Tag Archives: Engineering

Sewage reactor engineering, Stirred tank designs

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Over the past few years, I’ve devoted several of these essays to analysis of first-stage sewage treatment reactors. I described and analyzed the rotating disc reactor found at the plant is Holly here, and described the racetrack,“activated sludge” plug reactor found most everywhere else here. I also described a system without a primary clarifier found near Cincinatti. All of these were effective for primary treatment; soluble organics are removed by bio-catalyzed oxidation:

2 H-C-O-H + O2 –> CO2 + H2O.

A typical plant in Oakland county treats 2,000,000 gallons per day of this stuff, with the bio-reactor receiving liquid waste containing about 200 ppm of soluble and colloidal biomass. That’s 400 dry gallons for those interested, or about 3200 dry lbs./day. About half of this will be oxidized to CO2 and water. The rest (cell bodies) are removed with insoluble components, and applied to farmers fields or buried, or burnt in an incinerator.

There is another type of reactor used in Oakland County. It’s mostly used for secondary treatment, converting consolidated sludge to higher-quality sludge that can be sold or used on farms with less restriction, but it is a type of reactor used at the South Lyon treatment plant, for primary treatment. It is a Continually stirred tank reactor, or CSTR, a design that is shown in schematic below.

As of some years ago, the South Lyon system involved a single largish pond lined with plastic with a volume about 2,000,000 gallons total. About 700,000 gallons per day of sewage liquids went into the lagoon, at 200 ppm soluble organics. Air was bubbled through the liquid providing a necessary reactant, and causing near-perfect mixing of the contents. The aim of the plant managers is to keep the soluble output to the, then-acceptable level of 10 ppm; it’s something they only barely managed, and things got worse as the flow increased. Assume as before, a value V and a flow Q.

We will call the concentration of soluble organics C, and call the initial concentration, the concentration that enters,  Ci. It’s about 200 ppm. We’ll call the output concentration Co, and for this type of reactors, Co = C.  The reaction is first order, approximately, so that, if there were no flow into or out of the reactor, the concentration of organics would decrease at the rate of

dC/dt = -kC.

Here k is a reaction constant, dependent on temperature oxygen and cell content. It’s typically about 0.5/hour. For a given volume of tank the rate of organic removal is VkC. We can now do a mass balance on soluble organics. Since the rate of organic entry is QCi and the rate leaving by flow is QC. The difference must be the amount that is reacted away:

QCi – QC = VkC.

We now use algebra, to find that

Co = Ci/(1 + kV/Q).

V/Q is sometimes called a residence time; for the system. At normal flow, the residence time of the South Lyon system is about 2.8 days or 68.6 hours. Plugging these numbers in, we find that the effluent from the reactor leaves at 1/35 of the input concentration, or 5.7 ppm, on average. This would be fine except that sometimes the temperature drops, or the flow increases, and we start violating the standard. A yet bigger problem was that the population increased by 50% while the EPA standard got more stringent to 2 ppm. This was solved by adding another, smaller reactor, volume = V2. Using the same algebraic analysis, as above you can show that, with two reactors,

Co = Ci/ [(1 + kV/Q)(1+kV2/Q)].

It’s a touchy system, but it meets government targets, just barely, most of the time. I think it is time to switch to a plug-flow reactor system, as used in much of Oakland county. In these, the fluid enters a channel and is reacted as it flows along. Each gallon of fluid, in a sense moves by itself as if it were its own reactor. In each gallon, we can say that dC/dt = -kC. We can thus solve for Co in terms of the total residence time, where t again is V/Q. We can rearrange this equation and integrate: ∫dC/C = – ∫kdt. We then find that, 

      ln(Ci/Co) = kt = kV/Q

To convert 200 ppm sewage to 2 ppm we note that Ci/Co = 100 and that V = Q ln(100)/k = Q (4.605/.5) hours. An inflow of 1000,000 gallons per day = 41,667 gal/ hour, and we find the volume of tank is 41,667 x 9.21 = 383,750 gallons. This is quite a lot smaller than the CSTR tanks at South Lyon. If we converted the South Lyon tanks to a plug-flow, race-track design, it would allow it to serve a massively increased population, discharging far cleaner sewage. 

Robert Buxbaum, November 17, 2019

How tall could you make a skyscraper?

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Built in 1931, the highest usable floor space of the Empire State building is 1250 feet (381m) above the ground. In 1973, that record was beaten by the World Trade Center building 1, 1,368 feet (417 m, building 2 was eight feet shorter). The Willis Tower followed 1974, and by 2004, the tallest building was the Taipei Tower, 1471 feet. Building heights had grown by 221 feet since 1931, and then the Burj Khalifa in Dubai, 2,426 ft ( 739.44m):. This is over 1000 feet taller than the new freedom tower, and nearly as much taller than the previous record holder. With the Saudi’s beginning work on a building even taller, it’s worthwhile asking how tall you could go, if your only  limitations were ego and materials’ strength.

Burj Khalifa, the world’s tallest building, Concrete + glass structure. Dubai tourism image.

Having written about how long you could make a (steel) suspension bridge, the maximum height of a skyscraper seems like a logical next step. At first glance this would seem like a ridiculously easy calculation based on the math used to calculate the maximum length of a suspension bridge. As with the bridge, we’d make the structure from the strongest normal material: T1, low carbon, vanadium steel, and we’d determine the height by balancing this material’s  yield strength, 100,000 psi (pounds per square inch), against its density, .2833 pounds per cubic inch.

