Tag Archives: pressure

Rotating sail ships and why your curve ball doesn’t curve.

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The Flettner-sail ship, Barbara, 1926.

Sailing ships are wonderfully economic and non-polluting. They have unlimited range because they use virtually no fuel, but they tend to be slow, about 5-12 knots, about half as fast as Diesel-powered ships, and they can be stranded for weeks if the wind dies. Classic sailing ships also require a lot of manpower: many skilled sailors to adjust the sails. What’s wanted is an easily manned, economical, hybrid ship: one that’s powered by Diesel when the wind is light, and by a simple sail system when the wind blows. Anton Flettner invented an easily manned sail and built two ships with it. The Barbara above used a 530 hp Diesel and got additional thrust, about an additional 500 hp worth, from three, rotating, cylindrical sails. The rotating sales produced thrust via the same, Magnus force that makes a curve ball curve. Barbara went at 9 knots without the wind, or about 12.5 knots when the wind blew. Einstein thought it one of the most brilliant ideas he’d seen.

Force diagram of Flettner rotor (Lele & Rao, 2017)

The source of the force can be understood with help of the figure at left and the graph below. When a simple cylinder sits in the wind, with no spin, α=0, the wind force is essentially drag, and is 1/2 the wind speed squared, times the cross-sectional area of the cylinder, Dxh, and the density of air. Add to this a drag coefficient, CD, that is about 1 for a non-spinning cylinder. More explicitly, FD= CDDhρv2/2. As the figure at right shows, there is a sort-of lift in the form of sustained vibrations at zero spin, α=0. Vibrations like this are useless for propulsion, and can be damaging to the sail. In baseball, such vibrations are the reason knuckle balls fly erratically. If you spin the cylindrical mast at α=2.1, that is at a speed where the fast surface moves with the wind, at 2.1 times the wind speed, and the other side side moves to the wind, there is more force on the side moving to the wind (see figure above) and the ship can be propelled forward (or backward if you reverse the spin direction). Significantly, at α=2.1, you get 6 times as much force as the expected drag, and you no longer get vibrations. FL= CLDhρv2/2, and CL=6 at this rotation speed

Numerical lift coefficients versus time, seconds for different ratios of surface speed to wind speed, a. (Mittal & Kumar 2003), Journal of Fluid Mechanics.

At this rotation speed, α=2.1, this force will be enough to drive a ship so long as the wind is reasonably strong, 15-30 knots, and ship does not move faster than the wind. The driving force is always at right angles to the perceived wind, called the “fair wind”, and the fair wind moves towards the front as the ship speed increases. If you spin the cylinder at 3 to 4 times the wind speed, the lift coefficient increases to between 10 and 18. This drives a ship with yet force. You need somewhat more power to turn the sails, but you are also further from vibrations. Flettner considered α=3.5. optimal. Higher rotation speeds are possible, but they require more rotation power (rotation power goes as ω2, and if you go beyond α=4.3, the vibrations return. Controlling the speed is somewhat difficult but important. Flettner sails were no longer used by the 1930s when fuel became cheaper.

In the early 1980s, the famous underwater explorer, Jacques Cousteau revived the Flettner sail for his exploratory ship, the Alcyone. He used light-weight aluminum sails, and an electric motor for rotation instead of Diesel as on the Barbara. He claimed that the ship drew more than half of its power from the wind, and claimed that, because of computer control, it could sail with no crew. This latter claim was likely bragging. Even with today’s computer systems, people are needed as soon as something goes wrong. Still the energy savings were impressive enough that other ship owners took notice. In recent years, several ship-owners have put Flettner sails on cargo ships, as a right. This is not an ideal use since cargo ships tend to go fast. Still, it’s reported that, these ships get about 20% of their propulsion from wind power, not an insignificant amount.

And this gets us to the reason your curve ball does not curve: you’re not spinning it fast enough. You want the ball to spin at a higher rate than you get just by rolling the ball off your fingers. If you do this, α = 1 and you get relatively little sideways force. To get the ball to really curve, you have to snap your wrist hard aiming for α=1.5 or so. As another approach you can aim for a knuckle ball, achieved with zero rotation. At α=0, the ball will oscillate and your pitch nearly impossible to hit, or catch. Good luck.

Robert Buxbaum, March 22, 2023. There are also various Flettner airplane designs where horizontal, cylindrical “wings” rotate to provide lift, power too in some versions. The aim is high lift with short wings and a relatively low power draw. So-far, these planes are less efficient and slower than a normal helicopter.

