ENERGY CONVERSION ONE (Course 25741) CHAPTER NINE ….continued DC MOTORS AND GENERATORS.

ENERGY CONVERSION ONE (Course 25741)

CHAPTER NINE ….continuedDC 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


(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


• 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


(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


(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 speedAt 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.


• 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


• 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 :

F

ARFF N

FII *


• 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

00 n

n

E

E

A

A


• 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


• Magnetization Curve of typical 250 V dc motor Example -2


(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


(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


(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


• 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


• 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

A

ATA R

EVI


• 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


• 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


• 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


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 again8-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


• Effect of RF speed control on a shunt motor’s torque-speed (over motor’s normal operating range)


• Effect of RF speed control on a shunt motor’s torque-speed (over entire range from no-load to stall conditions)


• 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 torque-

speed 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


• 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


• 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


• Effect of armature voltage speed control


• 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:

• 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

indAT

K

R

K

V

2)(

ENERGY CONVERSION ONE (Course 25741) CHAPTER NINE ….continued DC MOTORS AND GENERATORS.

Documents

Transcript of ENERGY CONVERSION ONE (Course 25741) CHAPTER NINE ….continued DC MOTORS AND GENERATORS.