nirma university · web viewdc shunt motor (with pony brake) 3hp, 230v, 10.5a, 1500 rpm....

NIRMA UNIVERSITY

INSTITUTE OF TECHNOLOGY

ELECTRICAL ENGINEERING PROGRAMME

B. TECH. SEMESTER-VI (ELECTRICAL), JANUARY- 2020

EE603: Testing, Commissioning and Maintenance of Electrical Equipments (LPW)

INDEXSR.NO. TITLE

PAGE NO.DATE SIGN MARKSFROM TO

1. To determine the efficiency of a DC machine by direct load test.

2. To obtain the efficiency of a DC machine by Swinburne’s test.

3. To determine the efficiency of two identical DC shunt machines by Hopkinson’s test.

4. To obtain the efficiency two identical DC series machines by Field test.

5. To explicate out iron and friction losses of a DC shunt motor using Ward-Leonard method.

6. To determine efficiency of a 1-phase transformer by Sumpner’s test.

7. To obtain zero phase sequence impedance in 3-phase transformer.

8. Megger test on an induction motor and determination of no. of poles and full load speed of the given unlabelled old 3-phase induction motor without energizing.

EXPERIMENT NO: 1 DATE:

AIM: To determine the efficiency of a DC machine by direct load test.

MACHINE SPECIFICATION:

DC SHUNT MOTOR (with pony brake)

3HP, 230V, 10.5A, 1500 RPM

APPARATUS:

(1) Voltmeter (MC) 0 - 300V – 01 no.

(2) Ammeter (MC) 0 - 15A - 01 no.

THEORY:

The efficiency of small motors are determine by directly loading the motor. The motor

is loaded by means mechanical brake, eddy current brake or calibrated fan. The entire

output power is wasted. The efficiency of the motor is calculated directly by measuring

the power input and the power output. The common type of mechanical brake

employed in testing is the rope brake as shown in Fig.1. The load is regulated by the

weight carried by the scale span.

In case of dc generator, the generator is driven through a transmission

dynamometer and the entire output power is dissipated in resistors.

The actual load torque is given by, T=(W 1 −W 2) R Nm

Where,

The mechanical power output of the motor is given by, watt

Where, rad/sec

The electrical power input can be calculated by measuring the terminal voltage

'V' and the input current 'I'. Then electrical input is given by, and the

efficiency of the motor is given by, = 2π NT60VI

×100

The pulley to be used for motor output more than 2 HP, should be either air-cooled or

water cooled. For series motor there should be sufficient brake torque before the motor

is switched on to the supply mains.

For the temperature test, it is advantageous to use generator as a load and adjusting the

voltage of generator so that it can operate in parallel with the supply system. The

fluctuation in spring balance causes the error in this test.

PROCEDURE:

1. Connect the circuit as shown in Fig. 2.

2. Set the spring balance load to zero.

3. Run the motor at rated armature voltage and excitation.

4. Measure the speed at no load condition.

5. Gradually increase the mechanical load on motor pulley.

6. Note down the readings of ammeter, voltmeter, speed and spring balance.

7. Repeat the test for other load adjustment and record the readings.

8. Switch off the supply and disconnect the circuit.

OBSERVATION TABLE:

SR.NO.

ARMATURE VOLTAGE

V (V)

ARMATURE CURRENT

I (A)

LOAD W (kg)

SPEEDN (rpm)

1

2

3

4

5

Radius of the pulley = R = _______________ m

Calculation:

Torque delivered by the motor, Nm

= ____________Nm

Speed in rad/sec, rad/sec

= _________ rad/sec

Power output of the motor, watts

= __________ W

Power input to the,

= ______________ W

Efficiency of motor,

= _____________

Similarly, efficiency for all the readings can be determined.

RESULT TABLE:

Sr.No.

Torque(Nm)

Motor output

(W)

Motor speed(rpm)

MotorEfficiency

(%)1

2

3

4

5

Graph: Plot the graph of load versus efficiency

Conclusion:Comment on the graph and the results obtained it this test.

Quiz:

1. Why the brake test is not recommended for the machines with higher HP rating?

2. What measures needs to be taken before the test is performed on the DC series mo-

tor?

3. At which load the efficiency of the machine is the maximum?

4. What is the major difference between the direct and indirect testing?

5. Why the brake test is called as the direct test?


AIM: To obtain the efficiency of a DC machine by Swinburne's test


DC SHUNT MOTOR.

3HP, 230V, 13A, 1500 RPM

APPARATUS:

1) Panel mount digital Voltmeter (MC) 0 – 1000 V 01 no.

