Tag Archives: thermodynamics

Einstein’s theory of diffusion in liquids, and my extension.

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In 1905 and 1908, Einstein developed two formulations for the diffusion of a small particle in a liquid. As a side-benefit of the first derivation, he demonstrated the visible existence of molecules, a remarkable piece of work. In the second formulation, he derived the same result using non-equilibrium thermodynamics, something he seems to have developed on the spot. I’ll give a brief version of the second derivation, and will then I’ll show off my own extension. It’s one of my proudest intellectual achievements.

But first a little background to the problem. In 1827, a plant biologist, Robert Brown examined pollen under a microscope and noticed that it moved in a jerky manner. He gave this “Brownian motion” the obvious explanation: that the pollen was alive and swimming. Later, it was observed that the pollen moved faster in acetone. The obvious explanation: pollen doesn’t like acetone, and thus swims faster. But the pollen never stopped, and it was noticed that cigar smoke also swam. Was cigar smoke alive too?

Einstein’s first version of an answer, 1905, was to consider that the liquid was composed of atoms whose energy was a Boltzmann distribution with an average of E= kT in every direction where k is the Boltzmann constant, and k = R/N. That is Boltsman’s constant equals the gas constant, R, divided by Avogadro’s number, N. He was able to show that the many interactions with the molecules should cause the pollen to take a random, jerky walk as seen, and that the velocity should be faster the less viscous the solvent, or the smaller the length-scale of observation. Einstein applied the Stokes drag equation to the solute, the drag force per particle was f = -6πrvη where r is the radius of the solute particle, v is the velocity, and η is the solution viscosity. Using some math, he was able to show that the diffusivity of the solute should be D = kT/6πrη. This is called the Stokes-Einstein equation.

In 1908 a French physicist, Jean Baptiste Perrin confirmed Einstein’s predictions, winning the Nobel prize for his work. I will now show the 1908 Einstein derivation and will hope to get to my extension by the end of this post.

Consider the molar Gibbs free energy of a solvent, water say. The molar concentration of water is x and that of a very dilute solute is y. y<<1. For this nearly pure water, you can show that µ = µ° +RT ln x= µ° +RT ln (1-y) = µ° -RTy.

Now, take a derivative with respect to some linear direction, z. Normally this is considered illegal, since thermodynamic is normally understood to apply to equilibrium systems only. Still Einstein took the derivative, and claimed it was legitimate at nearly equilibrium, pseudo-equilibrium. You can calculate the force on the solvent, the force on the water generated by a concentration gradient, Fw = dµ/dz = -RT dy/dz.

Now the force on each atom of water equals -RT/N dy/dz = -kT dy/dz.

Now, let’s call f the force on each atom of solute. For dilute solutions, this force is far higher than the above, f = -kT/y dy/dz. That is, for a given concentration gradient, dy/dz, the force on each solute atom is higher than on each solvent atom in inverse proportion to the molar concentration.

For small spheres, and low velocities, the flow is laminar and the drag force, f = 6πrvη.

Now calculate the speed of each solute atom. It is proportional to the force on the atom by the same relationship as appeared above: f = 6πrvη or v = f/6πrη. Inserting our equation for f= -kT/y dy/dz, we find that the velocity of the average solute molecule,

v = -kT/6πrηy dy/dz.

Let’s say that the molar concentration of solvent is C, so that, for water, C will equal about 1/18 mols/cc. The atomic concentration of dilute solvent will then equal Cy. We find that the molar flux of material, the diffusive flux equals Cyv, or that

Molar flux (mols/cm2/s) = Cy (-kT/6πrηy dy/dz) = -kTC/6πrη dy/dz -kT/6πrη dCy/dz.

where Cy is the molar concentration of solvent per volume.

Classical engineering comes to a similar equation with a property called diffusivity. Sp that

Molar flux of y (mols y/cm2/s) = -D dCy/dz, and D is an experimentally determined constant. We thus now have a prediction for D:

D = kT/6πrη.

This again is the Stokes Einstein Equation, the same as above but derived with far less math. I was fascinated, but felt sure there was something wrong here. Macroscopic viscosity was not the same as microscopic. I just could not think of a great case where there was much difference until I realized that, in polymer solutions there was a big difference.

Polymer solutions, I reasoned had large viscosities, but a diffusing solute probably didn’t feel the liquid as anywhere near as viscous. The viscometer measured at a larger distance, more similar to that of the polymer coil entanglement length, while a small solute might dart between the polymer chains like a rabbit among trees. I applied an equation for heat transfer in a dispersion that JK Maxwell had derived,

where κeff is the modified effective thermal conductivity (or diffusivity in my case), κl and κp are the thermal conductivity of the liquid and the particles respectively, and φ is the volume fraction of particles. 

To convert this to diffusion, I replaced κl by Dl, and κp by Dp where

Dl = kT/6πrηl

and Dp = kT/6πrη.

In the above ηl is the viscosity of the pure, liquid solvent.

The chair of the department, Don Anderson didn’t believe my equation, but agreed to help test it. A student named Kit Yam ran experiments on a variety of polymer solutions, and it turned out that the equation worked really well down to high polymer concentrations, and high viscosity.

