Category Archives: Engineering

What drives the gulf stream?

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I’m not much of a fan of todays’ kids’ science books because they don’t teach science IMHO. They have nice pictures and a few numbers; almost no equations, and lots of words. You can’t do science that way. On the odd occasion that they give the right answer to some problem, the lack of math means the kid has no way of understanding the reasoning, and no reason to believe the answer. Professional science articles on the web are bad in the opposite direction: too many numbers and for math, hey rely on supercomputers. No human can understand the outcome. I like to use my blog to offer science with insight, the type you’d get in an old “everyman science” book.

In previous posts, I gave answers to why the sky is blue, why it’s cold at the poles, why it’s cold on mountains, how tornadoes pick stuff up, and why hurricanes blow the way they do. In this post, we’ll try to figure out what drives the gulf-stream. The main argument will be deduction — disproving things that are not driving the gulf stream to leave us with one or two that could. Deduction is a classic method of science, well presented by Sherlock Holmes.

The gulf stream. The speed in the white area is ≥ 0.5 m/s (1.1 mph.).

The gulf stream. The speed in the white area is ≥ 0.5 m/s (1.1 mph.).

For those who don’t know, the Gulf stream is a massive river of water that runs within the Atlantic ocean. As shown at right, it starts roughly at the end of Florida, runs north to the Carolinas, and then turns dramatically east towards Spain. Flowing east, It’s about 150 miles wide, but only about 62 miles (100 km) when flowing along the US coast. According to some of the science books of my youth this massive flow was driven by temperature according to others, by salinity (whatever that means), and yet other books of my youth wind. My conclusion: they had no clue.

As a start to doing the science here, it’s important to fill in the numerical information that the science books left out. The Gulf stream is roughly 1000 meters deep, with a typical speed of 1 m/s (2.3 mph). The maximum speed is the surface water as the stream flows along the US coast. It is about 2.5 metres per second (5.6 mph), see map above.

From the size and the speed of the Gulf Stream, we conclude that land rivers are not driving the flow. The Mississippi is a big river with an outflow point near the head waters of the gulf stream, but the volume of flow is vastly too small. The volume of the gulf stream is roughly

Q=wdv = 100,000 x 1000 x .5 =  50 million m3/s = 1.5 billion cubic feet/s.

This is about 2000 times more flow than the volume flow of the Mississippi, 18,000 m3/s. The great difference in flow suggests the Mississippi could not be the driving force. The map of flow speeds (above) also suggest rivers do not drive the flow. The Gulf Stream does not flow at its maximum speed near the mouth of any river.  We now look for another driver.

Moving on to temperature. Temperature drives the whirl of hurricanes. The logic for temperature driving the gulf stream is as follows: it’s warm by the equator and cold at the poles; warm things expand and as water flows downhill, the polls will always be downhill from the equator. Lets put some math in here or my explanation will be lacking. First lets consider how much hight difference we might expect to see. The thermal expansivity of water is about 2x 10-4 m/m°C (.0002/°C) in the desired temperature range). To calculate the amount of expansion we multiply this by the depth of the stream, 1000m, and the temperature difference between two points, eg. the end of Florida to the Carolina coast. This is 5°C (9°F) I estimate. I calculate the temperature-induced seawater height as:

∆h (thermal) ≈ 5° x .0002/° x 1000m = 1 m (3.3 feet).

This is a fair amount of height. It’s only about 1/100 the height driving the Mississippi river, but it’s something. To see if 1 m is enough to drive the Gulf flow, I’ll compare it to the velocity-head. Velocity-head is a concept that’s useful in plumbing (I ran for water commissioner). It’s the potential energy height equivalent of any kinetic energy — typically of a fluid flow. The kinetic energy for any velocity v and mass of water, m is 1/2 mv2 . The potential energy equivalent is mgh. Combine the above and remove the mass terms, and we have:

∆h (velocity) = v2/2g.

Where g is the acceleration of gravity. Let’s consider  v = 1 m/s and g= 9.8 m/s2.≤ 0.05 m ≈ 2 inches. This is far less than the driving force calculated above. We have 5x more driving force than we need, but there is a problem: why isn’t the flow faster? Why does the Mississippi move so slowly when it has 100 times more head.

To answer the above questions, and to check if heat could really drive the Gulf Stream, we’ll check if the flow is turbulent — it is. A measure of how turbulent is based on something called the Reynolds number, Re#, it’s the ratio of kinetic energy and viscous loss in a fluid flow. Flows are turbulent if this ratio is more than 3000, or so;

Re# = vdρ/µ.

In the above, v is velocity, say 1 m/s, d is depth, 1000m, ρ = density, 1000 kg/m3 for water, and  0.00133 Pa∙s is the viscosity of water. Plug in these numbers, and we find a RE# = 750 million: this flow will be highly turbulent. Assuming a friction factor of 1/20 (.05), e find that we’d expect complete mixing 20 depths or 20 km. We find we need the above 0.05 m of velocity height to drive every 20 km of flow up the US coast. If the distance to the Carolina coast is 1000 km we need 1000*.05m/20 = 1 meter, that’s just about the velocity-head that the temperature difference would suggest. Temperature is thus a plausible driving force for 0.5 m/s, though not likely for the faster 2.5 m/s flow seen in the center of the stream. Turbulent flow is a big part of figuring the mpg of an automobile; it becomes rapidly more important at high speeds.

World sea salinity

World sea salinity. The maximum and minimum are in the wrong places.

What about salinity? For salinity to work, the salinity would have to be higher at the end of the flow. As a model of the flow, we might imagine that we freeze arctic seawater, and thus we concentrate salt in the seawater just below the ice. The heavy, saline water would flow down to the bottom of the sea, and then flow south to an area of low salinity and low pressure. Somewhere in the south, the salinity would be reduced by rains. If evaporation were to exceed the rains, the flow would go in the other direction. Sorry to say, I see no evidence of any of this. For one the end of the Gulf Stream is not that far north; there is no freezing, For two other problems: there are major rains in the Caribbean, and rains too in the North Atlantic. Finally, while the salinity head is too small. Each pen of salinity adds about 0.0001g/cc, and the salinity difference in this case is less than 1 ppm, lets say 0.5ppm.

h = .0001 x 0.5 x 1000 = 0.05m

I don’t see a case for northern-driven Gulf-stream flow caused by salinity.

