16.50 Lecture 5: Non-Chemical Rockets, Optimum Exhaust Velocity

16.50 Lecture 5 Subjects: Non-Chemical rockets; Optimum exhaust velocity 1) Non-chemical rockets A shared characteristic of all non-chemical propulsion systems is that the energy and propellant mass are separate initially Chemical Chemical Energy mass . mc Non-chemical Energy Source Propulsive Mass . m 2 c 2 . mc . m There are several possible energy sources: 1) Solar a) Photovoltaic b) Solar thermal c) Solar pressure 2) Nuclear a) Fission b) Radioisotope c) Fusion? There are also many ways to bring the mass and energy together to produce thrust, but all behave according to the rocket equation. "V m final ! =e c mtot or breaking the final mass into its constituent parts, 1 "V m propsys + mstruct m pay ! =e c ! mtot mtot There are 2 general categories of systems, Thermal and Electrical, separable according to whether the energy is available in electrical or mechanical form, or only as thermal energy at some limiting temperature. A. Thermal Here the energy is used directly to heat the propellant, which is then expanded through a nozzle to produce thrust. Tc Energy C Propellant Now there is a chamber temperature Tc, limited by the energy source, and the exhaust velocity is given approximately by: c2 2" R c pTc = or c ! 2 c pTc ! T 2 " #1 M c What limits Tc? 1) Source Temperature e.g. Tsun = 6000°K 2) Materials So for these we generally want low M, e.g., H2 ! 2H. For a nuclear thermal rocket Tc ~ < 3000°K c < 8500 m/s H2 graphite Figures of merit are: a) Specific impulse, c; 2 b) Thrust per unit mass, F mengg B. Electrical If the energy is available in electrical form, then there is no limit in principle on c (other than the speed of light) and in practice we can achieve very high c with good efficiency by using any of a number of electrical accelerators. The system requirement is to produce a ΔV on a payload, mpay. In the absence of gravity "V m final ! loss, = e c , so why not make it very close to 1 by increasing c? To see the answer mtot we must analyze the whole system, and take account of the mass of the energy source. m elect P mpay m propellant m accel . m, c . m The total mass of the system can be broken out as: mtot = mpay + melect + mprop + meng + mstruct so that ratio of final mass to initial mass is "V m final m + melect + meng + mstruct ! = e c = pay mtot mtot The Figure of Merit for such a system is "V m pay m + meng + mstruct ! = e c ! elect mtot mtot Let us neglect meng + mstruct for the moment, compared to melect (or simply redefine mpay to include meng + mstruct, which makes sense for some missions). We know that: F = m˙ c and the power P is P ! m˙ c 2 Fc = 2 2 F melect , and an initial acceleration ao = . Then mtot P "V "V m pay ! ! #P # Fc =e c ! e =e c ! e mtot mtot 2mtot or finally in terms of the minimum number of dimensionless parameters, "V m pay ! # a "V c ) = e c ! ( e o )( mtot 2 "V Define a specific weight ! e " 3 (1) ! eao "V ) is determined by technology level (! e ) , the mission 2 requirement (ΔV) and how fast we want to achieve it (ao). So we should consider this relation a way to find copt to maximize mpay/mtot , given ΔV, ao and αe. Here the group ( Differentiating, m pay ) "V # $ a "V c 2 mtot = # e c + ( o )( ) =0 "V 2 " V !( ) c !( # a !V !V 2 "( !Vc )opt )opt e =( o ) (2) 2 c which we must solve for the optimum c/ΔV). For graphical presentation, let us eliminate the ! a "V group ( o ) between (2) and (1): 2 "V m pay ! "V (3) = e c (1 ! ) c mtot Equations (3) and (2) are represented below over a broad range of ΔV/c: ( 4 So we see that it only makes sense to choose !V c < 1 or > 1 for such systems. This is c !V !V > 1 the exponential is so small it outweighs the term representing melect. c !V Expanding the range 0 < ( )opt < 1, c because for Let us take a look at the meaning of these results: 1) If we choose ao and have given αe and ΔV, this gives us the ( maximum m pay . mtot !V )opt , and in turn the c 2) For given ΔV and αe, increasing ao (for a faster mission) increases ( m pay . mtot 3) For given ΔV and a0 , reducing αe (lighter power plant) increases !V )opt , which reduces c m pay . mtot Take an example: Suppose that the mission gives as a requirement ΔV = 104 m/s and technology enables αe = melect = 20 kg/kW = 0.020 kg/W P 5 Then ! eao "V = 100 ao where ao is in m/s2. 2 We can still choose how fast we want to do the mission, within limits. We know that the ! a "V = 1/e= .368. So for the assumed mission and technology, upper limit of e o 2 ! a "V !.368. This implies ao !.00368 m/s2 or 3.8x10-4 g's and for this maximum 100ao= e o 2 m pay m pay available acceleration, = 0, not a very useful result! Suppose we insist on = 0.5. mtot mtot !v ! a "v This gives ( )opt = .3, which in turn implies e o = .07. The acceleration is then ao = c 2 .07 = 7x10-4 m/s2 = 7x10-5 g's. This is only about 1/5 the maximum acceleration, but now 100 we have lots of payload. The time required to achieve the ΔV is ! "V "V m propellant m0 (1! e c ) c ! t= = = (1 ! e c ) m! F /c a0 and for our example, 10 4 / 0.3 t= (1" e"0.3 ) ! 1.23 ! 10 7 s. = 142 days 7 ! 10 "4 This type of propulsion requires patience! Note that for a H2, 02 rocket, 1 04 #V " m pay " m struct 4500 !e ".1 !.0084 !e c " mtot mtot So we would probably use 2 stages. But the main point is the very much smaller payload to total mass ratio of the chemical system. In addition, if the coasting period for a chemical rocket is very long, as in an interplanetary transfer, a continuous low thrust can in many cases accelerate the mission. 6