DC Machinery Fundamentals

7/30/2019 DC Machinery Fundamentals

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EEEB283 Electrical Machines and Drives DC Machinery Fundamentals

_____________________________________________________________________

1Dr. Ungku Anisa, UNITEN 2006

Chap. 8: DC Machinery Fundamentals

This chapter will firstly explain the principles of dc machine

operation using simple examples before considering some of thecomplications that occur in real dc machines.

8.1. A simple rotating loop between curved pole facesThe simplest rotating dc machine is shown below.

It consists of:

A rotating single loop of wire in a slot carved in acylindrical ferromagnetic core rotor

A pair of magnetic poles (north and south) statorThe magnetic field is supplied by the stator, i.e. N and S poles.

The air gap between the rotor and stator is of constant width

(i.e.gap is the same everywhere under the pole faces).

magnetic flux density is constant everywhereunder the pole surface.

From Chapter 1, we know thatirongap >> , hence

magnetic flux takes shortest possible path through

the air gap (i.e. perpendicular to the rotor surface).


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The voltage induced in a rotating loop

If the rotor is rotated, a voltage will be induced in the wire

loop. (Note: No voltage source applied to the rotor terminals.)

To determine the total voltage etot on the loop, each segment of

the loop (as shown in the figure above) has to be examined

separately and the resulting voltages summed up (see page 33).

The voltage on each segment is given by ( ) lBveind

rr

r

=

Hence, the total induced voltage in the loop:

=

+++==

edgespolethebeyond0

facespoleunder the2vBl

e

eeeeee

ind

addccbbatotind

When the loop rotates through 180,

segment ab is under the north pole instead of the southpole

direction ofeba and edc but magnituderemains constant

(8.1)

To and front view of the sim le dc machine


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The resulting voltageetotis shown in the figure below.

An alternative expression for eind that relates the behaviour of

the single loop to the behaviour of larger, real dc machines can

be obtained by examining the figure below:

The tangential velocity v of the loop edges can be expressed as v

= r. Therefore,

=

edgespolethebeyond0

facepoleunder the2 Blre

ind

Output voltage of the simple dc machine

Derivation of an alternative form of induced voltage equation.


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For a 2-pole machine, if we assume that the gap between the

poles is negligible (the gaps at the top and bottom of the

diagram) then the surface area of the pole can be written as

rlrl

AP

== 2

2

Therefore,

=

edgespolethebeyond0

facepoleunder the2

BAe

P

ind

Since the air gap flux density B is constant everywhere under

the pole faces, the total flux under each pole is:

BAP=

Thus, the final form of the voltage equation is:

=

edgespolethebeyond0

facepoleunder the2inde

In general, the voltage in any real machine will depend on the

same three factors:

the flux in the machine

the speed of rotation a constant representing the machine construction

Getting DC voltage out of the rotating loop

The voltage out of the loop is alternating between a constant

positive value and a constant negative value, i.e. ac voltage.

How can this machine be modified to produce a dc voltage?

(8.2)


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By using a mechanism called commutator and brushes

(see below)

two semicircular conducting segments (commutatorsegments) are added to the end of the loop

two fixed contacts (brushes) are placed at an angle suchthat at the instant the voltage in the loop is zero, the

contacts short-circuit the two segments

Every time voltage of the loop changes direction, the

contacts also switch connections and the output of

the contacts is always built up in the same way (as

shown below).

This connection-switching process is known as

.

Output voltage of a dcmachine with

commutator and brushes.


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The induced torque in the rotating loop

If the simple machine is connected to a battery, as shown below,

how much torque will be produced in the stationary loop

when the switch is closed and current is allowed to flow?

As before, the approach is to examine each segment of the loop

and then sum the effects of all segments (see page 34).

The force oneach segment is given by BliFrrr

= .

And the torque on the segment is given by

sinrFFrT ==r

r

, where is the angle between rr

and Fr

.

When the loop is beyond the pole edges, T= 0 (since Br

= 0).

The simple dc

machine

connection for

derivation of

induced torque

equation.

Front view ofthe machine

(the iron core

is omitted for

clarity).


