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, while a calculation shows that a minimal depth of about 10 kilometer is needed!
See also OSMOSIS: A MICROCOMPUTER LABORATORY TEACHER, a computer Simulation of the Levenspiel Fountain.


The operating principle of the diffusion machine is that 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 higher 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 hsolution is much higher than hH20 because the osmotic pressure, which increases hsolution, 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 e. g. in the (german) book Statistische Theorie der Wärme, Band 1: Gleichgewichtsphänomene by Wilhelm Brenig in chapter 2.2: For the Gibbs postulate (german: Gibbssche Gleichung/Gibbssche Fundamentalgleichung/Fundamentalgleichung der Thermodynamik), a description of systems at thermal equilibrium by canonical ensembles, is only possible when long-range forces like gravity can be neglected and therefore a system can 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: Es = Es1+Es2 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 Es12 = -G*m1*m2/r12 , so we have Es = Es1+Es2+Es12 . 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 Es12 << Es1 and Es12 << Es2 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, because they are an idealisation for the special case of short-range forces.
Von Laue worded this in 1906 as when a system consists of two subsystems with the entropies S1 = kB ln (W1) and S2 = kB ln (W2), then the entropy of the complete system is the sum S = S1+S2 only when W = W1 *W2. So the microstates of the subsystems must be independent. Source: M. von Laue, Zur Thermodynamik der Interferenzerscheinung, Ann. d. Phys. 20, S.365 (1906).
This is important because in theoretic thermodynamics the Second law of thermodynamics is only a result of the gibbs equation and not fundamental, no axiom. Therefore in the book from Brenig the second law (unequation 15.15) is only a result of the Gibbs postulate (equation 2.3).
This means that diffusion machines and similar machines are working in a region where the assumptions for the second law are not fulfilled and therefore the diffusion machines and similar machines can not collide with the second law; they are in disjoint areas.

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 hsolution, is given by the van't Hoff law :

Posmosis = 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 ρsolutionH20, decreases hsolution and the value of the hydrostatic pressure difference is (at a low vapor/gas density)

ΔPhydro = g * ( hsolutionsolution - hH20H20)

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, hcrit, where the pressures (Posmosis, ΔPhydro) do compensate each other exactly and above that height, hsolution is smaller than hH20:

Posmosis = ΔPhydro
Posmosis = g *hcrit *( ρsolution - ρH20) = g *hcrit *Δρ
hcrit = Posmosis / (g *Δρ) .

For ocean water / pure water hcrit 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 Posmosis. Therefore for low solute solutions hcrit does not depend on the concentration n'.
So examples for diffusion machines can be found at some places, e. g. at http://members.chello.at/karl.bednarik/OSMOPERP.PNG (with german text).

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 = (ρspheresolvent) *(4/3) *π *r3
= 2.1*10-25 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 cV = (4/3) *π *r3 / (4*r)3 = 0.0654 = 6.54 %. These microspheres are available 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 4.142*10-21 J.
The half density height, where we have N = 0.5 *N0, is at height h1/2 = -ln(0.5) *k*T/(m*g) .

That's why i'm using 20 nm Polystyrene spheres: The half density height is h1/2 = 0.693 * 4.142*10-21 / (2.1*10-25 *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 = 1.563*1022 m-3

So the osmotic pressure is:

Posmosis = 1.563*1022 * 4.142*10-21 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 -cV) *ρH20 + cV * ρPolystyrene = 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 hH2OH20solution = 10 m/1.00327 = 9.967 m water with the microspheres at the left side.
Because of the osmotic pressure, hsolution 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.


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 hcrit is proportional to the reciprocal value of g.
Another way is using charged particles, e. g. lead ions (Pb2+), and a vertical electric field, which adds the electric acceleration (E*q/m) to g. But because the oppositely charged ions, e. g. Cl-, are accelerated with the other sign, the electric field strength E must be height dependent, e. g. by using a spherical capacitor field.

