2 Magnetic Circuits

Indian Institute of Technology Guwahati

Department of Electronics & Electrical Engineering

1

EE 101

Electrical Sciences

Indian Institute of Technology Guwahati 2

Lectures 4-6

Magnetic Circuits

Electromagnetism

Dr. Govinda Bol Shrestha

Visiting Professor

Department of Electronics and Electrical Engineering


NOTATIONS AND CONVENTIONS

Quantity Symbol(s) Unit(s)

Magneto-motive force (mmf) F, (NI) A At

Magnetic Field Strength H A/m At/m

Magnetic Flux Wb

Magnetic Flux Density B T Wb/m2

Flux Linkage Wb Wb t

Inductance L, M H

Permeability , 0 H/m

Relative Permeability r

Reluctance H-1 At/Wb

3

Understanding of electromagnetic fields is important to clearly understand

the behavior of electric machines.

A: Ampere t: turns m: meter

Wb: Webers T: Tesla H: Henry


MAGNETIC INDUCTION

MAGNETIC FIELD INTENSITY (H)

A flow of electric current in a conductor creates a magnetic field

around the conductor as shown in Figure 3.1.

The direction of the magnetic field is defined by the Right Hand

Rule illustrated in Figure 3.1 (a).

The relationship between the magnetic field intensity (H) along a

closed path established by a current is specified by Ampere’s Law,

which is expressed as:

4

Figure 3.1 (a) Right Hand Rule (b) General closed loop for Ampere’s Law

I

P

H

(a)

H

I

(b)

H enccd I

This principle can be easily adopted in many practical situations. For example, when H is constant, (around a circular path)

H cos 0c C C

d Hd H d H


MAGNETIC FIELD PRODUCED BY A CONDUCTOR

Consider a long straight conductor carrying a current I as shown in

Figure 3.2 (a). The field intensity (H) will be same throughout the

circular path. Therefore,

When there are N conductors as shown in Figure 3.2 (b), each carrying a

current I, Then,

5

Figure 3.2 Magnetic field strength (H) around current carrying conductors

r

H

I

(a) Single conductor

r

H

I

(b) Multiple Copnductors

H

2

encc

c

enc

d I

H d H H r

I I

At/morA/m 2/ rIH

/ 2 A/m or At/mencI NI H NI r


MAGNETIC FLUX () AND FLUX DENSITY (B)

In a medium of free space (or air) the field intensity produces flux density given by:

B = 0H [Wb/m2 or T (Tesla)]

where,

0 (= 410-7 H/m), is the permeability of free space

The flux can then be calculated as:

when B is constant

It should be noted that both B and remain very small in the

medium of air because of the very small value of 0.

6

Figure 3.3 Magnetic field strength, flux, and flux density

dA

r

H B

I

,BABdAA


MAGNETIC MATERIALS

The magnetic flux or the flux density may be enhanced in a magnetic field by the use of magnetic materials. Consider a toroid of steel of radius r around the current carrying conductor as shown in Figure 3.4.

Then, at a point inside the toroid,

as before.

But,

B = 0 r H = H [Wb/m2 or T (Tesla)]

where,

0 is the permeability of free space

r is the relative permeability of steel, and

= 0 r, is the permeability of steel

r is near unity (1) for non magnetic materials but it can be very high (2000–6000) for Ferro-magnetic materials. Thus, the use of magnetic materials can enhance the flux density (B) and therefore the flux () by several magnitude of order for the same value of field intensity H.

7

r

HB

I

μr

Figure 3.4 Use of magnetic material

2/ rIH


MAGNETIC CIRCUITS IN ELECTRIC MACHINES

In order to obtain reasonably high values of flux density and the flux, most electromagnetic machines commonly utilize:

i. A structure of magnetic materials to utilize

the benefit of high permeability, and ii. Coils consisting of a large number of turns

to increase the total enclosed current.

Typical electromagnetic structures of a rotating

machine and a transformer are as shown.

In such structures, the flux density inside the

core structure will be several orders of magnitude

higher than in the surrounding space.

Therefore the flux outside the magnetic core can

conveniently be ignored when analyzing electromagnetic machines and magnetic structures.

8

I

V1

I1 I2

V2 N1 N2 E2 E1


TYPES OF MAGNETIC MATERIALS : FERRO-MAGNETISM

Non-magnetic materials

These are the material whose relative permeability is near unity.

