6.0 Basic DC and AC Motors

8/13/2019 6.0 Basic DC and AC Motors

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Purpose of “DC/AC Motor”

The purpose of a AC/DC Motor is to

Convert Electrical Energy into Mechanical Energy

Electricalenergy

AC/DCmotor

Mechanicalenergy



• There are two main types of electric motors

• There are direct current or DC and alternating

current or AC motors

• The reference of DC or AC refers to how the

electrical current is transferred through and

from the motor.



BASIC CONSTRUCTION

• Consist of stationary part (stator)

• Rotating part (rotor) connected to shaft that

couples the machine to its mechanical load

•Usually stator and rotor are made of iron tointensify the magnetic field



Armature and field windings

• A machine may contain several sets of

windings (commonly armature and field)• The purpose of the field winding is to set up

the magnetic field required to produce torque

• The armature windings carry currents thatvary with mechanical load

• In DC motor, the field winding is on the stator

while armature is on the rotor



ROTATING DC MACHINES

• Consist of

– Rotor (rotating part)

– Stator (stationary part)

– Brushes

– Commutator

– Shaft

– Field magnet



Induced EMF and commutation



• Rotation of the shaft

• When we supply the specified

voltage to a motor, it rotates the

output shaft at some speed.This rotational speed or angular

velocity, is typically measured

in revolutions/minute (rpm)

• Torque is the product of Force x Lever Arm

Length (Radius)

• Clockwise and Counter-Clockwise efforts are

distinguished by differences in sign (+ or -)

• The quantitative measure of the tendency of

a force to cause or change rotational motion

is called torque



Equivalent Circuit of the DC Motor

• The field circuit is represented by resistance RF and

inductance LF in series• Consider steady state operation in which current are

constant, and neglect the inductance because itbehaves as a short circuit for dc current

• Thus for DC field , F F F

I RV



• The voltage EA shown in the equivalent circuit

represents the average voltage induced in the

armature due to the motion of the conductors

relative to the magnetic field.

• The resistance RA is the resistance of the

armature windings plus the brush resistance.



DC MOTOR

Type Power

range

(hp)

Rotor Stator Comments and

applications

Wound field Shunt

connected

10-200 Armature

winding

Field

winding

Industrial applications,

grinding, machine tools

Series

connected

High torque at low

speed; dangerous if notloaded; drills,

automotive starting

motor

Compound

connected

Traction motors

Permanent

magnet field

1/20-

10

Armature

winding

Permanent

magnet

Servo applications,

machine tools,

computer peripherals,

automotive fans,

window motors



SHUNT CONNECTED DC MOTOR

• The field current is in parallel with the armature

• The field circuit consist of rheostat having aadjustable resistance (Radj) in series with field coilthat can be used to adjust motor speed

• If the adjustable resistance is increase whileholding the source voltage constant, the speedwould also increase

• Also, if the voltage source is increase the fieldcurrent could be hold constant to increase thespeed by increasing the value of adjustableresistance



• The machine is supplied by constant voltage

source, VT

• Has very high starting torque and draws very

large starting currents

• Usually, resistance inserted in series with

armature during starting to limit the current

to reasonable levels



Mechanical shaft

speed

Developed

torque

Induced

voltage

The armatureresistance



SEPARATELY EXCITED DC MOTORS

• Similar to shunt-connected motor except

different source are used for the armature and

field

• Separate as reason to be able to control speed

by varying one of these two sources



PERMANENT MAGNET MOTORS

• The field is supplied by magnets mounted on the statorrather than field coil

• Advantages; – No power required to establish the field-leading to better

efficiency – PM motor can be smaller than equivalent machine with

field winding

• Disadvantages; – The magnet can become demagnetized by overheating/

excessive armature current – Torque produced per ampere of armature current is

smaller



SERIES CONNECTED DC MOTORS

• The field winding is in series with armature

• Has moderate starting torque and startingcurrent

•Speed automatically adjust over a large range asthe load torque varies

• Because it slows down for heavier load, its outputpower is more nearly constant than other typesof motor

• Advantageous because the motor can operatewithin its maximum power rating for a widerange of load torque



• Ex: starter motor in automobiles, when engine is cold

the starter motor operate at lower speed, when

engine warm the starter spin faster.in either case, the

currrent drawn from battery remains withinacceptable limit



ADVANTAGES OF DC

MOTOR:

DISADVANTAGES OF DC

MOTOR:

– Ease of control – Deliver high starting

torque

– Near-linear

performance

– High maintenance – Large and expensive

(compared to induction motor)

– Not suitable for high-speed

operation due to commutatorand brushes

- Not readily available for use

at home



cos3 rmsrmsin V I P

mout out T P

60

2 mm n 7376.0

metersnewton pounds foot T T

746

wattshorsepower

P P

%100in

out

P

P

%100

load full

load full load noation speedregul



• The induced armature voltage is given by

•K is a machine constant that depends on thedesign parameters of the machine

• is the magnetic flux produced by each stator

pole

• is the angular velocity of the rotor

m A K E

m



• There are 3 types of power in DC motor which

include Pin, Pdev and Pout

A Adevmdev E I T P

LT in I V P

mout out T P

Radj

is a rheostat used to adjust motor speed



• The graph shows atorque/speed curve of atypical D.C. motor.

• Note that torque is

inversely proportional tothe speed of the outputshaft. In other words, thereis a tradeoff between howmuch torque a motor

delivers, and how fast theoutput shaft spins.

