1

A HAND BOOK FOR

DC MACHINES LAB

Electrical & Electronics Engineering

Department

2

DC MACHINES LAB MANUAL

(EEE DEPATMENT)

INDEX

SL.NO. NAME OF EXPERIMENTS PG.NO.

1 LOAD TEST ON DC SERIES MOTOR

2

2 LOAD TEST ON DC SHUNT MOTOR

6

3 LOAD CHARACTERISTICS OF DC SHUNT GENERATOR

10

4 OCC OF DC SHUNT GENERATOR

15

5 HOPKINSON'S TEST

21

6 SWINBURN'S TEST ON DC MACHINE

28

7 OC&SC TEST ON 1 PHASE TRANSFORMER

33

8

POLARITY & TRANSFORMATION RATIO TEST ON A 1

PHASE TRANSFORMER

39

9 OC&SC TEST ON 3 PHASE TRANSFORMER

42

10

LOAD CHARACTERISTICS OF DC COMPOUND

GENERATOR

46

11 SUMPNER'S TEST ON TWO 1 PHASE TRANSFORMER

50

12 SEPERATION OF LOSSES IN A 1 PHASE TRANSFORMER

55

3

EX. NO. : 1 DATE :

LOAD TEST ON DC SERIES MOTOR AIM:

1) To conduct load test on DC series motor and to determine the efficiency using

mechanical loading arrangement

2) To draw the following characteristics curves

Torque Vs Armature current (electrical characteristics)

Speed Vs Armature current

Speed Vs Torque (Mechanical characteristics)

Performance curve

APPARATUS REQUIRED:

Sl no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

MACHINES DETAILS:

THEORY:

Ina series motor the field winding is connected in series with the armature .The

field current and the armature current are the same.Series motor is a variable speed

motor. If the mechanical load on the motor increases, the armature current also

increases. Hence the flux in a series motor increases with increase in armature

current.

4

NαEb/ɸ

Taα ɸ Ia

ɸ α Ia , so Taα Ia2

ie, Starting torque of a series motor will be very high .

ifNαEb/ɸ ,WhenIa is very small speed become dangerously high, so a series motor

should never be started without somemechanical load on it, if there is no

mechanical load at the time of starting it may develop excessive speed and get

damaged due to heavy centrifugal force produced.

The load test is conducted by applying a spring balance load to brake drum of

machine. By tightening the belt over the pulley different load can be applied. The

spring balance reading S1 and S2 are noted and the effective radius ‘r’ of the

pulley is measured.

If N is the speed of motor, then Torque developed T= (S1-S2)rg Nm

The output power can be calculated as P=2πNT/60 Watts.

Efficiency of the motor=output power/input power.

PRECAUTIONS:

1. DC Series motor should notstart on no load condition.

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

PROCEDURE:

Connections are made as per the circuit diagram.

The precautions checked before starting the experiment.

A 220 V Dc supply is given to the circuit using two point starter.

Varying the load (Take readings up to rated current of the machine) by

tightening/releasing the belt over the pulley, the meter readings, spring

balance reading and speed are noted. After bring the load to initial position,

switch of the supply.

5

FORMULAE USED:

1.Torque(T)=9.81(S1-S2). R

Where, S1, S2→ spring balance readings in kg

R → Radius of brake drums in m

2. Output power (Pout) = 2πNT/60

Where, N=Speed in rpm

T=Torque in Nm

3. In put power (P in)= VL.IL Watts

VL=Input voltage in volts

IL=Input current in Amps

4. % Efficiency =( Pout/Pin)*100

CIRCUIT DIAGRAM

6

TABULAR COLUMN

S. No.

Input Voltage

VL (Volts)

In put current

IL (amps)

Speed N(rpm)

Spring balance Readings Torque

T(Nm)

Output Power

Pout (Watts)

Input Power

Pin (watts)

% η S1

(Kg) S2

(Kg) S1-S2

(Kg)

SAMPLE GRAPH :

Performance curve

RESULT:

7

EX. NO. :2

DATE :

LOAD TEST ON DC SHUNT MOTOR

AIM:

To conduct load test on DC shunt motor and to determine the efficiency using

mechanical loading arrangement

To Draw the following characteristics curves

Torque Vs Armature current (electrical characteristics)

Speed Vs Armature current

Speed Vs Torque (Mechanical characteristics)

Performance curve

APPARATUS REQUIRED:

Sl no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

MACHINES DETAILS:

THEORY:

Ina shunt motor the field winding is connected in parallel with the armature .The

field current and the armature current are not same. Shunt field windings are

having large number of turns of wire having high resistance. So shunt field current

is relatively small compared with the armature current. Applied voltage andRsh of

a shunt motor are constant, so field current Ish also will be constant.

8

Ish=V/Rsh

Hence, the flux in a shunt motor is constant.

N α Eb/ɸ

As ɸ is constant, speed of shunt motor will be constant.

Taα ɸ Ia

ɸ is constant, so Taα Ia

Sostarting torque of a shunt motor islow.

