Chapter4 dc motor

Mohd Rusllim Mohamed

Ext : 2080


BEE2123

ELECTRICAL MACHINES

Upon completion of the chapter the student should be able to: ◦ State the principle by which machines

convert mechanical energy to electrical energy. ◦ Understand the principle of DC generator

as it represents a logical behavior of dc motors. ◦ Discuss the operating differences between

different types of dc motors


Direct-current generators are not as common as they used to be, because direct current, when required, is mainly produced by electronic rectifiers. These rectifiers can convert the current of an ac system into direct current without using any moving parts as shown in Fig. 2.1.

Yet, still the DC generator is used in many plants to produce power needed to operate large dc motor.

Generator Motor


Fig 2.1 Full wave rectifier

../../Diagrams & Pictures/KIV Boxes/generatormotor.mov

A generator is a machine that converts mechanical energy into electrical energy by using the principle of magnetic induction (Fig. 2.2)

In this example, the ends of the wire loop have been connected to two sliprings mounted on the shaft, while brushes are used to carry the current from the loop to the outside of the circuit. (Details will be discussed later)


Fig 2.2 Principle of magnetic induction in DC machine

Recap : Whenever a conductor is moved within a magnetic field in such a way that the conductor cuts across magnetic lines of flux, voltage is generated in the conductor.

The AMOUNT of voltage generated depends on:

i. the strength of the magnetic field,

ii. the angle at which the conductor cuts the magnetic field,

iii. the speed at which the conductor is moved, and

iv. the length of the conductor within the magnetic field


The POLARITY of the voltage depends on the direction of the magnetic lines of flux and the direction of movement of the conductor.

To determine the direction of current in a given situation, the LEFT-HAND RULE FOR GENERATORS is used as shown in Fig. 2.3.


•thumb in the direction the conductor is being moved

•forefinger in the direction of magnetic flux (from north to south)

•middle finger will then point in the direction of current flow in an

external circuit to which the voltage is applied

Fig 2.3 Left Hand Rules

The simplest elementary generator that can be built is an ac generator. Basic generating principles are most easily explained through the use of the elementary ac generator. For this reason, the ac generator will be discussed first. The dc generator will be discussed later.

An elementary generator (Fig. 2.4)

consists of a wire loop mounted on the shaft, so that it can be rotated in a stationary magnetic field. This will produce an induced emf in the loop. Sliding contacts (brushes) connect the loop to an external circuit load in order to pick up or use the induced emf.


Fig 2.4 Elementary Generator

The pole pieces (marked N and S) provide the magnetic field. The pole pieces are shaped and positioned as shown to concentrate the magnetic field as close as possible to the wire loop.

The loop of wire that rotates through the field is called the ARMATURE. The ends of the armature loop are connected to rings called SLIP RINGS. They rotate with the armature.

The brushes, usually made of carbon, with wires attached to them, ride against the rings. The generated voltage appears across these brushes. (These brushes transfer power from the battery to the commutator as the motor spins – discussed later in dc elementary generator).


In Fig. 2.5, the armature winding has been left out so that it is easier to see the rotor action. The key thing to notice is that as the armature passes through the horizontal position, the poles of the electromagnet flip. Because of the flip, the north pole of the electromagnet is always above the axle so it can repel the field magnet's north pole and attract the field magnet's south pole.

It's this magnetic attraction and repulsion that causes the armature to rotate


N

s

Armature winding

has been left out

Fig. 2.5 Rotor being rotated

In Figure 2.6, an end view of the shaft and wire loop is shown. At this particular instant, the loop of wire (the black and white conductors of the loop) is parallel to the magnetic lines of flux, and no cutting action is taking place. Since the lines of flux are not being cut by the loop, no emf is induced in the conductors, and the meter at this position indicates zero.

This position is called the

NEUTRAL PLANE.


Fig. 2.6 00 Position (Neutral Plane)

In Figure 2.7, the shaft has been turned 900 clockwise, the conductors cut through more and more lines of flux, and voltage is induced in the conductor.

at a continually increasing angle , the induced emf in the conductors builds up from zero to a maximum value or peak value.

Observe that from 00 to 900, the black conductor cuts DOWN through the field. At the same time the white conductor cuts UP through the field. The induced emfs in the conductors are series-adding. This means the resultant voltage across the brushes (the terminal voltage) is the sum of the two induced voltages. The meter at position B reads maximum value.


