Assignments (Winter 2012)

8/12/2019 Assignments (Winter 2012)

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DEPARTMENT OF ELECTRICALAND COMPUTER ENGINEERING

COURSE ECSE-361POWER ENGINEERING

ASSIGNMENT #1SOLUTIONS

Prof: Anthony J. Rodolakis

Exercise #1

Given the following harmonic current waveforms)cos(400)(1 ttI

)15sin(25)(2 ttI

)15sin(25)30cos(4)(3 tttI

you are asked to determine:

The polar phasor form of these harmonic currents using)cos( t as reference framework

The cartesian (rectangular) form of the above determinedphasors.

The phasor form ofdt

dI

dt

dI

dt

dI 321 ,,

Solution

tI cos4001 , 02

4001 I (polar),

2

4000

2

400)0sin0(cos

2

4001 jjI

)1590cos(25sin252 ttI , 7552 I (polar),

83.429.1))75sin()75(cos(52 jjI

755302

4),15sin(25)30cos(4 33 IttI , (polar),

05.5928.7244.6743.3)966.0259.0(5)5.0866.0(828.23 jjjI

90084.2821 dt

dI, 907552

dt

dI, 9005.5928.73 dt

dI


2/28

Exercise #2

A single-phase circuit contains two parallel branches. The firstbranch is a series combination of a resistor 8R Ohms and aninductance of mHL 191.0 . The second branch is a capacitance

FC 530 capacitor. If the circuit is fed from a 50Hz source and

absorbs a total current of 10Amps, you are asked to determine:

The phasor form of the supply voltage

The phasor form of the individual branch currents

A phasor diagram showing all the above quantities

Solution

0.610)191.0)(50(2 3 jxjLjXL

ohms

0.6)10530)(50(2

116

jx

jCj

XC

ohms

)13.5375)(010( TOTTOTZIV , since

13.535.713.538

60

900.668(

900.6)68(

21

21

j

j

ZZ

ZZZTOT ohms

905.78

906

01021

2

1

ZZ

Z

II TOT

87.365.128

87.3610010

21

12

ZZ

ZII TOT


3/28

Exercise #3

It is given that a circuit is composed of two parallel branches and is

fed by an AC source of )cos(6502)( tte . The first branch is a

series combination of a resistor 301

R Ohms and aninductor mHL 6.1011 . The second branch is also a series

combination of a resistor 52 R Ohms and a capacitor of FC 221 .

If the voltage source frequency is assumed to be 60 Hz, you areasked to determine:

The active power consumed by every circuit element and theactive power consumed by the entire circuit

The reactive power consumed by every circuit element and thereactive power consumed by the entire circuit.

The active and the reactive power provided by the AC sourceand verify that apparent power is neither lost nor generated in

the circuit

The total circuit power factor and draw the power triangle of thecircuit, clearly indicating the totally consumed active, reactiveand apparent power.

Solution

302.3830)106.101)(60(230 31 jxjZ

ohms

125)10221)(50(2

15

62 j

xjZ

ohms

93.51360.1393.51652.48

0650

1

1

Z

VI A, 38.6750

38.6713

0650

2

2

Z

VI A

Therefore,32

30 103548.530)360.13( xPR W, 500,125)50( 2

5 RP W 00.0 CL PP W,

855,17500,12548,53 TOTP W32 108365.6)302.38()360.13( xQL lagging VARS

000,3012)50(

2

CQ leading VARS00.0530 QQR VARS,

4103164.2000,305.836,6 xQTOT (leading

VARS)Power from the source

37.52994.4438.675093.51360.132! IIIT A

162,23856.17))37.52sin()37.52(cos(994.44)(650( jIVS TOTSSOURCE kVA


4/28

Exercise #4

An installation is composed of two balanced Y-connected loadsthat are connected in parallel. Load #1 requires 15KW at a 0.6lagging power factor, while load #2 requires 10KVA at a 0.8leading power factor. Both loads are supplied by a balanced threephase ideal source of 480V. You are asked to determine:

The power triangle of every load

The power triangle of the load of the entire installation

The magnitude of the line current required by the entireinstallation

The rating of delta connected capacitor banks, connected at theservice entrance point of the installation, to raise the powerfactor of the installation to unity

The current that is expected to flow in each capacitor leg underthe above mentioned operating conditions.

