Lecture Notes EEE 360

Transcript Lecture Notes EEE 360

DC MACHINES

Dr. Abdulr-Razaq SH. Hadde

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.

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.

Construction

DC Machine Construction

Figure 8.1 General arrangement of a dc machine 5

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 . .

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.

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 .

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.

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

. 10

DC Machine Construction

I r_dc /2 Rotation Brush I r_dc Shaft I r_dc /2 Pole winding N 7 8 6 1 | 2 4 5 3 S Insulation Rotor Winding I r_dc Copper segment Figure 8.2 Commutator with the rotor coils connections .

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

N Rotation I r_dc /2 Brush I r_dc Shaft I r_dc /2 Pole winding 7 8 6 1 | 2 4 5 3 S Insulation Rotor Winding I r_dc Copper segment 13

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 from segment 1, 2 and exits 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.

S S

B v a 1

2 b v

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

)

2 b v a v

I r_dc N

N V dc V dc (b) Rotor current flow from segment 2 to 1 (slot

) 14

DC Generator Operation

• • • •

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.

v a B

N V

b v

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

) dc

B a

v 1

2 v

N V

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

) 16 dc

DC Generator Operation

• • • •

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.

v a B

N V

b v

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

) dc

B a

v 1

2 v

N V

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

) 17 dc

DC Generator Operation

• • • •

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 .

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

) S

v b a v 1

2 1

2 v a b B

I r_dc

v B

I r_dc N N V dc V dc (b) Rotor current flow from segment 2 to 1 (slot

) 18

DC Generator Operation

• • •

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 slot b .

b cut the flux lines entering the in This changes the polarity of the induced voltage in the coil.

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

v a B

N V

b v

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

) dc

B a

v 1

2 v

N V

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

) 19 dc

DC Generator Operation

• •

The simultaneously the commutator reverses its terminals, which assures that the output voltage (V dc ) 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.

v a B

N V

b v

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

) dc

B a

v 1

2 v

N V

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

) 20 dc

Generator

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

DC Generator Equivalent circuit

R f V f I f  max V brush R a E ag I ag V dc Load Mechanical power in Electrical power out •

Figure 8.12Equivalent circuit of a separately excited dc generator.

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 .

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:





I f

DC Generator Equivalent circuit

• •

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

E ag

 2

N r B



g v

The motor speed and flux equations are :

 

D g

2 





g D g

DC Generator Equivalent circuit

•

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

E ag

 2

N r B



g v

 2

N r B



  

D g

2   

N r





g D g

  

N r



 •

The equation is simplified.

E ag



N r



 

N r K

I f

 

K m I f

 26

DC Generator Equivalent circuit

• • •

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;

E ag



V dc



I ag R a



V brush

Motor

DC Motor Equivalent circuit

V brush R f R a Electrical power in  max V f I f E am I am V dc DC Power supply Mechanical power out • •

Figure 8.13

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

DC Motor Equivalent circuit

• •

The operation equations are: Armature voltage equation

V dc



E am



I am R a



V brush

The induced voltage and motor speed vs angular frequency

E am



K m I f

   2 

n m

DC Motor Equivalent circuit

• •

The operation equations are: The combination of the equations results in

K m I f

 

E am



V dc



I am R m P

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

out



E am I am T



P out

 

K m I am I f

Ideal Transformers

The ideal transformer: + I 1 V 1 IT N 1 N 2 I 2 V 2 + V 1 /N 1 = V 2 /N 2 N 1 I 1 = N 2 I 2 S 1 = V 1 I 1 * S 2 = V 2 I 2 * S 1 = S 2 k = V 2 /V 1 = N 2 /N 1 V 1 = V 2 /k I 1 = k* I 2 I 1 I 2 V 1 N 1 V 2 N 2 Z 2 Z 2 = V 2 /I 2 Z 2 ’ = V 1 /I 1 = (V 2 /k) / (I 2 *k) = Z 2 /k 2 Real transformer: I 1 R 1 jX 1 I 2 *k I c I m V 1 R 0 jX 0 V 1 N 1 N 2 R 2 V 2 jX 2 I 2 Load I 1 V 1

V 2 I 2

N 1 IT N 2 k = V 2 = V 1 I 1 N 2 N 1 R 1 + V 2 jX 1 V 1 I c R c I 2 *k I m jX m V 1 V 1 I 1 R 1 jX 1 I c R c N 1 N 2 I m jX m R 2 ’ jX 2 ’ R 2 jX 2 I 2 I 2 V 2 Load N 1 N 2

I 1

+ I 1 V 1 IT I 2 V 2



N 1

1 

N 2

+ R 1 jX 1 I 2 *k R 2 jX 2 I 2 V 1 I c R c I m jX m V 1 V 2 Load I 1 V 1 N 1 N 2 V 1 I 1 R 1 jX 1 I c R c I m jX m R 2 ’ jX 2 ’ I 2 N 1 N 2 Load V 2 I 2 R 2 & X 2 are referred to winding 1