Fig. 2.7 900 Position

After another 900 of rotation (Fig. 2.8), the loop has completed 1800

of rotation and is again parallel to the lines of flux. As the loop was turned, the voltage decreased until it again reached zero.

Note that : From 00 to 1800 the

conductors of the armature loop have been moving in the same direction through the magnetic field. Therefore, the polarity of the induced voltage has remained the same



As the loop continues to turn, the conductors again cut the lines of magnetic flux (Fig. 2.9).

This time, however, the conductor that previously cut through the flux lines of the south magnetic field is cutting the lines of the north magnetic field, and vice-versa.

Since the conductors are cutting the flux lines of opposite magnetic polarity, the polarity of the induced voltage reverses. After 270' of rotation, the loop has rotated to the position shown, and the maximum terminal voltage will be the same as it was from A to C except that the polarity is reversed.



After another 900 of rotation, the loop has completed one rotation of 3600 and returned to its starting position (Fig. 2.10).

The voltage decreased from its negative peak back to zero.

Notice that the voltage produced in the armature is an alternating polarity. The voltage produced in all rotating armatures is alternating voltage.



Observes ◦ The meter

◦ direction the conductors of the armature loop

◦ Direction of the

◦ current flow


Fig. 2.11 Output voltage of an elementary generator during one revolution

Since DC generators must produce DC current instead of AC current, a device must be used to change the AC voltage produced in the armature windings into DC voltage. This job is performed by the commutator.

The commutator is constructed from a copper ring split into segments with insulating material between the segments (See next page). Brushes riding against the commutator segments carry the power to the outside circuit.

The commutator in a dc generator replaces the slip rings of the ac generator. This is the main difference in their construction. The commutator mechanically reverses the armature loop connections to the external circuit.


Fig. 2.12 Commutator

The armature has an axle, and the commutator is attached to the axle. In the diagram to the right, you can see three different views of the same armature: front, side and end-on.

In the end-on view, the winding

is eliminated to make the commutator more obvious. You can see that the commutator is simply a pair of plates attached to the axle. These plates provide the two connections for the coil of the electromagnet.


Fig 2.13 Armature with commutator view

The diagram at the right shows how the commutator and brushes work together to let current flow to the electromagnet, and also to flip the direction that the electrons are flowing at just the right moment. The contacts of the commutator are attached to the axle of the electromagnet, so they spin with the magnet. The brushes are just two pieces of springy metal or carbon that make contact with the contacts of the commutator.

Through this process the commutator changes the generated ac voltage to a pulsating dc voltage which also known as commutation process.


Fig 2.14 Brushes and commutator

The loop is parallel to the magnetic lines of flux, and no voltage is induced in the loop

Note that the brushes make contact with both of the commutator segments at this time. The position is called neutral plane.


Fig. 2.15 00 Position (DC Neutral Plane)

As the loop rotates, the conductors begin to cut through the magnetic lines of flux.

The conductor cutting

through the south magnetic field is connected to the positive brush, and the conductor cutting through the north magnetic field is connected to the negative brush (Fig 2.16).

Since the loop is cutting lines

of flux, a voltage is induced into the loop. After 900 of rotation, the voltage reaches its most positive point.


Fig. 2.16 900 Position (DC)

As the loop continues to rotate, the voltage decreases to zero.

After 1800 of rotation, the conductors are again parallel to the lines of flux, and no voltage is induced in the loop.

Note that the brushes again make contact with both segments of the commutator at the time when there is no induced voltage in the conductors (Fig 2.17).



During the next 900 of rotation, the conductors again cut through the magnetic lines of flux.

This time, however, the conductor that previously cut through the south magnetic field is now cutting the flux lines of the north field, and vice-versa. (Fig 2.18).

Since these conductors are cutting the lines of flux of opposite magnetic polarities, the polarity of induced voltage is different for each of the conductors. The commutator, however, maintains the correct polarity to each brush.

The conductor cutting through the north magnetic field will always be connected to the negative brush, and the conductor cutting through the south field will always be connected to the positive brush.

Since the polarity at the brushes has remained constant, the voltage will increase to its peak value in the same direction.



As the loop continues to rotate (Fig 2.19), the induced voltage again decreases to zero when the conductors become parallel to the magnetic lines of flux.

Notice that during this 3600

rotation of the loop the polarity of voltage remained the same for both halves of the waveform. This is called rectified DC voltage.

