SME
Power Systems · Problem type
Droop: the frequency drop
Use this for “A generator with 5% droop picks up 0.1 pu more load on a 60 Hz system. How far does the frequency fall?”
Chart
f(0.1) = 59.7
This is an example. Type your own values to start a new problem.
Δf (Hz)
=
−
R
%
×
ΔP (pu)
×
fₙₒₘ (Hz)
f₁ (Hz)
=
fₙₒₘ (Hz)
+
(
Δf (Hz)
)
Formulas
Δf
=
−(
R
100
)
×
ΔP
×
fₙₒₘ
−0.3
=
−(
5
100
)
×
0.1
×
60
f₁
=
fₙₒₘ
+
Δf
59.7 = 60 + (−0.3)
Assumptions
•
The droop R is the percent the frequency falls from no load to full load.
•
The load change ΔP is per unit of the generator’s own rating; more load is positive.
•
Only the governor acts here; a slower control later brings the frequency back to rated.
Step-by-step
WE KNOW
R = 5%, ΔP = +0.1 pu, fₙₒₘ = 60 Hz
FIND
frequency change (Δf), new frequency (f₁)
Step 1 · Find frequency change (Δf)
Use Δf = −(R ÷ 100) × ΔP × fₙₒₘ
The droop is the fraction of rated frequency given up per unit of load picked up.
Δf
=
−(
5
100
)
×
0.1
×
60
Δf = −0.05 × 0.1 × 60
Δf = −0.005 × 60
Δf = −0.3 Hz
Step 2 · Find new frequency (f₁)
Use f₁ = fₙₒₘ + Δf
Add the change to the rated frequency.
f₁
=
60
+
(−0.3)
f₁
=
59.7
Hz
Check
Put the numbers back in. Both sides should match.
−0.3
=
−(
5
100
)
×
0.1
×
60
✓
59.7 = 60 + (−0.3) ✓
Topic check
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