1 dc machines dr. abdulr-razaq sh. hadde. 2 dc motor the direct current (dc) machine can be used as...

1

DC MACHINES

Dr. Abdulr-Razaq SH. Hadde

2

DC Motor

• The direct current (dc) machine can be used as a motor or as a generator.

• DC Machine is most often used for a motor. • The major advantages of dc machines are the

easy speed and torque regulation. • However, their application is limited to mills,

mines and trains. As examples, trolleys and underground subway cars may use dc motors.

• In the past, automobiles were equipped with dc dynamos to charge their batteries.

3

DC Motor

• Even today the starter is a series dc motor • However, the recent development of power

electronics has reduced the use of dc motors and generators.

• The electronically controlled ac drives are gradually replacing the dc motor drives in factories.

• Nevertheless, a large number of dc motors are still used by industry and several thousand are sold annually.

4

Construction

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DC Machine Construction

Figure 8.1 General arrangement of a dc machine

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DC Machines

• The stator of the dc motor has poles, which are excited by dc current to produce magnetic fields.

• In the neutral zone, in the middle between the poles, commutating poles are placed to reduce sparking of the commutator. The commutating poles are supplied by dc current.

• Compensating windings are mounted on the main poles. These short-circuited windings damp rotor oscillations. .

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DC Machines

• The poles are mounted on an iron core that provides a closed magnetic circuit.

• The motor housing supports the iron core, the brushes and the bearings.

• The rotor has a ring-shaped laminated iron core with slots.

• Coils with several turns are placed in the slots. The distance between the two legs of the coil is about 180 electric degrees.

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DC Machines

• The coils are connected in series through the commutator segments.

• The ends of each coil are connected to a commutator segment.

• The commutator consists of insulated copper segments mounted on an insulated tube.

• Two brushes are pressed to the commutator to permit current flow.

• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.

9

DC Machines

• The rotor has a ring-shaped laminated iron core with slots.

• The commutator consists of insulated copper segments mounted on an insulated tube.

• Two brushes are pressed to the commutator to permit current flow.

• The brushes are placed in the neutral zone, where the magnetic field is close to zero, to reduce arcing.

10

DC Machines

• The commutator switches the current from one rotor coil to the adjacent coil,

• The switching requires the interruption of the coil current.

• The sudden interruption of an inductive current generates high voltages .

• The high voltage produces flashover and arcing between the commutator segment and the brush.

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DC Machine Construction

|

Shaft

Brush

Coppersegment

Insulation

RotorWinding

N S

Ir_dcIr_dc/2

Rotation

Ir_dc/2

Ir_dc

12

3

4

5

6

7

8

Polewinding

Figure 8.2 Commutator with the rotor coils connections.

12

DC Motor Operation

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DC Motor Operation

• In a dc motor, the stator poles are supplied by dc excitation current, which produces a dc magnetic field.

• The rotor is supplied by dc current through the brushes, commutator and coils.

• The interaction of the magnetic field and rotor current generates a force that drives the motor

|

Shaft

Brush

Coppersegment

Insulation

RotorWinding

N S

Ir_dcIr_dc/2

Rotation

Ir_dc/2

Ir_dc

12

3

4

5

6

7

8

Polewinding

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DC Motor Operation

• Before reaching the neutral zone, the current enters in segment 1 and exits from segment 2,

• Therefore, current enters the coil end at slot a and exits from slot b during this stage.

• After passing the neutral zone, the current enters segment 2 and exits from segment 1,

• This reverses the current direction through the rotor coil, when the coil passes the neutral zone.

• The result of this current reversal is the maintenance of the rotation.

(a) Rotor current flow from segment 1 to 2 (slot a to b)

Vdc30

NS

Bv

v

a

b

1

2

Ir_dc

(b) Rotor current flow from segment 2 to 1 (slot b to a)

30NS Vdc

a

b

1

2

B

v v

Ir_dc

15

DC Generator Operation

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• The N-S poles produce a dc magnetic field and the rotor coil turns in this field.

• A turbine or other machine drives the rotor.

• The conductors in the slots cut the magnetic flux lines, which induce voltage in the rotor coils.

• The coil has two sides: one is placed in slot a, the other in slot b.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


17


• In Figure 8.11A, the conductors in slot a are cutting the field lines entering into the rotor from the north pole,

• The conductors in slot b are cutting the field lines exiting from the rotor to the south pole.

• The cutting of the field lines generates voltage in the conductors.

• The voltages generated in the two sides of the coil are added.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


18


• The induced voltage is connected to the generator terminals through the commutator and brushes.

• In Figure 8.11A, the induced voltage in b is positive, and in a is negative.

• The positive terminal is connected to commutator segment 2 and to the conductors in slot b.