The load test is conducted by applying a spring balance load to brake drum of

machine. By tightening the belt over the pulley different load can be applied. The

spring balance reading S1 and S2 are noted and the effective radius ‘r’ of the

pulley is measured.

If N is the speed of motor, then Torque developed T= (S1-S2)rgNm

The output power can be calculated as 2πNT/60 Watts

Efficiency of the motor=output power/input power.

PRECAUTIONS:

1. DC shunt motor should be started and stopped under no load condition.

2. Field rheostat should be kept in the minimum position.

3. 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 no load condition and minimum field rheostat position starter

resistance is gradually removed.

3. The motor is brought to its rated speed by adjusting the field rheostat.

4. Ammeter, Voltmeter readings, speed and spring balance readings are noted

under no loadcondition.

5. The load is then added to the motor gradually and for each load, voltmeter,

ammeter, springbalance readings and speed of the motor are noted(Take readings

up to rated current of the machine).

9

6. The motor is then brought to no load condition and field rheostat to minimum

position,then switch off the supply.

FORMULAE USED:

Torque (T) = 9.81 (S1-S2). r

Where, S1, S2→ Spring balance readings in kg

r→ Radius of brake drum in m

9.81→Constant to convert kg to Newton

2. Output Power (Pout)= 2πNT/60

Where N= Speed in rpm

3. In put power (Pin) = VL.IL Watts

VL=Input voltage in volts

IL=Input current in Amps

4. % Efficiency =( Pout/Pin)*100

CIRCUIT DIAGRAM:

10

TABULAR COLUMN:

S. No.

Input Voltage

VL (Volts)

If

Ia In put

current IL

(amps)

Speed N(rpm)

Spring balance Readings

Torque T(Nm)

Output Power

Pout (Watts)

Input Power

Pin (watts)

% η

S1

(Kg) S2

(Kg) S1-S2

(Kg)

SAMPLE GRAPH:

Performance curve

RESULT:

11

EX. NO. :3 DATE :

LOAD CHARACTERISTICS OF DC SHUNT GENERATOR

AIM:

To obtain internal and external characteristics of DC shunt generator.

APPARATUS REQUIRED:

S.No. Apparatus Specification Quantity

1 Ammeter

2 Voltmeter

3 Rheostats

4 Loading Rheostat

5 Tachometer

MACHINES DETAILS:

THEORY:

In a shunt generator, the field winding is connected in parallel with the armature

winding so that terminal voltage of the generator is applied across it.The shunt

field winding has many turns of fine wire having high resistance. Therefore, only a

part of armature current flows through shunt field winding and the rest flows

through the load.

For a DC Shunt GeneratorE0=V + IaRa

Ia= IL + Ish

Ish= V/Rsh

12

Where, E0 is the induced emf

V is the terminal voltage

Ia is the armature current

IL is the load current

Internal characteristics: it is the plot of the induced voltage E with armature

current. Drop in voltage is due to armature reaction which increases with increase

in load current.

External characteristics: It is the plot of terminal voltage with load current at

constant field resistance and speed. As the load on the machine is varied the

terminal voltage drops it is due to

1) The drop in voltage across the armature resistance, IaRa drop.

2) Armature reaction the air gap flux decreases which will reduced the

induced emf.

3) The drop in terminal voltage due to 1 & 2 results in decrease field

current which further reduces the induced emf.

PRECAUTIONS:

The field rheostat of motor should be at minimum position.

The field rheostat of generator should be at maximum position.

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

stopping.

PROCEDURE:

Connections are made as per the circuit diagram.

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.

13

Adjust the motor field rheostat to adjust the speed to rated speed of

generator.

Under no load condition, Ammeter and Voltmeter readings are noted, after

bringing the voltage to rated voltage by adjusting the field rheostat of

generator.

Load is varied gradually until rated current and for each load, voltmeter and

ammeter readings are noted.

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.

FORMULAE:

Eg = V + Ia Ra (Volts)

Ia = IL + If (Amps)

Eg : Generated emf in Volts

V : Terminal Voltage in Volts

Ia : Armature Current in Amps

IL : Line Current in Amps

If : Field Current in Amps

Ra : Armature Resistance in Ohms

14

CIRCUIT DIAGRAM:

PROCEDURE TO FIND Ra:

1. Connections are made as per the circuit diagram.

2. Supply is given by closing the DPST switch.

3. Readings of Ammeter and Voltmeter are noted.

4. Armature resistance in Ohms is calculated as Ra = (Vx1.5) /I

TABULAR COLUMN:

S.No. Voltage

V (Volts)

Current

I (Amps)

Armature

Resistance

Ra (Ohms)

15

S.No.

Field

Current

If (Amps)

Load

Current

IL

(Amps)

Terminal

Voltage

(V) Volts

Ia = IL + If

(Amps)

Eg=V + Ia

Ra (Volts)

SAMPLE GRAPH:

RESULT:

16

EX. NO. :4

DATE :

OPEN CIRCUIT CHARACTERISTICS OF DC SHUNT GENERATOR

AIM:

1) Toplot the OCC of a given DC Generator at rated speed & to find the critical

speed & critical resistance.

