Chapter 2 - AC Machinery Fundamentals PDF

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Chap. 2: AC Machinery Fundamentals AC Machines

1.

Synchronous machines

Induction machines

Magnetic field current supplied by separate DC source

Magnetic field current supplied by magnetic induction in field windings

A simple loop in a uniform magnetic field

Simplest AC machine - a loop of wire rotating in uniform magnetic field.

Front view

View of coil

AC machines terminology: Stator = stationary part of machine Rotor = rotating part of machine _____________________________________________________________________ 1 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Voltage induced in the rotating loop Rotor rotation will induce a voltage in the wire loop. To determine total voltage induced in loop etot Examine each segment separately using:

e ind=( v×B )⋅l Segment ab

(4.1)

 Wire velocity tangential to rotation path  B points to the right 

( v×B ) points into the page (right hand rule – thumb)

Segment bc



( v×B ) always perpendicular to l

 Since l is in the plane of the page, ( v×B ) is perpendicular to l for both portions of the segments. Therefore the voltage segment bc is zero. Segment cd

in the

 Wire velocity tangential to rotation path  B points to the right 

Segment da

e cb

( v×B ) points out of page

( v×B ) always perpendicular to l ( v×B )  Just as in segment bc, 

is perpendicular to l. Therefore the voltage in this segment

e ad

will be zero too.

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Hence, total induced voltage on loop:

e tot =eba +e cb + edc +e ad =vBl sinθ ab +vBl sin θcd ∴ etot =2 vBl sin θ

(4.2)

We know that if the loop is rotating with a constant angular velocity , then

θ=ωt and the tangential velocity v can be expressed as v =rω

Hence,

e tot =2 vBl sin ωt =2 ( rω ) Bl sin ωt =AB ωsin ωt ∴etot =φmax ωsin ωt

(4.3)

where: A = area of the loop = 2rl max = maximum flux through the loop

Generally, the induced voltage in real machines depends on:  flux in the machine (B or )  speed of rotation (v or )  machine constants (l or A)

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Torque induced in a current-carrying loop Assuming that the loop is at an arbitrary angle  to the magnetic field and a current i is flowing in it.

View of coil

Front view

This will cause torque to be induced on the loop. Again we will look at each segment separately. The force on each segment is given by:

F=i ( l×B )

(4.4)

and from this the torque is calculated as T = (F) (perpendicular distance) = Fr sin θ

Segment ab

(4.5)

 Current is into the page  B points to the right 

( l×B ) points downwards ∴ F=i ( l×B ) =ilB

Hence, Tab = F(r sin θab) = IlB (r sin θab) N.m _____________________________________________________________________ 4 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Segment bc



( l×B ) points parallel to each other ∴ F=i ( l×B ) =ilB

Angle between force & r = 0. Hence, Tbc = F (r sin 0) Tbc = 0 N.m Segment cd

 Current is out of the page  B points to the right 

( l×B ) points upwards. ∴ F=i ( l×B) =ilB

Hence, Tcd = F (r sin θ) Tcd = IlB (r sin θcd) N.m Segment da



( l×B ) points parallel to each other ∴ F=i ( l×B ) =ilB

Angle between force & r = 0. Hence, Tda = F (r sin θda) Tda = 0 N.m

Therefore, total induced torque on loop:

T ind =T ab +T bc + T cd +T da =rilB sin θab +rilB sin θcd ∴T ind =2rilB sin θ

(4.6)

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EEE4323 Electrical Machines – AC Machinery Fundamentals

(torque maximum when plane of loop parallel to magnetic field) Alternative torque expression: The current flowing in the wire loop will generate a magnetic flux density

B loop : |Bloop|=

μi G

(4.7)

where G = factor depending on loop geometry. Hence, since loop area A=2rl, by substituting (4.7) into (4.6):

T ind=2 rliB sin θ =A ( G/μ ) Bloop B s sin θ ∴T ind =kBloop Bs sin θ or

T ind =k ( Bloop ×Bs )

(4.8)

(4.9)

where k = AG/ . Generally, the torque in real machines depends on:  strength of the rotor magnetic field  strength of external (stator) magnetic field  angle between the two fields  Machine constants

2.

