Steady state and transient behaviour with governor

A turbine-generator with a speed governor is a closed loop: the governor senses speed and adjusts the gate so that mechanical input matches electrical load.

Model and block diagram

Rotor and load: $\Delta\Omega(s) = \dfrac{\Delta P_m - \Delta P_L}{2Hs + D}$

Governor with droop: $\Delta P_v(s) = \dfrac{1}{1+sT_g}\left[\Delta P_{ref} - \dfrac{\Delta\Omega}{R}\right]$

Turbine: $\Delta P_m = G_t(s)\,\Delta P_v$ with $G_t(s) = \dfrac{1 - T_w s}{1+0.5T_w s}$ for hydro.

 dPref +      +-------+ dPv +-------+ dPm
 ---->(+)---->|   1   |---->|turbine|---+
       ^ -    |1+sTg  |     | Gt(s) |   |
       |      +-------+     +-------+   v
    +-----+                   -dPL --->(+)
    | 1/R |                             |
    +-----+      +-----------+          |
       ^     dw  |     1     |          |
       +---------|  2Hs + D  |<---------+
                 +-----------+

Steady state behaviour

With $\Delta P_{ref} = 0$ and a step load $\Delta P_L$, the closed loop gives

$$\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D + \dfrac{G_t(s)}{R(1+sT_g)}}$$

Using the final value theorem ($G_t(0) = 1$):

$$\Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}} = \frac{-\Delta P_L}{\beta}$$

where $\beta = D + 1/R$ is the area frequency response characteristic. Since $1/R \gg D$ (e.g. $R = 0.05$ gives $1/R = 20$), the governor reduces the frequency error about 20 times compared with no governor.

• Speed-power relation: $\Delta P_m = \Delta P_{ref} - \Delta\omega/R$. This is the droop line: as load rises, speed falls in proportion.
• Units in parallel share a load change in proportion to $1/R$ (i.e. to rating, if all have the same per-unit droop).
• Changing $\Delta P_{ref}$ shifts the droop line up or down; this restores frequency to nominal (secondary control).

Transient behaviour

When a load $\Delta P_L$ is suddenly added:

1. Instant $t = 0^+$: the electrical output $P_e$ rises at once by $\Delta P_L$, because the generator must supply the connected load immediately. The mechanical input $P_m$ cannot change instantly (gate and water need time).
2. Deceleration: the deficit $P_e - P_m$ is supplied from the kinetic energy of the rotor, so speed starts to fall with slope $-\Delta P_L/2H$.
3. Load damping: as speed falls, frequency-sensitive load drops by $D\Delta\omega$, so $P_e$ comes down slightly from its initial jump.
4. Governor action: the speed fall is sensed; through the droop $1/R$ the governor opens the gate. For a hydro turbine, $P_m$ first dips (water inertia, non-minimum phase) and then rises.
5. Recovery: when $P_m$ exceeds $P_e$, the rotor accelerates and the fall in speed is arrested. After some oscillation, $P_m = P_e$ and speed settles at a new, lower value.

Steady state, from the block diagram ($s \to 0$):

$$\Delta\omega_{ss} = \frac{-\Delta P_L}{D + \frac{1}{R}}, \qquad \Delta P_{m,ss} = \Delta P_{e,ss} = \Delta P_L + D\,\Delta\omega_{ss}$$

 power
     |   Pe
 P0+ |   ._____
 dPL |   |     '--.________ Pm = Pe (final)
     |   |         _.-'
     |   |      _.'  Pm
  P0 |---'.   .'
     |     '.'  (hydro: Pm dips first)
     +---------------------------> t
 speed
     |----.
     |     \        .--------
     |      '._  _.'   dw_ss = -dPL/(D+1/R)
     |         ''
     +---------------------------> t

• Electrical power $P_e$: jumps by $\Delta P_L$ at once, then reduces slightly as frequency falls (load damping) and settles at $\Delta P_L + D\Delta\omega_{ss}$.
• Mechanical power $P_m$: starts at the old value (dips first in a hydro unit), rises with the governor and turbine time constants, overshoots slightly, and settles equal to $P_e$.
• Speed: falls, reaches a minimum (frequency nadir), recovers partly and settles at a lower value $\Delta\omega_{ss}$ fixed by the droop. With an isochronous governor ($R \to 0$) speed would return to nominal. Supplementary control (load reference change) is then used to bring the frequency back to 50 Hz.