Ferro-magnetic materials

These have very high relative permeability and experience very strong attractive force in magnetic fields.

Consider the magnetic circuit as shown in the following Figure.

The current I in the coil produces field strength H, which in turn produces flux density B and the flux .

The general nature of relationship between B and H for a common magnetic material is as shown.

9

I

N

B

H

Typical magnetization curve

Bm

Hm

B

H 0

Linear

Saturation

Knee


FERRO-MAGNETISM : MAGNETIZATION CURVE

The relationship between B and H can be explained using the notions of magnetic dipoles and magnetic domains. In the absence of any current in the coil, i.e., H, the dipoles are all randomly oriented and the magnetic domains cancel each other out, resulting in no net flux and flux density.

With the injection of current, i.e., application of H, the magnetic domains get aligned resulting in higher flux and flux density. Once all the domains are aligned, increase in I, i.e., H cannot produce further increase in B, and saturation occurs.

10

I

N

B

H


Bm

Hm

B

H 0

Linear

Saturation

Knee

e

e e e

e

e

B

B

B B

B

B

Dipoles

Magnetic Domains


MAGNETIZATION CHARACTERISTICS (B-H CURVE)

Variation of flux density B in Ferro-magnetic material for increasing values of H as shown in the figure is called the magnetization characteristics or the B-H curve.

Three important stages of the characteristics are indicated in the diagram.

Linear region, where B increases almost linearly with H

Knee region, where increase

in B slows down significantly,

due to reduction of domains

which can be aligned.

Saturation region, when B stops increases due to lack of domains which can be aligned and the curve practically flattens.

11


Bm

Hm

B

H 0

Linear

Saturation

Knee


TYPICAL MAGNETIZATION (B-H) CURVES

The B-H curves for three common magnetic materials are shown in Figure 3.11.

Note that different materials saturate at significantly different levels of flux density.

The initial sections of the curves (for lower values of B) are almost linear, where B=H with constant is valid for these linear regions of the curves.

The permeability µ changes rapidly after the knee point (i.e., in the saturation region.

12

Figure 3.11 Typical magnetization curves of three materials

0 200 400 600 800 1000 1200

Field intensity, H (A/m) F

lux D

ensi

ty, B

(T

)

1.4

1.2

1.0

0.8

0.6

0.4

0.2

Silicon steel

Cast steel

Cast iron

Unless otherwise stated, it will be assumed in these basic analyses of magnetic systems that they operate in the linear region.


Figure 3.12 Variation of permeability

1

Low

(increasing )

Saturation

H (At/m)

B

(T)

High

(decreasing )

H1

B1

H (At/m)

(H/m)

max : min

PERMEABILITY

Since, B = H, the permeability of the

material at a given level of flux density

can be obtained as the ratio of B/H

at each point of the magnetization

curve, as illustrated in Figure 3.12.

For example, at a point where, B = B1, and H = H1

= B1/H1 (H/m)

It can be seen that:

The value of increases with H to a max value and then decreases steadily after saturation sets in.

The value of remains approximately constant as indicated in red (within narrow limits) in the operating range of the flux density B.

13


MAGNETIC CIRCUITS

The flow of magnetic flux () in a magnetic circuit created by the current flowing in a coil may be analyzed as the flow of current in electric circuit. Consider the magnetic circuit shown in Figure 3.13 (a)

The following common assumptions will be adopted in the basic analysis of this circuit:

The flux is restricted to the magnetic material (which means there are no leakage and no fringing of flux).

The magnetic flux density (B) is uniform within the magnetic material, which is taken as the flux density along the mean path. (B=/A)

14

Figure 3.13 (a) An elementary magnetic circuit,

i

N A

ℓ


ANALYSIS OF MAGNETIC CIRCUITS

Applying Ampere’s law along the mean path,

Since, and

We get ,

So that flux

F = NI is called the magnetomotive force (mmf), analogous to emf. Note that this quantity depends purely on the electrical properties of the winding.

is called the Reluctance, analogous to Resistance.

Note that this is purely a property of the magnetic core material

and the structure.

or F= is called the Ohm’s law for the magnetic circuit, which may be represented and analyzed by drawing a magnetic equivalent circuit shown in Figure (b).

