Speed control by inertia and load damping

Without a governor, the gate (and so mechanical power $P_m$) stays fixed. When load changes, speed is held back only by two natural effects: the inertia of the rotating masses, which slows the rate of speed change, and the load damping, which reduces the load as frequency falls so that a new balance is reached.

Derivation of the transfer function

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,

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

Without a governor $\Delta P_m = 0$, so

$$\frac{\Delta\Omega(s)}{\Delta P_L(s)} = \frac{-1}{2Hs + D} = \frac{-1/D}{1 + sT}, \qquad T = \frac{2H}{D}$$

 dPm=0  +    +--------------+
 ------>(+)->|      1       |----+--> d(omega)
         ^ - |   2Hs + D    |    |
         |   +--------------+    |
       dPL

Response to a small step change in load

For a step $\Delta P_L(s) = \Delta P_L / s$:

$$\Delta\omega(t) = -\frac{\Delta P_L}{D}\left(1 - e^{-t/T}\right), \qquad T = \frac{2H}{D}$$

 d(omega)
   0 |-----.
     |      \
     |       '-.
     |          '--.______  -dPL/D
     +------------------------> t
           T = 2H/D

Role of inertia and load damping

• Inertia (H): At the first instant $P_m$ is unchanged, so the extra load is supplied from the kinetic energy of the rotor. Speed starts to fall at a rate $d\Delta\omega/dt = -\Delta P_L/2H$. Larger $H$ gives a slower fall and a larger time constant $T = 2H/D$. Inertia does not change the final value; it only slows the change.
• Load damping (D): As frequency falls, motor loads take less power ($D\,\Delta\omega$). The fall stops when this reduction equals $\Delta P_L$, giving steady-state deviation $\Delta\omega_{ss} = -\Delta P_L/D$. Larger $D$ gives a smaller final speed error and a faster settling.
• Since $D$ is small (1–2), the steady speed error is large: e.g. $\Delta P_L = 0.01$ pu and $D = 1$ gives a 1 % drop in frequency. So a governor is needed to restore speed; inertia and damping only make the system self-regulating for small changes.