ë - uotechnology.edu.iq Electrical...Parallel operation of two single -phase Transformers 3.1-Objective : To study the principle of parallel operation of two single phase transformers

-

Republic of Iraq

Ministry of Higher Education

and Scientific Research

University of Technology-Electromechanical

Department

1435 2014

September 14 2014 Lab. of AC Electrical Machines Electromechanical Eng. Dept

[2]

Experiment No.(1)

Open and Short - Circuit tests of a single - phase Transformer

A-Open Circuit test of a single -phase Transformer:

1.A.1-Objective: The purpose of this experiment is to find the iron losses from no-load test and to determine the magnetizing branch parameters.

1.A.2-Theory: The purpose of no load or open circuit test is to measure the iron losses and the components Iw and Im of the no load –current and hence find the parameters Ro and Xo of the equivalent circuit of the transformer.

For this purpose, one of the winding is open circuited and rated voltage at rated frequency is applied to the other winding. Generally in practice the H.V. winding is kept open and L.V. winding is excited and measurements are made on L.V. side. This is so because if the measurements are made on H.V. side the voltage to be applied and measured would be very large and no-load current Io will be very small. It is immaterial which winding is excited as long as normal voltage at normal frequency is applied since the iron losses and the fluxes will be same in both cases .The connection diagram for this test is shown in fig (1.3).

The equivalent-circuit of the transformer for the no-load condition is shown in fig(2.a) but since Io Z1 is very small, the circuit will be reduced to that shown in fig (1.1.a).

Let Vo, Io and Po be the reading of voltmeter, ammeter and wattmeter respectively. Neglect the small

12 RI o , copper losses. Hence from figure (1.1.b), we have


[3]

mm

Wm

ooWom

ooo

oW

IVX

IVR

IIII

IVPI

sin

cos

22

o

oo

o

Wo P

VIIIcos

The relation between WI and mI is shown in Fig.(1.2).

1.A.3-Procedure:

1- Connect the transformer as shown in fig(1.3). 2- Vary the input voltage from 0 to 125% rated value in steps, taking the readings at every step.

1.A.4-Calculation and Graphs.

1- Calculate Iw, Im and cos 0 for each value of Vo.

2- Plot Io , Iw, Im, Po and cos 0 against Vo. 3- Calculate the transformation ratio. 4- Calculate Rm and Xm for different values of Vo and find the average values.

1.A.5-Discussion: 1-why should the supply frequency be maintained constant ?

2- why generally the voltage is applied to the L.V. side ?

3- why is the P.F. of the transformer is small at no-load ?


[4]

Xm

Iw

Io

Z2

V2

I2

Fig.(1.1.a)

E1

Fig.(1.1.b)

V1

Om

Xm

Im

V1

E1

V2

Fig.(1.2)

Iw

RmRm

Im=Io SinOo

I1

Iw=Io CosOo

Io

Im

Oo

Io

Iw

Io

Z1

Im


[5]

Fig.(1.3) Open circuit test of single phase transformer

Transformer

w

Vi

HTLT

1- O

Io

Autotransformer

Po

V

Ii

V2

A

V1 V


[6]

B- Short - Circuit test of a 1-phase Transformer

1.B.1-Objective: The purpose of this experiment is to find the copper losses from short circuit test and hence estimate the efficiency for different load conditions and to determine the series impedance parameters.

1.B.2-Theory: The purpose of this test is to find the copper losses at full load and to find Req and Xeq of the transformer. For this purpose , one of the winding is short-circuited and a suitably small voltage is applied to the other winding so as to circulate full-load currents in the transformer windings. The applied voltage will be nearly that required to overcome the total impedance voltage drop of the windings and will be generally a few percent of rated value. Generally the L.V. winding is short-circuited and H.V. winding is excited. Because if the measurements are made on the L.V. side, the voltage would be very low and the current would be very high. It is immaterial which winding, is excited since the copper losses will be the same. The connection diagram for this test is shown in fig.(1.6).

Since the applied voltage is small the mutual flux is small and the magnetizing current and the core-losses may be neglected. The wattmeter reading will be equal to the total copper losses in both the windings corresponding to the current flowing in the circuit. Since the P.F. of the circuit is low, a low power factor wattmeter is to be used in the circuit. Fig (1.4.a) and (1.4.b) shows the eq. circuit of the transformer for the short-circuit condition and Fig(1.5) show the phasor diagram of Zeq .

Let Vsc, Isc and Psc be the readings of the voltmeter , Ammeter and wattmeter respectively.

Then Req = 2sc

sc

IP , eqZ =

SCIVsc

Xeq = eqReq 22Z

The above two tests can be used for indirect determination of efficiency and regulation of the transformer :

]Losses Total

Losses Total1[output


[7]

Regulation = %10002

sincos

V

IR eqIXeq

Where I = Secondary load current

02V = Secondary terminal voltage on no-load

P.F. angle of the load

Req = eq. resistance referred to secondary

Xeq = eq. reactance referred to secondary

1.B.3-Procedure: 1- Connect the transformer as shown in fig (1.6). 2-Supply H.V. side through on auto transformer. Increase the voltage in steps till 125% full load current flows on the H.V side, taking the readings at each step.

1.B.4-Calculation and Graphs: 1-Find the copper losses at full load .

2- Find Req and Xeq for each current and find average value. 3- Find e ciency at 0.8 pf. Lagging and at u.p.f for full-load condition. 4- Find the max. e ciency and current at 0.8 p.f and u.p.f loads .

1. B.5-Discussion: 1-why generally the voltage is applied to H.V.side?

2- Draw the exact equivalent circuit of the transformer ?