The voltage is pulsating. It does turn on and off, but it never reverses polarity. Since the polarity for each brush remains constant, the output voltage is DC.


Fig. 2.19 00 Position (DC Neutral Plane)

Observes ◦ The meter ◦ direction the

conductors of the armature loop

◦ Direction of the current flow


Fig. 2.20 Effects of commutation

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To increase the amount of output voltage, it is common practice to increase the number of turns of wire for each loop (Fig 2.20).

If a loop contains 20 turns of wire, the induced voltage will be 20 times greater than that for a single-loop conductor.

The reason for this is that each loop is connected in series with the other loops. Since the loops form a series path, the voltage induced in the loops will add.

In this example, if each loop has an induced voltage of 2V, the total voltage for this winding would be 40V (2V x 20 loops = 40 V).


Fig. 2.21 Effects of additional turns

When more than one loop is used, the average output voltage is higher and there is less pulsation of the rectified voltage.

Since there are four segments in the

commutator, a new segment passes each brush every 900 instead of every 1800.

Since there are now four commutator segments in the commutator and only two brushes, the voltage cannot fall any lower than at point A.

Therefore, the ripple is limited to the rise and fall between points A and B on the graph. By adding more armature coils, the ripple effect can be further reduced. Decreasing ripple in this way increases the effective voltage of the output.


Fig. 2.22 Effects of additional coils

Q.1Generators convert mechanical motion to electrical energy using what principle?

Q.2 What rule should you use to determine the direction of induced emf in a coil? Q.3 What is the purpose of the slip rings? Q.4 Why is no emf induced in a rotating coil when it passes through the neutral

plane? Q.5 What component causes a generator to produce dc voltage rather than ac

voltage at its output terminals?

Q.6 At what point should brush contact change from one commutator segment to the next?

Q.7 An elementary, single coil, dc generator will have an output voltage with how many pulsations per revolution?

Q.8 How many commutator segments are required in a two-coil generator?


The actual construction and operation of a practical dc generator differs somewhat from our elementary generators

Nearly all practical generators use electromagnetic poles instead of the permanent magnets used in our elementary generator

The main advantages of using electromagnetic poles are: (1) increased field strength and

(2) possible to control the strength of the fields. By varying the input voltage, the field strength is varied. By varying the field strength, the output voltage of the generator can be controlled.


Fig. 2.23 Four-pole generator (without armature)

If you ever have the chance to take apart a small electric motor, you will find that it contains the same pieces described above: two small permanent magnets, a commutator, two brushes, and an electromagnet made by winding wire around a piece of metal. Almost always, however, the rotor will have three poles rather than the two poles as shown in this article. There are two good reasons for a motor to have three poles: ◦ It causes the motor to have better dynamics. In a two-pole motor, if the

electromagnet is at the balance point, perfectly horizontal between the two poles of the field magnet when the motor starts, you can imagine the armature getting "stuck" there. That never happens in a three-pole motor.

◦ Each time the commutator hits the point where it flips the field in a two-pole motor, the commutator shorts out the battery (directly connects the positive and negative terminals) for a moment. This shorting wastes energy and drains the battery needlessly. A three-pole motor solves this problem as well.

It is possible to have any number of poles, depending on the size of the motor and the specific application it is being used in.


More loops of wire = higher rectified voltage In practical, loops are generally placed in slots of an iron core (Fig

2.24) The iron acts as a magnetic conductor by providing a low-reluctance

path for magnetic lines of flux to increase the inductance of the loops and provide a higher induced voltage. The commutator is connected to the slotted iron core. The entire assembly of iron core, commutator, and windings is called the armature (Fig 2.25). The windings of armatures are connected in different ways depending on the requirements of the machine.

Mohd Rusllim Mohamed Fig. 2.24 loops of wire are wound around slot in a metal core

Fig. 2.25 DC machine armature

Lap Wound Armatures ◦ are used in machines designed for low

voltage and high current ◦ armatures are constructed with large

wire because of high current ◦ Eg: - are used is in the starter motor of

almost all automobiles ◦ The windings of a lap wound armature

are connected in parallel (Fig 2.26). This permits the current capacity of each winding to be added and provides a higher operating current

◦ No of current path, C=2p ; p=no of poles


Fig. 2.26 Lap wound armatures

Wave Wound Armatures ◦ are used in machines designed for high

voltage and low current ◦ their windings connected in series (Fig

2.27) ◦ When the windings are connected in

series, the voltage of each winding adds, but the current capacity remains the same

◦ are used is in the small generator in hand-cranked megohmmeters

◦ No of current path, C=2


Fig. 2.27 Wave wound armatures

Frogleg Wound Armatures ◦ the most used in practical nowadays

◦ designed for use with moderate current and moderate armatures voltage

◦ the windings are connected in series parallel (Fig2.28).

