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Water vapor in a vertical pipe

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MTwater

MTwater

Student
Apr 18, 2024
1
Suppose I have a vertical pipe, its vertical height to be determined.

Suppose the bottom of the vertical pipe is open and dipped into water at atmospheric pressure.
A heater is installed in the bottom of the pipe so that water start boiling and vapor start to rise in the pipe.
Suppose water supply and amount of heat production have no practical limitations.

At the top of the vertical pipe, I install a vacuum pump with the intention to pump out water vapor that rises from the bottom of the pipe
Assume that the pipe is fully insulated thus no heat loss in the pipe.

What could become the maximum height of the vertical pipe so that the vacuum pipe will still be able to pump out the water vapor ?

Eric

 
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This forum is not for us to give you the answers to your homework problems.
 
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Unlimited.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
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The way I read this it can be reformulated to: what is the height of a column of water vapor at a constant temperature (T_sat @ 0 psig) such that the bottom of the column is at 0 psig and the very top is exactly 0 psia?

If I were assigned this problem I would use ideal gas law assumptions to determine some equation that gives the variation in pressure with height of a gas that varies in density due to gravity only, then plug in the pressures, MW, and temperature and solve for the height.
 
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At some point the atmosphere becomes so cold the pipe also cools and the vapor condenses and drops back down. Farther up the cooling is below the escape velocity of Earth's gravity so the water molecules just cascade back down, which is largely what keeps the vacuum of space from being an escape path.
 
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What could become the maximum height of the vertical pipe so that the vacuum pipe will still be able to pump out the water vapor ?
Click to expand...
As a WAG I suggest around the Clarke Belt.

Joking aside, How high must a column of steam be to exert a pressure equal to atmospheric pressure on the water surface.
Assuming that atmospheric pressure will force (or a vacuum will pull) water to a height of 32 Feet:
Unit weight of water/unit weight of steam X 32 feet.
Instrumentation 101. Manometer theory and calculations.
--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
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The molecular weight of water is 18, air is about 29. The temperature of the steam is 100C (373 Kelvin), while air is say 20C (293 Kelvin). So, the height of a column of steam that weighs the same as an equivalent column of air would be (29/18)x(373/293) or twice as high, because it is half the density.
 
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A commonly used value is the specific weight of water on Earth at 4 °C (39 °F), which is 62.43 pounds-force per cubic foot.
One cubic foot of steam weighs 0.03732 pounds.
One atmosphere (101.325 kPa or 14.7 psi) is also the pressure caused by the weight of a column of freshwater of approximately 10.3 m (33.8 ft).
Ratio between the weight of water and the weight of steam.
62.43 lb per cu ft /0.03732 lb per cu ft = 1670
1670 X 33.8 = 56,446 feet.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
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Except the density of a column of gas isn't constant, it varies with height.
 
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Except the density of a column of gas isn't constant, it varies with height.
Click to expand...
Back to the Clarke Belt I guess.
Can we consider an average density and take my last figure times 2?
That's 56,446 feet, not the Clarke Belt.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
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As stated, taking the limit as pressure that the vacuum pump can achieve goes to zero the height goes to infinity - LI's answer.

The question becomes how good is your vacuum pump which seems kind of practical for this sort of theoretical/thought experiment/homework problem. If you choose a vacuum you can get an actual number, e.g. for 1.47 psia (-27 in Hg) I get
133,000 ft
 
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