ee lab manual

Sreyas Institute Of Engineering and Technology

Electrical Engineering lab Manual

Dept of EEE Electrical Lab Manual 1

Experiment: 01

THEVENIN’S THEOREM

Aim: To determine the Thevenins theorem theoretically and practically.

Apparatus:

Theory:

“Any two terminal active linear network containing energy sources (generators) and resistances can be

replaced with an equivalent circuit consisting of a voltage source Vth in series with a resistance Rth. The

value of Vth is the open-circuit voltage between the terminals of the network and Rth is the resistance

measured between the terminals with all the energy sources eliminated (but not their internal

resistances).”

Circuit diagram

Voltmeter 0-40 dc 1 Num

Ammeter 0-200ma dc 2 Num




Procedure:

1) Connect the circuit as shown in the figure 1.

2) Measure the current through the load resistor in the linear circuit.

3) Calculate the Thevenin’s equivalent resistance of the circuit, when the source is set to zero.

Rth = (82||150) + 47=100 ohms.

4) Calculate the open circuit voltage across the terminals A & B which is equal to

the voltage across 150ohms resistor.

Vth= 15 X150=9.69V

82+150

5) Measure the voltage drop across 150ohms resistor after disconnecting terminals

A & C.

6) Find it to be equal to calculated value of Vth.

7) Now set the voltage to the obtained Vth in the Thevenin’s equivalent circuit using

variable power supply.

8) Measure the current through the load resistor in the Thevenin’s equivalent circuit.

9) Note that both current measured through the load resistor in the linear circuit

as well as in the equivalent circuit is same.

Repeat the above procedure for different values of resistors provided on the board.

Thus the Thevenin’s Theorem is proved.




Tabular column:

S.no Voltage applied[Vs]

Thevenins current[I]

Thevenins voltage[Vth]

Thevenins Resistance[Rth]

Equitant ckt load current[Il]

Results and Graph:

Viva Questions:

1) What is the statement of Thevenins theorem?

2) How you will calculate the Thevenins resistance?

3) Why Thevenins resister is connected in series with load resistance?

4) What is the aim of the Thevenins theorem?




Experiment: 02

NORTON’S THEOREMS

Aim: To determine the Norton’s theorem theoretically and practically.

Apparatus:

Theory:

“Any two terminal linear active network containing energy sources (generators) and resistance can be

replaced with an equivalent circuit consisting of a current source IN in parallel with a resistance RN. The

value of IN is the short-circuit current between the terminals of the network and RN is the resistance

measured between the terminals with all the energy sources eliminated (but not their internal

impedances).”

Circuit diagram






Procedure:

1) Connect the linear circuit as shown in the figure 2.

2) Measure the current through the load resistor in the linear circuit by connecting ammeter between A &

C.

3) Calculate the Norton’s equivalent resistance of the circuit.

RN= 82 X 150 + 47=100 ohms

82 + 150

4) Measure the Norton’s equivalent current which is the short circuit across the terminals A & B by

connecting current meter across A & B. This will be equal to 96.9mA.

5) Now connect the circuit as shown in Norton’s Equivalent Circuit where

RN=100 ohms and IN=96.9mA.

6) To get current source, after connecting circuit components Rload, RN

and points A & C shorted connect variable supply in series with the current meter in place of

current meter in place of current source shown. Adjust the voltage supply such that you read

96.9mA in the current meter.

7) Now, switch of the power, remove the meter short the positive terminal of

battery to terminal A. Remove the short between A & C and connect the current

meter between A & C. Note the current through RL. Observe it to be equal to the current

through RL measured in the linear circuit.

Repeat the above procedure for different values of resistors provided on the board.

Thus the Norton’s Theorem is proved.




Tabular column:

S.no Voltage applied[Vs] Load current[Il] Norton’s current[In]

Norton’s Resistance[Rth]

Equitant ckt load current[Il]

Results and Graph:

Viva Questions:

1) What is the statement of Norton’s theorem?

2) How you will calculate the Norton’s resistance?

3) How you will calculate the Norton’s current?

4) Why Norton’s resister is connected in parallel with load resistance?

5) What is the aim of the Norton’s theorem?




