-1
$\begingroup$

A hydropower plant uses a river to generate electrical energy. It has a drop of $\pu{107 m}$ and the flow rate of water is $\pu{275 m3 s-1}$. The flow rate of the water is the same in and out of the system. Determine whether the system is open or closed.

The answer is: the system is open.

I am a bit uncertain as to how to simplify the general energy balance for the system:

$$Q + W = \Delta H + \Delta E_\mathrm{k} + \Delta E_\mathrm{p}$$

I think that only $\Delta H$ can be removed from the equation since the enthalpy won't change. Is that correct?

$$Q + W = \Delta E_\mathrm{k} + \Delta E_\mathrm{p}$$

If so, how can I use that to calculate the maximum amount of electrical energy that can be generated?

$\endgroup$

closed as off-topic by Mithoron, airhuff, Todd Minehardt, Jan, paracetamol Nov 25 '17 at 14:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ This is a physics question. $\endgroup$ – Gert Nov 24 '17 at 17:32
  • $\begingroup$ This a question we were given in our chemical engineering class @Gert $\endgroup$ – J.Se Nov 24 '17 at 17:34
  • $\begingroup$ Maybe so, but there's no chemical content in it. Migrate it to physics.stackexchange. $\endgroup$ – Gert Nov 24 '17 at 17:36
  • $\begingroup$ I did and now it was put on hold as off-topic $\endgroup$ – J.Se Nov 24 '17 at 18:26
  • 2
    $\begingroup$ @paracetamol 'consider posting it there instead" is what should never be told under homework, I hope you see it now as OP admitted cross-posting of closed question... $\endgroup$ – Mithoron Nov 24 '17 at 20:15
2
$\begingroup$

The only three terms that need to be taken into account are $\dot{W}$, $\Delta \dot{E_p}$ and $\Delta \dot{E_k}$, as there's no heat involved. I use the 'dot' notation to indicate these are energies per unit of time (powers).

For potential energy power we have:

$$\Delta \dot{E_p}=-\dot{m}gH$$

Here $\dot{m}$ is the mass flow ($275\ \mathrm{kg/s}$).

For kinetic energy power we have:

$$\Delta \dot{E_k}=\frac12 \dot{m}v^2$$

$v$ is calculated from:

$$\dot{m}gH=\frac12 \dot{m}v^2$$ I.e.: $$v=\sqrt{2gH}$$ Now calculate $\Delta \dot{E_k}=\frac12 \dot{m}v^2$.

Then apply the energy (power) balance, as: $$\dot{W}=\Delta \dot{E_p}+\Delta \dot{E_k}$$

This is the power the plant extracts from the falling water, as work done on the turbines. This then converted to electrical energy.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.