A- Open circuit and load characteristic of self and separately excited D.C. shunt generatorB- Speed control of D.C shunt motorC- Load Characteristics of D.C. shunt motorD- Load Characteristics of D.C. series motorE- Load Characteristics of D.C. compound motorF- Load test on single phase transformer. G- Open circuit short circuit test on single phase transformer and separation of no load losses in single phase transformerH- Three phase transformer connectionI - Hopkinsons test on D.C. motoer -generator setJ- Sumpners test on transformersK- Swinburne's test L- Load characteristics of D.C. compound generator.

Ex.No: 1

OCC AND LOAD CHARACTERISTICS OF

DC SHUNT GENERATOR

AIM:

To conduct open circuit and load test on a Self-excited and Separately-excited (shunt) DC

generator and to determine the external and internal characteristics of it.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY1.

2.

3.

4.

5.

6.

Ammeter

Ammeter

Voltmeter

Rheostat

SPST switch

Tachometer

MC

MC

MC

(0 – 20A)

(0 – 2A)

(0 – 300V)

350, 1.4A

(0 –10000 RPM)

1

1

1

2

1

1

THEORY:

Open Circuit Test:

Open circuit characteristics may also called as No Load or Magnetic Characteristic. It show the

relation between the no load generated emf in the armature, Eo and the field or exciting current If at a

given fixed speed. It is just the magnetization curve for the material of the electromagnet. It’s shape is

practically same for all generators whether separately or self-excited.

Load Test:

Let us consider a self or separately excited dc generator giving its rated no load voltage ‘E o’ for

a certain constant field current. If there were no armature reaction and armature voltage drop, then this

voltage would have remained constant as shown in Fig by the dotted horizontal line I. But when the

generator is loaded, the voltage falls due to these two causes, thereby giving slightly drooping

characteristics. If we subtract from Eo the value of voltage drops due to armature reaction for different

loads, then we get the value of E, the emf actually induced in the armature under load conditions.

Curve II is plotted in this way and is known as the internal characteristic. The straight-line represents

the IaRa drop corresponding to different armature currents. If we subtract from E the armature drop

IaRa, we get terminal voltage V. Curve III represents the external characteristic.

PROCEDURE:

Open Circuit Test:

Connections are given as per the circuit diagram. Initially the SPST switch is kept open.

Supply is given and the motor is started with the help of a suitable starter. Emf generated due to

residual magnetism is noted from voltmeter when field current is zero. Then SPST switch is closed.

By varying the field rheostat in the generator side, different values of generated emf E o and field

current If are noted from voltmeter and ammeter. The same procedure is repeated for Separately-

excited DC shunt Generator.

Load Test:

Connections are given as shown in the circuit diagram. The motor is started with the help of dc

suitable starter and brought to its rated speed. Keeping the generator side DPST open, the field

rheostat in the generator side is adjusted for the rated voltage of the generator which is seen in the

voltmeter. Now the DPST switch is closed and the resistive load is put up on the generator step by

step. The terminal voltage, armature and load current values are noted down for each step from the

respective meters. Generator may be loaded upto 125%. Note that while taking each set of readings,

the field current is maintained constant as that for rated voltage (because due to heating, shunt field

resistance is increased). The same procedure is repeated for Separately-excited DC shunt Generator.

To Find Armature Resistance:

Connections are given are in the circuit diagram. Supply is given and the rheostat is varied and

the corresponding values of voltage and current are noted. From these values, dc armature resistance is

calculated. Then it may be multiplied by 1.3 to get hot armature resistance.

Open Circuit Test:

Open circuit characteristic is drawn by taking field current If along X-axis and generated

voltage Eo along Y-axis.

Load Test:

External characteristic is drawn taking load current IL along X-axis and terminal voltage V

along Y-axis. Internal characteristic is drawn by adding IaRa drop to external characteristic curve, Ia

along X-axis and E along Y-axis.

RESULT:

Thus the no-load test and load test are conducted on a self-excited and separately-excited DC

Shunt generator and the corresponding characteristics curves are drawn.

VIVA QUESTIONS:

1.What are self and separately excited generators?

2.Give the emf equation of a Dc generator.

3.What are the factors that may affect the voltage build up of a dc shunt

generator?

4.What is the procedure to shut down a generator.

5.Can we use the same machine as a generator and as a motor?

Ex.No: 2

SPEED CONTROL OF DC SHUNT MOTOR

AIM:

To draw the speed control characteristics of a dc shunt motor by

(i) Armature control method and

(ii) Field control method.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

5.

Ammeter

Ammeter

Voltmeter

Rheostat

Tachometer

MC

MC

MC

(0 – 2A)

(0 – 20A)

(0 – 300V)

350,1.4A

(0 – 10000 RPM)

2

1

1

2

1

THEORY:

We know back emf,

Eb= V- IaRa =ZNP / (A*60)

N = (V- IaRa) (A*60) / (ZP) rpm

N= K (V- IaRa) / rpm

N Eb /

From the above equation, it is clear that the variation of speed is possible in three different ways:

(i) Variation Flux (Or) Field (Flux) Control Method

In this method, speed can be increased above rated speed. In non-interpoles, it is in the ratio 2 :

1 and in interpoles maximum to minimum speed of 6 : 1 is fairly common.