If you balance these numbers, you calculate a height: 353,000 inches, 5.57 miles, but this is the maximum only for a certain structure, a wide flag-pole of T1 steel in the absent of wind. A more realistic height assumes a building where half the volume is empty space, used for living and otherwise, where 40% of the interior space contains vertical columns of T1 steel, and where there’s a significant amount of dead-weight from floors, windows, people, furniture, etc. Assume the dead weight is the equivalent of filling 10% of the volume with T1 steel that provides no structural support. The resulting building has an average density = (1/2 x 0.2833 pound/in3), and the average strength= (0.4 x 100,000 pound/in2). Dividing these numbers we get a maximum height, but only for a cylindrical building with no safety margin, and no allowance for wind.

H’max-cylinder = 0.4 x 100,000 pound/in2/ (.5 x 0.2833 pound/in3) = 282,400 inches = 23,532 ft = 4.46 miles.

This is more than ten times the Burj Khalifa, but it likely underestimates the maximum for a steel building, or even a concrete building because a cylinder is not the optimum shape for maximum height. If the towers were constructed conical or pyramidal, the height could be much greater: three times greater because the volume of a cone and thus its weight is 1/3 that of a cylinder for the same base and height. Using the same materials and assumptions,

The tallest building of Europe is the Shard; it’s a cone. The Eiffel tower, built in the 1800s, is taller.

H’max-cone = 3 H’max-cylinder =  13.37 miles.

A cone is a better shape for a very tall tower, and it is the shape chosen for “the shard”, the second tallest building in Europe, but it’s not the ideal shape. The ideal, as we’ll see, is something like the Eiffel tower.

Before speaking about this shape, I’d like to speak about building materials. At the heights we’re discussing, it becomes fairly ridiculous to talk about a steel and glass building. Tall steel buildings have serious vibration problems. Even at heights far before they are destroyed by wind and vibration , the people at the top will begin to feel quite sea-sick. Because of this, the tallest buildings have been constructed out of concrete and glass. Concrete is not practical for bridges since concrete is poor in tension, but concrete can be quite strong in compression, as I discussed here.  And concrete is fire resistant, sound-deadening, and vibration dampening. It is also far cheaper than steel when you consider the ease of construction. The Trump Tower in New York and Chicago was the first major building here to be made this way. It, and it’s brother building in Chicago were considered aesthetic marvels until Trump became president. Since then, everything he’s done is ridiculed. Like the Trump tower, the Burj Khalifa is concrete and glass, and I’ll assume this construction from here on.

let’s choose to build out of high-silica, low aggregate, UHPC-3, the strongest concrete in normal construction use. It has a compressive strength of 135 MPa (about 19,500 psi). and a density of 2400 kg/m3 or about 0.0866 lb/in3. Its cost is around $600/m3 today (2019); this is about 4 times the cost of normal highway concrete, but it provides about 8 times the compressive strength. As with the steel building above, I will assume that, at every floor, half of the volume is living space; that 40% is support structure, UHPC-3, and that the other 10% is other dead weight, plumbing, glass, stairs, furniture, and people. Calculating in SI units,

H’max-cylinder-concrete = .4 x 135,000,000 Pa/(.5 x 2400 kg/m3 x 9.8 m/s2) = 4591 m = 2.85 miles.

The factor 9.8 m/s2 is necessary when using SI units to account for the acceleration of gravity; it converts convert kg-weights to Newtons. Pascals, by the way, are Newtons divided by square meters, as in this joke. We get the same answer with less difficulty using inches.

H’max-cylinder-concrete = .4 x 19,500 psi/(.5 x.0866  lb/in3) = 180,138″ = 15,012 ft = 2.84 miles

These maximum heights are not as great as for a steel construction, but there are a few advantages; the price per square foot is generally less. Also, you have fewer problems with noise, sway, and fire: all very important for a large building. The maximum height for a conical concrete building is three times that of a cylindrical building of the same design:

H’max–cone-concrete = 3 x H’max-cylinder-concrete = 3 x 2.84 miles = 8.53 miles.

Mount Everest, picture from the Encyclopedia Britannica, a stone cone, 5.5 miles high.

That this is a reasonable number can be seen from the height of Mount Everest. Everest is rough cone , 5.498 miles high. This is not much less than what we calculate above. To reach this height with a building that withstands winds, you have to make the base quite wide, as with Everest. In the absence of wind the base of the cone could be much narrower, but the maximum height would be the same, 8.53 miles, but a cone is not the optimal shape for a very tall building.

I will now calculate the optimal shape for a tall building in the absence of wind. I will start at the top, but I will aim for high rent space. I thus choose to make the top section 31 feet on a side, 1,000 ft2, or 100 m2. As before, I’ll make 50% of this area living space. Thus, each apartment provides 500 ft2 of living space. My reason for choosing this size is the sense that this is the smallest apartment you could sell for a high premium price. Assuming no wind, I can make this part of the building a rectangular cylinder, 2.84 miles tall, but this is just the upper tower. Below this, the building must widen at every floor to withstand the weight of the tower and the floors above. The necessary area increases for every increase in height as follows:

dA/dΗ = 1/σ dW/dH.