Solution

2381521 LLTOT PPP kW, 1462021 LLTOT QQQ kVAR(lagging)

20)6.0tan(cos 111

LL PQ kVARS and 6)8.0tan(cos 1

21

LL PQ kVARS

lead

394.32)854.0)(480(3

23

cos3

TLL

TOT

LV

PI

kA, since

854.0)23

14(cos(tancos 1 T

Need 14kVARS to raise the pf of the installation to unity, thus

CAPCAPLLCAP IxIVQ )480(310143 3

, thus, AICAP 72.9


5/28





Exercise #1-Solution

1) Since NIHdl (Ampere) and for any distance rfrom the toroidcenter, we obtain

r

NIH

2 ,

r

NI

r

NIB r

220

The total flux through the cross-section will be

2/

2/ln

2

1

22

2/

2/

2/

2/

2/

2/

R

Rb

NIdr

rb

NIbdr

r

NIBbdr

R

R

R

R

R

R

,

with4

OUTIN DDR

,2

INOUT DD , and finally

2/

2/ln

2

2

R

Rb

N

I

N

IL

2) mAxxx

x

r

NIH /1032.29

10)350(

201612

2

3

3

, that yields

TxxxxxHB r237

0 1068.31032.291104 , that yields

32

7

42

0

2

/1040.51042

10)68.3(

2

1mJx

xx

xBWDENSITY

, that yields

JxxxxxVWWTOROIDDENSITYTOTAL

97.210)100)(50(10)350)((1040.5 632


6/28


1) Since by amperes law we have NIHdl , the magnetic circuitaround the core gives, due to the symmetry of the right and left paths

PATHRIGHTPATHRIGHTPATHCENTERPATHCENTER lHlHNIF (1)Also, since PATHRIGHTPATHLEFTPATHRIGHTPATHCENTER 2 , we obtain

PATHRIGHTPATHRIGHTCPATHRIGHTPATHCENTERPATHCENTERPATHCENTER BAAB 22

From the magnetization curve we obtain for TB PATHCENTER 5.0 that170 AmH PATHCENTER and for TB PATHRIGHT 25.0 ,

150 AmH PATHRIGHT .

Thus 11 10)72)(50(10)24)(70(400 xxIF , or AI 132.0

The magnetic permeability of the center leg will be705.0 PATHCENTERPATHCENTER HB , or

3101429.7 x , that yields

wbATxxxxx

x

A

lR

PATHCENTER

PATHCENTER

PATHCENTER /1025.510)88(101429.7

1024 343

1

The magnetic permeability of the right path will be5025.0 PATHRIGHTPATHRIGHT HB , or


wbAtxxxxx

x

A

lR

PATHRIGHT

PATHRIGHT

PATHRIGHT /105.2210)88(100.5

1072 342

1

Thus, wbAtxxR

RR LEGRIGHTLEGCENTERTOTAL /105.1610)25.1125.5(2

33

2) Similarly, for TB PATHCENTER 0.1

, we obtain TB PATHRIGHT 5.0

, densitiesthat yield 1900 AmH PATHCENTER and

170 AmH PATHRIGHT .

Thus, 11 10)72)(70(10)24)(170(400 xxI , or AI 228.0


3108824.5

x , that yields

wbATxxxxx

x

A

lR

PATHCENTER

PATHCENTER

PATHCENTER /1037.610)88(108824.5

1024 343

1



wbAtxxxxx

x

A

lR

PATHRIGHT

PATHRIGHT

PATHRIGHT /1018.1610)88(101429.7

1072 343

1

Thus, wbAtxxR


33

3) Reluctances of nonlinear magnetic cores are not constant but theydo vary depending on the operating point.


