Engineering Asked by Taako on July 7, 2021
I want to build a machine that pumps liquids (roughly density of water) out of an enclosed cannister. The cannister has for one inlet for air at the very top and has an outlet (near the top as well) that feeds a tube down to the bottom of the bottle. Both these in/out-lets are sealed with silicon and the one that allows air in has a one way valve for the air.
The outlet is connected to a manifold with valves that lets me choose when to allow liquid to flow out.
I want to use an air pump to pressurize the cannister and then control the liquid output (on/off, not flow rate) via the manifold valve.
In addition when the manifold is open the air pump will be constantly pumping to replace lost pressure.
I am considering a pump like this which claims to have an air flow rate of 3.5 liters/minute. Does this mean that if I keep the manifold open and pump I air at 3.5 liters/minute it will flow liquid out of the outlet at the same rate? Or is there something i’m not considering that is going to make it flow at a slower rate?
Such things that might be of concern is the the liquid is acting against gravity (not much, it’s being pumped up maybe 6 inches to the manifold) as well as the inner diameter of the fittings on the cannister (0.095 inch).
Pump in question:
https://smile.amazon.com/NW-5V-6VDC-Miniature-Vacuum-100KPa/dp/B078H8V563/ref=sr_1_2?dchild=1&keywords=air%20pump%205v&qid=1609264991&sr=8-2
And a diagram of my intended design:
https://i.imgur.com/f0w5H5j.png
Does this mean that if I keep the manifold open and pump I air at 3.5 liters/minute it will flow liquid out of the outlet at the same rate?
Short answer: No.
The 3.5 L/min rate is a flowrate of a compressible fluid. The keyword here is compressible. The true flow rate of the water-like liquid being driven by the air pressure (from your 3.5 L/min of airflow) that you will see will be roughly proportional to the difference in pressure that your air pump produces, and the pressure at the outlet of the manifold valve (looks like you would be using atmospheric), minus the height that you must pump the liquid (you described as ~6"), minus the loss of friction in the pipes/tubing.
We could actually try to calculate this using Bernoulli's equation + friction factor + physical properties (density & viscosity) of your fluid if provided. It looks like we already have the outlet pressure of the pump (from the link you provided).
Correct answer by ChemE mang on July 7, 2021
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