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ELECTRICAL MACHINES Class exercises 1

MAGNETIC CIRCUITS

1

Exercises on magnetic circuits – Ex #1 In the magnetic structure reported, the three iron paths A, B, C present the following reluctance values: ℜA = 250000 H-1, ℜB = 800000 H-1, ℜC = 400000 H-1. The airgap in the branch A presents a reluctance equal to ℜt = 2750000 H-1. The coils have NA = 200 e NB = 50 turns respectively.

Evaluate:  The inductances of the coils A and B. (LA = 12.2 mH, LB = 2.17 mH)  The value of the DC current in the coil A to have a flux equal to 0 in the path C, when IB= 5 A (IA = 4.688 A)

Exercises on magnetic circuits – Ex #2 The following circuit is given, with a part constituted by Permanent Magnet with wide hysteresis cycle, having a section Sm and lenght m. The remaining part is constituted by soft magnetic material; between the polar expansions an airgap is set with g = 0.2 cm and Sg = 10 cm2. air gap iron core

Sg, g permanent magnet

Sm, m Evaluate the minimum volume, the minimum section, the minimum lenght of Permanent Magnet in the case of material: KOERMAX 160, with Bg = 0.8 T (Vmagnet = 6.358 cm3, Sminimum = 17.8 cm2, minimum = 3.58 mm).

Exercises on magnetic circuits – Ex #2

KOERMAX 160  B = 450 mT  H = -356 kA/m

Exercises on magnetic circuits – Ex #3 The scheme of a differential relay is realized with a toroidal magnetic structure with average radius equal to 5 and circular section equal to 5 cm2. The two main coils have 10 turns and the relative permeability of the iron is equal to 5000. rm = 5 cm S = 5 cm2 µr = 5000 N = 10 turns

Evaluate: The number of turns NR of the release coil so that an unbalanced current of 30 mA at 50 Hz will induce at its terminals an RMS voltage equal to 2 V (Ns = 2122) The necessary unbalanced current to have, at 60 Hz, the same induced voltage at the terminals of the release coil (ΔI = 25 mA).

Exercises on magnetic circuits – Ex #4 The electromagnet has a fixed core (upper) with a coil supplied in DC current (500 turns). The mobil core has a mass of 8 kg and is maintained in rest position (y = 0,5 mm) by gravity. Both cores, of thickness s = 40 mm, are realized with iron having neglectable reluctance up to the saturation induction Bs=1.4 T.

Evaluate:  The autoinductance of the system (rest position and absence of saturation) (L0 = 1.005 H).  The maximum current in the coil before producing saturation (in rest position) (Imax = 2.23 A)  The minimum current to move the moving part from the rest position (Imin = 0.279 A)

Exercises on magnetic circuits – Ex #5 The electromagnet in figure is made by iron, with magnetic characteristics reported in the Table fe Sfe

I

Nturns

Fe = airgap = SFe = µ0 = Nturns =

airgap

32 1,5 4,2 4 π 10-7 250

cm mm cm2 H/m

B 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.55 1.6 1.65 1.7

H 40 47 56 67 82 102 131 175 248 380 700 1650 3500 4800 6250 8000 10000

PFe 0.16 0.26 0.37 0.52 0.67 0.84 1.02 1.33 1.45 1.74 2 2.55 3.2 3.35 3.7 4.1 4.4

1,8

B [T]

1,6 1,4 1,2 1 0,8 0,6 0,4 0,2 H [A/m]

0 0

2000

4000

6000

8000

10000

Evaluate the characteristic flux - current (excitation characteristic) and the value of the inductance of the winding

12000

Exercises on magnetic circuits – Ex #5 B

H

PFe

Φ

[T]

[A/m]

[W/kg]

0,3

40

0,4

MMFairgap

MMFTOT

I

Lsemplice

[Wb]

MMF Fe [Asp]

[Asp]

[Asp]

[A]

[H]

