DC GENERATORS
- A machine that converts mechanical energy to electrical energy
DIAGRAM
COUPLING
PRIME MOVER
(MECHANICAL)
DC GENERATOR
(ELECTRICAL)
Prime Mover
- A machine or equipment that drives the generator
TYPES OF PRIME MOVER
1. Motor
2. Turbine
COMMON SOURCES OF MECHANICAL ENERGY
1. Water/River – Hydroelectric
2. Steam – Thermal Plant (Geothermal, Coal Fired, Dentrothermal, Nuclear)
3. Wind - Windmill
4. Waves
5. Heat from the sun –Solar
6. Combustible Materials – Diesel/Gasoline, Gas Turbine
THREE IMPORTANT PARAMETERS IN INDUCING EMF
1. Conductor
2. Speed
3. Magnetic Flux
Basing on Faraday’s Principle
𝒆 = 𝜷𝒍𝒗 (𝒑𝒆𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓)
Where; 𝑒 = induced emf in volt
β = flux density in Tesla
𝑙 = length of conductor in m
𝑣 = velocity in m/s
Note: Tesla =
𝑊𝑒𝑏𝑒𝑟
𝑚2
; Gauss =
𝑀𝑎𝑥𝑤𝑒𝑙𝑙
𝑐𝑚2
1 weber = 1 x 108 max
1 maxwell = 1 magnetic line of force
Also:
𝒆=
𝜽
(𝒑𝒆𝒓 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓)
𝒕
Where; θ = flux in wb
t = time in cutting the flux
but: if we let N = no. of conductors in series
then,
𝜽
𝒆 = 𝑵( )
𝒕
Ex. A 30cm conductor placed to slide along a parallel metal bar spaced 30cm apart. The end of the railway is connected
to a voltmeter. A uniform magnetic field density of 100 gauss passes perpendicular through the rail and conductor. If the
conductor will move along the rail at a uniform velocity of 20cm in 0.1 sec, determine the voltage registered on the
voltmeter.
Given:
𝑀𝑎𝑥
𝑙 = 30𝑐𝑚; 𝛽 = 100 𝑔𝑎𝑢𝑠𝑠 ( 2 ) ; 𝑑 = 20𝑐𝑚; 𝑡 = 0.1 𝑠𝑒𝑐
𝑐𝑚
From the formula:
𝒆 = 𝜷𝒍𝒗
Substitute the given, take note 𝑣 =
𝑑
𝑡
𝒆 = (𝟏𝟎𝟎
𝑴𝒂𝒙
𝟐𝟎 𝒄𝒎
) (𝟑𝟎 𝒄𝒎) (
)
𝟐
𝒄𝒎
𝟎. 𝟏 𝒔𝒆𝒄
𝒆 = 𝟔𝟎𝟎, 𝟎𝟎𝟎
Converting to SI units, which is
𝑴𝒂𝒙
𝒔𝒆𝒄
𝑊𝑒𝑏𝑒𝑟
𝑠𝑒𝑐
𝒆 = 𝟔𝟎𝟎, 𝟎𝟎𝟎
𝑴𝒂𝒙 𝟏 𝑾𝒆𝒃𝒆𝒓
(
)
𝒔𝒆𝒄 𝟏𝒙𝟏𝟎𝟖 𝑴𝒂𝒙
𝒆 = 𝟎. 𝟎𝟎𝟔 𝑽 𝒐𝒓 𝟔 𝒎𝑽
PARTS OF A BASIC D.C. GENERATOR
1. Yoke or Frame or Casing – protects the entire machine
2. Armature core – made up of laminated sheet of iron (that has very low reluctance). It carries the conductors that are
responsible in cutting the flux.
3. Armature Winding – made up of copper wires or copper bars placed on the slots of the armature core.
4. Pole Core – made up of laminated sheet of iron bolted on the frame.
5. Field Winding – made up of copper wire usually smaller in size
6. Commutator – made up of copper brass insulated in between. It is used to rectify the A.C. signals to D.C. signals. Also
called as mechanical rectifier.
7. Brush – made up of carbon compound used to transmit the current to the load without twisting the wire.
COMMUTATOR ACTION IN RECTIFYING A.C. SIGNALS
Flemmings Right Hand Rule
-
The thumb represents the direction of motion of the conductor
The first finger represents field
The second finger represents current
TYPES OF WINDING
1. Lap Winding
- the windings are connected in parallel
2. Wave Winding
- the windings are connected in series
PARALLEL PATHS
- the path for the current generated
- represented by “a”
If lap winding, then
𝒂 = 𝒎𝒑
If wave winding, then
𝒂 = 𝟐𝒎
Where:
m = the number of conductors used
simplex: m = 1
duplex: m = 2
triplex: m = 3 etc…
p = number of magnetic poles
GENERATORS INDUCED VOLTAGE
𝑬=
Where: E = voltage in volts
S = speed in rpm
𝑺𝑷𝒁𝝓
𝟔𝟎𝒂
P = number of poles
Z = number of conductors
φ = magnetic flux in weber
a = number of parallel paths
Example 1: A four pole D.C. generator with simplex lap winding has 48 slots and 4 conductors per slot. How many coil
does it have?
Given
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
𝑇𝑦𝑝𝑒 𝑜𝑓 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑠𝑖𝑚𝑝𝑙𝑒𝑥 𝑙𝑎𝑝; 𝑠𝑙𝑜𝑡𝑠 = 48;
=4
𝑠𝑙𝑜𝑡
Take note, for every coil there 2 conductors, so
𝒄𝒐𝒊𝒍 = 𝟒𝟖 𝒔𝒍𝒐𝒕𝒔 (𝟒
𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝟏 𝒄𝒐𝒊𝒍
)(
)
𝒔𝒍𝒐𝒕
𝟐 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝒄𝒐𝒊𝒍 = 𝟗𝟔 𝒄𝒐𝒊𝒍𝒔
Example 2: A four pole D.C. generator with duplex wave winding has 48 slots and 4 conductors per slot. The flux per pole
is 2.5x106 maxwells and it runsat 1500rpm. What is the induced emf?
Given
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
𝑃 = 4; 𝑇𝑦𝑝𝑒 𝑜𝑓 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑑𝑢𝑝𝑙𝑒𝑥 𝑤𝑎𝑣𝑒; 𝑠𝑙𝑜𝑡𝑠 = 48;
= 4; 𝜙 = 2.5𝑥106 𝑚𝑎𝑥; 𝑆 = 1500𝑟𝑝𝑚
𝑠𝑙𝑜𝑡
Using the formula,
𝑺𝑷𝒁𝝓
𝑬=
𝟔𝟎𝒂
Substituting values with 𝑎 = 2𝑚, and converting φ to weber to conform to SI units
𝟒 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝟏 𝑾𝒆𝒃𝒆𝒓
(𝟏𝟓𝟎𝟎)(𝟒)(𝟒𝟖 𝒔𝒍𝒐𝒕𝒔) (
) (𝟐. 𝟓𝒙𝟏𝟎𝟔 𝑴𝒂𝒙) (
𝟖 𝑴𝒂𝒙)
𝒔𝒍𝒐𝒕
𝟏𝒙𝟏𝟎
𝑬=
𝟔𝟎(𝟐)(𝟐)
𝑬 = 𝟏𝟐𝟎 𝑽
TYPES OF DC GENERATOR ACCORDING TO THE TYPE OF EXCITER USED
1. Separately-Excited Generator
- the field winding is supplied from a separate source
Circuit Diagram:
+
𝐼𝑎
𝐼𝑓
𝑉𝑓
𝑅𝑎
𝐼𝐿
𝑅𝐿
𝑅𝑓
A
R
M
+
L
O
A
D
𝑉𝐿
𝐸
−
−
Where:
𝑉𝑓 = Field Voltage
𝐼𝑓 = Field Current
𝑅𝑓 = Field Resistance
𝐸 = Generated Voltage
𝑅𝑎 = Armature Resistance
𝐼𝑎 = Armature Current
𝑉𝐿 = Load Voltage
𝐼𝐿 = Load Current
c.2 Important terms in armature winding design
a. Pole pitch is the distance between two adjacent poles. It is also defined as the number of
armature slots per pole.