[8]

Isc

Zeq= Req + Xeq

Vsc

R2Isc X1

Fig.(1.4.b)

Xeq= X1+ X2

Fig.(1.4.a)

R1 X2

Zeq

2

Vsc

2

Fig.(1.5)

Oo

Req Xeq

Req

Xeq

Req= R1+ R2


[9]

A

Isc

Fig.(1.6) Short circuit test of single phase transformer

Vsc

wLTHT

A

I2

Transformer

Vi

Psc

1-Autotransformer

Ii

O

V


[10]

Experiment No.( 2 )

Load test of a single - phase Transformer

2.1-Objective:- Load test is used to determine the voltage regulation and efficiency of small transformers.

2.2-Theory :- The indirect method is used to determine the efficiency and regulation of large transformers, since it is not practical to carry a load test on these transformers. The load test can be used with small transformers, Since the provisions of test facilities and the dissipation of the load energy are available. The efficiency is determined directly from the measurements of input and output powers. i.e.

If the applied primary voltage is kept constant, then the terminal voltage will vary with the load and then regulation is given by:-

Regulation = %1000

0

2

22

V

VV

= %10002

sincos

V

IR eqIXeq

Where

V20 = secondary no load voltage

V2 =secondary load voltage

2.3-Procedure : 1-Connect the equipments as shown in g (2.1). Use current transformers, if necessary, for the measurement of power and power factor in the secondary circuit. The input voltage should be kept constant at 220V.

Note the secondary voltage V20 no-load and all instrument readings load.


[11]

2-Connect the resistive load and take readings of all instruments while the load current is increased in step up to 125o/o rated current.

3-Repeat (2) for inductive load with 0.8 and 0.9 p.f. lagging .

4-Repeat (2) for capaci ve load with 0.8 and 0.9 p.f. leading .

2.4-Calculations and graph: 1-plot Vt against I2 for all loads.

2-Calculate regulation for different p.f. at rated current I2

3-calculate and plot efficiency against I2 for all loads.

4-determine power factor for zero regulation.

2.5-Discussion: 1-Discuss the graphs and results of the experiment.

2- Draw the simpli ed phasor diagram for unity, 0.8 lagging and 0.8 leading power factor at rated secondary current.


[12]

I1

W1

LTIi

V

W2

V2

I2

LOAD

Autotransformer

AHT

Vi = 220 V A.C.

w

VV2o

w

O

V1

Fig(2.1) Load test of single phaseTransformer

A

Transformer

V

1-


[13]

Experiment No.( 3 )

Parallel operation of two single -phase Transformers

3.1-Objective : To study the principle of parallel operation of two single phase transformers

3.2-Theory: Transformers which are to operate in parallel are connected in parallel on both the primary and secondary sides main attention being given to the polarity of each. The following conditions must be established to give satisfactory operation, so that the load will be divided among the units in proportion to their rating and both reach their full load simultaneously:

1- primary winding of transformers should be suitable for the supply voltage and frequency . 2- The transformers should be properly connected with regard to polarity. 3- The voltage ratings of both primaries and secondaries should be identical. In other words the

transformers should have the same transformation ratio. 4- The equivalent impedance should be inversely proportional to the kVA ratings so that currents will

be proportional to their ratings. 5- The ratio of equivalent resistance to equivalent reactance should be same for each transformer. So

that all the currents will be in phase.

Any two single phase transformers will have the same polarity when their instantaneous terminal voltages are in phase, with this condition, voltmeter connected across similar terminals should read zero. if however the voltage measured is twice the normal voltage . Then the two transformers have the opposite polarity and one of them must be reversed.

Different R / X ratio or impedance ratios can be obtained in the case of two identical transformers by connecting a known external resistance or reactance in series with one transformer.

3.3- Procedure: 1- Connect two single phase transformers and the instruments as shown in the circuit diagram of

fig(3.1) . 2- Vary the load resistance and take readings of all instruments take one at rated currents. 3- Repeat 2 with small resistance connected in series with the primary of one of the transformers to

change R / X ratio.


[14]

3.4-Calculations and graph: 1. Plot P1 and P2 against load current according to the values obtained by the equal R / X ratios and

different R / X ratios. 2. Plot I1 and I2 against load current for both cases equal R / X ratios and different R / X ratio.

3. Plot cos 1 and cos 2 against load current for both cases.

3.5-Discussions: 1. Discuss the division of current and power between the two transformers when R / X ratios are

equal and when it is unequal . 2. Draw the phasor diagram for two transformers in parallel and explain it is features .


[15]

Tr.1

V2o V2V1

Use double rangevoltmeter

Switch

Ii

I2

w

V

AV

w

V

W1

Tr.2

Fig(3.1) Parallel operation of tow single phase Transformers

LOAD

A

VA.C.Supply V

AILI1

A

W2


[16]

Experiment No.( 4 )

BACK – TO- BACK TEST ON TWO SIMILAR single – PHASE TRANSFORMERS

4.1-Objective:

To determine the iron and copper losses of transformer and thus estimate the of the

transformer at different load conditions.

4.2-Theory:

Back -to Back tests on electrical machines permit two similar machines to be connected back to

back in such a way that one operates as a motor and the other as a generator. Full current at rated

voltage can circulate between the two machines and only the power to supply the full load losses are

provided by the mains. Such a test on two similar d.c. shunt machines is called “Hopkinson test” and on

two similar single- phase transformers “Sumpner test”.

Sumpner test is an indirect method of estimating the of a single- phase transformer. It is

applicable to only single- phase transformers and requires two similar single- phase transformers .

In the case of 3-ph transformer, heat-run test, known as “Delta / Delta heat run test”, can be conducted,

wherein the power fed from the mains will be only to supply the losses. Refer to connection diagram

shown in Fig-1, the two primaries are connected across the supply and the secondaries are connected in

opposition so that the voltage across the open ends of the secondaries is zero. With the secondaries

open-circuited, the primaries will draw a current Io which is the sum of the no-load currents of the two

transformers. The power measured by the wattmeter Wo is thus the total iron losses of the two

transformers. Since the two transformers are similar, the iron-loss of each transformer will be Wo/2 and

the no-load current of each transformer will be Io/2.

