Sheet2 Solved

8/12/2019 Sheet2 Solved

1/25

3.1 The rotor of Fig.3.25 is similar to that of Fig.3.2 (EX3.1) except that it has two

coils instead of one. The rotor is nonmagnetic and is placed in a uniform magnetic

field of magnitude 0B . The coil sides are of radius R and are uniformly spaced

around the rotor surface. The first coil is carrying a current1I and the second coil

is carrying a current 2I . Assuming that the rotor is 0.3m long , R=0.13m , and

0B =0.85 T , find the -directed torque as a function of rotor position for (a)

1I = 0 A and 2I = 5 A , (b) 1I = 5 A and 2I = 0 A , (c) 1I = 8 A and 2I = 8 A .

Fig.3 two-coil rotor for problem 3.1

Sol:

mNII .]cossin[1063.6 212

cos53.0)( Ta

sin53.0)( Tb

]cossin[2 210 IIRlBT

(a) = 0 A and = 5 A , (b) = 5 A and = 0 A , (c) = 8 A and = 8 A .

]cos8sin8[53.0)( Tc


2/25

3.2The winding current of the rotor of Problem 3.1 are controlled as a function of

rotor angle such that sin81 I A and sin82 I A .Write an expression

for the rotor torque as a function of the rotor position .

Sol

3.3 Calculate the magnetic stored energy in the magnetic circuit of Example 1.2.

Sol:

1.2=0.13(wb),I=10(A)

)(1310

13.01000H

I

NL

1.47

)(6502

1013

2

1 22 JLIW

cos8sin8 21 IandI

mNII .]cossin[1063.6 212

mN

mN

.5204.0

.]cos8sin8[1063.6 222

]cossin[2 210 IIRlBT


3/25

3.4 An inductor has an inductance which is found experimentally to be of form

0

0

1

2

xx

LL

where 0L =30mH, 0x =0.87mm,and x is the displacement of moveable

element.

Its winding resistance is measured and found to equal 110m.

a.The displacement x is held constant at 0.9mm, and the current is increased from 0

to 6.0A. Find the resultant magnetic stored energy in the inductor.

b. The current is then held constant at 6.0A, and the displacement is increased to

1.80mm.Find the corresponding change in magnetic stored energy.

Sol:

For mmx 9.0

)(5.29

87.0

9.01

302mHL

1.47

)(531.06105.292

1

2

1 232 JLIW

For mmx 8.1

)(6.19

87.0

8.11

302mHL

)(352.06106.192

1 23 JW

)(179.0531.0352.0 JW


4/25

3.5Repeat Problem 3.4, assuming that the inductor is connected to a voltage source

which increases from 0 to 0.4V (part a ) and then is held constant at 0.4 V(part b ). For both calculations, assume that all electric transients can be ignored.

Sol:

For a coil voltage of 0.4V, so coil current will equal AI 7.311.0

4.0

(a.)

If mHL 5.29 :

JouleLIWfld 202.07.3105.292

1

2

1 232

If mHL 6.19 :

JoulesLIWfld 134.07.3106.192

1

2

1 232

(b.)

)(068.0202.0134.0 JoulesWfld

3.6 The inductor of Problem 3.4 is driven by a sinusoidal current source of the form

tIti sin0 Where AI 5.50 and Hz50100 . With the displacement

held fixed at 0xx , calculate (a) the time-averaged magnetic stored energy ( fldW )

in the inductor and (b) the time averaged power dissipated in the winding

resistance.

Sol:

For mHLLxx 30, 00 . The rms. current is equal AIrms 89.32

5.5

(a.)

JoulesLIWfld 227.089.310302

1

2

1 232

(b.)

WRIP rmsdiss 63.111.089.3 22


5/25

3.11 The inductance of a phase winding of a three-phase salient-pole motor ismeasured to be of the form

mm LLL 2c o s)( 20

where m is the angular position of the rotor.

a. How many poles are on the rotor of this motor?b. Assuming that all other winding currents zero and that this phase is excited

by a constant currentI0, find the torque Tfld() acting on the rotor.

Sol:

(a): 2

(b):2

2222112

2

111fld iL2

1iiLiL

2

1W

m2

2

0m20

2

m

ldf

fld 2sinLI)2cosLL(2

Ia

d

dWT


6/25

3.13 Consider the plunger actuator of Fig. 3.29. Assume that the plunger isinitially fully opened(g = 2.25 cm) and that a battery is used to supply a

current of 2.5 A to the winding.

a. If the plunger is constrained to move very slowly (i.e., slowly comparedto the electrical time constant of the actuator), reducing the gap gfrom

2.25 to 0.20 cm, how much mechanical work in joules will be supplied

to the plunger.

b. For the conditions of part(a), how much energy will be supplied by thebattery(in excess of the power dissipated in the coil)?

