Lecture #06 September 17, 2015 The rocket equation Mass ratio and performance Structural and payload mass fractions Regression analysis Multistaging Optimal ΔV distribution between stages Trade-off ratios Parallel staging Modular staging 1 2015 David L. Akin - All rights reserved http://spacecraft.ssl.umd.edu
Derivation of the Rocket Equation Momentum at time t: Momentum at time t+δt: Some algebraic manipulation gives: M = mv Take to limits and integrate: Δm V e m v-v e m-δm v M =(m m)(v + v)+ m(v V e ) mfinal m initial m v = mv e dm m = Vfinal V initial dv V e v+δv 2
The Rocket Equation Alternate forms Basic definitions/concepts Mass ratio r m final or R m initial m initial m final r m final = e V Ve m ( initial ) mfinal v = V e ln = V e ln r Nondimensional velocity change Velocity ratio 3 m initial V V e
Rocket Equation (First Look) Mass Ratio, (M final /M initial ) 1 0.1 0.01 0.001 Typical Range for Launch to Low Earth Orbit 0 2 4 6 8 4 Velocity Ratio, (ΔV/Ve)
Sources and Categories of Vehicle Mass Payload Propellants Structure Propulsion Avionics Power Mechanisms Thermal Etc. 5
Sources and Categories of Vehicle Mass Payload Propellants Inert Mass Structure Propulsion Avionics Power Mechanisms Thermal Etc. 6
Basic Vehicle Parameters Basic mass summary Inert mass fraction Payload fraction m o = m pl + m pr + m in δ m in m o = λ m pl m o = Parametric mass ratio r = λ + δ m in m pl + m pr + m in m pl m pl + m pr + m in m o =initial mass m pl =payload mass m pr =propellant mass m in =inert mass 7
A Word About Specific Impulse Defined as thrust/propellant used English units: lbs thrust/(lbs prop/sec)=sec Metric units: N thrust/(kg prop/sec)=m/sec Two ways to regard discrepancy - lbs is not mass in English units - should be slugs Isp = thrust/weight flow rate of propellant - if Isp is in seconds, V e = g o I sp If the real intent of specific impulse is I sp = Ṫ m and T = ṁv e then I sp = V e 8
Rocket Equation (including Inert Mass) Payload Fraction, (M payload /M initial ) 1 0.1 0.01 0.001 Typical Range for Launch to Low Earth Orbit 0 2 4 6 8 Inert Mass Fraction δ 0 0.05 0.1 0.15 0.2 9 Velocity Ratio, (ΔV/Ve)
Limiting Performance Based on Inert Mass Asymptotic Velocity Ratio, (ΔV/Ve) 5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0 Typical Feasible Design Range for Inert Mass Fraction 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 Inert Mass Fraction, (M inert /M initial ) 10
Regression Analysis of Existing Vehicles Veh/Stage prop mass gross mass Type Propellants Isp vac isp sl sigma eps delta (lbs) (lbs) (sec) (sec) Delta 6925 Stage 2 13,367 15,394 StorabN2O4-A50 319.4 0.152 0.132 0.070 Delta 7925 Stage 2 13,367 15,394 StorabN2O4-A50 319.4 0.152 0.132 0.065 Titan II Stage 2 59,000 65,000 StorabN2O4-A50 316.0 0.102 0.092 0.087 Titan III Stage 2 77,200 83,600 StorabN2O4-A50 316.0 0.083 0.077 0.055 Titan IV Stage 2 77,200 87,000 StorabN2O4-A50 316.0 0.127 0.113 0.078 Proton Stage 3 110,000 123,000 StorabN2O4-A50 315.0 0.118 0.106 0.078 Titan II Stage 1 260,000 269,000 StorabN2O4-A50 296.0 0.035 0.033 0.027 Titan III Stage 1 294,000 310,000 StorabN2O4-A50 302.0 0.054 0.052 0.038 Titan IV Stage 1 340,000 359,000 StorabN2O4-A50 302.0 0.056 0.053 0.039 Proton Stage 2 330,000 365,000 StorabN2O4-A50 316.0 0.106 0.096 0.066 Proton Stage 1 904,000 1,004,000 StorabN2O4-A50 316.0 285.0 0.111 0.100 0.065 average 312.2 285.0 0.100 0.089 0.061 standard deviation 8.1 0.039 0.033 0.019 11
