Basic Ascent Performance Analyses

Transcription

1 Basic Ascent Performance Analyses Ascent Mission Requirements Ideal Burnout Solution Constant & Average Gravity Models Gravity Loss Concept Effect of Drag on Ascent Performance Drag Profile Approximation

2 Ascent Mission Requirements Basic objective: To leave spacecraft of given mass, safely at designated point, with necessary energy. Potential Energy: 1 st Component Characterized by altitude above earth s surface. Kinetic energy: 2 nd component characterized by the velocity magnitude of the spacecraft. Momentum: 3 rd component characterized by direction. Therefore, the ascent mission objectives can be stated in terms of the requirements on the position and velocity at the time of release, for a given payload.

3 Ascent Mission Problem As both the launch and terminal points are known a priori, the ascent mission design is essentially arriving at (1) vehicle and (2) flight path. The flight time is usually left to be determined as a part of the trajectory design solution, which is to be obtained from the applicable equations of motion. In most design exercises, simplified mathematical models are used for making predictions / design decisions, which are later verified through rigorous numerical simulations & experimental investigations.

4 Simplified Ascent Model Strategy Model simplification is carried out based on the overall accuracy requirements. As a first step, most solutions use idealized models to capture primary effects. Subsequently, these primary solutions are corrected for the secondary and tertiary physical effects. In this regard, primary physical effect in ascent mission is the propulsion, while secondary effect is gravity & aerodynamic forces are tertiary. In addition, there are possibilities of employing further simplification in description of these forces.

5 Ideal Burnout Performance Ideal burnout analysis is the first step that establishes the primary ascent mission performance and is evaluated under the assumptions of no gravity & atmosphere, flat non-rotating Earth and rectilinear trajectory. Objective of this analysis is to establish payload mass fraction/ Propellant features, from terminal conditions. Basic equation and the ideal solution for this case are, m( t) dv dm dv = mg ɺ 0Isp = dt m( t) g I ( V V0 ) ( Vb V0 ) g0i m sp b g0isp m = e = e m m sp

6 Ideal Burnout Example First stage of the Chinese Long March rocket has the following lift off parameters. m 0 = 79.4 T, m p = 60 T, I sp = 241 s, g 0 = 9.81m/s 2, Stage lift-off mass = 70 T. Determine ideal burnout velocity. Also, determine the propellant mass (keeping total lift-off mass of stage constant), if this velocity is to be increased by 10%. What would be its impact on the stage non-propellant mass ratio? What is max. possible increase in V id if min. non-propellant mass ratio permitted is 0.05? V id = = km/s; m p (+10% δv) = 62.5 T, (m s /m 01 ) (basic) = 0.14, (m s /m 01 ) (+10% δv) = 0.11, Max. δv = ~29%.

7 Effect of Gravity 1 Gravity reduces the ideal burnout velocity for a given mass fraction and, reduces the payload mass fraction, for a given burnout velocity. Its effect is maximum for vertical ascent case, as the weight of launch vehicle directly subtracts from the thrust generated. Basic equation governing this effect is, dv m( t) = mg ɺ 0Isp m( t) g( R, φ, λ) dt It can be seen that the gravity is a function of the trajectory, which, in turn, is available only after solution.

8 Effect of Gravity 2 As gravity is typically much smaller than propulsive forces, a reasonable first approximation can be obtained by assuming a constant sea level gravity and generating the solution for both velocity and altitude, as follows. dv mɺ dr d ( R ) E + h dh = g0 Isp gɶ ; = = V ( t) dt m dt dt dt The solution so obtained can then be corrected by determining the value of g at the burnout altitude and taking a suitably approximate value for the next cycle.

9 Effect of Gravity 3 In this case, the velocity solution is given by, m m V ( t) = g I ln gt ɶ ; V = g I ln gt ɶ sp b 0 sp b m( t) mb m V g I gt ɶ m 0 = ln ; Total Propellant Mass b 0 sp m0 m b p p It is seen that velocity solution now needs burn out time, which depends on burn profile. For a constant burn rate, burn time & burnout velocity can be obtained as, mp m m 0 p m( t) = m0 βt; tb = ; Vb ( t) = g0isp ln gɶ β ( m0 mp ) β

10 Effect of Gravity 4 The altitude solution is now obtained as follows. m0 h( t) = V ( t) dt = g0isp ln gt ɶ dt m ( t ) m 1 m βt h( t) = g I ln dt gt ɶ + h ; = x sp 0 m0 β t 2 m0 m g I 1 h( t) = ln x dx gt ɶ β sp 2 m g I β β β 1 h( t) = 1 t ln 1 t 1 t gt ɶ β m0 m0 m0 2 h b 0 0 sp 2 m0g0i sp = Λ Λ + Λ β [(1 )ln(1 ) ] 1 mp gɶ + h0 ; Λ = 2 β 2 m p m 0