𝑃𝑜𝑙𝑒 𝑝𝑖𝑡𝑐ℎ =
b. Winding Pitches
𝑛𝑜. 𝑜𝑓 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑠𝑙𝑜𝑡𝑠
𝑃
b. Commutator Pitch [Yc] is defined as the distance between two commutator segments to
which the two ends.
𝐹𝑜𝑟 𝑠𝑖𝑚𝑝𝑙𝑒𝑥 𝐿𝑎𝑝 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
𝐹𝑜𝑟 𝑠𝑖𝑝𝑙𝑒𝑥 𝑊𝑎𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
𝑦𝑐 = 2 − 1 = 1
𝑦𝑐 = 12 − 1 = 11
d. EMF Equation of DC Generator
Let,
Φ = Flux produced by each pole in weber (Wb)
P = number of poles in the DC generator.
N = speed of the armature conductor in rpm.
Z = total numbers of conductor
a = number of parallel paths
According to Faraday’s second law of electromagnetic induction, the magnitude of the induced
emf in each armature conductor is
𝑒=
b.1 Back Pitch [Yb] is measured in terms of coil sides. It is always odd number, since it is the
difference between odd and even number.
𝑦𝑏 = 14 − 1 = 13
b.2 Front Pitch [Yf] is defined as the distance measured between two coil sides which are
connected to some commutator segment.
𝐹𝑜𝑟 𝐿𝑎𝑝 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
𝐹𝑜𝑟 𝑊𝑎𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
𝑦𝑓 = 14 − 3 = 11
𝑦𝑓 = 27 − 14 = 13
b.3 Winding Pitch [Y] is defined as the distance between the starts of two consecutive coils
measure in terms of coil sides.
𝐹𝑜𝑟 𝐿𝑎𝑝 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
𝐹𝑜𝑟 𝑊𝑎𝑣𝑒 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,
pg. 5
𝑌 = 𝑦𝑏 − 𝑦𝑓
𝑌 = 𝑦𝑏 + 𝑦𝑓
𝑑∅ 𝑡𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑥
=
𝑑𝑡
𝑡𝑖𝑚𝑒
For one revolution of the conductor, total flux produced by all the poles
𝑡𝑜𝑡𝑎𝑙 𝑓𝑙𝑢𝑥 = ∅𝑃
And,
Time taken to complete one revolution
𝑡𝑖𝑚𝑒 =
Then, the induced emf of one conductor is
𝑒=
60
𝑁
∅𝑃 ∅𝑃𝑁
=
60
60
𝑁
Let us suppose there are Z total numbers of conductor in a generator, and arranged in such a
manner that all parallel paths are always in series.
Then, Z/a= number of conductors connected in series
We know that induced emf in each path is same across the line
Therefore,
1st edition MARADAZA2017
P = number of poles
Z = number of conductors
φ = magnetic flux in weber
a = number of parallel paths
Example 1: A four pole D.C. generator with simplex lap winding has 48 slots and 4 conductors per slot. How many coil
does it have?
Given
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
𝑇𝑦𝑝𝑒 𝑜𝑓 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑠𝑖𝑚𝑝𝑙𝑒𝑥 𝑙𝑎𝑝; 𝑠𝑙𝑜𝑡𝑠 = 48;
=4
𝑠𝑙𝑜𝑡
Take note, for every coil there 2 conductors, so
𝒄𝒐𝒊𝒍 = 𝟒𝟖 𝒔𝒍𝒐𝒕𝒔 (𝟒
𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝟏 𝒄𝒐𝒊𝒍
)(
)
𝒔𝒍𝒐𝒕
𝟐 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝒄𝒐𝒊𝒍 = 𝟗𝟔 𝒄𝒐𝒊𝒍𝒔
Example 2: A four pole D.C. generator with duplex wave winding has 48 slots and 4 conductors per slot. The flux per pole
is 2.5x106 maxwells and it runsat 1500rpm. What is the induced emf?
Given
𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟𝑠
𝑃 = 4; 𝑇𝑦𝑝𝑒 𝑜𝑓 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑑𝑢𝑝𝑙𝑒𝑥 𝑤𝑎𝑣𝑒; 𝑠𝑙𝑜𝑡𝑠 = 48;
= 4; 𝜙 = 2.5𝑥106 𝑚𝑎𝑥; 𝑆 = 1500𝑟𝑝𝑚
𝑠𝑙𝑜𝑡
Using the formula,
𝑺𝑷𝒁𝝓
𝑬=
𝟔𝟎𝒂
Substituting values with 𝑎 = 2𝑚, and converting φ to weber to conform to SI units
𝟒 𝒄𝒐𝒏𝒅𝒖𝒄𝒕𝒐𝒓𝒔
𝟏 𝑾𝒆𝒃𝒆𝒓
(𝟏𝟓𝟎𝟎)(𝟒)(𝟒𝟖 𝒔𝒍𝒐𝒕𝒔) (
) (𝟐. 𝟓𝒙𝟏𝟎𝟔 𝑴𝒂𝒙) (
𝟖 𝑴𝒂𝒙)
𝒔𝒍𝒐𝒕
𝟏𝒙𝟏𝟎
𝑬=
𝟔𝟎(𝟐)(𝟐)
𝑬 = 𝟏𝟐𝟎 𝑽
TYPES OF DC GENERATOR ACCORDING TO THE TYPE OF EXCITER USED
1. Separately-Excited Generator
- the field winding is supplied from a separate source
Circuit Diagram:
+
𝐼𝑎
𝐼𝑓
𝑉𝑓
𝑅𝑎
𝐼𝐿
𝑅𝐿
𝑅𝑓
A
R
M
+
L
O
A
D
𝑉𝐿
𝐸
−
−
Where:
𝑉𝑓 = Field Voltage
𝐼𝑓 = Field Current
𝑅𝑓 = Field Resistance
𝐸 = Generated Voltage
𝑅𝑎 = Armature Resistance
𝐼𝑎 = Armature Current
𝑉𝐿 = Load Voltage
𝐼𝐿 = Load Current
𝑅𝐿 = Load Resistance
Flux Analysis:
𝝓 = 𝒌 ∗ 𝑵 ∗ 𝑰𝒇
Where: k = proportionality constant
N = no. of turns (but if it is constant, then)
𝝓 = 𝒌′ ∗ 𝑰𝒇
Current Analysis:
𝑰𝒂 = 𝑰𝑳
Voltage Analysis:
𝑬 − 𝑰𝒂𝑹𝒂 − 𝑰𝑳 𝑹𝑳 = 𝟎
or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑽𝑳
or
𝑬 = 𝑰𝒂(𝑹𝒂 + 𝑹𝑳 ) ; 𝑬 = 𝑰𝑳 (𝑹𝒂 + 𝑹𝑳 )
Power Analysis:
𝑷𝒈 = 𝑬 ∗ 𝑰𝒂
and
𝑷𝑳 = 𝑽𝑳 𝑰𝑳
Where: Pg = Generated Power
PL = Load Power
Example 1: A DC Generator has no-load output voltage of 120V. Its armature circuit resistance is 0.95Ω and its field coil
is separately excited. If a load rated at 2kW, 115V is connected across the terminal, what power would be absorbed by
the load?