Water Towers, usually a good thing.

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Most towns have at least one water tower. Oakland county, Michigan has four. When they are sized right, they serve several valuable purposes. They provide water in case of a power failure; they provide increased pressure in the morning when people use a lot of water showering etc.; and they allow a town to use smaller pumps and to pump with cheaper electricity, e.g. at night. If a town has no tower, all these benefits are gone, but a town can still have water. It’s also possible to have a situation that’s worse than nothing. My plan is to show, at the end of this essay, one of the ways that can happen. It involves thermodynamic properties of state i a situation where there is no expansion headspace or excess drain (most towers have both).

A typical water tower — spheroidal design. A tower of the dimensions shown would contain about 1/2 million gallons of water.

The typical tower stands at the highest point in the town, with the water level about 170 feet above street level. It’s usable volume should be about as much water as the town uses in a typical day. The reason for the height has to do with the operating pressure of most city-level water pipes. It’s about 75 psi and each foot of water “head” gives you about 0.43 psi. You want pressures about 75 psi for fire fighting, and to provide for folks in apartment buildings. If you have significantly higher pressures, you pay a cost in electricity, and you start losing a lot of water to leaks. These leaks should be avoided. They can undermine the roads and swallow houses. Bob Dadow estimates that, for our water system the leakage rate is between 15 and 25%.

Oakland county has four water towers with considerably less volume than the 130 million gallons per day that the county uses. I estimate that the South-east Oakland county tower, located near my home, contains, perhaps 2 million gallons. The other three towers are similar in size. Because our county’s towers are so undersized, we pay a lot for water, and our water pressure is typically quite low in the mornings. We also have regular pressure excursions and that leads to regular water-boil emergencies. In some parts of Oakland county this happens fairly often.

There are other reasons why a system like ours should have water towers with something more like one days’ water. Having a large water reserve means you can benefit from the fact that electric prices are the lowest at night. With a days’ volume, you can avoid running the pumps during high priced, day times. Oakland county loses this advantage. The other advantage to having a large volume is that it gives you more time to correct problems, e.g. in case of an electric outage or a cyber attack. Perhaps Oakland thinks that only one pump can be attacked at one time or that the entire electric grid will not go out at one time, but these are clearly false assumptions. A big system also means you can have pumps powered by solar cells or other renewable power. Renewable power is a good thing for reliability and air pollution avoidance. Given the benefits, you’d expect Oakland county would reward towns that add water towers, but they don’t, as best I can tell.

Here’s one way that a water column can cause problems. You really need those pressure reliefs.

Now for an example of the sort of things that can go wrong in a water tower with no expansion relief. Every stand-pipe is a small water tower, and since water itself is incompressible, it’s easy to see that a small expansion in the system could produce a large pressure rise. The law requires that every apartment hose water system has to have expansion relief to limit these increases; The water tower above had two forms of reliefs, a roof vent, and an overflow pipe, both high up so that pressure could be maintained. But you can easily imagine a plumber making a mistake and installing a stand pipe without an expansion relief. I show a system like that at left, a 1000 foot tall water pipe, within a skyscraper, with a pump at the bottom, and pipes leading off at the sides to various faucets.

Lets assume that the pressure at the top is 20 psi, the pressure at the bottom will be about 450 psi. The difference in pressure (430 psi) equals the weight of the water divided by the area of the pipe. Now let’s imagine that a bubble of air at the bottom of the pipe detaches and rises to the top of the pipe when all of the faucets are closed. Since air is compressible, while water is not, the pressure at the bubble will remain the same as the bubble rises. By the time the bubble reaches the top of the pipe, the pressure there will rise to 450 psi. Since water has weight, 430 psi worth, the pressure at the bottom will rise to 880 psi = 450 + 430. This is enough to damage pump and may blow the pipes as well. A scenario like this likely destroyed the New Horizon oil platform to deadly consequences. You really want those pressure reliefs, and you want a competent plumber / designer for any water system, even a small one.

Robert Buxbaum, September 28- October 6, 2019. I ran for water commissioner is 2016.

Great waves, small circles, and the spread of ideas.

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Simplified wave motion, GIf by Dan Russel (maybe? I think?).

The scientific method involves looking closely at things. Sometimes we look closely for a purpose — to make a better mouse-trap, say. But sometimes it’s just to understand what’s happening: to satisfy curiosity, to understand the way the world works, or to answer a child. Both motivations bring positive results, but there is a difference in how people honor the product of these motivations. Scientific knowledge developed for curiosity is considered better; it tends to become the model for social understanding, and for art and literature. Meanwhile, science developed for a purpose is considered suspect, and often that suspicion is valid. A surprising amount of our knowledge was developed for war: for the purpose of killing people, destroying things, and occupying lands.

