Transcript Document

ENERGY CONVERSION ONE
(Course 25741)
CHAPTER NINE ….continued
DC MOTORS AND GENERATORS
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
• If a shunt dc motor has compensating windings
so that flux is constant regardless of load, &
motor’s speed & armature current known at any
one value of load , then speed at any other load
can be determined if IA is known at that load
• EXAMPLE-1: A 50 hp, 250 V, 1200 r/min dc
shunt motor with compensating windings,
• RA =0.06 Ω (including brushes, comp.
windings, & inter-poles). The field cct. has a
total resistance of Radj + RF of 50 Ω which
develops a no load speed of 1200 r/min. There
are 1200 turns per pole on shunt field winding
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
(a) Find speed of this motor when its input current
is 100 A
(b) Find speed of this motor when its input current
is 200 A
(c) Find the speed of this motor when its input
current is 300 A
(d) plot torque-speed characteristic of this motor
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
• EXAMPLE-1;SOLUTION:
• EA=K’φn, since IF is constant (VT & RF const.) & since
there are no A.R. φ would be constant
 relationships between speeds & internal
generated voltages of motor at 2 different load
conditions is: EA2/EA1=[K’φn2]/[K’φn1]
• constant K’ cancels, also φ canceled 
n2= EA2/EA1 . n1
• at no load IA is zero so EA1=VT=250 V
• While speed n1=1200 r/min
• If internal generated voltage at any other load is
determined, motor speed at the load can be
determined
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
(a) if IL=100 A then armature current :
IA=IL-IF=IL- VT / RF =100 – 250/ 50=95 A
 EA=VT-RAIA=250 –(95)(0.06)=244.3 V
Resulting speed:
n2=EA2/EA1xn1=244.3/250 x1200=1173 r/min
(b) IL=200 A  IA=200-250/50=195 A
EA=250-(95)(0.06)=238.3 V
n2= 238.3/250 x1200=1144 r/min
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
(c) if IL=300 A, then IA=IL-IF=300-250/50=295 A
EA = 250-(295)(0.06)=232.3 V
n2=232.3/250 x 1200=1115 r/min
(d) torque versus speed
At no load induced torque is zero
Pconv=EAIA=Tind ω
From this equation
Tind= EAIA / ω
IL=100 A 
Tind=(244.3)(95)/ [1173x1/60x2π]=190 N.m.
IL=200 A  Tind= 238.3x195/[1144x1/60x2π]=388 N.m.
IL=300 A  Tind=587 N.m.
TERMINAL CHARCATERISTIC of a
SHUNT DC MOTOR
• Torque – speed characteristic of motor
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• flux φ & EA of a dc machine is a nonlinear
function of mmf
 anything that changes mmf cause a nonlinear
effect on EA
• mmf should be used to determine EA & mmf
determined based on field current and A.R.
• magnetization curve is a direct plot of EA versus
IF for a given speed ω0  effect of variation in
field current can be determined directly from its
magnetization curve
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• If a machine has armature reaction, its flux will be
reduced with each increase in load. The total mmf in a
shunt dc motor is the field circuit mmf less the mmf
due to armature reaction (AR):
Fnet = NFIF – FAR
magnetization curves are expressed as plots of EA vs
field current, normally an equivalent field current is
defined that would produce the same output voltage
as the combination of all the mmf in the machine
The equivalent field current :
I
*
F
 IF
FAR

NF
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• one other effect must be considered when non linear analysis is
used to determine EA of a dc motor
• The magnetization curves for a machine are drawn for a
particular speed, usually the rated speed of the machine
• How can the effects of a given field current be determined if the
motor is turning at other than rated speed?
• The equation for the induced voltage in a dc machine when
speed is expressed as rev/min: EA = K’φn ,
• For a given effective field current, the flux in the machine is
fixed, so the EA is related to speed by
EA
n