2) Panel mount digital Ammeter (MC) 0 – 20 A 01 no.

3) Panel mount digital Ammeter (MC) 0 – 20 A 01 no.

4) Rheostat 366 Ω, 1.2 A. 01 no.

THEORY:

The efficiency of a dc motor can be determined by the following methods:

Direct loading method

Segregated-loss method

Regenerative method

In direct loading method a small error in the measurement of output or input introduces a

large error in the value of losses. It also consumes a lot of power for testing the machine. For

large size generators, it may become difficult or impossible to have such a large power

absorbing load at the site of testing. Hence it is advisable to adopt indirect method for the

determination of efficiency of the dc machines. Swinburne’s test (segregated-loss method) is

an indirect method of testing the efficiency of a motor or generator by measuring the losses.

Swinburne’s test is employed for dc shunt and compound motor. It is also known as `NO

LOAD TEST’. It is a simple indirect method in which losses are measured separately and

from the losses, the efficiency of dc machine can be determined at any desired load.

In this test the machine is run at no load at its rated voltage and rated speed.

Let I0 = no load current, Ish = shunt field current and V = rated voltage, then

No load armature current, I a 0=I 0−I sh

The total power input to armature = V I a0

The power drawn by the shunt field = I sh

As the output of the motor is zero, the power drawn at no-load supplies,

(i) Iron losses, (ii) mechanical losses (friction and windage losses), (iii) shunt field copper

losses, and (iv) armature copper losses at no load.

Hence

VI0 = Iron losses + mechanical losses (friction and windage losses) + Shunt field

copper losses + Armature copper losses at no load.

Here, iron losses, mechanical losses friction and windage losses), Shunt field copper losses

are practically constant at all loads, provided the applied voltage and the speed remain

constant. If the constant losses occurring in the motor be represented by WC, then

VI0 = WC + Armature copper losses at no load

Where, Armature copper loss at no load = I a 02 Ra

Ra is the armature winding resistance in ohms.

Thus the constant losses in the motor is given by,W C=V I a0−I a 02 Ra

The value of constant losses thus obtained can be used for calculating the efficiency of the

machine as a motor as well as generator at various loads.

Calculation of efficiency:

As a motor

Let current at full load drawn from the supply is IL Amps., the armature current is given by

I a=I L−I sh

Armature copper loss at full load = I a2 Ra

Constant losses at full load = W C

Total losses at full load = W C+ I a2 Ra

Input to the motor at full load =

Thus output of the motor = Input - losses = V I L−W C−I a2 Ra

Efficiency of motor at full load, % ηm=(V I L−W C−I a2 Ra

V I L)100

Similarly, efficiency at any load can be found.

As a generator

Let full load current fed by generator is IL Amps., the armature current is given by ,

Armature copper loss at full load =

Constant losses at full load = WC

Total losses at full load =

Output of the generator at full load =

Thus input of the generator = output + losses =

Efficiency of generator at full load, % ηg=( V I L

V I L+W C+ I a2 Ra

)100

Similarly, efficiency at any load can be worked out.

ADVANTAGES AND DISADVANTAGES OF SWINBURNE’S TEST

Advantages

1. As the power required during test for a large machine is less (only no load power),

this test is convenient and economical.

2. Because the constant losses are known, the efficiency of the dc machine at any load

condition, as a motor or generator can be determined.

Disadvantages

1. No account is taken of the change in iron losses from no-load to full-load. As load

changes the flux is distorted due to armature reaction which increases the iron losses.

2. This test does not give any indication of the satisfactory behavior of the machine

under loaded condition in respect of commutation and temperature rise.

PROCEDURE:


2. Apply the rated voltage and set the rated speed with the help of shunt field regulator.

3. Note down the readings of no-load current and field current.


5. Now connect the circuit as shown in Fig.2 for the measurement of armature

resistance.

6. Note down the armature resistance.

OBSERVATIONS:

TABLE -1: MEASUREMENT OF NO-LOAD POWER INPUT

SR.

NO.

INPUT

VOLTAGE

Vo (V)

NO-LOAD

CURRENT

Io (A)

FIELD

CURRENT

Ish (A)

ARMATURE

CURRENT

Ia = Io - Ish (A)

TABLE -2: MEASUREMENT OF ARMATURE RESISTANCE Ra

SR.

NO.

VOLTAGE

ACROSS

ARMATURE V

CURRENT IN

ARMATURE

I (Amp)

ARMATURE

RESISTANCE

R ()

AVERAGE

ARMATURE

RESISTANCE Ra ()

1

2

CALCULATIONS:

From no-load test-

Total input to the motor, = ____________ W

Armature current, = ____________ A

Armature cu-loss = = ____________ W

Thus the constant losses in the motor is given by,

W C=VI0−I a 02 Ra = __________ W

The value of constant losses thus obtained can be used for calculating the efficiency of

the machine as a motor as well as generator at various load.

