Lecture 24: Pressurization and Pump Cycles

16.512, Rocket Propulsion Prof. Manuel Martinez-Sanchez Lecture 24: Pressurization and Pump Cycles Thrust Chamber Pressurization 1. Introduction The most technically demanding part of a modern high pressure liquid propellant rocket is the chamber pressurization system. Consider for example the cluster of four RD-170 engines powering the Energia first stage. Their shared oxygen pump must deliver 1792 Kg/sec (1.63 m3) of LOX at 614 atm pressure to the combustors, for a power of 176,000 HP. The whole engine weighs 9500 Kg. Similarly, the SSME LH pump (see Fig. 3 in lecture 25) delivers 73 Kg/sec of LH (1.06 m3/sec) at 470 atm pressure, for a power of 76,000 HP. The SSME mass is 2870 Kg, of which the LH turbopump is 344 Kg. And these flow rates are both dwarfed by those of the old F1 Saturn engine, which swallowed a small river (3.5 m3/ sec) of LOX, albeit at a more modest pressure. Because of the importance of this system, we will devote this lecture to engine pressurization cycles and components. 2. Gas Pressurization Systems The simplest way to achieve the required thrust chamber pressure is to provide a small, high pressure gas reservoir, which, at firing time, pressurizes the propellant tanks. The tanks must then be thick enough to support the thrust chamber pressure, plus the injector drop. In addition, the gas reservoirs are also relatively heavy. Thus, the performance of a gas-pressurized rocket lags that of a pump-pressurized rocket (where the tanks can be much lighter), as shown in Fig. 1 (from Ref. 40). We notice here that the relative size of the gains due to a turbo pump system become overpowering for high ∆V rockets, but may not be worth the extra complexity for small ∆V Payload Weight/Gross Weight 200 0.40 Pump Fed Pressure Fed 100 0.20 0 0 10,000 20,000 30,000 Percent Improvement with Pump-Fed Engine 0.60 0 40,000 Mission Velocity (fps) Fig. 1. Payload ratio for typical gas pressurization and pump pressurization Thus, gas pressurization continues to play a role, particularly in small spacebased engines, such as monopropellant maneuvering and attitude control thrusters. A relatively large and sophisticated example is the Shuttle OMS/RCS system. If a gas pressurization system is adopted, a choice must be made between regulated or blow-down gas delivery. The regulated type is used most often, because it avoids chamber operation over an extended pressure range, which may be lead to stability problems. On the other hand, blow-down operation does result in a simpler and lighter system, with less gas inventory, and is standard in hydrazine monopropellant applications. Consider a regulated gas feed system, schematized in Fig. 2. Let PG ( t ) be the decreasing pressure in the gas reservoir, of constant volume VG and Pp the propellant tanks. The internal energy of the gas at some times t is UP + UG regulated tank pressure, where Vp ( t ) is the increasing ullage volume in both and the work of gas expansion in dt is Pp dVp . Therefore, integrating in (0,t), (assuming adiabatic expansion) UG + Up + Pp Vp = UGo (1) NOTE: If isothermal exp., UG + UP = UGo and then PG ( t ) VG + Pp VP ( t ) = PGo VG which can be re-written for an ideal gas as Pv ⎞ ⎛ ⎜ U sin g U = ⎟ γ − 1⎠ ⎝ PG ( t ) VG + γ Pp VP ( t ) = PGo VG (2) This relates PG (t) to VP (t) at any time during operation. If the final gas pressure is PGe (large enough to still overcome the regulator and injector drops), and if VTK represents the whole tankage volume, we obtain for the gas reservoir volume VG = γ Pp PGo − PGe VTK (3) (no γ for isothermal case) (more generally in between) The minimum mass of this gas reservoir, assumed spherical of radius RG, can be estimated by noting that its wall thickness tG must be tG = Pgo 2σGW RG (4) where σGW is the working stress of the wall material. If the density of this wall material is ρGW , we obtain for the gas tank mass MGTK = 3 ρGW 2 σGW γ Pp VTK ⎛P ⎞ 1 − ⎜ Ge ⎟ ⎝ PGo ⎠ (5) A similar calculation can be made for the mass of each of the propellant tanks. If, for simplicity, both, oxidizer and fuel tanks are assumed to be geometrically similar and made of the same material (a cylindrical body of length Lp, capped by hemispheres of radius Rp, with equal Lp/Rp for both ), then the propellant tank mass is MPTK = 3 ρPW 1 + Lp / Rp Pp VTK 2 σPW 1 + 3 Lp / Rp 4 REGULATOR PP PP High Pressure Gas FUEL OXIDIZER VP(t) PG(t) VG Fig. 2. Schematic of gas pressurization system = 0.2, γ = 5 3 o G the gas tank to that of the main propellant tanks, assume P / e PG We notice from (5) that the gas tank mass is nearly independent of its initial pressure: a higher PGo allows a smaller volume VG (Eq. 3), but requires thicker walls (Eq. 4). Values of PGo 5-10 times PP are common. To compare the mass of (He gas) and Lp Rp = 6. For equal material ratio ( ρ σ ), Eqs. (5) and (6) then give MGTK = 1.64 , indicating that the gas reservoir is likely to be the heavier MPTK component (although adding anti-slosh baffles, insulation, etc. may modify this conclusion). The mass of the propellant itself is Mp = VTK ρp , where the densities of oxidizer and fuel have been averaged to ρp . To compare to the minimum tankage mass, assume steel construction, with σ ρ ≅ 3.4 × 104 (m / sec)2 and Pp = 20 atm. Eq. (5) then gives MGTK / VTK = 184 Kg / m3 , and Eq. (3.6) gives MGTK / VTK = 112 Kg / m3 , for a total of 296 Kg / m3. This is to be compared to the mean propellant density; for N2O4-UDMH this is of the order of 1000 Kg/m3. Thus, not counting flanges valves, regulators, etc., the tankage mass is 30% of the propellant mass, a figure which is excessive for any ambitious mission. Advanced composite materials can reduce this mass significantly, however, Titanium is used mostly. An alternative method to reduce the required gas reservoir mass is to heat the pressurizing gas prior to injection into the propellant tank. This can be accomplished through heat exchanging, perhaps using the gas as a nozzle coolant, or by including in the gas a very small amount of oxygen and hydrogen, below the flammability limit, and then passing the gas through a catalytic bed reactor. In either case, assuming a gas temperature rise ∆ T is accomplished, a term MG cv ∆T needs to be added on the right hand side of Eq. (1), where (MG(t)) is the gas mass that has flowed through the heater. This yields eventually a modified expression for the gas reservoir volume: = K T V (7) o G T⎞ ⎟-P T ⎠ e G o G ⎛ P ⎜1 ⎝ p Δ + γP G V As an example, 1% O2 plus 2% H2 (by mole) in He yields upon reaction ∆T ≅ 230 K , which nearly cuts in half the required gas tank volume (and mass). One problem with heating methods is the potential for erratic variations in feed pressure if propellant sloshing cools the pressurizing gas in an unsteady manner.