Power output of a hydroelectric plant

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SUMMARY

The power output of a hydroelectric plant can be calculated using the formula for potential energy (PE) and its conversion to kinetic energy (KE). In the discussed scenario, water flows at a rate of 400 kg/s from a height of 30 meters, with gravitational acceleration at 9.8 m/s². Given that 80% of the lost potential energy is converted to electricity, the power output is determined to be 9408 Watts. This calculation confirms that the time factor is effectively one second for the purpose of power measurement in Watts.

PREREQUISITES
  • Understanding of potential energy (PE) and kinetic energy (KE) concepts
  • Familiarity with basic physics equations involving mass, gravity, and height
  • Knowledge of power calculation in terms of Joules per second (Watts)
  • Ability to perform dimensional analysis for unit consistency
NEXT STEPS
  • Study the principles of energy conversion in hydroelectric systems
  • Learn about the efficiency of energy conversion in different types of turbines
  • Explore advanced calculations involving varying flow rates and heights in hydroelectric plants
  • Investigate the environmental impacts and sustainability of hydroelectric energy production
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Students in physics, engineers working on renewable energy projects, and professionals involved in the design and optimization of hydroelectric power systems.

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Homework Statement



In a hydroelectric plant, water falls from a high elevation to a lower elevation,losing PE in the process/ In doing so, it turns turbines which generate electricity. In a particular hydroelectric plant, water flows at a rate of 400 kg/s. If 80% of the PE that the water loses gets converted to electricity, what is the power output of the dam, in watts?

Homework Equations



PE=mgy
PE lost=KE gained

The Attempt at a Solution



Known information:

m=400 kg/s (Right?)
g=9.8 m/s^2
y=30 meters

KE= .8(400 kg/s*9.8 m/s^2*30 m)=9408 Joules

That's easy enough, right? But don't a need a time to convert to Watts? Is the time just one second, so that my answer is 9408 Watts?

Thanks!
 
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You want to do a dimensional check here. Perhaps you are ending up with one/s left over... on the left hand side

Another thing you can do is calculate the amount of Joules in 100 seconds. Then divide by 100 to get the number of Joules/second, which is the power.
 
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KE= .8(400 kg/s*9.8 m/s^2*30 m)=9408 Joules

I don't really understand the phrasing of your first sentence. Do you mean that I have three/s total, but only two cancel out, so my answer is 9408 J/S, which is 9408 Watts?

Thank you!
 
Yesss
 
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Great! Thank you! That was too easy for the last problem in a chapter. 0.o
 

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