In a vacuum: steam sinks or evenly distributes?

[jrodatus]

Anybody know this one? I need to know for an evaporator design.

Because there's no buoyancy from air, I'd think it would sink (this answer was given on Yahoo), but if it sank then there would be a low pressure area left above, and so the vapor should rise to fill it, hence evenly distributing itself. No?

~Josh

 

[bbbeard]

Anybody know this one? I need to know for an evaporator design.

Because there's no buoyancy from air, I'd think it would sink (this answer was given on Yahoo), but if it sank then there would be a low pressure area left above, and so the vapor should rise to fill it, hence evenly distributing itself. No?

~Josh

Are the walls of the container adiabatic, or do they allow heat transfer? If the walls are cooler than the steam, some of it may condense into a liquid -- or not, depending on the temperature. Without condensation I would say the steam (which is a "vapor") behaves like a gas and fills the container.

BBB

 

[DaveC426913]

In vacuum, water's natural state is a vapour. Whether it condenses on the walls or not, it will vaporize again and distribute itself as a gas.

That being said, presumably there is some small gradient between ceiling and floor due to gravity balancing the natural vapour pressure.

 

[jrodatus]

Are the walls of the container adiabatic, or do they allow heat transfer? If the walls are cooler than the steam, some of it may condense into a liquid -- or not, depending on the temperature. Without condensation I would say the steam (which is a "vapor") behaves like a gas and fills the container.

Excellent! Thank you BBB. The walls are a copper pipe segment wound with an electric heating element, so hopefully there shouldn't be much condensation there. If my condenser coil is long enough w.r.t. to the evaporator, then the vapor should rise and flow around the bend...

It would be terribly nice to know how the presence of steam (kg/m^3) changes the pressure of the evaporator (" Hg). But I feel bad asking questions around here when I haven't contributed anything

EDIT: Wait... maybe I can figure that out using PV = nRT :)

~J

 

[bbbeard]

In vacuum, water's natural state is a vapour. Whether it condenses on the walls or not, it will vaporize again and distribute itself as a gas.

That being said, presumably there is some small gradient between ceiling and floor due to gravity balancing the natural vapour pressure.

I don't know what you mean by "In vacuum, water's natural state is a vapour" -- though I suspect it is just plain wrong. Pure water can exist in gaseous/vapor, liquid, and (various) solid forms, and the equilibrium composition depends on the thermodynamic state. The thermodynamic state in general depends on two degrees of freedom, e.g. pressure and temperature, temperature and specific volume, enthalpy and entropy, etc. But it is a misconception to think that water is vapor "in vacuum" (of course, if there is water vapor in a container, it is not a vacuum anyway).

Here's a thumbnail cartoon of the P-v diagram for water in the liquid-vapor-gas phases:

Imagine that the quantity of water and the volume of the container are such that the specific volume (=1/density) is in about the middle of the x-axis above. Then if the pressure (or temperature) is high enough, the state will be above the saturation line and will be all vapor. But if the pressure (temperature) is low enough the state will be below the line and will be a mixture of liquid and vapor.

BBB

 

[bbbeard]

It would be terribly nice to know how the presence of steam (kg/m^3) changes the pressure of the evaporator (" Hg). But I feel bad asking questions around here when I haven't contributed anything

EDIT: Wait... maybe I can figure that out using PV = nRT :)

~J

I think you need to acquaint yourself with the steam tables. For one thing, the ideal gas law is not such a great approximation for water in the vapor phase, though it will give answers that are the right order of magnitude... See the link in my previous post for a (very) brief intro to water thermodynamics...

Here's the NIST webbook if you need actual numbers...

BBB

 

[DaveC426913]

I don't know what you mean by "In vacuum, water's natural state is a vapour" -- though I suspect it is just plain wrong. Pure water can exist in gaseous/vapor, liquid, and (various) solid forms, and the equilibrium composition depends on the thermodynamic state.

Certainly. The phase diagram was where I drew my data from.

Tell me, have you actually looked to see what the temperature would have to be in order for water to be a stable liquid in a vacuum? Do you think it is anywhere near the temperature the OP is dealing with?