Here, A is the cross-sectional area of the building (square inches), H is height (inches), σ is the strength of the building material per area of building (0.4 x 19,500 as above), and dW/dH is the weight of building per inch of height. dW/dH equals  A x (.5 x.0866  lb/in3), and

dA/dΗ = 1/ ( .4 x 19,500 psi) x A x (.5 x.0866  lb/in3).

dA/A = 5.55 x 10-6 dH,

∫dA/A = ∫5.55 x 10-6 dH,

ln (Abase/Atop) = 5.55 x 10-6 ∆H,

Here, (Abase/Atop) = Abase sq feet /1000, and ∆H is the height of the curvy part of the tower, the part between the ground and the 2.84 mile-tall, rectangular tower at the top.

Since there is no real limit to how big the base can be, there is hardly a limit to how tall the tower can be. Still, aesthetics place a limit, even in the absence of wind. It can be shown from the last equation above that stability requires that the area of the curved part of the tower has to double for every 1.98 miles of height: 1.98 miles = ln(2) /5.55 x 10-6 inches, but the rate of area expansion also keeps getting bigger as the tower gets heavier.  I’m going to speculate that, because of artistic ego, no builder will want a tower that slants more than 45° at the ground level (the Eiffel tower slants at 51°). For the building above, it can be shown that this occurs when:

dA/dH = 4√Abase.  But since

dA/dH = A 5.55 x 10-6 , we find that, at the base,

5.55 x 10-6 √Abase = 4.

At the base, the length of a building side is Lbase = √Abase=  4 /5.55 x 10-6 inches = 60060 ft = 11.4  miles. Artistic ego thus limits the area of the building to slightly over 11 miles wide of 129.4 square miles. This is about the area of Detroit. From the above, we calculate the additional height of the tower as

∆H = ln (Abase/Atop)/ 5.55 x 10-6 inches =  15.1/ 5.55 x 10-6 inches = 2,720,400 inches = 226,700 feet = 42.94 miles.

Hmax-concrete =  2.84 miles + ∆H = 45.78 miles. This is eight times the height of Everest, and while air pressure is pretty low at this altitude, it’s not so low that wind could be ignored. One of these days, I plan to show how you redo this calculation without the need for calculus, but with the inclusion of wind. I did the former here, for a bridge, and treated wind here. Anyone wishing to do this calculation for a basic maximum wind speed (100 mph?) will get a mention here.

From the above, it’s clear that our present buildings are nowhere near the maximum achievable, even for construction with normal materials. We should be able to make buildings several times the height of Everest. Such Buildings are worthy of Nimrod (Gen 10:10, etc.) for several reasons. Not only because of the lack of a safety factor, but because the height far exceeds that of the highest mountain. Also, as with Nimrod’s construction, there is a likely social problem. Let’s assume that floors are 16.5 feet apart (1 rod). The first 1.98 miles of tower will have 634 floors with each being about the size of Detroit. Lets then assume the population per floor will be about 1 million; the population of Detroit was about 2 million in 1950 (it’s 0.65 million today, a result of bad government). At this density, the first 1.98 miles will have a population of 634 million, about double that of the United States, and the rest of the tower will have the same population because the tower area contracts by half every 1.98 miles, and 1/2 + 1/4 + 1/8 + 1/16 … = 1.

Nimrod examining the tower, Peter Breugel

We thus expect the tower to hold 1.28 Billion people. With a population this size, the tower will develop different cultures, and will begin to speak different languages. They may well go to war too — a real problem in a confined space. I assume there is a moral in there somewhere, like that too much unity is not good. For what it’s worth, I even doubt the sanity of having a single government for 1.28 billion, even when spread out (e.g. China).

Robert Buxbaum, June 3, 2019.

How long could you make a suspension bridge?

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The above is one of the engineering questions that puzzled me as a student engineer at Brooklyn Technical High School and at Cooper Union in New York. The Brooklyn Bridge stood as a wonder of late 1800s engineering, and it had recently been eclipsed by the Verrazano bridge, a pure suspension bridge. At the time it was the longest and heaviest in the world. How long could a bridge be made, and why did Brooklyn bridge have those catenary cables, when the Verrazano didn’t? (Sometimes I’d imagine a Chinese engineer being asked the top question, and answering “Certainly, but How Long is my cousin.”)

I found the above problem unsolvable with the basic calculus at my disposal. because it was clear that both the angle of the main cable and its tension varied significantly along the length of the cable. Eventually I solved this problem using a big dose of geometry and vectors, as I’ll show.

Vector diagram of forces on the cable at the center-left of the bridge.

Vector diagram of forces on the cable at the center-left of the bridge.

Consider the above vector diagram (above) of forces on a section of the main cable near the center of the bridge. At the right, the center of the bridge, the cable is horizontal, and has a significant tension. Let’s call that T°. Away from the center of the bridge, there is a vertical cable supporting a fraction of  roadway. Lets call the force on this point w. It equals the weight of this section of cable and this section of roadway. Because of this weight, the main cable bends upward to the left and carries more tension than T°. The tangent (slope) of the upward curve will equal w/T°, and the new tension will be the vector sum along the new slope. From geometry, T= √(w2 +T°2).

Vector diagram of forces on the cable further from the center of the bridge.

Vector diagram of forces on the cable further from the center of the bridge.