15

(a) An elementary magnetic circuit,

i

N A

ℓ

NI

(b) the magnetic equivalent circuit

H ,enccd I

H , & enccd Hl I NI Hl NI

= NI

B H

( / )

ANI NI FBA

A

/ A

/F


DEVELOPING MAGNETIC EQUIVALENT CIRCUITS

Magnetic equivalent circuits for more elaborate

magnetic circuits may be developed by adopting

the following procedure.

Coils represent sources, and the right hand rule should be used to specify the polarity of this source (i.e., the direction of flux).

Trace the mean path followed by the flux. Reluctance of various sections with different flux in them must be evaluated separately, as they will be required in the analysis of the circuit.

When the magnetic circuit consists of two or more closed loops, then Ohm’s law can be applied to each loop separately.

, for each closed loop i.

Magnetic circuits can be analyzed using magnetic equivalent circuits as long as the circuit remains linear i.e., the permeability μ remains constant.

16

i i iiiiHNI

(a) An elementary magnetic circuit,

i

N A

ℓ


EXAMPLE Draw the magnetic equivalent circuit for the magnetic circuit shown in Figure where the dimensions are in cm. The depth of the core is 10 cm and r of the material is known to be 1400. If the coil has 310 turns, estimate the current required in the coil to obtain a flux density of 0.8 T in the coil.

Solution - The cross-sectional areas of different sections are: Aab = Acd = 1010 = 100 cm2, Aac = Abd = 1015 = 150 cm2.

The lengths of various sections are:

The reluctance of various sections are:

17

15

N

i

25 15

10

25

10

a

c d

b

Figure

cm 352/10225,cm 402/15225 bdaccdab

1-

40

1-

40

T 13263101501400

35.0

,T 22736101001400

4.0

bdac

cdab


EXAMPLE Solution –

The equivalent circuit is drawn as shown in

Figure (b):

Total equivalent reluctance

e = 2(ab+ cd) = 72000 H-1

For a flux density of 0.8 T inside the coil,

ac = Bac Aac

= 0.8 150 10-4 = 0.012 Wb

Then,

Ni = ace=0.01272000=864 At

Therefore, i = 864/310=2.788 A

18

15

N

i

25 15

10

25

10

a

c d

b

Figure a

Figure (b). Magnetic equivalent circuit for Figure (a)

NI

eq

NI

ac bd

ab

cd


MAGNETIC CIRCUITS WITH AIR GAPS

Air gaps are integral part of magnetic circuits in various electric machines, as shown below.

Fringing of flux occurs in air gaps and the fringing flux often constitutes leakage flux. Although it is not easy to take account of leakage flux precisely, there are various approximate means of including their effects in the analysis. Leakage flux will be ignored for now.

The inclusion of the effects of air gaps in magnetic circuits will be illustrated with the following example.

19

Figure 3.17 (a) Magnetic circuit with air gap

i

g

I


MAGNETIC CIRCUITS WITH AIR GAPS - EXAMPLE An electromagnet of square cross section, shown in Figure 3.17 (a), has a coil of 1500 turns. The inner and outer radii of the core are 10 cm and 12 cm respectively and the air gap is 1 cm. If the current in the coil is 4 A and the relative permeability of the core material is 1200, determine the flux density in the circuit

Solution –

The equivalent circuit drawn as shown in

Figure (b) using the approach described earlier.

Cross section area Ac=Ag = 2 cm2 cm = 410-4 m2

Mean radius r (10+12)/2= 11 cm, Length of core = 2r – 1= 68.12 cm

eq = c+g= 21.023106 H-1

= (15004)/21.023106 = 2.8510-4 Wb

Bc=Bg= (2.8510-4)/(410-4) = 0.713 T

20

NI

g

c

(b) Magnetic equivalent circuit

1-6

47

1-6

47T 10 894.19

104104

01.0,T 10129.1

1041041200

6812.0

gc

Figure 3.17 (a) Magnetic circuit with air gap

i

g


FARADAY’S LAW (INDUCED EMF)

Consider a simple magnetic

circuit as shown Figure 3.20.

Input current i in the coil

of N turns establish a

flux in the core.

Flux Linkage (): When a flux passes through a coil of N turn, the flux is said to link the coil and,

= N is called the flux linkage of the coil.

Faraday’s law which is embedded in the Maxwell’s Equations states that a coil with flux linkage will have an induced voltage e in it given by:

The –ve sign indicates that the direction of the induced voltage will be such that it will tend to oppose the current/voltage (v) creating the flux.