Figure 3.29 Plunger actuator for Problem 3.12.

Sol:

(a): Work = Wfld(g = 0.2 cm) Wfld(g = 2.25 cm) = 46.7Joules

(b):


7/25

3.14

As shown in Fig.3.30 an N-turn electromagnet is to be used to lift a slab of iron of

mass M. The surface roughness of the iron is such that when the iron and the

electromagnet are in contact there is a minimum air gap of gmin=0.18mm in each

leg .the electromagnet cross-sectional area Ac=32cm and coil resistance is

2.8 .Calculate the minimum coil voltage which must be used to lift a slab ofmass 95kg against the force of gravity. Neglect the reluctance of the iron.

Sol:

9.895Kg

: gAcNL 2/20

NN 450

Ffid= 2

2

2

02

)4(2 ig

AcN

dg

dLi

g=9.8m/ 2

sec

931N Ffid imin

: mAAc

Ff id

N

gi 385)

2(

0

minmin

Vmin =385mA2.8=1.08V

Iron slab mass M

gg

Electromagnet

cross-sectional

Area Ac

N-turn winding


8/25

3.16 An inductor id made up of a 525-turn coil on a core of 14cm2cross - sectional

area and gap length 0.16mm the coil is connected directly to a 120-V 60-Hz

voltage source. Neglect the coil resistance and leakage inductance .Assuming

the coil reluctance to be negligible ,calculate the time-averaged force acting on

the core tending to close the air gap .How would this force very if the air-gap

length were doubled?

Sol:

g

AcNL

2

0 g

Li

dg

dLiFf id

22

22

Irms=Vrms/wL

NAcN

rmsV

Lg

IFf ld rms 115

022 222


9/25

3.17Fig. 3.31 shows the general nature of the slot-leakage flux produced by current I

in a rectangular conductor embedded in a rectangular slot in iron. Assume that

the iron reluctance is negligible and that the slot leakage flux goes straight

across the slot in the region between the top of the conductor and the top of the

slot.

(a)Derive an expression for the flux density Bs in the region between the top of

the conductor and the top of the slot.(b)Derive an expression for theslot-leakage sits crossing the slot above the conductor, in terms of the height xof the slot above the conductor, the slot width s, and the embedded length l

perpendicular to the paper.(c)Derive an expression for the force f created bythis magnetic field on a conductor of length l. In what direction does this force

act on the conductor?(d)When the conductor current is 850 A, compute theforce per meter on a conductor in a slot 2.5cm wide.

Sol:

part(a)

part(b)

NAAB

s

iBS

0

s

xlxlBAB sss

0


10/25

partc(c)

part(d)

s

li

dx

dWf

2

20

'

s

li

dx

dWf

2

2

0'

mNf 1.18


11/25

3.18A long, thin solenoid of radius r and height h is shown in Fig. 3.32. The magnetic

field inside such a solenoid is axially directed, essentially uniform and equal to

H=Ni/h. The magnetic field outside the solenoid can be shown to be negligible.

Calculate the radial pressure in newtons per square meter acting on the sides of

the solenoid for constant coil current i=Io.

Sol:

222

00

2

0'

22i

h

NrCoil

HW

2

0

2

00

0

'

Ih

Nr

dr

dWf

2

02

2

0

0 22Ih

N

hr

fP


12/25

3.22The two-winding magnetic circuit of Fig.3.36 has a winding on a fixed yoke and

a second winding on a moveable element .The moveable element is constrained

to motion such that the length of both air gaps remain equal.

a. Find the self-inductances of winding 1 and 2 in terms of the core dimension andthe number of turns

b. find the mutual inductance between the two winding.c. Calculate the coenergy Wfld(i1,i2)d. Find an expression for the force acting on the movable element as a function of

the

winding currents.

i1

i2Winding 1,N1 turns

Cross-sectionalarea A

Winding 2, N2 turnsMoveable element

+ _

+ _

Yoke

g0

g0

u

u

2

1


13/25

SOL:

(a)

0

02112

12

0

02111

11

2121111

2

0

0211

0

02

111

0

02211

2211

2

21

2

22

2

g

AuNNL

L

gAuNL

L

iLiL

ig

AuNNi

g

AuNN

g

AuiNiN

R

iN

iNiNiNF

1

1

0

02

222

22

1212222

1

0

0212

0

02

222

2

22

2

g

AuNL

L

iLiL

i

g

AuNNi

g

AuNN

2

(b)

2

0

02112

2i

g

AuNNL

(c)

222110

0

2121

2

2

2

2

2

1

2

1

0

0

0

212102

2

0

2

202

1

0

2

102112

2

222

2

111

'

