Table of Contents. I.)Introduction. II.)Lightweight propellant tanks. III.)Kerosene fuel and engines for the X-33/Venture star. IVa.)Aerodynamic lift applied to ascent to orbit. b.)Estimation of fuel saving using lift. V.)Kerosene fueled VentureStar payload to orbit.
I.) A debate among those questing for the Holy Grail of a reusable, single-stage-to-orbit vehicle is whether it should be powered by hydrogen or a dense hydrocarbon such as kerosene. Most concepts for such a vehicle centered on hydrogen, since a hydrogen/LOX combination provides a higher Isp. However, some have argued that dense fuels should be used since they take up less volume (equivalently more fuel mass can be carried in the same sized tank) so they incur less air drag and also since the largest hydrocarbon engines produce greater thrust they can get to the desired altitude more quickly so they also incur lower gravity drag loss. Another key fact is that for dense fuels the ratio of propellant mass to tank mass is higher, i.e., you need less tank mass for the same mass of propellant. This fact is explored in this report:
Single Stage To Orbit Mass Budgets Derived From Propellant Density and Specific Impulse. John C. Whitehead
32nd AIAA/ASME/SAE/ASEE Joint Propulsion ConferenceLake Buena Vista, FLJuly 1-3, 1996Whitehead notes that the propellant mass to tank mass ratio for kerosene/LOX is typically around 100 to 1, while for liquid hydrogen/ LOX it's about 35 to 1, which would result in a significantly greater dry mass for the hydrogen-fueled case just in tank weight alone. Based on calculations such as these Whitehead concludes the best option for a SSTO would be to use kerosene/LOX. The case for the X-33/VentureStar is even worse because the unusual shape of the tanks requires them to use more tank mass than a comparably sized cylindrical tank. This is discussed here:
Space Access Update #91 2/7/00. The Last Five Years: NASA Gets Handed The Ball, And Drops It. "...part of L-M X-33's weight growth was the "multi- lobed" propellant tanks growing considerably heavier than promised. Neither Rockwell nor McDonnell-Douglas bid these; both used proven circular-section tanks. X-33's graphite-epoxy "multi-lobed" liquid hydrogen tanks have ended up over twice as heavy relative to the weight of propellant carried as the Shuttle's 70's vintage aluminum circular-section tanks - yet an X-33 tank still split open in test last fall. Going over to aluminum will make the problem worse; X-
33's aluminum multi-lobed liquid oxygen tank is nearly four times as heavy relative to the weight of propellant carried as Shuttle's aluminum circular-section equivalent."Marshall Space Flight Center Lockheed Martin Skunk Works Sept. 28, 1999 X-33 Program in the Midst of Final Testing and Validation of Key Components.
X-33 Advanced Technology Demonstrator.
II.) I have proposed one possibility for lightweighting the X-33 tanks on this forum:
eFunda: Plate Calculator -- Simply supported rectangular plate with uniformly distributed loading.
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 spars 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. Another method is also related to the current design of having a honeycombed skin for the composite hydrogen tanks. Supposed we filled these honeycombed cells with fluid. It is known that pressurized tanks can provide great compressive strength. This is in fact used to provide some of the structural strength for the X-33 that would otherwise have to be provided by heavy strengthening members. This idea would be to apply fluid filled honycombed cells. However, what we need for our pressurized propellant tanks is *tensile strength*. A possible way tensile strength could be provided would be to use the Poisson's ratio of the honeycombed cells:
Poisson's ratio.
Chiral honeycomb.
III.) Any of these methods might allow you to reduce the weight of the tanks to be similar to that of cylindrical tanks and thus raise the payload to over 100,000 lbs. This would be for keeping the original hydrogen/LOX propellant. However, in keeping with the analyses that show dense propellants would be more appropriate for a SSTO vehicle I'll show that replacing the hydrogen-fueled engines of the X-33/ VentureStar with kerosene ones would allow the X-33 to actually now become an *orbital* craft instead of just suborbital, and the payload capacity of the VentureStar would increase to be comparable to that proposed for Ares V. The volume of the X-33 liquid hydrogen tanks was 29,000 gallons each and the liquid oxygen tank, 20,000 gallons, for a total of 78,000 gallons volume for propellant. This is 78,000gal*3.8 L/gal =3D 296,000 liters, 296 cubic meters. How much mass of kerosene/LOX could we fit here if we used these as our propellants? Typically the oxidizer to fuel ratio for kerosene/LOX engines is in the range of 2.5 to 2.7 to
- I'll take the O/F ratio as 2.7 to 1. The density of kerosene is about 806 kg/m^3 and we can take the density of liquid oxygen to be
Liquid Oxygen Propellant Densification Unit Ground Tested With a Large- Scale Flight-Weight Tank for the X-33 Reusable Launch Vehicle.
