Governing equations for a new compressible Navier-Stokes solver in general cylindrical coordinates

Σχετικά έγγραφα
HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

derivation of the Laplacian from rectangular to spherical coordinates

Approximation of distance between locations on earth given by latitude and longitude

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

Appendix to On the stability of a compressible axisymmetric rotating flow in a pipe. By Z. Rusak & J. H. Lee

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

4.6 Autoregressive Moving Average Model ARMA(1,1)

Appendix A. Curvilinear coordinates. A.1 Lamé coefficients. Consider set of equations. ξ i = ξ i (x 1,x 2,x 3 ), i = 1,2,3

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Areas and Lengths in Polar Coordinates

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

The Simply Typed Lambda Calculus

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Homework 8 Model Solution Section

Partial Differential Equations in Biology The boundary element method. March 26, 2013

Matrices and Determinants

Areas and Lengths in Polar Coordinates

C.S. 430 Assignment 6, Sample Solutions

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

DERIVATION OF MILES EQUATION FOR AN APPLIED FORCE Revision C

Second Order Partial Differential Equations

ES440/ES911: CFD. Chapter 5. Solution of Linear Equation Systems

Spherical Coordinates

EE512: Error Control Coding

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

Other Test Constructions: Likelihood Ratio & Bayes Tests

Κλινικοί Εκπαιδευτές: σύντομη περιγραφή ρόλου, κριτήρια επιλογής & πρόγραμμα εκπαίδευσης

Derivation of Optical-Bloch Equations

Section 8.3 Trigonometric Equations

Homework 3 Solutions

Example Sheet 3 Solutions

Math221: HW# 1 solutions

2 Προετοιμαςία & χοριγθςθ IV υγρϊν. Όροι Χρήςησ

Strain gauge and rosettes

Space-Time Symmetries

Eulerian Simulation of Large Deformations

Tutorial problem set 6,

CHAPTER 48 APPLICATIONS OF MATRICES AND DETERMINANTS

2. THEORY OF EQUATIONS. PREVIOUS EAMCET Bits.

Orbital angular momentum and the spherical harmonics

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

Solutions to Exercise Sheet 5

Parametrized Surfaces

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

w o = R 1 p. (1) R = p =. = 1

k A = [k, k]( )[a 1, a 2 ] = [ka 1,ka 2 ] 4For the division of two intervals of confidence in R +

Technical Information T-9100 SI. Suva. refrigerants. Thermodynamic Properties of. Suva Refrigerant [R-410A (50/50)]

DuPont Suva 95 Refrigerant

Statistical Inference I Locally most powerful tests

1 String with massive end-points

Concrete Mathematics Exercises from 30 September 2016

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Mean bond enthalpy Standard enthalpy of formation Bond N H N N N N H O O O

D Alembert s Solution to the Wave Equation

DuPont Suva 95 Refrigerant

6.3 Forecasting ARMA processes

PARTIAL NOTES for 6.1 Trigonometric Identities

Finite Field Problems: Solutions

Second Order RLC Filters

General 2 2 PT -Symmetric Matrices and Jordan Blocks 1

DuPont Suva. DuPont. Thermodynamic Properties of. Refrigerant (R-410A) Technical Information. refrigerants T-410A ENG

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 19/5/2007

Inverse trigonometric functions & General Solution of Trigonometric Equations

Variational Wavefunction for the Helium Atom

( ) 2 and compare to M.

Exercises to Statistics of Material Fatigue No. 5

Dr. D. Dinev, Department of Structural Mechanics, UACEG

Section 9.2 Polar Equations and Graphs

BSL Transport Phenomena 2e Revised: Chapter 2 - Problem 2C.4 Page 1 of 9

Reminders: linear functions

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

2 Composition. Invertible Mappings

ΠΣΤΥΙΑΚΗ ΔΡΓΑΙΑ. Μειέηε Υξόλνπ Απνζηείξσζεο Κνλζέξβαο κε Τπνινγηζηηθή Ρεπζηνδπλακηθή. Αζαλαζηάδνπ Βαξβάξα

MATH423 String Theory Solutions 4. = 0 τ = f(s). (1) dτ ds = dxµ dτ f (s) (2) dτ 2 [f (s)] 2 + dxµ. dτ f (s) (3)

