wherehfis the system losses. Ignoring minor losses, Manning’s formula
gives
hfQ^2 n^2 L/A^2 R4/30.0155Q^2.
Therefore
2 6000 0.879.81Q(1150.0115Q^2 )
givingQ9.2 m^3 s^1. Therefore the discharge per pump9.2/24.6 m^3 s^1.
The input to the motor 2 6/0.8514.12 MW. A 1 h operation of the
pump (i.e. an input of 14.12 MW h units) stores 9.2 60 60 3.31 104 m^3
of water.
TURBINE OPERATION
The input to the turbine
gQ(115 hf)/1000 kW.
Therefore
25 0000.919.81Q(115 0.0115Q^2 )
giving Q26.1 m^3 s^1. The generator output 25 0.8922.25 MW.
Therefore the maximum duration of turbine operation from 3.31 104 m^3
of stored water is 3.31 104 /26.1 60 60 0.352 h. Therefore the total
generated units (i.e. the output)22.250.3527.84 MW h, and the
overall efficiency of the system is 7.84/14.1255.5%.
Worked Example 12.6
The following data refer to a proposed hydroelectric power plant: tur-
bines, total power to be produced30 MW, normal operating speed 150
rev min^1 , net head available16 m; draft tube, maximum kinetic energy
at exit of draft tube1.5% of H, efficiency of draft tube85%, vapour
pressure3 m of water, atmospheric pressure10.3 m of water.
- What size, type, and number of units would you select for the pro-
posed plant?
- Starting from first principles, determine the turbine setting relative to
the tailrace water level.
Solution
For a low-head, high-discharge plant, Kaplan-type units are suitable.
542 HYDROELECTRIC POWER DEVELOPMENT
