Copyright 2018 Robert Clark
Multi-tubed Propellant Tanks.
One possibility would be to achieve the same lightweight tanks as cylindrical ones by using multiple, small diameter, cylindrical tubes.
A similar idea is described here:
OCTOBER 1, 2009 | MECHANICAL & FLUID SYSTEMS
Assemblies of Conformal Tanks
Space is utilized efficiently and sloshing is reduced.
https://www.techbriefs.com/component/content/article/tb/techbriefs/mechanics-and-machinery/5780
You could get the same volume by using varying lengths and diameters of the multiple cylinders to fill up the volume taken up by the tanks. The cylinders would not have to be especially small. In fact they could be at centimeter to millimeter diameters, so would be of commonly used sizes for tubes and pipes.
The weight of the tanks could be brought down to the usual 35 to 1 ratio for aluminum-lithium hydrogen/oxygen propellant to tank mass. Then the mass of the tanks on the X-33 would be 210,000 lbs/35 = 6,000 lbs, saving 9,200 lbs off the vehicle dry weight. This would allow the hydrogen-fueled X-33 to achieve its originally desired Mach 15 maximum velocity.
But note that since now the tanks are composed of cylindrically-shaped tubes, we no longer have the problem of the conformally-shaped carbon composite tanks failing. Then we could get in the range of 50 to 1 propellant to tank mass ratio by using carbon composites for the cylindrically-shaped tubes, and we could reduce the dry mass due to the tanks by an additional 2,000 lbs.
The same multi-tube approach applied to the full-scale VentureStar would allow it to significantly increase its payload carrying capacity. At a 35 to 1 aluminum-lithium ratio of propellant mass to tank mass for cylindrical tanks, the 1,929,000 lbs propellant mass would now require a mass of only 1,929,000/35 = 55,000 lbs for the tanks, a saving of 83,000 lbs off the original tank mass. This could go to extra payload, so from 45,000 lbs max payload to 128,000 lbs max payload.
But again we could now use carbon-composites for the cylindrical tubes. This would shave an additional 16,000 lbs off the weight of the tanks, and increase the payload now to 144,000 lbs.
An analogous possibility might be to use a honeycombed structure for the entire internal makeup of the tank. The X-33's carbon composite tank was to have a honeycombed structure for the skin alone. Using a honeycomb structure throughout the interior might result in a lighter tank in the same way as does multiple cylinders throughout the interior.
For the multi-tube approach and the honeycombed variant there would be a significant problem for maintenance however. As this is intended to be reusable, we would have difficulty examining all the interior parts of the tanks for cracks or punctures.
If you removed the many layers of the multi-tube method layer by layer for examination that would involve significant time and expense. Even more importantly by so many times having to physically move one of the layers of tubes you run an increased risk of damaging one of the tubes.
One possibility is that since there would be a gap space between one tube and a tube positioned diagonally to it, we could use this space to insert high resolution imaging equipment. It might also work to insert x-ray imaging devices.
Partitioned Propellant Tanks to Save Weight.
A different approach to getting near cylindrical-tube weight efficiency, might be to model the tanks, viewing them vertically, as conical but with a flat front and back, and rounded sides. Then the problem with the front and back naturally trying to balloon out to a circular cross section might be solved by having supporting flat panels at regular intervals within the interior.
The X-33 composite tanks did have support arches to help prevent the tanks from ballooning but these only went partially the way through into the interior. You might get stronger a result by having these panels go all the way through to the other side.
These would partition the tanks into portions. This could still work if you had separate fuel lines, pressurizing gas lines, etc. for each of these partitions and each got used in turn sequentially. A preliminary calculation based on the deflection of flat plates under pressure shows with the tank made of standard aluminum alloy and allowing deflection of the flat front and back to be only of millimeters that the support panels might add only 10% to 20% to the weight of the tanks, while getting similar propellant mass to tank mass ratio as cylindrical tank.
