Tag Archives: pressure

Change home air filters 3 times per year

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 pump

Hydrogen-power ed fuel cells provide a lot of advantages over batteries, e.g. for drones and extended range vehicles, but part of the challenge is compressing the hydrogen. On solution I’d proposed is a larger version of this steam-powered compressor, another is a membrane reactor hydrogen generator, and a few weeks ago, I wrote about an other clever innovative solutions: an electrochemical hydrogen pump. It was a fuel cell operating backwards, pumping was very efficient and compact, but the pressure was borne by the fuel cell membranes, so the pump is only suitable at low pressure differentials. I’d now like to describe a different, very clever hydrogen pump, one that operates by metallic hydride sorption and provides very high pressure.

Hydride sorption -desorption pressures vs temperature.

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

The basic metal hydride reaction is M + nH2 <–> MH2n. Where M is a metal or metallic alloy. While most metals will undergo this reaction at some appropriate temperature and pressure, the materials of interest are exothermic hydrides that undergo a nearly stoichiometric absorption or desorption reaction at temperatures near 1 atm, temperatures near room temperature. The plot at right presents the plateau pressure for hydrogen absorption/ desorption in several, common metal hydrides. The slope is proportionals to the heat of sorption. There is a red box shown for the candidates that sorb or desorb between 1 and 10 atmospheres and 25 and 100 °C. Sorbants whose lines pass through that box are good candidates for pump use. The ones with a high slope (high heat of sorption) in particular, if you want a convenient source of very high pressure.

To me, NaAlH4 is among the best of the materials, and certainly serves as a good example for how the pump works. The basic reaction, in this case is:

NaAl + 2H2 <–> NaAlH4

The line for this reaction crosses the 1 atm red line at about 30°C suggesting that each mol of NaAl material will absorb 2 mols of hydrogen at 1 am and normal room temperatures: 20-30°C. Assume the pump contains 100 g of NaAl (2.0 mols). We can expect it will 4 mols of hydrogen gas, about 90 liters at this temperature. If this material in now heated to 250°C, it will desorb most of the hydrogen (80% perhaps, 72 liters) at 100 atm, or 1500 psi. This is a remarkably high pressure boost; 1500 psi hydrogen is suitable for use filling the high pressure tank of a hydrogen-based, fuel cell car.

But there is a problem: it will take 2-3 hours to cycle the sober; the absorb hydrogen at low pressure, heat, desorb and cycle back to low temperature. If you only can pump 72 liters in 2-3 hours, this will not be an effective pump for automobiles. Even with several cells operating in parallel, it will be hard to fill the fuel tank of a fuel-cell car. The output is enough for electric generators, or for the small gas 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 supply them in research quantities.

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

At this point, I can imagine you saying that there is a simple way to make up for the low output of a pump with 100g of sorbent: use more, perhaps 10 kg distributed over 100 cells. The alloys don’t cost much in bulk, see chart above (they’re a lot more expensive in small quantities). With 100 times more sorbent, you’ll pump 100 times faster, enough for a fairly large hydrogen generator, like this one from REB. This will work, but you don’t get economies of scale. With standard, mechanical pumps give you a decent economy of scale — it costs 3-4 times as much for each 10 times increase in output. For this reason, the hydride sorption pump, though clever appears to be destined for low volume applications. Though low volume might involve hundreds of kg of sorbent, at some larger value, you’re going to want to use a mechanical pump.

Other uses of these materials include hydrogen storageremoval of hydrogen from a volume, e.g. so it does not mess up electronics, or for vacuum pumping from a futon reactor. I have sold niobium screws for hydrogen sorption in electronic packages, and my company provides chemical sorbers for hydrogen removal from air. For more of our products, visit www.rebresearch.com/catalog.html

Robert Buxbaum, May 26, 2017. 

The speed of sound, Buxbaum’s correction

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 udV/V = -dH = -TdS- VdP. It is now common to say that dS = 0 across the sound wave, and thus find that u2 = -V(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 u= -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

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?

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

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

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.

Joke re: SI pressure

Einstein, Newton, and the two Pascal brothers are playing hide and seek. Einstein has his eyes covered and is counting. The two Pascal bothers run and hide but Isaac Newton does not. He draws a square around him in the dust and stands waiting. When Einstein finishes counting he says, “I see you Sir Isaac standing there.” “No you don’t.” says Newton. “You see two Pascals: there’s one Newton in half a square meter area.