0,16

0,000126

12,80

358,10

370,90

1,48

0,0212

47

0,26

0,000168

15,04

477,46

492,50

1,97

0,0213

0,5

56

0,37

0,00021

17,92

596,83

614,75

2,46

0,0214

0,6

67

0,52

0,000252

21,44

716,20

737,64

2,95

0,0214

0,7

82

0,67

0,000294

26,24

835,56

861,80

3,45

0,0213

0,8

102

0,84

0,000336

32,64

954,93

987,57

3,95

0,0213

0,9

131

1,02

0,000378

41,92

1074,30

1116,22

4,46

0,0212

1

175

1,33

0,00042

56,00

1193,66

1249,66

5,00

0,0210

1,1

248

1,45

0,000462

79,36

1313,03

1392,39

5,57

0,0207

1,2

380

1,74

0,000504

121,60

1432,39

1553,99

6,22

0,0203

1,3

700

2

0,000546

224

1551,76

1775,76

7,10

0,0192

1,4

1650

2,55

0,000588

528

1671,13

2199,13

8,80

0,0167

1,5

3500

3,2

0,00063

1120

1790,49

2910,49

11,64

0,0135

1,55

4800

3,35

0,000651

1536

1850,18

3386,18

13,54

0,0120

1,6

6250

3,7

0,000672

2000

1909,86

3909,86

15,64

0,0107

1,65

8000

4,1

0,000693

2560

1969,54

4529,54

18,12

0,0096

1,7

10000

4,4

0,000714

3200

2029,23

5229,23

20,92

0,0085

fe Sfe

I Nturns

airgap

Exercises on magnetic circuits – Ex #5 Excitation characteristic with the given airgap 0,0008

Φ [Wb]

0,0007 0,0006 0,0005 0,0004 Flusso

0,0003 0,0002

Inductance of the winding with the given airgap

0,0001 Corrente [A] 0 0

5

10

15

20

25

30

35

0,025

Induttanza [H]

0,02

L [semplice]

0,015

0,01

0,005

Induzione B [T]

0 0

0,2

0,4

0,6

0,8

1

1,2

1,4

1,6

1,8

Exercises on magnetic circuits – Ex #5 Excitation characteristic with double airgap 0,0008

Φ [Wb]

0,0007 0,0006 0,0005 0,0004 Flusso 0,0003

Flusso 2 traf

0,0002

Inductance of the winding with double airgap

0,0001 Corrente [A] 0 0

5

10

15

20

25

30

35

0,025

Induttanza [H]

0,02 L [semplice] 0,015

L [traferro doppio]

0,01

0,005

Induzione B [T]

0 0

0,2

0,4

0,6

0,8

1

1,2

1,4

1,6

1,8

ELECTRICAL MACHINES Class exercises 2

TRANSFORMERS

11

Exercises on Transformers – Ex #1 A single phase transformer with the following plate data: SN = 66.7 kVA t = 11.5 kV/230 V vsc % = 5 % cosϕsc = 0.4 supplies a constant impedance load with the following plate data: Pload rated = 33.3 kW Vload rated = 290 V cosϕload = 0.85 (RL) The load is supplied with a line having a resistance of 0.1 Ω and a reactance of 0.2 Ω. If the transformer is supplied at its rated voltage, and neglecting the no load parameters, evaluate:     

the voltage at the load terminals (208 V) the current absorbed by the load (97.0 A) the line efficiency (0.948) the transformer efficiency (0.992) the total efficiency if the plant (0.940)

Exercises on Transformers – Ex #2 A single phase transformer, 33.3 kVA at 50 Hz, with transformer ratio 5.75 kV/230 V, presents the following characteristics: Psc % = 1.3 % Piron % = 1.2 % vsc % = 2.5 % The transformer is supplied at primary rated voltage, and connected to a load that absorbs a current equal to ¾ of the transformer rated current (load factor α = 0.75) with cosϕ = 0.8, evaluate: the

voltage drop from no load to load conditions (4 V) the efficiency of the transformer (0.968) the short circuit current at the secondary terminals (5793 A) the energy dissipated in a month of 25 working days, with the transformer always connected to the mains and with the following daily load profile:  7 hours at full load  8 hours at half load  9 hours at no load (396 kWh)

Exercises on Transformers – Ex #3 A three phase transformer, having a voltage ratio at no load equal to 1kV/250V, is tested at no load and at short circuit. No load test on the low voltage side: I20 = 7.5 A P0 = 375 W V20 = 250 V Short circuit test on the high voltage side: Vsc = 50 V Isc = 30 A

Psc = 750 W

Evaluate: the

the

equivalent circuit parameters reported at primary R’sc = 0.278Ω X’sc = 0.921Ω R’iron = 2667Ω

X’m = 309.9Ω

equivalent circuit parameters reported at secondary R”sc = 0.017Ω X”sc = 0.058Ω R”iron = 166.7Ω X”m = 19.4Ω

Exercises on Transformers – Ex #4 Two three phase transformers TA e TB present the following plate data: TA:

SN = 100 kVA t = 10kV/400V vsc % = 4.2 % cosϕsc = 0.35

TB:

SN = 120 kVA vsc % = 4 %

t = 10kV/400V cosϕsc = 0.35

The two transformers supply in parallel a three phase balanced load with constant impedance having the following data: Vload rated = 400 V cosϕload = 0.85 (RL) Pload rated = 180 kW Neglecting the no load parameters and supplying the primaries at rated voltage, evaluate:  the line voltage at the load terminals (388 V)  the current delivered by each transformer (IA = 131 A, IB = 165 A)  the short circuit current at the load terminals (7769 A)

