《航天推进 Space Propulsion》（英文版）Lecture 1b: Review of Rocket Propulsion

5. 522, Space Propulsic Prof. Manuel martinez-sanchez Lecture 1b: Review of Rocket Propulsion The only practical way to accelerate something in free space is by reaction. The idea is the same as in air breathing propulsion(to push something backwards)but in ockets the something"must be inside and is lost Here is a revealing derivation of the thrust equation for vacuum v(t) mt Mom()=M()v()+m(t)v()-c(e=constant M(t clt no drag, no interaction f molecules with ambient air d(Mom) 0 dM+ m(v-c) But dM.dv m=- mc call this thrust , f. Notice d(Mv) dt =F-vm=m(c-v) which can be (+)or(-) Using the same technique, the kinetic energy of the system rocket-jet is KE=SMV+5m(t[v(t)-c(t] dKE dv dt dM+1m(+c2-2vxc)=2mce dt 2 mcv So, if thermal (or electrical) energy is expended internally at the rate E, and converted to total kinetic energy with efficiency nth(or ne,) 16.522, Space P pessan Lecture 1b Prof. Manuel martinez Page 1 of 6

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 1 of 6 16.522, Space Propulsion Prof. Manuel Martinez-Sanchez Lecture 1b: Review of Rocket Propulsion The only practical way to accelerate something in free space is by reaction. The idea is the same as in air breathing propulsion (to push something backwards) but in rockets the “something” must be inside and is lost. Here is a revealing derivation of the thrust equation for vacuum: () () () ( ) ( ) ( ) t 0 Mom. t = M t v t + m t' v t' - c t' dt' = Constant ⎡ ⎤ ∫ ⎣ ⎦ i d Mom ( ) = 0 dt ( ) dv dM M +v +m v-c =0 dt dt i But dM dv m=- M =mc dt dt → i i call this thrust, F. Notice ( ) ( ) d Mv =F-vm=m c-v dt i i which can be (+) or (-) Using the same technique, the kinetic energy of the system rocket-jet is () () () t 2 2 0 1 1 KE = Mv + m t' v t' - c t' dt' 2 2 ⎡ ⎤ ∫ ⎣ ⎦ i So, ( ) dKE dv 1 dM 1 1 2 22 2 = Mv + v m v + c - 2vc m c dt dt 2 dt 2 2 + = i i m cv i −m i So, if thermal (or electrical) energy is expended internally at the rate E, i and converted to total kinetic energy with efficiency th e1. η (or ) η ; i.e., no drag, no interaction of molecules with ambient air

dKE dt Note that we counted both vehicle and wake Ke as produced by E, and this is unambiguous. If we try to define useful Propulsive work"as Fv=mcv, then we find Fv mvc 2V E is arbitrarily high! IfV> For this reason, nprop is not used for rockets. But it is still true that thrusting at high △ +cIn +v.cIn In the presence of external air, some modification is needed, leading to the well Ju+Ae(P -Pa)=mc defines c=u=Jet speed far from exhaust For finite Par in thermal rockets, increasing Ae increases ue(towards a limit To), but it eventually makes(Pe-P)A The best Ae is such as to make pe The thrust coefficient c is used to quantify the performance of nozzles. Starting from =mu+Ae(P-P) 16.522, Space P pessan Lecture 1b Prof. Manuel martinez Page 2 of 6

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 2 of 6 ( ) th d KE = E, dt η i then 2 th 1 E= mc 2 η i i jet kinetic power Note that we counted both vehicle and wake KE as produced by E i , and this is unambiguous. If we try to define “useful Propulsive work” as Fv = m cv, i then we find that the “propulsive efficiency” prop 2 th F.v mvc 2v = = 1 c E mc 2 η = η i i i is arbitrarily high! pr. c If v > , > 1 . 2 ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ η For this reason,ηprop is not used for rockets. But it is still true that thrusting at high speed increases kinetic energy more f 2 2 2 0 f0 0 f 1 1 m ∆ m v = m v + c ln - v 22 m ⎡ ⎤ ⎛ ⎞ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦ In the presence of external air, some modification is needed, leading to the well￾known formula ( ) F =mu +A P -P mc e ee a ≡ i i defines e c u  = Jet speed far from exhaust For finite Pa, in thermal rockets, increasing Ae increases ue (towards a limit e p0 max u 2c T ≅ ), but it eventually makes e ae (P - P )A negative. The best Ae is such as to make Pe = Pa. The thrust coefficient cF is used to quantify the performance of nozzles. Starting from ( ) F =mu +A P -P e ee a i (or finite Pe) 2 2 2 0 0 f 0 f f 1 m m = m c ln + v c ln 2m m ⎡ ⎤ ⎢ ⎥ ⎣ ⎦

