eeelab final latest

7/30/2019 Eeelab Final Latest

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ELECTRICAL AND ELECTRONICS ENGINEERING LAB MANUAL

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

PRAGATI ENGINEERING COLLEGE, SURAMPALEM 1

DPSTSWITCH

1. SWINBURNES TEST ON D.C. SHUNT MACHINE.(PREDETERMINATION OF EFFICIENCY OF GIVEN D.C.SHUNTMACHINE WORKING AS MOTOR AND GENERATOR)

AIM :

To conduct the Swinburnes (No- Load) test on the given D.C. shunt machine andpredetermine its efficiency vs. load curve treating it as (a) Motor and b) generator

APPARATUS:

S.NO NAME OF THE APPARATUS RANGE TYPE QUANTITY1 Ammeter 0-1A

0-2AMC 2No

2Voltmeter

0-300V0-30V

MC 2No

3 Rheostat 600/1A,300/1.5A

Wire wound 1

4 Tacho eter - Digital 1

CIRCUIT DIAGRAM :

(0-300V)MC

SWINBURNES TEST

300/1.5A

(0-1A)MC-Fuse10A

O O O

1L

2L

starter.pt3

L A F

A

AA

1F

2F

Z

ZZ

V220Supply.C.D

A A

V


2/24




5AFUSE

(0-2A)MC 300 /1.5A

30V (0-20V)MCD.C supply

DETERMINATION OF ARMATURE RESISTANCE BY DROP METHOD

THEORY:

In this test, the d.c shunt machine is run as a motor at the rated speed applying rated

voltage. The power input to the motor can be easily calculated from the readings of meters

and the constant losses are determined as below.

Power input to the motor =VI0, where

V =Rated applied voltage

I0 =Line current taken by the motor at no load

Ia=Armature current =I0 Ifwhere Ifis the field current.

Armature copper loss = (I0 If)2

R0, Where Ra is armature resistance in Ohms.

Field Copper loss =V If Wc = Constant losses =VI0 (Ia2

Ra) watts

The constant losses Wc include iron (hystresis and eddy current losses), mechanical(friction and windage) losses and the field copper loss.

The efficiency of the machine can be found out at a given load as below for (a) motorand (b) generator.(a) Motor : Suppose the machine is treated as a motor : Efficiency at full load is

found as below.

Output of the motor at rated full load =VIL (Ia2

Ra+Wc)

Efficiency at full load =(VIL (Ia2

Ra+Wc)) / VIL

The efficiency at full load is found as below.

1L

2L

A

AA

A

V


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Line current =IL / 2 at half full load.

Filed current (same as above) =If

Armature copper at full load Ia =(IL / 2) If

Armature copper losses at full load =(IL/2 If)2

Rawatts

At half full load =2/VI

WRI2

I

2

IV

2

Ca2

fLL

In the same way, described above, efficiency at full load, full load etc., can befound out.

(b)Generator: If the d.c. machine is treated as generator, efficiency is calculated asbelow. For full load,

Output of the generator =VIL, where V =rated voltage, IL =rated line current.

Wc, the constant losses are the same as determined above.

Armature current Ia =(IL +If), Ifis the same as before

Efficiency at full load =VIL / (VIL =Ia2

Ra+ Wc)

Efficiency at half full load is determined as below.

Armature copper loss at half full load = ((IL/2)+If)2

Ra

Efficiency at half full load =

ca

2

fLL

L

WRI2

I

2

VI

2

IV

Determination of Armature Resistance : The resistance of the armature can be determinedexperimentally by drop method using the circuit show in fig 2.2. Due allowances must egiven for the temperature rise under running condition. The revised armature resistance,

assuming a temperature rise of 500

is calculated.

Let the room temperature be =t00

Temperature under running condition = (t00

+500)

Let the temperature coefficient of copper be 0 =1/234.5.

Rt0 = R0 (1 +Rt0 0 ) Where Rt0 =Resistance at room temperature as determined above.

