O.C & S.C Tests on 1Φ Transformer

CYCLE -1

EXPERIMENT NO – 01

O.C & S.C TESTS ON 1Φ TRANSFORMER

AIM : To conduct O.C and S.C tests on the given the 1Φ Transformer to pre-determine efficiency and regulation for different loads and p.fs and comparing with the results obtained by conducting Direct load test.

APPARATUS REQUIRED : i) O.C and S.C tests : One (230V/115V, 2kVA) single phase transformer, one auto transformer, one (0-1)A and one (0-10)A a.c ammeters, one (0-150)V and one (0-30)V a.c voltmeters, one (0-1A, 0-150V, LPF) and one (0-10A, 0-30V, UPF) wattmeters and connecting wires. ii) Direct Loading : The above 1Φ transformer, one (0-10)A and one (0-20)A a.c ammeters, one (0- 150)V and one (0-300)V voltmeter, one (0-10A, 0-300V UPF) wattmeter, one water load and one lamp load and connecting wires. i) O.C.TEST:

WOC CIRCUIT DIAGRAM: A M L S1 P1 C 150V (0-1)A R A E HV 1Φ 50Hz SIDE 230V AC N (0-150)V V

C S2 P2

WOC = (0-1)A, (0-150)V, LPF Wattmeter

PROCEDURE : 1. The circuit is rigged up according to the diagram given above with the H.V side kept open.

2. The supply is switched on. By operating the auto-transformer, the rated voltage of 115V is applied to the L.V side of the transformer.

3. The corresponding reading of ammeter and wattmeter are noted and tabulated as under.

1 TABULAR COLUMN -1

Sl. Voc Ioc Woc x constant No Volts Amps Watts

1. 115

ii) S.C TEST :

W (0-10)A SC A M L P S A 1 1 HV LV

R C 30V E V 1Φ 50Hz 230V AC N (0-30)V

C P S 2 2

W = (0-10/20)A, (0-30)V, UPF Wattmeter SC PROCEDURE : 1. The connections are made as shown in the diagram above with the L.V side shorted.

2. With the auto-transformer position at zero, the supply is switched on. By carefully operating the auto transformer, the rated current of the H.V side of the transformer is circulated which is indicated by the ammeter. The rated current of H.V side can be calculated by using equation.

KVA x 103 2.0 x 103

ISC = ------= ------= 8.65 Amps 230 230

The corresponding readings of the voltmeter and wattmeter are noted and tabulated as under. 2 TABULAR COLUMN - 2

Sl. ISC VSC WSC x Constant No. Amps Volts Watts

1. 8.65

VERIFICATION BY DIRECT LOADING :

CIRCUIT DIAGRAM :

(0-10)A (0-20A) M L I I M L 1 2 P S A 1 1 A

C 300V C 150V R W W 1 2 V V 230V 1Φ 1 2 AC (0-300)V (0-150)V

P S 2 2

W = (0-10)A, (0-300)V, UPF Wattmeter 1

W = (0-20)A, (0-150)V, UPF Wattmeter 2 PROCEDURE:

1. The connections are made as shown in the diagram shown above.

2. With the load switch open, the supply switch is closed.

3. Using the autotransformer, the rated voltage of 230V is applied to the primary of the transformer and the no-load readings are noted.

4. The load switch is now closed. Using the water and lamp loads, the load current is increased till the wattmeter W2 indicates 250 Watts (0.25kVA at u.p.f) taking care to see that the primary applied voltage is 230V. This is achieved by using the auto-transformer. After making sure that the applied voltage is 230V and wattmeter W2 indicates 250 watts, the readings of all the meters are noted in tabular column-3.

5. The above step is repeated for different wattmeter (W2) readings indicated in the tabular column-3. At each step, the readings of all the meters are noted. 6. To verify the regulation calculated using O.C and S.C tests, the no-load readings, corresponding to V1=230 volts, are noted in tabular column-4. 3 7. Using the water load and lamp load, the secondary current of the transformer is made equal to it’s rated value (17.3 Amps) making sure that the primary applied voltage is 230V. The readings of all the meters are noted.

TABULAR COLUMN-3

Sl. W2 V1 V2 I1 I2 W1 % No. kVA Volts Volts Amps Amps kVA 1 0.0 230 2 0.25 __,,__ 3 0.5 __,,__ 4 1.0 __,,__ 5 1.5 __,,__ 6 2.0 __,,__

TABULAR COLUMN-4

Sl. I2 I1 V1 V2 % No. Amps Amps Volts Volts Regulation 1 0 230

2 17.3 230

CALCULATIONS:

1. O.C & S.C TESTS: a) Efficiency: The percentage efficiency can be calculated using the formula,

Output Output %  = ------x 100 = ------x 100 Input Output + Losses

X (KVA rating x 103 ) Cos  = ------x 100 3 2 X (KVA rating x 10 ) Cos  + WOC + X WSC where ‘X’ is the fraction of the full load KVA of the transformer at which efficiency is required to be calculated. cos  is the power factor of the load which can be assumed as 1, 0.8, 0.6 etc.

4 WOC is the iron loss of the transformer which is obtained from O.C test.

WSC is the full load copper losses of the transformer which is obtained from S.C test.

The efficiency can be calculated for X = 1/8, 1/4, 1/2, 3/4 and 1 (full load) and the results are tabulated as under, (The p.f of the load is assumed as 1.0 & 0.8).

Sl. Fraction Output Copper Iron Total CosΦ = 1.0 Cos Φ = 0.8 No. of F.L Losses Losses Losses Input %  Input %  ( X ) KVA Watts Watts Watts Watts Watts 1. 1/8 0.25

2. 1/4 0.5 __,,__

3. 1/2 1.0 __,,__

4. 3/4 1.5 __,,__

5. 1.0 2.0 __,,__

The graph of % v/s KVA output is plotted as shown.

%

cosΦ = 1.0

cosΦ = 0.8

0 0.5 1.0 1.5 2.0

Output in kVA

b) Percentage Regulation :

The percentage regulation of the transformer may be calculated using the formula

I2R02 Cos  + I2 X02 Sin  % Regulation = ------x 100 V2 Where, I2 = Rated secondary current = 17.3 Amps. V2 = Rated secondary voltage = 115 Volts.

Cos  = p.f of the load which can be assumed as 0.4, 0.6, 0.8 and 1. In the above equation, +ve sign is used if the p.f is lagging and –ve sign is used if the p.f is leading. R02 = Equivalent resistance of the transformer referred to secondary. 5 X02 = Equivalent reactance of the transformer referred to secondary.

Here, R02 and X02 can be calculated from the readings obtained during S.C. Test. But, as the S.C Test is conducted with the meters connected on H.V side (primary side), we can first calculate the equivalent resistance and reactance referred to primary.

WSC i.e., R01 = ------Ohms 2 I SC

VSC Also, Z01 = ------Ohms where Z01 is called the equivalent impedance referred to primary ISC

______2 2  X01 =  Z 01 - R 01 Ohms.

2 2 Therefore R02 = K R01 and X02 = K X01 where ‘K’ is the transformation ratio of the transformer given by

115 K = ------= 0.5 230 The results are tabulated as shown . Sl. p.f % Regulation No. 1 0.4 lag 2 0.6 lag 3 0.8 lag 4 u.p.f 5 0.8 lead 6 0.6 lead 7 0.4 lead The graph of % regulation V/s. p.f is drawn as shown below.

+VE REG.

0.4 0.6 0.8 1.0 0.8 0.6 0.4

PF (leading) PF (lagging)

-VE REG. 2. DIRECT LOADING : a) Efficiency :

The percentage efficiency under different loaded conditions are calculated using the equations, W2 % = ------x 100 W1 b) Regulation :

The % Regulation at F.L and u.p.f is calculated using the equation,

V2(NL) – V2 (FL) % Regulation = ------x 100 V2(NL)

The % efficiency and % Regulation calculated from O.C & S.C tests (for u.p.f) and by direct loading are tabulated, as shown below, to compare the results.