Experiment: 03

SUPERPOSITION THEOREM

Aim: Verification of the Superposition Theorems.

Apparatus:

Theory:

“In any bilateral linear network which contains more than one energy source, the current flowing

through any element is the vectorial sum of currents that are caused to flow in the element due to

individual energy sources at a time setting all other sources to zero.”

Circuit diagram:






Procedure:

1. Connect the circuit as shown in the circuit diagram figure (1).

2. Set V1=15V, for this connect fixed 15V supply.

3. Set V2=10V, for this adjust the variable supply to 10V.

4. Note the current (I) through E & F, when both V1 and V2 are applied.

5. For the same circuit apply voltage V1 and make sure that the V2 to be shorted (V2 =0) and

note down the current (I1) through E & F.

6. For the same circuit apply voltage V2 and make sure that the V1 to be shorted (V1 =0) and

note down the current (I2) through E & F.

7. The Superposition Theorem is verified i.e. I = I1 + I2.

Repeat the above procedure for different variable supplies.

Tabular column:

Applied Voltage V1 Applied Voltage V2 Current I1 Current I2 Current I3

12 5

10 5

8 5

6 5




Tabular column:


12 0

10 0

8 0

6 0


0 5

0 5

0 5

0 5

Results and Graph:

Viva Questions:

1) What is the statement of Superposition theorem? 2) How you will calculate the total resistance? 3) State the super position theorem? 4) What is meant by network? 5) What is meant by bilinear network? 6) Applications of superposition theorem?




Experiment: 04

RECIPROCITY THEOREM

Aim: Verification of the Reciprocity Theorems.

Apparatus:

Theory:

“In any bilateral linear network containing one or more generators, the ratio of the voltage(V)

introduced in one mesh to the current(I) in any second mesh is the same as the ratio obtained if the

position of voltage and current are interchanged, other emfs being removed.”

Circuit diagram:






Procedure:

1. Connect the circuit as shown in the circuit diagram figure (2).

2. Set the Voltage (say VI=10V) and connect it across A and B.

3. Connect C and D through current meter and note the current (I1).

4. Interchange the position of ammeter and voltmeter note the current (I2).

5. Observe that the both currents are same.

6. Calculate the ratio of voltage introduced between A& B to the current through C& D.

7. Calculate the ratio of voltage introduced between C& D to the current through A& B.

8. Note that the both ratios are equal. Thus the theorem is proved.

Repeat the same procedure for different voltages

Tabular column:

Voltage at AB Current at CD

12V

10V

08V

06V

04V

Results and Graph:

Current at AB Voltage at CD

12V

10V

08V

06V

04V




Experiment: 05

MAXIMUM POWER TRANSFER THEOREM

Aim: Verification of the Maximum Power Transfer Theorem.

Apparatus:

DRB 0-300 Ω 1 Num

Voltmeter 0-40 V Dc 1 Num

Ammeter 0-200ma Dc 1 Num

Theory:

“A resistive load will abstract maximum power from a network when the load resistance is equal

to the resistance of the network as viewed from the output terminals, with all the sources of emf

removed leaving behind their internal resistances.

Circuit diagram:




Procedure:

1. Connect the 5V to the terminals provided.

2. Connect the ammeter and voltmeter as shown in figure.

3. Calculate the power drawn by the circuit for different values of load resistors provided on

board and tabulate them.

PL=IL X VL

4. Observe that the maximum power is drawn when the load resistor is equal to the input

resistance.

5. Thus the Maximum Power Transfer Theorem is proved.

Tabular column:

S.No Load current IL P=I2R Resistance

30Ω

50Ω

70 Ω

90 Ω

100 Ω

120 Ω

140 Ω

160 Ω

180 Ω

200 Ω

Results and Graph:

Viva Questions:

1. State maximum power transfer theorem? 2. When the maximum power transfer theorem? 3. What is meant by power? Mention its unit? 4. What is meant by Energy? 5. What are the applications?




Experiment: 06

TWO PORT NETWORK PARAMETERS

Aim: To determine the two port network parameters.

