Advanced Computational Fluid and Aerodynamics 1 edition Paul G. Tucker Instructor Solution Manual, S

SOLUTION MANUAL COMPLETE

SOLUTIONS MANUAL FOR ADVANCED COMPUTATIONAL FLUID AND AERODYNAMICS

P. G. Tucker

University of Cambridge

1

INTRODUCTION AND USE OF THIS TEXT AS A CORE TO A CFD COURSE The problems in the text fall approximately into the following two categories: 1) Analytical type exercises related to numerical methods and their validation and 2) Code development exercises.

With regards to the second aspect, there are exercises that direct the reader to write both basic incompressible and compressible flow solvers. The latter is based around the Euler equations. The former is based around the Navier-Stokes equations and hence has additional diffusion terms.

Therefore, this incompressible solver is restricted to simple grid structures and has more of an emphasis on the finite difference method. The problems also direct the reader to generate meshes, produce validation solutions and also realistic initial flow guesses to support the above. The problems also direct the reader in the process of generating meshes for these exercises and appropriate boundary conditions along with the process of post processing solutions. The text can be used with a range of student abilities. For the computational exercises a full code can be developed and then areas deleted to provide a template. The extent of the template and the tasks that need to be completed in the template can be tailored to the typical class ability. Then the code can be validated through the exercises in the text and performance traits of the code can be explored through numerical exercises, as outlined in the problems in the text. Most of the detail needed to write the codes is given in the main text and the appendices in the text. Additional details are also given here.

A full compressible flow solver that could be used to form a template can be found in Tucker (2001).

At Cambridge University the CFD module is course work based. Such a course could be centered around either a compressible or incompressible solver or even the convection of a cylinder case (Problem 3, Chapter 4). Dealing with both would probably be too much for a typical undergraduate course. As well as writing the code, validating it and exploring its performance traits, other pieces of coursework are carried out that explore the students understanding of the material within the text of this book.

2

CHAPTER 2

2.1

Figure S.2.1. Schematic of channel flow geometry. !

Figure S.2.1 gives a schematic of channel flow geometry. We assume, !" = 0 since infinitely long, which from continuity equation gives 𝜕𝑣 =0 𝜕𝑦 and since 𝑣 0 = 𝑣 𝑢 = 0 implies 𝑣 = 0 Therefore the Navier-Stokes equations reduce to !! !

𝜇 !! ! = 0 !"

− !" = 0 ·.·

𝑝 = C (a constant)

.·.

𝑝 = 𝑝!

!! !

!"

𝜇 !! ! = 0 implies !" = 𝐶!

The differential equation can be solved by integration 𝑑𝑢 = 𝐶!

𝑑𝑦

to give 𝑢 = 𝐶! 𝑦 + 𝐶!

Boundary conditions are 𝑢 = 0 at 𝑦 = 0 .·. 𝐶! = 0 3

!

𝑢 = 𝑈 at 𝑦 = ℎ .·. 𝐶! = !

Therefore solution is 𝑢=

𝑦 𝑈 ℎ

𝑝 = 𝑝!

Note for compressible flow the solution is slightly different and given below. !

!

!

Steady flow − !" = 0, 𝐿! = ∞ − !" = 0 , 𝐿! = ∞ − !" = 0 𝑤 = 0 (given).

Therefore, equations reduce to 𝜕𝑢 𝜕!𝑢 𝜌𝜈 =𝜇 ! 𝜕𝑦 𝜕𝑦

(1)

𝜕𝑣 𝜕𝑝 𝜕!𝑣 1 𝜕 𝜕𝑣 𝜌𝜈 = − +𝜇 + 𝜇 𝜕𝑦 𝜕𝑦 𝜕𝑦 !

3 𝜕𝑦 𝜕𝑦

(2)

𝜕𝑤 𝜕!𝑤 = 𝜇 𝜕𝑦 𝜕𝑦 !

(3)

𝑤=0 𝜌𝜈

!"

The continuity equation combined with Taylor’s series can be used to demonstrate !" = 0 = 𝑣 ·.· .·.

!"

!"#

!"#

!"#

!"

!"

+ !" + !" + !" = 0 !"

!"# !"

= 0 or 𝜌 !" + 𝑣 !" = 01

!"

At wall (𝑦 = 0) 𝑣 = 0 so !" |!!! = 0 Taylor’s series for

about bottom plate at 𝑦 = 0 𝑣 𝑦 =𝑣 0 +

𝑦! 𝜕! 𝑣 | 𝑛! 𝜕𝑦 ! !!!

using this to give 𝑣 ℎ we have

1

Tempting to say at this point 𝑣 = 0 satisfies this equation.

4

(4)

ℎ! 𝜕 ! 𝑣 | 𝑛! 𝜕𝑦 ! !!!

𝑣 ℎ =𝑣 0 + Or

ℎ! 𝜕 ! 𝑣 | 𝑛! 𝜕𝑦 ! !!! which suggests if 𝑣 0 = 𝑣 ℎ = 0, 𝑣 everywhere zero. This satisfies Equation (4) !"

!! !

Since 𝑣 = !" = !! ! = 0, Equation (2) gives !" !"

= 0 and Equation (1) becomes !! !

!"

0 = 𝜇 !! !

implying !" = 𝐶!

which can be integrated to give 𝑑𝑢 = 𝐶!

𝑑𝑦

𝑢 = 𝐶! 𝑦 + 𝐶! since 𝑢 = 𝑈 at 𝑦 = ℎ and 𝑢 = 0 at 𝑦 = 0 we have 𝐶! = 0 and 𝐶! = 𝑈/ℎ and so 𝑢=

𝑈𝑦 ℎ

i.e a linear velocity profile.