Calculation of Efficiency

As a motor

Let full load current drawn from the supply is IL Amps, the armature current is given

by, = ___________ A

Armature copper loss at full load = = ___________ W

Constant losses at full load = WC = ____________ W

Total losses at full load = = ___________ W

Input to the motor at full load = = _____________ W

Thus output of the motor = Input - losses = V I L=W C−I a2 Ra =_______W

Efficiency of motor at full load % ηm=(V I L−W C−I a2 Ra

V I L)100 =_________

Similarly, efficiency at any load can be found.

As a generator

Let full load current fed by generator is IL Amps., the armature current is given by,

= _____________A

Armature copper loss at full load = = ____________ W

Constant losses at full load = WC = ___________ W

Total losses at full load = = ____________ W

Output of the generator at full load = = _____________ W

Thus input of the generator = output + losses = V IL+W C+ I a2 Ra = _______W

Efficiency of generator at full load % ηg=( V I L

V I L+W C+ I a2 Ra

)100 = ________

Similarly, efficiency at any load can worked out.

RESULT TABLES :

FOR MOTOR:

Sr.

No

.

Fraction

of load

Line

current

(A)

Armatur

e

cu-loss

(W)

Constant

losses

(W)

Total

losses

(W)

Motor

input

(W)

Motor

out put

(W)

Motor

Efficiency

(%)

1

2

3

4

5

6

FOR GENERATOR:

Sr.

No

.

Fraction

of load

Line

current

(A)

Armatur

e

cu-loss

(W)

Constant

losses

(W)

Total

losses

(W)

Generator

output

(W)

Generator

input

(W)

Generator

Efficiency

(%)

1

2

3

4

5

GRAPH : Plot the graph of efficiency v/s load current as a generator and as a

motor in the same graph paper.

RELEVANT IS :

1. IS 9320: 1979 Guide for testing of direct current machine.

2. IS 4889:1968 Method of determination of efficiency of rotating electrical machines.

CONCLUSION: (Discuss the following points)

1. Nature of the graph.

2. Give advantage of this test.

3. Comment on the efficiency determined by this method.

Quiz:

1. Why the magnetic and copper losses calculated by this method are less than the

actual values?

2. Show the losses taking place in the dc machines at different stage?

3. Why this test can not be performed on the dc series motor?

4. At what load the efficiency of the machines are the maximum?

5. What are the disadvantages of this method of testing?

EXPERIMENT NO: 3 DATE: AIM: To determine the efficiency of two identical DC shunt machines by Hopkinson’s

test.

MACHINE SPECIFICATIONS:

DC SHUNT MOTOR DC SHUNT GENERATOR

5 HP, 220V, 19A, 1500 RPM 3.2 kW, 230V, 14.5A, 1500 RPM

APPARATUS:

1. Panel mount digital Voltmeter (MC) 0 - 1000V 1 no.

2. Panel mount digital Voltmeter 0 - 1000 V 1 no.

3. Panel mount digital Ammeter (MC) 0 - 20A 2 nos.

4. Panel mount digital Ammeter (MC) 0 - 10A 2 nos.

5. Rheostat 366Ω, 1.2A 2 nos.

THEORY:

The efficiency obtained by Swinburne’s test is comparatively higher because, no

account is taken of the change in iron losses from no-load to full-load. As load changes

the flux is distorted due to armature reaction which increases the iron losses. Moreover

this test does not give any indication of the satisfactory behavior of the machine under

loaded condition in respect of commutation and temperature rise. These drawbacks can

be overcome if the efficiency of the machine is determined by Hopkinson’s test.

In this method, full load test can be carried out on two shunt machines, preferably

identical ones, without wasting their outputs. The two machines are mechanically

coupled to each other and connected electrically so that one of them runs as a motor

and the other as a generator. The mechanical output of the motor is supplied to the

generator and the electrical output of the generator is supplied to the motor. Hence the

majority of the motor input is supplied by the generator and remaining electrical power

is supplied by external power source.

In this test one of the DC machine is started as a motor by means of starter. Then, the

field of both the machines is adjusted such that the difference between the generator

emf and the motor emf becomes zero at the rated speed. There after by the use of

switch, generator emf is fed to the motor armature.

By adjusting either generator or motor field regulator, any load can be thrown on to the

machines. The net current drawn by the armature circuit of the motor from the supply

mains corresponds to the sum of iron losses, mechanical losses and armature circuit

copper losses of the two machines.