Apparatus:

Theory:

To represent the general nature of a network, it is normally represented by a rectangular box. If a

conductor is fastened to a node in the network and brought for access the end of the conductor is called

a terminal. Terminals are required for connecting driving force to the network, for connecting load or

making measurements.

The minimum number of terminals that are useful are two. One pair of terminals is given the name Port.

In the two port network shown in fig 1 we have four variable identified by two voltages and two

currents. We assume that the variables transform quantities and use V1 and I1 as variables at the input

port 1, V2 and I2 as the variables at the output port 2. Now only two of the four variables are

independent and the specific action of any two of them determines the remaining two.

Procedure:

PARAMETERS

The two port networks are characterized by parameters in terms of voltage and current

variables. The names of parameters are 1) Z-parameters, 2) Y-parameters,

3) H-parameters, 4) ABCD-parameters






1) Z-PARAMETERS

The Z-parameters are also called as Impedance parameters.Z11,Z12 ,Z21 ,Z22

are the Z-parameters of the two port network..

V1= Z11I1 + Z12I2

V2= Z21I1 + Z22I2

a) Input Impedance (Z11) Z11 is the input impedance when output is open circuited.

Z11 = V1/I1 when I2=0

To measure Z11 open circuit the output side of the two port network and apply voltage in

the input side of the two port network then measure the V1 and I1 and substitute in the above

equation.

b) Reverse Transfer Impedance (Z12) Z12 is the mutual or reverse transfer impedance when

input is open circuited.

Z12 = V1/I2 when I1=0

To measure Z12 open circuit the input side of the two port network and apply voltage in

the output side of the two port network then measure the V1 and I2 and substitute in the above

equation.

c) Forward Transfer Impedance (Z21) Z21 is the mutual or forward transfer impedance when

output is open circuited.

Z21 = V2/I1 when I2=0

To measure Z21 open circuit the output side of the two port network and apply voltage in

the input side of the two port network then measure the V2 and I1 and substitute in the above

equation.

d) Output Impedance (Z22) Z22 is the output impedance when output is open circuited.

Z22 = V2/I2 when I1=0




To measure Z22 open circuit the output side of the two port network and apply voltage

in the input side of the two port network then measure the V2 and I2 and substitute in the above

equation.

2) Y-PARAMETERS

The Y-parameters are also called as Admittance parameters.Y11,Y12 ,Y21 ,Y22

are the Y-parameters of the two port network..

I1= Y11V1 + Y12V2

I2= Y21V1 + Y22V2

a) Input Admittance (Y11) Y11 is the input admittance when output is short circuited.

Y11 = I1/V1 when V2=0

To measure Y11 short circuit the output side of the two port network and apply voltage


equation.

b) Reverse Transfer Admittance (Y12) Y12 is the mutual or reverse transfer admittance when

input is short circuited.

Y12 = I2/V1 when V2=0

To measure Y12 short circuit the output side of the two port network and apply voltage


equation.

c) Forward Transfer Admittance (Y21) Y21 is the mutual or forward transfer admittance when

output is short circuited.

Y21 = I1/V2 when V1=0

To measure Y21 short circuit the input side of the two port network and apply voltage in


equation.




d) Output Admittance (Y22) Y22 is the output admittance when output is short circuited.

Y22 = I2/V2 when V1=0

To measure Z22 short circuit the input side of the two port network and apply voltage in


equation.

3) H-PARAMETERS

The H-parameters are also called as Hybrid parameters. H11, H12, H21, H22

are the H-parameters of the two port network..

V1= H11I1 + H12V2

I2= H21I1 + H22V2

a) Input Impedance (H11) H11 is the input impedance when output is short circuited.

Z11 = V1/I1 when V2=0

To measure H11 short circuit the output side of the two port network and apply voltage


equation.

b) Reverse Voltage Ratio (H12) H12 is the mutual or reverse transfer voltage ratio when input is

open circuited.

H12 = V1/I2 when I1=0

To measure Z12 open circuit the input side of the two port network and apply voltage in


equation.

c) Forward Current Transfer Ratio (H21) H21 is the mutual or forward current transfer ratio

when output is short circuited.