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

Εργαστήριο Ανάπτυξης Εφαρμογών Βάσεων Δεδομένων. Εξάμηνο 7 ο

ΚΥΠΡΙΑΚΗ ΕΤΑΙΡΕΙΑ ΠΛΗΡΟΦΟΡΙΚΗΣ CYPRUS COMPUTER SOCIETY ΠΑΓΚΥΠΡΙΟΣ ΜΑΘΗΤΙΚΟΣ ΔΙΑΓΩΝΙΣΜΟΣ ΠΛΗΡΟΦΟΡΙΚΗΣ 6/5/2006

Problem Set 3: Solutions

CYLINDRICAL & SPHERICAL COORDINATES

Geodesic Equations for the Wormhole Metric

Differential equations

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

Higher Derivative Gravity Theories

On a four-dimensional hyperbolic manifold with finite volume

Notes on the Open Economy

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

ST5224: Advanced Statistical Theory II

Lecture 26: Circular domains

CRASH COURSE IN PRECALCULUS

Every set of first-order formulas is equivalent to an independent set

Fractional Colorings and Zykov Products of graphs

Congruence Classes of Invertible Matrices of Order 3 over F 2

Srednicki Chapter 55

2 Διενζργεια Διενζργεια υψθλοφ εκκενωτικοφ υποκλυςμοφ. Όροι Χρήςησ

CE 530 Molecular Simulation

Κάθε γνήσιο αντίγραφο φέρει υπογραφή του συγγραφέα. / Each genuine copy is signed by the author.

Transcript:

UNIVERSITY OF SOUTHAMPTON SCHOOL OF ENGINEERING SCIENCES AERODYNAMICS & FLIGHT MECHANICS GROUP Governing equations for a new compressible Navier-Stokes solver in general cylindrical coordinates by Richard D. Sandberg Report No. AFM-7/7 November 27 COPYRIGHT NOTICE - All right reserved. No parts of this publication may be reproduced stored in a retrieval system or transmitted in any form or by any means electronic mechanical photocopying recording or otherwise without the permission of the Head of the School of Engineering Sciences University of Southampton Southampton SO17 1BJ U.K. c School of Engineering Sciences University of Southampton

The aim of this report is to derive the governing equations for a new compressible Navier- Stokes solver in general cylindrical coordinates i.e. the streamwise and radial directions are mapped to general coordinates. A literature review revealed that simulations of complex cylindrical geometries have mostly been conducted using general curvilinear coordinates e.g. DeBonis & Scott 22. The method presented in this report is chosen over three-dimensional general curvilinear coordinates as it requires a smaller number of terms to be computed. Furthermore only two-dimensional metric terms need to be stored reducing the required allocated memory and therefore resulting in a more efficient code. 1 Governing equations in cylindrical coordinates The compressible Navier-Stokes equations consist of conservation of mass momentum and total energy. The flow is assumed to be an ideal gas with constant specific heat coefficients. All quantities are made dimensionless using the flow quantities at a reference location in the flow; here the free-stream/inflow location is used. The radius of the body is chosen as the reference length. The non-dimensionalization results in the following dimensionless parameters: Re = ρ u r µ M = u a P r = µ c p κ With z r θ denoting the streamwise radial and azimuthal directions respectively and u v w denoting the velocity components in z r θ directions respectively the non-dimensional compressible Navier Stokes equations in cylindrical coordinates are A = C = U t + A z + B r + 1 r ρu ρuu + p τ zz ρuw τ θz ρuh + q z uτ zz vτ rz wτ θz ρw ρuw τ θz w τ θr ρww + p τ θθ ρwh + q θ uτ θz vτ θr wτ θθ U = ρ ρu ρw ρe. C θ + 1 r D = 1 B = D = v + p τ rr w τ θr H + q r uτ rz vτ rr wτ θr v ρww τ rr + τ θθ 2w 2τ θr H + q r uτ rz vτ rr wτ θr 2 where the total energy is defined as E = T/γγ 1M 2 + 1/2u i u i with γ = 1.4 and the total enthalpy is H = E + p/ρ. 1

