U.S. patent application number 14/624370 was filed with the patent office on 2016-01-21 for coiled natural gas storage system and method.
This patent application is currently assigned to Other Lab, LLC. The applicant listed for this patent is Other Lab, LLC. Invention is credited to Samuel Eli Calisch, Tucker Gilman, Saul Griffith, Peter S. Lynn, Kevin Simon.
Application Number | 20160018057 14/624370 |
Document ID | / |
Family ID | 49514927 |
Filed Date | 2016-01-21 |
United States Patent
Application |
20160018057 |
Kind Code |
A1 |
Griffith; Saul ; et
al. |
January 21, 2016 |
Coiled Natural Gas Storage System and Method
Abstract
A system for storing natural gas comprises a plurality of
straight sections of tube. The plurality of straight sections of
tube are dense packed. The plurality of straight sections of tube
are configured to fill a designated volume.
Inventors: |
Griffith; Saul; (San
Francisco, CA) ; Lynn; Peter S.; (San Francisco,
CA) ; Calisch; Samuel Eli; (San Francisco, CA)
; Gilman; Tucker; (San Francisco, CA) ; Simon;
Kevin; (San Francisco, CA) |
|
Applicant: |
Name |
City |
State |
Country |
Type |
Other Lab, LLC |
San Francisco |
CA |
US |
|
|
Assignee: |
Other Lab, LLC
San Francisco
CA
|
Family ID: |
49514927 |
Appl. No.: |
14/624370 |
Filed: |
February 17, 2015 |
Related U.S. Patent Documents
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Application
Number |
Filing Date |
Patent Number |
|
|
13887201 |
May 3, 2013 |
|
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|
14624370 |
|
|
|
|
61642388 |
May 3, 2012 |
|
|
|
61766394 |
Feb 19, 2013 |
|
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Current U.S.
Class: |
220/581 |
Current CPC
Class: |
F17C 2201/0109 20130101;
F17C 2201/056 20130101; F17C 2209/221 20130101; F17C 2201/0138
20130101; F17C 2205/0107 20130101; F17C 2270/0168 20130101; F17C
2203/0673 20130101; F17C 2205/0111 20130101; F17C 2260/018
20130101; F17C 2201/0123 20130101; F17C 2205/0134 20130101; F17C
1/00 20130101; B60K 2015/03315 20130101; F17C 2209/2181 20130101;
F17C 2203/0636 20130101; F17C 2221/033 20130101; F17C 2201/054
20130101; F17C 2205/0115 20130101; F17C 1/005 20130101; F17C
2203/0646 20130101; F17C 2223/0123 20130101; B60K 15/03006
20130101; F17C 2205/0352 20130101 |
International
Class: |
F17C 1/00 20060101
F17C001/00 |
Claims
1. A system for storing natural gas, comprising: a tube that
defines a planar and rounded-rectangular body that coils from an
edge portion of the body to a central portion of the body, the body
including: a first tube-end disposed at the edge portion of the
body; a plurality of straight tube sections; a plurality of bend
sections coupled to and between the bend sections; and a second
tube-end disposed at the central portion of the body.
2. The system of claim 1, wherein the bend sections define a
90-degree bend between respective straight tube sections.
3. The system of claim 1, wherein the bend sections and straight
tube sections are defined is by a tube having the same
diameter.
4. The system of claim 1, wherein the rounded-rectangular body
defines a square rounded-rectangular body.
5. The system of claim 1, wherein each straight tube section is
disposed parallel to another straight tube section.
6. The system of claim 1, wherein each straight tube section is
disposed parallel to and contacting another straight tube section
along a respective length.
7. The system of claim 1, wherein a plurality of the bend sections
are nested within a curve of another bend section.
8. The system of claim 1, wherein the body defines a
central-orifice disposed at the central portion.
9. The system of claim 1, wherein the straight tube portions
comprise an inner liner with an outer braided layer.
10. The system of claim 1, wherein the straight tube portions and
bend portions are defined by a continuous unitary tube.
11. The system of claim 1, wherein the straight portions and bend
portions are defined by physically separable units that are coupled
together to define a contiguous body.
12. An assembly for storing natural gas, comprising a plurality of
the systems for storing natural gas of claim 1.
13. The assembly of claim 12, wherein the plurality of systems are
stacked in a layered configuration.
14. The assembly of claim 13, wherein a portion of the plurality of
systems are coiled in a clockwise configuration and wherein a
portion of the systems are coiled in a counter-clockwise
configuration.
15. The assembly of claim 12, wherein the plurality of systems are
stacked in a layered configuration having hexagonal packing.
16. A vehicle comprising a natural gas fuel-tank comprising the
system for storing natural gas of claim 1.
17. The vehicle of claim 16, wherein the system is disposed in a
chassis of the vehicle.
18. The vehicle of claim 16, wherein the vehicle comprises a truck
and the system is disposed in a bed of the truck.
Description
CROSS REFERENCE TO OTHER APPLICATIONS
[0001] This application is a continuation of, and claims priority
to, U.S. Non-Provisional patent application Ser. No. 13/887,201
entitled CONFORMING NATURAL GAS STORAGE filed May 3, 2013 which is
incorporated herein by reference for all purposes.
[0002] This application claims priority to U.S. Provisional Patent
Application No. 61/642,388 entitled CONFORMING ENERGY STORAGE filed
May 3, 2012 which is incorporated herein by reference for all
purposes.
[0003] This application claims priority to U.S. Provisional Patent
Application No. 61/766,394 entitled NATURAL GAS INTESTINE PACKED
STORAGE TANK filed Feb. 19, 2013 which is incorporated herein by
reference for all purposes.
BACKGROUND OF THE INVENTION
[0004] Since the 1990's heavy vehicles have been taking advantage
of compressed natural gas (CNG) engines. However, light vehicles,
such as passenger cars, still have yet to achieve widespread
adoption. Both private and public players began to identify
technological hurdles to CNG passenger vehicle growth. Industry
realized that if certain storage issues could be solved natural gas
offered incredible untapped opportunity. However, current CNG
storage solutions, both for integrated vehicles and converted
vehicles, are still bulky and expensive cylinder based systems. For
the integrated systems, various sized cylindrical tanks are
integrated into the vehicle chassis design. For the converted
vehicles, a big tank is placed in the trunk, eliminating storage or
spare tires.
BRIEF DESCRIPTION OF THE DRAWINGS
[0005] Various embodiments of the invention are disclosed in the
following detailed description and the accompanying drawings.
[0006] FIG. 1 is a diagram illustrating an embodiment of a cross
section at a middle of a tank.
[0007] FIG. 2 is a diagram illustrating an embodiment of a cross
section of a low number of tanks in a design.
[0008] FIG. 3 is a graph illustrating an embodiment of a finite
element analysis (FEA) and the analytical solution.
[0009] FIGS. 4A-4D are diagrams illustrating embodiments of tank
construction with straight tubes.
[0010] FIGS. 5A and 5B are diagrams illustrating embodiments of
brazing areas.