◦ Most large DC machines use frogleg wound armatures.


Fig. 2.28 Frogleg wound armatures

Most DC machines use wound electromagnets to provide the magnetic field.

Two types of field windings are used : ◦ series field, and

◦ the shunt field.


Series field windings ◦ are so named because they are connected in series with the

armature ◦ are made with relatively few windings turns of very large wire

and have a very low resistance ◦ usually found in large horsepower machines wound with

square or rectangular wire. The use of square wire permits the windings to be laid closer together, which increases the number of turns that can be wound in a particular space (Fig 2.29)

◦ Square and rectangular wire can also be made physically smaller than round wire and still contain the same surface area (Fig 2.30)

Fig. 2.29 square wire permits more turns than round wire in the same area Fig. 2.30 Square wire contains more

surface than round wire

Shunt field windings ◦ is constructed with relatively many turns of small

wire, thus, it has a much higher resistance than the series field.

◦ is intended to be connected in parallel with, or shunt, the armature.

◦ high resistance is used to limit current flow through the field.


When a DC machine uses both series and shunt fields, each pole piece will contain both windings (Fig 2.31).

The windings are

wound on the pole pieces in such a manner that when current flows through the winding it will produce alternate magnetic polarities.


Fig. 2.31 Both series and shunt field windings are contained in each pole piece

S – series field

F – shunt field


Winding

Lap

C=2p

Wave

C=2

Separately

Excited

Frogleg

Self excited

armature field

series shunt compound

Electric motors are everywhere! In a house, almost every mechanical movement that you see around you is caused by an DC (direct current) electric motor.

An electric motor is a device that transforms electrical energy into mechanical energy by using the motor effect.


DC motors consist of rotor-mounted windings (armature) and stationary windings (field poles). In all DC motors, except permanent magnet motors, current must be conducted to the armature windings by passing current through carbon brushes that slide over a set of copper surfaces called a commutator, which is mounted on the rotor.


Parts of an electric motor

The commutator bars are soldered to armature coils. The brush/commutator combination makes a sliding switch that energizes particular portions of the armature, based on the position of the rotor. This process creates north and south magnetic poles on the rotor that are attracted to or repelled by north and south poles on the stator, which are formed by passing direct current through the field windings. It's this magnetic attraction and repulsion that causes the rotor to rotate.

The greatest advantage of DC motors may be speed control. Since speed is directly proportional to armature voltage and inversely proportional to the magnetic flux produced by the poles, adjusting the armature voltage and/or the field current will change the rotor speed.

Today, adjustable frequency drives can provide precise speed control for AC motors, but they do so at the expense of power quality, as the solid-state switching devices in the drives produce a rich harmonic spectrum. The DC motor has no adverse effects on power quality.


Power supply, initial cost, and maintenance requirements are the negatives associated with DC motors

Rectification must be provided for any DC motors supplied from the grid. It can also cause power quality problems.

The construction of a DC motor is considerably more complicated and expensive than that of an AC motor, primarily due to the commutator, brushes, and armature windings. An induction motor requires no commutator or brushes, and most use cast squirrel-cage rotor bars instead of true windings — two huge simplifications.


Series motors connect the field windings in series with the armature.

Series motors lack good speed regulation, but are well-suited for high-torque loads like power tools and automobile starters because of their high torque production and compact size.


