machines manual 2016

7/24/2019 Machines Manual 2016

1/87

Electric Machines Lab EED, UET, Lahore 0


2/87

Sumpner Test / Back to Back Test of a Transformer


Experiment 1


Equipment

1. Two identical transformer

2. Two varaics

3. Two AC measuring panels

4. Calculator

Theoretical Background

The Sumpner test is a method of determining efficiency, regulation and heating under load

conditions. The open circuit and short circuit tests give us the equivalent circuit parameters

but cannot give heating information under various load conditions. The Sumpner test

provides copper loss, iron loss and heating information in a single test. In open circuit test,

there is no load on the transformer while in short circuit test also only fractional load gets

applied. In open circuit and short circuit tests, the loading conditions are absent and the

results are inaccurate. In Sumpner test, actual loading conditions are simulated hence the

results obtained are much more accurate. Thus Sumpner test is much improved method of

predetermining regulation and efficiency than open circuit and short circuit tests.

The Sumpner test requires two identical transformers. Both the transformers are connected to

the supply such that one transformer is loaded on the other. Thus power taken from the

supply is that much necessary for supplying the losses of both the transformers and there is

very small loss in the control circuit.

While conducting this test, the primaries of the two identical transformers are connected in

parallel across the supply V1. While the secondaries are connected in series opposition so that

induced EMFs in the two secondaries oppose each other. The secondaries are supplied from

another low voltage supply, connected in each circuit to get the readings. The connection

diagram is shown in the Figure below.


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.

T1 and T2 are two identical transformers. The secondaries of T1and T2 are connected in series

opposition. So EEF = EGH i.e. EMFs induced in two secondaries are equal but the secondaries

are connected such that E is connected to G and F is connected to H. Due to such series

opposition, two EMFs act in opposite direction to each other and cancel each other, so net

voltage in the local circuit of secondaries is zero, when primaries are excited by supply 1 of

rated voltage and frequency. There is no current flowing in the loop formed by two

secondaries. The series opposition can be checked by another voltmeter connected in the

secondary circuit as per polarity test. If it reads zero, the secondaries are in series opposition

and if it reads double the induced EMF in each secondary, it is necessary to reverse the

connections of one of the secondaries.

As per superposition theorem, if V2 is assumed zero then due to phase opposition no current

flows through secondary and both the transformers T1, T2 are as good as on no load, so open

circuit test gets simulated. The current drawn from source V1 in such case is 2Io where Io is no

load current of each transformer. The input power as measured by wattmeter W 1 thus reads

the iron losses of both the transformers.

Pi per transformer =W1/2 as T1, T2 are identical

Then a small voltage V2 is injected into the secondary with the help of low voltage

transformer, by closing the switch S. With regulation mechanism, the voltage V2 is adjusted

so that the rated secondary current I2 flows through the secondaries as shown. I2 flows from E

to F and then from H to G. The flow of I1 is restricted to the loop B A I J C D L K B and it

does not pass through W1. Hence W1 continues to read core losses. Both primaries and

secondaries carry rated current so short circuit test condition gets simulated. Thus the

wattmeter W2 reads the total full load copper losses of both the transformers.


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(Pcu full load) per transformer = W2/2

Key Point: Thus in the Sumpner test without supplying the load, full iron loss occurs in the

core while full copper loss occurs in the windings simultaneously. Hence heat run test can be

conducted on the two transformers. In open circuit and short circuit test, both the losses do

not occur simultaneously hence heat run test cannot be conducted. This is the advantage of

Sumpner test. From the test results the full load efficiency of each transformer can be

calculated as

Where Output = VA rating x cos 2

Key Point: As all the voltage, currents and powers are measured during the test, the

equivalent circuit parameters also can be determined. Hence the regulation at any load and

load power factor condition can be predetermined.

The only limitation is that two identical transformers are required.

Observations

No load voltage = Rated voltage = ________________V

No load current = 2 Io = ________________________A

No load power = Core losses of two transformer = W1 = _______________ Watt

Secondary side voltage (reduced voltage) = __________V

Rated current in secondaries = ____________________A

Power in secondaries -= Copper losses = ___________W

Iron loss of one transformer = _____________W

Copper loss of one transformer = ___________W

Total losses of one transformer = ___________W


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Calculate efficiency at full load at unity power factor

22

2

(VA rating) cos% 100 with cos 1

(VA rating) cos at full loadi cux

P P

= =

+ +

Calculate efficiency at half load at 0.8 power factor.

At half load, cos 2 = 0.8 and n = 1/2 = 0.5

2

2

2

2

n x (VA rating) cos% 100

n x (VA rating) cos ( at full load)

at half load= ( at full load) where n is the fraction of full load

i cu

cu cu

xP n P

P n P

=

+ +

Draw equivalent circuit and calculate all its parameters


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Comments

1. How you can determine the efficiency of a transformer?

2. Which winding is connected in parallel in this test?

3. How much voltage is applied on primary side while performing Sumpner test?

4. How much voltage is applied on secondary side while performing Sumpner test?

5. How secondaries of transformers are connected in this test?

6. How much current flows in primary and secondary winding while performing this

test?

7. What do you mean by phase opposition in reference to Sumpner test?

8. What is the necessary condition to be fulfilled by transformers in this test?


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8/87

Speed Control of a Three Phase Induction Motor


Vab Vbc Vca Average

Voltage

Speed in

RPM

Speed in

rad/sec

1

2

3

4

5

6

7

8

9

10

Frequency

Hz

Speed in

RPM

Speed in

rad/sec

Frequency

Hz

Speed in

RPM

Speed in

rad/sec

1 11

2 12

3 13

4 14

5 15

6 16

7 17

8 18

9 19

10 20


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Speed Control of a Three Phase Induction Motor


Plot graph between average voltage and speed of induction motor (Speed on y-axis)

Plot graph between supply frequency and speed of induction motor. (Speed on y-axis)

What is the synchronous speed?

Can we run induction motor above the synchronous machine?

What do you mean by slip? What are different methods to measure speed?

Briefly explain the method of speed measurement in this experiment.

What do you mean by an induction motor drive?

Draw internal block diagram of the induction motor drive

\


10/87

Voltage Regulation of a Transformer


Experiment 3

Voltage Regulation of a Single Phase Transformer with resistive, inductive and

capacitive load

Equipment

1. Two terminal boards consisting of voltmeter, ammeter and wattmeter

2. A varaic

3. A single phase transformer

4. Resistive load

5. Inductive load

6. Capacitive load

Procedure

1. First of all connect the primary of the transformer to the variable AC source available

in the Lab. This is so because we want to adjust the primary of the transformer at the

rated voltage and find the voltage regulation at the rated voltage. If WAPDA voltage

is less than the rated voltage of the transformer then set the primary voltage at 200 or

less but that should be maintained constant during the experiment.

2. At rated primary voltage there should be rated secondary voltage. If primary voltage

is different from rated then secondary will also be different from the rated. It is

recommended to set the rated voltage at secondary. Note down the value of secondary

voltage at no load.

3. Connect resistive load that consist of bulbs to the secondary of the transformer.

Connect an ammeter and a voltmeter across load to measure current and voltage

respectively.

4. Gradually vary the load by turning on the bulbs & take readings by noting down the

corresponding load currents (increasing Load currents) & voltages at that time. Take

about twelve readings.

5. Calculate the percentage regulation of the transformer at different loads by using the

formula

% Voltage Regulation = 100nl load

nl

V Vx

V


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Where Vnl is the secondary voltage at no load and VLoad is the secondary voltage with

load

6. Plot a graph between percentage regulation & load current.

7.