[0011] FIG. 6A is a diagram illustrating an embodiment of a
parameterization of a revolved curve.
[0012] FIG. 6B is a diagram illustrating an embodiment of a
differential element along the length of a tube.
[0013] FIG. 6C is a diagram illustrating an embodiment of a tapered
braid geometry.
[0014] FIGS. 7A-C are diagrams illustrating embodiments of bent
square spirals.
[0015] FIGS. 8A-C are diagrams illustrating embodiments of bent
square spirals.
[0016] FIGS. 9A-C are diagrams illustrating embodiments of bent
square spirals.
[0017] FIG. 10 is a diagram illustrating an embodiment of a group
of flat spirals filling a two-dimensional layer.
[0018] FIG. 11 is a diagram illustrating an embodiment of a
flexible elbow.
DETAILED DESCRIPTION
[0019] The invention can be implemented in numerous ways, including
as a process; an apparatus; a system; a composition of matter; a
computer program product embodied on a computer readable storage
medium; and/or a processor, such as a processor configured to
execute instructions stored on and/or provided by a memory coupled
to the processor. In this specification, these implementations, or
any other form that the invention may take, may be referred to as
techniques. In general, the order of the steps of disclosed
processes may be altered within the scope of the invention. Unless
stated otherwise, a component such as a processor or a memory
described as being configured to perform a task may be implemented
as a general component that is temporarily configured to perform
the task at a given time or a specific component that is
manufactured to perform the task. As used herein, the term
`processor` refers to one or more devices, circuits, and/or
processing cores configured to process data, such as computer
program instructions.
[0020] A detailed description of one or more embodiments of the
invention is provided below along with accompanying figures that
illustrate the principles of the invention. The invention is
described in connection with such embodiments, but the invention is
not limited to any embodiment. The scope of the invention is
limited only by the claims and the invention encompasses numerous
alternatives, modifications and equivalents. Numerous specific
details are set forth in the following description in order to
provide a thorough understanding of the invention. These details
are provided for the purpose of example and the invention may be
practiced according to the claims without some or all of these
specific details. For the purpose of clarity, technical material
that is known in the technical fields related to the invention has
not been described in detail so that the invention is not
unnecessarily obscured.
[0021] A system for storing natural gas is disclosed. The system
comprises a plurality of straight sections of tube. The plurality
of straight sections of tube are dense packed. The plurality of
straight sections of tube are configured to fill a designated
volume.
[0022] In 1890 Giuseppe Peano discovered a class of curves that
fill 2-dimensional space, a result which Hilbert extended to
3-dimensional cubes. It can be shown that such a curve can densely
fill any 2- or 3-dimensional (2D or 3D) shape. Based on these
insights, an Intestine Storage Concept is disclosed, a compressed
gas storage tank modeled after the human intestine, an example of a
high density curve efficiently filling an irregular volume.
[0023] A key insight here is that in the design of a cylindrical
tank the ratio of the tank mass to the contained gas mass is not
dependent on tank geometry. The mass of the material used, and thus
the bulk material cost is constant for a given pressure and
material yield stress. With no penalty paid (in material cost or
packing density) for moving to small diameter tubes, the ability is
gained to fit the tank in to any 3D shape desired--for example
embedded in the chassis of the vehicle. In the following, several
options for creating a compressed natural gas (CNG) tank with small
diameter tubing are disclosed. Also, how to pack that tubing into
desirable geometries for use in the automotive industry is
disclosed. Important design considerations are discussed: material
stress, diffusion through the tank, and possible tank failure due
to creep, fatigue, or thermal conditions.
[0024] It can be shown that it is possible to fill arbitrary shapes
with small tubes without a loss in mass, here we discuss possible
ways to pack these tubes. Because it is likely difficult, from a
manufacturing stand point, to use actual Hilbert curves as a
packing scheme alternative packing geometries are examined, and it
is considered how well they fit into space (conformability factor).
The conformability of a tank design (that is, the ratio of gas
volume contained in the tank to the volume of the smallest bounding
box of the tank) is a function of tube geometry, and packing
geometry. To calculate conformability factor exactly, a precise
definition of the packing scheme considered is needed, including
bends.
[0025] In various embodiments, the tank has an inlet coupler at one
end, the tank has multiple inlet couplers, the tank has an inlet
coupler after a taper from the one or more straight sections of
tube, the tank has an inlet coupler coupled to a reduced section of
tube, or any other appropriate manner for coupling to the tank for
access.
[0026] FIG. 1 is a diagram illustrating an embodiment of a cross
section at a middle of a tank. In the example shown, a plurality of
circular cross sections of the tubes are packed into a rectangular
bounding box. The packing densities for these packing schemes,
considered without boundary conditions, are well known: hexagonal
pack 102, with a packing density of
.pi. 6 3 ##EQU00001##
or about 91%; rectangular pack 100, with a packing density of
.pi./4 or about 78.5%; and, one of many binary packing schemes 104,
with two different radii tubes with packing density of .about.92%.
Binary pack 104 has tubes with a diameter ratio of {square root
over (2)}-1. A binary packing scheme increases packing efficiencies
and could be used to add additional longitudinal strength.
[0027] FIG. 2 is a diagram illustrating an embodiment of a cross
section of a low number of tanks in a design. In the example shown,
hexagonal and rectangular packing in a bounding box is shown for 1,
4, and 16 cross sectional tanks (e.g., hexagonal 200, 202, and 204,
and rectangular 210, 212, and 214). At low number of tanks
(circles) boundary conditions dominate the packing and rectangular
packing can be more efficient. As the ratio of radius to bounding
box gets larger, boundary conditions become less significant as the
packing efficiencies approach ideal. Since hexagonal has a greater
ideal packing efficiency, at some point (e.g., some radius, given a
particular bounding box) the hexagonally packed tubes will always
be better than the best rectangular packing. This question, what is
the radius at which hexagonal packing is always better, can only
truly be answered once a bounding box is chosen, since boundary
conditions are heavily dependent on bounding box. Once a bounding
box is chosen the maximum radius or tubes, and number of tubes can
be determined which will ensure highest packing efficiency.
[0028] For straight cylindrical tubes, far away from the ends, the
stress along the length remains constant. The magnitude of this
stress depends on the radius and wall thickness of the straight
lengths. Stress concentrations may also develop at tube ends,
bends, or fittings.
[0029] Polymers or metals can be extruded into tubes that can be
bent and packed into dense geometries. In order to develop an
understanding of the load bearing capabilities of these tubes, a
single length of cylindrical tubing without considering the end
conditions is examined. For geometries where t/r.ltoreq.1/20 we can
use thin-wall theory to calculate the principal stresses in a
cylindrical vessel. The stress is as follows: [0030] 1. Hoop Stress
(.sigma..sub.h). We assume this to be uniform across wall
thickness. [0031] 2. Longitudinal Stress (.sigma..sub.l). This
stress exists for cylinders with end caps. We assume this is
uniform across a section and sufficiently far from end caps. Stress
concentrations will need to be dealt with separately [0032] 3.
