1. CHAPTER 1
DC 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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2. 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.
coil
Slip ring:used to
generate ac signal
Carbon brush: used to
collect output voltage (ac
signal)
carbon brush
Figure 1.3: Physical Arrangement of AC Generating Method
2

3. 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.
Carbon
commutator:used to
brush generate dc signal
Carbon brush: used to
collect output voltage
(dc signal)
commutator
Figure 1.5 : Physical Arrangement of DC Generating Method
3

4. Figure 1.6: AC is converted into DC Signal through Commutation Process
1.4 EMF GENERATED ON DC MACHINE
The EMF generated:-
zNP
EMF = 2 φ
c60
where z = no of conductors in the armature circuit
c = no. of parallel path
N = speed in rpm
Φ = flux/pole (Wb)
P = no. of pole pair
Figure 1.7 : Permanent Magnet Dc Machine
4

5. 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
2 zPNφ
EMF =
c60
2 zP
K=
c60
Therefor ,
E1 = KN 1φ1
E 2 = KN 2φ 2
Constant field excitation:- i1 = i2 , ⇒ φ1 = φ 2
E1 KN 1φ1 N
= = 1
E 2 KN 2φ 2 N 2
E1 N 2 1000rpm(135V )
E2 = = = 158.8V
N1 850rpm
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
V1 = 140V , I F 1 = 2 A, N 1 = 900rpm
φαI F
(i) I F 2 = 2.5 A
E1 KN 1φ1 I
= = F1
E 2 KN 2φ 2 I F 2
E1 I F 2 140(2.5)
E2 = = = 175V
I F1 2
(ii) N 2 = 1000rpm, I F 2 = 2.2 A, V2 = ?
5

6. E1 KN 1φ1 NI
= = 1 F1
E 2 KN 2φ 2 N 2 I F 2
E1 N 1 I F 2 140(1000)(2.2)
E2 = = = 171.11V
N 2 I F1 (900)2
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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7. • 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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8. Machine Winding
Field winding Armature Winding
Separately excited **Self excited
-no direct connection -direct connection between
between armature circuit armature circuit and the
and the field circuit field circuit
Series Compound
Shunt excitation
excitation excitation
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.
•
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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9. (a) DC Series machine
Rf Rf
IL IL
Ra Ra
dc supply
DC loa d
Ia VT Ia
VT
1.0 k
1.0 k
1.0m 1.0m
5.0
+
+
Eg EC
-
-
Generator: Eg = VT + Ia(Ra + Rf) Motor: Ec = VT - Ia(Ra + Rf)
(b)DC Shunt machine
IL If IL
If
Ra Ra
DC loa d
1.0 k
dc supply
1.0 k
1.0 m
1.0 m
Ia Rf VT 5.0
Ia Rf
+ VT
+
Eg Ec
-
-
Generator: Eg = VT + IaRa Motor: Ec = VT - IaRa
(c) DC Compound machine
Rf2 Rf2
IL IF IL
Ra IF Ra
Ia
D Cload
Ia Rf1 VT Rf1 dc supply
1,0m
1,0
1,0m
DCM1
1,0m
+
VT
+
Eg Ec
D CM1
-
1,0m
1,0m
-
Generator: Eg = VT + Ia(Ra + Rf 2) Motor: Ec = VT - Ia(Ra + Rf 2)
Figure 1.11
** Eg = generated emf
Ec = counter emf
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10. 1.7 POWER FLOW DIAGRAM
Power flow diagram is normally represented as a fish bone.
Input Power = Output + Losses
Losses 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 losses
where Pout = VTIL
For DC Motor: Pin = Pout + Total losses
where Pin = VTIL
Total losses = Pca + Pcf + Pµ
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11. POWER FLOW DIAGRAM FOR DC GENERATOR
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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12. POWER FLOW DIAGRAM FOR DC MOTOR
DC SERIES
Pca
Pµ
Pm Pout
Pin=VTIL
Pcf
DC SHUNT
Pca
Pµ
Pm
Pin=VTIL Pout
Pcf
DC COMPOUND
Pca
Pcf2
Pin = VTIL Pm Pout
Pµ
Pcf1
1.8 MOTOR TORQUE
For load torque@shaft torque@net torque@output torque:
60 Pout
To =
2πN
For mechanical torque:
60 Pm
Tm =
2πN
For loss torque:
60 Pµ
TL =
2πN
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13. 1.9 EFFICIENCY
Generally efficiency is:
Pout
η=
Pin
For dc generator:
Pout VTIL
η= =
Pin VTIL + Losses
For dc motor:
Pout Pm − Pµ
η= =
Pin VTIL
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
0.08 ohm
IF IL 50 A
0.16 ohm
load
D CM1
1,0m
200 ohm
1,0m
500V 1,0m
-
Eg
+
Pµ = 520W
Pout = VI = 500(50) = 25000W
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14. 500V
IF = = 2.5 A
200Ω
I a = I F + I L = 2.5 + 50 = 52.5 A
E g = VT + I a ( Ra + R f )
E g = 500 + (52.5)(0.08 + 0.16) = 512.6V
Pin = Pout + Ploss
Ploss = Pca + Pcf + Pµ = (52.5 2 )(0.16) + (52.5 2 )(0.08) + (2.5 2 )(200) + 520 = 2431.5W
Pout 25000
η= = x100% = 91.13%
Pin 25000 + 2431.5
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 = 19.5A
N = 1400rpm
(i) Pm = E c I a
Ia = IL − IF
I L = 19.5 A
VT 150
IF = = =1
R F 150
I a = 19.5 − 1 = 18.5 A
E c = VT − I a Ra = 150 − 18.5(0.5) = 140.75V
Pm = E c I a = (140.75)(18.5) = 2603.88W
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15. OR :- Pm = Pin − Pca − Pcf
Pm = VT I L − I a Ra − I F RF
2 2
Pm = 150(19.5) − (18.5 2 )(0.5) − (12 )(150) = 2603.88W
(ii) Pout = Pm − Pµ = 2603.88 − 250 = 2353.88W
60 Pout 60(2353.88)
(iii) To = = = 16.06 Nm
2πN 2π (1400)
Pout P 2353.88
(iv) η= = out = = 80.47%
Pin VT I L 150(19.5)
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16. 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
(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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