Evaporation Preventing 2

[mwemag]

I am looking for a method and calculation parameters for preventing net evaporation of a hot static fluid in a closed box. The aim is to keep the fluid (or at least a certain amount of it) in its liquid state.

I'm not very familiar with thermodynamic calculations, but I suppose that the solution would be to create a equilibrium state in which the evaporation rate equals the condensation rate. This idea is based on the rule, that evaporation in a closed container will proceed until there are as many molecules returning to the liquid as there are escaping. At this point the vapor is said to be saturated, and the pressure of that vapor is called the saturated vapor pressure.

The inner space of the box is around 200 times larger than the volume of the fluid, (e.g. box volume 2 Liters, fluid volume 1 cl). The air in the box could also be any gas, whose pressure could be increased to around 30 atmospheres, or even higher if necessary. The constant temperature of the fluid would be a little below its boiling point at atmospheric pressure. It's evaporation rate at atmospheric pressure at the given temperature is around 1 g-cm2-s, vapor pressure at that temperature: ca. 50k Pa, and assuming the fluid having a flat surface.

The problem is that I don't know how to calculate the conditions for creating an equilibrium state in the given case. Actually every description of this topic I could find states that the saturated vapor pressure depends only on temperature, not on ambient gas pressure or volume sizes of container or fluid.

But in contrast to this rule, e.g. a small drop of water in a large closed vessel will probably evaporate completely because of the large air volume surroundig it, whereas a water volume of 3/4 of the vessel will likely reach an equilibrium state and therefore keep most of its volume liquid, indicating that the volume ratio indeed matters.

Second, by encreasing the pressure of the ambient air/gas the boiling temperature of any fluid will increase too, thereby also considerably reducing the evaporation rate of the fluid at lower temperatures, so ambient pressure is important too.

Obviously I have difficulties to understand the whole evaporation / saturated vapor pressure concept...

I appreciate any help.
Thanks

 

[StoneCold]

The liquid will evaporate at a given temperature until the partial pressure of the vapor is equal to the vapor pressure of the gas at that temperature. There are some assumptions in that but it is basically true. So figure out how many moles of gas are required to give you a partial pressure equal to the vapor pressure of the gas in the room. You need more moles of liquid than that to still have liquid after you have reached equilibrium.

Adjusting the total pressure will not help you. I would like to write more but I have to go.

Regards
Stonecold

 

[mwemag]

"until the partial pressure of the vapor is equal to the vapor pressure of the gas at that temperature"

For example, when I have water at constant 95°C with a corresponding vp of ca. 85 kPa and dry air as the initial fill gas in the container, having normal atmospheric pressure (100 kPa):

Would the evaporation saturate when the partial water-vapor pressure in the air reaches 85 kPa, establing a total air pressure of (85+100 kPa=) 185 kPa ?

And, to establish equilibrium conditions – do I have to find then how many moles of water vapor are required to add a pressure of 85 kPa to the given volume of dry air at atmospheric pressure and 95°C?

Thanks

 

[zdas04]

Why won't adjusting the total pressure help? The saturation curve for water in natural gas (see GPSA Field Data Book, figure 20-3) is a function of both temperature and pressure. For gas at 14.73 psia and 210F the saturation quantity for water vapor is 95,979 lbm/MMCF. If I raise the pressure to 1200 psig and keep the temp at 210F, the water content drops to 681 lbm/MMCF. So if I started with 100 gallons of water (call it 834 lbm), then at atmospheric pressure it would all be gone and the gas would be at about 5% RH. At 1,200 psig, I would still have a good portion of the water as liquid and the vapor space would be at 100% RH.

Seems like pressure matters a lot. If you look at it from the perspective of Henry's Law, increasing pressure has a significant effect on the amount of a gas (like your evaporation products) that are forced back into the liquid phase. As pressure goes up, the amount of a particular species of gas that can dissolve in a given liquid increases.

David

 

[IRstuff]

Partial pressure is affected by the total pressure, otherwise, antifreeze would boil in radiators and pressure cookers wouldn't work.

TTFN

FAQ731-376

 

[mwemag]

Total pressure definitely has a strong influence on the evaporation rate, since the different atmospheric pressures at different altitudes alter the boiling ponts of any liquid dramatically. And according to zdas04 it has a strong influence on the equilibrium conditions too.

The problem is how to integrate the total pressure factor into the calculation. The following formula for example is used to calculate the evaporation rate (in kg/m2*s):

(vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )

vp =Pa, ambient pp = Pa, molec. w = kg/mole, R = universal gas constant (8.3143), T = temp. in K

If the resulting value is zero then the equilibrium is established. But the overall pressure is not included in the formula, I don't know how to integrate it.

 

[zdas04]

It may be hard (in fact it really is hard), but that don't make it wrong. All of the psychometric calcs start with the assumption that pressure will change less than a few percentage points ever. If you can start with constant pressure, then the arithmetic gets considerably easier. Your original problem statement seemed to allow the pressure to change which has a much bigger impact in the real world than most researchers would like to admit.

I've been working on an empirical Evaporation=F(press,temp) equation for about 5 years now with mixed success. For temperatures under 100F, and pressures below 150 psia, my last equation was consistently within 15% of ASTM data. After several hundred thousand iterations in MathCad curve fitting programs, I generated a new equation last week that is pretty consistently 30% off for any pressure/temperature pair. I put it aside for at least a month before I'm going to start hunting for the latest flaw in my calcs (I didn't sleep this weekend working on the stupid thing).