Let, V = supply voltage

I1 = Current drawn from the supply

I2= Generator armature current

I3= Motor field current

I4= Generator field current

I1 + I2 = Current in armature of motor

Rg = Armature resistance of generator

Rm = Armature resistance of motor

The total power drawn by the armature circuit of the motor from the supply

= Watts

= (Iron losses and mechanical losses of two machines

+ generator and motor copper losses.)

= 2W 0+ I 22 Rg+ ( I1+ I 2 )2 Rm

so, W 0=12 (V I1−I 2

2 Rg−( I1+ I 2 )2 Rm)

Calculation of Efficiency

As a motor

Power input to the motor at full load =


Thus output of the motor = Input - losses

=

Efficiency of motor at full load

Using the above equations the efficiency at any load can be found.

As a generator

Power output of the generator at full load =


Thus input of the generator = output + losses =

Efficiency of generator at full load,

Using the above mentioned equation the effeiciency to any load can be found.

ADVANTAGES AND DISADVANTAGES OF HOPKINSON’S TEST

Advantages

i. Total power drawn from the supply is low.

ii. Both the machines carry full load current during testing and as such commutation

problems can be seen and final temperature rise can be checked.

Disadvantages

The only disadvantage of this test is the availability of two identical machines.

PROCEDURE:


2. Keep switch `s’ open

3. Start motor by the use of starter and brought up to its rated speed.

4. Set the reading of the voltmeter connected across the switch `s’ zero by increasing

the excitation of the generator.

5. Once the voltmeter across switch `s’ reads zero, close the switch `s’.

6. Now again increase the excitation of generator and note down all reading of

ammeters and voltmeter.

7. Repeat the step no.6 and tabulate the reading.


9. Now measure armature resistance of motor and generator by the use of multi-meter.

10. Note down the value of armature resistance.

OBSERVATION TABLE:

Sr.

No.

Input

voltage

V (V)

Current

drawn from

the supply

I1 (A)

Generator

armature

current

I2 (A)

Motor

field

current

I3 (A)

Motor armature

current

I1 + I2 (A)

Generator

field

current

I4 (A)

1.

2.

3.

4.

5.

Armature resistance of motor Rm = ____________ Ω

Armature resistance of generator Rg = ____________ Ω

CALCULATIONS:

Power input to the set = = ____________ W

Armature copper loss of the motor = = ____________ W

Armature copper loss of the generator = = ____________ W

Constant losses of each machine = ___________ W

Efficiency of motor

Shunt field cu loss of the motor = = __________W

Total losses of the motor = _______ W

Total power input to the motor

= ________

Efficiency of generator

Shunt field cu losses of generator = = _______ W

Total losses of generator = ______ W

Power output of the generator = ________ W

= _________

RESULT TABLE:

Sr.

No.

Motor Generator

Input

(W)

Copper

loss

(W)

Constant

loss

(W)

Efficienc

y

(%)

Output

(W)

Copper

loss

(W)

Constant

loss

(W)

Efficiency

(%)

1

2

3

4

5

6

GRAPH: Plot the graph of efficiency v/s output for a generator and as a motor in the

same graph paper.

RELEVANT IS :


2. IS 4889: 1968 Method of determination of efficiency of rotating

electrical machines.

CONCLUSION: (Discuss following points)

1. Discuss the shape of the graph.

2. Compare result with Swinburne’s test.

3. Comment on the efficiency determined by this test.

QUIZ:

1. Justify the assumptions of equal division of rotational losses.

2. Why this test is also known as regenerative test?

3. From field current value how will you say which machine works as generator and

which one works as a motor? Justify your answer.

4. Why in this test circuit one of the machines field winding current is not carried by

the starter?

5. Can this test is applicable for the test of the dc series machines? Justify your answer.


AIM: To obtain the efficiency two identical DC series machines by Field test.


D. C SERIES MOTOR D. C SERIES GENERATOR

3 HP, 230 V, 11.8 A, 1500 rpm 2.2 kW, 230 V, 11.8 A, 1500 rpm

APPARATUS :

1. Panel mount digital Voltmeter 0 - 300 V DC 02 nos.

2. Panel mount digital Voltmeter 0 - 30 V DC 02 nos.

3. Panel mount digital Ammeter 0 - 10 A DC 02 nos.

THEORY:

Swinburne’s method of testing is not applicable to a series machine, and

Hopkinson’s test on series motor is possible in case of small machines. This

method is useful in finding the efficiency of the DC series motor. Two identical series

motors are required in this test. Series motors which are mainly used for traction

work are easily available in pairs. The two machines are coupled mechanically.