Radial Stress (.sigma.r). Under the thin-walled assumption, the
radial Stress is insignificant compared to the longitudinal and
hoop stresses.
[0033] For hoop stress it is assumed that the stress in the
material is only due to the pressure inside the tank. The hoop
stress acts circumferentially around the cylinder. Since this force
is twice the longitudinal stress, it sets the required thickness of
the tank at a given pressure and material yield stress.
.sigma. h = .sigma. y fs = Pr t ##EQU00002## t = Prfs .sigma. y
##EQU00002.2##
[0034] with: Yield Stress (.sigma..sub.y), Pressure (P), Inner
Radius (r), Wall Thickness (t), Factor of Safety (fs)
[0035] The axial stress can be written as:
.sigma. a = P R 2 r - 1 ##EQU00003##
If we divide the hoop stress by the axial stress we get:
.sigma. h .sigma. a = R r + 1 ##EQU00004##
Again using a thin wall assumption: r.apprxeq.R, we can say:
.sigma. h .sigma. a 2 ##EQU00005##
which dictates that the hoop stress is about twice that of the
axial stress.
[0036] The tank mass is a function of tank pressure, volume and
material properties. Therefore, tank mass and bulk material cost
are not a function of tube radius. The equation for tank mass, for
a single cross section straight cylindrical tube is:
Mtank = .rho. V ( ( pfs .sigma. y + 1 ) 2 - 1 ) ##EQU00006##
Another important metric for CNG vehicles is overall tank energy
density. For a fixed internal volume, energy density is determined
from tank mass, where higher strength materials will drive down
mass, and increase energy density.
[0037] High strength fibers such as Carbon and ultrahigh molecular
weight polyethylene (UHMWPE) have greater strength-weight ratios
than metals, making them attractive materials for CNG applications.
By winding or weaving a fiber around a pressure vessel, the very
strong fibers significantly increase the burst pressure of a
vessel. Dry fiber reinforced vessels are made by weaving, braiding,
or winding high strength fibers around a pressure vessel.
[0038] By optimally orienting the fibers, it is possible to balance
the hoop and longitudinal stress evenly. The ideal angle to achieve
this is called the hose angle. This angle makes sure that material
strengths are being used the most effectively, which is valuable
when the strength-weight ratio is driving design decisions.
[0039] In some embodiments, a fiber reinforced composites are more
durable than dry fiber reinforcement because the matrix encases and
protects the fibers. Carbon, the strongest commercially available
fiber to date, is unfeasible in a dry application because the
fibers are too fragile. The drawback of fiber reinforced composites
is an increased manufacturing cost associated with laying the
fibers up in a matrix.
[0040] To confirm that adding bends does not compromise the
strength of the structure, the stresses in a pressurized bend
needed to be determined. A torus, or section of a bend, can be
geometrically represented as a circular cross-section of radius
r.sub.o revolved around a fixed axis distance R.sub.o from the
center of the circle. The hoop stress in a section of the torus can
be determined by applying conservation of momentum to two connected
cross-sections of tube. This results in a general equation for
pressure vessels that has a constant that needs to be determined By
examining the torus at the centerline where r=R.sub.o and
.alpha.=.pi./2, the constant can be found:
R o .sigma. h cos .pi. 2 - 1 2 cR o 2 = const ##EQU00007## const =
- cR o 2 2 ##EQU00007.2##
So, the equation for hoop stress in a torus is:
.sigma. h = P 2 t ( r 2 - R o 2 r cos .alpha. ) ##EQU00008##
where .alpha. is the tangential angle of the circular cross
section, c is substituted for
P t , ##EQU00009##
and r is the distance that the point is from the axis of rotation.
Alternatively:
r = Ro + ro cos .alpha. ##EQU00010## or : ##EQU00010.2## c .alpha.
s ( r 2 - R o 2 ) 2 r cos 2 .alpha. - .sigma. l r = - c cos .alpha.
##EQU00010.3##
using the fact that .alpha..sub.s=-1/r.sub.o for a circle, and
substituting in for r:
- c ( ( R o + r o cos .alpha. ) 2 - R o 2 ) 2 ( R o + r o cos
.alpha. ) r 0 cos 2 .alpha. - .sigma. l R o + r o cos .alpha. = - c
cos .alpha. ##EQU00011##
which simplifies to the equation of longitudinal stress in a
straight cylinder
.sigma. l = Pro 2 ##EQU00012##
These equations show that longitudinal stress remains unchanged,
that hoop stress increases minimally (e.g., 20% with a bend radius
of 3.5 tube radii), and that the bend stress can be kept below the
failure point of the tube by reducing the radius of the tubes at
the bend.
[0041] FIG. 3 is a graph illustrating an embodiment of a finite
element analysis (FEA) and the analytical solution. In the example
shown, the above equations are compared with the results from a FEA
in the plot above. The results of the FEA and analytical solutions
match within 1%. The results of the analysis show that with a 43%
reduction in tube radius at the bends, the same maximum hoop stress
is maintained with the same wall thickness. One major difficulty
with creating a multi-chamber high pressure tank is connecting the
chambers (e.g., tubes in this case) in a manner that does not
inhibitively decrease the working pressure or increase the
manufacturing cost. There are several options for constructing a
tank with many long straight tubes.
[0042] FIGS. 4A-4D are diagrams illustrating embodiments of tank
construction with straight tubes. In the example shown in FIG. 4A,
a plurality of straight sections of tube (e.g., 400) are all
connected at an end into a single chamber (e.g., 402). In various
embodiments, steel tubes are connected to a manifold (e.g., end
chamber) using brazing or welding, Aluminum or carbon fiber tubes
are connected using bonding, or any other appropriate connecting
manner. In the example shown in FIG. 4B, the tubes are connected in
rows to perpendicular tubes, which in turn are connected together
with a final, again perpendicular tube (not shown in FIG. 4B). In
the example shown in FIG. 4C, the tubes are individually connected
in a single long chamber using U fittings (e.g., 414 at the ends,
not visible in FIG. 4B) and T junctions (e.g., 410 and 412). In the
example shown in FIG. 4C, for an oddly shaped chamber the snaking
tube (e.g., tube 430) can be layered in to conform to the space.
The sections of straight tube are linked using a narrower tube
section that is bent in U shape 432 (e.g., the U fittings). In
cross section, the tubes are packed in a hexagonal packing (e.g.,
434 with U fittings shown at the ends) in order to achieve a
conformal tank to the space that efficiently uses the space. In
some embodiments, tube 430 is made from a single polymer chamber,
molded into shape, and over braided with a high strength fiber. In
some embodiments, the tank is bent into shape before impregnating
resin, or otherwise mechanically bound into shape. In the example
shown in FIG. 4D, a plurality of straight sections of tube have
sections of reduced radius. The sections of reduced radius form the
bends that enable the forming of the straight sections of tube into
a 3 dimensional shape that fills a designated volume (e.g., a
rectangular prism volume in FIG. 4D). Tube 440 shows a three
dimensional perspective view. Tube 442 shows a side view of
straight sections of tube with smaller diameter bent tubes between
the straight sections. Tube 444 shows an end view of the straight
sections of tube in a hexagonal dense pack. 448 indicates a region
that is enlarged and shown as region indicated by 446.
