1

VI. DC Machines

Introduction

• Separately-excited• Shunt• Series• Compound

DC machines are used in applications requiring a wide range of speeds by means of various combinations of their field windings

Types of DC machines:

2

DC Machines

Motoring

Mode

Generating

Mode

In Motoring Mode: Both armature and field windings are excited by DC

In Generating Mode: Field winding is excited by DC and rotor is rotated externally by a prime mover coupled to the shaft

Basic parts of a DC machine

1. Construction of DC Machines

3

Construction of DC Machines

Elementary DC machine with commutator.

Copper commutator segment and carbon brushes are used for:

(i) for mechanical rectification of induced armature emfs

(ii) for taking stationary armature terminals from a moving member

(a) Space distribution of air-gap flux density in an elementary dc machine;

(b) waveform of voltage between brushes.

Average gives us a DC voltage, Ea

Ea = Kg f r

Te = Kg f a= Kd If a

4

(a) Space distribution of air-gap flux density in an elementary dc machine;

(b) waveform of voltage between brushes.

Electrical Analogy

2. Operation of a Two-Pole DC Machine

5

Space distribution of air-gap flux density, Bfin an elementary dc machine;

One pole spans 180º electrical in space )sin(peakf BB

Mean air gap flux per pole: poleperavgpoleavg AB /

0 )sin( dABpeak

0 )sin( rdBpeak

Aper pole: surface area spanned by a pole

rBpeakpoleavg 2/ For a two pole DC machine,

Space distribution of air-gap flux density, Bfin an elementary dc machine;

One pole spans 180º electrical in space

)sin(peakf BB

Flux linkage a: )cos(/ poleavga N : phase angle between the magnetic axes of the rotor and the stator

0)( tt rwith 0= 0 )cos(/ tN rpoleavga

)sin(/ tNdt

de rpoleavgr

aa

0 )(1

dtteE aa

poleavgra NE /2

For a two pole DC machine:

6

rfga KE In general: Kg: winding factorf: mean airgap flux per poler: shaft-speed in mechanical rad/sec

602 r

rn nr: shaft-speed in revolutions per minute (rpm)

DC machines with number of poles > 2

120602rr

elecPnnP

f P: number of poles

rBP peakpoleavg 22

/

rfgpoleavgra KNP

E

/2

2

A four-pole DC machine

Schematic representation of a DC machine

7

Typical magnetization curve of a DC machine

Torque expression in terms of mutual inductance

d

dMii

d

dLi

d

dLiT fa

afa

af

fe 22

2

1

2

1

d

dMiiT faafe cosMM fa

afe iiMT ˆ

Alternatively, electromagnetic torque Te can be derived from power conversion equations

elecmech PP

aame IET mfga KE

amfgme IKT

afge IKT

8

In a linear magnetic circuit

affge IIKKT fff IKwhere

(a) separately-excited (b) series

(c) shunt (d) compound

Field-circuit connections of DC machines

9

Separately-excited DC machine circuit in motoring mode

mfga KE 2 a

CpK a

g

p : number of polesCa : total number of conductors in armature windinga : number of parallel paths through armature winding

ta VE

Te produces rotation (Te and m are in the same direction) 0and 0 ,0 memech ωTP

windage and frictionproducedpower output

poweroutput grossor power hanicalelectromec internal

wfoutmech PPP &

Separately-excited DC machine circuit in generating mode

ta VE

Te and m are in the opposite direction 0 and 0 ,0 memech ωTP

Generating mode– Field excited by If (dc)– Rotor is rotated by a mechanical prime-mover at m.– As a result Ea and Ia are generated

10

Separately-Excited DC Generator

rIEV aaat mfga KE where

LLt RIV also aL II where

3. Analysis of DC Generators

Terminal V-I Characteristics

Terminal voltage (Vt) decreases slightly as load current increases (due to IaRa voltage drop)

11

Terminal voltage characteristics of DC generators

Series generator is not used due to poor voltage regulation

Shunt DC Generator (Self-excited DC Generator)

– Initially the rotor is rotated by a mechanical prime-mover at mwhile the switch (S) is open.

– Then the switch (S) is closed.

12

When the switch (S) is closed

Ea= (ra + rf) If Load line of electrical circuit

Self-excitation uses the residual magnetization & saturation properties of ferromagnetic materials.