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Hence, the total induced torque in the loop is:

=

+++==

edgespolethebeyond0

facespoleunder the2rilBT

TTTTTT

ind

dacdbcabtotind

By employing the facts that rlAP

and BAP= , the torque

expression can be reduced to:

=

edgespolethebeyond0

facepoleunder the2 iT

ind

In general, the torque in any real machine will depend on thesame three factors:

the flux in the machine the current in the machine a constant representing the machine construction

(8.3)

(8.4)


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commutators


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8.2. Commutation in a simple four-loop DC machine(see page 35 for the detailed explanation)

The resulting terminal voltage as a function of time is shownbelow:

Note: this is a better approximation to a constant dc level thanthat produced by the single rotating loop of Section 8.1.

As the number of loops on the rotor increases, the

approximation to a perfect dc voltage gets better and better.

In summary:

Commutation is the process ofswitching the loop connections

on the rotor of a dc machine just as the voltage in the loop

switches polarity, in order to maintain an essentially constantdc output voltage.

Commutator segments = rotating segments to which the loops

are attached. Typically made of copper bars.

Brushes = stationary pieces that ride on top of the moving

segments. Typically made up of a mixture containing graphite,

so that they cause very little friction.

The resulting output voltage of the four-loop two-pole dc machine.


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8.3. Commutation and armature construction in real DCmachines

In real dc machines, there are several ways in which the loops

on the rotor (also called the armature) can be connected to itscommutator segments.

These different connections affect:

Number of parallel current paths within the rotor The output voltage of the rotor Number and position of the brushes riding on he

commutator segments

The rotor coils

Regardless of the way thewindings are connected to the

commutator segments, the rotor

windings consist ofdiamond-

shaped preformed coils which

are inserted into the armature

slots as a unit (see figure).

Each coil consists of a number

of turns (loops) of wire, each turn taped and insulated from

the other turns and from the rotor slot.

Each side of a turn is called a conductor and the number of

conductors on a machines armature is given by:

cCNZ 2=

whereZ = number of conductors on rotor

C = number of coils on rotor

Nc = number of turns per coil

Normally, a coil spans 180 electrical degrees. This means that

when one side is under the centre of a given magnetic pole, theother side is under the centre of a pole ofopposite polarity.

(8.8)

The shape of a

typical

preformedrotor coil.


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The physical poles may not be 180 mechanical degrees apart

but the magnetic field has completely reversed its polarity in

travelling from under one pole to the next.

The relationship between the electrical angle and mechanical

angle in a given machine is given by:

me

P

2=

where e = electrical angle, in degreesm = mechanical angle, in degreesP = number of magnetic poles on the machine

If a coil spans 180 electrical degrees, the voltages in theconductors on either side of the coil will be exactly the same in

magnitude but opposite in direction at all times. Such a coil is

called a full-pitch coil.

A fractional-pitch coil spans less than 180 electrical degrees,

and a rotor winding wound with fractional-pitch coils is called achorded winding.

The amount of chording in a winding is described by a pitchfactorp, defined by:

%100

180

coilofangleelectrical

=p

Most rotor windings are two-layer windings, meaning that sidesfrom two different coils are inserted into each slot. One side of

each coil will be at the bottom of its slot, and the other side will

be at the top of its slot.

(8.9)

(8.10)


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Connections to the commutator segments

There are a number of ways in which the rotor windings can be

connected to the commutator segments. The different winding

arrangements have different advantages and disadvantages.

Commutator pitch yc = distance (in number of segments)

between the commutator segments to which the two ends of acoil are connected.

Progressive winding (yc =1) =

the end of a coil (or a set number

of coils) is connected to acommutator segment ahead of

the one its beginning is

connected to.

Retrogressive winding ( yc =-1)

= the end of a coil (or a set

number of coils) is connected toa commutator segment behind

the one its beginning is

connected to.

If everything else is identical, the rotation direction of a

progressive-wound rotor is opposite to that of a retrogressive-

wound rotor.

Rotor (armature) windings are further classified according to the

plex of their windings:

1.A simplex rotor winding is a single, complete, closedwinding wound on a rotor.

2.A duplex rotor winding is a rotor with two complete andindependent sets of rotor windings. Each of the windingswill be associated with every other commutator segment:

one winding will be connected to segments 1, 3, 5, etc.,

A coil in a progressive winding. A coil in a retrogressive winding.


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and the other winding will be connected to segments 2, 4,

6, etc.

3.A triplex winding will have three complete andindependent sets of windings, each winding connected toevery third commutator segment on the rotor.

