:360]) figure (4) pl_wave_4_sparse = griddata(sp_vector,time_vector,pl_wave_4,xq,yq); meshc(xq/2/pi*360, yq, pl_wave_4_sparse) zlim([-50 50]); xlabel('Circumferential position, \alpha [mech. deg.]') ylabel('Time, t [s]') zlabel('f_4 (\alpha) [N]') set(gca,'XTick',[0:60:360]) figure (5) surface_superp_sparse = pl_wave_1_sparse + pl_wave_2_sparse + pl_wave_3_sparse + pl_wave_4_sparse + z_shift; meshc(xq/2/pi*360, yq, surface_superp_sparse) zlim([-50 50]); xlabel('Circumferential position, \alpha [mech. deg.]') ylabel('Time, t [s]') zlabel('f (\alpha) [N]') set(gca,'XTick',[0:60:360])

3) The MATLAB script is given in the question. The value of 513 is the location for the DC component in the shifted matrix. This function can be accomplished using the following MATLAB script: FFT2_raw_abs = abs(FFT2_raw); DC = FFT2_raw_abs (1,1); [DC_row, DC_col] = find(FFT2_abs_shifted == DC); FFT2_abs_shifted_cut = FFT2_abs_shifted (DC_row-10:DC_row+10,DC_col10:DC_col+10);

FFT2_raw = fft2(pl_wave_sum)/space_num/time_num; FFT2_shifted = fftshift (FFT2_raw); FFT2_abs_shifted = abs(FFT2_shifted); FFT2_abs_shifted_cut = FFT2_abs_shifted (513-10:513+10,513-10:513+10); FFT2_abs_shifted_cut_flipud = flipud (FFT2_abs_shifted_cut); 82

FFT2_abs_shifted_cut_flipud(FFT2_abs_shifted_cut_flipud 0. The phase angles in the third quadrant are the opposites of the corresponding ones in the first quadrant.

Q8. Solution 1) The contours for the sum waves of the four cases can be seen in Fig. 43. It can be seen by changing the phase angles, the locations of the peaks for the contours are shifted. However, the amplitudes of the

(a)

(b)

0.8

0.8 Time, t [s]

Time, t [s]

sum waves stay the same.

0.6 0.4

0.4 0.2

0.2 0

0.6

0

120

240

0

360

0

(d)

0.8

0.8 Time, t [s]

Time, t [s]

(c)

0.6 0.4

240

360

0.6 0.4 0.2

0.2 0

120

α [mech. deg.]

α [mech. deg.]

0

120

240

α [mech. deg.]

360

0

0

120

240

360

α [mech. deg.]

Fig. 43. Contours of the sum waves: (a) Case 1, (b) Case 2, (c) Case 3, and (d) Case 4.

83

2) The MATLAB script is provided below: %% 2D FFT analysis FFT2_raw = fft2(pl_wave_sum)/space_num/time_num; FFT2_shifted = fftshift (FFT2_raw); FFT2_abs_shifted = abs(FFT2_shifted); FFT2_abs_shifted_cut = FFT2_abs_shifted (513-5:513+5,513-5:513+5); FFT2_abs_shifted_cut_flipud = flipud (FFT2_abs_shifted_cut); FFT2_abs_shifted_cut_flipud(FFT2_abs_shifted_cut_flipud 0.01)*180/pi; % mech. deg. phase_angle_matrix(phase_angle_matrix==0) = NaN; phase_angle_matrix(abs(phase_angle_matrix)