Diffusion Machines

Some time ago i found a picture of a "Single Membrane Diffusion Machine":

Since the early 1700's this is also known as Daniel Bernoulli's "fluid energy" perpetual motion machine, described in the book "Perpetual Motion, the History of an Obsession" by Arthur W. J. G. Ord-Hume (copyright 1977 by George Allen Unwin; published by Barnes & Noble Press, Inc.).
It is also known as the "Octave Levenspiel's fountain" for ocean water / fresh water in the journal Scientific American, December 1971, page 101 and Science, 183, p. 157-160 (1974), but without a calculation and with the bug, that Levenspiel says that a depth of about 700 feet would be enough - calculation shows that a minimal depth of about 10 kilometer is needed!

The principle behind it is: The difference in density between the water and the solution causes a difference in hydrostatic pressure at the bottom of the solution column which is smaller than the difference in osmotic pressure. Thus, as shown, the right tube with the pure solvent will overflow and drip into the left tube with the solution. Hence the flow results.
In stationary mode, with a constant net flow, this flow is maintained by diffusion so that the concentration gradient in the left column is time-independent and without sedimentation. The diffusion moves the soluted particles in opposition to the flow of the solvent and the gravity so that there is no net movement of the soluted particles.

This is an interesting arrangement because when it is possible it converts ambient heat into potential energy, which can be used e. g. for power generation. But is it really possible and simple to implement?

From sub-marine springs it's known that the difference in density between the freshwater and the seawater can be used as a pressure gain, for pumping the fresh water from the see floor. Companies like Nymphea Water do use it to pump fresh water from sub-marine springs to the ocean surface. So the difference in hydrostatic pressure is simple and in use.

Because of the osmotic pressure a diffusion machine can only work when the osmotic pressure is smaller then the hydrostatic pressure difference which is caused by the density difference of the solution relative to the pure water; the solution must have a density higher than the pure water. If this is not fulfilled, the left side would always be higher and Δh would be negative.
It is clear that when it works, it's not easy to implement, because usually h_solution is much higher than h_H20 because the osmotic pressure, which increases h_solution, is usually many times stronger than the hydrostatic pressure difference. An example is ocean water: The osmotic pressure of ocean water is about 27 atm , but the density of ocean water is only 1.027 times the density of pure water; so the osmotic pressure is usually, in a laboratory, several thousand times higher than the hydrostatic pressure difference caused by the additional 2.7 % density !
This is the reason why in physics books it is always neglected that the solution has another density than the pure solvent, even when the barometric formula is used for calculating the boiling-point elevation based on the osmotic pressure, e. g. at http://www.sparknotes.com/chemistry/solutions/colligative/section1.html and http://www.chemprofessor.com/colligative.htm.
Another reason for this can be found in this book in chapter 2.2: For the Gibbs postulate, description of systems at thermal equilibrium by canonical ensembles and for all known proofs of the Second law of thermodynamics , it is necessary to neglect long-range forces like gravity, because with long-range forces a system can not be separated into independent subsystems:
If we take a system s and do split it into two patial systems s₁ and s₂, e. g. by impermeable walls, for the most simple systems the energy is an extensive quantity: E_s = E_s1+E_s2 and we can therefore use the canonical ensemble to describe the three systems. But when we take into account long-range forces, e. g. gravitation, we have a coupling term of E_s12 = -G*m₁*m₂/r₁₂ , so we have E_s = E_s1+E_s2+E_s12 . This means that quantities like the energies and entropies in such systems are coupled and therefore no extensive quantities in the strict sense; the Gibbs postulate is not fulfilled. Usually this can be neglected because we have E_s12 << E_s1 and E_s12 << E_s2 due to of the small gravitaional constant, shielding of electrical fields and the short-range of the magnetic fields. But in systems where the coupling can not be neglected, e. g. diffusion machines, things are not so simple; they can not be described by canonical ensembles.

Heights calculations

To simplify calculations only low solute concentrations will be used and calculations are done for no net flow mode, which means the H₂O flow over the bridge is stopped. The stationary mode with a constant H₂O flow will be calculated later (below) and is simple (linear) fluid dynamics .

