Mechanical energy/efficiency problem

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SUMMARY

The discussion centers on calculating the mechanical energy transferred to turbine blades from water flowing over a dam at a rate of 580 kg/s and falling 88 m. The speed of the water just before striking the turbine blades is calculated to be 41.5 m/s using the equation v=sqrt(2gh). The mechanical energy transferred to the turbine is determined by the efficiency of 55%, with the work in calculated as 500.704 kJ, leading to a work out of approximately 275.387 kJ.

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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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