DC MotorSlide3

7/29/2019 DC MotorSlide3

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DC Motor

Dr. Mohammed Moshiul Hoque


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Basic Principle

If a conductor is placed in a magnetic field andcurrent is allowed to flow through the conductor,the conductor will tend to move. A force isexerted on the conductor because it is carrying

current and it is located in a magnetic field The force (F) exerted on a conductor depends

upon the flux density of the magnetic field (B),the length of the conductor in the magnetic field

(L), and the magnitude of the current in theconductor (I).

F=0.886 BLI x 10-7 lb


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Basic Principle When its field magnets are excited and its

armature conductors are supplied with currentfrom the supply mains, they experience a forcetending to rotate the armature.

Armature conductors under N-pole areassumed to carry current downward (crosses)and those under S-poles, to carry current

upwards (dots). By applying Flemings LHR, the direction of the

force on each conductor can be found.

Each conductor experiences a force F whichtends to rotate the armature in anticlockwisedirection.

These forces collectively produce a drivingtorque which sets the armature rotating.

F

N SForce

Use LHR to determine the

direction of forceIndex: direction of flux from N to S poles

Middle: direction of current

Thumb: Direction of force


5/43

Back/Counter EMF

5-hp

240 V

20.4 A

Ra=0.97

Ohms Law, I = V/R

=240/0.97

=248A

~ appx. 10 times of the full loadcurrent (20.4A)

? ?Back/Counter EMF produced by the armature


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Back EMF

N S N S

(a) Force motor (LHR) (b) Motion generator (RHR)

-Generator principle: A voltage would be induced if the conductor cutlines of flux.

-The movement of a conductor through the magnetic field could be

performed manually, or by some external prime mover or By the

conductor itself.-When current flows through a conductor, a force is exerted on the

conductor that causes the conductor to move in the magnetic field.

-Movement of the conductor in the magnetic field produces

generator action, and a voltage is induced in the conductor in such a

direction as to oppose the applied voltage.


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Back/Counter EMF

In as much as twocurrents are inopposite directions,the two voltages are

in opposition.

Hence the voltageinduced is oppositeto the applied voltage

and therefore isreferred to theback/counter emf.

aR

bE

tVaI

Back/Counter

emf


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Eb can be acts as a governor

Eb depends on the N

N, Eb, Ia N , Eb, Ia

Ia, Torque will be large

Eb acts like a governor

It makes a motor self-regulating so that it drawsas much as current as is just necessary.

aR

bE

t

V

aI volt

bE

A

PZN


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Voltage Equation of a Motor

)(GeneratorE

(Motor)I

g

a

tVaI

aat

aabt

bta

RIV

RIEV

EVR

aR

bE

(Vt > Eb)

(Eg > Vt)

-Under normal condition, the back emf can never be greater than the terminal voltage

-This is understandable when it is realized that the applied voltage causes the

armature to rotate, which in turn produces the back emf.

-Eb=Vt, Ia=0, armature would cease to rotate, and Eb = 0.

@The back emf can never equal the applied voltage.

efficiency

motorhigher the,Ehigher theb

2

V

E

VI

IE

RIIEVI

b

a

ab

aaaba

Electrical input

Electrical equivalent

Mechanical power

developed

Armature cu

loss


10/43

Power Stages

All electrical power supplied to a motor is convertedinto mechanical power. Some of the power isdissipated in the field windings, and some in thearmature circuit.

The remainder is available for conversion intomechanical power.

The total power supplied to the motor = VtIL A portion of this power is consumed by the shunt

filed = Vt If. The difference between the power delivered to the

motor and the power consumed in the shunt field isthe power delivered to the armature.


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Power Stages

Total power

furnished to

motor = VtIL

Power toarmature =

Vt(IL-If)=VtIa

Power to

shunt field

= VtIf

Power

dissipated

in armatureresistance =

Ia2Ra

Electrical power

converted to

mechanical

power, Pm = VtIa Ia2Ra=EbIa

A

B

Iron/Friction Loss

Motor OutputC


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Power Stages

]I-I[IIVP

IIVP

IVIVP

fLaata

fLta

ftLta

The power delivered to the armature is not entirely converted

into mechanical power.

Part of this is dissipated in the armature resistance and

remainder is converted into mechanical power


13/43

Power Stages

]RI-V[EEIP

RIVIP

RIIVP

PowerMechanicaltoConvertedPowerElectrical

RIWarmature,inlossPowerIaVParmature,todeliveredPower

aatbbam

aatam

a2aatm

a

2

aa

ta

The electrical power converted into mechanical power is equal

to the product of the Back/counter emf and the armature current.

This mechanical power is not available outside of the motor, only

inside


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Condition for Maximum Power

aRaIaVImP

aR

aI

aVI

aIb

E

aRaIaIbEaVI

2

2

2

Gross mechanical power developed by a motor

2

2

02

0)2

(

tV

aR

aI

t

V

a

R

a

I

a

R

a

I

t

V

aR

aI

aI

t

V

adI

d

m

P

adI

d

2

2tV

2tV

tVknow,weAS

tV

bE

tV

b

E

tV

b

E

aRaIbE

Thus, gross mechanical power developed

by a motor is maximum when back emf is

equal to half of the applied voltage.


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Torque

Means turning or twistingmoment of a force about an axis.

If the conductor rotates in themagnetic field, then the currentthrough the conductor would

exert a force the conductor would tend to

rotate in a clockwise direction.

This tendency to produce rotation

is known as torque.Measured by the product of theforce and the radius at which thisforce acts.

