Air Pressure and Fluid Flow
Name: Chris H.
I have a bottle with two holes (one in the top one in the
bottom), I am performing an experiment showing how when the hole in the
top is covered (air hole) the water will not flow out the small hole in
the bottom. As there becomes less and less water in the bottle the time
delay between covering the hole at the top of the bottle and water
coming out the bottom increases - why?
The water column over the hole in the bottom of the bottle is held up by
the air pressure on the bottom of the column being greater than the air
pressure on top. The water over the parts of the bottom of the bottle
which are solid glass is held up by the glass pushing up on the
water. The force per unit area must be the same as the air pressure on
the water above the hole when the water is not flowing out.
If the hole on the top is open, the air pressure is (almost) the same on
the top and bottom of the water, so the water will fall with the
acceleration of gravity, ignoring viscosity, surface tension, and
friction between the water and the sides of the bottom hole.
However, when the top hole is closed, the air above the water must expand
to fill the increasing volume above the water. This, of course, reduces
the pressure of the air (pV = constant at constant temperature). Notice
that a 10% increase in the volume V produces a 10% decrease in the
pressure p. (Plug in some numbers if this is not obvious to you)
When the jar is almost filled with water, a 10% change in the volume of
the water occurs very quickly since the volume is small. Conversely, when
the jar is almost empty of water, losing the same amount of water produces
a much smaller percentage change in the volume (though the same magnitude
of volume change). Therefore it takes a longer time after the top hole is
covered for the water to stop flowing when the jar is almost full of water
then when there is less water in the jar.
I hope this is clear to you. Feel free to write another note if it is
Best, Dick Plano
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Update: June 2012