(ii) Armature Control Method

In this method, armature current is controlled by means of a rheostat in the armature circuit.

This method is used when speed below the no-load speed is required.

(iii) Voltage Control Method

It is obvious that changing the voltage applied to the motor can vary speed. The two commonly

employed types are (1) Multiple voltage control (2) Ward-Leonard System.

PROCEDURE:

Connections are given as shown in the circuit diagram and initially the field rheostat is kept in

minimum position and armature rheostat is kept in maximum position and armature rheostat is kept in

maximum position.

Armature Control Method:

Supply is given and the motor is started. The armature resistance is cut down and speed is

measured. If the motor runs below the rated speed, adjust the field rheostat and the motor is brought to

its rated speed. The field current, If is kept constant. Now the armature voltage (i.e. potential drop

across the armature) is varied by varying the armature rheostat and corresponding voltmeter, ammeter

(Ia) readings and speed are noted. The same procedure is repeated for another value of constant field

current If.

Field Control Method:

The motor is run at the rated speed as usual. Armature voltage and armature current is kept

constant. The field current is varied and the value of field current I f and speed are noted in the tabular

column. The same procedure is repeated for another constant value of armature current.

TABULATION:

GRAPH:

Armature Control Method:

Graph is drawn between speed and armature current taking armature current along X-axis and

speed along Y-axis.

Field Control Method:

Graph is drawn between speed and field current, taking speed along Y-axis and field current

along X-axis.

RESULT:

Thus the speed control of DC shunt motor is conducted and the speed control characteristics

curves are drawn.

VIVA QUESTIONS:

1.What is the equation governing the speed of a DC Motor?

2.What are all the factors affecting the speed of a DC motor?

3.Mention the methods of speed control of a Dc Motor.

4.Give the applications of DC Shunt Motor.

Ex.No: 3

LOAD TEST ON DC SHUNT MOTOR

AIM:

To conduct the load test on a given DC shunt motor and to draw the performance curves.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

5.

Ammeter

Voltmeter

Rheostat

Tachometer

Thread

MC

MC

0-20 Amps

0-300 Volts

350 Ohms, 1.4 Amps

(0 – 10000 RPM)

1

1

1

1

THEORY:

D.C shunt motor is one the field winding is connected in parallel with the armature winding.

The flux () is constant. So the torque (T) is proportional to armature current Ia.

The speed of DC shunt motor is constant. The torque, input power, output power and

percentage efficiency are calculated using the below formulae.

Torque T= (S1 ~ S2) R * 9.81 N-m

S1, S2 ---Spring Balance Readings (Kg)

R ---Radius of the brake drum (m)

Input Power Pi = VI watts

Output Power Po =2NT / 60 watts

Percentage Efficiency = (Output Power / Input Power)* 100

PROCEDURE:

The circuit connections are given as shown in the circuit diagram. Ensure that there is no load

on the brake drum. The supply is given by closing the DPST switch. The motor is started using the dc

3 point starter, and it is allowed to run at rated speed by adjusting the rheostat, which is connected to

the field circuit. After setting the rheostat it should not be altered. At no load condition, the input

voltage, current and speed are noted using voltmeter, ammeter and tachometer. Now the load on the

brake drum is increased upto full load gradually and the corresponding voltmeter ammeter readings

and speed are noted down. Then the load is gradually decreased and field rheostat is brought to the

minimum position and the supply is switched off. The torque, input power, output power and

percentage efficiency are calculated by using the formulae and tabulated.

RESULT:

Thus the load test on dc shunt motor is conducted and the performance characteristic curves are

drawn.

VIVA QUESTIONS:

1.What is a DC Shunt Motor?

2. Give the relationship between torque, flux and armature current.

3.Why do we require a DC starter?

4.How torque is developed in a Dc Motor?

5.What is meant by back emf?

Ex.No: 4

LOAD TEST ON DC SERIES MOTOR

AIM:

To conduct the load test on a given DC series motor and to draw the performance curves.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

Ammeter

Voltmeter

Tachometer

Thread

MC

MC

0-20 Amps

0-300 Volts

0 – 10000 rpm

1

1

1

1

THEORY:

The field winding is connected in series with the armature terminals. For a given current input

the starting torque developed by a series motor is greater than that developed by a shunt motor. Hence,

series motors are used where huge starting torques are necessary that is for street cars, cranes, hoists

and for electric railway operation. In addition to huge starting torque, there is another unique

characteristics of series motors which makes them especially desirable for traction work, that is when

a load comes on a series motor, it responds by increasing its speed and supplies the increased torque

with a small increase in current.

The torque, input power , output power , and % efficiency are calculated by using the

following formulae:

Torque T= (S1 ~ S2) R * 9.81 N-m

S1, S2 ---Spring Balance Readings (Kg)

R ---Radius of the brake drum (m)

Input Power Pi = VI watts

Output Power Po =2NT / 60 watts

Percentage Efficiency = (Output Power / Input Power)* 100

PROCEDURE:

The circuit connections are given as shown in the circuit diagram. Ensure that there is no load

on the brake drum. The DC series motor should always be started with some load. Otherwise the speed

will be enormous and fuse will blow out. Supply is given and the motor is started by using two-point

starter. At the initial load condition, voltmeter, ammeter and spring balance readings are noted. The

speed is also measured using tachometer. Now the load is increased gradually up to 125%. For various

load conditions. The voltmeter, ammeter and spring balance readings and speed are noted. The torque

input power , output power and percentage efficiency are calculated using the formulae and entered in

the tabulation.