Sl. Output %  by %  by No. kVA O.C & S.C tests direct loading 1 0 0 0 2 0.25 3 0.5 1. % Regulation at F.L., UPF by O.C & SC tests : 4 1.0 5 1.5 6 2.0 ______

2. % Regulation at F.L., UPF by direct loading : ______

CONCLUSIONS: (To be written by the student ) While drawing conclusions the points that should be considered are: a. Has the object of the experiment been achieved? b. Do the results substantiate the theory? If the results appear to contradict the theory, the probable reasons should be discussed. c. Proper interpretation of graph.

****** EXPERIMENT NO – 02

SUMPNER’S TEST ON 1Φ TRANSFORMERS

7 AIM : To conduct Sumpner’s test on two identical transformers and obtain a graph of % Efficiency V/s Load.

APPARATUS REQUIRED : Two identical 2 kVA 1Φ transformers, one (0-1) A and one (0-10) A a.c ammeters, one (0-30) V, one (0-300)V and one (0-600)V a.c voltmeters, one (0-1A, 0-300V) LPF wattmeter, one (0-10A, 0-30V) UPF wattmeter, one 15 amps single phase auto - transformer and connecting wires.

CIRCUIT DIAGRAM:

(0-1)A W 1 I M L P P 1 1 1 A

230V 50Hz AC C 300 V S S 1 1

V SEC PRI (0-300)V

N S V S 2 2

(0-30)V

P P 2 2

(0-10)AA A I M L 2

C 300V W V 2

E SPST (0-600)V

W = (0-1) A (0-300)V LPF Wattmeter 1 W = (0-10)A (0-300)V UPF Wattmeter 2

BED NO. NAME PLATE DETAILS

TRANSFORMER-1 TRANSFORMER-2

PROCEDURE:

1. The connections are made as shown in the circuit diagram.

8 2. With the SPST switch kept open and the auto-transformer at minimum position, the supply switch is closed.

3. The polarities of the secondary voltages of the transformers are checked using the voltmeter connected across SPST switch.

4. If the voltmeter indicates zero reading, the SPST switch is closed. If the voltmeter indicates twice the secondary voltage, then the secondary terminals of any one of the transformers are interchanged to make the voltmeter read zero and then SPST switch is closed.

5. Now, by operating the auto-transformer, the voltage applied to the secondary is increased gradually till they carry their rated current (8.7 amps for 2 kVA transformer). The reading of all the meters are noted and tabulated as under.

TABULAR COLUMN:

Sl. V1 I1 W1 x Constant V2 I2 W2 x Constant No. Volts Amps Watts Volts Amps Watts

1. 8.7

SPECIMEN CALCULATIONS: W1 x Constant Iron losses of each transformer = Wi = ------watts. 2

W2 x Constant Full load copper loss of each transformer = Wc = ------watts 2

The percentage efficiency of each of the transformers can be pre-calculated using the equation.

X (kVA rating x 103 ) Cos  %  = ------X 100 3 2 X (kVA rating x 10 ) Cos  + (Wi + X Wc)

Where X is the fraction of full load kVA at which % is to be predetermined and Cos  is the p.f of load. 1 1 1 3 The % is pre-determined for X = ------, ------, ------, ------, 1 and assumed load 8 4 2 4 p.f of 1.0 as well as 0.8 .

The results are tabulated as under.

9 Sl. Fraction Output Copper Iron Total cosΦ = 1.0 cosΦ = 0.8 No. of F.L (X) Losses Losses Losses Input %  Input %  kVA Watts Watts Watts Watts Watts 1. 1/8 0.25 2. 1/4 0.5 __,,__ 3. 1/2 1.00 __,,__ 4. 3/4 1.5 __,,__ 5. 1.0 2.0 __,,__

GRAPHS : The efficiency curves are drawn as shown below.

cosΦ = 1.0 % η cosΦ = 0.8

0 kVA OUTPUT

NOTE : The above test is also known as “Heat run test” and is mainly conducted to pre- determine the efficiency of the transformer as well as the temperature rise during loaded conditions.

******

EXPERIMENT NO – 03

LOAD TEST ON 1Φ INDUCTION MOTOR

10 AIM : To conduct load test on the given single phase induction motor and to plot it’s performance characteristics.

APPARATUS REQUIRED : A capacitor start, 1Φ induction motor, One (0-10A, 0-300V) UPF wattmeter, one (0-10)A a.c ammeter, one(0-300)V a.c voltmeter, and connecting wires.

CIRCUIT DIAGRAM :

W (0-10)A M L A M

C 300V

R F F 2 1 (0-300)V 230V 1Φ AC V

W = (0-10)A (0-300)V UPF Wattmeter

BED NO. NAME PLATE DETAILS

PROCEDURE : 1. The connections are made as shown in the figure.

2. With the SPST switch ‘S’ closed, the supply is switched on.

3. The SPST switch is opened once the motor starts running.

4. The no-load readings and the no load speed are noted.

5. The motor is loaded in steps of 1 Amp using the brake drum arrangement. At each step, the readings of meters, the speed and the spring balance readings F1 and F2 are noted. The 11 loading of the motor is continued till the current drawn by the motor equals it’s rated current.

The readings taken are tabulated as shown.

TABULAR COLUMN :

Sl. I W V N F1 F2 F = (F1- F2) No. Amps Watts Volts R.P.M Kgs. Kgs. Kgs. 1

2 5

3 6

4 7

5 8

CALCULATIONS :

If F1 and F2 are the spring balance readings and ‘r’ is the radius of the brake drum in ‘meters’, then the shaft torque is given by

Tsh = (F1 - F2) x r Kg - meters.

2  N Tsh  B.H.P output of the motor = ------where N is the speed of the motor at that 4500 load. The output of the motor in watts is then given by, output = B.H.P x 735.5 watts. But, the input to the motor = W watts. B.H.P x 735.5  % Efficiency of the motor = ------x 100 W 120 x f 120 x 50 If NS is the synchronous speed of the motor given by Ns = ------= ------= 1500 RPM P 4

Where f = frequency of the supply and P = number of poles of the motor and N is the actual speed, then the % slip can be calculated using the equation,

Ns - N % slip = ------x 100 Ns 12

W The p.f of the motor can be calculated using the equation, power factor = ------V x I The results are tabulated as under

Sl. I B.H.P % P.F % slip No. Amps output 1 2 5 3 6 4 7 5 8

GRAPHS :

PF % I % slip

%

%Slip

0 0.25 0.50 0.75 1.00 OUTPUT IN BHP

The graph of % slip, % efficiency, power factor and the line current V/s output in B.H.P is drawn as shown.

EXPERIMENT NO. 04

LOAD TEST ON 3Φ INDUCTION MOTOR

13 AIM : To conduct load test on the given three phase induction motor and to plot its performance characteristics.

APPARATUS REQUIRED : A three phase induction motor – D.C generator set, two current transformers, two (0-2)A (0-600)V UPF wattmeters, two (0-2)A a.c ammeters, one (0-25)A d.c ammeter, one (0-300)V d.c voltmeter, one (750 x 2) ohms rheostat, water load, and connecting wires.

CIRCUIT DIAGRAM:

C 25 (0-25)A + - M L A A 1 (750x2)Ω A A C 1 1 2 A A A Y-∆ 2 M L Z G R (0-2)A S W B + 1 T 1 415 V C 600V A V A 2 50Hz Y C (0-300)V R 1 3Φ C 600V B B T 1 2 _ AC B AA ZZ A E 2 B R M L

(0-2)A C 1 W 2 M L C 2 C 25

W1 & W2 = (0-2)A , (0-600)V, UPF, WATTMETER

BED NAME PLATE DETAILS NO. D.C Machine Induction Motor

PROCEDURE :

1. The connections are made as shown in the diagram.

2. With the secondary switches on C.Ts closed and DPST switch open, the supply switch is closed. 3. The induction motor is started using the star-delta starter.

4. Now, the d.c generator voltage is built up to its rated value of 220V by cutting–out (750 x2) ohms rheostat.

14 5. The secondary switches of the C.Ts are now opened and the no-load readings of all the meters and the no load speed are noted. If any of the wattmeters “Kick Back”, then the secondary switch of the corresponding C.T is closed and the current coil of that wattmeter is interchanged. The secondary switch of the C.T is now opened and the reading of wattmeter is noted as “-ve”.