Isentropic flow p o pr p=constant Pa=o Pe=P 0 A A and assuming ideal gas, 2 m=p u, At=Pox+1 R At RoTo with √RT F=PoA=+ non-dimensional Ce F=CPA Usually, CF-1.5-2 16.522, Space P pessan Lecture 1b Prof. Manuel martinez Page 3 of 6

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 3 of 6 Isentropic flow r p p ∝ r V p = constant and assuming ideal gas, 1 -1 tt t 0 0 t 2 2 m = u A = RT A +1 -1 γ ⎛ ⎞ ⎛⎞ ρρ γ ⎜ ⎟ ⎜⎟ ⎝ ⎠ ⎝⎠ γ γ i 0 0 0 P R T 0 t * P A m= c ⇒ i with ( ) * RT0 c= , Γ γ ( ) +1 2 2 -1 = +1 γ γ ⎛ ⎞ Γ γ ⎜ ⎟ ⎝ ⎠ γ e ee a 0 t * t0 0 u APP F=PA + - c AP P ⎡ ⎤ ⎛ ⎞ → ⎢ ⎥ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ non-dimensional cF So F=cPA F0 t Usually, F c 1.5 - 2 ∼

Specific impulse(inverse of specific fuel consumption) is defined as F mg We want high Isp(high c), since AV=cIn dt Using mo =mp +ms +m (mp =propellant, ms structure, m, payload) ms or m=moe-ms So, for given mp, ms, the higher c (Isp), the higher m; this dependence is at least linear( for small c,m=mp av -ms), but it becomes exponentially fast for high energy mIssions(》1) For chemical rockets c is limited by chemical energy/mass in fuel Cm=v2E In general, since u (Accounting for work of expansion leads to a replacement of e by H, the enthalpy. For ideal expansion, Brayton cycle, so nd H=CPTo(ideal gas) 2CpTo1 16.522, Space P pessan Lecture 1b Prof. Manuel martinez Page 4 of 6

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 4 of 6 Specific impulse (inverse of specific fuel consumption) is defined as * F sp F c c c I= = = g g m g i We want high Isp (high c), since 0 f dv dm m m =- c ∆V = c ln dt dt m → Using m =m +m +m , 0 PSL ( ) m = propellant, m = structure, m = payload P SL ∆V P c L S L0 S ∆V c m m = - m or m = m e - m e -1 − So, for given mP, mS, the higher c (Isp), the higher mL; this dependence is at least linear (for small LP S ∆V c ,m m -m c ∆V  ), but it becomes exponentially fast for high energy missions ( ∆V 1 c  ). For chemical rockets, c is limited by chemical energy/mass in fuel max c 2E. = In general, since 2 e th th 1 mu = E= mE 2 η η i ii e th u 2E = η (Accounting for work of expansion leads to a replacement of E by H, the enthalpy.) For ideal expansion, Brayton cycle, so -1 e 0 th P =1- P γ γ ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ η and H=cPT0 (ideal gas) -1 e e P0 0 P u = 2c T 1 - P γ γ ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎢ ⎥ ⎣ ⎦

This shows we want high Po/Pe. This means large area ratio., not necessarily high Po. In fact, for vacuum operation, higher Po simply means higher Pe, with no increase of ue (or Isp). In a closer look, higher Po can increase To by inhibiting dissociation-> higher Isp But the main reason to go to high Po is to reduce weight for a given thrust(also for boosting, where Pe cannot be much lower than Pa, so high Po/Pe means high Po) M4πR2tp 2πRtG=πR2po At 2R M4xR2P。Rp A=KR2 F2σcP。K M 2I PwR F = CPK R R F KC CoOk M (Kcn)。v Thus, for a given thrust level the engine mass scales like For a given total impulse(not thrust), it may be better to reduce Po, reduce thrust, operate longer. In boosters there is a complex tradeoff, involving gravity losses, drag penalties, improved Isp at high Po, etc. In general, for boosters it is found advantageous to go to high Po, limited mostly(in liquid rockets by turbomachinery For space engines the result is less clear, but they do tend to optimize at much lower The power of a rocket can be extremely high For the shuttle F=3000ton=3×10Nt, C=3300 m/sec 3×10×3.3×103=5×100wat=50GW! Thus, if one step in the power chain involves electrical power the engine is likely to be very heavy. Why then electrical? Because it breaks the ue limit, allowing any Isp, or, in other words, it gives very fuel efficient rockets. With EM forces one can increase ue almost arbitrarily however looking at the power requirements, Pro f a spa martine ssn Lecture 1b Page 5 of 6

