 |
 |
Fill and Evaporation Rates
Name: Robert
Status: Educator
Grade: 9-12
Location: WA
Country: United States
Date: December 2007
Question:
Hi, I am an Engineer at HP but I am volunteering at my son's
high school. I would like to structure a problem for the students which
would include a liquid substance that is 80% water going into a container
subject to evaporation. The question is: How long would it take to fill a
container of X mm if the liquid goes in at a specific rate given an
vaporation rate (of course there will be some precipitate remaining after
full evaporation)? What I would like to do is then have the students
construct an experiment to accelerate that behavior (at elevated temp and
low humidity) but I think I would need to do this at ambient first so I
could get an acceleration factor. Ideally, it would be great if I could
have a formula that equates volume as a function of time and temperature.
I thought I could use elementary differential equations for this but it
may actually be simpler than that. Any thoughts?
Replies:
I will guess there is probably a simpler way. The evaporation rate
depends on surface area, so if you are filling a cylindrical
container, you may be able simply to subtract a constant evaporation
rate from the fill rate to get the net fill rate.
If the evaporation rate also depends on the concentration of whatever
substance represents the other 20% of the liquid going in, then you will
need a differential equation. I will guess the evaporation rate will
depend on the fraction of molecules at the surface that are water, so
once the concentration has increased to the point that a precipitate is
being formed, the surface fraction will stop changing, and you'll be
back to simple subtraction.
Tim Mooney
We frequently receive inquiries about "rates" of evaporation in one form
on another. The problem is that "rates" of [evaporation, crystallization,
condensation, ...] the list is long are very difficult to measure
reproducibly because important variables (such as relative humidity) are
difficult to control. Example: if the relative humidity of the air is
100% it does not matter how fast you blow air across a sample, no water
will evaporate, so the solution will always become water-enriched. In
turn, the relative humidity is a sensitive function of the temperature
of the aqueous solution, and there is a "feedback" mechanism since the
preferential evaporation of some volatile component -- say acetone --
alters the partial pressure of water, and the evaporation causes a
reduction of the temperature solution which changes the values of all of
the above. All of these (and other) variables have "feed back" on the
conditions of evaporation.
While it may be an oversimplification, this is why it is very difficult
to predict weather. There are so many "feedback" variables operating I
fear that the differential equations will be non-linear. Way too involved
even for "state of the art" experts. Sorry, wish it were simpler.
Vince Calder
Robert,
You can make this situation somewhat simple, or you can make it quite
complicated. The more precise you want to be -- and I am not sure how precise
you would need to be for the exercise to be reasonably accurate (valuable?)
-- the more complicated. I am going to try to give you some considerations,
and you can decide (with experimental data) which can be safely neglected.
I am also going to give you the names of certain equations, but since you are
comfortable with differential equations, I am going to assume you can find
them and solve them -- but please feel free to email me back if you want
help with those steps.
First, you stated you want the container to have a liquid that is 80% water.
What do you want the other 20% to be? If you choose a soluble solid -- e.g.
sugar or salt for instance -- that would not evaporate, then you can
approximate the partial pressure exerted by the water using Raolt's law (the
mole fraction of the water times its partial pressure). When you know the
partial pressure of the water at its surface, you can make some simplifying
assumptions to calculate the mass transfer. For instance, assume you have a
cylindrical container and that the partial pressure of water at the exit
(top) of the container is ambient humidity. The partial pressure is a
function of temperature. You can then estimate the mass transfer of water by
diffusion from the surface to the top of the container by Fick's Law
(diffusion also is a function of temperature). If you are filling the vessel,
or waiting a long enough time that the length of the vapor column (from the
surface to the container top) changes, then the equations become more
challenging to solve analytically. Likewise, if you wait a long time, the
water will become depleted, its mole fraction will drop, and thus you have
another variable in your equations instead of a constant. I would recommend
a math software package such as Mathcad to solve them numerically.
If you want the other 20% to be a volatile liquid, such as ethanol,
methanol, etc. Now you have a different situation. Here you have a binary
mixture in equilibrium with the airspace above it. Each liquid exerts a
partial pressure that is a function of the mole fraction of each liquid, also
according to Raolt's law. I recommend you find what's called a T-XY diagram
for the mixture you select -- it tells you the vapor equilibrium
concentration at a given temperature and liquid composition. You can then
repeat the procedure above -- but if you run the experiment for a long
enough period of time, you are going to have the mole fractions of each
liquid changing along with the level of the fluid -- again, begging for a
numerical solution.
Both of these discussions neglect convection, and assume well-mixed liquids.
Neither may be true -- you could have depletion of more volatile substances
at the surface of the liquid (especially in the case of the sugar solution,
as its viscosity would inhibit mixing), and you could have convection
currents in the vapor phase causing higher mass transfer than diffusion
alone would predict. Especially if you are elevating temperature, these
convection currents would be more likely to develop. Also, as the liquid
evaporates, it carries energy with it. Water has a high heat of
vaporization, which means water that is left out to evaporate will be cooler
than ambient. This will affect the partial pressure at its surface as well
as diffusion.
So in summary, you have several conditions that are changing: the mole
fraction of each component, the level of the liquid, the temperature --
which affects the partial pressure of each component and the diffusion
coefficients of each component, and keeping in mind that the temperature at
the surface of the liquid will be lower than ambient. Very complicated, and
I'm not sure if all this complication is even useful, so I'm going to start
from the other side -- the experiment.