The rotating magnetic field

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Torque – produced to align rotor (loop) magnetic field with stator magnetic field. If stator magnetic field is rotated, torque will cause rotor to ‘chase’ the rotating stator magnetic field. How to create a rotating stator magnetic field? Use a set of three-phase windings displaced by 120 electrical around the machine circumference.

Currents have equal magnitude, phase difference = 120

rotating magnetic field is produced (constant magnitude)

When the following currents flow in the three coils:

iaa ' (t )=I M sin ωt A i bb ' (t )=I M sin( ωt −120° ) A i cc ' (t )=I M sin(ωt −240° ) A _____________________________________________________________________ 7 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Each coil will produce a magnetic field intensity:

H aa ' ( t )=H M sin ωt ∠ 0 ° Aturns/ m H bb ' ( t )=H M sin ( ωt−120 ° ) ∠120 ° Aturns / m H cc ' (t )=H M sin ( ωt−240 ° )∠ 240° Aturns / m Notice that the magnitude varies sinusoidally but the direction is always constant. The resulting flux densities are then given by:

B aa ' (t )=B M sin ωt ∠ 0° T B bb ' ( t )=B M sin ( ωt−120 ° )∠120 ° T B cc ' ( t )=B M sin ( ωt−240 ° )∠240 ° T where

B M =μH M .

In order to observe the net field rotation, let’s examine the flux densities at specific times:

ωt =90 °

ωt=0

B aa ' ( t ) =0 B bb ' ( t ) =B M sin (−120° ) ∠120 ° B cc ' (t ) =B M sin (−240°) ∠ 240°

B aa' ( t ) = B M sin 90o= B M ∠ 0o B bb' ( t ) =B M sin (−30° ) ∠120 ° B cc' (t )=B M sin (−150° ) ∠240 °

The total magnetic field:

The total magnetic field:

B net =Baa ' + Bbb ' + B cc ' ¿ 1. 5 B M ∠−90 °

B net =Baa ' + Bbb ' + B cc ' ¿ 1. 5 B M ∠0 °

Field magnitude constant, direction changes! _____________________________________________________________________ 8 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Proof of rotating magnetic field concept From previous observation, at any time t: The magnetic field will have same magnitude 1.5BM and rotates at angular velocity .

To prove this, let’s assume the following coordinate system:

x^

= horizontal unit vector

^y

= vertical unit vector

The total magnetic flux density in the stator is obtained from vectorial addition of the three magnetic field components, i.e.

B net =Baa ' ( t ) + Bbb ' ( t ) + Bcc ' ( t ) ¿ B M sin ωt ∠ 0 °+ B M sin ( ωt−120 ) ∠ 120 °+B M sin ( ωt−240 ) ∠ 240° By considering each x and y direction components and applying trigonometric identities:

B net = ( 1 .5 B M sin ωt ) x−( 1. 5 B M cosωt ) y=1 . 5 B M

(4.10)

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Field magnitude = 1.5BM at all times Direction =

counter clockwise (CCW) with velocity 

Relationship between electrical frequency and speed of magnetic field rotation In the three phase coils considered, the stator is said to be twopole and occurs in the following counter clockwise order: 1st pole

2nd pole

a – c’ – b - a’– c – b’ South

North

Poles complete one mechanical rotation in one electrical cycle of current.



The rotating magnetic field in the stator represented as moving north and south stator poles

Hence, the relation between the electrical and mechanical components are given by: _____________________________________________________________________ 10 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

θe = θm f e=f m ω e=ω m

(a) A simple four-pole stator winding. (b) The resulting stator magnetic poles. Notice that there are moving poles of alternating polarity every 90° around the stator surface.