5

2.2 Combine eqns (2.36) and (2.37) for steady flow for u velocity component (using λ = -2µ/3) ∂ ∂ ∂ ( ρuu) + ( ρ vu) + ( ρw φ ) ∂x ∂y ∂z ∂ ⎛⎛ 2 ⎞ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ = ⎜⎜ ⎜ 2µ − µ ⎟ ⎟⎟ + ⎜µ ⎟ + ⎜ µ ⎟+S ∂x ⎝ ⎝ 3 ⎠ ∂x ⎠ ∂ y ⎝ ∂ y ⎠ ∂z ⎝ ∂z ⎠ u

(A)

Combining (A) with eqns (2.38) ∂ ∂ ∂ ∂ ⎛⎛ 2 ⎞ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ( ρuu) + ( ρ vu) + ( ρ w w) = ⎜⎜ ⎜ 2µ − µ ⎟ ⎟⎟ + ⎜µ ⎟ + ⎜µ ⎟ ∂x ∂y ∂z ∂x ⎝ ⎝ 3 ⎠∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠

-

∂p ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂w ⎞ ∂ ⎛ ∂w ⎞ + ⎜λ ⎟ + ⎜µ ⎟ + ⎜λ ⎟ + ⎜µ ⎟ ∂x ∂x ⎝ ∂y ⎠ ∂y ⎝ ∂x ⎠ ∂x ⎝ ∂z ⎠ ∂z ⎝ ∂x ⎠

For a constant viscosity fluid ∂ ∂ ∂ 4 ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ( ρuu) + ( ρ vu) + ( ρ w w) = µ ⎜ ⎟ + µ ⎜ ⎟ +µ ⎜ ⎟ ∂x ∂y ∂z 3 ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠

-

∂p ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂w ⎞ ∂ ⎛ ∂w ⎞ + λ ⎜ ⎟ + µ ⎜ ⎟ + λ ⎜ ⎟ + µ ⎜ ⎟ ∂x ∂x ⎝ ∂y ⎠ ∂y ⎝ ∂x ⎠ ∂x ⎝ ∂z ⎠ ∂z ⎝ ∂x ⎠

or ∂ ∂ ∂ 4 ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ( ρuu) + ( ρ vu) + ( ρ w w) = µ ⎜ ⎟ + µ ⎜ ⎟ +µ ⎜ ⎟ ∂x ∂y ∂z 3 ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠

-

∂p ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂w ⎞ + λ +µ ⎜ ⎟ + λ +µ ⎜ ⎟ ∂x ∂x ⎝ ∂ y ⎠ ∂x ⎝ ∂z ⎠

(

)

(

)

or simplifying ∂ ∂ ∂ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ∂ ⎛ ∂u ⎞ ( ρuu) + ( ρ vu) + ( ρ w w) = µ ⎜ ⎟ + µ ⎜ ⎟ +µ ⎜ ⎟ ∂x ∂y ∂z ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ∂z ⎝ ∂z ⎠

-

⎛ 1 ⎞ ∂ ⎛ ∂w ⎞ ∂p ⎛ 1 ⎞ ∂ ⎛ ∂u ⎞ ⎛ 1 ⎞ ∂ ⎛ ∂v ⎞ + ⎜ ⎟µ ⎜ ⎟ + ⎜ ⎟µ ⎜ ⎟ + ⎜ ⎟µ ⎜ ⎟ ∂x ⎝ 3 ⎠ ∂x ⎝ ∂x ⎠ ⎝ 3 ⎠ ∂x ⎝ ∂y ⎠ ⎝ 3 ⎠ ∂x ⎝ ∂z ⎠

Which tidies to ∂ ∂ ∂ ∂2 u ∂2 u ∂2 u ( ρuu) + ( ρ vu) + ( ρ w w) = µ 2 +µ 2 + µ 2 ∂x ∂y ∂z ∂x ∂y ∂z

-

∂p ⎛ µ ⎞ ∂ ⎛⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂w ⎞ ⎞ + ⎜ ⎟ ⎜⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎟ ∂x ⎝ 3 ⎠ ∂x ⎝⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ ⎠

6

(B)

Dealing with left hand side we have ρu

∂ ∂ ∂ ∂ ∂ ∂ ∂2 u ∂2 u ∂2 u (u) + ρ v (u) + ρ w ( u)+ u ( ρu) + u ( ρ v) + u ( ρ w ) = µ 2 +µ 2 + µ 2 ∂x ∂y ∂z ∂x ∂y ∂z ∂x ∂y ∂z

-

∂p ⎛ µ ⎞ ∂ ⎛⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂w ⎞ ⎞ + ⎜ ⎟ ⎜⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎟ ∂x ⎝ 3 ⎠ ∂x ⎝⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ ⎠

The steady continuity equation (2.39) given below ∂ ∂ ∂ ( ρu) + ( ρ v) + ( ρ w) = 0 ∂x ∂y ∂z

Allows this the removal of the final three left hand side equation terms and hence the final answer ρu

∂ ∂ ∂ ∂2 u ∂2 u ∂2 u (u) + ρ v (u) + ρ w ( u) = µ 2 +µ 2 + µ 2 ∂x ∂y ∂z ∂x ∂y ∂z

-

∂p ⎛ µ ⎞ ∂ ⎛⎛ ∂u ⎞ ⎛ ∂v ⎞ ⎛ ∂w ⎞ ⎞ + ⎜ ⎟ ⎜⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎟ ∂x ⎝ 3 ⎠ ∂x ⎝⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎝ ∂z ⎠ ⎠

The derivation for the other 2 momentum equations is a direct repetition of the above for the respective equations.

7

2.3 Figure S.2.3.1 gives a schematic of the annulus geometry. Since it is infintely infinitely long,

∂ ∂z = 0 . Therefore the continuity equation reduces to 1 ∂ (ru) = 0 r ∂r

Or r

∂u +u=0 ∂r

u = 0 satisfies the above equation.

Figure S.2.3.1. Schematic of the annulus geometry.

Since u = w = ∂ ∂z = 0 the Navier-Stokes equations reduce to

0=-

0=

∂p ρ v 2 + ∂r r

1 ∂ ⎛ ∂v ⎞ v ⎜r ⎟ r ∂r ⎝ ∂r ⎠ r 2

(1)

As noted in the question, Equation (1) can be re-written as 0=

∂ c0 ∂r

c0 =

1 ∂vr r ∂r

Where (2)

Solving the differential (2) using integration

8

∫ dvr = c ∫ rdr 0

vr = c0

r2 + c1 2

(3)

Boundary conditions are

r = a; v = Ωa and r = b; v = 0 Application of these to Equation (3) gives

co a 2 Ωa = + c1 2

(4)

co b 2 0= + c1 2

(5)

2

Eliminating c1 from equations (4) and (5) gives

Ωa 2 =

co a 2 c0b 2 − 2 2

Therefore,

co =

2Ωa 2 b 2 Ωa 2 and c = − 1 a 2 − b2 a 2 − b2

Substitution of the above into Equation (3) gives

v=

1 ⎡ Ωa 2 r 2 b 2 Ωa 2 ⎤ − r ⎢⎣ a 2 − b 2 a 2 − b 2 ⎥⎦

or

Ωa 2 ⎡ r 2 − b 2 ⎤ v= r ⎢⎣ a 2 − b 2 ⎥⎦ Note the pressure distribution can be gained by substituting the above expression for v into

0=-

∂p ρ v 2 + ∂r r

9

2.4 W

R

ρ = Constant µ = Constant

!"

