Mechanical energy/efficiency problem

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Homework Help Overview

The problem involves calculating the speed of water falling over a dam and the mechanical energy transferred to turbine blades, considering efficiency. The context is within the subject area of mechanical energy and efficiency in fluid dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work in and work out, questioning how to express the rate of energy transfer. There is an exploration of the relationship between kinetic energy and the work done by the falling water.

Discussion Status

The discussion is active with participants seeking clarification on the calculations and concepts involved. Some have expressed understanding of the relationships between the variables, while others are still grappling with the definitions and calculations of work in and work out.

Contextual Notes

Participants note the importance of efficiency in the calculations and question how to express the results in terms of energy per second (J/s). There is also mention of the mass flow rate of water and its impact on the energy calculations.

Flatshoe
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Homework Statement



(i)Water flows over a dam at the rate 580kg/s and falls vertically 88m before striking the turbine blades. Calculate the speed of the water just before striking the turbine blades (SOLVED) and (ii) the rate at which mechanical energy is transferred to the turbine blades assuming 55% efficiency.

Homework Equations



Ek = 1/2mv^2 Mechanical efficiency = Work out/Work in

The Attempt at a Solution


Answer to part (i) v=sqrt2gh

v = 41.5 m/s


Mechanical efficiency = work out/work in

Work out = Ek1 - Ek2

Work in = Ek1 - Ek2/55%



Ok and I'm pretty much stuck. :(

Thanks in advance for any help guys!
 
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This problem looks familiar did you post it on some other message board?

As you stated the mechanical energy is the ratio of the work out to the work in. What's the work in? The efficiency is 55% so, 55% of the work in is returned.
 
Squeezebox said:
This problem looks familiar did you post it on some other message board?

As you stated the mechanical energy is the ratio of the work out to the work in. What's the work in? The efficiency is 55% so, 55% of the work in is returned.

I'm having trouble understanding how to calculate the work in... :(

Also it says calculate the "rate", so how would the answer be expressed? J/s?

Edit: I've never posted this anywhere before.
 
Flatshoe said:
I'm having trouble understanding how to calculate the work in... :(

Also it says calculate the "rate", so how would the answer be expressed? J/s?

Edit: I've never posted this anywhere before.

Work is Joules/s. 580 kg/s fall off the dam. So work in is the energy associated with the mass that falls each second. What's the energy associated with the mass?
 
Squeezebox said:
Work is Joules/s. 580 kg/s fall off the dam. So work in is the energy associated with the mass that falls each second. What's the energy associated with the mass?
Kinetic energy? 1/2mv^2 = 1/2(580)(41.552)^2 = 500704 J = 500.704 KJ

Is that the right calculation for the work in?
 
Flatshoe said:
Kinetic energy? 1/2mv^2 = 1/2(580)(41.552)^2 = 500704 J = 500.704 KJ

Is that the right calculation for the work in?

Yes, now how does that relate to work in and finally, work out.
 
Squeezebox said:
Yes, now how does that relate to work in and finally, work out.

Thanks, I understand it now!
 

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