Surface level winds in the Atlantic.

Surface level winds in the Atlantic. Trade winds in purple, 15-20 mph.

Now consider winds. The wind velocities are certainly enough to produce 5+ miles per hour flows, and the path of flows is appropriate. Consider, for example, the trade winds. In the southern Caribbean, they blow steadily from east to west slightly above the equator at 15 -20 mph. This could certainly drive a circulation flow of 4.5 mph north. Out of the Caribbean basin and along the eastern US coat the trade winds blow at 15-50 mph north and east. This too would easily drive a 4.5 mph flow.  I conclude that a combination of winds and temperature are the most likely drivers of the gulf stream flow. To quote Holmes, once you’ve eliminated the impossible, whatever remains, however improbable, must be the truth.

Robert E. Buxbaum, March 25, 2018. I used the thermal argument above to figure out how cold it had to be to freeze the balls off of a brass monkey.

Keeping your car batteries alive.

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Lithium-battery cost and performance has improved so much that no one uses Ni-Cad or metal hydride batteries any more. These are the choice for tools, phones, and computers, while lead acid batteries are used for car starting and emergency lights. I thought I’d write about the care and trade-offs of these two remaining options.

As things currently stand, you can buy a 12 V, lead-acid car battery with 40 Amp-h capacity for about $95. This suggests a cost of about $200/ kWh. The price rises to $400/kWh if you only discharge half way (good practice). This is cheaper than the per-power cost of lithium batteries, about $500/ kWh or $1000/ kWh if you only discharge half-way (good practice), but people pick lithium because (1) it’s lighter, and (2) it’s generally longer lasting. Lithium generally lasts about 2000 half-discharge cycles vs 500 for lead-acid.

On the basis of cost per cycle, lead acid batteries would have been replaced completely except that they are more tolerant of cold and heat, and they easily output the 400-800 Amps needed to start a car. Lithium batteries have problems at these currents, especially when it’s hot or cold. Lithium batteries deteriorate fast in the heat too (over 40°C, 105°F), and you can not charge a lithium car battery at more than 3-4 Amps at temperatures below about 0°C, 32°F. At higher currents, a coat of lithium metal forms on the anode. This lithium can react with water: 2Li + H2O –> Li2O + H2, or it can form dendrites that puncture the cell separators leading to fire and explosion. If you charge a lead acid battery too fast some hydrogen can form, but that’s much less of a problem. If you are worried about hydrogen, we sell hydrogen getters and catalysts that remove it. Here’s a description of the mechanisms.

The best thing you can do to keep a lead-acid battery alive is to keep it near-fully charged. This can be done by taking long drives, by idling the car (warming it up), or by use of an external trickle charger. I recommend a trickle charger in the winter because it’s non-polluting. A lead-acid battery that’s kept at near full charge will give you enough charge for 3000 to 5000 starts. If you let the battery completely discharge, you get only 50 or so deep cycles or 1000 starts. But beware: full discharge can creep up on you. A new car battery will hold 40 Ampere-hours of current, or 65,000 Ampere-seconds if you half discharge. Starting the car will take 5 seconds of 600 Amps, using 3000 Amp-s or about 5% of the battery’s juice. The battery will recharge as you drive, but not that fast. You’ll have to drive for at least 500 seconds (8 minutes) to recharge from the energy used in starting. But in the winter it is common that your drive will be shorter, and that a lot of your alternator power will be sent to the defrosters, lights, and seat heaters. As a result, your lead-acid battery will not totally charge, even on a 10 minute drive. With every week of short trips, the battery will drain a little, and sooner or later, you’ll find your battery is dead. Beware and recharge, ideally before 50% discharge

A little chemistry will help explain why full discharging is bad for battery life (for a different version see Wikipedia). For the first half discharge of a lead-acid battery, the reaction Is:

Pb + 2PbO2 + 2H2SO4  –> PbSO4 + Pb2O2SO4 + 2H2O.

This reaction involves 2 electrons and has a -∆G° of >394 kJ, suggesting a reversible voltage more than 2.04 V per cell with voltage decreasing as H2SO4 is used up. Any discharge forms PbSO4 on the positive plate (the lead anode) and converts lead oxide on the cathode (the negative plate) to Pb2O2SO4. Discharging to more than 50% involves this reaction converting the Pb2O2SO4 on the cathode to PbSO4.

Pb + Pb2O2SO4 + 2H2SO4  –> 2PbSO4 + 2H2O.

This also involves two electrons, but -∆G < 394 kJ, and voltage is less than 2.04 V. Not only is the voltage less, the maximum current is less. As it happens Pb2O2SO4 is amorphous, adherent, and conductive, while PbSO4 is crystalline, not that adherent, and not-so conductive. Operating at more than 50% results in less voltage, increased internal resistance, decreased H2SO4 concentrations, and lead sulfate flaking off the electrode. Even letting a battery sit at low voltage contributes to PbSO4 flaking off. If the weather is cold enough, the low concentration H2SO4 freezes and the battery case cracks. My advice: Get out your battery charger and top up your battery. Don’t worry about overcharging; your battery charger will sense when the charge is complete. A lead-acid battery operated at near full charge, between 67 and 100% will provide 1500 cycles, about as many as lithium. 

Trickle charging my wife's car. Good for battery life. At 6 Amps, expect this to take 3-6 hours.

Trickle charging my wife’s car: good for battery life. At 6 Amps, expect a full charge to take 6 hours or more. You might want to recharge the battery in your emergency lights too. 

Lithium batteries are the choice for tools and electric vehicles, but the chemistry is different. For longest life with lithium batteries, they should not be charged fully. If you change fully they deteriorate and self-discharge, especially when warm (100°F, 40°C). If you operate at 20°C between 75% and 25% charge, a lithium-ion battery will last 2000 cycles; at 100% to 0%, expect only 200 cycles or so.

Tesla cars use lithium batteries of a special type, lithium cobalt. Such batteries have been known to explode, but and Tesla adds sophisticated electronics and cooling systems to prevent this. The Chevy Volt and Bolt use lithium batteries too, but they are less energy-dense. In either case, assuming $1000/kWh and a 2000 cycle life, the battery cost of an EV is about 50¢/kWh-cycle. Add to this the cost of electricity, 15¢/kWh including the over-potential needed to charge, and I find a total cost of operation of 65¢/kWh. EVs get about 3 miles per kWh, suggesting an energy cost of about 22¢/mile. By comparison, a 23 mpg car that uses gasoline at $2.80 / gal, the energy cost is 12¢/mile, about half that of the EVs. For now, I stick to gasoline for normal driving, and for long trips, suggest buses, trains, and flying.