= R0(1 +(1/234.5)t0)

Rt0 +50 = R0

0

000

5.23450t1

Rt0 +50 = Rt0

0

00

000

t5.234

50t5.234ohms

Rt0+50 is called the hot resistance of the armature , which should be used for the armature

copper loss


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PROCEDURE :

1) Make connections as per the circuit diagram2) Keep the rheostat (speed regulator) in the field circuit in the minimum position.3) Switch on the D.C. supply and by means of the 3-point starter, gradually cut of the

starter resistance step by step and keep the starter handle in the ON position.4) By varying the rheostat in the field circuit increase the speed of the motor to its rated

value.5) Disconnect the supply and allow the machine to come to rest.6) Make connections as per the circuit diagram given in fig 2.27) Switch on the 30 V d.c. supply and adjust the armature current to its rated value.8)Take the readings of meters.9)Tabulate the results in tabular form 2.

Speed = Rated speed = 1500 RPM

TABULAR FORM:

S.No. VL IL If I = IL - If

TABULAR FORM:

ARMATURE RESISTANCE

S.No. Va I Ra


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CALCULATIONS AND GRAPHS:

As explained in theory, calculate the hot resistance of the armature. Also calculate theefficiencies of the d.c. machine treating it as (a) motor and (b) generator, at full, full, full and full loads and draw graphs.

(a) Efficiency vs. load as motor, and(b)Efficiency vs load as generator

RESULT:

QUIZ :

1. What are the disadvantages of this rest?2. Why the Swine Burnes test can not be used for D.C. series machines?3. Why do you need starter for the D.C. motors?4. What is the function of n0 volt and over load relays in the starter?5. Why is a 4 point starter preferred in some motors?6. What happens when the field circuit of a d.c. shunt motor is

accidentally open circuited while running?


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2. OC & SC TESTS ON SINGLE PHASE TRANSFORMER.(PREDETERMINATION OF EFFICIENCY AND REGULATION ATGIVEN POWER FACTOR AND DETERMINATION OF EQUIVALENTCIRCUIT)

AIM:To conduct the open circuit (no-load) and short circuit tests on the 1-phase

transformer and determine.a) Parameters of the equivalent circuit andb) Efficiency and regulation at various loads and power factors.

APPARATUS:S.NO NAME OF THE APPARATUS RANGE TYPE QUANTITY

1 Ammeter 0-10A0-10A

MIMI

2No

2 Voltmeter 0-150V0-75V

MIMI

2No

3 Watt Meter 2A,150V,60W10A,75V,30W

LPFUPF

2No

CIRCUIT DIAGRAM:

1 VARAIC230V/(0-270V)

1KVA

1KVA

10A,75V,30W

UPF Wattmeter

(0-300V)MI

(0-10A)MI

(0-2A)MI

SUPPLY.C.A,1

Hz50V230

SUPPLYCA

Hz

V

..,1

50

230

5A

A

B

E

C

A

V

fuse

fuse

TEST

CS ..

MI)A100(

HV LV

VV 115230

Ph

N

5A

LV

V

MI

V

)1500(

fuse

fuse

C

M L

V

C

M L

V

FTauto

HzV

/,1

50),2700/(230

C

B

WattmeterLPF

WVA )60,150,2(

TEST

CO ..

Ph

N

DPST A

A

V

HV

VV 230115

E

DPSTswitch

1 VARAIC230V/(0-270V)


7/24




THEORY:

The equivalent circuit of the 1-phase transformer referred to primary is givenbelow.

The approximate equivalent circuit is as given below.

R1 - Primary Resistance r2 - Secondary Resistancex1 - Primary Leakage reactance x2 - Secondary LeakageReactanceR01 - Equivalent shunt resistance referred to primary which represents Iron lossX01 - Equivalent shunt reactance referred to primary which represents the

magnetizing current.N1 - Number of Primary turns N2 -Number of secondary turns.V1 - Primary applied voltage V2 -Secondary terminal voltageV2 - Secondary terminal voltage referred to primaryE1 - Primary induced voltage.