As we continue from the center, there are more and more verticals, each supporting approximately the same weight, w. From geometry, if w weight is added at each vertical, the change in slope is always w/T° as shown. When you reach the towers, the weight of the bridge must equal 2T Sin Θ, where Θ is the angle of the bridge cable at the tower and T is the tension in the cable at the tower.

The limit to the weight of a bridge, and thus its length, is the maximum tension in the main cable, T, and the maximum angle, that at the towers. Θ. I assumed that the maximum bridge would be made of T1 bridge steel, the strongest material I could think of, with a tensile strength of 100,000 psi, and I imagined a maximum angle at the towers of 30°. Since there are two towers and sin 30° = 1/2, it becomes clear that, with this 30° angle cable, the tension at the tower must equal the total weight of the bridge. Interesting.

Now, to find the length of the bridge, note that the weight of the bridge is proportional to its length times the density and cross section of the metal. I imagined a bridge where the half of the weight was in the main cable, and the rest was in the roadway, cars and verticals. If the main cable is made of T1 “bridge steel”, the density of the cable is 0.2833 lb/in3, and the density of the bridge is twice this. If the bridge cable is at its yield strength, 100,000 psi, at the towers, it must be that each square inch of cable supports 50,000 pounds of cable and 50,000 lbs of cars, roadway and verticals. The maximum length (with no allowance for wind or a safety factor) is thus

L(max) = 100,000 psi / 2 x 0.2833 pounds/in3 = 176,500 inches = 14,700 feet = 2.79 miles.

This was more than three times the length of the Verrazano bridge, whose main span is ‎4,260 ft. I attributed the difference to safety factors, wind, price, etc. I then set out to calculate the height of the towers, and the only rational approach I could think of involved calculus. Fortunately, I could integrate for the curve now that I knew the slope changed linearly with distance from the center. That is for every length between verticals, the slope changes by the same amount, w/T°. This was to say that d2y/dx2 = w/T° and the curve this described was a parabola.

Rather than solving with heavy calculus, I noticed that the slope, dy/dx increases in proportion to x, and since the slope at the end, at L/2, was that of a 30° triangle, 1/√3, it was clear to me that

dy/dx = (x/(L/2))/√3

where x is the distance from the center of the bridge, and L is the length of the bridge, 14,700 ft. dy/dx = 2x/L√3.

We find that:
H = ∫dy = ∫ 2x/L√3 dx = L/4√3 = 2122 ft,

where H is the height of the towers. Calculated this way, the towers were quite tall, higher than that of any building then standing, but not impossibly high (the Dubai tower is higher). It was fairly clear that you didn’t want a tower much higher than this, though, suggesting that you didn’t want to go any higher than a 30° angle for the main cable.

I decided that suspension bridges had some advantages over other designs in that they avoid the problem of beam “buckling.’ Further, they readjust their shape somewhat to accommodate heavy point loads. Arch and truss bridges don’t do this, quite. Since the towers were quite a lot taller than any building then in existence, I came to I decide that this length, 2.79 miles, was about as long as you could make the main span of a bridge.

I later came to discover materials with a higher strength per weight (titanium, fiber glass, aramid, carbon fiber…) and came to think you could go longer, but the calculation is the same, and any practical bridge would be shorter, if only because of the need for a safety factor. I also came to recalculate the height of the towers without calculus, and got an answer that was shorter, for some versions, a hundred feet shorter, as shown here. In terms of wind, I note that you could make the bridge so heavy that you don’t have to worry about wind except for resonance effects. Those are the effects are significant, but were not my concern at the moment.

The Brooklyn Bridge showing its main cable suspension structure and its catenaries.

Now to discuss catenaries, the diagonal wires that support many modern bridges and that, on the Brooklyn bridge, provide  support at the ends of the spans only. Since the catenaries support some weight of the Brooklyn bridge, they decrease the need for very thick cables and very high towers. The benefit goes down as the catenary angle goes to the horizontal, though as the lower the angle the longer the catenary, and the lower the fraction of the force goes into lift. I suspect this is why Roebling used catenaries only near the Brooklyn bridge towers, for angles no more than about 45°. I was very proud of all this when I thought it through and explained it to a friend. It still gives me joy to explain it here.

Robert Buxbaum, May 16, 2019.  I’ve wondered about adding vibration dampers to very long bridges to decrease resonance problems. It seems like a good idea. Though I have never gone so far as to do calculations along these lines, I note that several of the world’s tallest buildings were made of concrete, not steel, because concrete provides natural vibration damping.

Why concrete cracks and why sealing is worthwhile

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The oil tanker Palo Alto is one of several major ships made with concrete hulls.

The oil tanker Palo Alto is one of several major ships made with concrete hulls.

Modern concrete is a wonderful construction material. Major buildings are constructed of it, and major dams, and even some ships. But under the wrong circumstances, concrete has a surprising tendency to crack and fail. I thought I’d explain why that happens and what you can do about it. Concrete does not have to crack easily; ancient concrete didn’t and military or ship concrete doesn’t today. A lot of the fault lies in the use of cheap concrete — concrete with lots of filler — and with the cheap way that concrete is laid. First off, the major components of modern concrete are pretty uniform: sand and rock, Portland cement powder (made from cooked limestone, mostly), water, air, and sometimes ash. The cement component is what holds it all together — cements it together as it were — but it is not the majority of even the strongest concretes. The formula of cement has changed too, but the cement is not generally the problem. It doesn’t necessarily stick well to the rock or sand component of concrete (It sticks far better to itself) but it sticks well enough that spoliation, isn’t usually a problem by itself.