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 :

QD = -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 hsolution, is the average concentration c'avg, we can calculate c'0 :

c'avg = c'0 *-D/v *(e-v/D *hsolution -1)/hsolution
c'0 = c'avg *hsolution *v/(D *(1 -e-v/D *hsolution))

This gives us e. g. the flow velocity v when we start with a given c'0/c'avg, hsolution, 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 *hsolution + 0.5*(v/D *hsolution)2 -1)/hsolution
c'avg/c'0 = -D/v *( -v/D *hsolution + 0.5*(v/D *hsolution)2)/hsolution
= 1 -0.5 *v/D *hsolution
v = 2 *D/hsolution *(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 another repulsion of the soluted particles which can boosst their diffusion (Debye-Hückel equation), but that also increases 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).
Another option is an electrical field parallel to the gravitational acceleration g, which is similar to a higher gravitational acceleration.

Membrane flux calculation

The diffusion flow, which is equal to the solvent (H2O) flow, is also equal to the solvent (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))

So for the bottom concentration we get

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

which can be easier formulated as

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.

Second example

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 = (ρspheresolvent) *(4/3) *π *r3 = 1.6*10-24 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 cV = (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 Nh = (1 m)2/(2*10-8 m)2=(5*107)2=2.515 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) = 1.04*10-13 m2/s, so we get a velocity of only 10-13 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, and also with the barotropic phenomenon (gravitational phase inversion) it is practical impossible to use such a Diffusion Machine for power generation of more than a few Picowatt!

Third example

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 like 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:

Middle of the cover from Understanding Engineering Thermo,Octave Levenspiel, 1996, ISBN-10: 0135312035
Cover of "Understanding Engineering Thermo",Octave Levenspiel, 1996, ISBN 0135312035.

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 (=hsolution). 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

PFountain = g * hsolutionsolution - Posmotic - g * hfresh_waterfresh_water
= 9.81 m/s2 *1.1*104 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)
ΔPtube = Q*(8 *ν *L) /(π *R4 ) = 9.75*10-5 m3/s *8 *8.90*10-4 Pas *1.1*104 m/(3.1416 m)4 = 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.

With the formulas for the density, compressibility and other data of water, which can be found here, the pressure difference, flux etc. can be calculated more precise.

Tuning of the Octave Levenspiel's Fountain

The power of this Octave Levenspiel's Fountain can be doubled by using an Osmotic Power Plant. Examples of osmotic power plants can be found e. g. in Norway:

and http://web.archive.org/web/20080117160950/http://www.waternet.com/news.asp?mode=4&N_ID=51017.

This combination is also called Double-Membrane Diffusion Machine:

Brownian capacitor-generator or brownian anti-pendulum

Some time ago i found an animated picture of a brownian capacitor generator, a special electrostatic generator (from Karl Bednarik):

The red part at the top is the positive part of a parallel plate type capacitor, the blue at the bottom the negative, the movably part between is a conducting plate driven by brownian motion and the orange and green part at the left is the output. The middle plate is fixed at the left side with an electrically isolated knuckle joint. The length, width and height is about 10 nm, so that the movably plate does show brownian motion.
The plates of the capacitor are isolated and therefore the electric charge of the parallel plate type capacitor is constant. The capacitor has a voltage of about the thermal voltage UT=kB*T/q, to avoid a too weak damping.
At the upper and the lower limit point the middle plate touches one output electrode and gets charged by the near capacitor plate, with the opposite electric charge. So the upper output electrode (orange) gets charged positive, the lower (green) negative.
The conducting parts (colored) may be made of graphite, graphene, metal or a (HT-)supercoductor.

I call it brownian anti-pendulum because it's similar to an electrostatic pendulum, also called Zamboni's pendulum, but a) the oscillation is against the direction of the electric field (so the electric field damps) and b) the pendulum does not touch the charged plates (red and blue), because the two output electrodes (orange and green) are the attachment points of the pendulum. Another reason why i call it anti-pendulum is that the motion, caused by the brownian motion is chaotic, and the friction drives this device while it damps a macroscopic pendulum.
It's an easy device to convert internal, brownian movements into electrical energy, but because of the necessary nanotechnology to build it, it's not easy to make.
Because this device uses the electroweak interaction as the long-range force to convert ambient heat into another energy it's similar to the diffusion machines above. The electroweak interaction is the reason why a reactance like the parallel plate type capacitor or a solenoid has no thermal noise: http://wiki.answers.com/Q/Why_REactance_does_not_add_thermal_noise.

