Tag Archives: drag

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

The mass of a car and its mpg.

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Back when I was an assistant professor at Michigan State University, MSU, they had a mileage olympics between the various engineering schools. Michigan State’s car got over 800 mpg, and lost soundly. By contrast, my current car, a Saab 9,2 gets about 30 miles per gallon on the highway, about average for US cars, and 22 to 23 mpg in the city in the summer. That’s about 1/40th the gas mileage of the Michigan State car, or about 2/3 the mileage of the 1978 VW rabbit I drove as a young professor, or the same as a Model A Ford. Why so low? My basic answer: the current car weighs a lot more.

As a first step to analyzing the energy drain of my car, or MSU’s, the energy content of gasoline is about 123 MJ/gallon. Thus, if my engine was 27% efficient (reasonably likely) and I got 22.5 mpg (36 km/gallon) driving around town, that would mean I was using about .922 MJ/km of gasoline energy. Now all I need to know is where is this energy going (the MSU car got double this efficiency, but went 40 times further).

The first energy sink I considered was rolling drag. To measure this without the fancy equipment we had at MSU, I put my car in neutral on a flat surface at 22 mph and measured how long it took for the speed to drop to 19.5 mph. From this time, 14.5 sec, and the speed drop, I calculated that the car had a rolling drag of 1.4% of its weight (if you had college physics you should be able to repeat this calculation). Since I and the car weigh about 1700 kg, or 3790 lb, the drag is 53 lb or 233 Nt (the MSU car had far less, perhaps 8 lb). For any friction, the loss per km is F•x, or 233 kJ/km for my vehicle in the summer, independent of speed. This is significant, but clearly there are other energy sinks involved. In winter, the rolling drag is about 50% higher: the effect of gooey grease, I guess.

The next energy sink is air resistance. This is calculated by multiplying the frontal area of the car by the density of air, times 1/2 the speed squared (the kinetic energy imparted to the air). There is also a form factor, measured on a wind tunnel. For my car this factor was 0.28, similar to the MSU car. That is, for both cars, the equivalent of only 28% of the air in front of the car is accelerated to the car’s speed. Based on this and the density of air in the summer, I calculate that, at 20 mph, air drag was about 5.3 lbs for my car. At 40 mph it’s 21 lbs (95 Nt), and it’s 65 lbs (295 Nt) at 70 mph. Given that my city driving is mostly at <40 mph, I expect that only 95 kJ/km is used to fight air friction in the city. That is, less than 10% of my gas energy in the city or about 30% on the highway. (The MSU car had less because of a smaller front area, and because it drove at about 25 mph)

The next energy sink was the energy used to speed up from a stop — or, if you like, the energy lost to the brakes when I slow down. This energy is proportional to the mass of the car, and to velocity squared or kinetic energy. It’s also inversely proportional to the distance between stops. For a 1700 kg car+ driver who travels at 38 mph on city streets (17 m/s) and stops, or slows every 500m, I calculate that the start-stop energy per km is 2 (1/2 m v2 ) = 1700•(17)2  = 491 kJ/km. This is more than the other two losses combined and would seem to explain the majority cause of my low gas mileage in the city.

The sum of the above losses is 0.819 MJ/km, and I’m willing to accept that the rest of the energy loss (100 kJ/km or so) is due to engine idling (the efficiency is zero then); to air conditioning and headlights; and to times when I have a passenger or lots of stuff in the car. It all adds up. When I go for long drives on the highway, this start-stop loss is no longer relevant. Though the air drag is greater, the net result is a mileage improvement. Brief rides on the highway, by contrast, hardly help my mileage. Though I slow down less often, maybe every 2 km, I go faster, so the energy loss per km is the same.

I find that the two major drags on my gas mileage are proportional to the weight of the car, and that is currently half-again the weight of my VW rabbit (only 1900 lbs, 900 kg). The MSU car was far lighter still, about 200 lbs with the driver, and it never stopped till the gas ran out. My suggestion, if you want the best gas milage, buy one light cars on the road. The Mitsubishi Mirage, for example, weighs 1000 kg, gets 35 mpg in the city.

A very aerodynamic, very big car. It's beautiful art, but likely gets lousy mileage -- especially in the city.

A very aerodynamic, very big car. It’s beautiful art, but likely gets lousy mileage — especially in the city.