Robert Buxbaum, January 4, 2018.

Hydrogen permeation rates in Inconel, Hastelloy and stainless steels.

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Some 20 years ago, I published a graph of the permeation rate for hydrogen in several metals at low pressure, See the graph here, but I didn’t include stainless steel in the graph.

Hydrogen permeation in clean SS-304; four research groups’ data.

One reason I did not include stainless steel was there were many stainless steels and the hydrogen permeation rates were different, especially so between austenitic (FCC) steels and ferritic steels (BCC). Another issue was oxidation. All stainless steels are oxidized, and it affect H2 permeation a lot. You can decrease the hydrogen permeation rate significantly by oxidation, or by surface nitriding, etc (my company will even provide this service). Yet another issue is cold work. When  an austenitic stainless steel is worked — rolled or drawn — some Austinite (FCC) material transforms to Martisite (a sort of stretched BCC). Even a small amount of martisite causes an order of magnitude difference in the permeation rate, as shown below. For better or worse, after 20 years, I’m now ready to address H2 in stainless steel, or as ready as I’m likely to be.

Hydrogen permeation data for SS 340 and SS 321.

Hydrogen permeation in SS 340 and SS 321. Cold work affects H2 permeation more than the difference between 304 and 321; Sun Xiukui, Xu Jian, and Li Yiyi, 1989

The first graph I’d like to present, above, is a combination of four research groups’ data for hydrogen transport in clean SS 304, the most common stainless steel in use today. SS 304 is a ductile, austenitic (FCC), work hardening, steel of classic 18-8 composition (18% Cr, 8% Ni). It shares the same basic composition with SS 316, SS 321 and 304L only differing in minor components. The data from four research groups shows a lot of scatter: a factor of 5 variation at high temperature, 1000 K (727 °C), and almost two orders of magnitude variation (factor of 50) at room temperature, 13°C. Pressure is not a factor in creating the scatter, as all of these studies were done with 1 atm, 100 kPa hydrogen transporting to vacuum.

The two likely reasons for the variation are differences in the oxide coat, and differences in the amount of cold work. It is possible these are the same explanation, as a martensitic phase might increase H2 permeation by introducing flaws into the oxide coat. As the graph at left shows, working these alloys causes more differences in H2 permeation than any difference between alloys, or at least between SS 304 and SS 321. A good equation for the permeation behavior of SS 304 is:

P (mol/m.s.Pa1/2) = 1.1 x10-6 exp (-8200/T).      (H2 in SS-304)

Because of the song influence of cold work and oxidation, I’m of the opinion that I get a slightly different, and better equation if I add in permeation data from three other 18-8 stainless steels:

P (mol/m.s.Pa1/2) = 4.75 x10-7 exp (-7880/T).     (H2 in annealed SS-304, SS-316, SS-321)

Screen Shot 2017-12-16 at 10.37.37 PM

Hydrogen permeation through several common stainless steels, as well as Inocnel and Hastelloy

Though this result is about half of the previous at high temperature, I would trust it better, at least for annealed SS-304, and also for any annealed austenitic stainless steel. Just as an experiment, I decided to add a few nickel and cobalt alloys to the mix, and chose to add data for inconel 600, 625, and 718; for kovar; for Hastelloy, and for Fe-5%Si-5%Ge, and SS4130. At left, I pilot all of these on one graph along with data for the common stainless steels. To my eyes the scatter in the H2 permeation rates is indistinguishable from that SS 304 above or in the mixed 18-8 steels (data not shown). Including these materials to the plot decreases the standard deviation a bit to a factor of 2 at 1000°K and a factor of 4 at 13°C. Making a least-square analysis of the data, I find the following equation for permeation in all common FCC stainless steels, plus Inconels, Hastelloys and Kovar:

P (mol/m.s.Pa1/2) = 4.3 x10-7 exp (-7850/T).

This equation is near-identical to the equation above for mixed, 18-8 stainless steel. I would trust it for annealed or low carbon metal (SS-304L) to a factor of 2 accuracy at high temperatures, or a factor of 4 at low temperatures. Low carbon reduces the tendency to form Martinsite. You can not use any of these equations for hydrogen in ferritic (BCC) alloys as the rates are different, but this is as good as you’re likely to get for basic austenitc stainless and related materials. If you are interested in the effect of cold work, here is a good reference. If you are bothered by the square-root of pressure driving force, it’s a result of entropy: hydrogen travels in stainless steel as dislocated H atoms and the dissociation H2 –> 2 H leads to the square root.

Robert Buxbaum, December 17, 2017. My business, REB Research, makes hydrogen generators and purifiers; we sell getters; we consult on hydrogen-related issues, and will (if you like) provide oxide (and similar) permeation barriers.

Change home air filters 3 times per year

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Energy efficient furnaces use a surprisingly large amount of electricity to blow the air around your house. Part of the problem is the pressure drop of the ducts, but quite a lot of energy is lost bowing air through the dust filter. An energy-saving idea: replace the filter on your furnace twice a year or more. Another idea, you don’t have to use the fanciest of filters. Dirty filters provide a lot of back-pressure especially when they are dirty.

I built a water manometer, see diagram below to measure the pressure drop through my furnace filters. The pressure drop is measured from the difference in the height of the water column shown. Each inch of water is 0.04 psi or 275 Pa. Using this pressure difference and the flow rating of the furnace, I calculated the amount of power lost by the following formula:

W = Q ∆P/ µ.

Here W is the amount of power use, Watts, Q is flow rate m3/s, ∆P = the pressure drop in Pa, and µ is the efficiency of the motor and blower, typically about 50%.

With clean filters (two different brands), I measured 1/8″ and 1/4″ of water column, or a pressure drop of 0.005 and 0.01 psi, depending on the filter. The “better the filter”, that is the higher the MERV rating, the higher the pressure drop. I also measured the pressure drop through a 6 month old filter and found it to be 1/2″ of water, or 0.02 psi or 140 Pa. Multiplying this by the amount of air moved, 1000 cfm =  25 m3 per minute or 0.42 m3/s, and dividing by the efficiency, I calculate a power use of 118 W. That is 0.118 kWh/hr. or 2.8 kWh/day.

water manometer used to measure pressure drop through the filter of my furnace. I stuck two copper tubes into the furnace, and attached a plastic hose. Pressure was measured from the difference in the water level in the hose.

