o.c and s.c test on single phase transformer aim: …

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O.C AND S.C TEST ON SINGLE PHASE TRANSFORMER

Aim: To determine the efficiency and regulation of a given single phase transformer by conducting the OC and SC test and also to draw its equivalent circuit Apparatus Required:

Sl No Equipment Rating Type Quantity

1. Volt meter 0-300V MI 1 No

0 – 25V MI 1 No

2. Ammeter 0-1A MI 1 No

0-15A MI 1 No

3. Watt Meter 0-150V, 0-1A 0.2 LPF 1 No

0-75V, 0-15A UPF 1 No

4. 1-Ø Transformer 3 KVA, 115/230V 1:2 Ratio 1 No

5. 1-Ø Auto Transformer 230V, 0-270V/15A 1 No

Theory: i). Open Circuit test: Consider a practical transformer on no load, i.e., secondary on open circuit. The primary will draw a small current I0 to supply the iron losses and a very small amount of copper loss in primary, hence the primary no load current I0 lags V1 by an angle Ø0 less than 900 and no load input power W0 = V1I0Cos Ø0. The no load primary current I0 can be resolved into two components IW & Iμ. IW is in phase with V1 and known as working/active/Iron loss component. IW = I0Cos Ø0 Iμ lags V1 by 900 and known as magnetizing component or lossless component. Iμ = I0Sin Ø0 Note: The no load primary copper loss I0

2R1 is very small and may be neglected. Therefore the no load input power W0 = Iron Loss. ii). Short Circuit test: This test is conducted to determine R01 (or R02), X01 (or X02) and full load copper losses of the Transformer. In this test, the secondary (usually the LV winding) is short circuited by a thick wire and a variable voltage is applied to the primary and gradually raised till full load current I1 flows through the primary. Under this condition the copper loss in the winding is equal to that of full load copper losses (since the current in the secondary is equal to its rated value). Since there is no output in the transformer under the short circuit conditions, the input power is all equal the losses of the Transformer. The voltage VSC is very small in this condition, the core loss is negligibly small, and hence the total loss can be considered as full load copper losses of the Transformer winding.



i). Open Circuit test: Circuit Diagram:

Procedure: 1) Connections are given as per circuit diagram 2) Switch on the power supply 3) With the help of Auto-Transformer, Apply voltage to HV side in steps (230V) 4) At each step note down Voltmeter, Ammeter and Wattmeter readings 5) After reaching maximum voltage of 230V on HV side, the supply is switched off

Tabular Column

Sl No Open Circuit Voltage (VOC)

Open Circuit Current (Im)

Iron loss (W)

1.

2.

3.

A Ph

1-P

H, 2

30V

, 50H

z,

AC

SU

PP

LY

V

0-1 A MI

0 V

230 V

0 V

115 V 1- Ph, 230V / 115V,

3 KVA Transformer

0 - 3

00

V MI

1-ph, 230V / 0-270V, 15A Auto transformer

150V, 1A, 0.2pf 1-ph watt meter

Op

en

Circ

uit

M L 5A

5A N

C V



ii) Short Circuit Test: Circuit Diagram:

Procedure: 1) Connection are given as per the circuit diagram 2) The mains switch on HV side is closed. 3) With the help of Booster -Transformer Current is injected in to HV winding in steps. 4) The voltmeter, ammeter and Wattmeter reading are noted down for each step in HV side. 5) After reaching full load current on Secondary side the supply is switched off. Tabular Column:

Sl No V1SC (V) I1SC (A) WSC ( W )

1.

2.

3.

Calculations:

a) Open Circuit Test:

1) No load Power Factor SCSC

SC

IV

WCos

0

Where WOC is Open circuit power in watts VOC is Open circuit voltage in volts I OC is Open circuit current in amps 2) Resistance to account for core loss

A Ph

1-P

H, 2

30V

, 50H

z,

AC

SU

PP

LY V

0-15 A MI

0 V

115 V

0 V

230 V 1- Ph, 230V / 115V,

3 KVA TRANSFORMER

0 - 7

5 V

MI

75V, 15A, UPF 1-ph watt meter

Sh

ort C

ircu

it M L

C V

15A

15A

1-Ph Auto Transformer 230V/ 0 - 230V

N



0CosI

VR

OC

OCC

3) Magnetizing Reactance

0SinI

VX

OC

OCM

b) Short Circuit Test Observations: 4) a) Equivalent winding Resistance referred to HV side

201

SCI

SC

I

WR

Where Wsc is short circuit power in Watts I SC is Short circuit current in Amps b) Equivalent winding Resistance referred to LV side

2

0102 KRR

Where K (Transformation Ratio) = ( V2 / V1 ) 5) a) Equivalent winding Impedance referred to H V side

SCI

VZ SC

1

01

b) Equivalent winding Impedance referred to LV side

2

0102 KZZ

6) a) Equivalent winding Reactance referred to HV side

2

01

2

0101 RZX

b) Equivalent winding Reactance referred to LV side

2

0102 KXX

Where K (Transformation Ratio) = ( V2 / V1 ) c) To find Efficiency:

Core loss or Constant loss (Wi)

Rated short circuit current I2 sc

KVA rating of transformer

Copper loss (Wsc) at full load Current

Output for other Pf = CosIV 22

Total power loss SCiT WWW

7) Output = ( X * KVA* COS ) in Watts



Where ‘X’ is Fraction of load KVA is Power rating of transformer

COS is Power Factor 8) Copper Losses = (X) 2 WSC in Watts Where Wsc Is Copper loss in short circuit condition Where ‘X’ is Fraction of load

9) Total power loss = (Copper Loss + Iron Loss) In Watts

10) 100%

LossesTotalOutput

OutputEfficiency (or)

100%

TL

L

WCosIV

CosIVEfficiency

c) Tabular Column to find Efficiency:

Rated Secondary Voltage ( V2 ) = …………… Volts

Constant Losses (Iron Loss) Wi = …………… Watts

Sl No:

Load Current (IL)

Wcu ( W ) Out Put =

V IL Cos ( W )

Total Losses (WT) = Wi + Wcu

% Efficiency

1 0 Amp 0 Watt 0.0 Wi=………. 0

Expected Graph:

i. Load Vs % Efficiency:



Load Current

% E

ffic

ienc

y

U.P.F

0.8 PF

d) To find Regulation: 100Re%2

022022

E

SinXICosRIgulationVoltage

At rated full load current and rated load voltage)

( +Ve for lagging and unity PF; - Ve for leading PF )

Tabular Column:

Expected Graphs:

ii. Power Factor Vs % Regulation:

Sl. No P. F

(Cos )

Lagging Pf % Regulation

Leading Pf % Regulation

1. 0

2. 0.2

3. 0.4

4. 0.6

5. 0.8

6. 1



PF

% R

egul

atio

n

Conclusion: The efficiency and regulation of a given single phase transformer is determined by conducting the OC and SC tests Viva Questions: 1. What is regulation of a transformer? 2. What are the different losses in a transformer? 3. What are the different types of transformers? 4. What is the efficiency of a transformer when compared with Induction motor? 5. What is the working principle of transformer? 6. What happens when DC is given to primary of a transformer? 7. Why the transformers are rated in KVA?