2) To plot the OCC at 3/4 th

rated speed.

3) To find the value of additional resistance to be added in field circuit in order to

make it as critical field resistance.

APPARATOUS REQUIRED:

S.No. Apparatus Range Type Quantity

1 Ammeter

2 Voltmeter

3 Rheostats

4 SPST Switch

5 Tachometer

MACHINES DETAILS:

THEORY :

The open circuit characteristic gives the variation of generated voltage(E) with

field current (If) at a constant speed.

17

E=ZɸN/60. P/A

ie, E α ɸN

When speed is constant, generated voltage is proportional to field flux. When the

field current is zero,a small residual flux is present in the machine and a small

voltage is generated even when field current is zero. As the field current is

increased the voltage increases linearly with field current. At some values of If,

magnetic circuit starts getting saturated. In saturation region, voltage increases

only slightly with increase in field current. So the generated voltage is directly

proportional to speed.

OCC of a generator shows the relation between the no load generated emfin the

generator & the field current in armature. If when the generator is driven at

constant rated speed, the field current is varied by adjusting the resistance

connected in the field. The field current is increased in suitable steps starting from

zero and the corresponding value of induced emf, E0is noted. The curve does not

start from the origin but from the point above the Y-axis, slightly above the origin.

This is due to the residual magnetism in the poles of generator at low value of field

current.

If we are given O.C.C. of a generator at a constant speed N1, then we can easily

draw the O.C.C. at any other constant speed N2.

18

Here we are given O.C.C. at a constant speed N1. It is desired to find

the O.C.C. at constant speed N2 (it is assumed that N1 < N2). For constant

excitation, E α N.

For If = OH, E1 = HC.

Therefore, the new value of e.m.f. (E2) for the same If but at N2

This locates the point D on the new O.C.C. at N2. Similarly, other points can be

located taking different values of If. The locus of these points will be the O.C.C.

at N2.

The maximum field circuit resistance (for a given speed) with which the shunt

generator would just excite is known as its critical field resistance.

The tangent of OCC at rated speed will represent the critical field resistance.

The critical speed of a shunt generator is the minimum speed below which it

fails to excite. Clearly, it is the speed for which the given shunt field resistance

represents the critical resistance.

In Fig curve 2 corresponds to critical speed because the shunt field resistance (Rsh)

line is tangential to it. If the generator runs at full speed N, the

19

new O.C.C. moves upward and the R'shline represents critical resistance

for this speed.

Speed Critical resistance

In order to find critical speed, take any convenient point C on excitation

axis and erect a perpendicular so as to cut RshandR'shlines at points B and

A respectively.

Then,

PRECAUTIONS:

The field rheostat of motor should be in minimum resistance position at the

time of starting and stopping the machine.

The field rheostat of generator should be in maximum resistance position at

the time of starting and stopping the machine.

SPST switch is kept open during starting and stopping.

PROCEDURE:

Connections are made as per the circuit diagram.

After checking minimum position of motor field rheostat, maximum position

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

gradually removed.

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

Voltmeter and ammeter readings are taken when the SPST switch is kept

open.

After closing the SPST switch, by varying the generator field rheostat,

voltmeter and ammeter readings are taken.

20

Field current is varied till generated voltage reaches 125% of the rated

voltage.

After bringing the generator rheostat to maximum position, field rheostat of

motor to minimum position, SPST switch is opened and DPST switch is

opened.

CIRCUIT DIAGRAM

CIRCUIT FOR MEASURING FIELD RESISTANCE

21

TABULAR COLUMN:

Sl

No.

Field

Current

If (Amps)

Armature

Voltage

Eo (Volts)

Measuring field Resistance

Sl

No.

Ammeter

reading(A)

Voltmeter

reading(v)

SAMPLE GRAPH :

RESULT:

22

EX. NO.:5

DATE :

HOPKINSON’S TEST

AIM:

To conduct Hopkinton’s test on a pair of identical DC machines to pre-determine

the efficiency of the machine as generator and as motor and to plot the efficiency

curve.

APPARATUS REQUIRED:

MACHINES DETAILS:

THEORY :

This test is called regenerative test or back to back test which can be carried out on

two identical d.c. machines mechanically coupled to each other. Thus the full load

test can be carried out on two identical shunt machines without wasting their

outputs. One of the machines is made to act as a motor while the other as a

generator. The mechanical output obtained from the motor drives the generator.

Sl

no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

23

Electrical output from the generator supplies part of input to the motor. The motor

is connected to the supply mains only to compensate for losses .In the absence of

losses, the motor-generator set would have run without any external power supply.

But due to losses, the generator output is not sufficient to drive the motor. Thus

motor takes current from the supply to account for losses.

The switch S is kept open.The first machine connected to supply will act as motor.