H21 = I2/I1 when V2=0




To measure H21 short circuit the output side of the two port network and apply voltage

in the input side of the two port network then measure the I2 and I1 and substitute in the above

equation.

d) Output Admittance (H22) H22 is the output impedance when output is open circuited.

H22 = I2/V2when I1=0

To measure H22 open circuit the input side of the two port network and apply voltage in


equation.

4) ABCD-PARAMETERS

In this system of parameters voltage and current at port 1 are expressed in terms of voltage and current

at port 2.

a) A is defined as the ratio of input voltage to the output voltage when the output is open

circuited.

A=V1/V2 when I2=0

b) B is defined as the ratio of input voltage to the output current when output is short

circuited.

B= V1/I2 when V2 =0

c) C is defined as the ratio of input current to the output voltage when the output is open

circuited.

C=I1/V2 when I2=0

d) D is defined as the ratio of input current to the output current when output is short

circuited.

D= I1/I2 when V2 =0

Results and Graph:




Experiment: 07

SERIES & PARALLEL RESONANCE

Aim: To determine series and parallel resonance

Apparatus:

Inductors of different value 3Num

Copacitor of different value 3Num

Resistors of different value 3Num

Circuit diagram:

Theory

The voltage across the inductor is VL = I XL

the voltage across the capacitor is VC = IXC

the voltage across the resistor is VR = IR

Phase relations among these voltages are shown in Figure 2. The voltage across the resistor is in

phase with the current. The voltage across the inductor leads the current by 90 degrees.




The voltage across the capacitor lags the current by 90 degrees. The total voltage across the

resistor, inductor and capacitor should be equal to the emf supplied by the generator.

Figure 2

From Figure 2 we can see that

If we divide both sides of this equation by current, we will get

E/I = Z = R2 + (XL - XC)

2,

Where (XL - XC) is called the total reactance and Z is called the impedance of the circuit.

We know that the capacitive reactance XC = 1/ωC, and the inductive reactance XL = ωL depend

on frequency. The value of frequency when

XL = XC, ωL = 1/ωC, or

ω= 1/√LC = ω0 = 2f0

The frequency f0 is called the resonance frequency of the circuit. At this frequency, the

impedance is smallest and the maximum value of the current (and the voltage across the resistor VR) can




be obtained. At this frequency, the circuit is said to be at resonance. At resonance, the current is in phase

with the generator voltage.

If we measure voltage across the resistor, depending on frequency, we will obtain a resonance

curve of the circuit as shown in Figure 3.

A resonance curve can be characterized by the resonance width ∆f, the frequency difference

between the two points on the curve where the power is half its maximum value or voltage is

Vmax/√LC = 0.707 Vmax

Figure 3

When the width is small compared with the resonance frequency, the resonance is sharp; that is,

the resonance curve is narrow. The circuit can be characterized by the quality factor

Q = f0/∆f.

If resistance is small and resonance is sharp, the quality factor is large. When the resistor is

large, the quality is small. Q is a measure of the rate at which energy is dissipated in the circuit if the AC

voltage source across the series circuit was removed.




PROCEDURE

1. Connect a series LCR circuit as shown in Figure 4

V1

R

C L

Figure 4

2. Calculate the resonance frequency of your circuit

3. Apply AC voltage to your circuit having the frequency being close to f0. Measure the peak-to-peak

voltage VR across your resistor. Also, measure the frequency using the oscilloscope (f = 1/T).

4. Decrease the frequency slowly and observe that VR decreases. Increase the frequency above f0

and observe that VR decreases. Estimate the frequencies where VR stops decreasing both above

and below f0. Your data will be collected between these two frequencies.

5. Take at least 10 readings of VR for frequencies below f0. Make sure that you take at least 5

readings in the vicinity of f0, since the data that is most meaningful is near f0, in the upper region

of the curve shown in figure 3.

6. Take 10 readings of VR for frequencies above f0. Make sure to take at least 5 readings in the

vicinity of the f0.




Tabular column:

FREQUENCY (Hz) VOLTAGE VR (V)

Parallel Resonance Circuit:

Theory

The parallel RLC circuit is very common in radio and TV circuits. The circuit analysis follows from

Kirchoff’s laws, but is not in many beginning texts. The common terminal voltage V is across each

element (see Fig. 5a), and the instantaneous current i is the algebraic sum of the instantaneous, iL, and

ic.

