III SEMESTER EE6365 ELECTRICAL ENGINEERING...

V+ TEAM

Anna University , Chennai

B.E. – ELECTRICAL AND ELECTRONICS ENGINEERING

III SEMESTER

EE6365 – ELECTRICAL ENGINEERING LABORATORY

LABORATORY MANUAL

V+ TEAM

CONTENT

S. No. Name of the experiment Marks Signature

1 Load test on a DC Shunt motor

2 Load test on a DC Series motor

3 Open circuit and load characteristics of a DC Shunt Generator

4 Open circuit and load characteristics of a DC Series Generator

5 Speed control of DC Shunt motor

6 Load test on a single phase transformer

7

Open circuit and short circuit tests on a single phase

transformer

8

Regulation of a three phase alternator by EMF and MMF

methods

9 V curves and inverted V curves of synchronous Motor

10 Load test on a three phase squirrel cage induction motor

11 Load test on single phase Induction Motor

12 Study of DC & AC Starters

V+ TEAM

LOAD TEST ON DC SHUNT MOTOR

AIM:

To conduct a direct load test on the given dc shunt motor and to plot the following performance

characteristics: 1) Efficiency vs Output power (2) Torque vs Output power (3) Speed vs Output Power (4)

Line current vs Output Power (5) Torque vs Speed.

NAME PLATE DETAILS:

Rated Voltage = Rated Current =

Rated Speed = Power rating =

EXCITATION:

Voltage = Current =

APPARATUS REQUIRED:

THEORY:

Load test on motors are performed to know about the efficiency, torque and speed characteristics,

which enable us to select an appropriate motor for an application. The torque equation of a DC Motor is

given by Ta = 0.159 (xPxZ / A) * Ia N-m, where =Flux per pole (Wb), P= No. of poles, Z= No. of armature

conductors, A = no. of parallel paths. As P, Z, A being constant, the above equation reduces to Ta=K Ia. In

a DC Shunt motor as is constant, the torque is directly proportional to armature current.

The speed of a DC motor is given by, N = K (V - IaRa) /. Since is constant, the speed is directly

proportional to (V-Ia Ra). As the load on the motor increases, the drop IaRa becomes negligible as `Ra ' is very

small and the speed is nearly constant.

Hence a DC shunt motor is considered as a constant speed motor. If a DC shunt motor is started on

load, it draws a heavy armature current which in turn will damage the machine itself. Hence DC shunt

motors are always started on no-load condition.

S. No. Apparatus Range & Type Quantity

1 DC Shunt Motor Set up - 1 No

2 Voltmeter (0-300 V), MC 1 No

3 Ammeter (0-2 A), MC

(0-20 A), MC

1 each

4 Rheostat 220 Ω, 1 A 1No

5 Tachometer, DPSTS - 1each

6 Connecting wires & Fuse - As Required

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PRECAUTIONS: (Not to be included in the Record)

1. Remove the fuse carriers before wiring and start wiring as per the circuit diagram.

2. Check the position of the rheostat as specified in the procedure.

3. The load on motor must be released initially.

4. Fuse calculations: As this is a load test, the required fuse ratings are 120% of the rated current of the

motor.

5. Replace the fuse carriers with appropriate fuse wires after the circuit connections are checked by the

Staff-in-charge.

PROCEDURE:

1. The circuit connections are made as per the circuit diagram.

2. Keeping the motor field rheostat in its minimum position and the starter in its OFF position the main

supply is switched ON by closing the DPSTS.

3. The motor is started using the three-point starter by slowly and carefully moving the starter handle

from its OFF to ON position.

4. The motor is brought to its rated speed by gradually adjusting the field rheostat and checked with the

help of a tachometer.

5. Under this no load condition, one set of readings namely, applied voltage (VL), armature current (Ia),

field current (If), the two spring balance readings (F1and F2) and motor speed (N) are noted down in

the tabular column.

6. The load on the motor is increased in steps gradually and at each step, all the meter readings and the

motor speed are recorded in the tabular column. The above procedure is repeated until the motor is

loaded to 120% of its rated current.

7. After the experiment is completed, the load on motor is gradually decreased to minimum and the

rheostat is brought back to its original position and then the main supply is switched OFF.

8. After completing the experiment, calculate the Torque, Armature current, Input power, Output

power and Efficiency using the formulae given.

9. The plots of Efficiency vs Output Power, Torque vs Output Power, Line Current vs Output Power

and Speed vs Output Power are plotted in the same graph sheet and also the Speed vs Torque

response is plotted in a separate graph sheet.

CIRCUIT DIAGRAM:



FUSE CALCULATION:

TABULAR COLUMN:

VL

(V)

IL

(A)

IF

(A)

F1

(Kg)

F2

(Kg)

F1 ~ F2

(Kg)

N

(rpm)

T

(Nm)

Ia

(A)

I/P

(W)

O/P

(W) %

(%)

FORMULAE USED:

1. The circumference of the brake drum (c) is measured and the radius of the drum (r) is calculated

using the expression r=c/2, R=r + t/2, t - thickness of the belt, Radius of the drum including

thickness of the belt, R= m.

2. The line current is calculated as IL =Ia+ If

3. The input power to the motor = VL*IL Watt.

4. The torque developed by the motor is given by T= 9.81* R * (F1~F2) N-m.

5. The output power of the motor =60

2 NT W

6. The efficiency of the motor = (Output Power/Input Power )* 100 %

MODEL GRAPH:



MODEL CALCULATIONS:

RESULT:

REVIEW QUESTIONS:

1. What is the need for a starter?

2. Why a DC shunt motor is called a constant Speed motor?

3. State few applications of DC shunt motor.

4. What is the role of commutator in a DC motor?

5. How to reverse the direction of rotation of DC shunt motor?



LOAD TEST ON DC SERIES MOTOR

AIM:

To find performance characteristics of a DC series motor by conducting the load test

APPARATUS REQUIRED:

S. No. Apparatus Range Type Quantity

1 DC Series Motor Set up - 1 No. 1

2 Voltmeter (0-300 V), MC 1 No. 2

3 Ammeter (0-20 A), MC 1 No. 1

4 Tachometer, DPSTS - 1each 5

5 Connecting wires & Fuse - As

Required

6

THEORY:

It is early observed that at no load the motor current and hence the flux per pole tends to

zero and as a consequence the motor speed tends to increase to infinite. This is a dangerous

situation and the centrifugal force will destroy the armature flux a series motor must never to

allowed to run at no load torque increases the motor speed drops heavily there by the KW load on

the motor of this type of speed torque characteristics and is ideally switched for function, cranes,

etc., Application were as during starting a large accelerating torque is demanded by the load is at

ways on the motor to there is no danger of under loading or no-loading.