[0043] FIGS. 5A and 5B are diagrams illustrating embodiments of
brazing areas. In the examples shown, given the maximum allowable
stress on a braze, how much overlap area is required to make a safe
join can be determined. In some embodiments, cylindrical thin
walled tubes are joined into other structures. The two cases of
interest in are: 1) joining tubes into circular holes as shown in
FIG. 5A (e.g., end caps or U fittings), or 2) joining tubes
perpendicularly into other thin walled tubes as shown in FIG. 5B.
In some embodiments, for joining a tube into a circular hole (e.g.,
FIG. 5A), the braze area, or the overlap area is given by
A.sub.b=2.pi. r.sub.o l, where r.sub.o is the tube outer radius
(also the circular hole radius) and l is the length of the
cylindrical interface (the thickness of the part with the circular
opening). The pressure area, or the area that contributes to the
sheer stress on the braze join, is given by A.sub.p=.pi.
r.sup.2.sub.i, where r.sub.i is the tube inner radius. The braze
area is easy to calculate, but will be a bit trickier with mitered
tubes. The wall thickness t=r.sub.o-r.sub.i is set by the material
properties at a given inner radius. Setting braze force equal to
pressure force, relating inner and outer radius, and solving for
braze length gives:
lc = rifb 2 .sigma. b ( 1 P - fm .sigma. y ) ##EQU00013##
[0044] For joining one tube into another calculating the braze area
becomes more difficult. The path length of the miter can be
calculated as:
C = 4 r .intg. 0 .pi. 2 ( 1 + sin 2 ( .theta. ) cos 2 ( .theta. ) a
2 - sin 2 ( .theta. ) ) .theta. ##EQU00014##
[0045] The integral finds the path length around a quarter miter
and r is the radius of the mitered tube (smaller tube), while a=R/r
the ratio of radii of the large tube to the small. The obvious
analogy with .pi. shows us that the `circumference` of the miter is
given as:
C = v ( a ) d ##EQU00015## where ##EQU00015.2## v ( a ) = 2 .intg.
0 .pi. 2 ( 1 + sin 2 ( .theta. ) cos 2 ( .theta. ) a 2 - sin 2 (
.theta. ) ) .theta. ##EQU00015.3##
and d is the diameter of the small tube. Using .nu.(a), the braze
area can be calculated as A.sub.b=2 .nu.(a)r.sub.o l, where r.sub.o
is the smaller tube outer radius and l is the length of the
interface (which will be the wall thickness of the larger end
tube). Now we can write the braze length can be written as:
l m = .pi. v ( a ) r i f b 2 .sigma. b ( 1 P + fm .sigma. y ) or
##EQU00016## l m = ( .pi. v ) lc ##EQU00016.2##
as R>>r, .nu..fwdarw..pi. and R.gtoreq.r the circular braze
is the worst case scenario.
[0046] Unlike liquid fuels, gas fuels can diffuse through
nonporous, seamless materials. Excessive diffusion is a concern
because lost fuel leads to lost money, shorter range, and explosive
gas mixtures. Additionally, diffusion linearly increases with both
increasing surface area and decreasing wall thicknesses. The
disclosed highly conformable design can achieve acceptable levels
of natural gas permeation with both polymer-matrix composites and
metal walls. This analysis is based on the solution-diffusion model
for permeation through non-glassy polymers. The model assumes two
different modes of transportation. The first, adsorption, is
linearly approximated by Henry's law of solubility. This model
describes the process of the permeate dissolving into and
subsequently out of the membrane. The second is diffusion, which is
treated as Fickian diffusion. This model approximates the transport
of the gas through the membrane directly. The combination of these
two modes can be linearly described by a diffusive gas permeability
coefficient, which is the product of the solubility coefficient and
the diffusion coefficient. This model only holds true when the
polymers are below their glass-transition temperature with little
or no pores. Given the small range of operating temperatures,
Permeability is assumed to be independent of temperature. It is
also assumed that permeability is independent of concentration and
pressure, both are common assumptions. The equation for
approximating this process is:
Q = P .DELTA. p A t ##EQU00017##
where Q is the volume flow (m.sup.3=sec) through the membrane,
.DELTA.p is the difference of the gases partial pressure between
the feed and permeate side of the membrane, t is the thickness of
the membrane, A is the surface area and P is the diffusive gas
permeability coefficient
P=D*S
Here D is the diffusion coefficient, and S is the solubility
coefficient. Permeability values are often expressed in a wide
range of units (barrer, cm.sup.3(STP)/(cm*s*mmHg),
mL(STP)*mm/(m.sup.2*day*atm), etc.). In some embodiments, SI units
are used for P (m.sup.3(STP)*m/(m.sup.2*s*Pa).
[0047] For the sake of simplifying the model, convection on the
permeate side of the membrane is assumed to be zero. This gives the
assumption that the mass fraction of methane on the permeate side
is approximately 1, causing the model to be dominated by pressure.
This is a reasonable assumption because the partial-pressure
differential across the pressure vessel is very large. In a real
system, even on the bottom of a fast-moving car, permeation rates
will only be slightly higher than in this simplified system.
Beginning by examining cylindrical tubes with no bends the surface
area A is calculated from the tank's volume (V), inner radius
(r.sub.i), and wall thickness (t):
A = 2 .pi. r avg L = V ( 2 r i + t r i 2 ) ##EQU00018##
[0048] Yielding the equation:
Q = P .DELTA. pV [ 2 tr i + 1 r i 2 ] ##EQU00019##
[0049] Three metrics can be used to determine if diffusion through
the tank walls is a limiting factor human safety, environmental
impact, and perceived comfort. Methane is explosive when it is
present in atmospheric concentrations between 5% and 15%. It is
unsafe to be in a space with a methane concentration of more than
1%. If the worst air change rate case for a sedan cabin of 3.5
m.sup.3 is used (1 changeover/hour); and it is assumed that every
bit of gas that escapes the tank goes directly into the car cabin,
it can be calculated if a design is acceptable.
[0050] Permeation also needs to be considered. the ANSI/CSA
NGV2-2007 16.12 Permeation Test requires that the steady state
permeation for the tank be no more than 0.25 cc hour per liter
water capacity of the container. ISO 11439:2000 has the same
requirement. Further, the global warming potential of the diffused
Methane should be considered in in relation to the global warming
potential of the Carbon Dioxide emitted from internal combustion
vehicles. According to the EPA (as of 1997), the average passenger
car consumes 550 gallons of gasoline, and emits 4,500 kg of
CO.sub.2 each year. Methane has 25 times the global warming
potential of CO2. The disclosed Methane gas intestine will need to
emit less than 180 kg of methane/year. Design choices can make it
so that greenhouse gas emissions due to diffusion from the small
radius tank will be much less than the greenhouse gas emissions
from an internal combustion engine.
