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Matoshri College of Engineering & Research Centre, Nashik.

Electrical Engineering Department

Matoshri Education Society’s

Matoshri College of Engineering & Research Centre, Eklahare, Nashik.


Certificate

This is to certify that,

Mr. / Miss:________________________

Roll No.:________________________

Exam Seat No.:_______________________

Class: S.E. Electrical

has completed the practical in Electrical Machines-I within the

premises of the institute for the academic year 2013-2014.

Staff HOD Principal



Matoshri Education Society’s

Matoshri College of Engineering & Research Centre, Eklahare, Nashik.


Index

Expt.

No. Title Date Page No. Remark

01 Open Circuit (OC) and Short Circuit (SC) test on a single phase Transformer.

02 Polarity test on a single phase Transformer.

03 Parallel operation of two single phase transformers.

04 To perform Sumpner’s or Back to Back test on two identical single phase Transformers.

05 Speed control of DC shunt motor.

06 Load test on D.C. shunt motor.

07 Load test on D.C. series motor.

08 Load test on 3Φ squirrel cage Induction Motor.

09 No load and blocked rotor test on three phase Induction Motor.

10 Effect of variation of rotor resistance on the performance of 3Φ slip ring Induction Motor.



Experiment No: 01

Date:

Title: Open Circuit (OC) and Short Circuit (SC) test on a single phase Transformer.

Roll No:



Experiment No: 01


Circuit diagram:



Experiment No: 01 Date:


After conduction of the experiment each student must answer following questions,

1) What are the objectives of the experiment?

2) What are the precautions to be taken while carrying out experiment?

3) What is the procedure of the experiment?

Aim: To determine,

1) The parameters of an equivalent circuit of a Transformer, 2) Efficiency and regulation of a Transformer.

By performing Open Circuit (OC) and Short Circuit (SC) test.

Apparatus:

Sr. No. Apparatus Specifications Qty.

01 Single phase Transformer 230V/115V, 2KVA 01

02 Single phase auto transformer 230V/0-270V, 10A 01

03 Voltmeter(AC) 0-300V 01

04 Ammeter(AC) 0-1A 01

05 Ammeter(AC) 0-10A 01

06 Wattmeter 1A/300V(LPF) 01

07 Wattmeter 10A/300V(UPF) 01

08 Connecting wires 1/18 ---

Theory:

Open Circuit (OC) or No Load Test:

The purpose of this test is to determine the shunt branch parameters of the equivalent

circuit of a transformer. One of the winding preferably low voltage is supplied at rated voltage

while other preferably high voltage winding is kept open circuited. The no load current I0 is very

small (2-6% of rated current) and R01 and X01are also small that V1 can be regarded as E1 by



neglecting the series impedance. This means that for all practical purposes the power input on no

load equals the core loss. W0 = Iron loss

The shunt branch parameters can easily be determined from the readings of V 0, I0 and W0.

W0 = V0 x I0 x cosΦ0 Iw = I0 cosΦ0 Im = I0 sinΦ0

R0 = 𝑉1

𝐼𝑤 X0 =

𝑉1

𝐼𝑚

These values are referred to the side from which the test is conducted and would easily be

referred to the other side using transformation ratio.

Short Circuit Test:

This test serves the purpose of determining the series parameters of transformer. This test is

usually conducted from the H.V. side of the transformer while L.V. side is short circuited as shown

in figure (2). Since the transformer resistances and leakage reactance’s are very small the supply

voltage required to circulate the short circuit current is 5-8% of the rated voltage. As a result the

exciting current under these conditions is very small, only about 0.1 to 0.5% of the full load

current. Thus the shunt branch of the equivalent circuit can be neglected. Since the transformer is

excited at very low voltage, the iron loss is negligible. Wsc = Copper loss

From the approximate equivalent circuit, the circuit parameters are computed as below.

Z01 = 𝑉𝑠𝑐

𝐼𝑠𝑐 = 𝑅01

2 + 𝑋012

Equivalent resistance R01 = 𝑊𝑠𝑐

(𝐼𝑠𝑐)2

Equivalent reactance X01 = 𝑍012−𝑅01

2

These values are referred to the side from which the test is conducted.



Precaution:

Auto Transformer should be at minimum voltage position at the time of closing & opening MCB.

Procedure:

I. Open Circuit (OC) Test:

1) Do the connections as per the circuit diagram.

2) Make sure that auto transformer is at it’s minimum (zero) value.

3) Switch ‘ON’ AC supply with the help of MCB.

4) With the help of auto transformer increase the voltage gradually till the voltmeter indicates

rated voltage of the winding which is connected to the supply side.

5) Note down corresponding voltmeter, ammeter, and wattmeter readings.

6) Auto transformer is again brought to minimum position.

7) Switch ‘OFF’ AC supply with the help of MCB.

II. Short Circuit (SC) Test:

1) Calculate rated primary and secondary current from KVA & voltage rating of transformer.

2) Do the connections as per the circuit diagram shown in figure (2).

3) Make sure that auto transformer is at it’s minimum (zero) value.


5) With the help of auto transformer increase the voltage gradually till ammeter indicates the

rated current flow through the windings.

6) Note down corresponding ammeter, voltmeter and wattmeter readings.

7) Auto transformer is again brought to minimum position.




Observation Table:



Vsc

(volts) Isc

(amps) Wsc

(watts)

8.7 A

Calculations:


Core loss, W0 =

W0 = V0 x I0 x cosΦ0

cosΦ0 =

Φ0 =

Iw = I0 cosΦ0

Im = I0 sinΦ0

R0 = 𝑉1

𝐼𝑤

X0 = 𝑉1

𝐼𝑚


Copper loss Wsc =

Wsc = Vsc x Isc x cos Φsc

cos Φsc =

Vo (volts)

Io (amps)

Wo (watts)

115 V



Φsc =

Z01 = 𝑉𝑠𝑐

𝐼𝑠𝑐 =

R01 = 𝑊𝑠𝑐

(𝐼𝑠𝑐)2 =

X01 = 𝑍012−𝑅01

2 =

Percentage Efficiency: for all loads and power factor

Output Power (X) x KVA rating x 1000 x cos

Efficiency, % = -------------------- = ------------------------------------------------

Input Power Output power + losses

(X) x KVA rating x 1000 x cos

= -------------------------------------------------------------

(X) x KVA rating x 1000 x cos + Wo +(X2) Wsc

Percentage Regulation: + = lagging and - = leading (X) x Isc (Ro1 cos Φsc Xo1sin Φsc) x 100 %R = --------------------------------------

V1

Where X is the load and it is 1 for full load, ½ for half load, ¾ load, ¼ load etc.. and the power

factor is, unity p.f., 0.8 p.f. lag and 0.8 p.f. lead

Result Table:

Sr. No. Load condition Power factor % Regulation % Efficiency

01 Full load Unity

02 Half load 0.8 lagging

03 Full load 0.8 lagging

04 Half load 0.8 leading

05 Full load 0.8 leading



Conclusion:



Questions:

1. What are the objectives of the experiment?

2. What are the specifications of transformer?

3. How measuring instruments (voltmeter, ammeter, wattmeter etc.) ratings are decided?

4. What is the procedure of experiment?

5. How current rating of winding is calculated when KVA rating and voltage ratings are

given?

6. What is the aim of Open Circuit test?

7. What is the aim of Open Circuit test?

8. What are the various equivalent circuit parameters of transformer?

9. How efficiency of transformer under various load condition is calculated?

10. How voltage regulation of transformer under various load condition is calculated?

11. What are the advantages of O.C. & S.C. test?

12. In open circuit test, why copper losses are neglected?

13. In short circuit test, why iron losses are neglected?

14. What is definition of transformer?

15. What are the types of transformer?

16. What is an e. m. f. equation of transformer?

17. Why laminated silicon steel stampings are used for making core of transformer?

18. When conducting short circuit test, rated voltage is applied to the transformer what

will happen?

19. Why to calculate equivalent circuit parameters of transformer?

20. What are the different losses that occur in transformer?

21. Why copper loss is called as variable loss?

22. Why iron loss is called as constant loss?

23. What do you mean by voltage regulation of transformer?

24. What is general range of transformer efficiency?

25. What is the condition for maximum efficiency?



Experiment No: 02

Date:

Title: Polarity test on a single phase Transformer.

Roll No:



Experiment No: 02


Circuit diagram:







2. What are the precautions to be taken while carrying out experiment?

3. What is the procedure of the experiment?

Aim: To confirm the terminal markings of windings of a single phase Transformer by polarity test.

Equipments:

Sr. No. Equipments Specifications Qty.

01 Single phase Transformer 230V/115V, 2KVA 01 02 Single phase auto transformer 230V/0-270V, 10A 01


04 Connecting wires 1/18 ----

Theory:

Similar polarity ends of the two windings of a transformer are those ends that acquire

simultaneously positive and negative polarity of emfs induced in them. These are indicated by the

dot convention. Usually the ends of the L.V. winding are labeled with small letter of the alphabet

and are suffixed 1 and 2 (a1, a2) while the H.V. winding ends are labeled by the corresponding

capital letter and are suffixed 1 and 2 (A1, A2).