21

Figure 3.20 (a) An elementary magnetic circuit, (b) the magnetic equivalent circuit

i

N A

ℓ

NI

dt

dN

dt

de


SINUSOIDAL EXCITATION OF MAGNETIC CIRCUITS-

Consider the magnetic circuit excited by a sinusoidal source shown in Figure 3.21. With a sinusoidal input current i, (which is due a sinusoidal voltage v) the mmf (Ni) and therefore the flux (=Ni/) produced in the core are also sinusoidal.

Let

Then,

Therefore, the rms value of the voltage is given by:

Thus, it is seen that the product Bmf is related to the input voltage and these two variables need to be treated jointly in the analysis of many magnetic circuits excited by sinusoidal sources. This observation will find useful applications in the analyses of magnetic circuits later.

22

tm sin

tfBNA

tfABN

tNdt

dNev

m

m

m

cos)(2

cos)2)((

cos

)(44.42/)(22/ fBNAfBNAVV mmm

i

N A

ℓ

e v

Figure 3.21 Sinusoidal excitation of magnetic circuit


Consider the magnetic circuit of Figure 3.20. A current i through the coil will produce a flux and therefore a flux linkage = N.

The inductance of the coil is defined as:

Since,

Thus, the inductance L is thus determined by the coil properties and the physical properties (dimensions) of the magnetic material

For linear magnetic circuits, reluctance is constant, and therefore the inductance L is constant for a given coil.

If the magnetic circuit consists entirely of Ferro-magnetic material, the B-H curve is hardly linear. Saturation often occurs, and do not remain constant so that L does not remain constant. If constant inductance L is desired, air gaps are often introduced in the magnetic circuit.

INDUCTANCE (SELF INDUCTANCE)

23

i

N A

ℓ

case)linear for (idi

dL

H 22

N

di

dLi

NNiNN

( / ).A


INDUCTANCE (SELF INDUCTANCE) For a magnetic circuit excited by ac source as shown in Figure 3.21, the

induced emf e can be expressed as:

This is the common form of Faraday’s Law

which is used in electric circuit analysis.

Often the negative sign is dropped by considering the direction of e separately.

If the resistance of the coil is negligibly small, then the input voltage v is completely balanced by the induced emf e, so that

If the resistance of the coil is incorporated in the analysis, then the input voltage is balanced by the drop in the resistance and the induced voltage in the inductor, so that:

24

i

N A

ℓ

dt

diL

dt

di

di

d

dt

de

dt

diLv

dt

diLriv


EXAMPLE The circular magnetic core shown in Figure 3.22 has a relative

permeability of 2200. The dimensions of the core are: r1=25 cm, r2=20 cm, and the cross section, A is circular. The coil has 102 turns. Calculate the inductance of the coil.

Solution

Mean radius

r = [(25+20)/2]10-2 = 0.225 m

Diameter of core section

= r1-r2 = 5 cm = 0.05 m

Cross section area A

= r2 = 0.0252 m2

Reluctance of core:

The magnetic equivalent circuit is as shown in Figure 3.22(b).

Inductance,

25

Figure 3.22 (a) Magnetic circuit (b) Magnetic Equivalent circuit

ℓ

A N

i

r1 r2

NI

eq

1-

27H 260435

025.01042200

225.02

A

H 04.0260435

10222

N

L


B-H CURVE -F Curve -i Curve

B-H Curve may be converted to -F Curve, or =i Curve simply by adjusting the x and y axis scales appropriately.

26


Bm

Hm

3

2

1

B

H 1 2 3 4

0

Linear

Saturation

Knee

3A

2A

A

3NA

2NA

NA

F l 2l 3l 4l

i l/N 2l/N 3l/N 4l/N

Convert to -F Curve,

- Multiply B axis by A

- Multiply H axis by l

Concert to =i Curve,

- Multiply axis by N

- Multiply F axis by 1/N


ENERGY IN MAGNETIC CIRCUITS Consider a lossless magnetic circuit shown in Figure 3.25.

If the current input is i at a voltage of v,

electrical energy input in time interval dt is:

But, ,

therefore,

Therefore the energy input to a magnetic circuit to establish flux in N turn coil can be written as:

This expression can be used to calculate the total electrical energy input ie, the stored energy in the system.

27

i

N A

ℓ

e v

Figure 3.25 Lossless magnetic circuit

idW vi dt ei dt

de

dt

idW ei dt i d

iW i d


MAGNETIC FIELD ENERGY -MAGNETIC STORED ENERGY

In a lossless system, without any output, the input energy will be stored as magnetic field energy , Wf .