If the secondaries are now connected to a variable low voltage source V2 and the injected voltage

is so adjusted to pass full load current through the secondaries, a condition similar to full-load

transformers are created but with the difference that the power drawn from the source Vo will be only

total iron-losses and the power drawn from the source V2 will be to supply the total copper-losses of the

two transformers. The injected voltage V2 is equal to the equivalent impedance drop in the transformers.

The reading of W2 will be the copper - losses of transformers and hence the copper-loss of each

transformer will be W2/2.


[17]

In practice, the full - load temperature-rise of the transformers can be studied by circulating the

full-load current through the secondaries for sufficient number of hours till the windings temperature

becomes steady. Thus this method enables to conduct “heat-run tests”, with power consumed equal to

sum of total losses of the two transformers only.

4.3-Procedure:1- Connect the two transformers as shown in fig (4.1), with the switch S1 being kept open.

2- Switch on the primary voltage source Vo and apply rated voltage across the two primaries. Take

the readings of Vo, Io and Wo.

3- Note the reading of voltmeter V3 across the switch S1.If the secondaries are properly

connected, the voltage a cross the switch S1 will be zero. Otherwise it will be twice the

secondary voltage. In which case the terminals of one of the secondaries are to be reversed

to obtain the voltage across the switch S1 equal to zero.

4- After making sure the voltage a cross switch S1=0, close the switch S and gradually increase the

secondary source voltage V2 in steps to circulate a secondary current. Note the reading of all

instruments in the secondary circuit I2, V2 and W2 at each step till I2 reads full load current

.Observe that the readings of Io, Vo and Wo will be unchanged and will remain constant as

recorded in step 1.

5- Temperature rise at full load can be calculated by “resistance method”, by allowing the full load

current to ow through for 3-4 hours and recording the ambient temperature and the

resistances of the windings before the start of the experiment and immediately after the

experiment. (the students are not required to do this part of the experiment but may discuss

this in this report).

4.4-Results and Graph:

1- Calculate the iron loss Pi and copper less Pcu of each transformer for full load condition.

2- Calculate the at 25%, 50%, 75%, 100% and 125% full load condi ons for unity p.f. and 0.8 p.f.

lagging and plot = f( Iload ) use the formula:


[18]

)(cos2

)(11

2 PPIVPP

cui

cui

lossesoutputlosses

where: V 2 = secondary rated voltage

I2= secondary current

cos = power factor

3- Calculate Req and Xeq using reading for full load condition as follows:

IV

IVZ eq

2

2

sec

sec

221

, I

W

IPReq 2

2

2

2sec

sec 22

RZX 2eq

2eqeq

4- Calculate the regula on for the full load upf, 0.8 pf lagging and 0.8 p.f leading use the formula:

VVVo

rogulation2

22Re

Where: V2o= secondary no load voltage

V r2 = secondary rated voltage

V2o is given by:

22

222

)sin()cos( XR eqeqoIVIVV

where: I= secondary rated current

+ sing for upf & lagging pfs

- sing for leading pfs

Regulation = %100sincos

2

22

VXR

o

eqeq II


[19]

4.5-Discussions:

1- What are the advantages and disadvantages of this test?

2- Why it is impractical to carry out actual load test on large rating transformers? 3- What will be the current in the primary windings :

a) When secondaries are not loaded?

b) When secondaries are loaded?


[20]

V2

V3

w

S1

P1

Tr.1

S1

V

V

Auto-transformer

W2

w

Io

Fig.(4.1) Back-Back test of tow single phase transformers

Wo

I2

A.C. Input voltage

Tr.2

A

P2 S2A.C. Output voltage

A


[21]

Experiment No.( 5 )

DELTA / OPEN DELTA LOSS TEST ON A 3-PHASE TRANSFORMER

5.1-Object:

To measure the iron and copper losses of a three –phase transformer. This test can also be used as

a heat – run test to determine the temperature rise of the transformer but this is beyond the scope of

this experiment.

5.2-Theory:

The primary and secondary of the transformer are delta – connected and the three phase rated

supply is applied to the primary with the secondary is open delta clearly the total power supplied will

then be the no-load (iron) loss of the transformer.

If the secondary delta is opened at one point no fundamental voltage appears at the open ends, though

third harmonic voltage will appear since such voltages for the three phase are added – up. If a single-

phase low voltage is now applied to the open secondary ends, a circulating current will flow in the delta

secondary and corresponding compensating currents will then flow in the primary delta. The full load

heat-run test can be carried out by circulating the full load current in the secondary windings for enough

time, and the total iron loss is thus given by :

21 WWWi

The total copper loss corresponding to a certain transformer secondary current is given by:

secWWcu

5.3-Procedure :

1- Connect the circuit as shown in gure(5.1). Apply the 3- phase rated voltage to the primary

winding, with no single – phase supply applied to the open-delta secondary windings.

2- Connect a voltmeter across the secondary terminals and measure the third – harmonic voltage.


[22]

3- Increase the single – phase secondary voltage ( secV ) step – by- step until ( secI ) reaches the

rated current. For each step record the readings of secsec,,2,1,1,0 VIWWVI and secW .

5.4-Results:

1- From the recorded readings, plot the iron and copper – loss against secI .

1- 2- Predict the e ciency characteris cs at u.p.f. and 0.8 p.f. then plot the e ciency against

percentage load.

=1-(Wi+Wc)/[o/p+Wi+Wc]

Percentage load=(Isec/Irat)*100

5.5-Conclusions:

1- Discus the efficiency curves above.