As a simple, first approximation to the above, you can take Dp = 0, since it’s much smaller than Dl and you can take Dl to equal Dl = kT/6πrηl as above. The new, first order approximation is:

D = kT/6πrηl (1 – 3φ/2).

We published in Science. That is I published along with the two colleagues who tested the idea and proved the theory right, or at least useful. The reference is Yam, K., Anderson, D., Buxbaum, R. E., Science 240 (1988) p. 330 ff. “Diffusion of Small Solutes in Polymer-Containing Solutions”. This result is one of my proudest achievements.

R.E. Buxbaum, March 20, 2024

Low temperature hydrogen removal

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Platinum catalysts can be very effective at removing hydrogen from air. Platinum promotes the irreversible reaction of hydrogen with oxygen to make water: H2 + 1/2 O2 –> H2O, a reaction that can take off, at great rates, even at temperatures well below freezing. In the 1800s, when platinum was cheap, platinum powder was used to light town-gas, gas street lamps. In those days, street lamps were not fueled by methane, ‘natural gas’, but by ‘town gas’, a mix of hydrogen and carbon monoxide and many impurities like H2S. It was made by reacting coal and steam in a gas plant, and it is a testament to the catalytic power of Pt that it could light this town gas. These impurities are catalytic poisons. When exposed to any catalyst, including platinum, the catalyst looses it’s power to. This is especially true at low temperatures where product water condenses, and this too poisons the catalytic surface.

Nowadays, platinum is expensive and platinum catalysts are no longer made of Pt powder, but rather by coating a thin layer of Pt metal on a high surface area substrate like alumina, ceria, or activated carbon. At higher temperatures, this distribution of Pt improves the reaction rate per gram Pt. Unfortunately, at low temperatures, the substrate seems to be part of the poisoning problem. I think I’ve found a partial way around it though.

My company, REB Research, sells Pt catalysts for hydrogen removal use down to about 0°C, 32°F. For those needing lower temperature hydrogen removal, we offer a palladium-hydrocarbon getter that continues to work down to -30°C and works both in air and in the absence of air. It’s pretty good, but poisons more readily than Pt does when exposed to H2S. For years, I had wanted to develop a version of the platinum catalyst that works well down to -30°C or so, and ideally that worked both in air and without air. I got to do some of this development work during the COVID downtime year.

My current approach is to add a small amount of teflon and other hydrophobic materials. My theory is that normal Pt catalysts form water so readily that the water coats the catalytic surface and substrate pores, choking the catalyst from contact with oxygen or hydrogen. My thought of why our Pd-organic works better than Pt is that it’s part because Pd is a slower water former, and in part because the organic compounds prevent water condensation. If so, teflon + Pt should be more active than uncoated Pt catalyst. And it is so.

Think of this in terms of the  Van der Waals equation of state:{\displaystyle \left(p+{\frac {a}{V_{m}^{2}}}\right)\left(V_{m}-b\right)=RT}

where V_{m} is molar volume. The substance-specific constants a and b can be understood as an attraction force between molecules and a molecular volume respectively. Alternately, they can be calculated from the critical temperature and pressure as

{\displaystyle a={\frac {27(RT_{c})^{2}}{64p_{c}}}}{\displaystyle b={\frac {RT_{c}}{8p_{c}}}.}

Now, I’m going to assume that the effect of a hydrophobic surface near the Pt is to reduce the effective value of a. This is to say that water molecules still attract as before, but there are fewer water molecules around. I’ll assume that b remains the same. Thus the ratio of Tc and Pc remains the same but the values drop by a factor of related to the decrease in water density. If we imagine the use of enough teflon to decrease he number of water molecules by 60%, that would be enough to reduce the critical temperature by 60%. That is, from 647 K (374 °C) to 359 K, or -14°C. This might be enough to allow Pt catalysts to be used for H2 removal from the gas within a nuclear wast casket. I’m into nuclear, both because of its clean power density and its space density. As for nuclear waste, you need these caskets.

I’ve begun to test of my theory by making hydrogen removal catalyst that use both platinum and palladium along with unsaturated hydrocarbons. I find it works far better than the palladium-hydrocarbon getter, at least at room temperature. I find it works well even when the catalyst is completely soaked in water, but the real experiments are yet to come — how does this work in the cold. Originally I planned to use a freezer for these tests, but I now have a better method: wait for winter and use God’s giant freezer.

Robert E. Buxbaum October 20, 2021. I did a fuller treatment of the thermo above, a few weeks back.

Weird thermodynamics near surfaces can prevent condensation and make water more slippery.

