lees-doroditsyn compressible boundary layer equations
TRANSCRIPT
Finite Difference Solutions to the Lees-DorodnitsynCompressible Boundary Layer Equations
Chelsea Onyeador
May 12, 2020
2.29 Final Project
1
Motivation
β’ Compressible, and specifically hypersonic, boundary layers can have complex velocity and temperature profiles
β’ Lees-Doroditsyn transformation results in decreased need for boundary layer scaling
β’ Boundary layer analysis is necessary to assess surface heating, skin friction, and external flow displacement effects
β’ Developing robust methods for modeling hypersonic boundary layers
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TSL L-D Coordinate Transformation
π = ΰΆ±0
π₯
πππ’πππ ππ₯ Ξ· =π’π
2πΰΆ±0
π¦
π ππ¦
πΆπβ²β² β² + ππβ²β² =2π
π’ππβ² 2 β
πππ
ππ’πππ
+ 2π πβ²ππβ²
ππβππ
πππβ²β²
ππ
ππ= 0
πΆ
πππβ²
β²
+ ππβ² = 2π πβ²ππ
ππ+πβ²π
βπ
πβπππ
β πβ²ππ
ππ+πππ’ππβπ
πβ²ππ’πππ
β πΆπ’π2
βππβ²β² 2
πΆ =ππ
ππππ, π =
β
βπ, and, πβ² =
π’
π’π
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Lees-Dorodnitsyn Equationsβ’ For 2D, Laminar, Compressible Boundary Layers
β’ Effectively Parabolic
β’ 5th Order system of coupled equations
β’ Introduce F, U, S, H, and Q to simplify into 5 - 1st order equations
β¦ πΉ = π =π
2π(stream function relation)
β¦ π =ππ
ππ=
π’
π’π(velocity relation)
β¦ π =ππ
ππ=
π2π
ππ2(shear stress relation)
β¦ π» = π =β
βπ=
π
ππ=
ππ
π(enthalpy/temperature relation)
β¦ π =ππ
ππ=
ππ»
ππ(heat transfer relation)
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Flow Parameters
FLUID MODELS
β’ Viscosity Sutherlandβs Law
β¦ π =ππππ
π
ππππ
32ππππ+ππππ
π+ππππ
β’ Calorically Perfect Gas
β¦ β = πππ
β¦ πΎ = 1.4
β’ Constant Pr
β¦ 0.71 (Van Driest), 0.75, or 1
β’ Isothermal wall
SPECIFIED PARAMETERS
β¦ ππππ₯
β¦ ππβ¦ ππ β βπβ¦ πππ π β π’π
β¦βπ€
βπβ βπ€
β¦ ππ = ππ
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Flow Parameters
FLUID MODELS
β’ Viscosity Sutherlandβs Law
β¦ π =ππππ
π
ππππ
32ππππ+ππππ
π+ππππ
β’ Calorically Perfect Gas
β¦ β = πππ
β¦ πΎ = 1.4
β’ Constant Pr
β¦ 0.71 (Van Driest), 0.75, or 1
β’ Isothermal wall
SPECIFIED PARAMETERS
β¦ ππππ₯
β¦ ππβ¦ ππ β βπβ¦ πππ π β π’π
β¦βπ€
βπβ βπ€
β¦ ππ = ππ
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Boundary Conditionsβ’ 5 Required for a 5th Order system:
β’ Wall:
β¦ πΉπ€πππ = 0
β¦ ππ€πππ = 0
β¦ πΊπ€πππ = πΊπ ππππππππ
β’ Edge:
β¦ πππππ = 1
β¦ πΊππππ = 1
β’ Initial condition @ π = 0
β¦ 1D Flat plate similarity solution (calculated with Newton-Raphson method)
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Residual Form
Fβ² β π = 0
πβ² β π = 0
πΆπ β² + πΉπ +2π
π’ππ» β π 2
ππ’πππ
+ 2πππΉ
πππ β π
ππ
ππ= 0
πΆ
πππ
β²
+ πΉπ β 2π πππ»
ππ+ππ»
βπ
πβπππ
β πππΉ
ππ+π»π’πβπ
πππ’πππ
+ πΆπ’π2
βππ 2 = 0
π»β² β π = 0
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πΉ, π, π, π», π are unknowns
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Finite Difference StencilBOX SCHEME
β’ Space-march in π direction
β’ 2nd Order
β’ Centered Difference in both directions
β’ Similar to Crank Nicholson
3-POINT BACKWARD DIFFERENCE
β’ Space-march in π direction
β’ 2nd Order
β’ Centered Difference in π
β’ 3-pt Backward Difference in π
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π
π +1
2π
π + 1
π +1
2
π + 1π
π
π β 2
π +1
2π
π + 1
π
π
π
π β 1
space marchspace march
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Box Stencil
π = ππ+12
π+12 =
1
4πππ+ ππ
π+1+ ππ+1
π+ ππ+1
π+1
πβ² = πβ²π+12
π+12 =
ππ+12
π+1β π
π+12
π
Ξπ=
12 ππ+1
π+1+ ππ
π+1β12 ππ+1
π+ ππ
π
Ξπ
ππ
ππ= α€ππ
πππ+12
π+12
=ππ+1
π+12 β π
π
π+12
Ξπ=
12 ππ+1
π+1+ ππ+1
πβ12 ππ