Let us start with pure water in a partially full cylindrical tube. Let us try
to model its evaporation. At the surface we have a partial pressure (a
function of temperature), and we assume ambient humidity at the exit of the
tube. Given Fick's law, we can find flux and with the cross sectional area,
mass transfer. You can then calculate an instantaneous volume change. As the
volume level drops, the distance through which evaporation occurs increases,
which decreases the mass transfer. So as the water evaporates, the rate of
evaporation slows down. Neglect everything else for the time being. Does
this data match your experiments where you leave out the tube for some
period of time? If yes, then you are done. If not, add in other variables
until you get the precision you desire. Go ahead and change your temperature
and humidity with this single-component case too if you like.
Now, add in your second component and repeat the above experiments. Again,
keep adding in variables until you get the desired precision. Which
variables? Try the ones that are easiest to measure. If you have a
temperature probe, stick it in the water. Compare it to the air temp. Have a
video camera? Record the evaporation over time, and gather very precise data
on the liquid level. These are just two examples.
Alright, I guess this has gotten long-winded enough. I hope I have given you
some food for thought in designing your experiment.
Burr Zimmerman
I agree with Vince Calder that this is a very difficult experiment to control very
precisely. However, this is an opportunity to make a try, see how it works,
and refine from there. Hey, this is what experimental science is all about!
Vince Calder alludes to a very tricky variable called 'latent heat of vaporization'
-- which is the energy it takes to vaporize a substance that is already at
its vaporization temperature (there is also a latent heat of fusion, the
energy to go from liquid to solid) to vaporize. This is relevant to this
discussion because the water at the surface will be *colder* than ambient
because as it evaporates, it cools (the vaporized water takes heat with it).
If you have ever drank chilled water from a terra cotta water jug, you have seen
this effect in action. The porous terra cotta allows some water to pass
though it and evaporate, chilling the remaining water. I heard somewhere
that Indians (from India -- not Native Americans) actually used the same
effect to make ice. It is the same reason damp clothes feel 'cool' -- water
is evaporating. Anyway, this is a very difficult thing to control -- or to
measure -- so it is something to keep in mind.
The reason this effect is important for you is that the vapor pressure of
the liquid is a function of temperature, and if your liquid is not
well-mixed, you will get less evaporation than you think because the surface
is cooler than ambient. Even in a temperature-controlled chamber, the
temperature at the liquid surface may be quite different than you expect.
Another variable you discussed was a mixture of two substances. You did not
say whether you meant two liquids, or one liquid and one solid. However,
both will affect evaporation rates. I will not go into too much detail here --
but please reply if you require more info -- but I would point you to
'Raolt's Law' to begin. As the more volatile liquid evaporates the remaining
substance will concentrate, and the rate of evaporation will change.
Calculating accurately with so many changing variables is very difficult to
do.
I can offer you a suggestion -- but please adjust for your own
circumstances. The goal is to create a setup that makes calculations easier.
With simpler calculations, you have an easier time matching up your results
with your calculations. I would use a very tall, narrow vessel (like a clear
plastic tube) with an aspect ratio of 10 or higher -- depends on the size of
your chamber. I would seal the bottom, fill with a small amount of water (~1
diameter in depth). I would put a magnetic mixing bar in the bottom to keep
the water well mixed -- make sure you do not turn up so high that it forms a
vortex. I would use deionized water to reduce the effect of solutes on the
evaporation rate.
The reason you want the water well mixed is to ensure it
is at the same temperature throughout instead of developing the cold surface
as previously mentioned -- as best as you can at least. The mixing bar will
add some heat as well, which may counteract the enthalpy of vaporization.
The long head space is to set up an unmixed diffusion column, which will
minimize the effect of wind or drafts. You can assume that the concentration
of water vapor at the liquid surface is the partial pressure of water at the
ambient temperature. You can assume the partial pressure at the top is the
ambient humidity. You can then calculate diffusion of water in air through
the length of the column (adjusting for the fact the column gets larger as
the water evaporates). Using Fick's law, you can calculate the flux of water
through the column, and then compare to experimental results. If you have
some probe thermistors, you could also mount one each near the surface and
near the bottom and measure water temperature directly instead of assuming
it is at ambient (be sure to calibrate each). If you need help with any of
the calculations, please let me know. Since you said you are comfortable with
differential equations, you may be able to do these without any problem.
As for different
runs, I would *not* hold one constant while you vary the other. It is better
to pick a set of conditions that span all variables' ranges. This is known
as 'design of experiments' -- you get higher quality data this way -- and
can do so running fewer experiments. For instance, if you have temperatures
A, B, and C and humidities 1, 2, and 3, you could run 9 experiments -- A1 A2
A3 B1 B2 B3 C1 C2 C3. That is known as a full factorial. You can run subsets
of that too depending on your time budget. If you picked one level to hold
for each variable, you might pick 2 and B -- and run A2, B2, and C2, and
then B1 and B3. That is five experiments. However, for 5 experiments, you may
get better data running A1, C3, A3, C1, and B2. If the effects at the
extremes are more important to know, this may be better. As you get into
more complex experiments, with many more variables, techniques such as this
become critical to get the most out of your experiments. If you want help
with this aspect, let me know, but I would suggest you Google things like
experimental design and full factorial / fractional factorial, Box Behnken,
central composite, Placket-Burman. Experimental design is a HUGE field
though, so be warned! (it is really interesting and useful too though). I
just came across this site which seems good at first glance:
http://www.itl.nist.gov/div898/handbook/pri/pri.htm, so maybe look at it.
Hope this helps,
Burr Zimmerman
Click here to return to the Engineering Archives
| |
Update: June 2012
|
|
