02 DC Machines

transcript

DC MachinesDC Machines

Two-pole DC machineTwo-pole DC machine

center of rotation

Horizontal view

Two-pole DC machineTwo-pole DC machine

Generated voltageGenerated voltage

NeFrom Faraday's law,

Consider a conductor rotating at n rpm in the field of p poles having a flux Фper pole.The total flux cut by the conductor in n revolutions is pФn;

The flux cut per second, giving the induced voltage e, is

Generated voltageGenerated voltageIf there is a total of z conductors on the armature, connected in a parallel paths,then the effective number of conductors in series is z/a,which produce the total voltage E in the armature winding.

Hence, for the entire winding,

mm wazp

thus,602

kwkE ama

2where,

Generated voltageGenerated voltage

1,2,2 azp

Therefore

ff ikAnd

mf wkiE

Induced torqueInduced torqueAssume no losses,

ame EiwT

mechanical power at armature = input electrical power

aae ikT

afe ikiT

Lap windingLap winding

• A two-layer winding in which each coil is connected in series to the adjacent coil.

• For a P-pole machines, a lap winding has P parallel paths between brushes.

• Lap winding is preferred for higher current applications.

Rotor winding Rotor winding

Wave windingWave winding

• Notice that the two ends of each coil are connected to commutator segments separated by the distance between poles.

• This configuration allows the series addition of the voltages in all the windings between brushes.

• This type of winding only requires one pair of brushes. • A wave winding has 2 parallel paths, regardless of the

number of poles.• Wave winding is preferred for higher voltage

applications.

Rotor Rotor

Question 1Question 1

The armature of a 120-V dc motor has a resistance of 1.5Ω and takes 4 A when operating at full load. Calculate

(a) the counter EMF produced by the armature, and

(b) the developed power by the armature.(c) efficiency of the motor, given the formula

%100in

DC motor circuitDC motor circuit

1.5 Ω

Answer : (a) 114 V; (b) 456 W

Question 2Question 2

A 60-kW four-pole generator has a lap winding placed in 48 armature slots, each slot containing six conductors. The pole flux is 0.08 Wb, and the speed of rotation is 1040 r/min.

a.Determine the generated voltage;

b.What is the current flowing in the armature conductors when the generator delivers full load?

Ans: (a) 399.36 V; (b) 37.56 A

Question 3 (p364) Example 8.1Question 3 (p364) Example 8.1

• When switch is closed at no load, current will flow into the rotor winding.

• Torque will be induced to give angular acceleration.

• Voltage is induced in the rotor winding.

• Rotor current will fall.

• Rotor will wind up in constant speed with zero torque (equilibrium).

Question 3 (p364)Question 3 (p364)

• When load torque of 10 Nm is applied, motor will slow down.

• Induced voltage decreases.

• Rotor current increases.

• Induced torque increases until equal to load torque (equilibrium) at lower speed.

• When assisting torque of 7.5 Nm is applied, motor will accelerate.

• Induced voltage increases.

• Rotor current reverses (generator).

• Opposite torque induced.

• Eventually opposite torque equal to assisting torque (equilibrium).

• When flux density were reduced to 0.2 T at no load, transient will occur and reach constant speed (zero torque ~equilibrium).

• Rotor current is zero.

• Speed ???.

Two-pole, 4-loop DC machineTwo-pole, 4-loop DC machine

• This machine has 4 complete loops buried in slots of rotor.

• The poles faces are curved to provide a uniform air-gap width and to give a uniform flux density everywhere under the faces.

Two-pole, 4-loop DC machineTwo-pole, 4-loop DC machineeEwt 4,0when

eEwt 2,45when

eEwt 4,90when

E, volts

wt45° 90° 135° 180° 225° 315°0°

CommutationCommutation

• It is the process of switching the loop connections on the rotor of a DC machine just as the voltage in the loop switches polarity, in order to maintain an essentially constant DC output voltage.