In determining the relative polarity of the two windings of a transformer the two windings

are connected in series across a voltmeter, while one of the windings is excited from the suitable

voltage source. If the polarities of the windings are as marked on the fig.(1), the voltmeter should

read V = V1-V2. If it reads (V1+V2) the polarity markings on one of the windings must be

interchanged.



Procedure:

1) Make connections as per circuit diagram.

2) Keep auto transformer at it’s minimum position (zero).


4) Increase voltage gradually upto 50% of rated voltage.

5) Measure the primary voltage V1 and voltage between A2 and a2 (V2).

6) If V1 > V2 the winding connections are series subtractive and the polarity markings are correct.

If V1 < V2 the winding connections are series additive and one of the polarity markings should be

reversed.

Observation Table:

Sr. No.

V1 (Volts)

V2 (Volts)

Winding Connection

Polarity Markings

1

2

Conclusion:

1. If voltmeter reading V2 is more than voltmeter reading V1, then windings are connected in

series additive.

2. If voltmeter reading V2 is less than voltmeter reading V1, then windings are connected in

series subtractive.



Questions:







32. What is an e.m.f. equation of transformer?


34. What is the necessity of polarity test on transformer?

35. If two transformers having opposite polarity are connected in parallel what will

happen?

36. How dot and cross are marked on transformer?

37. What do you mean by additive and subtractive polarity?



Experiment No: 03

Date:

Title: Parallel operation of two single phase transformers.

Roll No:



Experiment No: 03




Title: Parallel operation of two single phase transformers.



2. What are the precautions to be taken while carrying out experiment?


Aim: Parallel operation of two single phase transformers and study of their load sharing under

various conditions of voltage ratios and leakage impedance.

Apparatus:


01 Single phase Transformer 230V/115V, 2KVA 02 02 Single phase auto transformer 230V/0-270V, 10A 01


04 Ammeter(AC) 0-5A 02 05 Ammeter(AC) 0-10A 01 06 Lamp load 10A, 230V 01 07 Connecting wires 1/18 ----

Theory:

The need for operation of two or more transformers in parallel often arises due to:

1) Load growth, which exceeds the capacity of an existing transformer 2) Lack of space (height) for one large transformer A measure of security (the probability of two transformers failing at the same time is very less) 3) The adoption of a standard size of transformer throughout an installation

For supplying the load in excess of the rating of an existing transformer, a second transformer

may be connected in parallel with it. The primary windings are connected to the supply bus -bars

and secondary windings are connected to the load bus-bars.

There are certain definite conditions, which must be satisfied in order to avoid any local

circulating currents and to ensure that the transformers share the common load in proportion to

their KVA ratings.

http://electrical-engineering-portal.com/budgeting-and-financing-substation



The conditions for parallel operation are,

(1) Primary windings of the transformer should suitable for the supply system voltage and

frequency. If the transformers are operated at much lower frequencies, then the flux density

would be very high. The flux would saturate the core and the exciting current would be

abnormally high. The excessive iron losses will overheat the core. If the supply frequency is much

higher than for which transformers are rated. Then magnetization of the core will not be

sufficient.

(2) The terminals of the same polarities should be connected to each other while connecting

the primary and secondary windings in parallel. Paralleling of transformers with incorrect

polarities would result in dead short circuit and transformer windings would burn.

(3) The voltage ratings of both primary and secondary windings should be identical. In this

condition, it is not exactly satisfied; even then parallel operation is possible. But due to inequality

of the induced EMF in secondary’s, there will be, even on no load, some circu lating current

between them (and hence between the primary windings also), when secondary terminals are

connected in parallel. When secondary’s are loaded, this localized circulating current will tend to

produce unequal loading condition. Hence it may be impossible to take full KVA output from the

parallel connected group without one of the transformer becoming over heated.

4) The percentage impedances should be equally magnitude and have the same X/R ratio. If

this condition is not exactly satisfied i.e. impedances triangles are not identical in shape and size.

Parallel operation will still be possible but the power factors at which two transformers operates

will be different from the power factor of the common load therefore in this case the two

transformers will not share the load in proportion to their KVA ratings.

The currents carried by two transformers must be proportional to their KVA ratings. This is possible if their numerical impedances are inversely proportional to their KVA ratings and their percentage or per unit impedances are identical.

Precautions: 1. Auto transformer should be in zero position, before switching on the ac supply.

2. Polarity of the transformer connected in parallel should be same to avoid any local

circulating current through transformer winding. This can checked by observing reading of

voltmeter connected across switch.

Procedure: 1. Make the connections as per circuit diagram for two identical transformers with primaries

in parallel and secondary’s also in parallel to have equal ratio.

2. Make sure that all the lamps are in ‘OFF’ position.

3. Make sure that series single pole switch in the secondary circuit is ‘OFF’.



4. Switch ‘ON’ AC supply with the help of MCB.

5. Adjust the rated voltage of the primary winding of transformer.

6. Observe the voltmeters reading across the switch. If it is not zero, switch ‘OFF’ the supply

and connect secondary with their correct polarity again. Switch ‘ON’ the supply and

confirm that the voltmeter reads zero.

7. Close the switch to connect secondary in parallel.

8. Put on the load to adjust 70 % of full load current.

9. Note down all readings.

10. Switch ‘OFF’ the AC supply with the help of MCB.

11. Disconnect the circuit.

Observation Table:

Calculations: Let ZA and ZB be the parts of load impedances shared by each transformer.

Let IA and IB be the currents shared by each transformer.

ZB ZA IA = IT IB = IT ZA + ZB ZA + ZB

VA

ZA = = IA VB ZB = = IB

Sr.

No.

Voltage

Ratio

Impedance VA

volts

IA

amps

VB

volts

IB

amps

Total IT

amps

1 Equal Equal

2 Equal Equal

3 Equal Equal

4 Equal Equal



Result Table:

Sr.

No.

Observed Current

(IA) amps

Calculated Current

(IA) amps

Observed Current

(IB) amps

Calculated Current (IB)amps

1

2

3

4

Conclusion:

We can say that when two single phase transformer are connected in parallel, the

observed and calculated values of current are nearly equal.

Questions:

1. What are the objectives of experiment?





given?





10. When conducting short circuit test, rated voltage is applied to the transformer what

will happen?

11. What is the need of parallel operation of transformer?

12. What are the various conditions for parallel operation of transformer?

13. Can we connect two or more transformer in parallel having different KVA ratings?



14. What are the applications of parallel operation of transformers?

15. What are the factors which decide the load shared by the transformer?


happen?

17. What is the use of single pole switch?

18. How rating of voltmeter which is connected across switch is decided?

19. What are the advantages of parallel operation of transformer?



Experiment No: 04

Date:

Title: To perform Sumpner’s or Back to Back test on two identical single phase

Transformers.

Roll No:




Title: To perform Sumpners or Back to Back test on two identical single phase Transformers.





Aim: To perform Sumpners or Back to Back test on two identical single phase transformers & to

predetermine the efficiency and regulation of a given single phase Transformer and also to find

the parameters of the equivalent circuit.

Equipments:


01 Single phase Transformer 230V/230V, 2KVA 02

02 Single phase auto transformer 230V/0-270V, 10A 02

03 Voltmeter(AC) 0-300V (MI type) 02

04 Voltmeter(AC) 0-600V (MI type) 01

05 Ammeter(AC) 0-5A (MI type) 01

06 Ammeter(AC) 0-10A (MI type) 01

06 Wattmeter 5A, 300V-LPF 01

07 Wattmeter 10A, 300V-UPF 01

08 Single pole Switch (SPST) 10A, 230V 01

09 Connecting wires 1/18 ----

Theory:

While Open Circuit and Short Circuit tests on a transformer yield its equivalent circuit

parameters, these cannot be used for ‘heat run’ test wherein the purpose is to determine the

steady temperature rise if the transformer is fully loaded continuously. This is so because under

each of these tests the power loss to which the transformer is subjected is either the core loss or

copper loss but not both. The way out of this impasse without conducting the actual loading test is

Sumpners test which can only be conducted simultaneously on two identical transformers.



In conducting the Sumpners test the primaries of two transformers are connected in

parallel across the rated supply voltage (V1) while the two secondaries are connected in phase

opposition as shown in fig (1). Low voltage supply connected to secondary injects current into the

secondary circuit. As per the superposition theorem if V2 source is assumed to be shorted the two

transformers appear in open circuit to source V1 as their secondary’s are in phase opposition. The

current drawn from source V1 is 2I0 and power is 2P0 (twice the iron loss). When V1 is regarded

shorted the transformers are series connected across V2 and are short circuited on the side of

primaries. Therefore impedance seen at V2 is 2Z and when V2 is adjusted to circulate full load

current, the power fed in is 2Pc (twice the copper loss). Thus in the Sumpners test, the

transformers are not supplying any load, full iron loss occurs in their cores and full copper loss

occurs in their windings, net power input to the transformers being (2P0 + 2Pc).