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8/28






1) Since NIHdl (Ampere) and for any distance rfrom the toroidcenter, we obtain

r

NIH

2 ,

r

NI

r

NIB r

220

The total flux through the cross-section will be

2/

2/ln

2

1

22

2/

2/

2/

2/

2/

2/

R

Rb

NIdr

rb

NIbdr

r

NIBbdr

R

R

R

R

R

R

,

with4

OUTIN DDR

,2

INOUT DD , and finally

2/

2/ln

2

2

R

Rb

N

I

N

IL

2) mAxxx

x

r

NIH /1032.29

10)350(

201612

2

3

3

, that yields

TxxxxxHB r237

0 1068.31032.291104 , that yields

32

7

42

0

2

/1040.51042

10)68.3(

2

1mJx

xx

xBWDENSITY

, that yields

JxxxxxVWWTOROIDDENSITYTOTAL

97.210)100)(50(10)350)((1040.5 632


9/28


1) Since by amperes law we have NIHdl , the magnetic circuitaround the core gives, due to the symmetry of the right and left paths

PATHRIGHTPATHRIGHTPATHCENTERPATHCENTER lHlHNIF (1)Also, since PATHRIGHTPATHLEFTPATHRIGHTPATHCENTER 2 , we obtain

PATHRIGHTPATHRIGHTCPATHRIGHTPATHCENTERPATHCENTERPATHCENTER BAAB 22

From the magnetization curve we obtain for TB PATHCENTER 5.0 that170 AmH PATHCENTER and for TB PATHRIGHT 25.0 ,

150 AmH PATHRIGHT .

Thus 11 10)72)(50(10)24)(70(400 xxIF , or AI 132.0



wbATxxxxx

x

A

lR

PATHCENTER

PATHCENTER

PATHCENTER /1025.510)88(101429.7

1024 343

1



wbAtxxxxx

x

A

lR

PATHRIGHT

PATHRIGHT

PATHRIGHT /105.2210)88(100.5

1072 342

1

Thus, wbAtxxR


33

2) Similarly, for TB PATHCENTER 0.1

, we obtain TB PATHRIGHT 5.0

, densitiesthat yield 1900 AmH PATHCENTER and

170 AmH PATHRIGHT .

Thus, 11 10)72)(70(10)24)(170(400 xxI , or AI 228.0


3108824.5

x , that yields

wbATxxxxx

x

A

lR

PATHCENTER

PATHCENTER

PATHCENTER /1037.610)88(108824.5

1024 343

1



wbAtxxxxx

x

A

lR

PATHRIGHT

PATHRIGHT

PATHRIGHT /1018.1610)88(101429.7

1072 343

1

Thus, wbAtxxR


33

3) Reluctances of nonlinear magnetic cores are not constant but theydo vary depending on the operating point.


10/28


3322

1058.1001.05000

10)230(01.0 xx

xx

S

VRZRR

B

B

PUBASEPUHIGHTOTAL ohms

33

322

10529.01058.1005.05000

10)230(05.0 xxxxxSVXZXX

B

BPUBASEPUHIGHTOTAL

And for the low voltage side,

38.0)8.13/230(

1058.1001.02

3

xx

R LOWTOTAL ohms 9.1)8.13/230(

1058.1005.02

3

xx

X LOWTOTAL ohms

Magnetization branch

4.42419.44

10)8.13( 322

x

P

VR

OCTEST

OCTEST

LOWM ohms

91.913

3.15

10)8.13( 3

x

I

VZ

OCTEST

OCTEST

LOWM ohms

215.03.158.13

)9.44(cos

xIV

P

OCTESTOCTEST

OCTEST

OC lagging

747.14976.01.15))215.0(cos90cos( 1 xII OCMOCXM A

78.935747.14

108.13 3

X

I

VX

OCXM

OCTEST

LOWM ohms

At no load, using the cantilever equivalent circuit, we need

87.3632.3628.0108.13

4000

cos 3 xV

P

I LOADLOAD

LOAD

LOAD A

88.114329)9.138.0)(87.3632.362(108.13 3 jxV LOADNO V

%833.313800

1380014329

FULLLOADSECONDARY

FULLLOADAECONDARYNOLOADSECONDARY

V

VVVR

kWRIP TOTALLOADCU 88.4938.0)23.326(22

kWR

VP

TOTALM

OPERATING

IRON 375.484.4241

1432922

%6.97976.0

375.4888.494000

4000

IRONCUOUT

OUT

PPP

P


11/28

1

DEPARTMENT OF ELECTRICAL AND COMPUTER ENGINEERING

COURSE-ECSE-361-POWER ENGINEERING


Prof. A. J. Rodolakis

Exercise #1

For the linear machine shown below, the conducting bar whose two endsrest on the conducting rails is free to move along the rails with no frictionat all. If it is given that the bar length is 0.5m, the magnetic field density B is

