III. Electromechanical Energy Conversion

copper losses core losses(field losses) mechanical losses

dteiRdtidtiVdW telec 2

losses mechanical output energy mechanicalnet mechdW

losses field fldmechelec dWdWdW

fldmechelec dWdWdW

Schematic representation of an electromechanical energy conversion device

Differential energy input from electrical source:

For a lossless magnetic energy storage system:

Differrential energy balance eqn.:

1. Energy in Magnetic FielddteidWelec

idtdtdNdWelec

NiddWelec

ddWelec F

01

10

00

dd

W

dW

fld

elec

FF

F

Coenergy

10F dF fldW

11F fldfld WW

with 0 = 0 and F0 = 0

For a linear magnetic circuit:

1121 F fldfld WW

212

1 R

212

1PF

where R and P are the reluctance and permeance of the magnetic circuit respectively

2. Energy & Force in Singly-Excited E.M.C device

Electromechanical relay

• Assume mobile armature is initially at open position

• When the coil is excited by i(t), (t) is produced in M.C. and an electromagnetic force ffld is exerted on mobile armature tendingto align it with the densest part of M.F.

• When the armature moves

Armature Movement

• Slow motion case ( is gradually increased during motion)

• Fast motion case ( is constant during motion)

• Normal motion case

(i) Slow motion

42210001 AAAAddW fld F F

mechfldelec WWW

Shaded area gives Wmech

F is constant during motion

3110

AAdWelec F

43 AAWWW fldelecmech

(ii) Fast motion

is constant during motion

mechfldelec WWW

110

AdWelec F

43 AAWWW fldelecmech

Shaded area gives Wmech

4322100

01 F F AAAAAddW fld

(iii) Normal motion

mechfldelec WWW

Shaded area gives Wmech

• if dWmech > 0 then electical system does work on the armature

• if dWmech 0 then mechanical system does work on the armature

dxfdW fldmech

(a) Mechanical force in fast motion

mechfldelec dWdWdW

dxfdWd fldfld F

dxfddW fldfld F

dxx

xWd

xWxdW fldfld

fld

,,,

Note that

So

xW fld ,F

x

xWf fld

fld

,and

is independent of x

(a) Mechanical force in slow motion

Note that

xWxW fldfld F,FF,

xdWdxWd fldfld F,FF,

dxfddxWd fldfld FFdFF,

dxfxWd fldfld dFF,

So

FF,

xW fld

x

xWf fld

fld

F,and

F is independent of x

dx

xxW

dxW

xWd fldfldfld

,FF

F

,F,F

For a linear magnetic circuit

F21

fldfld WW 2221

21 PFR

1. When is constant during motion

dxf fld

dR221

2. When F is constant during motion

dxdL

dxf ifld

2

21

21 2

dPF

where R and P are the reluctance and permeance of the magnetic circuit respectively

Ex-1: Derive an expression for the electromagnetic force ffldproduced by the E.M.D. shown in the figure and then plot ffldagainst x.

Note: Initial gap length is c and i is constant during motion

Ex-2: For the electromagnet shown below:(a) For V = 120 Vdc, determine the energy stored in magnetic field and the lifting force produced (b) For V = 120 Vac at 60 Hz, determine the average lifting force

produced

Note: Initial gap length is 5 mm, internal resistance of the winding r = 6, N = 300, Ac = Ag = 6 x 6 = 36 cm2.

3. Rotating Machines

Torque

Te, Nm

Inertia

J, kg m2

Angular Speed, rad/s

Angular Displacement

, radRotational

Motion

Force

f, N

Mass

m, kg

Linear Speedv, m/s

Linear Displacement

x, mLinear Motion

,flde

WT

x

xWf fld

fld

,

F,flde

WT

x

xWf fld

fld

F,

fast motion

slow motion

For flux independent of during motion:

For mmf independent of during motion:

(a) Singly-excited device

0 tr

For a linear M.C.

ddLiTe

221

remech TP dtd

r

dtdLiPmech

221

Ex: The self inductance L in the figure as a function of rotor position is given by

L = L0 + L2 cos( 2)where the stator current is i(t) = Imsin(t)

(i) Obtain an expression for electromagnetic torque produced (instantaneous torque)

(ii) Let = rt + 0, find necessary condition for non-zero average torque

(b) Doubly-excited device

tiMtiL 2111

tiMtiL 1222

dtde 1

1

dt

de 22

2111 MiiL

1222 MiiL

dtd

ddMi

dtdiM

dtd

ddLi

dtdiLe

221

11

11

dtd

ddMi

dtdiM

dtd

ddLi

dtdiLe

112

22

22 1111 irev 2222 irev

rωii

LMML

dd

ii

dtd

LMML

ii

rr

vv

2

1

2

1

2

1

2

1

2

1

2

1

2

1

00

dtd

r

Terminal voltage equation (in matrix form):

2

1212211 v

viiivivTotal power

rTTTT

dd

dtdiviv

I

LI

ILIRIIVI 2211

Pcu

rrrelec ddLi

ddMii

ddLi

dtdiMi

dtdiiL

dtdiMi

dtdiiLririP

22

22112

11

22

222

11

112221

21 2

Pelec

Total electrical power input:

Copper loss Rate of increase in stored energy in MF + Electr.power converted to mech. en.

(in matrix form)

rωii

LMML

dd

ii

dtd

LMML

ii

rr

vv

2

1

2

1

2

1

2

1

2

1

2

1

2

1

00

Magnetic stored energy

2211 didiW fld

21222

211 2

121 iMiiLiL

rrrfld ddMii

ddLi

ddLi

dtdiMi

dtdiiL

dtdiMi

dtdiiLP

2122

212

11

22

222

11

11 21

21

dtdW

P fldfld

fldcuelecmech PPPP

rrrmech ddMii

ddLi

ddLiP

2122

212

1 21

21

rrrelec ddLi

ddMii

ddLi

dtdiMi

dtdiiL

dtdiMi

dtdiiLririP

22

22112

11

22

222

11

112221

21 2

remech TP

ddMii

ddLi

ddLiTe 21

222

121 2

121

I

LI

ddT T

e 21

In matrix form:

Ex1: Te ?

Ex2: Te ?