For the solvent the connected two columns are pascal's vases, because the membrane is penetrable for the solvent. So on both sides of the membrane the pressure of the solvent is the same. In the left column, the solution, the pressure is decreased by the osmotic pressure and increased by the additional density.

The osmotic pressure, which increases h_solution, is given by the van't Hoff law :

P_osmosis = n' *k *T/V

where n' is the number of solved particles, V the volume of the solution (left column) and k the Boltzmann constant.

The hydrostatic pressure difference of the two columns at the bottom, which is caused by the density difference ρ_solution -ρ_H20, decreases h_solution and the value of the hydrostatic pressure difference is (at a low vapor/gas density)

ΔP_hydro = g * ( h_solution *ρ_solution - h_H20 *ρ_H20)

At this point a qualitative analysis shows that it is clear that the diffusion machine works (if the right components are used): The osmotic pressure is independent from the heights, while the hydrostatic pressure difference increases with the heights linear. This means if the density of the solution is higher than the density of pure water/solvent, there is always a critical height, h_crit, where the pressures (P_osmosis, ΔP_hydro) do compensate each other exactly and above that height, h_solution is smaller than h_H20:

P_osmosis = ΔP_hydro
P_osmosis = g *h_crit *( ρ_solution - ρ_H20) = g *h_crit *Δρ
=>
h_crit = P_osmosis / (g *Δρ) .

For ocean water / pure water h_crit is about 10 kilometer, so ocean water and similar solutions are a bad choice for a diffusion machine. Another point is, that for such a high pressure of about 1000 Bar we have to take into account the different compressibility of the pure and the ocean water. Therefore for a diffusion machine a lower osmotic pressure is needed and this means bigger soluted particles, e. g. microspheres or macromolecules.
An important point is when we use low solute solutions, the density difference Δρ is proportional to the concentration, so as the osmotic pressure P_osmosis. Therefore for low solute solutions h_crit does not depend on the concentration n'.

First Example

To avoid high critical heights of several kilometer, i'm using microspheres instead of single atom or molecule solutes: Polystyrene spheres of diameter d = 2*r= 20 nm, with a density of 1.05 g/ml, water as solvent, a net mass per microsphere of
m = (ρ_sphere -ρ_solvent) *(4/3) *π *r³
= 2.1e-25 kg
and the microspheres in a concentration of one sphere per cube of edge length equal to two diameters, which means a volume concentration of c_V = (4/3) *π *r³ / (4*r)³ = 0.0654 = 6.54 %. These microspheres are availible e. g. from www.tedpella.com and www.emsdiasum.com .

An important point is that the microspheres are not to too big, because of the barometric formula :

c'(h) = c'₀ * e^-m*g*h/(k*T)

with m as the net weight per particle: Net density *(4/3) *π *r³, g the gravitational acceleration, h the height of the particles in the solution, k the boltzmann constant, the volume concentration c' and T the temperature. At 300 K, k*T is 4.142e-21 J.
The half density height, where we have N = 0.5 *N₀, is at height h_1/2 = -ln(0.5) *k*T/(m*g) .

That's why i'm using 20 nm Polystyrene spheres: The half density height is h_1/2 = 0.693 * 4.142e-21 / (2.1e-25 *9.81) m = 1393 m, which means that their density in the water is nearly constant in columns of a few tens meters; the diffusion is able to move the microspheres against gravity in columns of a few tens meters.
It's important that the barometric formula increases the osmotic pressure (at the membrane) but that this effect is less than 1 % (osmosis pressure increase). Therefore it can be neglected in this example.

The density of solute particles in my example is one sphere per cube with an edge length of two diameters:

n'/V = 1/(40 nm)³
= 1.563e22 m^-3

So the osmotic pressure is:

P_osmosis = 1.563e22 * 4.142e-21 Pa
= 64.7 Pa

which is equivalent to about 6.5 mm water height (mmH₂O).