Since iR is in phase with V, but iL lags V by 900, and ic leads V by 90

0, we can describe the

situation by the rotating vectors in Fig. 5b where I is the vector sum of IR, IL, and IC. Hence from Fig. 5b




Figure 5: A parallel LRC circuit.

Where , and .

Substituting these gives

Resonance still occurs for XL=XC but the total current, Irms, is then just the current through R and

is a MINIMUM.

Since V is the same for all the parallel elements, the relevant phase differences are between the

currents. To measure the total current Irms and the phase between it and the voltage Vrms, we will insert a

sampling resistor RS in series with the signal generator. See Figure 6.

Measure the voltage Vrms S across the sampling resistor RS by connecting the resistor ends to

scope channel 2 via a differential amplifier. Since the voltage and current are in phase across a resistor,

this signal Vrms S is proportional to the total current Irms.




At resonant frequency fr the currents from L and C will cancel since they are of equal magnitude

but (always) 1800 out of phase. Hence, at Fr the total current I will be just that thru R, i.e. I=IR, and V and

Vs (or I) will be in phase. The smaller the sampling resistor RS, the less disturbance its voltage drop Vrms

S will have on the net voltage Vrms applied across the parallel circuit. However, distortion in the generated

sine wave may bother when RS is small and large currents flow. Our suggested RS=10kΩ is a

compromise. Even so you may see a factor 2 change in Vrms as f varies; however we can compensate by

adjusting the signal generator amplitude to keep Vrms constant as the frequency changes.

V

RCL

Rs

Figure 6: A parallel LRC test circuit.

PROCEDURE:

1. Connect the circuit as shown in Fig. 6 and trigger the scope externally from the signal generator.

2. Set the frequency near fr adjust the signal generator amplitude for V in channel 1.

3. Vary the frequency f on both sides of fr by about 300 Hz. Adjust the signal generator amplitude to

keep Vrms (i.e. Y1) constant as f and Vrms S change. Since Irms Vrms S, plot Vrms S versus f. Note

the resonant minimum in Irms at fr.

Results and Graph:




Experiment:08

LOAD TEST ON DC SHUNT MOTOR

Aim: To determine the efficiency of a DC shunt motor by conducting brake test.

Apparatus:

Theory:

The precondition to be set for the load test on DC shunt motor is to run the motor at the rated

voltage and the rated speed. For Small motors the efficiency can be found directly by a brake test. The

loading arrangement done to the motor is that a brake drum is attached to the shaft of the motor and

spring balances are connected through which the brake drum is tightened so that the shaft is loaded. This

set is said to be called as applied mechanical load. The torque can be determined and speed is measured

from which the power out put can be calculated. The input to the motor is found by knowing the applied

voltage and load current. Hence the efficiency can be known

Let ‘S1’ and ‘S2’ are the spring balance readings.

The pull on the brake drum = 9.81 (S1-S2) Newton

Torque on the drum= Tsh = 9.81 (S1-S2) r Nw-m. Where ‘r’ is the radius of the drum

Motor power output = Tsh * 2π N/60 watts; where N is the rpm of the motor

Let input voltage and current are V and I

Power input to the motor is V*I

The efficiency = η = output/ input

Voltmeter 0 – 300V (MC) 1 No

Ammeter 0 – 30A (MC) 1 No


Tachometer 1 No




Procedure:

1. Give the connections as per the circuit diagram.

2. Start the motor using the starter.

3. Increase the load by tightening the brake band and note the observations of the ammeter,

voltmeter, tachometer and the applied loads S1and S2 till rated current is attained. Take at

least six readings

4. Unload the motor by slackening the brake band.

5. Switch off the supply to the motor by opening the DPST switch. Find the radius of the Brake

drum.




Precaution:

Before starting the experiment pour some water into the brake drum and also while doing the

experiment.

Stay away from the brake drum when switching off the motor.