Construction of DC motorConstruction of DC motor

Commutation problemsCommutation problems

1) Armature reaction

~ sparks at brushes (neutral-plane shift)

~ flux-weakening

2) L*di/dt voltages

~ sparks at brushes

Commutation Problems (1)Commutation Problems (1)

Solutions Solutions

• Interpoles / commutating poles~ Sparking due to Neutral-plane shift and L*di/dt voltages~ cheaper

• Compensating windings~ flux-weakening~ an expensive solution~ for very heavy, severe duty cycle motors

Types of DC motorsTypes of DC motors

• Separately excited DC motor

• Shunt DC motor

• Permanent magnet DC motor

• Series DC motor

• Compounded DC motor

Seperately excited DC motorSeperately excited DC motor

AAAT RIEV

EAVF VT

Shunt DC motorShunt DC motor

FAL III

FFF RIV

Constant if VT is constant

AAAT RIEV

Shunt DC motorShunt DC motor

• Equation gives straight line with a negative slope.

AAT RIwKV

Aind IK

Example 8.2Example 8.2

RAIL Prated = 50 hp

VT = 250 Vnm = 1200 rpm

With compensating windingsand interpolesRA = 0.06 ohmsRF = 50 ohmsnNL = 1200 rpmFind motor speed and induced torque when input

currents are 100 A, 200 A and 300 A.

ratings

• During no-load, IA = 0 A, EA = 250 V and nNL = 1200 rpm.

• When IL = 100 A, IA = ?

• PAG = EAIA, and PAG = torque * speed

Armature reactionArmature reaction

• If a motor has armature reaction, then as its load increases, the flux-weakening effects reduce its flux.

• As a result, motor speed tends to increase. wm

Induced torque

With AR

Without AR

Nonlinear analysis of shunt DC Nonlinear analysis of shunt DC motormotor

• Field current and armature reaction are the two principle contributors to the MMF.

• Effect of IF is determined from the magnetization curve.

• With armature reaction, the machine’s flux will be reduced with each increase in load.

• EA is directly proportional to wm when Ф is fixed.

• A 50-hp, 250-V, 1200 rpm, dc shunt motor without compensating windings has an armature resistance (including the brushes and interpoles) of 0.06 Ω. Its field circuit has a total resistance of 50 Ω, which produces a no-load speed of 1200 rpm. There are 1200 turns per pole on the shunt field winding, the armature reaction produces a demagnetizing mmf of 840 At at a load current of 200 A.

• Find the motor speed when input current is 200 A.

Effect of armature reactionEffect of armature reaction

• VT and RF were unchanged, thus IF and ФF is constant and EA α n.

• From the magnetization curve, when n=1200 rpm and IF*=4.3 A, EA=233 V.

• Actual EA=238.3 V

• Thus, n1=(238.3/233)*1200=1227 rpm

Speed control of shunt DC motorsSpeed control of shunt DC motors

• Adjusting RF and terminal voltage to the armature are two common ways to control the speed.

Changing the RChanging the RFF

• Let RF ↑;

• Since IF = VT / RF, IF ↓.

• Then Ф ↓, EA ↓ and IA ↑.

• But the increase in IA predominates over the decrease in Ф.

• Thus, torque ↑ and motor speeds up.

• When wm↑, EA↑, IA↓ and torque ↓.

• Finally, higher steady-state speed is reached.

Effect of REffect of RFF for normal operating for normal operating

rangerange

Induced torque

Larger RF

Smaller RF

Full-load torque

Warning about RWarning about RFF speed control speed control

• From equation above, when Ф↓ wNL ↑, and the slope gradient ↑.

• At very low speeds, the increase in IA is no longer enough to compensate for the decrease in Ф.

• Thus, torque ↓ and motor slows down.• Therefore, the results are not predictable for low

speeds.