0.33T with its direction as shown below, the series resistance is 0.5 Ohms,the battery source vol tage equals 120 Volts with the shown polarity and weassume that the bar has no inductance at all, you are asked to:

1. Determine the steady state bar speed magnitude and direction if aforce of 10 N is applied in the below shown direction.

2. Is this linear machine acting as a generator or as a motor?

Solution

1. For the shown external force of 10 N, the bar will reach steady state when

)5.0()33.0(10 IBIlFF EXTERNALLORENTZ , thus AI 6.60

In order for the Lorentz force to oppose the applied external force we needthe current to go down the bar. That yields

6.605.0

120

INDUCEDBARINDUCEDBARSOURCE

E

R

EVI . That yields

vBlvE INDUCEDBAR )5.0)(33.0(70.89 . That yields 6.543v m/s2. The machine is acting as a motor because it is overcoming the external

force by means of importing electrical energy from the source.


12/28


13/28

3

gaps. Then correct by taking into account fringing. (Fringing can beaccounted for, assuming that the cross section of the air-gap is a newsurface having an area determined by the formula

))(( glgwANEW with wl and g being the original area of the gapped

section and the air gap length respectively.

2. What will be the average value of the force holding the systemtogether if an alternating voltage of V100 (RMS) is applied across the

coil for a sufficient amount of time. Calculate this force by neglectingfringing in the air gaps. Then correct by taking into account fringingusing the same approach as in question 1 above.

Solution

1. DC excitation.

The total reluctance of the magnetic system, under the stated assumptions, willbe :

Since wbAtx

xAxxRxRxRxR GAPGAPGAPTOT /

103010422)(2)()()(

627

0

21

.

The system inductance will be

Hxxx

Ax

N

xR

NxL

TOT

556272

0

22 105343.3

2

100686.7

2

1030104250

2)()(

Therefore, without fringing

NxLdx

d

R

VxL

dx

dIxF 66.502)

)001.0(

105343.3()

5.7

40(

2

1)]([)(

2

1)]([

2

1)(

2

5222

and with fringing, since

Hx

x

x

xxL

2272 )030.0(0393.0

2

)030.0(104250)(

Nx

xxx

xLdx

d

R

VxL

dx

dIxFf

48.502)030.0()03.0(2

0393.0)5.7

40(

2

1

)]([)(2

1)]([

2

1)(

2

22

22

2. AC Exci tat ion

377602 x ,

ohmsjx

xLjRZ 63.60290.15324.135.7

105343.33775.7

5

)63.60377cos(54.62)63.60377cos(290.15

1002)( txt

xtI


14/28

4

Nx

txxLdx

dItF

2

522

)001.0(

105343.3)63.60377cos(54.62(

2

1)]([

2

1),001.0(

NttF )63.60377(cos8.1511),001.0( 2

NtdtT

F

T

AVERAGE 7562

7.511,1)()63.60377(cos8.1511

2 2/

0

2

and with fringing, since Hx

xxL

5107738.3

)(

ohmsjx

xLjRZ 20.62083.16227.145.7

107738.33775.7

5

)63.60377cos(2177.62)20.62377cos(20.62083.16

1002)(

txt

xtI

Nx

txxLdx

dIxFF 2

522

)001.0(

107738.3)63.60377cos(2177.62(

2

1)]([

2

1)(

NtxFF )63.60377(cos2.1459)( 2

NF AVERAGEF 730

Exercise #4

The back-end edges of two conducting plates, as shown below, areconnected through an ionized gas that is established at a distance of 1mfrom their front-end edges.