Therefore,

When the current is increased from 0 to I,

which is the area as shown.

If the current is subsequently reduced to zero, the energy will be returned to the system.

For linear system, B-H curve and so -i curve is linear, and the energy stored will be:

The stored energy can also be expressed as:

It should be noted that these expressions for stored magnetic field energy are for linear systems only.

28

fW i d

bcause, i

L

Wf

i

d

0 Energy Stored

I1

1

0

I

fW i d

i

Wf

Wi in Linear System I1

1

1 1 1 2

1 1 12 20 =

I

fW i d i Li

1 11 1 1 2

1 1 1 1 12 2 2 =

N i

fW N i


EXAMPLE The circular magnetic core shown in Figure 3.22 has a relative permeability of 2200. The dimensions of the core are: r1=25 cm, r2=20 cm, and the cross section, A is circular. The coil has 102 turns and a resistance of 4 Ω.

Calculate the magnetic field energy stored when connected to 10 V dc source using different approaches

Solution

From earlier example,

= 260435 H-1, and L = 0.4 H.

With 10 V dc source,

i =2.5 A, = Ni/ = 9.7910-4 Wb,

B = /A = 0.50 T, = N = 0.10 Wb t

Therefore,

or,

29

Figure 3.22 (a) Magnetic circuit (b) Magnetic Equivalent circuit

ℓ

A N

i

r1 r

2

NI

eq

2 4 21 12 2

(9.79 10 ) 200435 0.125 JfW

2 21 12 2

(2.5) 0.4 0.125 JfW i L


MAGNETIZATION CURVE: AN EXTENSION

Consider a simple magnetic circuit shown in Figure (a) excited by a variable input current.

When the field strength in a magnetic circuit is increased by increasing the current through the coil, the flux density increases as shown by the dark line in Figure (b) as discussed earlier.

If the field strength is now reduced by decreasing the current through the coil, the decrease in flux density does not retrace the path taken during the increasing filed strength but it traces a different path during decreasing field strength as shown by the red line in Figure (b). In fact some flux density Br is maintained even when the current and therefore the field strength is reduced to zero. This is called “Remnant” or “Retention”.

30

e v

i

N A

ℓ

(a) magnetic circuit

Bm

Hm

B

H 0

(b) Magnetization for increasing and decreasing current

Br


MAGNETIZATION CURVE: AN EXTENSION If a magnetic circuit is excited by AC source as shown, the input current and therefore the field strength follows a cycle, (i) increasing slowly to a maximum value (Hm) at c, (ii) decreasing slowly to zero at d, (iii) reversing the direction (i.e. getting negative) and increasing to a maximum value in the reverse direction (–Hm) at f, and (iv) finally reversing the direction to trace back to maximum positive value (Hm) at c.

A current flow in the opposite direction, i.e., a negative field-strength is required to reduce the flux density to zero. Such field strength is called the Coercive Force.

The path followed by the magnetization curve, and the retention and the coercive force are characteristics of specific magnetic materials and can vary significantly from material to material.

31

Bm

Hm

Br

Hc -Hm

-Bm

-Br

0

a

b

c

d

e

f

e v

i

N A

ℓ


HYSTERESIS LOOP

The path traced by the B-H curve for such a cycle of variation in the field strength is shown in the Figure and is called the Hysteresis Loop. The shape of this loop determines a part of magnetic losses.

The actual shape of the hysteresis loop depends on the magnetic material.

32

e v

i

N A

ℓ

Bm

Hm

Br

Hc -Hm

-Bm

-Br

0

a

b

c

d

e

f

Figure 3.27 (a) Magnetic circuit (b) Hysteresis loop


MAGNETIC LOSSES - HYSTERESIS LOSS

Magnetic losses, also called “core losses” or “iron losses”, consist of “Hysteresis loss” and “Eddy current loss”

For magnetic circuits excited by AC sources, the flux density in the core traverses a complete Hysteresis loop for each cycle of the input current as indicated in Figure 3.28. This process incurs a core loss known as “Hysteresis loss”.

Hysteresis loss per unit volume of the core can be shown to be equal to the area of the Hysteresis loop by tracing one full cycle of the loop and tallying the energy input during the process.