– when S is closed Ea = Er and If = If0– interdependent build-up of If and Ea continues– comes to a stop at the intersection of the two curve

as shown in the figure below

Solving for the exciting current, If

mga KE f fff IK

mda IKE f fgd KKK

where

where

Integrating with the electrical circuit equations

fff

ff irrdt

diLLiK aamd

Applying Laplace transformation we obtain

f0ffffff )()()( ILLsIrrssILLsIK aaamd

So the time domain solution is given by

tLL

Krr

a

mda

eIti

f

f

f0f )(

13

Let us consider the following 3 situations

tLL

Krr

a

mda

eIti

f

f

f0f )(

(iii) (ra + rf) < Kdm

(ii) (ra + rf) = Kdm

(i) (ra + rf) > Kdm

0)(lim f

tit

Two curves do not intersect.

0ff )( Iti

Self excitation can just start

Generator can self-excite

Self-excited DC generator under load

Ea= ra Ia + rf If

Load line of electrical circuit

Ia= If + IL Vt= Ea - ra Ia = rf If

Ea= (ra + rf) If + ra IL

14

Series DC Generator

)( saaat rrIEV mfga KE where

LLt RIV also saL III and

Not used in practical, due to poor voltage regulation

Compound DC Generators

ssaaat rIrIEV mfga KE where

LLt RIV also sL II and

(a) Short-shunt connected compound DC generator

sf IIIa and

15

(b) Long-shunt connected compound DC generator

saaat rrIEV mfga KE where

LLt RIV also fIII aL and sIIa and

Types of Compounding

(i) Cumulatively-compounded DC generator (additive compounding)

mmf field

series

mmf field

shunt

f

mmffield

sd FFF

for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)

sd f

(ii) Differentially-compounded DC generator (subtractive compounding)

sd FFF f

for linear M.C. (or in the linear region of the magnetization curve, i.e. unsaturated magnetic circuits)

sd f

Differentially-compounded generator is not used in practical, as it exhibits poor voltage regulation

16

Terminal V-I Characteristics of Compound Generators

Above curves are for cumulatively-compounded generators

Magnetization curves for a 250-V 1200-r/min dc machine. Also field-resistance lines for the discussion of self-excitation are shown

17

Examples

1. A 240kW, 240V, 600 rpm separately excited DC generator has an armature resistance, ra = 0.01 and a field resistance rf = 30. The field winding is supplied from a DC source of Vf = 100V. A variable resistance R is connected in series with the field winding to adjust field current If. The magnetization curve of the generator at 600 rpm is given below:

310300285260250230200165Ea (V)

65432.521.51If (A)

If DC generator is delivering rated load and is driven at 600 rpm determine:

a) Induced armature emf, Ea

b) The internal electromagnetic power produced (gross power)

c) The internal electromagnetic torque

d) The applied torque if rotational loss is Prot = 10kW

e) Efficiency of generator

f) Voltage regulation

2. A shunt DC generator has a magnetization curve at nr = 1500 rpm as shown below. The armature resistance ra = 0.2 , and field total resistance rf = 100 .

a) Find the terminal voltage Vt and field current If of the generator when it

delivers 50A to a resistive load

b) Find Vt and If when the load is disconnected (i.e. no-load)

18

Solution:

a) b)

NB. Neglected raIf voltage drops

3. The magnetization curve of a DC shunt generator at 1500 rpm is given below, where the armature resistance ra = 0.5 , and field total resistance rf = 100 , the total friction & windage loss at 1500 rpm is 400W.

a) Find no-load terminal voltage at 1500 rpm

b) For the self-excitation to take place

(i) Find the highest value of the total shunt field resistance at 1500 rpm

(ii) The minimum speed for rf = 100.

c) Find terminal voltage Vt, efficiency and mechanical torque applied to

the shaft when Ia = 60A at 1500 rpm.

19

Solution:

a) b) (i)

b) (ii)

• Series DC motor• Separately-excited DC motor• Shunt DC motor• Compound DC motor

DC motors are adjustable speed motors. A wide range of torque-speed characteristics (Te-m) is obtainable depending on the motor types given below:

4. Analysis of DC Motors

20

(a) Series DC Motors

DC Motors Overview

(b) Separately-excited DC Motors

21

(c) Shunt DC Motors

(d) Compound DC Motors

22

(a) Series DC Motors

DC Motors

)( saaat rrIEV

mfga KE

sa II

The back e.m.f:

Electromagnetic torque: afge IKT

Terminal voltage equation:

Assuming linear equation: sff IK

)( saaat rrIEV fg

saat

fg

am K

rrIV

K

E

sa II

afge IKT sff IK

2ade IKT

adfg IKK

asfge IIKKT

, mfga KE

ad

saatm IK

rrIV adfg IKK

saatmada rrIVIKE mfga KE

samd

ta rrK

VI

22

samd

tde

rrK

VKT

2 ade IKT

23

22

samd

tde

rrK

VKT

2

1

meT

thus

Note that: A series DC motor should never run no load!

m 0 eT overspeeding!