FORMULAE

1. Torque ( T )=S*9.81*R Nm

2. Input Power (Pi) = VL * IL watt

3. Output power (Po) = 2πNT/60 watt

4. Efficiency (η) = Output power / Input power X 100 %

where,

S - Load

R – Radius of brake drum

VL – Line Voltage

IL – Line current



N – Speed in rpm

T – Torque in Nm

CIRCUIT DIAGRAM

FUSE RATING NAME PLATE DETAILS

PRECAUTIONS:

1. DC series motor should be started and stopped with some load.

2. Brake drum should be cooled with water when it is under load.

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. After checking the load condition, DPST switch is closed and start using 2-point starter.

3. Ammeter, Voltmeter readings, speed and spring balance readings are noted on initial

condition.



4. The load is then added to the motor gradually and for each load, voltmeter, ammeter, spring

balance readings and speed of the motor are noted up to 125% of the rated current.

5. The motor is then brought to low load condition then DPST switch is opened.

TABULAR COLUMN:

Circumference of the Brake drum =_________m.

S. No.

Voltage

VL (V) Current

I L (A)

Spring Balance

Reading S=S1 S2

(Kg)

Speed

N

(rpm)

Torque

T (Nm)

Input

Power

Pi

(W)

Output

Power

Po

(W)

Efficiency

% S1 (Kg) S2 (Kg)

1

2

3

4

5

6

7

8

Radius of Brake Drum

Circumference = 2 π R =

Radius = R=Circumference / (2 π)

= _________m

MODEL CALCULATION



MODEL GRAPH

RESULT:



OPEN CIRCUIT AND LOAD CHARACTERISTICS OF

SELF EXCITED DC SHUNT GENERATOR

AIM:

To obtain open circuit characteristics and load characteristics of a self excited DC shunt

generator and find its critical resistance

APPARATUS REQUIRED:


1 DC Shunt Generator Set up - 1 No



(0-20) A MC 1 each

4 Rheostats 200 Ω/3 A

400 Ω/3 A 1 each

5 SPST & DPST switch - 1 No

6 Tachometer - 1 No


THEORY

A DC generator requires an excitation circuit to generate an induce voltage depending on

whether excitation circuit consumes power for the armature of the machine or from separately

require power supply. Generators may be classified as self excited or separately excited generators

respectively.

The induced emf in DC generators is given by the equation PфZN/60A volts. State P,Z,A

are constants the above equation are written as Eg= KфN. I f the speed of the generator also

maintained constant then Eg =Kф but the flux is directly proportional to the current Hence Eg

=K2If.From the above equation it is clear that the induced emf is directly propositional to the field

current. when speed maintained constant,. The plot between the induced emf and the field current is

known as open circuit characteristics of the DC generator.

The induced emf when the field current is zero is known as residual voltage. This emf is due

to the presence of a small amount of flux detained. In the field poles of the generator called residual

flux. Once the OCC is obtained parameters such as critical field resistance, critical speed and the

maximum voltage to which the machine can build up can be determined. If required the OCC at a

different speeds can also be obtained. .Critical speed is minimum speed below which the generator

shunt fails to excite.

FORMULAE:

Eg = V + Ia Ra (V)

Ia = IL + If (A) [for self Excited]

where



Eg : Generated EMF in V

V : Terminal Voltage in V

Ia : Armature Current in A

IL : Line Current in A

If : Field Current in A

Ra : Armature Resistance in Ohm

CIRCUIT DIAGRAM

OPEN CIRCUIT & LOAD TEST

OPEN CIRCUIT TEST

PRECAUTIONS:

At the time of starting and stopping the machine,

1. The field rheostat of motor should be in minimum resistance position.

2. The field rheostat of generator should be in maximum resistance position.

3. No load should be connected to generator.

PROCEDURE:


2. After checking minimum position of motor field rheostat, maximum position of generator

field rheostat, DPST switch is closed and starting resistance is gradually removed.

3. By adjusting the motor field rheostat, the motor is brought to rated speed.

4. By varying the generator field rheostat, voltmeter and ammeter readings are taken.

5. After bringing the generator rheostat to maximum position, field rheostat of motor to

minimum position, DPST switch is opened.



LOAD TEST

PRECAUTIONS:

1. The field rheostat of motor should be at minimum position.

2. The field rheostat of generator should be at maximum position.

3. No load should be connected to generator at the time of starting and stopping.

PROCEDURE:


2. After checking minimum position of DC shunt motor field rheostat and maximum position

of DC shunt generator field rheostat, DPST switch is closed and starting resistance is

gradually removed.


4. Connect the load by closing load DPST switch.

5. By adjusting the generator field rheostat, the generator is brought to rated voltage

6. Under no load condition, Ammeter and Voltmeter readings are noted.

7. Load is varied gradually and for each load, voltmeter and ammeter readings are noted.

8. Then the generator is unloaded and the field rheostat of DC shunt generator is brought to

maximum position and the field rheostat of DC shunt motor to minimum position, DPST

switch is opened.

TABULATION

Armature Resistance - Ra: _______ohm

OPEN CIRCUIT TEST

S. No. Field Current

If (A)

Armature Voltage

Eo (V)

1

2

3

4

5

6



LOAD TEST

MODEL CALCULATION

MODEL GRAPH

OC TEST

S. No.

Field

Current

If (A)

Load

Current

IL (A)

Terminal

Voltage

(V)

Ia = IL + If

(A)

Eg =V + Ia Ra

(V)

1

2

3

4

5

6

7



LOAD TEST



RESULT:



OPEN CIRCUIT AND LOAD CHARACTERISTICS OF

DC SERIES GENERATOR

AIM:

To obtain open circuit characteristics and load characteristics of a DC series generator and

find its critical resistance

APPARATUS REQUIRED:

Sl. No. Apparatus Range & Type Quantity

1 DC Series Generator Set up - 1 No



(0-20) A MC 1 each

4 Rheostats 200 Ω/ 3 A

400 Ω/3 A 1 each

5 SPST & DPST switch - 1 No

6 Tachometer - 1 No


THEORY

In a D.C. series generator the field winding is connected in series with the armature

winding. In this case the armature current flows through the field winding as well as the load. Since

the armature winding and the field winding are in series the armature current is the same as the

field current. The field winding has less number of turns of thick wire and hence its resistance is

low.

Ia = Ise = IL

The load characteristics of a D.C. series generator are plotted with the load current (IL) on

the X-axis and the Voltage (V) on the Y-axis. As in the case of the D.C. shunt generator there are

two types of load characteristics:

1. Internal characteristics – Induced emf E vs Load current IL. Here the drop is due to

armature reaction.