2- Comment on the value of the triple – frequency voltage. Explain why a fundamental frequency

voltage does not appear at the o.c. secondary terminals.

3- Comment on the suitability of experiment for prediction of efficiency.


[23]

Fig.(5.1) Delta / Open Delta Loss Test on a 3-phase transformer


[24]

Experiment No.( 6 )

OPEN AND SHORT CIRCUIT TEST ON 3 – PHASE INDUCTION MOTOR

6.1-Object:-

To determine the parameters of equivalent circuit of 3-phase induction motor.

6.2-Theory:-

The rotor of a.c motor does not receive electric power by conduction but by induction in exactly the

same way as secondary of a 2-winding transformer receives its power from primary. That is why such

motors are known as induction motors. In fact an induction motor can be treated as a rotating

transformer, i.e. one in which primary winding is stationary but the secondary is free to rotate.

The equivalent circuit of an induction motor as in the case of transformer the secondary value may be

transferred to the primary and vice versa.

Fig.(6.1.a) shows the equivalent circuit of an induc on motor where all values have been referred to the

stator.

The parameters of this equivalent circuit are determined by open and short circuit tests.

6.3-Open circuit (No- load) test:

Fig.(6.1.b) shows the circuit diagram for open circuit case of an induc on motor. The no load test is

carried out with different values of applied voltage, below and above the value of normal voltage. The

power input is measured by two wattmeter's ,the stator current (Io) by an ammeter and the applied

voltage(V1) by a voltmeter(V). Which are included in the circuit of g.(6.2).For the induc on motor is

running at light load, the p.f would be low i.e. less than (0.5) hence the total power input will be the

di erence of two wa meter reading (W1)and (W2) .

Let

Vo = applied voltage / phase

Io = motor current / phase


[25]

Po = watt meter reading i.e input in watt

Xm = magnetizing reactance

Rm = resistance whose losses correspond to the iron losses

For the separation of iron loss from the input no load power reading , we have that :

Wi = Po – mechanical loss

ocosIoVr 3 Wi

= Wi / 3Vr ooo

oW I

VPI cos

Rm = Vr phase/ Iw

ooWom IIII sin22

Xm = Vr phase/ Im Then the exciting impedance(Zm) for 3-

phase induction motor is:

Zm = Vo / Io

6.4-Short circuit test:

It is also known as locked – rotor test. Fig.(6.1.c) shows the equivalent circuit for short circuit test of an

induction motor. This test is used to find :-

1- Short –circuit current(Isc) with normal voltage applied to stator windings.

2- Power factor at short –circuit .

3-Total leakage reactance (Xeq) of motor and total resistance of the motor (Req) .

In this test, the rotor is locked and the rotor windings are short-circuited at slip-rings , if the motor has

wound rotor just as in the case of short-circuit test on a transformer.A reduced voltage at about (15 or


[26]

20 per cent of normal value) is applied to the stator terminals and so adjusted that full load current ows

in the stator.

As in this case (S =1), this means that R2/S is equal to R 2 where R 2 is the resistance of the rotor referred

to the stator side .

Because Is.c is much greater than the exciting current Io we can neglect, the magnetizing branch and

now the equivalent circuit reduces to that shown in g.(6.1.c) It is evident that Isc = I 2 and the phase

value of this current is

Is.c = Vs.c / Zeq

Zeq = Vs.c / Is.c , Req = Ps.c /3(I s.c) , Xeq = ( Zeq2 - Req2)

If we known R1, then 2 =Req – R1

If we do not known, we can assume R1= 2 =Req 2

Since X1 can only be calculate by design data, we can assume X1=X2=Xeq / 2

The power required by the motor at short circuit(Psc) is consumed by the stator and rotor cupper

losses neglecting a small Pfe( iron losses) because of small test voltage during the short circuit test.

)R(3Re3 2122 RIqI SCSC Psc = Pcu1 + Pcu2 =

Assuming that the stator and rotor leakage fluxes are similar in a medium with constant permeability,

the reactance is almost constant, Xeq = const.

Similarly Req = const. , therefore Zeq = const. and hence the relation ship Isc = f( Vsc ) is almost a linear

one .

= Req / Zeq = f( Vsc ) shown a constant value. SCcos Correspondingly 6.5-Test

procedures:

1- Connect the circuit as shown in g.(6.4)


[27]

2- If the motor is a slip ring induction motor, open the rotor winding when at rest and supply a.c. voltage

to the stator. Measure the voltage appearing across a pair of slip rings of the rotor.

3-Close the circuit of rotor winding, then Start the motor on no load. Take the readings of all instruments

while the voltage is reduced in steps from 120% to 20% rated voltage.

4- Lock the rotor securely so that it can not rotate, then short circuit the rotor terminals. Supply the

stator with a reduced voltage. Take readings of all instruments while the voltage is reduced from that

which gives about ( 1.25) I rated ll that which gives( 0.2) I rated (reading should be taken quickly to

avoid over heating because of lack of ventilation).

5- Measure the stator and rotor resistances (if it is a slip ring rotor, rotor resistance can be measured, if it

is a squirrel cage motor, it cannot be measured) by using voltmeter-ammeter and a low dc. Voltage. To

obtain the effective a.c.value of resistance multiply dc. Resistance by a factor k=1.2.

6.6-Graph and calculation:-

as func on of V1 on one graph sheet as in g(6.3) ocos 1- Plot Po , Io and

2- Find the following from no load test : Rm , Xm , Iw , Im

3- From short circuit test data find Zeq , Req , and Xeq for the rated current.

. sheetgraph oneon Vsc offunction as cos and o 4- Plot Isc , Psc ,

5- Draw the equivalent circuit of the induction motor ,showing the value of the parameters of the

circuit.

6- To find the iron losses and mechanical losses ,draw the variation of no load losses with the applied

voltage ,i.e.

