Hydroelectric Power Plant Efficiency Calculation

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Homework Help Overview

The discussion revolves around the calculation of efficiency in a hydroelectric power plant, specifically focusing on the relationship between the mass of falling water and the electrical power generated. Participants are examining the equations related to energy and power in this context.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the mass of falling water to the electrical power generated using an equation. Some participants question the use of mass versus the rate of mass flow in the equation, suggesting a need for clarification on the terms used.

Discussion Status

Participants have confirmed the correctness of the original poster's numerical result while engaging in a discussion about the appropriate terms to use in the equations. There is an exploration of different interpretations regarding the definitions of energy and power in the context of the problem.

Contextual Notes

There is a focus on ensuring the correct application of terms in the equations, with some ambiguity regarding whether energy or power is being referenced. This may affect the interpretation of the original poster's approach.

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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