The molecular stress tensor components are τ zz = 2µ 2 u z v r 1 w r θ + v 3 τ rr = 2µ u z + 2v r 1 w r θ + v 4 τ θθ = 2µ u z v w r + 21 r θ + v 5 τ rz = µ u Re r + v 6 z τ θz = µ w Re z + 1 u 7 r θ τ θr = µ 1 v Re r θ w + w. 8 r The heat-flux vector components are q z = q r = q θ = µ T P rγ 1M 2 Re z 9 µ T P rγ 1M 2 Re r 1 µ 1 T P rγ 1M 2 Re r θ 11 where the Prandtl number is assumed to be constant at P r =.72. The molecular viscosity µ is computed using Sutherland s law c.f. White 1991 setting the ratio of the Sutherland constant over freestream temperature to.36867. To close the system of equations the pressure is obtained from the non-dimensional equation of state p = ρt /γm 2. 2 Transformation to general cylindrical coordinates In order to allow for complex geometries the r z plane is mapped to general coordinates ξ η. Hence streamwise and radial derivatives need to be expressed in terms of the new variables. By using the chain rule the following expressions can be derived z r = 1 r J η = 1 z J η ξ r ξ ξ + z ξ η η = rη ξ r ξ η 12 = zξ η z η ξ 13 where J is the determinant of the coordinate transformation. For conciseness the metric terms are abbreviated as e.g. r η = r so that J = z η ξr η z η r ξ and the asterisk denotes a metric term already divided by J. 2

Following Anderson 1995 the r and z derivatives of equation 1 can be transformed as follows provided that the grid transformation does not vary in time U t = 1 J ξ Ar η Bz η + η Ar ξ + Bz ξ 1 C r θ 1 r D 14 Unfortunately it is not straightforward to plug in the vectors A and B into 14 as they contain terms that are pre-multiplied by 1/r. As r = rξ η when taking ξ and η derivatives of A and B the chain rule needs to be applied. To illustrate this procedure the derivation of the radial momentum equation is chosen as example. Here we only focus on the derivatives in the ξ and η directions r η v + p τ rr z η ξ r ξ v + p τ rr z ξ. 15 η The only term containing 1/r is the stress tensor component τ rr. Therefore the equation is separated into r η v + p 2µ u ξ z + 2v z η 1 2µ w r ξ r θ + v z η r ξ v + p 2µ u η z + 2v z ξ + 1 2µ w r η r θ + v z ξ. 16 Applying the chain rule to the last term of each row results in 1 2µ w ξ r θ + v z η = 1 2µ w r ξ θ + v z η 1 2µ w η r θ + v z ξ = 1 2µ w r η θ + v z ξ 1 r 2 r ξz η 2µ 1 r 2 r ηz ξ 2µ w θ + v w θ + v Substituting these two expressions into 16 yields r η v + p 2µ u ξ z + 2v z η r r ξ v + p 2µ u η z + 2v z ξ r 1 2µ w r ξ θ + v z η + 1 2µ w r η θ + v z ξ + 1 2µ w r 2 θ + v r ξ z η r η z ξ. 17 }{{} J The last term of 17 occurs because of the partial derivative with respect to r of τ rr. In the case of streamwise derivatives z we get r ξr η r η r ξ = hence 1/r 2 terms cancel. The resulting governing equations are now summarized. For brevity the inner derivatives are not written in terms of ξ η however the streamwise and radial derivatives need to be computed with equations 12 and 13. 3

2.1 Continuity equation ρ t = 1 J ξ ρur η z η + η ρur ξ + z ξ 1 ρw r θ r 18 2.2 Streamwise momentum equation ρu t = 1 { ρuu + p 2µ 2 u J ξ z v r ρuu + p 2µ 2 u η z v r + 1 2µ w r ξ θ + v r η 1 2µ r η 1 ρuw τ θz 1 r θ r r η z η r ξ z ξ } w θ + v r ξ 19 2.3 Radial momentum equation t = 1 { r η J ξ r ξ η 1 2µ w r ξ θ + v z η + 1 2µ w r 2 θ + v 1 r θ w τ θr 1 r v + p 2µ u z + 2v z η r v + p 2µ u z + 2v z ξ r + 1 } 2µ w r η θ + v z ξ v ww 2µ Re v r 1 w r θ + v 2 4