With water of density 1.000 g/ml the solution has a density of (1 -c_V) *ρ_H20 + c_V * ρ_Polystyrene = 1.00327 g/ml, so with a column of 10 m water at the right side, the different compressibilities of the solution and the pure water are doubtless neglectable, and we get h_H2O *ρ_H20/ρ_solution = 10 m/1.00327 = 9.967 m water with the microspheres at the left side.
Because of the osmotic pressure, h_solution gets increased by 6.5 mm, so the real height difference Δh is (33 -6.5) mm = 26.5 mm. The gravity induced column height difference is 5 times higher than the osmotic pressure induced height difference!
The 26.5 mm difference is not much, and 10 m columns are big, but it's easy to see and enough for a clear proof of concept.
When we have a water flow over the bridge, this causes a concentration gradient in the left column, but that does not influence the hydrostatic pressure difference and even when the osmotic pressure at the membrane would be increased by a solvent flow by 100 %, the osmotic pressure remains less than half the hydrostatic pressure difference; the hydrostatic pressure difference always dominates.
But this example is hard to implement: To avoid capillary action a column width of several mm should be used and the 6.54 % solution of 20 nm spheres of Polystyrene for this column does cost some thousand euros.

Tuning

In the example above the effect of only 26.5 mm height difference is not great, but because of the high half density height of the 20 nm microspheres, you can use microspheres with a density difference (to the solvent) ten times higher which gives a ten times higher hydrostatic pressure difference, thus 32 cm height difference and that's enough for a water mill!
An example is PTFE (Teflon) instead of polystyrene: Polystyrene has a density difference to water of 0.05 g/ml, while PTFE has about 1.2 g/ml!

A further way of tuning is using a higher acceleration g, because you can use the diffusion machine e. g. in a centrifuge and h_crit is proportional to the reciprocal value of g.

A power bottleneck is the semipermeable membrane and slow diffusion, so the membrane should be made wider than shown in the figure. A way to reduce this bottleneck would be a high ambient temperature, because the viscosity is smaller at higher temperatures and diffusion is faster at higher temperatures. Another way to reduce the bottleneck semipermeable membrane is using low viscosity fluids or superfluids.

Tuning example

A simple tuning example is using hexane instead of water, because hexane does not dissolve polystyrene and has lower viscosity and density.
At 25°C the density of hexane is 0.659 g/ml and the (dynamic) viscosity is only 0.320 mPas, while water has a (dynamic) viscosity of 0.891 mPas.
Because hexane has a smaller density than water, the density difference of the polystyrene microspheres to the solvent is increased from 0.05 g/ml to 0.409 g/ml.
So with hexane instead of water the viscosity is about three times smaller and the density difference to the polystyrene microspheres is about eight times higher.

Detailed Theory

Diffusion flow calculation

The diffusion flow in the left column is the first bottleneck and equal to the H₂O flow in stationary mode.
This flow through the horizontal area A is given by Fick's first law :

Q_D = -A *D*dc'/dh

with the diffusion constant D of the microspheres, given by the Einstein Relation :

D = k *T/(6 *π *η *r)

with the dynamic viscosity η = ν * ρ, the density ρ and kinematic viscosity ν.
From this formula we see that the temperature should be high and viscosity and radius should be small.
Another point is the density of the spheres, which should be high, to get a small critical height.

In stationary mode, in the solution we have a vertical solute flux upwards, caused by diffusion, and an opposed vertical flux downwards that is caused by the solvent flow. Because of the stationarity, these two fluxes do have the same amount of flux J and do compensate each other:

J = -D*dc'/dh = v *c'

with the solvent velocity v. Because of the stationarity, the net solute flow is zero and that's the reason why we have a net solvent flow.
So for the velocity of the solvent flux, which is equal to the diffusion velocity, the last equation gives us:

v = -D/c' *dc'/dh

which is a concentration dependent velocity, because the velocity v is proportional to the relative concentration gradient dc'/(c'*dh) .
The most important result of this is that the concentration gradient dc'/dh can't be constant, because with a linear concentration gradient the upper low-density half would be faster than the lower high-density half but for stationarity they must have the same speed.
Therefore the next step is to find a gradient dc'(h)/dh, which gives a constant velocity v.
For the first approximation of the flow we neglect the volume of the soluted particles because we assume low solute concentrations. This approximation is the exact result for a left column, which narrows from the bottom to the top a little, so that the velocity of the solvent is really constant.
The constant velocity gives us the vertical concentration c'(h):

v = -D/c'(h) *dc'(h)/dh = const
=>
dc'(h)/dh = -v/D *c'(h)
=>
c'(h) = c'₀ *e^{-v/D *h}