Precautions:

. 1) Auto transformer should be in zero position, before switching on the ac supply.

2) Transformer should be operated under rated values.

Procedure:

1) Do the connections as per circuit diagram.

2) Make sure that both the auto-transformers are their minimum (zero) position.

3) Switch ‘ON’ AC supply.

4) Apply rated primary voltage V1 and keep voltage injected in secondary at zero.

5) A voltmeter is connected across the secondary and with the secondary supply off i.e. switch S is

kept open. The voltmeter reading is noted.

6) If the reading of voltmeter reads higher voltage, the terminals of any one of secondary coil is

interchanged in order that voltmeter reads zero.

6) Note down the readings of V1, I1 and W1.

7) The secondary is now switched on and SPST switch is closed with variac of second auto

transformer is zero.

8) After switching on the second auto transformer, variac is adjusted so that full load rated

secondary current flows.

9) Note down the readings of V2, I2 and W2.



10) Switch ‘OFF’ AC supply.

Observation Table:

Calculations:

W1

Core loss of each transformer, W0 = ----- Watts

2

W2

Full load copper loss of each transformer, Wc = ------ Watts.

2

No load primary current, I0 = I1/2

Equivalent circuit parameters:

W0

W0 = V1I0 Cos o o = Cos-1

---------

V1 I0

Iw = I0 Coso Iμ = I0 sino

Ro = V1 / Iw Xo = V1 / Iμ

Ro2 = Wc / (I2)2

Zo2 = V2 / 2*I2

Xo2 = Zo22 – Ro2

2

Equivalent Circuit:

Sr. No.

Condition V1

(volts) I1

(amps) W1

(watts) V2

(volts) I2

(amps) W2

(watts)

1. Without injection of voltage from secondary

0 0 0

2. With injection of voltage from secondary



Percentage Efficiency: for all loads and power factor

Output Power (X) x KVA rating x 1000 x cos

Efficiency, % = -------------------- = ------------------------------------------------ x100

Input Power Output power + losses

(X) x KVA rating x 1000 x cos

= -------------------------------------------------------------

(X) x KVA rating x 1000 x cos + Wo +(X2) Wc

Percentage regulation:

1.Unity power factor: % R = I2 Ro2 Coso __________ x100 V2/2

2. Lagging power factor: % R = I2 (Ro2 Coso + Xo2Sino) ______________________ X 100 V2 /2

3. Leading power factor:

http://1.bp.blogspot.com/-OZXzwuswN24/TZpqZ3vTMyI/AAAAAAAAAUA/tqExe_tSYCI/s1600/3.JPG



% R = I2 (Ro2 Coso - Xo2Sino) _____________________ X 100 V2 /2

Conclusion:

Thus the efficiency and regulation of a given single phase transformer is calculated by

conducting back-to-back test and the equivalent circuit parameters are also found out.

Questions:






given?






happen?

30. What is the use of single pole switch?

31. How rating of voltmeter which is connected across switch is decided?

32. What are the advantages of back to back test?

33. Reading shown by the wattmeter which is connected in primary circuit indicates what?

34. Reading shown by the wattmeter which is connected in secondary circuit indicates

what?

35. Why back to back test is called regenerative test?

36. Can we perform back to back test on two different KVA rating transformer?



Experiment No: 05

Date:

Title: Speed control of DC shunt motor.

Roll No:



Experiment No: 05


Circuit diagram:

Specifications:i) Output power=3HP

ii) Rated voltage=220V

iii) Rated current=12A

iv) Rated speed=1500RPM

v) Insulation class=B









Aim: To obtain speed control of DC shunt motor by

a. Varying armature voltage with field current constant.

b. Varying field current with armature voltage constant.

Equipment: DC shunt motor

Specifications:i) Output Power=3HP ii) Rated voltage=220V iii) Rated current=12A iv) Rated speed=1500RPM v) Insulation class=B Name of manufacturer: BENN Electricals

Apparatus: Sr. No.

Apparatus Specifications Qty.

01 DC shunt motor 3HP, 220V, 12A, 1500RPM 01

02 Three point starter For 3HP 01

03 Ammeter(DC) 0-1A (Moving Coil type) 01

04 Voltmeter(DC) 0-300V (Moving Coil type) 01

05 Rheostat 300ohm,1.2A 01

06 Rheostat 110ohm, 3A 01

07 Tachometer 0-3000RPM 01

08 Connecting wires 1/18 ---

Theory:

DC shunt motor is a constant speed motor. It is used in constant speed drives while in

lathes, centrifugal pumps etc. but in some applications speed control is required. The speed of DC

motor can be expressed by following relationship:



N α V−IaRa

Φ

Therefore, speed of DC motor can be regulated by change of Φ, Ra or V in other words by, i)

Field (flux) control

ii) Rheostatic control or armature voltage control.

iii) Voltage control methods

I. Field (flux) control method:

The speed is inversely proportional to field (flux), hence field current.

i.e. N α1

Φ α

1

𝐼𝑓

Since the flux is produced by the field current, control of speed by this method is obtained

by control of the field current. In the shunt motor, this is done by connecting a variable resistance

in series with the shunt field winding. The resistance is called shunt field regulator.

The shunt field current is given by, 𝐼𝑠ℎ = 𝑉

𝑅𝑠ℎ+𝑅𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙

The connection of external resistance in the field reduces the field current and hence the

flux Φ is also reduced. The reduction in flux will result in an increase in the speed. Consequently,

the motor runs at a speed higher than normal speed. For this reason, this method of speed control

is used to give motor speeds above normal speed.

Since voltage across motor remains constant it continues to deliver constant output. This

characteristic makes this method suitable for fixed output loads. The performance curve of DC

shunt motor is field current (If) Vs speed (N) is shown n figure.



Merits:

1) Good working efficiency, compact controlling equipment & capability of minute speed control.

2) The speed is not affected by load & speed control can be performed effectively even at light loads.

Demerits:

1) Inability to obtain speeds below rated speed.

2) Inability to operate at high speeds because of armature reaction & commutation difficulties.

II. Armature voltage (Rheostatic) control method:

This method consists of obtaining reduced speeds by the insertion of external resistance in

series with armature circuit. When value of external resistance increases, voltage drop across

external resistance increases, hence voltage across armature decreases and speed of motor also

decreases. This method is used when speeds below rated speed is required.

Merits:

1) Simplicity & easy of connections.

Demerits:

1) Variable resistance is used which dissipates large amount of heat resulting in low efficiency of machine & high operating cost.

2) Inability to obtain speeds above rated speed.



Precautions:

1) Field Rheostat should be kept in the minimum resistance position at the time of starting and

stopping the motor.

2) Armature Rheostat should be kept in the maximum resistance position at the time of start ing

and stopping the motor.

Procedure:



2) Check the maximum position of armature rheostat and minimum position of field rheostat.

3) Switch ‘ON’ DC supply with the help of MCB.

4) Start the DC shunt motor with help of 3 point starter.

5) Armature voltage is fixed to various values and for each fixed value, by adjusting the field

rheostat, speed is noted for various field currents.

6) Reduce field current step by step & note down speed for different field currents by tachometer.

II. Armature voltage control method:

1) Field current is fixed to various values and for each fixed value, by varying the armature

rheostat, speed is noted for various voltages across the armature.

2) Bring field rheostat to minimum position and armature rheostat to maximum position.

3) Switch ‘OFF’ DC supply with the help of MCB.



Observation Table:


II. Armature voltage (Rheostatic) control method:

Sr. No.

If1= amps If2= amps Va (volts) N (rpm) Va (volts) N (rpm)

1

2

3

4

5

6

7

8

Sr. No.

Va1= volts Va2= volts If (amps) N (rpm) If (amps) N (rpm)

1

2

3

4

5

6

7

8



Conclusion:

The variation of speed with armature voltage & field (flux) has been studied. The curves

are shown in graphs. The speed of DC shunt motor is directly proportional to armature voltage. (N

α Va) & inversely proportional to field current. (N α1/ Φ α1/If) Thus to increase speed, the

armature voltage should be increased or field currents should be decreased.

Questions:

1. What is the title of experiment?


3. What are the specifications of machine?

4. How measuring instruments (voltmeter, ammeter, etc.) ratings are decided?


6. What should be the position of field winding rheostat initially? And why?

7. If field winding of D.C. shunt motor is open circuited when motor is running what will

happen?

