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Speed droop in sharing common load

Short answer

Speed droop provides each unit a definite speed-load line, ensuring stable load division. The load change share ratio equals (P_r/R) of the units; equal per-unit droop divides load proportionally to ratings, preventing instability.

Explain the importance of speed droop in sharing common load by two alternator operating in parallel with suitable block diagram and derivation.

Speed droop is the drop in speed (frequency) of a governed unit as its load goes from no load to full load, expressed in per unit of rated speed. It is defined by

R=Δf/f0ΔP/Pr(pu)R = \frac{\Delta f / f_0}{\Delta P / P_r} \quad \text{(pu)}

A 4 % droop means the frequency falls by 4 % (2 Hz at 50 Hz) from no load to full load. Droop gives every governor a definite speed–load line, so parallel units share a load change in a fixed, stable ratio. Without droop (isochronous governors), parallel units fight each other and the load division is undefined.

Block diagram (two units on a common bus)

   +-->[1/R1]--+
   |           v -
   |  dPref1->(+)->[Gov1]->[Turb1]--dPm1--+
   |                                      v
   |                        -dPL ------->(S)--+
   |                                      ^   |
   |  dPref2->(+)->[Gov2]->[Turb2]--dPm2--+   |
   |           ^ -                            v
   +-->[1/R2]--+                    [1/(2Hs+D)]
   |                                          |
   +--------------- d(omega) -----------------+

S = summing point: ΔPm1+ΔPm2−ΔPL\Delta P_{m1} + \Delta P_{m2} - \Delta P_L.

Both units see the same frequency deviation Δf\Delta f because they are synchronised.

Derivation

In steady state, the governor of unit ii obeys its droop line:

ΔPi=ΔPref,i−1RiΔff0Pri\Delta P_i = \Delta P_{ref,i} - \frac{1}{R_i}\frac{\Delta f}{f_0}P_{ri}

With fixed references (ΔPref,i=0\Delta P_{ref,i} = 0):

ΔP1=−Pr1R1Δff0,ΔP2=−Pr2R2Δff0\Delta P_1 = -\frac{P_{r1}}{R_1}\frac{\Delta f}{f_0}, \qquad \Delta P_2 = -\frac{P_{r2}}{R_2}\frac{\Delta f}{f_0}

The total change must supply the load change:

ΔPL=ΔP1+ΔP2=−Δff0(Pr1R1+Pr2R2)Δff0=−ΔPLPr1R1+Pr2R2\begin{aligned} \Delta P_L &= \Delta P_1 + \Delta P_2 = -\frac{\Delta f}{f_0}\left(\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}\right) \\ \frac{\Delta f}{f_0} &= \frac{-\Delta P_L}{\frac{P_{r1}}{R_1} + \frac{P_{r2}}{R_2}} \end{aligned}

and the ratio of sharing is

ΔP1ΔP2=Pr1/R1Pr2/R2\frac{\Delta P_1}{\Delta P_2} = \frac{P_{r1}/R_1}{P_{r2}/R_2}

If R1=R2R_1 = R_2 (same per-unit droop), ΔP1/ΔP2=Pr1/Pr2\Delta P_1/\Delta P_2 = P_{r1}/P_{r2}, i.e. load is shared in proportion to ratings.

 f                     f
 f0|\ unit 1           |\  unit 2
   | \                 |  \
 f'|--*-----common f'--|---*
   |   \               |    \
   +----+---> P1       +-----+---> P2
       P1'                  P2'

Importance

  • Gives a unique, stable division of load among parallel units.
  • A smaller droop makes a unit "stiffer", so it takes a bigger share of any change.
  • Equal per-unit droop on all units makes each carry load proportional to its rating, avoiding overload of small units.
  • The frequency drop needed is small (typically 4–5 % droop), and the load reference setting is then used to bring frequency back to nominal.

✓ Answer checked by Claude Opus 5.5 (high effort)

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