1st Semester 1427-1428H
Final Exam
Time: 3.0 hrs.
..
King Saud University
College of Engineering
Electrical Engineering Department
EE339-Electric Machines
Answer All Questions:
Question1
(a) Explain in details, how can we get the series and shunt parameters of the approximate
equivalent circuit of single-phase transformer in the lab. And explain how can we get the
following conditions?
i. Maximum efficiency at constant value of terminal voltage and load power factor.
ii. Maximum efficiency at constant value of terminal voltage and load current.
iii. Negative voltage regulation.
Solution:
series and shunt parameters of the approximate equivalent circuit of single-phase
transformer in the lab by performing the following two tests.
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Short-Circuit Test.
This test is performed by short-circuiting one winding and applying rated current to the other
m
winding, as shown in Fig.1a. In the equivalent circuit of for the transformer, the impedance of
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ta
the excitation branch (shunt branch composed of Rc1 and X m1 ) is much larger than that of the
M
series branch (composed of Req1 and Req1 ). If the secondary terminals are shorted, the high
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impedance of the shunt branch can be neglected. The equivalent circuit with the secondary
A
short-circuited can thus be represented by the circuit shown in Fig.3.14b. Note that since
D
r.
2
2
is small, only a small supply voltage is required to pass rated current through
Z eq1 = Req
1 + X eq1
the windings. It is convenient to perform this test by applying a voltage to the high-voltage
winding.
As can be seen from Fig.3.14b, the parameters Req1 and X eq1 can be determined from the
readings of voltmeter, ammeter, and wattmeter. In a well designed transformer, R1 = a 2 R2 = R2′
and X l1 = a 2 X l 2 = X l′2 .
Note that because the voltage applied under the short-circuit condition is small, the core
losses are neglected and the wattmeter reading can be taken entirely to represent the copper
losses in the windings, represented by I12 Req1 .
-3-
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m
ta
El
M
Fig.1 Short-circuit test. (a) Wiring diagram for short-circuit test. (b). Equivalent circuit at
r.
A
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short-circuit condition.
D
No-Load Test (Or Open-Circuit Test)
This test is performed by applying a voltage to either the high-voltage side or low-voltage
side, whichever is convenient. Thus, if a 1100/ 110 volt transformer were to be tested, the
voltage would be applied to the low-voltage winding, because a power supply of 110 volts is
more readily available than a supply of 1100 volts.
A wiring diagram for open circuit test of a transformer is shown in Fig.2a. Note that the
secondary winding is kept open. Therefore, from the transformer equivalent circuit of the
equivalent circuit under open-circuit conditions is as shown in Fig.2b. The primary current is the
exciting current and the losses measured by the wattmeter are essentially the core losses. The
equivalent circuit of Fig.2 b shows that the parameters Rc1 and Xm1 can be determined from the
voltmeter, ammeter, and wattmeter readings.
Note that the core losses will be the same whether 110 volts are applied to the low-voltage
winding having the smaller number of turns or 1100 volts are applied to the high-voltage
-4-
winding having the larger number of turns. The core loss depends on the maximum value of flux
in the core.
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(a)
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(b)
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Fig.2 No-load (or open-circuit) test. (a) Wiring diagram for open-circuit test. (b) Equivalent
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r.
A
circuit under open circuit
(i) For constant values of the terminal voltage V2 and load power factor angle
maximum efficiency occurs when:
dη
=0
dI 2
η=
(1)
V2 I 2 cos ϕ 2
V2 I 2 cos ϕ 2 + Pc + I 22 Req 2
Then condition for maximum efficiency is:
Pc = I 22 Req 2
(2)
That is, core loss = copper loss. For full load condition,
Pcu , FL = I 22, FL Req 2
(3)
-5-
ϕ 2 , the
(ii) For constant values of the terminal voltage V2 and load current I 2 , the maximum
efficiency occurs when:
dη
=0
dϕ 2
If this condition is applied to effeciency equation, the condition for maximum
efficiency is
ϕ 2 = 0 Then, cos ϕ 2 = 1
that is, load power factor = 1
Therefore, maximum efficiency in a transformer occurs when the load power factor is
unity (i.e., resistive load) and load current is such that copper loss equals core loss.