)2.6()(

3

,)(

2

12

OOLOSS

LOSS

FiginshownasVVVand

RIPP

whereVfP

r


[28]

6.7-Discussions :-

1- Give the slip value of the motor at the following speed :

Stands ll , 20% of synchronous speed and synchronous Speed .

2- Calculate the frequency of the rotor currents and voltages at above speeds

3- Upon what factors the following quantities depend (Psc , P f +W , Xm , X1 and Ior.


[29]

FIG(6.1) Equivalent circuit of three- phase IM


[30]

V1 (volt )

Po

CosO

V1 rated

Io

FIG(6.3)-Variation of no load quantities with v1= applied voltage

CosO


[31]

Three-ph

ase A.

C.Supp

ly V

W1

W2

S

w

Fig(6.4) Circuit connection for IM tests

Stator

R

Autotransformer

T

AwI1

Rotor


[32]

Experiment No.( 7 )

Load Test on a 3- Phase Induction Motor

7.1-Object:

To conduct the load test and find the efficiency, slip , power factor , input current , input power

,rotor copper loss …etc as a function of output power .

7.2-Theory:

The net input power to the rotor i.e. the power transferred from stator to rotor ( P12) is stator

input P1 - stator copper losses Pcu1 - stator iron losses Pi1 . i.e. P12 = P1- Pcu1- Pi1 .

The net output power of the motor i.e. the shaft output P2= P12 –rotor copper losses Pcu2 - rotor

iron losses Pi2 which is negligibly small at running Condition - Pmech .i.e. P2= P22- Pcu2- Pi2-

Pmech(Pmech=mechanical losses or rotational losses) .Since the rotor current can not be measured in

squirrel cage motor, rotor copper losses can be calculated as (rotor input power * slip).

The stator iron losses at rated voltage and mechanical losses friction and windage losses can be

obtained by conducting no-load test at various input voltage Vi . and plotting the graph of input power

on no-load as a function of Vi2 (refer to expt.No 9 for details).

If the torque of the motor shaft can be measured during the load test the output power can be

easily calculated as: P2 = T

602 N

where T = torque in N.m. and =

N= rotor speed r.p.m

If there is facility to measure the torque and the motor is to be loaded by means of a DC generator

the shaft output of the motor can be approximately estimated by one of the following methods:

a) Find the overall ( ) of the IM-DC Gen. Set :

powerInputMotorInductionPowerOutputGeneratorDC

set =


[33]

= IM * dc Gen.

Assuming of IM = of DC. Gen.

IMtoInputOutputGenDC .

set = IM * DC. G = 2 =

InputIMOutputGenDC .

IM = DC. G =

.GenDCofOutput = Output of IM = Input of IM *

b) Assume an of DC generator say 90% ,

9.0.GenDCofOutput

Then output IM = input of DC.Gen.=

9.0

.GenDCofPout Or, P2(IM) =

7.3-Procedure:

1- Connect the circuit as shown in g (1)

2- Switch on the supply and the motor should start immediately .If the direction of rotation is

wrong, interchange any two supply leads allow the motor to run for ten minutes at 25% over

load to warm up its windings.

3- Reduce the load to zero in steps take readings of voltage, current, power input, speed and slip as

a function of load. Record all results


[34]

7.4-Calculation:

1- Find the efficiency from the output power and input power .

2- Plot the graph of speed slip efficiency and power factor to a common base of output power

3- Plot the graph of torque current and output power to a common base

of slip.

4- Plot Torque - Slip characteristics .

7.5-Conclusion:

1- what is the idea of interchanging two lead if the direction of rotation is wrong ?

2- explain the construction of cage I.M.

3- explain the importance of slip measurement and give full load slip of the motor.

Fig. ( 1)


[35]

Experiment No.( 8 )

Operation of 3- phase I . M. Under

Unbalanced condition

8.1-THEORY:-

The unbalanced opera on of 3-phase I.M from a single phase supply is not a common operating

condition but it may be considered during an emergency when there is a loss of one phase of three

phase supply or it might be connected intentionally due to the lack of a near by three phase supply .The

unbalanced condition of operation leads to a large vibration which reduce the life of motor and produce

hum .The unbalanced operation reduces the motor torque, capability and efficiency. To prevent burning

the motor , it is not allowed to run for a pre longed period when the unbalanced in voltage is more than

5% for the same, motor is disconnected from the source whenever single phasing occurs unless the

single phasing is always accompanied by a light load.

The unbalanced system can be solved by the use of symmetrical components. Thus unbalanced

stator voltages can be analyzed into positive, naegatve and zero sequence components.

The positive sequence components are responsible for producing the forward torque component. The

negative sequence components are responsible for producing the backward torque component. The zero

sequence components if present will not affect the torque or mechanical output. For the delta

connection of stator winding, zero sequence components of current can be eliminated since there is no

return path through the neutral of a 3-phase system.

(In = 3 Io since In = 0 then Io = 0)

The effect of the negative sequence voltage is thus seen to be reduction of the torque and mechanical

output. Because of the backward by rotating magnetic field. Produced torque has negative sign because

the synchronous speed is (- S) and the slip is (2-s) while in the positive sequence has positive sign and

speed ( S) and the slip is (s). Torque/speed characteris cs for 3 and 1- is shown in

Fig (1) and g (2) shown T/slip characteris c of posi ve and nega ve components.

T= Tp - Tn


[36]

Pin= W1 +W2

Po/p (G) =VL IL , Po/p (M) =Po/ (G) = VL IL/0.9

= Po/p /Pin

T=Po/p / =Po/p /2 60N

S= (Ns - N)/Ns

P.f =Pin /Vi Ii (single phase)

Or p.f = Pin / 3 Vi Ii (three phase)

8.2-Experiment:

If the motor windings are connected as star and one of the supply-phases is out the motor is

opera ng as shown in g(3-a), also if the motor windings are connected as delta and one of the supply

phases blown the motor is operating as shown in g(3-b). In both cases the motor is equivalent to the

single phase motor which it will have all the motor characteristics.