Tabular column:

(“r” is the radius of the brake drum)

S no. V

(Volt)

IL

(Amp)

N

(rpm)

S1

(Kg)

S2

(Kg)

W=S1-S2

(Kg)

T= 9.81

W r

(Nm)

Output=

2πNT/60

(watts)

Input

VIL (Watts)

η =

output/

Input (%)




Expected graph:

Results and discussion: (to be written in the main laboratory)

viva Questions:

1. What are the methods for finding the efficiency? 2. What are the basic requirements to conduct the load test? 3. Compare the load characteristics for different types of DC motors. 4. If two motors are required to drive a common load, how will they share the total load?




Experiment: 09

MAGNETISATION CHARACTERISTICS OF SEPAERTELY EXCITED DC GENERATOR

Aim: To obtain the open circuit magnetization characteristics (OCC) of a DC shunt generator and to find the

following

a) Maximum Voltage built up b) critical field resistance c) critical speed

d) residual magnetism

Apparatus::

Theory:

This characteristic is also called as no load characteristic through which the generator performance

parameters can be determined. This characteristic gives the value of maximum voltage the generator can

give and to avoid the failure of excitation, the field winding resistance value which is called as critical field

resistance can be determined. Also, the above parameters at various speeds can be determined. The basic

set up for determining the above parameters is that the generator is run on separately excited condition.

The basic requirement is that the prime mover ,the motor, is run at its rated speed and by varying the

generator excitation in steps, the generator voltage is noted and the procedure is repeated in forward as

well as reverse direction so that an observation can be made such the both the induced values will not be

same.

Voltmeter 0 – 300V (MC) 1 No


Field regulator 600Ω /1.7A 1 No




Circuit diagram:

Procedure:

1. Make the connections as per the circuit diagram. Keep the field regulator in the generator field circuit in the maximum resistance position.

2. Start the motor with the starter and note the speed at which the MG set runs… 3. Keep the field switch open and note the voltmeter reading. 4. Close the field switch and note the voltmeter and ammeter readings. 5. Increase the field current by reducing the field regulator resistance in steps and note the

voltmeter and ammeter readings at each step. 6. Once rated voltage of the generator is reached start decreasing the field current in steps and

note the voltmeter and ammeter readings at each step. 7. Bring the generator field regulator to the maximum resistance position and open the field

switch. 8. Switch off the supply to the motor.

Precautions: While changing the field current to and fro motion must be avoided, i.e. either increase or

decrease continuously.




Tabular column:

Sample calculations:

S.No If( Gen filed

current)

Induced voltage

Vg

( Increasing)

Vg

( decreasing)

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.




Plot the graph of no load EMF (E0) Vs Field Current (If). Draw a tangent to the ascending curve. The

slope of the tangent gives the critical resistance.

Results and Graph: (to be written in the main laboratory)

viva Questions:

a) What are the merits of separate excitation b) If the generator does not build up voltage, what are the possible causes/ c) How are two generators operated in parallel? d) What are the different methods of excitation?




Experiment: 10

SPEED CONTROL OF DC SHUNT MOTOR

Aim: To control the speed of a DC shunt motor by armature control method and field control

method.

Apparatus:

Voltmeter 0 – 300V (MC)and 0 -15V (MC) 1 No

Ammeter 0 – 2A (MC)and 0 – 5A (MC) 1 No

Rheostat 40 Ω 5A and 600Ω 1.7A 1 No

Theory:

Speed control methods:

The basic speed control methods are flux control methods, and Armature control method.

The advantages of flux control method are that it provides smooth and easy control and above rated

speed is possible. As the field winding resistance is high, the field current is small and hence the power

loss in the external resistance is very small which makes the method more economical and efficient. As

filed current is small, the size of the resistance required is small. The disadvantages of flux control method

are that below the rated speed is not possible and at high speeds, the commutation problems are severe.

The advantages of armature control methods smooth speed control below the normal speed is possible

and the disadvantage is as the entire armature current passes through the external resistance, there is

huge power loss. The armature current is more than the field current and the rheostat required is of large

size and large power loss results. The above rated speed is not possible by this method.




Circuit Diagram

Armature Voltage control Method

Connect the circuit as per the circuit diagram given.

Run the motor using starter.

