B.Tech. Pt. II

EXPERIMENT No. 1

OPEN CIRCUIT & SHORT CIRCUIT TESTS

AIM: Perform open circuit and short circuit test on a two winding transformer and hence determine its equivalent circuit parameter. Calculate its % regulation at full load U.P.F., 0.8 lagging P.F., 0.8 leading P.F. and Z.P.F. lagging, Z.P.F. leading.

SPECIFICATIONS OF TRANSFORMER: Name Plate rating of transformer under test.

Calculate the following quantities:

Full load current referred to primary (high Voltage) side =

Full load current referred to secondary (Low Voltage) side =

Turn ratio =

Record the resistances of Primary and Secondary windings using Digital Multimeter.

APPARATUS REQUIRED:

ApparatusNumberRangeRemarks

For Open Circuit Test-from Low Voltage side

Ammeters11/2A > 5-10% of

Voltmeter1150- 300V

LPF Wattmeter1150 - 300V / 5A>

For Short Circuit Test-from High Voltage side

Ammeters110A >

Voltmeter130V

Wattmeter175V /10A

THEORY: A transformer can be easily analyzed using lumped parameter approach, in which the transformer is represented by a black-box known as equivalent circuit. If the equivalent circuit parameter is known then the input output relation can be easily established. A typical equivalent circuit of transformer is shown in Figure. A transformer works on the basis of the mutual flux coupling the two coils. In order to minimize the leakage flux the magnetic flux is allowed to flow in the path provided by the ferromagnetic core. This core is subjected to alternating magnetization in which the direction of flux alternates at the supply frequency and due to this certain amount of power is wasted in Hysteresis and eddy current losses.

Using Text Book write:

The e.m.f equation of a two winding transformer.

Derive equivalent circuit of a two winding transformer.

Define core loss-Hysteresis & Eddy current losses.

Define referred value.

DETERMINATION OF EQUIVALENT CIRCUIT PARAMETERS:

At no load the current is assumed to flow in the parallel combination of and or,

Where real component of current

Similarly, imaginary component of

Input power,

Where =resistance of low voltage winding and = core lossThe input power is the wattmeter reading of LPF wattmeter when rated voltage (Low voltage side) at rated frequency is applied on low voltage side of transformer, keeping high voltage side open. The e.m.f or the voltage appearing across parallel branches of equivalent circuit is,

Where =leakage reactance of low voltage side.

Thus,

If drop in winding is assumed to be zero, then

Core loss = Wattmeter reading or

e.m.f. =Applied voltage or

Thus using open circuit test, the two equivalent circuit parameters can be calculated as,

.

These values are referred to low voltage side. The rated copper loss is normally taken as wattmeter reading during short circuit test at rated current, when a feeble voltage is applied on high voltage side of transformer, keeping its low voltage side short circuited. Transformers series equivalent circuit parameters referred to high voltage side are

Thus,

These parameters are referred from high voltage side. The equivalent circuit parameters referred to high voltage side are obtained by obtained referred values of from high voltage side.

Thus,

Similarly,

The parameters, which are referred to high voltage side remains unchanged. The approximate equivalent circuit diagram is shown in Figure.

Here, transformer resistance referred to HV side,

And transformer leakage reactance referred to HV side,

Similarly equivalent circuit of transformer referred to low voltage side are

POWER EFFICIENCY: It is assumed that KVA rating of transformer is ; Rated Copper loss or the wattmeter reading under short circuit test is ; Rated Core Loss or wattmeter reading under Open Circuit test is . The power efficiency of a two winding transformer operating at fraction of load and at a power factor is,

% Power Efficiency,

Calculate % efficiency at different fraction of load and at two different power factors.

OBERVATIONS:

Resistance of High voltage winding, =

Resistance of low voltage windings, = Open Circuit Test (High voltage side Open)Short Circuit Test ( Low Voltage side Shorted)

Voltage VoRated LV, 110V Voltage VSC5-10% of rated voltage

CurrentIoACurrentISCRated High Voltage side current, 6.8A

WattageWoWWattageWSCW

PHASOR DIAGRAMS: Draw to the scale phasor diagrams at ZPF Lag /Lead, UPF and 0.8 P.F. Lag/Lead.

Use approximate equivalent circuit for V= VHigh; I = IHigh; REX1=Rexh & XEX1=Xexh. Choose a suitable scale. Measure the length of phasor E in terms of Volt.% REGULATION Further, calculate the no load voltage E for different power factors.

For a lagging power factor angle, E can be written as,

Thus calculate the % regulation (UP) & (DOWN) using the expressions given below:

How will you calculate the above for leading power factor? Compare different quantities in Tabular form..

P.F.No Load Voltage E% Regulation% Power Efficiency

CalculatedMeasuredUPDOWN

ZPF Lag.

0.8 Lag.

UPF

0.8 Lead

ZPF Lead

TABLE: POWER EFF. At 0.8 P.F. LAGGING & 0.8 P.F. LEADING

S.No.FractionInputConstant lossVariable

LossTotal

Loss% Efficiency

(W)(W)(W)(W)

10

20.1

30.2

161.5

PROCEDURE:

Note the name plate rating of transformer, decide the range of instruments.