Ea

Rf

M VT (dc

supply)

Ra ia

)( faaaT RRiEV

La iinote :

KIKKE aa 21


P

Pout Pin= VTiL

Pca=ia2Ra

Pcf=ia2Rf

Pm

P is normally given

Pin = Pout + total losses

Where,

Pca =armature copper loss

Pcf =field copper loss

P=stray, mech etc

in

out

mm

oo

P

PEfficiency

N

Ptorquemechanicalfor

N

Ptorqueloadoutputfor

N

P

,

2

60,

2

60,/

2

60

Pm= Ea ia

Example 1:

A dc machine in Figure 1 is consumed a 6.5kW when the 12.5 A of armature current is passing thru the armature and field resistance of 3.3 and 2.0 respectively. Assume stray losses of 1.2kW. Calculate

a) terminal voltage, VT

b) back emf, Ea

c) net torque if the speed is at 3560rpm

d) efficiency of the machine

[520V, 453.75V, 12N-m, 68.8%]


Ea

Rf

M VT (dc

supply)

Ra ia

Figure 1

6.5 , 12.5 ,

3.3 , 2.0 ,

1.2

in a

a f

P kW I A

R R

P kW

).

sin ,

6.5520

12.5

in T L

L a

inT

L

a P V I

ce I I

P kV V

I

). ( )

520 12.5(3.3 2.0)

453.75

a T a a fb E V I R R

V

60).

2

,

453.73(12.5) 5672

5672 1.2 4472

60(4472)12

2 (3560)

oo

o m m a a

m

o

o

Pc

N

P P P P E I

P W

P k W

Nm

). % 100%

4472100%

6.5

68.8%

o

in

Pd

P

k

Abu Zaharin Ahmad

Example 2: A 600V 150-hp dc machine in

Figure 2 operates at its full rated load at 600 rpm. The armature and field resistance are 0.12 and 0.04 respectively. The machine draws 200A at full load. Assume stray losses 1700W. Determine

a) the armature back emf at full load, Ea

b) developed power and developed torque

c) assume that a change in load results in the line current dropping to 150A. Find the new speed in rpm and new developed torque. {Hint: Ea=K}


Ea

Rf

M VT (dc

supply)

Ra ia

Figure 2

[Ea=568V, P=113.6kW, T=1808Nm, n=608.45rpm, T=1017Nm]

600 , 200 ,

0.12 , 0.04 ,

1700 , 600

T a

a f

V V I A

R R

P W n rpm

). ( )

600 200(0.12 0.04)

568

a T a a fa E V I R R

V

60).

2

568(200) 113.6

60(113.6 )1808

2 (600)

mm

m a a

o

Pb

N

P E I kW

kNm

2 2

1 1

2 22 1 2 1

1 1

2

2

2 2

).

tan

600 150(0.12 0.04)

576

576600 608.5

568

576(150)

86.4

a

a

a a

a a

a

m a a

E Kc

E K

Since K is cons t

E Eor n n

E E

E

V

n rpm

E I

Nm

Abu Zaharin Ahmad

Shunt motors use high-resistance field windings connected in parallel with the armature.

Varying the field resistance changes the motor speed.

Shunt motors are prone to armature reaction, a distortion and weakening of the flux generated by the poles that results in commutation problems evidenced by sparking at the brushes.

Installing additional poles, called interpoles, on the stator between the main poles wired in series with the armature reduces armature reaction.


Ea VT (dc

supply)

Ra ia

if

Rf M

iL

)( aaaT RiEV

ffT

faL

RiV

iiinote

:


P

Pout Pin=VTiL

Pca=ia2Ra

Pcf=if2Rf

Pm

P is normally given


Where,



P=stray, mech etc

in

out

mm

oo

P

PEfficiency

N


N


N

P

,

2

60,

2

60,/

2

60

Pm= Ea ia

Example : ◦ A voltage of 230V is applied to armature of a

machines in Fig 3 results in a full load armature currents of 205A. Assume that armature resistance

is 0.2. Find the back emf, net power and torque by assuming the rotational losses are 1445W at full load speed of 1750rpm.

[189V, 37.3kW, 203.5Nm]


230 , 205 ,

0.2 , 1445 , 1750

T a

a

V V I A

R P W n rpm

).

230 205(0.2)

189

a T a aa E V I R

V

60).

2

60(37.3 )203.5

2 (1750)

oo

o

Pc

N

kNm

) ,

189(205) 38.745

38.745 1445 37.3

o m m a a

m

o

b P P P P E I

P kW

P k kW

Abu Zaharin Ahmad

the concept of the series and shunt designs are combined.