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It is a fundamental of science that that the properties of every pure one-phase material is totally fixed properties at any given temperature and pressure. Thus for example, water at 0°C is accepted to always have a density of 0.998 gm/cc, a vapor pressure of 17.5 Torr, a viscosity of 1.002 centipoise (milliPascal seconds) and a speed of sound of 1481 m/s. Set the temperature and pressure of any other material and every other quality is set. But things go screwy near surfaces, and this is particularly true for water where the hydrogen bond — a quantum bond — predominates.

its vapor pressure rises and it becomes less inclined to condense or freeze. I use this odd aspect of thermodynamics to keep my platinum-based hydrogen getter catalysis active at low temperatures where they would normally clog. Normal platinum catalysts are not suitable for hydrogen removal at normal temperatures, eg room temperature, because the water that forms from hydrogen oxidation chokes off the catalytic surface. Hydrophobic additions prevent this, and I’d like to show you why this works, and why other odd things happen, based on an approximation called the  Van der Waals equation of state:

{\displaystyle \left(p+{\frac {a}{V_{m}^{2}}}\right)\left(V_{m}-b\right)=RT} (1)

This equation described the molar volume of a pure material, V_{m}, of any pure material based not the pressure, the absolute temperature (Kelvin) and two, substance-specific constants, a and b. These constants can be understood as an attraction force term, and a molecular volume respectively. It is common to calculate a and b from the critical temperature and pressure as follows, where Tc is absolute temperature:

{\displaystyle a={\frac {27(RT_{c})^{2}}{64p_{c}}}}, {\displaystyle b={\frac {RT_{c}}{8p_{c}}}.} (2 a,b)

For water Tc = 647 K (374°C) and 220.5 bar. Plugging in these numbers, the Van der Waals gives reasonable values for the density of water both as a liquid and a gas, and thus gives a reasonable value for the boiling point.

Now consider the effect that an inert surface would have on the effective values of a and b near that surface. The volume of the molecules will not change, and thus b will not change, but the value of a will change, likely by about half. This is because, the number of molecules surrounding any other molecule is reduced by about half while the inert surface adds nothing to the attraction. Near a surface, surrounding molecules still attract each other the same as before, but there are about half as many molecules at any temperature and pressure.

To get a physical sense of what the surface does, consider using the new values of a and b to determine a new value for Tc and Pc, for materials near the surface. Since b does not change, we see that the presence of a surface does not affect the ratio of Tc and Pc, but it decreases the effective value of Tc — by about half. For water, that is a change from 647 K to 323.5K, 50.5°C, very close to room temperature. Pc changes to 110 bar, about 1600 psi. Since the new value of Tc is close to room temperature, the the density of water will be much lower near the surface, and the viscosity can be expected to drop. The net result is that water flows more readily through a teflon pipe than through an ordinary pipe, a difference that is particularly apparent at small diameters.

This decrease in effective Tc is useful for fire hoses, and for making sailing ships go faster (use teflon paint) and for making my hydrogen removal catalysts more active at low temperatures. Condensed water can block the pores to the catalyst; teflon can forestall this condensation. It’s a general trick of thermodynamics, reasonably useful. Now you know it, and now you know why it works.

Robert Buxbaum August 30, 2021

The remarkable efficiency of 22 caliber ammunition.

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22 long rifle shells contain early any propellant.

The most rifle cartridge in the US today is the 22lr a round that first appeared in 1887. It is suitable to small game hunting and while it is less–deadly than larger calibers, data suggests it is effective for personal protection. It is also remarkably low cost. This is because the cartridge in almost entirely empty as shown in the figure at right. It is also incredibly energy efficient, that is to say, it’s incredibly good at transforming heat energy of the powder into mechanical energy in the bullet.

The normal weight of a 22lr is 40 grains, or 2.6 grams; a grain is the weight of a barley grain 1/15.4 gram. Virtually every brand of 22lr will send its bullet at about the speed of sound, 1200 ft/second, with a kinetic energy of about 120 foot pounds, or 162 Joules. This is about twice the energy of a hunting bow, and it will go through a deer. Think of a spike driven by a 120 lb hammer dropped from one foot. That’s the bullet from a typical 22lr.

The explosive combustion heat of several Hodgdon propellants.

The Hodgdon power company is the largest reseller of smokeless powder in the US with products from all major manufacturers, with products selling for an average of $30/lb or .43¢ per grain. The CCI Mini-Mag, shown above, uses 0.8 grains of some powder 0.052 grams, or about 1/3¢ worth, assuming that CCI bought from Hodgdon rather than directly from the manufacturer. You will notice that the energies of the powders hardly varies from type to type, from a low of 3545 J/gram to a high of 4060 J/gram. While I don’t know which powder is used, I will assume CCI uses a high-energy propellant, 4000 J/gram. I now calculate that the heat energy available as 0.052*4000 = 208 Joules. To calculate the efficiency, divide the kinetic energy of the bullet by the 208 Joules. The 40 grain CCI MiniMag bullet has been clocked at 1224 feet per second indicating 130 foot pounds of kinetic energy, or 176 J. Divide by the thermal energy and you find a 85% efficiency: 176J/ 208 J = 85%. That’s far better than your car engine. If the powder were weaker, the efficiency would have to be higher.

The energy content of various 22lr bullets shot from different length barrels.

I will now calculate the pressure of the gas behind a 22lr. I note that the force on the bullet is equal to the pressure times the cross-sectional area of the barrel. Since energy equals force times distance, we can expect that the kinetic energy gained per inch of barrel equals this force times this distance (1 inch). Because of friction this is an under-estimate of the pressure, but based on the high efficiency, 85%, it’s clear that the pressure can be no more than 15% higher than I will calculate. As it happens, the maximum allowable pressure for 22lr cartridges is set by law at 24,000 psi. When I calculate the actual pressure (below) I find it is about half this maximum.

