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A hydropower plant uses a river to generate electrical energy. It has a drop of $107m$ and the flow rate of water is $275$ $m^3$/$s$. The flow rate of the water is the same in and out of the system.

I was asked to determine whether the system is open or closed. Ans: open

And to simplify the general energy balance for the system

$Q + W = ΔH + ΔE_k + ΔE_p$

here is where I am a bit uncertain as to how to simplify it, I think that only $ΔH$ can be removed from the equation since the enthalpy won't change. Is that correct?

And so if $Q + W =ΔE_k + ΔE_p$ how can I use that to calculate the maximum amount of electrical energy that can be generated?

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  • $\begingroup$ Why won't the formula for (gravitational) potential energy do $E=m*g*h$? That'd yield 289MW. $\endgroup$ – Bart Jan 24 '18 at 23:20
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You can eliminate Q and Ek as well. This is because the heat transfer between the water and ambient is negligible in hydro-electricity(for steam power plants Q might be significant) . Also, at the the reservoir/dam where water is stored, the water usually has negligent speed, thus low value of K.E. The potential energy of water is the significant term, with it being equal to mgh or Qρgh in the rate form; where Q is discharge or volume flow rate of water. This potential energy(p.e) is responsible for the work done W by the turbine runner. The electrical work generated would be some factor times the p.e, because of losses such as penstock losses, losses incurred in shaft, runner etc.

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What's wrong with P = QgH * density ?

P = Power, Q is flow rate, g = gravity, density of the fluid water and H is height...

And most water power engineers take gravity as 10 for a first stab - the error gets "lost" in the efficiency later...

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