02 dc machines
TRANSCRIPT
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DC MachinesDC Machines
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Two-pole DC machineTwo-pole DC machine
a
bc
d
l
r
eind
+ −
center of rotation
ab
dc
Horizontal view
N S
w
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Two-pole DC machineTwo-pole DC machine
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Two-pole DC machineTwo-pole DC machine
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Generated voltageGenerated voltage
dtd
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
60np
e
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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,
aznp
E60
mm wazp
En
w
2
thus,602
azp
kwkE ama
2where,
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Generated voltageGenerated voltage
1,2,2 azp
mwE 2
Therefore
mwazp
E 2
ff ikAnd
mf wkiE
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Induced torqueInduced torqueAssume no losses,
ame EiwT
mechanical power at armature = input electrical power
aae ikT
afe ikiT
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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.
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Rotor winding Rotor winding
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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.
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Rotor Rotor
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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
out
P
P
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DC motor circuitDC motor circuit
120 V
1.5 Ω
+
eind
-
4 A
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Answer : (a) 114 V; (b) 456 W
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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
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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).
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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.
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Question 3 (p364)Question 3 (p364)
• 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).
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Question 3 (p364)Question 3 (p364)
• 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 ???.
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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.
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Two-pole, 4-loop DC machineTwo-pole, 4-loop DC machineeEwt 4,0when
eEwt 2,45when
eEwt 4,90when
2e
4e
E, volts
wt45° 90° 135° 180° 225° 315°0°
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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.
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Construction of DC motorConstruction of DC motor
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Commutation problemsCommutation problems
1) Armature reaction
~ sparks at brushes (neutral-plane shift)
~ flux-weakening
2) L*di/dt voltages
~ sparks at brushes
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Commutation Problems (1)Commutation Problems (1)
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Commutation Problems (2)Commutation Problems (2)
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Commutation Problems (3)Commutation Problems (3)
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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
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Types of DC motorsTypes of DC motors
• Separately excited DC motor
• Shunt DC motor
• Permanent magnet DC motor
• Series DC motor
• Compounded DC motor
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Seperately excited DC motorSeperately excited DC motor
AAAT RIEV
EAVF VT
RA
RF
LF
IA
IL
AL II
wKEA
IF
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Shunt DC motorShunt DC motor
VT
RF
IF
IAEA
RAIL
FAL III
FFF RIV
Constant if VT is constant
AAAT RIEV
wKEA
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Shunt DC motorShunt DC motor
• Equation gives straight line with a negative slope.
Aind
T RK
wKV
AAT RIwKV
Aind IK
indAT
K
R
K
Vw
2)(
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Example 8.2Example 8.2
VT
RF
IF
IAEA
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
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• During no-load, IA = 0 A, EA = 250 V and nNL = 1200 rpm.
• When IL = 100 A, IA = ?
• PAG = EAIA, and PAG = torque * speed
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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
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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.
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Example 8.3Example 8.3
• 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.
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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
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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.
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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.
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Effect of REffect of RFF for normal operating for normal operating
rangerange
wm
Induced torque
Larger RF
Smaller RF
Full-load torque
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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.
indAT
K
R
K
Vw
2)(
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Effect of REffect of RFF for entire range for entire range
wm
Induced torque
Larger RF
Smaller RF
Full-load torque
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Example 8.4Example 8.4
• 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)
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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
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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.
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Effect of armature voltage speed Effect of armature voltage speed controlcontrol
wm
Induced torque
Larger VA
Smaller VA
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Example 8.5Example 8.5
• 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?
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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
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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.
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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
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Power and torque limits versus Power and torque limits versus speedspeed
nm
Max. torque Max. power
nmnbasenbase
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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
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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.
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Hysteresis lossHysteresis loss
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Different types of magnetic Different types of magnetic materialsmaterials
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Example 8.6Example 8.6
• 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
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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)
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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)