[0051] Although Methane is an odorless gas, Methane suppliers often
add other chemicals to Methane such as hydrogen sulfide to make the
gas smell noticeable. There is a concern that diffused methane will
smell before health concerns become a risk. If consumers can smell
the gas, they will be distrustful of the technology, limiting its
market viability. This threshold will need to be examined based on
common aromatic gases that are added to methane. The effects of
adding fiber reinforcement to polymers is not entirely understood.
It is generally understood that wet layup FRPs have between 2 and
10 times the permeability of a pure polymer barrier. This effect
increases as more layers are added to the composite. One
explanation is that the wet layup process with fibers introduces
voids that expedite diffusion. If this is the case, then vacuum
layups or other layup techniques could reduce voids, reducing
permeation. A factor of ten increase in permeability could rule out
FRPs as a design option.
[0052] In some embodiments, a design goal is a variable radius
cylinder which starts out with a large radius r0 and shrinks to a
small radius r1<r0. The mean curvatures at the ends are 1=r0 and
1=r1. When the radius starts to shrink, the curvature in the axial
direction changes from zero to positive, increasing the mean
curvature. Since the amount of bending force is roughly
proportional to mean curvature, higher mean curvature means a
higher pressure can be resisted.
[0053] Unfortunately, eventually the taper must become concave in
the axial direction in order to meet up with the smaller cylinder,
causing a decrease in the mean curvature which must be countered by
a smaller radius. This raises the question of the fastest possible
taper that stays within a given mean curvature bound.
[0054] Let z be the distance along the axial direction, r=r(z) the
variable radius and define:
E=1+R.sup.2.sub.z
The mean curvature is given by
H = 1 2 r E - r zz 2 E 3 / 2 ##EQU00020##
The mean curvature of the large cylinder is 1/2r.sub.0, so the
space of curves satisfying H>1/2r.sub.0 needs to be understood.
Extremal curves where equality holds are therefore solutions to
H = 1 2 r 0 ##EQU00021## 1 2 r E - r zz 2 E 3 / 2 = 1 2 r 0
##EQU00021.2## r 0 E - r zz rr 0 = rE 3 / 2 ##EQU00021.3## r zz rr
0 + rE 3 / 2 - r 0 E = 0 ##EQU00021.4##
[0055] This equation appears impossible to solve analytically, but
is easy to treat numerically. One solution option is a convex shape
such as a partial spherical cap. Another solution option is to use
single constant mean curvature curve with mean curvature larger
than 1/2r.sub.0, chosen such that the ratio between the minimum and
maximum radii is r.sub.0/r.sub.1. There are likely further options,
so further criteria are needed to pick out the best one.
[0056] Examining the extreme cases of objects with no curvature in
one direction establishes that having a constant mean curvature
does not guarantee that stress is maintained as constant. Instead,
a pair of coupled equations generated from examining differential,
axi-symmetrical elements can be used to create curves of
continuously varying stress when revolved and pressurized.
Performing a transformation from cylindrical coordinates to
cartesian coordinates yields equations below in [0053], obtaining a
first and second piola-kirchhoff stress in relation to the cauchy
stress tensor in the `undeformed` material.
[0057] FIG. 6A is a diagram illustrating an embodiment of a
parameterization of a revolved curve. In the example shown, s
represents the length travelled along the curve. The function, r(s)
represents the radius of the revolved curve, and z(s) represents
the z position of the revolved curve. Function a(s) represents the
angle with respect to the z-axis. Function .alpha..sub.s(s)
represents the derivative of the angle with respect to s at the
location s along the curve.
[0058] FIG. 6B is a diagram illustrating an embodiment of a
differential element along the length of a tube. In the example
shown, a tube has radius r.sub.0 on the left side and r.sub.1 on
the right side. P is the internal pressure. The angle .alpha..sub.0
is the angle with respect to the axis line of the differential
element as it meets the tube with radius r.sub.0. The stress as the
differential element meets the tube with radius r.sub.0 is
.sigma..sub.0, and the stress as the differential element meets the
tube with radius r.sub.1 is .sigma..sub.1.
[0059] FIG. 6C is a diagram illustrating an embodiment of a tapered
braid geometry. In the example shown, .theta..sub.b is the braid
angle, T is the tension in the fibers, and l is a length in the
axial direction.
P=J.gradient.F.sup.-T
S=JF.sup.-1.sigma.F.sup.-T
[0060] F=.gradient.p
p=[r(s)cos(.theta.),r(s)sin(.theta.),z(s)]
J=det F
[0061] In the equations above, p is the curve parameterized by s
(the length travelled along the curve), and J is the Jacobian.
Since the Jacobian is 1,
P=FS
The fact that the first piola-kirchhoff tensor represents the
cylindrical stresses on a cartesian plane can be used to apply the
divergence theorem in cylindrical coordinates to get the
equation:
- cn = .gradient. sa P = P s s + r s r P s + 1 r P .theta. .theta.
##EQU00022##
Where c is the ratio internal pressure/wall thickness, and n is the
normal vector
n=[cos(.alpha.)cos(.theta.),cos(.alpha.)sin(.theta.),-sin(.alpha.)cos(.t-
heta.)]
This gives us the system of equations:
- c [ cos ( .alpha. ) cos ( .theta. ) cos ( .alpha. ) sin ( .theta.
) - sin ( .alpha. ) cos ( .theta. ) ] = S s ss [ sin ( .alpha. )
cos ( .theta. ) sin ( .alpha. ) sin ( .theta. ) cos ( .alpha. ) ] +
S ss [ .alpha. s cos ( .alpha. ) cos ( .theta. ) + 1 r sin 2 (
.alpha. ) cos ( .theta. ) .alpha. s cos ( .alpha. ) sin ( .theta. )
+ 1 r sin 2 ( .alpha. ) sin ( .theta. ) - .alpha. s sin ( .alpha. )
- 1 r cos ( .alpha. ) sin ( .alpha. ) cos ( .theta. ) ] + S aa r [
- cos ( .theta. ) sin ( .theta. ) 0 ] ##EQU00023##
which can be simplified by evaluating the system at .theta.=0
- c [ cos ( .alpha. ) 0 - sin ( .alpha. ) ] = S s ss [ sin (
.alpha. ) 0 cos ( .alpha. ) ] + S ss [ .alpha. s cos ( .alpha. ) +
1 r sin 2 ( .alpha. ) 0 - .alpha. s sin ( .alpha. ) - 1 r cos (
.alpha. ) sin ( .alpha. ) ] + S aa r [ - 1 0 0 ] ##EQU00024##
This system of differential equations can be simplified by
canceling the S.sup.ss.sub.s terms to yield the relationship:
S ss .alpha. s cos ( .alpha. ) - S aa r = - c cos ( .alpha. )
##EQU00025##
which defines the relationship between axial and hoop stress for
any axi-symmetric pressure vessel. Another equation for S.sup.ss
can be derived from applying conservation of momentum to a
differential element of varying radius:
2.alpha.r.sub.1 cos(.alpha..sub.1).sigma.|.sub.s1-2.pi.r.sub.0
cos(.alpha..sub.0).sigma.|.sub.s0-2.pi.c.intg..sub.s.sub.0.sup.s.sup.1r(s-
)sin(.alpha.(s)ds=0
Substituting sin(.alpha.(s))ds=dr and evaluating the integral, the
following relationship is found:
rS.sup.ss cos(.alpha.)-cr.sup.2/2=const
Since any physical pressure vessel must eventually have a radius of
zero to close the volume, the constant must always be equal to
zero. Therefore, the following relationship is derived for axial
stress:
S ss = Pr 2 t cos .alpha. ##EQU00026##
Therefore, a closed form solution for S-as well:
S aa = .sigma. hoop = Pr t cos ( .alpha. ) + Pr 2 .alpha. s 2 t cos
2 ( .alpha. ) ##EQU00027##
The hose angle for any geometry is then.