2. External characteristics – Terminal Voltage V vs Load current IL. Here the drop is due

to armature and series field resistance.

The Voltage equation of a D.C. series generator is given by

V = E – Ia (Ra + Rse)

The load characteristics are shown in the model graph. It will be noticed that a series

generator has rising voltage characteristic i.e. with increase in load, its voltage is also increased, but



it is seen that at high loads, the voltage starts decreasing due to excessive demagnetizing effects of

armature reaction. In fact, terminal voltage starts decreasing as load current is increased as shown

by the dotted curve and for a particular high value of load current the terminal voltage is reduced to

zero.

FORMULAE:

Eg = V + Ia Ra (V)

Ia = IL + If (A) [for self Excited]

where

Eg : Generated EMF in V

V : Terminal Voltage in V

Ia : Armature Current in A

IL : Line Current in A

If : Field Current in A

Ra : Armature Resistance in Ohm

OPEN CIRCUIT TEST

PRECAUTIONS:

1. The field rheostat of motor should be in minimum resistance position.

2. The field rheostat of generator should be in maximum resistance position.

3. No load should be connected to generator.

4. Generator field should be separately excited.

PROCEDURE:


2. After checking minimum position of motor field rheostat, maximum position of generator

field rheostat, DPST switch is closed and starting resistance is gradually removed.


4. By varying the generator field rheostat, voltmeter and ammeter readings are taken.

5. After bringing the generator rheostat to maximum position, field rheostat of motor to

minimum position, DPST switch is opened.



LOAD TEST

PRECAUTIONS:

1. The field rheostat of motor should be at minimum position.

2. No load should be connected to generator at the time of starting and stopping.

PROCEDURE:

1. Connections are given as shown in the circuit diagram.

2. The DC supply is switched ON and the DC shunt motor (prime mover) is started using the 3-

point starter. The motor is brought to its rated speed by adjusting its field rheostat and the same

is checked with the help of a tachometer.

3. The load DPST is now closed and the loading rheostat is switched on in steps and at each step

the motor speed is maintained constant by adjusting the motor field rheostat and then the

terminal voltage (VL) and the load current (IL) are noted down.

4. The procedure is continued until the load current is equal to 120% of the rated current of the

generator. After the experiment is completed the load on the generator is gradually decreased to

minimum, bring motor field rheostat to minimum position and then the supply is switched OFF.

5. The resistances of the armature and the series field winding of the generator are found by giving

low voltage supply and connecting a voltmeter and ammeter.

6. The external and internal characteristics of the given DC series generator are plotted.

TABULATION

Armature Resistance - Ra: _______Ω

OPEN CIRCUIT TEST

S. No. Field Current

If (A)

Armature Voltage

Eo (V)

1

2

3

4

5



6

LOAD TEST

S. No. Load Current

IL= Ia = If (A)

Terminal Voltage

Vt (V)

Ia *(Ra+Rse)

(V)

Eg =Vt + Ia (Ra+Rse)

(V)

1

2

3

4

5

6

7

MODEL CALCULATION



MODEL GRAPH

OC TEST & LOAD TEST

RESULT:



SPEED CONTROL OF A DC SHUNT MOTOR

AIM:

To control the speed of the given D.C. shunt motor by

i) Armature - control method and

ii) Field/Flux - control method and plot the corresponding characteristics

NAME PLATE DETAILS:

Rated output =

Rated voltage =

Rated armature current =

Rated field current =

Rated speed =

APPARATUS REQUIRED:


1 DC Shunt Motor Set up - 1 No

2 Voltmeter (0-300 V) MC 1 No

3 Ammeter (0-2 A) MC

(0-5 A) MC 1 each

4 Rheostats (50 Ω/8 A)

(200 Ω/3 A) 1 each

5 Tachometer - 1 No

6 DPSTS - 1 No


THEORY:

The speed of a dc shunt motor can be expressed as N α (Eb/). From the above relation, it is clear that

the motor speed is directly proportional to the back emf (Eb) in the armature and is inversely proportional to

the flux Ф. In armature control method, the motor speed is controlled by adjusting the back emf (armature

voltage), keeping the flux constant. In the field control method, the flux (field current) is varied keeping the

back emf constant. These methods are widely used for controlling the speed of shunt as well as compound

motors.

The armature control method is generally used for varying the speed below the rated speed of the

machine. Flux control or Field control method is used for varying speeds above the rated speed of the

machine.

PRECAUTIONS: (Not to be included in the record)

1. Remove the fuse carriers and start wiring as per the circuit diagram.

2. Fuse rating calculation: Since this is no load test, the required fuse rating is only 20% of the rated

current of the motor.

3. Before switching ON the supply, the armature rheostat is kept at maximum resistance position and

field rheostat at minimum resistance position.

4. Replace the fuse carriers with appropriate fuse wire after the circuit connections.



PROCEDURE:


2. Keeping the armature rheostat in maximum resistance position and the field rheostat in minimum

resistance position, the main supply to the circuit is switched ON.

3. The motor is started using 3-point starter by slowly and carefully moving the starter handle from its

OFF to ON Position.

4. Armature rheostat is decreased slowly and brought to minimum resistance position.

5. Speed of the motor is adjusted to rated speed (1500 rpm) by increasing the field resistance.

(i)ARMATURE CONTROL METHOD:

1. Now one set of readings namely the field current (If1), Armature voltage (Va), armature current (I a)

and the motor speed (N) are noted down in the tabular column.

2. Keeping the field current constant at the above value (If1) the armature rheostat is gradually

increased in steps & at each step the armature voltage (Va), armature current(Ia) & the corresponding

speeds are noted down.

3. This procedure is continued until the armature rheostat is brought completely to its maximum

resistance position.

4. Now by adjusting the field rheostat, the field current is set to another fixed value (If2)

5. Armature rheostat is brought back to minimum resistance position and the steps 5-7 are repeated and

another set of readings are tabulated.

6. After the experiment is completed, armature rheostat is brought back to maximum resistance

position is brought to minimum resistance position and the main supply is switched OFF.

7. After completing the experiment, Eb is calculated.

8. Using the data, plot of Speed vs Armature voltage is drawn in the graph sheet.

(ii)FLUX CONTROL METHOD:

1. Armature rheostat is kept in its minimum resistance position and motor is brought to rated speed

(1500 rpm) by adjusting the field rheostat.

2. The armature voltage (Va1), armature current (I a1) is noted and keeping this constant, the field

current is varied in steps by adjusting the field rheostat and at each step the field current (If) and the

corresponding speeds are noted down in the tabular column.