The change in kinetic energy per inch of barrel is calculated as the change in 1/2 mv2, where m is the mass of the bullet and v is the velocity. There is a web-site with bullet velocity information for many brands of ammunition, “ballistics by the inch”. Data is available for many brands of bullet shot from gun barrels that they cut shorter inch by inch; data for several 22lr are shown here. For the 40 grain CCI MiniMag, they find a velocity of 862 ft/second for 2″ barrel, 965 ft/second for a 3″ barrel, 1043 ft/second for a 4″ barrel, etc. The cross-section area of the barrel is 0.0038 square inches.

Every 22 cartridge has space to spare.

Based on change in kinetic energy, the average pressure in the first two inches of barrel must be 10,845 psi, 5,485 psi in the next inch, and 4,565 psi in the next inch, etc. If I add a 15% correction for friction, I find that the highest pressure is still only half the maximum pressure allowable. Strain gauge deformation data (here) gives a slightly lower value. It appears to me that, by adding more propellant, one could make a legal, higher-performance version of the 22lr — one with perhaps twice the kinetic energy. Given the 1/3¢ cost of powder relative to the 5 to 20¢ price of ammo, I suspect that making a higher power 22lr would be a success.

Robert Buxbaum, March 18, 2021. About 10% of Michigan hunts dear every year during hunting season. Another 20%, as best I can tell own guns for target shooting or personal protection. Just about every lawyer I know carries a gun. They’re afraid people don’t like them. I’m afraid they’re right.

magnetic separation of air

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As some of you will know, oxygen is paramagnetic, attracted slightly by a magnet. Oxygen’s paramagnetism is due to the two unpaired electrons in every O2 molecule. Oxygen has a triple-bond structure as discussed here (much of the chemistry you were taught is wrong). Virtually every other common gas is diamagnetic, repelled by a magnet. These include nitrogen, water, CO2, and argon — all diamagnetic. As a result, you can do a reasonable job of extracting oxygen from air by the use of a magnet. This is awfully cool, and could make for a good science fair project, if anyone is of a mind.

But first some math, or physics, if you like. To a good approximation the magnetization of a material, M = CH/T where M is magnetization, H is magnetic field strength, C is the Curie constant for the material, and T is absolute temperature.

Ignoring for now, the difference between entropy and internal energy, but thinking only in terms of work derived by lowering a magnet towards a volume of gas, we can say that the work extracted, and thus the decrease in energy of the magnetic gas is ∫∫HdM  = MH/2. At constant temperature and pressure, we can say ∆G = -CH2/2T.

The maximum magnetization you’re likely to get with any permanent magnet (not achieved to date) is about 50 Tesla, or 40,000 ampere meters. At 20°C, the per-mol, magnetic susceptibility of oxygen is 1.34×10−6  This suggests that the Curie constant is 1.34 ×10−6 x 293 = 3.93 ×10−4. Applying this value to oxygen in a 50 Tesla magnet at 20°C, we find the energy difference, ∆G is 1072 J/mole = RT ln ß where ß is a concentration ratio factor between the O2 content of the magnetized and un-magnetized gas, C1/C2 =ß

At room temperature, 298K ß = 1.6, and thus we find that the maximum oxygen concentration you’re likely to get is about 1.6 x 21% = 33%. It’s slightly more than this due to nitrogen’s diamagnetism, but this effect is too small the matter. What does matter is that 33% O2 is a good amount for a variety of medical uses.

I show below my simple design for a magnetic O2 concentrator. The dotted line is a permeable membrane of no selectivity – with a little O2 permeability the design will work better. All you need is a blower or pump. A coffee filter could serve as a membrane.bux magneitc air separator

This design is as simple as the standard membrane-based O2 concentrator – those based on semi-permeable membranes, but this design should require less pressure differential — just enough to overcome the magnet. Less pressure means the blower should be smaller, and less noisy, with less energy use.  I figure this could be really convenient for people who need portable oxygen. With current magnets it would take 4-5 stages or low temperatures to reach this concentration, still this design could have commercial use, I’d think.

On the theoretical end, an interesting thing I find concerns the effect on the entropy of the magnetic oxygen. (Please ignore this paragraph if you have not learned statistical thermodynamics.) While you might imagine that magnetization decreases entropy, other-things being equal because the molecules are somewhat aligned with the field, temperature and pressure being fixed, I’ve come to realize that entropy is likely higher. A sea of semi-aligned molecules will have a slightly higher heat capacity than nonaligned molecules because the vibrational Cp is higher, other things being equal. Thus, unless I’m wrong, the temperature of the gas will be slightly lower in the magnetic area than in the non-magnetic field area. Temperature and pressure are not the same within the separator as out, by the way; the blower is something of a compressor, though a much less-energy intense one than used for most air separators. Because of the blower, both the magnetic and the non magnetic air will be slightly warmer than in the surround (blower Work = ∆T/Cp). This heat will be mostly lost when the gas leaves the system, that is when it flows to lower pressure, both gas streams will be, essentially at room temperature. Again, this is not the case with the classic membrane-based oxygen concentrators — there the nitrogen-rich stream is notably warm.