.theta. hose = arctan 2 + r .alpha. s cos ( .alpha. )
##EQU00028##
[0062] The above equation demonstrates that a line which is
constrained to have .alpha..sub.s=0 at the junction with a cylinder
will have a continuously varying braid angle without any
discontinuities. Any curve which has a continuously varying
curvature (e.g., with no discontinuity in the change of curvature),
tangential angle, and radius and satisfies .alpha.(0)=0,
.alpha..sub.s(0)=0, .alpha.(end)=0, and .alpha..sub.s(end)=0 when
revolved around the axis of symmetry, will allow for the
construction of a taper with continuously varying braid angle.
[0063] This taper is created by establishing the necessary boundary
conditions and generating a curve that matches either the hoop or
axial stress of the larger cylinder according to the coupled
ordinary differential equations (ODEs) below. That curve is then
revolved around a central axis and subsequently braided over to
create a composite without any discontinuities in braid angle.
- c [ cos ( .alpha. ) 0 - sin ( .alpha. ) ] = S s ss [ sin (
.alpha. ) 0 cos ( .alpha. ) ] + S ss [ .alpha. s sin ( .alpha. ) -
1 r cos ( .alpha. ) sin ( .alpha. ) 0 - .alpha. s sin ( .alpha. ) -
1 r cos ( .alpha. ) sin ( .alpha. ) ] + S aa r [ - 1 0 0 ]
##EQU00029##
In some embodiments, a taper to a reduced radius section of a tube
has a continuously varying curvature. In some embodiments, the
continuously varying curvature enables the targeting of an
optimized braid angle for the overbraid. For example, the optimized
braid angle helps maintain the desired shape of the tube (and the
braid) under pressure after being bent into desired shape from a
straight shape (e.g., the braid does not later move or deform
significantly). In some embodiments, this enables a dry braid to be
used. In some embodiments, a taper targets optimization of the
internal volume of the bend without exceeding a stress maximum of
the bend.
[0064] Working with a braiding machine will allow building
prototypes of both a snake design and a spiral. The spiral design
requires tapering of the braid, and bending the braid before laying
it in a mold and curing it. Below a braid is described that reduces
diameter and maintains hose angle. The hose angle is defined here
to be the fiber angle that equalizes hoop and axial stress. A
prototype of a square spiral can be achieved using a triaxial braid
with helical third axis.
[0065] With full control over the mandrel feed rate (v) and carrier
angular velocity (.omega.) it is possible to create a tapered braid
that maintains stress equilibrium in the fibers over the taper.
This builds on the definition of the hose (or ideal) angle
established above. A definition of hose angle is developed for
tapered braids. In a braided pressure vessel, the axial force and
hoop force should be in equilibrium. This is achieved by adjusting
the braid angle (.theta..sub.b) such that the portion of hoop force
and axial force in the braid fibers is equally carried by the
tension in the fiber. Since the hoop force is a function of cone
radius, angle, and curvature we expect that the braid angle will be
as well. In a traditional, cylindrical, pressure vessel this angle
takes on a single value (arctan {square root over (2)}) and is
called the `hose` angle (the braid angle of a braided fire hose).
To find the `hose` angle for a cone we assume square cross-section
braid fibers with thickness t. The axial stress balance for the
fibers is given by:
T sin .theta..sub.h=.sigma..sub.axial tan .theta..sub.h
The hoop stress balance for the fibers is:
T cos .theta..sub.h=.sigma..sub.hoop
dividing and solving for .theta..sub.h we get:
.theta. h = arctan .sigma. axial .sigma. hoop ##EQU00030##
[0066] Braiders will be able to use this equation to dynamically
set machine parameters and keep that braid angle ideal throughout
the braiding process. Although it is likely that braiders will use
this relationship to determine the parameters for their specific
machine, a general way is developed here to translate between ideal
braid angle and angular velocity of carriers (.omega.), and take-up
speed (.nu.). A basic equation for braid angle would be:
.theta..sub.b=arctan(.omega.R/.nu.)
[0067] Solve for .omega./.nu. to get a relationship between braid
radius and the ratio of carrier angular velocity to take-up
speed.
v .omega. = 1 R .sigma. axial .sigma. hoop ##EQU00031##
[0068] Given a function R(z), that is the shape of the mandrel
along the z axis, an equation is obtained for the takeup speed at a
fixed carrier frequency as a function of mandrel position.
[0069] The cover fraction (c.sub.f) is the fraction of the surface
area of a composite braid that is covered by fiber. In pressure
vessels, having a cover factor that is too small will result in
excessive radial shear stresses acting on the diamond shaped
interstices in between the fibers of a biaxial weave. Those
interstices are triangular in a tri-axial weave. When the shear
forces exceed the shear strength of the matrix, the structure fails
as the interstices `blow out.` A geometric function of the minimum
acceptable cover factor is defined assuming ideal braid angle and
isotropic matrix. The blow-out force on a unit diamond (the space
defined by the area between the middle of the yarn for four
interlacing fibers) is equal to the area in between the fibers
which is defined as:
A = 2 tan .theta. ( 2 .pi. R n - w 2 cos .theta. ) ##EQU00032##
[0070] Where A is the area between the fibers, .theta. is the braid
angle, R is the radius of the tube, n is the number of carriers,
and w is the width of the carriers. This force is applied in shear
across the diamond. Due to scaling laws (surface area vs. edge
length), the largest shear stress will be experienced at the
boundary of the interstitial diamond. The perimeter of the area is
defined by the fiber geometry as:
l = 4 sin .theta. ( 2 .pi. R n - w 2 cos .theta. ) ##EQU00033##
Since stress is the ratio between force and area:
.tau. yield > force area = PA lt ##EQU00034## .tau. yield > P
cos .theta. 2 t ( 2 .pi. R n - .omega. 2 cos .theta. )
##EQU00034.2## .tau. yield > P ( 4 .pi. R cos .theta. - .omega.
