Hydroelectric Power Plant Efficiency Calculation

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SUMMARY

The discussion focuses on the calculation of efficiency in hydroelectric power plants using the equation .80(1/2)mv² = Ee, where v represents the velocity of falling water and Ee denotes the electrical power generated. The correct approach involves using the mass flow rate (dm/dt) instead of mass (m) to accurately reflect the relationship between energy and power. The equation can be adjusted to .80(1/2)(dm/dt)v² = dE/dt when considering energy, confirming that the original equation is valid when Ee is defined as power.

PREREQUISITES
  • Understanding of basic physics concepts related to energy and power
  • Familiarity with hydroelectric power generation principles
  • Knowledge of calculus, specifically derivatives for mass flow rate
  • Ability to manipulate equations involving kinetic energy
NEXT STEPS
  • Study the principles of energy conservation in hydroelectric systems
  • Learn about the derivation and application of the mass flow rate in fluid dynamics
  • Explore the relationship between energy and power in electrical systems
  • Investigate efficiency calculations in various types of power plants
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Students in physics or engineering, professionals in renewable energy sectors, and anyone involved in the design or analysis of hydroelectric power systems.

BrainMan
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Homework Statement



upload_2016-10-18_20-24-9.png


Homework Equations

The Attempt at a Solution



What I attempted to do is

.80(1/2)mv2 = Ee

Where v is the velocity of the falling water, m is the mass of the falling water, and Ee is the electrical power generated by the electric plant.

So then I solved for m since you are given the Ee and the velocity and I got 3.86 x 10e5 kg/s.
 

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Your number is correct. To be strictly correct you should use dm/dt in your equation instead of m.
 
kuruman said:
Your number is correct. To be strictly correct you should use dm/dt in your equation instead of m.
But then it would be dE/dt on the other side, assuming E stands for energy, not power.
The OP equation can be justified by prefixing it with "in the time in which a mass m of water passes through".
 
haruspex said:
But then it would be dE/dt on the other side, assuming E stands for energy, not power.
If E stands for energy, .80(1/2)(dm/dt)v2 = dE/dt is correct.
If Ee stands for power (as indicated by OP in attempt at solution), .80(1/2)(dm/dt)v2 = Ee is correct.
 

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