Robert E. Buxbaum, October 11, 2017. I find thermodynamics wonderful, both as science and as an analog for society.

Heraclitus and Parmenides time joke

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From Existential Commics

From Existential Comics; Parmenides believed that nothing changed, nor could it.

For those who don’t remember, Heraclitus believed that change was the essence of life, while  Parmenides believed that nothing ever changes. It’s a debate that exists to this day in physics, and also in religion (there is nothing new under the sun, etc.). In science, the view that no real change is possible is founded in Schrödinger’s wave view of quantum mechanics.

Schrödinger's wave equation, time dependent.

Schrödinger’s wave equation, time dependent.

In Schrödinger’s wave description of reality, every object or particle is considered a wave of probability. What appears to us as motion is nothing more than the wave oscillating back and forth in its potential field. Nothing has a position or velocity, quite, only random interactions with other waves, and all of these are reversible. Because of the time reversibility of the equation, long-term, the system is conservative. The wave returns to where it was, and no entropy is created, long-term. Anything that happens will happen again, in reverse. See here for more on Schrödinger waves.

Thermodynamics is in stark contradiction to this quantum view. To thermodynamics, and to common observation, entropy goes ever upward, and nothing is reversible without outside intervention. Things break but don’t fix themselves. It’s this entropy increase that tells you that you are going forward in time. You know that time is going forward if you can, at will, drop an ice-cube into hot tea to produce lukewarm, diluted tea. If you can do the reverse, time is going backward. It’s a problem that besets Dr. Who, but few others.

One way that I’ve seen to get out of the general problem of quantum time is to assume the observed universe is a black hole or some other closed system, and take it as an issue of reference frame. As seen from the outside of a black hole (or a closed system without observation) time stops and nothing changes. Within a black hole or closed system, there is constant observation, and there is time and change. It’s not a great way out of the contradiction, but it’s the best I know of.

Predestination makes a certain physics and religious sense, it just doesn't match personal experience very well.

Predestination makes a certain physics and religious sense, it just doesn’t match personal experience very well.

The religion version of this problem is as follows: God, in most religions, has fore-knowledge. That is, He knows what will happen, and that presumes we have no free will. The problem with that is, without free-will, there can be no fair judgment, no right or wrong. There are a few ways out of this, and these lie behind many of the religious splits of the 1700s. A lot of the humor of Calvin and Hobbes comics comes because Calvin is a Calvinist, convinced of fatalistic predestination; Hobbes believes in free will. Most religions take a position somewhere in-between, but all have their problems.

Applying the black-hole model to God gives the following, alternative answer, one that isn’t very satisfying IMHO, but at least it matches physics. One might assume predestination for a God that is outside the universe — He sees only an unchanging system, while we, inside see time and change and free will. One of the problems with this is it posits a distant creator who cares little for us and sees none of the details. A more positive view of time appears in Dr. Who. For Dr. Who time is fluid, with some fixed points. Here’s my view of Dr. Who’s physics.  Unfortunately, Dr. Who is fiction: attractive, but without basis. Time, as it were, is an issue for the ages.

Robert Buxbaum, Philosophical musings, Friday afternoon, June 30, 2017.

A clever, sorption-based, hydrogen compressor

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Hydrogen-powered fuel cells provide weight and cost advantages over batteries, important e.g. for drones and extended range vehicles, but they require highly compressed hydrogen and it’s often a challenge compressing the hydrogen. A large-scale solution I like is pneumatic compression, e.g. this compressor. One would combine it with a membrane reactor hydrogen generator, to fill tanks for fuel cells. The problem is that this pump is somewhat complex, and would likely add air impurities to the hydrogen. I’d now like to describe a different, very clever hydrogen pump, one that suited to smaller outputs, but adds no impurities and and provides very high pressure. It operates by metallic hydride sorption at low temperature, followed by desorption at high temperature.

Hydride sorption -desorption pressures vs temperature.

Hydride sorption -desorption pressures vs temperature, from Dhinesh et al.

The metal hydriding reaction is M + nH2 <–> MH2n. Where M is a metal or metallic alloy and MH2n is the hydride. While most metals will undergo this reaction at some appropriate temperature and pressure, the materials of practical interest are exothermic hydrides that is hydrides that give off heat on hydriding. They also must undergo a nearly stoichiometric absorption or desorption reaction at reasonable temperatures and pressures. The plot at right presents the plateau pressure for hydrogen absorption/ desorption in several, exothermic metal hydrides. The most attractive of these are shown in the red box near the center. These sorb or desorb between 1 and 10 atmospheres and 25 and 100 °C.

In this plot, the slope of the sorption line is proportional to the heat of sorption. The most attractive materials for this pump are the ones in the box (or near) with a high slope to the line implying a high heat of sorption. A high heat of sorption means you can get very high compression without too much of a temperature swing.