n ) 4 tn ##EQU00034.3##
Where P is the pressure inside the vessel, and t is the thickness
of the epoxy Including the geometric definition of coverage
factor:
c f = 1 - ( 1 - .omega. n 4 .pi. R cos .theta. ) 2 ##EQU00035## or
##EQU00035.2## .omega. n = 4 .pi. R cos .theta. ( 1 - 1 - cp )
##EQU00035.3##
[0071] and using the equation for the optimal winding angle for the
section of a cylindrical pressure vessel that is at a constricted
radius relative to the widest section of tube:
.theta. hose = arctan 2 + r .alpha. s cos ( .alpha. )
##EQU00036##
[0072] Where R.sub.o is the maximum radius of the pressure vessel,
and .alpha. is the angle of the taper. The function of a minimum
acceptable cover factor at a given radius reduction and the ideal
braid angle is given by:
c f = 1 - ( 1 - .tau. yield nt RP .pi. cos .theta. ) 2 ##EQU00037##
or ##EQU00037.2## c f > 1 - ( 3 + R .alpha. s cos .alpha. ) (
.tau. yield nt RP .pi. ) 2 ##EQU00037.3##
[0073] As long as the above inequality is true, within a factor of
safety for .tau..sub.yield, the material will not fail by having a
blowout. The actual cover factor for the tapered braid can be
calculated as above. It is assumed that the number of carriers (n)
and the yarn width (.omega.) are constant over the braid. At the
largest radius (R.sub.o) the cover factor is C.sub.o and the braid
angle is .theta..sub.o. It is convenient to define two
constants:
A = C 0 1 - ( 1 - B R 0 cos .theta. 0 ) where B = n .omega. / ( 4
.pi. ) ##EQU00038##
The cover factor is then
c f = A ( 1 - ( 1 - B R cos .theta. ) 2 ) ##EQU00039##
[0074] In some embodiments, the taper is braided using the `hose`
angle, .theta..sub.h, and the ideal taper is:
c f = A ( 1 - ( 1 - B R 3 + R .alpha. s cos .alpha. ) 2 )
##EQU00040##
[0075] Once the nominal radius is chosen, and the reduced radius
(as well as values for the number of carriers and yard width), the
inequality governing the risk of blowout is as follows:
c f = A ( 1 - ( 1 - B R 3 + R .alpha. s cos .alpha. ) 2 ) > 1 -
( 3 + R .alpha. s cos .alpha. ) ( .tau. yield nt RP .pi. ) 2
##EQU00041##
[0076] In some embodiments, a braided fiber based tank is
constructed by over braiding a polymer liner. For example, the
polymer liner is created as a single long tube with reduced
sections that will bend when the tank is folded into shape. In some
embodiments, a plurality of straight sections of tube comprise a
single long tube with reduced radius sections. In some embodiments,
the reduced radius sections comprise bent sections (e.g., where the
reduction has considered the desired bending, wall thickness,
spacing of the larger tubes, to make the desired packing without
deformation of the bend section or mechanical failure when
bending). In some embodiments, this liner can be formed as a single
unit blow molded from a single die or with a set of discrete length
dies. In some embodiments, the plurality of straight sections of
tube comprise multiple long tubes with reduced radius sections that
are joined together (e.g., sealed together). In some embodiments,
the liner sections can be molded separately and connected with a
smaller diameter tubes and sealed into a larger assembly as shown
in FIG. 4C and FIG. 4D. The length of the chambered sections can be
algorithmically determined to `program` the final shape, which can
be any desired shape. The liner is then braided over with a
braiding machine in a single long length. In some embodiments, the
plurality of straight sections of tube have overbraided fibers. In
some embodiments, the overbraided fibers comprise a single long
overbraid. In various embodiments, after braiding the tank is
folded into shape and held together with a resin, over wrapped
fibers, woven bag or mechanically constructed box, or any other
appropriate holding manner. The liner acts both as a mandrel for
braiding and as a barrier for gas diffusion. The shape of the
reduction is created such that the ideal braid angle can be
maintained over the entire length of the tank without
discontinuities in the required angle.
[0077] All metal, type I, tanks are the lowest cost option for CNG
storage. Metals manufacturing has a longer and more developed
history than that of composites. Inexpensive and efficient methods
for forging and forming metal geometries are common, readily
accessible, and sufficiently vetted for safety and long term
viability.
[0078] In some embodiments, the type I tank is a fully
brazed-T-pulled designed. The fully brazed tank uses straight
lengths of tube with tapered (swaged) ends. The ends are brazed
onto a ball pulled tube in a bulk brazing process. Layers could be
hexagonally packed and connected again using a ball pulled
tube.
[0079] In some embodiments, the type I tank includes swaged
straight tubes that are brazed onto U joints. Similar to the above
design the joints will be easier to braze or weld. In some
embodiments, it is possible to swage straight lengths of tube at
the ends, tapering and increasing wall thickness enough to allow
internal or external national pipe threads (NPT). Once threads are
made hydraulic hose or small diameter thick walled tubing can be
used to connect to the ends.
[0080] In some embodiments, it is possible to swage or otherwise
form a tube so that the center of the tube has a reduced diameter.
With special tooling a mandrel bender could be used to bend this
tube into a snake. The snake would be a single layer of a larger,
hexagonally packed. In some embodiments, it is possible to swage
straight lengths of tube at the ends, tapering and increasing wall
thickness enough to allow internal or external NPT threads. Once
threads are made hydraulic hose or small diameter thick walled
tubing can be used to connect to the ends.
[0081] In some embodiments, it is possible to swage or otherwise
form a tube so that the center of the tube has a reduced diameter.
With special tooling a mandrel bender could be used to bend this
tube into a snake. The snake would be a single layer of a larger,
hexagonally packed.
[0082] In some embodiments, in order to utilize current bending
technology with minimum custom tooling a single diameter tube could
be bent into a spiral. Spirals can stack hexagonally to allow for
better packing density. One disadvantage of the spiral tank is that
to achieve maximum packing density, the turns of the spiral need to
be touching, which does not leave enough room for many tube-bending
dies. By making one layer of a tank out of two (or more) spirals
that have 1 diameter spacing between the turns and interlacing them
at a 180 degree offset, you can achieve similar packing density to
a tight spiral with relaxed manufacturing constraints.
[0083] In some embodiments, in order to achieve sufficient energy
density for light vehicle applications, the tank must weigh <100
kg. To achieve this, high yield strength steels or aluminums are
used. Because of complexities with joining aluminum alloys in some
embodiments steel is selected. In some embodiments, brazing options
are eliminated because of the inconsistency introduced from that
number of brazed joints, as well as the difficulty in fixturing the
assembly for batch brazing. A similar argument holds against
mechanical fittings. In some embodiments, a bent snake is selected
because of the lack of fittings or joins, but forming the reduced
sections is challenging.
[0084] In some embodiments, a bent square spiral is a good design
because of the convenience of having only quarter bends with
relatively large bend radius. In various embodiments, bent square
spirals are used for a tank where the bent square spirals comprise
one of the following: a single long tube square spiral closely
packed, a single sparse square spiral, a double-sparse square
spiral, or any other appropriate spiral.