Connect circuit diagram as shown in Figure for Open circuit Test keeping high voltage side kept Open (why?). Apply rated voltage corresponding to low voltage winding. Note

Connect circuit diagram as shown in Figure for Short circuit Test keeping low voltage side short circuited (why?). Apply rated current of high voltage winding. Note. Note that very small voltage is required during short circuit test.

Measure the winding resistances of high voltage and low voltage windings respectively using a Digital Multi meter.

PLOT:

% Efficiency versus fraction of load for two different power factors

% regulation versus load current at unity power factors

PRECAUTIONS:

1. Choose correct rating instruments required for different tests.

2. During short circuit test, apply reduced voltage such that only rated current flows. Excessive voltage may cause melting of insulation of shorted wire and subsequent fire.

B.Tech. EE Pt. II

EXPERIMENT No. 2

LOAD TEST ON A SINGLE PHASE TRANSFORMER

AIM: Perform load test on transformer and plot % regulation, % efficiency versus load current.

SPECIFICATIONS OF TRANSFORMER: Name Plate rating of transformer under test.

COMPUTER PROGRAM: Obtain the equivalent circuit parameters referred to low voltage side from the earlier experiment. Write a computer program for finding % regulation (UP & DOWN) and % efficiency at any power factor and any load current.

APPARATUS REQUIRED:

ApparatusNumberRange

Ammeters215A & 10A

Voltmeter2150- 300V

Wattmeter1150 - 300V / 15A

THEORY: The load test on a two winding can be performed from either side. When a transformer is subjected to electrical loading the voltage on the input side takes place. For a 220V/11V, it is convenient to perform load test from LV side, because the input voltage can be kept constant during the experiment.

For the primary winding, we have,

And for secondary winding referred to primary side,

When an a.c. voltage of magnitude V1 is applied across the primary winding, it results in an induced e.m.f. equal .According to Lenzs law, this e.m.f. opposes the applied voltage and results in a flow of current,

Here it must be clearly understood that the quantity is the leakage reactance of primary winding. It should not be confused with the quantity , where being the self inductance of primary winding.

% REGULATION Further, calculate the no load voltage E for different power factors.

For a lagging power factor angle, E can be written as,

Thus calculate the % regulation (UP) & (DOWN) using the expressions given below:

How will you calculate the above for leading power factor?

POWER EFFICIENCY: It is assumed that KVA rating of transformer is ; Rated Copper loss or the wattmeter reading under short circuit test is ; Rated Core Loss or wattmeter reading under Open Circuit test is . The power efficiency of a two winding transformer operating at fraction of load and at a power factor is,

% Power Efficiency,

The power efficiency can also be calculated using experimental observation.

PROCEDURE:

Note the name plate rating of transformer, decide the range of instruments.

Connect circuit diagram as shown in Figure for load test. Choose a resistive load. Initially do not switch ON any load. Keep the voltage on LV side equal to its rated value.

Slowly change the load switches.

Do you observe any change in input voltage? If yes correct the input voltage using auto transformer.

Record different voltages, currents ant wattmeter.

Repeat the experiment for capacitive and inductive loads.

OBERVATIONS: Using developed computer program, calculate no load voltage , % regulations and % efficiency For different input currents and power factor.

S.No.Low Voltage sideHV sideE.M.FP.F.% Efficiency% Regulation

VIIIWIVoIoWoEILVHVExperimentCalculationExperimentCalculation

10

2

3

10

PLOT: For different types of loads, viz resistive, inductive and capacitive.

% Efficiency versus load currents

% regulation versus load current

RESULT & DISCUSSIONS:

Discuss the graphs for different types of loads. Why % regulation is negative for capacitive loads

PRECAUTIONS:

3. Choose correct rating instruments required for different tests.

4. Do not apply high voltage on LV windings.

EXPERIMENT No. 3

CHARACTERISTICS OF SELF EXCITED DC GENERATORSAIM:(a) Experimental determination of magnetizing characteristics (Open Circuit

Characteristics or OCC) of Self-excited DC generators

(b) Perform load tests on DC Generators & draw the Load Characteristics of

Self-excited DC Generators. (Only for Electrical Students)

SPECIFICATIONS OF THE MACHINE: Name Plate rating of machine under test.

APPARATUS REQUIRED:

ApparatusNumberRange

D.C. Ammeters21/2A & 30-0-30A

D.C. Voltmeter1300V

Tachometer1

THEORY: Write essential conditions of self excitation of dc generator. Define residual magnetism, critical field resistance and critical speed etc. from your text book

CIRCUIT DIAGRAM:

Self-excited DC Generator:

PROCEDURE:

1) Connect as shown in the Fig 1 for self-excited dc generator.

2) Run the generator at the rated speed using starter of dc motor.

3) Maintain its speed constant throughout the experiment using motor field resistance.