Bottom line is that if pressure can change significantly, then the equilibrium conditions will be a function of both pressure and temperature.

Sorry.

David

 

[mwemag]

The total pressure of a gas mixture is always the sum of the individual partial pressures. If the original gas has a pressure of 1 atmosphere (?100 kPa) and the added vapor pressure (needed to establish equlibrium) is 85 kPa, then the total pressure is 185 kPa.

Assuming now to increase the total pressure by 20 times, then the partial vapor pressure should be 20 times higher too, but as it is the saturated vapor it won't increase (since it depends only on temperature and any additional vapor would condense). It will still have 85 kPa – but now with only the 20th part of molecules needed to form that pressure.

If this assumption is correct, then the fluid volume part which is lost through evaporation before the equilibrium state is 20 times smaller with the 20 atmospheres condition than with the 1 atmosphere condition.

I hope I'm right...

 

[zdas04]

Let's say that you've increased the pressure by sqeezing the box instead of adding more stuff.

The original gas may still 54% of the vapor phase, or it may not be. I've been doing some work with Henry's Law constants this week and I've found that doubling the pressure tends to increase the solubility of a gas in water by about twice. For example, CO2 has a Henry's law constant that is 1/4 that of O2. So if I increase the pressure 20 times, the amount of CO2 that now disolves at constant pressure is 4 times the amount of O2 that disolves, so in this case if you started with 54% CO2 and 46% O2, you might end up with 40% CO2 and 60% O2 in the vapor space.

David

 

[IRstuff]

You might want to play at:

http://webbook.nist.gov/cgi/fluid.c...nit=m/s&VisUnit=uPa*s&STUnit=N/m&RefState=NBP

TTFN

FAQ731-376

 

[mwemag]

zdas04, actually the total gas pressure in the device will increase already because of the gas heated by the hot temperature of the fluid.

The empirical value shows that the original pressure of the fill gas in the cold state at RT, around 5 atmospheres, will increase in the hot state to around 25 atmospheres immediatly after the liquid reaches it's constant temperature. I am designing the application to run with at a much higher temperature (around 85 percent higher), therefore the aim is to estimate the equilibrium conditions / evaporation rate for that higher Temperature. Without empirical data I can only approximately estimate that the total pressure in the hot state with the higher T would be in the range 35-45 atmospheres.

Of course this approximate estimation will not allow me to calculate the evaporation rate, but it seems (as the vapor saturation in the 14.73 vs. 1200 psia example of your first answer shows) that a good part of the original liquid mass can be prevented from evaporation just by this higher overall pressure.

I tried to recalculate your example with metric volume units and saturated vapor pressure instead of saturation quantity in lbm/MMCF without success (unable to find even a definition for lbm/MMCF). It probably would help if you could particularize your example, eventually with data on the container-fluid volume ratio too.

Thanks a lot.

 

[zdas04]

Sorry about the units confusion, but it is all I have. "lbm" means "pounds mass of water vapor" and "MMCF" means "millions of cubic feet of natural gas adjusted to 60F and 14.73 psia (standard temperature and pressure in the US Oil & Gas industry)".

ASTM D 1142 has a high quality technical description of the process of evaporation (I think I paid $40 for it last week at the ASTM web page, but I'm not positive of the price). It presents a two constant equation that relates temperature and pressure to evaporation:

W=A/p+B

Where "W" is the water vapor in lbm/MMCF (there's that term again). "A" and "B" are fixed for a given temperature (and ASTM provides a table of these values from -40F to +460F. "p" is absolute pressure in psia.

I'm sure there is a metric version of this (probably in mg/m^3 or something like that), but I had no reason to look for it.

David

David Simpson, PE
MuleShoe Engineering

http://www.muleshoe-eng.com

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

The harder I work, the luckier I seem

 

[mwemag]

Thanks!
I should have concentrated on psychometric descriptions much earlier...

By putting "saturation quantity" into play, it was easy to do a simple demonstrative calculation with an astonishing result:

The saturation quantity of water vapor in air at 95°C and normal atm. pressure is 961.88 kg/m3 (metric), which is around 0.5 grams per Liter (=1 dm3 =1000 cm3). The same mass in form of liquid water at that temp has a volume of around 0.5 cm3. Therefore a volume of only 0.5 cm3 liquid water is enough to fill a volume of 1 dm3 with saturated vapor.

Now, let's say I want to sacrifice one half of the original water volume by evaporation to get equilibrium, thus saving the other half. So I duplicate the liquid water to 1 cm3, this volume being the initial quantity. 0.5 cm3 of the water will evaporate at the designated temperature, reaching saturated pressure in the above gas volume, and the remaining 0.5 cm3 liquid water will be "save" from evaporation.

If this calculation is right, then the problem seems to be solved even at normal atm. pressure, without the need to increase the overall pressure at all.

Even though I admit having difficulties to imagine that a water layer with a thickness of only (!) 0.1 mm (= 1 cm3) evenly distributed on the bottom of a 10x10x10 cm container would not completely evaporate at constant 95°C in practice...

How plausible is the above estimation?

 

[IRstuff]

It would evaporate IFF it were exposed to unsaturated air.

TTFN

FAQ731-376