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.

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 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 s_{1} and s_{2},
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*m_{1}*m_{2}/r_{12} ,
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, 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 S_{1} = k_{B} ln (W_{1}) and S_{2} = k_{B} ln (W_{2}), then the
entropy of the complete system is the sum S = S_{1}+S_{2} only when
W = W_{1} *W_{2}. 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.

To simplify calculations, only low solute concentrations will be used and
calculations are done for no net flow mode, which means the H_{2}O flow over the
bridge is stopped. The stationary mode with a constant H_{2}O 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'.

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).

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 *r^{3}

= 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
c_{V} = (4/3) *π *r^{3} / (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 *r^{3},
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 *N_{0}, 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 * 4.142*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*10^{22} m^{-3}

So the osmotic pressure is:

P_{osmosis} = 1.563*10^{22} * 4.142*10^{-21} Pa
= 64.7 Pa

which is equivalent to about 6.5 mm water height (mmH_{2}O).

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.

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.

Another way is using charged particles, e. g. lead ions (Pb^{2+}), 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.

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.

The diffusion flow in the left column is the first bottleneck and equal to the
H_{2}O 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 *hsolution} -1)/h_{solution}

=>

c'_{0} = c'_{avg} *h_{solution} *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}, 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 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.

The diffusion flow, which is equal to the solvent (H_{2}O)
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 *R^{4} * 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).

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

P = Q *ρ *Δh *g

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)}]^{h}_{0} = 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 *r^{3}
= 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
c_{V} = (4/3) *π *r^{3} / (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}/(2*10^{-8} m)^{2}=(5*10^{7})^{2}=2.5^{15} 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/m^{3} * 10 nm) =
1.04*10^{-13} m^{2}/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!

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:

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 (=h

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 m

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

= 9.81 m/s

= +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 *R

=>

ΔP

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/s

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.

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:

http://www.statkraft.com/energy-sources/osmotic-power/

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:

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
U_{T}=k_{B}*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.

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 U

So the original nanoscopic version is busted not because of the second law of thermodynamics but because of the elementary electric charge.

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 U

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, k

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

The middle plate voltage is -U

At the upper attachment point C

-U

= -U

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

-U

= -U

So the output voltage, the open-circuit voltage difference between the output electrodes, is U

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.

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:

I

The factor 2 takes into account the two attachments per cycle. So with U

I

= 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

P

= 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!

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.

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