3. This procedure is continued until the speed reaches 1800 rpm.

4. After obtaining a set of readings corresponding to Va1, the field rheostat is brought back to its initial

position.

5. Now armature rheostat is adjusted in such a way that the armature voltage is fixed at another

constant value of Va2 and armature current (I a2)

6. The steps 15 to 17 are repeated again to obtain another set of readings corresponding to Va2.

7. After the experiment is completed, field rheostat is brought to minimum resistance position and the

main supply is switched OFF.

8. Using the data, Plot of Speed vs field current is drawn in the graph sheet.



CIRCUIT DIAGRAM:

FUSE CALCULATION:

TABULAR COLUMN:

ARMATURE CONTROL METHOD FLUX CONTROL METHOD

Eb = Va - IaRa(eff)

If

(A)

Va

(V)

Ia

( A )

Eb

(V)

N

(rpm)

If1=

If2=



MODEL CALCULATION:

MODEL GRAPH:

RESULT:

REVIEW QUESTIONS:

1. Which method of speed control is used for controlling the speed of the motor above its rated speed?

Give reason.

2. Which method of speed control is used for controlling the speed of the motor below its rated speed?

Give reason.

Va

(V)

If

(A)

N

(rpm)

Va1=

Ia1=

Eb1=

Va2=

Ia2=

Eb2=



3. Explain the reasons for the shape of the graphs obtained.

4. Write the methods to control the speed of a DC series motor.

5. What are the factors affecting speed of a dc motor?

LOAD TEST ON A SINGLE PHASE TRANSFORMER

AIM:

To conduct a direct load test on the given single phase transformer and also to determine the

efficiency and the regulation at different load conditions

NAME PLATE DETAILS:

KVA rating =

Rated H.V side Voltage =

Rated L.V side Voltage =

APPARATUS REQUIRED:


1 Single Phase Transformer 230/115 V 1 No

2 Single Phase Autotransformer 220/(0-270 V) 1 No

3 Voltmeter AC (0-150 V), MI

(0-300 V), MI 1each

4 Ammeter AC (0-10 A), MI 1 No

5 Wattmeter (20A,150 V, UPF) 1 No

6 Resistive load - 1 No

7 DPST Switch - 1 No

8 Connecting Wires & Fuse - As Required

THEORY:

Direct load test is conducted to determine the efficiency characteristics and regulation characteristics

of the given transformer. Transformer is a static device having copper windings and iron core. Eddy current

losses and hysteresis losses that together called as iron or constant losses takes place in the iron core. The

variable losses or copper losses which are variable with load current takes place in the windings. These

losses decide the efficiency of the transformer under different load conditions. An ideal transformer is

supposed to give constant secondary voltage irrespective of the load current. But practically, the secondary

voltage decreases as the transformer is loaded due to primary and secondary impedance drops. Since these

drops are dependent on load current, this variation in terminal voltage is found using direct loading.

PRECAUTIONS: (Not to be included in the record)

1. Remove the fuse carriers and start wiring as per the circuit diagram.



2. Fuse rating calculation: Since this is load test, the required fuse rating is only 120% of the rated

current of the motor.

3. Keep the autotransformer in its minimum position at the time of starting..

4. Replace the fuse carriers with appropriate fuse wire after the circuit connections.

PROCEDURE:

1. The connections are made as per the circuit diagram.

2. Keeping the autotransformer in its minimum position and the DPST switch on load side in open

position, the main supply is switched ON.

3. By slowly and carefully operating the autotransformer the rated voltage (115V) is applied to the L.V

side of the transformer.

4. Under this no-load condition, one set of readings namely VH.V, IH.V, WH.V, VL.V and WL.V, are

recorded in the tabular column.

5. The DPST switch on the load side is now closed and the load is increased in gradual steps and at

each step primary voltage is maintained at no load voltage and then all meter readings are noted

down in the tabular column.

6. The procedure is continued until the current on the H.V side is equal to its full load value.

7. After the experiment is completed, the load is decreased to its minimum, the auto transformer is

brought to the minimum position and the supply is switched OFF.

8. After completing the experiment, Output Power, Input Power, Efficiency and Regulation are

calculated.

9. Using the data, plot of % Regulation vs Output Power and % Efficiency vs Output Power are drawn

in the graph sheets.

CIRCUIT DIAGRAM:

FUSE CALCULATION:



TABULAR COLUMN:

S. No.

LOAD

INPUT OUTPUT

(%) %Regulation VLV

(V)

ILV

(A)

WLV

(W)

VHV

(V)

IHV

(A)

WHV

(W)

FORMULAE USED:

1. The Output Power of the transformer = VH.V * IH.V on the H.V side

2. The Input Power of the transformer = WL.V = Wattmeter reading on the L.V side

3. % Efficiency (%η) = (OUTPUT / INPUT) * 100

4. % Regulation = *100

MODEL GRAPH:

MODEL CALCULATIONS:

RESULT:

REVIEW QUESTIONS:



1. Define – Regulation of a transformer

2. What is the effect of load power factor on regulation of Transformer?

3. What is the condition for maximum efficiency?

4. Determine the % load at which maximum efficiency occurred for the given 1-Φ transformer?

5. Explain mutual induction principle.

OPEN CIRCUIT AND SHORT CIRCUIT TESTS ON A SINGLE PHASE TRANSFORMER

AIM:

To conduct the open circuit test and short circuit test on a single-phase transformer and hence

predetermine the efficiency and regulation. And also to draw the equivalent circuit referred to H.V side and

to plot performance characteristics: (i)%η vs Output Power(ii) % Regulation vs Power factor.

NAME PLATE DETAILS:

KVA rating =

Rated H.V side Voltage =

Rated L.V side Voltage =

APPARATUS REQUIRED:


1 Single Phase Transformer 115/230 V 1 No.

2 1Ф Autotransformer 230/(0-270) V 1 No.

3 Voltmeter AC (0-150 V),MI, (0-75 V),MI 1 each

4 Ammeter AC (0-10 A),MI, (0-2 A),MI 1 each

5 Wattmeter ( 150V,2.5A,LPF), (75V,10A,UPF) 1 each

6 DPST Switch - 1 No.


THEORY:

The actual performance characteristics of transformers can be obtained by conducting a direct load test

on them. When this has to be performed on large rating transformers, the loads of the required size may not

be available, and the power consumed during this test will be very large as the transformers are loaded up to

120% of their rated capacity.

The time required to perform such a test is also more. The same performance characteristics can be

obtained by comparatively easier methods, which are known as indirect methods or predetermination

techniques. To predetermine the efficiency and regulation of transformers, the open circuit test [to determine

the core loss] and short circuit test [to determine the full load copper loss (variable loss)] are carried out.