For fully developed flow !" = 0 and so from continuity equation: 1 𝜕 𝜕𝑤 𝑟𝑢 + =0 𝑟 𝜕𝑟 𝜕𝑧 1 𝜕 𝜕𝑢 𝑟𝑢 = 0 or 𝑟 + 𝑢 = 0 implying 𝑢 = 0 𝑟 𝜕𝑟 𝜕𝑟 And so the Navier-Stokes equations reduce to the following, noting v = 0 also: µ

1 𝜕 𝜕𝑤 𝑟 𝑟 𝜕𝑟 𝜕𝑟

=

𝜕𝜌 𝜕𝑧

(1)

!"

For fully developed flow !" = 𝐶 and we can write: 𝜕 𝑟

𝜕𝑤 1 𝜕𝜌 = 𝜕𝑟 µ 𝜕𝑧

𝑟 𝑑𝑟

Which on integration gives 𝑟 !" !"

𝜕𝑤 1 𝜕𝜌 = 𝑟² + 𝐶! 𝜕𝑟 2µ 𝜕𝑧

(2)

= 0 at r = 0 so 𝐶! = 0 and so (2) becomes: 𝜕𝑤 1 𝜕𝜌 = 𝑟 𝜕𝑟 2µ 𝜕𝑧

Which can be integrated again: 𝜕𝑤 =

1 𝜕𝜌 2µ 𝜕𝑧

𝑟 𝑑𝑟

To give: 𝑤=

1 𝜕𝜌 𝑟² + 𝐶!

4µ 𝜕𝑧

At r = R, w = 0 so:

10

(3)

1 𝜕𝜌 𝑅² 4µ 𝜕𝑧

𝐶! = − Therefore: 𝑤= =

1 𝜕𝜌 !

1 𝜕𝜌 ! 𝑟 − 𝑅 4µ 𝜕𝑧 4µ 𝜕𝑧 1 𝜕𝜌 𝑟² − 𝑅² 4µ 𝜕𝑧

(4)

!

𝑄 = 2𝜋

𝑊𝑟 𝑑𝑟

(5)

!

Substituting (4) into (5) gives: 2𝜋 𝜕𝜌 𝑄= 4µ 𝜕𝑧

!

𝑟 ! − 𝑅²𝑟 𝑑𝑟

!

Which yields Poiseuille’s Law: ! !"

! ∆!

𝑄 = − !" !" 𝑅! (or − !" ! 𝑅! )

(6)

𝑄 = 𝜋 𝑅² 𝑊

(7)

We can also write:

Equating (6) and (7) gives the following pressure gradient expression: 𝜕𝜌 8𝑊µ = − 𝜕𝑧 𝑅²

(8)

This can be substituted into equation (4) to give: 𝑤 = 2𝑊 1 −

𝑟 ! 𝑅

(9)

Derivation of fraction factor expression from the above is as follows: 𝑓=

1

𝜏! = − µ

𝜏!

2 𝜌 𝑊² 𝜕𝑤 𝜕𝑟 𝑟 = 𝑅

11

(10) (11)

Substituting (9) into (11) 𝜏 = −2 𝑊 µ 𝜏=

𝜕 𝑟² 1− 𝜕𝑟 𝑅²

2 𝑊 µ 𝜕𝑟² 4𝑊µ𝑟 = 𝑅² 𝜕𝑟 𝑅²

𝜏! is given by setting r = R: 4𝑊µ 𝑅

(12)

𝑓=

8𝑊µ 𝑅 𝜌 𝑊²

(13)

𝑅𝑒 =

2𝑅𝜌𝑊 µ

(14)

16 𝑅𝑒

(15)

𝜏! = Substituting (12) into (10) gives:

Combining (13) and (14) gives: 𝑓=

12

2.5 Show that when Equation (2.47). is transformed to a rotating coordinate system the following extra terms arise 𝑆!,!"#$#%&' = 𝑆!,!"#"$%&#'( + 2𝜌𝜈Ω! + Ω!! 𝑟𝑆!,!"#$#%&' = 𝑆!,!"#"$%&#'( − 2𝜌𝑢Ω! !"# !"#$%

!"# !"#$

Need to make the substitution v = v + Ωr

(A)

The solution is yielded by following the equations (2.58) – (2.72) given in the text. Alternatively, substitution (A) can be used and viscous terms ignored to simplify matters.

13

2.6 Equation (2.47) for steady flow (with no extra source terms) and the corresponding source for tangential momentum is given below2 1 ∂ 1 ∂ ∂ ( ρ ruv) + ( ρ vv) + ( ρ w v) r ∂r r ∂θ ∂z 1 ∂ ⎛ ∂v ⎞ 1 ∂ ⎛ 1 ∂v ⎞ ∂ ⎛ ∂v ⎞ = ⎜ r Γϕ r ⎟ + ⎜ Γϕθ ⎟ + ⎜ Γϕ z ⎟ r ∂r ⎝ ∂r ⎠ r ∂θ ⎝ r ∂θ ⎠ ∂z ⎝ ∂z ⎠ -

+

1 ∂p µ v ρuv 2 ∂ - 2 + 2 (µ u) r ∂θ r r r ∂θ

⎞ 1 ∂ ⎛ ∂w ⎞ 1 ∂ ⎛ ∂w ⎞ 1 ∂ ⎛ ∂u ⎞ 1 ∂ ⎛ ∂ ⎜ µ r ⎟ + 2 ⎜ λ (r u) ⎟ + ⎜λ ⎟ + ⎜µ ⎟ 2 r ∂θ ⎝ ∂z ⎠ r ∂z ⎝ ∂θ ⎠ ⎠ r ∂r ⎝ ∂θ ⎠ r ∂θ ⎝ ∂r

To make axisymmetric remove all derivatives with respect to theta as per below 1∂ ∂ ( ρ ruv) + ( ρ wv) r ∂r ∂z 1∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞ µ v ρuv = ⎜ r Γϕ r ⎟ + ⎜ Γϕ z ⎟ - 2 r ∂r ⎝ ∂r ⎠ ∂z ⎝ ∂z ⎠ r r

For constant density fluid (making use of the axisymmetric form of the continuity equation) we can write

ρv ∂

∂w ρru ∂v ∂v (ru) + ρv + + ρw r! ∂r!!#!!! ∂" z r ∂r ∂z $ zero from continuity

=

1 ∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞ µ v ρuv ⎜ r Γφ r ⎟ + ⎜ Γφ z ⎟ - 2 r ∂r ⎝ ∂r ⎠ ∂z ⎝ ∂z ⎠ r r

Note the axisymmetric continuity equation is ρ ∂ ∂w (ru) + ρ =0 r ∂r ∂z Hence we have

2

Note -

v ∂µ r ∂r

term is ignored. This term is erroneous.