Inert Mass Fractions for Existing LVs Inert Mass Fraction 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00 0 1 10 100 1,000 10,000 Gross Mass (MT) LOX/LH2 LOX/RP-1 Solid Storable 12
Regression Analysis Given a set of N data points (x i,y i ) Linear curve fit: y=ax+b find A and B to minimize sum squared error N error = (Ax i + B y i ) 2 Analytical solutions exist, or use Solver in Excel Power law fit: y=bx A error = i=1 Polynomial, exponential, many other fits possible i=1 N [A log(x i )+B log(y i )] 2 13
Solution of Least-Squares Linear Regression (error) A A N x 2 i + B i=1 N =2 (Ax i + B y i )x i =0 i=1 N x i i=1 (error) B N x i y i =0 i=1 N =2 (Ax i + B y i )=0 A i=1 N x i + NB i=1 N y i =0 i=1 A = N x i y i x i yi N x 2 i ( x i ) 2 B = yi x 2 i x i xi y i N x 2 i ( x i ) 2 14
Regression Analysis - Storables Inert Mass Fraction 0.10 0.08 0.06 0.04 0.02 0.00 R 2 = 0.0541 1 10 100 1,000 Gross Mass (MT) 15
Regression Values for Design Parameters Vacuum Ve (m/sec) Inert Mass Fraction. Max ΔV (m/sec) LOX/LH2 4273 0.075 11,070 LOX/RP-1 3136 0.063 8664 Storables 3058 0.061 8575 Solids 2773 0.087 6783 16
Revised Analysis With ε Instead of δ m in m in + m pr r = + =) = r r = m pl + m in m pl + m pr + m in r = m pl m pr +m in + m in m pr +m in m pl m pr +m in + m pr+m in m pr +m in r = + +1 where m pl m in + m pr 17 = r 1 r
Economy of Scale for Stage Size Stage#Inert#Mass#Frac6on#ε# 0.20# 0.18# 0.16# 0.14# 0.12# 0.10# 0.08# 0.06# 0.04# 0.02# 0.00# 1# 10# 100# 1,000# 10,000# Stage#Gross#Mass,#MT# Storable#MER# LOX/RPD1# Storables# LOX/LH2# Cryo#MER# 18
Stage Inert Mass Fraction Estimation LOX/LH2 =0.987 (M stage hkgi) 0.183 storables =1.6062 (M stage hkgi) 0.275 19 Note: storables also pertains to LOX/ hydrocarbon
The Rocket Equation for Multiple Stages Assume two stages V 1 = V e1 ln V 2 = V e2 ln Assume V e1 =V e2 =V e ( mfinal1 m initial1 ( mfinal2 m initial2 ) ) = V e1 ln(r 1 ) = V e2 ln(r 2 ) V 1 + V 2 = V e ln(r 1 ) V e ln(r 2 )= V e ln(r 1 r 2 ) 20
Continued Look at Multistaging There s a historical tendency to define r 0 =r 1 r 2 V 1 + V 2 = V e ln (r 1 r 2 ) = V e ln (r 0 ) But it s important to remember that it s really V 1 + V 2 = V e ln (r 1 r 2 ) = V e ln m final1 m initial1 m final2 m initial2 And that r 0 has no physical significance, since m final1 = m initial2 r 0 = m final2 m initial1 21
Multistage Inert Mass Fraction Total inert mass fraction Convert to dimensionless parameters δ 0 = m in,1 + m in,2 + m in,3 m 0 δ 0 = m in,1 m 0 0 = 1 + 2 1 + 3 2 1 General form of the equation n stages j 1 0 = + m in,2 m 0,2 m 0,2 m 0 22 j=1 = m in,1 m 0 + m in,2 m 0 + m in,3 m 0,3 m 0,3 m 0,2 m 0,2 m 0 j =1 + m in,3 m 0