BRAKE TEST ON 3-PHASE SLIP RING INDUCTION MOTOR

Aim: To perform the Brake test on the given 3-Ø Induction motor and to obtain the performance characteristics of the motor. Apparatus Required:

Sl No Equipment Rating Type Quantity

1 Auto Transformer 415/ 0-470V, 10A 1No

2 Ammeter (0-10A) MI 1No

3 Voltmeter (0-300/600V) MI 1No

4 Wattmeter 300/600V, 0-5/10A

UPF 1No

Nameplate details:

1. Rated Voltage

2. Rated Current

3. Speed

4. Type of Excitation

5. Power Theory:

Slip ring induction motor is a type of induction motor in which the rotor is provided with

3 phase double layer distributed winding consisting of coils as used in alternators. The

rotor is wound for as many poles as the number of stator poles and is always wound 3

phase even when the stator is wound two phase.

The three phase are starred internally. The other three winding terminals are

brought out and connected to three insulated slip rings mounted on then shaft with

brushes resting on them. These brushes are further externally connected to a three phase

star connected rheostat. This makes possible the introduction of additional resistance in

the rotor circuit during the starting period for increasing the starting torque of the motor

and for changing its speed torque /current characteristics. When running under normal

conditions the slip rings are automatically short circuited by means of a metal collar

which is pushed along the shaft and connections all the rings together.. Next the brushes

are automatically lifted from the slip rings to reduce the frictional losses and the wear and

tear.

In the wound rotor type the rotor slots accommodate an insulated winding similar

to that used on the stator. The rotor winding is uniformly distributed and is usually

connected in star. The three leads from the star connection ate then connected to three

slip rings of collector rings mounted on but insulated from the shaft. Carbon brushes

pressing on the slip rings allow external resistors to be inserted in series with the rotor



winding for speed and starting torque control. Actually the wound type rotor of induction

motor costs more and requires increased maintenance it is therefore only used where

i) The driven load requires speed control

ii) High starting torque is required.

Since the rotor is wound with polyphase windings and carrier slip rings it is called

slip ring induction motor or wound rotor

STATOR

R

Y B

R

Y

B

●

●

●

●

●

● ●

●

● || || || || || || || || || ||

|| || || || ||

TPST Switch

V

A

0-600V MI

0-10A MI 600V,10A,UPF M L

C V

10 A

M

C V

600V,10A,UPF L

10 A

10 A

●

●

●

R

Y

B

● ●

●

●

●

●

V

S1

S2

S3

R1

R2

R3

ROTO

R

ROTOR RESISTANCE STARTER

R

B

Y

Y



STATOR

R

Y

B

R Y B

● ●

● ●

●

●

●

● ● || || || || || || || || || ||

|| || || || ||

TPST Switch

V

A

0-600V MI 0-10A MI

600V,1

0A,UPF

M

L V

10 A

M C

V

600V,1

0A,UPF

L

10 A 10 A

● ● ● R Y B

● ● ● ● ● ●

V

S1 S2 S3 R1 R2 R3 ROTOR

ROTOR RESISTANCE STARTE

R

R

B

Y

Y



Procedure: 1. Make the connections as per circuit diagram. 2. Ensure variac (Auto Transformer) position in zero output voltage and switch on 3-Φ mains

supply 3. Increase the voltage by using variac till rated voltage is impressed across stator winding. 4. Take no load readings of wattmeter, ammeter, voltmeter, speed and spring balance readings. 5. By increasing load on the motor by tightening brake drum in steps, & take the readings of all the

readings. 6. Repeat step-5 till 100-115% of rated current is reached. 7. Replace the load on motor, reduce voltage & switch off mains supply. Tabular Column:

SL

No

VL

V

IL

A W1

watt W2

watt

S1

kg

S2

kg

Speed ‘N’ in

RPM

Input (w)

Torque ( T ) in

N-m

Output

%Efficiency

1

Calculations: Radius of the Brake drum r = …….. Mtrs Torque (T) = (S1- S2) r.g N.mtr

Power Output = wattsNT

60

2

%Efficiency = (Output / Input) X 100.

Expected Graphs: 1. % Efficiency Vs Output Power 2. Speed Vs Output Power in 3. Torque Vs Output Power 4. Load Current Vs BHP.

% Efficiency

Torque

Load Current

X

Y

O

Out Put Power in BHP

% E

ffic

ienc

y

Load

Cur

rent

Torq

ue

Spe

ed

Speed



Conclusion:

Performance characteristics of the 3-Ø Induction Motor is obtained by conducting Brake Test

Viva Questions:

1. What is the working principle of 3 – phase induction motor? 2. What are the different types of 3 – phase induction motor? 3. As the load increases, what happens to the speed of the motor? 4. Why 3 – phase induction motor is widely used for industrial applications? 5. What are the various losses in induction motors? 6. Why induction motor is called as rotating transformer?



REGULATION OF 3-Ø ALTERNATOR BY

EMF AND MMF MEHOD

Aim: To predetermine the regulation of 3-Ø alternator by Synchronous Impedance Method Apparatus Required:

Si. No Equipment Rating Type Quantity

1. Volt meter 0-600V MI 1 No

2. Ammeter 0-2A MC 1 No

0-5A MI 1 No 3. Tubular Rheostat 0 - 290 / 2.8A Wire Wound 2 No

4. Tachometer 0 – 10K rpm Analog 1 No

5. S.P.S.T Switch 5A 1 No

Name Plate Details: D.C Shunt Motor 3-Ø Alternator:

1. Rated Line Voltage

2. Full Load Line Current

3. Rated Speed

4. No of Poles

5. Frequency Fuse Rating: For DC Shunt Motor and 3-ØAlternator: 125% of rated full load current

Theory: The regulation of a synchronous generator (Alternator) is the rise in voltage at the terminals when the load is reduced from full load rated value to zero, speed and field current remaining constant. The voltage regulation depends upon the power factor of the load. There are various methods for determination of regulation one of them is Synchronous Impedance Method is based on the concept of replacing the effect of armature reaction by a fictitious reactance.



Procedure: (i) Open Circuit Test: 1. Connection’s are made as per the circuit diagram 2. Keep the potential divider in minimum output position 3. DC supply is given to the Motor & Excitation circuit 4. Start the motor by gently & steadily pushing the handle from start to run position 5. By Adjusting the field regulator of the motor set the speed of Alternator to rated Speed 6. Now vary the field excitation of alternator in steps and note down the voltmeter Reading (Eg) and ammeter reading as (If) 7. From the above data draw the curve between the Field current Vs Generated EMF, Which is OCC. Tabular Column:

Sl No Field Current

If in ( A )

Generated Voltage ‘Eg’ in ( V ) / Line

Generated Voltage ‘Eg’ in ( V ) / Ph

1.

(ii)Short Circuit Test: 1. After the completion of O.C.C keep the potential divider in minimum Output position 2. Close TPST knife Switch, and vary the field current in steps such that the ammeter Reads short circuit current 3. Note down the readings of field current (If) and short circuit Current as (Isc) 4. Continue the procedure till the Ammeter reads I fL 5. Graph is plotted as SCC and Regulation is calculated from the graph.