Energy input when the flux density

changes from –Br to Bm is:

= area abcdea

33

m

r

B

B HdB

Bm

Hm

Br

Hc -Hm

-Bm

-Br

0

a

b

c d

e

Figure 3.28 Hysteresis loop and Hysteresis loss


HYSTERESIS LOSS Energy released when the flux density changes from Bm to Br is:

= area cdec

Net energy input in each cycle, i.e., energy loss per cycle is,

w = 2 area abcea

= area of the hysteresis loop

The shape/area of the hysteresis loop

and therefore the hysteresis loss depends

on the core material.

Hysteresis loss is commonly estimated using the empirical formulae given by Steinmetz formulae :

where,

Bm is the maximum flux density in the core, f is the frequency of the source, Kh is a constant, and n is the Steinmetz index (1.52.5) for common core materials)

34

Bm

Hm

Br

Hc -Hm

-Bm

-Br

0

a

b

c d

e

Figure 3.28 Hysteresis loop and Hysteresis loss

r

m

B

B HdB

n

h h mP K B f


EDDY CURRENT LOSS

When a time varying flux is created in a magnetic core, it will produce induced voltage in any perceivable closed path as shown in Figure (a). The direction of these “Eddy” currents will be such that they tend to oppose the original flux responsible for such currents.

The eddy currents flowing in these

closed paths incur power losses, which

are called “Eddy current losses”.

The magnitude of the “Eddy current”

and therefore the “Eddy loss” will

depend on (i) the magnitude of the

induced voltage in each closed path,

and (ii) the resistance of such closed

paths. Therefore, it is very difficult to quantify Eddy losses.

35

Eddy current and eddy current paths

Core

B

i

Eddy

Current

(a) In solid core material


EDDY CURRENT LOSS

Eddy current loss (Pe) in a magnetic circuits excited by a sinusoidal source is commonly estimated using the empirical formulae:

where, Bm is the maximum flux density in the magnetic core,

f is the frequency of the source, and

Ke is a constant.

Eddy currents and therefore eddy losses are reduced by using laminations of high resistivity core material, instead of solid core material. This increases (i) the effective length of eddy current paths, and (ii) the resistance of these eddy current paths. Reduced “Eddy currents” lead to reduced “Eddy current losses”.

36

Eddy current and eddy current paths

i

B

Eddy

Currents

Laminated

Core

(b) In laminated core material

2 2

e e mP K B f


SINUSOIDAL EXCITATION OF MAGNETIC CIRCUITS – A REVISIT

Both Eddy current loss as well as Hysteresis loss depend on frequency f and the maximum flux density Bm. It should be noted that these two variables are not fully independent.

For a magnetic circuit excited by a sinusoidal input it was shown earlier that

Thus, it is seen that the product Bmf is jointly related to the input voltage, and these two variables cannot vary independently for a given input voltage.

37

/ 2 =2 ( ) / 2 = 4.44 ( ) m m mV V NA B f NA B f


EXAMPLE An electromagnet is known to have hysteresis loss of 180 W when excited by 60 Hz, 120 V source. If the Steinmetz index of the core material is 1.6, estimate the hysteresis loss when the electromagnet is connected to 120 V, 50 Hz source.

Solution

Let Bm1 and Bm2 be the maximum flux densities under the two conditions.

Then, Ph1=KhBm11.6 f1 , and Ph2=KhBm2

1.6 f2 = ?

Therefore, ,

and Ph1= 180 W, f1 = 60 Hz, and f2 = 50 Hz

The voltage equation under the two conditions yield,

V1 = 120 V = K Bm1 f1 , and V2= 120 V = K Bm2 f2

Bm1f1 = Bm2 f2 Bm2/Bm1 = f1/f2 = 60/50 = 1.2

Therefore,

38

1.6

2 2 2

1.6

1 1 1

h m

h m

P B f

P B f

1.61.62 2

2 11.6

1 1

501.2 180 200.8

60

mh h

m

B fP P W

B f


PRINCIPLES OF ELECTROMECHANICAL ENERGY CONVERSION

Electromechanical devices (machines) convert electrical energy into mechanical energy and vice versa. Most of these devices utilize magnetic field as a medium.

Conservation of energy has to be satisfied by all these processes.

, where,

Wi is the input (electrical) energy,

Wo is the output (mechanical) energy,

Wf is the field or stored energy,

Wl is the energy lost in the system.

The flow of energy in the process is as shown in the Figure. This process is reversible except for the losses.