Exercises on Transformers – Ex #5 Two three phase transformers TA e TB supply in parallel a load made with three equal impedances (star connected): ZL = 0.56 + j 0.56 Ω. The transformers present the following plate data: TA:

SN = 160 kVA V’’sc = 29.1 V

t = 20kV / 400V Psc = 3200 W

TB:

SN = 100 kVA V’’sc = 28.8 V

t = 20kV / 396V Psc = 1800 W

Neglecting the no load parameters and supplying the primaries at rated voltage, evaluate:   

the no load circulating current (12.4 A) the current absorbed by the load (276.8 A) the line voltage at the load terminals (379.7 V)

ELECTRICAL MACHINES Class exercises 3

ASYNCHRONOUS MACHINE

17

Exercises on Asynchronous Machine – Ex #1 A three phase asynchronous motor with the following nameplate data is given: VsN = 380 V IsN = 15.8 A f = 50 Hz PN = 7.5 kW 4 poles nN = 1450 rpm Rs = 0.73 Ω

Isc Rsc Locked rotor (short circuit) test data: Psc = 1100 W Vsc = 80 V cosϕsc = 0.5 R'r Xsc Neglecting the no load parameters, evaluate: Xds=X'dr   

The starting torque (79.5 Nm) The maximum torque slip (0.281) The maximum torque (136.8 Nm)

15,87713 A 1,454545 Ohm 0,724545 Ohm 2,519347 Ohm 1,259673 Ohm

Istart

75,41638 A

Tstart

78,70428 Nm

S_Tmax Tmax

0,27623 137,084 Nm

Exercises on Asynchronous Machine – Ex #2 Nameplate data of a three phase asynchronous motor: PN = 7.5 kW VsN = 380 V IsN = 14.1 A 4 poles sN% = 3.33 % ηN = 0.941 Is0 % = 46 % Cstart / CN = 2.5 Istart / IsN = 7.46 Considering the simplified single phase equivalent circuit: Evaluate:  The motor rated torque (49.4 Nm)  The required supply frequency to obtain a rotating magnetic field at 2500 rpm (83.33 Hz) Neglecting the mechanical and iron losses and also the magnetizing current at start (Ir,start / IsN = 7.46), evaluate the following parameters:  The magnetizing reactance X0 (33.7 Ω)  The total leakage reactance Xdt (1.064 Ω)  The stator and rotor resistance Rs and Rr (Rs = 0.6 Ω, Rr = 0.58 Ω).

Exercises on Asynchronous Machine – Ex #3 During a test at rated load, a 6 poles three phases asynchronous motor supplied at 380V/50Hz absorbs a power = 23,2 kW with cosϕ = 0.88 at the speed of 960 rpm. The no load test has given the following results: no load power P0 = 1,1 kW, no load current I0 = 13.5A. The stator phase resistance Rs = 0,27 Ω, while the mechanical losses are 1/3 of the no load losses. Evaluate:  the absorbed current in rated load test conditions (40 A)  the power delivered by the motor during the test (20098 W)  the efficiency of the motor during the test (0.886)  the torque delivered by the motor during the test (200 Nm)  the no load torque, neglecting the no load slip (3.5 Nm).

Exercises on Asynchronous Machine – Ex #4 A three phase asynchronous motor, 4 poles, star connected, has the following rated data: VsN = 380 V IsN = 31 A nN = 1460 rpm Rs =0.238 Ω PN = 15 kW In the locked rotor test, at rated current, the following data are measured: Psc = 1960 W Vsc = 74.82 V cosϕsc = 0.488 In the no load test, at rated voltage, the following data are measured : P0 = 630 W cosϕ0 = 0.067 Pfv = 260 W Is0 = 14.3 A Evaluate:  the equivalent circuit parameters, taking into account the stator voltage drop in the no load conditions and considering Xds = Xdr (in the locked rotor test, neglect the magnetizing parameters Riron and Xm) (Rr = 0.442 Ω, Xds = Xdr = 0.608 Ω, Riron = 593.4 Ω, Xm = 14.73 Ω)  the rated torque of the motor (98.1 Nm)  the starting torque, neglecting Riron and Xm (209.33 Nm)  the rotating magnetic field speed at 20 Hz (600 rpm).

Exercises on Asynchronous Machine – Ex #5 A compressor presents the following rated data: Pn = 10 kW @ n = 1460 rpm, Cstart = 230 Nm. Three motors are available: A) PN = 15 kW VsN = 380 V f = 50 Hz IsN = 31.5 A 4 poles Psc = 2.4 kW cosϕsc = 0.51 Rs = 0.14 Ω B)

PN = 20 kW Psc = 2.5 kW

IsN = 55 A

6 poles

C)

PN = 15 kW VsN = 380 V f = 50 Hz IsN = 31 A Psc = 1.96 kW cosϕsc = 0.488 Rs = 0.238 Ω

4 poles

VsN = 380 V cosϕsc = 0.4

f = 50 Hz Rs = 0.1 Ω

Neglecting the no load parameters in the equivalent circuit, evaluate:  Which of the three motors can be adopted to supply the compressor (A)  The current absorbed by the adopted motor, at the motor-compressor speed of 1450 rpm (10.77 A).