4

)2(4

4

2

442

1

2

1

iNiNg

Au

iiNNiNiNg

Au

g

iiNNui

g

ANui

g

ANuiiLiLiLW fld


14/25

(d)

2221120

0

2

0

2

22110021

0

'

'

4

)4(

40)4(,

iNiNg

Aug

iNiNAugii

g

WW

fld

fld


15/25

3.24Two windings, one mounted on a stator and the other on a rotor,have self-and

mutual inductances of

L11=4.5H L22=2.5H L12=2.8cos H

where is the angle between the axes of the windings. The resistances of the

windings may be neglected. Winding 2 is short-circuited, and the current in

winding 1 as a function of time is i1=10sinwt A.

a. Derive an expression for the numerical value in newton-meters of theinstantaneous torque on the rotor in terms of the angle.

b. Compute the time-averaged torque in newton-meters where 045 c. If the rotor is allowed to move ,will it rotate continuously or will it tend to cone to

rest ? If the latter, at what value of 0

(1)mN)sin(2cos(2wt))-78.5(1

cos2sinsin2,2

cos2-1wtsin

cossin)(sin314cossin14.3

,sin10

cos12.1)5.2

cos8.2()(

0

2

sin8.2

2

1

2

1

)(

:

2

22

1

1

111

22

122

2221212

21

'

2112

2

222

2

111

'

fld

fld

fld

fld

fld

T

wt

wtiT

wti

iiH

HiL

Li

iLiL

mNiiW

T

iiLiLiLW

a

sol

mNT

b

fld

5.78)1(45

)(

0

00

fld

fld

270or90

0d

Tdand0T

,)(

c


16/25

3.25

Fig 3.37

A loudspeaker is made of a magnetic core of infinite permeability and circular

symmetryas shown in Fig 3.37a and b. The air-gap length g is much less than the radius

0r of the central core. The voice coil is constrained to move only in the x direction and is

attached to the speaker conewhich is not shown in the figure. A constant radial magnetic

field is produced in the air gap by a direct current in coil 1 11 Ii . An audio-frequency

signal tIi cos22

is then applied to the voice coil. Assume the voice coil to be ofnegligible thickness and composed of N2turns uniformly distributed over its height h.

Also assume that its displacement is such that it remains in the air gap hlx 0

aCalculate the force on the voice coilusing the Lorentz Force LawEq.3.1

bcalculate the self-inductance of each coil.

ccalculate the mutual inductance between the coils. (HintAssume that current is

applied to the voice coiland calculate the flux linkages of coil 1. Note that these

flux linkages very with the displacement x)

dcalculate the force on the voice coil from the coenergy'

fidW .

Sol:

ag


17/25

bg

AN

Ag

N

R

NL

g

g

0

2

0

22

g

NlrL

2

10011

2

)3

2(

2 220022

hxl

g

NrL

c

)

2

(2 2100

12 lh

x

g

NNrL

d

2121002

2

2

2002112

2

222

2

111

2

2

1

2

1ii

g

NNri

g

NriiLiLiL

dx

dwfld

tIiIi cos, 2211

tIIg

NNrtIg

Nrwfld cos2cos 212100222

2

200


18/25

3.29

Fig 3.40

Figure 3.40 shows a circularly symmetric system in which a moveable plunger

constrained to move only in the vertical directionis supported by a spring of spring

constant K=5.28 N/m. The system is excited by a samarium-cobalt permanent-magnetin the shape of a washer of outer radius R3inner radius R2and thickness tm . The

system dimension are

R1=2.1cm R2=4cm R3=4.5cm h=1cm g=1mm tm=3mm

The equilibrium position of the plunger is observed to be x=1.0mm

aFind the magnetic flux density Bg in the fixed gap and Bx in the variable gap

bCalculate the x-directed magnetic force pulling down on the plunger

cThe spring force is of the form xXKfspring 0 . Find 0X

ola gg HB 0 xx HB 0 )( cmRm HHB

005.1 R mKAHc 712

'

T

RRhtRgRhx

tHRB

R

m

mcg 0562

)(22

)(

2

2

2

3

2

101

10


19/25

TBR

hB gx 5 3 5.0

2

1

b

NAB

iLi

RR

htRgRhx

NhRBhRN

R

m

g

)(

22

22

2

2

2

3

2

10

1

22

101

N

RR

htRgRhx

tHhR

RR

htRgRhx

NihR

dX

dLi

f

R

m

mc

R

m

fld 0158.0

)(

22

2

)(

22

2

2 2

2

2

2

3

2

101

22

10

2

2

2

2

3

2

101

22

102

c

xXKf 0

mmK

fxX 20


20/25

3.30The plunger of a solenoid is connected to a spring. The spring force is given by

xaKf 9.00 ,wherex is the air-gap length. The inductance of the solenoid isof the form axLL /10 , and its winding resistance isR.

