Newtonian Analysis of Rarified Flows

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1 Atmospheric Regimes on Entry Basic fluid parameters Definition of Mean Free Path Rarified gas Newtonian flow Continuum Newtonian flow (hypersonics) SphereConeAero so ware 2012 David L. Akin - All rights reserved 1

2 Atmospheric Regimes on Entry 2

3 Basic Fluids Parameters M Mach Number = v a a speed of sound = γrt R = m 1 ordered energy random energy = 2 mv2 1 2 m v2 g 3 = v2 3RT = γ 3 Re Reynold s number = Re = ṁv τa = ρav2 µ v L A = ρvl µ v 2 a 2 = γ 3 M 2 inertial force viscous force

4 Random vs. Ordered Energy 4

5 More Fluid Parameters K Knudsen number K = number of collisions with body number of collisions with other molecules K = λ L λ mean free path L vehicle characteristic length 5

6 Estimating Mean Free Path Assume: All molecules are perfect rigid spheres Each has diameter σ, mass m, and velocity Consider a cube with side length L containing N molecules N/6 molecules are traveling in each direction ±X ±Y ±Z v 6

7 Consider Collisions in +Z Direction number of potential +Z collisions v v 2 v n(+z) = 1 6 N πσ2 L L 3 = 1 6 N πσ2 L 2 frequency of +Z collisions Area = πσ 2 f(+z) = n(+z) t = f(+z) = πρσ2 v 3m π 6 N σ2 L 2 L 2 v 7

8 Consider Collisions in +X Direction v v 2 v frequency of +X collisions f(+x) = n(+x) 2πρσ2 v = t 6m f( X) =f(+y )=f( Y )=f(+x) f( Z) =0 Total frequency of collisions f = π 3 ( ) ρσ2 v m 8

9 Mean Free Path λ = v f = m/σ 2 π 3 ( )ρ 1 ρ at sea level: λ = m at 100 km: λ =0.3 m 1 ft 9

10 Newtonian Flow Mean free path of particles much larger than spacecra --> no appreciable interaction of air molecules Model vehicle/ atmosphere interactions as independent perfectly elastic collisions α α 10

11 Newtonian Analysis mass flux = (density)(swept area)(velocity) ρ A sin(α) A dm dt =(ρ)(a sin α)( ) α 11

12 Momentum Transfer Momentum perpendicular to wall is reversed at impact Bounce momentum is transferred to vehicle Momentum parallel to wall is unchanged sin(α) F F = dm dt = ρ A sin α(2 sin α) =2ρ 2 A sin 2 α 12

13 Lift and Drag L = F cos α =2ρ 2 A sin 2 α cos α D = F sin α =2ρ 2 A sin 3 α c L = c D = L 1 2 ρ 2 A = 4 sin2 α cos α D 1 2 ρ 2 A = 4 sin3 α L D = cos α α D F L sin α = cot α 13

14 Flat Plate Newtonian Aerodynamics Angle of Attack (deg) Lift coeff. Drag coeff. L/D 14

15 Example of Newtonian Flow Calculations Consider a cylinder of length l, entering atmosphere transverse to flow da = rdθdl dṁ = ρda cos θ = ρ cos θrdθdl df = dṁ =2ρ 2 cos 2 θrdθdl dd = df cos θ =2ρ 2 cos 3 θrdθdl dl = df sin θ =2ρ 2 cos 2 θ sin θrdθd df r θ dl dd 15

16 Integration to Find Drag Coefficient Integrate from θ = π 2 π 2 D = + π 2 π 2 l 0 =2ρ 2 r dd =2ρ 2 r + π 2 + π 2 π 2 l 0 cos 3 θdθdl cos 3 θdθ = 8 3 ρ 2 r π 2 By definition, D = 1 2 ρ 2 Ac D and, for a cylinder A =2rl ρ 2 rc D = 8 3 ρ 2 r = c D =

17 Continuum Newtonian Flow (Hypersonics) Air molecules predominately interact with shock waves Effect of shock wave passage is to decelerate flow and turn it parallel to vehicle surface α Shock wave 17

18 Continuum Newtonian Flow (Hypersonics) Treat hypersonic aerodynamics in manner similar to previous Newtonian flow analysis All momentum perpendicular to wall is absorbed by the wall α 18

19 Mass Flux (unchanged) mass flux = (density)(swept area)(velocity) ρ A sin(α) A dm dt =(ρ)(a sin α)( ) α 19

20 Momentum Transfer Momentum perpendicular to wall is absorbed at impact and transferred to vehicle Momentum parallel to wall is unchanged sin(α) cos(α) F = dm dt = ρ A sin α( sin α) =ρ 2 A sin 2 α F 20

21 Lift and Drag L = F cos α = ρ 2 A sin 2 α cos α D = F sin α = ρ 2 A sin 3 α c L = c D = L 1 2 ρ 2 A = 2 sin2 α cos α D 1 2 ρ 2 A = 2 sin3 α α L D = cos α sin α = cot α D F L 21

22 Modified Newtonian Flow Coefficient of pressure in classical Newtonian flow c p =2sin 2 (α) Coefficient of pressure in modified Newtonian flow c p = c pmax sin 2 (α) Cp(max) is the pressure coefficient behind a normal shock at flight conditions c pmax = P shock P 1 2 ρ v 2 22

23 Maximum Coefficient of Pressure c pmax = 2 γm 2 (γ + 1) 2 M 2 γ γ 1 1 γ +2γM 2 4γM 2 2(γ 1) γ +1 1 as M c pmax (γ + 1) 2 4γ γ γ 1 4 γ +1 c pmax for γ =1.4 c pmax 2 for γ =1 23

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