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Well I have to be WRONG first Billy. The product you linked to is for propane tanks and Propane is stored at a very low 180 psi. You were talking about LNG tanks that operate at 3,600 psi. ...
Arthur
Yes, currently only few cars and trucks have NG adsorbed as a liquid on high-surface to volume materials. The first hit the road in Feb 2007 with a rectangular tank operating at 500 psi. In a decade there should be many more on the roads than EVs.
Fact that currently most flat tanks are filled with LPG not LNG is no defense for you telling me: I lacked engineering skill, was suggesting an impractical "unobtainium" flat tank that could not be built, and ridiculing me in many posts.
YOU ARE WRONG. You should apologize, but I know you will not - you never admit error or apologize.
“… Ground corncobs, a plentiful agricultural by-product in Missouri and surrounding Corn Belt states, are the main ingredient in a new technology that may soon power
natural-gas-powered vehicles on the country's highways. … The initial goal of ANG {Company licensed by MU} is to utilize the high-surface-area carbon in
natural gas tanks for ground transportation vehicles within the next year. …”
From:
http://munews.missouri.edu/news-releases/2009/0513-suppes-pfeifer-natural-gas-license.php
PS you could make a flat tank that operates at 3600 psi without the filler and its liquid film but
there is no need for it with the MU “corn cob” technology. – I have more carefully analysed one (while working on a potential patent application) and even including the fillets it has 8.5% more volume per unit volume of wall material used. What follows is the first draft, but not finished when I learned the flat tanks are commercially available.
Analysis:
Consider a unit length of a long, round, gas filled cylinder, with ID of “D” and walls “t” thick everywhere. Assume the “safety margin” is in t. Assume its axis is the z-axis of a (X, Y, Z) Cartesian coordinate system with the y-axis vertical. This cylinder is but one of many adjoining parallel cylinders whose axes are all in the ZX plane. Their intersections with the XY plane are illustrated below, but the one centered on the z a-axis has been made bold. Our attention will focus on it initially.
…OOO
OOOO … But for reasons soon to be explained, I will illustrate these cross sections with: …CCCCCCC …
I call the “t-thick” semicircle in the XY plane with y > 0 “upper” (or “U”) and that with y < 0 “bottom” (or “B”) and the two, zero height, “t-wide” bands on the x-axis R & L, (for right & left). Thus: the first R band with x > 0 extend between the points (D/2, 0) and ([D/2] + t, 0).
Note that the pressure induced tension in the “t-thick” semicircles, U & B, is strictly vertical where they join the x-axis bands. I.e. all along all of the R & L bands, the tension force is strictly vertical. Thus, IF there were no pressure difference horizontally across it, a vertical flat slab of height H, and t thickness, parallel to the z-axis, could have always existed separating the two semicircles, U & B, but the tension in semicircles U & B would increase. The reason why there is no horizontal pressure force across theses flat, H tall, slabs is that the gas pressure on both sides is the same. (All these “vertically stretched” tubes are interconnected in a manifold at one end and individually closed at the other end.)
The “first vertical slab” R with x > 0 “belongs” to the first tube to the right of the one centered on the z-axis, not to that z-axis tube. Thus each tube, except one at the extreme right of the set of tubes, has only the three sides: B, L & U, not four “belonging” to it. This is the fundamental reason why the set can contain more gas per unit of wall mass; however H must be greater than D for there to be significantly less material used per unit of gas contained. Now I will address some of the other details and show this gain in material use efficiency mathematically.
For example, at the top wall U (and the bottom of wall B), the tension is still exactly horizontal but increased compared to the round tube. Thus the thickness there would need to be T = t(1+H/D), including the constant safety factor, to keep this flat sided tube from splitting horizontally. If the top U and bottom B were flat, instead of semicircles, they would have a tendency to bow outward but if T >> D this short thick “flat bridge” of length (t/2+D+t/2) spanning the gap D wide would not flex much when the walls are made of very stiff material. Note each tube’s bridge is part of a large flat sheet and only conceptually considered to be (D +t) wide (and T thick) for the analysis of one tube separately.
For a numerical example, taking pi as 3.14, let H = 3.14D, then the flat tops of U & B are “bridges” of thickness 4.14t spanning a gap of only D. If D = 1cm and t = 0.1cm, then the flat bar or bridge is 0.414 cm thick and only 1.0 cm long. Such a thick, short, bar would not bend much as the pressure is increased. It would probably be ripped off the vertical slabs, which are in this numerical case only 0.1 thick before the bending significantly as the failure mode. These t-thick divisions between the adjoining tubes are the “tension webs” holding the large top and bottom sheets (U & B) together.
Proper design of the ends of these tension webs where they join to the large flat sheet (U & B) that closes each of the D wide tubes is a complex computer modeling problem, which models the transformation of the purely vertical stress in the vertical tension web slabs to a purely horizontal stress at the midpoints of horizontal “bridges” U & B spanning the gap D between the tension web divisions between the individual tubes.
To facilitate this 90 degree turn in the stress direction, there would be no square corners where the tension web slabs join the horizontal bridge bars U & B. Instead “fillets” of approximately 0.2 cm radius would exist. Then the gap directly exposed to the gas pressure spanned by the top and bottom bridge bars would be only 0.6 cm when D = 1.0 cm. I assume that a bar with thickness of 0.414 cm spanning a gap of only 0.6 cm surely would have negligible bending. I. e. bending a bar of very stiff material significantly whose thickness is more than 2/3 of its length by uniform pressure applied to one side is not likely to be a problem.