RS-84.
Bringing launch costs down to earth. "Three federally funded projects are underway to develop new rocket engines that can make it more affordable to send payloads into orbit."
With 307,000 kg kerosene/LOX fuel and 24,800 kg dry weight, the mass ratio would be 13.4. According to the Astronautix page, the sea level Isp of the RS-84 would be 301 s, and the vacuum 335 s. Take the average Isp as 320s. The total Isp for a rocket to orbit including gravity and air drag losses is usually taken to be about 9,200 m/s. Then an average exhaust velocity of 3200 m/s and mass ratio of 13.4 would give a total delta-v of 8,300 m/s. Even if you add on the 462 m/ s additional velocity you can get for free by launching at the equator this would not be enough for orbit. So for the X-33 I'll look at the cases of the lighter for its thrust NK-33, used as a trio. Note that though not designed to be a reusable engine to make, say, 100 flights, all liquid fuel rocket engines undergo extensive static firings during testing so the NK-33 probably could make 5 to 10 flights before needing to be replaced.The NK-33 is almost legendary for its thrust to weight ratio of 136. According to the Astronautix page its weight is 1,222 kg , with a sea level Isp of
297 sec and a vacuum Isp of 331:NK-33.
Drag: Loss in Ascent, Gain in Descent, and What It Means for Scalability. Thursday 2008.01.10 by gravityloss * Ariane A-44L: Gravity Loss: 1576 m/s Drag Loss: 135 m/s * Atlas I: Gravity Loss: 1395 m/s Drag Loss: 110 m/s * Delta 7925: Gravity Loss: 1150 m/s Drag Loss: 136 m/s * Shuttle: Gravity Loss: 1222 m/s Drag Loss: 107 m/s * Saturn V: Gravity Loss: 1534 m/s Drag Loss: 40 m/s (!!) * Titan IV/Centaur: Gravity Loss: 1442 m/s Drag Loss: 156 m/s
IV.a) It's capability of reaching orbit and possibly even with a small payload could be increased with another additional factor. Among the questors for a SSTO vehicle, the idea to use wings for a horizontal landing has been derided because of the view they were just dead weight on ascent and you need to save as much weight off the empty weight of the vehicle as possible to achieve orbit. However, the key fact is that wings or a lifting body shape can reduce the total delta- v for orbit by using aerodynamic lift to supply the force to raise the vehicle to a large portion of the altitude of orbit rather than this force being entirely supplied by the thrust of the engines. This fact is discussed on page 4 of this report:
AIAA 2000-1045 A Multidisciplinary Performance Analysis of a Lifting-Body Single- Stage-to-Orbit Vehicle. Paul V. Tartabini, Roger A. Lepsch, J. J. Korte, Kathryn E. Wurster
38th Aerospace Sciences Meeting & Exhibit. 10-13 January 2000 / Reno, NV "One feature of the VentureStar design that could be exploited during ascent was its lifting body shape. By flying a lilting trajectory, it was possible to significantly decrease the amount of gravity losses, thereby improving vehicle performance and payload capability. Yet increasing the amount of lift during ascent generally required flight at higher angles-of-attack and resulted in greater stress on the vehicle structure. Accordingly, the nominal trajectory was constrained to keep the parameter q-_ below a 1500 psf-deg structural design limit to ensure that the aerodynamic loads did not exceed the structural capability of the vehicle. The effect of this trajectory constraint on vehicle performance is shown in Fig. 3. There was a substantial benefit associated with using lift during ascent since flying a non-lifting trajectory resulted in a payload penalty of over 1000 lbs compared to the nominal case."Then the gravity losses could be further reduced by flying a lifting trajectory, which would also increase the payload capability by a small percentage.
