Mechanical energy/efficiency problem

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Water flows over a dam at a rate of 580 kg/s, falling 88 meters, resulting in a speed of 41.5 m/s just before hitting the turbine blades. The mechanical efficiency of the system is 55%, meaning only 55% of the input energy is converted to output work. The work in is calculated using kinetic energy, yielding approximately 500.704 kJ for the falling water. The discussion revolves around understanding how to calculate work in and out, with clarification on expressing the rate of energy transfer in Joules per second (J/s). Overall, participants successfully clarify the calculations needed to solve the problem.
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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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