Repeat all the above steps for inductive and capacitive load.

Observations

1. Resistive load

Sr.

#

Primary

Voltage Vp

(V)

Kept

constant

Load

Current

IL

(A)

Secondary

(Load) Voltage

VL

(V)

No load

Secondary

Voltage

Vnl

(V)

Percentage

Voltage

Regulation

100nl L

nl

V Vx

V

1

2

3

4

5

6

7

8

9

10


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2. Inductive Load

Sr.

#

Primary

Voltage Vp

(V)

Kept

constant

Load

Current

IL

(A)

Secondary

(Load) Voltage

VL

(V)

No load

Secondary

Voltage

Vnl

(V)

Percentage

Voltage

Regulation

100nl L

nl

V Vx

V

1

2

3

4

5

6

7

8

9

10


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3. Capacitive load

Sr.

#

Primary

Voltage Vp

(V)

Kept

constant

Load

Current

IL

(A)

Secondary

(Load) Voltage

VL

(V)

No load

Secondary

Voltage

Vnl

(V)

Percentage

Voltage

Regulation

100nl L

nl

V Vx

V

1

2

3

4

5

6

7

8

9

10


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Draw phasor diagram for secondary side of transformer with resistive load

Draw phasor diagram for secondary side of transformer with inductive load

Draw phasor diagram for secondary side of transformer with capacitive load.

Plot graph between load current and % voltage regulation for resistive load (%VR ony-axis)

Plot graph between load current and % voltage regulation for inductive load (%VR on

y-axis)

Plot graph between load current and % voltage regulation for capacitive load (%VR

on y-axis)

What relationship can be observed in the graphs? Which graph has more slope and

why?

What can be done to keep % Regulation as minimum as possible?

What is the best ideal Voltage Regulation value and why? Explain

What difference is observed between resistive and capacitive load voltage regulation.


15/87

Equivalent Circuit of a Transformer


Experiment 4

Equivalent circuit of a single phase transformer using open circuit and short circuit test

Equipment

1. Two terminal boards consisting of voltmeter, ammeter and wattmeter

2. A varaics

3. A single phase transformer


It is possible to experimentally determine the values of the inductances and resistances in the

transformer model. An adequate approximation of these values can be obtained with only two

tests, the open-circuit test and the short-circuit test. In the open-circuit test, one transformer

winding is open-circuited, and the other winding is connected to full rated line voltage. Look

at the equivalent circuit in Figure.

Under the conditions described al1 the input current must be flowing through the excitation

branch of the transformer. The series elements, Rpand Xpare too small in comparison to Rc

and XM to cause a significant voltage drop, so essentially all the input voltage is dropped

across the excitation branch. The open-circuit test connections are shown in Figure. Full line

voltage is applied to one side of the transformer, and the input voltage, input current, and


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input power to the transformer are measured. (This measurement is normally done on the low

voltage side of the transformer, since lower voltages are easier to work with.) From this

information, it is possible to determine the power factor of the input current and therefore

both the magnitude and the angle of the excitation impedance.

The easiest way to calculate the values of Rcand XMis to look first at the admittance of the

excitation branch. The conductance of the core-loss resistor is given as

1c

c

GR

=

and the susceptance of the magnetizing inductor is given as

1M

M

BX

=

Since these two elements are in parallel, their admittances add, and the total excitation

admittance is

1 1

E c M

E

c M

Y G jB

Y jR X

=

=

The magnitude of the excitation admittance (referred to the side of the transformer used for

the measurement) can be found from the open-circuit test voltage and current:

ocE

oc

IY

V=

The angle of the admittance can be found from a knowledge of the circuit power factor. The

open-circuit power factor (PF) is given by

cosoc

oc oc

PPF

V I= =

1cosoc

oc oc

P

V I

=

The power factor is always lagging for a real transformer, so the angle of the current always

lags the angle of the voltage by degrees. Therefore, the admittance YEis


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( )1cos

ocE

oc

ocE

oc

IY

V

IY PF

V

=

=

By comparing two equations for YE , it is possible to determine the values of Rc and XM

referred to the connected side directly from the open-circuit test data.

In short-circuit test, the secondary winding terminals of the transformer are short-circuited,

and the primary winding terminals are connected to a variable voltage source. The input

voltage is adjusted until the current in the short-circuited windings is equal to its rated value.

(Be sure to keep the primary voltage at a safe level. It would not be a good idea to burn out

the transformer's windings while trying to test it.) The input voltage, current, and power areagain measured. Since the input voltage is so low during the short-circuit test, negligible

current flows through the excitation branch. If the excitation current is ignored, then the

entire voltage drop in the transformer can be attributed to the series elements in the circuit.

The magnitude of the series impedances referred to the primary side of the transformer is

scsc

sc

VZ

I=

The power factor of the current is given by

cossc

sc sc

PPF

V I= =

and is lagging. The current angle is thus negative, and the overall impedance angle is

positive: Therefore,

1

cossc

sc sc

P

V I

=

The series impedance ZSEis equal to

eRSE q eqZ jX= +

2 2p p(R a ) (X a )SE s sZ R j X= + + +

It is possible to determine the total series impedance referred to any side by using this

technique, but there is no easy way to split the series impedance into primary and secondary


18/87



components. Fortunately, such separation is not necessary to solve normal problems. Note

that the open-circuit test is usually performed on the low-voltage side of the transformer, and

the short-circuit test is usually performed on the high voltage side of the transformer, so Rc

and XMare usually found referred to the low-voltage side, and Reqand Xeqare usually found

referred to the high-voltage side. All of the elements must be referred to the same side (either

high or low) to create the final equivalent circuit.

Circuit diagram for open circuit test

Connect the circuit as shown in the figure. Use 220 V tap on the primary side.

Observations

No load primary voltage Voc= _______________ (It should be rated voltage of the primary

side)

No load primary current Ioc

= _________________


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No load Power Poc= ________________________ (Iron losses of the transformer)

No load power factor = cosoc

oc oc

PPF

V I= = =___________________

No load power factor angle = 1cosoc

oc oc

P

V I

=

=________________

Iw(current through Rc) = cosocI =____________

Im(current through Xm) = sinocI =____________

occ

w

VR

I= =__________

ocm

m

VX

I= =__________

Alternative way

ocE

oc

IY

V= =________________

ocE

oc

IYV

= =_____________

E c MY G jB= =_____________

As1 1

E

c M

Y jR X

= so1

c

c

RG

= =___________ and1

M

M

XB

= =_____________

Circuit diagram for short circuit test

Connect the circuit as shown in figure. Make sure at start the applied voltage is zero.

Gradually and carefully increase the voltage on the primary side such that rated current flows

in primary side.


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Observations

Short circuit: Primary voltage Vsc= _______________ (It should be very small voltage)

Short circuit: Primary current Isc= ________________ (It should be the rated current)

Short circuit: Secondary current= _________________ (Slightly lower than above)

Cu-Losses: Power Psc= ________________________

Short circuit power factor = cossc

sc sc

PPF

V I= = =___________________

Short circuit power factor angle = 1cossc

sc sc

P

V I

=

=________________

scE

sc

VZs

I= =__________________

scE

sc

VZs

I= =________________

eRSE q eqZ jX= + =______________

eR q =________________________ & eX q =________________________

Or2

eRsc

q

sc

P

I= =___________________

22e eX Rq SE qZ= =___________


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Xeq=_________________________

Draw the equivalent circuit referred to the primary side and clearly mention all the

values

Draw the equivalent circuit referred to the secondary side and clearly mention all the

values (a=Is/Ip in short circuit test) a=____

How we can minimize the no-load current of the transformer?