We're not answering a hypothetical physics exam question here, we're addressing the OP's question.

 

[bbbeard]

Certainly. The phase diagram was where I drew my data from.

Tell me, have you actually looked to see what the temperature would have to be in order for water to be a stable liquid in a vacuum? Do you think it is anywhere near the temperature the OP is dealing with?

We're not answering a hypothetical physics exam question here, we're addressing the OP's question.

I think the OP did not mean that the system was literally evacuated, because that assumption would be inconsistent with the presence of water vapor. I think he meant that there was no air in the system. Given that the system has no air in it, but has an unspecified volume at an unspecified temperature and contains an unspecified amount of water, then I think it is grossly misleading to throw in statements like "In vacuum, water's natural state is a vapour."

And yeah, I'd say I'm pretty familiar with the steam tables. Are you? Riddle me this:

A container has a fixed volume of 1 cubic meter. It contains 1 kilogram of water and no other substance. The system is heated until it is in equilibrium with a uniform temperature of 100 C. What thermodynamic state is the water in? Can you determine the pressure and if so, what is that pressure? Is there any liquid water? If so, how much of the water is liquid and how much is vapor?

BBB

 

[klimatos]

Anybody know this one? I need to know for an evaporator design.

Because there's no buoyancy from air, I'd think it would sink (this answer was given on Yahoo), but if it sank then there would be a low pressure area left above, and so the vapor should rise to fill it, hence evenly distributing itself. No?

~Josh

No.

Firstly, gas molecules will randomly distribute themselves in an available space, they will not evenly distribute themselves. This is the basic premise of statistical thermodynamics. The probabilities will be evenly distributed, the molecules will not.

Secondly, gases do not "fill" the available space. This model of expanding gas molecules passed out of usage almost two-hundred years ago with the general adoption of kinetic gas theory. At NTP they take up about 0.1% of the available space. Even with liquids and solids, the molecules do not "fill" all of the space.

Thirdly, you use "steam" in the title (T>100°C) and "vapor" in the text (T<100°C). Which do you mean? Knowing the temperature is important.

Taking your title as representative of your intention, the steam molecules will be randomly distributed subject to gravity. Since your title makes no mention of liquid water, we must assume that none exists at the ambient temperature and pressure.

 

[Bill_K]

A container has a fixed volume of 1 cubic meter. It contains 1 kilogram of water and no other substance. The system is heated until it is in equilibrium with a uniform temperature of 100 C. What thermodynamic state is the water in? Can you determine the pressure and if so, what is that pressure? Is there any liquid water? If so, how much of the water is liquid and how much is vapor?

Oh what the heck. For saturated water vapor at 100^oC, pressure is 101.42 kPa. Specific volume of the vapor is 1.6720 m³/kg. I therefore need 0.6 kg of vapor. The remaining 0.4 kg is liquid and occupies negligible volume. (I hope this wasn't a homework question!)

 

[bbbeard]

Oh what the heck. For saturated water vapor at 100^oC, pressure is 101.42 kPa. Specific volume of the vapor is 1.6720 m³/kg. I therefore need 0.6 kg of vapor. The remaining 0.4 kg is liquid and occupies negligible volume. (I hope this wasn't a homework question!)

Correct [though actually my steam tables says v_g=1.6729 m³/kg], and no, this wasn't a homework problem. I was just trying to get Dave to acknowledge that his premises were faulty.

Since I've been posting here, several times now visitors have posted questions which require a thermodynamic answer, and other folks jump in with "rules of thumb" which are really very questionable and misleading. I try to stamp out these errata when I can.

 

[bbbeard]

Thirdly, you use "steam" in the title (T>100°C) and "vapor" in the text (T<100°C). Which do you mean? Knowing the temperature is important.

And speaking of errata, typically the term "vapor" applies to the gaseous phase of a substance below its critical point, not the boiling point at standard atmospheric pressure. For water the critical point is 374°C (647 K). "Steam" used colloquially includes vapor and condensed liquid droplets, but engineering practice refers to any gaseous water, either above or below the critical temperature, as "steam". The steam tables certainly don't draw a line at 100 degrees and refer to >100°C as steam and <100°C as vapor.

BBB