chapter 1 dc machines new
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CHAPTER 1DC MACHINES
1.1 INTRODUCTION
Energy is needed in different forms:
Light bulbs and heaters - electrical energy Fans and rolling miles - mechanical energy Need for energy converters
Figure 1.1
DC generator – supply current to the load DC motor – need current from the supply AC electric supply - AC machines (synchronous and asynchronous) DC electric supply - DC machines
Advantages of DC Machine (i) Adjustable motor speed over wide ranges(ii) Constant mechanical output (torque)(iii)Rapid acceleration or deceleration(iv) Responsive to feedback signals
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Figure 1.2 : DC machine Construction
1.2 GENERATION OF AC SIGNAL
Generator needs something to rotate or move itself. It is called the prime mover. The shaft of the prime mover will be coupled to the shaft of the generator so that the
generator can rotate once the prime mover rotates. In a big system, they normally used 3-phase induction motor. But, in small scale machine, they can use a mechanism to rotate the generator which
then generated electricity.
carbon brushFigure 1.3: Physical Arrangement of AC Generating Method
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coil
Slip ring:used to generate ac signalCarbon brush: used to collect output voltage (ac signal)
Figure 1.4 : Emf generated at various instant (angle position)
1.3 GENERATION OF DC SIGNAL
The generating of DC signal can be done by replacing the slip ring with commutator so that rectification process can happen.
Commutator can generate dc signal. This is done through rectification or commutation process; which converts ac signal
into dc mechanically. Therefore a commutator is called a mechanical rectifier.
Figure 1.5 : Physical Arrangement of DC Generating Method
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commutator
Carbon brush
commutator:used to generate dc signalCarbon brush: used to collect output voltage (dc signal)
Figure 1.6: AC is converted into DC Signal through Commutation Process
1.4 EMF GENERATED ON DC MACHINE
The EMF generated:-
where z = no of conductors in the armature circuitc = no. of parallel pathN = speed in rpmΦ = flux/pole (Wb)P = no. of pole pair
Figure 1.7 : Permanent Magnet Dc Machine
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Example 1
If the no-load voltage of a separately excited generator is 135 V at 850 rpm, what will be the voltage if the speed is increased to 1000 rpm? Assume constant field excitation.
Solution
Constant field excitation:-
Example 2
A separately excited generator has no-load voltage of 140 V when the field current is adjusted to 2 A. The speed is 900 rpm. Assume a linear relationship between the field flux and current. Calculate:(i) the generated voltage when the field current is increased to 2.5 A. Assume N1=N2.
(ii) the terminal voltage when the speed is increased to 1000 rpm with the field current set at 2.2 A.
Solution
(i)
(ii)
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1.5 DC MACHINE CONSTRUCTION
rotor/armature pole
Figure 1.8 : DC Machine Construction
DC motor principles DC motors consist of rotor-mounted windings (armature) at the rotor side and
stationary windings (field poles) at the stator side.
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In all DC motors, except permanent magnet motors, current must be conducted to the armature windings by passing current through carbon brushes that slide over a set of copper surfaces called a commutator, which is mounted on the rotor.
The commutator bars are soldered to armature coils. The brush/commutator combination makes a sliding switch that energizes particular
portions of the armature, based on the position of the rotor. This process creates north and south magnetic poles on the rotor that are attracted to
or repelled by north and south poles on the stator, which are formed by passing direct current through the field windings.
It's this magnetic attraction and repulsion that causes the rotor to rotate.
Figure 1.9 : DC Machine Construction
1.6 MACHINE WINDINGS
Machine winding can be divide into 2:-(i) armature winding (rotor side)(ii) field winding (stator side)
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Figure 1.10: Winding Connection in DC Machine
SELF-EXCITED FIELD WINDING
In self-excited dc machine, there are three types of excitation method namely:(i) Series Excitation: the field winding is connected in series with the armature
circuit.(ii) Shunt Excitation: the field winding is connected in parallel with the
armature circuit.(iii) Compound Excitation: the field winding are connected in series and parallel
with the armature circuit.
The schematic diagrams for the three types of these machines are illustrated in Figure 1.11.
The difference between dc motor and dc generator is in terms of the current direction.
(a) DC Series machine
Machine Winding
Field winding Armature Winding
**Self excited-direct connection between armature circuit and the field circuit
Separately excited-no direct connection between armature circuit and the field circuit
Series excitation
Shunt excitation Compound excitation
In dc generator: armature current, Ia is supplied by the armature.In dc motor: armature current, Ia is received by the armature.