Ignoring losses, the energy balance equation reduces to:

39

i o fW W W W

Energy balance in electro-mechanical devices

Lost Energy

Wl

Stored Energy

Wf

Input Energy

Wi

Ooutput Energy

Wo

i o fW W W


COUPLED CIRCUITS - MUTUAL INDUCTANCE

Consider the magnetic circuit with two coils of self inductances L1 and L2 shown in Figure 3.24(a). Let a current i1 in Coil 1 produces flux 1 and Coil 2 is left open. A portion 21 of 1 links both coil 1 as well as coil 2, while the portion 11 links only coil 1.

Then, , and,

where,

is called the mutual inductance of Coil 2 with respect to Coil 1.

Such circuits are called coupled circuits and the mutual inductance

makes power transfer from one coil to the other.

40

+

e1

_

+

e2

_

N1 N2

i1

1

11

21

Figure 3.24(a) Magnetic circuit with two coils

11 1

die L

dt

21 21 121 2 2

1

2 21 1 21 1 121

1 1

d d die N N

dt di dt

dN di d di diM

di dt di dt dt

21 2121 2

1 1

d dM N

di di


COUPLED CIRCUITS - MUTUAL INDUCTANCE

Similarly, if coil 2 is supplied with a current, coil 1 is left open, then we will have induced voltage in coil 1 due to coupling.

where,

is the mutual inductance of Coil 1 with respect to Coil 2.

It can be shown that M12 = M21, which is usually written M which is called the mutual inductance between the two coils.

It can further be shown that,

The ratio, is called the co-efficient of coupling, which is less than 1.

Then,

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12 1212 1

2 2

d dM N

di di

212 12

die M

dt

1 2M L L

1 2/ ,k M L L

1 2 ,M k L L


COUPLED CIRCUITS – DOT CONVENTION

Dot convention makes use of a dot placed at one end of each coupled coil.

Current entering dotted terminal of one coil produces a voltage which is sensed positive at the dotted terminal of the second coil.

Then the magnetic circuit shown here can be represented by the schematic diagram shown below it.

Then for the dot positions and the polarities as shown, the induced voltage in the second coil can be easily be written as:

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+

e1

_

+

v1

_

+

e2

_

N1 N2

i1

1

11

21

Figure 3.24(a) Magnetic circuit with two coils

12

div M

dt



Various possible combinations of dots and the polarities of the induced emfs are shown in the diagram.

It may be noted that current entering non-dotted terminal of one coil produces a voltage which is sensed positive at the non- dotted terminal of the second coil.

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Dot convention can be used to determine the polarities of induced voltages when both of the coupled coils carry current.

For circuit (a), the equations for v1 and v2 can be written as:

For circuit (b), the equations for v1 and v2 will be:

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1 21 1

2 12 2

- 0

- 0

di div L M

dt dt

di div L M

dt dt

1 21 1

2 12 2

- 0

+ 0

di div L M

dt dt

di div L M

dt dt


EXAMPLE

For the circuit shown, find the current I2.

Solution:

The equations for the two loops are:

Using the parameters of the circuit and =10, the equations in steady state form can be written as:

Solving, I2 = 0.1724-16.7

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1 21 1 1

2 12 2 2 2

- 1 0

+ 0, 400

di div i L M

dt dt

di div L M v i

dt dt

1 1 2

2 2 1

10 0 - I 1 10 1 I 10 9 I 0

400I + 10 100 I 10 9 I 0

j j

j j


ENERGY IN COUPLED CIRCUITS

For the coupled circuit shown in (a),

The total energy stored in the magnetic

field can be shown to be,

However, if the current enters into the dotted terminal in one coil, while the other current leaves a dotted terminal as in figure (b), then the total stored energy becomes.

The energy equation is valid for any instantaneous currents at an instant,

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1 12 2

1 1 2 2 1 22 2W L I L I MI I

1 12 2

1 1 2 2 1 22 2W L I L I MI I

1 12 2

1 1 2 2 1 22 2( ) ( ) ( ) ( ) ( )w t L i t L i t Mi t i t


EXAMPLE

For the circuit shown in (b),

L1= 3.2 H, L2= 5 H, k = 0.5

i1= 4 cos (100t-60) A, i2= 2 cos (100t-60) A

Find the values of v1, v2, and the total stored energy at t=0.

(-762 V, -173.2 V, 4.9 J)

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Thank you!

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2 Magnetic Circuits