Thus second machinewhich is acting as generator will act as a load to the first

machine. The speed of motor is adjusted to normal value with the help of the field

rheostat. The voltmeter reading is observed. The voltage of the generator is

adjusted by its field rheostat so that voltmeter reading is zero. This will indicate

that the generator voltage is having same magnitude and polarity of that of supply

voltage. This will prevent heavy circulating current flowing in the local loop of

armatures on closing the switch. Now switch S is closed. The two machines can be

put into any load by adjusting their field rheostats. The generator current I2 can be

adjusted to any value by increasing the excitation of generator or by reducing the

excitation of motor. The various reading shown by different ammeters are noted

for further calculations.

The input to the motor is nothing but the output of the generator and small

power taken from supply. The mechanical output given by motor after supplying

losses will in turn drive the generator.

PRECAUTIONS:

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

starting and stopping the machine.

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

of starting and stopping the machine.

24

3. SPST switch should be kept open at the time of starting and stopping the

machine.

PROCEDURE:

1. Connections aremade as per the circuit diagram.

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

position of field rheostat of generator, opening of SPST switch, DPST switch is closed

and starting resistance is gradually removed.

3. The motor is brought to its rated speed by adjusting the field rheostat of the

motor.

4. The voltmeter V1 is made to read zero by adjusting field rheostat of generator

and SPST switch is closed.

5. By adjusting field rheostats of motor and generator, various Ammeter readings,

voltmeter readings are noted. (Care should be taken to avoid current exceeding

rated values of motor and generator)

6. The rheostats and SPST switch are brought to their original positions and DPST

switch is opened

25

CIRCUIT DIAGRAM:

DETERMINATION OF ARMATURE RESISTANCE

26

TABULAR COLUMN :

AS MOTOR

sl.no.

V

(volt)

I1

(A) I2(A) I3(A)

Motor

armature

copper

loss

Field

loss

stray

losses/2W)

Total

Losses

(W)

Out

put

power

Input

power efficiency

AS GENERATOR

sl.no.

V

(volt) I1(A) I2(A) I3(A)

Motor

armature

copper

loss

Field

loss

stray

losses/2W)

Total

Losses

(W)

Out

put

power

Input

power efficiency

27

FORMULAE USED

Input Power = VI1watts

Motor armature cu loss = (I1+ I2)2 Ra watts

Generator armature cu loss = I22 Ra watts

Total Stray losses W = V I1 - (I1+I2)2 Ra + I22 Ra watts.

Stray loss per machine = W/2 watts.

AS MOTOR:

Input Power = Armature input + Shunt field input

= (I1+ I2) V + I3V = (I1+I2+I3) V

Total Losses = Armature Cu loss + Field loss + stray loss

= (I1 + I2)2 Ra + VI3 + W/2 watts

Efficiency % = Input power – Total Losses / input power

AS GENERATOR:

Output Power = VI2 watts

Total Losses = Armature Cu loss+ Field Loss + Stray loss

= I22 Ra + VI4 + W/2 watts

Efficiency % = Output power / Output Power+ Total Losses

28

MODEL GRAPH :

RESULT:

29

EX. NO. :6 DATE :

SWINBURNE’STEST

AIM:

To conduct Swinburne’s test on DC machine to determine efficiency when

working as generator and motor without actually loading the machine. Also plot

efficiency curves.

APPARATU REQUIRED:

Sl.

No

Name of the

Apparatus Range Type Quantity

MACHINE DETAILS:

THEORY :

The power that a dc machine receives is called input power & power it gives out is

called output power.

ie ,The efficiency of a dc machine,

Efficiency= output/input

Output=Input-Losses & Input=Output+ Losses

Swinburne’s test is a no load test conducted on a DC Machine which run as motor

at noload& losses of the machine are determined. Once the losses of machine are

known,its efficiency at any desired load can be determined in advance. This test

only applicable in machines that flux is constant at all loads. In this test, the dc

shunt machine is run as a motor on no load with supply voltage adjusted to rated

voltage. The speed of the motor is adjusted to rated speed with the help of field

rheostat.

Let

30

IL → No load line current

Ia0 → No load armature current

V → input voltage

Ish → Shunt field current

I a02. Ra → No load armature copper loss

At no load condition,

No load input power (Pin) =V.IL

No load input current(Iao) =IL-Ish

No load power input to the armature =V.Iao =V.(IL-Ish)

Since the output of the motor is zero,the no load input power to the armature

supplies iron loss,frictionloss,windage loss & armature copper loss.

No load armature loss (WCU) = (Iao)2.Ra

Constant loss (Wc)= Input to motor-Armature copper loss

=Pin-Wcu

=V.IL-Iao2Ra

=V.IL-(IL-Ish)2.Ra

EFFICIENCY OF MOTOR

Input power (Pout) = VIL

Ia+Ish = IL

Armature copper loss (Wcu)

Constant loss(Wc )

Input=Output +Losses

V(Ia+Ish)=Output+Wc+Ia2Ra

Ia2Ra-VIa+Output+Wc-VIsh=0

Ia =V± V2− 4Ra(output +Wc− VIsh)/ 2Ra

Wcu = Ia2Ra=(IL-Ish)

2Ra

Wc =Pin-Wcu

=Pin -(IL-Ish)2Ra

31

Total losses = Wcu+Wc

%Efficiency = Pout/Pin * 100

= (Pin-total losses)/Pin* 100

EFFICIENCY OF GENERATOR

Output power (P out) = VIL

Armature current (Ia) = IL+Ish

Armature copper loss (Wc) = Ia2Ra=(IL+Ish)

2Ra

Total losses = Wcu+Wc=(IL+Ish)2Ra+Wc

%Efficiency = Pout/Pin * 100

=Pout/(Pout+total losses) * 100

PRECAUTIONS: The field rheostat should be in the minimum position at the time of starting and

stopping the motor

PROCEDURE:

1. Connections are made as per the circuit diagram.

2. After checking the minimum position of field rheostat, DPST switch is closed

and starting resistance is gradually removed.