8.3-Procedures:

1. connect the cct as shown in g (4)

2. Switch on the supply (three-phase) and the motor should start immediately with very light load,

the generator operating at no load. (in a single phase supply the motor cannot be starting

because it has no starting torque in this case special arrangement needed).

(the load on the 3-phase motor in this case is its losses and the generator losses).

3. Switch on the switch S1 then take the reading as shown in the table as a function of load, current

should be not higher than the rated current of the motor.

I1 W 1 W 2 V1 N IL VL S T

8.4-Calculation


[37]

1. find the M

2. Plot in comparison with the readings of 3-phase operation N, S, and with Po/p M.

3. Plot in comparison with the readings of 3-phase operation T, I1 and Po/p M with speed.

Fig (1) T/S characteristic 3- and 1-

3-

1-

T

S


[38]

Fig (2) T/slip characteristic of positive and negative components.


[39]

Fig (3)

S1 I=I1+I2

I

I2

I2

I S1

I1

I

I

V

W

W1

A

M G

A

V

(a)

(b)

S1

S2

S3


[40]

Fig (4)


[41]

Experiment No.( 9 )

Speed Control of Three Phases I. M. by Variation of Rotor Resistance

9.1-General:

Motor speed may be varied by putting variable resistance back into the rotor circuit. This reduces

rotor current and speed. The high starting torque available at zero speed, the down shifted break down

torque, is not available at high speed. See R2 curve at 90% Ns, below. Resistors R0 R1 R2 R3

Increase in value from zero. A higher resistance at R3 reduces the speed further. Speed regulation

is poor with respect to changing torque loads. This speed control technique is only useful over a range of

50% to 100% of full speed. Speed control works well with variable speed loads like elevators and printing

presses.

Wound rotor induction motor qualities.


[42]

• Excellent starting torque for high inertia loads.

• Low starting current compared to squirrel cage induction motor.

• Speed is resistance variable over 50% to 100% full speed.

• Higher maintenance of brushes and slip rings compared to squirrel cage motor.

9.2-Experimental Program:

1- connect the machine as shown in fig.1

2- Start the 3- induction motor at R2.

3- Gradually load the induction motor by the dc generator coupled its shaft. Take readings for

various loads

4- Adjust the rotor resistance of 3- induction motor to= R2’ repeat (3)

5- Adjust the rotor resistance of 3- induction motor to= R2” repeat (3)

9.3-Graphics:

Plot N=f (T) for three values of rotor resistance of 3- induction motor R2, R2’, R2” on one graph

sheet

9.4-Conclusion and discussion:

1- Explain why pole changing method of speed control can not be used for slip ring induction

motor?

2- Why vf

is kept constant in modern industrial application of speed control on 3- induction

motor?

3- Give a comparison among speed control methods of 3- induction motor.


[43]

Figure (1)

Circuit Diagram

M M

V

A

G

A

V


[44]

Experiment No.( 10 )

SPEED CONTROL OF 3- PHASE INDUCTION MOTOR

10.1-General: The poly phase induction motors are constant speed motors i.e. the speed is almost constant as

the load torque increase up to rated value. The running speed of an induction motor is:-

SP

fSnn sr

11201

s

rs

nnnS

Where:-

rn rotor speed in r.p.m sn synchronous speed in r.p.m f supply frequency in Hz p number of poles s slip

Since the varia on of the slip S from no load to full load is very small (about 0.05), for constant frequency and constant number of poles, the speed of an induction motor is almost constant. The torque of an induction motor is given by:-

SR

XXS

RR

Vn

Ts

2

221

22

1

21 .3.

602

1 rated tourque

Also we have:-


[45]

221

221

22

1.

602

3XXRR

RVn

Ts

st starting tourque

221

211

21

max2

.

602

3

XXRR

Vn

Ts

rated tourque

Slip at which maximum torque occurs : 221

21

2max

XXR

RS

Where V1, R1, X1, X2, and R2' are all phase values.

Hence the speed torque characteristics can be modified by:-

1. By varying the supply voltage. (Figure 2) 2. Varying the stator parameters (R1, X1). (Figure 3) 3. Varying the rotor parameters (R2, X2). (Figure 4)

The last method of speed control can be used in the case of slip ring induction motors.

Another method of speed control in the case of slip ring induction motor is by injection of voltage into rotor circuit by an external frequency converter or by special construction as in the case of Schrage motor ( gure 5). These methods require addi onal machine or special construc on.

Another method of controlling the speed of squirrel cage induction motor is by changing the poles, i.e. the stator winding may be connected for different pole number and hence the synchronous speed of the motor can be changed, (Figure 6). (This method is not possible in slip ring induc on motor. Why? Discuss that. This method has its own limitation:-

1. The speed can be controlled only in steps. 2. its possible to have only two different synchronous speed for one winding 3. In cascade control we can have speeds corresponding to P1, P2, and P1 ± P2 pole numbers

but this method requires two machines mechanically coupled. With the development of semiconductor devices, it is now possible to change the supply frequency (F) continuously so that the speed of induction motors can be continuously varied by varying the frequency of voltage applied to the stator winding. This method has found a great use in the modern industrial application and now the induction motors are gradually replacing the variable speed D.C. drive in industry.

The object of this experiment is to study the speed control of an induction motor by varying the frequency of the applied voltage. For this purpose an alternator is connected to a variable speed drive


[46]

(Schrage motor in our case) and the frequency of the alternator voltage is varied by changing the speed of the Schrage motor speed by changing the brush position. The magnitude of the alternator voltage can be controlled by controlling the alternator field excitation, but at constant excitation this voltage is proportional to the frequency.