D

A

O

L

1V1

2V1E

CI WI

OI

2

2

2

1 rN

N

2

2

2

1 xN

N

1r 1x

01R 01X

D

A

O

L

1V1

2V

2

2

2

111 r

N

NrR

2

2

2

111 x

N

Nxx

CI WI

OI

01R01

X


8/24




O.C. Test :

Suppose the readings of all meters in this test are given below.

Applied Voltage = V1 reading of wattmeter =W0

Current drawn =I0IC, IW, R01 are calculated from the above readings as below.Where IC, IW are the magnetizing Component, Iron loss (working) component of no loadcurrent respectively.

V1I0Cos 0 =W0 ; 0 =No-Load p.f. angle.

IC =I0 Cos 0 =W0 / V1I0 sin 0 = )CosII( 022

02

0

R01 = V1 / IC , X01 =V1 /Im.The reading of the wattmeter gives the iron loss of the transformer for the rated

terminal voltage applied in the test.

Iron Loss =W

S.C. Test:In this test with the secondary / L.V. winding short circuited, a reduced voltage is

applied to the H.V. winding in this till full load current flows. Let following be the readingsobtained.

Applied Voltage = Vs Watt meter reading =WsCurrent Drawn =Is

R1, X1and Z1are calculated from the readings as below.

R1

=Ws/I

s

2; Z

1=V

s/I

s; X

1= 2

1

2

1 RZ

Copper Loss at full load i.e., Ifull =Ws

EFFICIENCY & REGULATION:

Efficiency of the transformer at a given load I2 and Power factor cos 2 is given

))I/I(WWCosIV(/)CosIV( 2full2si222222

Regulation of the transformer at a given load I2and power factor Cos 2 is foundfrom the formula.

% Regulation =((E2 V2)/E2) x 100, Where E2and V2are related by the equation E22

=(V2

Cos 2 +I2 R2)2

+(V2 Sin 2 +I2X2)2; Where E2=Secondary induced voltageV2=Secondary terminal voltage

The approximate formula of the above is given asE2 V2=I2R2 Cos 2 +I2 X2 sin 2 .Note: For all lagging loads 2 is positive and for leading loads 2 is negative in the aboveformulae.


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TABULAR FORM:a) O.C. Test :

S.No VOC IOC WOC

b) S.C. Test :

S,No VSC ISC WSC

PROCEDURE:

1) Make connection as per the circuit diagram shown in fig. 1 (a) for the O.C. test.2) Apply rated voltage to the primary of the transformer and note the readings of all

meters.3) Make connections as per the circuit diagram shown in fig. 1(b) for the S.C. test.4) By means of the variac, apply a reduced voltage till full load current flows in the

winding. Note readings of all meters.5) Evaluate R01, X01, R1 and X1 and insert the values in the equivalent circuit.6) Find the regulation and efficiency of the Transformer at, , and 1of full load

current and at power factors of 0.8 lag, unity and 0.8 leading.7) Draw the approximate equivalent circuit and insert the values calculated as discussed

in theory.8) (i) Taking Load current on the x axis draw efficiency vs load current graph at

various power factors.Taking load current on the x-axis draw the regulation vs Loadcurrent curve for various power factors.

RESULT:

QUIZ:

1. Why is the Iron Loss negligible in the S.C. test? Why is the copper Lossnegligible in the O.C. Test.

2. Under what conditions do you get negative regulation for transformer?3. Using the approximate formula find the power factor for zero regulation?4. Why are percentage values used more commonly for the resistance, reactance

and impedance of the transformer instead of absolute values?5. What is the significance of leakage reactance in the transformer?