The trajectory I'll use to illustrate this will first be straight- line at an angle up to some high altitude that still allows aerodynamic lift to operate. At the end of this portion the vehicle will have some horizontal and vertical component to its velocity. We'll have the vertical component be sufficient to allow the vehicle to reach 100 km, altitude. The usual way to estimate this vertical velocity is by using the relation between kinetic energy and potential energy. It gives the speed of v =3D sqrt(2gh) to reach an altitude of h meters. At 100,000 m, v is 1,400 m/s. Now to have orbital velocity you need 7,800 m/s tangential, i.e., horizontal velocity. If you were able to fly at a straight-line at a constant angle to reach 7,800 m/s horizontal velocity and 1,400 m/s vertical velocity and such that the air drag was kept at the usual low
100 to 150 m/s then you would only need sqrt(7800^2 + 1400^2) =3D 7,925 m/s additional delta-v to reach orbit. Then the total delta-v to orbit might only be in the 8,100 m/s range. Note this is significantly less than the 9,200 m/s delta-v typically needed for orbit, including gravity and air drag. The problem is with usual rocket propulsion to orbit not using lift the thrust vector has to be more or less along the center-line of the rocket otherwise the rocket would tumble. You can gimbal the engines only for a short time to change the rocket's attitude but the engines have to be then re-directed along the center line. However, the center line has to be more or less pointing into the airstream, i.e., pointing in the same direction as the velocity vector, to reduce aerodynamic stress and drag on the vehicle. But the rocket thrust having to counter act gravity means a large portion of the thrust has to be in the vertical component which means the thrust vector has to be nearly vertical at least for the early part of the trip when the gross mass is high. Then the thrust vector couldn't be along the center line of a nearly horizontally traveling rocket at least during the early part of the trip. However, using lift you are able to get this large upwards vertical component for the force on the rocket to allow it to travel along this straight-line. A problem now though is that at an altitude short of that of space, the air density will not be enough for aerodynamic lift. Therefore we will use lift for the first portion of the trajectory, traveling in a straight-line at an angle. Then after that, with sufficient vertical velocity component attained to coast to 100 km altitude, we will supply only horizontal thrust during the second portion to reach the 7,800 m/s horizontal velocity component required for orbital velocity.IV.b) How much fuel could we save using a lifting straight-line portion of the trajectory? I'll give an example calculation that illustrates the fuel savings from using aerodynamic lift during ascent. First note that just as for aircraft fuel savings are best at a high L/D ratio. However, the hypersonic lift /drag ratio of the X-33/ VentureStar is rather poor, only around 1.2, barely better than the space shuttle:
AIAA-99-4162 X-33 Hypersonic Aerodynamic Characteristics. Kelly J, Murphy, Robert J, Nowak, Richard A, Thompson, Brian R, Hollis NASA Langley Research Center Ramadas K. Prabhu Lockheed Martin Engineering &Sciences Company
This explains the low increase in payload, about 1,000 lbs., less than .5% of the vehicle dry weight, by using a lifting trajectory for the VentureStar. However, some lifting body designs can have a lift/ drag ratio of from 6 to 8 at hypersonic speeds:
Waverider Design.
- gMcos(=CE=B8). We'll set L =3D gMcos(=CE=B8). Then the force along the stra= ight- line is Fx =3D T - gMsin(=CE=B8) - gMcos(=CE=B8)/(L/D). As with the calcula= tion for the usual rocket equation, divide this by M to get the acceleration along this line, and integrate to get the velocity. The result is V(t) =3D Ve*ln(M0/Mf) -g*tsin(=CE=B8) - g*tcos(=CE=B8)/(L/D), wit= h M0 the initial mass, and Mf, the mass at time t, a la the rocket equation. If you make the angle =CE=B8 (theta) be shallow, the g*tsin(=CE= =B8) term will be smaller than the usual gravity drag loss of g*t and the (L/D) divisor will make the cosine term smaller as well. I'll assume the straight-line path is used for a time when the altitude is high enough to use the vacuum Ve of 331s*9.8 m/s^2 =3D 3244 m/s. According to the Astronautix page, 3 NK-33's would have a total vacuum thrust of 4,914,000 N and for an Isp of 331s, the propellant flow rate would be 4,914,000/(331x9.8) =3D 1,515 kg/sec. I'll use the formula: V(t) =3D Ve*ln(M0/Mf) - g*tsin(=CE=B8) - g*tcos(=CE=B8)/(L/D) , to calculate the velocity along the inclined straight-line path. There are a couple of key facts in this formula. First note that it includes