(b) Separately-excited DC Motors

aaat rIEV

mfga KE The back e.m.f:

Electromagnetic torque: afge IKT

Terminal voltage equation:

Assuming linear equation: fff IK

24

, mfga KE

aaat rIEV

fg

aemfgt K

rTKV

afge IKT

2fg

aem

fg

t

K

rT

K

V

efg

a

fg

tm T

K

r

K

V2

elm TK 0

No load (i.e. Te = 0) speed: fg

t

K

V

0

Slope: 2fg

al

K

rK

very small!

Slightly dropping m with load

(c) Shunt DC Motors

aaat rIEV

mfga KE The back e.m.f:

Electromagnetic torque: afge IKT

Terminal voltage equation:

Assuming linear equation: fff IK

25

, mfga KE

aaat rIEV

fg

aemfgt K

rTKV

afge IKT

2fg

aem

fg

t

K

rT

K

V

efg

a

fg

tm T

K

r

K

V2

elm TK 0

No load (i.e. Te = 0) speed: fg

t

K

V

0

Slope: 2fg

al

K

rK

very small!

Slightly dropping m with load

Same as separately excited motor

elm TK 0

Note that: In the shunt DC motors, if suddenly the field terminals are disconnected from the power supply, while the motor was running,overspeeding problem will occur

m 0 f

mfga KE Ea is momentarily constant, but f will decrease rapidly.

so overspeeding!

26

Motor Speed Control Methods

Shaft speed can be controlled byi. Changing the terminal voltageii. Changing the field current (magnetic flux)

(a) Controlling separately-excited DC motors

afge IKT

aaat rIEV

i. Changing the terminal voltage

elm TK 0fg

t

K

V

0

et TV , 0

27

afge IKT

ii. Changing the field current

elm TK 0fg

t

K

V

0

eTI , , 0ff

(linear magnetic circuit)fff IK

Shaft speed can be controlled byi. Adding a series resistanceii. Adding a parallel field diverter resistanceiii. Using a potential divider at the input (i.e. changes the

effective terminal voltage)

(b) Controlling series DC motors

28

i. Adding a series resistance

)( tsaata rrrIVE

fg

am K

E

afge IKT

Ea drops, Ia stays the same

For the same Te, f is constant

but m drops since Ea = Kg f m..

New value of the motor speed, m is given by

mfga KE

, r mt aE

For the same Te produced

ii. Adding a parallel field diverter resistance

fg

am K

E

sff IK

Is drops i.e. Is < Ia.

Ea remains constant,

When series field flux drops, the motor speed m = Ea / Kg f should rise, while driving the same load.

mfga KE

, , , r mfd as II

For the same Te produced, Ia increases

)||( dsa

ata rrr

EVI

sds rrr ||

When we add the diverter resistance

29

iii. Using a potential divider

This system like the speed control by adding series resistance as explained in section (i) where rt RTh and Vt VTh.

tTh VRR

RV

21

2

Let us apply Thévenin theorem to the right of Vt

21 || RRRTh

)( ThsaaTha RrrIVE

fg

am K

E

afge IKT

Ea drops rapidly, Ia stays the same

For the same Te, f is constant

but m drops rapidly since Ea = Kg f m..

mfga KE

, V mt aE

For the same Te produced

New value of the motor speed, m is given by

If the load increases, Te and Ia increases and Ea decreases, thus motor speed m drops down more.

Ex1: A separately excited DC motor drives the load at nr = 1150 rpm.

a) Find the gross output power (electromechanical power output) produced by the DC motor.

b) If the speed control is to be achieved by armature voltage control and the new operating condition is given by:

nr = 1000 rpm and Te = 30 Nm

find the new terminal voltage Vt while f is kept constant.

30

Soln:

a) Gross output power is given by

For Ia = 36 A

Thus

V11237036125 .aaTa rIVE

aame IETP conv

kW03436112conv . aame IETP

Note that induced torque is found to be Te = 33.4 Nm

b) New gross output power is found to be

kW1436010002302222conv2 ./ aame IETP

As Ea = Kg f m and f is constant

V 1971121150

10001

1

22 . a

r

ra E

n

nE

2

1

2

1

2

1 r

r

m

m

a

a

n

n

E

E

A332197

143

2

22 .

..

k

E

PI

a

conva

Then, new armature current is found to be

Consequently, the new terminal voltage is given by

V10937033219722 ...aaat rIEV

31

Ex2: A 50 hp, 250 V, 1200 rpm DC shunt motor with compensating windings has an armature resistance (including the brushes, compensating windings, and interpoles) of 0.06 . Its field circuit has a total resistance Radj + RF of 50 , which produces a no-load speed of 1200 rpm. The shunt field winding has 1200 turns per pole.

a) Find the motor speed when its input current is 100 A.

b) Find the motor speed when its input current is 200 A.

c) Find the motor speed when its input current is 300 A.

d) Plot the motor torque-speed characteristic.

