Assuming an absolute vacuum (-0.1 MPa) is created with a pump, what diameter of water droplets can be drawn in?
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In multi-effect evaporation, vacuum is usually applied in the last effect, and it is water vapor that is removed. Today, while designing a vacuum tank, this question came to mind: assuming a vacuum is created using a pump (at -0.1 MPa), what diameter would an ideal spherical water droplet need to have in order to overcome gravity and be drawn upward?
There shouldn’t be any water droplets drawn out, right? ! The reason why water droplets are drawn in is due to air currents – they are carried along by stronger air currents, rather than pressure pushing the water droplets upward! If it’s caused by pressure, then it’s not a “water droplet,” but rather a “water column”!
As the air current rises, it carries small liquid droplets with it; but what size are the droplets that can be carried upward? Assume the upward air flow velocity is 1.5 m/s.
This should be related to the form of the liquid – there is a big difference between a liquid surface and liquid droplets – and it must also have a significant impact on the diameter of the inlet. It should be possible to calculate this precisely.
This flow rate is essentially the speed required for gravity separation; at a speed of 1.5 m/s, droplets with a size of 100um to 200um or larger can basically be removed through gravity separation. As for smaller droplets, they are expelled along with the air flow.