Republic of Iraq
Ministry of Higher Education
and Scientific Research
-
University of TechnologyElectromechanical
Department
2014
1435
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 V o, Io and Po be the reading of voltmeter, ammeter and wattmeter respectively. Neglect the small
I o2
R 1 , copper losses. Hence from figure (1.1.b), we have
[2]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Po
Vo
IW
I o2
Im
Rm
X
I o cos
m
cos
I W2
o
I o sin
o
V
IW
V
Im
IW
I o
o
I oV
Po
o
The relation between IW and I m 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 V o.
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 ?
[3]
September 14 2014
I1
Z1
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Z2
I2
Io
Io
Iw
V1
E1
Rm
Io
Im
Iw
Xm
V2
V1
Fig.(1.1.a)
Xm
V2
Fig.(1.1.b)
E1
Iw=Io CosOo
Rm
Im
Iw
Oo
Io
Im=Io SinOo
Im
Fig.(1.2)
[4]
Om
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Po
Ii
Io
V1
HT
A
w
Vi
LT
V
V
1- O Transformer
Autotransformer
Fig.(1.3) Open circuit test of single phase transformer
[5]
V2
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 R eq 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 corelosses 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 =
Xeq =
Psc
I sc
2
Z 2 eq
,
V
Z eq = sc I
SC
R 2 eq
The above two tests can be used for indirect determination of efficiency and regulation of the
transformer :
[1
Total Losses
]
output Total Losses
[6]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Regulation =
Where
IReq cos
IX eq sin
V2
100%
0
I = Secondary load current
V2 = Secondary terminal voltage on no-load
0
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 ?
[7]
September 14 2014
Isc
R1
Lab. of AC Electrical Machines Electromechanical Eng. Dept
X1
R2
X2
Isc
Vsc
Req
Xeq
Vsc
Fig.(1.4.a)
Fig.(1.4.b)
Req= R1+ R2
Zeq
Xeq
Xeq= X1+ X2
2
Zeq=
Oo
2
Req + Xeq
Req
Fig.(1.5)
[8]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Isc
Ii
Psc
A
w
HT
LT
I2
Vi
Vsc
V
A
Autotransformer
1- O Transformer
Fig.(1.6) Short circuit test of single phase transformer
[9]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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
then regulation is given by:Regulation =
=
V2
V2
0
V2
will vary with the load and
100%
0
IReq cos
IX eq sin
V2
100%
0
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 V 20 no-load and all instrument readings load.
[10]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
[11]
Vi = 220 V A.C.
Ii
[12]
Autotransformer
V1
V
w
I1
A
HT
1- O Transformer
LT
V
V2o
W2
w
Fig(2.1) Load test of single phaseTransformer
W1
I2
A
V
V2
LOAD
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
[13]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
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 .
[14]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Ii
V1
A.C.Supply
W1
Tr.1
A
I1
A
w
V
V
IL
A
V2o
V
V2
LOAD
Switch
W2
Tr.2
I2
w
A
V
V
Use double range
voltmeter
Fig(3.1) Parallel operation of tow single phase Transformers
[15]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 V 2 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.
[16]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
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 V o, 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 V 2 in steps to circulate a secondary current. Note the reading of all
instruments in the secondary circuit I2, V 2 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:
[17]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
1
where:
losses
output losses
( Pi
1
V 2I
2
cos
P )
(P P
cu
i
cu
)
V 2 = secondary rated voltage
I2= secondary current
cos
= power factor
3- Calculate Req and Xeq using reading for full load condition as follows:
Z
eq
X eq
1 V sec
2 I sec
V
2I
,
2
R
eq
2
P
2I
W
sec
2
sec
2
2
I
2
2
2
2
Zeq
R eq
4- Calculate the regula on for the full load upf, 0.8 pf lagging and 0.8 p.f leading use the formula:
V
Re gulation
V
2o
V
2 r
2o
Where: V2o= secondary no load voltage
V
2r
= secondary rated voltage
V2o is given by:
V
2o
(V cos
I2
R
eq
)2
(V sin
I2
X
eq
where: I= secondary rated current
+ sing for upf & lagging pfs
- sing for leading pfs
Regulation =
I 2 Req cos
I2
V
X
eq
sin
100%
2o
[18]
)2
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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?
[19]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Io
Wo
w
A
Tr.1
A.C. Output voltage
P1
Tr.2
S1
V3
V
P2
S2
S1
W2
w
V
V2
A
I2
Auto-transformer
A.C. Input voltage
Fig.(4.1) Back-Back test of tow single phase transformers
[20]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 singlephase 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 :
Wi W1 W2
The total copper loss corresponding to a certain transformer secondary current is given by:
Wcu Wsec
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.