GRAPHS:

The performance characteristics curves are drawn as

1. Output power VS Speed

2. Output power VS Current

3. Output power VS Torque

4. Output power VS Efficiency

RESULT:

Thus the load test on a DC Series motor is conducted and the performance characteristic curves are drawn.

VIVA QUESTIONS:

1. What is the special character of series motor?

2. Why the series motor is started with some load?

3. How can you give the load for the motor?

4. Give some applications of series motor?

5. Why the traction motors are series motors?

EX.NO: 5

LOAD TEST ON DC COMPOUND MOTOR

AIM

To conduct the load test on given DC compound motor and draw the performance curves.

APPARATUS REQUIRED

S.NO ITEM TYPE RANGE QUANTITY

1.2.3.

AmmeterVoltmeterTachometer

MCMC

(0 – 20A)(0 – 300V)

(0 – 10000 RPM)

111

CIRCUIT DIAGRAM

THEORY

Compound motor has both shunt and series field windings. If series field opposes shunt filed

then the motor is said to be differentially compound. If the series flux is in the same direction of shunt

flux then the motor is said to be cumulative compound. The characteristic of such motors lies in

between those of shunt and series field windings. The torque, input power, output power and

efficiency are calculated by using the following formulae:

Torque T= (S1 ~ S2) R X 9.81 N-m

S1, S2 ---Spring Balance Readings (Kg)

R ---Radius of the brake drum (m)

Input Power Pi = VI watts

Output Power Po =2NT / 60 watts

Percentage Efficiency = (Output Power / Input Power) X 100

Differential Compound Motors

Series field opposes the shunt field, flux is decreased as load is applied, this result in constant

speed. Due to this there is a decrease in rate at which the motor torque increases with load. It is not in

common use.

Cumulative Compound Motors

Such machines are used in drawing heavy machine tools. Due to shunt winding speed will not

become excessively high, but due to the series field it will be able to take heavy load. In conjunction it

is used where sudden temporary loads as in rolling mills.

PROCEDURE

The circuit connections are given as per the circuit diagram. Ensure that there is no load on the

brake drum. The supply is given by closing the DPST switch. The motor is started using four point

starter and is allowed to run at rated speed by adjusting the rheostat, which is connected to the field

circuit. After setting the rheostat it should not be altered. At no load condition input voltage current

and speed are noted using voltmeter, ammeter and tachometer. Now the load on the brake drum is

increased gradually up to 125% of full load and corresponding voltmeter, ammeter readings and speed

are noted. Then load is decreased and field rheostat is brought to minimum position and supply is

switched off. The torque input power and percentage efficiency are calculated using the formulae and

tabulated.

GRAPH

The performance characteristics curve are drawn as

1. Output power Vs Speed

2. Output power Vs Current

3. Output power Vs Torque

4. Output power Vs Efficiency

Taking Output Power along X-axis and current, torque, speed and efficiency along Y-axis.

RESULT

Thus the load test on DC compound motor is conducted and the performance curves are drawn.

VIVA QUESTIONS

1. What are the two types of compound motor?

2. What is the difference between the compound motor and shunt motor?

3. What is the difference between cumulative compound and differential compound?

4. Which compound motor can be applied for sudden temporary loads?

5. Give some applications of compound motor?

6. Which starter is used to start the compound motor?

Ex No: 6

LOAD TEST ON A SINGLE PHASE TRANSFORMER

AIM:

To conduct load test on a given single-phase transformer and to draw the performance

characteristic curves.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1. Voltmeter MI (0 – 300V) 1

2. Voltmeter MI (0 – 150V) 1

3. Ammeter MI (0 – 5A) 1

4. Ammeter MI (0 – 10A) 1

5. Wattmeter UPF (10A,150V) 1

6. Wattmeter UPF (5A,300V) 1

7. 1 phase transformer 1 KVA 1

8. Lamp Load - - 1

THEORY:

Efficiency:

Transformer is a static device by which electric power is transformed from one circuit to

another circuit at the same frequency. Hence there are no friction or windage losses. So the losses are

1. Core Or Iron Loss: It includes both hysteresis and eddy current losses. Because the core flux

remains constant for all loads, the core loss is same for all loads.

2. Copper Loss: This loss is due to Ohmic resistance of the transformer winding. These losses may

be found out by conducting open circuit test and short circuit test. Further the test result will be

useful in predetermining the efficiency of higher rating transformers. But for small transformers a

direct loading method can be employed where input and output powers can be measured. The

efficiency of a good transformer will be in the range of 95 to 98%.

Regulation:

Regulation is an important parameter by which the performance of a transformer can be

assessed. The lesser the value, the better the transformer because every transformer should keep its

secondary terminal voltage as constant as possible under all conditions of load.

