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111/19/2011 ELCT708 DC Machine

Direct Current (DC) Machines

Dr. Mostafa Soliman

DC Machine

Part no. 2

DC Motor Characteristics

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DC Motors

Equivalent circuit.

The equivalent circuit of DC Motors (and Generators) has two

components:

Armature circuit: it can be represented by a voltage source and aresistance connected in series (the armature resistance). The armature

winding has a resistance, Ra.

The field circuit: It is represented by a winding that generates themagnetic field and a resistance connected in series. The field windinghas resistance Rf.

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DC Motor Equivalent circuit

Equivalent circuit of a separately excited dc motor

Equivalent circuit is similar to the generator only the currentdirections are different.

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DC Motor Equivalent circuit

The operation equations are: Armature voltage equation

brushaaat VRIEV

The induced voltage (counter or back emf) and motor

speed vs angular frequency

fa IKE 60

2 n

f

aabrusht

aabrushtaf

KI

RIVV

RIVVEIK

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Power Flow and Losses in

DC Motors

Pin = VtIL

I2R Losses

(Copper losses)

Core Losses

(Both cores)Vt = Terminal voltage

IL= Line current

Mechanical

Losses

Stray Losses

Pout

P(developed or converted) = EaIa

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The relationship between

the induced EMF and torque The developed power (EaIa) is the converted power from the

electrical domain to the mechanical domain.

afad

ff

aaad

daa

IIKIKT

IK

IKIET

TIE

circuit)magneticlinear(assuming

)(

Shaft torque or output torque is:

lossmechanicallosscorelossmechanicallosscore PPIEPPPPT aadloadload

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7/29711/19/2011 ELCT708 DC Machine

Classification of DC Motors

Separately Excited and Shunt Motors- Field and armature windings are either connected separate or in

parallel.

Series Motors- Field and armature windings are connected in series.

Compound Motors

- Has both shunt and series field so it combines features of series andshunt motors.

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8/29811/19/2011 ELCT708 DC Machine

Shunt DC Motors

The armature and field windings are connected in parallel. Almost constant speed operation.

By KVL around the outer loop:

aatfa

ff

a

brushaata

RIVnIkE

constII

kE

VRIVE

so,

neglectedisif,

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9/29911/19/2011 ELCT708 DC Machine

Torque developed by shunt motor

T

Ik

R

Ik

V

RIk

TIkV

Ik

TI

RIkV

RIEV

f

a

f

t

a

f

ft

f

a

aat

aaat

2

forsolving

so

but

so

Very important:

The field circuit has not to be open at any time, otherwise, the speed will increase

rapidly and the motor will be destroyed.

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Torque developed by shunt motor

IfVtandIf(hence ) are

constant, speed is

directly proportionalto the torque.

TIKR

IkV

f

a

f

t2

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Starting of Shunt DC Motors At the starting of a DC motor, Ea = 0, so:

To limitIa, a resistance is inserted in series with Ra

then removed after the development ofEa.

currenthighydangerousl0a

t

a

ata

R

V

R

EVI

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Series DC Motors The armature and field winding are connected in series.

High starting torque.

Ia = If

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Torque developed by series

motorBy KVL around the loop:

T

k

RR

k

Tk

V

k

TI

k

RR

kI

VRRIVIkIkE

IkIIkT

IIRRIVE

sat

asa

a

t

saatafa

afa

af

saata

1or

but

)(

)(

thatnoteweand

)(

2

Series motor cannot be

started with no-load

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Separately Excited DC Motors The armature winding supplies the load.

The field winding is supplied by a separate DC source whosevoltage is variable.

Good speed control.

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Compound DC Motors

Both series and shunt fieldsexist.

The series field produces s.

The series field produces sh.

If both s and sh are in the

same direction

cumulatively compound DC

motor.

If both s and sh are in

opposite direction

differentially compound DC

motor.

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Compound DC Motors

Differentially compound

Cumulatively compound:

Has higher starting torque than the

shunt motor but much lower than the

starting torque in case of series

motor.

Differentially Compound:

Unstable operation.

Not easy to start.

Not used commonly.

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Comparison of DC MotorsShunt Motors: Constant speed motor (speed regulation is very good).

Adjustable speed, medium starting torque. (TStart = 1.4 TFL) can be started

with no-load.

Applications: centrifugal pump, machine tools, blowers fans, etc.

Series Motors: Variable speed motor which changes speed drastically from one

load condition to another. It has a high starting torque. Must be started with

load.

Applications: hoists, electric trains, conveyors, elevators, electric cars, etc.

Compound motors: Variable speed motors. It has a high starting torque and the

no-load speed is controllable unlike in series motors. can be started with no-

load.

Applications: Rolling mills, sudden temporary loads, heavy machine tools, etc

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Speed Control of DC Motors

Speed can be controlled by varying:

1) Armature circuit resistance using an externalresistanceRA Ext.