Take Note: No-load output voltage = Generated Voltage, E
Given:
E=120V; Ra=0.95Ω; Rated Power=2000W; Rated Voltage=115V; PL=?
Using the formula,
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑽𝑳
V=115 cannot be used for VL because this is just the stated rated voltage/ideal voltage but in reality it will change. The
only thing constant is the load resistance RL, therefore instead of using VL, its equivalent formula will be used which is
𝑉𝐿 = 𝐼𝐿 𝑅𝐿 .
In solving for 𝑅𝐿 , the formula for power will be used which is 𝑃𝐿 =
𝑉𝐿2
,
𝑅𝐿
so
𝑽𝟐𝑳 𝟏𝟏𝟓𝟐
𝑹𝑳 =
=
= 𝟔. 𝟔𝟏𝟐𝟓Ω
𝑷𝑳 𝟐𝟎𝟎𝟎
Going back to the formula,
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑽𝑳
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝑳 𝑹𝑳
But current is the same, therefore
𝑬 = 𝑰𝒂 (𝑹𝒂 + 𝑹𝑳 )
Substituting values,
𝟏𝟐𝟎 = 𝑰𝒂 (𝟎. 𝟗𝟓 + 𝟔. 𝟔𝟏𝟐𝟓)
𝑰𝒂 = 𝟏𝟓. 𝟖𝟕 𝑨
But take note 𝐼𝑎 = 𝐼𝐿 . Also, in solving for 𝑃𝐿 , we cannot use the formula P=VI because of the voltage, so the formula
used will be
𝑷𝑳 = 𝑰𝟐𝑳 𝑹𝑳
𝑷𝑳 = (𝟏𝟓. 𝟖𝟕𝟐 )(𝟔. 𝟔𝟏𝟐𝟓)
𝑷𝑳 = 𝟏, 𝟔𝟔𝟓. 𝟒 𝑾
Example 2: If a no-load voltage of a separately excited DC Generator is at 110V at 1350rpm, what would be the voltage if
the speed increases to 1600rpm? And if it is decreased to 1100rpm?
Given
𝐸1 = 110𝑉 @ 𝑆1 = 1350 𝑟𝑝𝑚; 𝐸2 =? @ 𝑠2 = 1600 𝑟𝑝𝑚; 𝐸3 =? @ 𝑆3 = 1100 𝑟𝑝𝑚
For this problem, since the generator will be the same even if the speed is changed, the following ratio and proportion
formula can be used,
𝑬𝟏 𝑺𝟏
=
𝑬𝟐 𝑺𝟐
Solving for 𝐸2
𝟏𝟏𝟎 𝟏𝟑𝟓𝟎
=
𝑬𝟐
𝟏𝟔𝟎𝟎
𝑬𝟐 = 𝟏𝟑𝟎. 𝟑𝟕 𝑽
Solving for 𝐸3
𝟏𝟏𝟎 𝟏𝟑𝟓𝟎
=
𝑬𝟑
𝟏𝟏𝟎𝟎
𝑬𝟑 = 𝟖𝟗. 𝟔𝟑 𝑽
2. Self-Excited Generator
- the source of the field circuit is from the generated voltage of the armature
TYPES OF SELF-EXCITED GENERATOR
A. Shunt Generator
- the field circuit is in parallel with the armature.
Circuit Diagram:
+
𝐼𝑎
𝑅𝑎
𝐼𝑠ℎ
𝑅𝑠ℎ
+
A
R
M
𝐼𝐿
𝑅𝐿
L
O
A
D
𝑉𝐿
𝐸
−
−
Where: 𝐼𝑠ℎ = Shunt current
𝑉𝑠ℎ = Shunt Voltage
Current Analysis:
𝑰𝒂 = 𝑰𝒔𝒉 + 𝑰𝑳
Voltage Analysis:
𝑬 − 𝑰𝒂𝑹𝒂 − 𝑰𝑳 𝑹𝑳 = 𝟎
Or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑽𝑳
Power Generated:
𝑷𝒈 = 𝑬 ∗ 𝑰𝒂
and
𝑷𝑳 = 𝑽𝑳 𝑰𝑳
Example 1: The emf induced in the armature at 400kW, 250V shunt generator is 258.866V when the terminal voltage
and load current are at rated values and the shunt field current is 12A. The armature resistance including the brushes is
0.0055Ω. Find,(a) Terminal Voltage, (b) Power Generated, (c) Power Output
Solution:
Since the problem stated specifically that the generator voltage and current are at rated values, then we can directly use
𝑉𝐿 = 250 𝑉 𝑎𝑛𝑑 𝑃𝐿 = 400 𝑘𝑊
For question a, terminal voltage means the voltage at the load side, therefore
𝑻𝒆𝒓𝒎𝒊𝒏𝒂𝒍 𝑽𝒐𝒍𝒕𝒂𝒈𝒆 = 𝑽𝑳 = 𝟐𝟓𝟎 𝑽
For question b,
𝑷𝒈 = 𝑬 ∗ 𝑰𝒂
For Ia,
𝑰𝒂 = 𝑰𝑳 + 𝑰𝒔𝒉
𝑰𝒂 =
𝑰𝒂 =
𝑷𝑳
+ 𝑰𝒔𝒉
𝑽𝑳
𝟒𝟎𝟎, 𝟎𝟎𝟎
+ 𝟏𝟐
𝟐𝟓𝟎
𝑰𝒂 = 𝟏, 𝟔𝟏𝟐 𝑨
So,
𝑷𝒈 = (𝟐𝟓𝟖. 𝟖𝟔𝟔)(𝟏𝟔𝟏𝟐)
𝑷𝒈 = 𝟒𝟏𝟕, 𝟐𝟗𝟏. 𝟗𝟗𝟐 𝑾
For question c, Power output means the power at the load side, therefore
𝑷𝒐𝒘𝒆𝒓 𝑶𝒖𝒕𝒑𝒖𝒕 = 𝑷𝑳 = 𝟒𝟎𝟎, 𝟎𝟎𝟎 𝑾
Example 2: A 4-pole shunt generator has a simplex lap connected armature with 728 conductors and a flux per pole of
25𝑚Wb. The generator supplies two hundred fifty 110V, 75W incandescent lamps. The field and armature resistances
are 110Ω and 0.025Ω respectively. Determine the speed of this generator.