SN

r

+

F

Center of Rotation


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Torque

P = T watt = 2N/60 [if N is in r. p. m] P =T x 2N/60 = NT x 2 /60 P = NT/9.55 watt


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Armature Torque of a Motor

P = Ta x [Ta armature torque running at rps] Pa = Ta x2N We know, electrical power converted into

mechanical power, P = Eb Ia watt Pa = PTa x2N = Eb Ia = Z N x (P/A) x Ia (**)

m-NZI0.159a

A

P

A

PZIT

aa

a

.2

1


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(1)aIaT

m-NaZI0.159

A

P

a

1. In case of a series motor, is directly proportional to Ia

(1)Ta Ia. Ia Ia22. For shunt motore, is practically constant(1)Ta K. Ia Ia


21/43


(**) Ta x2N = Eb Ia

mNI

9 .55 aa

N

IE

N

IET

N

IET

abab

a

ab

a

.2

60

602

2


22/43

Shaft Torque

The whole of the armature torque is not

available for doing useful work-

because a certain percentage of it is required for

supplying iron & friction losses in the motor.

The torque which is available for doing useful

work is known as shaft torque.

It is so called because it is available at the shaft.


23/43

Shaft Torque

mNN

outputT

N

output

2T

2N/6

output

2

output

T

2TOutput

wattToutput

sh

sh

sh

sh

sh

55.9

60

Tlost = Ta-Tsh

Due to ironand friction

losses


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Torque and Speed

(2)K

RIV

K

E

N

(1)NKE

aatb

b

, N & Ta

It cannot be so because torque tends to produce rotation.oIf torque increases, motor speed must increase rather

decrease.

oStrange ??????????


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Torque and Speed

1. A reduction in flux would reduce the back emf

Eb = KN [, Eb]2. The reduction in back emf would produce an

increase in armature current

3. A reduction in flux would reduce the torque and anincrease in the armature current would increasethe torque

a

bt

a

R

EVI

{Eb, Ia}

T = KIa[, T; Ia, T]


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Torque and Speed

Since the increase in the armature current isgreater than the decrease in flux, the resultingtorque would increase.

4. The increased torque will increase the speed ofthe machine and therefore, the magnitude of theback emf will increase [T, N , E ].

5. The increase in the speed and back emf bringsabout a reduction in the armature current and

torque. Reduction is such as to meet therequirements of the new conditions at a newconstant speed.


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Motor Characteristics

Torque and Armature Current (Ta vs. Ia)

Speed and Armature Current (N vs. Ia)

Speed and Torque (N vs. Ta)


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Ta vs. Ia (Series Motor)

TaIa Field windings carry the

armature current (Ia = If)

Up to saturation, Ia Before Saturation,TaIa Ia. Ia = Ia2 At light loads, Ia and hence is small. But Iaincreases,Ta increases as the square

of the current.


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Ta vs. Ia (Series Motor)

After Saturation, isalmost independent of Ia

TaIa= Ia Characterizes becomes

straight line

Tsh < Ta due to stray losses

Use where huge startingtorque (hoists and electrictrains)


31/43

N vs. Ia (Series Motor)

K

EN

b

bE

N

Ia, , N, variesinversely

When load is heavy, Ia is

large. N is low

When Ia falls to a small

value, N

(2)K

RIV

K

EN aatb


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N vs. Ta (Series Motor)

Also known as

mechanical

characteristics.

Speed is high,

torque is low and

vice-versa.

TaIa

b

EN

, N, TaN , Ta


34/43

Ta vs. Ia (Shunt Motor)

is constant

Since a heavy stating

load will need a heavy

starting current, shunt

motor should never be

started on heavy load.

TaIa Ia


35/43

N vs. Ia (Shunt Motor)

is constant, NEb Eb is also practically constant,

N is constant (most cases)

Eb & with Ia/IL Eb decreases slightly more

than so that the whole thereis some decrease in N Suitable: driving shafting, m/c

tools, lathes, wood-workingm/c

bEN


36/43

N vs. Ta (Shunt Motor)

N is constant with Ta

Ta

TaIa IaNEb


37/43


38/43

Compassion of Shunt and Series

Motors

Self Study

Page (842-843)

Book: B. L. Thereja


39/43

Problem

A 4-pole, 240 A, wave connected shunt motor gives 11.19KW when running at 1000 rpm and drawing armature &field currents of 50 A and 1.0 A respectively. It has 540conductors. Its resistance is 0.01. Assuming a drop of 1volt per brush. Find (a) total torque (b) useful torque (c)useful flux/pole (d) rotational losses, and (f) efficiency

Given,

P = 4

Vt = 240 V

Po = 11.19 KW = 11190 W

N = 1000 rpm

Ia = 50 AIf= 1 A

Z = 540

Ra = 0.1

VB = 1 V/brush = 1* 2 = 2 V

A = 2 (wave connected)

T t = ?

Tsh = ?

u = ?Pr = ? = ?


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41/43

Problem

W460

11190-685011,

PLoss,Rotational

W685011,

350-12000

100250-12000

PDeveloped,Power

WPloss,Brush

WP

W(d)

R

d

B

cu

0

22

100502

2501.050

000,1250240

PP

lossesP

IV

RI

IVP

d

in

aB

aa

atin

91.4%

P

PNow,

WPhave,We

W

51240

III

Pinput,Total(e)

in

o

o

faL

in

10012240

11190100

11190

12240

][)150(240

LtIV


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DC MotorSlide3