8. Why D.C. shunt motor is called as constant speed motor?

9. How the field winding and armature winding are connected?

10. Comment on i. Field winding current vs speed, ii) Armature voltage vs speed curve.

11. What are the applications of D.C. shunt motor?

12. What is the necessity of starter?

13. What is back emf in D.C. motor?

14. What is current equation of D.C. shunt motor?

15. What is voltage equation of D.C. shunt motor?

16. What is speed equation of D.C. shunt motor?

17. What is power equation of D.C. shunt motor?

18. How field winding and armature winding of D.C. shunt motor is identified?

19. What are the different methods of speed control of D.C. shunt motor?

20. What are the advantages of flux control method?

21. What are the disadvantages of flux control method?

22. What are the advantages of armature voltage control method?

23. What are the advantages of armature voltage control method?



Experiment No: 06

Date:

Title: Load test on D.C. shunt motor.

Roll No:



Experiment No.: 06


Circuit diagram:

Specifications:i) Rated output power=3HP













Aim: To conduct load test on D.C. shunt motor & to find efficiency.

Equipment: DC shunt motor with Brake Drum Arrangement (BDA).

Specifications:i) Rated output power=3HP





Name of manufacturer: BENN Electricals

Apparatus:

Sr. No.


01 DC shunt motor with BDA 3HP, 220V, 12A, 1500RPM 01

02 Three point starter For 3HP 01

03 Ammeter(DC) 0-10A 01

04 Voltmeter(DC) 0-300V 01

05 Rheostat 300ohm,1.2A 01


07 Connecting wires 1.5 Sq.mm. ---



Theory:

The DC shunt motor has two windings namely armature winding and field winding which

are connected in parallel.

Current equation,

I=Ia + Ish

Where, I = Total current, Ia = Armature current, Ish = Field winding current

Voltage equation,

The voltage applied across armature has

1) To overcome the back emf (Eb)

2) To supply armature ohmic drop

V=Eb +IaRa

Where, V = Supply voltage, Eb = back emf, Ra = Armature winding resistance

N α 1/Φ,

In DC shunt motor, field winding is supplied from a constant voltage so that the field current is

constant. As field current is constant, flux produced by field winding is constant. Since flux is

constant so speed is also approximately constant & it produces moderate torque. The various

applications are,

1. Drills & milling machines

2. Lathe & centrifugal pumps

3. Reciprocating pumps & wood working tools

So for choice of motor according to the user’s requirement we should know the

performance characteristics of DC shunt motor which we find by conducting load test on it. The

three important characteristics of DC shunt motor are

1. Ta Vs Ia (Torque verses armature currents):

In DC shunt motor speed is constant (though at heavy loads Φ decreases some what due to

increased armature reaction) since Ta α ΦIa But Φ=constant. So Ta α Ia. Nature of this curve is

shown in fig. It is practically a straight line passing through origin. Tsh is slightly less than Ta. Since a



heavy starting load will need a heavy starting current, shunt motor never be started on heavy

load.

2. N Vs Ia (Speed verses armature current):

If Φ is assumed to be constant, the N α Eb is also constant at a particular load conditions

(Eb= V-IaRa) but strictly speaking both Eb & Φ decreases with increasing load, so there is some

decrease in speed. The drop varies from 5-15% of full load speed being dependant on saturation,

armature reaction & brush position, hence actual speed curve is slightly dropping as shown by

dotted line in fig. But for all practical purposes shunt motor is taken as a constant speed motor.

3. N Vs Ta (Speed verses Torque):

Earlier we found the Ta Vs Ia & N Vs Ia characteristics & taking the help of these two curves

we can find that N Vs Ta.

Precautions:

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

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

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



Procedure:


2) Set field rheostat to it’s minimum (zero) resistance position & do not put any load on the machine.


4) Start the DC shunt motor with help of 3 point starter.

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

6) Note down readings of voltmeter, ammeter, tachometer at the no load conditions.

7) The load is then added to the motor gradually and for each load, voltmeter, ammeter, spring balance readings and speed of the motor are noted.

8) Take 7 or 8 readings for different loads till full load conditions.

9) The motor is then brought to no load condition and field rheostat to minimum position.


Observation Table:

Sr. No.

Supply voltage Total

current Spring balance Speed

V (volts) I (amps) S1(kg) S2(kg) N(rpm) 1

2

3

4

5

6

7

8



Radius of pulley, r = 0.1125m OR Circumference (cm) r= ------------------- m (by measuring circumference ‘r’ can be calculated) 100 x2

Calculation:

1. Input power, Pin = V x I ------watts

2. Shaft torque, Tsh = 9.81 x r x (S1-S2) ------N. m

3. Output power, Pout = 2 x π x N x Tsh

_____________ -------watts

60

4. Efficiency of DC shunt motor, % η = Pout

__________ x 100

Pin

Result Table:

Conclusion:

Sr. No.

Supply voltage

Total current

Input power

Shaft torque

Output power

Efficiency

V (volts) I (amps) Pin

(watts) Tsh (Nm) Pout

(watts) % η

1

2

3

4

5

6

7

8



Questions:





41. What are the different types of d.c. motors?

42. Why D.C. shunt motor is called as constant speed motor?


44. Comment on i. Efficiency curve, ii. Armature current vs torque curve, iii) Armature

current vs speed curve, iv) Torque vs speed curve

45. What are the applications of D.C. shunt motor?


47. What are the various types of D.C. motor starters?


49. What is current equation of D.C. shunt motor?

50. What is voltage equation of D.C. shunt motor?

51. What is speed equation of D.C. shunt motor?

52. What is power equation of D.C. shunt motor?

53. What is the expression for torque developed?

54. Write the condition for maximum power developed by d.c. motor.

55. What is the relationship between back emf, speed of rotation and flux per pole?

56. How field winding and armature winding of D.C. shunt motor is identified?

57. How radius of pulley can be measured?

58. Explain the principle of operation of a DC motor

59. How does the back emf in a d.c. motor make the motor self regulating?

60. How can we reverse the direction of a d.c. shunt motor?



Experiment No: 07

Date:

Title: Load test on D.C. series motor.

Roll No:



Experiment No.: 07


Circuit diagram:

Specifications:i) Output Power=3HP













Aim: To conduct load test on D.C. series motor & to find efficiency.

Equipment: DC series motor with Brake Drum Arrangement (BDA).

Specifications:i) Rated output Power=3HP





Name of manufacturer: BENN Electricals

Apparatus:

Sr. No.


01 DC series motor with BDA 3HP, 220V, 12A, 1500RPM 01

02 Two point starter For 3HP 01

03 Ammeter(DC) 0-10A 01

04 Voltmeter(DC) 0-300V 01


06 Connecting wires 1.5 Sq.mm. ---



Theory:

The series motor provides high starting torque and is able to move very large shaft loads when it is first energized. Figure shows the wiring diagram of a series motor. From the diagram you can see that the field winding in this motor is wired in series with the armature winding. This is the attribute that gives the series motor its name. Since the series field winding is connected in series with the armature, it will carry the same amount of current that passes through the armature. For this reason the field is made from heavy -gauge wire that is large enough to carry the load. Since the wire gauge is so large, the winding will have only a few turns of wire. In some larger DC motors, the field winding is made from copper bar stock rather than the conventional round wire used for power distribution. The square or rectangular shape of the copper bar stock makes it fit more easily around the field pole pieces. It can also radiate more easily the heat that has built up in the winding due to the large amount of current being carried.

Electrical diagram of series motor.

The amount of current that passes through the winding determines the amount of torque the

motor shaft can produce. Since the series field is made of large conductors, it can carry large

amounts of current and produce large torques.

Current equation,

I = Ia = Ish

Voltage equation,

The voltage applied across armature has

1) To overcome the back emf (Eb)

2) To supply armature ohmic drop

3) To supply series field winding drop

V=Eb + Ia (Ra+Rse)



1) Ta Vs Ia (Torque verses armature currents):

The equation for torque in DC motor is given by, T α φ. Ia Flux φ for series motor is proportional to armature current Ia. Thus T α (Ia)2 . The relationship between torque and armature current is therefore a parabola. With increase in Ia, the flux increases linearly but due to saturation of the magnetic core, beyond certain magnitude of Ia, the increase in flux is very negligible. Thus T is proportional to the square of Ia up to the saturation point beyond which T varies linearly with Ia. This is shown in figure. From torque load characteristics it is seen that a series motor when stared with load develops a very high starting toque. Hence series motor are used in applications where high starting torque is required such as in electric trains, hoists, trolleys etc.

1. N Vs Ia (Speed verses armature current):

The speed of DC motor can be expressed by following relationship:

N α V−IaRa

Φ

It is seen that speed is inversely proportional to flux φ. For Dc series motor, flux is proportional to

Ia. Thus V is constant; N is inversely proportional to Ia. Thus speed versus armature current

characteristics is therefore rectangular hyperbola as shown in figure.

It is seen from characteristics that the speed decreases as the load on the motor increases.

