Tuesday, 19 February 2008


Air or Steam lift pump video on YouTube. This is used for feeding a steam Babington, using Air, Steam, or even Propane to lift waste vegetable oil to the ball.

An air lift pump can lift a liquid to a height above the surface of the liquid equal to about 2/3 of the depth at which the air is injected into the bottom of the vertical pipe.

The capacity of the air-lift pump depends largely on the percentage of submergence of the foot piece; that is, the greater the submergence of the foot piece below the water level in the discharge pipe, the greater the volume (column) of water the pump can deliver per unit of time. However, the deeper the foot piece is submerged, the greater the compressed air pressure must be to lift the column of water.


In air-lift pump operation, compressed air has to be regulated correctly. The amount of compressed air should be the minimum needed to produce a continuous flow. Too little air results in liquid being discharged in spurts, or not at all. Too much air causes an increase in the volume of discharge but at lower discharge pressure. If air is increased still further, discharge volume begins to decrease

Some details for water pumping..

Sizing the air lift pump
The flowrate through an air lift pump is proportional to the flowrate of the air powering it. The literature reports air lift pump flow rates of 20 to 2,000 gpm and lifts to more than 700 ft.
An empirical calculation attributed to the Ingersoll Rand Co. correlates the flow of air with that of water.

Va = 0.8 Ll/(C log10{(Ls + 34)/34]
Where Va = volume of free air (cu. ft.) needed per gallon of water
Ls = length of the submerged section (ft.)
Ll = length of the lift section (ft.)
C = constant that depends on Ll (see Table 1)
Another relevant variable is the relationship between Ls and Ll. Functional air lift designs exhibit a curious non-linear phenomenon. The ratio of submerged length to total length, Ls/(Ll + Ls), runs about 0.6 when the lift is only around 20 ft., but decreases to about 0.4 when the lift is 500 ft.

The last relevant variable is the air pressure needed to make the device operate. This depends, of course, on the specific gravity of the fluid. The depth that corresponds to one psi is inversely proportional to the specific gravity. For water, one psi corresponds to 2.31 ft. of depth. If one ignores the friction losses in the line, the applicable relationship is:

P = (Ls * sg/2.31)
P = required gas pressure (psig)
sg = specific gravity of the fluid (dimensionless)

No comments:

Post a Comment