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

4. The armature current, field current and voltage readings are noted.

5. The field rheostat is then brought to minimum position DPST switch is opened. PROCEDURE FOR DETERMINING THE RESISTANCE:

1. Connections are made as per the circuit diagram.

2. Supply is given by closing the DPST switch.

3. Readings of Ammeter and Voltmeter are noted.

4. Armature resistance in Ohms is calculated as Ra = (Vx1.5) /I

CIRCUIT DIAGRAM

32

TABULAR COLUM:

33

TABULAR COLUM

SAMPLE GRAPH

RESULT :

34

EX. NO. :7

DATE :

O.C AND S.C TESTS ON SINGLE PHASE TRANSFORMER

AIM:

To conduct open circuit and short circuit tests on the given 2.5KVA

transformerand predetermine the following:-

1. Equivalent circuit as referred to low voltage side

2. Equivalent circuit a referred to high voltage side

3. Efficiency curve at 0.8 pf lag and lead

4. Regulation at unity, lagging and leading power factors

APPARATUS REQUIRED:

Sl no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

MACHINES DETAILS

THEORY:

Open Circuit test

This test is usually conducted on the low voltage side side of the transformer. It is

conducted to determine the core loss(iron loss or no load loss). The low voltage

side of the transformer is supplied at rated voltage with the high voltage side left

open. The current, voltage and power on the input side is noted. Since the no-load

primary current is small(2-10% of the rated current) the copper losses in the

primary winding can be neglected and the power loss read by the wattmeter is the

core loss of the transformer. Since the flux linking with the core is constant at all

loads, the core loss remains same for all loads. The parameters R0 and X0 (the

shunt branch) are determined using this test.

35

Input power, W0 = V0I0cos0

cos0 = P0 / V0 I0

Iw = I0cos0, I= I0sin0

R0 = V0/ Iw , X0= V0 / I

V2/V1 (HV/LV)

R01=R0K

2 , X0

1=X0K

2

Iw1=Iw.K ,I

I

Short Circuit test

The short circuit test is conducted to determine the full load copper loss and the

equivalentresistance and leakage reactance referred to the winding in which the test

is conducted. The test is conducted on the high voltage side with the low voltage

side short circuited by a thick conductor. A low voltage just enough to circulate the

rated current of the transformer is supplied to the transformer. The voltage supplied

is usually only 5-10% of the normal supply voltage and so the flux linking with the

core is small. Thus core losses can be neglected and the wattmeter reading gives

the full load Cu loss of the transformer

WSC=Isc2R02,

R02=WSC/Isc2

Z02=Vsc/Isc

X02=

R01=R02/ K2

X01=X02/.K2

I2=KVA*100/V2

36

PROCEDURE

Open Circuit test

1. Connections are made as shown in the connection diagram 1.

2. The high voltage side is left open. The supply is switched on with the

autotransformer in theMinimum position.

3. The autotransformer is gradually varied till the voltmeter reads the rated voltage

of the Primary side of the transformer. The corresponding ammeter and wattmeter

readings are noted down.

Short Circuit test

1. Connections are made as shown in the diagram 2.

2. The low voltage side is short circuited. Supply is switched on with the

autotransformer in theminimum position.

3. The autotransformer is grad-dually varied till the ammeter reads the rated

current of the Transformer on the high voltage side.

Regulation of the transformer (which gives the variation of the secondary terminal

voltage from no load to full load expressed as a percentage of the secondary

terminal voltage with the primary voltage held constant) is then calculated using

the approximate formula at various power factors and half the full load, Regulation

curve is then plotted.

Rated current =Rated volt Amperes of transformer/ Rated voltage on high voltage

side

CIRCUIT DIAGRAM

OPEN CIRCUIT TEST

37

SHORT CIRCUIT TEST

TABULAR COLUMN

OC TEST

V0 (volt) I0 (Ampere) W0(watt)

SC TEST

Vsc (volt) Isc(Ampere) Wsc(watt)

38

REGULATION

Lagging power factor Leading power factor

sl.no. cosɸ Regulation sl.no. cosɸ Regulation

Efficiency from Transformer Tests F.L. Iron loss =W o from open-circuit test

F.L. Cu loss =Wc from short-circuit test

Total F.L. losses = Wo+WSC

We can now find the full-load efficiency of the transformer at any p.f. without

actually loading the transformer.

