Is this the Maximum Power Generated from a Small Hydro Plant?

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SUMMARY

The maximum power generated from the small hydro plant, based on the calculations provided, is 148,687.5 Watts. This figure is derived from the potential energy (Ug) and kinetic energy (K) of the water flowing over a 10m waterfall with a flow rate of 1.5 m/s. The calculations assume a 25% efficiency in energy conversion, resulting in a total energy output of 594,750 J/s before efficiency is applied. The final power output confirms the effectiveness of the hydro plant design under the specified conditions.

PREREQUISITES
  • Understanding of gravitational potential energy (Ug = mgh)
  • Knowledge of kinetic energy calculations (K = 0.5mv²)
  • Familiarity with fluid dynamics and flow rates
  • Basic principles of energy efficiency in mechanical systems
NEXT STEPS
  • Research the principles of energy conversion efficiency in hydroelectric systems
  • Learn about the impact of friction and turbulence on energy calculations in fluid dynamics
  • Explore advanced calculations for hydroelectric power generation, including variable flow rates
  • Investigate the design and optimization of turbine systems for small hydro plants
USEFUL FOR

Engineers, renewable energy developers, and students studying mechanical engineering or environmental science will benefit from this discussion, particularly those focused on hydroelectric power generation and efficiency optimization.

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Homework Statement



A small hydro plant is being built. A nearby river is 4m wide and 1m deep. The water flow is 1.5m/s over the brink of a 10m high waterfall. The turbine company advertises 25% efficiency in converting the potential energy of the water into electric energy. What's the power you can generate?

Homework Equations



Ug = mgh
K= 0.5mv2


The Attempt at a Solution



Waterfall.jpg


Since V = 1.5 m/s, I found the volume of water passing over the brink of the waterfall, every second:
Volume/ sec = LWH = 1.5(4)(1) = 6m3/ sec

Convert to mass:
1m3 = 1000 kg
6m3 = 6000 kg

Since I did the beginning calculations for "every second", I wrote the rest of the values as "per second" too

At the top of the waterfall, there is grav.potential and kinetic energy:
Ug = mgh = 6000(9.8)(10) = 588 000 J / s
AND
K = 0.5mv2 = 0.5(6000)(1.5)2 = 6750 J/s
SO
Etop = 6750 + 588 000 = 594 750 J/s


I'm not sure if it is right to assume that this system is conservative because there's a lot of friction in flowing water, but I did anyway...
At the bottom of the waterfall, there is kinetic energy. Assuming a conservative system:
Ebottom = Etop
Ebottom = 594 750 J/s

% efficiency is 25%, so:
25% of 594 750 = 148687.5 J/s = 148687.5 W = POWER
 
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I just need someone to check the answers because I don't know if they're right.
 

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