Voltage regulation is obtained from the following equation
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Vno − load − Vload
*100
Vload
m
%reg =
ta
The vector diagram for the transformer in differerent operation conditions is
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shown in the following figure. It is clear that the only way to have negative
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voltage regulation is when a Vno−load < Vload and this only happened with leading
A
power factor. So we get negative regulation in leading power factor only.
r.
V1
D
(iii)
ϕ
V2′
I 2′
I 2′ Z eq1
I 2′ Req1
V1
I 2′ Z eq1
I 2′
I 2′ X eq1
(a)
I 2′ X eq1
I 2′ Req1
V2′
-6-
(b)
V1
I 2′
I 2′ X eq1
I 2′ Z eq1
I 2′ Req1
ϕ
V2′
(c)
Vector diagram for transformer for different power factor (a) lagging PF (b) Unity PF
(c) Leading PF.
A
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(b) A single-phase, 10kVA, 2000/200 V, 60 Hz distribution transformer has the
following characteristics:
Core loss at full voltage=120W, Copper loss at half load =80W
V. Determine the efficiency of the transformer when it delivers full load at 0.8 power
factor lagging.
VI. Determine the rating at which the transformer efficiency is a maximum.
Determine the efficiency if the load power factor is 0.9
VII.
Determine the rating of transformer at 92% efficiency and 0.8 power factor.
VIII.
The transformer has the following load cycle:
No load for 6 hours
66% full load for 10 hours at 0.8 PF
85% full load for 8 hours at 0.9 PF
Determine the all day efficiency of the transformer
Solution:
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r.
(i) Pout = 10 * 0.8 = 8kW
Pcore = 120W
Pcu
80
= x 2 Pcu , FL =
= 320W
2
Pcu , FL
1
⎛ ⎞
⎜ ⎟
⎝ 2⎠
8000
η=
*100 = 94.74%
8000 + 120 + 320
(ii) Maximum efficiency occurs at Pcore = Pcu
2
Then Pcu = 120 = xmax
Pcu , FL
xmax =
Pcore
=
Pcu , FL
120
= 61.23%
320
0.6123 *10000 * 0.9
*100 = 95.83%
0.6123 *10000 * 0.9 + 120 + 120
8000 * x
= 92%
(iii) η =
8000 * x + 120 + 320 * x 2
8000 * x = 0.92 8000 * x + 120 + 320 * x 2
η=
(
)
294.4 x 2 − 640 x + 110.4 = 0
-7-
640 ± 640 2 − 4 * 294.4 *110.4
x=
= 1.087 ± 0.898
2 * 294.4
Then x = 1.985 refused or x = 0.189
(iv) E24 = 0 + 10 * 0.66 * 0.8 *10 + 8 * 0.9 * 0.85 *10 = 114kWh
Ecore = 120 * 24 *10 −3 = 2.88kWh
Ecu = 320 * 0.66 2 *10 + 320 * 0.852 * 8 = 3.244kWh
E24
Then η all _ day =
*100
E24 + Ecore + Ecu
114
η all _ day =
*100 = 94.9%
114 + 2.88 + 3.244
Question2
(a) What effect does armature reaction have on the torque-speed characteristics of a shunt DC
motor? Can the effect of armature reaction be serious? What can be done to remedy this
problem?
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If the shunt DC motor is load increases the flux weakening effects reduce its flux. The effect of a flux
reduction is to increase the motor’s speed characteristics as shown in the following figure.
ωm
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With AR
D
r.
A
No AR
Tind
Yes, the effect of armature reaction can be serious on shunt DC motor because as we explain before
when the flux in a motor is decreased, its speed increases. But increasing the speed of a motor can
increase its load, resulting in more flux weakening. It is possible for some shunt DC motors to reach
a runaway condition as a result of flux weakening.
We can remedy the effects of this problem by using interpole or compensating winding
(b) DC machine (12 kW, 100 V, 1000 rpm) is connected to a 100 V dc supply and is
operated as a dc shunt motor. At no-load condition, the motor runs at 1000 rpm and the
armature takes 6 amperes.