[21]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
3- Increase the single – phase secondary voltage ( Vsec ) step – by- step until ( I sec ) reaches the
rated current. For each step record the readings of I 0,V1,W1,W2, I sec,Vsec and Wsec .
5.4-Results:
1- From the recorded readings, plot the iron and copper – loss against I sec .
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.
[22]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Fig.(5.1) Delta / Open Delta Loss Test on a 3-phase transformer
[23]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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
[24]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
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
Wi
3Vr Io cos
o
Po
Vo
IW
= Wi / 3Vr
I o cos
o
Rm = Vr phase/ Iw
Im
I o2
IW2
I o sin
o
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
[25]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
2
2
3I SC Re q 3I SC ( R1 R 2 )
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.
procedures:
1- Connect the circuit as shown in g.(6.4)
[26]
cos
SC
Correspondingly
6.5-Test
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
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)
cos
o
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.
and cos
.
o
as function of Vsc on one graph sheet
5- Draw the equivalent circuit of the induction motor ,showing the value of
4- Plot Isc , Psc ,
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.
PLOSS
PLOSS
and V
f (V ) , where
2
PO
3 I O R1
(
V 2
)
Vr
as shown in Fig ( 6 . 2 )
[27]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 ,
[28]
Xm , X1
and Ior.
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
FIG(6.1) Equivalent circuit of
three- phase IM
[29]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
CosO
Po
Io
CosO
V1 (volt )
V1 rated
FIG(6.3)-Variation of no load quantities with v1=
applied voltage
[30]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
W1
Three-phase A.C.Supply
I1
w
A
V
R
S
w
W2
T
Stator
Autotransformer
Fig(6.4) Circuit connection for IM tests
[31]
Rotor
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 = P 1- 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= P 22- Pcu2- Pi2Pmech(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 V i2 (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:
2 N
60
where
N= rotor speed
P2 =
T = torque in N.m. and
T
=
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 :
DC Generator Output Power
Induction Motor Input power
[32]
set
=
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
=
IM
*
dc Gen.
Assuming of IM = of DC. Gen.
DC Gen. Output
Input to IM
DC
Gen. Output
IM Input
Output of
b) Assume an
Output of
P
set
out
DC Gen.
=
IM
=
IM
=
*
DC. G
DC. G
=
2
=
=
Output of IM = Input of IM *
of DC generator say 90% ,
DC Gen.
Then output IM = input of DC.Gen.=
0.9
of
DC Gen.
Or,
0.9
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
[33]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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)
[34]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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
[35]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
Pin= W1 +W2
Po/p (G) =V L IL
, Po/p (M) =Po/ (G)
= VL IL/0.9
= Po/p /Pin
T=Po/p /
=Po/p /2
N
60
S= (N s - 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
W1
W2
V1
N
IL
8.4-Calculation
[36]
VL
S
T
September 14 2014
1. find the
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
T
3-
1-
S
Fig (1) T/S characteristic 3- and 1-
[37]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Fig (2) T/slip characteristic of positive and negative components.
[38]
September 14 2014
I1
Lab. of AC Electrical Machines Electromechanical Eng. Dept
S1
I
S1
I
I=I1+I2
I
(a)
I2
I
I2
(b)
Fig (3)
W
1
A
A
V
V
M
S1
G
S3
W
S2
[39]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Fig (4)
[40]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
[41]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
• 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
v
is kept constant in modern industrial application of speed control on 3f
motor?
3- Give a comparison among speed control methods of 3- induction motor.
[42]
induction
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
A
A
V
V
MM
Figure (1)
Circuit Diagram
[43]
G
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
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:-
ns 1
nr
S
120 f
1
P
ns
S
S
nr
ns
Where:-
n r rotor speed in r.p.m
ns 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:-
T
1
.
ns
2
R1
60
3V1 2
R2
S
.
2
X1
X2
Also we have:-
[44]
2
R2
S
rated tourque
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
V1 2 R 2
3
.
T st
n s R1 R 2 2
X1 X 2
2
60
V12
3
.
Tmax
2
ns 2 R
R12
X1 X 2
2
1
60
starting
2
rated tourque
S max
Slip at which maximum torque occurs :
tourque
R2
R 12
X1
X2
2
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
[45]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
(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
E
V ph
2.22 f Z ph
Where
f
Supply frequency
Flux per pole
Z ph
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
E1
V
i.e. 1 ratio constant, so that flux approximately remains constant.
f
f
Let K
f
where (f) is the operating frequency and (f1) the normal rated frequency for which
f1
the machine is designed (i.e. 50 Hz). The synchronous speed is now ( Kns , the applied voltage KV1
and all the reactance KX . Substituting these parameters in torque equation-1 and simplifying we
have.