What causes the swirl of tornadoes and hurricanes

Some weeks ago, I presented an explanation of why tornadoes and hurricanes pick up stuff based on an essay by A. Einstein that explained the phenomenon in terms of swirling fluids and Coriolis flows. I put in my own description that I thought was clearer since it avoided the word “Coriolis”, and attached a video so you could see how it all worked — or rather that is was as simple as all that. (Science teachers: I’ve found kids love it when I do this, and similar experiments with centrifugal force in the class-room as part of a weather demonstration).

I’d like to now answer a related question that I sometimes get: where does the swirl come from? hurricanes that answer follows, though I think you’ll find my it is worded differently from that in Wikipedia and kids’ science books since (as before) I don’t use the word Coriolis, nor any other concept beyond conservation of angular momentum plus that air flows from high pressure to low.

In Wikipedia and all the other web-sits I visited, it was claimed that the swirl came from “Coriolis force.” While this isn’t quite wrong, I find this explanation incomprehensible and useless. Virtually no-one has a good feel for Coriolis force as such, and those who do recognize that it doesn’t exist independently like gravity. So here is my explanation based on low and high pressure and on conservation of angular momentum.  I hope it will be clearer.

All hurricanes are associated with low pressure zones. This is not a coincidence as I understand it, but a cause-and-effect relationship. The low pressure center is what causes the hurricane to form and grow. It may also cause tornadoes but the relationship seems less clear. In the northern hemisphere, the lowest low pressure zones are found to form over the mid Atlantic or Pacific in the fall because the water there is warm and that makes the air wet and hot. Static air pressure is merely the weight of the air over a certain space, and as hot air has more volume and less density, it weighs less. Less weight = less pressure, all else being equal. Adding water (humidity) to air also reduces the air pressure as the density of water vapor is less than that of dry air in proportion to their molecular weights. The average molecular weight of dry air is 29 and the molecular weight of water is 18. As a result, every 9% increase in water content decreases the air pressure by 1% (7.6 mm or 0.3″ of mercury).

Air tends to flow from high pressure zones to low pressure zones. In the northern hemisphere, some of the highest high pressure zones form over northern Canada and Russia in the winter. High pressure zones form there by the late fall because these regions are cold and dry. Cold air is less voluminous than hot, and as a result additional hot air flows into these zones at high altitude. At sea level the air flows out from the high pressure zones to the low pressure zones and begins to swirl because of conservation of angular momentum.

All the air in the world is spinning with the earth. At the north pole the spin rate is 360 degrees every 24 hours, or 15 degrees per hour. The spin rate is slower further south, proportionally to the sine of the latitude, and it is zero at the equator. The spin of the earth at your location is observable with a Foucault pendulum (there is likely to be one found in your science museum). We normally don’t notice the spin of the air around us because the earth is spinning at the same rate, normally. However the air has angular momentum, and when air moves into into a central location the angular speed increases because the angular momentum must be conserved. As the gas moves in, the spin rate must increase in proportion; it eventually becomes noticeable relative to the earth’s spin. Thus, if the air starts out moving at 10 degrees per hour (that’s the spin rate in Detroit, MI 41.8° N), and moves from 800 miles away from a low pressure center to only 200 miles from the center, the angular momentum must increase four times, or to 40 degrees per hour. We would only see 30 degrees/hr of this because the earth is spinning, but the velocity this involves is significant: V= 200 miles * 2* pi *30/360 = 104 mph.

To give students a sense of angular momentum conservation, most science centers (and colleges) use an experiment involving bicycle wheels and a swivel chair. In the science centers there is usually no explanation of why, but in college they tend to explain it in terms of vectors and (perhaps) gauge theories of space-time (a gauge is basically a symmetry; angular momentum is conserved because space is symmetric in rotation). In a hurricane, the air at sea level always spins in the same direction of the earth: counter clockwise in the northern hemisphere, clockwise in the southern, but it does not spin this way forever.