The water manometer I used to measure the pressure drop through the filter of my furnace. I stuck two copper tubes into the furnace, and attached a plastic tube half filled with water between the copper tubes. Pressure was measured from the difference in the water level in the plastic tube. Each 1″ of water is 280 Pa or 0.04psi.

At the above rate of power use and a cost of electricity of 11¢/kWhr, I find it would cost me an extra 4 KWhr or about 31¢/day to pump air through my dirty-ish filter; that’s $113/year. The cost through a clean filter would be about half this, suggesting that for every year of filter use I spend an average of $57t where t is the use life of the filter.

To calculate the ideal time to change filters I set up the following formula for the total cost per year $, including cost per year spent on filters (at $5/ filter), and the pressure-induced electric cost:

$ = 5/t + 57 t.

The shorter the life of the filter, t, the more I spend on filters, but the less on electricity. I now use calculus to find the filter life that produces the minimum $, and determine that $ is a minimum at a filter life t = √5/57 = .30 years.  The upshot, then, if you filters are like mine, you should change your three times a year, or so; every 3.6 months to be super-exact. For what it’s worth, I buy MERV 5 filters at Ace or Home Depot. If I bought more expensive filters, the optimal change time would likely be once or twice per year. I figure that, unless you are very allergic or make electronics in your basement you don’t need a filter with MERV rating higher than 8 or so.

I’ve mentioned in a previous essay/post that dust starts out mostly as dead skin cells. Over time dust mites eat the skin, some pretty nasty stuff. Most folks are allergic to the mites, but I’m not convinced that the filter on your furnace dies much to isolate you from them since the mites, etc tend to hang out in your bed and clothes (a charming thought, I know).

Old fashioned, octopus furnace. Free convection.

Old fashioned, octopus furnace. Free convection.

The previous house I had, had no filter on the furnace (and no blower). I noticed no difference in my tendency to cough or itch. That furnace relied for circulation on the tendency for hot air to rise. That is, “free convection” circulated air through the home and furnace by way of “Octopus” ducts. If you wonder what a furnace like that looks like here’s a picture.

I calculate that a 10 foot column of air that is 30°C warmer than that in a house will have a buoyancy of about 0.00055 psi (1/8″ of water). That’s enough pressure to drive circulation through my home, and might have even driven air through a clean, low MERV dust filter. The furnace didn’t use any more gas than a modern furnace would, as best I could tell, since I was able to adjust the damper easily (I could see the flame). It used no electricity except for the thermostat control, and the overall cost was lower than for my current, high-efficiency furnace with its electrical blower and forced convection.

Robert E. Buxbaum, December 7, 2017. I ran for water commissioner, and post occasional energy-saving or water saving ideas. Another good energy saver is curtains. And here are some ideas on water-saving, and on toilet paper.

Forced diversity of race is racist

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Let me browse through some thoughts on efforts to address endemic racism. I’m not sure I’ll get anywhere, but you might as well enter the laboratory of my mind on the issue.

I’d like to begin with a line of the bible (why not?) “‘Do not pervert justice; do not show partiality to the poor or favoritism to the great, but judge your neighbor fairly.” (Lev. 19:15). This sounds good, but in college admissions, I’ve found we try to do better by showing  favoritism to the descendants of those who’ve been historically left-out. This was called affirmative action, it’s now called “diversity”.  

In 1981, when I began teaching chemical engineering at Michigan State University, our department had race-based quotas to allow easier admission to the descendants of historically-disadvantaged groups. All major universities did this at the time. The claim was that it would be temporary; it continues to this day. In our case, the target was to get 15% or so black, Hispanics and American Indians students (7 in a class of 50). We achieved this target by accepting such students with a 2.0 GPA, and not requiring a math or science background; Caucasians required 3.0 minimum, and we did require math or science. I’m not sure we helped the disadvantaged by this, either personally or professionally, but we made the administration happy. The kids seemed happy too, at least for a while. The ones we got were, by and large, bright. To make up for the lack of background we offered tutoring and adjusted grades. Some diversity students did well, others didn’t. Mostly they went into HR or management after graduation, places they could have gone without our efforts.

After some years, the Supreme court ended our quota based selection, saying it was, itself racist. They said we could still reverse-discriminate for “diversity,” though. That is, if the purpose wasn’t to address previous wrongs, but to improve the class. We changed our literature, but kept our selection methods and kept the same percentage targets as before.

This is a popular meme about racism. It makes sense to me.

This is a popular meme about racism.

The only way we monitored that we met the race-percent target was by a check-box on forms. Students reported race, and we collected this, but we didn’t check that black students look black or Hispanic students spoke Spanish. There was no check on student honesty. Anyone who checked the box got the benefits. This lack of check bread cheating at MSU and elsewhere. Senator Elizabeth Warren got easy entry into Harvard and Penn, in part by claiming to be an Indian on her forms. She has no evidence of Indian blood or culture Here’s Snopes. My sense is that our methods mostly help the crooked.

The main problem with is, I suspect, is the goal. We’ve decided to make every university department match the state’s racial breakdown. It’s a pretty goal, but it doesn’t seem like one that helps students or the state. Would it help the MSU hockey squad to force to team to racially match the state; would it help the volleyball team, or the football team?  So why assume it helps every academic department to make it’s racial makeup match the state’s. Why not let talented black students head to business or management departments before graduation. They might go further without our intervention.

This is not to say there are not racial inequalities, but I suspect that these diversity programs don’t help the students, and may actually hurt. They promote crookedness, and divert student attention from achieving excellence to maintaining victim status. Any group that isn’t loud enough in claiming victim status is robbed of the reverse-discrimination that they’ve been told they need. They’re told they can’t really compete, and many come to believe it. In several universities, we gone so far as to hire “bias referees” to protect minorities from having to defend their intellectual views in open discussion. The referee robs people of the need to think, and serves, I suspect, no one but a group of powerful politicians and administrators — people you are not supposed to criticize. On that topic, here is a video of Malcolm X talking about the danger of white liberals. Clearly he can hold his own in a debate without having a bias referee, and he makes some very good points about white liberals doing more harm than good.

