Transient response of turbine-generator with governor

With a speed governor, the turbine-generator responds to a sudden load increase by first giving up stored kinetic energy and then, through governor action, increasing the water flow until mechanical input again equals electrical output.

System model

Inertia (rotating mass): for small changes, in per unit, the swing equation is

$$2H\frac{d\,\Delta\omega}{dt} = \Delta P_m - \Delta P_e$$

where $H$ = inertia constant (s), $\Delta\omega$ = speed deviation (pu). Taking the Laplace transform,

$$\Delta\Omega(s) = \frac{1}{2Hs}\big[\Delta P_m(s) - \Delta P_e(s)\big]$$

Load damping: a composite load has frequency-insensitive parts (lighting, heating) and frequency-sensitive parts (motors). So

$$\Delta P_e = \Delta P_L + D\,\Delta\omega$$

where $\Delta P_L$ = non-frequency-sensitive load change and $D$ = load damping constant (% change in load per % change in frequency, typically 1–2).

Substituting,

$$\begin{aligned} 2Hs\,\Delta\Omega(s) &= \Delta P_m(s) - \Delta P_L(s) - D\,\Delta\Omega(s) \\ \Delta\Omega(s) &= \frac{\Delta P_m(s) - \Delta P_L(s)}{2Hs + D} \end{aligned}$$

The governor with speed droop $R$ gives

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

and the hydraulic turbine gives

$$\frac{\Delta P_m(s)}{\Delta P_v(s)} = \frac{1 - T_w s}{1 + 0.5T_w s}$$

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

Transient response to a sudden load increase

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.

Conclusion: inertia limits the initial rate of fall, the governor limits the depth of the fall and brings $P_m$ up to match $P_e$, and the droop leaves a small steady frequency error which is removed by changing the load reference.