The air that’s sucked into the hurricane become heated and saturated with water. As a result, it becomes less dense, expands, and rises, sucking fresh air in behind it. As the hot wet air rises it cools and much of the water rains down as rain. When the, now dry air reaches a high enough altitude its air pressure is higher than that above the cold regions of the north; the air now flows away north. Because this hot wet air travels north we typically get rain in Michigan when the Carolinas are just being hit by hurricanes. As the air flows away from the centers at high altitudes it begins to spin the opposite direction, by the way, so called counter-cyclonally because angular momentum has to be consevered. At high altitudes over high pressure centers I would expect to find cyclones too (spinning cyclonally) I have not found a reference for them, but suspect that airline pilots are aware of the effect. There is some of this spin at low altitudes, but less so most of the time.

Hurricanes tend to move to the US and north through the hurricane season because, as I understand it, the cold air that keeps coming to feed the hurricane comes mostly from the coastal US. As I understand it the hurricane is not moving as such, the air stays relatively stationary and the swirl that we call a hurricane moves to the US in the effective direction of the sea-level air flow.

For tornadoes, I’m sorry to say, this explanation does not work quite as well, and Wikipedia didn’t help clear things up for me either. The force of tornadoes is much stronger than of hurricanes (the swirl is more concentrated) and the spin direction is not always cyclonic. Also tornadoes form in some surprising areas like Kansas and Michigan where hurricanes never form. My suspicion is that most, but not all tornadoes form from the same low pressure as hurricanes, but by dry heat, not wet. Tornadoes form in Michigan, Texas, and Alabama in the early summer when the ground is dry and warmer than the surrounding lakes and seas. It is not difficult to imagine the air rising from the hot ground and that a cool wind would come in from the water and beginning to swirl. The cold, damp sea air would be more dense than the hot, dry land air, and the dry air would rise. I can imagine that some of these tornadoes would occur with rain, but that many the more intense?) would have little or none; perhaps rain-fall tends to dampen the intensity of the swirl (?)

Now we get to things that I don’t have good explanation for at all: why Kansas? Kansas isn’t particularly hot or cold; it isn’t located near lakes or seas, so why do they have so many tornadoes? I don’t know. Another issue that I don’t understand: why is it that some tornadoes rotate counter cyclonicly? Wikipedia says these tornadoes shed from other tornadoes, but this doesn’t quite seem like an explanation. My guess is that these tornadoes are caused by a relative high pressure source at ground level (a region of cold ground for example) coupled with a nearby low pressure zone (a warm lake?). My guess is that this produces an intense counter-cyclonic flow to the low pressure zone. As for why the pressure is very low in tornadoes, even these that I think are caused by high pressure, I suspect the intense low pressure is an epee-phenomenon caused by the concentration of spin — one I show in my video. That is, I suspect that the low pressure in the center of counter-cyclonic tornadoes is not the cause of the tornado but an artifact of the concentrated spin. Perhaps I’m wrong here, but that’s the explanation that seems to fit best with the info I’ve got. If you’ve got better explanations for these two issues, I’d love to hear them.

How and why membrane reactors work

Here is a link to a 3 year old essay of mine about how membrane reactors work and how you can use them to get past the normal limits of thermodynamics. The words are good, as is the example application, but I think I can write a shorter version now. Also, sorry to say, when I wrote the essay I was just beginning to make membrane reactors; my designs have gotten simpler since.

Above, for example, is a more modern, high pressure membrane reactor design:  72 tube reactor assembly; high pressure. The area at right is used for heat transfer. Normally the reactor would sit with this end up, and the tube area filled or half-filled with catalyst, e.g. for the water gas shift reaction, CO + H2O –> CO2 + H2. According to normal thermodynamics, the extent of reaction for this will not be affected by pressure once it reaches equilibrium, only by temperature. If you want the reaction to go reasonably to completion, you have to operate at low temperatures, 250- 300 °C, and you have to cool externally to remove the heat of reaction. In a membrane reactor, you can operate at much higher temperatures and you don’t have to work so hard to remove heat. The trick is to operate with the reacting gas at high pressures, and to extract hydrogen at lower pressures. With a large enough difference between the reacting pressure and the extract pressure, you can achieve high extents (high conversions) at any temperature.

Here’s where we sell membrane reactors; we also sell catalyst and tubes.