• The negative terminal is connected to segment 1 and to the conductors in slot a.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


19


• When the coil passes the neutral zone: – Conductors in slot a are

then moving toward the south pole and cut flux lines exiting from the rotor

– Conductors in slot b cut the flux lines entering the in slot b.

• This changes the polarity of the induced voltage in the coil.

• The voltage induced in a is now positive, and in b is negative.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


20


• The simultaneously the commutator reverses its terminals, which assures that the output voltage (Vdc) polarity is unchanged.

• In Figure 8.11B – the positive terminal is

connected to commutator segment 1 and to the conductors in slot a.

– The negative terminal is connected to segment 2 and to the conductors in slot b.

30NS Vdc

Bv

v

a

b

Ir_dc


30NS Vdc

a

b

1

2vv

B

Ir_dc


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Generator

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DC Generator Equivalent circuit

• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.

• This induced voltage is represented by a voltage source.

• The stator coil has resistance, which is connected in series.

• The pole flux is produced by the DC excitation/field current, which is magnetically coupled to the rotor

• The field circuit has resistance and a source

• The voltage drop on the brushes represented by a battery

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• Figure 8.12 Equivalent circuit of a separately excited dc generator.

RfRa

Vbrush

VdcEagVf

max

IfIag

Load

Mechanicalpower in

Electricalpower out

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• The magnetic field produced by the stator poles induces a voltage in the rotor (or armature) coils when the generator is rotated.

• The dc field current of the poles generates a magnetic flux

• The flux is proportional with the field current if the iron core is not saturated:

fag IK1

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• The rotor conductors cut the field lines that generate voltage in the coils.

• The motor speed and flux equations are :

vBNE grag 2

2gDv ggag DB

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• The combination of the three equation results the induced voltage equation:

• The equation is simplified.

agrggrg

grgrag NDBND

BNvBNE

222

fmfragrag IKIKNNE 1

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• When the generator is loaded, the load current produces a voltage drop on the rotor winding resistance.

• In addition, there is a more or less constant 1–3 V voltage drop on the brushes.

• These two voltage drops reduce the terminal voltage of the generator. The terminal voltage is;

brushaagdcag VRIVE

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Motor

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DC Motor Equivalent circuit

• Figure 8.13 Equivalent circuit of a separately excited dc motor• Equivalent circuit is similar to the generator only the current directions are

different

RfRa

Vbrush

VdcEamVf

max

IfIam

Mechanicalpower out

Electricalpower in

DC Powersupply

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DC Motor Equivalent circuit• The operation equations are:• Armature voltage equation

brushaamamdc VRIEV

The induced voltage and motor speed vs angular frequency

fmam IKE mn 2

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DC Motor Equivalent circuit• The operation equations are:• The combination of the equations results in

The current is calculated from this equation. The output power and torque are:

mamdcamfm RIVEIK

amamout IEP fammout IIKP

T

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Ideal Transformers

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I1 I2 I1 I2

V1 V2

V1 V2N1 N2

IT+ +- -

I1 I2

Z2V1 V2

N1 N2

The ideal transformer: V1/N1 = V2/N2 N1I1 = N2I2 S1 = V1I1* S2 = V2I2* S1 = S2 k = V2/V1 = N2/N1 V1 = V2/k I1 = k* I2 Z2 = V2/I2 Z2’ = V1/I1 = (V2/k) / (I2*k) = Z2/k2

Real transformer:

I2I1

N2N1

R1jX1 R2 jX2

LoadV1 V2Ic

R0 jX0

Im

V1

I2*k

34

N2N1 IT

V2

+

-

V2V1

V1

V2 =

N1

N2k =

V1

I1 I2R1 jX1

Ic

Rc jXm

R2 jX2

Load

Im

N1 N2

I1 R1 jX1

Ic

Rc jXm

R2’ jX2’

Im

N1 N2

V1

I2

I2*k

35

I1 I2 I1 I2

V1 V2

V1 V2N1 N2

IT+ +- -

I2I1

N2N1

R1jX1 R2 jX2

LoadV1 V2Ic

Rc jXm

Im

V1

I2*k

I1 R1 jX1

Ic

Rc jXm

R2’ jX2’

Im

N1 N2

V1

I2

Load

R2 & X2 are referred to winding 1

1

2

1

2

N

N

V

Vk

36

I1 Req jXeq I2

V1

N1 N2

Load

Req = R1 + R2’

Xeq = X1 + X2’Neglecting excitation current

N1 N2

I2jXeqI1

V1 Load

V2

V2

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Determination of Req and Xeq from short-circuit test. Vsc is normally equal to five percent of the rated voltage of winding #1, or: Vsc = .05 x V1rated Measure Isc, Vsc, Psc Psc = /I2sc/* Rs1 Rs1 = Psc / (/I2sc/) Zsc1 = /Vsc1///Isc1/= /Req + jXeq/ = /Rs1 + jXs1/ Xs1 = Zsc2 - Rs2 Vsc = .05 x V2rated Measure Isc, Psc, Vsc The same procedure as before however, the short-circuit impedance is referred to

winding #2.