To me, NaAlH4 appears to be the best of the materials. Though I have not built a pump yet with this material, I’d like to. It certainly serves as a good example for how the pump might work. The basic reaction is:

NaAl + 2H2 <–> NaAlH4

suggesting that each mol of NaAl material (50g) will absorb 2 mols of hydrogen (44.8 std liters). The sorption line for this reaction crosses the 1 atm horizontal line at about 30°C. This suggests that sorption will occur at 1 am and normal room temperature: 20-25°C. Assume the pump contains 100 g of NaAl (2.0 mols). Under ideal conditions, these 100g will 4 mols of hydrogen gas, about 90 liters. If this material in now heated to 226°C, it will desorb the hydrogen (more like 80%, 72 liters) at a pressure in excess of 100 atm, or 1500 psi. The pressure line extends beyond the graph, but the sense is that one could pressure in the neighborhood of 5000 psi or more: enough to use filling the high pressure tank of a hydrogen-based, fuel cell car.

The problem with this pump for larger volume H2 users is time. It will take 2-3 hours to cycle the sober, that is, to absorb hydrogen at low pressure, to heat the material to 226°C, to desorb the H2 and cycle back to low temperature. At a pump rate of 72 liters in 2-3 hours, this will not be an effective pump for a fuel-cell car. The output, 72 liters is only enough to generate 0.12kWh, perhaps enough for the tank of a fuel cell drone, or for augmenting the mpg of gasoline automobiles. If one is interested in these materials, my company, REB Research will be happy to manufacture some in research quantities (the prices blow are for materials cost, only I will charge significantly more for the manufactured product, and more yet if you want a heater/cooler system).

Properties of Metal Hydride materials; Dhanesh Chandra,* Wen-Ming Chien and Anjali Talekar, Material Matters, Volume 6 Article 2

Properties of Metal Hydride materials; Dhanesh Chandra,* Wen-Ming Chien and Anjali Talekar, Material Matters, Volume 6 Article 2

One could increase the output of a pump by using more sorbent, perhaps 10 kg distributed over 100 cells. With this much sorbent, you’ll pump 100 times faster, enough to take the output of a fairly large hydrogen generator, like this one from REB. I’m not sure you get economies of scale, though. With a mechanical pump, or the pneumatical pump,  you get an economy of scale: typically it costs 3 times as much for each 10 times increase in output. For the hydride pump, a ten times increase might cost 7-8 times as much. For this reason, the sorption pump lends itself to low volume applications. At high volume, you’re going to want a mechanical pump, perhaps with a getter to remove small amounts of air impurities.

Materials with sorption lines near the middle of the graph above are suited for long-term hydrogen storage. Uranium hydride is popular in the nuclear industry, though I have also provided Pd-coated niobium for this purpose. Materials whose graph appear at the far, lower left, titanium TiH2, can be used for permanent hydrogen removal (gettering). I have sold Pd-niobium screws for this application, and will be happy to provide other shapes and other materials, e.g. for reversible vacuum pumping from a fusion reactor.

Robert Buxbaum, May 26, 2017 (updated Apr. 4, 2022). 

Alcohol and gasoline don’t mix in the cold

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One of the worst ideas to come out of the Iowa caucuses, I thought, was Ted Cruz claiming he’d allow farmers to blend as much alcohol into their gasoline as they liked. While this may have sounded good in Iowa, and while it’s consistent with his non-regulation theme, it’s horribly bad engineering.

At low temperatures ethanol and gasoline are no longer quite miscible

Ethanol and gasoline are that miscible at temperatures below freezing, 0°C. The tendency is greater if the ethanol is wet or the gasoline contains benzenes

We add alcohol to gasoline, not to save money, mostly, but so that farmers will produce excess so we’ll have secure food for wartime or famine — or so I understand it. But the government only allows 10% alcohol in the blend because alcohol and gasoline don’t mix well when it’s cold. You may notice, even with the 10% mixture we use, that your car starts poorly on the coldest winter days. The engine turns over and almost catches, but dies. A major reason is that the alcohol separates from the rest of the gasoline. The concentrated alcohol layer screws up combustion because alcohol doesn’t burn all that well. With Cruz’s higher alcohol allowance, you’d get separation more often, at temperatures as high as 13°C (55°F) for a 65 mol percent mix, see chart at right. Things get worse yet if the gasoline gets wet, or contains benzene. Gasoline blending is complex stuff: something the average joe should not do.

Solubility of dry alcohol (ethanol) in gasoline. The solubility is worse at low temperature and if the gasoline is wet or aromatic.

Solubility of alcohol (ethanol) in gasoline; an extrapolation based on the data above.

To estimate the separation temperature of our normal, 10% alcohol-gasoline mix, I extended the data from the chart above using linear regression. From thermodynamics, I extrapolated ln-concentration vs 1/T, and found that a 10% by volume mix (5% mol fraction alcohol) will separate at about -40°F. Chances are, you won’t see that temperature this winter (and if you you do, try to find a gas mix that has no alcohol. Another thought, add hydrogen or other combustible gas to get the engine going.

Robert E. Buxbaum, February 10, 2016. Two more thoughts: 1) Thermodynamics is a beautiful subject to learn, and (2) Avoid people who stick to foolish consistency. Too much regulation is bad, as is too little: it’s a common pattern: The difference between a cure and a poison is often just the dose.

The speed of sound, Buxbaum’s correction

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Ernst Mach showed that sound must travel at a particular speed through any material, one determined by the conservation of energy and of entropy. At room temperature and 1 atm, that speed is theoretically predicted to be 343 m/s. For a wave to move at any other speed, either the laws of energy conservation would have to fail, or ∆S ≠ 0 and the wave would die out. This is the only speed where you could say there is a traveling wave, and experimentally, this is found to be the speed of sound in air, to good accuracy.