Also for any load equal to x x full-load,

Note that iron loss remains the same at all loads.

Voltage Regulation The voltage regulation of a transformer is the arithmetic difference (not phasor

difference) between the no-load secondary voltage (0V2) and the secondary

voltage V2 on load expressed as percentage of no-load voltage i.e.

39

The +ve sign is for lagging p.f. and ve sign for leading p.f.

EQUIVALENT CIRCUIT

RESULT:

40

EX. NO. :8

DATE :

POLARITY & TRANSFORMATION RATIO TEST ON SINGLE

PHASE TRANSFORMER

AIM:

To check the polarity of single phase transformer & find the transformation ratio.

APPARATUS REQUIRED:

Sl no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

MACHINES DETAILS:

THEORY:

Transformation Test

The transformation ratio ‘K’ is equal to the number of turns in the secondary

divided by the number of turns in the primary OR it is the ratio of induced emf in

the secondary to induced emf in the primary.

ie, K=N2/N1 =E2/E1

41

Polarity Test

In a transformer one terminal of primary side is positive with respect

toother at every instant. for the secondary side also one terminal of primary side is

positive with respect to other at every instant. If the polarity of terminals is not

marked, polarity test is performed to determine the terminals which have some

instant polarity. The polarity of the transformer winding is important for practical

operation.

The voltmeter reads V1+V2 , The transformer is said to posses additive polarity.

The voltmeter reads V1-V2 , The transformer is said to posses subtractive polarity.

PROCEDURE:

Connections are made as per the circuit diagram.

Autotransformer is kept in minimum position & supply is given.

Vary the autotransformer & take different readings till it exceeds the rated

value of voltage 110V.

Note the voltmeter reading V1,V2& V3.If V3=V1+V2, then it is additive

polarity otherwise it is subtractive polarity.

Bring the autotransformer to minimum position & switch off the supply.

CIRCUIT DIAGRAM

ADDITIVE POLARITY

42

SUBSTRACTIVE POLARITY

TRANSFORMATION RATIO TEST

TABULAR COLUMN

Additive polarity

Sl no. V1 V2 V3

Substractive polarity

Sl no. V1 V2 V3

Transformation ratio test

Sl no. V1 V2 K=V2/V1

RESULT:

43

EX. NO. :9

DATE :

OC & SC TEST ON THREE PHASE TRANSFORMER

AIM:

1)To determine the perphase equivalent circuit parameter referred to LV side &

HV side.

2) To predetermine the efficiency & regulation for different load condition and to

plot the curves.

APPARATOUS REQUIRED

THEORY:

A three-phase transformer can be constructed by having three primary and three

Secondary windings on a common magnetic circuit. The three single-phase core

type transformers, each with windings (primary and secondary) on only one leg

Have their unwound legs combined to provide a path for the returning flux. The

Primaries as well as secondaries may be connected in star or delta. If the primary is

energized from a 3-phase supply, the central limb (i.e., unwound limb) carries the

fluxes produced by the 3-phase primary windings. Since the Since the phasor sum

of three primary currents at any instant is zero, the sum of three fluxes passing

through the central limb must be zero. Hence no flux exists in the central limb and

it may, therefore, be eliminated. This modification gives a three leg core type 3-

phase transformer. In this case, any two legs will act as a return path for the flux in

the third leg. For example, if flux is in one leg at some instant, then flux is /2 in

the opposite direction through the other two legs at the same instant. All the

connections of a 3-phase transformer are made inside the case and for delta-

Sl. No.

Name of the Apparatus

Range Type Quantity

44

connected winding three leads are brought out while for star connected winding

four leads are brought out.

PROCEDURE:

OC TEST :

1. Connections are made as per the circuit diagram.

2. Three phase auto transformer is placed in minimum position & supply is

given by DPST switch is closed and starting resistance is gradually removed.

3. Adjust the autotransformer to get the rated voltage which is read by

voltmeter which is equal to 230V.

4. Note the readings, sum of of the readings of two wattmeter’s to give core

loss.

5. set the auto transformer to its initial position & switch off the supply.

SC TEST:

1. Connections are made as per the circuit diagram.

2. Three phase auto transformer is placed in minimum position & supply is

given by DPST switch is closed and starting resistance is gradually removed.

3. Adjust the autotransformer to get the rated current which is read by

Ammeter in high voltage side which is equal to rated current.

4. Note the readings, sum of of the readings of two wattmeter’s to give copper

loss.

5. set the auto transformer to its initial position & switch off the supply.