M

20 A

20 A

+

-

Z

ZZ

A

AA

L A F 3 - Point

Starter

22

0V

, DC

Su

pp

ly

DPST Switch

A F

1

F 2

B TPST Switch

A 0

- 5A

MI

+ -

0 - 2A MC

Potential Divider 0 - 290 ohms / 2.8A

0 - 2

90 o

hm

s / 2.8

A

V

0- 6

00

V M

I

R

Y

3 - Ø Alternator

N

R

Y B

R

Y

B



Tabular Column:

Sl No Field Current

‘ If ’ ( A )

Short circuit Current ‘ISC’ ( A )

1.

III). Determination of Armature Circuit Resistance (Ra / Ph):

Procedure: 1. Connect the circuit as shown in figure 2. Vary the rheostat in steps & note the readings of Ammeter and Volt meter 3. Resistance per phase = Volt meter reading / Ammeter reading 4. Take two or three readings at approximately equal intervals of ammeter Readings from 0-5A and compute the average value for Ra Tabular Column:

Sl No Va ( V ) I L ( A ) Ra = ( Va / IL )

1.

2.

Avg Ra = …………….. Sample Calculations: Synchronous Impedance Method:

22 DAODOAEO

sc

ocs

I

VZ at constant I f

(or) BC

ACZS

Where AB = Rated open circuit Voltage at a field current / Ph AC = Rated short circuit Current at that field current / Ph

22

aSS RZX

2

2

Saaa XIVSinRIVCosE

A V + -

0 - 6 Ω / 5A A + - 0 - 20A MC

+

-

DC

15

V S

up

ply

0 - 1

5V

MC

R or Y or B Phase

Neutral

DPST SWITCH



+ve for lagging pf -ve for leading pf Eo = No load voltage developed in armature E = No load voltage after allowing for armature reaction V = Rated terminal voltage

Armature resistance Ra (DC) = ……………

Armature resistance Ra (AC) = 1.6 Ra in DC = ……………..

Synchronous impedance Zs = ( Eo / Isc) = ………………

100Re%

V

VEgulation o

Phasor Diagram:

V

Ø

A

C

D

O

B

I a X

s

I aRa

Eo

Graph: 1. Field current If Vs Generated EMF / Ph 2. Field current If Vs Short circuit Current Isc

X

Y

Gene

rate

d EM

FEg

/ Ph

Shor

t circ

uit C

urre

nt I

SC

Field Current If

O

A

B

C



Conclusion: Regulation of the 3-Ø Alternator has been determined by Synchronous Impedance Method Viva Questions: 1. What is the working principle of alternator? 2. Give the difference between A.C. Generator and D.C Generator. 3. Why alternators are rated in KVA? 4. What are advantages of stationary armature? 5. Is the efficiency of the alternator is determined by direct loading? 6. Give the e.m.f equation of an Alternator?



PARALLEL OPERATION OF TWO SINGLE PHASE TRANSFORMERS Aim: To operate two transformers in parallel and to determine the relation of sharing of loads to their impedances. Apparatus Required:

S. No. APPARATUS TYPE RANGE QUANTITY

1. 1-ф Transformers 3-KVA, 230/110V 2

2 1-ф variac 10 A, 230/(0-230)V 1

3. Voltmeter MI (0-300)V 3

4. Ammeter MI (0-15/30)A 3

5. 1-ф Load Resistive 3 KVA 1

Theory: As the load on the transformer is increased to nearly the double, it is necessary to replace the existing transformer with that of double capacity or to operate another transformer in parallel with the existing one. To operate two transformers in parallel, it is essential that the following conditions should be satisfied: 1. Their polarities must be same. 2. Their impedance voltages must be same. 3. Their voltage ratios must be equal. Procedure: A.] Short-Circuit test: 1. Make the connections are as per the circuit diagram. 2. Ensuring the Variac in minimum position, switch-ON the single phase a.c. mains supply. 3. Adjust the variac such that full-load current flows through the primary. 4. Note and tabulate the values of current, voltage and power. 5. Evaluate the impedance of the transformer, and hence, calculate its percentage impedance voltage. 6. Repeat the similar procedure for the second transformer.



A Ph

1-P

H, 2

30V

, 50H

z,

AC

SU

PP

LY V

0-15 A MI

0 V

115 V

0 V

230 V 1- Ph, 230V / 115V,

3 KVA TRANSFORMER

0 - 7

5 V

MI

75V, 15A, UPF 1-ph watt meter

Sh

ort C

ircu

it

M L

C V

15A

15A

1-Ph Auto Transformer 230V/ 0 - 230V

N

A

0-15 A MI

V

0 - 3

00 V M

I

0 V

230 V

0 V

115 V

1- Ph, 230V / 115V, 3 KVA Transformer

Ph

1-P

H, 2

30V

, 50H

z,

AC

SU

PP

LY

A

0-30 A MI

1-ph, 230V / 0-270V, 28A Auto transformer

30A

30A N

0 V

230 V

0 V

115 V

1- Ph, 230V / 115V, 3 KVA Transformer

A 0-1 5A MI

6KW,220V

V

0 - 3

00 V

SPST SWITCH



B.] Parallel operation: 1. Make the connections as per the circuit diagram. 2. Ensuring the minimum position of the variac, switch-ON the single phase ac

mains supply is, keeping the switch open. 3. Adjust the variac to the rated voltage and ensure the voltage across the switch is

to be zero. If the voltmeter doesn’t read zero, interchange the polarities and repeat the steps 2 & 3.

4. Measure the readings of both the transformers on the H.V. and L.V. sides. 5. Increase the load in steps and repeat the step-4 till 50% of the full-load of both

the transformers is reached. 6. Tabulate the readings. Tabular Column: A.] Short-Circuit Test: Transformer-1 Transformer-2

I1 V1 Z1

B.] Parallel Operation:

S.No. Load I1 I2 I I1Z1 I2Z2 Theoritical I1 I2

Result: The parallel operation of the two transformers is studied and their load sharing is observed. Viva Questions: 1. what should be the rating of two transformers for parallel operation 2. what is the advantage of parallel operation of transformers 3. with the help of parallel operation can we calculate the efficiency 4. What are the different losses in a transformer 5. If the regulation is less then the transformer is a good transformer or not 6. Define regulation of a transformer 7. what is the effect on losses of a transformer with variation supply frequency

I2 V2 Z2



SCOTT CONNETION AIM: To obtain a balanced two-phase supply from 3-phase system by using Scott connection. NAME PLATE DETAILS: Main Transformer: Rating = ………………KVA Primary volts = ……………. V Secondary volts =………….V Teaser Transformer: Rating = ………………KVA Primary volts = ……………. V Secondary volts =………….V APPARATUS:

S.No. Equipment Type Range Quantity

1. Ammeter

2. Voltmeter

3. Variac

. THEORY: In order to supply power to two-phase electric furnaces, to interlink three-phase and two-phase systems, three-phase to two-phase conversion or vice-versa is essential. The common type of connection used to achieve the above conversion is “Scott Connection”. Two single-phase transformers of identical rating with suitable tappings provided on primaries of both are required for this connection. Transformer A _ _ _ 50% tapping and is called the Main Transformer. Transformer B _ _ _ 86.6% tapping and is called the Teaser Transformer. The voltage across the primary, CO of Teaser transformer will be 86.6% of the voltage across the primary AB of a main transformer (refer to fig). The neutral point of the three-phase system will be on the Teaser transformer, such that the voltage between O & N is 28.8% of the applied voltage. Thus the neutral point divides the Teaser primary winding, CO in the ratio 1:2. The phasor diagram of voltages across the primaries and seconderies has shown in the fig. The voltage across the two seconderies a1a2 and b1b2 should be same in magnitude but in phase quadrature, which may be verified experimentally by recording the voltage across the seconderies Va1a2, Vb1b2 and the voltage a2b2 will a1 and b1 connected together. The voltage Va1a2 and Vb1b2 will be in phase quadrature, if the following relationship holds well between the three voltages.