Ea VT (dc

supply)

Ra ia

if

Rf1 M

iL Rf2

)( 2faaaT RRiEV

1

:

ffT

faL

RiV

iiinote


P is normally given


Where,



P=stray, mech etc

in

out

mm

oo

P

PEfficiency

N


N


N

P

,

2

60,

2

60,/

2

60

Pm= Ea ia

P

Pout Pin=VTiL

Pca=ia2Ra

Pcf1=if2Rf1

Pm

Pcf2=ia2Rf2

There is no direct connection between the armature and field winding resistance

DC field current is supplied by an independent source ◦ (such as battery or another generator or prime

mover called an exciter)

nKniKC

pnZE fffa

60

2

fff RiV

aaaT RiEV

Where p= no of pole pair

n= speed (rpm)

Z=no of conductor

=Flux per pole (Wb)

C= no of current/parallel path

=2p (lap winding)

=2 (wave winding)

KVL:

Circuit analysis:

La iinote :

Ea

Ra La ia

M

Rf

VT Vf Lf

If

PMDC is a dc motor whose poles are made of permanent magnets.

Do not require external field circuit, no copper losses

No field winding, size smaller than other types dc motors

Disadvantage: cannot produce high flux density, lower induce voltage

Torque –speed characteristic for shunt and separately excited dc motor


a

ff

a

ff

a

a

a

aa

aa

R

nIK

R

IVK

excitedseparatelyassame

n

E

R

EV

n

IE

IEtorqueDeveloped

22

,

2

2

,

22

a

ff

R

IVKc

2

Starting

torque

a

ff

R

nIKslope

2

22

nNL n n

m

=0 n=0

By referring to the Torque –speed characteristic for shunt and separately excited dc motor

note that, there are three variables that can influence the speed of the motor,

V

If Ra

Thus, there are three methods of controlling the speed of the shunt and separately excited dc motor,

i. Armature terminal – voltage speed control

ii. Field speed control

iii. Armature resistance speed control


a

ff

a

ff

R

nIK

R

IVK

22

22

Variables

i. Armature resistance speed control - Speed may be controlled by changing Ra

- The total resistance of armature may be varied by means of a rheostat in series with the armature

- The armature speed control rheostat also serves as a starting resistor.

- From -n characteristic,


a

ff

a

ff

start

R

nIKslope

R

IVKc

2

2

22Will be changed

Torque –speed characteristic


Ra1

nNL n n1

m

Ra2

Ra3

Ra1 < Ra2 < Ra3

n2 n3

Advantages armature resistance speed control: i. Starting and speed control functions may be combined in

one rheostat ii. The speed range begins at zero speed iii. The cost is much less than other system that permit

control down to zero speed iv. Simple method

Disadvantages armature resistance speed control : i. Introduce more power loss in rheostat ii. Speed regulation is poor (S.R difference nLoaded & nno

loaded) iii. Low efficiency due to rheostat


ii. Field Speed Control - Rheostat in series with field winding (shunt or separately

ect.) - If field current, If is varied, hence flux is also varied

- Not suitable for series field

- Refer to -n characteristic,

- Slope and nNL will be changed




If1 < If2 < If3

1 < 2 < 3

nNL3 n n1

m

n2 n3 nNL2

nNL1

Base speed

Advantages field speed control: i. Allows for controlling at or above the base speed

ii. The cost of the rheostat is cheaper because If is small value

Disadvantages field speed control : i. Speed regulation is poor (S.R difference nLoaded & nno

loaded)

ii. At high speed, flux is small, thus causes the speed of the machines becomes unstable

iii. At high speed also, the machines is unstable mechanically, thus there is an upper speed limit


iii. Armature terminal – voltage speed control - Use power electronics controller

- AC supply rectifier

- DC supply chopper

- Supply voltage to the armature is controlled

- Constant speed regulation

- From -n characteristic,

- C and nNL will be change

- Slope constant




nNL1 n n1

m

V1 < V2 < V3

n2 n3 nNL1 nNL3

Advantages armature terminal voltage speed control: i. Does not change the speed regulation

ii. Speed is easily controlled from zero to maximum safe speed

Disadvantages armature terminal voltage speed control : i. Cost is higher because of using power electronic

controller


A 120v shunt motor has the following parameters: Ra = 0.40 Ω, Rf= 120 Ω and rotational 240W. On full load the input current is 19.5A and the motor runs at 1200rpm. Determine:

a) Developed power

b) Output power

c) Output torque

d) Efficiency


Given VT = 120 V, Ra = 0.40 Ω, Rf= 120 Ω , Pu= 240 W, IL= 19.5 A, nr=1200 rpm

a) Developed power

• 1st find If =VT/Rf

• Ia=IL- If • Ea= VT – IaRa

• Pm= EaIa b) Output power

Pout = Pm- Pu

c) Output torque

To=60Po /2n d) Efficiency

Pin = VTIL

=(Po/Pin) *100


460 V dc motor drives as 50 hp and at 900 rmp. The stunt field resistance as 57.5 Ω and the armature resistance is 0.24 Ω of the motor efficiency is 82%, determine

a) the rotational loss

b) the new developed power if the motor speed is reduced to 750 rpm using armature resistance speed control

Note : 1 horse power = 746watt.