At a very low load, the speed is dangerously high. Thus if a series motor is allowed to run at a very

light load or at no load, its speed will become much higher than its normal speed which may cause

damage to the motor. For this reason series motor are never started on no load.



Precautions:

1. The motor should be started and stopped with load.

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

Procedure:


2) Put load on the machine.


4) Start the DC series motor with help of 2 point starter.

5) For various loads, note down reading of voltmeter, ammeter, tachometer & spring balance.

6) Bring the load to initial position.


Observation Table:

Radius of pulley, r = 0.1125 m (given) OR Circumference (cm) r = ------------------- meter (by using this equation, r can be calculated) 100 x2

Sr. No.

Supply voltage Current Spring balance Speed

V (volts) I=Ia=If (amps) S1(kg) S2(kg) N(rpm)

1

2

3

4

5

6

7

8



Calculation:

5. Input power, Pin = V x I ------watts

6. Shaft torque, Tsh = 9.81 x r x (S1-S2) ------N.m

7. Output power, Pout = 2 x π x N x Tsh _____________ -------watts 60

8. Efficiency of DC series motor, % η = Pout __________ x 100 Pin

Result Table:

Conclusion:

Thus load test on DC series motor is conducted and its efficiency is determined.

Sr. No.

Supply voltage

Current Input power

Shaft torque

Output power

Efficiency

V (volts) I (amps) Pin

(watts) Tsh (Nm) Pout

(watts) % η

1

2

3

4

5

6

7

8



Questions:

24. What is the title of experiment?



27. How measuring instruments (voltmeter, ammeter, etc.) ratings are decided?


29. If field winding of D.C. series motor is open circuited when motor is running what will

happen?


31. Comment on i. Armature current vs speed, ii) Armature torque vs speed curve.

32. What are the applications of D.C. series motor?


34. What type of starter is used for series motor? And Why?


36. What is current equation of D.C. series motor?

37. What is voltage equation of D.C. series motor?

38. What is speed equation of D.C. series motor?

39. What is power equation of D.C. series motor?

40. Whether D.C series is motor is started on no-load? Justify.

41. Why torque of a D.C series motor is high?

42. Why D.C series motor is called a variable speed drive?

43. What are the applications of D.C series motor?

44. Mention the methods to control the speed of the D.C series motor.



Experiment No: 08

Date:

Title: Load test on 3Φ squirrel cage Induction Motor.

Roll No:




Title: Load test on 3Φ squirrel cage Induction Motor.





Aim: To perform load test on 3Φ squirrel cage Induction Motor and plot the performance

characteristics.

Equipment: 3- Squirrel cage Induction Motor.

Specifications: i) Rated Voltage: 415 Volts

ii) Full load current: 4.7 Amps

iii) Rated speed: 1480 RPM

iv)Rated output: 3 HP (2.2 KW)

v) No. of poles: 4

vi)Frequency: 50 Hz

Apparatus:

Sr. No.


1 3 Φ squirrel cage Induction Motor with BDA.

3 Φ, 3HP, 415V, 50Hz, 1480RPM, 4pole

1

2 3 Φ Auto transformer 10A, 440V/0-470V 1 3 Voltmeter(AC) 0-600V 1 4 Ammeter(AC) (0-5A) 1 5 Wattmeter 5A,600V 2 6 Tachometer 0-3000RPM 1 7 Connecting wires 1.5 sq.mm. ---

Theory:

Induction motor is a machine which converts AC electrical energy into mechanical energy.

In this motor the rotor does not receive electric power by conduction but by induction inexactly as

the secondary of 2 – winding transformer receives its power from the secondary. That is why such

motors are known as rotating transformer. When you give three phase supply to the three phase



stator winding then it is called as three phase induction motor. It is widely used in practice due to

simple, robust construction, cheapest & highly efficient.

In case of slip ring Induction Motor by adding external resistance in rotor circuit high

starting torque & smooth speed variation can be obtained. This is not possible in case of squirrel

cage induction motor, because rotor bars are permanently shorted by means of end ring. To check

performance of small squirrel cage induction motor, direct loading method is used where

induction motor are attached directly by pulley & break arrangement. Hence this method is called

direct loading method.

The effect of applying load on the speed, slip, stator current, power factor, efficiency and

torque are discussed below:

Effect on speed:

When the induction motor is on no load the speed is slightly below the synchronous speed. The

current due to induced emf in the rotor is responsible for torque production required at no load,

as the load is increased the rotor speed is slightly reduced. The emf induced in the rotor and hence

the current increases to produce higher torque required until the torque is equal to the torque

required by the load on the motor.

Effect on slip:

Synchronous speed depends upon of frequency of stator supply voltage and number of poles for

which that motor winding is made. Therefore if poles and frequency are constant, synchronous

speed is constant. Thus with increase in load on the motor, rotor speed decreases, slip will

increase. %slip = (Ns – N /Ns) *100

Effect on stator current:

Current drawn by the stator is determined by two factors. One component is the magnetizing

current required to maintain the rotating field. The second component produces a field which is

equal and opposites to that formed by the rotor currents. The rotor current increases with loads,

the stator current will also therefore increases with load. Power factor of an induction motor on

no load is very low because of the high value of magnetizing current. With load the power factor

increases because the power component of the current is increased.

Effect on torque:

The torque will increases with increase in loads, with increase in output.

Performance characteristics of Induction motor,

a) Speed power characteristics:

1) 3Φ squirrel cage Induction Motor operates like D.C. shunt motor at substantially constant

speed from no load to full load.

2) As rotor of induction motor never runs at synchronous speed, it always operates with certain

amount of slip.



3) As rotor impedance is very small, only slight change in speed is enough to cause large change

in rotor current to develop torque necessary to drive load.

4) Percent slip is less than one at no load while full load it is 2 to 5%.

5) Small change in % slip or speed from no load to full load indicates why squirrel cage induction

motor is constant speed.

b) Power factor characteristics:

1) Because of air gap between rotor & stator, the reluctance of magnetic circuit is high induction

motor draws large magnetizing current to produce required flux in air gap.

2) Current drawn by induction motor operating at no load is largely magnetising current so no load

current lags behind applied voltage by very large angle & thus power factor of induction motor at

no load is quite low.

3) As load is increased, rotor current increases, equivalent stator current increases, angle between

stator voltage & equivalent stator current decreases & hence power factor increases.

c) Efficiency characteristics:

1) At no load slip is small &rotor speed is nearly equal to synchronous speed. Hence rotor & stat or

currents are small enough to develop torque that meets fixed losses (core, friction & windage

losses).

2) η = Output power _________________

Input power

At no load, Output power =0, η=0

3) As the mechanical load on motor increases, the speed drop due to retarding effect on load

torque is developed.

4) Then increases slip hence rotor and current & corresponding stator current increases to develop

torque that meets increased load requirement.

Now, P out =2πNT/60

5) At all load; there are variable losses in stator & rotor in addition to fixed losses.

6) At light load, efficient is quite low because fixed losses are relatively large part of input power.

7) With increase in load, output power increases hence efficiency also increases rapidly &

becomes maximum at fixed losses = copper losses. Beyond this point copper losses become

relatively large causing efficiency to decrease.



d) Starter current characteristics:

At no load or about 30 to 40 % of rate current with increasing load slip also increases linearly

with increase in stator current.

e) Torque characteristics:-

1) With increasing load, speed falls; slip increases (T α S in stable region)

2) A point will reach where maximum torque will be developed.

3) Further increasing load causes drop in speed. The slip increases, the drawing torque

decreasing (T α S) in unstable region & rotor will stop ultimately.

Applications:

1.Squirrel cage induction motor having moderate starting torque and constant

speed characteristics preferred for driving fans, blowers, water pumps, grinders, lathe

machine, printing machines and drilling machines.

2. Slip ring induction motors can have high starting torque as high as maximum

torque. Hence they are preferred for lifts, hoists, elevators, cranes, compressors

Precautions:

1) 3-phase autotransformer should be at minimum voltage position.

2) There should be no-load at the time of starting (Loosen the belt on the brake drum)

3) Brake drum should be filled with water.

Procedure:


2) Make sure that their is no load on machine & auto transformer is at it’s minimum (zero) value.


4) With the help of auto transformer increase the voltage gradually till voltmeter indicates rated

voltage. Note down no load reading.

5) Increase the load on motor gradually & note down corresponding readings.

6) Repeat step 5 till full load is reached.

7) Decrease the load; bring auto-transformer to its minimum voltage position.


Any case wattmeter starts indicating negative reading (below 0) stops. Interchange

wattmeter connections either interchange M & L or interchange C & V.



Restart motor at no load condition.

Adjust load & consider power reading after coil is reversed to be –ve

Observation Table:

Sr. No.

