SME
Power Systems · Problem type
Swing frequency
Use this for “A generator with H = 5 s runs at δ₀ = 30° with Pₘₐₓ = 2 pu on a 60 Hz grid. Find its swing frequency.”
Chart
A steady oscillation at ωₙ = 8.081 rad/s: with ζ = 0 nothing takes energy out, so the swing never shrinks
This is an example. Type your own values to start a new problem.
Kₛ (pu)
=
Pₘₐₓ (pu)
×
cos(
δ₀
°)
(
ωₙ (rad/s)
)
2
=
2π
×
f (Hz)
×
Kₛ (pu)
÷
(2
×
H (s)
)
fₙ
Swings per second
calculated · Hz
T
Swing period
calculated
Formulas
Kₛ
=
Pₘₐₓ
×
cos(
δ₀
)
1.7321 = 2 × cos(30°)
ωₙ
=
√(2π
×
f
×
Kₛ
2 ×
H
)
8.0806
=
√(2π
×
60
×
1.7321
2 × 5
)
fₙ
=
ωₙ
2π
1.2861
=
8.0806
2π
T
=
1
fₙ
0.7776
=
1
1.2861
Assumptions
•
Small swings about δ₀: the curve Pₘₐₓ sin δ is replaced by its slope there, Kₛ = Pₘₐₓ cos δ₀ per radian.
•
Damping is ignored, so the rotor angle rocks at ωₙ without dying away.
•
Powers are per unit on the machine’s base, so H is in seconds on that base.
Step-by-step
WE KNOW
Pₘₐₓ = 2 pu, δ₀ = 30°, H = 5 s, f = 60 Hz
FIND
synchronizing power (Kₛ), swing frequency (ωₙ), swings per second (fₙ), swing period (T)
Step 1 · Find synchronizing power (Kₛ)
Use Kₛ = Pₘₐₓ × cos(δ₀)
The
slope
of
Pₘₐₓ
sin
δ
at
δ₀
:
how
much
more
power
a
small
extra
angle
sends.
Kₛ = 2 × cos(30°)
Kₛ = 2 × 0.866
Kₛ = 1.73205 pu
Step 2 · Find swing frequency (ωₙ)
Use ωₙ = √(2π × f × Kₛ ÷ (2 × H))
The
swing
equation
(
2
H
2π
f
)Δδ″
=
−KsΔδ
is
a
spring
and
mass:
ωₙ
2
=
stiffness
inertia
.
ωₙ
=
√(2π
×
60
×
1.73205
2 × 5
)
ωₙ
=
√(
652.968
10
)
ωₙ
=
√
65.2968
ωₙ = 8.0806 rad/s
Step 3 · Find swings per second (fₙ)
Use fₙ = ωₙ ÷ (2π)
One swing is 2π radians of ωₙ t.
fₙ
=
8.0806
2π
fₙ = 1.28607 Hz
Step 4 · Find swing period (T)
Use T = 1 ÷ fₙ
The period is the time for one swing.
T
=
1
1.28607
T
=
0.7776
s
Check
Put the numbers back in. Both sides should match.
1.73205 = 2 × cos(30°) ✓
8.0806
=
√(2π
×
60
×
1.73205
2 × 5
)
✓
1.28607
=
8.0806
2π
✓
0.7776
=
1
1.28607
✓
Topic check
See what you know: 5 questions on this topic
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