Rocket Performance MARYLAND U N I V E R S I T Y O F. Ballistic Entry ENAE Launch and Entry Vehicle Design

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1 Rocket Performance Parallel staging Modular staging Standard atmospheres Orbital decay due to drag Straight-line (no gravity) entry based on atmospheric density David L. Akin - All rights reserved

2 Parallel Staging Multiple dissimilar engines burning simultaneously Frequently a result of upgrades to operational systems General case requires brute force numerical performance analysis 2

3 Parallel-Staging Rocket Equation Momentum at time t: M = mv Momentum at time t+δt: (subscript b =boosters; c =core vehicle) M = (m m b m c )(v + v) + m b (v V e,b ) + m c (v V e,c ) Assume thrust (and mass flow rates) constant 3

4 Parallel-Staging Rocket Equation Rocket equation during booster burn V = V e ln mfinal m initial = V min,b +m e ln in,c + m pr,c +m 0,2 m in,b +m pr,b +m in,c +m pr,c +m 0,2 where χ= fraction of core propellant remaining after booster burnout, and where V e = V e,bṁ b +V e,c ṁ c ṁ b +ṁ c = V e,bm pr,b +V e,c (1 χ)m pr,c m pr,b +(1 χ)m pr,c 4

5 Analyzing Parallel-Staging Performance Parallel stages break down into pseudo-serial stages: Stage 0 (boosters and core) V 0 = V e ln Stage 1 (core alone) V 1 = Subsequent stages are as before ( ) min,b +m in,c +χm pr,c +m 0,2 m in,b +m pr,b +m in,c +m pr,c +m 0,2 V e,c ln m in,c +m 0,2 m in,c + m pr,c +m 0,2 5

6 Parallel Staging Example: Space Shuttle 2 x solid rocket boosters (data below for single SRB) Gross mass 589,670 kg Empty mass 86,183 kg Ve 2636 m/sec Burn time 124 sec External tank (space shuttle main engines) Gross mass 750,975 kg Empty mass 29,930 kg Ve 4459 m/sec Burn time 480 sec Payload (orbiter + P/L) 125,000 kg 6

7 Shuttle Parallel Staging Example V e = V e,b = 2636 m sec χ = V e,c =4459 m sec = (1, 007, 000) (721, 000)(1.7417) 1, 007, , 000(1.7417) =2921 m sec V 0 = 2921 ln V 1 = 4459 ln 862, 000 3, 062, 000 =3702m sec 154, , 700 =6659m sec V tot =10, 360 m sec 7

8 Modular Staging Use identical modules to form multiple stages Have to cluster modules on lower stages to make up for nonideal ΔV distributions Advantageous from production and development cost standpoints 8

9 Module Analysis All modules have the same inert mass and propellant mass Because δ varies with payload mass, not all modules have the same δ! Use module-oriented parameters ε Conversions m in m in + m pr ε = δ 1 λ σ m in m pr σ = δ 1 δ λ 9

10 Rocket Equation for Modular Boosters Assuming n modules in stage 1, r 1 = n(m in )+m o2 n(m in + m pr )+m o2 = nε + mo2 m mod n + m o2 m mod If all 3 stages use same modules, n j for stage j, where r 1 = n 1ε + n 2 + n 3 + ρ pl n 1 + n 2 + n 3 + ρ pl ρ pl m pl ; m mod = m in + m pr m mod 10

11 Example: Conestoga 1620 (EER) Small launch vehicle (1 flight, 1 failure) Payload 900 kg Module gross mass 11,400 kg Module empty mass 1,400 kg Exhaust velocity 2754 m/sec Staging pattern 1st stage - 4 modules 2nd stage - 2 modules 3rd stage - 1 module 4th stage - Star 48V (gross mass 2200 kg, empty mass 140 kg, V e 2842 m/sec) 11

12 Conestoga 1620 Performance 4th stage V V 4 = V e4 ln m f4 = 2842 ln m o4 Treat like three-stage modular vehicle; M pl =3100 kg = m in m mod = = pl = = 3104 m sec m pl = 3100 m mod = n 1 = 4; n 2 = 2; n 3 =1 12