4.All armature with more than one set of windings are calledmultiplex windings.

Finally, armature windings are classified according to thesequence of their connections to the commutator segments:

Lap winding Wave winding Frog-leg winding (combines lap and wave windings in a

single rotor)

Note: for individual characteristics, advantages and

disadvantages of these windings, please refer to Chapman

textbook page 493 502.

8.4. Problems with commutation in real machinesIn practice, the commutation process is not as simple as

described theoretically in Sections 8.2 and 8.3. There are twomajor effects that disturb the commutation process:

I. Armature reactionII.L di/dt voltagesThis section explores the nature of these problems and the

solutions employed to mitigate their effects.


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I. Armature reactionIn real DC machines, the magnetic field is provided by field

windings on the stator, i.e. no magnets.

If the magnetic field windings of a dc machine are connected

to a power supply and the rotor of the machine is turned by

an external source of mechanical power, then

a voltage will be induced in the conductors of the

rotor.

This voltage will be rectified into a dc output by the action ofthe machines commutator.

When a load is connected to the terminals of the machine:

current will flow in its armature windings. this current produces

which will distort the original magnetic field from the

machines poles

Armature reaction = of the flux in the

machine as the load is increased.

It causes two serious problems in real dc machines.

Problem 1: Neutral-plane shift

Magnetic neutral plane = the plane within the machine where

the velocity of the rotor is exactly parallel to the magnetic

flux lines, so that in the conductors in the plane.

To understand the problem of neutral plane shift, the two-pole

dc machine shown below is employed.


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The flux is distributeduniformly under the polefaces.

The rotor windings havevoltages built up as shown on

the left.

Hence, the neutral plane isexactly vertical.

Now suppose a load is connected such that the machine actsas a generator (i.e. no power supplied to rotor windings).

Current flows out of thepositive terminal of the

generator, as shown on the

left.

This current flow produces a magnetic field from the rotorwindings, as shown in picture (c). (direction: right-hand rule)..

This rotor magnetic field affects the original magnetic fieldfrom the poles that produced the generators voltage in the

first place.

In some places under thepole surfaces, it subtractsfrom the pole flux.

In other places, it adds to

the pole flux.


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The overall result is that themagnetic flux in the air gap of

the machine is skewed as

shown on the left.

Notice that the place on the rotor where induced voltage in aconductor would be zero (i.e. the neutral plane) has

.

In a dc generator magnetic neutral plane shifts

the direction of rotation.

In a dc motor magnetic neutral plane shifts

to the direction of

rotation (due to reverse in rotor currentdirection, hence flux add and subtract

at opposite corners from that shown in

picture (d) above).

Note: amount of shift depends on the machine load.

What is the effect of neutral-plane shift?

The brushes must short out commutator segments just atthe moment when the voltage across them is zero.

If the brushes are set to short out conductors in the verticalplane, then the voltage between the segments is zero until

the machine is loaded.

When the machine is loaded, the neutral plane shifts andthe brushes short out commutator segments withfinite

voltage across them.


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The result is:

Current flow circulating between the shorted segments Large sparks at the brushes when the current path is

interrupted as the brush leaves the segment

End result: at the

brushes.

This is a very serious problem. It leads to:

Drastically reduced brush life. Pitting of the commutator segments. Greatly increased maintenance costs.

Notice that this problem cannot be fixed by placing the

brushes over the full-load neutral plane, as it will then causesparks at no load.

In extreme cases, the neutral-plane shift can lead to flashover

in the commutator segments near the brushes.

(The air near the burshes in a machine is normally ionized as aresult of sparking on the brushes. Flashover occurs when the

voltage of adjacent comutator segments gets large enough to

sustain an arc in the ionized air above them. If flashover occurs,the resulting arc can even melt the commutators surface.)

Problem 2: Flux weakening

To understand flux weakening, refer to the magnetisation curve

shown below:


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Most machines operate near the saturation point (knee ofcurve).

At locations on the pole surface where rotor mmfadds to thepole mmf, only a small increase in flux occurs.

But at locations on the pole surface where rotor mmfsubtracts from the pole mmf, a decrease influx occurs.

The net result:

The total average flux under the entire pole face is

decreased (see figure below).


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Flux weakening causes problems in both generators and motors.

In generators, the effect is reduced voltage supplied by the

generator for any given load.