In which case the PF of transformer is higher, open circuit test or short circuit test?

Why?

Teacher Signature: ________________

Date:_______________


22/87

Speed Control of a DC motor


Experiment 5

Speed Control of Separately Excited and Shunt DC Motor

Equipment

1. Terminal Board

2. DC machine

3. Tacho-meter

4. Two varaics

5. Variable resistance

Speed Control of Separately Excited Motor

Voltage Control Method

In this method of speed control field winding current is kept constant by keeping the field

voltage constant as separate supply is used to excite field winding. The applied armature

voltage is changed from 0 to 160V and the corresponding speed variation is noted down.

Speed of DC motor is directly proportional to the armature voltage. Field Current is kept

constant at 0.6 A. (Field current varies from machine to machine)

Field Current Control Method

For field control method, armature voltage is kept constant at nearly 120-140V and the field

current Ifis changed by varying the resistance added in series with the field winding. Field

current Ifis changed such that the speed varies between 500 rpm and 1800 rpm. Speed of the

DC motor is inversely proportional to the field current.

Speed Control of Shunt Motor

In this motor only one supply is used to energize field and armature winding. Field and

armature winding are connected in parallel.

Voltage Control Method

In this method of speed control field winding current is kept constant with the help of variable

resistance connected in series with field winding because the field voltage also changes with


23/87



change of armature voltage. Armature voltage is changed from 50 to 150 V such that only

armature current changes and field current remains constant and corresponding speed is noted.

Field Current Control Method

For field control method, armature voltage is kept constant at nearly 120-140V and the field

current Ifis changed by varying the resistance added in series with the field winding. Field

current Ifis changed such that the speed varies between 500 rpm and 1800 rpm. Speed of the

DC motor is inversely proportional to the field current.

Procedure

1.

Connect armature winding and field winding with separate supplies. Use volt-meter and

ammeter of the terminal board to measure the voltage and current of respective

winding. First turn on the field supply and set its voltage nearly 100V. Then turn on the

armature supply and gradually increase the armature voltage and note down the speed at

different voltages.

2. Then keep armature winding voltage and field winding voltage constant and change the

field current using the variable resistance connected in series with the field winding.

Note down the speed of the machine and record in the table.

3. To turn off your machine always switch off field circuit first and then switch off the

armature supply. Reverse procedure can cause machine to accelerate quickly which can

result in mechanical damage and due to reduced back EMF the armature current will be

very high which can result in electrical damage.

4. Now connect armature and field windings in parallel and repeat the above procedure.

Only one supply will be used in this case.


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Observations

1. Voltage Control Method for separately excited machine

Sr. # Armature Voltage (V)

Va

Speed (rpm)

N

Speed (rad/sec)

W=2N/60

Field Current (A)

If(constant)

1

2

3

4

5

6

7

8

9

10

11

12

Draw Equivalent circuit used for above experiment.


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2. Field Current Control Method for separately excited machine


Va(constant)

Speed (rpm)

N

Speed (rad/sec)

W=2N/60

Field Current (A)

If

1

2

3

4

5

6

7

8

9

10

11

12



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3. Voltage Control Method for shunt type machine


Va

Speed (rpm)

N

Speed (rad/sec)

W=2N/60

Field Current (A)

If(constant)

1

2

3

4

5

6

7

8

9

10

11

12



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4. Field Current Control Method for shunt type machine


Va(constant)

Speed (rpm)

N

Speed (rad/sec)

W=2N/60

Field Current (A)

If

1

2

3

4

5

6

7

8

9

10

11

12

Draw Equivalent circuit used.


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Plot the graph between speed and field current for separately excited and shunt machine

(speed on y-axis and field current on x-axis).Clearly indicate the scale and units.

Plot the graph between speed and armature voltage for separately excited and shunt

machine (speed on y-axis and armature voltage on x-axis).Clearly indicate the scale and

units.

Comments

1. What characteristics do you observe from the speed vs armature voltage graph? Can

this method be used above rated armature voltage?

2. What characteristics do you observer from the speed vs field current graph? What will

happen if you reduce field current to minimum value?

3. What can be other methods to change the speed of Shunt Motor?

Instructor Signature:__________________


29/87

Magnetic Characteristics of a DC generator


Experiment 6

Magnetic Characteristics of a DC Shunt and Separately Excited Generator

Equipment

1. Terminal Board

2. DC motor generator set

3. Tacho-meter

4. Two varaics

5. Variable resistance

Procedure

1. Make connection for one machine as motor. Motor connection can be of shunt

type or of separately excited type. It is your choice.

2. Connect a voltmeter at armature terminal of second coupled machine which will

be used as a generator. As we are going to perform the test on a shunt generator,

field and armature are kept in parallel.

3. Connect field winding of DC generator in parallel to armature winding such that

there is a variable resistance and ammeter in series with field winding. Select the

meter of proper range (should not be more than 1A as field current is usually very

small)

4. Run your first machine as a motor by applying armature voltage such that

machine runs at 1500 rpm (you can set it at any value in the range of 100 to 160V

but must be kept constant)

5. Now gradually vary the variable resistance in the field of DC shunt generator.

Observe the reading of the voltmeter connected across armature winding. Voltage

must be induced in armature winding. Keep in mind that there is no load

connected with armature.

6. Now gradually change the variable resistance such that the field current increases

and note down the armature voltage. Please select such an interval that you can

record at least 10 readings over the entire range of variable resistance.

7. Once readings are obtained with increasing field currents, now decrease the

currents with same values and note down the armature voltage. You will observe


30/87



that armature voltage will be higher than the case when you were increasing field

current and this is because of residual magnetic field.

8. Note the readings in table and draw the magnetic characteristics.

9. Repeat the experiment for separately excited generator.

Observations

Sr

#

Field Current

If(mA)

Voltage Generated Vg

(V)

Ascending Filed Current Descending Field Current

1 0

2

3

4

5

6

7

8

9

10

11


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Why there is a difference between the ascending and descending values?

Which generator has the highest no load voltage and why?

Define magnetic characteristics

Is there any voltage generated at zero field current. If yes then what is the value and

why?

The drawn graph is between field current and generated voltage, explain how it

resemble magnetization curve?

On which factor the magnitude of induced EMF depends upon?

What is the saturation value of field current in your graph?

What do you mean by critical resistance?

What is the value of critical resistance in this experiment?


32/87

Load Test of a DC Series Motor


Experiment 7


Equipment

1. Terminal Board

2. DC series machine and DC generator

3. Tacho-meter

4. Load Unit


The circuitry of this motor includes an armature in series with an inductive field winding. The

relationship between speed & torque for a series motor can be given by

This equation shows that a high torque is obtained at low speed and a low torque is obtained

at high speed. This is a special characteristic feature of DC Series Motor. Due to this feature

DC Series Motor are used in such applications where large starting torque is required. It is

also notable in this motor that T I a2 which means that DC series will produce

unidirectional torque both for AC and DC supply. Thus it is a universal motor which can be

operated both on DC as well as AC supply. DC series motor has a dangerously high starting

current if it is directly connected to the DC power supply because of very low EMF at start.

The starting current is limited either by inserting a resistance or using a low DC terminal

voltage at start. DC Series motor has a very poor speed control as it is very slow for heavy

load and fast for lighter loads. If no load is applied, the motor may over-speed and destroy

itself. This feature makes this motor unsafe for using with belts for a brake as belt may cause

motor to be destroyed.


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Data

Without any connection take a DMM and connect it across different winding of the machines

and measure the ohmic resistance.