14

ρw

∂v ∂v ∂ ⎛ ∂v ⎞ 1 ∂ ⎛ ∂v ⎞ ρuv µ v + ρu = - 2 ⎜µ ⎟ + ⎜r µ ⎟ ∂z ∂r ∂z ⎝ ∂z ⎠ r ∂r ⎝ ∂r ⎠ r r

Multiplying through by r and rearranging, we obtain: ρw

∂ ∂v ∂ ⎛ ∂rv ⎞ ∂ ⎛ ∂v ⎞ v rv + ρur = ⎜µ + ⎜ r µ ⎟ − ρuv − µ ⎟ ∂z ∂r ∂z ⎝ ∂z ⎠ ∂r ⎝ ∂r ⎠ r

( )

Now

∂ ∂v rv = r + v ∂r ∂r

( )

Therefore:

ρu

∂ ∂v rv = ρur + ρuv ∂r ∂r

( )

Hence:

ρw

∂ ∂ ∂ ⎛ ∂(rv) ⎞ ∂ ⎛ ∂v ⎞ v rv + ρu rv = ⎜ µ + r µ − µ ∂z ∂r ∂z ⎝ ∂z ⎟⎠ ∂r ⎜⎝ dr ⎟⎠ r

( )

( )

This is an equation of the standard form for the angular momentum (rv) except for the last two terms.

Let T=

∂ ⎛ ∂v ⎞ v rµ ⎟ − µ ⎜ ∂r ⎝ ∂r ⎠ r

Now

∂ ∂φ rφ = r + φ ∂r ∂r

( )

Put

φ = rµ

∂v ∂r

Therefore, ∂ ⎛ 2 ∂v ⎞ ∂ ⎛ ∂v ⎞ ∂v r µ ⎟ = r ⎜ rµ ⎟ + rµ ⎜ ∂r ⎝ ∂r ⎠ ∂r ⎝ ∂r ⎠ ∂r

Therefore: T=

1 ∂ ⎛ 2 ∂v ⎞ ∂v v r µ ⎟ −µ −µ ⎜ r ∂r ⎝ ∂r ⎠ ∂r r

and 15

T=

1 ∂ ⎛ 2 ∂v ⎞ ∂v v r µ ⎟ −µ −µ ⎜ r ∂r ⎝ ∂r ⎠ ∂r r

Since

∂ ∂v rv = r + v ∂r ∂r

( )

and

∂v 1 ∂ v = rv − ∂r r ∂r r

( )

then ⎛1 ∂ 1⎡ v⎞⎤ ∂v v T = ⎢ r 2µ ⎜ rv − ⎟ ⎥ − µ − µ = r⎣ r⎠⎦ ∂r r ⎝ r ∂r ⎤ 1 ∂ ⎛ v⎞ 1 ∂ ⎡ ∂ ∂v v = rv ⎥ − rµ ⎟ − µ − µ . ⎢r µ ⎜ r ∂r ⎣ ∂r ∂r r ⎦ r ∂r ⎝ r ⎠

( )

( )

Let

S=−

1 ∂ v ∂v r µv − µ − µ r ∂r r ∂r

( )

Since µ is constant, we have:

µ ⎛ ∂v ⎞ v ∂v S = − ⎜ r + v⎟ − µ − µ = r ⎝ ∂r r ∂r ⎠ ⎛ v ∂v ⎞ v ∂v = −2µ − 2µ = −2µ ⎜ + ⎟ = r ∂r ⎝ r ∂r ⎠ = −2

µ ∂ rv . r ∂r

( )

The full equation becomes:

ρw

⎤ µ ∂ ∂ ∂ ∂ ⎛ ∂rv ⎞ 1 ∂ ⎡ ∂ rv + ρu rv = ⎜ µ + r µ rv rv ⎢ ⎥−2 ∂z ∂r ∂z ⎝ ∂z ⎟⎠ r ∂r ⎣ ∂r r ∂r ∂ ⎦

( )

( )

( )

Note that the last two terms can be combined as:

1 ∂ ⎡ 3 ∂ ⎛ v⎞⎤ ⎢r µ ⎜ ⎟ ⎥ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ since this is same as

16

( )

(2a)

1 ∂ ⎡ 3 ∂ ⎛ rv ⎞ ⎤ ⎢r µ ⎜ 2 ⎟ ⎥ = r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ =

1 ∂ ⎡ 3 ⎛ 1 ∂(rv) ⎞ rv ⎤ −2 2 ⎥ = ⎢r µ ⎜ 2 ⎟ r ∂r ⎣ r ⎦ ⎝ r ∂r ⎠

=

1 ∂ ⎡ ∂(rv) ⎤ 2 ∂ ⎛ 3 rv ⎞ r µ ⎟= ⎢r µ ⎥− r ∂r ⎣ ∂r ⎦ r ∂r ⎜⎝ r3 ⎠

=

1 ∂ ⎡ ∂(rv) ⎤ 2µ ∂ rv ⎢r µ ⎥− r ∂r ⎣ ∂r ⎦ r ∂r

( )

Therefore:

ρw

∂ ∂ ∂ ⎛ ∂(rv) ⎞ 1 ∂ ⎡ 3 ∂ ⎛ v ⎞ ⎤ rv + ρu rv = ⎜ µ + ⎢r µ ⎜ ⎟ ⎥ ∂z ∂r ∂z ⎝ ∂z ⎟⎠ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦

( )

( )

(2b)

i.e. we end up with the equation below ρu

∂( rv ) ∂( rv ) ∂( rv ) ⎞ 1 ∂ ⎛ 3 ∂ ⎛ v⎞⎞ ∂ ⎛ + ρw = r Γφ r ⎜ ⎟ ⎟ + Γφ z ⎜ ⎜ ∂r ∂z r ∂r ⎝ ∂r ⎝ r ⎠ ⎠ ∂z ⎝ ∂z ⎟⎠

(C)

This is a non-standard on account of the form of the radial diffusion term, but has the (slight) advantage over (2a) in that there is no source term. Using a conservative discretization scheme, there should be no difficulties in ensuring equality of the boundary fluxes, i.e. that torques are balanced in the converged solution.