The concentration at the bottom, c'₀, is needed for calculating the osmotic pressure at the membrane and can be calculated by integration from the bottom to the surface. Because the integral over c'(h) from the bottom to the top, divided by the height h_solution, is the average concentration c'_avg, we can calculate c'₀ :

c'_avg = c'₀ *-D/v *(e^{-v/D *h_solution} -1)/h_solution
=>
c'₀ = c'_avg *h_solution *v/(D *(1 -e^{-v/D *h_solution}))

This gives us e. g. the flow velocity v when we start with a given c'₀/c'_avg, h_solution, and D.
Because the diffusion velocity v is usually small, especially for microspheres and macromelecules, we get a good first approximation of by using "only" the first three summands of the taylor series of the exponential function:

c'_avg = c'₀ *-D/v *(1 -v/D *h_solution + 0.5*(v/D *h_solution)² -1)/h_solution
=>
c'_avg/c'₀ = -D/v *( -v/D *h_solution + 0.5*(v/D *h_solution)²)/h_solution
= 1 -0.5 *v/D *h_solution
=>
v = 2 *D/h_solution *(1 - c'_avg/c'₀) .

For calculating the membrane pressure difference, caused by the flow, we need the solvent flow Q instead of the velocity v:

Q = dV/dt = A *v

with dV as the volume of the liquid, poured in the time dt, and A as the cross sectional area of the left column, that is filled by the solvent.

If the soluted particles are charged with the same sign, e. g. polystyrene microspheres in water, than we have an one another repulsion of the soluted particles which boosts their diffusion but also the osmotic pressure because of the oppositional charged ions in the solvent.
When the charged microspheres are in a colloid crystal the microspheres there do only brownian motion at their lattice site; they do not diffuse. So a diffusion machine with a colloid crystal is strictly speaking a brownian motion machine and not a diffusion machine.
But colloid crystals are very sensitive, and in first order their effective density is only the density of the solvent because the lattice sites are fixed by the container (column).
Maybe a ferrofluid in a homogeneous magnetic field would be much better.

Membrane flux calculation

The diffusion flow, which is equal to the H₂O flow, is also equal to the water flow through the membrane, which is the second bottleneck. The flow through the membrane can be calculated with the Poiseuille's law :

Q = dV/dt = T *π *R⁴ * ΔP/(8 *ν *L)

with dV as the volume of the liquid, poured in the time dt, L as the length of the cylindric tube (hole), R the internal radius of the cylindric tube (hole), ΔP the pressure difference between the two ends, ν the dynamic fluid viscosity and T as the number of (cylindric) tubes (holes) in the membrane.
This is the relation between flow and pressure difference between both sides of the membrane for a membrane of width L and with T cylindric tubes (holes).

Power calculation

The power of the solvent flow over the bridge is flow times density times height difference times gravitational acceleration:

P = Q *ρ *Δh *g

Barometric formula density calculation of the solved particles

As mentioned above, the solved particles are concentrated on the bottom, described by the barometric formula. For calculating a diffusion machine we need to know the density on the bottom, at the membrane, for a given average concentration c'_avg, for calculating the osmotic pressure at the membrane.
The average concentration is given by integration over the concentration, divided by the height h:

c'_avg = 1/h *c'₀ *[ -k*T/(m*g) *e^-m*g*h/(k*T)]^h₀ = c'₀/h * k*T/(m*g) *(1 -e^-m*g*h/(k*T))

c'₀ = c'_avg * m*g*h/(k*T *(1 -e^-m*g*h/(k*T)))

c'_avg/c'₀ = k*T/(m*g*h) *(1 -e^-m*g*h/(k*T)) .