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 5 of 6 This shows we want high P0/Pe. This means large area ratio e * A A , not necessarily high P0. In fact, for vacuum operation, higher P0 simply means higher Pe, with no increase of ue (or Isp). In a closer look, higher P0 can increase T0 by inhibiting dissociation higher Isp. But the main reason to go to high P0 is to reduce weight for a given thrust (also for boosting, where Pe cannot be much lower than Pa, so high P0/Pe means high P0). Roughly 2 w 2 0 F0 t M 4 R t 2 R t = R p F cPA π ρ  πσ π 0 R t=p 2σ 2 0 w 2 2 t F 0 M 4 R PRρ A = K R F 2 c P K R π σ  w 2 F 0 F F 0 M2 F R F = c P K R R = F Kc c P K π ρ σ  ( ) w 32 32 0 F M2 1 = F K c P π ρ σ Thus, for a given thrust level, the engine mass scales like 0 1 P . For a given total impulse (not thrust), it may be better to reduce P0, reduce thrust, operate longer. In boosters, there is a complex tradeoff, involving gravity losses, drag penalties, improved Isp at high P0, etc. In general, for boosters it is found advantageous to go to high P0, limited mostly (in liquid rockets) by turbomachinery. For space engines the result is less clear, but they do tend to optimize at much lower P0. The power of a rocket can be extremely high. For the Shuttle 7 F 3000 ton = 3 ×10 Nt ,  c 3300 m/sec  1 11 mc Fc 3×10 3.3 10 =5 10 watt =50 GW!! 2 7 3 10 2 22 = = ×× × i Thus, if one step in the power chain involves electrical power, the engine is likely to be very heavy. Why then electrical? Because it breaks the ue limit, allowing any Isp, or, in other words, it gives very fuel efficient rockets. With EM forces one can increase ue almost arbitrarily; however looking at the power requirements

FC P F ac MM 2nc 2nc 0.5 watt/Kg-2 Kg/watt= 2000 Kg/KW 50waKg→2×102→20 100 (m/sec2)10 So, with reasonable mass/power tios for electric power one gets very low accelerations 01 0.001 ELECTRICAL If one pushes Isp in an electric thruster, Mp is reduced for given av, but Ms increases due to the high P, unless a is very low(which may be impossible because of mission duration constraints). Thus, an optimum Isp is found to exist(w 2000 sec, depending on conditions) Reference ez-Sanchez, M. and J E. Pollard. Spacecraft Overview. "Journal of Propulsion and Power 14, no. 5(September-October 1998): 688-699 [A publication of the American Institute of Aeronautics and Astronautics, Inc.] Pro f a spa martine ssn Lecture 1b

16.522, Space Propulsion Lecture 1b Prof. Manuel Martinez-Sanchez Page 6 of 6 c c 1 1 Fc P mc 2 2 = = η η i c c P F c ac M M2 2 = = η η 0.5 watt/Kg 2 Kg/watt = 2000 Kg/KW → -2 50 watt/Kg 2 10 20 →× → So, with reasonable mass/power ratios for electric power one gets very low accelerations If one pushes Isp in an electric thruster, MP is reduced for given ∆V, but Ms increases due to the high P, unless a is very low (which may be impossible because of mission duration constraints). Thus, an optimum Isp is found to exist (~2000 sec, depending on conditions). Reference Martinez-Sanchez, M., and J. E. Pollard. “Spacecraft Electric Propulsion – An Overview.” Journal of Propulsion and Power 14, no. 5 (September–October 1998): 688-699. [A publication of the American Institute of Aeronautics and Astronautics, Inc.]