What causes problem is that the strength of concrete is strongly affected (decreased) by having lots of sand, aggregate and water. The concrete used in sidewalks is as cheap as possible, with lots of sand and aggregate. Highway and wall concrete has less sand and aggregate, and is stronger. Military and ship concrete has little sand, and is quite a lot stronger. The lowest grade, used in sidewalks, is M5, a term that refers to its compressive strength: 5 Mega Pascals. Pascals are European (Standard International) units of pressure and of strength. One Pascal is one Newton per square meter (Here’ a joke about Pascal units). In US (English) units, 5 MPa is 50 atm or 750 psi.

Ratios for concrete mixes of different strength.

Ratios for concrete mixes of different strength; the numbers I use are double these because these numbers don’t include water; that’s my “1”.

The ratio of dry ingredients in various concretes is shown at right. For M5, and including water, the ratio is 1 2 10 20. That is to say there is one part water, two parts cement, 10 parts sand, and 20 parts stone-aggregate (all these by weight). Added to this is 2-3% air, by volume, or nearly as much air as water. At least these are the target ratios; it sometimes happens that extra air and water are added to a concrete mix by greedy or rushed contractors. It’s sometimes done to save money, but more often because the job ran late. The more the mixer turns the more air gets added. If it turns too long there is extra air. It the job runs late, workers will have to add extra water too because the concrete starts hardening. I you see workers hosing down wet concrete as it comes from the truck, this is why. As you might expect, extra air and water decrease the strength of the product. M-10 and M-20 concrete have less sand, stone, and water as a proportion to cement. The result is 10 MPa or 20 MPa strength respectively.

A good on-site inspector is needed to keep the crew from adding too much water. Some water is needed for the polymerization (setting) of the concrete. The rest is excess, and when it evaporates, it leaves voids that are similar to the voids created by having air mix in. It is not uncommon to find 6% voids, in commercial concrete. This is to say that, after the water evaporates, the concrete contains about as much void as cement by volume. To get a sense of how much void space is in the normal concrete outside your house, go outside to a piece of old concrete (10 years old at least) on a hot, dry day, and pour out a cup of water. You will hear a hiss as the water absorbs, and you will see bubbles come out as the water goes in. It used to be common for cities to send inspectors to measuring the void content of the wet (and dry) concrete by a technique called “pycnometry” (that’s Greek for density measurement). I’ve not seen a local city do this in years, but don’t know why. An industrial pycnometer is shown below.

Pyncnometer used for concrete. I don't see these in use much any more.

Pycnometer used for concrete. I don’t see these in use much any more.

One of the main reason that concrete fails has to do with differential expansion, thermal stress, a concept I dealt with some years ago when figuring out how cold it had to be to freeze the balls off of a brass monkey. As an example of the temperature change to destroy M5, consider that the thermal expansion of cement is roughly 1 x 10-5/ °F or 1.8 x10-5/°C. This is to say that a 1 meter slab of cement that is heated or cooled by 100°F will expand or shrink by 10-3 m respectively; 100 x 1×10-5 = 10-3. This is a fairly large thermal expansion coefficient, as these things go. It would not cause stress-failure except that sand and rock have a smaller thermal expansion coefficients, about 0.6×10-5 — barely more than half the value for cement. Consider now what happens to concrete that s poured in the summer when it is 80°F out, and where the concrete heats up 100°F on setting (cement setting releases heat). Now lets come back in winter when it’s 0°F. This is a total of 100°F of temperature change. The differential expansion is 0.4 x 10-5/°F x 100°F =  4 x10-4 meter/meter = 4 x10-4 inch/inch.

The force created by this differential expansion is the elastic modulus of the cement times the relative change in expansion. The elastic modulus for typical cement is 20 GPa or, in English units, 3 million psi. This is to say that, if you had a column of cement (not concrete), one psi of force would compress it by 1/3,000,000. The differential expansion we calculated, cement vs sand and stone is 4×10-4 ; this much expansion times the elastic modulus, 3,000,000 = 1200 psi. Now look at the strength of the M-5 cement; it’s only 750 psi. When M-5 concrete is exposed to these conditions it will not survive. M-10 will fail on its own, from the temperature change, without any help needed from heavy traffic. You’d really like to see cities check the concrete, but I’ve seen little evidence that they do.

Water makes things worse, and not only because it creates voids when it evaporates. Water also messes up the polymerization reaction of the cement. Basic, fast setting cement is mostly Ca3SiO5

2Ca3SiO5 + 6 H2O –> 3Ca0SiO2•H2O +3Ca(OH)2•H2O.

The former of these, 3Ca0SiO2•H2O, forms something of a polymer. Monomer units of SiO4 are linked directly or by partially hydrated CaO linkages. Add too much water and the polymeric linkages are weakened or do not form at all. Over time the Ca(OH)2 can drain away or react with  CO2 in the air to form chalk.

concrete strength versus-curing time. Slow curing of damp concrete helps; fast dry hurts. Carbonate formation adds little or no strength. Jehan Elsamni 2011.