6. The induction motor is now indirectly loaded by loading the d.c generator in steps of about 3 amps. At each step, the readings of all the meters and the speed are noted. The loading of the induction motor is continued till the current drawn by the motor equal it’s rated value. The readings are tabulated in the tabular column as shown.

CALCULATIONS :

The input to the induction motor = (W1 + W2) watts

Output of the D.C Generator Output of the induction motor = ------Efficiency of the Generator

VDC X IDC = ------nG

The efficiency of the generator “G’’ at an output can be found from the known efficiency graph of the d.c generator which is drawn using the following data.

Output In Watts 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000

G 0 0.48 0.75 0.82 0.88 0.9 0.92 0.91 0.9 0.88 0.86

Output of I.M in watts The B.H.P output of I.M = ------735.5

The percentage efficiency of the I.M can be calculated using equation

Output of I.M in watts %  = ------x 100 Input to I.M in watts

The shaft torque of the I.M can be found using the equation

B.H.P x 4500 T = ------Kg-mts where N is speed of the I.M in RPM. 2  N

120 x f If Ns is the synchronous speed of the Induction Motor given by Ns = ------P 15 (where f = frequency of the supply and P = no. poles of the induction motor) and N is the actual

speed, then the % slip can be calculated using the equation

NS - N % Slip = ------x 100. NS

120 x 50 Here, Ns = ------= 1500 RPM 4 The p.f of the induction motor can be calculated using the equations.

___ W1 - W2  = tan –1  3 x ------and p.f = cos  W1 + W2

The results obtained are tabulated as shown.

Sl. IDC VDC I1 x I2 x W1 x W2 x Speed B.H.P %  Torque % Power No Amps Volts C.T C.T Const. Const N T Slip Factor Ratio Ratio x C.T x C.T R.P.M kg-mt Amps Amps Ratio Ratio Watts Watts 1 0

2 3

3 6

4 9

5 12

5 15

7 18

GRAPHS :

The graphs of B.H.P Vs % efficiency, % slip, Torque and p.f and Speed Vs Torque are drawn as shown . 16

PF % T % slip

%

%Slip

0 0.25 0.50 0.75 1.00 OUTPUT in BHP

****** CYCLE -2

EXPERIMENT : 05

NO-LOAD AND BLOCKED ROTOR TESTS ON 3-PHASE INDUCTION MOTOR AND PERFORMANCE EVALUATION USING ‘ CIRCLE DIAGRAM’. C 25

3Φ A C AIM : To conduct no-load and block rotor tests on the given three phase Induction motor1 to2 obtain the circle diagram to pre-determine AUTO the performance of the I.M. M L T R W A (0-2)A V (0-600)V APPARATUS REQUIRED : A threeA phase1 Induction Motor, Two current transformers, Two (0- 2)A (0-600)V UPF wattmeters, TwoN (0-2)A (0-100)V M LPF L wattmeters, Two (0-2)A a.c ammeters, R S One (0-5)A d.c ammeter one (0-12)VF d.c voltmeter, Lamp Load, 600 One 3 phase auto-transformer and 415V 3Φ C A C connecting wires. O 2 1 50 Hz AC Y R C 600 B B M 1 2 E CIRCUIT B DIAGRAM : R A M L W (0-2)A 1 M L 17

C 25

W & W = (0-2)A, (0-600)V LPF Wattmeter 1 2

BED NAME PLATE DETAILS NO. INDUCTION MOTOR

PROCEDURE :

I. NO-LOAD TEST :

1. The connections are made as shown in the diagram.

2. With the auto-transformer at the minimum position, the supply is switch on.

3. By operating the auto-transformer, the rated voltage of 415V is applied to the stator of the induction motor.

4. All the meter readings are noted down. If any of the wattmeter “KICK BACK”, then the secondary switch of the corresponding C.T is closed and the current coil of the wattmeter is interchanged. The secondary switch of the C.T is now opened and the reading of the wattmeter is noted as ‘-ve’.

II. BLOCKED ROTOR TEST :

1. The circuit diagram remains the same .In this test except that the L.P.F wattmeters are replaced by U.P.F wattmeters and (0-600)V voltmeter by (0-100)V voltmeter. 18 2. With the auto-transformer at minimum position, the supply switch is closed.

3. With the rotor of the I.M blocked (held stationary manually), the voltage applied to the stator is increased gradually till the I.M draws it’s rated current. The readings of the voltmeter and wattmeters are noted.

NO-LOAD TEST :

I1 + I2 Sl. Vo I1 x I2 x Io = W1 x Const. x W2 x Wo = W1 + W2 No. Volts C.T C.T ------C.T.Ratio Const. x Watts Ratio Ratio 2 Watts C.T.Ratio Amps Amps Amps Watts

1 415

BLOCKED ROTOR TEST :

I1 + I2 Sl. I1 x C.T I2 x C.T ISC = ------VSC W1 x Const. x W2 x Const. x WSC = W1 + W2 No. Ratio Ratio 2 C.T Ratio C.T Ratio Watts Amps Amps Amps Volts Watts Watts

1 7.5

(0-5)A A A 1 2 + - A III) MEASUREMENT OF STATOR RESISTANCE

+ + - V V CIRCUIT DIAGRAM : Sl. I V Rdc = ------220V DC (0-15)V No. I - Amps Volts Ohms 19 1 2 2 3 3 4

Average Rdc = Ohm 1. The connections are made as shown.

2. Using the lamp load, different currents are passed through any one phase of the stator of the Induction motor. For each of the current, the voltage drops across the phase winding is measured and noted. 3. VThe a.c resistance per phase (R1) is calculated using the equation R1 = 1.4 R d.c .

H CONSTRUCTION OF THE CIRCLE DIAGRAM :

K The stator of the given Induction motor is delta connected and hence the current Io and Is.c measured above are the line currents. M A 1. From the set of readings taken during no-load test, the angle o (angle between applied

voltage V and the no-load phase current Io /  3) is found using the equation,

__ W1 - W2

tan o =  3 x ------W1 + W2 P __

! As shown in figure the vector Io /  3 represented by O O is drawn, by selecting a suitable scale, at an angle Φ lagging behind the applied voltage V. Φ o SC

N E OUTPUT LINE

Φ Q R 0 TORQUE LINE F

I O' S C 0

0 T G

20 L

2. From the set of readings taken during the blocked rotor test, the angle Is.c Is.c (angle between applied voltage Vs.c and s.c phase current ------) is found using the equation,  3 W1 - W2

tan  s.c =  3 x ------W1 + W2

The s.c current / phase ISN, when rated voltage V is applied, is calculated using the equation, I s.c V ISN = ------x ------Amps.

 3 V s.c

This current ISN is then represented by a vector OA making an angle s.c lagging V as shown.

3. The point O! and A are joined and the line is bisected at right angles . Let this bisector cut the horizontal line drawn from O! at C.

4. With C as center and O!C as radius, a semicircle is drawn as shown . This passes through the point A. O! A is the output line.

5. The length AG in cms represent the s.c power input/phase with rated voltage V, given by

W S.C V 2 WSN = ------[------] watts . 3 Vs.c 21 WSN  Per phase power scale = ------Watts / cm. AG

6. Using the power scale, the point E, which fixes the torque line is located by taking the length EF equal to s.c stator copper loss/phase with normal voltage, given by

2 I SN x R1 EF = ------cms Power scale

CALCULATIONS :

(a) The line current, slip, efficiency, p.f and torque at Full Load can be calculated using the following procedure.

The line GA is extended to H such that

Rated output of I.M / Phase in Watts 3.7 x 103/3 AH = ------= ------cms. Power Scale Power Scale

From H, a line parallel to the output line AO! is drawn to obtain the full load operating point ‘P’. From the point P a perpendicular PT is drawn to the horizontal axis. From the circle diagram, the % efficiency can be calculated using the equation,

PQ % efficiency at F.L = ------x 100 PT

Rotor Cu loss QR Similarly, slip at F.L = ------x 100 = ------x 100 Rotor input PR

PT p.f at F.L = ------OP

Full load line current =  3 x OP x Current scale

Full load torque = 3 (PR x Power Scale) syn. Watts = X (say)

X = ------Kg – mts where NS = 1500 r.p.m

2  NS ------x 9.81 60

22 (b) The max. power output and max. torque can be calculated by drawing tangents parallel to the output and torque line respectively. Let these tangents touch the circle at M and K as shown.