Note you might not need to have a partitioned tank, with separate fuel lines, etc., if the panels had openings to allow the fuel to pass through. These would look analogous to the wing ribs in aircraft wings that allow fuel to pass through. You might have the panels be in a honeycomb form for high strength at lightweight that still allowed the fuel to flow through the tank. Or you might have separate beams with a spaces between them instead of solid panels that allowed the fuel to pass through between the beams
We'll view the X-33 hydrogen tanks standing vertically as conical with flattened front and back and rounded sides. This report on page 19 by the PDF file page numbering gives the dimensions of the X-33 hydrogen tanks as 28.5 feet long, 20 feet wide and 14 feet high:
Final Report of the X-33 Liquid Hydrogen Tank Test Investigation Team.
https://web.archive.org/web/20120127103443/http://alpha.tamu.edu/public/jae/misc/tankreport.pdf
Call it 9 meters long, 6 meters wide, and 4.3 meters deep for this calculation. I'll simplify the calculation by approximating the shape as rectangular, i.e., uniformly 6 meters wide. Note that the rounded portions of the sides, top, and bottom will be considered separately. I'll call the vertical length of each section x, and the bulkhead thickness h. Since the length of the tank is 9m, the number of sections is 9/x.
Typically propellant tanks are pressurized in the 20-40 psi range. I'll take it as 30 psi; call it 2 bar, 2x10^5 Pa. Referring to the drawing of the tank, each bulkhead takes part in supporting the internal pressure of the two sections on either side of it. This means for each section the internal pressure is supported by one-half of each bulkhead on either side of it, which is equivalent to saying each bulkhead supports the internal pressure of one section.
The force on each section is the cross-sectional area times the internal pressure, so 6m*x*(2*10^5 Pa), with x as in the diagram the vertical length of each section. The bulkhead cross-sectional area is 6m*h, with h the thickness of the bulkheads. Then the pressure the bulkheads have to withstand is 6m*x*(2*10^5 Pa)/6m*h = (2*10^5 Pa)*x/h.
The volume of each bulkhead is 6m*h*4.3m. The density of aluminum-lithium alloy is somewhat less than aluminum, call it 2,600 kg/m^3. So the mass of each bulkhead is (2,600 kg/m^3)*6m*h*4.3m = 67,080*h. Then the total mass of all the 9/x bulkheads is (9/x)*67080*h = 603,720*(h/x).
Note that additionally to the horizontal bulkheads shown there will be vertical bulkheads on the sides. These will have less than 1/10 the mass of the horizontal bulkheads because the length of each section x will be small compared to the width of 6m, and will have likewise small contribution to the support of the internal pressure.
The tensile strength of some high strength aluminum-lithium alloys can reach 700 MPa, 7*10^8 Pa. Then the pressure the bulkheads are subjected to has to be less than or equal to this: (2*10^5 Pa)*x/h <= 7*10^8 Pa, so x/h <= 3,500, and h/x => 1/3,500. Therefore the total mass of the bulkheads = 603,720*(h/x) => 172.5 kg. Note we have not said yet how thick the bulkheads have to be only that their total mass is at or above 172.5 kg, for one of the twin rear tanks. It's twin would also require 172.5 kg in bulkhead mass. The third, forward, tank had about 2/3rds the volume of these twin rear tanks so I'll estimate the bulkhead mass it will require as 2/3rds of 172.5 kg, 115 kg. Then the total bulkhead mass would be 460 kg, about 15% of the 3,070 kg tank mass I calculated for the reconfigured X-33.
This is for the bulkheads resisting the outwards pressure of the sections. Notice I did not calculate the pressure inside the tank on the bulkheads from the propellant on either side. This is because the pressure will be equalized on either side of the bulkheads. However, we will have to be concerned about the pressure on the rounded right and left sides of the tank, and the rounded top and bottom of the tank, where the pressure is not equalized on the outside of the tank.
Before we get to that, remember the purpose of partitioning the tank was to minimize the bowing out of the front and back sides from the internal pressure. Consider this page then that calculates the deflection of a flat plat under a uniform load:
eFunda: Plate Calculator -- Clamped rectangular plate with uniformly distributed loading.
This calculator computes the maximum displacement and stress of a clamped (fixed) rectangular plate under a uniformly distributed load.
http://www.efunda.com/formulae/solid_mechanics/plates/calculators/CCCC_PUniform.cfm
In the data input boxes, we'll put 200 kPa for the uniform load, 6 meters for the horizontal distance, .3 m, say, for the vertical distance, and 6 mm for the thickness of the plate. For the vertical distance x I'm taking a value proportionally small compared to the tank width, but which won't result in an inordinate number of partitioned sections of the tank. For the thickness I'm taking a value at 1/1000th the width of the tank, which is common for cylindrical tanks. For the material specifications for aluminum-lithium we can take the Young's modulus as 90 GPa. Then the calculator gives the deflection as only 2.35mm, probably adequate.