Solution:
To determine the speed, the only formula that can be used is,
𝑺𝑷𝒁𝝓
𝑬=
𝟔𝟎𝒂
Solving for E,
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑽𝑳
But,
𝑰𝒂 = 𝑰𝑳 + 𝑰𝒔𝒉
𝑷𝑳 𝑽𝒔𝒉
+
𝑽𝑳 𝑹𝒔𝒉
𝟐𝟓𝟎(𝟕𝟓) 𝟏𝟏𝟎
𝑰𝒂 =
+
𝟏𝟏𝟎
𝟏𝟏𝟎
𝑰𝒂 =
𝑰𝒂 = 𝟏𝟕𝟏. 𝟒𝟓
So,
𝑬 = (𝟏𝟕𝟏. 𝟒𝟓)(𝟎. 𝟎𝟐𝟓) + 𝟏𝟏𝟎
𝑬 = 𝟏𝟏𝟒. 𝟐𝟗
Therefore, solving for the speed with 𝑎 = 𝑚𝑝
𝟏𝟏𝟒. 𝟐𝟗 =
𝑺(𝟒)(𝟕𝟐𝟖)(𝟎. 𝟎𝟐𝟓)
𝟔𝟎(𝟏)(𝟒)
𝑺 = 𝟑𝟕𝟔. 𝟕𝟖 𝒓𝒑𝒎
B. Series Generator
- the field winding is connected in series with the armature.
Circuit Diagram:
𝑅𝑠𝑒
+
𝐼𝑎
𝐼𝑠𝑒
𝑅𝑎
𝐼𝐿
𝑅𝐿
+
A
R
M
L
O
A
D
𝑉𝐿
𝐸
−
−
Where: 𝐼𝑠𝑒 = Series Field Current
𝑅𝑠𝑒 = Series Field Resistance
Current Analysis:
𝑰𝒂 = 𝑰𝒔𝒆 = 𝑰𝑳
Voltage Analysis:
𝑬 − 𝑰𝒂𝑹𝒂 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝑳 𝑹𝑳 = 𝟎
or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝑳 𝑹𝑳
or
𝑬 = 𝑰𝒂 (𝑹𝒂 + 𝑹𝒔𝒆 + 𝑹𝑳 ); 𝑬 = 𝑰𝒔𝒆 (𝑹𝒂 + 𝑹𝒔𝒆 + 𝑹𝑳 ); 𝑬 = 𝑰𝑳 (𝑹𝒂 + 𝑹𝒔𝒆 + 𝑹𝑳 )
Power Generated:
𝑷𝒈 = 𝑬 ∗ 𝑰𝒂
and
𝑷𝑳 = 𝑽𝑳 𝑰𝑳
Example: A series generator is used to supply a series lighting system that draws a total current of 4.5A. if the armature
and the series field resistances are 0.1Ω and 0.2Ω respectively. Calculate the induced emf if the terminal voltage is 600V
with a diverter resistance of 0.3Ω connected across the field circuit.
Solution:
Diverter resistance is just a resistor in parallel with the series field resistance.
Rx 𝑅𝑠𝑒
+
𝐼𝑠𝑒
𝐼𝑎
𝑅𝑎
𝐼𝐿
𝑅𝑑
𝑅𝐿
+
A
R
M
L
O
A
D
𝑉𝐿
𝐸
−
−
Solving for the total resistance between the parallel connection of the series field resistance and the diverter resistance,
Rx,
(𝑹𝒔𝒆 )(𝑹𝒅 )
𝑹𝒙 =
𝑹𝒔𝒆 + 𝑹𝒅
𝑹𝒙 =
(𝟎. 𝟐)(𝟎. 𝟑)
𝟎. 𝟐 + 𝟎. 𝟑
𝑹𝒙 = 𝟎. 𝟏𝟐
From this, the equivalent circuit becomes,
𝑅𝑥
+
𝐼𝑥
𝐼𝑎
𝑅𝑎
𝐼𝐿
𝑅𝐿
A
R
M
+
L
O
A
D
𝑉𝐿
𝐸
−
−
From this, the circuit is just the same as an ordinary series generator. Take note that the induced emf in question is the
generated voltage, E, and recall also that terminal voltage is the load voltage VL. From the problem, it was given that the
load draws current of 4.5A, therefore, the load current is 4.5A. From this, solving for the generated voltage
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝑳 𝑹𝑳
Instead of using 𝐼𝐿 𝑅𝐿 , we will use the given load voltage of 600V. Also, 𝑅𝑠𝑒 will now become 𝑅𝑋 . Lastly, since current is
the same for series generator,
𝑬 = 𝑰𝑳 (𝑹𝒂 + 𝑹𝑿 ) + 𝑽𝑳
𝑬 = (𝟒. 𝟓)(𝟎. 𝟏 + 𝟎. 𝟏𝟐) + 𝟔𝟎𝟎
𝑬 = 𝟔𝟎𝟏 𝑽
C. Compound Generator
- a shunt generator with additional series field winding
Types of Compound Generator
1. Long-Shunt Compound Generator
- the series field is connected in series with the armature
Circuit Diagram:
𝑅𝑠𝑒
+
𝐼𝑎
𝐼𝑠𝑒
𝑅𝑎
𝐼𝐿
𝐼𝑠ℎ
𝑅𝐿
𝑅𝑠ℎ
+
A
R
M
L
O
A
D
𝑉𝐿
𝐸
−
−
Current Analysis:
𝑰𝒂 = 𝑰𝒔𝒆 = 𝑰𝒔𝒉 + 𝑰𝑳
Voltage Analysis:
𝑬 − 𝑰𝒂𝑹𝒂 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝑳 𝑹𝑳 = 𝟎
or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝑳 𝑹𝑳
or
𝑬 = 𝑰𝒂(𝑹𝒂 + 𝑹𝒔𝒆) + 𝑽𝑳 ; 𝑬 = 𝑰𝒔𝒆(𝑹𝒂 + 𝑹𝒔𝒆) + 𝑽𝑳
2. Short-Shunt Compound Generator
- the series field is in series with the load
Circuit Diagram:
𝑅𝑠𝑒
+
𝐼𝑎
𝐼𝑠𝑒
𝑅𝑎
𝐼𝐿
𝐼𝑠ℎ
+
A
R
M
𝑅𝑠ℎ
𝑅𝐿
L
O
A
D
𝑉𝐿
𝐸
−
−
Current Analysis:
𝑰𝒔𝒆 = 𝑰𝑳
and
𝑰𝒂 = 𝑰𝒔𝒉 + 𝑰𝑳
or
𝑰𝒂 = 𝑰𝒔𝒉 + 𝑰𝒔𝒆
Voltage Analysis:
𝑬 − 𝑰𝒂𝑹𝒂 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝑳 𝑹𝑳 = 𝟎
or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝑳 𝑹𝑳
or
𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝒔𝒆(𝑹𝒔𝒆 + 𝑹𝑳 ); 𝑬 = 𝑰𝒂𝑹𝒂 + 𝑰𝑳 (𝑹𝒔𝒆 + 𝑹𝑳 )
GENERATORS EFFICIENCY
- the ratio of the power output to that of the power input
%𝑬𝒇𝒇 =
𝑷𝒐
∗ 𝟏𝟎𝟎%
𝑷𝒊
or
%𝑬𝒇𝒇 =
𝑷𝒐
∗ 𝟏𝟎𝟎%
𝑷𝒐 + 𝑷𝑳𝑶𝑺𝑺𝑬𝑺
Types of Power Losses
1. Electrical Losses
- losses due to its windings and brush contact
a) Copper Loss, Pcu
a.1) Armature Loss, Pa
𝑃𝑎 = 𝐼𝑎2 𝑅𝑎
a.2) Shunt Field Loss, Psh
𝑃𝑠ℎ = 𝐼𝑠ℎ2 𝑅𝑠ℎ
a.3) Series Field Loss, Pse
𝑃𝑠𝑒 = 𝐼𝑠𝑒 2 𝑅𝑠𝑒
a.4) Diverter Loss, Pd
𝑃𝑑 = 𝐼𝑑2 𝑅𝑑
a.5) Brush Contact Loss, Pb
𝑃𝑏 = 𝐼𝑏 2 𝑅𝑏
2. Mechanical Losses
- sometimes called as CONSTANT LOSS or STRAY POWER LOSS or ROTATIONAL LOSS
- it is constant provided that the speed and flux are also constant
a) Core Loss, Pco
a.1) Eddy Current Loss, Pe
𝑃𝑒 = 𝑘𝑒 𝑆 2 𝜙 2
a.2) Hysteresis Loss, Ph
𝑃ℎ = 𝑘ℎ 𝑆(𝛽𝑚)1.6
Where: βm = maximum flux density
b) Friction and Windage Loss, Pfw
- due to friction, ventilation bearing, vibration etc.