10/24




3. BRAK E TEST ON 3-PHASE INDUCTION MOTOR.(DETERMINATION OF PERFORMANCE CHARACTERISTICS)

AIM:To conduct the brake test on the given 3-phase induction motor and plot its

performance characteristics.APPARATUS:

S.NO NAME OF THE EQUIPMENT TYPE RANGE QUANTITY

1) Ammeter MI (0-10A) 12) Voltmeter M (0- 00V) 13) Wattmeter UPF 10A/600V/1500W 24) Tachomeer digital - 1

CIRCUIT DIAGRAM:1

MI

)100(

STARTER

DELTASTAR

UPF

WVA 1500,600,10/5

fuse

fuse

fuse

2W2W

2V

1V

B

1L

2L

TPST

3L

2U

1U

LI

R

plysup.C.A

,Hz50

V415,3

2S1S

MI

V)6000( 1V

2VMI

V)6000(

VC

M L

2W

1V

2V

ROTOR

1U

2U

STATOR

W

C

M L

V

UPF

WVA 1500,600,10/5

Y


11/24




THEORY:Brake test in a direct method of testing. It consists of applying a brake to a water

cooled pulley mounted on the shaft of the motor. A rope is wound round the pulley and its

two ends are attached to two spring balances S1and S2. The tension of the rope can be

adjusted with the help of swivels. Then,The force acting tangentially on the pulley = (S1 S2) Kgs.

If R1 is the pulley radius, the torque at the pulley,

Tsh=(S1 S2) R kg. Mt.

If w is the angular velocity of the motor.W =2N/60, Where N is the speed in RPM.

Motor output =Tsh x w =2N (S1 S2) kg.mt.wt=9.81 x 2N (s1 S2) R watts.

Where R=radius of the fully =0.0115m

The motor input can be measured directly as in the circuit diagram 6. For finding theperformance characteristics, the speed of the motor can also bemeasured by a tachometer.TABULAR COLOUMN:

S.No S1(Kg) S2(Kg) W1(w) W2(w) VL(v) IL(a) N(rpm) T n-m S=slip Input Output =output/input

MODEL CALCULATIONS:Slip= Ns-N/ Ns NS= Synchronous Speed=1500 rpm

T=9.81*(S1-S2)*R where R=radius of the fully

Input =W1+W2,

Output =T*W

Efficiency=Output/input


12/24




PROCEDURE:1. Make the connections as per the circuit diagram shown in above.2. Loosen the rope of the break drum such tat S1=S2=0.3. Close the switch S and apply the rated 3-phase a.c. supply to the motor. Note

the readings of all meters.4. Gradually increase the load by tightening the rope and note down the readingsof all meters and tabulate the results as shown below.

5. Starting from no-load, take the readings as the line current is increased from, , , and 1 of its full value.

6. The output and input of the motors, Efficiency, Torque and slip can becalculated and the performance characteristic.a) Load vs. Efficiencyb) Load vs. Speedc) Load vs. Torqued) Load vs. Slip and speed are draw

RESULT:

QUIZ:

1. What are the types of starters generally used for squired cage inductionmotors?

2. How is the supply voltage related to the starting Torque?3. In what respects slipping I.M. superior to squirrel cage?4. What is the value of rotor resistance, which gives maximum starting torque?5. For what value of slip, do you get maximum running torque in I.M?


13/24




4. REGULATION OF 3 ALTERNATOR BY SYNCHRONOUSIMPEDANCE METHOD

AIM:

To determine the regulation of the given 3-phase alternator by SynchronousImpedance method and

.CIRCUIT DIAGRAM:

Fig (a)

THEORY:

In this experiment the 3-phase alternator is driven by the D.C. shunt motor at ratedspeed. The field is excited by a variable d.c. voltage output drawn from a rectifier.

The regulation of the alternator is determined by synchronous impedance method

from the O.C.C. and S.C.C tests.

O.C.Test:

The open circuit characteristic (O.C.C.) or the magnetizing characteristic of thealternator is obtained by plotting the variation of induced voltage E as the field current isgradually increased. The shape of the O.C.C. is as shown in Fig. (b). This curve actuallyshows a small voltage at zero field current due to the residual magnetism. But this isneglected and not shown. In the initial stages when the field current is small O.C.C. is a

1L

fuse

SupplCD

V

.