With four “fillets” filling in all corners of each “vertically stretched” tube the volume of a 1cm unit of length along the z-axis times the cross section area of these four 0.2 cm radius “fillets” which is area of a square 0.4 cm on an edge minus the area of a circle 0.4 cm in diameter. Or 0.16 – 3.14 [(0.2)^2] = [0.16 – (3.14x0.04)] = 0.0344 square centimeters. This 0.0344 cc must be added to 1cm length times the rectangular cross section area and subtracted from 1 cm long unit of rectangular contained area’s volume.
The 1cm length of rectangular wall volume is 1 times twice the top bar cross section area + 1Ht or:
2x0.414x(1+0.1) + (3.14x1)x0.1 = 0.828x1.1 + 0.314 = 0.9108 + 0.314 = 1.2248 cc
Thus the total material volume of one centimeter length of the “stretched tube” is 1.2248 + 0.0344 = 1.2592 cc.
One centimeter length of the rectangular tube’s contained volume is: HD = 3.14cc but the 0.0344cc of the fillets is subtracted from this to get the volume of gas contained in each centimeter of the vertically stretched tube’s length. I.e. that 1cm long gas volume is 3.1056 cc.
Thus the gas volume to containing material volume ratio for the stretched tube is 3.1056 / 1.2592 = 2.46633
The ID radius r of a circular tube which also holds 3.1056 cc of gas per centimeter of length is r^2 = 3.1056 / 3.14 = 0.9890440 so r = 0.99451cm or slightly less than D = 1cm. As the wall thickness required is directly proportional to the diameter, the walls are only 0.099451cm thick, also less than t = 0.1cm of the vertically stretched tube. The outer radius of the round walls is 0.99451 +2x0.099451 = 1.193412 whose square is: 1.424232. Thus the difference between the area of ID and OD circles is 3.14(1.4242322 – 0.9890440) =3.14x0.4351882 = 1.3649095 which is significantly greater than the 1.2592 cc of material required by the vertically stretched tube to hold 2.46633 cc of gas per centimeter of length.
SUMMARY: The tube with vertical stretch of 3.14 cm spaced 1cm apart is more efficient in the use of wall material by the factor of 1.3649095 / 1.2592 = 1.0852 or 8.5% more efficient even with the four “fillets” added to the material used and decreasing the gas volume storage capacity; however, the two extreme edges of this set of parallel “stretched tubes” have not been considered. They could be semicircles in cross section with ID of D = 3.14cm. The right most edge tube would have four sides as its vertical wall closes the vertically stretched “C shape” tube adjoining it as illustrated below:
…………….CCCCCCCCCCCCCCCCCCCCCCD
The left most extreme tube, however, like all the others has only three walls previously called B, L, & U but for this tube they are all just one semicircle like a mirror image of the extreme right, “D shaped,” edge tube illustrated above. Because pressure outside of these two “D-shaped tube is atmospheric their walls must 3.14t thick.
Recall t was the wall thickness of a 1 cm diameter circular tube, which included the desired safety factor. The wall of a 3.14cm circular tube would be 3.14 times thicker to have the same safety factor as must these two extreme edge tubes curved semicircular walls. The vertical wall of the “D-shaped” right most tube illustrated above is, however, only t thick, like any other “tension web divider” as it has no pressure difference across it. There are two fillets inside each of the “D-shaped extreme tubes to facilitate the smooth transition of the vertical “tension web” divider stress vertical wall into the outer wall, even thought for these two extreme tube the outer wall is “D-shaped.”
These two D-shaped extreme edge tube if consider to have been joined have the same efficiency as 3.14cm diameter tube, except for the fact there is a 0.1cm bat across the diameter. When there are about 100 parallel tubes in this “flat gas storage” tank the overall efficiency might drop to only 8.4% better than the circular storage tube. 102 of these tubes would be a “flat tank” about 110 + 7= 117cm wide and 2x0.414 + 3.14 = 3.528 cm tall (That is less than 1.4 inches tall - very suitable for mounting on a large truck’s roof as its fuel tank. It could be wider and as long as the truck’s roof and be the roof. It could be thicker for more fuel stored. I.e. D > 3.14 cm)
If D were greater than the assumed 3.14cm then the efficiency relative to any larger circular tube storing the same amount of gas, or any combination of smaller circular tube storing that amount of gas, would be even greater. This is because the efficiency (gas volume to required wall volume) of circular tube tanks does not depend upon their diameter. E.g. if a circular tube’s diameter is made10 times larger, it will hold 100 times more gas but its walls must be 10 times thicker and the circumference is 10 times greater so the wall volume is also 100 times greater with no improvement in the gas volume to wall volume ratio.
Greater material efficiency than a more conventional pressurized gas storage tank is not the main commercial advantage. This “flat tank” could be the roof of an 18 wheeler truck holding CNG fuel for it with much less wind resistance than round tanks sitting on top of the truck or the roof of an intra-city bus, which also need to keep wind resistance as low as possible.
