The right size pipe for falling water

There are a lot of engineering calculators out there and when the problem I'm dealing with isn't a simple product of units (for which I use wolfram alpha)

I make good use of them.

However, I recently came across a problem for which I struggled to calculate an answer and for which I struggled to find calculators.

Given a reservoir of height h how thick a pipe is required to extract P power.

The energy in the water is a simple calculation that comes from mgh (mass times acceleration due to gravity (9.81m/s^2) times the height the mass is raised).

However, we don't want to take the whole reservoir we simply want to power something.

Water through a pipe travels as a result of pressure differences between the ends of the pipe. Water pressure at a depth is independent of the total volume of water and is the same at the sides of a given depth as it is forcing downwards.

This can be calculated by working out the weight of the water atop an area at a depth.

The wight of a volume of water is easy to calculate thanks to metric 1m^3 of water wights 1 tonne. 1 tonne exerts a force of 1000kg*9.81m/s^2 but to keep things simple you can talk about kgf kilograms of force. This makes our calculations slightly easier than working in newtons.

1kgf = 9.81N

So the pressure of water on the surface of the reservoir is 0 (relative pressure to the latent atmosphere, you can't have the surface of the water, or any other liquid bordering a vacuum.).

There are multiple ways to try to work out the speed of the water, and therefore volume flow of water.

The speed of the water (assuming a frictionless pipe) will be the same as the speed of the water falling from the height freely.

v = at

We know the acceleration that's just 9.81m/s/s

But how long will it take?

We know the distance is 10m.

$\\v = at \\\int v\ dv = \int at\ dv \\s = vt+ \frac{1}{2}at^2\\\frac{ds}{dt} = \frac{d^2 s}{dt^2 t} + v_0 \\s = \int \frac{d^2s}{dt^2 t} dt = \frac{1}{2}at^2 + v_0 t + c$

in this case, the constant is the starting distance which we can set to zero.

We can also set the starting velocity to zero since the water starts at rest.

$s = \frac{1}{2}at^2$

since we are claiming to know the displacement s

$\\s = \frac{1}{2}at^2\\2s/a = t^2$

The energy contained in the water follows 1/2mv^2.

The power is just the energy over time.

But what is the mass?

The mass of the water flowing per second is a function of the velocity and the cross-section of the pipe.

$\\\frac{m}{t} = Av \\m = Avt \\E = \frac{1}{2}mv^2 \\E = \frac{1}{2} (Avt) v^2 \\E = \frac{1}{2} A v^3 \ t \\$

$v = at \\\\t = \sqrt{\frac{2s}{a}} \\v = a \sqrt{\frac{2s}{a}} \\v = \sqrt{2sa} \\P = \frac{1}{2} A v^3 \\P = \frac{1}{2} A ({2sa})^{\frac{3}{2}} \\$

And that is our formula for the power for a given pipe size. Rearranging we can get the A for a given power

$P = \frac{1}{2} A ({2sa})^{\frac{3}{2}} \\\\A = \frac{2P}{({2sa})^{\frac{3}{2}}} \\A = 2P({2sa})^{\frac{2}{3}}$

Head of damn
Power
Area of pipe

There is another way to do this which doesn't require all the suvat but didn't initially get trigged as a direction.

Rather than using suvat to calculate the velocity we can use potential energy.

Energy in a mass of water raised h is mgh

Energy in a mass of water flowing is 1/2mv^2.

Therefore we can calculate the velocity as the conversion of this gravitational potential energy into kinetic energy.

$\frac{1}{2}mv^2 = mgh \\\frac{1}{2}v^2 = gh \\v^2 = 2gh \\\\v = \sqrt{2gh} \\$

This looks slightly different but only because we have put g as the acceleration and h as the displacement.

From the velocity and target power, we can get the area in the same way as before.

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The right size pipe for falling water

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