Solution of 4th

Year B.Tech., 8th

Sem, RTU Exam -2014

Sub: Electric Drives and Their Control

Q. 1 (a) Explain how the load equalization reduces the fluctuations in motor torque and

speed? (8)

[first Mid-term question set by shiv shanker sharma ]

[Rajasthan University 2005, 2006, RTU 2010, 2012]

Ans. Load Equalization:- Some loads are such that there is a heavy load on the motor for a

short duration and a long no load running time. In such cases it is not practical to use a motor

drive of full load rating as it would increase cost as well as no load losses, in such cases we

use a fly wheel connected to motor shaft to equalize the load. During light load speed of

wheel increases thus it stores kinetic energy during no load or light load ,when heavy load

comes for short duration it is met by motor developed torque and dynamic torque developed

as a result of reduction of speed of flywheel

Motro torque speed curve

For successful operation of the flywheel system the motor speed vary with torque and

assuming a linear motor speed- torque curve in the region of interest

Hence o ro

r

TT

where Tr= rated torque and T instantaneous torque

Because of slow response due to inertia motor can be assumed to be in electrical equilibrium

during transient operation of the motor-load system.

The torque of the system is given as (1 )m mt

tt t

LT T c Te

so

max min(1 )

hh

m m

h

t tt t

LT T c T e and min max(1 )h

h

m m

h

t tt t

LT T c T e

A. 1 (b) Rated torque

Tr = 50 1000 60

2 980

= 487.456 N-m

Tmax = 2XTr = 2X487.456 = 974.912 N-m Tmin = 200N-m

Tlh = 1000+200 = 1200 N-m th = 10 sec

2 1000 2 980

,60 60

o r

J= min

max

( )log ( )

hr

lho re

lh

tT

T T

T T

= 487.456 10

1200 2002 (1000 980)log ( )

1200 974.91e

=

= 3593.645 Kg-m2

Weight of flywheel = J/r2 = 3593.645/(0.9)2 = 4436.59Kg

(ii) Let the time taken be t1 sec

Tmax = 974.91 N-m , Tmin = 700 N-m , Tl = 200 N-m

J = max

min

( )log ( )

lr

lo re

l

tT

T T

T T

3593.645 = 487.4562 974.91 200

(1000 980) log ( )60 700 200

l

e

t

= 5186.19Kg-m2

Hence Moment of inertia of the flywheel = 5176.19Kg-m2

OR

Q. 1 (a) [Also in Assingment for first Mid-term set by shiv shanker sharma]

[RTU 5 years]

Q. 1 (b) [Also in Assingment for first Mid-term set by shiv shanker sharma]

Ans. Steady State Characteristics of Different Types of Motors and Loads

Unit II

Q.2(a)

Q.2(b) Since torque is proportional to armature current

Ia2 =80% of Ia

Ia2 = 0.8X200 = 160A

E1 = V-Ia1Ra

=220 200X0.06 = 208V

E2 = N2/N1XE1 = 700/800X208 = 182 V

Internal voltage of the variable voltage source

=182-160(0.06 + 0.04) = 166V

Q.2 a) Speed control of DC motor When a dc motor drives a load between no-load and full-load, the IR drop due to armature resistance

i

the speed regulation is poor. even for a fixed setting of the rheostat. In etlect. the I R drop across the

rheostat increases as the armature current increases. This produces a substantial drop in speed with

increasing mechanical load.

Armature speed control using a rheostat Schematic diagram of a shunt motor including the field

rheostat

Field speed control According to Eq. 1 we can also vary the speed of a dc motor by varying the field flux (I). Let LIS now

keep the armature voltage constant so that the numerator in Eq 1 is constant. Consequently, the motor

speed now changes in inverse proportion to the flux (P: if we increase the tlux the speed will drop,

and vice versa. This method of speed control is frequently used when the motor has to run above its

rated called base speed. To control the flux (and hence, the speed), we connect a rheostat Rr in series

with the field (Fig 3) To understand this method of speed control, suppose that the motor in Fig. 4 is

initially running at constant speed. The counter-emf E" is slightly less than the armature supply

voltage E" due to the IR drop in the armature. If we suddenly increase the resistance of the rheostat,

both the exciting current I, and the flux will diminish. This immediately reduces the cemf Eo,

causing the armature current I to jump to a much higher value. The current changes dramatically

because its value depends upon the very small difference between and despite the weaker field. the

motor develops a greater torque than before. It will accelerate until is again almost equal to Es

Clearly. to develop the same with a weaker flux. the motor must turn faster. We can therefore raise the

motor speed above its nominal value by introducing a resistance in series with the field. For shunt-

wound motors. this method of speed control enables high-speed/base-speed ratios as high as 3 to 4.