13 Constellation 1620 Performance (cont.) r 1 = n 1 + n 2 + n 3 + pl = n 1 + n 2 + n 3 + pl r 2 = n 2 + n 3 + pl = n 2 + n 3 + pl r 3 = n 3 + pl = n 3 + pl V 1 = 1814 m sec ; V 2 = 2116 m sec V 3 = 3223 m sec ; V 4 = 3104 m V total = 10, 257 m sec sec = = =0.3103

14 Discussion about Modular Vehicles Modularity has several advantages Saves money (smaller modules cost less to develop) Saves money (larger production run = lower cost/ module) Allows resizing launch vehicles to match payloads Trick is to optimize number of stages, number of modules/stage to minimize total number of modules Generally close to optimum by doubling number of modules at each lower stage Have to worry about packing factors, complexity 14

15 OTRAG

16 Modular Example Let s build a launch vehicle out of seven Space Shuttle Solid Rocket Boosters M in =86,180 kg M pr =503,500 kg ε m in m in + m pr = σ m in m pr = Look at possible approaches to sequential firing 16

17 Modular Sequencing - SRB Example Assume no payload All seven firing at once - ΔV tot =5138 m/sec sequence - ΔV tot =9087 m/sec sequence - ΔV tot =9175 m/sec sequence - ΔV tot =9250 m/sec sequence - ΔV tot =9408 m/sec sequence - ΔV tot =9418 m/sec Sequence limited by need to balance thrust laterally 17

18 Atmospheric Density with Altitude Ref: V. L. Pisacane and R. C. Moore, Fundamentals of Space Systems Oxford University Press,

19 Energy Loss Due to Atmospheric Drag Drag D 1 2 v2 Ac D Drag acceleration a d = D m = m c D A v2 2 Ac D m <== Ballistic Coefficient a d = v2 2 orbital energy E = µ 2a de dt = µ da 2a 2 dt 19

20 Energy Loss Due to Atmospheric Drag Since drag is highest at perigee, the first effect of atmospheric drag is to circularize the orbit (high perigee drag lowers apogee) de drag dt = a d v v 2 circ = µ a de drag dt = v2 2 µ a de drag dt = µ a 2 µ a = µ 3 2 a 2 20

21 Derivation of Orbital Decay Due to Drag Set orbital energy variation equal to energy lost by drag = o e µ da 2a 2 dt = 2 µ a da dt = µa 3 2 h hs a = h + r E = da dt = dh dt dh dt = µ (h + r E ) o e h hs 21

22 Derivation of Orbital Decay (2) This is a separable differential equation... 1 re + h e h hs dh = µ o dt h re + h e h hs dh = h o 1 µ o t t o dt Assume 1 re h r E + h h o e h hs dh = re for r E h µ o (t t o ) 22

23 Derivation of Orbital Decay (3) h s e h hs e h o hs re = µ o (t t o ) e h hs h(t) = h s ln e h o hs Note that some variables typically use km, and others are in meters - you have to make sure unit conversions are done properly to make this work out correctly! e h o hs = 23 µre o (t t o ) h s µre o (t t o ) h s

24 Orbit Decay from Atmospheric Drag Altitude (km) β=500 β=1500 β= Time (sec) 24

25 Time Until Orbital Decay e h hs e h o hs = µre h s o (t t o ) To find the time remaining (t o =0) until the orbit reaches any given critical altitude, some algebra gives t(h crit )= h s p e h o hs µre o h crit e hs t(h crit ) / 25

26 Decay Time to r=120 km Altitude (km) β=500 β=1500 β= Decay Time (yrs) 26

27 (no lift) s = distance along the flight path dv dt = g sin D m dv dt = dv ds ds dt = V dv ds = 1 2 d(v 2 ) ds D mg horizontal v, s 1 2 d(v 2 ) ds = g sin D m Drag D 1 2 v2 Ac D 1 2 sin 2 d(v 2 ) ds = g sin v 2 2m Ac D d(v 2 ) dh = g sin v 2 2m Ac D ds dh ds = dh sin 27