However, we still have to consider what happens to the rounded sides and the bottom and top. Look at the last figure on this page:
Thin-Walled Pressure Vessels.
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
http://www.efunda.com/formulae/solid_mechanics/plates/calculators/CCCC_PUniform.cfm
In the data input boxes, we'll put 200 kPa for the uniform load, 6 meters for the horizontal distance, .3 m, say, for the vertical distance, and 6 mm for the thickness of the plate. For the vertical distance x I'm taking a value proportionally small compared to the tank width, but which won't result in an inordinate number of partitioned sections of the tank. For the thickness I'm taking a value at 1/1000th the width of the tank, which is common for cylindrical tanks. For the material specifications for aluminum-lithium we can take the Young's modulus as 90 GPa. Then the calculator gives the deflection as only 2.35mm, probably adequate.
However, we still have to consider what happens to the rounded sides and the bottom and top. Look at the last figure on this page:
Thin-Walled Pressure Vessels.
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/pressure_vessel.cfm
It shows the calculation for the hoop stress of a cylindrical pressure vessel. The calculation given is 2*s*t*dx = p*2*r*dx, using s for the hoop stress. This implies, s = p*r/t, or equivalently t = p*r/s. So for a given material strength s, the thickness will depend only on the radius and internal pressure.
However, what's key here is the same argument will apply in the figure if one of the sides shown is flat, instead of curved. Therefore in our scenario, the rounded sides, top and bottom, which we regard as half-cylinders, will only need the thickness corresponding to a cylinder of their same diameter, i.e., one of a diameter of 4.3m.
So the rounded portions actually require a smaller thickness than what would be needed for a cylinder of diameter of the full 6m width of the tank.
This means the partitioned tank requires material of somewhat less mass than a cylindrical tank of dimension the full width of the tank plus about 15% of that mass as bulkheads.
The new high strength metal alloys might also save further on this weight. However, we now have to consider the Young's modulus of the alloys, because of the deflection of the plates calculation, and not just the tensile strength.
A Key Advantage of Partitioned Tanks.
A Key Advantage of Partitioned Tanks.
There is an another advantage of using partitioned tanks in addition to the weight savings. A problem with weight growth of the X-33/VentureStar arose in regards to the size of the wings. For stability reasons, you would want the center of gravity (CG) to remain ahead of the center of pressure (CP) during the entire flight. But as the propellant is burned off, the propellant mass near the front will be decreased and this will increase the effect of the heavy engines at the rear on the moving the CG rearward. To deal with this problem during the X-33 development, the wings kept getting larger and larger. But this cancels out the advantage the X-33 had in its dry mass in its original design in not needing heavy wings.
The original X-33/VentureStar was supposed to look like this:
But in the later incarnations, it looked like this:
The added wing size was to move the CP rearward to keep the CG ahead of it.
Partitioned propellant tanks are nothing new actually in aerospace. They are quite commonly used on jet airliners to deal with the problem of CG shift as fuel is burned off:
Concorde.
Balancing by Fuel-Pumping.
The Concorde Tank-Schematic:
"1 + 2 + 3 + 4 are the Collector-Tanks, feeding the engines directly. Usually they feed there counterpart engines – but they can be cross-switched to feed more and/or other engines at the same time.
5 + 7 and 8 + 6 are the Main-Transfer Tanks, feeding the 4 Collector-Tanks. Initially 5 + 7 are active. If those are empty 6 + 8 take over (or must be activated from the Engineering Panel!).
5a + 7a are Auxiliary-Tanks (to 5 and 7).
9 + 10 are the Trim-Tanks for balancing forward
11 is the Trim-Tank for balancing afterward"
Then the partitioned tanks could solve two problems of the dry mass of the X-33/VentureStar: weight growth in the tanks and in the wings.
Bob Clark
1 comment:
The thin-wall vessel stress formula (similar for spheres and cylinders) is "exact" if the "r" in the formula is the inside radius. It does not apply to tapered cylinders. That leads to a 3-D stress state.
Consider the air mattress. Segments of cylindrical tubes joined together with a tension-only perforated membrane at each joint, and segments of hemispherical ends. If you double the thickness of the membrane for the joint piece, and extend it tapered one membrane thickness into each membrane, it is strong enough. This was indicated by 3-D finite element analysis, for isotropic material (metals). Probably doesn't fully apply to your shapes,
though. Use EB welding in vacuo for full weld efficiency.
-- GW
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