Rotating Seal ?= stuffing box (Pierce Nichols) ----------------------------- A stuffing box is the contraption that passes the prop shaft on a power boat through the hull. It has to let the shaft rotate freely while sealing out the water. It is similar to this -- the rotor must rotate freely, and the liquid must stay sealed within the piping. This is an area for study and tweaking. It is certainly not a show-stopper. Chilled Propellants (H. Spencer) ------------------- At Space Access 96, it was pointed out that the improvement in density is not the only gain from chilled propellants, and indeed it may not be the biggest gain. The other thing that chilling does is to reduce the vapor pressure of the propellant. The main reason you have to pressurize the tanks of a pump-fed vehicle is to keep the pumps from cavitating (or cavitating too much, anyway). Reducing the vapor pressure reduces the tendency for pumps to cavitate, which means you can reduce tank pressure... and since tanks are primarily pressure vessels, that means lighter tanks. The Rocket Equation ------------------- Delta V = Isp*g*ln(Mi/Mf) where Isp is the specific impulse, g is earth gravity (if you're using seconds for Isp) Mi is initial mass and Mf is burnout mass. This still won't account for gravity losses. Density of LOx -------------- LOx is 1140 kg/m^3 (at -182C) Density of hydrocarbon fuels (lb/gal) ---------------------------- average actual Mil Spec 0F 100F AvGas 6.1 5.7 6.0 JP-4 6.7 6.4 6.5 JP-5 7.2 6.8 6.8 JP-8 - - 6.7 5.7 lb/gal == 683 kg/m^3 7.2 lb/gal == 863 kg/m^3 According to Henry Spencer: [BEGIN] Typically the density changes are small -- but still potentially useful -- unless you cool to cryogenic temperatures, which is workable for only a few fuels. (Propane is the canonical example of a fuel that benefits, going from 0.59 at its boiling point to 0.72 at LOX temperatures.) At LOX temperatures, RP-1 is a rock-hard solid. Its density is of purely academic interest unless you have some very unusual engine concept... If you want hydrocarbons that are still liquid at LOX temperatures, you have to resort to propane or something even lighter. Propane is still liquid there, although only just. Methane is still a gas, although again only just. Those are the two popular choices for cryogenic hydrocarbons. RP-1 freezes at about 225K (about 405R). [END] And: [BEGIN] Here are densities for hydrocarbons at 30 psia (typical of a tank interior) and at T=175.68 R (the temperature of NBP LOX at 30 PSI): (T=175.68 R = -175.55 C) Propane: 45.0 lb/ft3 (720 kg/m^3) Methane: 28.09 lb/ft3 (448 kg/m^3) Ethane: 40.19 lb/ft3 (644 kg/m^3) For RP-1, I have only an ESTIMATE based on the work of Dr. Joeseph Scharrer: RP-1 at 175.68 R density = 58.77 lb/ft3 (941 kg/m^3) RP-1 at 70F density = 50.28 lb/ft3 (805 kg/m^3) Anyone have any better data? Jim Glass [END] So I guess I need to rework my calculations. Calculation ----------- Assume we have a column of liquid with cross sectional area C and length L rotating about one end. (Such as a propellant feed line running through a rotor blade.) The accelleration at some point is: a = v^2/r where v is the linear velocity and r is the distance from the center of rotation. The velocity at some point is: v = 2(pi)rw where w is the rotational speed (in revolutions per second) The acceleration at some point is therefore: a = (2(pi)rw)^2/r = 4(pi)^2(w^2)r The mass of an infinitesimal fraction of the column is: dm = C(rho)(dr) where rho is the density of the fluid. The total centripetal force accellerating the column is thus: f = ma f = INTEGRAL[0,L,((rho)C4(pi)^2(w^2)(r)(dr))] f = C(rho)2(pi)^2(w^2)(L^2) So if we desire a propellant pressure P at the outer end of the column, we need to adjust the rotational speed such that the force exerted by the end of the column is equal to the centripetal acceleration. Thus the pressure P (in N/m^2) times the cross