220

L A F

A

AA

Z

ZZ

A7.1/300

A

AV

X

X

S

S

S

XX

fI

R

B Y

2L

switch

TPST

outputcdVariable ..XX

fuse

fuse

RECTIFIE

fuse

MC)A20(

M)A100(

MI)V6000(

1F 2F

DPST

switch

..230

CAV

Ph

N

DPST

switch


14/24




straight line since the field is unsaturated. With higher field currents E increases slowly andfinally reaches a fixed due to magnetic saturation.

S.C. Test:In this the machine is run at constant speed and the armature terminals are short-

circuited. A reduced excitation is applied to the field. The S.C.C. is obtained by plottingarmature current I on the y-axis and the field current on the x-axis. This is characteristic is astraight line as shown in fig. (b).

MODEL GRAPH & CALCULATION:

Regulation of alternator at a given load current Iaand p.f. Cos is an index of the drop of

voltage with load. It is given by

% Regulation =((E-V)/V) x 100 -------- (1)Where E is the induced voltage in the armature / phase.V is the rated terminal voltage of the armature / phase.The phasor diagram of the alternator is given in fig. (c)

Fig. (b)

E2 =(V COS +IaRa)2

+V sin +IaXs)2

- - - - - - - (2)

Where E and V are phase values, and

Ia = Armature current / phase

Ra = Armature resistance / phase.

Xs = Synchronous reactance / phase.

Note: For lagging power factor is positive and for leading power factors is negative.

Regulation by synchronous impedance method:Suppose it is required to find the synchronous impedance and synchronous reactance

of the alternator at a load current of Ia.

voltsinEph

.C.C.O

.C.C.S

V

1E

aI

faI 2fI CurrenField


15/24




For the armature current Ia, find the field current Ifarequired from the S.S.C. From

the O.C.C. find E1corresponding to Ifa. Then Zathe synchronous impedance is given by

Zs= E1/Ia.

Xs, the synchronous reactance of the alternator is given byXs = 2a

2s RZ , Where Ra is the a.c. resistance of the armature.

Knowing Ra and Xs, for a given rated terminal voltage V, Regulation at any given loadcurrent and power factor can be found from equation (1) and (2).

Fig. c PHASOR DIAGRAM OF ALTERNATOR

Corresponding to Ifr, find the induced voltage E from the O.C.C. then regulation at thegiven load current and p.f. is given by

% regulation =(E V)/E) x 100

TABULAR COLOUMNS:

a) O.C.TEST:S.NO. LINE VOLTAGE(VL) PHASE VOLTAGE(VPh)=(VL) /3

b) S.C.TEST:S.NO. FIELD CURRENT(If) ARMATURE CURRENT(Ia)

aI

saXI

aaRI

saZI

E


16/24




PROCEDURE:

1. Make connections as per the circuit diagram for obtaining the O.C.C. ormagnetization curve. Keep the switch S is open.

2. Start the D.C. motor by means of the 3-point starter and adjust the speed of theset to the rated value.

3. Switch on the rectifier and supply the field current to the field terminals of thealternator.

4. Starting with zero field current, gradually vary the field current and obtain thecorresponding reading of voltmeter connected across the armature. Continuethis till you get nearly a constant voltage in the armature.

5. Plot the O.C.C. taking Ifon the x-axis and E-phase on the y-axis. This will beobtained as shown in fig (b).

6. For obtaining he S.C.C. make connections as per the circuit diagram in fig (a).Replace all the meters with meters of new ranges.

7.

Start the set by means of the 3-pt starter of the d.c. motor and adjust speed tothe rated value by the field regulator.

8. Close the switch S. Switch on the rectifier. Starting with zero field current,gradually increase the field current and note the corresponding armaturecurrent. Continue this till full load current flows through the armature.

9. The d.c. resistance of the armature is found using the drop method. Allow20% extra fro the skin effect and find the Rac.

10. Calculate Xs the synchronous reactance of the armature at a given Ia and findthe regulation from equations (1) and (2) as explained in theory.

11. Draw the regulation vs. load curve graphs at (a) 0.8 lag (b) unity and (c) 0.8lead power factors using the above method.