Keep the field current constant and vary the terminal voltage by varying the resistance in the armature circuit in steps of 5V

Note down the terminal voltage and speed.




Field Control Method

Keep the terminal voltage constant and vary the field current by changing the resistance connected in the field circuit in steps.

Note the field current and speed

Tabular column:

Armature control Method Field Control Method

S.No. Vg

(Volts)

N

(Speed)

S.No. If N

(Speed)

1.

1.

2. 2.

3. 3.

4 4

5 5

6 6

7 7

8 8

9 9

10 10




Graphs to be obtained are

Draw the graph of (a) Speed Vs armature voltage (b) Speed Vs Field current

Sample graphs:

Results and discussion: (to be written in the main laboratory)

Viva Questions:

1. Compare the efficiency of Generator and Motor .Comment on the result.

2. Can the Swinburne’s test be performed on a series motor?

3. Why is it not possible to get higher speeds with armature voltage method?

4. Can lower speeds be obtained by using field control method?

5. What are the disadvantages of armature and field control methods?




Experiment: 11

SWINBURNE’S TEST ON DC SHUNT MOTOR

Aim: To control the speed of a DC shunt motor by armature control method and field control

method.

Apparatus:

Swinburne’s test or no load test is indirect method of testing the dc motors in which the flux

remains practically constant .In this test, actually without loading the motor the motor losses and hence

the efficiency at different loads can be predetermined. In this method, the motor is run on no load at its

rated voltage. At the starting, the motor is run at rated speed and the noload armature current Ia is noted.

The power fed tpo the motor is V(Ia + Ish).

The field copper loss is Ish2 Rsh and armature copper loss Ia

2 Ra.

The total copper losses are Ish2 Rsh + Ia

2 Ra.

The constant losses are Wconst = V(Ia + Ish).- Ish2 Rsh + Ia

2 Ra

These constant losses are remaining constant even it is loaded.

If the full load current is IL , the armature current flowing through is IL- Ish,

Full load copper losses are Ish2 Rsh + Ia

2 Ra

The constant losses are already known.

Hence the total losses are Constant losses + Full load copper losses.

The input to the motor if the machine is motor is V IL

The output of the motor is input _ losses

The efficiency of the motor is the ratio output to the input.

Voltmeter 0 – 300V (MC)and 0 -15V (MC) 1 No

Ammeter 0 – 2A (MC)and 0 – 5A (MC) 1 No

Rheostat 40 Ω 5A and 600Ω 1.7A 1 No




The advantage of this test is the efficiency of the machine either motor or generator can be

predetermined at any load without conducting the actual test. The disadvantage is that the constant

losses are not really constant but varies from no-load to full load and the efficiency computed can

reliable.

Circuit Diagram




Swineburne’s method:

S.No. Iao

(amps)

Ifo

(amps)

V

(volts)

N

Rpm

1

For finding armature resistance

S. No. V (volts) I (amperes) Ra=V/I(Ω)




Determination of constant losses:

No load current ILo = Iao+ Ifo

No load input = V* ILo

W const= Constant losses = V * ILo – Iao2 Ra

Assume suitable values of the load current

EFFICIENCY AS A MOTOR

SNo.

V

IL

If

Ia=

IL- If

I2R

losses

Pcu=

Ia2 Ra

Const.

losses

W const

Total

Losses

Pcu+ Pcon

Input

V IL

Output = input

– total losses

η = output

/input

(%)




EFFICIENCY AS A GENERATOR

Results and Graph: (to be written in the main laboratory)

Viva Questions:

1. Compare the efficiency of Generator and Motor .Comment on the result.

2. Can the Swinburne’s test be performed on a series motor?

3. Why is it not possible to get higher speeds with armature voltage method?

4. Can lower speeds be obtained by using field control method?

5. What are the disadvantages of armature and field control methods?

SNo. V

IL If

Ia=

IL+- If

Armature

coppeer

losses

Pcu= Ia2 Ra

W

const

Total

Losses

Pcu+ Pcon

watts

Output

= VIL

watts

Input =

output

+total

losses

watts

η =

output

u/input

(%)




