Power output of a hydroelectric plant

Click For Summary

Homework Help Overview

The discussion revolves around calculating the power output of a hydroelectric plant based on the potential energy lost by falling water. The original poster presents a scenario involving water flow rate and the conversion of potential energy to electrical energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between potential energy and kinetic energy, questioning the need for time in the power calculation. There is also a discussion about dimensional analysis and the conversion of energy to power.

Discussion Status

Some participants have offered guidance on dimensional checks and alternative methods for calculating power. There appears to be an ongoing exploration of the concepts involved, with no explicit consensus reached on the final answer.

Contextual Notes

The original poster is working within the constraints of a homework assignment, which may impose specific rules on the approach to the problem. There is a mention of a potential misunderstanding regarding units and the calculation of power output.

Medgirl314
Messages
568
Reaction score
2

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!
 
Physics news on Phys.org
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.
 
  • Like
Likes   Reactions: 1 person
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
 
  • Like
Likes   Reactions: 1 person
Great! Thank you! That was too easy for the last problem in a chapter. 0.o
 

Similar threads

Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
4
Views
3K
Replies
20
Views
4K
  • · Replies 13 ·
Replies
13
Views
5K