28 (2) Exponential atmosphere = o e h hs d o = e h hs dh h s = o e o h hs dh h s = o dh h s dh = hs d sin 2 sin 2 d(v 2 ) d d(v 2 ) dh = g sin v 2 2m Ac D v 2 = g sin 2 h s Ac D m d(v 2 ) d = 2gh s 28 + h sv 2 sin Ac D m

29 (3) Let m c D A Ballistic Coe cient d(v 2 ) d h s sin v2 = 2gh s Assume mg D to get homogeneous ODE Use d(v 2 ) d h s sin v2 = 0 v 2 as integration variable d(v 2 ) v 2 = h s sin d v v e d(v 2 ) v 2 = h s sin 0 d v e = velocity at entry 29

30 (4) Note that the effect of ignoring gravity is that there is no force perpendicular to velocity vector constant flight path angle γ straight line trajectories ln v2 v 2 e = 2 ln v v e = h s sin v v e = exp h s 2 sin v v e = exp h s o 2 sin o Check units: m kg m 3 kg m 2 30

31 Earth Entry, γ=-60 v/v e / o Beta=100 kg/m^

32 What About Peak Deceleration? To find n max, set d dt d 2 v dt 2 = 1 2 d 2 v dt 2 = 1 2 n dv dt = dv dt v2 2 = d2 v dt 2 = 0 2v dv dt 2 2 v v2 d dt + v 2 d dt = 0 = 0 2 v 3 = v 2 d dt 2 v = d dt 32

33 Peak Deceleration (2) From exponential atmosphere, d dt = From geometry, dh dt = v sin d dt = o h s e v h s sin 2 v = h hs dh dt = dh h s dt 2 v = v sin h s d dt Remember that this refers to the conditions at max deceleration nmax = hs sin 33

34 Critical β for Deceleration Before Impact At surface, = o Value of at which vehicle hits crit = oh s sin ground at point of maximum deceleration How large is maximum deceleration? dv dt = v2 2 dv = v2 dt max 2 dv = n v2 max dt max 2 sin h s = 1 2 v 2 h s sin Note that this value of v is actually v nmax 34

35 Peak Deceleration (3) From page 14, v nmax v e dv dt max = 1 2 v v e = exp 35 h s 2 sin h s = exp sin 2 sin h s 2 v e e 1 2 = e 1 2 h s sin = v2 e sin 2h s e Note that the velocity at which maximum deceleration occurs is always a fixed fraction of the entry velocity - it doesn t depend on ballistic coefficient, flight path angle, or anything else! Also, the magnitude of the maximum deceleration is not a function of ballistic coefficient - it is dependent on the entry trajectory (v e and γ) but not spacecraft parameters (i.e., ballistic coefficient).

36 Terminal Velocity Full form of ODE - d v 2 d h s sin v2 = 2gh s At terminal velocity, v = constant v T h s sin v2 T = 2gh s v 2 T = 2g sin 36

37 Cannon Ball γ= diameter sphere, c D =0.2, V E =6000 m/sec Iron Aluminum Balsa Wood Weight 40 lb 15.6 lb 14.5 oz β (kg/m 2 ) ρmd (kg/m 3 ) hmd (m) ,300 32,500 Vimpact (m/s) * Vterm (m/sec) *Artifact of assumption that D 37 mg

38 Atmospheric Density with Altitude Pressure=the integral of the atmospheric density in the column above the reference area = f(h) P o = Z 1 o gdh = o g Z 1 o e h hs dh = o gh s he h hs = o gh s [0 1] i 1 o Earth: o = kg m 3 ; h s = 7524m; P o = o gh s P o (calc) = 90, 400 Pa; P o (act) = 101, 300 Pa o, P o 38

39 Nondimensional Ballistic Coefficient v v e = exp h s o 2 sin o Po =exp 2 g sin Let b = g (Nondimensional form of ballistic coe cient) o h s P o Note that we are using the estimated value of P o = o gh s, not the actual surface pressure. v 1 = exp v e 2 sin o o crit = oh s sin crit = 1 sin 39

40 Entry Velocity Trends, γ=-90 Density Ratio Velocity Ratio