Va2b2 = √ (Va1a2 ²+Vb1b2 ²)

The behavior of the above circuit can be studied experimentally, at equal loading on

the two seconderies with Upf load → (If the two seconderies of main and teaser

transformers carry equal currents, then the current flowing in the primary windings on three-phase side will also be equal) PROCEDURE 1. Connect the circuit as shown in the fig. 2. Ensure that the switches S1 and S2 are open 3. Adjust the 3-phase variac for minimum voltage at its output 4. Switch on the A.C. supplies and apply the rated voltage across the primaries of the

transformers 5. Record the voltages V1, V2 and V3 and verify that the output is balanced two-phase

supply 6. Switch off the A.C. supplies and remove the dotted connection of the two secondaries

and the voltmeter V3. Adjust the variac to minimum output 7. Switch on the A.C. supply again. Adjust the output voltage of the variac as per the rated

voltage of the primaries of the transformer. 8. Close the switches S1 and S2 to load both the secondaries. Adjust equal loading

conditions also. 9. Switch off the load from both secondaries and adjust the variac, so that its output

voltage is minimum and then switch off the supply.



● B ● ● ●

415V, 3-, 50Hz, AC Supply

● Y ● ● ●

● R ● ● ●

V

V

V

0 – 150 V,MI

0 – 150V,MI

0 – 300 V,MI

0

0

230V

115V Main Transformer

3KVA,230V/115V

50%

0 0

115V

Teaser Transformer 3 KVA,230V / 115V

86.6%

V

0–600 V,MI

CIRCUIT DIAGRAM:SCOTT CONNECTION


|| || || || || || || || || || || || || || || || ||

T P S

T

10 A

10 A

10 A



0

0

230V

115V

Main Transformer

3 KVA,230V / 115 V

50%

0 0

115V

Teaser Transformer 3 KVA,230V / 115V

86.6%

V

0–600 V,MI

CIRCUIT DIAGRAM:SCOTT CONNECTION


● B ● ● ●

|| || || || || || || || || || || || || || ||


● Y ● ● ●

● R ● ● ●

T P S

T

20 A

10 A

20 A

A

0–15A ,MI

3KW,220V

A

0–15A ,MI

3KW,220V



TEST READINGS

For balanced two-phase supply

S.No. V1 (V) V2 (V) V3 (V)

Under loaded conditions

S.No. Im (A) It (A) I1 (A) I2 (A) I3 (A) V1 (V) V2 (V)

RESULT Hence 3-phase to 2-phase conversion is obtained by Scott connection. VIVA QUESTIONS: 1. What is a teaser transformer? 2. What are the advantages of Scott Connection? 3. What do you mean by Tertiary Winding? 4. What is the percentage of winding at which Scott connection is made? 5. What are the applications of Scott Connection?



SLIP TEST

AIM: To determine Xd i.e direct-axis synchrounous reactance and quadrature axis synchronous reactance xq of a salient pole synchronous machine by slip test APPARATUS:

THEORY:

Using slip test ,slip test, xd and xq can be determined. The synchronous machine is driven by a separate motor at a speed slightly different from synchronous speed.the field winding is left open and positive sequence balanced voltages of reduced magnitude and of rated frequency are impressed the armature terminals.under these conditions ,the relative velocity between the field poles and therotating armature mmf wave is equal to the difference between synchronous speed and the rotor speed i.e slip speed. A small A>C voltage across the open field winding indicates that the field poles and rotating MMF wave, are revolving in the same direction and that is what is required in the slip test.if the field poles revolve in a direction opposite to the rotating MMF wave, negative sequence reactance would be measured. At one instance, when the peack of armatureMMF wave is in line with the field pole or direct axis, the reluctance offered by the small air gap is minimum.at this instant the impressed terminal voltage per phase divided by the corresponding armature current per phase divided by the corresponding armature current per phase give d axis synchronous reactance . After on e-quarter of slip cycle, the peack of armature MMF wave acts on the interplar or q-axis of the magnetic circuit and the reluctance offered by long airgap is maximum.at this instant, the ratio of armature terminal voltage per phase to the corresponding armature current per phase, gives q-axis synchronous reactancexq.

S.No. NAME OF APPARATUS

RANGE TYPE QUANTITY

1 3-q alternator with prime mover

KVA 3.5,VOLTS-415,RPM:1500,AMP:4.85

salient 1

2 D.C ammeter (0-2.5)amp MC

3 A.C ammeter (0-2.5)amp MI 1

4 A.C voltmeter (0-75)volts MI 1

5 3-Q variac 400/(0-470)volts - 1

6 TPST switch -------- --- 1

7 synchroscope ----- analog 1

8 tachometer (0-2000)RPM digital 1

9 rheostat 270 ohm,1.2Amps --- 1



Oscillograms of armature current ,terminal voltage and the induced EMF in the O.C field winding are shown in the figure.a much larger slip than would be used in practice, has been shown in fig. when the armature MMF wave is along the direct axis , the armature flux passing through O.f winding is maximum. Therefore the induced field EMF i.e dqa/dt is zero.when armature MMF wave is along q-axis the armature flux linfing the field winding is zero.therefore the induced field emf dqa/dt is maximum.thus the q-axis can be located on the oscillogram . waveforms of voltage across open field and the armature current varies cyclically at twice the slip frequency. PROCEDURE: 1. connect the circuit as shown in the circuit diagram. 2. before starting the motor ensure d.c motor field rheostat is in minimum position.then

swith on the dc main supply and then by dragging the starter handle swith on the motor. 3. after setting the speed of the prime mover to slightly less than the synmcronous speed,

the phase sequence of the emf generated is cheked by the phase sequence indicater. 4. switch on 3 phase 415volts ac supply and chek the phase sequence using phase sequence

indicater.ensure that the phase sequence of the supply coincides with that of the armature output voltage.

5. once the phase sequence is ensured to be the same, the field circuit instant ,close TPST switch .

6. now by varying the 3 phase variac in steps the values of minimum and maximum voltages and currents are to be noted down for ech step.