(i) Find the value of the resistance of the shunt field control rheostat Radj .
(ii) Find the rotational losses at 1000 rpm.
(iii) Find the speed, electromagnetic torque, and efficiency of the motor when rated
current flows in the armature.
(1)
Consider that the air gap flux remains the same as that at no load.
-8-
(2)
Consider that the air gap flux is reduced by 5% when rated current
flows in the armature because of armature reaction.
(iv) Find the starting torque if the starting armature current is limited to
150% of its rated value.
(I) Neglect armature reaction.
(II) Consider armature reaction, If(AR) = 0.16 A.
Solution:
(i) No load, I A = 6 A
E A = Vt − I A R A = 100 − 6 * 0.1 = 99.4V
From magnetizing reactance (attached figure), to generate
E A = 99.4V at 1000 rpm required I F = 0.99 A
VF 100
=
= 10 A
I F 0.99
= 101 − 80 = 21Ω
Then RF = RFC + RFW =
RFC = 101 − RFW
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m
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r.
E A _ FL
88
*1000 = 885.31rpm
ω NL =
E A _ NL
99.4
D
ω FL =
A
E A _ FL K AΦ FL ω FL ω FL
=
=
E A _ NL K AΦ NLω NL ω NL
M
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ta
Protational = E A I A = 99.4 * 6 = 596.4W
(iii) The motor is loaded and I A = I A rated = 120 A
(1) No armature reaction, that is Φ NL = φ FL
E A _ NL = 99.4V
E A _ FL = VT − I A R A = 100 − 120 * 0.1 = 88V
y
(ii) At no load the electromagnetic power developed is lost
as rotational power
885.31
* 2π = 92.71 Rad / sec .
60
E I
88 *120
T= A A=
= 113.9 N .m
92.71
ωm
Pout = E A I A − Protational = 10560 − 596.4 = 99636.6W
ωm =
Pin = V T IT = VT (I A + I F ) = 100 * (120 + 0.99) = 12 099W
Pout 9963.6
=
*100 = 82.35%
12099
Pin
(2) with the armature reaction, Φ FL = 0.95 * Φ NL
E A _ FL K AΦ FL ω FL ω FL
=
=
E A _ NL K AΦ NLω NL ω NL
η=
-9-
88
ω
= 0.95 * FL
99.4
1000
∴ ω FL = 931.91 rpm
Note that the speed increases if the flux decreases because
of armature reaction.
931.91
* 2π = 97.59 Rad / sec .
60
88 *120
T=
= 108.21 N .m
97.59
P
9963.6
η = out =
*100 = 82.35% (assuming rotational
Pin 12 099
loss do not change with speed.
ωm =
(iv)
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(II) I F = 0.99 A when I A = 180 A
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∴ K AΦ = 0.949V / Rad / sec .
I A = 1.5 *120 = 180 A
Tstart = 0.949 *180 = 170.82 N .m
1000
* 2π
60
M
E A _ NL = 99.4 = K AΦωm = K AΦ
y
(I) If the armature reaction is neglected, the flux
condition under load can be obtained from the noload condition.
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r.
A
I F (eff ) = I F − I F ( AR ) = 0.99 − 0.16 = 0.83 A
From magnetizing curve the corresponding generator
voltage is
E A = 93.5V (= K AΦωm ) at 1000 rpm
93.8
93.5
K AΦ =
=
= 0.893V / Rad / sec
ωm 1000 * 2π / 60
Tstart = 0.893 *180 = 160.71N .m
Question 3
(a) In 3-phase wound rotor induction motor, prove that the maximum torque is constant
for any value of external resistance
The mechanical torque from three-phase induction motor can be obtained from the
following equation:
Tmech
Vth2
R2′
=
*
*
ω syn (Rth + R2′ / s )2 + ( X th + X 2′ )2 s
1
- 10 -
An expression for maximum torque can be obtained by setting dT / ds = 0 .