Robert Buxbaum, November 5, 2017. In a related problem, black folks are arrested too often. I suggest rational drug laws. Some financial training could help too.

How Tesla invented, I think, Tesla coils and wireless chargers.

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I think I know how Tesla invented his high frequency devices, and thought I’d show you, while also explaining the operation of some devices that develop from in. Even if I’m wrong in historical terms, at least you should come to understand some of his devices, and something of the invention process. Either can be the start of a great science fair project.

physics drawing of a mass on a spring, left, and of a grounded capacitor and inception coil, right.

The start of Tesla’s invention process, I think, was a visual similarity– I’m guessing he noticed that the physics symbol for a spring was the same as for an electrical, induction coil, as shown at left. A normal person would notice the similarity, and perhaps think about it for a few seconds, get no where, and think of something else. If he or she had a math background — necessary to do most any science — they might look at the relevant equations and notice that they’re different. The equation describing the force of a spring is F = -k x  (I’ll define these letters in the bottom paragraph). The equation describing the voltage in an induction coil is not very similar-looking at first glance, V = L di/dt.  But there is a key similarity that could appeal to some math aficionados: both equations are linear. A linear equation is one where, if you double one side you double the other. Thus, if you double F, you double x, and if you double V, you double dI/dt, and that’s a significant behavior; the equation z= at2 is not linear, see the difference?

Another linear equation is the key equation for the motion for a mass, Newton’s second law, F = ma = m d2x/dt2. This equation is quite complicated looking, since the latter term is a second-derivative, but it is linear, and a mass is the likely thing for a spring to act upon. Yet another linear equation can be used to relate current to the voltage across a capacitor: V= -1/C ∫idt. At first glance, this equation looks quite different from the others since it involves an integral. But Nicola Tesla did more than a first glance. Perhaps he knew that linear systems tend to show resonance — vibrations at a fixed frequency. Or perhaps that insight came later. 

And Tesla saw something else, I imagine, something even less obvious, except in hindsight. If you take the derivative of the two electrical equations, you get dV/dt = L d2i/dt2, and dV/dt = -1/C i . These equations are the same as for the spring and mass, just replace F and x by dV/dt and i. That the derivative of the integral is the thing itself is something I demonstrate here. At this point it becomes clear that a capacitor-coil system will show the same sort of natural resonance effects as shown by a spring and mass system, or by a child’s swing, or by a bouncy bridge. Tesla would have known, like anyone who’s taken college-level physics, that a small input at the right, resonant frequency will excite such systems to great swings. For a mass and spring,

Basic Tesla coil. A switch set off by magnetization of the iron core insures resonant frequency operation.

Basic Tesla coil. A switch set off by magnetization of the iron core insures resonant frequency operation.

resonant frequency = (1/2π) √k/m,

Children can make a swing go quite high, just by pumping at the right frequency. Similarly, it should be possible to excite a coil-capacitor system to higher and higher voltages if you can find a way to excite long enough at the right frequency. Tesla would have looked for a way to do this with a coil capacitor system, and after a while of trying and thinking, he seems to have found the circuit shown at right, with a spark gap to impress visitors and keep the voltages from getting to far out of hand. The resonant frequency for this system is 1/(2π√LC), an equation form that is similar to the above. The voltage swings should grow until limited by resistance in the wires, or by the radiation of power into space. The fact that significant power is radiated into space will be used as the basis for wireless phone chargers, but more on that later. For now, you might wish to note that power radiation is proportional to dV/dt.

A version of the above excited by AC current. In this version, you achieve resonance by adjusting the coil, capacitor and resistance to match the forcing frequency.

A more -modern version of the above excited by AC current. In this version, you achieve resonance by adjusting the coil, capacitor and resistance to match the forcing frequency.

The device above provides an early, simple way to excite a coil -capacitor system. It’s designed for use with a battery or other DC power source. There’s an electromagnetic switch to provide resonance with any capacitor and coil pair. An alternative, more modern device is shown at left. It  achieves resonance too without the switch through the use of input AC power, but you have to match the AC frequency to the resonant frequency of the coil and capacitor. If wall current is used, 60 cps, the coil and capacitor must be chosen so that  1/(2π√LC) = 60 cps. Both versions are called Tesla coils and either can be set up to produce very large sparks (sparks make for a great science fair project — you need to put a spark gap across the capacitor, or better yet use the coil as the low-voltage part of a transformer.

power receiverAnother use of this circuit is as a transmitter of power into space. The coil becomes the transmission antenna, and you have to set up a similar device as a receiver, see picture at right. The black thing at left of the picture is the capacitor. One has to make sure that the coil-capacitor pair is tuned to the same frequency as the transmitter. One also needs to add a rectifier, the rectifier chosen here is designated 1N4007. This, fairly standard-size rectifier allows you to sip DC power to the battery, without fear that the battery will discharge on every cycle. That’s all the science you need to charge an iPhone without having to plug it in. Designing one of these is a good science fair project, especially if you can improve on the charging distance. Why should you have to put your iPhone right on top of the transmitter battery. Why not allow continuous charging anywhere in your home. Tesla was working on long-distance power transmission till the end of his life. What modifications would that require?

Symbols used above: a = acceleration = d2x/dt2, C= capacitance of the capacitor, dV/dt = the rate of change of voltage with time, F = force, i = current, k = stiffness of the spring, L= inductance of the coil, m = mass of the weight, t= time, V= voltage, x = distance of the mass from its rest point.

Robert Buxbaum, October 2, 2017.

Estimating the strength of an atom bomb

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As warfare is a foundation of engineering, I thought I’d use engineering to evaluate the death-dealing power of North Korea’s atomic/hydrogen bomb, tested September 3, 2017. The key data in evaluating a big bomb is its seismic output. They shake the earth like earthquakes do, and we measure the power like earthquakes, using seismometers. I’ve seen two seismographs comparing the recent bomb to the previous. One of these, below, is from CTBTO, the Center for Test Ban Treaty Oversight, via a seismometer in western Kazakhstan (see original data and report).

Seismic output of all North Korean nuclear tests.