7. evaluate the values of direct axis and quadrature axis reactance using respective formulae. 8. draw the vector diagram and calculate regulation of alternator for different values of

power factor and load current



220V DC

Supply

● ● ●

● ● ● || || || || || || || || || || || ||


B

Z

ZZ

0-5A MI

V 0-600V MI

X

XX

415/ 0-470V,15A 3-φ Dimmerstat

L A F

V

0-75 V MI

18 Ω,

12 A U

W

● R ● ● || || || || || || || || || || || || || || ||


● Y ● ●

● B ● ●

N ●

CIRCUIT DIAGRAM

● ● ● +

3 Pt starter

V

A

NAME PLATE DETAILS:

D P S T

T P S T

10 A

10 A

M

A

AA

10 A

10 A

10 A

270Ω,

1.2 A



FORMULAE: Max armature terminal voltage per phase Xd=---------------------------------------------------- Minimum armature current per phase Minimum armature terminal voltage per phase Xq=------------------------------------------------------ Maximum armature current per phase TABULAR COLUMN:

S.No I(min) amps

I(max) amps

V( min) volts

V(max) volts

Xd (ohms)

Xq (ohms)

LAOD QUANTITY .8 power factor Unity power factor

0.6 power factor 0.4 power factor

Leading lagging leading lagging leading lagging

Full load

Half load

1/4th load

RESULT: The values of quadrature axis and direct axis reactance have been found out. the regulation of salient pole machine has been pre-determined for different values of load and power factor. Viva Questions: 1. What is meant by Quadrature and Direct axis reactance? 2. What is meant by two axis reactance theory 3. Whaty is the relation between Xd &Xq for non salient pole machine 4. Why this test is called as Slip test 5. What are the advantages of Slip Test? 6. Which type of power plants Salient pole Alternators are used.



SUMPNER’S TEST (BACK TO BACK TEST)

AIM: To conduct sumpner’s test on two identical single phase transformers and to determine the efficiency at differentloads. NAME PLATE DETAILS OF TRANSFORMER Rating=______KVA Primary volts=_____V Secondary volts=______V APPARATUS

THEORY: The two primaries are connected in parallel to the supply across the rated voltage, and the secondaries are connected in series opposition so that their resultant voltage is ‘zero’. Any desired current can now be circulated in the windings by injecting a voltage into the secondary circuit. It is also called sometimes s back to back test from the way the connections are made. Current in the two secondaries is rated full load current of the transformer. Thus heat run test can be conducted on the transformers without actually loading them and hence steady state temperature rise in the transformer can be estimated. The current drawn by the primaries is twice the no-load current of each transformer. The wattmeter (W1) connected in the circuit of the primaries measures the total core loss of both the transformers. Thus iron losses of each transformers = Wo/2. Similarly, wattmeter (W2) connected in the secondary circuit measures the total full load copper losses of the two transformers. Hence full load copper losses of each transformer = Wcu/2.Where Wcu is the wattmeter reading (W2) when full load current is flowing in the secondary circuit. A low voltage hardly 5-8% of the rated value is applied across the secondaries for full load current to flow.

S.No. Equipment Type Range Quantity

1 Wattmeter 2 Ammeter 3 Voltmeter 4 Variac 5 Resistive load



0-300V,MI

0-2A,MI

0-75V,MI

0-20A,MI

0-600V,MI

0 230 V 230 V 0

115 V 115V

3KVA

115/230V

230/0-270V Variac

300V,10A,LPF

230/ 0-270V Variac

C V

150V,10A,UPF

M L

M L

C V

CIRCUIT DIAGRAM: SUMPNER’S TEST

230V,

1-φ,50Hz

AC Supply

● ●

● ● || || || || || || || || || || || ||

N

Ph

230V,

1-φ,50Hz

AC Supply

● ●

● ● ● || || || || || || || || || || || ||

N

Ph

A

A

V

V

V

● ●

NAME PLATE DETAILS:

D P S

T

D P S

T

5A

5 A

15 A

●

spst

230V/0-270V5A

230V/0-270V,28A



PROCEDURE: 1. Make the connections as per the circuit diagram. 2. Ensure that the switches S2 and S3 are open. 3. Energize the primaries by closing the switch S1 4. Observe the reading of voltmeter V2, which should be zero for correct connections of the secondaries. In case the voltmeter reads twice the rated voltage of each transformer, open the switch S1 and interchange the connections at the secondary terminals of one of the transformers. Close the switch S1 and verify the reading in voltmeter V2 now reads zero. 5. Adjust the setting of the variac to give zero output voltage. 6. Replace the voltmeter V2 by low range voltmeter. 7. Close the switch S3 and then S2 8. Adjust the output voltage of the variac, so that the current flowing in the secondaries is full load secondary current of each transformer. 9. Record the readings of all the instruments connected in the primary and secondary circuits. Only one set of readings is sufficient to calculate the efficiency at different loads. 10. Switch off the supply to primary and secondary circuits. TEST READINGS:

S.No PRIMARY SIDE SECONDARY SIDE

V0 (V) I1 (A) W1 (W) V2 (V) I2 (A) W2 (W)

CALCULATIONS: Iron loss of each transformer W0 = W1/2 =____ W No-load current of each transformer I0 = I1/2 = ____ A. Full load copper losses of each transformer Wsc = W2/2 = ______ W. Injected voltage of each transformer Vsc = V2/2 = ______V. Efficiency at full load and power factor = cosΦ: Output = KVA * cosΦ Total losses of each transformer under full load = W1/2 + W2/2 = W0 + Wsc INPUT = OUTPUT + TOTAL LOSSES Efficiency (η) = (output / input ) * 100 % =_______ Similarly efficiency at different loads can be calculated.



Tabular Form: Precautions: 1. Even if the voltmeter reads zero at the first instance, it is advisable to check the reading of

voltmeter by interchanging the connections at the secondary terminals of one of the transformers

2. Ensure that the variac should be at zero position 3. LV range voltmeter must be connected in secondary side after checking the phase

opposition Result: Efficiency of the given two transformers are found using Sumpner’s test. Viva Questions: 1. What is the other name for sumpner’s test? 2. What are the advantages of Sumpner’s test when compared with the other tests? 3. Define regulation of a transformer 4. What are the disadvantages of Sumpners Test? 5. Sumpener’s test is a direct test or indirect test

S.No LOAD % η



REGULATION OF 3-ф ALTERNATOR USING ZPF METHOD

Aim: To determine the regulation of 3-ф alternator by zero power factor method. Apparatus Required:

S.No. APPARATUS TYPE RANGE QUANTITY

1. Synchronising Board - - 1

2. Ammeters MI MC

(0-10)A (0-1.5)A

1 1

3. Voltmeter MI (0-666)V 1

4. Wattmeter UPF (0-5/10)A, 250V 1

5. Tachometer Digital (0-9999) r.p.m. 1

6. Rheostats - 1500Ω, 1.2 A 250 Ω, 1.2 A

1 1

7. Connecting Wires - - As per requirement

Theory: This method is based on the separation of armature leakage reactance drop and the armature reaction reactance. Hence, it gives more accurate results. It is a graphical method to find out the regulation of the given three phase alternator. The experimental data required is No-load Curve and the ZPF curve, also called the wattles load characteristic curve. It is a curve of terminal voltage against the excitation, when the armature is delivering full-load current at zero power factor. The construction of the graph to obtain regulation using the zpf method is as follows: 1. Take a point b on the zpf curve as shown. 2. Draw bk equal to bo (o is the origin). Ob’ is the field current at rated armature current

under short-circuits conditions. 3. Through k, draw kc parallel to oc’ to meet OCC at c. 4. Drop a perpendicular from c to meet bk at a. The Δabc is called the Potier triangle. 5. The length ac gives the leakage reactance drop, IaXal and ab is the armature reaction mmf

at rated current, Xp. For cylindrical rotor machine, Potier reactance Xp is approximately equal to leakage reactance, Xal. In salient-pole machine, Xp may be larger than 3 times of Xal.