DC compound motor (First Machine)

Resistance of armature winding (A1-A2) =__________

Resistance of field winding (F1-F2) =___________

Resistance of series field (D1-D2)=___________

Write down the name plate data of machines

DC separately excited / shunt motor (Second Machine)

Resistance of armature winding (A1-A2) =__________

Resistance of field winding (F1-F2) =___________

Write down the name plate data of machines

Procedure

1. Setup consists of two DC machines which can be configured for different types.

Configure one machine as a DC series motor i.e. connects its series field winding (D1-

D2) in series with armature winding (A1-A2). For excitation use the supply with currentrating above 3A. Connect the circuit as shown in the circuit diagram.


34/87



2. Configure second machine as a separately excited DC generator. Use the supply of 2A

rating for the excitation of field winding (F1-F2) and connect load with the armature

winding.

3. Use voltmeter and ammeter measure the load voltage and current. Power can be

calculated by taking the product of two quantities. (You can use load unit or the bulbs for

the load).

4. For speed measurement use techo-generator and brake control unit.

5. Run DC series motor by gradually increasing the voltage up to 100V (Do not set voltage

above the rated value). Select a voltage for the operation of DC series motor and that

must be kept constant throughout the experiment. If voltage changes because of the load

then adjust it with the help of varaics.

6. Set the excitation voltage of separately excited DC generator at rated value. Observe the

voltage across the armature terminals/load terminal. You can use brake control unit to

apply the load on the DC generator.

7. Vary the load and record load power, torque and speed. If you are using the bulbs as a

load then record load voltage, load current and calculate the power. With the help of

readings calculate torque and efficiency.

8. Now draw the graph between the following quantities.

Output Power v/s Torque

Output Power v/s Speed

Output Power v/s Efficiency

Output Power on the x-axis and all other quantities on the y-axis.


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Connection Diagram


36/87


Electric Machines Lab EED, UET, Lahore

Sr. # Motor

Terminal

Voltage

Vt(V)

Motor

Current

Ia

Input

Power

P=VtIa

Motor

BEMF

Eb

Speed

N

(rpm)

Speed

w

(rad/sec)

Load

Voltage

VL

Load

Current

IL

Load

Power

Pout

VLIL

In

1

2

3

4

5

6

7

8

9

10


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Plot graph between speed and output power. (power on x-axis)

Plot graph between Induced EMF and output power (power on x-axis)

Plot graph between shaft torque and output power (power on x-axis)

Plot graph between efficiency and output power (power on x-axis)

What are the applications of series motor? (Mention at least five applications)

Is it necessary to maintain the constant input voltage while performing load test?

Can we run DC series motor at no load? If not, Why?

Why applied voltage does not remain constant in this experiment?

What is the important precaution in this experiment?

Is series motor a constant speed motor?

Why speed of a series motor decrease so drastically with load?

How does the torque of a series motor vary with load current? Write down the exact

expression.


38/87

Load Characteristics of the DC Shunt Generator


Experiment 8

Load Characteristics of a DC Shunt and Separately Excited Generator

Equipment:

1. Brake Control Unit SM26712. Drive Machine, DC Machine SM26413. Tacho-generator SM26314. Power Supply Unit SM26315. Terminal Board SM26356. Test Machine, DC Machine SM26417. Digital Multi-meter (DMM)8. Measuring Unit9. Shunt Regulator10.

Load Resistor SM2676

Background:

Shunt Generator is also called self-excited generator as field winding is connected across the

armature so that armature voltage can supply the field current. Some residual magnetism

must exist in the magnetic circuit of the generator. Because of this residual magnetism small

voltage appears across the armature terminal even when the field current is disconnected. On

connecting the field (a resistance in series), armature voltage increases which in turn increase

the field current which again builds up the voltage, so this process continues until a specific

voltage is reached. In actual cases, this build up follows approximately the magnetization

curve. Field Circuit resistance is important for voltage stabilization. If the resistance slope

coincides with somewhat linear portion of the magnetization curve it results in unstable

voltage situation. If resistance is greater than this, build up is insignificant. And for lower

values of resistance, generator will build up higher voltages. Shunt Generator uses a field

having suitable impedance across the armature such that the armature voltage supplies the

necessary field current, although in some cases, separate external voltage may also be

applied. Use a field current rheostat to adjust the current in field winding and to adjust the

armature terminal voltage. For simplicity, we assume magnetic linearity & find that

f fK I = (0.1)

For a DC shunt generator the voltage equation can be written as:

t a aE V I R= + (0.2)


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Where E is the induced EMF, Iais the armature current and Vtis the terminal/load voltage.

The induced EMF is given as:

b b f f f E K K K I KI = = = (0.3)

Speed relation can be formulated as:

t a at a a f t a a

f

V I RE V I R KI V I R

KI

+= + = + = (0.4)

Asa a

I R is very small therefore we can write above equation as:

t

f

V

KI = (0.5)

Thus speed varies inversely with field current. Speed also varies inversely with load. For load

torque T we have the following relation:

1am m

EI

T T = (0.6)

Important characteristics of a shunt generator are:

1.

Residual Magnetism is present in the magnetic field and armature system.

2. Field is wound such as to aid the residual magnetism.

3. Field circuit resistance is less than the critical field current resistance.

Procedure

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

2. Adjust the value of Rf (Shunt Regulator) and proper direction of field current or

direction of rotation of prime mover to aid the residual magnetism so that voltage

generation takes place.

3. Power up the circuit with rheostat (Load Resistor) at zero i.e. at no load and note the

voltage and field current for zero load current.

4. Now attach the load resistance to the generator with the resistance set to maximum.

Note the load voltage and field current.


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5. For next reading slowly decrease the resistance which increases the load current.

(Maximum load current is 2.2A). You can change the resistance value of the load as

prescribed in diagram to make your load current near 2A.

6. In this way take 10 readings of load current, field current & load voltage. (Take

readings at equal intervals)

7. Calculate the Armature current by adding both the load & field currents for each

reading.

8. Calculate the total induced emf produced by using the Raequal to 8.

9. Now draw the graph between voltage and load current. Voltage on the y-axis and load

current on the x-axis. Also draw a graph between E (Total emf induced) and the

armature current (E on the y-axis and Iaon the x-axis).

10.Repeat the experiment for separately excited generator

Precautions

1. Maximum allowable voltage to be generated by generator is 160V.

2. Maximum speed of the DC Motor acting as prime mover for the generator is 1500

rpm.

3. Take extra care while measuring field current by DMM. Use proper terminals and

range otherwise you may damage the DMM.

4. Use measuring unit or DMM supplied to measure load current and load voltage. Dont

use any other device to measure these.

5. If shunt regulator is not available, just dont use it. It is being used over here for

adjustment of field current to change the magnetic flux inside your machine.


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Observations

Circuit Diagram


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Plot graph between voltage and load current

Plot graph between E (EMF induced) and the armature current


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Load test of a three phase induction motor


Experiment 9

Components of voltage drop in a DC shunt generator

Equipment:

1. Brake Control Unit SM26712. Drive Machine, DC Machine SM26413. Tacho-generator SM26314. Power Supply Unit SM26315. Terminal Board SM26356. Test Machine, DC Machine SM26417. Digital Multi-meter (DMM)8. Measuring Unit9. Shunt Regulator10.

Load Resistor SM2676

Voltage Drop Components

There are three components of voltage drop in DC shunt generator which are as follows:

1. Voltage drop due to armature resistance: The armature current Iawhich is the sum of

the field current If& the Load Current ILflows through the armature thus due to the

armature resistance IaRavoltage drop due occurs.