Seismic output, to scale, of all declared DPNK nuclear tests as observed from IMS station AS-59 in Western Kazakhstan

North Korea’s previous bomb, exploded 9 September 2016, was reported to be slightly more powerful than the ones we dropped on Hiroshima and Nagasaki, suggesting it was about 20 kilotons. According to CTBTO, it registered 5.3 on the Richter scale. The two tests before that appear somewhat less powerful, perhaps 7-10 kilotons, and the two before that appear as dismal failures — fizzles, in atomic bomb parlance. The MOAB bomb, by comparison, was 9 Tons, or 0.009 kiloTons, a virtual non-entity.

To measure the output of the current bomb, I place a ruler on my screen and measure the maximum distance between the top to bottom wiggles. I find that this bomb’s wiggles measure 5 cm, while the previous measures 5 mm. This bomb’s wiggles are ten times bigger, and from this I determine that this explosion registered 6.3 on the Richter scale, 1.0 more than the previous — the Richter scale is the logarithmic measure of the wiggle amplitude, so ten times the shake magnitude  is an addition of 1.0 on the scale. My calculation of 6.3 exactly matches that of the US geological survey. The ratio of wiggle heights was less on the, NORSAR seismometer, Norway, see suggesting 5.8 to 5.9 on the Richter scale. The European agencies have taken to reporting 6.1, an average value, though they originally reported only the 5.8 from NORSAR, and a bomb power commensurate with that.

We calculate the bomb power from the Richter-scale measure, or the ratio of the wiggles. Bomb power is proportional to wiggle height to the 3/2 power. Using the data above, ten times the wiggle, this bomb appears to be 10^3/2 = 31.6 times as powerful as the last, or 31.6 x 20kTon = 630kTon (630,000 tons of TNT). If we used the European value of 6.1, the calculated power would be about half this, 315 kTons, and if we used the NORSAR’s original value, it would suggest the bomb had less than half this power. Each difference of 0.2 on the Richter scale is a factor of two in power. For no obvious reason we keep reporting 120 to 160 kTons.

NORSAR comparison of North Korean bomb blasts

NORSAR comparison of North Korean blasts — suggests the current bomb is smaller; still looks like hydrogen.

As it happens, death power is proportional to the kiloton power, other things being equal. The bombs we dropped on Hiroshima and Nagasaki were in the 15 to 20 kTon range and killed 90,000 each. Based on my best estimate of the bomb, 315 kTons, I estimate that it would kill 1.6 million people if used on an industrial city, like Seoul, Yokohama, or Los Angeles. In my opinion, this is about as big a bomb as any rational person has reason to make (Stalin made bigger, as did Eisenhower).

We now ask if this is an atom bomb or a hydrogen-fusion bomb. Though I don’t see any war-making difference, if it’s a hydrogen bomb that would make our recent treaty with Iran look bad, as it gave Iran nuclear fusion technology — I opposed the treaty based on that. Sorry to say, from the seismic signature it looks very much like a hydrogen bomb. The only other way to get to this sort of high-power explosion is via a double-acting fission bomb where small atom bomb sets off a second, bigger fission bomb. When looking at movies of Eisenhower-era double-acting explosions, you’ll notice that the second, bigger explosion follows the first by a second or so. I see no evidence of this secondary-delay in the seismic signature of this explosion, suggesting this was a hydrogen bomb, not a double. I expect Iran to follow the same path in 3-4 years.

As a political thought, it seems to me that the obvious way to stop North Korea would be to put pressure on China by making a military pact with Russia. Until that is done, China has little to fear from a North Korean attack to the south. Of course, to do that we’d likely have to cut our support of NATO, something that the Germans fear. This is a balance-of-power solution, the sort that works, short of total annihilation. It was achieved at the congress of Vienna, at the treaty of Ghent, and by Henry Kissinger through détente. It would work again. Without it, I see the Korean conflict turning hot again, soon.

Robert Buxbaum, September 11, 2017.

Activated sludge sewage treatment bioreactors

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I ran for water commissioner of Oakland county in 2016, a county with 1.3 million people and eight sewage treatment plants. One of these plants uses the rotating disk contractor, described previously, but the others process sewage by bubbling air through it in a large tank — the so-called, activated sludge process. A description is found here in Wikipedia, but with no math, and thus, far less satisfying than it could be. I thought I might describe this process relevant mathematics, for my understanding and those interested: what happens to your stuff after you flush the toilet or turn on the garbage disposal.

Simplified sewage plant: a plug-flow reactor with a 90+% solids recycle used to maintain a high concentration of bio-catalyst material.

Simplified sewage plant: a bubbling, plug-flow bio-reactor with 90% solids recycle and a settler used to extract floc solids and bio-catalyst material.

In most of the USA, sanitary sewage, the stuff from your toilet, sink, etc. flows separately from storm water to a treatment plant. At the plant, the sewage is first screened (rough filtered) and given a quick settle to remove grit etc. then sent to a bubbling flow, plug-flow bioreactor like the one shown at right. Not all cities use this type of sludge processes, but virtually every plant I’ve seen does, and I’ve come to believe this is the main technology in use today.

The sewage flows by gravity, typically, a choice that provides reliability and saves on operating costs, but necessitates that the sewage plant is located at the lowest point in the town, typically on a river. The liquid effluent of the sewage, after bio-treatment is typically dumped in the river, a flow that is so great more than, during dry season, more than half the flow of several rivers is this liquid effluent of our plants – an interesting factoid. For pollution reasons, it is mandated that the liquid effluent leaves the plant with less than 2 ppm organics; that is, it leaves the plant purer than normal river water. After settling and screening, the incoming flow to the bio-reactor typically contains about 400 ppm of biomaterial (0.04%), half of it soluble, and half as suspended colloidal stuff (turd bits, vegetable matter, toilet paper, etc). Between the activated sludge bio-reactor and the settler following it manage to reduce this concentration to 2 ppm or less. Soluble organics, about 200 ppm, are removed by this cellular oxidation (metabolism), while the colloidal material, the other 200 ppm, is removed by adsorption on the sticky flocular material in the tank (the plug-flow tank is called an oxidation ditch, BTW). The sticky floc is a product of the cells. The rate of oxidation and of absorption processes are proportional to floc concentration, F and to organic concentration, C. Mathematically we can say that

dC/dt = -kFC

where C and F are the concentration of organic material and floc respectively; t is time, and k is a reaction constant. It’s not totally a constant, since it is proportional to oxygen concentration and somewhat temperature dependent, but I’ll consider it constant for now.