Still, it strikes me that Mach’s assumptions may have been too restrictive for short-distance sound waves. Perhaps there is room for other sound speeds if you allow ∆S > 0, and consider sound that travels short distances and dies out far from the source. Waves at these, other speeds might affect music appreciation, or headphone design. As these waves were never treated in my thermodynamics textbooks, I wondered if I could derive their speed in any nice way, and if they were faster or slower than the main wave? (If I can’t use this blog to re-think my college studies, what good is it?)

I t can help to think of a shock-wave of sound wave moving down a constant area tube of still are at speed u, with us moving along at the same speed as the wave. In this view, the wave appears stationary, but there is a wind of speed u approaching it from the left.

Imagine the sound-wave moving to the right, down a constant area tube at speed u, with us moving along at the same speed. Thus, the wave appears stationary, with a wind of speed u from the right.

As a first step to trying to re-imagine Mach’s calculation, here is one way to derive the original, for ∆S = 0, speed of sound: I showed in a previous post that the entropy change for compression can be imagines to have two parts, a pressure part at constant temperature: dS/dV at constant T = dP/dT at constant V. This part equals R/V for an ideal gas. There is also a temperature at constant volume part of the entropy change: dS/dT at constant V = Cv/T. Dividing the two equations, we find that, at constant entropy, dT/dV = RT/CvV= P/Cv. For a case where ∆S>0, dT/dV > P/Cv.

Now lets look at the conservation of mechanical energy. A compression wave gives off a certain amount of mechanical energy, or work on expansion, and this work accelerates the gas within the wave. For an ideal gas the internal energy of the gas is stored only in its temperature. Lets now consider a sound wave going down a tube flow left to right, and lets our reference plane along the wave at the same speed so the wave seems to sit still while a flow of gas moves toward it from the right at the speed of the sound wave, u. For this flow system energy is concerned though no heat is removed, and no useful work is done. Thus, any change in enthalpy only results in a change in kinetic energy. dH = -d(u2)/2 = u du, where H here is a per-mass enthalpy (enthalpy per kg).

dH = TdS + VdP. This can be rearranged to read, TdS = dH -VdP = -u du – VdP.

We now use conservation of mass to put du into terms of P,V, and T. By conservation of mass, u/V is constant, or d(u/V)= 0. Taking the derivative of this quotient, du/V -u dV/V2= 0. Rearranging this, we get, du = u dV/V (No assumptions about entropy here). Since dH = -u du, we say that u2 dV/V = -dH = -TdS- VdP. It is now common to say that dS = 0 across the sound wave, and thus find that u2 = -V2 (dP/dV) at const S. For an ideal gas, this last derivative equals, PCp/VCv, so the speed of sound, u= √PVCp/Cv with the volume in terms of mass (kg/m3).

The problem comes in where we say that ∆S>0. At this point, I would say that u2 = -V(dH/dV) = VCp dT/dV > PVCp/Cv. Unless, I’ve made a mistake (always possible), I find that there is a small leading, non-adiabatic sound wave that goes ahead of the ordinary sound wave and is experienced only close to the source caused by mechanical energy that becomes degraded to raising T and gives rise more compression than would be expected for iso-entropic waves.

This should have some relevance to headphone design and speaker design since headphones are heard close to the ear, while speakers are heard further away. Meanwhile the recordings are made by microphones right next to the singers or instruments.

Robert E. Buxbaum, August 26, 2014

If hot air rises, why is it cold on mountain-tops?

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This is a child’s question that’s rarely answered to anyone’s satisfaction. To answer it well requires college level science, and by college the child has usually been dissuaded from asking anything scientific that would likely embarrass teacher — which is to say, from asking most anything. By a good answer, I mean here one that provides both a mathematical, checkable prediction of the temperature you’d expect to find on mountain tops, and one that also gives a feel for why it should be so. I’ll try to provide this here, as previously when explaining “why is the sky blue.” A word of warning: real science involves mathematics, something that’s often left behind, perhaps in an effort to build self-esteem. If I do a poor job, please text me back: “if hot air rises, what’s keeping you down?”

As a touchy-feely answer, please note that all materials have internal energy. It’s generally associated with the kinetic energy + potential energy of the molecules. It enters whenever a material is heated or has work done on it, and for gases, to good approximation, it equals the gas heat capacity of the gas times its temperature. For air, this is about 7 cal/mol°K times the temperature in degrees Kelvin. The average air at sea-level is taken to be at 1 atm, or 101,300  Pascals, and 15.02°C, or 288.15 °K; the internal energy of this are is thus 288.15 x 7 = 2017 cal/mol = 8420 J/mol. The internal energy of the air will decrease as the air rises, and the temperature drops for reasons I will explain below. Most diatomic gases have heat capacity of 7 cal/mol°K, a fact that is only explained by quantum mechanics; if not for quantum mechanics, the heat capacities of diatomic gases would be about 9 cal/mol°K.