The flux-weakening effect can be more serious for motors:

when flux is decreased, its speed but increase in motor speed can increase its load, resulting

in more flux weakening (hence, speed increase further)

it is possible for some shunt dc motors to reachcondition, where motorspeed just keeps

on increasing until the machine is disconnected from the

power line or until it destroys itself.


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II.L di/dt voltagesThe L di/dt voltages occur in commutator segments that are

being by the brushes.

Also sometimes known as inductive kick.

To understand this problem, refer to the figures belowrepresenting a series of commutator segments (a, b, c, d ..) and

the conductors (or coils) connected to them.

Assuming that the current in the brush = 400A, the current in

each path is 200A (since current split into two coils).

When a commutator segment is shorted out, the current flow

through the commutator segment must reverse.

How fast must this reversal occur?

Assuming the machine is rotating at 800 rpm and that there are50 commutator segments (a reasonable number for a typical

motor), each commutator segment moves under a brush andclears it again in t= 0.0015s.

Calculations:

800 rpm = 83.78 rad/s

1 segment takes up 0.126 rad ( = 50 / 2 )

Time to clear 1 segment = 0.126 / 83.78 = 0.0015s


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Therefore, the rate of change in current with respect to time in

the shorted loop must average:

With even a tiny inductance L in the loop, a very significant

inductive voltage kickv = L di/dtwill be induced in the shorted

commutator segment.

This high voltage naturally causes sparking in the brushes ofthe machine, resulting in the same arcing problems as caused

by the neutral-plane shift.

Solution to the problems with commutation

The following three approaches have been developed to partiallyor completely correct the problems of armature reaction and L

di/dtvoltages:

I. Brush shiftingThis method attempts to stop the sparking at the brushes caused

by the neutral-plane shifts andL di/dteffects.

A/s667,266s0015.0

)200(200=

=

dt

di

The current reversal in the

coil undergoing

commutation as a function

of time for both ideal and

real commutation, with thecoil inductance taken into

account.

ideal

commutation

real

commutation

- 200 A


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Since the neutral plane shifts, hence shift the brushes (to the

new neutral plane position) to stop the sparking.

But there are several serious problems associated with this

method:

The neutral plane moves with changes in load and the shiftdirection reverses when going from motor to generatoroperation.

Even though brush sparking is stopped, brush shiftingaggravates the flux-weakening effect of armature reaction

in the machine. This is true because of two effects:

i. The rotor mmf now has a vector component that opposesthe mmf from the poles (see figure below).

ii. The change in armature current distribution causes theflux to bunch up even more at the saturated parts of the

pole faces.

However, this method is obsolete.

Before brush shifting (i.e.

brushes over the vertical lane)

After brush shifting (i.e. brushes

over the shifted neutral lane)


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It is only used in very small motors where other better solutions

are not economical.

II. Commutating poles or interpolesBasic idea:

If the voltage in the wires undergoing commutation can be

made zero, then there will be no sparking at the brushes.

Method:

Small poles calledcommutating poles or interpoles are placed

midway between the main poles anddirectly

over the conductors being commutated.

The flux provided by the commutating poles will exactly

the voltage in the coils undergoing

commutation. Therefore,

there will be no sparking at the brushes.

The commutation poles do not change the operation of themachine because they are so small and only have effect on the

few conductors undergoing commutation.

Hence, armature reaction under the main pole faces is

unaffected, i.e.

flux weakening in the machine is unaffected by

the commutating poles (i.e. flux weakening problem isnot resolved).

How is cancellation of voltage in the commutator segments

accomplishes for all values of load?

This is done by connecting the interpole windings in series

with the windings on the rotor.


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As the load increases

The rotor current (IA) increases

Magnitude of neutral plane shift

and

size ofL di/dt effects increases

Increases voltage in conductors undergoing commutation

(which was supposed to be zero)

BUT

Interpole flux increases as well

(due to series connection, i.e. IA, Iinterpole)

Gives larger voltage in the conductor which

opposes voltage due tothe neutral-plane shift

The net result is:

Commutating pole cancels the neutral-plane

shift andL di/dt effects over a broad range of loads.

Note: Interpoles work for both motor and generator

operation.

A dc machine

with interpoles


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What polarity must the flux in the interpoles be?

Interpoles must induce voltage in the conductors undergoing

commutation that is opposite to the voltage caused by the

neutral-plane shift andL di/dteffects.