As shown in the figure above, the process relies on a high recycle of floc (solids) to increase the concentration of cells, and speed the process. Because of this high recycle, we can consider the floc concentration F to be a constant, independent of position along the reactor length.

The volume of the reactor-ditch, V, is fixed -it’s a concrete ditch — but the flow rate into the ditch, Q, is not fixed. Q is high in the morning when folks take showers, and low at night. It’s also higher — typically about twice as high — during rain storms, the result of leakage and illegal connections. For any flow rate, Q, there is a residence time for a bit of sewage flowing through a tank, τ = V/Q. We can now solve the above equation for the value of τ for an incoming concentration C° = 400 ppm, an outgoing concentration Co of 2 ppm. We integrate the equation above and find that:

ln (C°/Co) = kFτ

Where τ equals the residence time, τ = V/Q. Thus,

ln (C°/Co) = kFV/Q.

The required volume of reactor, V, is related to the flow rate, Q, as follows for typical feed and exit concentrations:

V = Q/kF ln( 400/2) = 5.3 Q/kF.

The volume is seen to be dependent on F. In Oakland county, tank volume V is chosen to be one or two times the maximum expected value of Q. To keep the output organic content to less than 2 ppm, F is maintained so that kF≥ 5.3 per day. Thus, in Oakland county, a 2 million gallon per day sewage plant is built with a 2-4 million gallon oxidation ditch. The extra space allows for growth of the populations and for heavy rains, and insures that most of the time, the effluent contains less than 2 ppm organics.

Bob Martin by the South Lyon, MI, Activated Sludge reactor

Bob Martin chief engineer the South Lyon, MI, Activated Sludge plant, 2016. His innovation was to control the air bubblers according to measurements of the oxygen content. The O2 sensor is at bottom; the controller is at right. When I was there, some bubblers were acting up.

As you may guess, the activated sludge process requires a lot of operator control, far more than the rotating disk contractor we described. There is a need for constant monitoring and tweaking. The operator deals with some of the variations in Q by adjusting the recycle amount, with other problems by adjusting the air flow, or through the use of retention tanks upstream or downstream of the reactor, or by adding components — sticky polymer, FeCl3, etc. Finally, in have rains, the settler-bottom fraction itself is adjusted (increased). Because of all the complexity. sewer treatment engineer is a high-pay, in demand, skilled trade. If you are interested, contact me or the county. You’ll do yourself and the county a service.

I’d mentioned that the effluent water goes to the rivers in Oakland county. In some counties it goes to the fields, a good idea, I think. As for the solids, in Oakland county, the solid floc is concentrated to a goo containing about 5% solids. (The goo is called unconsolidated sludge) It is shipped free to farmer fields, or sometimes concentrated to more than 5% (consolidated sludge), and provided with additional treatment, anaerobic digestion to improve the quality and extract some energy. If you’d like to start a company to do more with our solids, that would be very welcome. In Detroit the solids are burned, a very wasteful, energy-consuming process, IMHO. In Wisconsin, the consolidated sludge is dried, pelletized, and sold as a popular fertilizer, Milorganite.

Dr. Robert Buxbaum, August 1, 2017. A colleague of mine owned (owns?) a company that consulted on sewage-treatment and manufactured a popular belt-filter. The name of his company: Consolidated Sludge. Here are some sewer jokes and my campaign song.

A rotating disk bio-reactor for sewage treatment

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One of the most effective designs for sewage treatment is the rotating disk bio-reactor, shown below. It is typically used in small-throughput sewage plants, but it performs quite well in larger plants too. I’d like to present an analysis of the reactor, and an explanation of why it works so well.

A rotating disc sewage reactor.

A rotating disk sewage reactor; ∂ is the thickness of the biofilm. It’s related to W the rotation rate in radians per sec, and to D the limiting diffusivity.

As shown, the reactor is fairly simple-looking, nothing more than a train of troughs filled with sewage-water, typically 3-6 feet deep, with a stack of discs rotating within. The discs are typically 7 to 14 feet in diameter (2-4 meters) and 1 cm apart. The shaft is typically close to the water level, but slightly above, and the rotation speed is adjustable. The device works because appropriate bio-organisms attach themselves to the disk, and the rotation insures that they are fully (or reasonably) oxygenated.

How do we know the cells on the disc will be oxygenated? The key is the solubility of oxygen in water, and the fact that these discs are only used on the low biological oxygen demand part of the sewage treatment process, only where the sewage contains 40 ppm of soluble organics or less. The main reaction on the rotating disc is bio oxidation of soluble carbohydrate (sugar) in a layer of wet slime attached to the disc.

H-O-C-H + O2 –> CO2 + H2O.

As it happens, the solubility of pure oxygen in water is about 40 ppm at 1 atm. As air contains 21% oxygen, we expect an 8 ppm concentration of oxygen on the slime surface: 21% of 40 ppm = 8 ppm. Given the reaction above and the fact that oxygen will diffuse five times more readily than sugar at least, we expect that one disc rotation will easily provide enough oxygen to remove 40 ppm sugar in the slime at every speed of rotation so long as the wheel is in the air at least half of the time, as shown above.

Let’s now pick a rotation speed of 1/3 rpm (3 minutes per rotation) and see where that gets us in terms of speed of organic removal. Since the disc is always in an area of low organic concentration, it becomes covered mostly with “rotifers”, a fungus that does well in low nutrient (low BOD) sewage. Let’s now assume that mass transfer (diffusion) of sugar in the rotifer slime determines the thickness of the rotifera layer, and thus the rate of organic removal. We can calculate the diffusion depth of sugar, ∂ via the following equation, derived in my PhD thesis.

∂ = √πDt

Here, D is the diffusivity (cm2/s) for sugar in the rotifera slime attached to the disk, π = 3.1415.. and t is the contact time, 90 seconds in the above assumption. My expectation is that D in the rotifer slime will be similar to the diffusivity sugar in water, about 3 x 10-6 cm2/s. Based on the above, we find the rotifer thickness will be ∂ = √.00085 cm2 = .03 cm, and the oxygen depth will be about 2.5 times that, 0.07 cm. If the discs are 1 cm apart, we find that, about 14% of the fluid volume of the reactor will be filled with slime, with 2/5 of this rotifer-filled. This is as much as 1000 times more rotifers than you would get in an ordinary constantly stirred tank reactor, a CSTR to use a common acronym. We might now imagine that the volume of this sewage-treatment reactor can be as small as 1000 gallons, 1/1000 the size of a CSTR. Unfortunately it is not so; we’ll have to consider another limiting effect, diffusion of nutrients.