45

CIRCUIT DIAGRAM:

OC Test

SCTest

46

TABULAR COLUMN

OC TEST

V0 (volt) I0 (Ampere) W0(watt)

SC TEST

Vsc (volt) Isc(Ampere) Wsc(watt)

SAMPLE CALCULATION

OCTEST

NoLoad current/phase = I0

No Load Voltage / Phase =V0/√3

No Load out put /phase =W0/3

We have , V0I0cos ϕ0 =W0

No Load Powerfactor, cos ϕ0 = W0/V0I0

ϕ0=cos-1

(W0/V0I0)

Working component of current ,Iw =I0cos ϕ0

Magnetising component of current, Im =I0sin ϕ0

Shunt branch Resistance , R0 =V0/Iw

Shunt branch Reactance , X0 = V0/Im

Transformation Ratio, K =V2/V1

R0H =R0* K2 , X0H= X0* K

2

47

SC TEST

SC Load current/phase = Isc

SC Load Voltage / Phase =Vsc/√3

SC out put /phase =Wsc/3

We have ,VscIsccosϕsc =Wsc

No Load Powerfactor, cosϕsc = Wsc/VscIsc

Φsc=cos-1

(Wsc/Vsc

Z02 =VSC/Isc

R02 =Wsc /Isc2

X02=√(Z022-R02

2)

Transformation Ratio, K =V2/V1

R0L=R02/ K2

X0L =X02/ K2

Result

48

EX. NO. :10

DATE :

LOAD CHARACTERISTICS OF DC COMPOUND GENERATOR

AIM:

To conduct a load test on the given dc compound generator & plot its internal &

external characteristics when working in cumulative & differential mode.

APPARATOUS REQUIRED:

MACHINE DETAILS:

THEORY:

The compounding of a machine is done by adding a few coils in series with the

armature coil circuit of a shunt machine. Compound generator, both series and

shunt excitation are combined. The shunt winding can be connected either across

the armature only (short-shunt connection S) or across armature plus series field

(long-shunt connection G). The compound generator can be cumulatively

compounded or differentially compounded generator.

When a compound generator has its series field flux aiding its shunt

field flux, the machine is said to be cumulative compound. When the series field is

connected in reverse so that its field flux opposes the shunt field flux, the

generator is then differential compound. The easiest way to build up voltage in a

compound generator is to start under no load conditions. At no load, only the shunt

field is effective. When no-load voltage build up is achieved, the generator is

loaded. If under load, the voltage rises, the series field connection is cumulative. If

the voltage drops significantly, the connection is differential compound.

Sl. No.

Name of the Apparatus

Range Type Quantity

49

External characteristic:

The series excitation aids the shunt excitation. The degree of compounding

depends upon the increase in series excitation with the increase in load current.

(i) If series winding turns are so adjusted that with the increase in load

current the terminal voltage increases, it is called over-compounded

generator. In such a case, as the load current increases, the series field

m.m.f. increases and tends to increase the flux and hence the generated

voltage. The increase in generated voltage is greater than the IaRa drop so

that instead of decreasing, the terminal voltage increases .

(ii) If series winding turns are so adjusted that with the increase in load

current, the terminal voltage substantially remains constant, it is called

flat-compounded generator. The series winding of such a machine has

lesser number of turns than the one in over-compounded machine and,

therefore, does not increase the flux as much for a given load current.

Consequently, the full-load voltage is nearly equal to the no-load voltage.

(iii) If series field winding has lesser number of turns than for a flatcompounded

machine, the terminal voltage falls with increase in load

current Such a machine is called under-compounded generator.

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.

50

PROCEDURE:

1. Connections are made as per the circuit diagram.

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.

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

after bringing the voltage to rated voltage by adjusting the field rheostat

of generator.

4. Load is varied gradually and for each load, voltmeter and ammeter

readings are noted.

5. 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.

6. The connections of series field windings are reversed the above steps are

repeated.

7. The values of voltage for the particular currents are compared and then

the differential and cumulative compounded DC generator is concluded

accordingly.

PROCEDURE FOR MEASURING Ra:

1. Connections are made as per the circuit diagram.

2. Supply is given by closing the DPST switch.

3. Readings of Ammeter and Voltmeter are noted.

4. Armature resistance in Ohms is calculated as Ra = (Vx1.5) /I

51

CIRCUIT DIAGRAM

TABULAR COLUMN

FORMULAE:

Eg = V + Ia Ra (Volts)

Ia = IL + If (Amps)

Eg : Generated emf in Volts

V : Terminal Voltage in Volts

Ia : Armature Current in Amps

IL : Line Current in Amps

If : Field Current in Amps

Ra : Armature Resistance in Ohms

RESULT:

S.No. Cumulatively Compounded Differentially Compounded

V (Volts) IL (Amps) V (Volts) IL (Amps)

52

EX. NO. :11

DATE :

SUMPNER’S TEST ON A PAIR OF 1-Ф TRANSFORMER

AIM:

To conduct OC & SC tests on the given 1- Transformer and to calculate its

1) Equivalent circuit parameters

a). Referred to H.V side

b). Referred to L.V side

2) Efficiency at various loads.

3) Regulation at various power factors

4) Maximum Efficiency.