*both* the gravity and air drag. Secondly, note that though using aerodynamic lift generates additional, large, induced drag, this is covered by the fact that the L/D ratio includes this induced drag, since it involves the *total* drag. I'll take the time along the straight-line path as 100 sec. Then Mf =3D 328,700kg -100s*(1,515 kg/s) =3D 177,200 kg. After trying some examples an angle of 30=C2=BA provides a good savings over just using the usual non- lifting trajectory. Then V(t) =3D 3244*ln(328,700/177,200) - 9.8*100(sin (30=C2=BA) + cos(30=C2=BA)/5) =3D 1,345 m/s. Then the vertical component of= this velocity is Vy =3D 1,135*sin(30=C2=BA) =3D 672.3 m/s and the horizontal, Vx= =3D 1,135*cos(30=C2=BA) =3D 1,164.5 m/s. To compare this to a usual rocket trajectory I'll calculate how much fuel would be needed to first make a vertical trip to reach a vertical speed of 672.3 m/s subject to gravity and air drag, and then to apply horizontal thrust to reach a 1,164.5 m/s horizontal speed. The air drag for a usual rocket is in the range of 100 m/s to 200 m/s. I'll take the air drag loss as 100 m/s for this vertical portion. Then the equation for the velocity along this vertical part including the gravity loss and the air drag loss would be V(t) =3D 3244*ln(M0/Mf) - 9.8*t - 100 m/s, where M0 =3D328,700 kg and Mf =3D 328,700 - t(1,515). You want to find the t so that this velocity matches the vertical component in the inclined case of 672.3 m/s. Plugging in different values of t, gives for t =3D 85 sec, V(85) =3D 680 m/s. Now to find the horizontal velocity burn. Since this is horizontal there is no gravity loss, and I'll assume this part is at very high altitude so has negligible air drag loss. Then the velocity fomula is V (t) =3D 3244*ln(M0/Mf). Note in this case M0 =3D 328,700 - 85*1,515 =3D 199,925 kg, which is the total mass left after you burned off the propellant during the vertical portion, and so Mf =3D 199,925 - t*1,515. Trying different values of t gives for t =3D 40, V(40) =3D 1,171.5 m/s. Then doing it this way results in a total of 125 sec of fuel burn, 25 percent higher than in the aerodynamic lift case, specifically 25s*1,515 kg/s =3D 37,875 kg more. Or viewed the other way, the aerodynamic lift case requires 20% less fuel over this portion of the trip than the usual non-lift trajectory. With a 307,000 kg total fuel load, thus corresponds to a 12.3% reduction in the total fuel that would actually be needed. Or keeping the same fuel load, a factor 1/.877 =3D 1.14 larger dry mass could be lofted, which could be used for greater payload. For a reconfigured X-33 dry mass of 21,700 kg, this means 3,038 kg extra payload. Remember though this is for our imagined new X-33 lifting shape that is able to keep a high L/D ratio of 5 at hypersonic speed, not for the current X-33 shape which only has a hypersonic L/D of 1.2. With the possibility of using morphing lifting body or wings with high hypersonic L/D ratio allowing a large reduction in fuel requirements to orbit, this may be something that could be tested by amateurs or by the "New Space" launch companies.V.) Now for the calculation of the payload the VentureStar could carry using kerosene/LOX engines. The propellant mass of the VentureStar was
1,929,000 lbs. compared to the X-33's 210,000 lbs., i.e., 9.2 times more. Then its propellant tank volume would also be 9.2 times higher, and the kerosene/LOX they could contain would also be 9.2 times higher, or to 9.2*307,000 =3D 2,824,400 kg. We saw the VentureStar dry mass was 257,000 lbs, 116,818 kg, with half of this as just the mass of the LH2/LOX tanks, at 138,000 lbs, 62,727 kg. However, going to kerosene/LOX propellant means the tanks would only have to be 1/100th the mass of the propellant so only 28,244 kg. Then the dry mass would be reduced to 82,335 kg. We need kerosene/LOX engines now. I suggest the RS-84 be completed and used for the purpose. You would need seven of them to lift the heavier propellant load. They weigh about the same as the aerospike engines on the current version of the VentureStar so you wouldn't gain any weight savings here. To calculate how much we could lift to orbit I'll take the average Isp of the RS-84 as 320. Then if we took the payload as 125,000 kg the total liftoff mass would be 2,824,400 + 82,335 + 125,000 =3D 3,031,735 kg, and the ending dry mass would be 207,335 kg, for a mass ratio of 14.6. Then the total delta-v would be 3200ln(14.6) =3D 8,580 m/s. Adding on the 462 m/s equatorial speed brings this to 9042 m/s. With the reduction in gravity drag using a lifting trajectory this would suffice for orbit.Bob Clark