2. Voltage drop due to armature reaction: Reaction between the armature field (due to

current Ia) and the main field (due to If) gives rise to two fields at some angle to each

other. Due to all this there is a shift in the magnetic axis and redistribution of charges

takes place in some of the conductors thus producing a significant voltage drop. This

voltage drop is named as the voltage drop due to armature reaction.

3. Voltage drop due to field current reduction: When the voltage drop due to the above

mentioned reasons take place, the terminal voltage decreases which causes the field

current to decrease further as field current is given by /f t fI V R= .Due to field current

decrease, the magnetic field strength decreases, thus less voltage is induced.

Load Characteristics in DC Generator

Following characteristics are observed for different types of DC Generator:


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Characteristics A - Ideal Characteristics

Characteristics B - Keeping the Ifconstant

Characteristics C - Accounts for all the three voltage drops

Procedure

1. Connect the circuit for the shunt generator using DC shunt motor as the prime mover.

Include the shunt regulator resistance in the field winding circuit with the milli-

ampere Meter.

2. Adjust the value of shunt regulator so that the field current has a value of 45mA. Note

down the no-load Voltage.

3. Add a variable resistive load of 200.

4. Start increasing the load current i.e. decreasing the resistance and note down the

values of load current IL, field current If,terminal voltage Vtfor different values of the

load current. The terminal voltage recorded has a drop for all the three components of

voltage drop as mentioned in the above text. Make the speed of the prime mover

constant at every set of readings to 1500 rpm by changing the applied voltage of the

motor.


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5. After taking the above set of readings, another set of readings is to be taken for the

same load currents ILbut making the field current constant at 45mA at every reading

by varying the shunt regulator included in the field circuit. In this way the terminal

voltage recorded do not account the drop due to field current Ifdecrease as the field

current is leveled at every reading.

6. In order to calculate the voltage drop due to armature reaction, the voltage drop due to

IaRais subtracted from the second set of readings terminal voltage for the same load

current IL. All these calculations are done in the third table.

7. Now draw the graph for all the three characteristics as shown in the above figure (2)

with load current on the x-axis and terminal voltage on the y-axis.

Precautions

DC Motor acting as prime mover for the generator should not be over powered or

underpowered so that it goes beyond its prescribed speed. A suitable operating speed should

be around 1500 rpm.

Observations

No Load Voltage = VtA= _______ V

Table 1

(Accounts for all the three voltage drop components)


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Table 2

(Accounts for the two voltage drop components i.e. armature reaction and IaRadrop as the

field current Ifis kept constant for every value)

Table 3

(Calculation for all three components of voltage drop)

Which component of the voltage drop is the largest and why?Armature Reaction

increases with load or not? Why?

Which of the following drops would be present in separately excited generator and

series Generator? Discuss with reason?

Write all types of Losses occurring in a DC Generator?


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Connection Diagram


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Experiment 10

Load Test of a Three Phase Induction Motor

Objective

To determine the load characteristics of an induction motor and to plot a graph between the

following quantities:

Output power and efficiency

Speed and efficiency

Power factor (cos ) and efficiency

Reactive power and efficiency

Torque and efficiency

Induction Motor

Principle

In ac motors the rotor does not receive the electrical power by conduction but by induction in

the same way as in case of secondary of a two-winding transformer receives its power from

primary. Thats why such motors are called as induction motors. In fact the induction motorcan be treated as rotating transformer i.e. one in which primary winding is stationary but the

secondary is free to rotate.

Where a poly phase electrical supply is available, the three-phase (or poly phase) AC

induction motor is commonly used, especially for higher-power motors. The phase

differences between the three phases of the poly phase electrical supply create a rotating

electromagnetic field in the motor. Through electromagnetic induction, the rotating magnetic

field induces a current in the conductors in the rotor, which in turn sets up a counterbalancing

magnetic field that causes the rotor to turn in the direction the field is rotating. The rotor

always rotates slower than the rotating magnetic field produced by the poly phase electrical

supply; otherwise, no counterbalancing field will be produced in the rotor.

Construction

An induction motor essentially consists of two main parts; a stator and a rotor.


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Stator

The stator is the outer body of the motor which houses the driven windings on an iron core.

In a single speed three phase motor design, the standard stator has three windings, while a

single phase motor typically has two windings. The stator core is made up of a stack of round

pre-punched laminations pressed into a frame which may be made of aluminum or cast iron.

The laminations are basically round with a round whole inside through which the rotor is

positioned. The inner surface of the stator is made up of a number of deep slots or grooves

right around the stator. It is into these slots that the windings are positioned. The arrangement

of the windings or coils within the stator determines the number of poles that the motor has.

A standard bar magnet has two poles, generally known as north and south. Likewise, an

electromagnet also has a north and a south pole.

As the induction motor stator is essentially like one or more electromagnets depending on the

stator windings, it also has poles in multiples of two. i.e. 2-pole, 4-pole, 6-pole etc. Greater

the no of poles slower will be the speed and vice versa. The stator winding ,when supplied

with 3 phase currents ,produces a magnetic flux which is of constant magnitude but revolves

at synchronous speed (given by Ns=120f/P).This revolving magnetic flux induces an emf in

the rotor by mutual induction.

The winding configuration, slot configuration and lamination steel all have an effect on the

performance of the motor. The voltage rating of the motor is determined by the number of

turns on the stator and the power rating of the motor is determined by currents in winding, the

losses which comprise copper loss and iron loss, and the ability of the motor to dissipate the

heat generated by these losses. The stator design determines the rated speed of the motor and

most of the full load, full speed characteristics.

Rotor

The Rotor comprises a cylinder made up of round laminations pressed onto the motor shaft,

and a number of short-circuited windings. The rotor windings are made up of rotor bars

passed through the rotor, from one end to the other, around the surface of the rotor. The bars

protrude beyond the rotor and are connected together by a shorting ring at each end. The bars

are usually made of aluminum or copper, but sometimes made of brass. The position relative

to the surface of the rotor, shape, cross sectional area and material of the bars determine the

rotor characteristics. Essentially, the rotor windings exhibit inductance and resistance, and


50/87



these characteristics can effectively be dependent on the frequency of the current flowing in

the rotor. A bar with a large cross sectional area will exhibit a low resistance, while a bar of a

small cross sectional area will exhibit a high resistance.

Likewise a copper bar will have a low resistance compared to a brass bar of equal

proportions. Positioning the bar deeper into the rotor, increases the amount of iron around the

bar, and consequently increases the inductance exhibited by the rotor. The impedance of the

bar is made up of both resistance and inductance, and so two bars of equal dimensions will

exhibit different A.C. impedance depending on their position relative to the surface of the

rotor. A thin bar which is inserted radially into the rotor, with one edge near the surface of the

rotor and the other edge towards the shaft, will effectively change in resistance as the

frequency of the current changes. This is because the A.C. impedance of the outer portion of

the bar is lower than the inner impedance at high frequencies lifting the effective impedance

of the bar relative to the impedance of the bar at low frequencies where the impedance of

both edges of the bar will be lower and almost equal. The rotor design determines the starting

characteristics.

Main types of rotor

1-Squirrel cage rotor

2-Phase wound or wound rotor

Squirrel Cage Rotor

Most common AC motors use the squirrel cage rotor, which will be found in virtually all

domestic and light industrial alternating current motors. The squirrel cage takes its name from

its shape - a ring at either end of the rotor, with bars connecting the rings running the lengthof the rotor. It is typically cast aluminum or copper poured between the iron laminates of the

rotor, and usually only the end rings will be visible. The vast majority of the rotor currents

will flow through the bars rather than the higher-resistance and usually varnished laminates.