Portland limestone cement strength versus curing time. Slow curing and damp helps; fast dry hurts. Carbonate formation adds little or no strength. Jehan Elsamni 2011.

Ca(OH)2 + CO2 → CaCO3 + H2O

Sorry to say, the chalk adds little or no strength, as the graph at right shows. Concrete made with too much water isn’t very strong at all, and it gets no stronger when dried in air. Hardening goes on for some weeks after pouring, and this is the reason you don’t drive on 1 too 2 day old concrete. Driving on weak concrete can cause cracks that would not form if you waited.

You might think to make better concrete by pouring concrete in the cold, but pouring in the cold makes things worse. Cold poured cement will expand the summer and the cement will detach from the sand and stone. Ideally, pouring should be in spring or fall, when the temperature is moderate, 40-60°F. Any crack that develops grows by a mechanism called Rayleigh crack growth, described here. Basically, once a crack starts, it concentrates the fracture forces, and any wiggling of the concrete makes the crack grow faster.

Based on the above, I’ve come to suspect that putting on a surface coat can (could) help strengthen old concrete, even long after it’s hardened. Mostly this would happen by filling in voids and cracks, but also by extending the polymer chains. I imagine it would be especially helpful to apply the surface coat somewhat watery on a dry day in the summer. In that case, I imagine that Ca3SiO5 and Ca(OH)2 from the surface coat will penetrate and fill the pores of the concrete below — the sales pores that hiss when you pour water on them. I imagine this would fill cracks and voids, and extend existing CaOSiO2•H2O chains. The coat should add strength, and should be attractive as well. At least that was my thought.

I should note that, while Portland cement is mostly Ca3SiO5, there is also a fair amount (25%) of Ca2SiO4. This component reacts with water to form the same calcium-silicate polymer as above, but does so at a slower rate using less water per gram. My hope was that this component would be the main one to diffuse into deep pores of the concrete, reacting there to strengthen the concrete long after surface drying had occurred.

Trump tower: 664', concrete and glass. What grade of concrete would you use?

Trump tower: 664′, concrete and glass. What grade of concrete would you use?

As it happened, I had a chance to test my ideas this summer and also about 3 years ago. The city inspector came by to say the concrete flags outside my house were rough, and thus needed replacing, and that I was to pay or do it myself. Not that I understand the need for smooth concrete, quite, but that’s our fair city. I applied for a building permit to apply a surface coat, and applied it watery. I used “Quickrete” brand concrete patch, and so far it’s sticking OK. Pock-holes in the old concrete have been filled in, and so far surface is smooth. We’ll have to see if my patch lasts 10-20 years like fresh cement. Otherwise, no matter how strong the concrete becomes underneath, the city will be upset, and I’ll have to fix it. I’ve noticed that there is already some crumbling at the sides of flags, something I attribute to the extra water. It’s not a problem yet, but hope this is not the beginning of something worse. If I’m wrong here, and the whole seal-coat flakes off, I’ll be stuck replacing the flags, or continuing to re-coat just to preserve my reputation. But that’s the cost of experimentation. I tried something new, and am blogging about it in the hope that you and I benefit. “Education is what you get when you don’t get what you want.” (It’s one of my wise sayings). At the worst, I’ll have spent 90 lb of patching cement to get an education. And, I’m happy to say that some of the relatively new concrete flags that the city put in are already cracked. I attribute this to: too much sand, air, water or air (they don’t look like they have much rock): Poor oversight.

Dr. Robert E. Buxbaum. March 5, 2019. As an aside, the 664 foot Trump Tower, NY is virtually the only skyscraper in the city to be built of concrete and glass. The others are mostly steel and glass. Concrete and glass is supposed to be stiffer and quieter. The engineer overseeing the project was Barbara Res, the first woman to oversee a major, NY building project. Thought question: if you built the Trump Tower, which quality of concrete would you use, and why.

A logic joke, and an engineering joke.

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The following is an oldish logic joke. I used it to explain a conclusion I’d come to, and I got just a blank stare and a confused giggle, so here goes:

Three logicians walk into a bar. The barman asks: “Do all of you want the daily special?” The first logician says, “I don’t know.” The second says, “I don’t know.” The third says, “yes.”

The point of the joke was that, in several situations, depending on who you ask, “I don’t know” can be a very meaningful answer. Similarly, “I’m not sure.”  While I’m at it, here’s an engineering education joke, it’s based on the same logic, here applied:

A team of student engineers builds an airplane and wheel it out before the faculty. “We’ve designed this plane”, they explain, “based on the principles and methods you taught us. “We’ve checked our calculations rigorously, and we’re sure we’ve missed nothing. “Now. it would be a great honor to us if you would join us on its maiden flight.”

At this point, some of the professors turn white, and all of them provide various excuses for why they can’t go just now. But there is one exception, the dean of engineering smiles broadly, compliments the students, and says he’ll be happy to fly. He gets onboard the plane seating himself in the front of the plane, right behind the pilot. After strapping himself in, a reporter from the student paper comes along and asks why he alone is willing to take this ride; “Why you and no one else?” The engineering dean explains, “You see, son, I have an advantage over the other professors: Not only did I teach many of you, fine students, but I taught many of them as well.” “I know this plane is safe: There is no way it will leave the ground.”heredity cartoon

Robert Buxbaum, November 2i, 2018.  And one last. I used to teach at Michigan State University. They are fine students.

magnetic separation of air

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

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

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

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

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

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

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

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

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

A very clever hydrogen pump

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I’d like to describe a most clever hydrogen pump. I didn’t invent it, but it’s awfully cool. I did try to buy one from “H2 Pump,” a company that is now defunct, and I tried to make one. Perhaps I’ll try again. Here is a diagram.