Assuming the output and load power factor, the efficiency at different loads is computed.



The regulation on full load for different assumed load power factors can also be computed. The data

obtained from these tests are also useful to find the equivalent circuit parameters. The results obtained from

these tests are almost closer to the actual values obtained by direct load test.

PRECAUTION: (Not to be included in the Record)

1. Remove the fuse carriers and start wiring as per the circuit diagram

2. Keep the autotransformer in its minimum position at the time of starting

3. Fuse calculations: (i) For O.C test -- fuse rating -- 20% of rated current of L.V side

(ii) For S.C test -- fuse ratings -- 120% of rated current of H.V side

4. Replace the fuse carrier with appropriate fuse wires after the circuit connections are checked by the

Staff-in-charge

PROCEDURE:

O.C TEST:

1. The connections are made as per the circuit diagram.

2. Keeping the H.V winding open and the autotransformer in its minimum position, the main supply is

switched ON.

3. By slowly and carefully adjusting the autotransformer, the rated voltage (115V) is applied to the L.V

winding of the transformer.

4. Under this condition the ammeter (Io), Voltmeter (Vo) and Wattmeter (Wo) readings are noted down.

5. After the experiment is completed, the autotransformer is slowly brought back to its minimum

position and then the main supply is switched OFF.

6. After completing the experiment, constant loss, efficiency and regulations are calculated.

7. Using the data, Plot of % Regulation Vs Power Factor is drawn in the graph sheets.

S.C TEST:


2. Short circuiting the L.V winding and keeping the autotransformer in its minimum position, the main

supply is switched ON.

3. By slowly and carefully adjusting the auto transformer, the rated current IHV is circulated through the

H.V winding, (Note: H.V

.V

x1000ratingKVA VHI )

4. Under this condition, the ammeter (Isc), the voltmeter (Vsc) and the Wattmeter (Wsc) readings are

noted down.

5. After the experiment is completed, the autotransformer is brought back to its minimum position

and main supply is switched OFF.

6. After completing the experiment, copper loss, output power, efficiency and regulations are

calculated.



7. Using the data, Plot of % Efficiency vs Output Power and regulation graph(%reg vs pf) is drawn in

the separate graph sheets.

CIRCUIT DIAGRAM:

OPEN CIRCUIT TEST:

FUSE CALCULATION:

SHORT CIRCUIT TEST:



FUSE CALCULATION:

TABULATION:

TEST VOLTAGE

(V)

CURRENT

(A)

POWER

(W)

O.C. TEST

(On L.V side)

VO =

IO=

WO=

S.C. TEST

(On H.V side)

VSC =

ISC =

WSC=

FORMULAE USED:

I. To Obtain the Equivalent Circuit Parameters w.r.t H.V Side:

1. From the O.C test the constant loss (Iron loss) is noted Wc = Wo = ________ Watt.

2. From the S.C test the full load copper loss is noted WF.L = Wsc = _______ watt.

For a transformer, the equivalent circuit parameters can be determined either with respect to H.V

side or with respect to L.V side. If the parameters are estimated on the H.V side the resulting equivalent

circuit is called H.V side equivalent circuit of the transformer. From the O.C test Ro and Xo are calculated

using the following expressions,

I

V

w

0

).(0 VLR I

V

m

0

).(0 VLX

Where,Iw = I0 Coso , Im = I0 Sino

0=

00

01

IV

WCos



Since these values are calculated with respect to L.V side (because O.C test is conducted on the L.V

side), the equivalent values of `Ro' and `Xo' as referred to H.V side are determined as

K

R

2

0(L.V)

).(0 VHR K

X

2

0(L.V)

).(0 VHX

Where K = (secondary voltage) / (primary voltage)

K=115/230 for a step down operation; K = 230/115 for a step up operation.

Since we are assuming a step down operation K= 115/230 = 0.5.

I

Wsc

2

sc

).( VHTR , I

Vsc

sc

).( VHTZ R - Z T(HV)2

T(HV)2

)(HVTX

RT(H.V) and XT(H.V) are the total equivalent resistance and reactance of the transformer as referred to

the H.V side whose values are calculated from the S.C test.

Now the H.V side equivalent circuit is drawn and the parameters values are mentioned in the circuit.

II. TO PREDETERMINE THE EFFICIENCY:

The percentage efficiency is then predetermined for different load conditions for a specified load

power factor using the expression,

Output power = x*KVA*cos*1000 Watt

Copper loss = x2 *WSC Watt

Total loss = Core loss (Wo) + Copper loss watt

Input power = Output Power + Total loss Watt

% Efficiency = (Output Power/Input Power) 100 x %

Where `x' is the fraction of the full load which is 0.25 for 25% load, 0.5 for 50% load, 0.75 for 75% load, 1.0

for full load and 1.25 for 125% load and cos is the load p.f (assumed as 0.8 lag).

The efficiency values so calculated are entered in the tabular column as shown below.

III. TO PREDETERMINE THE PERCENTAGE REGULATION:

% Regulation = 100*)X (R T(HV)T(HV).

HV

VH

V

SinCosI , Where ‘+’ for lagging power factor, ‘-‘for

leading power factor, IH.V = Rated current on H.V side, VH.V = Rated voltage on H.V side

TABULATION:

TO PREDETERMINE THE EFFICIENCY:

% OF

LOAD

OUTPUT

POWER

COPPER

LOSS

TOTAL

LOSS

INPUT

POWER EFFICIENCY

X (W) (W) (W) (W) (%)

0.2

0.4

0.6



0.8

1.0

1.2

TO PREDETERMINE THE % REGULATION:

POWER FACTOR % Reg. for lagging P.F % Reg. for leading P.F

0.2

0.4

0.6

0.8

1

MODEL GRAPHS:

REGULATION CURVE: EFFICIENCY CURVE:

EQUIVALENT CIRCUIT REFERRED TO HV SIDE:

Output power (w)

%

η



MODEL CALCULATION:

RESULT:

REVIEW QUESTIONS:

1. Why is O.C test conducted on the L.V side and S.C test on the H.V side?

2. Define – Regulation in a transformer

3. Why is the regulation graph not passing through the origin?

4. Write the condition for maximum efficiency.

5. What are the parameters that are calculated from O.C and S.C tests?

REGULATION OF A THREE PHASE ALTRENATOR BY EMF AND MMF METHODS

AIM:

To conduct OC and SC tests on a given 3-Φ alternator and hence to predetermine the regulation by

(i) EMF method (ii) MMF method.