11 Constant g Burnout Example First stage of the Chinese Long March rocket has the following lift off parameters. m 0 = 79.4 Tons, m p = 60 Tons, I sp = 241 s, g 0 = 9.81m/s 2, Payload mass = 9.4 Tons, β = 600 kg/s, R E = 6400 km Determine the burnout velocity and altitude. Next, estimate the change in gravity due to altitude and obtain the corrected values of these parameters. V g = km/s; h b = 79.7 km g b = g 0; g av = g 0 V g = = km/s; h b = 80.3 km

12 Impact of Gravity on Ascent Mission It is known that for spherically symmetric gravitational potential, gravitational field is conservative i.e. total mechanical energy is conserved. However, this is strictly true only if the mass is also conserved. In the context of ascent mission, while in a global sense the combustion products represent the mass lost due to propellant burning, in practical sense this mass is useless and therefore energy associated with this is also useless. Non-conservative nature of Gravitational force is a result of this reality.

13 Gravity Energy Loss Formulation Practically, final payload mass is the only mass that we are interested in and hence we define Ideal kinetic energy density in terms of burnout mass of vehicle as, E = E = 1 V ' ( m0 m p ) 2 Here, m p is the total propellant mass and V 0 is the ideal burnout velocity. In the presence of gravity, the total specific energy at burnout can be expressed as, E 1 E = = V + gh ɶ ' b 2 g g g ( m0 mp ) 2

14 Gravity Energy Loss Estimation In a practical sense, the difference between E 0 and E g is the loss of mechanical energy, that can be attributed to the presence of gravity. The loss represents the energy that is imparted to the un-burnt propellant by the currently burning propellant mass & potential energy of combustion products. Thus, currently burning propellant imparts energy to un-burnt propellant & combustion products which is lost in subsequent time intervals. It can also be seen that the longer one carries the propellant, the higher is likely to be the loss.

15 Gravity Energy Loss Example First stage of the Chinese Long March rocket has the following lift off parameters. m 0 = 79.4 Tons, m p = 60 Tons, I sp = 241 s, g 0 = 9.81m/s 2, Payload mass = 9.4 Tons, β = 600 kg/s. Determine the ideal and gravity based total energies and obtain the energy loss due to gravity. If β is 1200 kg/s, what is the energy loss? What do you think will happen if all propellant is burnt instantaneously? E 0 = 5.55x10 6 ; E g = 3.54x10 6 ; Loss = ~36% (β = 600) E 0 = 5.55x10 6 ; E g = 4.46x10 6 ; Loss = ~20% (β = 1200) E 0 = 5.55x10 6 ; E g = 5.55x10 6 ; Loss = 0% (β = )

16 Effect of Drag Drag in rockets is about an order of magnitude lower than gravity and is tertiary nonlinear effect. A simplified linearized drag model based on the ideal trajectory solution can be used to predict its effect. In many cases, a constant average drag acceleration, based on the total energy loss, gives reasonable estimate of the rocket performance in the presence of drag. Under vertical motion assumption, the equation is, dv g0isp dm D( t) g0isp dm = gɶ = gɶ ad ( t) dt m( t) dt m( t) m( t) dt

17 As the aim is to determine total energy loss due to drag, concepts of drag acceleration / energy is considered adequate for preliminary estimates. In this context, a triangular approximation can be used to capture the gross effect. Drag Profile Approximation dv ( t ) 1 = dm g I 0 sp ( a g Davg +ɶ) dt m( t) dt

18 Drag Acceleration Example First stage of Chinese Long March rocket has following lift off parameters. m 0 = 79.4 T, m p = 60 T, I sp = 241 s, g 0 = 9.81m/s 2, Payload mass = 9.4 T, Burnout mass = 19.4 T, β = 600 kg/s, C D0 = 1.0, S r = π m 2. a t = 50s. Determine the impact of drag on the burnout performance. (V g = 2.35 km/s, h g = 79.7 km). h 50s = 14.1 km, V 50s = 631 m/s, ρ 50s = kg/m 3 D 50s = kn, m 50s = 49.4 Tons, a D50s = 2.83 m/s 2 a Davg = 1.42 m/s 2, V D =2.21 km/s, h D =72.0 km What would happen ifβ=1200 kg/s?

19 Summary Basic Ascent performance of a rocket can be arrived at using simplified trajectory models. Gravity causes a significant performance loss, which can be reduced by higher burning rates. Consideration of gravity also necessitates the definition of propellant burn profile, which impacts the burnout performance. Aerodynamic drag represents an energy loss, which increases significantly with higher burning rates. This can be reduced by burning propellant at slower rate within atmosphere and at faster rate once outside of it.

20 Atmospheric Density Values Altitude (km) Density (kg/m 3 )