RESULT:

QUIZ:

1. How is armature reaction considered in the determination of regulation ofalternator?

2. Why synchronous impedance is called fictitious impedance? Is it constant forall field currents?

3. The regulation determined by synchronous impedance method is calledpessimistic method, . Why?

4. Do you get zero or negative regulation for some loads? Explain.5. What are the errors in the two methods of finding regulation?


17/24




5. BRAK E TEST ON D.C.SHUNT MOTOR.

AIM :To conduct the brake test on the given D.C. Shunt motor and draw its performance

curves.

APPARATUS:

S.No Name of the equipment type range Quantity1) Ammeter MC (0-1A),(0-20A) 22) Voltmeter MC (0-300V) 13) Rheostat Wirewound 300/1.5A 1

4) Tachometer digital - 1

CIRCUIT DIAGRAM:

(0-20A)MC+

(0-1A)MC 300/1.7 A

A(0-300V)MC

AA

Brake-load-

ArrangementCIRCUIT DIAGRAM FOR BRAKE TEST

DPSTSwitch

1L

2L

L A F

M

Z

Starterpt.3

V

A

Y

1S 2S

A

Z Z

2F1F

Supply

C.D

V220


18/24




THEORY:The brake test is a direct test on the d.c. motor. Therefore, the performance

characteristics correspond to the actual performance of the motor under running conditions.

This test can be used only for small motors. Because for large machines, dissipation of heat

produced on the pulley is a problem.It utilizes a brake drum fixed on the shaft of the motor as shown in fig 7.2. A rope is

wound round the pulley and its two ends are attached to two springs balances S1 and S2.

The tension of the rope can be adjusted with the help of handles on the frame. The force

acting tangentially on the pulley is equal to the difference between the two readings of the

spring balances S1and S2.

Let the radius of the pulley =R meters

Shaft torque Tsh=(S1 S2) R x 9.81 N M

Motor Output =TshX w = TshX 2N/60 whereW is the angular velocity of rotor in rad / sec. N is the speed of the motor in RPM.

Evidently, the output of the motor is utilized in overcoming the mechanical friction between

the pulley and the belt and heat is produced. Cooling of the pulley is therefore required to

dissipate this heat. The following performance curves can be drawn in this test.

1) Efficiency vs BHP2) Speed vs BHP3)Torque vs Ia.

MODEL GRAPHS:

N

KW

Efficiency

NSpeed aT

shT

N

'T'Torque

'N'Speed

aI


19/24




PROCEDURE:

1) Make connections as per the circuit diagram2) Start the machine by means of the 3 point starter and adjust the speed of the

motor to its rated value.3) Take the readings of all meters at no-load. Also note down the readings of

spring balances when the rope is removed from the pulley. Now put the rope

around the pulley and load the motor by gradually tightening it.

4) At each load, note the readings of meters, ad the spring balances. Cool thebrake drum by pouring cold water in the pulley when the motor in on load.

Continue this process till full load is reached. Stop the machine and tabulate

the results as below. Measure the radius of the pulley.

TABLER FORM:

S.No VL IL IF IA S1(Kg) S2(Kg) N(rpm) T n-m Input Output =output/input

CALCULATIONS:VL =Line voltage

IA =IL - IF

T=9.81*(S1-S2)*R where R=radius of the fully

Input =VL*IA watts

Output =T*W where W=2N/60 ,N=Speed of the motor in r.p.m.

Efficiency=Output/input


20/24




GRAPH:Calculate the efficiency of the machine and torque at each load current and plot the graphs

a) Efficiency vs BHP

b) Speed vs. BHP

c) Torque vs. Armature current, Ia

RESULT:

QUIZ :

1. Why the iron losses are constant in a d.c. shunt motor ?2. What is the application of d.c. shunt motors?3. Why d.c. shunt motor is called constant speed motor?4. How do the hysteresis and eddy current losses depend on the speed?5. Why are the armature and pole cores laminated?6. What is the advantage of this test over the Swine Bernes test ?


