King Saud University
College of Engineering
Electrical Engineering Department
EE331-Electromechanical Energy Conversion II
Test #1
2nd Semester
1431/32H
Time: 60 min.
Question1 (10 Marks)
A 13.8 kV 10 MVA 0.8-PF-lagging 60Hz two-pole star connected steam turbine
generator has synchronous reactance of 18 Ω per phase and an armature resistance of 2
Ω per phase. This generator is operating in parallel with a large power system (infinite
bus).
(a) (2 Marks) What is the magnitude of E A at rated conditions?
(b) (2 Marks) What is the torque angle of the generator at rated conditions?
(c) (3 Marks) If the field current is constant, what is the maximum power possible out
of this generator? How much reserve power or torque does this generator have at full
load?
(d) (3 Marks) At absolute maximum power possible, how much reactive power will this
generator be supplying or consuming? Sketch the corresponding phasor diagram.
(Assume I F is still unchanged).
Question2 (15 Marks)
A 20-MVA 12.2-kV 0.8-PF-lagging Y-connected synchronous generator has a
negligible armature resistance and a synchronous reactance of 8.18 Ω . The generator is
connected in parallel with a 60-Hz, 12.2-kV infinite bus that is capable of supplying or
consuming any amount of real or reactive power with no change in frequency or
terminal voltage.
(a) (3 Marks) What is the internal generated voltage Ea of this generator under rated
conditions?
(b) (3 Marks) Suppose that the generator is initially operating at rated conditions. If the
flux is decreased by 15 percent, what will the new armature current I a be?
(c) (4 Marks) Suppose that the generator is initially operating at rated conditions. If the
flux is increased by 15 percent, what will the new armature current I a be?
(d) (5 Marks) Draw the phasor diagram of the generator in cases (a), (b) and (c).
King Saud University
College of Engineering
Electrical Engineering Department
EE331-Electromechanical Energy Conversion II
Test #1
2nd Semester
1431/32H
Time: 90 min.
Question1 (10 Marks)
A 13.8 kV 10 MVA 0.8-PF-lagging 60Hz two-pole star connected steam turbine generator has
synchronous reactance of 18 Ω per phase and an armature resistance of 2 Ω per phase. This
generator is operating in parallel with a large power system (infinite bus).
(i) (2 Marks) What is the magnitude of E A at rated conditions?
(ii) (2 Marks) What is the torque angle of the generator at rated conditions?
(iii) (3 Marks) If the field current is constant, what is the maximum power possible out of this
generator? How much reserve power or torque does this generator have at full load?
(iv) (3 Marks) At absolute maximum power possible, how much reactive power will this
generator be supplying or consuming? Sketch the corresponding phasor diagram.
(Assume I F is still unchanged).
Solution:
(a) The phase voltage of this generator at rated conditions is
Vφ =
13800 V
= 7967V
3
The armature current per phase at rated conditions is:
S
10 *106
IA =
=
= 418 A
3 * VT
3 * 13800
Therefore, the internal generated voltage at rated conditions is
E A = Vφ + RA * I A + jX s I A
(
)
E A = 7967∠0o + (1.5 + j12) * 418∠ − 36.87o = 12040∠17.6o V
The magnitude of E A is 12040V
(b) The torque angle of the generator at rated conditions is δ = 17.6°.
(c) Ignoring R A , the maximum output power of the generator is given by:
Pmax =
3Vφ E A
XS
=
3 * 7967 *12040
= 24MW
12
The power at maximum load is 8 MW, so the maximum output power is three times the full load
output
power.
(d) The phasor diagram at these conditions is shown below:
Under these conditions, the armature current is
IA =
E A − Vφ
R A = jX S
=
12040∠90o − 7967∠0o
= 1194∠40.6o A
1.5 + j12
The reactive power produced by the generator at this point is
Q = 3Vφ I A sinθ = 3 * 7967 *1194* sin(0 − 40.6) = −18.6MVAR
The generator is actually consuming reactive power at this time.
Question1 (15 Marks)
A 20-MVA 12.2-kV 0.8-PF-lagging Y-connected synchronous generator has a negligible
armature resistance and a synchronous reactance of 8.18 Ω . The generator is connected in
parallel with a 60-Hz, 12.2-kV infinite bus that is capable of supplying or consuming any
amount of real or reactive power with no change in frequency or terminal voltage.
(i) (3 Marks) What is the internal generated voltage Ea of this generator under rated
conditions?
(ii) (3 Marks) Suppose that the generator is initially operating at rated conditions. If the flux is
decreased by 15 percent, what will the new armature current I a be?
(iii) (4 Marks) Suppose that the generator is initially operating at rated conditions. If the flux is
increased by 15 percent, what will the new armature current I a be?
(iv) (5 Marks) Draw the phasor diagram of the generator in cases (i), (ii) and (iii).
(i) I a =
S
3V L
=
20 * 10 3
3 * 12.2
= 946.48 ∠ − cos −1 (0.8) = 946.48 ∠ − 36.87 A
E∠δ = Vφ ∠0 + I a ∠θ * jX s
Vφ =
12200
∠0 = 7043.67∠0V
3
E∠δ = 7043.67∠0 + 946.48 ∠ − 36.87 * j8.18 = 13238.76 ∠27.89V
(ii) φ 2 = 0.85φ1 and due to Ea αφ
∴ Ea 2 = 0.85 Ea1 = 0.85 *13238.76 = 11252.946V
P=
3Vφ E a
Xs
sin δ , But
3Vφ
Xs
= consta nt , Then PαEa sin δ
Then, Ea1 sin δ1 = Ea 2 sin δ 2
Then, 13238 .76 * sin (27.89 ) = 11252 .946 sin δ 2
∴I a 2 =
Ea 2 ∠δ 2 − Vφ ∠0
jX S
11252.946∠33.39 −
=
j8.18
∴δ 2 = 33.39o
12200 o
∠0
3
= 809.74∠ − 20.79 A
(iii) φ3 = 1.15φ1 and due to Ea αφ
∴ Ea 3 = 1.15 Ea1 = 1.15 *13238.76 = 15224.574V
P=
3Vφ Ea
Xs
sin δ , But
3Vφ
Xs
= consta nt , Then PαEa sin δ
Then, Ea1 sin δ1 = Ea 3 sin δ 3
Then, 13238.76 * sin(27.89) = 15224.574sin δ 3
∴δ 3 = 24 o
12200 o
15224
.574∠24−
∠0
Ea2∠δ2 −Vφ∠0
3
∴Ia2 =
=
=1130
.125∠−47.95A
jXS
j8.18
(iv)
Line of constant power
δ3
δ1
θ2
Ea2
Ea1
Ea3
δ2
θ1
Line of constant power
θ3
Ia2
Ia1
Ia3