Broader speed ranges tend to produce instability and poor commutation. Under certain abnormal

conditions, the may drop to dangerously low values. For example, if the exciting current of a shunt

motor is interrupted accidentally, the only flux remaining is that due to the permanent magnetism in

the poles. This flux is so small that the motor has to rotate at a dangerously high speed to induce the

required emf. Safety devices are introduced to prevent such runaway conditions.

Q.2 b) Let the flux at 200V be 1

Then flux at 175V be 2

2 = 175/200X1

Since the load is constant

Ia22 = Ia11

Ia2 = 1/2Ia1 = 10.5/0.875 =11.4A

E1= V1 Ia1Ra = 200-10.5X0.5 =195V

E2 = V2 Ia2Ra = 175 -11.4X0.5 = 169.33V

E N

E1/E2 = 1N1/2N2

N2 = E2/E1 X1/2 XN1 =169.3/195 X1/0.875 X2000= 1923 rpm

Unit -III

Q.3 a) Starting an induction motor

Direct on line starting

For motors over a few kW, however, it is necessary to assess the eVect on the supply system before

deciding whether or not the motor can be started simply by switching directly onto the supply. If

supply systems were ideal (i.e. the supply voltage remained unaVected regardless of how much

current was drawn) there would be no problem starting any induction motor, no matter how large. The

problem is that the heavy current drawn while the motor is running up to speed may cause a large

drop in the supply system voltage, annoying other customers on the same supply and perhaps taking it

outside statutory limits.

Star/delta (wye/mesh) starter

This is the simplest and most widely used method of starting. It provides for the windings of the

motor to be connected in star (wye) to begin with, thereby reducing the voltage applied to each phase

to 58% of its DOL value. Then, when the motor speed approaches its running value, the windings are

switched to delta (mesh) connection. The main advantage of the method is its simplicity, while its

main drawbacks are that the starting torque is reduced (see below), and the sudden transition from star

to delta gives rise to a second shock albeit of lesser severity to the supply system and to the load.

For star/delta switching to be possible both ends of each phase of the motor windings must be brought

out to the terminal box. This requirement is met in the majority of motors, except small ones which

are usually permanently connected in delta.

Autotransformer starter

A 3-phase autotransformer is usually used where star/delta starting provides insuYcient starting

torque. Each phase of an autotransformer consists of a single winding on a laminated core. The mains

supply is connected across the ends of the coils, and one or more tapping points (or a sliding contact)

provide a reduced voltage output, as shown in Figure The motor is Y connected to the reduced voltage

output, and when the current has fallen to the running value, the motor leads are switched over to the

full voltage. If the reduced voltage is chosen so that a fraction a of the line voltage is used to start the

motor, the starting torque is reduced to approximately 2 times its DOL value, and the current drawn

from the mains is also reduced to a2 times its direct value. As with the star/delta starter, the torque per

ampere of supply current is the same as for a direct start.

The switchover from the starting tap to the full voltage inevitably results in mechanical and electrical

shocks to the motor. In large motors the transient over voltages caused by switching can be enough to

damage the insulation, and where this is likely to pose a problem a modified procedure known as the

Korndorfer method is used. A smoother changeover is achieved by leaving part of the winding of the

autotransformer in series with the motor winding all the time.

Resistance or reactance starter

By inserting three resistors or inductors of appropriate value in series with the motor, the starting

current can be reduced by any desired extent, but only at the expense of a disproportionate reduction

in starting torque.

For example, if the current is reduced to half its DOL value, the motor voltage will be halved, so the

torque (which is proportional to the square of the voltage see later) will be reduced to only 25% of

its DOL value. This approach is thus less attractive in terms of torque per ampere of supply current

than the star/delta method. One attractive feature, however, is that as the motor speed increases and its

effective impedance rises, the volt drop across the extra impedance reduces, so the motor voltage rises

progressively with the speed, thereby giving more torque. When the motor is up to speed, the added

impedance is shorted-out by means of a contractor. Variable-resistance starters (manually or motor

operated) are sometimes used with small motors where a smooth jerk free start is required, for

example in textile lines

Solid-state soft starting

This method is now the most widely used. It provides a smooth build-up of current and torque, the

maximum current and acceleration time are easily adjusted, and it is particularly valuable where the

load must not be subjected to sudden jerks. The only real drawback over conventional starters is that

the mains currents during run-up are not sinusoidal, which can lead to interference with other

equipment on the same supply.