For a generator,

neutral-plane shifts in the direction of rotation. conductors undergoing commutation have the same

voltage polarity as the pole they just left.

to cancel this effect:Interpoles must be of the samepolarity as the

next upcoming main pole in a generator.

For a motor,

reverse in rotor current direction compared to generatormode

neutral-plane shifts opposite to the direction of rotation. conductors undergoing commutation have the same

voltage polarity as the pole they are approaching.

to cancel this effect:Interpoles must be of the same polarity as the

previous main pole in a motor.

Determining the required polarity

of an interpole. The flux from the

interpole must produce a voltage

that opposes the existing voltage in

the conductor.


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Use of commutating poles:

very common they correct sparking problems at low cost

BUT flux-weakening problem is till present!

III. Compensating windingsFlux-weakening problem can be very serious in very heavy,

severe duty cycle motors.

Compensating windings are:

placed in slots carved in the faces of the poles parallelto the rotor conductors.

Connected in series with the rotor windings such thatload changes will change the currents in the compensating

windings.

Figures below shows the basic concept of compensating

windings.

1. Here, the pole flux is shownby itself.

2. The rotor flux and compensatingwinding flux is shown.

3. The sum of the fluxes is just equal to the original pole flux

by itself.


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A more careful development of the compensating winding effect

is shown below:

Mmf due to compensating windings is equal and oppositeto the mmf due to the rotor at every point under the polesurface.

The resulting net mmf is just the mmf due to the poles.Flux in the machine is unchanged regardless of load.

The flux and magnetomotive forces in a dc machine with compensating

windings.


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The major disadvantage of compensating windings:

expensive must be machined into pole faces does not cancelL di/dt effects also requires interpoles

8.5. Internal generated voltage and induced torqueequations for real DC machines

Internal generated voltage equation

The voltage out of the armature of a real machine is:

( )

( )vBla

ZE

E

A

A

=

= conductoreachonvoltage

pathcurrentper

conductorsofnumber

whereZ= total number of conductor

a = number of current paths

The velocity of each conductor in the rotor can be expressed as v= r, where r is the radius of the rotor.

Therefore,

( )Blra

ZE

A=

Since the area per pole PrlAP

2= , hence the total flux per

pole in a P pole machine is:

BP

rlBA

P

2==

The voltage out of the armature of a real dc machine is:

(8.11)

(8.13)

(8.12)

( )

KE

a

ZP

P

rlB

a

ZPrlB

a

ZE

A

A

=

=

==

2

2

2


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Induced torque equation

The torque on the armature of a real machine is:

( ) ( )

=

=

a

lBrIZT

T

A

ind

indconductoreachontorqueconductorsofnumber

The flux per pole in the machine can be expressed as:

( )

P

rlBBA

P

2==

Therefore, torque on the armature of a real dc machine is:

( )

Aind

AAind

IKT

Ia

ZPIrlB

a

ZT

=

==2

8.6. The construction of DC machinesA simplified diagram of a dc machine is shown below.

(8.14)

(8.15)


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The physical structure of the machine consists of two parts:

o Stator, consists of :- the frame,which provides physical support.- pole pieces, which project inward and provides path

for the machine magnetic flux.- pole shoes, i.e. the ends of the pole pieces that

spread out near the rotor surface for even flux

distribution.- pole face, i.e. exposed surface of a pole shoe.

o RotorAir gap = distance between stator and rotor.

There are two principal windings on a dc machine:

o Armature windings windings in which voltage isinduced (i.e. rotor windings in the dc machine)

o Field windings windings that produce the mainmagnetic flux in the machine (i.e. stator windings in thedc machine)

Note: Since the armature windings are on the rotor, a dc

machines rotor is sometimes called an armature.

8.7. Power flow and losses in DC machinesThe efficiency of a dc machine:

%100%100%100

=+

==in

lossin

lossout

out

in

out

P

PP

PP

P

P

P


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Losses in DC machines

1.Electrical or copper losses (I2R loss):Armature loss:

AAARIP2

=

Field loss :FFF

RIP2=

2.Brush losses power loss across the contact potential at themachine brushes.

Brush drop loss: ABDBD IVP =

3.Core losses hysteresis and eddy current losses occurring inthe metal of the motor.

