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Microtek
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If the bubbles were dispersed in the water it would not be neccessary to actively remove them, as long as the flow rate in the system was
significantly greater than the speed with which the bubbles rise. You would then just pump them out along with the rest.
It would be interesting to do the experiment with a 20 m long strong garden hose. If it was filled with water and the middle part was raised 10 m so
one end was 1 m higher than the other ( without allowing any of the water to leak out ). What would happen when both ends of this siphon were released
? I'm inclined to think it would work like any other siphon, but maybe the under pressure at the top will disrupt the flow. Still, if this is the
case, forcing water into one end should provide enough forced flow I think.
[Edited on 9-2-2006 by Microtek]
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unionised
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1 atm = 103 KPa
10 m head of sea water is equivalent to
9.81 *10 * about 1020 = 100.0 KPa
At the top of the "syphon" the pressure is only about 1/33 of an atmosphere.
On a warm day the water will boil- never mind the dissolved gases.
Let's calculate them anyway. Air disolves in water at 30C to the extent of about 16 ml/l (measured at NTP).
Drop the pressure and the volume increases while the solubility drops. In moles/Litre the solubility is proportional to the presure so 32/33 of that
air bubbles out and- because the pressure has dropped- it occupies 33 times as much space. It takes up over half a litre. We are not talking about a
few little bubbles here- we are talking about nearly as much air as water and the problem gets worse if you use water that is cold enough not to boil.
[Edited on 12-2-2006 by unionised]
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Microtek
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That is all true in a static situation, but I think the result will be very different when when water is allowed to flow into the low pressure zone (
or indeed when it is forced in there by a pump ).
Anyway, I don't think this discussion is really going anywhere at this point, but if I decide to conduct the experiment I'll let you know how it turns
out.
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