When a transformer is loaded with a constant primary voltage, the secondary voltage decreases

(assuming lagging power factor loads) because of its internal resistance and leakage reactance.

Let the change in secondary terminal voltage from no load to full load is (VNL- VFL). If this change is divided by VNL , then the regulation is known as ‘DOWN’. If this change

is divided by VFL, then it is known as ‘UP’.

% Regulation down = (VNL- VFL)*100 / VNL

% Regulation down = (VNL- VFL)*100 / VFL

Normally, we are interested in finding regulation ‘down’.

As we know that there are

1. Resistive load

2. Inductive load

3. Capacitive load

4. Combination of 1,2 and 3 loads are possible.

The regulation characteristic curve will be as shown in the model graph, where it is drawn between

power factor and % regulation. But for resistive loads, it is enough to draw the same between output

power and % regulation.

i) Performance efficiency = (Output power / Input power)* 100

ii) Percentage regulation ‘down’ = (VNL- VFL)*100 / VNL

PROCEDURE:

Primary and secondary circuit connections are made as shown in the circuit diagram. AC supply

is given to the primary winding of the transformer by closing the DPST switch. At no load condition,

the voltmeter, ammeter and wattmeter readings corresponding to the primary and terminal voltage of

the secondary of the transformer are noted. Now the SPST switch is closed and load is applied to the

transformer in steps. For every step of increase in load, the primary current, voltage, power and the

secondary current, terminal voltage and power are noted. The transformer is loaded up to 125% and

then the load is decreased. The supply is switched off. From these values, the percentage efficiency

and percentage regulation are calculated and the corresponding characteristic curves are drawn.

GRAPH:i) Percent load KVA Vs Percentage efficiencyii) Output power Vs Percentage regulation

RESULT:

Thus the load test on a single-phase transformer is conducted and the characteristic curves aredrawn.

VIVA QUESTIONS:

1. On what principle the transformer operates?

2. What is the efficiency of the transformer?

3. What is the role of core in a transformer?

4. What are the losses in the transformer?

Ex.No: 7(A)

OPEN CIRCUIT AND SHORT CIRCUIT TESTS

ON A SINGLE PHASE TRANSFORMER

AIM:

To conduct open circuit and short circuit tests on a single phase transformer and hence

determine the equivalent circuit parameters. Also to determine the percentage efficiency and

regulation at any desired load.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

5.

6.

7.

Ammeter

Ammeter

Voltmeter

Voltmeter

Wattmeter

Wattmeter

Auto transformer

MI

MI

MI

MI

UPF

LPF

0 – 2A

0 – 10A

0 – 300V

0 – 60V

10A, 150V

5A, 300V

1

1

1

1

1

1

1

THEORY:

Equivalent Circuit:

The equivalent circuit of a transformer is as shown in the figure. It consists of two circuits,

(1) magnetizing circuit consisting of Ro and Xo

(2) Working circuit consisting of R01, X01 and load ZL. The above four quantities namely Ro , Xo,

R01and X01 are known as parameters or constant of transformer. Ro and Xo can be found out

with the help of open circuit (or) no-load test and R01and X01 by short circuit test. Once we

know these quantities, the performance of the transformer such as efficiency, regulation etc can

be found out at any desired load and power factor.

Open Circuit Test:

As the name implies, the secondary (High Voltage side) is left open. Primary is given with

rated normal voltage. With normal voltage, normal flux will be set up in the core. Hence normal iron

losses will occur. As the primary current is small, (Usually 2 to 10% of rated load current) Copper loss

will be negligibly small in primary and nil in secondary (it being open).

Hence power consumed under open circuit is practically core loss and which is same for all loads. The

voltmeter, ammeter and wattmeter readings are noted down. Then

W = V1 Io coso ; coso = W/(V1 Io)

I = I osino ; I= I coso

Xo = V1/ I ; Ro = V1/ I

Short Circuit Test:

This is an economical method for determining the following:

1) Equivalent impedance (Z01 or Z02) , Leakage reactance ( X01 or X02 ) and total resistance(R01 or

R02) of the transformer as referred to the winding in which the measuring instruments are

placed.

2) Copper loss at full load (at and any desired load). This loss is used in calculating the efficiency

of the transformer.

3) Knowing Z01 or Z02 the total voltage drop in the transformer as referred to primary or secondary

can be calculated and hence regulation of the transformer is determined.

Now rated full load current is made to flow in the circuit by applying a small percentage of normal

voltage by means of an auto-transformer. Because of this, core losses are very small with the result

that wattmeter represents full load copper loss for the whole transformer.

If Vsc is the voltage required to circulate rated load currents, then

Z01 = Vsc / I1 ; Also Wsc = I12R01 R01 = Wsc / I1

2

X01 = (Z012 - R01

2)

Subsequently, it is evident that copper loss depends on current and iron loss on voltage. Hence

total transformer loss depends on volt-ampere (VA) and not on phase angle between voltage and

current. (I.e.) it is independent of load power factor. That is why rating of transformer is in KVA and

not in KW. The

Percentage regulation = ((xI1 / V1) (R01 cos0 X01 sin0)) x 100

Percentage Efficiency = output power/input power

Output power = ((x) * full load KVA * 1000 * cos)

Input power = output power – losses

Losses = Iron loss + Copper loss

Copper loss = Variable losses Wcu = x2 * Wsc

Where x is the fractional load.

