Mark Borgerson wrote:
In article ,
says...
Mark Borgerson wrote:
In article ,
says...
snip
snip
How do you get 33' as 1/2 of the diffusion path.
A quick thumbnail guesstimation at where equilibrium would likely be
reached. I didn't take the time to calculate the exact heights.
I think there will be
about 33 feet of water in the column on each side
Then I think you would be wrong, unless your columns are significantly
longer than that, probably more like 50+ feet.
---to provide the
weigth that pulls the pressure down. That would leave only about
7 feet of water vapor path on each side of the column.
There is no vacuum to hold the water up - the vacuum is what you are
trying to *create*. The water columns will drop until there is an
equilibrium point reached between the external atmospheric pressure, the
height (weight as you state) of the water column, and the pressure in
the headspace (the U-tube). The water columns *must* retreat, or the
headspace stays at atmospheric pressure. If the tubes are long enough,
and the initial column heights are high enough, then when you reach
equilibrium, you'll have close to a vacuum and close to 33' water column
heights. And a lot more empty headspace than you started with.
I see the problem. I am assuming that you completely fill a 40 foot
tube with water using a pump capable of providing about 16-20 PSIG.
That fills the tube completely with water--at which point you
close the tube (with a one-way valve).
Uhmm, a manual valve is a manual valve. A "one-way" valve is a
checkvalve, and you wouldn't need to close it.
When you release the
pressure at the bottom end, the water falls to the point where the
weight of the water column is one atm (about 14.7PSIA) minus the
vapor pressure of water at 20deg C. The vapor pressure of water
at 20C is about 17.5mmHg, or about 2.3% of the 760mmHg standard
atmosphere.
Since a mercury has a density 13.6, the column of water will
be 13.6 * (760- 17.6)mm high. That's 10.1m high, or
about 33.12 feet high. In a 40-foot tube, that would leave
about 7 feet of water vapor at the top of the tube and 33 feet
of water below the vapor.
Ahh, no. See below...
Use the ideal gas law: PV=nRT
For our evacuation purposes, nRT is a constant (#moles is constant, R
doesn't change, and assume constant temperature), so if you start with a
volume of 1 liter, and a pressure of 14.7 psia, and you want to reduce
that pressure to 1.47psia, then you need a 10-fold volume increase. You
want to reduce it to 0.147psia? then you need a 100-fold initial-volume
increase.
What is the 1 liter to which you refer?
It is an example, for illustration purposes. It's the headspace (i.e.
the amount of volume *not* filled with water, prior to closing the valve
and letting the water columns 'fall').
The point is, whatever your starting headspace volume is, to get
anywhere near a vacuum, the *VOLUME* of the headspace must increase 100
fold. That is the relevance of the ideal gas law. If you start with a
100ml headspace, then to get a decent vacuum, the water columns have to
drop to a point where the headspace is 10L. *AT THAT POINT* you have
sufficient vacuum to support a water column of around 30'. But the
columns have dropped significantly to achieve that vacuum, and thus the
columns must be much higher, as must the initial water column height.
The *only* way the headspace volume increases is if the water columns
drop significantly, and the only way significant vacuum is created is if
the sealed volume increases tremendously.
This is not a closed system---the tube is open to a reservoir at
atmospheric pressure at the bottom.
I'm assuming that you start with a head space (or initial volume)
of zero. You then simply have to evaporate enough water to fill
the top of the tube with water vapor to the point where vapor
pressure + water weight = 1ATM.
I'm not sure that 'diffusion' is the proper term for the motion
of the water vapor. After all, the heat engine is providing
water vapor on one side and condensing it on the other---so there
is a net mass flow and probably a small pressure differential to
move the vapor.
Well, diffusion is the primary mechanism. What happens when your 'heat
engine' creates water vapor? It doesn't just immediately condense on
the other side. It creates pressure on the heating side, which does two
things. One, it drives both the water columns *downward*, and it raises
the boiling point on the seawater side (it does, however, make
condensation on the fresh side more efficient as well). You can't look
at this as a static system where the pressure stays the same or the
column heights stay the same. It's a dynamic system, and will reach an
equilibrium point with the columns much lower than the initial starting
point, and the headspace pressure much higher.
Well, not much higher----only about 17.5 mmHG higher. But that IS
a lot higher than zero! ;-)
No, a lot higher. You're confusing vapor pressure with the "steam"
pressure during distillation. Huge difference. Vapor pressure is the
countervailing force (fighting condensation as it were) on the FRESH
water side of the system (and the seawater side). Vapor pressure in
both columns will be about the same, so you have to boil the seawater
column to get any significant vapor transfer. This results in *much*
higher pressure, which lowers the columns and increases the vapor path
and....
And don't forget, there will also be significant evaporation (due to low
partial pressures) on the freshwater side that will be in equilibrium
with (and in opposition to) the condensation process. It's not as simple
a system as it seems.
That's why this system *will* work, but it must work very slowly.
Still (pun intended), you need a lot of heat to provide the energy
to evaporate the water or it will soon cool to the point where
its vapor pressure is reduced and the process slows drastically.
My 'guess' would be that the system would end up operating around
4-5psia when equilibrium is reached, which would require a temp of about
60°C (140°F) to maintain boiling.
AHA!, you're assuming a much higher operating temperature than me.
Yes, because you're mistaking the amount of "vacuum" you'll have
available when the system reaches equilibrium.
I was assuming something on the order of 20 to 25C.
Then you are assuming an almost perfect vacuum, which can't happen since
the boiling Must significantly raise the headspace pressure.
You're going
to have to add to your energy budget the heat necessary to raise
the water temperature from 20C to 60C, then.
If the equilibrium pressure is really 1/3ATM, then there will be
about 20 feet of water in the 40-foot tube and 20 feet of vapor.
If you're going to work at those temperatures and pressures, you
probably need only a 22-foot tube.
You need to get back to the gas law to see where this error lies. You
have to *create* the vacuum. That requires a HUGE increase in volume
for whatever the initial headspace is. For this to happen you need a
much longer tube to start with.
Keith Hughes