• Relationship between force and stored energy in magnetic systems

Electrical Machines I

Week 5-6: Singly Excited Systems

RECALL… REMEMBER…. !!

• Energy stored in linear and non linear systems

• Relationship between force and stored energy in magnetic systems

• Energy and Co-energy

Singly Excited Magnetic Field System: Force and torque

Linear system Rotational system

involves an object moving from one

point to another in a straight line involves an object rotating about an axis

Force: Torque:

Distance (displacement): Distance (displacement):

Velocity:

Mass: Moment of inertia:

Singly Excited Magnetic Field System: Force and torque

Linear system Rotational system

Newton’s law of motion: Newton’s law of motion:

Power:

Opposing force due to spring and friction for

example

Opposing force due to load torque and friction

for example

Singly Excited Magnetic Field System:

Previously we have calculated the force acting on a “plunger” as a function of the system variables as follows:

dx i dL

F

²

2

= 1

x F

W

∂

= ∂

It is our objective today is to calculate the force or torque acting on a “single excited reluctance machine” in terms of system variables

A reluctance motor is an ac synchronous motor whose reluctance changes as a function of angular displacement . Owing to its constant speed operation, it is commonly used in electric clocks, record players, and other precise timing devices

The word synchronous actually mean “existing or occurring at the same time” and in electric machines mean that the motor rotates with a speed that is related to the

supply frequency

Singly Excited Magnetic Field System: Reluctance motor

Single phase, 2 pole reluctance motor

What is the required torque to be able to make

this motor rotate in terms

of system variables?

: Mechanical speed and is NOT EQUAL to 2π

I

ωm

V_s Rotor

axis

Stator magnetic

axis Rotor

Excitation(field) winding

Singly Excited Magnetic Field System: Reluctance motor

x F

W

∂

= ∂

Q ∂ θ

= ∂

∴

W

T

(a) (b) (c) (d)

min max

max , L , ℜ

ϕ ϕ ⇓ , L _⇓ , ℜ _⇑ ϕ _min , L _min , ℜ _max ϕ _max , L _max , ℜ _min

2

2 1 Li W

=

Q

We will study the rotor movement from 0 to 360˚ and started to plot the inductance and reluctance variation with time

Singly Excited Magnetic Field System: Reluctance motor

When the magnetic axes of the rotor and the stator are at right angles to each other (the quadrature or q- axis position), the reluctance is maximum, leading to a minimum inductance. As the rotor rotates with a

uniform mechanical speed &_' the inductance goes through maxima and minima

Maximum inductance

= Minimum reluctance

Minimum inductance

= Maximum reluctance

One electrical cycle of ≡ half one mechanical cycle of

≡ ⁾₊^* since this is a two pole machine

Singly Excited Magnetic Field System: Reluctance motor

,_- ,_. / ,₀ 1

If we try to write the general equation of this inductance as a function of theta it will be in the form of:

, 2 ,

/ ,

345 12

Offset=shift Peak inductance value ,₁ ,_. ,₀

1

, 2 -. 7 , _. / , ₀ / -. 7 , _. , ₀ 345 12

2θ is because 2 complete rotations

are made in the 360˚

θ

∂

= ∂

∴ L

i T

²

2 1

θ

∂

= ∂

e

T W Q

( )

2

1 L i W

= θ Q

Singly Excited Magnetic Field System: Reluctance motor

, 2 -. 7 , _. / , ₀ / -. 7 , _. , ₀ 345 12

( ) ^{( )} _ ^

  −

−

=

∴ * 2 sin 2 θ

2 1 2

1 ₂

q d

e

i L L

T

This is a

“mechanical angle”

Replace the variable / 8 Where 8 is the initial position of the rotor’s magnetic axis with respect

to the stator’s magnetic axis.

( ⁻ ) ^{( ω + δ )}

−

=

∴ T

i L

L

sin 2

t 2

1 ₂

Singly Excited Magnetic Field System: Reluctance motor

Since the current is sinusoidal and can be expressed as:

t I

i = _max cos ω

( ) ^{( ) ( )}

( )

( ) ^{sin( )}

sin cos

sin 2

2 cos 2 1

cos 1

2 2 cos

1

2

2 2 max

b a

b a

b a

a a

t t

L L

I

T

− +

+

= +

=

+

−

=

∴

Q Q

δ ω

ω

If you assume it as a “sin”

function its totally correct as well. It is preferred that we assume that the supply voltage is

“sinusoidal” and since the coil is inductive, current waveform will lag the sinusoidal voltage by 90˚.

This means that the current will be a “cosine” function

( ) ⁽ _{( (} ⁾ ₎ ^{( (} ₎ ^{) )} _



 





+ +

+

−

−

−

− +

−

=

∴ ω ω δ

δ ω

ω δ

ω

2 2

sin 5

. 0

2 2

sin 5

. 0 2

sin 4

1 2

max t t

t t

L t L

I T

m s

m s

m q

d e

Singly Excited Magnetic Field System: Reluctance motor

This “big” equation implies that for this reluctance machine to be able to rotate the average value of the torque terms of this equation MUST NOT

equal to zero

1. First term: average is zero as it is periodical so no torque is produced by this term 2. Second term: if we are able to make , this term will be re-written as

This condition will produce an average torque

( 2 δ )

sin 5 .

0 −

−

( ) ⁽ _{( (} ⁾ ₎ ^{( (} ₎ ^{) )} _



 





+ +

+

−

−

−

− +

=

∴ ω ω δ

δ ω

ω δ

ω

2 2

sin 5

. 0

2 2

sin 5

. 0 2

sin 4

1 2

max t t

t t

L t L

I T

m s

m s

m q

d e

Singly Excited Magnetic Field System: Reluctance motor

3. Third term: if we are able to make , this term is discard (same as the second term but with a negative sign which does not imply)

( ) ^{[ sin ( ) 2 δ ]}

8

1 2

max

e

I L L

T = − −

∴

Thus the condition which will successfully drive this motor will be represented as

Given that the machine

&_' &₅ Maximum torque

Singly Excited Magnetic Field System: Reluctance motor

( ) ^{[ sin ( ) 2 δ ]}

8

1 2

max

e

I L L

T = − −

∴

&_' &₅

Average torque will be equal to zero if

,_. ,₀

This condition happens if the rotor is cylindrical

(not salient)

Maximum torque occurs when 59: 1; <, ; >7^° The torque sign depends on the value of δ,

• if δ is (+ve), T will be (-ve) in this case, the torque is opposite to the direction of rotation. i.e. the machine acts as generator.

• if δ is (-ve), T will be (+ve) in this case, the torque is same direction of rotation. i.e. the machine acts as motor.