Solution:

At no load, the armature current is zero and therefore Ea = VT = 250 V.

a) Since the input current is 100 A, the armature current is

250100 95

50T

A L F LF

VI I I I A

R

Therefore 250 95 0.06 244.3A T A AE V I R V

and the resulting motor speed is:

22 1

1

244.31200

2501173A

A

En n m

Erp

32

b) Similar computations for the input current of 200 A lead to n2 = 1144 rpm.

c) Similar computations for the input current of 300 A lead to n2 = 1115 rpm.

d) At no load, the torque is zero. The induced torque is A Aind

E I

For the input current of 100 A:2443 95

1902 1173 / 60ind N m

-

For the input current of 200 A:

For the input current of 300 A:

2383 195388

2 1144 / 60ind N m

-

2323 295587

2 1115 / 60ind N m

-

The torque-speed

characteristic of the motor is:

Ex3: A 100 hp, 250 V, 1200 rpm DC shunt motor with an armature resistance of 0.03 and a field resistance of 41.67 . The motor has compensating windings, so armature reactance can be ignored. Mechanical and core losses may be ignored also. The motor is driving a load with a line current of 126 A and an initial speed of 1103 rpm. Assuming that the armature current is constant and the magnetization curve is

a) What is the motor speed if the field

resistance is increased to 50 ?

b) Calculate the motor speed as a function

of the field resistance, assuming a

constant-current load.

c) Assuming that the motor next is

connected as a separately excited and

is initially running with VA = 250 V,

IA = 120 A and at n = 1103 rpm while

supplying a constant-torque load,

estimate the motor speed if VA is

reduced to 200 V.

33

Soln:

a) For the given initial line current of 126 A, the initial armature current will be

Therefore, the initial generated voltage for the shunt motor will be

After the field resistance is increased to 50 Ω, the new field current will be

shunt

1 1 1

250126 120

41.67A L FI I I A

1 1 250 120 0.03 246.4A T A AE V I R V

2

2505

50FI A

The values of EA on the magnetization curve are directly proportional to the flux. Therefore, the ratio of internal generated voltages equals to the ratio of the fluxes within the machine. From the magnetization curve, at IF = 5A, EA1 = 250V, and at IF = 6A,

EA1 = 268V. Thus:

b) A speed vs. RF characteristic is shown on the

right

The ratio of the two internal generated voltages is

2 2 2 2 2

1 1 1 1 1

A

A

E K n

E K n

Since the armature current is assumed constant, EA1 = EA2 and, therefore

1 12

2

nn

1 1 1 12

2 2

2681103 1187

250A

A

n E nn rpm

E

34

and since the flux is constant

Since the both the torque and the flux are constants, the armature current IA is

also constant. Then

c) For a separately excited motor, the initial generated voltage is

Since

1 1 1A T A AE V I R

2 2 2 2 2

1 1 1 1 1

A

A

E K n

E K n

2 12

1

A

A

E nn

E

2 22 1

1 1

200 120 0.031103 879

250 120 0.03T A A

T A A

V I Rn n rpm

V I R

separately excited

Ex4: A 250-V series dc motor with compensating windings. and a total series

resistance ra + rs of 0.08 . The series field consists of 25 turns per pole.

with the magnetization curve (at 1200 rpm) shown below.

a) Find the speed and induced torque of

this motor for when its armature current

is 50 A.

b) Calculate the efficiency of the motor

35

Soln:

a) To analyze the behavior of a series motor with saturation. pick points along the operating curve and find the torque and speed for each point. Notice that the magnetization curve is given in units of magnetomotive force (ampere-turns) versus Ea for a speed of 1200 r/min. so calculated Ea values must be compared to the equivalent values at 1200 r /min to deternine the actual motor speed.

For Ia = 50 A

Since Ia = Is = 50 A. the magnetomotive force is

V 24608.050250 saaTa rrIVE

turns-A12505025 ss NIF

From the magnetization curve at F = 1250 A-turns. Ea0 = 80 V. To get the

correct speed of the motor.

rpm 3690120080

2460

0 r

a

ar n

E

En

b) Efficiency is given by the ratio of the output and input power values. Thus,

To find the induced torque supplied by the motor at that speed, recall that

Pconv = Ea Ia = Tem. Therefore,

Nm 8316036902

50246.

/

m

aae

IET

%.% 49810050250

50246100100100

tt

aa

tt

me

in

out

IV

IE

IV

T

P

P