Effect of REffect of RFF for entire range for entire range

Induced torque

Larger RF

Smaller RF

Full-load torque

• A 100-hp, 250-V, 1200 rpm shunt dc motor has an armature resistance of 0.03 Ω and a field resistance of 41.67 Ω. The motor has compensating windings, so armature reaction can be ignored. Mechanical and core losses may be neglected. The motor is assumed to be driving a load with a line current of 126 A and an initial speed of 1103 rpm. Assume that the armature current is constant.

• What is the motor speed if the field resistance is raised to 50 Ω? (use fig 8.30)

Effect of changing REffect of changing RFF

• In this case, EA is constant since IA is constant. So n α 1/Ф.

• Initial speed is 1103 rpmWhen RF=41.67Ω, IF=6 A, and from magnetization curve, EA=268

• When RF=50 Ω, IF=5 A and EA=250• Since EA α Ф on curve, Ф1/Ф2 = EA1/EA2 =

268/250 = 1.076• Thus n2 = (Ф1/Ф2)*n1 = 1187 rpm

Changing the armature voltageChanging the armature voltage

• No changing to the VF.

• When VA↑, IA↑ torque ↑ and wm ↑.

• Then, EA↑ and causes IA↓ and torque ↓.

• Higher steady-state speed is reached.

Effect of armature voltage speed Effect of armature voltage speed controlcontrol

Induced torque

Larger VA

Smaller VA

• The motor in ex 8.4 is now connected separately excited. The motor is initially running with VA=250 V, IA=120 A and n=1103 rpm, while supplying a constant-torque load.

• What will the speed of this motor be if VA is reduced to 200 V?

Effect of changing VEffect of changing VAA

• Flux is constant, so EA α n.

• Initial conditions: VA=250 V, IA=120 A, n=1103 rpm

• So EA0=246.4 V

• VA is changed to 200 V with constant torque, thus IA is constant

• EA1=196.4 V

• Thus n1=(196.4/246.4)*1103=879 rpm

Safe operating rangesSafe operating ranges

• RF control is good for speeds above base speed but not for speeds below base speed.

• Armature voltage control is good for speeds below base speed but not for speeds above base speed.

Torque and power limitsTorque and power limits

• The limiting factor is the heating of armature conductors.

• For armature voltage control at low speed, Ф is constant, thus,

• The maximum torque is independent of wm. And,

max,max AIK

wP maxmax

Power and torque limits versus Power and torque limits versus speedspeed

Max. torque Max. power

nmnbasenbase

High speed power and torque limitHigh speed power and torque limit

• During speeds equal and above rated speed, RF control is used.

• Pmax is maintained from overloading.

• When RF↑, Ф↓, IA↑↑ (overloading) and w↑.

• Thus, torque limit is decreased when speed increases.

wP max

The PM DC motorThe PM DC motor

• No field losses, smaller size

• Low Ф & torque, i.e. lower induced torque per ampere of IA than shunt type.

• Risk of demagnetization

• Behave like shunt type, except that the flux is fixed.

• Speed control by VA and RA control.

Hysteresis lossHysteresis loss

Different types of magnetic Different types of magnetic materialsmaterials

• A 250-V series DC motor with compensating windings, and a total of series resistance RA + RS of 0.08Ω. The series field consists of 25

TutorialsTutorials

1. A dc motor operates at 1680rpm when drawing 28A from a 230V supply. The armature resistance is 0.25Ω and other losses are negligible.

a) Calculate the no-load speed if IA = 0A at no load.

b) Determine the developed power under loaded conditions.

c) Determine the torque developed under the given load.

(1733rpm,6244W, 35.5Nm)

2. A 240V shunt motor has an armature resistance of 0.25Ω. Under load, the armature current is 24A. Suppose the flux is suddenly decreased by 2.5%, what would be the immediate effect on the developed torque?

(1.93 times)

3. For Q2, determine the new steady-state speed after the field has been decreased. Assume the motor was operating at 640rpm before the field was adjusted.

(656rpm)

02 DC Machines

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