10.2-Speed control by variation of supply voltage: The induced e.m.f (back e.m.f) equation of induction motor is

phph ZfVE 22.2

Where

f Supply frequency

Flux per pole

phZ Number of series conductors per phase.

Hence if only (f) is varied keeping applied voltage constant, the air gap flux varies. If f is decreased from rated value, increase and the machine get saturated and the magnetizing current increase.

Hence it is customary to keep the air gap flux at its normal value. For this purpose we must keep

fVei

fE 11 .. ratio constant, so that flux approximately remains constant.

Let 1ffK where (f) is the operating frequency and (f1) the normal rated frequency for which

the machine is designed (i.e. 50 Hz). The synchronous speed is now ( sKn , the applied voltage 1KV

and all the reactance KX . Substituting these parameters in torque equation-1 and simplifying we have.

KSR

XXSRR

VnT

Ks

2

221

22

21

1.3

.60

2

1

may be K>1 or K>2 It may be seen that this expression is of the same as the original torque equation, but resistances have become larger by a factor K. similar results can be obtained for starting torque, maximum torque and maximum slip.


[47]

221

221

221

.

602

3

XXKR

KR

KR

V

nT

sst

221

211

21

max

2

.

602

3

XXKR

KR

Vn

Ts

The slip at which maximum torque occurs becomes:-

221

21

2

max

XXKR

KR

S

Typical speed torque curves for four different frequencies are shown in gure (6), the slip at which maximum torque occurs becomes larger as the operating frequency decreases and the maximum torque gets reduced slightly. The starting torque increases for small reductions in frequency but attains a maximum and then decreases with the further reduction in frequency.

The reduction in developed torque at low frequencies is partly due to the apparent increase in resistances of the machine and also due to the air gap flux remains constant while V/f constant, which the above torque equa on (equa on 2) neglects. This reduc on in air gap ux which results from the voltage drop in the stator impedance will obviously spend on the input current and hence on the load of the motor.

10.3-Experimental program: 1. Connect the machine as shown in figure -1 2. Start the Schrage motor, adjust its speed for rated frequency and adjust the D.C.

excitation for the alternator for rated voltage motor through the auto transformer. 3. Gradually load the induction motor by the D.C. generator which coupled with its shaft.

Keep the speed of the Schrage motor constant. Adjust the excitation of the alternator to compensate for the voltage drop of the alternator. Take all reading for various loads

4. Adjust the frequency of alternator to f=1.1 fr, the voltage change proportionately repeat 3.


[48]

5. Adjust the frequency to f=0.9 fr, repeat 3. 6. Adjust the frequency to f= 0.8 fr, repeat 3.

10.4-Graphs:

Plot Tfn for the four frequencies i.e. (1.1 fr, fr, 0.9 fr, and 0.8 fr) on one graph sheet

10.5-Conclusion and discussions: 1. Explain why pole changing method of speed control can not be used for slip ring induction

motor?

2. Why f

V is kept constant in this expression?

3.Draw the Tfn curve for:-

Only V1 is changed Only P is changed Only f is changed


[49]

VA

IM

Schrage motor

Test Motor

Auto-Transformer

Three phase suply

W2

W1

Load Generator

G

DC SUPPLY

ALTERNATOR

Fig.(1)- Tested machine connection diagram


[50]

Fig(2)- speed- torque c/c with stator voltage control

T(N.m)

N(rpm)

0.6V1 0.7V1 0.8V1 0.9V1 V1

Fig(3)- speed- torque c/c with stator Impedance control T(N.m)

N(rpm)

Series x

Series R

Motor with R1 and X1


[51]

16R2

12R2

8R2

4R2

2R2

R2

Fig(4) a- speed torque c/c of slip -ring IM with rotor resistance

T(N.m)

N(rpm)


[52]

b- speed torque c/c of slip -ring IM with rotor impedance control

T(N.m)

N(rpm)

R2

With series inductive reactance

With series resistance

With series capacitance


[53]

Fig(6) a- Constant torque operation speed torque c/c with poles changing motor

T(N.m)

N(rpm)

12-poles

6-poles

Fig(5) speed torque c/c with injection of voltage into rotor

T(N.m)

N(rpm)

Ej in phase opposition E2

Ej=0

Ej in phase with E2

N1 Ns

N2


[54]

Fig(6) b- Constant torque operation speed torque c/c with HP driver

T(N.m)

N(rpm)

12-poles

6-poles


[55]

T(N.m)

N(rpm)

K=0.2

K=0.5

K=0.75

K=1

T(N.m)

N(rpm)

Fig.(7) Speed –Torque C/C with V/f constant


[56]


Synchronous Motor Operation

11.1-Theory:

A 3-phase synchronous motor is not self starting i.e. it does not start when the rotor is excited by dc

and stator is applied with 3-phase voltage in order to start a synchronous motor the following methods

are usually used

a) Using a Pony Motor: A separate small dc motor is used to bring the speed of the synchronous motor to near about

synchronous speed and then the drive motor is disconnected and the rotor is connected to dc supply

b) Using the Synchronous induction motor: Using a wound rotor induction motor to start as an induction motor (rotor is shorted through

external resistance) and run as a synchronous motor by connecting the rotor to dc supply.

c) Using Damper Windings: In the pole shoes of the salient poles of the synchronous motor, thus obtaining the starting torque

by induction motor action.

The last method is used in this lab. As all the synchronous motors used are of salient pole construction

and have been provided with damper Windings for starting purpose.

11.2-General:

The variation of the excitation for a constant load and constant applied voltage, cause the armature

current to change and the graph of aI eI i.e. aI =f )( eI will be in the form of V g. (1) thus the

armature current aI required by the synchronous motor can be brought into phase or to lagging

or leading with respect to the terminal voltage by suitable adjustment of the excitation current. It

leads the terminal voltage at over excita on and lags at under excita on g. (2) This behavior is very

important, since it enables the motor to be used for power factor correction.