PROCEDURE:

Open Circuit Test

The connections are given as shown in the circuit diagram. Keeping the secondary at no load

(i.e. left open), apply the rated primary voltage by closing the DPST switch. Note down the readings of

ammeter, voltmeter and wattmeter. From these values, the exciting branch parameters (R0, X0) of the

equivalent circuit are calculated using the formulae appearing in the theory.

Short Circuit Test

The connections are made as shown in the circuit diagram. Keeping the secondary winding

shorted, the voltage applied to the primary by autotransformer is adjusted for rated primary current.

The corresponding voltmeter and wattmeter readings are taken. Take care to see that current does not

exceed rated value, I1. From these values R01,X01 of the transformer are determined using the formulae

appearing in the theory and marked all parameters on the equivalent circuit. Then efficiency and

regulation are found out using the formulae and entered in the tabulation.

GRAPH:

Graph is drawn between % load KVA and % efficiency taking along X-axis and Y-axis

respectively. Also a graph is drawn between power factor and % regulation taking % regulation along

Y axis and power factor along X-axis.

RESULT:

Thus the open circuit and short circuit test is conducted on a single-phase transformer and the

equivalent circuit parameters are found out. Hence percentage of regulation and efficiency are

determined at various loads and power factors.

VIVA QUESTIONS:

1. Does the frequency of transformer change under any circumstances?

2. Why transformer rating is in KVA?

3. What will happen when the primary of a transformer is connected to dc supply?

4. Name the losses in a transformer.

5. Does the transformer draw any current when its secondary is open?

6. What do you understand by predetermination?

Ex.No: 7(B)

SEPARTION OF NO LOAD LOSSES IN A SINGLE PHASE TRANSFORMER

AIM:

To determine the no load losses i.e. hystrersis and eddy current loss of a single-phase transformer.

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QTY1 Transformer I Phase 1 KVA 12 Ammeter MI 0 – 2A 13 Voltmeter MI 0 – 150V 14 Wattmeter LPF 150V,5A 15 Tachometer ` 0 – 10000 Rpm 16 Rheostat 350, 1.4A

600, 1.2A11

CIRCUIT DIGRAM:

THEORY: Core losses are of two types. First one is hysteresis loss another one is eddy current loss. Hysteresis loss is due to magnetic property of the core. Hystersis loss is reduced by using high silicon content core materiel. Eddy current loss is when ever flux linking the core change an emf is induced in the core. As the core forms a closed circuit current flows through the core. That is treated as loss. Laminating the core reduces eddy current loss.

PROCEDURE:

Connections are given as per the circuit diagram. Field rheostat motor is kept at minimum position and that of generator at maximum position. The motor is brought to its rated speed using the field rheostat. At rated speed corresponding reading of voltmeter, ammeter and wattmeter are noted. For different speeds the readings are tabulated.

RESULT:

Thus the no load losses of a single-phase transformer are separated in to eddy current and hysteresis losses.

VIVA QUESTIONS:

1. What are types of core losses?2. How will you reduce eddy current losses?3. How will you reduce hysteresis losses?

Ex.No: 8

3 PHASE CONNECTIONS IN TRANSFORMER

AIM:

To connect the three given single phase transformers for 3 phase operation, with primary side

in delta and secondary side in star and also.

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QTY1 Testing Transformer 12 Voltmeter MI 0 – 600V 2

THEORY:

Instead of using a 1 phase transformers for 3 phase operation. This facilities flexibility in

connection. Depending on the requirements, the three units can be connected for star/star, delta/ delta,

delta/star, star/delta operation. With three separate units reliability of service also increases as if one of

the transformers gets damaged the remaining two can be operate in open delta, even though with

reduced capacity. Starting and finishing ends of primary ends of primary and secondary winding of all

of the 1 phase transformers are usually marked, so as to ensure proper connections.

The load on the secondary side may be balanced side may be balanced or unbalanced. If load is

balanced, current flowing through the neutral will be zero, when secondary side is in star with neutral

connection. However if the load is unbalanced then out of balance current flows through the neutral.

PROCEDURE:

1. POLARITY TEST:

This is done on a 1 phase transformer to test the polarity of the terminals. For this, the connections

are given as shown in the figure. When the power is supplied on the voltmeter shows the

difference in voltage between the primary and secondary windings of the transformer.

2. ADDITIVE TEST:

This test is done on a 1 phase transformer to test polarity of the terminals. For this, the connections

are given as shown in the figure. When the power is supplied on the voltmeter shows the addition

of voltage of primary and secondary winding of the transformer.

3. Y-Y CONNECTION:

In this type of connection all the primary winding of the similar transformer are connected in Y

design and similarly the secondaries are too Y connected. Two voltmeter are used in which one is

connected between R&V of input and other between two secondary terminals. The power is

switched on and the corresponding line voltage are noted down.

4. Y- CONNECTION:

In this primaries are star connected and secondaries are delta connected. The power is switched on

and corresponding line voltage is noted down.