The iron losses and friction losses are made equal by connecting two field’s in

series with the motor armature and coupling them together.

One machine is started as a motor with the help of starter. It is desirable to

start the set when the load is connected to Generator, otherwise it results

dangerous speed. After this at full load the readings are noted.

Let, Vs = supply voltage

I1 = motor current

Vg = terminal voltage of generator

Ig = load current

Rm = motor armature resistance

Rg = generator armature resistance

Rse = series resistance of each field.

The power intake of the whole set =

Power output of the set =

Total losses in the set,

Armature and field copper loss

Total stray loss of the set

Stray loss of the each machine

Motor efficiency:

Motor input = 1 1V I

Motor losses = o cumot semotW W W = 21o m seW R R I

Motor efficiency

Generator efficiency:

Generator output =

Generator losses = =

Generator efficiency

ADVANTAGES AND DISADVANTAGES OF FIELD TEST

Advantages

1. The main advantage of this test is, it is applicable to series machines.

Disadvantages

1. Output of the generator is entirely wasted on dead load.

2. Small inaccuracy in the measurement of the motor input or generator output produces

large error in their difference for traction motors as they are of very high efficiency.

PROCEDURE:


2. Apply load on the generator first and then start the motor at test using starter.

3. Change the load of generator and take the readings of ammeters and

voltmeters.

4. Measure the armature and field resistance with the help of multimeter.

5. Disconnect the circuit.

OBSERVATION TABLE :

Sr.

No

.

Motor

armature

current

I1 (A)

Supply

voltage

Vs (V)

Voltage

across motor

series field

Vsem (V)

Voltage across

generator series

field

Vseg (V)

Generato

r

volts

Vg

(V)

Generator

load current

Ig (A)

1.

2.

3.

4.

5.

6.

CALCULATIONS :

Input power to the set = = _________ W

Output power of the set = = ________ W

Motor armature copper loss = = _______ W

Generator armature copper loss = = _______ W

Motor and the generator field copper loss = = ______ W

Total loss of the set = ________ W

Stray loss for both machines = 2 2 21 1m g g sem segW I R I R I R R

= _______ W

Efficiency for the motor:

Motor input = 1 1segV V I = ________ W

Motor losses = 21o m semW R R I = _________ W

Motor efficiency,

21 1 1

1 1

% 100seg o m semm

seg

V V I W R R I

V V I

= _________

Efficiency for the generator:

Generator Input = g gV I = _________ W

Generator losses = = _________ W

Generator efficiency = %= ________

RESULT TABLE :

Sr.

No

.

Motor Generator

Input

(W)

Losses

(W)

Efficiency

(%)

Output

(W)

Losses

(W)

Efficiency (%)

1

2

3

4

5

GRAPH : Plot the graph of output V/ s efficiency for motor and generator on the

same graph paper.

CONCLUSION: Discuss the usefulness of the test in all respect.

Quiz:

1. Why it is not advisable to start series motor without load?

2. What are the assumptions taken for this test?

3. Why the value of field resistance in series motor is less than the field resistance of

the shunt motor?

4. What are the advantages of this method?

5. Why this test is not called the regenerative test?


AIM: To explicate out iron and friction losses of a DC shunt motor using Ward-

Leonard method.


DC MOTOR DC GENEARTOR

5 HP, 220 V, 19 A, 1500 rpm 3.2kW, 220 V, 14.5 A, 1500 rpm

APPARATUS:

1. Panel mounted digital Voltmeter 0 - 300 V (D.C) 01 No.

2. Panel mounted digital Ammeter 0 - 3 A (D.C) 01 No.

3. Panel mounted digital Ammeter 0 - 10 A (D.C) 01 No.

4. Rheostat 366 , 1.2 A 02 Nos.

5. Rheostat100 , 5 A 01 No.

THEORY:

By the use of Swinburne’s test, Hopkinson’s test and field test, we can find the

efficiency at D.C machine as well as the total stray losses. But we can not separate out

the iron losses (i.e. hysteresis loss and eddy current losses) and friction and windage

losses. Using this method we can separate out the above losses.

In this test the main motor of set is run at no-load. The no-load losses are provided by

the generator. At different excitation current the speed is maintained constant by

changing the applied voltage to the motor. The smooth variation required in applied

voltage to keep speed constant is provided by Ward-Leonard set. The speed is kept

constant to keep mechanical (friction and windage) losses constant. Then total iron and

friction losses are measured by subtracting the Cu losses from the total power taken by

the motor when running on no-load. It is desirable to feed the field current from

separate supply.