The plunger is initially stationary at position ax 9.0 when a dc voltage of

magnitude V0is applied to the solenoid.

a. Find an expression for the force as a function of time required to hold theplunger at position 2/a

b. If the plunger is then released and allowed to come to equilibrium, find theequilibrium positionX0. You may assume that this position falls in the range

aX 00 .

Sol :

part (a):

ax 9.0 00 1.0/1 LaxLL

/0)( te

R

Vti , RL /

/2

2

00

0

2

/0

0

2

/0

2

2

2

/1

2

2

t

t

t

fld

eR

V

a

L

a

Le

R

V

dx

axLde

R

Vdx

dLif

part (b):

X02

0

0

0

0

02

9.09.0

R

V

aK

La

K

faX


21/25

3.31 Consider the solenoid system of Problem 3.30. Assume the following parameter

values:

cmNKRcmamHL /5.35.12.20.4 00

The plunger has mess KgM 2.0 . Assume the coil to be connected to a dc

source of magnitude 4A. Neglect any effects of gravity.

a. Find the equilibrium displacementX0.b. Write the dynamic equations of motion for the system.c. Linearize these dynamic equations for incremental motion of the system

around its equilibrium position.

d. If the plunger is displaced by an incremental distance from itsequilibrium positionX0 and released with zero velocity at time 0t , find(i)

The resultant motion of the plunger as a function of time, and (ii) The

corresponding time-varying component of current induced across the coilterminals.

Sol :

part (a):

AIi 40

Na

Lif 45.1

102.22

10416

2 2

30

2

cmKfaX 56.15.345.12.29.09.0 00

part (b):

Nxxdt

xd

xaKfdt

xdM

5.348.52.29.05.345.12.0

9.0

2

2

02

2

dt

dxv

dtdx

a

LRI

dtdLIRIv

182.06

0000

part (c):

)(0 txXx , )(0 txVv


22/25

xdt

xd

xdt

xd

5.17

5.348.52.0

2

2

2

2

dt

xdv

dt

dxv

182.0

182.06

part (d)

mttx c o s)(

sec/18.45.17 rad Vttv sin76.0)(


23/25

3.32 The solenoid of Problem 3.31 is now connected to a dc voltage source of

magnitude 6V.

a. Find the equilibrium displacement 0X .

b. Write the dynamic equations of motion for the system.

c. Linearize these dynamic equations for incremental motion of the system

around its equilibrium position.

Sol:

(a)

AR

VI 400

3.31 cm.X 5610 .(b)

dtdiRV 0

flux linkage iL(x )

dynamic equation:

dt

dx)i

a

L(

dt

di)

a

x(LiR

dt

dx

dx

dLi

dt

di)

a

x(LiR

dt

dx

dx

dL

idt

di

LiR

dt

dLi

dt

diLiR

(Li)dt

diRV

00

0

0

1

1

or

dt

dx

i.dt

di

x).(i.

dt

dx)i

.(

dt

di)

.

x(i.

18204550110451

22

104

221104516

3

33

( 40)


24/25

and

x)a.(Ka

Li

x)a.(Kfdt

xd

M

x)a.(Kdt

xdMf

902

90

90

00

2

02

2

02

2

or

x...dt

xd.

x)..(..

i

dt

xd.

53936109091020

229053222

10420

3

2

2

32

2

2

(c)

The equation can be linearized by letting txXx ' 0 and tiIi ' 0 .

The result is

dt

dxI

a

L

dt

di

a

XLRi

''

000

0 10

or

dt

dx.

dt

di.i.

'' 72801051510 3

and

'''

xKia

LI

d t

xdM 0

00

2

2

or

''

'

xidt

xd350727.02.0 2

2

()

()


25/25

3.33 Consider the single-coil rotor of Example 3.1. Assume the rotor winding to be

carrying a constant current of AI 8 and the rotor to

. have a moment of inertia 201250 mkg.J

a. Find the equilibrium position of the rotor. Is it stable?

b. Writer the dynamic equations for the system.

c. find the natural frequency in hertz for incremental rotor motion around this

equilibrium position.

Sol:

(a)

Rotor current =8A

Torque sin0TT

mN....RlIBT 0048030050020822 00

The stable equilibrium position will be at 0 .

(b)

Tdt

dJ sin02

2

(c)

The incremental equation of motion is

Tdt

dJ 02

2

and the natural frequency is

s cr a dJ

T/62.0

0125.0

0048.00

Corresponding to a frequency of

Hzf

f

099.02

62.0

2

2

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Sheet2 Solved