Multistage Payload Fraction Total payload fraction (3 stage example) λ 0 = m pl m 0 = m pl m 0,3 m 0,3 m 0,2 m 0,2 m 0 Convert to dimensionless parameters λ 0 = λ 3 λ 2 λ 1 Generic form of the equation λ 0 = n stages j=1 λ j 23
Effect of Staging 0.25 Inert Mass Fraction δ=0.15 Payload Fraction 0.20 0.15 0.10 0.05 1 stage 2 stage 3 stage 4 stage 0.00 0 1 2 3 4 5 Velocity Ratio (ΔV/Ve) 24
Effect of ΔV Distribution 70 1st Stage: Solids 2nd Stage: LOX/LH2 Normalized Mass (kg/kg of payload) 60 50 40 30 20 10 0 2000 3000 4000 5000 6000 7000 8000 9000 10000 Stage 2 ΔV (m/sec) Total mass Stage 2 mass Stage 1 mass 25
ΔV Distribution and Design Parameters 0.45 0.4 0.35 0.3 1st Stage: Solids 2nd Stage: LOX/LH2 0.25 0.2 delta_0 lambda_0 lambda/delta 0.15 0.1 0.05 0 2000 3000 4000 5000 6000 7000 8000 9000 10000 Stage 2 ΔV (m/sec) 26
Lagrange Multipliers Given an objective function y = f(x) subject to constraint function z = g(x) Create a new objective function y = f(x)+ [g(x) z] Solve simultaneous equations y x = 0 y = 0 27
Optimum ΔV Distribution Between Stages Maximize payload fraction (2 stage case) subject to constraint function Create a new objective function λ o = λ 0 = λ 1 λ 2 =(r 1 δ 1 )(r 2 δ 2 ) V total = V 1 + V 2 ( e V 1 V e,1 δ 1 )( e V 2 V e,2 δ 2 ) + K [ V 1 + V 2 V total ] Very messy for partial derivatives 28
Optimum ΔV Distribution (continued) Use substitute objective function max (λ o ) max [ln (λ o )] Create a new constrained objective function ln (λ o )=ln (r 1 δ 1 )+ln (r 2 δ 2 )+K [ V 1 + V 2 V total ] Take partials and set equal to zero [ln (λ o )] r 1 =0 [ln (λ o )] r 2 =0 [ln (λ o )] K =0 29
Optimum ΔV Special Cases Generic partial of objective function [ln (λ o )] r i = 1 r i δ i + K V e,i r i =0 Case 1: δ 1 =δ 2 V e,1 =V e,2 r 1 = r 2 = V 1 = V 2 = V total 2 Same principle holds for n stages r 1 = r 2 = = r n = V 1 = V 2 = = V n = V total n 30
Sensitivity to Inert Mass ΔV for multistaged rocket where V tot = n stages k=1 V k = n k=1 m o,k = m pl + m pr,k + m in,k + m f,k = m pl + m in,k + nx j=k+1 V e,k ln nx j=k+1 m o,k m f,k (m pr,j + m in,j ) (m pr,j + m in,j ) 31
Finding Payload Sensitivity to Inert Mass Given the equation linking mass to ΔV, take and solve to find m pl m in,k ( V tot ) m pl = ( Vtot )=0 This equation shows the trade-off ratio - Δpayload resulting from a change in inert mass for stage k (for a vehicle with N total stages) dm pl + ( V tot) m in,j dm in,j =0 32 k j=1 V e,j ( N l=1 V e,l ( ) 1 m o,j 1 m f,j ) 1 m o,l 1 m f,l
Trade-off Ratio Example: Gemini-Titan II Initial Mass (kg) Stage 1 Stage 2 150,500 32,630 Final Mass (kg) 39,370 6099 Ve (m/sec) 2900 3097 dm -0.1164-1 33
Payload Sensitivity to Propellant Mass In a similar manner, solve to find m pl m pr,k = ( Vtot )=0 k j=1 V e,j N l=1 V e,l ( This equation gives the change in payload mass as a function of additional propellant mass (all other parameters held constant) ( ) 1 m o,j ) 1 m o,l 1 m f,l 34
Trade-off Ratio Example: Gemini-Titan II Initial Mass (kg) Stage 1 Stage 2 150,500 32,630 Final Mass (kg) 39,370 6099 Ve (m/sec) 2900 3097 dm -0.1164-1 dm 0.04124 0.2443 35