5. -Y CONNECTION:

In this type of connection primary winding of all the three transformers is connected in and all

secondaries are connected in Y. the voltmeter connected across the lines and the reading in noted

down.

6. - CONNECTION:

In this type of connection, the primaries of the 3 transformers are connected in . The voltmeters

are connected across the corresponding voltage are noted down.

GRAPH:

RESULT:

The input and output voltage of primary and secondary line voltage of 3-phase transformer

using Y-Y, Y-, -, -Y connection was found out.

VIVA QUESTIONS:

1. What are the types of 3 phase transformers according to winding connections?

2. Which type of three-phase transformer is used in power transmission?

Ex.No. 9

HOPKINSON’S TEST

AIM:

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QTY1 Voltmeter MC 0 – 300V 22 Ammeter MC 0 – 20A

0 – 2A22

3 Rheostat 350, 1.4A 24 Tachometer 0 – 10000 Rpm 1

Theory:

By 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 and are so adjusted

electrically that one of them runs as a motor and the other as a generator. The mechanical output of the

motor drives the generator and the electrical output of generator is used in supplying the greater part

of input to the motor. If there were no losses in the machines, they would have run without any

external power supply. But due to these losses, generator output is not sufficient to drive the motor and

vice-versa. The losses are supplied either by an extra motor which is belt-connected to the motor

generator set or as suggested by kapp, electrically from the supply mains.

The two shunt machines are connected in parallel. They are to begin with , started as unloaded

motors. Then, the field of one is weakened and that of the other is strengthened so that the former runs

as a motor and the latter as a generator.

Procedure:

Motor is started up from the supply mains with the help of a starter whereas main switch SPST

of the generator is kept open. Its speed is adjusted to normal value by means of its field regulator.

Generator voltage is read on Voltmeter. The voltage of the generator is adjusted by its field regulator

until voltmeter reads zero, there by showing that its voltage is the same, both in polarity and

magnitude as that of the main supply. There after, SPST is closed to parallel the machines. By

adjusting the respective field regulators, any load can now be thrown on to the machines. Generator

current I can be adjusted to any desired value by increasing the excitation of generator or by reducing

the excitation of motor and the corresponding values of different ammeters are read.

Formulae:

% = (I1/ (I1+I2))

For Generator:

Armature cu loss = (I1+I3)2 RaField cu loss = I3 V

For Motor:

Armature cu loss = (I1+I2-I4)2 Ra

Stray losses:

(W) = VI2 – [ (I1+I3)2 Ra + VI3 + (I1+I2-I4)2Ra+VI4]

Total losses (Wg) = (W/2)+(I1+I3)2Ra+VI3

Efficiency g = VI1/(VI1+Total losses)

Total losses Wm = (W/2) + (I1+I2-I4)2Ra+VI4

Efficiency m = [V(I1+I2)-Total loss]/V(I1+I2)

Ex.No: 10

SUMPNER'S TEST

AIM:

To conduct Sumpner's test on two similar single-phase transformer and to obtain efficiency and

regulation at various loads.

APPARATUS REQUIRED:

S.NO APPARATUS TYPE RANGE QTY1 Autotransformer2 Ammeter3 Voltmeter4 Wattmeter5 SPST

CIRCUIT DIGRAM:

THEORY:If low voltage is applied to the second side with the help of an autotransformer, full load

current can be circulated through the windings and this current is limited by the equivalent

impedance's of the two transformers referred to secondary side. Hence the wattmeter connected to

secondary side tough it may not be a problem to conduct load test on a small transformer, it is difficult

to perform load test on large transformers, as it results in wastage of large amount of energy. It will

also be difficult to get a suitable load at required power factor capable of absorbing the energy. Hence

sumpner's test is a suitable method to predetermine the efficiency and regulation at any load, at any

power factor, provided two identical transformers are available. The test is useful to estimate the

temperature rise also. Here power is drawn from the supply only for meeting the losses. The load

condition is simulated.

The primaries of the identical transformer is connected in series with their polarities in

opposition. The primary circuit when energized draws the magnetizing and core loss current of the two

transformers (I1). However since the secondary emfs are equal and opposite, no current flow through

the secondary circuit. Hence the transformers behave as if their secondaries are open circuited. So if an

ammeter and wattmeter (W1) are connected on the side it will read the no load current and core losses

of the two transformers. The voltmeter reading V2, corresponds to the voltage across the two

windings. Thus the sumpner's test enables to perform open circuit test simultaneously. The transformer

can also be tested for temperature rise economically with out loading.

PROCEDURE:

Connections are made as show in figure. Keep the SPST switch 'S' on the secondary side open.

See that the auto transformer on the primary side and secondary side are set in 'zero' output position.

Switch on the supply on the primary side and vary the autotransformer applying normal voltage.

Check the polarity if the transformer A and B. In case of correct polarity the reading of the voltmeter

connected between the free terminal of the secondary, i.e. V2 will be zero. If it read double the

voltage, it indicates incorrect polarities. If polarities are correct proceed with the experiment. If the

polarities are not correct, switch off the supply and secondary terminal connection are inter changed so

as to correct the polarity.