Example: With the half density height as total height h we have:

c'_avg/c'₀ = 1/ln(2) *(1 -e^-ln(2)) = 0.7212

which means the density on the bottom is 1/0.7212=1.386 times the average density and at the half density height = h the density is 1.386/2=0.693 times the average density.
An interesting point is that these formulas do show that at constant c'_avg, which means that the hydrostatic pressure difference is linear with h, the osmotic pressure on the bottom, which is proportional to c'₀/c'_avg, does increase faster than the hydrostatic pressure difference. This means that at too great heights the hydrostatic pressure difference gets smaller than the osmotic pressure and the diffusion machine does not work at too great heights!

For salts the density calculation is more complicated because the two sorts of solved particles (cation and anion) do have the same charge with different sign but generally a different density. Because of the Coulomb's law and therefore strong attraction of the two sorts of ions it's clear that the concentration of both must nearly be the same. Therefore we can assume that we have both ions with the same concentrations and with their arithmetic mean density as their effective density. Because of the hydration shell it's not easy to estimate the effective density of an ion in water, but we can get a good approximation by using the density of pure water and the density of salt water.

Note that here the c'₀ and the c'(h) from the barometric formula are caused by gravitation and they are independent from the c'₀ and the c'(h) which are caused by the diffusion flow (see above). To keep things easy, in my examples i'm using parameters which ensure that the c'₀/c'_avg from the barometric formula is less than one percent and therefore neglectable.

In this example we use the following parameters: The solution of hexane has a width, depth and height of 1 m (cube of edge length 1 m). The pure hexane column has also a depth of 1 m, but a width of only 1 cm, because the bottlenecks are the diffusion in the solution and the membrane, so the friction in the hexane column is negligible. The solution consists of hexane and polystyrene microspheres of diameter d = 2*r= 20 nm, with a density of 1.05 g/ml, a net mass per microsphere of
m = (ρ_sphere -ρ_solvent) *(4/3) *π *r³
= 1.6e-24 kg
and the microspheres in a concentration of one sphere per cube of edge length equal to two diameters, which means a volume concentration of c_V = (4/3) *π *r³ / (4*r)³ = 0.0654 = 6.54 %. The concentration at the membrane (bottom) is c'(0) = c'₀ = c'_avg *1.5.
The ambient temperature is 25°C.
For the gravitation induced c'_avg/c'₀ we get 263.9 *(1 -e^-1/263.9) = 0.9981 which means the density gradient which is caused by gravitation is about three hundred times smaller than the density gradient which is caused by the flow. Therefore it is clear that we can neglect the barometric formula in this example.
The membrane has an area of 1 m² and one 10 nm hole, with a length of 20 nm, per square of edge length two diameters, so we have N_h = (1 m)²/(2e-8 m)²=(5e7)²=2.5e15 holes in the membrane.

With these parameters we can start by calculating the flow of the hexane through the left column, this gives the pressure difference at the membrane and finally the height of the pure hexane h_hexane and power P. For equilibrium, the flow of the hexane has to be throttled by the throttle, but that's easy to implement, e. g. with a thin tube.

The Einstein Relation gives a diffusion constant of D = 1.38*10^-23 J/K *300 K / (6 * 3.1415 * 0.32 mPa/s * 658 kg/m³ * 10 nm) = 1.04e-13 m²/s, so we get a velocity of only 1e-13 m/s which is really slow, because this gives a hexane flow of much less than one liter per year!

So even with Osmium instead of Polystyrene, which gives a 56 times greater concentration gradient, it is practical impossible to use such a Diffusion Machine for power generation of more than a few Picowatt!