 Max. torque output = 3 (KL x power scale) syn. Watts. = X (say)

X = ------Kg – mt

2  NS ------x 9.81 60 Max. power output = 3 (MN x power scale) watts.

Starting torque AE ------= ----- F.L torque PR

NOTE : It is recommended that the scale of the current vectors should be so chosen that the diameter is more than 20 cms in order that the performance data of the motor may be read with reasonable accuracy from the circle diagram.

******

EXPERIMENT NO. 06

PARALLEL OPERATION OF TWO DISSIMILAR 1-PHASE TRANSFORMERS

AIM : To connect two single phase transformers of different ratings in parallel and to study how they share a common load.

APPARATUS REQUIRED :

23 1. Load Test: One 2 kVA and one 3 kVA single phase 1:1 transformers, one (0-10)A one (0-15)A and one (0-25)A A.C ammeters, one (0-300)V and one (0-600)V A.C voltmeters, Three (0- 10/20A, 0-300V) U.P.F Wattmeters, one water load, one choke coil and connecting wires. 2. S.C Tests: one Auto-transformer, one (0-30)V voltmeter, one (0-10)A one (0-15)A A.C ammeters, One (0-10/20A, 0-300V) U.P.F Wattmeter and connecting wires.

CIRCUIT DIAGRAM:

(0-10)A (0-25)A 2 kVA M L I I M L A A L A

P S 1 1 C 300V C 300V W W A L R V 230V 1Φ V (0-300)V L AC N

P S 2 2

(0-20)A

3 kVA A M L I W = (0-10)A, (0-300)V UPF Wattmeter B A

W = (0-20)A, (0-300)V UPF Wattmeter A B W = (0-20)A, (0-300)V UPF Wattmeter L P S 1 1 V C 300 V W B

K (0-600)V

P S 1 1

NAME PLATE DETAILS BED NO. TRANSFORMER-A TRANSFORMER-B

PROCEDURE :

1. The connections are made as shown in the diagram given.

2. With the SPST switch open, the supply is switched on..

3. The reading of the voltmeter connected across switch ‘K’ is observed. If it reads zero, which means that the two secondary e.m.f’s are opposing, the switch ‘K’ is closed. If it reads twice the secondary rated voltage, the secondary terminals of one of the transformers are interchanged and then the switch ‘K’ is closed. 24 4. The no load readings are then noted. The load switch is now closed and the total load on the transformers is increased in steps of 3 amps using both resistive and reactive loads. At each step, the readings of the various meters are noted. The total load is increased till the current supplied by any one of the transformers reaches it’s rated value. The readings and the results are tabulated in the tabular column shown.

NOTE : The reactive load (choke coil) is used in order to get lower p.fs.

SHORT CIRCUIT TESTS :

W (0-10)A SC A M L S P A 1 1

R C 30V E V 1Φ 50Hz 230V AC N (0-30)V

C S P 2 2

W = (0-10/20)A, (0-30)V, UPF Wattmeter SC Short circuit test is conducted on each of the two transformers separately with the meters connected on the secondary and the readings are tabulated as under.

TRANSFORMER ‘A’ (2 kVA) TRANSFORMER ‘B’ (3 kVA)

Sl. ISC VSC WSC Sl. ISC VSC WSC No. Amps Volts Watts No. Amps Volts Watts 1 8.69 1 13.0

TABULAR COLUMN :

! ! ! ! ! ! Sl. IL VL WL IA -A IB -B WA WB IA -A IB -L WA WB No. Amps Volts Watts Amps Amps Watts Watts Amps Amps Watts Watts 1 0 2 3 3 6 4 9 5 12

25 6 15 7 18

CALCULATIONS :

I. TRANSFORMER ‘A’ (2kVA) :

From the s.c test readings, the following can be calculated.

ZA = equivalent impedance of transformer ‘A’ refereed to it’s VSC secondary = ------Ohms ISC RA = equivalent resistance of transformer ‘A’ refereed to it’s WSC secondary = ------Ohms 2 I SC  XA = equivalent reactance of transformer ‘A’ refereed to it’s

2 2 secondary =  Z A - R A Ohms

 ZA = (RA + j XA) Ohms = ZA  θA.

! The current IA to be shared by transformer ‘A’ when the total load current is IL is given by

ZB θ B ! ! IA -A = IL-L x ------where ZB  θ B is the equivalent impedance

ZA θA + ZB θB

of the transformer ‘B’ referred to it’s secondary which can be calculated using a similar procedure

explained above. L is the p.f angle of the load which can be calculated using the equation,

WL ! ! Cos L = ------. The current IA -A is compared with IA -A where A is given by VL IL

Cos A = ------VL IA

The Volt-amps to be shared by transformer ‘A’ is given by

ZBθB ! ! QA -A = QL -L x ------Volt-amps where QL is the total load Volt-amps

ZAθA + ZBθB

given by VLIL. The power to be supplied by the transformer ‘A’ is then given by

26 ! ! ! WA = QA Cos A watts .

II. TRANSFORMER ‘B’ (3 kVA):

! ! In a similar way, the current IB and the power WB to be shared by the transformer ‘B’ are calculated using the equations,

ZAA

! ! IB -L = IL -L x ------Amps.

ZAA + ZBB

ZAA ! ! QB -B = QL-L x ------Volt-amps

ZAA + ZBB ! ! !  WB = QB Cos QB Watts.

! ! ! ! The calculated values of IA , WA and IB ,WB are compared with the actual values IA,WA and IB ,

WB respectively as shown in the tabular column.

******

EXPERIMENT NO – 07

SCOTT CONNECTION USING 1Φ TRANSFORMERS

AIM : To convert a Three Phase System into two independent single phase systems using two Single Phase Transformers and to study their performance.

APPARATUS REQUIRED : One single phase 2 kVA transformer with 86.6% tapping in primary and another single phase 2kVA transformer with 50% tapping in primary, Two(0-10)A (0-

27 300)V UPF wattmeters, Two (0-10)A a.c ammeters, Two (0-300)V and One(0-600)V a.c voltmeters, One 3 Auto transformer, Two water or lamp loads and connecting wires.

CIRCUIT DIAGRAM:

(0-10)A (0-10)A W I 1 1 I M L R A A

P S 3Φ 1 1 C 300 415 V 3Φ AC A V TEASER V SUPPLY U (0-300)V TR. (0-300)V1 T 86.66% R O (0-10)A T P S A 2 2 Y R I A Y A I N 2 B P 50% P S 2 1 F MAIN TR. S 1 O V R (0-10)A 2 M E (0-300)V R C 300 A I B S 2 M L

W (0-10)A 2 W1 & W2 = (0-10)A , (0-300)V, UPF WATTMETER

BED NO. NAME PLATE DETAILS

PROCEDURE : 1. The connections are made as shown in figure. 2. With the auto-transformer at minimum position, the three phase supply is switched on. 3. Using the auto-transformer, the voltage applied to the primary of the Scott connected transformers is increased till the secondary voltages become 230V. 4. The no-load readings are noted.

28 5. The secondary’s are now loaded equally in steps of about 2 amps. At each step the readings of all the meters are noted. 6. The loads on the secondary’s are now removed.

7. One of the secondary’s is now loaded to fixed value (say I1 = 4 amps). 8. The load on the other secondary is increased in steps of about 2 amps. At each step, the readings of all the meters are noted and tabulated as under.

Sl. IR IY IB W1 W2 I1 I2 V1 V2 % E No. Amps Amps Amps Watts Watts Amps Amps Volts Volts Q 1 0 0 U A 2 2 2 L

3 4 4 L O 4 6 6 A D 5 8 8 U 1 4 0 N E Q 2 4 2 U A 3 4 4 L

4 4 6 L O 5 4 8 A D

CALCULATION :

As the loads applied to the secondary are resistive, the p.f is unity. Therefore, the total output of the secondary’s = (V1 I1 + V2 I2) W

The corresponding input to the primary is given by (W1 + W2) watts.

Therefore, the efficiency of the Scott connected transformer

29 V1 I1 + V2 I2 = ------x 100 W1 + W2

The % is calculated for both equal and unequal load conditions and a graph of % V/s total output in watts is drawn as shown below.