In fig.1 curve A represents the input to the armature excluding armature Cu-loss. Curve

B is a horizontal line drawn from the point of intersection between curve A and Y axis.

It represents the mechanical losses. These losses are normally constant at a constant

speed. The total stray losses at an excitation OP is represented by PR. PQ represents the

frictional losses and QR represents iron losses.

For a given flux, the power wasted in magnetic hysteresis is directly proportional to the

frequency of flux reversals, and hence to the speed. The power wasted in eddy currents,

on the other hand, increases in proportion to the square of the frequency and hence also

of the speed. If, then, the total iron losses be known for a given flux at two different

values of speed, hysteresis loss can readily be separated from the eddy current loss. For

this the test is repeated either one half or twice normal speed, whichever is the more

convenient. The results are than plotted in terms of E / n as shown in Fig.2. Taking the

case when the second test is at half normal speed,

Core loss at normal speed

Core loss at half speed, for same flux

Hence, and

Thus if, as shown in Fig. 2, the total iron loss curve be represented by A, and curve B

represents twice the total loss at half speed, PL being the total iron loss at normal speed

and PN twice the loss at half speed for the same flux, then LN is one half of the eddy

current loss at full speed. Hence, on constructing a third graph C by subtracting from

each ordinate of B an amount NM = LN, the ordinates PM of this third graph will give

the hysteresis loss at full speed, and the value of ML will give the eddy current loss.

Having thus separated out the losses for one particular speed over the whole range of

working flux, the losses at any other speed are at once obtained by taking the hysteresis

loss to be proportional to the speed, and the eddy loss proportional to the square of the

speed.

PROCEDURE:

1. Connect the circuit as shown in Fig.3 keeping switch S1 open.

2. Start the motor of Ward-Leonard set and adjust its excitation to give the rated speed.

3. Adjust the field current of generator to zero.

4. Excite the field circuit of the motor to be tested and then close the switch S1.

5. Adjust speed of the motor under test to its rated value by varying the output voltage of

generator. This can be done by varying Ward-Leonard generator field current.

6. Measure the terminal voltage and armature current.

7. Change the excitation current of motor to be tested keeping its speed constant as

mentioned in step 5 and take several readings.

8. Repeat the procedure for half speed of motor under test.

CALCULATIONS:

Normal Speed Reading No. _______

N = ________ rpm, n = _______ rps

Armature resistance Ra = __________

Field current If = _________ A

Armature current Ia = ________ A

Terminal voltage V = _________ V

Armature input = __________ W

Armature copper loss = ________ W

Iron and friction losses = _________ W

Induced emf = __________ V

= _________ volts/rps

Friction and windage losses Pfw1 = __________ W

Iron losses = __________ W

Half the Normal Speed Reading No. _______

N = ________ rpm, n = _______ rps

Armature resistance Ra = __________

Field current If = _________ A

Armature current Ia = ________ A

Terminal voltage V = _________ V

Armature input = __________ W

Armature copper loss = ________ W

Iron and friction losses = _________ W

Induced emf = __________ V

= _________ volts/rps

Friction and windage losses Pfw2 = __________ W

Iron losses = __________ W

= _________ W

= __________ W

OBSERVATION TABLE:

Sr.

No.

Speed

N (rpm)

Armature current Ia

(A)

Armature Voltage

(V)

Field Current If

(A)

CALCULATION TABLE :

NORMAL SPEED :

Sr.

No.

Input Power

P = VIa

(W)

Armature Cu

Losses Ia2Ra

(W)

Iron & Friction

Losses

Po = P - Ia2 Ra

(W)

Induced

E.M.F

E = V - IaRa

(V)

E / n

volts / rps

Iron Losses

Pi = P0 - Pfw

(W)

HALF THE NORMAL SPEED :

RESULT TABLE:

Sr.

No

.

Speed

(rpm)

Armatur

e

Input

(W)

Armature

Copper

Loss (W)

Iron

& FW

Losse

s

(W)

FW

Losses

(W)

Iron

Losses

(W)

Induce

d

E.M.F

E (V)

E / n

(volts/rps)

Eddy

Curren

t Loss

(W)

Hysteresis

Loss (W)

GRAPH :

(A) If Vs P01 & P02, (B) E / n Vs Pi1 & 2Pi 2 for both speeds.

Construct a third graph by subtracting each ordinate of graph B from that of graph A

at a particular E / n ratio.