CIRCUIT DIAGRAM: REGULATION OF ALTERNATOR USING ZPF METHOD

● R

● Y

● B

● ● || || || || || || || || || || || ||

● ●

● ●



0-5A, MI

300V,10A,UPF

U

W

M L

C V

V

●

● ●

● ●

● ●

|| || || || || || || || || || || ||

A

V

A

0-300 V ,MI

10 A

220V DC

Supply

● ● ●

● ● ● || || || || || || || || || || || ||

10 A

D P S

T

360Ω

1.2A

● ● ●

A

M

18 Ω,12A

145 Ω

2.8A

F

FF

●

AA

10 A

10 A

10 A

T P S T

T P S T

10 A

220V DC

Supply

● ● ● || || || || || || || || ||

D P S T

+

+

+ 0-1A

+ X

XX



Procedure: 1. Make the connections as per the circuit diagram. 2. Ensure the minimum resistance position in field of the motor(prime-mover), switch-ON the DC motor supply mains. 3. Pull the starter handle to ON position and adjust the speed to its rated value. 4. Ensure the maximum resistance position in 3- ф alternator field circuit and switch-ON the excitation of alternator and adjust its voltage to its rated value. 5. Ensure that the synchronizing switch is in open position and switch-ON the 3- ф mains supply and observe the following: a) All the 6 bulbs glow and become dark at the same time ( indicates that the phase sequence of the alternator and the source are same) b) Adjust the voltage of the alternator equal to the supply voltage. c) Adjust the speed of the alternator such thatall the bulbs glow and become dark slowly (indicates that the frequency of both the supplies are same). d) Observe the instant thet the bulbs’ filament has no glow , i.e. dark and close the switch(synchronizing switch). This is called the dark lamp synchronization of alternator with the supply source. e) Under floating condition, adjust the field of dc motor and 3- ф alternator such that the wattmeter reads zero watts at rated current and note the corresponding field current, armature current. f) Draw the Potier triangle, and hence, find the regulation of the alternator. Tabular Column: S.C Test O.C.Test

Ia If Va If

Z.P.F. Test:

Ia If V

Result: The regulation of the 3- ф alternator has been pre-determined using ZPF method. Viva Questions: 1.Define regulation of an alternator 2.Why emf method is called as pessimistic metrhod 3.What is the disadvantage of emf method 4.What is meant by synchronous impedance 5.Is synchronous impedance constant with respect to Excitation 6.What is the other name for mmf method



Equivalent circuit of 1- ф induction motor AIM: To determine the equivalent circuit parameters of a single phase induction motor from no-load and blocked rotor test. NAME PLATE DETAILS: 1-Phase induction motor Power = _______ Watts Voltage = ________ Volts Current = ________Amps Speed = _________ R.P.M APPARATUS:

S.No Equipment Type Range Quantity

1. Wattmeter

2. Ammeter

3. Voltmeter

4. Variac

5. Tachometer

THEORY: No-load and blocked rotor tests are performed on 1-phase induction motor to determine its parameters of equivalent circuit. Equivalent circuit in figure is drawn on the basis of double field revolving theory, in which the iron loss component has been neglected. The motor consist of a stator winding, represented by its resistance R1 and leakage reactance X1 and two imaginary rotors, generally called as forward and backward rotors. Each rotor has been assigned to the actual rotor values in terms of stator. Exciting branch has been shown with exciting reactance only with one-half of the total magnetizing reactance assigned to each rotor. If the forward rotor operates at a slip S, then the backward rotor has a slip of (2-S). The complete parameters of the equivalent circuit can be calculated from the following steps.

1. Measurement of A.C. resistance of stator main winding: The D.C. resistance of main winding of stator i.e., Rdc is measured by V-I method at full load current the effective value of resistance is taken 1.3 times Rdc.

2. Magnetizing reactance Xm from no-load test: At no-load cos Ф0 = W0 / (V0 * I0) Now voltage across a and b is Vab = (V0 – I0) / (Ф0 * (r1 + r2¹/4) + J * (x1 + x2¹/2)) Thus Xm = 2 * (Vab/ I0) 3. PARAMETERS FROM BLOCKED ROTOR TEST Ze = Vsc/Isc Re = Wsc/I²sc Xe = √ (Ze² - Re²) = X1¹ + X2¹



Moreover X1 can be taken equal to X2¹ i.e., X1 = X2¹ = ½ Xe Thus X2f = X 2b = ½ X2¹ Similarly R2f = R2b = ½ R2¹ FROM NO-LOAD TEST: Under no-load, motor runs at rated voltage and frequency. S is extremely small under no-load and then the equivalent circuit can be simplified as 1. Term R2¹/ 2S concerning to forward rotating field becomes quite large and can be treated as infinite. 2. Term R2¹/ (2(2-S)) reduces equal to R2¹/4 which is much smaller than Xm/2 and as such the exciting branch of backward rotating field may be considered to be open as shown in the figure. PROCEDURE: FOR NO-LOAD TEST 1. Connect the circuit diagram as shown in the figure. 2. Ensure that the motor is unloaded and the variac is set at zero position. 3. Switch ON the supply and increase the voltage gradually, till the rated voltage is reached, thus the motor runs at rated speed under no-load. 4. Record the readings of all the meters connected in the circuit. 5. switch off the supply after reducing the voltage to zero FOR BLOCKED ROTOR TEST 6. Change the ranges of all the instruments, which are suited for the test in figure. 7. Block the rotor either by tightening the belt or by hand. 8. Switch ON the A.C. supply and increase the voltage gradually, so that the current drawn by the motor under blocked rotor condition is equal to the full load current of the motor. 9. Record the readings of all the meters connected in the circuit. 10. Switch off the supply fed to the motor. 11. Measure the resistance per phase of the stator winding.