APPARATUS REQUIRED:

MACHINES DETAILS:

THEORY :

In this test load test is conducted without actual loading.Sumpner’s test which can

onlybe conducted simultaneously on two identical transformers. In conducting the

Sumner’stestthe primaries of the two transformers are connected in parallel across

the rated voltagesupply (V1), while the two secondary’s are connected in phase

opposition. As per thesuperposition theorem, if V2 source is assumed shorted, the

Sl

no Name of Apparatus Type Specification Qty

1 Ammeter

2 Voltmeter

3 Rheostat

4 Tachometer

53

two transformers appear in open circuit to source V1 as their secondary are in

phase opposition and therefore no currentcan flow in them. The current drawn

from source V1 is thus 2I0 (twice the no-load current of each transformer) and

power is 2P0 (= 2P0, twice the core loss of each transformer). When V1 is regarded

as shorted, the transformers are series-connected across V2 and are short-

circuitedon the side of primaries. Therefore, the impedance seen at V2 is 2Z and

whenV2 is adjusted tocirculate full-load current (Ifl), the power fed in is 2Pc (twice

the full-load copper-loss of eachtransformer). Thus in the Sumpner’s test while the

transformers are not supplying any load,full iron-loss occurs in their core and full

copper-loss occurs in their windings; net powerinput to the transformers

being(2Po+2Pc).The heat run test could , therefore, be conducted onthe two

transformers, while only losses are supplied.

For each trans former the results are Voltage =V1 , Current = I0 /2 , Core losses = P0 /2

Voltage =Vsc/2 , Current = Isc, Copper losses = Psc/2

P0 = Pi (iron-loss)

P0 = V1 I0cos0

cos0 = P0 / V1 I0

Iw= I0cos0, I= I0sin0

R0 = V1/ Iw, X0 = V1 / I.

Vsc=Voltage, Isc= Current ,Psc= Power (Copper loss)

54

CIRCUIT DIAGRAM

PROCEDURE

(1) Connections are done as per the circuit diagram.

(2) By using the variac rated voltage is made to apply across the low voltage sideof

the transformer.

(3) Before closing the DPST switch the reading of the voltmeter connected across

DPSTswitch must be zero.

(4) By using the variac in H.V side rated current is made to flow in the circuit.

(5) At this instant note down all the meter readings.

(6) By using above tabulated readings the efficiency and regulation of the

transformersare calculated.

55

TABULAR COLUMNS

56

SAMPLE GRAPHS:

RESULT:

57

EX. NO. :12

DATE :

SEPARATION OF NO LOAD LOSSES IN A SINGLE PHASE

TRANSFORMER

AIM:

To separate no load losses of a transformer in to eddy current loss and hysteresis

loss.

APPARATUS REQUIRED:

THEORY:

It is seen that the core losses of transformer includes,

1. Hysteresis loss

2. Eddy current loss

For a given volume and thickness of laminations, these losses depend on the

operating frequency, maximum flux density in the core and the voltage.

The hysteresis loss is given by Steinmet'z relation,

Ph = Kh Bm1.67

f v watts

i.e. Ph = A Bm1.67

f watts ..............(1)

where A = constant assuming constant voltage

The eddy current loss is given by,

Pe = Ke Bm2 f

2 t

2 watts

i.e. Pe = B Bm2 f

2 watts .........(2)

where B = constant for given thickness t of core

Thus the total core loss becomes,

Pi = Ph + Pe = A Bm1.67

f +B Bm2 f

2 .........(3)

Sl. No.

Name of the Apparatus

Range Type Quantity

58

Practically conduct two tests on transformers at two different frequencies

f1 and f2, keeping maximum flux density in the core same. The results are to be

used in the equations (1), (2) and (3) to obtain the constants A and B. Thus the core

losses i.e. iron losses can be separated into hysteresis and eddy current losses.

PRECAUTIONS:

1. The motor field rheostat should be kept at minimum resistance position.

2. The alternator field rheostat should be kept at maximum resistance position.

PROCEDURE:

1. Connections are given as per the circuit diagram.

2. Supply is given by closing the DPST switch.

3. The DC motor is started by using the 3 point starter and brought to rated speed

by adjusting itsfield rheostat.

4. By varying the alternator filed rheostat gradually the rated primary voltage is

applied to thetransformer.

5. The frequency is varied by varying the motor field rheostat and the readings of

frequency arenoted and the speed is also measured by using the tachometer.

6. The above procedure is repeated for different frequencies and the readings are

tabulated.

7. The motor is switched off by opening the DPST switch after bringing all the

rheostats to theinitial position.

59

CIRCUIT DIAGRAM:

TABULAR COLUMN:

S.No. Speed

N (rpm)

Frequency

f (Hz)

Voltage

V (Volts)

Wattmeter

reading

Watts

Iron loss

Wi (Watts)

Wi / f

Joules

60

FORMULAE USED:

1. Frequency, f =(P*NS) / 120 in Hz P = No.of Poles & Ns = Synchronous speed

in rpm.

2. Hysteresis Loss Wh = A * f in Watts A = Constant (obtained from graph)

3. Eddy Current Loss We = B * f2

in Watts B = Constant (slope of the tangent

drawn to the curve)

4. Iron Loss Wi = Wh + We in Watts Wi / f = A + (B * f)

Here the Constant A is distance from the origin to the point where the line cuts

the Y- axis in the graph between Wi / f and frequency f. The Constant B is

Δ(Wi / f ) / Δf

MODEL GRAPH:

RESULT:

61