CONCLUSION :

Hysteresis loss at N1 rpm = _________ W

Eddy current loss at N1 rpm = ___________ W

RELEVANT IS:


2. IS 4889: 1968 Method of determination of efficiency of rotating electrical machines.

QUIZ:

1. Discuss the effect of speed on the hysteresis and eddy current losses?

2. If the readings are available at the rated and twice the rated speed than what will be

the equations of the hysteresis and eddy current losses at rated speed?

3. Discuss the factors on which the hysteresis and eddy current losses depend?

4. Why the hysteresis and eddy current losses are called iron losses?

5. If the load on the motor is reduced to half of their rated value, then what will happen

to their iron losses?


AIM : To determine efficiency of a 1-phase transformer by Sumpner’s test.


Single phase transformer

4.5 kVA, 230 V / 115 V, 50 Hz.

APPARATUS: 1. Panel mounted digital Ammeter 0-20 A (M.I.) 01 no.

2. Panel mounted digital Ammeter 0-20 A (M.I.) 01 no.

3. Panel mounted digital Ammeter 0-20 A (M.I.) 01 no.

4. Panel mounted digital Ammeter 0-20 A (M.I) 01 no.

5. Wattmeter 300 V/10 A, 0 - 300 W 01 no.

6. Wattmeter 30 V/ 20 A, 0 - 1500 W 01 no.

7. Auxiliary transformer 230 / 20 V 01 no.

8. Single phase variac 20 A 01 no.

9. Single pole switch 01 no.

THEORY :

This test provides data for finding regulation, efficiency, and heating under load

condition and is employed only when two identical transformers A and B are available.

These transformers are connected in parallel on both side. Their e.m.f are in opposition

round the close circuit formed by the two primaries and two secondaries. Disregarding

the effect of the auxiliary transformer, the wattmeter W1 will read the total core loss of

the two transformers together. A voltmeter across the switch Sw should read zero,

indicating complete equality of secondary voltages and phase opposition, and when it is

closed wattmeter W1 should be unaffected.

A circulating e.m.f. is introduced into the primary circuit by means of the secondary of

an auxiliary transformer C energized from a suitable AC supply of any convenient

frequency, it need not be the same as that of the main supply to avoid unequal loading.

The e.m.f. necessary is the product of the circulating current and the impedance in

primary turns. The e.m.f. and the circulating current are associated with wattmeter W2

consequently records the I2R loss. The current is adjusted by control of voltage applied

for a full value of the circulating current, the full load I2R is expanded in the winding of

the transformer under test and the cores are energized to normal induction density.

PROCEDURE:

1. Connect the circuit as shown in the circuit diagram.

2. Keep the switch Sw open.

3. Switch on the supply to both primary windings and noted down the readings of

current drawn, supply voltage and wattmeter W1 reading. (The voltmeter across the

switch Sw should read 0 due to complete equality of secondary voltages and phase

opposition of both secondaries.

4. Close the switch Sw.

5. Inject the voltage in the secondary by means of auxiliary transformer C up to full

load secondary current I2.

6. Note down the readings of W1, W2 etc.

7. Calculate the individual efficiency for different load condition at power factor 0.8

and unity and comment on performance.

OBSERVATION TABLE :

Sr No. Supply

Voltage

Vs (V)

No load

current I0

(A)

Total iron

loss W1

(W)

Current

I1

A)

Current

I2

(A)

Total Cu

loss W2

(W)

1.

2.

3.

4.

5.

6.

7.

8.

RESULT TABLE:

Sr.

No.

Output

VsI2cos

(Watts)

Total losses

W1 + W2

(Watts)

Input

o/p + loss

(Watts)

Individual

Efficiency

(%)

P.F.

= cos

Load

1. 1.0 1/4 F.L.

2. 1.0 1/2 F.L.

3. 1.0 3/4 F.L.

4. 1.0 F.L.

5. 0.8 1/4 F.L.

6. 0.8 1/2 F.L.

7. 0.8 3/4 F.L.

8. 0.8 F.L.

CONCLUSION:

QUIZ:

1. How many transformers are needed in this test? Why?

2. What is the purpose of using an auxiliary transformer in this test?

3. State the advantages of this test over other methods of determining the efficiency of

transformer.

4. What is the roll of wattmeter on the primary side of the transformer?

5. Comment upon the reading of wattmeter, connected in the secondary circuit of the

transformers.


AIM: To obtain zero phase sequence impedance in 3-phase transformer.

APPARATUS:1. Three-phase transformer2. Variac3. 0-10 A MI meter4. 0-300 V MI meter5. 230 V, 50 Hz mains supply

WRITE DOWN NAME PLATE DETAILS OF TRANSFORMER:

THEORY:As per IS 2026 - 1972, there are three types of tests to be carried out on transformers.