21/24




1.MAGNETIZATION CHARACTERISTICS OF D.C. SHUNTGENERATOR , DETERMINATION OF CRITICAL FIELDRESISTANCE & CRITICAL SPEED

AIM:To draw the magnetization characteristics of the given D.C.Shunt generator anddetermine the 1) critical resistance 2) critical speed

APPARATUS:

S.NO NAME OF THEAPPARATUS

RANGE TYPE QUANTITY

1 Ammeter (0-1A) MC 12 Voltmeter (0-300V) MC 13 Rheostat 1200/0.5A,

300/1.5AWire wound 2

4 Tach eter - Digital 1

CIRCUIT DIAGRAM:

Fuse

10A F1

3 pt Starter

+

+

(0 30

(0 1 A )

+L1

L2

L

AA

Z

Z

Z

Z

SupplCD

V

.

220

2F

DPST

switch

V

AA1/600

A1.5/300A

A F

A

MAGNETISATION CHARACTERISTICS OF D.C SHUNT GENERATOR

1200/0.5 A


22/24




THEORY:Magnetization characteristic of DC shunt generator is defined as The flux Vs

Field current graph at constant speed. Since, the induced e.m.f. in a dc shunt generator isdirectly proportional to the flux , at constant speed, induced e.m.f. Vs field current can bedefined as Magnetization characteristic

The magnetization characteristic can be divided into the following sections

ab: The induced voltage oa corresponds to the value at zero field current. This value is equalto the induced voltage due to residual magnetism. In the section ab, the flux produced andhence the induced voltage increases in proportion to the field current and therefore thecharacteristic is a straight line.

bc: At b called the knee point, the core of the field starts getting saturated and the increase in

flux or induced voltage is not in proportional to the field current thereafter.At the point C, the core is almost saturated and any further increase in field current doesnot produce a change in the flux or EMF.

The critical resistance of the field winding is defined as the max. resistance of the fieldabove which the generator fails to excite at a given speed. This is obtained by drawing atangent to the magnetization or open circuit characteristics of the machine (o.c.c) shown infig.

The slope of the straight line od drawn in fig represent a particular value of the fieldresistance. With this reistance increases the slope and the point d goes lower and finallyod becomes a tangent. This corresponds to the tangent to the magnetization characteristic.Thus the resistance of the field corresponding to the gives the critical resistance.

For the given resistance of the shunt field corresponding to od, suppose the speed of themachine is decreased. Evidently all the points in the magnetization move lower and for aparticular speed, od becomes a tangent to it. This speed is called speed of the generator.

PROCEDURE:

1. Make connections as per the circuit diagram in fig 1.12. By means of the 3-point starter, start the set and adjust the speed to the rated value.3. With the switch S open, note the reading of voltmeter (E). This gives the voltage inducedin the armature due to residual magnetism.4.Close the switch S and gradually increase the field current by moving the jockey.At each

value of field current If, note the corresponding changes in the induced voltage in thearmature (E) , ensuring constancy of speed throught.5.Cotinue the procedure given in 4 above, till the induced voltage becomes nearly constant.6.Repeat the procedure from 2 to 5 with decreasing values of field current.7. Tabulate the results below.

Rated Speed =1500rpm


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TABULAR FORM:

S.No If(Amps)increasing

Eo(V)

MODEL GRAPH:

TYPICAL PLOT OF E.M.F E VS IF OF A DC SHUNT GENERATOR


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PRAGATI ENGINEERING COLLEGE SURAMPALEM 24

CALCULATIONS AND GRAPHS:

1.Plot E Vs Ifboth for increasing and decreasing values of field currents.2. Find the critical resistance of the field and the critical speed as explained in theory.

RESULT:

QUIZ:

1. State three causes for d.c shunt generator to fail to excite.

2. What is armature reaction? What is its effect in a dc generator.3. What is the function of interpoles.4. What type of connection is used(lap or wave) for

(a) High current and low voltage(b) Low current and high voltage

5. In a dc generator, the load current is 100amps. If the number of poles is 4, what is thecurrent in the armature conductors if it isa) Lap woundb) Wave wound.

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