sectional area C must equal the force calculated above. CP = C(rho)2(pi)^2(w^2)(L^2) thus: P = (rho)2(pi)^2(w^2)(L^2) so the rotational speed is : w = SQRT(P/2(rho)(pi)^2(L^2)) w = SQRT(P/2(rho))/(pi)L and the tip-speed is : v = 2(pi)Lw v = 2(pi)L(SQRT(P/2(rho)))/(pi)L v = 2 SQRT(P/2(rho)) Which is interesting, since it does not depend on the length of the blades, but only on the target pressure and propellant density. The SSME operates at roughly 3000 PSI, which is 20.68e6 N/m^2. To achieve this pressure with liquid oxygen with a density of 1140 kg/m^3, the tips must be moving at 190 m/s. Suppose we desire a chamber pressure of 3000 psi. According to Jim Glass, propane at -175C has a density of 720 kg/m^3, which implies a tip-velocity of 240 m/s. RP-1 at 21C has a density of 805 kg/m^3, and thus implies a tip-velocity of 227 m/s. This analysis ignores the fact that the motion of the fuel is not circular, but is actually a spiral. Thus acceleration of the fuel decreases as the outward flow rate increases, but I'll neglect this term for the time being, until I have some estimates for engine size and propellant flow rate. It also assumes that the fluids are incompressible. The speed of sound at sea level is ~343 m/s. If the blade tips are moving at 202 m/s, then the craft can climb vertically at 277 m/s (Mach 0.8) before the tips go supersonic. Of course, the speed of sound changes with altitude (temperature). This also neglects any horizontal motion to improve the efficiency of the rotor. The Enstrom F28 helicopter has a main rotor speed range of 313-385 RPM, and has an engine speed range of 2750-2900 RPM. According to http://www.ai.mit.edu/projects/cbcl/heli/aero/rotational_vel.html (which is based on US Army Field Manual 1-51) a reasonable tip speed for an actual helicopter is 397 kts (204 m/s) Another factor which needs to be calculated is the amount of thrust needed to tangentially accelerate the fuel as it flows outward along the blade. The fuel changes tangential velocity from 0 to ~200 m/s as it traverses the blade length. Hudson stated that he currently (4/96) intends to use an electric motor to counteract bearing friction, as I have also been considering. For the full scale vehicle, this may require a considerable amount of power, since the entire vehicle (sans rotor) must be supported by the thrust bearing, and the rotating propellant coupling may generate substantially more friction than an ordinary bearing. Selection of airfoil shape is tricky. The center of pressure is typically at the quarter chord point for a subsonic airfoil, but at the half chord point for a supersonic airfoil. Also, if a non-symmetric airfoil is selected, the center of pressure changes with angle of attack. These factors can create large twisting moments on the rotor blade, especially when the tips are supersonic, but the root isn't. Since the body of the craft will not have any tremendous aerodynamic function, it can be essentially a perfect Sears-Haack body. I think. Probably one end will need to be blunt for reentry, a-la ICBM. I Need to learn more hypersonics! Thurst-to-weight (Jim Kingdon) ---------------- From some data Marcus posted a while ago (and my own calculations) the SSME T/W ratio is 67. From http://www.pwfl.com/rl10.htm the RL10A-4-1 T/W ratio is 60. Note that the RL-10 T/W ratio and ISP have been improving in recent years (rather dramatically, to my point of view). Disclaimer: comparing T/W ratios is difficult. There is always some dispute about how much of the plumbing, mounting hardware, etc., to include in the weight. LOX safety (H. Spencer) ---------- People underestimate the hazards of LOX. For example, if you're using aluminum plumbing, you have to be careful to get rid of *all* particulate contamination in the LOX, because otherwise, when the LOX goes around a corner fast and particles slam into the wall... the aluminum ignites. {bummer!}