2.4 Azimuthal momentum equation ρw t = 1 J { ξ η ρuw µ Re w r η z w r ξ w µ Re ρuw µ w µ Re z Re r 1 µ u v r ξ Re θ r η θ w z η + 1 r η 1 µ v r 2 Re θ w 1 ρww + p τ θθ 1 2w 2τ θr r θ r w z η r w z ξ 21 } µ u v Re θ r ξ θ w z ξ 2.5 Energy equation ρe t { ρuh + q z u 2µ ξ 2 u z v r 2 v r u z w z = 1 vτ rz w µ r η 22 J Re H + q r uτ rz v 2µ w µ w z η Re r ρuh + q z u 2µ 2 u η z v vτ rz w µ w r ξ r Re z H + q r uτ rz v 2µ 2 v r u w µ w z ξ z Re r + 1 µ u 2 w r ξ Re 3 θ + v w u r η θ µ 2 w v Re 3 v θ + v w θ w z η 1 µ u 2 w r η Re 3 θ + v w u r ξ θ µ } 2 w v Re 3 v θ + v w θ w z ξ + 1 µ 2 w v r 2 Re 3 v θ + v w θ w 1 ρwh + q θ uτ θz vτ θr wτ θθ 1 H + q r uτ rz vτ rr wτ θr r θ r 5

3 Vector form of general cylindrical coordinates The equations in general cylindrical coordinates can be written as where  r = U t = 1 J { µ Re u 2 3 1 r  = ˆB = 2µ ξ ξ Ârη ˆBz η + Âr ξ + η ˆBz ξ Âr r η ˆB r z η + η  r r ξ + ˆB r z ξ } + 1 r C θ 1 r D 1 r ˆB 2 rr 23 ρuh + q z u 2µ ρu ρuu + p 2µ 2 u v z r ρuw µ Re 2 u v z H + q r uτ rz v 2µ w + θ v µ Re w u θ θ + v w u θ ˆB rr = w z r vτrz w µ Re ρuv τ rz v + p 2µ u + 2 v z r w µ w Re r u + 2 v z r ˆBr = µ Re w v µ Re 2µ w + θ v µ v Re θ w w z w µ w Re r 2µ µ Re θ w 2 3 v w θ + v w + θ v v θ w 2 3 v w θ + v w v. θ w In order to make the new code most efficient a good compromise between the number of arithmetic operations and the number of three dimensional arrays stored needs to be found. The emphasis here lies on reducing the number of stored arrays to a minimum while minimizing the number of repeated operations. One choice is to simply store all nine velocity derivatives. However for this option the number of operations is high as the stress tensor components constantly need to be reassembled. The most efficient choice of 6

arrays to store appears to be the following set 1 τ zz = 2µ 2 u z v 2 τ rz = µ u r Re r + v z 4 τ rr = 2µ 2 v r u 5 τ θr = µ w z Re r 7 τ 2 = µ u 8 τ 3 = µ v Re θ Re θ w. 3 τ θz = µ w Re z 6 τ 1 = 2µ w θ + v Using these eight arrays in addition to the primitive variables and the heat-flux vector  ˆB Âr ˆB r and ˆB rr can be assembled. In order to evaluate C and D the following relations can be used τ θz = τ θz + 1 r τ 2 τ θr = τ θr + 1 r τ 3 τ rr = τ rr 1 r τ 1 τ θθ = τ zz τ rr + 2 r τ 1. 24 Finally all vectors required for 23 can be written as ρu ρuu + p τ zz  = ˆB = v + p τ rr ρuw τ θz w τ θr ρuh + q z u τ zz vτ rz w τ θz H + q r uτ rz v τ rr w τ θr τ 1  r = ˆBr = ˆBrr = τ 2 uτ 1 wτ 2 C = References τ 1 τ 3 vτ 1 wτ 3 τ 1 τ 3 wτ 3 vτ 1 ρw ρuw τ θz + 1τ r 2 w τ θr + 1τ r 3 ρww + p τ zz τ rr + 2τ r 1 ρwh + q θ u τ θz + 1τ r 2 v τ θr + 1τ r 3 w τ zz τ rr + 2τ r 1 D = v ρww τ rr 1 r τ 1 + τ zz τ rr + 2 r τ 1 2w 2 τ θr + 1 r τ 3 H + q r uτ rz v τ rr 1 r τ 1 w τ θr + 1 r τ 3. Anderson J. 1995 Computational Fluid Dynamics The Basics with Applications. McGraw-Hill. DeBonis J. & Scott J. N. 22 Large-eddy simulation of a turbulent compressible round jet. AIAA J. 4 7 1346 1354. White F. M. 1991 Viscous Fluid Flow. McGraw Hill. 7