300V,5A,,LPF

(0-300V)MI

0-5A,MI

S1

S2

M1

M2

230/0-270V 1-φ variac

C M L

C V

Centrifugal switch

● ●

230V,

1-φ,50Hz

AC

Supply

● ●

● ● ● || || || || || || || || || || || ||

V

A

● N

CIRCUIT DIAGRAM: EQUIVALENT CIRCUIT OF A SINGLE PHASE INDUCTION MOTOR

NAME PLATE DETAILS:

10 A

10 A

D P S T

75V,10A,,uPF

(0-75V)MI

0-10A,MI

S

1

S2

M1

M2

230/0-270V 1-φ variac

C M L

C V

Centrifugal switch

● ●

230V,

1-φ,50Hz

AC

Supply

● ●

● ● ● || || || || || || || || || || || ||

V

A

●

CIRCUIT DIAGRAM: BLOCKED ROTOR TESTON A SINGLE PHASE INDUCTION MOTOR

NAME PLATE DETAILS:

10 A

10 A

D P S T

L



TEST READINGS NO-Load test

S.No. V0 I0 W0

Blocked rotor test

S.No. Vsc Isc Wsc

Ra test

S.No. V I Ra = V/I

CALCULATIONS: 1. Zf = Rf + J Xf = J (Xm/2) [ (R2¹/2S) + J (X2¹/2)] / (R2¹/2S) + [ (X2¹/2) + (Xm/2)] = _______ohms Where Zf is the forward impedance. 2. Zb = Rb + J Xb = J (Xm/2) [ (R2¹/2(2-S)) + J (X2¹/2)] / (R2¹/2(2-S)) + [ J(X2¹/2) + (Xm/2)] = _________ohms Where Zb is the backward impedance. 3. Zt = Zf + Zb + Z1 {where Z1 = R1 + JX1}= ________ ohms 4. Current drawn by the motor at above slip = I1 = V/Zt = _______ Amps 5. Cos Ф = Rt / Zt 6. Voltage across the backward rotor = Ef = I1 * Zf = _______volts 7. Impedance of the rotor = Z3 = √ [(R2¹/2S) ² + (X2¹/2S) ²] = ____ohms I3 = Ef/Z3 = ______ Amps T = I3² [R2¹/2S] in sync-watts 8. Voltage across backward rotor = Eb = I1 * Zb Z5 =√ [(R2¹/2(2-S)) ² + (X2¹/2) ²] = _____volts I5 = Eb/ 2S = _______Amps Tb = I2² [(R2¹/2(2-S))] = _____sync-watts. 9. Net torque (T) = Tf – Tb sync-watts Mechanical output = Pm = Tnet (1-S) sync-watt P0 = Pm – friction and windage losses = _______watts η = (P0/VI cos Ф) * 100



GRAPH: Draw a graph of efficiency Vs output

η

Output RESULT: Efficiency of single phase induction motor is calculated using equivalent circuit parameters of 1-phase induction motor. VIVA QUESTIONS: 1. State the conditions, under which no-load test is performed? 2. Which theory is commonly used for the analysis of induction motor? 3. What is the slip of forward and backward rotors? 4. What is the phase displacement in space between the two windings? 5. How the phase splitting can be increased between the two windings? 6. How is the starting winding disconnected from the supply?



CIRCLE DIAGRAM OF 3- SQUIRREL CAGE INDUCTION MOTOR

AIM:

To conduct the no-load and blocked rotor test on 3-phase squirrel cage induction motor and to predetermine the performance using circle diagram. APPARATUS REQUIRED:

S.No NAME OF THE APPARATUS RANGE TYPE QUANTITY

1. 3- squirrel cage induction motor

3.0HP, 1410RPM, 415V, 4.9amp

Cage 1 No.

2. 3- auto transformer 415/(0-470) volts 1 No.

3. Ammeter (0-5) Amps MI 1 No.

4. Voltmeter (0-150/300/600) volts MI 1 No.

5. Wattmeter (0-150) V, 10amps,UPF

MI 2 No.

6. Wattmeter (0-600) V, 10amp,LPF MI 2 No.

THEORY:

“The locus of the tip of the stator current phasor of an induction motor lies on

a semicircle and moves on it as the load on the motor increased from the no-load to full-

load value”. This locus is known as circle diagram of the induction motor. The complete set

of performance characteristics can be predetermined from the diagram.

Using no-load test and blocked rotor test, the circle diagram for 3-

squirrel cage induction motor can be plotted.

(1) NO-LOAD TEST: this test is performed to determine no-load current Io,

no-load power factor coso , windage losses and friction losses Pwf, no-load core losses Pi, no-load power input Po ,no-load resistance Ro and reactance Xo. More over it also reveals any mechanical fault, noise, etc…. The no-load test is performed with different values of applied voltage below and above the rated voltage, while the motor is running under no-load. The power input for all these three-phase Po, voltage Vo and current Io and measured by wattmeter voltmeter and ammeter. Since motor is running with out load, the power factor would be less than 0.5,hence total power input is equal to the difference of the two readings of the wattmeter. at no-load .power input is equal to



core losses, stator copper losses and windage, friction losses. core losses occur only in the stator as the slip is extremely small and so the frequency of the rotor current is low as .05Hz.the magnitude of no-load current in an induction motor is about 30-40 percent of full load current because of the air gap. so the stator copper losses at no-load needs to be accounted for this can be estimated by determining stator resistance The mechanical power developed corresponding to windage and friction losses only and the rotor circuit can, therefore be considered to be open. This is also evident from the magnitude of rotor circuit resistance. The resistance becomes very high because the slip at no-load is extremely small and soothe rotor circuit is practically open at no-load.

If windage and friction losses so determined, the stator copper losses are subtracted from power input at no-load, core losses can be determined. Knowing the total core losses, no-load current ,applied voltage ,the value of magnetizing component of no-load current, no-load resistance and reactance can be determined.

(2)BLOKED ROTOR TEST: this test is performed to determine, the short circuit current with normal applied voltage to stator, power factor on short circuit, total equivalent resistance and reactance of the motor as referred to stator. in this circuit the rotor is held firmly and stator is connected across supply of variable voltage. This test is just equivalent to short circuit test on transformer,. The connection diagram remains the same as in the case if no-load test.



CIRCUIT DIAGRAM: NO LOAD TEST ON 3- INDUCTION MOTOR

V

A

0-600V MI -- NL

0-10A MI

R

Y

B

●

R

Y B

STATOR


●

●

●

●

●

● ● ● || || || || || || || || || || ||

0-150V MI -- BR


N 415 V/0-470 V,3-Φ AUTO TRANSFORMER

●

10 A

10 A

10 A

T P S

T ROTOR

●

600V,10A,LP

M L

C V

600V,10A,LPF--NL

L

V

M

C

V

A

0-600V MI -- NL

0-10A MI

R

Y

B

●

R

Y B

STATOR


●

●

●

●

●

● ● ● || || || || || || || || || || ||

0-150V MI -- BR


N 415 V/0-470 V,3-Φ AUTO TRANSFORMER

●

10 A

10 A

10 A

T P S T

ROTOR

●

600V,10A,LP

M L

C V

600V,10A,LPF--NL L

V

M

C

CIRCUIT DIAGRAM: BLOCKED ROTOR TEST ON 3- INDUCTION MOTOR



PROCEDURE:

(A): NO-LOAD TEST:

(1): connect the circuit as shown in the circuit diagram.

(2): left the secondary of the transformer opened and using autotransformer apply rated voltage at the primary of the transformer.

(3): induction motor is made to rotate freely and then note the values of voltmeter, ammeter and wattmeter.

