# Ocean currents and fluid pressure

## Main Question or Discussion Point

If we imagine a fluid travelling through a pipe as being a warm
current of sea-water moving over a cold denser layer of sea-water,
with the air immediately above the warm current forming one wall of
the pipe, would the pressure
of the warm current decrease as the current travels faster,and if so,
would less water evaporate from the warm current as it flows faster?
I ask this because oceanographers are trying to find a mechanism by
which a change in the rate of flow of ocean currents can cause more
seawater to evaporate and thus a warming of the Earth's atmosphere
(apparently water vapour is a better greenhouse gas than carbon
dioxide).

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From observing my two 10-gal freshwater aquariums, I notice that the uncovered one evaporates faster than the one fully covered even when both are exposed to the same 15 watts light heating fluorescent lamps and constant flow of warm water over cold water thru 200 gal per hour filtration system.

This phenomenon seems to imply that evaporation is connected to atmospheric pressure rather than water pressure.

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The covered aquarium allows some water vapour to condense back into the
main body of water.Atmospheric pressure may be the dominant factor but the pressure of the warm surface current in the sea (in which salt will also have an effect on evaporation - a salty solution has a higher boiling point than a pure liquid) will have an effect.The relative velocites of air/surface current and of cold water under surface current/ surface current contribute to evaporation from the warm surface current.A faster moving surface current would be expected to take more water vapour out of the atmosphere
by virtue of the current having a its pressure lowered,and the surface current
would absorb more water from the cold layer beneath it too.

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kurious said:
The covered aquarium allows some water vapour to condense back into the
For the 4 years since I had the tanks, I never notice any condensation. As a matter of fact, the covered one has a higher ambient temperature therefore it should have evaporated more instead it hardly evaporate at all. Usually I have to add about 1 gal for open top tank every 3 weeks or so but never add any water to the covered one during the same time period.

Yet the plants of java fern in both tanks are growing like crazy and the fishes are all doing very well.

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LURCH
Do you have a means of testing the atmospheric humidity of the air inside the tyop of the covered tank, and comparing it to that of the air just above the uncovered tank? I think humidity may be playing a much bigger role than air pressure in the differing evaporation rates.

The humidity is that of the room, in winter, more evaporation for both tanks. In summer, less evaporation. The covered tank is not really completely covered, there are a lot of holes along the topside for tubings and cable to in and out. But we know that pressure is always maximum when normal to the surface area, in this case, the pressure is vertically oriented in every case.

Perhaps the extra water vapour in the Earth's atmosphere results from heating by
the magma under the crust.Currents that change their speed probably change direction too and maybe come into contact with hotspots.There are also hydrothermal vents in the ocean .Perhaps the deep currents can encourage these to release more heat by changing the pressure on the sea floor.The sea floor would have to be very sensitive to pressure changes though.
If you like the sea read "Mapping the deep" by Robert Kunzig.Great book combining exploration biology,chemistry,geology and oceanography.That's where I got the idea for the original question from ( chapter on thermohaline circulation).

The question is a little hard for me to follow, but as far as the fish tanks go the coverd one should have a slower air exchange than the open one. Meaning that the air trapped in the coverd tank can only absorb vapor at a rate perportional to its air exchange rate (everything else being equal). Assuming that the air being exchanged is of lesser humidity. I would think that the velocity and volume of air moving across the surface is a big factor in evaporation.

So if we have one tank that has a standard open top, and another one shaped more like a bottle with a small opening at the top, I assume that the bottle woule evaporate more slowly everything else being equal.

Property changes of real fluids (solid, liquid or vapors) can be determined only by the difference of a final property value and an initial value. But for ideal gases, properties can be found directly without knowing the initial and final values.

A general relation for all ideal gases can be derived from the equation of state given by

$$\frac {p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$$

Boyle's law for constant temperature and Charles' law for constant pressure.