The front-edges are connected to a current source supplying the

conducting plates with 1000A. If the length of the front edges of bothplates is 100mm and the distance between the plates is 10mm, you areasked to determine.1. The inductance of the conducting loop formed by the plates and the

ionized gas assuming the magnetic field to be uniform between theplates and zero elsewhere.

2. The force acting on the ionized gas column.3. The pressure, per unit surface, on the surface of the ionized column.


15/28

5

Solution

1. Applying Amperes law on any conducting plate yields that 11 2yHIdlH with 1H and y being the magnetic field produced by the current in the plate

and the length of the plate edges respectively. Thus, mAy

IH /

21 .

Since there is another plate as well carrying the return loop current there willbe a similar magnetic filed, yielding a total magnetic filed intensity of

mAy

I

y

I

y

IHHHT /

2221 . Since the total reluctance seen by the

magnetic flux iszx

yR

0 with z and x the distance to the ionized gas column

and the distance between the plates respectively. Thus, the inductance of

the loop will be Hxx

xx

zx

y

R

NL

8

3

371

0

2

10410100

)1010)(1(104)(

2. The force on the ionized gas column will be

Nxxxy

xI

y

xz

dz

dIzL

dz

dIF

272

0

2

0

22 102100

10)104(1000

2

1

2

1)(

2

1)]([

2

1

3. The pressure on the column will be given by

2

6

2

/8.621010010

102mN

xx

x

A

FP


16/28

1



ASSIGNMENT #5

SOLUTIONS


Exercise #1The magnetic path of a two-pole reluctance motoring device can be approximated by

the analytical expression wbAtxR /)2cos5.15.2(1006.5 4 .The devices magnetic core

is excited by a 15-turn coil that can be assumed to have negligible resistance. If a 60Hz, potential difference of 110V RMS is applied across the coils terminals, you areasked to determine:

1. The maximum value of magnetic flux produced in the machine air-gaps2. The angular velocity of rotor rotation that allows the machine to produce an

average unidirectional torque.3. The maximum average electromechanical torque the machine can possibly

produce.4. The mechanical power output of the machine under maximum average torque

production conditions.

Solution

1.mfNE 44.4 , mWb

xxm 527.27

156044.4

110

2. Since )2cos()( cba RRRR and tt mcos)(

)]2cos([)(cos2

1)]([)(

2

1)( 222

cbam RRR

d

dtR

d

dtT

2sin)(cos)]2cos[)(cos2

1)( 2222 camcam RRtRR

d

dtT

2sin)2cos1(2

1)( 2 cam RRtT , and if 0 tm

)222sin(2

1)222sin(

2

1)22[sin(

2

1)( 000

2 tttttRRT mmmcam

There is an average torque only when sradm /377

3. )2sin(

2

1

2

1)( 0

2 camAVERAGE RRT and camMAXAVERAGE RRT 2

4

1)( for

045

with NxxxxRRT camMAXAVERAGE 38.145.11006.510)527.27(4

1

4

1)45( 4622

4. kWxTP mMAXAVERAGE 42.537738.14


17/28


18/28

3

Exercise #3The wire loop shown below is under the influence of a uniform magnetic field of densityB=0.5T, rotates counter-clockwise with an angular speed of 103 rad/s and has a radiusof 0.1m and a stack length of 0.5m. You are asked to determine:

1. The voltage induced around the wire loop, as a function of time. If this inducedvoltage is an AC voltage, determine its frequency, its crest and its RMS value.

2. The current that will flow in the loop, as a function of time, if a 5 Ohm resistor isconnected at the 2 ends of the loop (ignore loop inductance).3. The torque, in direction and magnitude, as a function of time that will be exerted

on the wire loop when current is allowed to flow in the 5 ohm resistor, themaximum possible torque exerted on the loop and the position at which thishappens.

4. The maximum flux that cuts across the wire loop and the position at which thishappens.

5. The average electric power consumed in the 5 ohm resistor?6. The average mechanical power that is needed to rotate the loop at 103 rad/sec,

when the loop is closed through the 5 ohm resistor.

7. The relationship between the electrical and the mechanical power.8. If the wire loop acts as a generator or a motor when closed through the 5 ohmresistor.