Waves provide a wonderful example of science exploration that was developed mostly for curiosity, and so they have become models of social understanding and culture — far more so than the atom bomb and plague work discussed previously.

Waves appear magical: You poke a pond surface with a stick, and the influence of that poke travels, as if by magic, to all corners of the pond. Apparently the initial poke set off something, and that sets off something else, and we’ve come to use this as a model for cultural ideas. Any major change in music, art, or cultural thought is described as a wave (and not as a disease). The sense of wave is  that a small push occurs, and the impact travels across a continent and across an ocean. The Gifs above and below shows how this happens for the ordinary wave — the one with a peaked top. As shown, the bits of water do not move with the wave. Instead they just circulate in a small circle. The powerful waves that crosses an ocean are composed of many small circles of water rolling in the general direction of the wave. With ideas too, I think, one person can push a second, and that second a third, each acting in his or her own circle, and a powerful transmission of ideas results. Of course, for a big wave, you need a big circle, but maybe not in cases of reflection (reflected waves can add, sometimes very destructively).

simplified wave movement

In the figures I’ve shown, you will notice that the top of the circle always moves in the same direction as the top of the wave. If the wave moves to the right, the circle is clockwise. There are also Rayleigh waves. In these, the top of the wave is not peaked, but broad, with little indents between ripples. For Rayleigh wave the motion is not circular, but elliptical, and the top of the ellipse moves in the opposite direction to that of the wave. These waves go slower than the normal waves, but they are more destructive. Most of the damage of earthquakes is by the late-arriving Rayleigh waves.

If regular waves are related to fast-moving ideas, like rock n roll, Rayleigh waves might be related to slower-traveling, counter-intuitive ideas, paradigm shifts: Religions, chaos, entropy, feminism, or communism. Rayleigh waves are mostly seen in solids, and the destructive power of counter-intuitive ideas is mostly seen in rigid societies.

Then there are also pressure waves, like sound, and wiggle waves (transverse waves). Pressure waves travel the fastest, and work in both solids and liquids. Wiggle waves travel slower (and don’t travel in liquids). Both of these involve no circles at all, but just one bit of material pushing on its neighbor. I think the economy works this way: bouncing springs, for the most part. Life is made up of all of these, and life is good. The alternative to vibration, I should mention, is status. Status is a form of death. There is a certain sort of person who longs for nothing more than an unchanging, no-conflict world: one government and one leadership. Avoid such people.

Robert Buxbaum, February 10, 2019

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.

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). 

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

The future of steamships: steam

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Most large ships and virtually all locomotives currently run on diesel power. But the diesel  engine does not drive the wheels or propeller directly; the transmission would be too big and complex. Instead, the diesel engine is used to generate electric power, and the electric power drives the ship or train via an electric motor, generally with a battery bank to provide a buffer. Current diesel generators operate at 75-300 rpm and about 40-50% efficiency (not bad), but diesel fuel is expensive. It strikes me, therefore that the next step is to switch to a cheaper fuel like coal or compressed natural gas, and convert these fuels to electricity by a partial or full steam cycle as used in land-based electric power plants. 

Ship-board diesel engine, 100 MW for a large container ship

Diesel engine, 100 MW for a large container ship

Steam powers all nuclear ships, and conventionally boiled steam provided the power for thousands of Liberty ships and hundreds of aircraft carriers during World War 2. Advanced steam turbine cycles are somewhat more efficient, pushing 60% efficiency for high pressure, condensed-turbine cycles that consume vaporized fuel in a gas turbine and recover the waste heat with a steam boiler exhausting to vacuum. The higher efficiency of these gas/steam turbine engines means that, even for ships that burn ship-diesel fuel (so-called bunker oil) or natural gas, there can be a cost advantage to having a degree of steam power. There are a dozen or so steam-powered ships operating on the great lakes currently. These are mostly 700-800 feet long, and operate with 1950s era steam turbines, burning bunker oil or asphalt. US Steel runs the “Arthur M Anderson”, Carson J Callaway” , “John G Munson” and “Philip R Clarke”, all built-in 1951/2. The “Upper Lakes Group” runs the “Canadian Leader”, “Canadian Provider”, “Quebecois”, and “Montrealais.” And then there is the coal-fired “Badger”. Built in 1952, the Badger is powered by two, “Skinner UniFlow” double-acting, piston engines operating at 450 psi. The Badger is cost-effective, with the low-cost of the fuel making up for the low efficiency of the 50’s technology. With larger ships, more modern boilers and turbines, and with higher pressure boilers and turbines, the economics of steam power would be far better, even for ships with modern pollution abatement.

