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I have a turbine which produces 50MW, the water falls down from three meters height.

How much water does flow per second?

Please do not answer this question, but rather give an explanation how I can calculate it myself.

Tim
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  • How is this about mathematics? – copper.hat Jan 09 '19 at 21:41
  • @copper.hat I can't quite follow what your question is. Do you mean why I posted it on Mathematics? As I didn't see a Physics SE. And physic belongs to math. The math is calculating the waterflow. – Tim Jan 09 '19 at 21:44
  • And what is the difficulty? It is not about math. It is a physics homework question. – copper.hat Jan 09 '19 at 21:45
  • I don't know how I can get the water flow out of generated Power and fall height. – Tim Jan 09 '19 at 21:46
  • What power would be generated if the water flowed at $x$ $m^3/s$? This is basic physics. You need the density of water and the acceleration due to gravity. (Using the usual improbable & unelaborated assumptions of homework questions, presumably in this case the assumption is that all the potential energy is transformed into usable work. Impossible, of course.) – copper.hat Jan 09 '19 at 21:46
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    Mhm. 1 m^3 = 1000 Liters = 1000kg so W=1000*3=3000 So if I understand right, 50'000'000 / 3000=16666.6.. Does this mean a waterflow of 16'666L/s or 16qm of water per seconde? – Tim Jan 09 '19 at 21:51
  • Doing a dimensional analysis is probably your best bet. – Cuhrazatee Jan 09 '19 at 22:00
  • @TylerKharazi What does this mean? I'm not that good at physics. Could you maybe provide a (simple) formula? Thanks – Tim Jan 09 '19 at 22:00
  • This is essentially what you were doing in the comment I upvoted. You should look at the units of power, which are Watts, and the units of Watts are kg m^2/(s^3). Often times in physics, you should look at the units of the problem, they can provide a great intuition for what the final solution ought to look like. Also, FYI there is a physics stack exchange, https://physics.stackexchange.com/. – Cuhrazatee Jan 09 '19 at 22:04
  • @TylerKharazi Thanks! So was my calulation correct? – Tim Jan 09 '19 at 22:06
  • If your answer does not include $g$ or the density of water $\rho$ then it cannot be correct. – copper.hat Jan 09 '19 at 22:09

1 Answers1

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This is physics, not math.

If a mass $m$ of water falls through a height $h$ then the potential energy change is $mgh$.

If $\dot{m}$ of water passes through a height $h$ then the potential energy change per unit time is $\dot{m}gh$.

If water has an average density of $\rho$, and the volume flow rate is $\dot{V}$ then the mass flow rate is $\dot{m} = \rho \dot{V}$.

If all the potential energy is converted to usable work $P$ in the turbine then we have $P = \rho \dot{V} gh$.

You have $P,h,g,\rho$. Compute $\dot{V}$.

Addendum:

$\rho$ is the density of water, we usually take $\rho = 1000\ kg/m^3$.

$g$ is the acceleration due to gravity, we usually take $g = 9.81 \ m/s^2$.

In the above, $h=3 \ m$, $P= 50 \times 10^6 \ W$.

The dot as in $\dot{m}$ usually means the rate of change with respect to time. So, $\dot{V}$ is the change of volume per unit time, or flow rate.

copper.hat
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