Real Transformer Ideal Transformer

Req jXeq

Vsc

Isc

Vsc

jXeqReqIsc

VscVsc

Isc

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The per-unit system: pu = Actual Quantity / Base Value of Quantity Name plate rating: An electric element has a nameplate rating which specifies the safe operating voltage, current and power for the element. Example: A light bulb is rated at 120V and 50Watts. Base Values: The base values are normally selected equal to base values. Pb = Qb = Sb = S1(rated)

Vb = VL-N(rated) Zb = Vb/Ib = Vb / (Sb/Vb) = Vb2/Sb Yb = 1/Zb Ppu = PA/Pb Qpu = QA/Qb Spu = SA/Sb Vpu = VA/Vb Ipu = IA/Ib Zpu = ZA/Zb

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The nameplate rating for transformers, generators and machines are given in percent (which is the same as pu, i.e. percent% = pu) using the device power rating as Sb and voltage rating as Vb. Example: A 3 generator is rated 300 MVA, 20 kV and ten percent reactance. Compute the impedance of the machine in ohms. Sb = 300 MVA/3 = 100 x 106 MVA Vb = 20 kV/ 3 = 11.56 x 103 V Zb = Vb2/Sb = (11.56 x 103)2 / 100 x 106 = 1.34 Zgen = j.1 x (1.34) = j.134

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Rs’ jXs’

V1(rated) V2(rated)

Rs jXs

V1(rated) V2(rated)

N1 N2 N1 N2

I1I1 I2

S(rated) S(rated)LV HV

a = N1

N2=

V1(rated)

V2(rated)

P.U. Equivalent Circuit of a Transformer. Select: Vb1 = V1(rated) Sb = S(rated) Select: Vb2/Vb1 = V2(rated) / V1(rated) or Vb2 = (V2(rated) / V1(rated))Vb1 Zb1 = Vb1/Ib1 = Vb1

2/Sb Sb = Vb1Ib1 Zb2 = Vb2/Ib2 = Vb2

2/Sb Ib2 = Sb/ Vb2 Zb1 / Zb2 = Vb1

2/ Vb22 = Vb1

2/( V2(rated) / V1(rated))2 Vb12 = 1/(1/a)2 = a2

Zb1 = Zb2 / k2

Zpu1 = Rs+jXs/Zb1 = Rs+jXs/ (Zb2/k2 ) = (Rs*k2) +(jXs*k2)/Zb2 = R’s+jX’s/ Zb2 = (Rs*k2 + jXs*k2)1/ Zb2 or Zpu1 = Zpu2

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Rs jXs

V1(rated) V2(rated)

N1 N2

Vb1 = V1(rated) Sb = S(rated 1 phase) Vb2 = V2(rated)

Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb

Zpu

Vb1 Vb2

Zpu = Rs+jXs/Zb1

Vb1 Vb2

I1 pu

V1 pu

Rs/a2 + jXs/a2

V2(rated)

N1 N2

Sb = S1(rated) Vb2 = V2(rated)

Zb1 = (Vb1)2/Sb Zb2 = (Vb2)2/Sb

Zb2 = Zb1*k2

Zpu

Vb1

Vb2

Sb

Vb1

I1 pu

V1 pu V2 pu

Vb2

I2 pu

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Transformation between two bases:

Selection 1: Sb1 = SA Vb1 = VA

Then:Zb1 = (Vb1)2 / Sb1 Zpu1 = ZL/ Zb1

Selection 2: Sb2 = SB Vb2 = VB


Zpu2 / Zpu1 = ZL / Zb2 x Zb1 / ZL = Zb1 / Zb2

= (Vb1)2 / Sb1 x Sb2 / (Vb2)2

Zpu2 = Zpu1 (Vb1 / Vb2)2 (Sb2 / Sb1)

“1” = old“2” = new

Zpu (new) = Zpu (old) (Vb (old) / Vb (new))2 (Sb (new) / Sb (old))

Transformation between two bases:

Selection 1: Sb1 = SA Vb1 = VA


Selection 2: Sb2 = SB Vb2 = VB


Zpu2 / Zpu1 = ZL / Zb2 x Zb1 / ZL = Zb1 / Zb2

= (Vb1)2 / Sb1 x Sb2 / (Vb2)2

Zpu2 = Zpu1 (Vb1 / Vb2)2 (Sb2 / Sb1)

“1” = old“2” = new

Zpu (new) = Zpu (old) (Vb (old) / Vb (new))2 (Sb (new) / Sb (old))

ZL

1 dc machines dr. abdulr-razaq sh. hadde. 2 dc motor the direct current (dc) machine can be used as...

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