Electrolytic membrane H2 pump

Electrolytic membrane H2 pump

This pump works as the reverse of of a PEM fuel cell. Hydrogen gas is on both sides of a platinum-coated, proton-conducting membrane — a fuel cell membrane. As in a PEM fuel cell, the platinum splits the hydrogen molecules into H atoms. An electrode removes electrons to form H+ ions on one side of the membrane; the electrons are on the other side of the membrane (the membrane itself is chosen to not conduct electricity). The difference from the fuel cell is that, for the pump you apply a energy (voltage) to drive hydrogen across the membrane, to a higher pressure side; in a fuel cell, the hydrogen goes on its own to form water, and you extract electric energy.

As shown, the design is amazingly simple and efficient. There are no moving parts except for the hydrogen itself. Not only do you pump hydrogen, but you can purify it as well, as most impurities (nitrogen, CO2) will not go through the membrane. Water does permeate the membrane, but for many applications, this isn’t a major impurity. The amount of hydrogen transferred per plate, per Amp-second of current is given by Faraday’s law, an equation that also shows up in my discussion of electrolysis, and of electroplating,

C= zFn.

Here, C is the current in Amp-seconds, z is the number or electrons transferred per molecule, in this case 2, F is Faraday’s constant, 96,800, n is the number of mols transferred.  If only one plate is used, you need 96,800 Amp-seconds per gram of hydrogen, 53.8 Amp hours per mol. Most membranes can operate at well at 1.5 Amp per cm2, suggesting that a 1.1 square-foot membrane (1000 cm2) will move about 1 mol per minute, 22.4 slpm. To reduce the current requirement, though not the membrane area requirement, one typically stacks the membranes. A 100 membrane stack would take 16.1 Amps to pump 22.4 slpm — a very manageable current.

The amount of energy needed per mol is related to the pressure difference via the difference in Gibbs energy, ∆G, at the relevant temperature.

Energy needed per mol is, ideally = ∆G = RT ln Pu/Pd.

where R is the gas constant, 8.34 Joules per mol, T is the absolute temperature, Kelvins (298 for a room temperature process), ln is the natural log, and Pu/Pd is the ratio of the upstream and downstream pressure. We find that, to compress 2 grams of hydrogen (one mol or 22.4 liters) to 100 atm (1500 psi) from 1 atm you need only 11400 Watt seconds of energy (8.34 x 298 x 4.61= 11,400). This is .00317 kW-hrs. This energy costs only 0.03¢ at current electric prices, by far the cheapest power requirement to pump this much hydrogen that I know of. The pump is surprisingly compact and simple, and you get purification of the hydrogen too. What could possibly go wrong? How could the H2 pump company fail?

One thing that I noticed went wrong when I tried building one of these was leakage at the seals. I found it uncommonly hard to make seals that held even 20 psi. I was using 4″ x 4″ membranes so 20 psi was the equivalent of 320 pounds of force. If I were to get 200 psi, there would have been 3200 lbs of force. I could never get the seals to stay put at anything more than 20 psi.

Another problem was the membranes themselves. The membranes I bought were not very strong. I used a wire-mesh backing, and a layer of steel behind that. I figured I could reach maybe 200 psi with this design, but didn’t get there. These low pressures limit the range of pump applications. For many applications,  you’d want 150-200 psi. Still, it’s an awfully cool pump,

Robert E. Buxbaum, February 17, 2017. My company, REB Research, makes hydrogen generators and purifiers. I’ve previously pointed out that hydrogen fuel cell cars have some dramatic advantages over pure battery cars.

just water over the dam

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Some months ago, I posted an engineering challenge: figure out the water rate over an non-standard V-weir dam during a high flow period (a storm) based on measurements made on the weir during a low flow period. My solution follows. Weir dams of this sort are erected mostly to prevent flooding. They provide secondary benefits, though by improving the water and providing a way to measure the flow.

A series of weir dams on Blackman Stream, Maine. These are thick, rectangular weirs.

A series of compound, rectangular weir dams in Maine.

The problem was stated as follows: You’ve got a classic V weir on a dam, but it is not a knife-edge weir, nor is it rectangular or compound as in the picture at right. Instead it is nearly 90°, not very tall, and both the dam and weir have rounded leads. Because the weir is of non-standard shape, thick and rounded, you can not use the flow equation found in standard tables or on the internet. Instead, you decide to use a bucket and stopwatch to determine the flow during a relatively dry period. You measure 0.8 gal/sec when the water height is 3″ in the weir. During the rain-storm some days later, you measure that there are 12″ of water in the weir. Give a good estimate of the storm-flow based on the information you have.

A V-notch weir, side view and end-on.

A V-notch weir, side view and end-on.

I also gave a hint, that the flow in a V weir is self-similar. That is, though you may not know what the pattern will be, you can expect it will be stretched the same for all heights.