If polarities are found correct after checking the SPST switch 'S' is closed. Switch on the

supply on the secondary side with the help of the autotransformer. Sufficient voltage is applied to

allow rated current through the secondary windings. Note down the reading of the voltmeters,

ammeters and wattmeter's on the primary on the primary and secondary side. Reduce the voltage and

supply on the primary side is switch off. Correction of equivalent circuit drawing to be given.

TABULATION:

GRAPH:

RESULT:

Thus the transformation ratio of different three phase connections are checked and determined.

Ex.No:11

SWINBURNE’S TEST

AIM

To predetermine the performance characteristics of a dc shunt motor by conducting

Swinburne’s test at any desired load.

APPARATUS REQUIRED

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

Ammeter

Voltmeter

Rheostat

Tachometer

MC

MC

0-10 Amps

0-300 Volts

350 Ohms, 1.4 Amps

2

2

2

1

CIRCUIT DIAGRAM

THEORY

It is a simple method in which losses are measured separately and from their knowledge,

efficiency at any desired load can be predetermined. This test is applicable only to these machines in

which flux is practically constant. i.e., shunt and compound wound machines.

The machine is run as a motor on no load at its rated voltage, i.e., voltage stamped on the

nameplate. The speed is adjusted to the rated speed with the help of shunt field rheostat.

By knowing no load current Io and shunt field current If no load armature current Iao can be

calculated as,

Iao = Io - If

Let V= Supply Voltage

No load input power = V Io

No load input to armature = V (Io - If)

Power input to shunt field = V If

No load Power input to armature supplies the following:

1. Iron losses in core

2. Friction loss

3. Windage loss and

4. Armature copper loss = (Iao)2 Ra

Hot armature resistance is to be used. For that Ra = 1.3 Rdc (approx) relation may be used. If

we subtract armature copper loss from no load input power VIo, we get constant loss Wc = VIo - (Iao)2

Ra.

Knowing the constant losses, its efficiency and performance at any other can be determined as

given below.

Let, I = Load current at which efficiency is required, then,

Armature Current, Ia = I - If , if the machine is motoring;

And Ia = I + If ; if the machine is generating.

PROCEDURE

The connections are made as shown in the circuit diagram. The machine is run as a motor

under no load at its rated voltage. The speed is adjusted to rated speed using the field rheostat. The

input voltage, no load current and field current are noted. Calculations are done to the motor and

generator.

To Find Armature Resistance Ra

The connections are made as shown in the circuit diagram. The rheostat is adjusted and the

values of armature current and voltage are noted. The average dc resistance Rdc is obtained. The hot

armature resistance Ra= 1.3Rdc (approx) is calculated and used.

MODEL CALCULATION:

GRAPH

Graph is drawn by taking Output power along X-axis and Efficiency along Y-axis.

RESULT

Thus Swinburne’s Test is conducted and performance characteristics are obtained for the

machine running as a motor and generator.

VIVA QUESTIONS

1. What is the purpose of conducting Swinburne’s test?

2. What are the losses in dc machine?

3. What are the advantages and disadvantages of Swinburne’s test?

4. What happens if the direction of current at the terminals of a series motor is reversed?

5. A dc motor fail to start when switched on what could be the possible reasons?

Ex.No 12

LOAD TEST ON DC COMPOUND GENERATOR

AIM;

To conduct load characteristics on a given DC compound generator and plot is characteristics.

APPARATUS REQUIRED:

Sl. No APPARATUS TYPE RANGE Qty

1 Voltmeter MC 0 – 300V 1

2 Ammeter MC 0 – 20A 1

3 Rheostat 350, 1.4A 1

4 Loading Rheostat 3Kw 1

5 Tachometer 0 – 10000 Rpm 1

CIRCUIT DIAGRAM

THEORY:

A shunt generator is unsuitable where constancy of terminal voltage is essential, because .its

terminal voltage decreases as the load on it increase. This decrease in V is particularly objectionable

for lighting circuit where even slight change in the voltage makes an apprectionable change in the

candlepower of the incandescent lamps. A shunt generator may be made to supply substantially

constant voltage by adding to it a few turns joined in series with either armature or load. These turns

are so connected as to aid the shunt turns when the generator supplies load. As the load current

increase, the current there by increasing the flux. Due to the increase in flux, induced emf also

increase. By adjusting the number of series turn, this increase in emf can be made to balance the

combined voltage drop in the generator due to armature reaction and the armature drop. Hence, V

remains practically constant, which means that the field current is also almost unchanged. The decree

in terminal voltage is due to three cases, first two are neutralized by the series field ampere-turns and

the third one, there fore does not occur.

If the series field ampere turns such as to produce the same voltage at rated load as no-load,

then the generator is flat-compounded. It should be noted, however, that even

in the case of a flat-compounded generator, the voltage is not constant from no-load to rated load. At

half the load the voltage is actually greater than the rated voltage.

If the series filed amp-turns are such that the rated-load voltage is less than the no load voltage, then

the generator is under-compounded, if rated-load voltage is less than the no-load voltage, then the

generator is under-compounded but such generator are seldom used.

For short distances such as in hotels and office buildings, flat-compound generators are used

because the loss of voltage over small lengths of the feeder is negligible. But when it is necessary to

maintain a constant voltage then an over-compounded generator, which combines the functions of a

generator and a booster, is invariably used.