Experiment: 12

OPEN CIRCUIT AND SHORT CIRCUIT TESTS ON A SINGLE PHASE TRANSFORMER

Aim:

1. To predetermine the efficiency of a single phase transformer, 2. To obtain the equivalent circuit of the transformer and 3. to find the regulation of the transformer

Apparatus:

Voltmeter 0 – 150 V and 0 – 60V (MI) 1 No

Ammeter 0 – 1A (MI)and 0 – 15A (MI) 1 No

Wattmeter (LPF) 1/2A,300V 1 No

Wattmeter (LPF) 10/20A,75V 1 No

Transformer 230/115V, 3 KVA 1 No

Theory:

These two tests open Circuit test (O.C test) and Short Circuit Test( S.C test )are performed to find

the no load magnetizing loss and full load copper loss of a single phase transformer. The open

circuit test gives the iron loss. The test parameters give the core values of core resistance and

reactance is determined. The short circuit test gives the copper loss and in turn gives winding

resistance and reactance is determined. The equivalent circuit of a transformer can be deduced

from the results of these tests. From the OC and SC tests the performance of the transformer at

different power factors and load conditions can be determined. The maximum efficiency can be

determined. .




Circuit Diagram:




Procedure:

OPEN CIRCUIT TEST (O.C.TEST)

1. Make the connections as per the circuit diagram, the 230V winding of the transformer is kept open

2. Apply the rated voltage,i.e.115V through the auto transformer. 3. Note down the voltmeter Voc, ammeter Ioc and wattmeter Woc readings and tabulate as

shown . The wattmeter used in the OC test should be low power factor wattmeter, since it must be able to

measure power at low power factor at which the transformer works on no load,

SHORT CIRCUIT TEST (S.C TEST)

1. make the connections as per the circuit diagram and keep the 115V winding of the transformer short circuited

2. Apply the low voltage to the 230V side through the auto transformer and increase the voltage gradually till the full load current (13A) flows in the 230V winding.

3. Note down the voltmeter, ammeter and wattmeter readings and tabulate as shown

Tabular column:

OC Test:

V sc (Volts) I sc (amperes) P sc (watts)




SC Test:

To draw Equivalent circuit:

P0 = Iron Loss = I0 V0 Cos Ø0

Cos Ø0 = P0/( V0 X I0), Ø0 = Cos-1 P0/( V0 X I0)

R0 = V0 /( I0 Cos Ø0) = V0 / Iw

X0 = V0 /( I0 Sin Ø0) = V0 /Iµ

Psc = Copper Loss = I2sc X R01

R01= Psc/ I2sc

Z01 =Vsc/Isc

Vo Volts) Io (amperes) Po (watts)




X01=

Sample Graph:

Load at which max efficiency occurs is the same whatever the power factor, However numerical value of

“η” decreases with decrease in P.F

TO CALCULATE THE EFFICIENCY AT U.P.F/0.8 PF/0.6 PF

SI.No Load

Current

IL

Iron

Loss

P0

Copper

Loss

Psc

Total

Loss P0+

Psc

Output

KVA X

P.f

Input =

Output-

Losses

η=Output/Input




Plot the graph Output Vs

1) Efficiency 2) Iron Loss 3) Cu.Loss

And from the graph find the condition for efficiency to be maximum

To calculate the percentage regulation at UPF/0.8 lag/0.8 lead

% Regulation = 1

01011 )sin(

V

XCosRI

(or)

2

02022 )sin(

V

XCosRI

Where V1 = Primary Rated voltage

I1 = Rated Primary Current




Positive sign for Lagging Power factor

Negative sign for Leading Power Factor

Results and Graph:

Viva Questions:

1) Explain why the wattmeter reading in O.C Test is taken as Iron Loss?

2) Explain why the wattmeter reading in S.C Test is taken as Copper Loss?

3) What re the uses of transformers, explain with example?

4) Why the efficiency of the transformer is high as compared to the electrical motor?

5) What are the materials used for making the core and winding of the transformer?

6) Explain why those materials are used?

7) What do you understand by an Auto-transformer?

8) Why transformer rating is in KVA not KW.

9)What is the all day efficiency of a transformer.

ee lab manual

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