Q.3 b) For synchronous speed Ns = 120f/P = 120X50/6 = 1000

s = 104.72 rad/sec

For regenerative braking

Sm =

1

2

2 2

1 1 2( )

r

r x x

=

2

10.242

1 4

1

22

1 221 1 2( )

VphI

rr x x

s

=2

2

400

345.5

11 4

0.242

A

22 22

max

13 45.53

( 0242)244.5

104.72ms

rI

sT N m

Maximum over loading torque may hold = 244.5 N-m speed at which totque is maximum

= (1-sm)x1000 = 1242 rpm

OR

Q.3 a) control arrangements for inverter-fed drives

The variable-voltage autotransformer can be replaced by three sets of thyristors connected back-to

back. as shown in Fig. The sets are called valves. To produce rated voltage across the motor the

respective thyristors are fired with a delay e equal to the phase angle lag that would exist if the motor

were directly connected to the line. Fig. shows the resulting current I and line-to-neutral voltage for

phase A. The valves phases Band C are triggered the same way. Except for an additional delay of

1200 and 240

0 respectively.

Fig 1 Fig 2

To reduce the voltage across the motor, the firing angle e is delayed still more. For example, to obtain

50 percent rated voltage, all the pulses are delayed by about 100. The resulting distorted voltage and

current waveshape for phase A are pictured very approximately. The distortion increases the losses in

the motor compared to the autotransformer method. Furthermore, the power factor is considerably

lower because of the large phase angle lag e. Nevertheless, to a first approximation, the torque-speed

characteristics shown in Fig. 2 still apply.

Due to the considerable PR losses and lower power factor, this type of electronic speed control is only

feasible for motors rated below 20 hp. Small hoists are also suited to this type of control, because they

operate intermittently. Consequently, they can cool off during the idle and light-load periods.

Q.3 b) Single-phasing

If one line of a 3-phase line is accidentally opened, or if a fuse blows while the 3-phase motor is

running, the machine will continue to run as a single phase motor. The current drawn from the

remaining two lines will almost double, and the motor will begin to overheat. The thermal relays

protecting the motor will eventually trip the circuit-breaker, thereby disconnecting the motor from the

line. The torque-speed curve is seriously affected when a 3-phase motor operates on single phase. The

breakdown torque decreases to about 40% of its original value. and the motor develops no starting

torque at all. Consequently, a fully loaded 3-phase motor may simply stop if one of its lines is

suddenly opened. The resulting locked-rotor current is about 90% of the normal 3-phase LR current. It

is therefore large enough to trip the circuit breaker or to blow the fuses. The torque-speed curve is

seriously affected when a 3-phase motor operates on single phase. The breakdown torque decreases to

about 40% of its original value. and the motor develops no starting torque at all. Consequently, a fully

loaded 3-phase motor may simply stop if one of its lines is suddenly opened. The resulting locked-

rotor current is about 90% of the normal 3-phase LR current. It is therefore large enough to trip the

circuit breaker or to blow the fuses

Unit IV

Q.4 a)

Q.4 (b) STATIC ROTOR RESISTANCE CONTROL

The rotor resistance control of a wound-rotor induction motor is inefficient because the speed

reduction is obtained by wasting slip power in extemal resistors. It has, however, advantages of low

cost, a good power factor, and a high torque-to-current ratio for a wide range of speed, including

starting and braking. The variable frequency control is the only other method of induction motor

speed control which gives a high torque-to-current ratio. However, it is very expensive. Hence, rotor

resistance control finds application in drives requiring low cost and a high torque-to-current ratio,

such as low-power excavators, crane hoists, and so on.

Instead of mechanically varying the resistance, the rotor circuit resistance can be varied statically by

using the principIe of a chopper. This gives stepless and smooth variation of resistance and

consequently of motor speed. As shown in figure 1, the slip frequency ac rotor voltages are converted

into de by a 3-phase diode bridge and applied across an external resistance R. The self-commutated

semiconductor switch S, connected in parallel with R, is operated periodically with a period T and

remains on for an interval ton in each period The effective value of resistance R changes from R to O

as to changes from O to T. The filter inductor L, is provided to minimize the ripple in current Id' A

high ripple in Id produces high harmonic content in the rotor, increasing copper losses and causing

derating of the motor. The filter inductor also helps in eliminating discontinuous conduction at light

loads. As in the case of a de motor, discontinuous conduction makes the speed regulation poor. The

main contributor to the ripple is the diode bridge and not the semiconductor switch, because it

operates at a sufficiently high frequency. The diode bridge output voltage Vd changes from its

maximum value at standstill to nearly 5 percent of the maximum value at near rated motor speed. If

switch S is realized using a thyristor, reliable commutation can only be obtained either by using a

bulky commutation capacitor or an auxiliary source for charging the commutation capacitor. Hence, a

thyristor is not suitable for this application. Because induction motors are usually dsigned with a

stator-to-rotor tums ratio greater than 1, the voltage Vd is small.