Voltage Current Wattmeter Spring balance Speed

V (volts) I (amps) W1(watts) W2 (watts) S1(kg) S2(kg) (rpm)

1

2

3

4

5

6

7

8

Radius of pulley=

Calculations:

1. Input power, Pin = W1 + W2 -------watts

2. Shaft torque, Tsh = 9.81 x r x (S1-S2) -------Nm

3. Output power, Pout = 2 x 𝜋 x N x Tsh ______________ -------watts 60

4. Efficiency, %η = Pout ________ x 100 Pin

5. P er c en t ag e s l i p = (N s -N )/ N s *1 0 0

Ns = synchronous speed in rpm

N = Actual speed of rotor in rpm

6. Power factor = (W1+W2)/√3VI



Result Table:

Sr. No.

Input power

Shaft torque

Output power

Efficiency Slip Power factor

Pin (watts)

Tsh (Nm)

Pout (watts)

%η %s cos∅

1

2

3

4

5

6

7

8

Conclusion:

Questions: Q.1 What are the objectives of the experiment? Q.2 What are the precautions to be taken while performing experiment?

Q.3 What is the procedure of the experiment?



Q.4 What are the specifications of Induction motor?

Q.5 Whether two wattmeter’s measures input power or output power?

Q.6 What are the advantages of two wattmeter method to measure power?

Q.7 How to calculate multiplying factor of wattmeter? What is the value in this experiment?

Q.8 Why one of the wattmeter shows negative reading for first few readings? And how to take

such negative reading?

Q.9 How to calculate output power in this experiment?

Q.10 How to calculate shaft torque in this experiment?

Q.11 How to calculate efficiency of an Induction motor?

Q.12 How to calculate radius of pulley if it is not given?

Q.13 What is working principle of induction motor?

Q.14 What are the main parts in induction motor?

Q.15 Why stator and rotor core is made up of from Silicon Steel Stampings?

Q.16 What are the types of Induction Motor?

Q.17 What is synchronous speed of induction motor? How it is calculated?

Q.18 Can N=Ns? What will happen when N=Ns?

Q.19 What is slip speed and slip?

Q.20 Why frequency of rotor emf & rotor current is very small? And how it is calculated?

Q.21 What are the different losses that occurs in induction motor? Where it occurs?

Q.22 Draw power flow diagram of induction motor?

Q.23 Draw torque-speed characteristic of induction motor? And show stable, unstable operating

region, maximum torque, starting torque, slip at start, slip at running condition.

Q.24 What is torque equation of Induction Motor? What is the condition for maximum torque?

Q.25 What are the advantages of squirrel cage induction motor?

Q.26 What are the disadvantages of squirrel cage induction motor?

Q.27 What are the applications of squirrel cage induction motor?

Q.28 What is the difference between squirrel cage and slip ring induction motor?

Q.29 Wheather this is direct loading test or indirect loading test?

Q.30 What are the drawbacks of direct loading test?

Q.31 Explain how power factor of induction motor varies with load?

Q.32 What is the general range of efficiency of induction motor?

Q.33 If no load current of induction motor is compared with no load current of transformer for

same capacity, in which case it will be more? Why?

Q.34 Can we add external resistance in series with rotor winding in case of squirrel cage induction

motor?

Q.35 What are the different methods to improve starting torque of squirrel cage induction motor?

Q.36 What is the necessity of starter for three phase induction motor?

Q.37 What are the different types of starters used for squirrel cage induction motor?

Q.38 Which type of starter is used in this experiment?



Experiment No: 09

Date:

Title: No load and blocked rotor test on three phase Induction Motor.

Roll No:




Title: No load and blocked rotor test on a three phase Induction Motor.





Aim: To perform no load and blocked rotor test on a three phase Induction Motor and

i) To determine parameters of equivalent circuit.

ii) To determine performance of induction motor from circle diagram.

Equipment: 3- Squirrel cage Induction motor.

Specifications: i) Rated Voltage: 415 Volts

ii) Full load current: 4.7 Amps

iii) Rated speed: 1480 RPM

iv) Rated output: 3 HP

v) No. of poles: 4

vi) Frequency: 50 Hz

Apparatus:

Sr. No.


1 3 Φ squirrel cage Induction motor with BDA.

3 Φ, 3HP, 415V, 50Hz, 1480RPM, 4pole

1

2 3 Φ Auto transformer 10A, 440V/0-470V 1 3 Voltmeter(AC) 0-300-600V 1 4 Ammeter(AC) (0-5A) 1 5 Wattmeter 5A,150-600V 2 6 Tachometer 0-3000RPM 1 7 Connecting wires 1.5 sq.mm. ---

Theory: No-Load Test: Balanced voltages are applied to the stator terminals at the rated frequency with the rotor uncoupled from any mechanical load. Current, voltage and powerare measured at the motor input. The losses in the no-load test are those due to core losses, winding losses, windage and friction.



Blocked Rotor Test: The rotor is blocked to prevent rotation and balanced voltages are applied to

the stator terminals at a frequency of 25 percent of the rated frequency at a voltage where the

rated current is achieved. Current, voltage and power are measured at the motor input.

Disadvantage: 1. Speed control of Induction motor is not simple & cheap as compared to dc shunt motor. 2. Just like dc motor, its speed decreases with increasing load. 3. Its starting torque is somewhat inferior to that of dc shunt motor. Distinguish Feature of an Induction motor: In case of dc shunt motor the supply is given to field winding & armature winding. But in case of Induction motor, ac supply is given to only one winding of stator & not to rotor. The rotor of such motor receives power by induction in exactly same way as secondary of two winding transformer. Construction:

1. Laminated core stator carrying polyphase winding. 2. Laminated rotor core carrying either cage winding with shaft wounded slip ring. 3. A stiff shaft to preserve very short air gap. 4. A frame from stator housing covers bearing & terminal box. In this, cage of frame is not a

part of magnetic circuit as in d.c machine. 5. Non salient pole construction used for all polyphase motor.

Working Principle of Induction motor: Induction motor works on the principle of Electromagnetic Induction, when 3Φ supply is

given to the Stator winding a rotating magnetic field is produced. This field produces an effect on rotating pole around stator which is stationary. So there is relative motion between stator & rotating magnetic field. Hence an emf is induced in rotor. This emf drives current through rotor conductors. Thus, rotor conductors produced average magnetic field. Now, there are two fluxes present, namely rotating magnetic field & rotor flux. These fluxes produce necessary torque which results in rotating action of Induction Motor. Equivalent circuit: An Induction Motor has been shown equivalent to transformer. The only difference is that the mechanical load on secondary of I.M. is replaced by equivalent electrical load of resistance RL is given by,



All the values shown are per phase values. The circuit is similar to series R-L circuit. The

reactance X1e is fixed while the total resistance R1e + (R2'(1-s)/s) is variable. This is because the slip

s varies as load varies. The voltage across the parallel exciting branch is V1. Hence we can write the

expression for the rotor current referred to stator as,

I2r' = V1/√((R1e + RL )2 + X1e

2)

Where RL' = R2'(1-s)/s = Variable equivalent load resistance R1e = R1 + R2' = Equivalent resistance of motor referred to stator X1e = X1 + X2' = Equivalent reactance of motor referred to stator Dividing and multiplying by,

Equivalent circuit parameters:

W0

W0 = 3 Vo Io cos o o = cos-1 ------------

3 V0 I0

Iw = I0 coso Iμ = I0 sino

Ro = Vo / Iw Xo = Vo / Iμ

Ro1 = Wsc / (Isc)2 Zo1 = Vsc / Isc

Xo1 = Zo12 – Ro12

Procedure to draw circle diagram:

By using the data obtained from the no load test and the blocked test, the circle diagram can be drawn using the following steps. 1. Draw the line by taking the current on X axis, voltage on Y axis. 2. From the no load test, find out the no load current (Io) and draw the vector OA with the magnitude of Io from the origin by suitable current scale, which lags behind the voltage by an angle Фo, Фo = Cos-1(Wo / √3VoIo)

http://3.bp.blogspot.com/-4x7235CVNXM/TmgD544Q8zI/AAAAAAAABZM/ZThOF5ZuCTg/s1600/ABB111.jpeg



3. From the current ISC find out ISN (short circuit current corresponding to the normal voltage) through the formula ISN= ISC (VO/VSC). Draw the OB vector with the magnitude of ISN from the origin by the same current scale, which lags the voltage V by an angle ФSC.

Where, ФSC = Cos-1(Wsc / √3VscIsc) 4 . Jo i n t h e p o i n t s B an d A t o g et t h e o u t p u t l i n e . 5. Draw the parallel line for the X axis from point A and for the Y axis from point B upto the X axis (point E) let both the lines intersects at point D. 6. Then draw the bisector for the output line and extend it to the line AD, let the point of intersection be C. 7. By keeping the point C as centre draw a semicircle with the radius CA 8. Let EB be the line of total loss (ED (constant loss) +DB (variable loss)) 9. In the line DB locate the point G to separate the stator and rotor copper losses Rotor Cu loss/Stator Cu loss = Wsc-3Isc2R1/ 3Isc2R1 Where R1= stator resistance in ohm Variable loss = Stator Cu loss + Rotor Cu loss 10. To get the torque lines join the points A and G.