c) Stray Load Loss, PSL
- flux distortion
- 1% of the output for machines 150kW and over
Example 1: A 250kW, 230V compound generator is delivering 800A at 230V. the shunt field current is 12A. Armature
resistance is 0.007Ω and series field resistance is 0.002Ω. The stray power loss is 5500W and the generator is connected
long shunt. Determine the full-load efficiency at rated voltage.
Given:
𝐼𝐿 = 800𝐴;
𝑉𝐿 = 230𝑉;
𝐼𝑠ℎ = 12𝐴;
𝑅𝑎 = 0.007Ω; 𝑅𝑠𝑒 = 0.002Ω; 𝑃𝑚𝑒𝑐ℎ = 5500; %𝑒𝑓𝑓 =?
Solving first electrical losses due to all the resistance of the windings starting with the shunt field winding, but, since the
shunt field resistance is not given, then the following formula will be used
𝑷𝒔𝒉 = 𝑽𝒔𝒉 𝑰𝒔𝒉
But, the shunt field resistance is in parallel with the load, therefore 𝑉𝑠ℎ = 𝑉𝐿
𝑷𝒔𝒉 = 𝑽𝑳 𝑰𝒔𝒉
𝑷𝒔𝒉 = (𝟐𝟑𝟎)(𝟏𝟐)
𝑷𝒔𝒉 = 𝟐𝟕𝟔𝟎 𝑾
Next is the series field winding, since the series field winding resistance is given, the following formula can be used,
𝑷𝒔𝒆 = 𝑰𝒔𝒆𝟐 𝑹𝒔𝒆
But,
𝑰𝒔𝒆 = 𝑰𝑳 + 𝑰𝒔𝒉
𝑰𝒔𝒆 = 𝟖𝟎𝟎 + 𝟏𝟐
𝑰𝒔𝒆 = 𝟖𝟏𝟐
So,
𝑷𝒔𝒆 = (𝟖𝟏𝟐)𝟐 (𝟎. 𝟎𝟎𝟐)
𝑷𝒔𝒆 = 𝟏, 𝟑𝟏𝟖. 𝟔𝟖𝟖 𝑾
Last is for the armature winding, since the armature resistance is given, the following formula can be used,
𝑷𝒂 = 𝑰𝒂𝟐 𝑹𝒂
But, armature is in series with the series field, therefore
𝑷𝒂 = 𝑰𝒔𝒆𝟐 𝑹𝒂
𝑷𝒂 = (𝟖𝟏𝟐)𝟐 (𝟎. 𝟎𝟎𝟕)
𝑷𝒂 = 𝟒, 𝟔𝟏𝟓. 𝟒𝟎𝟖 𝑾
Solving for the efficiency of the generator,
%𝑬𝒇𝒇 =
%𝑬𝒇𝒇 =
𝑷𝒐
∗ 𝟏𝟎𝟎%
𝑷𝒐 + 𝑷𝑳𝑶𝑺𝑺𝑬𝑺
(𝟖𝟎𝟎)(𝟐𝟑𝟎)
∗ 𝟏𝟎𝟎%
(𝟖𝟎𝟎)(𝟐𝟑𝟎) + 𝟓𝟓𝟎𝟎 + 𝟐𝟕𝟔𝟎 + 𝟏𝟑𝟏𝟖. 𝟔𝟖𝟖 + 𝟒𝟔𝟏𝟓. 𝟒𝟎𝟖
%𝑬𝒇𝒇 = 𝟗𝟐. 𝟖𝟒%
Example 2: A short-shunt compound generator supplies a load of 50A at a terminal voltage of 250V. the armature, series
field and shunt field resistances are 0.05Ω, 0.03Ω and 250Ω respectively. Calculate the over-all efficiency if the constant
loss is 325W.
Given:
𝐼𝐿 = 50𝐴;
𝑉𝐿 = 250𝑉;
𝑅𝑎 = 0.05Ω; 𝑅𝑠𝑒 = 0.03Ω; 𝑅𝑠ℎ = 250Ω; %𝑒𝑓𝑓 =?