Given VT = 460 V , Po= 50 hp (due to word of drives as otherwise rated output) Ra = 0.24 Ω , Rf= 57.5 Ω , nr= 900 rpm, = 82% a) Pu?

• Po = 50hp x 746 W • =(Po/Pin) *100, hence Pin • Pin = VTIL, hence IL • If =VT/Rf

• Ia=IL- If • Ea= VT – IaRa

• Pm= EaIa • Pu = Pm- Pout


Ea=KKn ◦ Ea1 Kn1

◦ Ea2 Kn2

◦ For Ea1 / Ea2 Kn1/Kn2 ,then Ea2

Note that for armature resistance speed control Torque remains the same

𝜏1 = 𝜏2

60𝑃

𝑚

2𝜋𝑛1

= 60𝑃

𝑚2

2𝜋𝑛1

60𝐸

𝑎1𝐼

𝑎1

2𝜋𝑛1

= 60𝑃

𝑚2

2𝜋𝑛2


There are two factors affecting the performance of dc machine 1. Armature reaction

2. Armature inductance

Definition of armature reaction: 1. It is the term used to describe the effects of the armature

mmf on the operation of a dc machine as a "generator" no matter whether it is a generator or motor.

2. It effects both the flux distribution and the flux magnitude in the machine.

3. The distortion of the flux in a machine is called armature reaction

Two effects of armature reaction: 1. Neutral Plane Shift 2. Flux Weakening

Effect on flux distribution: Neutral plane shift ◦ When current is flowing

in the field winding, hence a flux is produced across the machine which flows from the North pole to the South pole.

◦ Initially the pole flux is uniformly distributed and the magnetic neutral plane is vertical

Effect on flux distribution: Neutral plane shift ◦ effect by the air gap on the

flux field causes the distribution of flux is no longer uniform across the rotor.

◦ There are two points on the periphery of the rotor where B= 0.

Effect on flux distribution: Neutral plane shift ◦ when a load connected to the

machines a resulting magnetic field produced in the armature

◦ If the armature is rotated at a speed by an external torque each armature coil experiences a change in flux t as it rotates.

◦ A voltage is generated across the terminals of each winding according to the equation e = t

Effect on flux distribution: Neutral plane shift ◦ Both rotor and pole fluxes

(flux produced by the field winding and the flux produced by the armature winding) are added and subtracted together accordingly

◦ The fields interact to produce a different flux distribution in the rotor.

◦ Thus, the flux on the middle line, between the two field poles, is no longer zero.

Effect on flux distribution: Neutral plane shift

The combined flux in the machine

has the effect of strengthening or weakening the flux in the pole. Neutral axis is therefore shifted in the direction of motion.

The result is current flow circulating between the shorted segments and large sparks at the brushes. The ending result is arcing and sparking at the brushes.

Solution to this problem: ◦ placing an additional poles on the

neutral axis or mid-point that will produce flux density component, which counter-acts that produced by the armature.

Effect on flux magnitude: Flux Weakening

Most machine operate at

saturation point When the armature reaction

happen, at location pole surface: ◦ The add of rotor mmf to pole

mmf only make a small increase in flux

◦ The subtract of rotor mmf from pole mmf make a large decrease in flux.

◦ The result is the total average flux under entire pole face is decreased.

◦ This is called Flux Weakening

d –flux decrease under subtracting section of poles

When rotor turns, thus we have inductance value, e1 = L(di/dt). Lat say current ia1.

That means, we have ability to store energy

If the machine is turn ‘off’, thus, e1 will decreased. This will affect the current as well. Say ia2.

When the machine is turn ‘on’ again, it will produce e2 while e1 is still inside. The current now is reversed direction from previous (decreasing) current.

Thus, it will cause sparking resulting the same aching problem caused by neutral plane shift.

Chapter4 dc motor

Engineering

Transcript of Chapter4 dc motor