π+1+ ππ
π
Ξπ
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π
π +1
2π
π + 1
π +1
2
π + 1π
π
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3pt - Backward Stencil
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π = ππ
π+12 =
1
2πππ+ ππ
π+1
πβ² = πβ²π
π+12 =
πππ+1
β πππ
Ξπ
ππ
ππ= α€ππ
πππ
π+12
=
32ππ
π+12 β 2π
πβ1
π+12 +
12ππβ2
π+12
2Ξπ=
32 ππ
π+1+ ππ
πβ 2 ππβ1
π+1+ ππβ1
π+12 ππβ1
π+1+ ππβ1
π
4Ξπ
π β 2
π +1
2π
π + 1
π
π
π
π β 1
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3pt - Backward Stencil
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π = ππ
π+12 =
1
2πππ+ ππ
π+1
πβ² = πβ²π
π+12 =
πππ+1
β πππ
Ξπ
ππ
ππ= α€ππ
πππ
π+12
=
32ππ
π+12 β 2π
πβ1
π+12 +
12ππβ2
π+12
2Ξπ=
32 ππ
π+1+ ππ
πβ 2 ππβ1
π+1+ ππβ1
π+12 ππβ1
π+1+ ππβ1
π
4Ξπ
π = πππ’π = π’ππ = π’π ππβπ = βππ = βπ ππ
α€πβπππ
=πβπππ
π
=πβπππ
ππ
α€ππ’πππ
=ππ’πππ
π
=ππ’πππ
ππ
π β 2
π +1
2π
π + 1
π
π
π
π β 1
Analytical Evaluations:
Finite difference approximations:
12
Newton-Raphson Solverβ’ Necessary for Non-linear system
β’ 10 β Diagonal Sparse Matrix
β’ Analytically calculated jacobian
β’ Utilized MATLAB built in matrix solver
β’ Quadratic convergence
β’ Convergence criteria based on magnitude of max residual
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Matrix pattern
13
Newton vs. MATLAB FSolveβ’ Compared the Newton-Raphson Method against MATLABβs nonlinear solver
β’ Fsolve is a quasi-Newton method
β’ Assumed both solution methods had similar accuracies
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Newton-Raphson MATLAB FSolve
Runtime 6-7 sec 116 sec
# of Iterations 3-4 2
Implementation Complexity ~650 lines ~125 lines
# of Function Evaluations 525 243
(Solved for 30 x 30 system)
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Selected Results
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Error Analysisβ’ L2 Error Analysis of πΏβ π , displacement
thickness
β’ Compared to solution with fine resolution
β’ Roughly 2nd order convergence, as expected
β’ Comparable performance of Box Scheme and 3-pt scheme
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πΏβ = ΰΆ±0
ππ
1 βΟ
Οππ
ππ¦
ππππ = ΰΆ±
0
ππ
1 βΟ
Οππ
π» 2π
π’πππππ
16
Conclusionsβ’ Newton-Raphson (in this case) is significantly preferable to FSolve
β’ Stability is not a significant concern for the backwards difference scheme
β’ Oscillations are not present in the box scheme
β’ There is no clear advantage to using the box scheme instead of the 3-pt backwards difference
β’ Further sampling of the solution space may be necessary to determine overall behavior of solution methods
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Thank youQuestions?
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Appendix
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π = ππ+12
π’π = π’ππ+12
= π’π ππ+12
βπ = βππ+12
= βπ ππ+12
α€πβπππ
=πβπππ
π+12
=πβπππ
ππ+12
α€ππ’πππ
=ππ’πππ
π+12
=ππ’πππ
ππ+12
Analytical Evaluations:
π = πππ’π = π’ππ = π’π ππβπ = βππ = βπ ππ
α€πβπππ
=πβπππ
π
=πβπππ
ππ
α€ππ’πππ
=ππ’πππ
π
=ππ’πππ
ππ
Analytical Evaluations:
Chapman Rubesin factorβ’ πΆ = πΆ π
β’ πΆπ+
1
2
π+1
2 = πΆ ππ+
1
2
π+1
2
β’ πΆπ+
1
2
π+1
2 =1
4πΆ ππ
π+ πΆ ππ+1
π+ πΆ ππ
π+1+ πΆ ππ+1
π+1
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Similarity Equations
πΆπβ²β² β² + ππβ²β² = 0
πΆ
πππβ²
β²
+ ππβ² + πΆπ’π2
βππβ²β² 2 = 0
Fβ² = π
πβ² = π
πΆπ β² + πΉπ = 0
π»β² = π
πΆ
πππ
β²
+ πΉπ + πΆπ’π2
βππ 2 = 0
π = π»
π = πΉ
2 - Coupled 2nd and 3rd Order Diff. Eqns 5 - Coupled 1st Order ODEs (Drela Notation)
Where πβ² =ππ
ππ
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