Solution1. The voltage across the coil will be

)103sin(15.5)103sin()5.0)(5.0)(103)(1.0(22)( tttBSINRLtELOOP

Frequency of AC voltage Hzf 39.162

103

2

Effective voltage VE

E MAXRMS 64.32

15.5

2

2. If a 5 Ohm resistor is connected across there will be a current in the loop

)103sin(2783.0)103sin(03.15

)103sin(15.5)()( tt

t

R

tEtI LOOPLOOP

3. For the direction of B shown, the produced torque due to the current flow will be ofthe counter-clockwise direction and oppose the rotation

)103(sin0515.0)103sin()5.0)(5.0)](103sin(03.1)[1.0(2sin2)( 2 ttttRILBtT ,


19/28


20/28

5

For st 00417.0 :

cos)00417.0377sin(283.6cossin)101.0(2

)1000)(10(104)0,(

3

7 xtx

xB

cos283.6cos)90sin(283.6)0,( 0 B

For st 0125.0 :

cos)0125.0377sin(283.6cossin)101.0(2

)1000)(10(104)0,(

3

7 xtx

xB

cos283.6cos)270sin(283.6)0,( 0 B ,

i.e. a mirror image of the previous spatial wave.

2. The resultant stator MMF spatial wave will be at t=1/4 of the supply cycle calculatedas follows:

)00417.0()00417.0()00417.0()00417.0( CBATOT BBBB , with

The components of the three space vectors along the axis of phase a-a being:

At t=0

0.0)0.0( AB , )120sin(283.6)0.0( BB , )120sin(283.6)0.0( CB , yielding240120 )120sin(283.6)120sin(283.6)0.0(

jj

TOT eeB

)]240sin()240[cos(2

3(283.6)]120sin()120[cos(

2

3(283.6)0.0( jjBTOT

]2

35.0[

2

3(283.6]

2

35.0[

2

3(283.6)0.0( jjBTOT

283.62

3

2

3

2

3283.62)0.0( jjBTOT , i.e. the total stator field space vector sits at

90 behind the phase a magnetic axis.

At t=0.00416 s (1/4 of the supply cycle), or 090t , yielding

283.6)90sin(283.6)0.0( 0 jA eB ,120120 )30sin(283.6)12090sin(283.6)0.0( jjB eeB ,240240 )210sin(283.6)12090sin(283.6)0.0( jjC eeB

Yielding a total value

240120 )2

1(283.6)

2

1(283.6283.6)0.0( jjTOT eeB

)]240sin()240)[cos(21(283.6)]120sin()120)[cos(

21(283.6283.6)0.0( jjBTOT

]2

3

2

1)[

2

1(283.6]

2

3

2

1[

2

1(283.6283.6)0.0( jjBTOT

283.62

3)0.0( TOTB , i.e. the total stator field space vector sits at the phase a

magnetic axis.


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26/28

1





Exercise #1

A three-phase, Y-connected, 2-pole, 60Hz induct ion motor has a rotorspeed equal to its synchronous speed and its rotor is open circuited. Youare asked to:

1. Determine the frequency of the rotor-induced voltage if the stator issupplied by a 25Hz source.2. Determine the nature of the rotor-induced voltage if a dc source is

connected between the 2 stator phase leads and the third is leftopen.

Solution

1. Since the rotor runs at synchronous speed it will run at 3600RPM. If thestator is supplied by a frequency of 25Hz then (assuming the same direction

of rotation as the rotor) we will have that 4.1

1500

36006025

xs ,

Hzxsff sr 35254.1 . Note: The solution assumes that the rotor rotates

at the 60 Hz synchronous speed.

2. )cos()120cos((cos2

)120cos(2

cos2

)( DCNININI

F

)2

3sincos

2

3(

2120sinsin120coscos(cos

2)(

NINIF

)30cos(2

3)30cos(

2

3)sin

2

1cos

2

3(

2

3)(

NININIF , a co-

sinusoidal spatial distribution. The rotor windings run past this stationary

field with a speed of 3600 RPM, and a sinusoidal 60Hz voltage will beinduced in it.

