We use gravity flow to transfer leachate (contaminated water) from a tank to an open pond on different elevations. Head difference, between the level in  the tank and the dischage pipe end into pond, is 4-6 metres. Horizontal distance bewteen the tank and the pond is about 1000 metres. Pipe diameter 63 mm poly. Pipe goes through a valley between the tank and the pond that is about 2 metres below the discharge point. System suffers from air locks each time we go to use. (Pipe is drained between uses as it forms part of another system). Bleeding the air from the low point restores the flow.

Can someone explain the fluid mechanics of why an air lock stops flow in this type of situation, what head difference would stop this occuring?

In a comment related to that of Chemical, the Froude number can be used to determine if the liquid velocity is sufficient to push the gas bubble ahead of it to clear the pipeline.

For fluid flow in pipes, the Froude number is easily calculated as,

Fr = u/sqrt(g*D)

where u is the fluid velocity, g is gravity, D is the pipe diameter, and the "sqrt" represents the square root.

First, consider horizontal pipes.  If the liquid velocity is too low, then a separated flow geometry will be present with the liquid flowing underneath the gas.  At high liquid velocities, the liquid will flow as a plug and push the gas in front of it to clear the line.  As the liquid Froude number approaches or exceeds unity, the liquid velocity will be sufficient to obtain the plug flow geometry and push the gas in front of it.

For the pipe describe by mikeellsmnore, the diameter is .063 m and gravity is 9.81 m/s.  Thus, to obtain a Froude number > 1, the liquid velocity must be greater than,

u > Fr*sqrt(g*D) = 1.0*sqrt(9.81*.063) = 0.79 m/s, or 2.6 feet per sec.

Next, for a downward flowing vertical pipe, the gas bubble will tend to rise due to buoyancy.  To avoid the collection of gas at the piping high point (and the potential for vapor lock), the liquid velocity should be greater than the velocity of the rising gas bubble.  In other words u liquid > u bubble<filter

The bubble rise velocity can be estimated by,

u bubble = 0.345*sqrt(g*D)

If this expresion is rearranged, we get back the expression for Froude number:

(u bubble)/sqrt(g*D) = Fr = 0.345

So, to clear the bubble from the vertical pipe, the Froude number should exceed 0.3.  This is similar to the statement made by chemical.

In your case, if you are starting from a zero flow condition, then the question is what is required to start the fluid moving?  

To start the fluid moving, the pressure in the tank plus the static head of the water between the tank and the gas pocket (i.e., the driving pressure) must be greater than the pressure in the gas pocket.  The pressure in the gas pocket will be equal to the pressure at the discharge end of the pipe  (I assume the pipe discharges above the surface of the pond water?) plus the static head of the water column between the gas pocket and the pond (i.e., the back preesure).

At first glance, the piping geometry you described should provide enough driving pressure to get the system flowing.  Hoevever, the problem may be that the gas pocket extends beyond the "valley" and is present farther up the piping upstream of the valley.  In this case, the static head of water between the tank and the gas pocket will be reduced and the driving pressure will be insufficient to overcome the back pressure.

I am not familiar with the valving procedure you use when refilling the pipe system.  Perhaps the procedure can be modified to reduce the penetration of the gas pocket between the tank and the valley, thereby maximizing the driving pressure.

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