E A0
n0
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• EXAMPLE-2:
• A 50 hP, 250V, 1200r/min DC shunt motor
without compensating windings has an
armature resistance (including the brushes and
interpoles) of 0.06 .
• Its field circuit has a total resistance Radj + RF of
50 , which produces a no-load speed of 1200
r/min. There are 1200 turns per pole on the
shunt field winding, and the armature reaction
produces a demagnetising mmf of 840 A.turns
at a load current of 200A. magnetization curve
of this machine is shown next
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• Magnetization Curve of typical 250 V dc motor
Example -2
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
(a) Find the speed of this motor when its input
current is 200A.
(b) This motor is essentially identical to the one in
Example 1, except for the absence of
compensating windings.
How does its speed compare to that of the
previous motor at a load current of 200A?
(c) calculate & plot torque-speed characteristic of
motor
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
(a) IL=200 A IA=IL-VT/RF=200-250/50=195 A
EA=250-195x0.06=238.3 V
at this current A.R. demagnetizing mmf is 840
A.T.s so effective shunt field current is:
IF*= IF- FAR/NF = 5.0-840/1200 =4.3 A
from magnetization curve, effective field current
produce EA0 of 233 V at a speed of n0=1200
r/min
However since voltage is 238.3 V, shows actual
operating speed of motor is:
EA/EA0 n0 =238.3/233 x 1200 =1227 r/min
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
(b) at 200 A load in Example 1, motor’s speed
was n=1144 r/min in this example motor’s
speed is 1227 r/min
Note: speed of motor with A.R. is higher than
speed of motor with no A.R. This relative
increase in speed is due to flux weakening in
machine with A.R.
(c) for this item, (since there is no information
about FAR in other load currents) it is assumed
that FAR varies linearly with load current, &
using MATLAB speed-torque characteristic
obtained
NONLINEAR ANALYSIS of a SHUNT
DC MOTOR
• Torque-speed characteristic of motor with
armature reaction
SPEED CONTROL of SHUNT DC
MOTOR
•
•
Two common methods:
1- Adjusting the field resistance RF (and thus
the field flux)
2- Adjusting the terminal voltage applied to the
armature
Less common method:
3- Inserting a resistor in series with the
armature circuit
SPEED CONTROL of SHUNT DC
MOTOR
• Changing the Field Resistance
• If the field resistance increases, field current
decreases (IF↓ = VT/RF↑), and as the field current
decreases, flux decreases as well.
• A decrease in flux causes an instantaneous decrease
in the internal generated voltage EA↓ (=Kφ↓ω), which
causes a large increase in the machine’s armature
current since
VT  E A 
I A 
RA
SPEED CONTROL of SHUNT DC
MOTOR
• Induced torque in a motor is given by ind =KφIA
since flux in machine decreases while current IA
increases, which way does the induced torque
change?
Look at
this example: 
• armature
current flow is
IA=(250V-245V)/
0.25Ω= 20A
SPEED CONTROL of SHUNT DC
MOTOR
• What happens in this motor if there is a 1%
decrease in flux?
• If the flux decrease by 1%, then EA must
decrease by 1% too, because EA = Kφω
Therefore, EA will drop to:
EA2 = 0.99 EA1 = 0.99 (245) = 242.55 V
• armature current must then rise to
IA = (250-242.55)/0.25 = 29.8 A
• Thus, a 1% decrease in flux produced a 49%
increase in armature current
SPEED CONTROL of SHUNT DC
MOTOR
• back to original discussion, the increase in
current predominates over the decrease in flux
so, ind>load , the motor speeds up
• However, as the motor speeds up, EA rises,
causing IA to fall. Thus, induced torque ind
drops too, and finally ind equals load at a higher
steady-sate speed than originally
• Summarizing behaviour:
1- increasing RF causes IF (=VT/RF) to decrease
2- decreasing IF decrease φ
SPEED CONTROL of SHUNT DC
MOTOR
3 – Decreasing φ lowers EA(=Kφω)
4 - Decreasing EA increases IA (=VT-EA)/RA
5- increasing IA increases Tind (=KφIA), with change in IA
dominant over change in flux
6-increasing Tind makes Tind>Tload & speed ω increases
7-increase in ω, increases EA= Kφω again
8-increasing EA decreases IA
9-Decreasing IA decreases Tind until Tind=Tload at a higher
ω
• Effect of increasing RF on O/P characteristic of a shunt
motor shown in next figure
SPEED CONTROL of SHUNT DC
MOTOR
• Effect of RF speed control on a shunt motor’s
torque-speed (over motor’s normal operating
range)
SPEED CONTROL of SHUNT DC
MOTOR
• Effect of RF speed control on a shunt motor’s
torque-speed (over entire range from no-load to
stall conditions)
SPEED CONTROL of SHUNT DC
MOTOR
• According to equation of speed presented before:
(a) no-load speed is proportional to reciprocal of flux in
motor
(b) while slope of the curve is proportional to reciprocal
of flux squared
• Therefore a decrease in flux causes slope of torquespeed to become steeper
• over this range, an increase in field resistance
increases motor’s speed
• For motors operating between no-load & full-load
conditions, an increase in RF may reliably be expected
to increase operating speed
SPEED CONTROL of SHUNT DC
MOTOR
• In previous figure (b) terminal characteristic of motor over full
range from no-load to stall shown
• In figure can see at very slow speeds, an increase in field
resistance will actually decrease speed of motor
• This effect occurs because at very low speeds, the increase in
armature current caused by decrease in EA not enough to
compensate for decrease in flux in induced torque field
resistance
• Some small dc motors used for control purposes operate at
speeds close to stall conditions
• For these motors, an increase in RF might have no effect or it
might even decrease speed of motor
• Since the results are not predictable, field resistance speed
control should not be used in these types of dc motors. Instead,
the armature voltage method should be employed
SPEED CONTROL of SHUNT DC
MOTOR
• CHANGING ARMATURE VOLTAGE
• This method involves changing the voltage applied to
the armature of the motor without changing the
voltage applied to the field
• If the voltage VA is increased, then the IA must rise [ IA
= (VA ↑ -EA)/RA]. As IA increases, the induced torque
ind =KφIA↑ increases, making ind > load , and the
speed of the motor increases
• But, as the speed increases, the EA (=Kφω↑)
increases, causing the armature current to decrease
This decrease in IA decreases the induced torque,
causing ind = load at a higher rotational speed
SPEED CONTROL of SHUNT DC
MOTOR
• Effect of armature voltage speed control
SPEED CONTROL of SHUNT DC
MOTOR
• Inserting a Resistor in Series with the Armature
Circuit
• If a resistor is inserted in series with the armature
circuit, the effect is to drastically increase the slope of
the motor’s torque-speed characteristic, making it
operate more slowly if loaded. This fact can be seen
from the speed equation:
VT
RA



2 ind
K ( K )
• The insertion of a resistor is a very wasteful method of
speed control, since the losses in the inserted resistor
are very large. For this reason, it is rarely used