2)IFcan be varied by using an external resistanceRadj inseries withRF to control the flux, hence the speed.

3) The applied voltage to the armature circuit resistance,if the motor is separately excited

f

aat

kI

RIV

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Separately DC Motor Speed Control

MCTR 908 Electric Drives 19

There are two main regions to control the speed of the separately excited DC motors.

Region 1: for speeds up to base speed Called Constant Torque Region.

Armature voltage is variable while keeping the field current constant.

Region 2: for speeds higher than the base speed Called Constant Power Region.

Field current is variable while the armature voltage is constant (at rated value).

S l DC M S d C l

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Armature Voltage Control (Constant Torque Region):

1. This method implies changing the voltage applied to the armature of the motor without

changing the voltage applied to its field. Therefore, the motor must be separately excitedto use armature voltage control.

2. This control method is used for speeds from zero up to the base speed.

3. This method cannot be used for speed higher than the base speed because the voltage

should not exceed the rated voltage.

Armature Voltage Control


4. In this region, the torque is kept

constant (at maximum value) andas a result,Ia and ,Ifare constant

also. (TdIaIf)

S l DC M S d C l

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Armature Voltage Control (Constant Torque Region):

a. Increasing the armature voltage VA increases the armature current (Ia = (VA - Ea)/Ra);

b. Increasing armature currentIa

increases the induced torque Tind

(Tind

= KIa);

c. Induced torque Tindis now larger than the load torque Tloadand, therefore, the speed

increases.

d. Increasing speed increases the internal generated voltage (EA = K);

e. IncreasingEa decreases the armature currentIa

f. DecreasingIa decreases the induced torque until Tind= Tloadat a higher speed

(balanced condition)



S l DC M S d C l

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Field Weakening Control (Constant Power Region):

1. This method is used to control the speed for values higher than the base speed.

2. In this region, the armature voltage and current are kept constant.

3. The power developed by the motor (Td) remains constant.

4. In order to keep the power constant, the torque has to be decreased as it will be inversely

proportional to the speed (Td1/).

Field Weakening


5. In this region, the torque decreases

with increasing the speed and as aresult, If will decrease asIa is

constant (TdIaIf)

Field Weakening


S l DC M S d C l

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Field Weakening Control (Constant Power Region):

a. Increasing field resistanceRf decreases the field current (If= Vf/Rf);

b. Decreasing field currentIf decreases the flux ;

c. Decreasing flux decreases the internal generated voltage (Ea = K);

d. DecreasingEa increases the armature current (Ia = (VTEa)/Ra);

e. Changes in armature current dominate over changes in flux; therefore, increasingIa

increases the induced torque (Tind= KIa);

f. Increased induced torque is now larger than the load torque Tloadand, therefore, the

speed increases;

g. Increasing speed increases the internal generated voltageEa;

h. IncreasingEa decreases the armature currentIa

i. DecreasingIa decreases the induced torque until Tind= Tloadat a higher speed .

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Example 1

Q1) A 240 V, shunt DC motor takes an

armature current of 20 A when running at

960 rpm (full load). The armature resistanceis 0.2 . Determine the no load speed if the

no load armature current is 1 A.

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Example 1

rpm45.975960*236

8.239.).(

.).(

.).(

.).(

.).(

constantisIassuming

V8.2392.0*1240load)(no

V2362.0*20240load)(full

f

lnnlFn

lnn

lFE

lnE

E

E

ERIV

c

c

c

c

caat

loadfullload,no F.l.n.l.

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Example 2

Q2) A 120 V shunt motor has the following

parameters: Ra = 0.4, RF = 120 and

rotational (core, mechanical and stray) losses are240 W. On full load, the line current is 19.5 A and

the motor runs at 1200 rpm, find:

The developed power

The output power, and

The output torque.

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Example 2

N.m67.141200

60

2

1843

)(

watt18432402083(b)

watt1.20835.18*6.112*V6.112)4.0(*)5.18(120

A5.1815.191120

120

)(

out

out

devout

aadev

aata

af

fLa

P

Tc

lossesrotationalPP

IEPRIVE

IAI

IIIa

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Example 3

Q1) A DC series motor is operated at fullload from a 240 V supply at a speed of

600 rpm. The Ea is found to be 217.2V at a line current of 38 A, find:

a) The armature resistance assuming theseries field resistance is 0.2 .

b) Find the no-load speed given that theno-load current is 1 A.

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Example 3

rpm151,25

V6.2396.0*1240

:loadnoAt(b)

4.0

6.038

2.217240)(*)(

2

11

22

1

2

nIn

In

E

E

E

R

RR

RRIEVa

a

a

a

a

a

a

fa

faaat

Date post:	03-Apr-2018
Category:	Documents
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