If we repeat the winding pattern twice, we have a four pole machine: a – c’ – b - a’– c – b’- a – c’ – b - a’– c – b’ Poles complete 1/2 a mechanical rotation in 1 electrical cycle of current. Hence, the electrical and mechanical components related by:

θe =2θ m f e=2 f m ω e=2 ω m

Therefore, in general, for a P (number of pole) pole stator,

_______________________

P θe = θ m 2 P f e= f m 2 11 Updated 16 February 2011

(4.12)

(4.13) _____________________

EEE4323 Electrical Machines – AC Machinery Fundamentals

P ω e= ω m 2

f m=

(4.14)

nm

n

60 , and m is the mechanical speed of the Since magnetic field in revolutions per minute, thus

f e= 3.

P n 120 m

(4.15)

Magnetomotive force and flux distribution in AC machines

In the previous section, the following assumptions were made:  flux produced inside ac machine is in free space  direction of flux density is perpendicular to plane of coil (given by right hand rule) Not true is real AC machines! Because of ferromagnetic rotor in centre of machine with a small airgap between the stator and the rotor. The rotor can be: (a) cylindrical (nonsalient-pole) (b) salient-pole

Only cylindrical rotors are considered in this course. _____________________________________________________________________ 12 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

In the machine,

ℜgap >> ℜstator , ℜrotor

Flux density vector B takes shortest possible path across the air gap. To produce sinusoidal voltage in machine Requires

Sinusoidal flux density vector B magnitude Occurs only when Magnetising intensity H (and mmf, F) is sinusoidal along air gap surface

Cylindrical rotor with sinusoidally varying air gap flux density

MMF or H and B as a function of angle in the air gap _____________________________________________________________________ 13 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

How to achieve sinusoidal mmf along airgap? Distribute the turns of winding that produces mmf in the slots, Vary number of conductors in each slot in sinusoidal manner.

For example:

Distributed stator winding designed to produce a sinusoidally varying air gap flux density

The mmf distribution resulting from the winding, compared to an ideal distribution.

The number of conductors in each slot is:

nC =N C cos α

(4.16)

where NC = number of conductors at an angle 0. _____________________________________________________________________ 14 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Increasing number of slots and making them closely spaced gives a better approximation of sinusoidal distribution of mmf. In practise, the winding distribution of equation (4.16) is not possible due to the finite number of slots and integral number of conductors possible in each slot.

4.

Induced voltage in AC machines

The rotating field can induce voltages in the three-phase windings. The induced voltage in a coil on a two-pole stator

Figure above shows a rotating rotor with a sinusoidally distributed magnetic field in the centre of a stationary coil.

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Assumptions:  air gap flux density vector magnitude, sinusoidally with mechanical angle  direction of

|Bgap|

varies

B gap always radially outward

The magnitude of the air gap flux density vector at a point around the rotor is given by:

|Bgap| =B M cosα

(4.17)

Where  = angle measured from the direction of peak rotor flux density. Since the rotor is rotating within the stator at an angular velocity

m, then

|Bgap|

around the stator is:

|Bgap|=B M cos ( ω m t−α )

(4.18)

The voltage induced in a moving coil inside a stationary field is e ind =( v×B )⋅l (4.19) Where:

v

B l

= velocity of the wire relative to the magnetic field (from rotor moving)

= magnetic flux density vector = length of conductor in the magnetic field

However, we have a stationary coil in a moving field. Hence, to use (4.19) we must be in a frame where magnetic field appears to be stationary, i.e. _____________________________________________________________________ 16 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

If we sit on the magnetic fields so that the field appears to be stationary, the sides of the coil will appear to go by at an apparent velocity of vrel. Stator coil in reality is not moving and the magnetic field is from rotor.