Consider the diffusive mass transfer of sugar from a 1,000,000 gal/day flow (43 liters/sec). Assume that at some point in the extraction you have a concentration C(g/cc) of sugar in the liquid where C is between 40 ppm and 1 ppm. Let’s pick a volume of the reactor that is 1/20 the normal size for this flow (not 1/1000 the size, you’ll see why). That is to say a trough whose volume is 50,000 gallons (200,000 liters, 200 m3). If the discs are 1 cm apart, this trough (or section of a trough) will have about  4×10^8 cm2 of submerged surface, and about 9×10^8 total surface including wetted disc in the air. The mass of organic that enters this section of trough is 44,000 C g/second, but this mass of sugar can only reach the rotifers by diffusion through a water-like diffusion layer of about .06 cm thickness, twice the thickness calculated above. The thickness is twice that calculated above because it includes the supernatant liquid beyond the slime layer. We now calculate the rate of mass diffusing into the disc: AxDxc/z = 8×10^8 x 3×10-6 x C/.06 cm = 40,000 C g/sec, and find that, for this tank size and rotation speed, the transfer rate of organic to the discs is 2/3 as much as needed to absorb the incoming sugar. This is to say that a 50,000 gallon section is too small to reduce the concentration to ln (1) at this speed of disc rotation.

Based on the above calculation, I’m inclined to increase the speed of rotation to .75 rpm. At this speed, the rotifer-slime layer will be 2/3 as thin 0.2 mm, and we expect an equally thinner diffusion barrier in the supernatant. At this faster speed, there is now 3/2 as much diffusion transfer per area because the thinner diffusion barrier, and we can expect a 50,000 liter reactor section to reduce the initial concentration by a fraction of 1/2.718 or C/e. Note that the mass transfer rate to the discs is always proportional to C. If we find that 50,000 gallons of tank reduces the concentration to 1/e, we find that we need 150,000 gallons of reactor to reduce the concentration of sugar from 40 ppm to 2 ppm, the legal target, ln (40/2) = 3. This 150,000 gallons is a remarkably small volume to reduce the sBOD concentration from 40 ppm to 2 ppm (sBOD = soluble biological oxygen demand), and the energy use is small too if the disc bearings are good.

The Holly sewage treatment plant is the only one in Oakland county, MI using the rotating disc contacted technology. It has a flow of 1,000,000 gallons per day, and has a contactor trough that is 215,000 gallons, about what we’d expect though their speed is somewhat higher, over 1 rpm and their input concentration is likely lower than 40 ppm. For the first stage of sewage treatment, the Holly plant use a vertical-draft, trickle-bed reactor. That is they drizzle the sewage-liquids over a slime-coated packing to reduce the sBOD concentration from 200 ppm to before feeding the flow to the rotating discs. My sense of the reason they don’t do the entire extraction with a trickle bed is that the discs use far less energy.

I should also add that the back-part of the disc isn’t totally useless oxygen storage, as it seems from my analysis. Some non-sugar reactions take place in the relatively anoxic environment there and in the liquid at the bottom of the trough. In these regions, iron reacts with phosphate, and nitrate removal takes place. These are other important requirements of sewage treatment.

Robert E. Buxbaum, July 18, 2017. As an exercise, find the volume necessary for a plug flow reactor or a stirred tank reactor (CSTR) to reduce the concentration of sugar from 40 ppm to 2 ppm. Assume 1,000,000 gal per day, an excess of oxygen in these reactors, and a first order reaction with a rate constant of dC/dt = -(0.4/hr)C. At some point in the future I plan to analyze these options, and the trickle bed reactor, too.

If the wall with Mexico were covered in solar cells

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As a good estimate, it will take about 130,000 acres of solar cells to deliver the power of a typical nuclear facility, 26 TWhr/year. Since Donald Trump has proposed covering his wall with Mexico with solar cells, I came to wonder how much power these cells would produce, and how much this wall might cost. Here goes.

Lets assume that Trump’s building a double wall on a strip of land one chain (66 feet) wide, with a 2 lane road between. Many US roads are designed in chain widths, and a typical, 2 lane road is 1/2 chain wide, 33 feet, including its shoulders. I imagine that each wall is slanted 50° as is typical with solar cells, and that each is 15 to 18 feet high for a good mix of power and security. Since there are 10 square chains to an acre, and 80 chains to a mile we find that it would take 16,250 miles of this to produce 26 TWhr/year. The proposed wall is only about 1/10 this long, 1,600 miles or so, so the output will be only about 1/10 as much, 2.6 TWhr/year, or 600 MW per average daylight hour. That’s not insignificant power — similar to a good-size coal plant. If we aim for an attractive wall, we might come to use Elon Musk’s silica-coated solar cells. These cost $5/Watt or $3 Billion total. Other cells are cheaper, but don’t look as nice or seem as durable. Obama’s, Ivanpah solar farm, a project with durability problems, covers half this area, is rated at 370 MW, and cost $2.2 Billion. It’s thus rated to produce slightly over half the power of the wall, at a somewhat higher price, $5.95/Watt.

Elon Musk with his silica solar panels.

Elon Musk with his, silica-coated, solar wall panels. They don’t look half bad and should be durable.

It’s possible that the space devoted to the wall will be wider than 66 feet, or that the length will be less than 1600 miles, or that we will use different cells that cost more or less, but the above provides a good estimate of design, price, and electric output. I see nothing here to object to, politically or scientifically. And, if we sell Mexico the electricity at 11¢/kWhr, we’ll be repaid $286 M/year, and after 12 years or so, Republicans will be able to say that Mexico paid for the wall. And the wall is likely to look better than the Ivanpah site, or a 20-year-old wind farm.

As a few more design thoughts, I imagine an 8 foot, chain-link fence on the Mexican side of the wall, and imagine that many of the lower solar shingles will be replaced by glass so drivers will be able to see the scenery. I’ve posited that secure borders make a country. Without them, you’re a tribal hoard. I’ve also argued that there is a pollution advantage to controlling imports, and an economic advantage as well, at least for some. For comparison, recent measurement of the Great Wall of China shows it to be 13,170 miles long, 8 times the length of Trump’s wall with China.

Dr. Robert E. Buxbaum, June 14, 2017.