Payload Sensitivity to Exhaust Velocity Use the same technique to find the change in payload resulting from additional exhaust velocity for stage k m pl V e,k = ( Vtot )=0 ( ) k j=1 ln mo,j m f,j N l=1 V e,l ( ) 1 m o,l 1 m f,l This trade-off ratio (unlike the ones for inert and propellant masses) has units - kg/(m/sec) 36
Trade-off Ratio Example: Gemini-Titan II Initial Mass (kg) Stage 1 Stage 2 150,500 32,630 Final Mass (kg) 39,370 6099 Ve (m/sec) 2900 3097 dm -0.1164-1 dm 0.04124 0.2443 dm (kg/m/sec) 37 2.87 6.459
Parallel Staging Multiple dissimilar engines burning simultaneously Frequently a result of upgrades to operational systems General case requires brute force numerical performance analysis 38
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 39
Parallel-Staging Rocket Equation Rocket equation during booster burn V = V e ln m final m initial = V e ln m in,b +m in,c + m pr,c +m 0,2 m in,b +m pr,b +m in,c +m pr,c +m 0,2 χ where = 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 40
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 = V e,c ln 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 m in,c +m 0,2 m in,c + m pr,c +m 0,2 41
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 Isp 269 sec Burn time 124 sec External tank (space shuttle main engines) Gross mass 750,975 kg Empty mass 29,930 kg Isp 455 sec Burn time 480 sec Payload (orbiter + P/L) 125,000 kg 42
Shuttle Parallel Staging Example V e,b = gi sp,e =(9.8)(269) = 2636 m sec V e = χ = 480 124 480 =0.7417 2636(1, 007, 000) + 4459(721, 000)(1.7417) 1, 007, 000 + 721, 000(1.7417) V 0 = 2921 ln 862, 000 3, 062, 000 =3702m sec 154, 900 V 1 = 4459 ln 689, 700 =6659m sec V tot =10, 360 m sec 43 V e,c =4459 m sec =2921 m sec
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 44
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 Introduce two new parameters Conversions m in m in + m pr = ε = δ 1 λ 45 m in m mod σ = σ m in m pr δ 1 δ λ
Rocket Equation for Modular Boosters Assuming n modules in stage 1, If all 3 stages use same modules, n j for stage j, r 1 = where ρ pl n(m in )+m o2 n(m in + m pr )+m o2 = r 1 = n 1ε + n 2 + n 3 + ρ pl n 1 + n 2 + n 3 + ρ pl m pl m mod ; m mod = m in + m pr 46 nε + mo2 m mod n + m o2 m mod
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) 47
Conestoga 1620 Performance 4th stage Δ 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 = 1400 11400 =0.1228 pl = 900 + 140 900 + 2200 = 3104 m sec m pl = 3100 m mod 11400 =0.2719 n 1 = 4; n 2 = 2; n 3 =1 48
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 49 4 0.1228 + 2 + 1 + 0.2719 4+2+1+0.2719 2 0.1228 + 1 + 0.2719 2+1+0.2719 1 0.1228 + 0.2719 1+0.2719 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 =0.5175 =0.4638 =0.3103
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 50
OTRAG - 1977-1983 51
Today s Tools Mass ratios Estimation of vehicle masses from v and inert mass fractions (both and ) Regression analysis Staging calculations Optimization of v distribution between stages Trade-off ratios Parallel staging calculations Modular vehicle calculations 52