4.Mechanical losses losses associated with mechanicaleffects, i.e. friction and windage losses.

5.Stray losses (or miscellaneous losses) losses that cannotbe placed in any of the above categories.

The power flow diagram

For a dc motor:

Pout = Tappm

T


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8.1 A simple rotating loop between curved pole facesDetailed segment analysis to obtain the voltage induced in a

simple rotating loop

Segment ab:

Velocity vab tangential to rotation path

=edgespolethebeyond,0

poleunder the,surfacetoBr


facepoleunder the,pageintopositivevBle

ba

Segment bc:

direction of( )Bv rr either into or out of page ( ) lBv rrr 0=cbe Segment cd:

Velocity vcdtangential to rotation path




facepoleunder the,pageofoutpositivevBledc

Segment da:

( ) lBv

rr

r

0=ade


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Detailed segment analysis to obtain the torque induced in a

simple rotating loop

Segment ab:Current directed out of the page




facepoleunder the,directionmotionotangent tilBF

ab

r


facepoleunder the,ckwisecounterclorilBTab

Segment bc:

Current flowing from upper left to lower right ( ) BlBliFbc rrrrr toparallelissince0== 0=bcT Segment cd:

Current directed into the page




facepoleunder the,directionmotionotangent tilBFcdr


facepoleunder the,ckwisecounterclorilBTcd

Segment da:Current flowing from upper left to lower right ) BlBliFda rrrrr toparallelissince0== 0=daT


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8.2 Commutation in a simple four-loop DC machineCommutation = process ofconverting the ac voltages

and currents in the rotor of a dc machine to dc voltages and

currents in its terminals.

A simple four-loop, two-pole dc machine (see figure below) has:

Four complete loops buried (in a special manner) in slotscarved in the laminated steel of the rotor.

Pole faces that are curved to provide uniform flux densityeverywhere under the faces.

The unprimed end of each loop is the outermost wirein each slot.

The primed end of each loop is the innermost wire inthe slot directly opposite.

The dc machine and its windings connection to the machines

commutator are shown below:

Notice that:

At t = 0, in the dc machine:

The 1, 2, 3 and 4 ends of the loops are under the northpole face and the voltage in each of the loop ends is given

by:

LoopStretches between

commutator

segments:

1 a and b

2 b and c

3 c and d

4 dand a

A four-loop two-pole dc machine at

time t= 0.


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( ) pageofoutpositivevBllBveind ==

rr

r

The 1, 2, 3 and 4 ends of the loops are under the southpole face and the voltage in each of the loop ends is givenby:

( ) pagetheintopositivevBllBveind ==

rr

r

The overall result is shown below whereby each coilrepresents

one side (or

conductor)of a loop.

The total voltage at the brushes of the machine is:== 0henw4 teE

where e = vBl = induced voltage on one side of a loop.

Notice that there are two parallel paths for current throughthe machine. The existence oftwo or moreparallel paths

for rotor current is a common feature of all commutationschemes.

(8.5)

The voltage on the

rotor conductors of a

four-loop two-pole

dc machine at timet= 0.


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Figure below shows the machine at time t = 45:

Loops 1 and 3 have rotated into the gap between thepoles, so the voltage across each of them is zero.

At this instant the brushes of the machine are shorting outcommutator segments ab and cd.

Only two loops 2 and 4 are under the pole faces, so theterminal voltageEis given by:

== 54henw2 teE (8.6)

The same four-

loop two-pole dc

machine at time

t= 45.

The voltage on the

rotor conductors of

the dc machine at

time t= 45.


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The rotor continues to turn another 45, i.e. t = 90:

The 1, 2, 3 and 4 ends of the loops are under the northpole face and the voltage in each of the loop ends is given

by:

( ) pageofoutpositivevBllBveind

==rr

r

The 1, 2, 3 and 4 ends of the loops are under the southpole face and the voltage in each of the loop ends is givenby:

( ) pagetheintopositivevBllBveind

==rr

r

The overall result is shown above on the right. Theterminal voltage of the machine is:

Note: The voltages on loops 1 and 3 have reversed between

t = 0 and t = 90, but their connections have also reversed.Hence, the total voltage is still built up in the same direction as

b f

(8.7)== 09when4 teE

The same two-pole dc machine at time t= 90 with the voltages onthe conductors.

DC Machinery Fundamentals

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