Isochronous governor and speed droop model

An isochronous governor keeps the turbine speed constant at its set value for all loads ("isochronous" = same speed). It uses integral control: the gate keeps moving until the speed error is zero.

Operation

 w_ref +   error  +------+ dY +-------+ dPm
 ---->(+)-------->| K/s  |--->|turbine|---+
       ^ -        +------+    +-------+   |
       |                                  v
       |                         -dPL -->(+)
       |      +-----------+               |
       +------| 1/(2Hs+D) |<--------------+
        speed +-----------+

1. A load increase makes $P_e > P_m$, so speed falls.
2. The speed error is integrated: $\Delta Y = -\dfrac{K}{s}\Delta\Omega$. The gate opens and $P_m$ rises.
3. The rotor accelerates back. The integrator stops only when the error is zero, so speed returns exactly to $\omega_{ref}$, with $\Delta P_m = \Delta P_L$.

Why not for several units in parallel

• Parallel units run at one common frequency. Each isochronous governor tries to hold the frequency at its own set point.
• Set points are never exactly equal, so the units fight: the one with the higher set point keeps opening and the other keeps closing, until one is at full load and the other at zero or motoring.
• Integral control has no fixed relation between speed and load, so the load sharing is not defined and is unstable. Only one unit in a system can be isochronous; others need droop.

Mathematical model of a governor with speed droop

Droop is obtained by feeding the gate position back through a gain $R$ around the integrator (steady-state feedback):

                 +------+  dY
 -dw ---->(+)--->| K/s  |---+---> to turbine
           ^ -   +------+   |
           |     +-----+    |
           +-----|  R  |<---+
                 +-----+

With no change in load reference:

$$\Delta Y(s) = \frac{K}{s}\left[-\Delta\Omega(s) - R\,\Delta Y(s)\right]$$ $$\begin{aligned} s\,\Delta Y &= -K\Delta\Omega - KR\,\Delta Y \\ \Delta Y(s)(s + KR) &= -K\,\Delta\Omega(s) \\ \frac{\Delta Y(s)}{\Delta\Omega(s)} &= -\frac{K}{s + KR} = -\frac{1}{R}\cdot\frac{1}{1 + sT_G} \end{aligned}$$

where $T_G = \dfrac{1}{KR}$ is the governor time constant. Including the load reference,

$$\Delta Y(s) = \frac{1}{1 + sT_G}\left[\Delta P_{ref}(s) - \frac{1}{R}\Delta\Omega(s)\right]$$

Steady state ($s \to 0$): $\Delta Y = \Delta P_{ref} - \Delta\omega/R$, i.e. $R = -\Delta\omega/\Delta Y$ (pu). This is the speed droop characteristic: speed falls linearly as output rises. A 5 % droop means a 5 % speed drop gives a 100 % change in gate (output). It gives a unique and stable load sharing between parallel units in proportion to $1/R$.