% EQUAL LOADS

UNEQUAL LOADS

OUTPUT IN WATTS

Vector diagram for u.p.f loads

V 1 I R

KI = 1.15KI 2 1

I 1

I I B Y

S S 1 1 I V 2 2 C TEASER B TRANSFORMER PRI 0 SEC SECONDARY 0 V (b) TO SHOW 90 PHASE SHIFT BETWEEN1 THE TWO SECONDARY

VOLTAGES V1 AND V2 : (0-300)V (0-600)V

S S 2 2 V 3 V =√2V V =V 3 3 S S 1 1

MAIN V TRANSFORMER 2 V =V 3 SECONDARY 30 (0-300)V

S S 2 2

Fig (a) Fig (b) Fig (c) PROCEDURE :

1. The no-load voltages V1 and V2 across the two secondary’s of the Scott connected transformers are first noted and the supply is switched off.

2. The secondary’s are then connected in series a shown in fig (b). The supply switch is again closed (without disturbing the position of the Auto transformer) and the total voltage V3 is noted.

TABULAR COLUMN :

Sl. V1 V2 V3 No. Volts Volts Volts

CALCULATIONS :

If V1 = V2 = V (say)

It will be seen that V3 =  2V. This can happen only when both V1 and V2 are equal and displaced by 900. This is as shown in the vector diagram.

31 ******

EXPERIMENT : 08

SPEED CONTROL OF 3 PHASE INDUCTION MOTOR BY STATOR VOLTAGE AND ROTOR RESISTANCE CONTROL

AIM : To conduct load test on the given 3 phase slip-ring induction motor under a) Two different voltages applied to the stator b) Two different rotor resistances and to compare the performances.

APPARATUS REQUIRED : One slip ring induction motor with variable rotor resistor, two current transformers, two (0-2)A (0-600)V UPF wattmeters, one (0-10)A a.c ammeter, one(0- 600)V a.c voltmeter, brake-drum arrangement and connecting wires.

CIRCUIT DIAGRAM:

C 25 3Φ C 25 3Φ M L A R A M L A1 R A 1 U U (0-600)V T (0-600)V 415 V 3Φ T (0-2)A A 415 V 3Φ O (0-2)A A V AC O V AC T M L R T M L R R R C 600V A B C R R C 600V A2 B2 C2 A 2 2 2 A W R F F N A 1 W R F1 F1 N A 1 1 1 S S F F Y O (0-10)A B STATOR C ROTOR Y R Y O (0-10)A B1 STATOR C1 ROTOR Y R R 1 1 R W M 2 W B M C 600V 2 B E C 600VA E M L A R M L R B B (0-2)A (0-2)A

PROCEDURE : M L M L C 25 C 25 W 1 & W1 = (0-2)A (0-600)V UPF WATTMETER

BED NO. NAME PLATE DETAILS

32 a) STATOR VOLTAGE CONTROL :

1. The connections are made as shown in the diagram. The external rotor resistance connected to the slip ring I.M is made equal to zero by short circuiting the rotor terminals marked R Y & B.

2. With the 3 auto-transformer at minimum position and C.T secondary switches closed, the supply switch is closed.

3. Using the 3 auto-transformer, the rated voltage of 400V is applied slowly to the stator of the slip ring I.M.

4. The C.T secondary switches are now opened and the no-load readings of all the meters are noted. If any of the wattmeters “KICK BACK”, then the secondary switch of the corresponding C.T is closed and the current coil connections of the wattmeter are inter changed. The secondary switch is then opened and the reading of that wattmeter is taken as – ve.

5. The I.M is now loaded, using break drum arrangement, in steps of about 1 amp. At each step, the readings of all the meters are noted and entered in the tabular column-1 shown. The loading of the I.M is continued till it draws it’s rated current. NOTE : When the load on the I.M is increased, the voltage applied to the I.M decreases due to auto-transformer. Before noting the readings, at each step, the applied voltage is maintained constant at 400V.

6. The mechanical load on the I.M is now reduced to zero and the voltage applied to I.M is reduced by about 10% (say) of the rated voltage i.e, to 360V using the auto-transformer.

7. The procedure given in step 5 is now repeated and the readings of all the meters are noted in Tabular column-2 shown.

8. The mechanical load on the I.M is now reduced and the voltage applied to the I.M is reduced to zero. The supply switch is then opened. b) ROTOR RESISTANCE CONTROL :

1. The short circuit placed at the rotor terminals marked RYB is now removed and the Star connected variable resistor is connected across the rotor windings.

2. Using the handle provided on the variable resistor, about 25% of the external rotor resistance /phase is included in the rotor circuit.

3. The procedure given in steps 2 to 5 of stator voltage control is now repeated and the readings of all meters are noted in tabular column- 3 shown.

4. The mechanical load on the I.M is now reduced and the voltage applied to the I.M is reduced to zero and the supply switch is then opened. 33 TABULAR COLUMN-1 : R OTOR RESISTANCE R/Phase = 0 Ω

Sl. VRY IY W1 x W2 x F1 F2 F = Speed BHP % Torque Slip Power No. Volts Amps Const Const kgs. kgs. (F1- N T factor x C.T x C.T F2) RPM kg.mts Ratio Ratio kgs. Watts Watts 1 400

2 400

3 400

4 400

5 400

6 400

TABULAR COLUMN-2 : ROTOR RESISTANCE R / phase = 0 Ω

Sl. VRY IY W1 x W2 x F1 F2 F = Speed BHP % Torque Slip Power No. Volts Amps Const Const kgs. kgs. (F1- N T factor x C.T x C.T F2) RPM kg.mts Ratio Ratio kgs Watts Watts

34 1 360 .

2 360

3 360

4 360

5 360

6 360

TABULAR COLUMN-3 : ROTOR RESITANCE R / phase = 25%

Sl. VRY IY W1 x W2 x F1 F2 F = Speed BHP % Torque Slip Power No. Cons Const (F1-F2) N T factor Amps t x C.T x C.T Ratio Ratio Volts Watts Watts kgs. kgs. kgs. RPM kg.mts 1 400

2 400

3 400

4 400

5 400

6 400

SPECIMEN CALCULATIONS :

If F1 and F2 are the spring balance readings and ‘r’ is the radius of the brake drum in ‘meters’, then the shaft torque is given by

Tsh = (F1-F2) r Kg-mts.

2π NTsh B.H.P output of the I.M = ------where N is the speed of the motor at that load. 4500 The output of the I.M, is then given by output = B.H.P x 735.5 watts.

But, the input to the motor = (W1 + W2) watts.

B.H.P x 735.5 35  % efficiency of the I.M = ------x 100 W1 + W2

120xf 120 x 50 If NS is the synchronous speed of the motor given by NS = ------= ------= 1500 RPM P 4 (where f is the frequency of the supply, P is the number of poles of the motor) and N is the actual speed, then the slip can be calculated using the equation, NS - N Slip = ------NS The p.f of the motor can be calculated using the equations,

W1 – W2  = tan-1 3 (------) and p.f = cos  W1 + W2

The results are tabulated in the respective tabular columns.

GRAPHS :

The graphs of Speed v/s Torque and ; p.f , % v/s B.H.P are drawn for both stator voltages and rotor resistances as shown.

SPEED N PF %  V=400V % (V=400V)

% (V=360V)

V=360V PF (V=400V)

PF (V=360V) ROTOR RESISTANCE CONTROL

0 0 TORQUE (kg-m) BHP

SPEED N PF %  R = 0Ω % ( R = 0Ω)

R=25% % ( R=25%)

PF (R =25%)

PF (R=0 Ω)

FL 36

0 0 TORQUE (kg-m) BHP The following observations can be made from the above tests and the graphs drawn.