Solving first electrical losses due to all the resistance of the windings starting with the series field winding since it is just
in series with the load,
𝑷𝒔𝒆 = 𝑰𝒔𝒆𝟐 𝑹𝒔𝒆
Since it is in series with the load, then 𝐼𝑠𝑒 = 𝐼𝐿
𝑷𝒔𝒆 = 𝑰𝟐𝑳 𝑹𝒔𝒆
𝑷𝒔𝒆 = (𝟓𝟎)𝟐 (𝟎. 𝟎𝟑)
𝑷𝒔𝒆 = 𝟕𝟓 𝑾
Next is the shunt field winding,
𝑷𝒔𝒉 = 𝑰𝒔𝒉𝟐 𝑹𝒔𝒉
𝑽𝒔𝒉 𝟐
𝑷𝒔𝒉 = (
) 𝑹𝒔𝒉
𝑹𝒔𝒉
𝑷𝒔𝒉 = (
𝑷𝒔𝒉 = [
𝑽𝑳 + 𝑽𝒔𝒆 𝟐
) 𝑹𝒔𝒉
𝑹𝒔𝒉
𝑽𝑳 + (𝑰𝒔𝒆 )(𝑹𝒔𝒆 ) 𝟐
] 𝑹𝒔𝒉
𝑹𝒔𝒉
𝟐
𝑽𝑳 + (𝑰𝒔𝒆 )(𝑹𝒔𝒆 )
𝑷𝒔𝒉 = [
] 𝑹𝒔𝒉
𝑹𝒔𝒉
𝟐
𝟐𝟓𝟎 + (𝟓𝟎)(𝟎. 𝟎𝟑)
𝑷𝒔𝒉 = [
] (𝟐𝟓𝟎)
𝟐𝟓𝟎
𝑷𝒔𝒉 = 𝟐𝟓𝟏. 𝟓 𝑾
Last is for the armature winding,
𝑷𝒂 = 𝑰𝒂𝟐 𝑹𝒂
But
𝑰𝒂 = 𝑰𝒔𝒉 + 𝑰𝒔𝒆
𝑰𝒂 =
𝑰𝒂 =
𝑽𝑳 + (𝑰𝒔𝒆 )(𝑹𝒔𝒆 )
+ 𝑰𝒔𝒆
𝑹𝒔𝒉
𝟐𝟓𝟎 + (𝟓𝟎)(𝟎. 𝟎𝟑)
+ 𝟓𝟎
𝟐𝟓𝟎
𝑰𝒂 = 𝟓𝟏. 𝟎𝟎𝟔 𝑨
so,
𝑷𝒂 = (𝟓𝟏. 𝟎𝟎𝟔)𝟐 (𝟎. 𝟎𝟓)
𝑷𝒂 = 𝟏𝟑𝟎. 𝟎𝟖 𝑾
Therefore, solving for the efficiency of the generator,
%𝑬𝒇𝒇 =
%𝑬𝒇𝒇 =
𝑷𝒐
∗ 𝟏𝟎𝟎%
𝑷𝒐 + 𝑷𝑳𝑶𝑺𝑺𝑬𝑺
(𝟓𝟎)(𝟐𝟓𝟎)
∗ 𝟏𝟎𝟎%
(𝟓𝟎)(𝟐𝟓𝟎) + 𝟑𝟐𝟓 + 𝟕𝟓 + 𝟐𝟓𝟏. 𝟓 + 𝟏𝟑𝟎. 𝟎𝟖
%𝑬𝒇𝒇 = 𝟗𝟒. 𝟏𝟐%
DC MOTORS
- A machine that converts electrical energy to mechanical energy
Speed Characteristics of DC Motors
𝑺 = 𝒌𝒃
𝑬𝒃
𝝓
Where: S = Speed in rpm
kb = proportionality constant
Eb = Back emf or Counter emf
= as the armature rotates, it also induces its own voltage
Φ = flux per pole
Torque Characteristics of DC Motors
Torque – turning moment of rotating body
Consider a shaft,
𝑟
𝐹
𝑻 = 𝑭 ∗ 𝒓 (𝑵. 𝒎)
Where: T = Torque in Newton-meter
F = Force in Newton
r = radius in meters
HORSEPOWER DEVELOPED IN A ROTATIONAL BODY
𝑯𝑷 =
𝟐𝝅𝑺𝑻
𝟒𝟒𝟕𝟔𝟎
Where: HP = mechanical power in horsepower
S = speed in rpm
T = torque in N-m
Note:
𝟏 𝑯𝑷 = 𝟕𝟒𝟔 𝑾
From this, a variation of a formula for torque can be developed
𝑻 = 𝒌𝒕 𝝓𝑰𝒂
Where: T = torque
kt = proportionality constant
φ = flux per pole
𝐼𝑎 = armature current
TYPES OF DC MOTORS
A. Shunt Motor
+
𝐼𝑠
𝐼𝑎
𝑅𝑎
𝐼𝑠ℎ
𝑉𝑠
𝑅𝑠ℎ
+
𝐸𝑏
A
R
M
−
−
Where:
Vs = Supply Voltage
Is = Supply Current
Current Analysis:
𝑰𝒔 = 𝑰𝒔𝒉 + 𝑰𝒂
Voltage Analysis:
𝑽𝒔 − 𝑰𝒂𝑹𝒂 − 𝑬𝒃 = 𝟎
or
𝑽𝒔 = 𝑰𝒂𝑹𝒂 + 𝑬𝒃
Power Analysis:
𝑷𝒔 = 𝑽𝒔𝑰𝒔
and
𝑷𝒅 = 𝑬𝒃𝑰𝒂
Where: Ps = Supply Power in Watts
Pd = Power Developed in the armature
Example: A 50Hp, 550V shunt motor has an armature resistance of 0.36Ω when operating at rated load speed while the
armature takes 75A. What resistance should be inserted in the armature circuit to obtain 20% reduction in flux and
when developing 70% of the rated torque?
Given:
𝑃 = 50𝐻𝑝;
𝑉𝑆 = 550𝑉;
𝑅𝑎 = 0.36;
𝐼𝑎1 = 75𝐴;
𝑅𝑠 =?; 𝜙2 = 0.8𝜙1 ; 𝑇2 = 0.7𝑇1
This problem has two scenarios where the first scenario is the standard shunt motor and the second scenario has an
added resistance to the armature. First, recall the characteristics of Torque
𝑻 = 𝒌𝒕 𝝓𝑰𝒂
But,
𝑻𝟐 = 𝟎. 𝟕𝑻𝟏
𝒌𝒕 𝝓𝟐 𝑰𝒂𝟐 = 𝟎. 𝟕(𝒌𝒕 𝝓𝟏 𝑰𝒂𝟏 )
But, constant remains constant as long as it’s the same motor and also from the problem 𝜙2 = 0.8𝜙1 , thus
𝒌𝒕 (𝟎. 𝟖𝝓𝟏 )𝑰𝒂𝟐 = 𝟎. 𝟕(𝒌𝒕 𝝓𝟏 𝑰𝒂𝟏 )
𝑰𝒂𝟐 =
𝟎. 𝟕
(𝑰 )
𝟎. 𝟖 𝒂𝟏
𝟎. 𝟕
(𝟕𝟓)
𝟎. 𝟖
= 𝟔𝟓. 𝟔𝟐𝟓𝑨
𝑰𝒂 𝟐 =
𝑰𝒂𝟐
With the resistance inserted, the circuit becomes like this,
+
𝐼𝑠
𝐼𝑎2
𝑅𝑎
𝐼𝑠ℎ
𝑉𝑠
𝑅𝑠ℎ
+
𝐸𝑏2
A
R
M
−
−
Using the formula,
𝑽𝒔 = 𝑰𝒂𝑹𝒂 + 𝑬𝒃
But this time, since there is a resistance in series with the armature, the formula becomes,
𝑽𝒔 = 𝑰𝒂𝟐 (𝑹𝒂 + 𝑹𝒔) + 𝑬𝒃𝟐
This formula now needs 𝐸𝑏2 to be solved, recall the characteristics of motor speed,
𝑺 = 𝒌𝒃
𝑬𝒃
𝝓
Since the motor will still be the same even if there is an added resistance, and the problem did not state any changes to
the motor speed after adding the resistance, it is safe to assume that 𝑆2 = 𝑆1. From this, we can have a ratio and
proportion approach,
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃 𝝓
𝟐
Since it is just the same motor, constant will stay the same. Add in also the scenario for flux which is 𝜙2 = 0.8𝜙1 , the
formula now becomes
𝑬𝒃𝟏
𝝓𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃
