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.

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.

since we are claiming to know the displacement s

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.

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

Head of damn
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.

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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