Very low voltages at very high currents are typical in the bars and end rings; high efficiency

motors will often use cast copper in order to reduce the resistance in the rotor.

In operation, the squirrel cage motor may be viewed as a transformer with a rotating

secondary - when the rotor is not rotating in sync with the magnetic field, large rotor currents


51/87



are induced; the large rotor currents magnetize the rotor and interact with the stator's

magnetic fields to bring the rotor into synchronization with the stator's field. An unloaded

squirrel cage motor at synchronous speed will consume electrical power only to maintain

rotor speed against friction and resistance losses; as the mechanical load increases, so will the

electrical load - the electrical load is inherently related to the mechanical load. This is similar

to a transformer, where the primary's electrical load is related to the secondary's electrical

load. This is why, as an example, a squirrel cage blower motor may cause the lights in a

home to dim as it starts, but doesn't dim the lights when its fan-belt (and therefore mechanical

load) is removed. Furthermore, a stalled squirrel cage motor(overloaded or with a jammed

shaft) will consume current limited only by circuit resistance as it attempts to start. Unless

something else limits the current (or cuts it off completely) overheating and destruction of the

winding insulation is the likely outcome. Virtually every washing machine, dishwasher,

standalone fan, record player, etc. uses some variant of a squirrel cage motor.

Wound Rotor

An alternate design, called the wound rotor, is used when variable speed is required. In this

case, the rotor has the same number of poles as the stator and the windings are made of wire,

connected to slip rings on the shaft. Carbon brushes connect the slip rings to an external

controller such as a variable resistor that allows changing the motor's slip rate. In certain

high-power variable speed wound-rotor drives, the slip-frequency energy is captured,

rectified and returned to the power supply through an inverter.

Compared to squirrel cage rotors, wound rotor motors are expensive and require maintenance

of the slip rings and brushes, but they were the standard form for variable speed control

before the advent of compact power electronic devices. Transistorized inverters with variable

frequency drive can now be used for speed control, and wound rotor motors are becoming

less common. (Transistorized inverter drives also allow the more-efficient three-phase motors

to be used when only single-phase mains current is available, but this is never used in

household appliances, because it can cause electrical interference and because of high power

requirements.)

This type of motor is becoming more common in traction applications such as locomotives,

where it is known as the asynchronous traction motor.The speed of the AC motor is


52/87



determined primarily by the frequency of the AC supply and the number of poles in the stator

winding, according to the relation:

Ns = 120 f / p

where

Ns = Synchronous speed, in revolutions per minute

F = AC power frequency

p = Number of poles per phase winding

Slip

Actual rpm for an induction motor will be less than this calculated synchronous speed by an

amount known asslip, that increases with the torque produced. With no load, the speed will

be very close to synchronous. When loaded, standard motors have between 2-3% slip, special

motors may have up to 7% slip, and a class of motors known as torque motorsare rated to

operate at 100% slip (0 RPM/full stall).

The slip of the AC motor is calculated by:

S = (Ns Nr) / Ns

where

Nr = Rotational speed, in revolutions per minute.

S = Normalized Slip, 0 to 1.

Of all ac machines the poly phase induction motor is the one which is extensively used forvarious kinds of industrial drives.

Rotating magnetic Field

The fundamental principle of operation of AC machine operation is that if a 3-phase set of

currents, each of equal magnitude and displaced by 120 (a balanced 3-phase system) flows

in a 3-phase winding, then it will produce a rotating magnetic field of constant magnitude.To

produce a rotating magnetic field, the following conditions must be met:


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The supply must be poly phase

There must be angular displacement between the axis of the coils for the a, b, and c

phases

The magnetic field produced by the armature windings rotates at a speed which is given by

P

fN

120=

where f is the frequency and P is the number of poles in the machine.

In Pakistan, where the line frequency is 50 Hz, the maximum attainable speed of an AC

motor (i.e. with a 2 pole machine) is

rpmN 30002

)50(120max ==

Torque

As seen above, the 3-phase stator windings set up a rotating magnetic field. The flux

produced passes through the air-gap, sweeps past the rotor surface and so cuts the rotor

conductors which are yet (at startup) stationary. Due to the relative speed between the

rotating flux and the stationary conductors, an emf is induced in the latter according to

Faradays law of electromagnetic induction and this is dynamically induced emf. The

frequency of the induced emf is the same as the supply frequency. Its magnitude is

proportional to the relative velocity between the flux and the conductors and its direction is

given byFlemings Right-hand rule. Since the rotor conductors form a closed circuit, rotor

current is produced whose direction as given by Lenzs law is such as to oppose the cause

producing it. In this case, the cause which produces the rotor current is the relative velocity

between the rotating flux and the stationary conductors. Hence to reduce the relative speed,

the rotor starts to rotate in the same direction as that of the fluxand tries to catch up with it.

Percentage Slip

In practice, the rotor never succeeds in catching up with the stator field. If it really did so,

then there would be no relative speed between stator and rotor, hence no rotor emf, no rotor

current and therefore no torque to maintain rotation. That is why; the rotor runs at a speed

which is always less than the speed of the stator field. The difference in speed depends on the


54/87



load on the motor. The difference between the synchronous speed NSand the actual speed Nr

of the rotor divided by the synchronous speed is called the percentage slip. Mathematically,

100%

=

S

rS

N

NNSlip

Losses in an Induction Motor

The power input to an induction motor, Pin, is in the form of three-phase electric voltages and

currents. The first loss in the machine is I2R losses in the stator windings (the stator copper

loss). Then some amount of power is lost as hysteresis and eddy currents loss in the stator

(stator core losses). The power remaining at this point is transferred to the rotor of the

machine across the air gap between the stator and the rotor. This power is called the air-gap

power of the machine. After the power is transferred to the rotor, some of it is lost as I2R

losses (the rotor copper loss), and the rest is converted from electrical to mechanical form.

Finally, friction and windage losses and stray losses are subtracted. The remaining power is

the output of the motor, Pout.

The core losses in the induction motor come partially from the stator circuit and partially

from the rotor circuit. Since an induction motor normally operates at a speed near

synchronous speed, the relative motion of the magnetic fields over the rotor surface is quite

slow, and the rotor core losses are very tiny compared to the stator core losses.

The higher the speed of an induction motor, the higher its friction, windage and stray losses

will be there. These three categories of losses are sometimes lumped together and called

rotational losses. The total rotational losses of a motor are often considered to be constant

with changing speed, since the component losses change in opposite directions with change

in speed

Advantages of Induction Motor

It has very simple and extremely rugged, almost unbreakable construction (especially

squirrel cage type).

Its cost is low and is very reliable.

It has sufficiently high efficiency.

It requires minimum of maintenance.


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It starts up from rest and needs no extra starting motor and has not to be

synchronized .its starting arrangement is simple specially for squirrel cage type

motor.

Disadvantages

Its speed cannot be varied without sacrificing some of its efficiency.

Just like a dc shunt motor its speed decreases with increase in load.

Its starting torque is somewhat inferior to that of the dc shunt motor.

Block Diagram

Connect your setup according the following block diagram.

Apparatus Required

Following apparatus is required for this experiment

Terminal Board, Measuring unit, Induction motor, DC generator and break control unit

Procedure

Set the apparatus as shown in the block diagram above.

Vary the line voltage, VL, and for each different value note the values of line current,

IL, the total true 3-phase power, WT, and the total reactive power, Q, using measuring

unit.

Measure the total power using the concept of 2-wattmeter method of 3-phase

measurement.