ROTOR RESISTANCE CONTROL :

1. Speed of the slip ring induction motor can be controlled by inserting additional resistances in the rotor circuit.

2. P.F of the I.M under loaded conditions improves with increase in rotor resistance.

3. For a given torque, the slip at which the I.M works is proportional to the rotor resistance.

4. The starting torque can be increased with increase in rotor resistance.

5. Max. Torque is independent of rotor resistance.

6. Efficiency of the M/c decreases with increase in rotor resistance.

******

CYCLE -3

EXPERIMENT NO – 09

CONNECTION OF THREE 1Φ TRANSFORMERS IN STAR - DELTA

37 AIM : To conduct load test on three 1 transformers connected in Star-Delta to determine efficiency and regulations.

APPARATUS REQUIRED : Three similar, 2kVA, 230V/230V, 1 transformers, one (0-10)A and one (0-20)A a.c ammeters, two (0-300-600)V ac voltmeters, two (0-20)A (0-300)V UPF wattmeters, one (0-10)A (0-500)V UPF double element wattmeter, one 15A, 3 auto- transformer, one 3 resistive load and connection wires.

CIRCUIT DIAGRAM :

W 1 W i 3 Φ M L

415 V A (0-500)V P C 1 3Φ AC U S S V 1 2 V T SUPPLY (0-300)V O T 3 Φ R P P R 2 2 A A L A Y N P S S O 2 2 1 S A B F P P S S D O 1 1 1 2 R (0-10)A (0-20)A M E R C

M L W i 2 W i = (0-10)A (0-600)V DOUBLE ELEMENT UPF WATTMETER

W 1 & W 2 = (0-20)A (0-300)V UPF WATTMETERS

BED NO. NAME PLATE DETAILS

PROCEDURE :

1. The connections are made as shown in the circuit diagram taking care to see that all similar terminals are connected to obtain Star point on the primary side and dissimilar terminals are connected to obtain Delta connection on the secondary side.

2. With the auto- transformer position at zero, the supply switch voltage is closed. Using the auto- transformer, a line voltage of VP=400V (which correspond to a phase voltage of 230V) is applied to the primary of the star-delta connected 3 transformer.

38 3. The readings of all the meters are noted.

4. The star-delta transformer is now loaded, using the balanced 3 resistive load, in steps of about 3 amps. At each step, the readings of all the meters are noted and tabulated as shown. The load

on the transformer is increased till the load current (line current) IS is equal to 3 x 8.7 = 15 amps.

NOTE :

1) While using the 3 balanced resistive load, three switches are closed (vertically) every time to increase the load current.

2) When the transformer is loaded, the primary voltage VP (applied voltage) will decrease slightly due to Auto-transformer. Using the auto-transformer, the primary voltage brought back to 400V at each step of loading and then the readings of all the meters are noted.

TABULAR COLUMN :

Sl. VP IP Wi IS VS W1 W2 % % Reg No. Volts Amps Watts Amps Amps Watts Watts 1 400 0 0 0 2 400 3 400 4 400 5 400 6 400

NOTE : As the 3 balanced load is resistive, p.f=1 and hence W1 and W2 always indicate equal readings.

CALCULATIONS :

The % under different loaded conditions can be calculated using the equation, output W1 + W2 % = ------x 100 = ------x 100 Input Wi

The % regulation at different load currents, can be calculated using the equation,

VS (no load) – VS (load) % Regulation = ------x 100 VS (no load)

Where VS (no load) is the secondary voltage VS when IS=0 and VS (load) is the secondary voltage VS noted for different values of IS.

GRAPH :

39 A graph of % v/s output in KW is drawn as shown.

% 

0 1 2 3 4 5 6

OUTPUT IN kW

****** EXPERIMENT NO. 10

LOAD TEST ON INDUCTION GENERATOR

AIM : To run the given 3-phase squirrel cage induction motor as an induction generator and then to conduct load test to obtain it’s performance characteristics.

APPARATUS REQUIRED : One squirrel cage induction motor – D.C machine set, Two current transformers, Two (0-2)A, (0-600)V UPF wattmeters, one (0-10)A a.c ammeter, One (0-25)A d.c ammeter, One (0-300)V d.c voltmeter, one 750  rheostat and connecting wires.

(0-25)A 25 C - +

A1 CIRCUIT DIAGRAM L M : A1 C2 750Ω

A2 CI A F L M R 600 C + Y-∆ A ` B1 (0-300V) D 220V C1 Y A2 + - DC

(0-10)A S A2 AA FF B 600 w2 C T B1 B2

W1 L M A B2 R G/M T 40 L M E C1 R

25 C C2 V A V

W1 & W2 = (0-2)A, (0-600)V UPF WATTMETERS

BED NAME PLATE DETAILS NO. INDUCTION MOTOR D.C MACHINE

PROCEDURE :

1. The connections are made as shown in the diagram.

2. With the DPST switch open and 750  rheostat in cut-in position, 3 phase supply switch is closed.

3. The induction motor is started using the star-delta starter.

4. The d.c supply voltage is now measured and the voltage across d.c generator is made equal to the d.c. supply voltage by cutting-out the 750  rheostat.

5. After making sure that the polarities of the two d.c supplies are same, the D.P.S.T switch is closed.

6. The no-load speed and the readings of all the meters are noted after opening the C.T secondary switches. If any of the wattmeters “KICK BACK”, then the secondary switch of the corresponding C.T is closed. The current coil connections of the wattmeter are interchanged. The secondary switch of the C.T then opened and the reading of the wattmeter is taken as –ve.

7. The d.c machine, which is now floating across the d.c supply is made to run as a motor by under-exciting it’s field. This is done cutting-in the 750  rheostat.

8. The motor current is made equal to 3A using the 750  rheostat. The readings of all the meters are noted.

9. The speed of the set is now increased, in steps, to get motor currents given in the tabular column. At each step, the readings of all the meters are noted. The loading of the induction 41 generator is continued, by increasing it’s speed, till the current supplied by it reaches it’s rated value.

10. The speed of the set is now reduced gradually by cutting-out the 750  rheostat till the d.c ammeter indicates zero. The DPST and TPST switches are then opened.

CALCULATIONS :

Input to the Induction generator = output of the D.C motor = input to the d.c motor x M

= VDC x IDC x M watts.

The efficiency of the d.c motor at different inputs can be found from the known input V/s %

M characteristics which is drawn from the following data.

Input in 0 500 1000 1500 2000 3000 4000 5000 Watts

% M 0 0.4 0.6 0.74 0.8 0.87 0.85 0.82

Output of the Induction generator = (W1 + W2) watts

(W1 + W2)  %  of the Induction generator = ------x 100

VDC x IDC x M

The p.f of the induction generator can be found from the following equation. __ W1 - W2 tan  =  3 ------and p.f = cos . W1 + W2 If IL is the average line current, then the power current (in phase current) and the wattless current (Quadrature current) of the line current can be calculated using the equations,

Power current = IL cos  and Watt less current = IL sin 

The readings and the results are tabulated as shown.

TABULAR COLUMN

42 Sl. IDC VDC N IY=IL W1 x Const x W2 x Const x W =(W1+W2) % p.f IL x IL x Ratio Ratio Cos  Sin  No. Amps Volts RPM Amps Watts Watts Watts Amps Amps

1 0

2 3

3 6

4 9

5 12

6 15

7 18

GRAPHS:

As the functions of the induction generator, in practice, is completely determined by the speed at which it is driven the performance curves are drawn by employing the speed as independent variable. The graph of speed V/S % , p.f, Total current, power Current and watt less current are drawn as shown.

TOTAL CURRENT

WATTLESS CURRENT TOTAL POWER WATTLESS CURRENT PF CURRENT %  CURRENT POWER CURRENT

% 

1500 SPEED

CONCLUSIONS: (To be written by the student ) While drawing conclusions the points that should be considered are: 43 a. Has the object of the experiment been achieved? b. Do the results substantiate the theory? If the results appear to contradict the theory, the probable reasons should be discussed. c. Proper interpretation of graph.

******

EXPERIMENT : 11

NO-LOAD AND BLOCKED ROTOR TESTS ON 3 - PHASE INDUCTION MOTOR - PERFORMANCE EVALUATION USING EQUIVALENT CIRCUIT

AIM : To conduct no-load and block rotor tests on the given three phase Induction motor to obtain it’s equivalent circuit and to Pre-determine the performance of the I.M.

APPARATUS REQUIRED : A three phase Induction Motor, Two current transformers, Two (0-2)A (0-600)V u.p.f wattmeters, Two (0-2)A (0-75)V u.p.f wattmeters, Two (0-2)A a.c ammeters, One (0-5)A d.c ammeter one (0-12)V d.c voltmeter, Lamp Load, One 3 phase auto- transformer and connecting wires.

