Steady state operation of hydropower plant

In steady state, a hydro unit runs at constant speed with constant water flow and gate opening, and the mechanical power from the turbine exactly equals the electrical power plus losses ($P_m = P_e$).

 Reservoir (head H)
 ~~~~~~~ |___tunnel___[surge tank]
                         \ penstock (Q)
                          \
               gate/  +---------+   +-----+
             governor>| turbine |===| gen |==> Pe
                      +---------+   +-----+
                          | tailrace

Power relation: the hydraulic power converted is

$$P = \eta\, \rho\, g\, Q\, H_{net}$$

where $Q$ = discharge (m³/s), $H_{net}$ = gross head minus head losses in tunnel and penstock, $\eta$ = overall efficiency of turbine and generator.

Steady state conditions:

• Water velocity in the penstock is constant, so there is no water hammer and the surge tank level is steady.
• Gate (or needle) opening is fixed at the value set by the governor for the present load.
• Speed is constant at synchronous value; frequency is steady. The accelerating power $P_m - P_e = 0$.
• Generator voltage is held by the AVR at its set value.

Change of operating point: if the load changes, the governor senses the speed change and moves the gate until a new steady state is reached where $P_m = P_e$ again (with a small frequency change set by the droop). The power output is proportional to $Q$, so for a fixed head the gate opening sets the power.