)( mT

A

dev K V R

K T



• Motor characteristics are frequently given as

two points on this graph:

– The stall torque represents the point on the graph

at which the torque is a maximum, but the shaft is

not rotating

– The no load speed is the maximum output speed

of the motor (when no torque is applied to theoutput shaft)

K

V T

T

A

V R

K



• Several methods can be used to control the

speed of dc motors:

– Vary the voltage supplied to the armature circuit

while holding the field constant.

– Vary the field current while holding the armature

supply voltage constant.

– Insert resistance in series with the armaturecircuit.



A certain 5-hp three-phase induction motor operates from

a 440-Vrms (line to line) three-phase source and draws a

line current of 6.8-Arms at a power factor of 78% lagging

(cos θ = 0.78) under rated full load conditions.

The full load speed is 1150 rpm. Under no-load conditions,the speed is 1195 rpm, and the line current is 1.2 Arms at a

power factor of 30% lagging.

Determine :

(a) Power loss(b) Efficiency

(c) The input power with no-load

(d) Speed regulation



A group of students have been assigned to carry out an

experiment using a new dc motor through Pout= 37.285 kW

that operates from a 220-Vrms dc source with losses of 3350

W under rated full-load conditions. The full-load speed is

1150 rpm. Under no-load conditions, the speed is 1200rpm.

Determine :

(a) Source current

(b) Efficiency with full load(c) Speed regulation



A 50-hp shunt-connected dc motor has the machine constant

KΦ=2.228. The dc supply voltage is VT=240V, the armatureresistance is RA= 0.065Ω, the field resistance is RF=10Ω, and

adjustabler esistance is Radj=14Ω. At speed of nm=1200 rpm, the

rotational loss is Prot=1450 W. If this motor drives a hoist that

demands a torque of Tout=250 Nm independent of speed,

determine:

(a) Filed current

(b) Trot and Tdev

(c) Armature current and voltage

(applying Kirchhoff’s voltage law) (d) Angular velocity of rotor

(e) Motor speed

(f) Efficiency

With an aid of the equivalent circuit diagram.



Phase - defines the type of electrical powerbeing supplied to the motor

Each phase is displace 120°



Electro-Magnets

Stator

Rotor



Time 5

Time 4 Time 6

Time 1 Time 7

Time

0° 60° 120° 180° 240° 300° 360°

A

C

B

Time 2

Time 3 Time 5 Time 7

N S

B1C2

A1

A2

B2 C1

N S

NS

B1

A1

A2

B2 C1

N

B1C2

A1

A2

B2 C1

B1C2

A1

A2

B2 C1

B1C2

A1

A2

B2 C1

B1C2

A1

A2

B2 C1

B1C2

A1

A2

B2 C1

S

N

S

N

N

N

S

S

N

N

S

S

N

N

S

S

N

N

S

SC2

S



f is Applied Frequency

P is magnetic poles that rotate at synchronous speed

Synchronous Speed - The speed of the stator’s magnetic field rotation

P

f

n s120



Synchronous Speed (60 Hz) = 7200

No of Poles

P ns

2 36004 1800

6 1200

8 900

10 720

12 600

Synchronous speed versus number of poles for f=60 Hz



s

m s

n

nn s

•Motor slip, s is defined to be the relative speed as a

fraction of synchronous speed

•Slip s varies from 1 when rotor is stationary to 0 whenthe rotor turns at synchronous speed

sn

is mechanical speed of the rotormn

is synchronous speed



O rpm 900 rpm 1750

Motor

Speed

1800

Sync

Speed

0.0278

Slip



• In figure, the reflected resistance is split intotwo parts as follows;

• The power delivered, Pdev to the resistance isthe part that is converted to mechanical form.

• The equivalent circuit shown is one of thethree phase in AC motor, thus the totaldeveloped power is

r r

r

R s

s

R s

R

'

1

'

'

2)'('1

3 r r dev I R s

s P



• The power delivered to the rotor resistance R’r

is converted to heat• Generally we refer to I2R losses as copper

losses. The total copper losses in the rotor is

• The stator copper loss is

• The input power from the three phase source

is

• is the power factor

2)'('3 r r r I R P

23 s s s I R P

cos3 s sin V I P

cos



• Part of the developed power is lost to friction

and windage.

• Another loss is core loss due to hysteresis and

eddy currents.

• The output power is developed power minus

rotational loss;

• Efficiency of the machine

rot devout P P P

%100in

out

P

P



• The developed torque is

• The power Pag

that crosses the air gap into the

rotor is delivered to the rotor resistances

m

dev

dev

P T

devr ag P P P

2)'('13 r r ag I R s

P

ag dev P s P )1(



•

• We also have . Using thisequation to substitute in above equation, we

get;

• For speed to increase from a standing start,the initial torque or starting torque producedby the motor must be larger than the torquerequired by the load

m

ag

dev

P sT

)1(

sm s )1(

s

ag dev P T

E l 2 I d ti t



Example 2: Induction-motor

performance

• A certain 30-hp four pole 440-V-rms 60-Hzthree phase delta-connected induction motorhas

• Under load, the machine operates at 1746

rpm and has rotational losses of 900W. Findthe power factor, the line current, the outputpower, copper losses, output torque andefficiency

0.2 s X

2.1 s

R

50m X 8.0'r X

6.0'r R



• Factors to be considered; – Electrical sources available

– Output power required

– Load torque versus speed

– The service-life requirements – Efficiency

– Speed regulation

– Starting current

– Desired operating speed – Acceptable frequency of maintenance

– Ambient temperature



1. Describe briefly operation of DC motor withthe aid of sketch diagram

2. Describe shunt-connected, separately-excitedand series-connected type of DC motor withthe aid of sketch diagram

3. How speed can be controlled in shuntconnected motor?

4. Determine the synchronous speed for motor

6.0 Basic DC and AC Motors

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