From: glhurst@onr.com (Gerald L. Hurst)
Newsgroups: sci.environment,sci.engr,sci.engr.chem,sci.chem,sci.misc,
	sci.research
Subject: Re: Fresh water from sea water - new economical process
Date: 10 Jan 1996 07:39:39 GMT

In article <4csu4a$aa@ixnews8.ix.netcom.com>, ferrad@ix.netcom.com(Adrian Ferramosca )
says:

>He won't give up, will he ...
>
>In <4cqcs9$bbo@geraldo.cc.utexas.edu> glhurst@onr.com (Gerald L. Hurst) writes:
>>
>>In article <4cq1aa$93a@cloner3.netcom.com>, ferrad@ix.netcom.com(Adrian
>Ferramosca ) says:
>>>
>>>I agree, if you're talking steady state.  But it does occur, and the
>>>only explanation I can put forward is that the kinetics of the process
>>>dictate that the energy of the system gets temporarily shifted towards
>>>the departing gas phase.  Eventually, if left to reach steady state, it
>>>will distribute evenly, and return to the 'can' and cause the ice to
>>>melt.
>>
>>You're not going to beat thermodynamics with a kinetic stick. If
>>"the the energy of the system gets temporarily shifted towards the
>>departing gas phase," then you have harnessed Maxwell's demon and
>>perpetual motion is just around the corner. All you have to do is
>>let that energy segregated in the escaping gas flow through a heat
>>engine on it's way back to the can.
>
>No, I'm not saying that you get it all for free, but unsteady state
>effects can shift equilibrium.  Nothing ever occurs at steady state.
>If you can harness the articial separation of energy, you can get in
>this case the required product, and do whatever you want with the
>higher energy stream.  You will obviously not get back anywhere near
>the energy put in, but the idea here is to keep the low enthalpy
>product, we don't need the high enthalpy one!  Let the product warm
>itself up of its own accord.

Sorry you think I'm being stubborn. It's just that I seem to have
persistent doubts about spontaneous processes which result in a
net decrease in entropy, even if they occur only for a few minutes.

Let's look at your process in its simplest form. You have a can
of carbonated salt water similar to a salty diet Coke. You chill
it down to its freezing point at say -2 degC.  Let's assume we're
dealing with one liter of solution with a rather high (compared
to Coke) equilibrium CO2 pressure of 4 atmospheres.  Let's
further make the (incorrect) assumption that when we pop the lid,
the CO2 is going to bubble off instantly leaving us with a residual
CO2 pressure of one atmosphere.

The solubility of CO2 in a liter of water is about is about  1.85
liters/atmosphere and a one-molar slt concentration will vave
little effect on this value.  If the CO2 pressure drops from 4 atm to
1 atm absolute, the Amount of CO2 released  at about -2 deg will
be about

1,85liter/atm*3atm*1mole/22.2liter = 0.25 moles CO2

The enthalpy change for CO2(Aq) -->CO2(g) is about -4.85 Kcal/mole

hence the volatilizing CO2 should absorb about 4.85Kcal/mole*0.25mole
= 1.25 kcal.

Water requires the loss of about 0.080 Kcal/g to freeze, hence the 
amount of ice created by the escaping CO2 should be about

1.25Kcal/0.080Kcal/g =  15.6 g or about 1.6% of the original water.

Now let's look at the contribution of water vapor in removing heat 
from the solution. The vapor pressure of water at -2 degC is about 
4mm. If  the vapor entrained by the escaping CO2 reaches saturation 
then the 0.25 moles of the former should carry with it approximately 
4/760*0.25 = 1.3exp-3 moles of water vapor. Multiplying this value 
by the ratio of the heat of fusion to the heat of vaporization yields
1.3exp-3moles*540cal/g/80cal/g = 8.8exp-3 moles of ice or about 
0.16grams, an insignificant amount even compared with the merely 
small (15.6 grams) amount produced by the CO2.

At this stage we see that the expected thermodynamic changes can only
account for enough energy transport to freeze about 1.6 percent of 
the water.  Clearly an economical process will require freezing at 
least the bulk of the fluid, let us say certainly no less than 50 
percent or in the current example no less about 500 grams.

If the carbon dioxide and water vapor could leave the solution at 
a higher temperature than the liquid we could theorize that by 
some quirk of kinetics the warm gas could carry away the necessary 
amount of heat. This is only a pipe dream though because in order 
for it to happen the net sum of q/T for the liquid and gas phases 
would have to be negative, that is the entropy of the liquid would 
go down more than the entropy of the gas went up.  Unless we are 
to break the second law, it is a no go.

The alternative, which has been suggested by the inventor, is to 
have more energy removed by water vapor (since we have much more 
water than CO2). If the water is to carry off more heat, it has 
to be vaporized to a partial pressure greater than its equilibrium 
vapor pressure of 4mm and maintained at -2 deg (or slightly lower 
as the solution becomes more concentrated) at least until it is no 
longer in thermal contact with the liquid.  In other words we have 
to produce a mix of CO2 and water vapor at -2 deg or less which is 
in a metastable supercooled state with respect to the water vapor.

The amount of water vapor in the CO2 required to carry away the
latent heat of fusion of the freezing water  is readily calculated 
by multiplying the 500 g of ice desired by the ratio of the heat 
of fusion to the heat of vaporization

500g*80cal/g/540cal/g = 74 g of water vapor  or 74g/18g/mole = 4.1 
moles.

Our product gas the consists of 4.1 moles of water vapor mixed with
only 0.25 moles of  CO2, all at -2 degC.

Ignoring the relatively minor CO2 content, we have an astounding 91
liters of ice cold super-cooled steam - the same as if we had been
able to cool 125 liters of regular steam through 102-103 degC without
it condensing.

Imagine now that as expected this cold steam is unstable and reverts
to its natural state of water and hot steam.  As the cold steam 
condenses it forms water which is heated at the rate of  18 
cal/deg/mole by the vapor which is surrendering about 9.72 kcal/mole. 
The process goes on until 

[(4.1-N)*8.2cal/deg+N*1cal/deg]*102deg=9720*Ncal, 

that is until the amount of heat provided by N condensing moles out 
of 4.1 provides enough heat to warm up both the condensed (N) and
uncondensed (4.1-N ) moles of water to 100degC. The only new data 
here are the heat capacities of liquid water (18cal/deg/mole) and 
water vapor (Cp = 8.2 cal/deg/mole)

If I did the math right, N (the amount of water condensing) is only 
0.33 moles so the final equilibrium mixture consists of  O.33 moles 
(6 grams) of boiling hot water and 4.1-0.33 = 3.8 moles (68 grams, 
116 liters) of live steam.

These calculations are only approximate, but they lead me to believe
that if the proposed desalination process works as described, the
next cold Coke you open may blister your face with a blast of steam.

If you find any mistakes in the math, be advised that I was just
testing you.

Jerry (Ico)