Water turbine response to step gate change

The water turbine transfer function relates a small change in gate opening to the change in mechanical power output. Due to the inertia of water in the penstock, power first moves in the wrong direction when the gate is moved.

Derivation of the transfer function

Assumptions: lossless turbine, incompressible water, rigid penstock, and small changes about an operating point.

Velocity of water in the penstock: $U = K_u\,G\sqrt{H}$, where $G$ = gate opening and $H$ = head at the gate.

Turbine power: $P_m = K_p\,H\,U$.

Linearising and normalising by initial values ($U_0, G_0, H_0, P_{m0}$), with bar for per-unit change:

$$\begin{aligned} \Delta\bar U &= \tfrac{1}{2}\Delta\bar H + \Delta\bar G \\ \Delta\bar P_m &= \Delta\bar H + \Delta\bar U \end{aligned}$$

Eliminating $\Delta\bar H = 2(\Delta\bar U - \Delta\bar G)$:

$$\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$$

Acceleration of the water column (length $L$, area $A$): the net force from the head change accelerates the water mass,

$$\rho L A\,\frac{d\Delta U}{dt} = -A\rho g\,\Delta H \;\Rightarrow\; T_w\,\frac{d\Delta\bar U}{dt} = -\Delta\bar H$$

where the water starting time is

$$T_w = \frac{L\,U_0}{g\,H_0}$$

In Laplace form: $T_w s\,\Delta\bar U = -2(\Delta\bar U - \Delta\bar G)$, so

$$\frac{\Delta\bar U}{\Delta\bar G} = \frac{1}{1 + \tfrac{1}{2}T_w s}$$

Substituting in $\Delta\bar P_m = 3\Delta\bar U - 2\Delta\bar G$:

$$\frac{\Delta\bar P_m(s)}{\Delta\bar G(s)} = \frac{3}{1 + 0.5T_w s} - 2 = \frac{1 - T_w s}{1 + 0.5T_w s}$$

Response to a unit step change in gate opening

For a unit step $\Delta\bar G(s) = 1/s$:

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

Initial value ($s \to \infty$): $\Delta\bar P_m(0^+) = \dfrac{-T_w}{0.5T_w} = -2$

Final value ($s \to 0$): $\Delta\bar P_m(\infty) = 1$

Time response:

$$\Delta\bar P_m(t) = 1 - 3e^{-2t/T_w}$$

 dPm
  1 |                 ____________
    |             _.-'
  0 |----.     _-'
    |    |   .'
    |    |  /   time constant Tw/2
 -2 |    |_/
    +----+-------------------------> t
       gate step at t = 0

Explanation: when the gate opens suddenly, the flow cannot increase at once because the water column has inertia. The larger opening with the same flow lowers the pressure (head) at the turbine, so power first drops to $-2$ times the gate change. As the water accelerates, flow and power rise, and power settles at $+1$ times the gate change with time constant $T_w/2$. Similarly, closing the gate first gives a power rise (with pressure rise, i.e. water hammer).

This behaviour (zero in the right-half plane) is why hydro governors are given a large transient droop and a slower response.