# Power generation with a long, horizontal pipe, using atmospheric pressure differences

Straight to the point.

If I float a pipe starting from a point in the ocean that has the atmospheric pressure of 101 kPa (Call it Point A) all the way to another geographic position on the ocean with an atmospheric pressure of 96 kPa (point B). Lets say the distance between the two points is 1000 km. The air temperature at point A is 28 °C while the air temp at point is 18 °C.

If you opened both ends of the pipe,would the atmospheric pressure cause air to flow from Point A to Point B? (factoring in reasonable friction co-eff for a pipe.)

Can you please provide factors that would hinder the flow of air!

Could such a pipe, practically be used to generate a flow of air to generate electrical power?

• Why do you need a pipe? The ocean is open. Wind will blow without it. – mmesser314 May 24 '16 at 13:37
• You should do a back-of-the-envelope calculation of the impedance of a 1000km pipe to air flow... – Jon Custer May 24 '16 at 13:43

Poiseuille's law will tell you that a pipe of 0.1 m diameter will achieve a flow velocity of around $v=0.1\mathrm{~m/s}$, which is actually more than I expected. For pipes a bit wider than this, the flow will be turbulent (Re>2200), for which you can't apply Poiseuille. For turbulent flow, you can look into the Darcy-Weisbach equation.

The interesting part is to what extent pipe-flow formulas can be applied to such a system.

One thing to consider is the influence of the Earth's rotation. Your pipe is horizontal, but you don't state whether it's aligned East-West or North-South and at what latitude. The Coriolis acceleration is $a=-2\Omega\times v$, where $\Omega={}$7.3e-5 rad/s is the rotation speed of the Earth. This acceleration is always perpendicular to the pipe, no matter the orientation, so we can forget about it.

The next one is the centrifugal acceleration, $a=\Omega^2 R$, where $R$ is the distance to the Earth's axis of rotation. With $R\sim 10^6\mathrm{~m}$, this could be somewhat significant ($a=0.005~\mathrm{m/s^2}$), but you have stated that the tube will follow the sea surface, which is already on a surface line of constant gravitational potential. So, no effect here either.

The difference in air temperature is not relevant either, also because the pipe is on a constant-potential surface; density differences will not lead to buoyant forces.

So, I don't think there are significant forces other than the pressure difference that would affect the flow rate.

And electricity generation you can forget about. The kinetic energy flow in an unrestricted pipe is $P=\pi\rho D^2 v^3/8$, where $\rho$ is the density of air (1.3 kg/m3) and $D$ the pipe diameter. That's $P=4~\mathrm{\mu W}$. And if you put a turbine in the tube, the flow speed will decrease even further.

In general, power-generation schemes where the pressure difference is a given (and available for free) do not benefit from attempts to channel the flow. The kinetic energy of the fluid mass passing through an otherwise unrestricted pipe is a hard upper limit for the amount of power that you can harvest. Any restriction in the pipe, such as nozzles, funnels, and turbines, can only decrease that power, never increase. Unfortunately, every now and then, someone comes up with variations of "funnel the wind into a pipe", tries to patent it and develop it into a product, only to discover that they cannot beat the efficiency of an unshrouded wind turbine for any definition of "efficiency" that matters for practical purposes.

• I hope you don’t mind answering some follow up questions. How would you go about calculating the flow velocity if you had to increase the pipe diameter greater than 0.1m? Would it at all be possible to produce meaningful kinetic energy if you could increase the pipe diameter up to possible sizes (say max 5.0m), as well as including a nozzle to increase the velocity? Side Note: This is the closest example to the system that I could find. The difference being this example doesn’t use a model that factors out hydrostatic pressure. coldenergyllc.com/ACMPatent.pdf – Travtheguy May 25 '16 at 4:45
• Answer updated. – Han-Kwang Nienhuys May 25 '16 at 7:39
• Thanks again for your feedback. I fully understand what you are saying. I hope you at least found the topic interesting. Keep up the good work. – Travtheguy May 25 '16 at 9:19