Nuclear steam boilers can be very compact

Nuclear steam boilers can be very compact

Steam powered ships can burn fuels that diesel engines can’t: coal, asphalts, or even dry wood because fuel combustion can be external to the high pressure region. Steam engines can cost more than diesel engines do, but lower fuel cost can make up for that, and the cost differences get smaller as the outputs get larger. Currently, coal costs 1/10 as much as bunker oil on a per-energy basis, and natural gas costs about 1/5 as much as bunker oil. One can burn coal cleanly and safely if the coal is dried before being loaded on the ship. Before burning, the coal would be powdered and gassified to town-gas (CO + H2O) before being burnt. The drying process removes much of the toxic impact of the coal by removing much of the mercury and toxic oxides. Gasification before combustion further reduces these problems, and reduces the tendency to form adhesions on boiler pipes — a bane of old-fashioned steam power. Natural gas requires no pretreatment, but costs twice as much as coal and requires a gas-turbine, boiler system for efficient energy use.

Todays ships and locomotives are far bigger than in the 1950s. The current standard is an engine output about 50 MW, or 170 MM Btu/hr of motive energy. Assuming a 50% efficient engine, the fuel use for a 50 MW ship or locomotive is 340 MM Btu/hr; locomotives only use this much when going up hill with a heavy load. Illinois coal costs, currently, about $60/ton, or $2.31/MM Btu. A 50 MW engine would consume about 13 tons of dry coal per hour costing $785/hr. By comparison, bunker oil costs about $3 /gallon, or $21/MM Btu. This is nearly ten times more than coal, or $ 7,140/hr for the same 50 MW output. Over 30 years of operation, the difference in fuel cost adds up to 1.5 billion dollars — about the cost of a modern container ship.

Robert E. Buxbaum, May 16, 2014. I possess a long-term interest in economics, thermodynamics, history, and the technology of the 1800s. See my steam-pump, and this page dedicated to Peter Cooper: Engineer, citizen of New York. Wood power isn’t all that bad, by the way, but as with coal, you must dry the wood, or (ideally) convert it to charcoal. You can improve the power and efficiency of diesel and automobile engines and reduce the pollution by adding hydrogen. Normal cars do not use steam because there is more start-stop, and because it takes too long to fire up the engine before one can drive. For cars, and drone airplanes, I suggest hydrogen/ fuel cells.

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.

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

My steam-operated, high pressure pump

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Here’s a miniature version of a duplex pump that we made 2-3 years ago at REB Research as a way to pump fuel into hydrogen generators for use with fuel cells. The design is from the 1800s. It was used on tank locomotives and steamboats to pump water into the boiler using only the pressure in the boiler itself. This seems like magic, but isn’t. There is no rotation, but linear motion in a steam piston of larger diameter pushes a liquid pump piston with a smaller diameter. Each piston travels the same distance, but there is more volume in the steam cylinder. The work from the steam piston is greater: W = ∫PdV; energy is conserved, and the liquid is pumped to higher pressure than the driving steam (neat!).

The following is a still photo. Click on the YouTube link to see the steam pump in action. It has over 4000 views!

Mini duplex pump. Provides high pressure water from steam power. Amini version of a classic of the 1800s Coffee cup and pen shown for scale.

Mini duplex pump. Provides high pressure water from steam power. A mini version of a classic of the 1800s Coffee cup and pen shown for scale.

You can get the bronze casting and the plans for this pump from Stanley co (England). Any talented machinist should be able to do the rest. I hired an Amish craftsman in Ohio. Maurice Perlman did the final fit work in our shop.

Our standard line of hydrogen generators still use electricity to pump the methanol-water. Even our latest generators are meant for nom-mobile applications where electricity is awfully convenient and cheap. This pump was intended for a future customer who would need to generate hydrogen to make electricity for remote and mobile applications. Even our non-mobile hydrogen is a better way to power cars than batteries, but making it mobile has advantages. Another advance would be to heat the reactors by burning the waste gas (I’ve been working on that too, and have filed a patent). Sometimes you have to build things ahead of finding a customer — and this pump was awfully cool.