Transcript DC Generator - UniMAP Portal

BASIC ELECTRICAL TECHNOLOGY
DET 211/3
Chapter 7
Direct Current (DC) Motor
&
Direct Current (DC) Generator
DC Motor
There are four major types of DC motor in general
use:
• Separately excited DC motor
• Shunt DC Motor
• Series DC Motor
• Compounded DC Motor
Speed Regulations
• DC motors are often compared by their speed regulations.
• Speed Regulations (SR) of the a motor is defined by
SR 
nl  fl
x 100%
fl
n nl  n fl
SR 
x 100%
n fl
The Equivalent Circuit of a DC Motor
RA
Armature circuit represented by voltage source, EA and a
resistor RA.
The brush voltage drop represented by battery, Vbrush opposing
the direction of current flow in the machine.
The field coils, which produce the magnetic flux are represented
by inductor LF and resistor RF.
The separate resistor, Radj represents an external variable
resistor used to control the amount of current in the field circuit.
The Equivalent Circuit of a DC Motor
The brush voltage drop is often only a very tiny fraction of
the generated voltage in a machine so the voltage drop
may be left out or approximately included in the value of
RA.
The internal resistor in the field coils is sometimes lump
together with the variable resistor, and the total is called
RF.
The Equivalent Circuit of a DC Motor
The internal generated voltage in DC motor is
E A  K
The induce torque developed by DC motor is
ind  KI A
These two equations, the KVL equation of the armature circuit and
the machine’s magnetization curve are all the tools necessary to
analyze the behavior and performance of the dc motor.
Separately Excited and Shunt DC motors
VF
IF 
RF
VT  E A  I A R A
IL  IA
Separately excited DC motor
VT
IF 
RF
VT  E A  I A R A
IL  IA  IF
Shunt DC motor
Example
A 50hp, 250V, 1200rpm DC shunt motor with
compensating windings has an armature resistance of
0.06Ω. Its field circuit has a total resistance of 50Ω, which
produces a no load speed of 1200rpm. There are 1200
turns per pole on the shunt field winding. Find:
1.
2.
3.
4.
the speed of this motor when its input current is 100A
the speed of this motor when its input current is 200A
the speed of this motor when its input current is 300A.
the induced torque of this motor for above conditions.
Solution
VT  250V
n  1200rpm
2n 2 x 1200


 125.67 rads 1
60
60
E A  K
The relationship between the speeds and internal
generated voltages of the motor at two different load
conditions is
E A1  K1
(1)
EA2  K2
(2)
The flux is constant and
no armature reaction
Solution
E A1 K1

E A 2 K2
At no load, n  1200rpm
IA = 0A
E A1  VT  250V
1)
VT
250
IA  IL  IF  IL 
 100 
 95A
RF
50
Solution
EA at this load will be
EA  VT  IA R A  250  95(0.06)  244.3V
The resulting speed of this motor is
E A1 K1

E A 2 K2
E A 2 1 244.3x125.67
2 

 122.8rads 1
E A1
250
60 60 x122.8
n2 

 1173rpm
2
2
Answer
2)
IA = 195A,
EA = 238.3V,
n2 = 1144rpm
3)
IA = 295A,
EA = 232.3V,
n2 = 1115rpm
Solution
4)
Pconv  E A I A  ind
E A IA
ind 