APPARATUS REQUIRED:

S. No. Apparatus Range &Type Quantity

1 Alternator Set up - I No.

2 Rheostat 200 Ω/3 A 2 Nos.

3 Voltmeter AC (0-600 V), MI I No.

4 Ammeter AC (0-10 A), MI I No.

5 Ammeter DC (0-2 A), MC I No.

6 Tachometer - 1No.

7 DPST Switch and TPST Switch - 1 each


THEORY:

Loading an alternator causes its terminal voltage to drop or rise depending upon (i) Magnitude of

load (ii) Nature of load. For a pure resistive load it drops by 8-12% below no-load value while for a

lagging p.f. load the drop is 25-50% below no load value and it is 20-30% higher for leading p.f. loads. The

reasons are 1) Armature resistance 2) Armature winding leakage reactance and 3) Armature reaction.



Electromotive force (EMF) and Magnetomotive force (MMF) methods are used to predetermine the

regulation of non-salient pole alternators. In emf method, the effect of armature reaction is represented as a

fictitious reactance Xar for each phase of the alternator. In mmf method effect of armature leakage reactance

is replaced by additional armature reaction. MMF method is more accurate.

Direct load Test is not preferred due to the absence of large sized loads and the enormous power

wastage involved in testing. Voltage regulation is defined as a percentage of rated voltage when load

current is reduced to zero suddenly by throwing off the load keeping If and speed constant.


1. Remove the fuse before starting wiring.

2. Fuse rating calculation: Since this is no load test, the required fuse rating is only 20% of the rated

current of the alternator.

3. Keep Motor field Rheostat in minimum resistance Position.

4. Keep the potential divider for alternator field in minimum voltage position.

5. Check that the TPST on alternator side is open.

PROCEDURE:

1. The circuit connections are given as shown in the circuit diagram.

2. Keeping the motor field rheostat in the indicated position and with the TPSTS open, the motor

supply is switched ON, by closing DPSTS1.

3. Motor is started using the 3-point starter by moving the handle from OFF to ON position and the

motor is brought to its rated speed by adjusting the rheostat in the motor field circuit.

4. Supply is switched on to the field winding of alternator by closing the DPSTS2.

O.C. TEST:

1. Using the 200 ohm potential divider, current in field circuit is increased in steps of 0.1A and at each

step the alternator induced voltage indicated by voltmeter and the corresponding field current (If) are

noted in tabular column.

2. This procedure is continued until the alternator voltage is 120% of its rated voltage.

3. After completing O.C. Test, the potential divider and motor field rheostat are brought to its

minimum position.

4. After completing the experiment, calculate Synchronous Impedance, Synchronous Reactance &

Regulation using the formulae given.

5. Using the data, Plot the graph between Eo Vs If.

S.C. TEST:

1. The alternator terminals are short circuited by closing TPST switch through an ammeter.

2. The rated current is made to flow through the armature of the stator windings by carefully adjusting

220 ohms potential divider from the minimum position.

3. After completing the experiment, calculate the Load current, Field Current and Regulation.

4. Using the data, Plot the graph Isc vs If and % Regulation vs Power Factor for both the EMF and

MMF methods.

CIRCUIT DIAGRAM:



FUSE CALCULATION:

MOTOR

ALTERNATOR

TABULATION:

O.C. TEST S.C TEST

S. No.

If

(A)

Vg(1-1)

(V)

Vg (ph)

(V)

S.C TEST



S. No. If

(A)

ISC

(A)

FORMULAE USED:

1. Synchronous Impedance (ph),ZS = (Open circuit voltage/phase) / (Short circuit current/phase)

(For the same field current)

2. Synchronous Reactance (Ph), XS= [(ZS) 2-(Ra) 2]1/2

EMF Method:

E0=(Vcos Φ +IaRa)2+(VSin Φ)±IaXS)21/2

E0 = Induced EMF per Phase, V=Rated voltage per phase, Ra = Armature resistance in Ω, Ia = Armature

current in A; ‘+’ for lagging p.f. load. ‘-‘for leading p.f. load.

100E

Reg % 0 XV

V

MMF Method:

From the O.C.C. graph, find (1) If1 - Field current required to produce rated voltage per phase.

(2) If2 - Field current required to produce rated current per phase during

S.C. test.

If = If12+If2

2-2If1If2cos(90±Φ)1/2

Where ‘+’ for lagging p.f. load, ‘-‘for leading p.f. load.

Now determine Vo corresponding to I from graph.

100E

Reg % 0 XV

V

EMF METHOD:

S. No. P.F

(lag)

Eo

(V)

Reg.

(%)

P.F

(lead)

Eo

(V)

Reg.

(%)

1.

2.

3.

4.

5.

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

MODEL CALCULATION: (EMF method)



MODEL GRAPH:

MMF METHOD:

S. No. P.F.

(lag)

If

(A)

E0

(V)

Reg.

(%)

P.F.

(lead)

If

(A)

Reg.

(%)

E0

(V)

1.

2.

3.

4.

5.

0.2

0.4

0.6

0.8

1.0

0.2

0.4

0.6

0.8

1.0

MODEL CALCULATION: (MMF method)



RESULT:

REVIEW QUESTIONS:

1. Define – Regulation

2. What is meant by pessimistic method?

3. Which method is called as optimistic method?

4. What are the advantages of EMF and MMF method?

5. List out the various methods used to predetermine the regulation.

V AND INVERTED V CURVES OFTHREE PHASE

SYNCHRONOUS MOTOR

AIM:

To draw V and inverted V curves for given three phase synchronous motor

APPARATUS REQUIRED:

S. No. Name of the apparatus Range Type Quantity

1 Voltmeter (0 – 600 V) M. I. 1

2 Ammeter (0 – 10 A) M. I. 1

3 Ammeter (0 – 2 A) M. C. 1

4 Wattmeter 600 V, 10 A UPF 1

5 Tachometer (0 – 10000

rpm) Digital 1

6 Three phase Auto

Transformer (0 – 470 V) - 1



7 Rheostat 950 Ω, 0.8 A - 1

CIRCUIT DIAGRAM:

PRECAUTION

1. All the switches should be kept open at the time of starting the experiment.

2. The potential divider in the field circuit of synchronous motor should be kept at minimum

potential position.

PROCEDURE


2. Close the T. P. S. T. switch.

3. The auto transformer is varied gradually to start the motor.

4. The auto transformer is adjusted till the voltmeter reads the rated voltage of the synchronous

motor.

5. Close the D. P. S. T. switch and increase the field current.

6. At no load condition, increase the field current in steps and note down the corresponding

armature current.