The second example showed that a simple diffusion machine is impractical because of the very low power - it is even hard to show that they work because they must be shielded from perturbations like temperature differences and temperature changes!
Maybe there is a chance to get more power by using superfluids, but that also seems to be impractical.
So it only makes sense to calculate an Octave Levenspiel's fountain as the next example: A more than 10 km long vertical pipe which is in the ocean, open at the top at the ocean surface (sea level), and closed at the bottom with a semiperable membrane, which is only permeable for water molecules.
Because the diffusion is done by the ocean, the Octave Levenspiel's fountain does not has the main bottleneck of a "simple Diffusion Machine", as shown in the first picture. Another advantage is that the Octave Levenspiel's fountain is more simple: we only have to calculate (or investigate) the osmotic pressure of the ocean water at the depth of the tube bottom (=h_solution). The compressibility and different compressibilities of sea and fresh water can be neglected for a first approximation because the osmotic pressure is independent of the compressibility and in first order fresh water and sea water have the same compressibility.

To keep things simple, the third example is a low cost version: The tube bottom is placed on the ocean ground at Mariana Trench, about 11 km below sea level, with a tube of diameter 1 m, and as semiperable membrane we use a reverse osmosis membrane with a reverse osmosis flow of about 16 m³ per hour (= 4.4 l/s) at a pressure difference (outside - inside the tube bottom) of about 50 bar.
An example of such a membrane is the membrane of an AP 100K from www.ampack1.com .
Due to Poiseuille's law the flow is proportional to the hydrostatic pressure difference minus the osmotic pressure, so we can easily calculate the flow for other pressure differences than 50 bar.

With an average density of ocean water/fresh water of 1.0/1.028 kg/l, an osmotic pressure of about 28 atm at 11 km below sea level and the tube filled up to sea level with fresh water, we have a pressure difference across the membrane, from the ocean to the inside of the tube, of

P_fountain = g * h_solution *ρ_solution - P_osmotic - g * h_{fresh_water} *ρ_{fresh_water}
= 9.81 m/s² *1.1e4 m *28 kg/m³ -2.8 MPa
= +221 kPa .

So without a (net) flow across the membrane, the fresh water surface is 22 m above sea level, which is enough for a nice fountain. If we would take into account the compressibility of water, the densities and differences would be a little higher, but this calculation is good enough for a first approximation.
With the fresh water surface at sea level, we have a flow of

4.4 l/s * 221 kPa/5 MPa = 0.195 l/s .

And the pressure loss in the 11 km long tube is

Q = dV/dt = π *R⁴ * ΔP/(8 *ν *L)
=>
ΔP_tube = Q*(8 *ν *L) /(π *R⁴ ) = 9.75e-5 m³/s *8 *8.90e-4 Pas *1.1e4 m/(3.1416 m⁴) = 2.2 mPa

which is less than 0.001 % and therefore negligible.

Because this system is linear (no turbulence), the maximum power of the fountain is at half of the calculated flow and at half of the maximum height:

p = m*g*h/t = g*h *dm/dt = g*h *Q * ρ = 9.81 m/s² *11 m *0.5 *0.195 l/s * 1.028 kg/l = 11 W

So the Octave Levenspiel's fountain works, but only with very long tubes and with a maximum power in the region of ten Watt. That's more than ten orders of magnitude more power above simple diffusion machines, but much too less for an economical power generation.

Tuning of the Octave Levenspiel's fountain

The power of this Octave Levenspiel's fountain can be doubled by using an Osmotic Power Plant: http://www.waternet.com/news.asp?mode=4&N_ID=51017.
This is also called Double-Membrane Diffusion Machine:

The materials for a diffusion machine must produce a hydrostatic pressure gradient which is greater than the osmotic pressure gradient.
This means at the bottom (h=0):

Δρ *g > P₀ *m*g/(k*T) *e^-m*g*h/(k*T)

Because of h=0 we have e^-m*g*h/(k*T)=1, so:

Δρ *g > P₀ *m*g/(k*T)
=>
Δρ > P₀ *m/(k*T)

So the density difference must be relative high (and positive), but that's trivial.
This criterion is enough for a double membrane Diffusion Machine, which uses only the different pressure gradients.
For a single membrane version such materials are only candidates, because here the integral over the difference of the pressure gradients must be greater than P₀.