Lets consider a volume of this air at this standard condition, and imagine that it is held within a weightless balloon, or plastic bag. As we pull that air up, by pulling up the bag, the bag starts to expand because the pressure is lower at high altitude (air pressure is just the weight of the air). No heat is exchanged with the surrounding air because our air will always be about as warm as its surroundings, or if you like you can imagine weightless balloon prevents it. In either case the molecules lose energy as the bag expands because they always collide with an outwardly moving wall. Alternately you can say that the air in the bag is doing work on the exterior air — expansion is work — but we are putting no work into the air as it takes no work to lift this air. The buoyancy of the air in our balloon is always about that of the surrounding air, or so we’ll assume for now.

A classic, difficult way to calculate the temperature change with altitude is to calculate the work being done by the air in the rising balloon. Work done is force times distance: w=  ∫f dz and this work should equal the effective cooling since heat and work are interchangeable. There’s an integral sign here to account for the fact that force is proportional to pressure and the air pressure will decrease as the balloon goes up. We now note that w =  ∫f dz = – ∫P dV because pressure, P = force per unit area. and volume, V is area times distance. The minus sign is because the work is being done by the air, not done on the air — it involves a loss of internal energy. Sorry to say, the temperature and pressure in the air keeps changing with volume and altitude, so it’s hard to solve the integral, but there is a simple approach based on entropy, S.

Les Droites Mountain, in the Alps, at the intersect of France Italy and Switzerland is 4000 m tall. The top is generally snow-covered.

Les Droites Mountain, in the Alps, at the intersect of France Italy and Switzerland is 4000 m tall. The top is generally snow-covered.

I discussed entropy last month, and showed it was a property of state, and further, that for any reversible path, ∆S= (Q/T)rev. That is, the entropy change for any reversible process equals the heat that enters divided by the temperature. Now, we expect the balloon rise is reversible, and since we’ve assumed no heat transfer, Q = 0. We thus expect that the entropy of air will be the same at all altitudes. Now entropy has two parts, a temperature part, Cp ln T2/T1 and a pressure part, R ln P2/P1. If the total ∆S=0 these two parts will exactly cancel.

Consider that at 4000m, the height of Les Droites, a mountain in the Mont Blanc range, the typical pressure is 61,660 Pa, about 60.85% of sea level pressure (101325 Pa). If the air were reduced to this pressure at constant temperature (∆S)T = -R ln P2/P1 where R is the gas constant, about 2 cal/mol°K, and P2/P1 = .6085; (∆S)T = -2 ln .6085. Since the total entropy change is zero, this part must equal Cp ln T2/T1 where Cp is the heat capacity of air at constant pressure, about 7 cal/mol°K for all diatomic gases, and T1 and T2 are the temperatures (Kelvin) of the air at sea level and 4000 m. (These equations are derived in most thermodynamics texts. The short version is that the entropy change from compression at constant T equals the work at constant temperature divided by T,  ∫P/TdV=  ∫R/V dV = R ln V2/V1= -R ln P2/P1. Similarly the entropy change at constant pressure = ∫dQ/T where dQ = Cp dT. This component of entropy is thus ∫dQ/T = Cp ∫dT/T = Cp ln T2/T1.) Setting the sum to equal zero, we can say that Cp ln T2/T1 =R ln .6085, or that 

T2 = T1 (.6085)R/Cp

T2 = T1(.6085)2/7   where 0.6065 is the pressure ratio at 4000, and because for air and most diatomic gases, R/Cp = 2/7 to very good approximation, matching the prediction from quantum mechanics.

From the above, we calculate T2 = 288.15 x .8676 = 250.0°K, or -23.15 °C. This is cold enough to provide snow  on Les Droites nearly year round, and it’s pretty accurate. The typical temperature at 4000 m is 262.17 K (-11°C). That’s 26°C colder than at sea-level, and only 12°C warmer than we’d predicted.

There are three weak assumptions behind the 11°C error in our predictions: (1) that the air that rises is no hotter than the air that does not, and (2) that the air’s not heated by radiation from the sun or earth, and (3) that there is no heat exchange with the surrounding air, e.g. from rain or snow formation. The last of these errors is thought to be the largest, but it’s still not large enough to cause serious problems.

The snow cover on Kilimanjaro, 2013. If global warming models were true, it should be gone, or mostly gone.

Snow on Kilimanjaro, Tanzania 2013. If global warming models were true, the ground should be 4°C warmer than 100 years ago, and the air at this altitude, about 7°C (12°F) warmer; and the snow should be gone.

You can use this approach, with different exponents, estimate the temperature at the center of Jupiter, or at the center of neutron stars. This iso-entropic calculation is the model that’s used here, though it’s understood that may be off by a fair percentage. You can also ask questions about global warming: increased CO2 at this level is supposed to cause extreme heating at 4000m, enough to heat the earth below by 4°C/century or more. As it happens, the temperature and snow cover on Les Droites and other Alp ski areas has been studied carefully for many decades; they are not warming as best we can tell (here’s a discussion). By all rights, Mt Blanc should be Mt Green by now; no one knows why. The earth too seems to have stopped warming. My theory: clouds. 

Robert Buxbaum, May 10, 2014. Science requires you check your theory for internal and external weakness. Here’s why the sky is blue, not green.