56/87



Note the values of torque, T, the output power, Pout, and the motor speed, Nr, are

measured from the brake control unit (BCU) for each value of line voltage.

Calculate the values of power factor, cos , % slip and the efficiency, from the

formulas given. Plot the graphs as mentioned in the objective.

Name Plate Data of Induction Motor

Name Plate Data of DC Machine


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Synchronous speed = 1500rpm

No. VL

(V)

IL

(A)

WT

(W)

Q

(VAR)

cos Nr

(rpm)

% Slip Torque

(N.m)

Pout

(W)

Formulas

100%

=

S

rS

N

NNSlip ----------------------(1)

LL

T

IV

W

3

cos = --------------------------------(2)

From the above readings plot the following graphs. Also give the comments of three lines

for each plot.

Plot o/p versus efficiency

Plot reactive power versus efficiency

Plot speed versus efficiency

Plot power factor versus efficiencyPlot torque versus efficiency


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Plot torque slip curve

Answer the following questions in your note book.

1. What is the purpose of load test of induction motor?

2.

Write down the expression for calculating the three phase active, reactive and apparentpower. Draw Power diagram. How we can calculate power factor?

3. What is power factor of induction motor at no load and why?

4. Why does power factor increases with load?

5. What type of motor was used in experiment, slip ring or squirrel?

6. Is induction motor is a variable speed motor?

7. Why does motor damage due to over loading.

8.

Can power factor of induction motor be leading?


59/87

Load test of a DC Series Generator


Experiment 11

Load Test of a DC Series Generator

Equipment

1. Brake Control Unit SM26712. Drive Machine, DC Machine SM26413. Tacho-generator SM26314. Power Supply Unit SM26315. Terminal Board SM26356. Test Machine, DC Machine SM26417. Measuring Unit

Procedure

1. Make connection of DC motor to run as a DC shunt motor that will act as prime

mover for DC series generator.

2. Connect armature winding (A1-A2) and series field winding (D1-D2) in series with

each other.

3. Connect load with DC series generator through measuring unit such that load voltage,

current and power in measured.

4.

Run DC shunt motor near to rated speed of DC series machine

5. Gradually increase the load on DC series generator and take at least ten different

readings with-out violating the ratings of the machine.

6. If you do not get any output voltage then try again by reversing the terminals of field

winding.

Precautions

1.

Take the first reading at no-load.

2. Do not exceed the Load Current value above 3A to be on the safe side.

3. Maximum speed of the DC Shunt Motor acting as prime mover for the generator is

1500 rpm.


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Circuit Diagram


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Observations

Speed of Machine = _____________________

Load

Voltage

(V)

Load

Current

(A)

IaRa

drop

Ia Rs

drop

Induced

EMF

(V)

Load

Power

(Watt)

InputPower

(Watt)

Approximate

Efficiency

1

2

3

4

5

6

7

8

9

1

0

Draw the following graphs

1. Load voltage vs load power (load voltage on y-axis)

2. Induced EMF and load current (Induced EMF on y-axis)

Why is there a difference between the Terminal Voltage & the Induced emf?

Why is it so that the terminal voltage starts increasing when the load is increased?


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Calculation of Xdand Xqof a Synchronous Machine


Experiment 12

Calculation of Xdand Xqand no. of poles of a Synchronous Machine

Objective

To calculate Xd and Xq in a 3-phase salient pole synchronous machine by applying single

phase supply to only two windings.

Synchronous Motor

Synchronous motors are AC motors that have a field circuit supplied by an external DC

source. They convert AC electrical power to mechanical power. It is electrically identical to

an alternator or AC generator. Some characteristics of the synchronous motor are:

It runs either at synchronous speed or not at all i.e. while running it maintains a

constant speed. The only way to change its speed is to vary the supply frequency.

It is not self-starting. It has to be run up to synchronous or near synchronous speed by

some means before it can be synchronized to the supply.

It is capable of being operated under a wide range of power factors both lagging and

leading.

Production of Torque

In a synchronous motor, a three-phase set of stator currents produces a rotating magnetic

field, BS. The field current, IF of the motor produces a steady-state magnetic field, BR.

Therefore, there are two magnetic fields present in the machine, and the rotor field will tend

to line up with the stator field, just as two bar magnets will tend to line up if placed near each

other. Since the stator magnetic field is rotating, the rotor magnetic field will constantly try ocatch up. Larger the angle between the two magnetic fields, greater the torque on the rotor of

the motor. The basic principle of a synchronous motor operation is that the rotor chases the

rotating stator magnetic field around in a circle, never catching up with it.

Speed of the Synchronous Motor

The rotor (which is initially unexcited) is speeded up to synchronous or near synchronous

speed by some arrangements and then excited by the DC source. The moment this

synchronously rotating rotor is excited, it is magnetically locked into position with the stator


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i.e. the rotor poles are engaged with the stator poles and both run synchronously in the

same direction. It is because of this interlocking of stator and rotor poles that the motor has

either to run synchronously or not at all. The synchronous speed is given by the usual

relation:

P

fN

S

120=

However, this engagement is not very rigid. As the load on the motor is increased, the rotor

progressively tends to fall back in phase by some angle but it still continues to run

synchronously.

Types of the synchronous Motor

There are basically two types of synchronous motors depending on the shape of the rotor. The

stator is same for all AC machines.

1. Cylindrical or non-salient

The rotor is cylindrical in shape and has slots in the outer periphery to support the field

windings. This motor is usually designed for a small number of poles. Maximum poles used

are four. According to equation (1), the smaller the number of poles the greater the speed,

therefore this type of motor has high speed applications.

2. Salient pole

In this type of motor, the shaft is prepared separately and the poles are designed separately.

The two are then bolted together. Therefore, with this arrangement we can have as many

numbers of poles as we like and according to equation (1), the speed of this motor is therefore

lesser than cylindrical type motor. However, this motor is more powerful than the cylindrical

machine.


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Figure-1: Types of synchronous motors

Q-axis and d-axis

The line parallel to the direction of magnetic flux of field windings is called the direct

axis, D-axisor polar axis

The line perpendicular to the D-axis is quadrature axis or Q-axisor interpolar axis.

The angle between the two axes is always 90E.In Figure-1, it is clear that the air gap

between the stator and rotor is uniform in the case of cylindrical rotor machine whereas non-

uniform in the case of salient-pole machine. Hence, along the d-axis the air gap is minimum,and along the q-axis it is maximum in salient-pole machine. Now, since inductance is

proportional to flux which is further proportional to inductive reactance, therefore, with the

change of position of the rotor the reactance of the armature windings varies. The maximum

reactance is called Xd and the minimum is called Xq.

Defination of Xdand Xq

Xd: It is the reactance of the armature winding when axis of armature winding coincides

with the D-axis of the rotor winding

Xq: It is the reactance of the armature winding when axis of armature winding coincides

with the Q-axis of the rotor winding

Hence the conclusion is that in cylindrical machine Xd= Xq, whereas for salient-pole Xd>

Xq.


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Block Diagram

Figure-2: Block Diagram

Apparatus

Terminal Board

Measuring unit

Synchronous motor

Procedure

Set the apparatus as shown in block diagram.

Apply a single phase low value AC supply to only one winding. Since 1-phase does

not produce a rotating magnetic field, therefore the rotor will not rotate.

Rotate the rotor slowly by hand and note the current variation in the armature

winding.

Record the minimum and maximum values.

Find the values of reactance of the windings by dividing the voltage by the current

assuming the resistance of the windings to be negligible.