T
1
ns 1
2
.K
60
3V12
R
R2
S
.
2
X1
X2
2
R2
KS
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.
[46]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
V 12
3
T st
2
T max
ns
60
.
R1
K
R2
K
R2
K
2
X1
X2
2
V 12
3
.
ns
R1
2
60 2
K
R1
K
2
X1
X2
2
The slip at which maximum torque occurs becomes:-
R2
K
S max
R1
K
2
X1
X2
2
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.
[47]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 n
f T 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?
V
is kept constant in this expression?
f
3.Draw the n f T curve for:-
2. Why
Only V1 is changed
Only P is changed
Only f is changed
[48]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Three phase suply
Auto-Transformer
W1 A
ALTERNATOR
Test Motor
V
IM
G
Load Generator
Schrage motor
W2
DC SUPPLY
Fig.(1)- Tested machine connection diagram
[49]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
N(rpm)
0.6V1
0.7V1
0.8V1
0.9V1
V1
Fig(2)- speed- torque c/c with stator voltage
control
T(N.m)
N(rpm)
Series R
Motor with
R1 and X1
Series x
Fig(3)- speed- torque c/c with stator Impedance
control
[50]
T(N.m)
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
N(rpm)
R2
2R2
4R2
8R2
12R2
16R2
Fig(4) a- speed torque c/c of slip ring IM with rotor resistance
[51]
T(N.m)
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
N(rpm)
R2
With series
resistance
With series
inductive
reactance
With series
capacitance
b- speed torque c/c of slip -ring IM
with rotor impedance control
[52]
T(N.m)
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
N(rpm)
N1
Ns
Ej in phase
with E2
N2
Ej=0
Ej in phase
opposition
E2
Fig(5) speed torque c/c with
injection of voltage into rotor
T(N.m)
N(rpm)
6-poles
12-poles
Fig(6) a- Constant torque operation speed
torque c/c with poles changing motor
[53]
T(N.m)
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
N(rpm)
6-poles
12-poles
Fig(6) b- Constant torque operation
speed torque c/c with HP driver
[54]
T(N.m)
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
N(rpm)
T(N.m)
N(rpm)
K=1
K=0.75
K=0.5
K=0.2
T(N.m)
Fig.(7) Speed –Torque C/C with V/f constant
[55]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Experiment No.( 11 )
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
armature current
I a Ie
i.e.
I a =f ( I e )
will be in the form of V g. (1) thus the
I a 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.
[56]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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 I e , and then
smoothly reduce it. The variation of the excitation current is limited by the value of armature current
Ia
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
I a =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.
[57]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
11.3.2.1-Procedure:
1- By changing the dc generator load and excitation current obtain
Ia
Ir
And unity power
factor.
2- Reduce the motor load smoothly till the no load condition while keeping the armature current
constant (
I ar
Ia
I r ) by increasing the excita
I a sin
on current, Record all reading to table 2 where
.
11.4-Calculation and Graphs:
1- From part A plot
cos
= f (I e ) ,
I a =f ( I e ) , for P=0, P=0.25Pr, and P=0.5Pr on the same
graph.
2- From B plot I a , I ar ,
cos
and
I e = 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).
[58]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
[59]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Experiment No.( 12 )
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.
[60]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
e
180
0
A
Min boost
B
B
135
A
0
B
90
A
0
B
45
0
Max boost
[61]
A
September 14 2014
B
Lab. of AC Electrical Machines Electromechanical Eng. Dept
0
0
a
A
AO= input voltage , AB= output voltage
OB=Regulator voltage
FIG(1).
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 :
[62]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
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.
[63]
Lab. of AC Electrical Machines Electromechanical Eng. Dept
September 14 2014
A
R
V
L
G
C
M
DC
AC
DC
LOAD
A
Fig.(1)-Alternator cct diagram
2
[64]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
Table(1)
Speed Constant=
rpm
Ia=Constant corresponding to rated voltage at no load=
PF
Unity PF
Steps
Ia
Amp.
0.7 Lag
V
Ia
0.7 Lead
V
Ia
V
1
2
3
4
5
6
Table(2)
Speed Constant=
rpm
V=Constant =
volt
PF
Unity PF
Steps
Ia
0.7 Lag
Ie
Ia
0.7 Lead
Ie
1
2
3
[65]
Ia
Ie
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
4
5
6
[66]
September 14 2014
Lab. of AC Electrical Machines Electromechanical Eng. Dept
[67]