7. The potential divider is brought to the minimum potential position.



8. Repeat the same procedure for different load conditions.

9. Reduce the load on the motor.

10. Reduce the field current to zero value.

11. Reduce voltage by varying auto transformer.

12. Open all the switches.

TABULAR COLOUMN

S. No. Ia (A) If (A)

WATTMETER READING POWER

FACTOR W1 (W) W2 (W) W1 + W2 (W)

1

2

3

4

5

6

7

8

9

10

11

12

13

14

GRAPH

Field current, If Vs Armature current, Ia

Field current, If Vs Power factor, cosф



RESULT

Thus the V and inverted V curves of the given synchronous motor have been drawn.

LOAD TEST ON THREE PHASE INDUCTION MOTOR

AIM:

To conduct the direct load test on a given 3-phase induction motor and plot the performance

characteristics of the machine.

NAME PLATE DETAILS:

Rated voltage =

Rated power =

Rated current =

Frequency =

Rated speed =

APPARATUS REQUIRED:


1 3 Ф Induction Motor - 1 No.

2 3 Ф Autotransformer 415 V (0 – 470 V), 1 No.



12.4 KVA

3 Voltmeter AC (0-600 V), MI 1 No.

4 Ammeter AC (0-10 A), MI 1 No.

5 Wattmeter 600 V, 10 A , UPF

( Double Element) 1 No.

6 Tachometer, TPST Switch - 1 each

7 Connecting wires & fuse - As Required

THEORY:

Squirrel cage induction motors are so called because of the rotor construction, which is the most

rugged construction. The rotor conductors are heavy bars of copper, Aluminium that are permanently short-

circuited. The rotor slots are given a slight skew for quieter operation and to prevent the locking tendency of

the rotor. The direct load test is conducted on the squirrel cage induction motor to plot its performance

characteristics under loading condition. This is more accurate than the predetermination techniques as the

latter doesn’t take into account the effect of factors such as temperature, which cause significant change in

its operation.


1. Remove the fuse carrier before starting wiring

2. Fuse rating calculation: Since this is load test, the required fuse rating is only 120% of the rated

current of the motor

3. Before switching on the supply ensure the motor in on no load condition and the autotransformer is

in the minimum position

4. Replace the fuse carriers with appropriate fuse wires after the circuit connections are checked by the

staff in charge

PROCEDURE:

1. The connections are given as shown in circuit diagram.

2. The 3Ф ac supply is switched ON to the motor using the starter.

3. Under this load condition, one set of readings of the ammeter (IL), voltmeter (VL), wattmeter (W), spring

balance and the speed (N) of motor are noted down.

4. Now the mechanical load on motor is increased in regular steps in such a way that the current drawn by

the motor increases in steps of 1A.

5. At each step of loading, the entire meter readings are noted down in the tabular column.

6. This procedure is continued until the current drawn by the motor equals 120% of its rated value.

7. After the experiment is completed, the main supply is switched OFF.

8. After completing the experiment, Torque, Output Power, Power Factor, % Slip and % efficiency are

calculated by using the given formulae.

9. Using the obtained data, the plot of % efficiency Vs Output power, .% Slip vs Output power,

Speed vs Output power, power factor vs Output power, Line current vs Output power and

Slip vs torque.

CIRCUIT DIAGRAM:



FUSE CALCULATION:

TABULAR COLUMN:

S. No.

Line

voltage

VL

(V)

Line

current

IL

(A)

Input

power

(W)

Speed

(RPM)

Spring balance reading

Torque

(Nm)

Output

power

(W)

P.F % %Slip F1

(Kg)

F2

(Kg)

F1 ~ F2

(Kg)



MODEL GRAPHS:

FORMULAE USED:

1. Torque = 9.81*R*(F1~F2 ) N-m

R – Radius of the brake drum including belt thickness. (m)

F1, F2 – spring balance readings in kg.

2. Output Power = 2πNT/60 W

T – Torque in N-m

N – Speed in rpm.

3. Power Factor = Input Power / 3 VL* IL Watt

4. % Efficiency = (Output/Input)* 100 %

5. % Slip = [(Ns-N)/Ns]*100%

MODEL CALCULATION:



RESULT:

REVIEW QUESTIONS:

1. What is Skewing?

2. What is cogging?

3. What is crawling?

4. Define – Slip

LOAD TEST ON SINGLE PHASE INDUCTION MOTOR

AIM:

To draw load characteristics of a single phase induction motor by conducting the load test

APPARATUS REQUIRED:

S. No. Apparatus Range Type Quantity

1 Single phase induction motor - - 1

2 Single phase auto transformer (0 – 270 V) 1

3 Voltmeter (0-300) V MI 1



4 Ammeter (0-10 ) A MI 1

5 Wattmeter 300 V, 10 A, UPF 1

6 Connecting wires - - As required

THEORY:

Constructional of this motor is more or less similar to a poly phase induction motor, except

that its stator is provided with a single phase winding. A centrifugal switch is used in some type of

motor in order to cut out a winding, used in some type of motor, in order to cut out a winding, used

in some type of motors for starting squirrel cage rotor, when fed from a single phase only

alternating one which alternates along one phase axis only. Now, alternating or pulsating flux

acting on a stationary squired cage rotor cannot produce rotation that is why a single phase motor is

not self starting.

FORMULAE

1. Torque ( T )=S*9.81*R Nm

2. Output power (Po) = 2πNT/60 watts

3. Efficiency (η) = Output power / Input power X 100 %

4. Slip S = (Ns – Nr) / Ns * 100 %

5. Synchronous speed Ns = 120 f / P rpm

6. Power factor cos Φ = Pin / (VL *IL)

where,

R – Radius of brake drum.

VL – Line Voltage

IL – Line current

N – Speed in rpm

Nr - Rated speed in rpm

T – Torque in Nm

CIRCUIT DIAGRAM



FUSE RATING NAME PLATE DETAILS

PRECAUTIONS:

1. Motor should be started and stopped under no load condition.

2. Brake drum should be cooled with water when it is under load.

PROCEDURE:


2. The DPST switch is closed. The autotransformer is adjusted to get rated voltage and

corresponding no load readings are noted down.

3. Gradually increase the load upto the rated current and for each load the corresponding meter

readings are tabulated

4. Then load is removed and autotransformer reduced to zero. Then DPST switch opened.

TABULAR COLUMN:

Circumference of the Brake drum =_________m.