Differentiating above equation with respect to slip s and equating the results to zero
gives the following condition for maximum torque:
R2′
2
= Rth2 + ( X th + X 2′ )
STmax
∴ STmax =
R2′
Rth2 + ( X th + X 2′ )2
The maximum torque per phase fromis:
Tmax =
1
2ω syn
*
Vth2
2
Rth + Rth2 + ( X th + X 2′ )
This equation shows that the maximum torque developed by the induction machine is
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independent of the rotor circuit resistance which is clear from the following figure:
Torque speed characteristics for varying R2 .
(b) A three-phase, 100 kVA, 460 V, 60 Hz, eight-pole induction machine has the following
equivalent circuit parameters:
R1 = 0.07 Ω, R2′ = 0.05 Ω, X 1 = X 2′ = 0.2 Ω, and X m = 6.5 Ω
(i)
Derive the Thevenin equivalent circuit for the induction machine.
(ii)
If the machine is connected to a three-phase, 460 V, 60 Hz supply, determine the
starting torque, the maximum torque the machine can develop, and the speed at which the
maximum torque is developed.
- 11 -
(iii) If the maximum torque is to occur at start, determine the external resistance required
in each rotor phase. Assume a turns ratio (stator to rotor) of 1.1.
Solution:
(i) From Thevenin Equivalent Circuit
Xm
6.5
V1 =
* 265.6 = 257.7V
X1 + X m
0.2 + 6.5
( j 6.5)* ( j 0.2 + .07 ) = 0.06589 + j 0.1947Ω
RTh + jX Th =
0.07 + j 0.2 + j 6.5
VTh =
(ii)
[
Tst =
3 * 257.7 2 * 0.05
94.25 * (0.06589 + 0.05)2 + (0.19467 + 0.2 )2
] = 624.7 N .m
3 * 257.7 2 * 0.05
= 2267.8 N .m
2⎤
2
⎡
2 * 94.25 * 0.06589 + 0.06589 + (0.19467 + 0.2 )
⎢⎣
⎥⎦
0.05
STmax =
= 0.1249
2
2
0.06589 + (0.19467 + 0.2 )
ns = (1 − 0.1249 ) * 900 = 787.5rpm
Speed in rpm for which max torque occurs - = 1 − ST
max
r2
r12
+ ( x1 + x2 )
2
α r2
M
∴ STmax = k * r2
STmax
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ta
STmax =
El
(iii)
)
m
(
al
Tmax =
STstart
r2
r2 at start
S
1* 0.05
∴ r2 at start = start * r2 =
= 0.4Ω
r2 start
0.1249
D
r.
A
li
=
∴ r2
external
at start =
(0.4 − 0.5) = 0.243Ω
1.2 2
Question 4
(a) State the required condition for interconnecting synchronous generator to infinite bus.
Then explain how can you perform the synchronization process in the lab?
The required condition for interconnecting synchronous generator to infinite bus are:
1- Both have same phase sequence
2- Both have same frequency
3- Both have same voltage
4- No phase-shift between voltage waveforms.
The following figure shows a wiring diagram for the parallel operation of two alternators.
The following steps have to be performed before
Step 1: Alternator B is driven at or near its rated speed, and the field current is raised to a
level at which its no-load voltage is nearly equal to that of the grid. The no-load voltage
- 12 -
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is checked by placing a voltmeter between any two lines of the incoming alternator, as
shown in the figure when the circuit breaker is in the open position.
Step 2: To verify the phase sequence, three lamps are connected asymmetrically as
shown. When the phase sequence of alternator B is the same as that of alternator A (or
the grid), lamp L1 is dark while the other two lamps glow brightly. If the phase sequence
is not proper, the three lamps become bright or dark simultaneously.
Step 3: When the phase sequence is proper and the frequency of the incoming generator
is exactly equal to that of the grid, lamp L1 stays dark while the other lamps stay bright.