Short of buying a lighter car, you have few good options to improve gas mileage. One thought is to use better grease or oil; synthetic oil, like Mobil 1 helps, I’m told (I’ve not checked it). Alternately, some months ago, I tried adding hydrogen and water to the engine. This helps too (5% -10%), likely by improving ignition and reducing idling vacuum loss. Another option is fancy valving, as on the Fiat 500. If you’re willing to buy a new car, and not just a new engine, a good option is a hybrid or battery car with regenerative breaking to recover the energy normally lost to the breaks. Alternately, a car powered with hydrogen fuel cells, — an option with advantages over batteries, or with a gasoline-powered fuel cell. 

Robert E. Buxbaum; July 29, 2015 I make hydrogen generators and purifiers. Here’s a link to my company site. Here’s something I wrote about Peter Cooper, an industrialist who made the first practical steam locomotive, the Tom Thumb: the key innovation here: making it lighter by using a forced air, fire-tube boiler.

What’s Holding Gilroy on the Roof

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We recently put a sculpture on our roof: Gilroy, or “Mr Hydrogen.” It’s a larger version of a creepy face sculpture I’d made some moths ago. Like it, and my saber-toothed tiger, the eyes follow you. A worry about this version: is there enough keeping it from blowing down on the cars? Anyone who puts up a large structure must address this worry, but I’m a professional engineer with a PhD from Princeton, so my answer is a bit different from most.

Gilroy (Mr Hydrogen) sculpture on roof of REB Research & Consulting. The eyes follow you.

Gilroy (Mr Hydrogen) sculpture on roof of REB Research & Consulting. The eyes follow you. Aim is that it should withstand 50 mph winds.

The main force on most any structure is the wind (the pyramids are classic exceptions). Wind force is generally proportional to the exposed area and to the wind-speed squared: something called form-drag or quadratic drag. Since force is related to wind-speed, I start with some good statistics for wind speed, shown in the figure below for Detroit where we are.

The highest Detroit wind speeds are typically only 16 mph, but every few years the winds are seen to reach 23 mph. These are low relative to many locations: Detroit has does not get hurricanes and rarely gets tornadoes. Despite this, I’ve decided to brace the sculpture to withstand winds of 50 mph, or 22.3 m/s. On the unlikely chance there is a tornado, I figure there would be so much other flotsam that I would not have to answer about losing my head. (For why Detroit does not get hurricanes or tornadoes, see here. If you want to know why tornadoes lift things, see here).

The maximum area Gilroy presents is 1.5 m2. The wind force is calculated by multiplying this area by the kinetic energy loss per second 1/2ρv2, times a form factor.  F= (Area)*ƒ* 1/2ρv2, where ρ is the density of air, 1.29Kg/m3, and v is velocity, 22.3 m/s. The form factor, ƒ, is about 1.25 for this shape: ƒ is found to be 1.15 for a flat plane, and 1.1 to 1.3 a rough sphere or ski-jumper. F = 1.5*1.25* (1/2 *1.29*22.32) = 603 Nt = 134 lb.; pressure is this divided by area. Since the weight is only about 40 lbs, I find I have to tie down the sculpture. I’ve done that with a 150 lb rope, tying it to a steel vent pipe.

Wind speed for Detroit month by month. Used to calculate the force. From http://weatherspark.com/averages/30042/Detroit-Michigan-United-States

Wind speed for Detroit month by month. Used to calculate the force. From http://weatherspark.com/averages/30042/Detroit-Michigan-United-States

It is possible that there’s a viscous lift force too, but it is likely to be small given the blunt shape and the flow Reynolds number: 3190. There is also the worry that Gilroy might fall apart from vibration. Gilroy is made of 3/4″ plywood, treated for outdoor use and then painted, but the plywood is held together with 25 steel screws 4″ long x 1/4″ OD. Screws like this will easily hold 134 lbs of steady wind force, but a vibrating wind will cause fatigue in the metal (bend a wire often enough and it falls apart). I figure I can leave Gilroy up for a year or so without worry, but will then go up to replace the screws and check if I have to bring him/ it down.

In the meantime, I’ll want to add a sign under the sculpture: “REB Research, home of Mr Hydrogen” I want to keep things surreal, but want to be safe and make sales.

by Robert E. Buxbaum, June 21, 2013