Exercise #2

A 4-pole, three-phase, Y-connected induct ion motor is rated 5HP and isconnected across a three-phase, 60-Hz, 220 V (line to line RMS) AC supply.The per=phase induction motor equivalent circuit parameters were


27/28

2

determined from tests and are: Rs=0.48Ohms, Xs=j0.80Ohms,Rr=0.42Ohms and Xr=j0.80Ohms, while Xm=j30.00Ohms and Rm=infinite(subscripts s and r refer to stator and rotor respectively, subscript m refersto magnetization parameters and all rotor quantities are already referred tostator). For this machine, and at an operating slip equal to 4%, determine

1. The RMS value of the current the machine draws from the supplymains.2. The power factor the motor operates at. Is this a lagging or a leading

power factor?3. The machine mechanical output power and electromagnetic torque.

Solution

1. ohmsjjjs

ZR 35.453.108.05.1008.04.0

42.008.

42.0

ohmsZZ

ZZZ

RM

RM

C

2.2370.915.7154.32

35.94315

903035.453.10

)9030)(35.453.10(

ohmsjjjZZZ CRTOTAL 20.26446.10621.439.9)281.391.8()8.048.0(

AZ

VI

TOTOAL

S

TOTAL 20.26134.1220.26446.10

127

2. 897.0)20.26cos(cos lagging

3. kWIVP TOTALS 147.4)897.0)(314.12)((127(3cos33

Stator loss, kWRIP STOTLOSSS 212.0)48.0()314.12(33 22

Gap power, kWPPP LOSSSGAP 935.3212.0147.43

Mechanical power,

Ax

RZZ

ZIRIP R

RM

MTOTAL

RRLOSSR 868.1154.32

30134.12][ 22

kWPPP LOSSRGAPMECH 778.342.0)868.11(3935.3 2

Electromagnetic torque

NmxP

TM

MECH

ELEC 8.20181

10778.3 3

since sradn

M /181)1728(22

3600

60

)04.01(2

602

,

Exercise #3An asynchronous, 4-pole, three-phase 50Hz, motor is Y-connected, and

operates at 380V. It is known that its stator leakage impedance 41 jZs is

equal to its rotor leakage impedance when the motor is at standstill. Themagnetizing reactance of the motor is 20 ohms. You are asked todetermine:

1. The maximum possible electromagnetic torque (pullout torque) themachine can develop


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2. The shaft mechanical speed when the motor delivers the abovemaximum electromagnetic torque

3. The motor slip when the machine produces an electromagnetictorque equal to half its pull -out torque

Solution

1. Using the exact equivalent circuit, we obtain

Vj

j

XXjR

jXVV

MSS

MSTHEV 4.218.183

6.8702.24

904400

)204(1

200220

)(

)204(1

)41(20

)(

)(

j

jj

XXjR

jXRjXjXRZ

MSS

SSM

THEVTHEVTHEV ohms

36.369.036.78343 jZTHEV ohms

The synchronous angular speed is (4-poles)

sradP

fSYN /157

4

)50(44

NmXXRR

VT

RTHEVTHEVTHEV

THEV

SYN

MAX 67.39)(2

31

22

2

2. Since 135.03923.7

1

)436.3(69.0

1

)( 2222

RTHVTHEV

R

MAX

XXR

Rs

that yields a mechanical speed of

sradsSYNm /8.135)135.01(157)1( or RPM1300)135.01(1500

3. If we denote by *s the slip at which MAXELECTR TT2

1

22

2

22

2

)()(

31

)(2

31

2

1

RTHEVR

THEV

THEVR

SYNRTHEVTHEVTHEV

THEV

SYN XXs

RR

V

s

R

XXRR

V

That yields

])()[(2

1])([

2 2222RTHEV

R

THEVRTHEVTHEVTHEV XXs

RRXXRR

s ,

])436.3()

1

69.0[(2

1

])436.3(69.069.0[

2 2222

ss

The quadratic yields 2 solutions, namely 53.01

s and 035.02

s . The smaller

slip is the solution.Note:The solution provided accounted for the magnetization impedance ofthe machine to illustrate the solution in the general case.

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Assignments (Winter 2012)