Divide (A) by r: this is a dangerous practice since we are dividing both sides by zero as r → 0 we have:

ρw

∂ ⎛ v⎞ u ∂v ∂ ⎛ ∂ ⎛ v ⎞ ⎞ 1 ∂ ⎛ ∂v ⎞ uv v + ρ = ⎜ µ ⎜ ⎟ ⎟ + 2 ⎜ rµ ⎟ − ρ 2 − µ 3 ⎜ ⎟ ∂z ⎝ r ⎠ r ∂r ∂z ⎝ ∂z ⎝ r ⎠ ⎠ r ∂r ⎝ ∂r ⎠ r r

Now ∂ ⎛ v ⎞ 1 ∂v v = − ∂r ⎜⎝ r ⎟⎠ r ∂r r 2

Hence:

ρ

u ∂v ∂ ⎛ v⎞ uv = ρu ⎜ ⎟ + ρ 2 r ∂r ∂r ⎝ r ⎠ r

17

Therefore:

ρw

∂ ⎛ v⎞ ∂ ⎛ v ⎞ ∂ ⎡ ∂ ⎛ v ⎞ ⎤ 1 ∂ ⎛ ∂v ⎞ uv v + ρ u = ⎢ µ ⎜ ⎟ ⎥ + 2 ⎜ r µ ⎟ − 2ρ 2 − µ 3 ⎜ ⎟ ⎜ ⎟ ∂z ⎝ r ⎠ ∂r ⎝ r ⎠ ∂z ⎣ ∂z ⎝ r ⎠ ⎦ r ∂r ⎝ ∂r ⎠ r r

Let T=

1 ∂ ⎛ ∂v ⎞ v rµ ⎟ − µ 3 2 ⎜ r ∂r ⎝ ∂r ⎠ r

Since ∂ ⎛ v ⎞ 1 ∂v v = − , ∂r ⎜⎝ r ⎟⎠ r ∂r r 2 r

∂v ∂ ⎛ v⎞ = r2 ⎜ ⎟ + v ∂r ∂r ⎝ r ⎠

Hence:

⎤ 1 ∂⎡ ∂ ⎛v ⎞ v T = 2 ⎢r 2µ ⎜ ⎟ + µv ⎥ − µ 3 ∂r ⎝ r ⎠ ⎥⎦ r r ∂r ⎢⎣ 1 ∂⎡ ∂ ⎛ v ⎞⎤ 1 ∂ v = 2 ⎢r 2µ ⎜ ⎟⎥ + 2 µv − µ 3 ∂r ⎝ r ⎠⎥⎦ r ∂r r ∂r ⎢⎣ r

( )

=

1 ∂⎡ ∂ ⎛ v ⎞⎤ 1 ∂ v r ⋅r µ µv − µ 3 ⎢ ⎜ ⎟⎥ + 2 2 ∂r ⎝ r ⎠⎥⎦ r ∂r r ∂r ⎢⎣ r

=

1⎡ ∂⎛ ∂ ⎛ v⎞⎞ ⎤ 1 ∂ ⎛ v⎞ 1 ∂ v r r µ µv − µ 3 = ⎢ ⎥+ 2 r µ ⎜ ⎟ + 2 ⎜ ⎟ 2 ⎜ ⎟ ∂r ⎝ r ⎠ r ∂r r ⎢⎣ ∂r ⎝ ∂r ⎝ r ⎠ ⎠ ⎥⎦ r r

=

1 ∂ ⎡ ∂ ⎛ v ⎞ ⎤ µ ∂ ⎛ v ⎞ µ ∂v v + 2 −µ 3 = ⎢r µ ⎜ ⎟ ⎥ + 2 ⎜ ⎟ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ r ∂r ⎝ r ⎠ r ∂r r

=

1 ∂ ⎡ ∂ ⎛ v ⎞ ⎤ µ ∂ ⎛ v ⎞ µ ⎛ ∂v v ⎞ r µ + − = ⎢ ⎥+ r 2 ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦ r ∂r ⎜⎝ r ⎟⎠ r 2 ⎜⎝ ∂r r ⎟⎠

=

1 ∂ ⎡ ∂ ⎛ v⎞⎤ µ ∂ ⎛ v⎞ µ ∂ ⎛ v⎞ r µ + + r = ⎢ ⎥ r 2 ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦ r ∂r ⎜⎝ r ⎟⎠ r 2 ∂r ⎜⎝ r ⎟⎠

=

1 ∂ ⎡ ∂ ⎛ v ⎞ ⎤ 2µ ∂ ⎛ v ⎞ r µ ⎢ ⎥+ r 2 ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦ r ∂r ⎜⎝ r ⎟⎠

( )

( )

18

Therefore the full equation reads:

ρw

∂ ⎛ v⎞ ∂ ⎛ v⎞ ∂ ⎡ ∂ ⎛ v⎞⎤ 1 ∂ ⎡ ∂ ⎛ v⎞⎤ µ ∂ ⎛ v⎞ uv + ρ u = µ + r µ − 2ρ 2 ⎢ ⎥ ⎢ ⎥+2 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ∂z ⎝ r ⎠ ∂r ⎝ r ⎠ ∂z ⎣ ∂z ⎝ r ⎠ ⎦ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ r ∂r ⎝ r ⎠ r

(3)

i.e. we have ρu

∂ ⎛ v⎞ ∂ ⎛ v⎞ 1 ∂ ⎛ ∂ ⎛ v⎞⎞ ∂ ⎛ ∂ ⎛ v⎞⎞ µ ∂ ⎛ v⎞ uv r Γφ r ⎜ ⎟ ⎟ + Γ φ z ⎜ ⎟ ⎟ −2 ⎜ ⎟ + ρ w ⎜⎝ ⎟⎠ = ⎜ ⎟ − 2ρ 2 ∂r ⎝ r ⎠ ∂z r r ∂r ⎜⎝ ∂r ⎝ r ⎠ ⎠ ∂z ⎜⎝ ∂z ⎝ r ⎠ ⎠ r ∂r ⎝ r ⎠ r

19

(D)

2.7 (A)

Tangential velocity v

ρw

∂v ∂v ∂ ⎛ ∂v ⎞ 1 ∂ ⎛ ∂v ⎞ uv v + ρu = µ ⎟+ rµ ⎟ − ρ − µ 2 ⎜ ⎜ ∂z ∂r ∂z ⎝ ∂z ⎠ r ∂r ⎝ ∂r ⎠ r r

(I)

(II)

(III)

(IV)

(V)

(VI)

At the axis of symmetry, v = 0, therefore (I) and (III) are zero.

Let 𝑣 = 𝑎𝑟 + 𝑏𝑟 ! + 𝑐𝑟 ! +…

→ 𝑣/𝑟 = 𝑎 + 𝑏𝑟 + 𝑐𝑟 ! +…

𝑢 = 𝑑𝑟 + 𝑒𝑟 ! + 𝑓𝑟 ! +…

→ 𝑢/𝑟 = 𝑑 + 𝑒𝑟 + 𝑓𝑟 ! +…

!" !"