CIRCUIT DIAGRAM :

C 25

3Φ A C 1 2 AUTO M L T R W A (0-2)A V (0-600)V A 1 N M L R S F 600 415V 3Φ C A C O 2 1 50 Hz AC Y R C 600 B B M 1 2 E B R A M L W (0-2)A 1 M L

C 25

W & W = (0-2)A, (0-600)V LPF Wattmeter 1 2

BED NAME PLATE DETAILS NO.

INDUCTION MOTOR

44 NO-LOAD TEST :

1. The connections are made as shown in the diagram.

2. With the auto-transformer at the minimum position, the supply is switch on.

3. By operating the auto-transformer, the rated voltage of 415V is applied to the stator of the induction motor. All the meter readings are noted down. If any of the wattmeter “KICK BACK”, then the secondary switch of the corresponding C.T is closed and the current coil of the wattmeter is interchanged. The secondary switch of the C.T is now opened and the reading of the wattmeter is noted as ‘-ve’.

BLOCKED ROTOR TEST :

1. The circuit diagram remains the same for this test except that the l.p.f wattmeters are replaced by u.p.f wattmeters and (0-600)V voltmeter by (0-100)V voltmeter.

2. With the auto-transformer at minimum position, the supply switch is closed.

3. With the rotor of the I.M blocked (held stationary manually), the voltage applied to the stator is increased gradually till the I.M draws its rated current. The readings of the voltmeter and wattmeters are noted.

NO-LOAD TEST :

I1 + I2 Sl. Vo I1 x I2 x Io = W1 x Const. x W2 x Wo = W1 + W2 No. Volts C.T C.T ------C.T.Ratio Const. x Ratio Ratio 2 Watts C.T.Ratio Watts Amps Amps Amps Watts

1 415

BLOCKED ROTOR TEST :

I1 + I2 Sl. I1 x C.T I2 x C.T ISC = ------VSC W1 x Const. x W2 x Const. x WSC = W1 + W2 No. Ratio Ratio 2 C.T Ratio C.T Ratio 45 Amps Amps Amps Volts Watts Watts Watts

1 7.5

MEASUREMENT OF STATOR RESISTANCE

CIRCUIT DIAGRAM :

(0-5)A A A 1 2 + - A V Sl. I V Rdc = ------No. I + + - V Amps Volts Ohms 220V DC (0-15)V 1 2 - 2 3 3 4

Average Rdc = Ohm 1. The connections are made as shown.

2. Using the lamp load, different currents are passed through any one phase of the stator of the Induction motor. For each of the current, the voltage drops across the phase winding is measured and noted. 3. The a.c resistance per phase (R1) is calculated using the equation R1 = 1.4 R d.c .

R =R +R ' X =X +X ' EQUIVALENT CIRCUIT : 01 1 2 01 1 2

I / √ 3 I ' 0 2 I I W μ

(1-s) R X R ' = R ' x 0 0 L 2 V =415V s 46 The complete equivalent circuit of the Induction motor is as shown above where

R1 = Stator resistance / Phase 1 R2 = Rotor resistance / Phase referred to stator R01 = Total resistance / Phase referred to stator X1 = Stator leakage reactance / Phase 1 X2 = Rotor leakage reactance / Phase referred to stator X01 = Total reactance / Phase referred to stator R0 = Exciting shunt resistance per phase X0 = Exciting shunt reactance resistance 1 RL = Mechanical load equivalent resistance referred to stator.

I0 I0

 we know that IW = ---- cos o and I = ---- sin 0 where 3 3

__ W1 - W2 –1 0 = tan  3 ------W1 + W2 V0 V0

 R0 = ------ and X0 = ------

IW I  Wsc Wsc 1  R01 = ------= ------= R1 + R2 2 2 3 (Isc / Ph) 3 (Isc /  3)

1  R2 = (R01 - R1) 

Vsc  3 Vsc

Also Zo1 / phase = ------= ------

Isc /  3 Isc ______2 2  X01 =  Z 01 - R o1

Hence, the equivalent circuit of the Induction motor can be drawn. 47 TO DETERMINE %  , B.H.P, TORQUE & P.F FROM EQUIVALENT CIRCUIT:

The above quantities at different slips can be found as follows. Let the % slip be 3. 1 - S 1 – 0.03 1 1 1  RL = R2 ( ------) = R2 ( ------) S 0.0.3

0 V1/ph 415 0 1  I2 /ph = ------= ------Amps. 1 Z total / ph R01 + RL + J X01

I0- o 1  I1/ph = I2 /ph + I0/ph = I1 - (say) where I0/ph = ----- 3  p.f at 3 % slip = Cos 

Now, input to the I.M =  3 VL IL Cos  watts. __ __

= 3 x 415 x IL Cos  where IL = 3 I1/ph

1 2 1 output of I.M = 3(I2 /ph) x RL watts.

output Hence %  = ------x 100 can be found Input

Speed at 3% slip = N = NS (1-S) = 1500 (1-0.03) = 1455 r.p.m

output B.H.P. at 3% slip = ------735.5

4500 B.H.P Torque at 3% slip = ------x ------Kg. Mts. 2 N Hence, the performance of the I.M can be pre-determined using it’s equivalent circuit.

48 ******

VIVA QUESTIONS a) TRANSFORMERS

1. What is a transformer?

2. Explain the principle of operation of transformer?

49 3. What are the losses in a transformer and in which part of the transformer do they occur?

4. What is the material commonly used for transformer core?

5. What are the important properties of the core material?

6. Why the transformer core is laminated?

7. What are the two components of the core loss?

8. How is the hysterisis loss affected by a change in flux density?

9. On what factors does a) hysterisis loss b) eddy current loss depend? Why can these losses be

assumed constant?

10. What are the drawbacks of a solid core used for transformer? Under which conditions of

load, a-ve regulation can be used?

11. What are the various causes of voltage drop in a transformer?

12. Why are transformer windings divided into several coils?

13. What purposes are served by placing transformer in oil filled tanks?

14. Distinguish between power and distribution transformers?

15. What is the effect of leakage flow in a transformer?

16. What useful information is obtained from open circuit test?

17. Explain the function of active and reactive components of the no load current of a

transformer?

18. Why is the wave shape of magnetizing current of a transformer non sinusoidal?

19. How can the equivalent impedance of a transformer be determined?

20. What does equivalent resistance of a transformer mean? How it may be calculated in

primary turns and secondary turns?

21. Give reasons for the following,

i) The open circuit test on a transformer is conducted at rated voltage.

ii) Only a low voltage is applied to a transformer during short circuit test.

50 iii) Usually the low voltage winding is short-circuited and the high voltage winding

open circuited for the open circuit test.

iv) Usually the low voltage winding is short-circuited and the high voltage winding

is excited for the short circuit test.

v) During the open circuit test, the copper losses are negligible and during short

circuit test the iron losses are negligible.

vi) The secondary of a current transformer should not be kept open.

22. A transformer is rated 50Hz, 230V/115. What will happen if 460V, 50Hz supply is

connected to the primary?

23. what will happen if the above transformer is connected to 230V, 25 Hz supply/

24. An o.c test is conducted on a 230V/115V transformer with LV side excited. What will be

the value of oc test parameters referred to HV side?

25. During the s.c test the wattmeter shows 125 watts when 20 amps current is flowing through

short circuited secondary. What will be the wattmeter reading when 40 amps current is

flowing through short circuited secondary?

26. What do you understand by regulation of a transformer?

27. Distinguish between regulation and efficiency of a transformer? Show how power factor

affects both of them?

28. what are the conditions for maximum efficiency?

29. Two similar transformers have a F.L regulation of 4% and 6%respectively. Which of the

above transformers is better?

30. Is the voltage regulation at 0.6 lag power factor better or worse than that at 0.8 lagging p.f

for the same KVA output? Give reasons for your answer.

31. What is the main advantage of ‘Back to Back’ test of transformers?

51 32. What is the condition to be satisfied by the two transformers to be tested using ‘Back to

Back’ test?

33. Is the primary side balanced with a balanced load on the secondaries of Scott connected

transformers?