(4): make input zero by reducing the voltage using variac and switch off the main supply.

(B): BLOKED ROTOR TEST:

(1): connections are made as per the circuit shown in the figure 2.

(2): by tightening the handles of the load, Blok the rotor of the induction motor tightly. And then apply rated current by varying variac voltage.

(3): note the values of ammeter, voltmeter and wattmeter.

(4): adjust the autotransformer to its minimum position and switch off the main supply.

(5): calculate the resistance of the stator by ammeter-voltmeter method.

(6) Calculate the values required and plot the circle diagram for a given induction motor.

(7): construction of circle diagram should be explained clearly, by explaining all the parameters.

TABULAR COLOUMN:

1. NO-LOAD TEST:

No-load voltage No-load current Wattmeter 1 Wattmeter 2 Total power

2.BLOKED ROTOR TEST:

Bloked voltage Blocked current Wattmeter 1

Wattmeter 2

Total power



PRECAUTIONS:

1. The autotransformer should be kept at its minimum position when not required. 2. In blocked rotor test only rated current is applied and not rated voltage. 3. When blocked rotor test is performed values should be taken quickly. 4. In no-load test the rotor should be made to move freely. 5. In no-load LPF wattmeter and at blocked rotor UPF wattmeter should be used. 6. Accurate values should be noted with out any errors.

RESULT:

Circle diagram of a squirrel cage induction motor is plotted and its construction is explained.

Viva Questions 1. What is the torque line in the Circle Diagram? 2. In comparison with transformer Induction motor is having high efficiency or low efficiency 3. What are the advantages of circle diagram? 4. What is the assumption made for drawing a circle diagram? 5. Is SC and OC test is Possible for both type of Induction motors 6. What are the different starting methods of a 3 phase induction motor? 7. What is the value of slip for an induction generator?



V AND INVERTED V CURVES OF A SYNCHRONOUS MOTOR Aim: To determine the variation of armature current and power factor of a synchronous motor with excitation. Apparatus:

S.No NAME OF THE APPARATUS

TYPE RANGE QUANTITY

1. Ammeter MC 0-15A 0-2A

1 1

2. Voltmeter MC 0-300V 1

3. Ammeter MI 0-5A 1

4. Voltmeter MI 0-600V 1

5. Wattmeter Dynamometer 300V/5A/UPF 1

6. Tachometer Digital 0-9999rpm 1

7 Load Resistive 3KW 1

8 Synchronizing Board ----------- --------- 1

Theory

If the load remains constant the load angle δ remains unchanged and with changes of

field current the back emf developed in the armature changes. If the field excitation is

increased, back emf increases and if the field current is decreases the armature induced

emf decreases. Let it be assumed that the field excitation is gradually increased keeping

the load on the motor unchanged.

Let

V= Rated voltage/phase

Eb=Back emf/phase

δ = load or torque angle

Ia= Armature current

Φ = power factor angle

Active component of current = Ia cos Φ

Power input/ phase = V Ia cos Φ

As long as the load on the motor remainbs unchanged not only the power angle

remains fixed in magnitude, but the power input to the motor is constant.

VIa cos Φ= contant

But Vis also constant, Therefore for fixed load we have

Ia cos Φ = constant.

That is the active component of the armature is of constant magnitude.

Let the field current increase, this increases the back emf from phasor diagram it can be

seen that the resultant voltage phasor Er moves to the left lagging behind it by the fixed

angle θ is the armature current phasor Ia..



Since the active component ofg the current is fixed in magn9itude, we have Ia cos

Φ1 = Ia cos Φ =OP. But it can be seen that as Er moves leftwards, andgle φ progressively

decreases. Hence cos Φ increases and Ia decreases.

For a certain excitation it so happens that cosφ becomes zero so that Ia is in phase with

The P.F is unity is termed as normal excitation.

When a synchronous motor is under excited it is evident that the motor draws lagging

current and in over excited draws leading current.

V curves of the motor can be drawn at different loads as also on no-load. The

field excitation which causes minimum armature current of different loads is not same.

As the field excitation is gradually increased the motor PF which is lagging increases and

at normal excitation it becomes equal to unity. For further increase of field excitation the

PF which is now leading progressively decreases.

The graphical plot of power factor Vs field current is an Λ curve

Procedure

1. Make the connections as per the ckt diagram

2. Ensuring the minimum resistance in the field circuit of DC motor supply is switched

on for the DC motor.

3. Using 3 pointer starter the motor is started and brought to rated speed.

4. The alternator is brought to rated voltage

5. Keeping the synchronizing switch open 3Ф supply to the alternator is ON, and the

following observations are made on synchronizing board

i) All the six bulbs are glowing and becoming dark at a time

6. The voltage of alternator is adjusted to supply voltage

7. The speed of alternator is adjusted such that bulbs glow and become dark slowly.

8. At the instant when tyh bulb has no glow (dark) switch is closed.

9. The supply is supplied to alternator and DC motor, this condition is called floating

condition which machine driving the other is not known.

10. DC motor supply is switched off so that it can act as generator and alternator will acts

as synchronous motor.

11. The excitation of synchronous motor is at UPF. If the excitation is increased then it

acts as leading PF else acts as lagging PF

12. With out load on DC generator If values for both lag and lead PF is verified and noted

the Ia, V and PF

13. Generator is loaded and above step is repeated.

14. A graph is plotted between If Vs Ia and PF Vs If



Result

V & Λ Curves are plotted for Synchronous Machine Viva Questions:

1. What is the principle operation of a Synchronous Motor?

2. What is the application of Synchronous Motor?

3. What is advantage of plotting the V & Λ Curves

4. What is Load Angle?

5. The active component of armature current is having const magnitude or varies

6.what happens to the pf if the field current is gradually increased



LIST OF EXPERIMENTS

S.No

Name of the Experiment

1.

OC & SC Test on single phase Transformer

2.

Brake Test on Three phase Slip ring Induction motor

3.

Circle diagram of a Squirrel cage Induction motor

4.

Regulation of Three phase Alternator by MMF & EMF method

5.

Parallel operation of two single phase Transformers

6.

Scott connection

7.

Sumpner’s Test

8.

Regulation of Alternator using ZPF method

9.

V & inverted V curves of a Synchronous motor

10.

Slip Test

11.

Equivalent circuit of single phase Induction motor



20 A

220V DC Supply

● ● ●

● ● ● || || || || || || || || || || || || || ||

20 A

D P S T

+ L A F

● R

● Y

● B

● ● || || || || || || || || || || || ||

● ●

● ●



0-5A, MI

300V,10A,UPF

U

W

M L

C V

V

●

● ●

● ●

● ●

|| || || || || || || || || || || ||

A

V

A

0-300 V ,MI

270Ω

1..2A

A

M

18

Ω,1

2A

145 Ω

2.8A

F FF

AA

10 A

10 A

10 A

T P S T

T P S T

5 A

220V DC

Supply

● ● ● || || || || || || || || ||

D P S T

+

+ 0-1A

+ X

XX

A +

-

spst

0-20A A

v + v

- 0-300V

loa

d

-

o.c and s.c test on single phase transformer aim: …

Documents