To find out maximum quantities:

11. Draw the tangent to the semicircle in such a way that it should be parallel to the output line. Let the point of tangent be H. 12. Join the points H and C that will be perpendicular to the output line then draw a line parallel to the Y axis from H to output line. The point at where the parallel line meets the output line is names as H’. 13. Here the HH’ will be the maximum output power. 14. Draw the tangent to the semicircle in such a way that should be parallel to the torque line. Let the point of tangent be I. 15. Join the points I and C that will be perpendicular to the torque line then draw a line parallel to the Y axis from I to torque line. The point at where the parallel line meets the output line is named as I’. 16. Here the II’ will be the maximum t orque line. 17. Draw the tangent to the semicircle in such a way that should be parallel to the X axis. Let the point of tangent be J. 18. Join the points J and C and extend the same up to X axis. The point at where the line meets the input line is named as J'. 19. Here the JJ’ will be the maximum input power. 20. From the circle diagram find maximum input power, maximum torque, maximum output power, rotor Cu loss, stator Cu loss and slip. To find the total input power on short circuit with normal voltage PSN by following formula, PSN= Wsc (Vrated / Vsc)and value of line BE equal to PSN, so power for 1 cm is equal to ratio between PSN and length of EB in cm. To find out full load quantities:

21. Extend the line EB from B to K such that BK = Output power (from the name plate details of motor) / Power scale. 22. Draw the parallel line to output line AB, which cuts the semicircle at point L. (near by Y xis) 23. Draw the parallel line for Y axis from point L to X axis (point Q). Then join O and L.



Procedure to find out full load quantities and maximum quantities from circle diagram:

1. F i xed l o s s = D E*P o w er S ca l e (wat t s ) 2. S t at o r Cu l o s s = DG * P o wer S ca l e(w at t s ) 3. Ro t o r Cu l o s s = G B * P o wer S c a l e ( wat t s ) 4. Maximum torque = *HH’*Power Scale+ / *2ЛN/60+ N-m 5. Maximum Output power = II’ * Power Scale (watts) 6. Maximum Input power = JJ’ * Power Scale (watts) 7. Maximum Efficiency = Maximum Output power / Maximum input power 8. Full load current = OL * Power Scale (amps) 9. Full load Power factor = Cos (angle between OL and Y axis) 10. Full load torque = *LN’ * Power Scale+ / *2ЛN/60+ N-m 11. Full load output power = LM * Power scale (watts) 12. Full load input power = LQ * power Scale (watts) 13. Full load efficiency = Full load output power / Full load input power 14. Full load Stator Cu loss = NP * Power Scale (watts) 15. Full load rotor Cu loss = MN * Power Scale (watts) 16. Full load rotor input = LN * Power scale (watts) 17. Full load slip = Full load rotor Cu loss / Full load rotor input 18. Full load speed =synchronous speed * (1-Slip) 19. Starting Torque = *BG * Power Scale+ / *2ЛN/60+ N-m

Procedure:

I. No load test:




2) Make sure that there is no load on the motor & auto transformer is at it’s minimum (zero)

value.


4) With the help of auto transformer increase the voltage gradually till voltmeter indicates

rated voltage of the motor.

5) Note down corresponding readings of ammeter, voltmeter & wattmeter’s.

6) With the help of auto transformer reduce the voltage to zero.


II. Blocked rotor test: 1) Do the connections as per circuit diagram.

2) Make sure that rotor of motor is blocked with the help of mechanical loading

arrangement. And position of auto transformer is at it’s minimum (zero) value.


4) With the help of auto transformer increase the voltage gradually till rated current flow

through the stator winding of the motor. (Rated current is mentioned on nameplate)

5) Note down corresponding readings of ammeter, voltmeter & wattmeter’s.

6) With the help of auto transformer reduce the voltage to zero.


Observation Table: I.No Load Test:

Sr. No.

Rated voltage

No load Current

Wattmeter Readings Total Power Wo=W1+W2

Vo (volts) Io ( amps) W1 (watts) W2 (watts) (watts)

1 415V

II.Blocked Rotor Test:

Sr. No.

Reduced Voltage

Rated Current

Wattmeter Readings Total Power

Wsc=W1+W2 Vsc(volts) Isc (amps) W1 (watts) W2 (watts) (watts)

1 4.7A



Questions:




4) What are the specifications of induction motor?

5) How measuring instruments (voltmeter, ammeter, wattmeter etc.) ratings are selected?

6) What is the aim of open circuit test?

7) What is aim of blocked rotor test?

8) While performing open circuit test rated voltage is applied. why?

9) While performing blocked rotor test rated current is allowed to flow through the winding.

Why?

10) What are the various equivalent circuit parameters of induction motor?

11) What is the use of equivalent circuit parameters of induction motor?

12) Draw approximate equivalent circuit of induction motor?

13) What is circle diagram of induction motor?

14) What is the use of circle diagram of induction motor?

15) What is the procedure to draw circle diagram of induction motor?

16) How to separate rotor copper loss & stator copper loss?

17) Which are the various quantities can be calculated from circle diagram?

18) How to calculate full load current, full load slip, full load power factor, full load efficiency?

19) How to calculate maximum torque, maximum output power?



Experiment No: 10

Date:

Title: Effect of variation of rotor resistance on the performance of 3Φ slip ring

Induction Motor.

Roll No:




Title: Effect of variation of rotor resistance on the performance of 3Φ slip ring Induction Motor.





Aim: To plot torque slip characteristics (in normal operating region) of 3Φ slip ring Induction

Motor for various values of rotor resistance inserted in rotor circuit.

Equipment: 3- slip ring Induction Motor.

Specifications: i) Rated output power: 5 HP

ii) Rated voltage: 415 Volts

iii) Rated current: 7.5 Amps

iv) Rated speed: 1410 RPM

v) Supply frequency: 50 Hz

vi) Type: Slip ring

vii) Connection of winding: Υ / Υ

viii) Rating: Continuous duty (S1)

ix) Class of insulation: B

Name of the manufacturer: BENN Electricals

Apparatus:

Sr. No.


1 3 Φ slip ring Induction Motor with BDA.

3 Φ, 5HP, 415V, 7.5A, 50Hz, 1410RPM

1

2 3 Φ Auto transformer 10A, 440V/0-470V 1 3 Rotor resistance starter For 5 HP 1 4 Voltmeter(AC) 0-600V 1 5 Ammeter(AC) 0-10A 1 6 Wattmeter 10A,600V 2 7 Tachometer 0-3000RPM 1 7 Connecting wires 1.5 sq.mm. ---



Theory:

3Φ Induction Motor according to the rotor construction can be classified into two types,

1) Squirrel cage Induction Motor

2) Slip ring Induction Motor

The construction of slip ring induction motor:

1) The stator construction of 3 Φ slip ring Induction Motor is similar to squirrel cage Induction

Motor i.e. a laminated stator core is slotted on periphery carrying polyphase distribution type

winding.

2) The rotor consists of 3 Φ Y-connected double layer distribution type winding similar to that of

stator, wound with same no. of poles like stator.

3) The 3 ends of 3 Φ Y-connected winding are permanently connected to the slip rings & brush

assembly mounted on shaft.

Features of slip ring Induction Motor:

1) The possibilities of adding an external resistance in series with each phase of rotor winding is an

important feature of this motor.

2) By this arrangement the value of rotor resistance per phase can be controlled which in turn

helps into controlling starting torque.

3) Such added resistances are cut out gradually & then finally removed from rotor circuit in normal

running condition.

Advantages of slip ring Induction Motor:

1) High starting torque.

2) Low starting current.

3) Speed can be varied due to additional external resistance in rotor circuit.

Disadvantages:

1) Construction is complicated & delicate.

2) Due to slip rings & brushes, frequent maintenance is necessary which increases maintenance

cost.

3) Rotor Cu loss is high hence efficiency is less.

Applications:



1) Slip ring Induction Motor is preferred for loads which require high starting torque.

2) They are used in cranes, hoists, elevators, compressors, crusher etc.

Effect of variations of rotor resistance on performance of slip ring Induction Motor as

follows,

a) Effect of rotor resistance on starting current,

Rotor current under running condition is given by,

𝑰𝟐 =s . E2

(𝑅2)2 + (𝑠.𝑋2)2

At start, s=1

𝑰𝟐 = E2

(𝑅2)2 + (𝑋2)2 =

E2

𝑍2

Now, with increase in rotor resistance, rotor impedance increases which reduces rotor current and

also starting current of stator.

b) Effect of rotor resistance on starting power factor,

Power factor cosΦ2 under rotor running condition is given by,

cos Φ2 = R2

(𝑅2)2 + (𝑋2)2 =

R2

𝑍2

Thus, with increase in rotor resistance, rotor impedance increases and hence power factor also

increases.

c) Effect on starting torque,

Torque developed in Induction Motor depends upon following three factors,

i) Rotor current, I2 at slip ‘s’

ii) Rotor P.F, cos Φ2 at slip ‘s’

iii) Flux (Φ)

T α Φ2 I2 cos Φ2

T α K. E2.I2.Cos Φ2 *Φ2= E2K]

Substituting values of I2 & Cos Φ2

T =K . E2 𝑥s .E 2

(𝑅2)2+(𝑠.𝑋2)2 𝒙 R2

(𝑅2)2+(𝑠.𝑋2)2



𝑇 = 𝐾 .𝑠.𝑅2 .𝐸2

2

𝑅22 + 𝑠2 .𝑋2

2

This is a general torque equation of induction motor under running condition.