34. Is it possible to get a balanced 3 phase supply from a 2 phase supply in Scott connected

transformers?

35. State with the help of connection and vector diagram how a 2 phase supply can be obtained

from a 3phase supply?

36. Why is the condition of correct polarity the most important and most essential condition for

parallel operation of transformers?

37. What is the effect of

i) Unequal voltage ratio

ii) Unequal p.u impedances

iii) Unequal x / R ratio on the parallel operation

38. List the advantages and disadvantages of 3 phase transformers?

39. What are the distinguishing features of Y-Y,Δ-Y,Y- Δ AND Δ- Δ in a3 phase connection of

transformers. Compare their advantages and disadvantages?

40. How can the problems of unbalanced voltages and third harmonic currents overcome in Y-Y

connection?

41. Discuss the advantages and disadvantages and applications of auto transformers. Compare it

with a two winding transformer. b) INDUCTION MOTORS

42. Explain the principle of operation of the polyphase induction motor?

43. What is the essential difference between squirrel cage and slip ring induction motors?

44. Instead of squirrel cage rotor, if a solid iron cylinder is used will the motor run?

52 45. Explain the double cage and deep bar rotors in induction motors?

46. What is the synchronous speed for 50Hz supply and a four pole motor?

47. What is the relative speed of the rotor w.r.t stator field at synchronous speed?

48. what is meant by %slip?

49. What is the slip at stand still?

50. What is `crawling’ of 3 phase induction motor?

51. What is cogging?

52. How is cogging is eliminated?

53. Give the reasons:

i) The speed of an induction motor can not be equal to synchronous peed

ii) The reactance of rotor varies widely between starting and running condition

iii) The induction motor is most commonly used motor

iv) Induction motor can be called a generalized transformer

v) In a 3 phase induction motor, the rotor core loss is negligible

54. The rotating field of the stator and rotor are stationary w.r.t to each other. Justify the

statement?

55. In a slip ring induction motor if the rotor circuit is open will the motor work?

56. Why does the pf improve when the induction motor is loaded?

57. While performing the load test on induction motor, do you reach the maximum torque?

58. An induction motor is running with some load. Explain what happens when one phase goes

out?

59. Why the starters are are required for the induction motors?

60. What are the various methods of starting of induction motors?

61. What are the different methods of speed control of induction motors?

62. What is the effect of decreasing the supply voltage on induction motor?

53 63. What is the effect of introducing resistance in the rotor circuit of an induction motor? In

which type of the motor it is possible under what conditions is it done and why?

64. How do you mark the stable operation region on the circle diagram?

65. Explain synchronous watts as a unit of torque?

66. Name the different types of I phase induction motors?

67. What is induction generator? Discuss its principle of operation and what are its limitations?

68. What are the conditions for the induction motor to run as an induction generator?

QUESTION BANK

A) TRANSFORMERS

1. Conduct OC and SC test on the given single phase transformer to pre-determine its efficiency at F.L and ………………p.f lag/lead.

2. By conducting suitable tests on the given single phase transformer, obtain the equivalent circuit referred to either primary or secondary and mark the values on the diagram.

3. Find the power factor at maximum and minimum regulation of the given single phase transformer by conducting suitable test on the transformer. 54 4. Conduct suitable experiment on the given single phase transformer to find the iron loss and copper loss at ……..…. F.L and hence find it’s efficiency and regulation at F.L and ……... … p.f lag/lead.

5. Obtain the equivalent circuit of the given single phase transformer and also pre-determine it’s efficiency at ……... F.L and ______p.f lag/lead

6. Conduct suitable tests on the given single phase transformer and find the load at which the efficiency is maximum.

7. Obtain the equivalent circuit of the given single phase transformer and hence determine the secondary terminal voltage at F.L and ……....pf lag/lead.

8. Obtain the graph of % regulation V/s p.f of the given single phase transformer by conducting suitable tests.

9. Connect the given two dissimilar transformers in parallel and determine how a load of ………… amps/watts, at any p.f other than UPF, is shared by each transformer. Verify analytically.

10. Connect the given two dissimilar transformers in such a way that they share a common load and show how a load of ……... Amps/watts, at any p.f other than UPF, is shared between them. Verify analytically.

11. Conduct back-to-back test on the given two identical single phase transformers and pre- determine the efficiency of each transformer at ………load and………pf lag/lead.

12. Conduct Sumpner’s test on the given two identical transformers and pre-determine the efficiency of each transformer at …….load and …p.f lag/lead.

13. Conduct load test on the Scott connected transformers to show that i) Primary currents are balanced when secondary currents are equal. ii)Primary currents are unbalanced when secondary currents are unequal.

14. Obtain two single phase supplies from a 3 phase supply using two single phase transformers and show that the single phase voltages differ by 900.

15. Conduct load test on the Scott connected transformers and determine the efficiency of the set when a) Both secondaries carry F.L current at u.p.f b) One secondary carries F.L current with the other carrying ½ F.L current, both at u.p.f.

16. Conduct different polarity tests on the given single phase transformer to demonstrate both subtractive and additive polarities.

17. Conduct suitable experiments on the given single phase transformer to pre-determine it’s efficiency and regulation at F.L and u.p.f . Verify the same by conducting any other test.

55 18. Conduct suitable experiments on the given single phase transformer to pre-determine it’s efficiency and regulation at ½ F.L and u.p.f. Verify the same by conducting load test on the transformer.

19. Connect the given three single phase transformers in  -  and conduct load test on the same to obtain i) % at ½ F.L and u.p.f ii) % Regulation at F.L and u.p.f

20. Connect the given three single phase transformers in  - . Conduct load test on the same and draw a graph of % v/s output at u.p.f.

B) INDUCTION MACHINES

21. Conduct load test on the given 3 phase squirrel cage Induction motor and determine the %, p.f and torque when running at a slip of ………percent.

22. Conduct load test on the given 3 phase squirrel cage Induction motor to obtain the following graphs i) % v/s B.H.P ii) p.f v/s B.H.P

23. By conducting a suitable experiment on the given 3 phase squirrel cage Induction motor, determine the % and % slip when it is delivering ………..percent of it’s rated output.

24. Conduct load test on the given 3 phase squirrel cage Induction motor to obtain the following graphs i) % slip v/s B.H.P ii) Line current v/s B.H.P

25. By conducting suitable experiments on the given 3 phase Induction motor, obtain it’s circle diagram. Hence predetermine the following i) % and slip at F.L and ii) Pull-out torque.

26. Conduct suitable experiment on the given 3 phase induction motor to predetermine i) The ratio, Max. torque/FL torque ii) Starting torque

27. By conducting suitable experiment on the given 3 phase Induction motor, obtain it’s circle diagram. Hence pre-determine the following i) Line current, torque and p.f at F.Lii) Maximum power output.

28. You are given a 3 phase Induction motor. How do you proceed to obtain it’s equivalent circuit? Hence predetermine the % and torque when running at ………… percent slip.

29. Run the given 3 phase Induction motor as an Induction generator to obtain the following graphs i) % vs Speed ii) p.f vs Speed.

30. Run the given 3 phase Induction motor as an Induction generator and determine experimentally the p.f at which the generator works when it is running at a slip of ………. percent.

31. Conduct load test on the given 1 phase Induction motor to obtain the following graphs i) % vs B.H.P ii) p.f vs B.H.P

56 32. By conducting a suitable experiment on the given 1phase Induction motor, determine the %, p.f and B.H.P when running at a slip of ………….. percent.

33. Conduct a suitable experiment on the given 1 phase Induction motor, and hence determine the %, % slip and p.f when delivering …………. percent of it’s rated output.

34. Conduct a suitable experiment on the given 1 phase Induction motor to obtain the following graphs i) % Slip vs B.H.P ii) Line current vs B.H.P

35. Conduct load test on the given 3 phase slip ring I.M under two different applied voltages to obtain it’s performance characteristics in each case. Comment on it’s performance.

36. Conduct load test on the given 3 phase slip ring I.M for two values of rotor resistances to obtain it’s performance characteristics in each case. Comment on it’s performance

37. Conduct load test on the given 3 phase slip ring IM under the following conditions to obtain it’s speed v/s torque characteristics. 1. Under two different applied voltages 2. Under two different rotor resistances.