As

= 𝑎 + 2𝑏𝑟 + 3𝑐𝑟 ! +…

𝑟 → 0, 𝐼𝐼 → 𝜌𝑑𝑟𝑎 and 𝑉 → 𝜌𝑑𝑟𝑎

therefore, II and V are equal but opposite. Both terms ∝ 𝑟.

IV:

VI:

!"

𝑟 !" = 𝑎𝑟 + 2𝑏𝑟 ! + 3𝑐𝑟 ! + …

→

! !

!"

!

! !"

!"

𝑟 !" = 𝑎 + 4𝑏𝑟 + 9𝑐𝑟 ! + ...

! !"

𝑟 !" = ! + 4𝑏 + 9𝑐𝑟 +…

i.e. 𝐼𝑉 → ∞ as 𝑟 → 0

!

!

i.e. 𝑉𝐼 → ∞ as 𝑟 → 0

!!

= ! + 𝑏 + 𝑐𝑟 +…

Therefore, as 𝑟 → 0, IV and VI both → ∞ in such a way that the difference tend to a constant (3b): however, the terms are treated differently (one is a diffusion term and one is a source term), so it is likely that a net source term arises from the two combined, giving rise to irregular behaviour near an axis of symmetry.

(B)

Angular momentum, rv ⎤ µ ∂ ∂ ∂ ∂ ⎛ ∂rv ⎞ 1 ∂ ⎡ ∂ ρw rv + ρu rv = ⎜ µ + r µ rv rv ⎢ ⎥−2 ∂z ∂r ∂z ⎝ ∂z ⎟⎠ r ∂r ⎣ ∂r r ∂r ⎦

( )

(I)

( )

(II)

( )

(III)

(IV)

( )

(V)

𝑟𝑣 = 0 at the axis, therefore terms I and II do not matter.

II:

𝑟𝑣 = 𝑎𝑟 ! + 𝑏𝑟 ! + 𝑐𝑟 ! +... ! !"

𝑟𝑣 = 2𝑎𝑟 + 3𝑏𝑟 ! + 4𝑐𝑟 ! +… !

𝜌𝑢 !" 𝑟𝑣 = 𝜌(𝑑𝑟 + 𝑒𝑟 ! + 𝑓𝑟 ! +. . . )(2𝑎𝑟 + 3𝑏𝑟 ! + 4𝑐𝑟 ! +…)

20

Therefore, as 𝑟 → 0, 𝐼𝐼 → 2𝜌𝑑𝑟 ! i.e. 𝐼𝐼 ∝ 𝑟 ! (no problem) Hence II = 0 at axis of symmetry and IV and V have therefore to balance.

IV:

𝑟𝑣 = 𝑎𝑟 ! + 𝑏𝑟 ! + 𝑐𝑟 ! +... !

𝑟𝑣 = 2𝑎𝑟 + 3𝑏𝑟 ! + 4𝑐𝑟 ! +…

!" !

𝑟 !" 𝑟𝑣 = 2𝑎𝑟 ! + 3𝑏𝑟 ! + 4𝑐𝑟 ! +… !

!

!"

𝑟 !" 𝑟𝑣

= 4𝑎𝑟 + 9𝑏𝑟 ! + 16𝑐𝑟 ! +…

! !

!

= 4𝑎 + 9𝑏𝑟 + 16𝑐𝑟 ! +…

𝑟 !" 𝑟𝑣 ! !"

V:

(constant as 𝑟 → 0)

!

− ! 2𝑎𝑟 + 3𝑏𝑟 ! + 4𝑐𝑟 ! +. . . = −4𝑎 − 6𝑏𝑟 − 8𝑐𝑟 ! +. .. (constant as 𝑟 → 0)

Therefore, as 𝑟 → 0, terms IV and V are equal and opposite (and constant). The difference is proportional to 𝑟! .

Again, the problem arises that the diffusion and source terms are treated differently, so that a net source term may arise from the combination, but at least the net source term will be small.

(C) ρu

∂( rv ) ∂( rv ) ∂( rv ) ⎞ 1 ∂ ⎛ 3 ∂ ⎛ v⎞⎞ ∂ ⎛ + ρw = r Γφ r ⎜ ⎟ ⎟ + Γφ z ∂r ∂z r ∂r ⎜⎝ ∂r ⎝ r ⎠ ⎠ ∂z ⎜⎝ ∂z ⎟⎠

v = ar + br 2 + cr 3 +… u = dr + er 2 + fr 3 +… st

The only new term is the 1 right hand side below

T=

1 ∂ ⎡ 3 ∂ ⎛ v⎞⎤ ⎢r ⎥ r ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦

v = a + br + cr 2 r ∂ ⎛ v⎞ = b + 2cr ∂r ⎜⎝ r ⎟⎠ r3

∂ ⎛ v⎞ = br 3 + 2cr 4 ⎜ ⎟ ∂r ⎝ r ⎠

21

∂ ⎡ 3 ∂ ⎛ v⎞⎤ 2 3 ⎢r ⎥ = 3br + 8cr ⎜ ⎟ ∂r ⎣ ∂r ⎝ r ⎠ ⎦ 1 ∂ ⎡ 3 ∂ ⎛ v⎞⎤ 2 ⎢r ⎥ = 3br + 8cr ∝r ⎜ ⎟ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ (D)

Angular velocity, v/r ∂ ⎛ v⎞ ∂ ⎛ v⎞ ∂ ⎡ ∂ ⎛ v⎞⎤ 1 ∂ ⎡ ∂ ⎛ v⎞⎤ µ ∂ ⎛ v⎞ uv ρ w ⎜ ⎟ + ρu ⎜ ⎟ = ⎢ µ ⎜ ⎟ ⎥ + − 2ρ 2 ⎢r µ ⎜ ⎟ ⎥ + 2 ⎜ ⎟ ∂z ⎝ r ⎠ ∂r ⎝ r ⎠ ∂z ⎣ ∂z ⎝ r ⎠ ⎦ r ∂r ⎣ ∂r ⎝ r ⎠ ⎦ r ∂r ⎝ r ⎠ r (I)

(II)

(III)

(IV)

(V)

v/r is not zero on an axis of symmetry, so none of the terms are zero at r = 0.

v = a + br + cr 2 + ... r ∂ ⎛ v⎞ = b + 2cr + ... ∂r ⎜⎝ r ⎟⎠

𝐼𝐼 → 𝜌 𝑑𝑟 + 𝑐𝑟 ! 𝑏 + 2𝑐𝑟 = 𝜌𝑏𝑑𝑟 +…

i.e. ∝ 𝑟!