Change the voltage to different values to obtain a set of readings and calculate the

average values of Xd and Xq.

The maximum and minimum current occurs when the rotor passes two consecutive

poles. Using this fact and the mark on the machines back, determine the number of

poles by observing the number of variations in current during one complete

revolution (i.e. 360)


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Note the Name Plate data of synchronous machine

Observations

No. of

observation

Voltage, V

(V)

Minimum

current, Id(A)

Maximum

current, Iq(A)

dd I

VX =

qq I

VX =

1

2

3

Average value of Xd= _______

Average value of Xq= _______

No. of poles of the machine =_______


67/87



Rotate the shaft of the machine slowly such that you can take 12 readings in rotation and note

down the following readings.

Angle Voltage, V

(V)

Minimum

current, Id

(A)

Maximum

current, Iq

(A)d

dI

V

X = q

qI

V

X =

Ld Lq

0

30

60

90

120

150

180

210

240

270

300

330

360

30


68/87



Plot d-axis and q-axis inductances from above table

Answer the following questions

What is d and q axis of a synchronous machine?

What is Xd and Xq?

What is the advantage of finding Xd and Xq?

Which one is greater Xd or Xq? Explain Why?

Why this experiment is done at low voltages?

At what speed the machine is run in this experiment?

Why Id is smaller as compared to Iq?


69/87

Equivalent Circuit of a Three Phase Induction Motor


Experiment 13


Equipment

1. Terminal Board

2. Measuring unit

3. Three phase induction motor

4. Variable three phase supply

5. Calculator


To find out the equivalent circuit of a three phase induction machine normally three tests are

performed on the induction motor called open circuit test, blocked rotor test and DC test for

stator resistance. The equivalent circuit of an induction motor is a very useful tool for

determining the motors response to changes in load. R1, R2, X1, X2 and Xm are calculated

using the above mentioned tests. The equivalent circuit is shown in figure below.

This circuit is modified to represent rotor copper losses and converted power as


70/87



No Load Test

The no-load test of an induction motor measures the rotational losses of the motor. In this test

the motor is run without any external mechanical load on it. It spins freely and the only load

on the motor is the friction and windage loss. Therefore, all the input power is consumed by

mechanical losses. The slip is very low, possibly as small as 0.01 or less, because the speed

of the rotor would be very much near synchronous speed. The no-load power input, Wo, is

measured by two wattmeter and the no-load current, Io, by an ammeter and Voby a voltmeter.

In this test, the rotor copper losses are negligible because the current I2is extremely small, so

they may be neglected. The stator copper lossesare given by:

1

2

13 RIPSCL = --------------------- (1)

so the input power must equal

Pin= 12

13 RI + Prot

Where Prot is the rotational lossesof the motor and are the sum of core losses, friction and

windage losses, and stray losses. Hence, if R1 is known (from the DC test or stator-

resistance test) and the input is known then the rotational losses of the machine can be

determined.

In addition to this, the power factor at no load, o, can also be calculated from the following

formula:

OO

OO

IV

W

3cos =

Blocked Rotor Test

This test is also known as locked-rotor test.This test is used to find:

Short-circuit current with reduced voltage applied to stator

Power factor on short circuit

In this test, the rotor is locked or blocked, so that it cannot move. In the case of a slip-ring

type rotor, the rotor windings are short-circuited at the slip-rings. A reduced voltage is

applied to the motor, and is so adjusted that full load current flows in the stator. The resulting


71/87



voltage, current, and power are measured. Since the rotor is not moving, slip =1 and so the

rotor resistance referred to the stator, which is given by R2/s, becomes R2, which is a small

value. The power factor is found from the relation:

SS

SS

IV

W

3cos =

Apparatus Setup:

Name Plate Data of Induction Motor

Rated Stator Voltage = __________V

Rated Stator Current =__________A

Type of connection =_______________

Stator Resistance Measurement

Find out the resistance of a stator winding with the help of DMM. If both terminals of a phase

winding are available externally then you can directly find per phase stator resistance R 1but

if machine is internally Y-connected and only one terminal of a phase winding is available

then measure the resistance between two phases and divide it by 2 to get the per phase

resistance of the stator circuit.


72/87


73/87



Take the average voltage of the above three RMS line voltages

Average RMS terminal Voltage Vfl = ____________Volts

Line 1 current: Ia= ____________ Ampere

Line 2 current: Ib= ____________ Ampere

Line 3 current: Ic = ____________ Ampere

Average line current: I2 = (Ia+ Ib+ Ic)/3 = ____________ Ampere

Input Power = Pin,br=__________ W

Calculations for equivalent circuit:

No load Test:

Stator resistance R1 = ____________

Average line current from no-load test: IL,oc = I1 = ____________ Ampere

Average RMS terminal no-load line voltage = VL = ____________Volts

Phase voltage = ___________3

Lp

VV Volts= =

No-load Impedance = 11

___________p

nl mV

Z X XI

= = + =

Stator copper losses = 21 13 __________SCLP I R Watt = =

No-load rotational losses = r , , _________________________P _________ot in nl SCL nlP P Watt = = =

No load power factor =1

,___________

3nl

L

Pin nlPF

I V= =

No load impedance angle = 1cos ( ) ____________nl nlPF = =

Approximate Voltage across magnetizing branch Vm= Vp I1R1 = ___________V

(It has been assumed that there is no voltage drop across X1as X1is unknown)

Current through Xm: Im= I1sin( nl )=_____________ Ampere

Current through Rc: Ic= I1cos( nl )=_____________ Ampere


74/87



Approximate value of Xm=m

______________I

mV=

Approximate value of Rc= ______________m

c

V

I=

Parallel combination of Xmand Rc: _____________m c

m

m c

jX RZ

jX R= =

+

Approximate value of X1= Znl-Xm= ___________

Or X1=Znl-imag(Zm)=_________________

Blocked Rotor test:

,

,,

2

,

____________

____________3

____________

____________

L fl

L flp fl

in br

V V

VV V

I A

P W

=

=

=

=

2 2

, , 2 1 2 22

, 2 12 22 2 , 2 1 2

22

3 33

3 33

in br SCL RCL in br

in br in br

P P P P I R I RP I R

I R P I R RI

= + = +

= =

2, 2 1

22

2

3_______________

3

in br P I RR

I

= =

Full load Impedance =

, 2 2

1 2 1 22 ____________ ( ) ( )

p fl

fl

V

Z R R X XI= = = + + +

Approximate X2:2 2

2 1 2 1( ) __________flX Z R R X= + =


75/87



Draw the equivalent circuit of Induction of motor with values of R1, X1, R2, X2and Xm

calculated above.

Thevenin Equivalent Circuit:

1 1

2 21 1

1

1

2

1

1

_______________

_____________( )

___________

______________

____________

mth p

m

mth p

m

mth p

m

th

mth

m

jXV V

R jX jX

XV V V

R X jX

XV V V

X XX X

XR R

X X

= =+ +

= =

+ +

+ =

=

+

Draw the Thevenin Equivalent Circuit:


76/87



Calculation of Torque:

( )

( )

22

2 22 2

2max

2 22

2

max2 2

2

______

______

120________

2_________ / sec

60

3__________________ .

( ) ( )

________________( )

3____

2 ( )

s

ss

thstart

s th th

th th

th

s th th th

f Hz

P poles

fN rpm

P

Nw rad

V RT N m

w R R X X

RsR X X

VT

w R R X X

=

=

= =

= =

= =+ + +

= =+ +

= =

+ + +

______________ .N m

Draw Torque speed Characteristics of Induction motor and indicate starting and pull-

out torque.

machines manual 2016

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