4) Initially keep the load switch OFF. Whether the generator develops voltage? If No, why? If Yes, go to Step No.8

5) Switch OFF the prime mover.

6) Interchange either connections of armature circuit or connections of field circuit of generator.

7) Repeat Step-2. Whether the generator develops voltage? If Yes why?

8) Vary field current of generator in steps and record the developed voltages as in Table 1.

9) Record developed voltage of generator, when generator field is open, i.e., Ifg=0.0

10) Plot Open circuit voltage versus field current.

11) Adjust the open circuit voltage at 200V at 1000 RPM. Increase resistive load in steps. (Whether the speed falls? If Yes why?)

12) Maintain the speed at constant value through out the experiment. Note the load current and terminal voltage for different loads as in Table 2.

13) Now switch OFF prime mover and put a short circuit across the self excited dc generator. Run prime mover at required speed and record current (VL=0.0). Plot the load characteristics of self excited shunt generator. Repeat 1-8 at two different speeds.

.

PLOT GRAPHS:

(a) Magnetizing characteristics of dc Generators:

Field current (on X- axis) Versus Open Circuit Voltage (on Y- axis)

(b) Load Characteristics:

Load current (on X- axis) Vs Terminal Voltage (on Y- axis) for

NOTE: Obtain internal characteristics of the DC Generator by adding resistance drop to the external characteristics.

RESULTS: Comment on graphs and torque of motorDISCUSSIONS: Effect of speed, armature reaction, field resistance, residual magnetism, type of load, type of generators and commutation etc.

PRECAUTIONS:

1) While obtaining load characteristics of separately excited dc generator, the armature terminal should never be shorted. Since field is already connected to dc supply, it will cause excessive armature current to flow thereby blowing of supply fuses and damaging armature windings, brushes and commutator.

OBSERVATIONS: -

Speed = Constant

Magnetizing Characteristics of dc generators

Self excited DC Generator

Increasing field currentDecreasing field current

S.No.Open Circuit

Voltage (V)Field Current

(A)S.No.Open Circuit

Voltage (V)Field Current (A)

1.01.

2.2.

15.15

Load Characteristics of Self excited dc Generators: Speed = Constant Ifg = Constant

S.No.Terminal VoltageLoad Current

Output Power Po=V.ASpeed

NMotor Torque

60*Po/ (2 ( N)

(Volt)(Amp)(Watt)(RPM)(N-m)

1.0

2.

15

160

Determination of Armature resistance: Draw your own circuit diagram or use digital multi meter.

B.Tech. EE Pt. II

EXPERIMENT 4

SPEED CONTROL OF d.c. SHUNT MOTOR

AIM: - Study different methods of speed control of a dc shunt motor.

SPECIFICATIONS OF THE MACHINE: Name Plate rating of machine under test.

APPARATUS REQUIRED:

ApparatusNumberRange

D.C. Ammeters25-0-5A & 1/2A

D.C. Voltmeter1300V

Tachometer1

Digital Multimeter1

THEORY: The speed of a dc shunt motor mainly depends on field flux or field current and the voltage appearing across armature of dc motor. This can be understood from the basic equations of dc motor. The back e.m.f. appearing across the armature can be given as

E b = 2p Z N / 60 (2a)

= K N

and also from circuit point of view,

Eb= V- Ia RaWhere

2a = total number of parallel pathN = speed of operation

2p = total number of polesIa = armature current

= Flux per poleV = applied voltage

Z = Number of conductorsRa = armature resistance

Combining above equations, we get

Speed

N= Eb / K = (V- Ia Ra) / K

Since the flux is directly proportional to field current If , under low saturation of the magnetic circuit,

N Eb

&

N 1 / If

It is also clear that the speed of a dc motor depends on design parameters, which can done only at design level.

There are basically two experimental methods of speed control of a dc motor.

(a) Armature Voltage Control

(b) Field Flux Control

Note: The resistance Ra shall be rated to carry a current at least 20% of rated motor current continuously. Its resistance is to be such that Ia.Ra is at least 50% of applied voltage. (why?)

PROCEDURE:

1) Keep maximum armature resistance and minimum field resistance.

2) Field flux control: In the circuit shown the motor is now started with the help of a starter. Slowly decrease the armature resistance Ra to minimum possible value.

3) Adjust the field current If such that the speed N is 20-30% below the rated speed of the motor.

4) Vary the field current in small steps and note the corresponding speed until the speed is increased to 15-20% above the rated value.

5) Plot speed N versus field current If (Ra is not changed)

6) For a new value of Ra held constant. Repeat steps 1 to 5.7) Armature Control: Adjust field current If such that the speed N is nearly rated speed.

8) Increase the resistance Ra in steps and note the corresponding voltage across the armature Eb and the speed. The speed should go down to 50% rated value.

9) Repeat steps 7-8 for a new value of If held constant.

10) Plot speed N versus Armature voltage Eb (If =constant)

RESULTS & CONCLUSIONS:

Discuss the shape of characteristics.

What have you learned from this experiment?

What do you conclude from this experiment?PRECAUTIONS:_1275999396.unknown

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