S. No

Voltage

VL (V) Current

I L (A)

Input

Power

Pin (W)

Spring Balance

Reading S=S1 S2

(Kg)

Speed

N

(rpm)

Torque

T

(Nm)

Output

Power

Po (W)

Efficiency

% SLIP

S

Power

Factor

Cos Φ S1 (Kg) S2 (Kg)

1

2



3

4

5

6

7

8

Radius of Brake Drum

Circumference = 2 π R =

Radius = R=Circumference / (2 π)

= _________meter

MODEL CALCULATION



MODEL GRAPH

RESULT:



STUDY OF DC AND INDUCTION MOTOR STARTERS

AIM:

To study the different kinds of D.C motor starters

THEORY:

The value of the armature current in a D.C shunt motor is given by

Ia = (V – Eb )/ Ra

Where V = applied voltage.

Ra = armature resistance.

E b = Back emf.

In practice the value of the armature resistance is of the order of one ohms and at the instant of

starting the value of the back emf is zero volts. Therefore under starting conditions the value of the armature

current is very high. This high inrush current at the time of starting may damage the motor. To protect the

motor from such dangerous current the DC Motors are always started using starters.

The types of D.C motor starters are

i) Two point starters ii) Three point starters iii) Four point starters.

The functions of the starters are i) It protects from dangerous high speed. ii) It protects the motor from

overloads.

i) TWO POINT STARTERS:

It is used for starting D.C. series motors which has the problem of over speeding due to the loss of

load from its shaft. Here for starting the motor the control arm is moved in clock-wise direction from its

OFF position to the ON position against the spring tension. The control arm is held in the ON position by the

electromagnet E. The exciting coil of the hold-on electromagnet E is connected in series with the armature

circuit. If the motor loses its load, current decreases and hence the strength of the electromagnet also

decreases. The control arm returns to the OFF position due to the spring tension, thus preventing the motor

from over speeding. The starter also returns to the OFF position when the supply voltage decreases

appreciably. L and F are the two points of the starter which are connected with the motor terminals.

ii) THREE POINT STARTER:

It is used for starting the shunt or compound motor. The coil of the hold on electromagnet E is

connected in series with the shunt field coil. In the case of disconnection in the field circuit the control arm

will return to its OFF position due to spring tension. This is necessary because the shunt motor will over

speed if it loses excitation. The starter also returns to the OFF position in case of low voltage supply or

complete failure of the supply. This protection is therefore is called No Volt Release (NVR).

Over load protection:

When the motor is over loaded it draws a heavy current. This heavy current also flows through the

exciting coil of the over load electromagnet ( OLR). The electromagnet then pulls an iron piece upwar6.ds

which short circuits the coils of the NVR coil. The hold on magnet gets de-energized and therefore the

starter arm returns to the OFF position, thus protecting the motor against overload. L, A and F are the three

terminals of the three point starter.



iii) FOUR POINT STARTER:

In a four point starter arm touches the starting resistance, the current from the supply is divided into

three paths. One through the starting resistance and the armature, one through the field circuit, and one

through the NVR coil. A protective resistance is connected in series with the NVR coil. Since in a four point

starter the NVR coil is independent of the of the field ckt connection, the d.c motor may over speed if there

is a break in the field circuit. A D.C motor can be stopped by opening the main switch. The steps of the

starting resistance are so designed that the armature current will remain within the certain limits and will not

change the torque developed by the motor to a great extent.



AUTO – TRANSFORMER STARTING:

An auto transformer starter consists of an auto transformer and a switch as shown in the fig. When

the switch S is put on START position, a reduced voltage is applied across the motor terminals. When the

motor picks up speed, say to 80 per cent of its normal speed, the switch is put to RUN position. Then the

auto-transformer is cut out of the circuit and full rated voltage gets applied across the motor terminals.

The circuit diagram in the fig is for a manual auto-transformer starter. This can be made push button

operated automatic controlled starter so that the contacts switch over from start to run position as the motor

speed picks up to 80% of its speed. Over-load protection relay has not been shown in the figure. The switch

S is air-break type for small motors and oil break type for large motors. Auto transformer may have more

than one tapping to enable the user select any suitable starting voltage depending upon the conditions.

Series resistors or reactors can be used to cause voltage drop in them and thereby allow low voltage

to be applied across the motor terminals at starting. These are cut out of the circuit as the motor picks up

speed.

STAR- DELTA METHOD OF STARTING:

The starter phase windings are first connected in star and full voltage is connected across its free

terminals. As the motor picks up speed, the windings are disconnected through a switch and they are

reconnected in delta across the supply terminals. The current drawn by the motor from the lines is reduced to

as compared to the current it would have drawn if connected in delta. The motor windings, first in star and

then in delta the line current drawn by the motor at starting is reduced to one third as compared to starting

current with the windings delta-connected.

In making connections for star-delta starting, care should be taken such that sequence of supply connections

to the winding terminals does not change while changing from star connection to delta connection.

Otherwise the motor will start rotating in the opposite direction, when connections are changed from star to

delta. Star-delta starters are available for manual operation using push button control. An automatic star –

delta starter used time delay relays (T.D.R) through which star to delta connections take place automatically

with some pre-fixed time delay. The delay time of the T.D.R is fixed keeping in view the starting time of the

motor.

FULL VOLTAGE OR DIRECT –ON-LINE STARTING:

When full voltage is connected across the stator terminals of an induction motor, large current is

drawn by the windings. This is because, at starting the induction motor behaves as a short circuited

transformer with its secondary, i.e. the rotor separated from the primary, i.e. the stator by a small air-gap.

At starting when the rotor is at standstill, emf is induced in the rotor circuit exactly similar to the

emf induced in the secondary winding of a transformer. This induced emf of the rotor will circulate a very

large current through its windings. The primary will draw very large current from the supply mains to

balance the rotor ampere-turns. To limit the stator and rotor currents at starting to a safe value, it may be

necessary to reduce the stator supply voltage to a low value. If induction motors are started direct-on-line

such a heavy starting current of short duration may not cause harm to the motor since the construction of

induction motors are rugged. Other motors and equipment connected to the supply lines will receive reduced

voltage. In industrial installations, however, if a number of large motors are started by this method, the

voltage drop will be very high and may be really objectionable for the other types of loads connected to the

system. The amount of voltage drop will not only be dependent on the size of the motor but also on factors



like the capacity of the power supply system, the size and length of the line leading to the motors etc. Indian

Electricity Rule restricts direct on line starting of 3 phase induction motors above 5 Hp.

ROTOR RESISTANCE STARTER:

In a slip-ring (wound rotor) induction motor, resistance can be inserted in the rotor circuit via slip

rings ,so as to increase the starting torque. The starting current in the rotor winding is

Where Rext = Additional resistance per phase in the rotor circuit.





RESULT:

Thus the starters used of three phase induction motor were studied.



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