Slip,

𝑠 =𝑁𝑠 − 𝑁

𝑁𝑠

Where, s = slip Ns = synchronous speed in rpm Nr = rotor speed in rpm At start, speed N = 0. Slip,

𝑠 =𝑁𝑠 − 0

𝑁𝑠= 1

Starting Torque,

𝑇𝑠𝑡 = 𝐾.𝑅2 .𝐸2

2

𝑅22 + 𝑋2

2

As discussed above, rotor current I2 decreases while rotor power factor Cos Φ2 increases at start

with increased rotor resistance.

As the effect of increased power factor predominates over decreased rotor current. Thereby,

increasing starting torque with increased rotor resistance.

d) Effect of Rotor Resistance on torque slip & speed torque characteristics,

Effect of addition of external resistance on torque slip & speed torque characteristics in

running conditions is given by,

𝑇 = 𝐾 .𝑠.𝑅2 .𝐸2

2

𝑅22 + 𝑠2 .𝑋2

2

Also the slip at given speed ‘N’ is given by,

𝑠 =𝑁𝑠 − 𝑁

𝑁𝑠

Case I:- When N = Ns ,s=0 In this case torque is zero, hence torque slip curve starts from S=0 as

shown in figure & speed torque characteristics starts from N=Ns as shown in figure.

Case II: Effect of the rotor resistance in the low slip region of characteristics. when speed ‘N’ is

very near to synchronous speed, the slip ‘s’ is very low (𝑠2 .𝑋22 ) is negligibly in comparison with 𝑅2 .

Torque,

𝑇 𝛼 𝑠.𝑅2

𝑅22 α

𝑠

𝑅2



Thus, in low slip or stable region, torque is inversely proportional to R2 for given value of slip ‘s’

i.e. if Rotor resistance is increased the torque developed decreases for same value of slip or speed.

Now, to produce the same torque speed falls increasing slip & compensation the reduce in torque.

Thus, by adding external resistance in rotor circuit, speed below normal value can be obtained.

Case III: Effect of rotor resistance in the high slip region.

As slip increases (that is speed decreases with increase in load), the term 𝑠2 .𝑋22 becomes large, so

that 𝑅22 may be neglected in comparison with𝑠2 .𝑋2

2

𝑇 𝛼𝑠.𝑅2

𝑠2.𝑋22 α

𝑅2

𝑠 .𝑋22

Thus the torque is inversely proportional to slip towards standstill conditions. The torque slip

characteristic is represented by a rectangular hyperbola.

e) Effect of Rotor resistance on max torque: Maximum torque is obtained by putting S=Sm in

torque eqn,

Max torque(Tm) = K*E22/2*X2

It is seen that variation in Rotor resistance does not change the magnitude of maximum torque

but nearly changes the value of slip at which it occurs, large the rotor resistance greater is value of

slip at which max torque occurs.

Max torque can be obtained at start if value of S= Sm =R/X2 = i.e. making rotor resistance equal

to its reactance by adding external resistance.

f) Effect of increasing rotor resistance on P out & η:

Addition of external high resistance in rotor circuit is helpful in achieving high starting

torque but such high resistance will increase rotor copper loss, which will decrease power output

& efficiency.

Hence, such an added resistance should be gradually & then finally removed from rotor

circuit under normal running condition of motor. Thus, good performance at start & in running

condition is obtained.

Precautions:



1) 3-phase autotransformer should be at minimum voltage position.

2) There should be no-load at the time of starting (Loosen the belt on the brake drum)

3) Brake drum should be filled with water.

Procedure:


2) Make sure that their is no load on machine, auto transformer is at it’s minimum (zero) value &

rotor resistance starter should be at it’s maximum value.


4) With the help of auto transformer increase the voltage gradually till voltmeter indicates rated

voltage. Note down no load reading.

5) Increase the load on motor gradually upto full load & note down corresponding readings.

6) Decrease the resistance in the rotor circuit with the help of rotor resistance starter and repeat step 5.

7) Decrease the load; bring auto-transformer to its minimum voltage position.


Any case wattmeter starts indicating negative reading (below 0) stops. Interchange

wattmeter connections either interchange M & L or interchange C & V.

Restart motor at no load condition.

Adjust load & consider power reading after coil is reversed to be –ve

Observation Table:



Rotor position

Sr. No.

Voltage Current Wattmeter Spring balance Speed

V (volts)

I (amps)

W1 (watts)

W2 (watts)

S1 (kg) S2 (kg) N (rpm)

R1

1

2

3

4

5

R2

6

7

8

9

10

R3

11

12

13

14

15

R4

16

17

18

19

20

Radius of pulley=



Calculations:

7. Input power, Pin = W1 + W2 -------watts

8. Shaft torque, Tsh = 9.81 x r x (S1-S2) -------Nm

9. Output power, Pout = 2 x π x N x Tsh ______________ -------watts 60

10. Efficiency, %η = Pout ________ x 100 Pin

11. P er c en t ag e s l i p , %s = (N s -N ) - - - - - - - - - - x1 0 0 N s

Ns = synchronous speed in rpm

N = Actual speed of rotor in rpm

12. Power factor, cosφ = (W1+W2)/√3VI



Result Table:

Rotor position

Sr. No.

Input power Shaft

torque Output power

Efficiency Slip P.F.

Pin (watts) Tsh (Nm) Pout

(watts) %η %s cosφ

R1

1

2

3

4

5

R2

6

7

8

9

10

R3

11

12

13

14

15

R4

16

17

18

19

20



Conclusion:

Questions: Q.1 What are the objectives of the experiment? Q.2 What are the precautions to be taken while performing experiment?

Q.3 What is the procedure of the experiment?

Q.4 What are the specifications of Induction motor?

Q.5 Whether two wattmeter’s measures input power or output power?

Q.6 What are the advantages of two wattmeter method to measure power?

Q.7 How to calculate multiplying factor of wattmeter? What is the value in this experiment?

Q.8 Why one of the wattmeter shows negative reading for first few readings? And how to take

such negative reading?

Q.9 How to calculate output power in this experiment?

Q.10 How to calculate shaft torque in this experiment?

Q.11 How to calculate efficiency of an Induction motor?

Q.12 How to calculate radius of pulley if it is not given?

Q.13 What is working principle of induction motor?

Q.14 What are the main parts in induction motor?

Q.15 Why stator and rotor core is made up of from Silicon Steel Stampings?

Q.16 What are the types of Induction Motor?

Q.17 What is synchronous speed of induction motor? How it is calculated?

Q.18 Can N=Ns? What will happen when N=Ns?

Q.19 What is slip speed and slip?

Q.20 Why frequency of rotor emf & rotor current is very small? And how it is calculated?

Q.21 What are the different losses that occurs in induction motor? Where it occurs?

Q.22 Draw power flow diagram of induction motor?



Q.23 Draw torque-speed characteristic of induction motor? And show stable, unstable operating

region, maximum torque, starting torque, slip at start, slip at running condition.

Q.24 What is torque equation of Induction Motor? What is the condition for maximum torque?

Q.25 What are the advantages of slip ring induction motor over squirrel cage induction motor?

Q.26 What are the disadvantages of slip ring induction motor over squirrel cage induction motor?

Q.27 What are the applications of slip ring induction motor?

Q.28 What is the difference between squirrel cage and slip ring induction motor?

Q.29 Wheather this is direct loading test or indirect loading test?

Q.30 What are the drawbacks of direct loading test?

Q.31 Explain how power factor of induction motor varies with load?

Q.32 What is the general range of efficiency of induction motor?

Q.33 If no load current of induction motor is compared with no load current of transformer for

same capacity, in which case it will be more? Why?

Q.34 Explain effect of adding external resistance in series with rotor winding on starting t orque,

power factor, efficiency and speed?

Q.35 What are the different methods to improve starting torque of slip ring induction motor?

Q.36 What is the necessity of starter for three phase induction motor?

Q.37 What are the different types of starters used for slip ring induction motor?

Q.38 Which type of starter is used in this experiment?

Q.39 Wheather rotor winding is connected in star or delta?

Q.40 Can we connect rotor winding of slip ring induction motor in delta? Why?

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