I 1 R eq jX eq I 2 V 1 V 1 N 1 N 2 R eq = R 1 + R 2 ’ X eq = X 1 + X 2 ’ I 1 jX eq Neglecting excitation current I 2 N 1 N 2 V 2 V 2 Load Load

Determination of Req and Xeq from short-circuit test. Isc Req jXeq Vsc Vsc Real Transformer Ideal Transformer Vsc is normally equal to five percent of the rated voltage of winding #1, or: Vsc = .05 x V 1 rated Measure Isc, Vsc, Psc Psc = /I 2 sc/* Rs 1 Rs 1 = Psc / (/I 2 sc/) Zsc 1 = /Vsc 1/ //Isc 1/ = /Req + jXeq/ = /Rs 1 + jXs 1 / Xs 1 =



Zsc 2 - Rs 2 Isc Vsc Isc Req jXeq Vsc Vsc = .05 x V 2 rated Measure Isc, Psc, Vsc The same procedure as before however, the short-circuit impedance is referred to

winding #2.

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 = S 1



(rated) Vb = V L-N(rated) Zb = Vb/Ib = Vb / (Sb/Vb) = Vb 2 /Sb Yb = 1/Zb Ppu = P A /Pb Qpu = Q A /Qb Spu = S A /Sb Vpu = V A /Vb Ipu = I A /Ib Zpu = Z A /Zb

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 10 6 MVA Vb = 20 kV/



3 = 11.56 x 10 3 V Zb = Vb 2 /Sb = (11.56 x 10 3 ) 2 / 100 x 10 6 = 1.34



Zgen = j.1 x (1.34) = j.134

 39

I 1 Rs jXs I 2 I 1 V 1 (rated) V 2 (rated) LV N 1 N 2 S (rated) HV N 1 a = N 2 P.U. Equivalent Circuit of a Transformer. V 1 (rated) = V 1 (rated) V 2 (rated) N 1 N 2 S (rated) Rs’ jXs’ V 2 (rated) Select: Vb 1 = V 1 (rated) Sb = S (rated) Select: Vb 2 /Vb 1 = V 2 (rated) / V 1 (rated) or Vb 2 = (V 2 (rated) / V 1 (rated) )Vb 1 Zb 1 = Vb 1 /Ib 1 = Vb 1 2 /Sb Sb = Vb 1 Ib 1 Zb 2 = Vb 2 /Ib 2 = Vb 2 2 /Sb Ib 2 = Sb/ Vb 2 Zb 1 / Zb 2 = Vb 1 2 / Vb 2 2 = Vb 1 2 /( V 2 (rated) / V 1 (rated) ) 2 Vb 1 2 = 1/(1/a) 2 = a 2 Zb 1 = Zb 2 / k 2 Zpu 1 = Rs+jXs/Zb 1 = Rs+jXs/ (Zb 2 /k 2 ) = (Rs*k 2 ) +(jXs*k 2 )/Zb 2 = R’s+jX’s/ Zb 2 = (Rs*k 2 + jXs*k 2 )1/ Zb 2 or Zpu 1 = Zpu 2

Rs jXs V 1 (rated) V 2 (rated) Vb 1 = V 1 (rated) N 1 N 2 Sb = S (rated 1 phase) Vb 2 = V 2 (rated) Zb 1 = (Vb 1 ) 2 /Sb Zb 2 = (Vb 2 ) 2 /Sb Vb 1 I 1 pu V 1 pu Zpu Vb 1 Vb 2 Vb 1 Zpu = Rs+jXs/Zb 1 Rs/a 2 + jXs/a 2 Sb Vb 1 V 2 (rated) N 1 N 2 Sb = S 1 (rated) Vb 2 = V 2 (rated) Zb 1 = (Vb 1 ) 2 /Sb Zb 2 = (Vb 2 ) 2 /Sb Zb 2 = Zb 1 *k 2 V 1 I pu 1 pu Vb 2 Zpu Vb 2 Vb 2 I 2 pu V 2 pu

Sb 1 = S A Vb 1 = V A Zb 1 = (Vb 1 ) 2 / Sb 1 Zpu 1 = Z L / Zb 1 Sb 2 = S B Vb 2 = V B Zb 2 = (Vb 2 ) 2 / Sb 2 Zpu 2 = Z L / Zb 2 Zpu 2 / Zpu 1 = Z L / Zb 2 x Zb 1 / Z L = Zb 1 / Zb 2 = (Vb 1 ) 2 / Sb 1 x Sb 2 / (Vb 2 ) 2 Zpu 2 = Zpu 1 (Vb 1 / Vb 2 ) ) 2 (Sb 2 / Sb 1 ) ) “1” = old “2” = new Zpu (new) = Zpu (old) (Vb (old) / Vb (new) ) 2 (Sb (new) / Sb (old) ) ) Z L