alibaba - 29-7-2007 at 23:19
hi, newbie here wants to ask a question:
i wanted to know what would happen if i have an enclosed plastic canister (bottle/tube) 40 % full of pure Methylene Chloride. what would be the effect
if i dip this tube in halfway in warm water with temerature56 C( 39.8C Methylene Chloride's boiling point):
which of the following will occur:
1. the liquid will boil and vaporise to the upper part of the tube where it is cooler and condense?
2. the boiling liquid will build up alot of pressure and blow open my bottle?
3. something else?
this is for a project which i intend to make but i wanted to ask those chemists out there who are familiar before i do something stupid?
thank you in advance
smuv - 30-7-2007 at 00:52
You can calculate the vapor pressure of methylene chloride at that temperature, and determine if your container can hold the calculated pressure.
This will give a hint as to what will happen.
In terms of plastic containers...be careful as methylene chloride can soften/dissolve many plastics.
What are you planning to do? You have peaked my interest.
not_important - 30-7-2007 at 01:18
It depends. If the heat input exceeds the heat loses, you'll condense less vapour than you generate and pressure will rise. What happens next depends
on a lot of things. As the pressure rises the boiling point goes up. If the heat source is at or very close to the 1 atmosphere(*) boiling point of
DCM, the pressure will rise to slightly above one atmosphere; there are tables and charts around that will tell you the vapour pressure at various
temperatures that you can use to figure out how much the pressure will rise.
(*) - more properly the pressure at the time of sealing the system, plus pressure generated by the expansion of the gases as the system gets warmer.
Condensation removes heat from the system, if it condenses fast than it boils - more heat is lost than is put into the system - then the pressure
stabilises at very slightly above the vapour pressure of DMC at the temperature of the heat source. If condensation can't keep up, the pressure rises
until it matches that vapour pressure of DMC at the heat source's temperature, allowing for the effect of condensation, or the container bursts.
If your bottle is plastic, the failure of it is more likely as most plastics quickly lose strength with increasing temperature, and DMC is a decent
solvent or softening agent for many plastics.
Knowing the temperature of the heat source, the vapour pressure of DMC at that temperature, and the bursting strength of the container will get you
most of the way to the answer. Again, if there is more heat lost through condensation than is put in from the heat source, the pressure will rise
very little - mostly from the PV = nRT expansion of the gases in the system when it was sealed. If you are counting on passive heat loses to the
surrounding air, it is likely that condensation will not keep up with the heat input.
evil_lurker - 30-7-2007 at 02:18
More than likely the plastic container would get melted pretty quick.
But if it was glass, and the container was sealed, you would observe a pressure rise followed by condensation on the glass.
changes
alibaba - 30-7-2007 at 23:09
ok, supposedly i change my container with a glass tube. fill it up say 30% with the liquid and dip it halfway of its height into hot water exceeding
the boiling point? would this lead to the liquid vapourizing in the upper part of the glass tube which is cooler since the bottom half is in thw hot
water? if this tube was not straight but a curved shape with a 90degree angle halfway - see the improvised drawing below
---------------
. ------------
...
...
...
...
will this lead to the liquid from the vertical part vapourising and condensing in the upper flat/horizontal/ part of the glass tube . i hope i have
clearly outlined my question.
thank you
woelen - 30-7-2007 at 23:49
Yes, you may expect that some liquid will vaporize and form droplets on the horizontal part of the glass tube, assuming of course, that at least
initially, this part of the glass tube is cooler than the part, which is immersed in the hot water.
Actually, this makes a nice demonstration (I have done a similar thing myself with bromine, which forms beautiful condensing droplets on cooler places
of glass). If the tube is tilted a little bit, such that the horizontal part is directed somewhat downwards (but not touching the hot water), then you
even may see a drop of liquid collecting in the second "leg" of the tube.
alibaba - 31-7-2007 at 00:53
actually i was hoping to achieve an effect similar to the "drinking bird" where the liquid from one end of the tube will move to the other where it is
cooler. what changes must be made to the configuration to achieve such an effect? logically the entire liquid from the submerged part of the tube
should evaportae and condense in the other horizontal/even downward/ part when held long enough in the heated water?
suggestions?
thank you
aspiration
alibaba - 31-7-2007 at 01:02
would it help to partially evacuate the air to help the liquid move up?
woelen - 31-7-2007 at 01:45
I doubt that all of the liquid will go up in a reasonable time. While the liquid condenses, the tube will heat up. Also, heat will be transferred
from the hot water surface to the piece of the tube, which is above the surface. Another effect is increase of vapor pressure in the tube, which will
make evaporation harder. So, I expect that some liquid will go up, but also quite some liquid will remain in the vertical leg of the tube.
You only can get (almost) all liquid in the upper part, if the temperature difference is maintained. That will require active cooling of the higher
part, and/or local heating of the lower part, immersed in the water.
alibaba - 31-7-2007 at 01:54
since th temperature of the water in which the tube is submerges is 50-60C, this will solve the issue of maintaining a temp difference with the room
temp of about 30c- or so i think.
could this be the solution?
thank you
woelen - 31-7-2007 at 02:33
I doubt that this solves the issue of maintaining temperature difference. The part of the tube, in which liquid condenses will become hot as well, due
to warm vapor going upwards, and due to the vapor to liquid transition. Quite a lot of heat is produced in this process (exactly the same amount of
heat is produced, as what was needed to evaporate the liquid).
I really think that you will need some form of active cooling in order to have all liquid go upwards, otherwise I only expect part of the liquid to go
upwards.
not_important - 31-7-2007 at 05:30
The "drinking bird" does not depend on moving the fluid by evaporation, but by using the vapour pressure difference to force the fluid through the
connecting tube tube, changing the center of gravity. When the the bird 'drinks', the water moistens it's head, resulting in slight cooling through
evaporation. You can make the bird 'drink' without water by positioning a small lamp so as to _slightly_ warm part of the lower body.
There's very little transfer of fluid via vapour, it's the pressure differential that counts. It's much easy to accomplish this than to distill a
large percentage of the liquid from one buld to the other.
Let's see, OK, here's sites on the D.B.
historic
http://www.backstreet.demon.co.uk/oddstuff/drinkingbirds/dri...
scientific
http://nicholnl.wcp.muohio.edu/DingosBreakfastClub/DippyBird...
Even with DCM you'll need to transfer a lot of heat to move enough of it to change the center of gravity enough, so both active heating and cooling
are likely to be needed. Making the "head" from metal with heatsink fins could help reduce the need to active cooling; getting a shape that will
properly retain liquid until it starts to tip may be a problem. If you're thinking of driving the liquid back and forth through distillation, rather
than through gravity return, then the design needs to be symmetrical and becomes more difficult.
[Edited on 31-7-2007 by not_important]