𝟎. 𝟖(𝝓𝟏 )
𝑺𝟏 = 𝒌𝒃
𝑺𝟏 𝑬𝒃𝟏
=
𝑺𝟐 𝑬𝒃𝟐
𝟎. 𝟖
But since 𝑆2 = 𝑆1 , then
𝟏=
𝑬𝒃𝟏
𝑬𝒃𝟐
𝟎. 𝟖
𝑬𝒃𝟐 = 𝟎. 𝟖𝑬𝒃𝟏
Solving first for 𝐸𝑏1 ,
𝑽𝒔 = 𝑰𝒂𝟏 𝑹𝒂 + 𝑬𝒃𝟏
𝟓𝟓𝟎 = (𝟕𝟓)(𝟎. 𝟑𝟔) + 𝑬𝒃𝟏
𝑬𝒃𝟏 = 𝟓𝟐𝟑𝑽
Solving for 𝐸𝑏2 ,
𝑬𝒃𝟐 = 𝟎. 𝟖𝑬𝒃𝟏
𝑬𝒃𝟐 = 𝟎. 𝟖(𝟓𝟐𝟑)
𝑬𝒃𝟐 = 𝟒𝟏𝟖. 𝟒
Solving for 𝑅𝑠
𝑽𝒔 = 𝑰𝒂𝟐 (𝑹𝒂 + 𝑹𝒔) + 𝑬𝒃𝟐
𝟓𝟓𝟎 = (𝟔𝟓. 𝟔𝟐𝟓)(𝟎. 𝟑𝟔 + 𝑹𝒔) + 𝟒𝟏𝟖. 𝟒
𝑹𝒔 = 𝟏. 𝟔𝟒𝟓Ω
B. Series Motor
𝑅𝑠𝑒
+
𝐼𝑠
𝐼𝑠𝑒
𝐼𝑎
𝑅𝑎
𝑉𝑠
+
𝐸𝑏
−
A
R
M
−
Current Analysis:
𝑰𝒔 = 𝑰𝒔𝒆 = 𝑰𝒂
Voltage Analysis:
𝑽𝒔 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝒂𝑹𝒂 − 𝑬𝒃 = 𝟎
or
𝑽𝒔 = 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝒂𝑹𝒂 + 𝑬𝒃
or
𝑽𝒔 = 𝑰𝒔(𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃; 𝑽𝒔 = 𝑰𝒔𝒆(𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃; 𝑽𝒔 = 𝑰𝒂(𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃
Power Analysis:
𝑷𝒔 = 𝑽𝒔𝑰𝒔
and
𝑷𝒅 = 𝑬𝒃𝑰𝒂
Example: A 400V series motor working with an unsaturated field is taking 60A and running at 840RPM. If the armature
and series field resistances are 0.04Ω and 0.06Ω respectively, at what speed will the motor run when developing half-full
load torque?
Given:
𝑉𝑠 = 400𝑉;
𝐼𝑎 = 60𝐴;
𝑆1 = 840 𝑟𝑝𝑚; 𝑅𝑎 = 0.04Ω;
𝑅𝑠𝑒 = 0.06Ω; 𝑆2 =?; 𝑇2 = 0.5𝑇1
Recall the torque characteristics,
𝑻 = 𝒌𝒕 𝝓𝑰𝒂
But,
𝑻𝟐 = 𝟎. 𝟓𝑻𝟏
𝒌𝒕 𝝓𝟐 𝑰𝒂𝟐 = 𝟎. 𝟓(𝒌𝒕 𝝓𝟏 𝑰𝒂𝟏 )
The problem did not state that there are changes in flux but flux is directly proportional to the field winding current and
with the change in torque there will be changes in the armature current. Considering the motor connected in series, a
change in armature current will also mean a change in field winding current, therefore a change will also occur in the
field winding flux. Recall,
𝝓 = 𝒌′ 𝑰𝒇
For series motor, 𝐼𝑓 = 𝐼𝑠𝑒, but also 𝐼𝑠𝑒 = 𝐼𝑎, thus
𝝓 = 𝒌′ 𝑰𝒂
Substituting flux to the formula from torque,
𝒌𝒕 (𝒌′ 𝑰𝒂𝟐 )𝑰𝒂𝟐 = 𝟎. 𝟓(𝒌𝒕 (𝒌′ 𝑰𝒂𝟏 )𝑰𝒂𝟏 )
Constant remains the same, thus
(𝑰𝒂𝟐 )𝟐 = 𝟎. 𝟓(𝑰𝒂𝟏 )𝟐
𝑰𝒂𝟐 = √𝟎. 𝟓(𝑰𝒂𝟏 )𝟐
𝑰𝒂𝟐 = (√𝟎. 𝟓)(𝑰𝒂𝟏 )
𝑰𝒂𝟐 = (√𝟎. 𝟓)(𝟔𝟎)
𝑰𝒂𝟐 = 𝟒𝟐. 𝟒𝟐𝟔𝑨
Since the problems asks for the motor speed after the change in torque, the same ratio and proportion can be used
therefore,
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃 𝝓
𝟐
Solving first for 𝐸𝑏1 ,
𝑽𝒔 = 𝑰𝒂𝟏 (𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃𝟏
𝟒𝟎𝟎 = (𝟔𝟎)(𝟎. 𝟎𝟔 + 𝟎. 𝟎𝟒) + 𝑬𝒃𝟏
𝑬𝒃𝟏 = 𝟑𝟗𝟒𝑽
Solving for 𝐸𝑏2 ,
𝑽𝒔 = 𝑰𝒂𝟐 (𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃𝟐
𝟒𝟎𝟎 = (𝟒𝟐. 𝟒𝟐𝟔)(𝟎. 𝟎𝟔 + 𝟎. 𝟎𝟒) + 𝑬𝒃𝟐
𝑬𝒃𝟐 = 𝟑𝟗𝟓. 𝟕𝟓𝟕𝟒𝑽
Going back to the ratio and proportion
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃
𝝓𝟐
𝑬𝒃𝟏
𝒌′ 𝑰𝒂𝟏
𝑬𝒃
𝑺𝟐 = 𝒌𝒃 ′ 𝟐
𝒌 𝑰𝒂𝟐
𝑺𝟏 = 𝒌𝒃
𝟑𝟗𝟒
𝟖𝟒𝟎
𝟔𝟎
=
𝟑𝟗𝟓. 𝟕𝟓𝟕𝟒
𝑺𝟐
𝟒𝟐. 𝟒𝟐𝟔
𝑺𝟐 = 𝟏, 𝟏𝟗𝟑. 𝟐𝟓 𝑹𝑷𝑴
C. Compound Motor
1. Long-Shunt Compound Motor
𝑅𝑠𝑒
+
𝐼𝑠
𝐼𝑠𝑒
𝐼𝑎
𝑅𝑎
𝐼𝑠ℎ
𝑉𝑠
𝑅𝑠ℎ
+
𝐸𝑏
−
A
R
M
−
Current Analysis:
𝑰𝒔𝒆 = 𝑰𝒂
so
𝑰𝒔 = 𝑰𝒔𝒉 + 𝑰𝒔𝒆
Or
𝑰𝒔 = 𝑰𝒔𝒉 + 𝑰𝒂
Voltage Analysis:
𝑽𝒔 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝒂𝑹𝒂 − 𝑬𝒃 = 𝟎
or
𝑽𝒔 = 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝒂𝑹𝒂 + 𝑬𝒃
or
𝑽𝒔 = 𝑰𝒔𝒆(𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃; 𝑽𝒔 = 𝑰𝒂(𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃
Power Analysis:
𝑷𝒔 = 𝑽𝒔𝑰𝒔
and
𝑷𝒅 = 𝑬𝒃𝑰𝒂
2. Short-Shunt Compound Motor
𝑅𝑠𝑒
+
𝐼𝑠
𝐼𝑠𝑒
𝐼𝑎
𝑅𝑎
𝐼𝑠ℎ
𝑉𝑠
𝑅𝑠ℎ
+
𝐸𝑏
A
R
M
−
−
Current Analysis:
𝑰𝒔 = 𝑰𝒔𝒆 = 𝑰𝒔𝒉 + 𝑰𝒂
Voltage Analysis:
𝑽𝒔 − 𝑰𝒔𝒆𝑹𝒔𝒆 − 𝑰𝒂𝑹𝒂 − 𝑬𝒃 = 𝟎
or
𝑽𝒔 = 𝑰𝒔𝒆𝑹𝒔𝒆 + 𝑰𝒂𝑹𝒂 + 𝑬𝒃
or
𝑽𝒔 = 𝑰𝒔𝑹𝒔𝒆 + 𝑰𝒂𝑹𝒂 + 𝑬𝒃
Power Analysis:
𝑷𝒔 = 𝑽𝒔𝑰𝒔
and
𝑷𝒅 = 𝑬𝒃𝑰𝒂
Example 5: A 550V long-shunt compound motor has an armature resistance of 0.815Ω and series field resistance of
0.15Ω. the full-load speed is 1900rpm when the armature current is 22A. At what speed will the motor operate at noload if the armature current drops to 3A with a corresponding drop in flux to 90% of the full-load value?