(i) Routine test (ii) Type tests (iii) Commissioning tests (iv) Special tests

Routine tests are performed at manufacturer premises to see whether the desired

specifications are met or not. These tests are performed on each and every product

manufactured. For a transformer, these are

Measurement of winding resistance

Checking ratio, polarity and phase relationship

Measurement of impedance voltage

Measurement of full load losses

Measurement of no load current and no load losses

Type tests are performed by the manufacturer at their premises on a sample product of a

lot. These tests are not performed on each product manufactured. The purpose of these

tests is to check whether the design specifications are achieved or they need certain

alteration. Type tests include

Temperature rise test

Impulse voltage withstand test for oil immersed type transformer only.

Special tests are performed at the manufacturer's premises in presence of the purchaser.

The purchaser should state about it at the time of ordering. They are performed at extra

cost to the purchaser. The followings are the special tests for transformer.

Measurement of zero phase sequence impedance

Impulse voltage withstands test including chopped wave.

PROCEDURE:1) Carry out megger test for the windings first.

2) Make the connections as shown in circuit diagram.

3) Keep the secondary terminals open.

4) Adjust the test voltage so that 10 to 20 % of rated current flows through the

windings of the transformer or through the circuit.

5) Record the readings.

6) Calculate the zero phase sequence impedance.

OBSERVATION TABLE:Sr.No.

Input voltage – V (V)

Current I (A)

Zo = 3 V/I Average Zo Zo p.u.

12345678

CONCLUSION:

QUIZ:1. Draw a complete specification for a 3-phase power transformer, which is to be

intended to purchase.2. State the possible hazards, if commissioning tests are not performed carefully. 3. What is an impulse wave? How is it specified?4. State the importance of zero sequence impedance.5. Explain the positive, negative and zero phase sequence current.6. Out of positive phase sequence, negative phase sequence and zero phase sequence

impedance, which impedance is high and why?

EXPERIMENT NO.: 8 DATE:

AIM: Megger test on an Induction Motor and determination of no. of poles and full load speed of the given unlabelled 3-phase induction motor without energizing or connecting it across the supply.

APPARATUS:1. Galvanometer ( 5-0-5 V) range (one)2. Screw driver - (one)3. Megger (one)

WRITE DOWN NAME PLATE DETAILS OF INDUCTION MOTOR:

THEORY & PROCEDURE:

DETERMINATION OF NO.OF POLES AND SPEED:

This method is useful in case there is no nameplate on body of an old induction motor.

The motor stator and rotor being magnetic material, can retain residual flux or

magnetism. If a galvanometer is connected across the two terminals of one phase or

phase to phase, and the shaft is rotated by hand for 10 revolutions, the galvanometer

will show corresponding no. of oscillations. These no. of oscillations depend upon the

no. of pair of pole of machine.

Say, the oscillations measured found to be 40.

No. of pairs of poles = No. of oscillations per revolution

= 40/10

= 4

So, No. of poles = 8

Synchronous speed, Ns = 120*50/8 = 750 rpm

Considering full load slip = 4 %,

Full load speed N = Ns (1 - 0.04) = 750*0.96 = 720 rpm

MEGGER TEST:

Before commissioning of the motor, megger test is to be carried out on the motor. Due

to the moisture or vapour, the insulation of the winding might have been deteriorated.

Without testing the insulation, if motor is energized, winding may get spark over and

burnt. Hence, the large motors are never started without its insulation test or megger

test, if motor has remained idle for long time.

In insulation test, the insulation between conductor of each phase and earth is checked

as well as the insulation between conductors of two phases is checked with the help of

megger.

Tabulate the readings taken in the observation table.

OBSERVATIONS FROM TEST 1:

No. of oscillations for 10 revolutions of rotor =

No. of oscillations per revolution of rotor =

Synchronous speed, Ns =

If slip, s =

Actual speed N = Ns (1-s) =

OBSERVATION TABLE FOR TEST 2:

Sr. No. Insulation resistance between each phase of winding and body ( earth ) - M

Insulation resistance between two phases of winding - M

A B C A & B B & C C & A1.2.3.

CONCLUSION:

QUIZ:1. State the different types of meggers available.2. Explain the principle of working of the megger.3. Can we measure the insulation with the help of multimeter? Why?4. Check the no. of oscillations obtained in one revolution and illustrate its reasons.5. As a thumb rule, what should be the I.R of the winding? Discuss the effect of

moisture on quality of insulation?

nirma university · web viewdc shunt motor (with pony brake) 3hp, 230v, 10.5a, 1500 rpm....

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