𝑉𝐼 → −2𝜌𝑑𝑎 constant i.e. ∝ 𝑟 ! - terms are unequal v = a + br + cr 2 + ... r

IV:

∂ ⎛ v⎞ = b + 2cr + ... ∂r ⎜⎝ r ⎟⎠ r

∂ ⎛ v⎞ = br + 2cr 2 + ... ⎜ ⎟ ∂r ⎝ r ⎠

∂ ⎡ ∂ ⎛ v⎞⎤ ⎢r ⎥ = b + 4cr + ... ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦ 1 ∂ ⎡ ∂ ⎛ v⎞⎤ b ⎢r ⎥ = + 4c + .. r ∂r ⎣ ∂r ⎜⎝ r ⎟⎠ ⎦ r 𝐼𝑉 → ∞ as 𝑟 → 0 V:

! !

!!

𝑏 + 2𝑐𝑟 = ! + 4𝑐 +… i.e.

𝑉 → ∞ as 𝑟 → 0

22

(VI)

Terms IV and V are both infinite at the axis of symmetry and both are positive. There is nothing in the rest of the equation to balance these terms (this is a consequence of dividing equation (1) by r as 𝑟 → 0).

23

2.8 Momentum Equations (three-dimensional, variable fluid properties) 𝜌

𝑇!" = −𝑝𝛿!" + 𝜇 !!

!

𝐷𝑢! 𝜕𝑇!" = + 𝐵𝑜𝑑𝑦 𝐹𝑜𝑟𝑐𝑒 𝐷𝑡 𝜕𝑥!

𝜕𝑢! 𝜕𝑢! 2 2 𝜕𝑢! 𝜕𝑢! 2 + − 𝛿!" ∇ ∙ 𝒗 − 𝜌𝑢𝛿!" + 𝜇 ! + − 𝛿 ∇∙𝒗 𝜕𝑥! 𝜕𝑥! 3 3 𝜕𝑥! 𝜕𝑥! 3 !" !

!

.·. 𝜌 !"! = − !! 𝑝 + ! 𝜌𝑘 𝛿!" + !! 𝜇!"" !

!

!!! !!!

!!

!

+ !!! − ! 𝛿!" ∇ ∙ 𝒗 + 𝐵𝑜𝑑𝑦 𝐹𝑜𝑟𝑐𝑒 !

Cartesian 2 𝜕𝑢! 2 𝑇!! = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − ∇∙𝒗 3 𝜕𝑥! 3 2 𝜕𝑢! 2 𝑇!! = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − ∇∙𝒗 3 𝜕𝑥! 3 2 𝜕𝑢! 2 𝑇!! = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − ∇∙𝒗 3 𝜕𝑥! 3 𝑇!" = 𝜇!""

𝜕𝑢! 𝜕𝑢! + 𝜕𝑥! 𝜕𝑥!

𝑇!" = 𝜇!""

𝜕𝑢! 𝜕𝑢! + 𝜕𝑥! 𝜕𝑥!

𝑇!" = 𝜇!""

𝜕𝑢! 𝜕𝑢! + 𝜕𝑥! 𝜕𝑥!

𝜕𝑢! 𝜕𝑢 ≡ 𝜕𝑥! 𝜕𝑥

𝜕𝑢! 𝜕𝑣 ≡ 𝜕𝑥! 𝜕𝑟

𝜕𝑢! 1 𝜕𝑤 𝑣 ≡ + 𝜕𝑥! 𝑟 𝜕𝜃 𝑟

𝜕𝑢! 𝜕𝑢 ≡ 𝜕𝑥! 𝜕𝑟

𝜕𝑢! 𝜕𝑣 ≡ 𝜕𝑥! 𝜕𝑥

𝜕𝑢! 1 𝜕𝑢 ≡ 𝜕𝑟! 𝑟 𝜕𝜃

𝜕𝑢! 𝜕𝑤 ≡ 𝜕𝑥! 𝜕𝑥

𝜕𝑢! 1 𝜕𝑣 𝑤 ≡ − 𝜕𝑥! 𝑟 𝜕𝜃 𝑟

𝜕𝑢! 𝜕𝑤 ≡ 𝜕𝑥! 𝜕𝑟

Note, in what follows a non-standard notation is used for velocity where u, v and w are used to represent the axial, radial and tangential velocity components, respectively.

24

Polars 2 𝜕𝑢 2 𝑇!! = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − ∇∙𝒗 3 𝜕𝑥 3 2 𝜕𝑣 2 𝑇!! = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − ∇∙𝒗 3 𝜕𝑟 3 2 2 𝜕𝑤 𝑣 2 𝑇!! = −𝑝 + 𝜌𝑘 + 𝜇!"" +2 − ∇∙𝒗 3 𝑟 𝜕𝜃 𝑟 3 𝑇!" = 𝜇!""

𝜕𝑢 𝜕𝑣 + 𝜕𝑟 𝜕𝑥

𝑇!" = 𝜇!""

1 𝜕𝑢 𝜕𝑤 + 𝑟 𝜕𝜃 𝜕𝑥

𝑇!" = 𝜇!""

1 𝜕𝑣 𝑤 𝜕𝑤 − + 𝑟 𝜕𝜃 𝑟 𝜕𝑟

Standard-form 𝜌

𝜕𝜙 𝜕 𝜕 1 𝜕 𝜕𝜙 + 𝜌𝑢! − 𝛤!,! + 𝑟 𝜌𝑣! − 𝛤!,! 𝜕𝑡 𝜕𝑥 𝜕𝑥 𝑟 𝜕𝑟 𝜕𝑟

+

𝛤!,! 𝜕𝜙 1 𝜕 𝜌𝑤! − 𝑟 𝜕𝜃 𝑟 𝜕𝜃

+ 𝑆!

Axial (u): 𝜕𝑇!! 1 𝜕 1 𝜕 + 𝑟𝑇!" + 𝑇 𝜕𝑥 𝑟 𝜕𝑟 𝑟 𝜕𝜃 !" 𝜕 2 𝜕 𝜕𝑢 2 = − 𝑝 + 𝜌𝑘 + 𝜇!"" 2 − 𝛻∙𝒗 𝜕𝑥 3 𝜕𝑥 𝜕𝑥 3 1 𝜕 𝜕𝑢 𝜕𝑣 1 𝜕 1 𝜕𝑢 𝜕𝑤 + 𝑟𝜇!"" + + 𝜇!"" + 𝑟 𝜕𝑟 𝜕𝑟 𝜕𝑥 𝑟 𝜕𝜃 𝑟 𝜕𝜃 𝜕𝑥 Hence: 𝛤!,! = 2𝜇!"" 𝛤!,! = 𝜇!"" 𝛤!,! = 𝜇!"" 𝑆! = −

𝜕 2 1 𝜕 𝜕𝑣 1 𝜕 𝜕𝑤 𝜕 2 𝑝 + 𝜌𝑘 + 𝑟𝜇!"" + 𝜇!"" − 𝜇 𝛻∙𝒗 𝜕𝑥 3 𝑟 𝜕𝑟 𝜕𝑥 𝑟 𝜕𝜃 𝜕𝑥 𝜕𝑥 3 !""