Figure 1 Static rotor resistance control of wound-rotor induction motor.

Hence, a transistor is suitable for low-power drives. A GTO may be employed for ratings beyond the

capability of transistors. The self-commutation capability of these devices ensures reliable

commutation at all operating points and makes the semiconductor switch compact. An alternative

static rotor resistance control circuit is obtained by using either a 6-pulse or 3-pulse controlled

rectifier instead of the diode bridge and semiconductor switch S. The power consumed by R is then

controlled by controlling the rectifier firing angle. As the firing angle is increased from O to the

maximum, the effective rotor resistance increases from R to a maximum value, controlling the speed.

OR

Q.4 a)

Static Kramers Drives

Q.4 b) Operation with a current source

The variable frequency supply for speed control of an induction motor can be a voltage source' or a

current source.

Operation at a Fixed Frequency

The equivalent circuit of figure 1 is applicable. The only difference is that the motor is now fed by a

current source I, instead of a voltage source Y. The motor input current will be independent of motor

parameters and the terminal voltage V will change due to the change in the motor impedance. The

input current I, is shared between the rotor impedance and the magnetizing reactance Xm For low

values of s, the rotor current is small and the magnetizing current 1m is nearly equal to L. Since I, is

usually much higher than the normal magnetizing current, the motor operates under saturation for low

values of slip. Therefore, the motor should be analyzed taking the saturation into account. The

nonlinear relationship between E and Im is obtained experimentally. The following equations can be

written from the equivalent circuit.

2'

'2 '2 2rr r

RX I E

s

(1)

2'

' ' 2 '2 2 2( )r r m r s mR

X X I I Xs

(2)

m m mE I X (3)

Subtracting equation (1) frorn (2) and then substituting frorn equation (3) gives

A suitable value (less than Is) is assumed for 1mfor a given L; E and Xm are obtained from the

magnetization characteristic; 1; is calculated from equation (3); s is evaluated from equation (2), and

then T and V are obtained frorn equations (5) and (9), respectively.

Unit V

Q.5 a)

Q.5 b) when the motor operates as a variable speed drive motor utilizing a variable frequency supply,

it can be regeneratively braked and all the K.E. returned to the mains. As in an induction motor,

regeneration is possible if the synchronous speed is less than the rotor speed. The input frequency is

gradually decreased to achieve this at every instant. The K.E. of the rotating parts is returned to the

mains. The braking takes place at constant torque. With a CSI and cyclo-converter regenerative is

simple and straight forward .

Speed-torque characteristic with a fixed frequency supply

OR

Q.5 a) Cylindrical rotor motor operation from a current source- In figure , the equivalent circuit and

the phasor diagram of synchronous motor have been redrawn, taking the armature current as the

reference vector. The angle ' between the phases Ir and I, is related to by the following equation:

Now,

Power developed is

And torque

An alternative expression for torque is obtained by substituting from equation (1) into equation (2),

giving

From equations (3) or (4), the pull-out torque is

The pull-out torque can be increased by increasing either I, or Ir. For given values of I, and Ir. the

torque is a function of the angle '. Hence ' is defined as the torque angle. A torque versus torque

angle curve is shown in figure. The motoring operation is obtained for and the braking operation is

obtained for a given value of motoring torque is obtained at

two values of '-that is, and (7T-6). Similarly, a given braking torque is available at two values of

' -that is, (7T+ 82) and (37T- 82). Figure shows the locus of I:n as ' is varied from 0 to 7T with Ir

and I, constant. Phasors, I, and V are shown for two values of ' which yield the same value of

torque. The following may be noted:

1. When ' is less than 7T/2, the machine always operates at a lagging power factor. However, when 8'

is greater than 7T/2, the machine may operate at either a lagging or a leading power factor.

2. For ' greater than 7T/2, I:n and V have lower values and the power factor is higher.

3. Because of the large value of I:n for ' less than 7T/2, the machine may operate under heavy

saturation.

For these reasons, during the motoring operation, the preferred range of ' is 7T/2

discussion ofsection 8.1.5 about the braking and multiquadrant operations of voltage source inverter

fed induction motor drives is also applicable to voltage source inverter fed synchronous motor drives,

except that the changeover from motoring to braking, or vice versa, in a synchronous motor drive is

obtained by a change in torque angle 8 rather than in frequency