PROCEDURE:

1. Connections are given as per the circuit diagram.

2. The motor is started by a three-point starter and brought to its rated speed by varying the

filed rheostat of the motor.

3. Then the generator field is varied and brought to its rated voltage and now the no load

readings are noted.

4. Now, the load switch is switched on and the resistive load is added gradually.

5. At each step of loading the generator speed is maintained constant.

6. For each step of loading the readings are noted from the ammeter, voltmeter and are

tabulated.

7. The loading is continued till the generator is loaded to its rated load.

8. After reaching the full load, the load is cut down slowly and brought to no load and the

motor is switched off.

MODEL CALCULATION:

Ia – Ise = Ir +IL

Where Ia = Armature Current in Amps

Ir = Shunt Field Current in Amps

IL = Load Current in Amps

Ise = Series field current in amps

Eg = V+IaRa = Generated EMF in volts

V = Terminal voltage in Volts

Ra = Armature Resistance in ohms

RESULT

The load test on compound generator is conducted and its characteristics are plotted.

1.9 VIVA QUESTIONS

1.What is compound generator?

2.What are types of compound generators?

3.What are the types of compound generators according to field winding connection?

INFERENCES & APPLICATION

MODEL GRAPH:

1. Over compound generator

2. Flat compound generator

3. Under compound generator

4. Shunt generator

5. Differential compound generator.

TABULATION:

S.No V Volts If Amps IL Amps Ia = If+IL Amps Eg = V+IaRa

1

2

3

4

5

6

7

8

9

10

Ex.No: 1

OCC AND LOAD TEST ON SEPARATELY EXCITED DC SHUNT GENERATOR

AIM:

To conduct open circuit and load test on a separately excited (shunt) DC generator and to

determine the external and internal characteristics of it.

APPARATUS REQUIRED:

S.No ITEM TYPE RANGE QUANTITY

1.

2.

3.

4.

5.

6.

7.

Ammeter

Ammeter

Voltmeter

Rheostat

DPST switch

SPST switch

Tachometer

MC

MC

MC

1

2

1

2

1

1

THEORY:

Open Circuit Test:

Open circuit characteristics may also called as NO LOAD or MAGNETIC

CHARACTERISTIC. It show the relation between the no load generated emf in the armature, Eo and

the field or exciting current If at a given fixed speed. It is just the magnetization curve for the material

of the electromagnet. It’s shape is practically same for all generators whether separately or self-

excited.

Load Test:

Let us consider a self or separately excited dc generator giving its rated no load voltage ‘E o’ for

a certain constant field current. If there were no armature reaction and armature voltage drop, then this

voltage would have remained constant as shown in Fig by the dotted horizontal line I. But when the

generator is loaded, the voltage falls due to these two causes, thereby giving slightly drooping

characteristics. If we subtract from Eo the value of voltage drops due to armature reaction for different

loads, then we get the value of E, the emf actually induced in the armature under load conditions.

Curve II is plotted in this way and is known as the internal characteristic. The straight-line represents

the IaRa drop corresponding to different armature currents. If we subtract from E the armature drop

IaRa, we get terminal voltage V. Curve III represents the external characteristic.

PROCEDURE:

Open Circuit Test:

Connections are given as per the circuit diagram. Initially the SPST switch is kept open.

Supply is given and the motor is started with the help of a suitable starter. Emf generated due to

residual magnetism is noted from voltmeter when field current is zero. Then SPST switch is closed.

By varying the field rheostat in the generator side, different values of generated emf E o and field

current If are noted from voltmeter and ammeter.

Load Test:

Connections are given as shown in the circuit diagram. The motor is started with the help of dc

suitable starter and brought to its rated speed. Keeping the generator side DPST open, the field

rheostat in the generator side is adjusted for the rated voltage of the generator which is seen in the

voltmeter. Now the DPST switch is closed and the resistive load is put up on the generator step by

step. The terminal voltage, armature and load current values are noted down for each step from the

respective meters. Generator may be loaded upto 125%. Note that while taking each set of readings,

the field current is maintained constant as that for rated voltage (because due to heating, shunt field

resistance is increased).

To Find Armature Resistance:

Connections are given are in the circuit diagram. Supply is given and the rheostat is varied and

the corresponding values of voltage and current are noted. From these values, dc armature resistance is

calculated. Then it may be multiplied by 1.3 to get hot armature resistance.

TABULATION:

GRAPH:

Open Circuit Test:

Open circuit characteristic is drawn by taking field current If along X-axis and generated

voltage Eo along Y-axis.

Load Test:

External characteristic is drawn taking load current IL along X-axis and terminal voltage V

along Y-axis. Internal characteristic is drawn by adding IaRa drop to external characteristic curve, Ia

along X-axis and E along Y-axis.

RESULT:

Thus the no-load test and load test is conducted on a separately-excited DC generator and the

corresponding characteristics curves are drawn.

VIVA QUESTIONS:

1.What is separately excited generators?

2.Give the emf equation of a DC generator.

3.What are the factors that may affect the voltage build up of a dc shunt

generator?

4.What is the procedure to shut down a generator.

5.Can we use the same machine as a generator and as a motor?