[57]

11.3-Experimental Investigation:

11.3.1- A) V – Curve of the Synchronous Motor:

To obtain the V-curve it is recommended to start with maximum excitation current eI , and then

smoothly reduce it. The variation of the excitation current is limited by the value of armature current aI

which should not exceed 1.2 mes Irated i.e. 1.2Ir. It is easier to evaluate the motor load by the required

power input from the supply. It must be noted that the required power input to the motor is not strictly

constant; it is different values of the armature current due to the variation of motor losses.

11.3.1.1- Procedure:

1. Connect the circuit diagram as show in g. (3)

2. Keeping the field circuit open and start the motor, when the motor has attained a speed near

about synchronous speed, close the field circuit with no load on the motor. adjust the field

excitation such that aI =1.2Ir . Reduce the excitation current and observe the armature current

and input power according to table 1

3. Repeat 2 above for P=0.25Pr, and P=0.5Pr the synchronous motor is loaded by the dc machined

coupled to it. The dc machine is operated as a generator and loaded by a load rheostat.

11.3.2- B)Regulation of the reactive power by Active power Variation:

The synchronous machine can be used simultaneously as a loaded motor or as a capacitor to

regulate the Reactive power of the supply.

The aim of this experiment is to show the armature current (its reactive component), the excitation

current and the power factor as a function of the input power at constant armature current applied

voltage.


[58]

11.3.2.1-Procedure:

1- By changing the dc generator load and excitation current obtain ra II And unity power

factor.

2- Reduce the motor load smoothly till the no load condition while keeping the armature current

constant ( ra II ) by increasing the excita on current, Record all reading to table 2 where

sinaar II .

11.4-Calculation and Graphs:

1- From part A plot cos = f )( eI , aI =f )( eI , for P=0, P=0.25Pr, and P=0.5Pr on the same

graph.

2- From B plot aI , arI , cos and eI = f (P) on the same graph.

11.5-Discussion:

1- Explain V curve by using Phasor diagram.

2- How the regulating characteristics can be obtained from V curve?

3- Comment on the motor operation as a phase advancer (capacitor).


[59]


[60]


ALTERNATOR OPERATING CHARACTERISTICS

12.1-Theory :

The main operating characteristics of an alternator are the A) External B) Regulating characteristics .

12.1-A - External Characteristics

It is the terminal voltage as a function of the armature current i.e V= f(Ia)

At constant speed and constant field current and constant power factor.

At no- load the terminal voltage is equal to the no- load induced voltage E. When the load is increasing

the armature- reaction effect is also increasing –cross magnetizing for pure resistance, demagnetizing

for pure inductive load and magnetizing for pure capacitive load for intermediate power factors the

terminal voltage decrease for resistance – inductive load (V < E) and will increase for resistance –

capacitive loading i.e

(V > E) , the regulation therefore is :

% Reg = E – Vr / Vr

Where

E = no – load terminal voltage

Vr = full – load terminal voltage

And it is positive value for lagging power factors and negative value for leading power factors.


[61]

e B

B

0

B

0

B

0

0

180

135

90

45

A

A

A

Min boost

Max boost

A


[62]

12.1-B - Regulating characteristics

It is the excitation current Ie as function of the load current Ia at constant power factor and speed .

i.e Ie = f( Ia ) when N , Vr and cos are constant . This characteristics shows to that extent the excitation

current has to be regulated in order to keep the terminal voltage constant always.

12.2-Procedure:

1- Connect the circuit as shown in g.(2)

2- Start the d.c motor and run at the rated speed adjust the excitation of the alternator to generate the

rated voltage at zero armature current keep the excitation current and speed constant and increase the

load from

Cos = 1. (2) Cos = 0.7 0 -125% ) full- load ( 4 or 5) steps for (1) (

Lagging and (3) Cos = 0.7 leading . Record all readings according to table 1 .

3- Repeat 2 above but keep the terminal voltage constant always by changing the excitation current

while the load is changing for (1) cos =1

(2) cos = 0.7 lagging (3) cos = 0.7 leading , keep the speed constant.

Record all readings according to table 2 .

12.3-Graphs and Calculation :

a

0 B 0 A

AO= input voltage , AB= output voltage OB=Regulator voltage

FIG(1).


[63]

1- From (2) plot V = f( Ia ) for di erent power factors. And calculate the voltage regula on at rated

current .

2- From (3) plot Ie = f( Ia ) for di erent power factors. Find the excita on current ra os i.e Ie = Ier /

Ieo (in relative units)

Where

Ier the excitation current at rated voltage and rated armature current

Ieo the excitation current at rated voltage and zero armature current

12.4-Discussion :

1- What do you understand by ( armature - reaction )? Discuss the effect of armature reaction in an

alternator when is loaded with Resistive , Inductive , Capacitive loads.

2- Draw the phasor diagram of the armature with leading. Lagging and unity power factors.

3- Discuss the importance of the external and regulating characteristics.

4- Give the value of voltage regulation for different power factors at full- load current and interpret their

values.


[64]

DC

AC

Fig.(1)-Alternator cct diagram

A

A

LOAD

V

DCCM

R

L G

2


[65]

Table(1)

Speed Constant= rpm

Ia=Constant corresponding to rated voltage at no load= Amp.

0.7 Lead 0.7 Lag Unity PF PF

V Ia V Ia V Ia Steps

1

2

3

4

5

6

Table(2)

Speed Constant= rpm

V=Constant = volt

0.7 Lead 0.7 Lag Unity PF PF

Ie Ia Ie Ia Ie Ia Steps

1

2

3


[66]

4

5

6


[67]

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