[0085] FIGS. 7A-C are diagrams illustrating embodiments of bent
square spirals. In the example shown in FIG. 7A, bent square spiral
700 spirals inwards in a clockwise direction. The tube has 90
degree bends and is bent with no space between the tube as it
spirals inward. Bent square spiral 702 spirals inwards in a counter
clockwise direction. The tube has 90 degree bends and is bent with
no space between the tube as it spirals inward. The alternating
spirals (e.g., bent square spiral 700 and bent square spiral 702)
are stacked to be hexagonally dense packed. In the example shown in
FIG. 7B, a cross section view (e.g., similar to cut along B-B) is
shown for eight alternating spirals stacked to achieve a hexagonal
packing. In the example shown in FIG. 7C, a perspective view of the
stack of alternating bent square spirals is shown. In some
embodiments, straight sections are connected with 90.degree. bends
and are configured such that the bends make a spiral.
[0086] FIGS. 8A-C are diagrams illustrating embodiments of bent
square spirals. In the example shown in FIG. 8A, bent sparse square
spiral 800 spirals inwards in a counter clockwise direction. The
tube has 90 degree bends and is bent with a single tube space
between the tube as it spirals inward. Bent sparse square spiral
802 spirals inwards in a counter clockwise direction. The tube has
90 degree bends and is bent with a single tube space between the
tube as it spirals inward. The alternating spirals (e.g., bent
sparse square spiral 800 and bent sparse square spiral 802) are
nested within each other to form a single layer of a stack. In the
example shown in FIG. 8B, a cross section view (e.g., similar to
cut along A-A) is shown for eight alternating nested spirals
stacked to achieve a hexagonal packing. In the example shown in
FIG. 8C, a perspective view of the stack of alternating bent nested
square spirals is shown. In some embodiments, straight sections are
connected with 90.degree. bends and are configured such that the
bends make a spiral.
[0087] FIGS. 9A-C are diagrams illustrating embodiments of bent
square spirals. In the example shown in FIG. 8A, bent double sparse
square spiral 900 has two tubes that spiral inwards in a counter
clockwise direction. The two tubes have 90 degree bends and are
bent with a double tube space between the two tubes as they spiral
inward. Bent double sparse square spiral 802 has two tubes that
spiral inwards in a counter clockwise direction. The two tubes have
90 degree bends and are bent with a double tube space between the
two tubes as they spiral inward. The alternating double spirals
(e.g., bent double sparse square spiral 900 and bent double sparse
square spiral 902) are nested within each other to form a single
layer of a stack. In the example shown in FIG. 9B, a cross section
view (e.g., similar to cut along AA-AA) is shown for six
alternating nested spirals stacked to achieve a hexagonal packing.
In the example shown in FIG. 8C, a perspective view of the stack of
alternating bent nested double square spirals is shown. The
double-sparse spiral has two uniquely bent tubes per half layer,
with a total of four per layer. To achieve hexagonal packing units,
four unique bent tubes are used, as layers stack with an
alternating pattern, A B A B. The alternating layers are sized so
that the layers stack to achieve a hexagonal dense pack. In some
embodiments, the flat tank layers can then be layer into the
chassis of a car or truck bed. In some embodiments, straight
sections are connected with 90.degree. bends and are configured
such that the bends make a spiral. In some embodiments, the double
sparse spiral can be extended to a sparse of an arbitrary number of
sparse spirals. In some embodiments, the plurality of straight
sections of tube comprise a bent sparse square spiral of multiple
spirals in the same layer (e.g., double or 2, triple or 3, 4, 5, 6,
etc.). In some embodiments, the bent sparse square spiral is nested
with a second bent sparse square spiral of multiple spirals to form
a sparse square layer. In some embodiments, the sparse square layer
is stacked with another sparse square layer in a dense pack
configuration.
[0088] FIG. 10 is a diagram illustrating an embodiment of a group
of flat spirals filling a two-dimensional layer. In the example
shown, a flat space is filled with 4 larger spirals (e.g., spiral
1000, spiral 1002, spiral 1004, and spiral 1006) and 4 smaller
spirals (e.g., spiral 1010, spiral 1012, spiral 1014, and spiral
1016). The large spirals and the small spirals shown are double
sparse nested square spirals. A three dimensional volume can also
be filled with stacks of groups of flat spirals to fill any
arbitrary volume efficiently.
[0089] In some embodiments, hexagonally packed fiber tube systems
are used for tanks. A traditional carbon fiber structure can be
made from straight carbon fiber reinforced polymer (CFRP) tubes
with glued in end-caps. These end-caps might consist of "U" bends
that join two adjacent tubes together. The straight tubes are
pultruded in a continuous process, and the end caps made in a batch
molding process. The wall thickness of the end of the pultruded
tube is given a slight taper, which is matched by a slight taper in
the end-cap. These elbows can be either composite or metal
hydraulic hose.
[0090] In some embodiments, instead of depending on bond strength
to hold the end-caps in place, it is possible to use external CFRP
bands or through rods that run parallel to the pultruded tubes and
take the full axial load, holding the end-caps in against the
internal pressure. For this case, the pultruded tube need have
circumferential fibers, and so can have a thinner wall thickness,
the axial fiber requirement being shifted to the external bands or
through rods.
[0091] In some embodiments, stacked spiral composite structures can
be made by wet pultruding the carbon fiber weave over an ethylene
vinyl alcohol (EVOH) liner and laying it into a two part spiral
mold with a breather cloth. The EVOH liner is pressurized, and a
vacuum is pulled over the mold to drive out any air, off-gas, and
excess resin. The spiral tank layers is stacked and connected using
a carbon spine that runs along the outer edge of the tank.
[0092] In some embodiments, a modular design of a long carbon fiber
(CF) tube is laid into a square spiral. For example, a tube
diameter is between 0.6''-2'', and a bounding box is about
3'.times.3'. The spiral tanks are stacked hexagonally to form
larger volumes.
[0093] In some embodiments, the densest packing with the most
flexibility is to stack tubes hexagonally and simply connect the
ends.
[0094] FIG. 11 is a diagram illustrating an embodiment of a
flexible elbow. In the example shown, connecting the ends of tubes
uses a flexible elbow. Internal tube 1100 comprises a flexible
internal rubber bend. Internal tube 1100 sits against internal
liner 1102 of tube 1104. With a flexible elbow the straight lengths
are braided over (e.g., using composite overwrap 1106) and bent
before curing. In various embodiments, the flexible elbows comprise
reduced sections of a plastic liner, attached sections of hydraulic
hose, a composite rubber tube, or include any other appropriate
elbow configuration.
[0095] In some embodiments, using a continuous internal liner as a
mandrel, the entire tank is braided in a single process. The shape
of the tank is `programmed` into the liner such that reduced
sections is bent into elbows after braiding.
[0096] In various embodiments, the tank is used to store other
gases than natural gas and conform to a volume--for example, air,
oxygen, carbon dioxide, nitrogen, or any other appropriate gas.
[0097] Although the foregoing embodiments have been described in
some detail for purposes of clarity of understanding, the invention
is not limited to the details provided. There are many alternative
ways of implementing the invention. The disclosed embodiments are
illustrative and not restrictive.
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