Detailed Theory

Capacity calculation for the original nanoscopic device

A middle plate with a length and height of 10 nm and a medium distance of 3 nm to a capacitor plate has a capacity of about

C = epsilon*A/d = 8.8542e(-12-8-8+8)/0.3 F = 2.95e-19 F

This means with one extra electron the plate has a voltage of Ue = e/C = 1.602e-19/2.95e-19 V = 0.543 V. The absolute value is much higher than the thermal voltage UT=kB*T/e, which means that the capacity calculation shows that the original version can not work, because the elementary electric charge is too big, because the probability that the middle plate gets charged to the voltage U at a limit stop is about exp(-U/UT). Another point is that the parallel plate type capacitor must have a voltage (several times) higher than the middle plate voltage which means the damping by the electric field is so weak, that also a charge transfer from one output electronde to the other is very unlikely.
So the original nanoscopic version is busted not because of the second law of thermodynamics but because of the elementary electric charge.

Capacity calculation for the microscopic device

Because the original nanscopic device is too small, the device may work if it is bigger, which means microscopic, with a length, width and height of about 1 µm, which is 100 times larger then the original. To prevent freezing of the middle plate, e. g. by Van der Waals force, the device should have an add-on like a repulsive magnetic force caused by permanent magnets, because graphite and graphen are so strong diamagnets that they can be used for magnetic levitation over permanent magnets: http://de.wikipedia.org/wiki/Levitation_%28Technik%29#Magnetische_Levitation.
Another traditional way is to use the middle plate as a physical pendulum, with a pendulum frequency of about

f = sqrt(g/l)/(2pi) = sqrt(9.81/5e-7)/(2*3.1416) *1/s = 705 Hz.

A middle plate with a length and height of 1 µm and a medium distance of 0.3 µm to a capacitor plate has a capacity of about

C = epsilon*A/d = 8.8542e(-12-6-6+6)/0.3 F = 2.95e-17 F

This means with one extra electron the plate has a voltage of Ue = e/C = 1.602e-19/2.95e-17 V = 5.43 mV. This means that when the middle plate is charged with the thermal voltage -kB*T/e of about 25 mV it has about five extra electrons or holes. So the 1 µm plate is big enough to make the device work.

Pendulum motion calculation for the microscopic device

The second criterion for a working microscopic device is the amplitude, because it must be big enough to reach the output electrodes and small enough that the brownian motion is enough to drive the device.
With a maximum deflection of the lower end of 10°, which is 1 µm * sin(10°) = 174 nm, the height increase of the center of mass is 1/2 * 1 µm *(1-cos(10°)) = 7.6 nm = h.
With graphen we have a lattice constant of 2.46 A, which gives about 12 atomic mass units per (2.46 A)² and (1 µm / 2.46 A)² = 1.65e7 carbon atoms. So the mass of the 1 µm graphen square is m = 12*u*n = 12*1.66e-27 *1.65e7 = 3.3e-19 kg.
So the potential energy, m*g*h is 3.3e-19 kg *9.81 m/s^2 *7.6e-9 m = 2.46e-26 J and much lower than the thermal energy at one degree of freedom, kB*T = 1.38e-23 *300 J = 4.14e-21 J, and this means the browninan motion is strong enough to drive the device.

Voltage calculation

The middle plate is electrically a dual capacitor, with capacitance C+ to the positive (red) plate and capcitance C- to the negative plate (blue). When the output current is zero, this is also called capacitive divider. Both capacitor plates have the same absolute value of voltage, UCAP/2. So for calculating the output voltage and current we have to get these two capacities at the attachment points and simply apply the capacitor equation U=Q/C for both capacities.

The device starts symmetrical with a neutral middle plate in the middle. Because of Q=0 we have U=Q/C=0 independent of C+ and C-, which are equal here, with value C0.
The middle plate voltage is -UCAP/2 + UCAP * C+/(C+ +C-) = 0 V.

At the upper attachment point C+ is about two times higher and C- is about the half, so we have C+ = 2 * C0, C- = 1/2 * C0, or C+ = 4 * C-. The voltage of the middle plate is the same as the upper output electrode voltage:
-UCAP/2 + UCAP * C+/(C+ +C-)
= -UCAP/2 + UCAP * C+/(C+ + 1/4 * C+) = -UCAP/2 + UCAP * 1/(1 + 1/4) = 0.3 * UCAP.

While moving to the lower attachment point the middle plate is isolated.

While moving to the lower attachment point the middle plate is isolated.