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1.0
k
1.0m +-
DC
loa
d
Ra
Eg
Ia
RfIL
VT
1.0
k
1.0m +-
5.0
Ra
EC
Ia
RfIL
dc supplyVT
Generator: Eg = VT + Ia(Ra + Rf) Motor: Ec = VT - Ia(Ra + Rf)
(b)DC Shunt machine
1.0
k 1
.0m
+-
DC
lo
ad
Ra
Eg
Ia Rf
IfIL
VT
1.0
k 1
.0m
+-
5.0
Ra
Ec
Ia Rf
IfIL
dc supplyVT
Generator: Eg = VT + IaRa Motor: Ec = VT - IaRa
(c) DC Compound machine
1,0m
1,0
m 1
,0m
+-
DC
M1
DC
load
RaILIF
Rf2
Rf1VT
Eg
Ia 1,0m
1,0
m 1
,0m
+-
DC
M1 1,0
RaILIF
Rf2
Rf1dc supply VTEc
Ia
Generator: Eg = VT + Ia(Ra + Rf 2) Motor: Ec = VT - Ia(Ra + Rf 2)
Figure 1.11
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** Eg = generated emf Ec = counter emf
1.7 POWER FLOW DIAGRAM
Power flow diagram is normally represented as a fish bone.
POWER FLOW DIAGRAM FOR DC GENERATOR
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Input Power = Output + LossesLosses can be divided into 2:-
(i) Copper losses (Armature copper loss , Pca) and (Field copper loss,Pcf)(ii) Iron losses, Pµ (friction, stray, windage,mechanical losses)
For DC Generator: Pin = Pout + Total losseswhere Pout = VTIL
For DC Motor: Pin = Pout + Total losseswhere Pin = VTIL
Total losses = Pca + Pcf + Pµ
POWER FLOW DIAGRAM FOR DC MOTOR
DC SERIES Pµ Pcf
Pin Pout= VTIL
Pca
DC SHUNT Pµ Pcf
Pin Pout= VTIL
Pca
DC COMPOUND Pµ Pcf1
Pin Pout= VTIL
Pca Pcf2
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1.8 MOTOR TORQUE
For load torque@shaft torque@net torque@output torque:
For mechanical torque:
For loss torque:
DC SERIES Pca Pµ
Pm Pout Pin=VTIL
Pcf
DC SHUNT Pca Pµ PmPin=VTIL Pout
Pcf
DC COMPOUND Pca Pcf2
Pin = VTIL Pm Pout Pµ Pcf1
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1.9 EFFICIENCY
Generally efficiency is:
For dc generator:
For dc motor:
Example 3
A short-shunt compound generator delivers 50 A at 500 V to a resistive load. The armature, series field and shunt field resistances are 0.16, 0.08 and 200 Ω, respectively. Calculate the generated EMF and armature current, if the rotational losses are 520 W, determine the efficiency of the generator.
Solution
1,0m
1,0
m 1
,0m
+- DC
M1
load
0.16 ohm
0.08 ohm
200 ohm 500V
Eg
50 AILIF
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Example 4
A 150 V shunt motor has the following parameters:Ra = 0.5 Ω, Rf = 150 Ω and rotational loss 250 W. On full load the line current is 19.5 A and the motor runs at 1400 rpm. Determine:(i) the developed power/developed mechanical power(ii) the output power(iii) the output torque(iv) the efficiency at full load
Solution
IL = 19.5AN = 1400rpm
(i)
OR :-
(ii)
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(iii)
(iv)
Tutorial 1
1. A 300 V compound motor has armature resistance 0.18 Ω, series field resistance 0.3 Ω and shunt field resistance 100 Ω. The rotational losses are 200 W. On full load the line current is 25 A and the motor runs at 1800 rpm. Determine:
(i) the developed mechanical power(ii) the output power
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(iii) the output torque(iv) the efficiency at full load
2. A 120 V series motor has 0.2 Ω field resistance. On full load, the line current is 16.5 A. The output power is 1500 W and rotational loss is 150 W. Find the value of armature resistance.
3. Briefly explain the difference between motor and generator of DC machine.
4. A compound DC motor rated at 415 V, 6 HP, 2000 rpm has armature resistance 0.18 Ω, series field resistance 0.3 Ω and shunt field resistance 100 Ω. The rotational losses are 200 W. The full load line current is 40 A.
(i) Find the developed mechanical power.(ii) Find the output power.(iii) Find the load torque.(iv) Find the efficiency of the motor.(v) Draw the power flow diagram for this type of motor.
5. A DC series generator delivers 100 kW at 10 kV to a load. The armature resistance is 20 Ω and the field resistance is 50 Ω. Calculate:
(i) the generated emf, Eg
(ii) the input power if the stray and friction losses are 400 W.
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