At IL = 100A
ind
244.3x 95

 189 Nm
122.8
At IL = 200A
ind  388Nm
At IL = 300A
ind  587 Nm
Series DC Motor
VT  E A  I A (R A  R s )
I L  I A  IS
Compounded DC Motor
A compounded DC motor is a motor with both a shunt
and a series field.
Current flowing into
magnetomotive force.
a
dot
produces
a
positive
If current flows into the dots on both field coils, the
resulting magnetomotive forces add to produces a larger
total magnetomotive force. It is called cumulative
compounding.
Equivalent circuit of compounded DC motor
VT  E A  I A (R A  R s )
IA  IL  IF
VF
IF 
RF
Long shunt connection
The net magnetomotive force
Fnet  F F  FSE - FAR
Effective shunt field current
I*F  I F 
N SE
F
I A  AR
NF
NF
Short shunt connection
Applications of DC motor types
Separately-excited dc motor applications:
i) Golf cars (buggy)
ii) Forklift
iii) Aerial lift equipment
DC Generator
There are five major types of DC generators:
1. Separately excited generator. In a separately excited generator, the
field flux is derived from a separately power source independent of
the generator itself.
2. Shunt generator. In a shunt generator, the field flux is derived by
connecting the field circuit directly across the terminals of the
generator.
3. Series generator. In a series generator, the field flux is produced by
connecting the field circuit in series with the armature of the
generator.
4. Cumulatively compounded generator. In a cumulatively compounded
generator, both a shunt and a series field are present, and their
effects are additive.
5. Differentially compounded generator. In a differentially compounded
generator, both a shunt and a series field are present, but their effects
are subtractive.
DC Generator
DC generators are compared by their voltages, power
ratings, efficiencies, and voltage regulations.
Voltage regulation, VR is defined by
Vnl  Vfl
VR 
x 100%
Vfl
Separately Excited Generator
IL  IA
IF 
VF
RF
VT  E A  I A R A
Example
If no load voltage of a separately-excited dc
generator is 135V at 850 r/min, what will be
the voltage if the speed is increased to 1000
r/min? Assume constant field excitation
Solution
V1  135V
n1  850rpm
n2  1000rpm
V2  ? V
Constant field excitation means; if1 = if2 or constant
flux; 1 = 2
E A1 Kn1 n1


E A 2 Kn 2 n 2
n2
EA 2 
EA1
n1
1000
(
)135  158.8V
850
Shunt DC Generator
IA  IF  IL
VT  E A  I A R A
Series DC Generator
I A  IS  I L
VT  E A  I A (R A  R S )
The Cumulatively Compounded DC Generator
Total Magnetomotive force
Fnet  F F  FSE - FAR
N F I*F  N F I F  N SE I A  FAR
I*F  I F 
Cumulatively compounded dc generator with a long
shunt connection
N SE I A FAR

NF
NF
IA  IF  IL
VT  E A  I A (R A  R S )
VT
IF 
RF
Cumulatively compounded dc generator with a short
shunt connection
Differentially Compounded DC Generator
With a long shunt connection
Fnet  F F  FSE - FAR
N I  N F I F  N SE I A  FAR
*
F F
Equivalent shunt field
current,
I eq  
N SE I A FAR

NF
NF
N SE I A FAR
I  IF 

NF
NF
*
F
Example
A short-shunt compound generator delivers 50A at
500V to a resistive load. The armature, series field
and shunt field resistance are 0.16, 0.08 and 200,
respectively. Calculate the armature current if the
rotational losses are 520W, determine the efficiency
of the generator
Solution
Pu  520W
If 
Pout  500Vx 50A  25000W
500
 2.5A
200
Ia  If  IL  2.5A  50A  52.5A
Armature Copper Loss:
Pca  (Ia )2 Ra  (52.5)2 (0.16)  441W
Series Field Copper Loss: Pcf 2
Shunt Field Copper Loss:
 ( Ia )2 Rf 2  (52.5)2 (0.08)  220.5W
Pcf 1  ( If )2 Rf 1  (2.5)2 (200)  1250W
Friction + Stray + windage + etc:
So,Total Losses =
Pu  520W
( 441  220.5  1250  520)  2431.5W
Continued…
Efficiency, η =

Pout
Pout

Pin
Pout  Totallosse s
25000
 0.9113 @ 91.13%
25000  2431.5
Assignment 6
QUESTION 1
The equivalent circuit of the separately-excited dc motor
Figure above shows fixed field voltage VF of 240V and armature
voltage VA that can be varied from 120 V to 240 V. What is the
no-load speed of this separately-excited dc motor when Radj =
175 and
a) VA = 120V, b) VA = 180V, c) VA = 240V ?
Assignment 6
QUESTION 2
The equivalent circuit of the shunt dc motor
a) If the resistor Radj is adjusted to 175, what is the rotational
speed of the motor at no-load conditions?
b) Assuming no armature reaction, what is the speed of the
motor at full load? What is the speed regulation of the motor?
The magnetization curve for the dc motor of Question 1 and Question
2. This curve was made at a constant speed of 1200 r/min.