While moving to the lower attachment point the middle plate is isolated.

At the lower attachment point C- is about two times higher and C+ is about the half, so it's the same than as at the upper attachment point but with changed signs and the voltage of the lower output electrode is:
-UCAP/2 + UCAP * C+/(C+ +C-)
= -UCAP/2 + UCAP * 1/4 *C-/(1/4 * C- + C+) = -UCAP/2 + UCAP * 1/(1 + 1/4) = -0.3 * UCAP.

So the output voltage, the open-circuit voltage difference between the output electrodes, is U0 = 0.6 * UCAP.

While moving to the upper attachment point the middle plate is isolated and it's symmetric to the case when the middle plate moves to the lower attachment point.

Current and power calculation

In the chapter before the current is zero, so the first operating point is at the open-circuit voltage. Because this voltage generator is linear we only need a second operating point to get the internal resistance to get the maximum output power with a load resistance which is equal to the internal resistance. So we take the short circuit current as the second operating point. At each attachment point the voltage difference between the middle plate and the output electrode is, with the voltages of the last chapter, 0.3 * UCAP. So the charge exchange at each attachment point is Q0 = Ca *Ua with the capacity of the middle plate at an attachment point Ca = C+ +C- = 2.5 C0, and Ua as the open-circuit voltage of the output electrode, which has an absolute value of 0.3 * UCAP.

The last step is to put the charge exchange at each attachment point together with the pendulum frequency f to get the short circuit current:
Is = f * 2 * Q0.
The factor 2 takes into account the two attachments per cycle. So with UCAP = 25 mV we have
Is = f * 2 * 2.5 C0 * 0.3 * UCAP
= 705 * 5 * 2.95e-17 0.3 * 0.025 A = 7.8e-16 A
in first approximation because the pendulum has a resonance frequency of 705 Hz but moves chaotic and is damped by the electric field.
So the maximum output power is
Pmax = 1/4 * U0 * Is
= 0.25 * 0.025 * 7.8e-16 W = 4.9e-18 W.
So at first look it's the same as with the diffusion machines, we get nearly no power out if it. But the calculated anti-pendulum is much smaller. We can put 1e18 into one cubic meter and therefore get a higher power density of about five Watt per cubic meter, while the diffusion machines are in the region of a picowatt per cubic meter!

Tuning the anti-pendulum

To get higher capacities and a higher capacity ratio it's nearby to use the capacitor plates with an angle a little greater than zero:

Macroscopic model

With some parts from a DIY store and some cheap electronic components it's easy to make a (quick and dirty) macroscopic model which gets driven by two disc magnets in series between two air-core coils as anti-helmholtz coils at the pendulum frequency:

For low-pass filtering the output i use a 10 µ F/100 V foil capacitor which is soldered parallel to a 1 GOhm resistor to eleminate polarisation voltages and electrostatic voltages which may come from the environment. To reduce electostatic charging in the room the relativ air humidity is 50 ±5 % and to reduce thermoelectric voltage i used only thermoelectric voltage low-grade tin-solder with 20 % Cd (Sn Cd 20).

For driving the coils, each consisting of two parallel cheap 250324 coils, i use a dual port parallel port card whith a direct DA conversion, with a 75 Ohm resistor from each of the 12 output pins to the summary point:

In the picture you can see a switch with two capacitors for low-pass filtering, against harmonics. I also added the 5 input pullup resistors.
BTW: Such an output can also be used to drive a relay (with 5 V / 110 Ohm input) or a white 1 W LED directly.
The device driver is the simple Linux user-space program sine_9bit2.c.

The shown first version of the model has the bug that i used wood as insulating mounting plate, but wood is no good insulator at high resistance applications. So i had to make a workaround and use (blue) plastic as insulation at the left output electrode.
The RLC meter showed a capacity variance of only factor 3, because it's a quick and dirty version and it's necessary to left some space for the two disc magnets.
This model works at 1.219 Hz and with a capacitor voltage of 36 V it shows an output voltage of 10 mV with a 10 MOhm load resistor, 1 V with 1 GOhm and so on.
The output voltage changes it's sign when the capacitor voltage sign is changed and the output voltage is zero when the pendulum movement is blocked or if the coil current is switched off or when the capacitor voltage is zero. So leakage currents can be neglected (at the output) and the model works as calculated above. The next step is to make a microscopic original.