Any small mismatch in the frequency forces the three lamps to go from dark to bright in
a sequential order. In addition to the lamps to check the condition for synchronism, a
device called a synchroscope is also connected across one of the phases. The
synchroscope is shown connected across phase a in. The synchroscope measures the
phase angle between the a phase of the incoming alternator B and the grid. When the two
frequencies are the same and the phase sequence is proper, the phase difference between
the two a phases must be zero. This corresponds to the vertically up position of the
synchroscope’s pointer. The slow clockwise rotation of the synchroscope’s pointer
indicates that the phase a of alternator B is mowing ahead of phase a of the grid. In other
words, the frequency (and therefore the speed) of alternator B is slightly greater than that
of the infinite bus, and vice versa for the counterclockwise rotation. The speed is altered
by controlling the mechanical input to the alternator.
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Step 4: Alternator B is ready to be put into service by closing the circuit breaker when (a)
the line voltage of incoming alternator B is equal to that of the infinite grid, (b) lamp L1 is
dark while the other two lamps are bright, and (c) the synchroscope pointer is pointing
vertically up (zero phase-difference position).
Once the circuit breaker is closed, alternator B is on line. At this time, it is neither
receiving nor delivering power. This is referred to as the floating stage of the alternator
A synchroscope.
(b) A three-phase 5
kVA, 208 V,
four-pole, 60 Hz, star connected synchronous machine has negligible stator winding
resistance and a synchronous reactance of 8 ohms per phase at rated terminal voltage.
- 13 -
The machine is first operated as a generator in parallel with a three-phase, 208 V, 60 Hz
power supply.
(I) Determine the excitation voltage and the power angle when the machine is delivering
rated kVA at 0.8 PF lagging. Draw the phasor diagram for this condition.
(II) If the field excitation current is now increased by 20 percent (without changing the
prime mover power), find the stator current, power factor, and reactive kVA
supplied by the machine.
(IV) With the field current as in (I) the prime mover power is slowly increased. What is
the steady-state (or static) stability limit? What are the corresponding values of the
stator (or armature) current, power factor, and reactive power at this maximum
power transfer condition?
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r.
A
208
= 120V / Phase
3
Stator current at rated kVA
5000
IA =
= 13.9 A
3 * 208
φ = −36.9 o for lagging power factor 0.8
(I) VT =
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The per-phase equivalent circuit for the synchronous generator is shown in the following
figure(a).
E F = VT ∠0o + I A * jX s
E F = 120∠0o + 13.9∠ − 36.9o * 8∠90o = 206.9∠25.5o
The power angle is
δ = 25.5o
(II) The new excitation voltage is:
E F′ = 1.2 * 206.9 = 248.28V
Because power transfer remains same
VT E F
V E′
sin δ = T F sin δ ′
XS
XS
E F sin δ = E F′ sin δ ′
E
sin 25.5
sin δ ′ = F sin δ =
E F′
1.2
δ ′ = 21o
The stator current is
- 14 -
E F − VT 248.28∠21o − 120∠0o
=
= 17.86∠ − 51.5o A
o
jX S
8∠90
Power factor= cos 51.5 = 0.62 Lag
Reactive kVA= 3 | VT | * | I A | sin 51.5 = 5.03kVA
IA =
⎛ 120 * 248.28
120 2 ⎞⎟
Q = 3⎜
cos 21 −
*10 − 3 = 5.03kVAR
⎜
⎟
8
8 ⎠
⎝
The maximum power transfer occurs at
Pmax =
δ = 90o
3VT E F 3 * 206.9 *120
=
= 9.32kW
XS
8
y
E F − VT 206.9∠90o − 120∠0o
IA =
=
= 29.9∠30.1o A
o
XS
8∠90
Power factor =cos 30.1=0.865 leading
The stator current and power factor can also be obtained by drawing the phasor diagram for
maximum power transfear condition. The phasor diagram is shown in the figure above.
Because δ = 90o , E F is the voltage drop I A * X S and the current phasor I A is in the quadrature with
I A * X S , then
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r.
A
= 29.9 A
82
From the two triangles abc and abd
∠bac = ∠adb = φ
ab
120
tan φ =
=
= 0.58
ad 206.9
φ = 30.1o
Power factor =cos 30.1o =0.865 lead
m
206.9 2 + 120 2
D
IA =
al
| I A * X S |2 =| E F |2 + | VT |2
- 15 -