The total voltage induced in the coil will be the sum of voltages induced in each of the four sides. Assume that rotor moves counter clockwise (B moves CCW, hence vrel foes CW)

Segment ab

  = 180o  B directed radially outward 

( v×B ) points out of page (in the direction

of l) e ba=( v×B )⋅l =vBl =−v [ B M cos ( ω m t−180 ° ) l] (-ve sign : voltage is built up with a polarity opposite to the assumed polarity)

Segment bc



( v×B ) is perpendicular to l e cb= ( v×B )⋅l=0 (nothing induced in l)

Segment cd

  = 0o  B directed radially outward 

( v ×B ) points out of range e dc =( v ×B )⋅l =vBl =v [ B M cosω m t ] l

Segment da



( v ×B ) always perpendicular to l

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EEE4323 Electrical Machines – AC Machinery Fundamentals

e ad= ( v×B)⋅l =0

Hence, total induced voltage on the coil:

e tot =e ba +e dc =−vB M l cos ( ω m t−180 °) +vB M l cos ω m t ∴ etot =2 vB M l cos ω m t

(4.20)

Since v =rωm , therefore

∴ eind =2 vB M l cos ω m t=2 rωm B M l cos ω m t

(4.21)

Finally, the flux passing through the coil is given by

φ=2 rlB M

(4.22)

Hence, the induced voltage can be expressed as

e ind=φω cos ωt Note: ω m=ω e=ω

(4.23)

since it is a two-pole stator.

Or if the stator coil has NC turns, then the total induced voltage of the coil is

e tot =N c [ φω cos ωt ]

(4.24)

Remember: This derivation goes through the induced voltage in the stator when there is a rotating magnetic field produced by the rotor.

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EEE4323 Electrical Machines – AC Machinery Fundamentals

Induced voltage in a three-phase set of coils If now we have three coils each having NC turns placed around the rotor magnetic field, Voltages induced in each coil will have: Same magnitude but having a phase difference of 120o

The resulting voltages in each of the coil are:

e aa ' (t )=N C φω cos ωt V

(4.25a)

e bb ' (t )=N C φω cos ( ωt −120 ° ) V

(4.25b)

e cc ' (t )=N C φω cos ( ωt −240 ° ) V

(4.25c)

Hence, Generates a uniform magnetic

A three-phase set of currents flowing in the stator

field in the stator windings

& A uniform rotating magnetic field produced by the rotor

Generates a 3 phase voltage in the stator windings

Referring back to the induced voltage derived earlier… _____________________________________________________________________ 19 Updated 16 February 2011

EEE4323 Electrical Machines – AC Machinery Fundamentals

Maximum induced voltage achieved when sin (…) = 1 Hence, the peak voltage in any phase of the three-phase stator is: Emax =N C φω =2 πN C φf (4.26) since ω=2 πf . Finally, the rms voltage at any phase of the three-phase stator:

Erms =

E max 2 πN c φf = =√ 2 πN c φf √2 √2

(4.27)

Note: this is the induced voltage at each phase, as for the lineline voltage values, it will depend upon how the stator windings are connected, whether as Y or .

5.

Induced torque in an AC machine

Figure below shows a simplified ac machine with a single coil of wire mounted on the rotor. The stator flux density distribution: B S ( α ) =B S sin α

(4.28)

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EEE4323 Electrical Machines – AC Machinery Fundamentals

The torque produced in the rotor is obtained by analysing the force and torque on each of the two conductors separately: Induced force

Torque on conductor

T ind ,1=( r×F ) =rFsin 90 ° T ind ,1=rIlB

F=i ( l×B ) =ilBS sin α

Conductor 1

(direction as shown)

(counter clockwise)

T ind ,2=( r×F ) =rFsin 90 ° T ind ,2=rIlB

F=i ( l×B ) =ilBS sin α

Conductor 2

(direction as shown)

(counter clockwise)

Therefore, the total torque on the loop:

T =2 rIlB sin α

(4.29)

However, there are two known facts: 1. Current i flowing in the rotor produces a magnetic field on its own. Direction of field: Right Hand Rule (Thumb) Magnitude of its magnetising intensity:

|H R|=Ci

2. Angle between the peak stator flux density rotor magnetising intensity Furthermore,

H R

B S and peak

is .

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