The upshot of this hint is that, there is one, fairly constant flow coefficient, you just have to find it and the power dependence. First, notice that area of flow will increase with height squared. Now notice that the velocity will increase with the square root of hight, H because of an energy balance. Potential energy per unit volume is mgH, and kinetic energy per unit volume is 1/2 mv2 where m is the mass per unit volume and g is the gravitational constant. Flow in the weir is achieved by converting potential height energy into kinetic, velocity energy. From the power dependence, you can expect that the average v will be proportional to √H at all values of H.

Combining the two effects together, you can expect a power dependence of 2.5 (square root is a power of 0.5). Putting this another way, the storm height in the weir is four times the dry height, so the area of flow is 16 times what it was when you measured with the bucket. Also, since the average height is four times greater than before, you can expect that the average velocity will be twice what it was. Thus, we estimate that there was 32 times the flow during the storm than there was during the dry period; 32 x 0.8 = 25.6 gallons/sec., or 92,000 gal/hr, or 3.28 cfs.

The general equation I derive for flow over this, V-shaped weir is

Flow (gallons/sec) = Cv gal/hr x(feet)5/2.

where Cv = 3.28 cfs. This result is not much different to a standard one  in the tables — that for knife-edge, 90° weirs with large shoulders on either side and at least twice the weir height below the weir (P, in the diagram above). For this knife-edge weir, the Bureau of Reclamation Manual suggests Cv = 2.49 and a power value of 2.48. It is unlikely that you ever get this sort of knife-edge weir in a practical application. Be sure to measure Cv at low flow for any weir you build or find.

Robert Buxbaum, vote for me for water commissioner. Here are some thoughts on other problems with our drains.

More French engineering, the Blitzkrieg motorcycle.

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There’s something fascinating that I find in French engineering. I wrote a previous essay about French cars, bridges, and the Eiffel tower. Here’s a picture or two more. Things I wanted to include but didn’t. First here’s a Blitzkrieg Vespa motorcycle; the French built some 800 of these from 1947 to 1962 and used them in Vietnam and Algeria. What’s remarkable is how bizarrely light and unprotected it is. It’s a design aesthetic that follows no one, and that American engineers would not follow.

French Blitzkrieg Vespa used in Vietnam

French Blitzkrieg Vespa used in Vietnam; cannon range is 4.5 miles.

The key engineering insight that allows this vehicle to make sense is that recoil-less rifles are really recoil-less if you design them right. Thus, one can (in theory) mount them on something really light, like a Vespa. Another key (French) insight is that a larger vehicle may make the soldier more vulnerable rather than less by slowing him down and by requiring more gasoline and commissariat services.

Americans do understand the idea of light and mobile, but an American engineers idea of this is a jeep or an armored truck; not a Vespa. From my US engineering perspective, the French went way overboard here. The French copy no one, and no one copies the French, as they say. Still, these things must have worked reasonably well or they would not have made 800 of them over 15 years. A Vespa is certainly cheaper than a Jeep, and easier to transport to the battle zone….

Robert Buxbaum, February 18, 2016. The Italians have a somewhat similar design aesthetic to the French: they like light and cheap, but also like maneuverable and favor new technology. See my blog about a favorite Fiat engine.

Weird flow calculation

Here is a nice flow problem, suitable for those planning to take the professional engineers test. The problem concerns weir dams. These are dams with a notch in them, somethings rectangular, as below, but in this case a V-shaped notch. Weir dams with either sort of notch can be used to prevent flooding and improve the water, but they also provide a way to measure the flow of water during a flood. That’s the point of the problem below.

A series of weir dams on Blackman Stream, Maine. These are thick, rectangular weirs.

A series of weir dams with rectangular weirs in Maine.

You’ve got a classic V weir on a dam, but it is not a knife-edge weir, nor is it rectangular or compound as in the picture at right. Instead it is nearly 90°, not very tall, and both the dam and weir have rounded leads. Because the weir is of non-standard shape, thick and rounded, you can not use the flow equation found in standard tables or on the internet. Instead, you decide to use a bucket and stopwatch to determine that the flow during a relatively dry period. You measure 0.8 gal/sec when the water height is 3″ in the weir. During the rain-storm some days later, you measure that there are 12″ of water in the weir. The flow is too great for you to measure with a bucket and stopwatch, but you still want to know what the flow is. Give a good estimate of the flow based on the information you have.

As a hint, notice that the flow in the V weir is self-similar. That is, though you may not know what the pattern of flow will be, you can expect it will be stretched the same for all heights.

As to why anyone would use this type of weir: they are easier to build and maintain than the research-standard, knife edge; they look nicer, and they are sturdier. Here’s my essay in praise of the use of dams. How dams on drains and rivers could help oxygenate the water, and to help increase the retention time to provide for natural bio-remediation.

If you’ve missed the previous problem, here it is: If you have a U-shaped drain or river-bed, and you use a small dam or weir to double the water height, what is the effect on water speed and average retention time. Work it out yourself, or go here to see my solution.

Robert Buxbaum. May 20-Sept 20, 2016. I’m running for drain commissioner. send me your answers to this problem, or money for my campaign, and win a campaign button. Currently, as best I can tell, there are no calibrated weirs or other flow meters on any of the rivers in the county, or on any of the sewers. We need to know because every engineering decision is based on the flow. Another thought: I’d like to separate our combined sewers and daylight some of our hidden drains.