As example the parameters from the first example are used:

3.27 kg/m³ > 64.7 Pa * 2.1e-25 kg / 4.142e-21 J = 0.00328 kg/m³

So in the first example the hydrostatic pressure gradient is about three orders of magnitude greater than the barometric formula induced osmotic pressure gradient.

Another result of the criterion is that it gives us a criterion for the temperature:

T > P₀ *m/(k*Δρ)

but because the osmotic pressure is proportional to T, it does NOT mean that the Diffusion Machine always works if the temperature is high enough.

Another result from the criterion is:

m < k*T*Δρ/P₀

This means the mass m of the solute particles should not be too high.
That's clear because a too great mass causes sedimentation and before sedimentation the barometric formula concentrates the solute (and therefore P₀) at the bottom stronger than linear.

The last result is:

P₀ < k*T*Δρ/m

This means that the osmotic pressure must be small enough; the solute particles should not be too small and therefore, for the constant volume concentration, not too numerous.

It is also possible to use a double-membrane diffusion machine with solved particles which do have a density that is lower than the solvent density.

Other sorts of diffusion machines

A much smaller diffusion machine can be made by using vapor/gas that is more dense than the liquids, so that we have osmosis upside down:
barotropy.html

The idea of using reverse osmosis, which is driven by diffusion and the density difference between the inside and the outside of a long vertical column, has also been used for water in air.
It is described in the book from David E. H. Jones, "The Inventions of Deadalus", Chapter "The desert waterer" on pages 72-73, ISBN 0716714132. It was first published in the New Scientist, 25 May 1978.
It is also described in the german version "Zittergas und schräges Wasser", Chapter "Wasser in der Wüste" on pages 279-281, ISBN 3871447684, but the translator has lost the density rho in some lines, so the calculations are a little incomplete in the german version.
An interesting point is that the water molecule weight is lower than the oxygen and nitrogen molecule weight, which means that the barometric formula gives a water density in air which increases with the height (at thermal equilibrium and very high pressure, because of the buoyancy in liquids, very dense gases and supercritical gases, more correctly known as supercritical fluids).

A more simple version of a diffusion machine is metal in a gravity field: An isolated vertical metal cable in a plasma, for example a wolfram (W) cable isolated with e. g. ThO₂ in a Cs (or H) plasma at 2000 to 3000 K, at thermal equilibrium. By the gravity the inside the metal free movably electrons are pulled down with the gravity force m_e*g and in stationary equilibrium without current this causes an electric field of strength E_g, which pulls the electrons up so that there is a floating (no net movement) of the electrons:

m_e *g = E_g *q_e.

By measuring this field, on accelerated conductors, C. R. Tolman and T. D. Steward have measured the electron mass in 1916 (Bergmann-Schaefer: Lehrbuch der Experimentalphysik, Band II, 6. Auflage, 1971, P. 483-484).
Such a field can not come into existence in a plasma because an electric field would be exhausted by the free movable negative and positive ions.
When the metal cable in the plasma is not isolated at the ends, the electric field in the cable causes a current from the lower cable end, through the plasma, to the upper cable end. The current of this current generator can be used, as any other electric current, to supply an electric circuit.
But technically this is hard to do because the gravity on earth causes an electric field of only E_g = g *m_e/q_e = 55.8 pV/m in metals.

By using a ceramic proton conductor instead of a metal, protons instead of electrons and a hydrogen plasma instead of a Cs plasma, also at thermal equilibrium (at about 3000 K), we get an about two thousand times stronger electric field with of -102 nV/m on earth.
Because the proton has another charge sign, the electric field has the opposite direction of the electric field caused by the electrons. This can be used for combining isolated W cables and ceramic proton conductors to a battery.

But the rub in all kinds of Diffusion Machines is that their maximum output power is usually very small, about a few picowatt or less.
This is a consequence of the Second law of thermodynamics . Another perception is that if Diffusion Machines would have a maximum output power of several kilowatt or more the Second law of thermodynamics would be missing, so we should be happy that practicable Diffusion Machines in first approximation do not work and that exact calculation does show that their maximum output power is much less than a kilowatt.

Last Change: 2008-11-05