MACCHINE ELETTRICHE Class exercises 4

DC MACHINE

23

Exercises on DC Machine – Ex #1 A separately excited DC motor, with rated power equal to 10 kW, moves a mechanical load at the speed of 600 rpm, while supplied at 110 V and absorbing a current of 60 A with efficiency equal to 0.75. The armature resistance is equal to 0.1 Ω. Neglecting the excitation losses, evaluate:  the power delivered by the motor (4950 W)  the armature joule losses (360 W)  the mechanical and iron losses (together) (1290 W)  the supply voltage required to double the motor speed when the absorbed current is equal to 50 A (consider the flux to be constant) (213 V).

Exercises on DC Machine – Ex #2 A permanent magnet DC motor is given, with the following data:  Torque constant kT = 0.8 Nm/A  armature resistance Ra = 0.75 Ω If the motor is supplied with a voltage equal to Va =200 V, evaluate the absorbed and delivered power in the following rotating conditions (no mechanical and iron losses): 1. ωr = 230 rad/s (Pin = 4267 W, Pout = 3919 W, Motor) 2. ωr = 280 rad/s (Pin = 7168 W, Pout = 6400 W, Brake) 3. Evaluate the rotating speed of the motor when it is lifting a 500 kg load; the mechanical reduction (supposed ideal) allows to lift the load of 2 cm every complete revolution of the motor (232 rad/s).

Exercises on DC Machine – Ex #3 A separately excited DC motor, supplied at the voltage of Va = 600V and Vexc = 600V delivers at the shaft a power equal to Pout = 25 kW, while absorbing a current equal to Ia = 46 A. The armature resistance is Ra = 0,64 Ω, and the total excitation resistance is Rexc = 460 Ω.

Evaluate: the losses in the motor (Pja = 1354W, Pjecc = 783W, PFe + Pav = 1246W) the absorbed no load current I0 (2.1 A) the efficiency when the mechanical output power is equal to 25 kW (0.88) the thermal resistance machine-ambient (Rth) if the steady state overtemperature in continuous service (S1) is equal to 100°C when the delivered power is 25 kW(0.029°C/W)

Exercises on DC Machine – Ex #4 A separately excited DC motor for railway traction presents the following data: Vrated = 2000 V Irated = 850 A Prated = 1600 kW Istart = 1100 A Ra = 0.138 Ω nN = 990 rpm nmax = 1650 rpm The excitation current is electronically regulated to be always: Iecc = 0.235 Ia. Evaluate:  the rated torque of the motor (15433 Nm)  the armature supply voltage of the motor at 500 rpm and rated torque (1068 V)  the delivered torque when the motor absorbs a current equal to 500 A (5340 Nm). During braking, the machine works as a generator, disconnected from the supply and connected to a braking resistance; the excitation current is maintained at the value before braking; evaluate:  the required resistance to brake with a maximum current equal to Istart (consider that the machine was working at rated voltage and current) (≥ 1.573 Ω).

Exercises on DC Machine – Ex #5 A series excited DC motor is supplied at the rated voltage of 1000 V. The motor is started with the insertion of a series resistance equal to 2.2 Ω, delivering a starting torque equal to 7000 Nm and absorbing a current equal to 400 A. With the starting resistance disconnected, the motor delivers in rated conditions a torque equal to 3600 Nm. Evaluate:  the parameters of the equivalent circuit ( Rexc – Ra – Kt ) (in this case Rexc = Ra) ( (Ra = Recc = 0.15Ω, Kt = 43.75 10-3 Nm/A2). 

the rotation speed (695 rpm)



the required supply voltage to increase the speed value of 10%, under the hypothesis that the resistent torque is proportional to the speed (1143 V).

Exercises on DC Machine – Ex #6 A locomotor that must be supplied by the line voltage VL = 3000V is equipped with 12 series excited DC motors with the following nameplate data: VN = 1000 V IN = 350 A PN = 300 kW Ra = 0.15 Ω Rexc = 0.25 Ω nN = 1000 rpm At start all the motors are connected in series; evaluate:  the required starting resistance Rstart to limit the starting current at the value of 500 A (1.2 Ω)  the speed of the motors when, disconnected the starting resistance, the absorbed current is equal to the rated one (128 rpm). To obtain the maximum speed of the locomotor, the motors are connected to get 4 parallels of the 3 motors in series; evaluate:  the speed of the motors when, with the starting resistance inserted, the motors absorb the rated current (350 rpm).