SME
Power Systems
Grid stability
Use this for “Find the critical clearing time for a generator with H = 5 s delivering 1 pu with Pₘₐₓ = 2 pu.”
Chart
δ₀ = 30°, δcr = 79.56°, δₘₐₓ = 150°; A₁ = 0.865, A₂ = 0.865: equal, so δcr is the critical angle
This is an example. Type your own values to start a new problem.
Pₘ
Mechanical power
example · pu
Pₘₐₓ
Peak power sent
example · pu
H
Inertia constant
example
f
System frequency
example · Hz
δ₀
Starting angle
worked out (not typed) · °
δcr
Critical clearing angle
worked out (not typed) · °
δₘₐₓ
Largest swing angle
worked out (not typed) · °
tcr
Critical clearing time
calculated
Formulas
δ₀
=
sin⁻¹(
Pₘ
Pₘₐₓ
)
30
=
sin⁻¹(
1
2
)
δcr
=
cos⁻¹((π
−
2
×
δ₀
×
π
180
)
×
sin(
δ₀
°)
−
cos(
δ₀
°))
79.562
=
cos⁻¹((π
−
2
×
30
×
π
180
)
×
sin(30°)
−
cos(30°))
δₘₐₓ
=
180°
−
δ₀
150 = 180° − 30
tcr
=
√(4
×
H
×
(δcr
−
δ₀
)
×
π
180
÷
(2π
×
f
×
Pₘ))
0.2142
=
√(4
×
5
×
(79.562
−
30)
×
π
180
÷
(2π
×
60
×
1))
Assumptions
•
The fault drops the power sent to 0 until it is cleared; the same line returns after, with the same Pₘₐₓ.
•
Damping is ignored: the rotor speeds up over A₁ and must give it all back over A₂.
•
Powers are per unit on the machine’s base, so H is in seconds on that base.
•
During the fault the angle grows as δ₀ + (πfPₘ ÷ (2H))t², δ in radians.
Step-by-step
WE KNOW
Pₘ = 1 pu, Pₘₐₓ = 2 pu, H = 5 s, f = 60 Hz
FIND
starting angle (δ₀), critical clearing angle (δcr), largest swing angle (δₘₐₓ), critical clearing time (tcr)
Step 1 · Find starting angle (δ₀)
Use δ₀ = sin⁻¹(Pₘ ÷ Pₘₐₓ)
Before
the
fault
the
power
sent
matches
the
turbine:
Pₘₐₓ
sin
δ₀
=
Pₘ.
δ₀
=
sin⁻¹(
1
2
)
δ₀
=
30°
Step 2 · Find critical clearing angle (δcr)
Use δcr = cos⁻¹((π − 2 × δ₀ × π ÷ 180) × sin(δ₀°) − cos(δ₀°))
Setting
A₁
=
A₂
gives
cos
δcr
=
(π
−
2
δ₀
)
sin
δ₀
−
cos
δ₀
,
with
δ₀
in
radians
inside
the
bracket.
δcr
=
cos⁻¹((π
−
2
×
30
×
π
180
)
×
sin(30°)
−
cos(30°))
δ₀
=
30
×
π
180
=
0.5236
rad
cos δcr = (π − 2 × 0.5236) × sin 30° − cos 30° = 0.1812
δcr = cos⁻¹(0.1812)
δcr = 79.562°
Step 3 · Find largest swing angle (δₘₐₓ)
Use δₘₐₓ = 180° − δ₀
The
swing
can
go
only
as
far
as
where
the
sine
falls
back
to
Pₘ:
180°
−
δ₀
.
δₘₐₓ = 180° − 30
δₘₐₓ = 150°
Step 4 · Find critical clearing time (tcr)
Use tcr = √(4 × H × (δcr − δ₀) × π ÷ 180 ÷ (2π × f × Pₘ))
With
no
power
out,
the
angle
grows
as
δ₀
+
(
πfPₘ
2
H
)
t
2
:
solve
for
the
time
to
reach
δcr.
tcr
=
√(4
×
5
×
(79.562
−
30)
×
π
180
÷
(2π
×
60
×
1))
tcr
=
√(
17.3004
376.991
)
tcr
=
√
0.04589
tcr = 0.2142 s
Check
Put the numbers back in. Both sides should match.
30
=
sin⁻¹(
1
2
)
✓
79.562
=
cos⁻¹((π
−
2
×
30
×
π
180
)
×
sin(30°)
−
cos(30°))
✓
150 = 180° − 30 ✓
0.2142
=
√(4
×
5
×
(79.562
−
30)
×
π
180
÷
(2π
×
60
×
1))
✓
Topic check
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