Given:
𝑉𝑠 = 550𝑉;
𝑅𝑎 = 0.815Ω; 𝑅𝑠𝑒 = 0.15Ω; 𝑆1 = 1900𝑟𝑝𝑚; 𝐼𝑎1 = 22𝐴;
𝑆2 =?; 𝐼𝑎2 = 3𝐴;
𝜙2 = 0.9𝜙1
For this problem, the same ratio and proportion will be used,
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃 𝝓
𝟐
Solving first for 𝐸𝑏1 , since this problem is a long-shunt compound motor, then
𝑽𝒔 = 𝑰𝒂𝟏 (𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃𝟏
𝟓𝟓𝟎 = (𝟐𝟐)(𝟎. 𝟏𝟓 + 𝟎. 𝟖𝟏𝟓) + 𝑬𝒃𝟏
𝑬𝒃𝟏 = 𝟓𝟐𝟖. 𝟕𝟕 𝑽
Solving for 𝐸𝑏2 ,
𝑽𝒔 = 𝑰𝒂𝟐 (𝑹𝒔𝒆 + 𝑹𝒂) + 𝑬𝒃𝟐
𝟓𝟓𝟎 = (𝟑)(𝟎. 𝟏𝟓 + 𝟎. 𝟖𝟏𝟓) + 𝑬𝒃𝟐
𝑬𝒃𝟐 = 𝟓𝟒𝟕. 𝟏𝟎𝟓 𝑽
Back to the ratio and proportion formula,
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃 𝝓
𝟐
𝑬𝒃
𝑺𝟏 = 𝒌𝒃 𝝓 𝟏
𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃
𝟎. 𝟗(𝝓𝟏 )
𝑺𝟏 𝑬𝒃𝟏
=
𝑺𝟐 𝑬𝒃𝟐
𝟎. 𝟗
𝟏𝟗𝟎𝟎
𝟓𝟐𝟖. 𝟕𝟕
=
𝟓𝟒𝟕. 𝟏𝟎𝟓
𝑺𝟐
𝟎. 𝟗
𝑺𝟐 = 𝟐, 𝟏𝟖𝟒. 𝟑𝟏 𝒓𝒑𝒎
MOTOR STARTING
- at starting, the armature of the motor draws high current due to zero back emf. To limit the current, a rheostat is
inserted/ connected in series with armature circuit.
Example: A shunt motor inserted with a starting resistor
𝑅𝑠
+
𝐼𝑠
𝐼𝑎
𝑅𝑎
𝐼𝑠ℎ
𝑉𝑠
𝑅𝑠ℎ
+
𝐸𝑏
−
A
R
M
−
Current Analysis:
𝑰𝒂 =
Where:
𝑽𝒔
𝑹𝒂 + 𝑹𝒔
Rs = Starting Resistance
Note: After the motor developed its rated speed/torque or the motor starts to rotate, the rheostat must set back to
zero/minimum value so the motor characteristics is not affected
Example 1: A shunt motor is running at the speed of 720RPM and draws 50A from 500V source. The armature and field
resistance are 0.4 & 250Ω respectively. Determine;
a. starting resistance to maintain the torque
b. ohmic value of the rheostat to reduce the speed to 540RPM.
Solution:
a. to maintain the torques means to maintain the armature current, thus the given 50A must be maintained. Therefore,
𝟓𝟎 =
𝟓𝟎𝟎
𝟎. 𝟒 + 𝑹𝒔
𝑹𝒔 = 𝟗. 𝟔Ω
b. for the speed to drop to 540rpm, the starting resistor is still connected but the resistance is adjusted, with this solving
for 𝑅𝑠 should follow the same ratio and proportion formula. Solving first for 𝐸𝑏1 during when the circuit has no
resistance connected,
𝟓𝟎𝟎 − (𝟓𝟎)(𝟎. 𝟒) − 𝑬𝒃𝟏 = 𝟎
𝑬𝒃𝟏 = 𝟒𝟖𝟎
Solving for 𝐸𝑏2 , using the ratio and proportion formula,
𝑬𝒃𝟏
𝝓𝟏
𝑬𝒃𝟐
𝑺𝟐 = 𝒌𝒃 𝝓
𝟐
𝑺𝟏 = 𝒌𝒃
Since the problem did not state that there changes in flux or torque, it is safe to assume that they are the same,
therefore the resulting formula will be,
𝑺𝟏 𝑬𝒃𝟏
=
𝑺𝟐 𝑬𝒃𝟐
𝟕𝟐𝟎 𝟒𝟖𝟎
=
𝟓𝟒𝟎 𝑬𝒃𝟐
𝑬𝒃𝟐 = 𝟑𝟔𝟎
Solving for 𝑅𝑠,
𝑽𝒔 − 𝑰𝒂𝟐 (𝑹𝒂 + 𝑹𝒔) − 𝑬𝒃𝟐 = 𝟎
Since it was assumed that the torque is the same, therefore the armature current will still be the same with the starting
resistance, thus
𝟓𝟎𝟎 − (𝟓𝟎)(𝟎. 𝟒 + 𝑹𝒔) − 𝟑𝟔𝟎 = 𝟎
𝑹𝒔 = 𝟐. 𝟒Ω