[dropcap3]1[/dropcap3]Hydraulic Power Calculation
To calculate hydraulic power you need to measure Head (meters) and Flow (liters per second).
The greater the head and flow the greater the power.
[titled_box title=”Where did this formula come from?”]
The energy released by a falling body of water = weight x distance
And weight is the downward force the water exerts which = mass x acceleration due to gravity.
And mass = density x volume
Therefore put those all together and you get:
Hydraulic Power = Joules per second = Watts = Density x Volume x acceleration due to gravity x distance.
And density of water = 1000 kg per cubic meter.
And volume of water = Q = cubic meters (m3) (use cubic meters, not liters per second because density of water is being measured in terms of cubic meters)
And acceleration due to gravity = 9.81 meters per second squared (m⁄s2)
If you assemble all this together in the formula again
Hydraulic Power = Watts = 1000 (kg⁄m3) x Q (m3) x 9.81 (m⁄s2) x H (m)
Watts = 9810 QH
kilowatts = 9.81QH, where Q is cubic meters per second and H is meters of head.[/titled_box]
Note: Head referred to above is the vertical distance between upper and lower water levels (and there are some rules for proper measurement of this distance). However, the turbine may not be installed at the current upper water level height. If it’s installed lower then the useful head is less. This can be referred to as the “Net Head” or “Useful Head”.
Therefore we have to recalculate the hydraulic power available to the turbine using useful head.
Net Hydraulic Power Available to Turbine= Useful Head (meters) x Flow (lps) x 9.81 = Watts