Radial (v): 25

𝜕𝑇!" 1 𝜕 1 𝜕 𝑇!! + 𝑟𝑇!! + 𝑇!" − 𝜕𝑥 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕 𝜕𝑢 𝜕𝑣 = 𝜇!"" + 𝜕𝑥 𝜕𝑟 𝜕𝑥 1 𝜕 2 𝜕𝑣 2 + −𝑟 𝑝 + 𝜌𝑘 + 𝑟𝜇!"" 2 − 𝛻∙𝒗 𝑟 𝜕𝑟 3 𝜕𝑟 3 1 𝜕 1 𝜕𝑣 𝑤 𝜕𝑤 + 𝜇!"" − + 𝑟 𝜕𝜃 𝑟 𝜕𝜃 𝑟 𝜕𝑟 𝜇!"" 2 𝜕𝑤 2𝑣 2 − + − 𝛻∙𝒗 𝑟 𝑟 𝜕𝜃 𝑟 3

!

−

−𝑝 + ! 𝜌𝑘 𝑟

Hence: 𝛤!,! = 𝜇!"" 𝛤!,! = 2𝜇!"" 𝛤!,! = 𝜇!"" !

Call −𝑝 + ! 𝜌𝑘

𝜎: 1 𝜕 𝜎 𝜕𝜎 𝜕 2 𝑟𝜎 − = = −𝑝 + 𝜌𝑘 𝑟 𝜕𝑟 𝑟 𝜕𝑟 𝜕𝑟 3 !

!

Also ∇ ∙ 𝒗 terms give − !" ! 𝜇!"" ∇ ∙ 𝒗

𝑆! = −

𝜇!"" 𝜕𝑤 𝜕 2 𝜕 𝜕𝑢 𝑣 𝜕 2 𝑝 + 𝜌𝑘 + 𝜇!"" − 2𝜇!"" ! − 𝜇!"" 𝛻 ∙ 𝒗 − 2 ! 𝜕𝑟 3 𝜕𝑥 𝜕𝑟 𝑟 𝜕𝑟 3 𝑟 𝜕𝜃 1 𝜕 𝜕 𝑤 + 𝜇!"" 𝑟 𝑟 𝜕𝜃 𝜕𝑟 𝑟

𝑤! +𝜌 𝑟

Circumferential (w):

26

𝜕 1 𝜕 1 𝜕 𝑇!" 𝑇!" + 𝑟𝑇!" + 𝑇!! + 𝜕𝑥 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕 1 𝜕𝑢 𝜕𝑤 1 𝜕 !

1 𝜕𝑣 𝑤 𝜕𝑤 = 𝜇!"" + + ! 𝑟 𝜇!"" − + 𝜕𝑥 𝑟 𝜕𝜃 𝜕𝑥 𝑟 𝜕𝑟 𝑟 𝜕𝜃 𝑟 𝜕𝑟 1 𝜕 2 2 𝜕𝑤 2𝑣 2 + − 𝑝 + 𝜌𝑘 + 𝜇!"" + − ∇∙𝒗 𝑟 𝜕𝜃 3 𝑟 𝜕𝜃 𝑟 3 𝜕 𝜇!"" 𝜕𝑢 𝜕 𝜕𝑤 1 𝜕 𝜕𝑣 = + 𝜇!"" + ! 𝑟𝜇!"" 𝜕𝑥 𝑟 𝜕𝜃 𝜕𝑥 𝜕𝑥 𝑟 𝜕𝑟 𝜕𝜃 1 𝜕 ! 𝜕 𝑤 1 𝜕 2 1 𝜕 𝜕𝑤 + 𝑟 𝜇!!! ( ) − 𝑝 + 𝜌𝑘 + !

2𝜇!"" 𝑟 𝜕𝑟 𝜕𝑟 𝑟 𝑟 𝜕𝜃 3 𝑟 𝜕𝜃 𝜕𝜃 1 𝜕 1 𝜕 2 + !

2𝜇!"" 𝑣 + (− ∇ ∙ 𝒗) 𝑟 𝜕𝜃 𝑟 𝜕𝜃 3 Multiply equation by (r) to obtain one for (rw): Hence: 𝛤!,! = 𝜇!"" 𝛤!,! = 𝜇!"" 𝛤!,! = 2𝜇!"" 𝑆! = −

𝜕 2 𝜕 𝜕𝑢 1 𝜕 𝜕𝑣 2 𝜕 𝑝 + 𝜌𝑘 + 𝜇!"" + 𝑟𝜇!"" + 𝜇 𝑣 𝜕𝜃 3 𝜕𝑥 𝜕𝜃 𝑟 𝜕𝑟 𝜕𝜃 𝑟 𝜕𝜃 !"" 𝜕 2 − 𝜇 ∇∙𝒗 𝜕𝜃 3 !""

27

2.9 The procedure for this computational exercise is detailed in the text. Note, although MATLAB might be convenient it is probably non-optimal for the later computational exercises. For these a language such as FORTRAN is preferable.

2.10 The procedure for this computational exercise is detailed in the text.

28

CHAPTER 3

3.1 Please see equations (1)-(3) of Thompson et al. (1974) and accompanying text for the full equations.

The general form of the solution is given below. We can state the chain rule of differential calculus as below to transform derivatives between the physical and computational planes. 𝜕 = 𝜕𝑥!

!

!!!

𝜕! = 𝜕𝑥! 𝜕𝑥!

𝜕 𝜕𝜉! 𝜕𝜉! 𝜕𝑥! !

!

!!! !!!

𝜕 ! 𝜕𝜉! 𝜕𝜉! 𝜕𝜉! 𝜉! 𝜕𝑥! 𝜕𝑥!

Note, in the equation above there are no cross terms i.e. it is explicitly assumed that there is an orthogonal system (the grid generated will be orthogonal). The above expands out to 𝜕! 𝜕 ! 𝜕𝜉! ! 𝜕! 𝜕𝜉! 𝜕𝜉! 𝜕 ! 𝜕𝜉! ! = + 2 + 𝜕𝜉! 𝜕𝜉! 𝜕𝑥! 𝜕𝑥! 𝜕𝑥!! 𝜕𝜉!! 𝜕𝑥! 𝜕𝜉!! 𝜕𝑥! and 𝜕! 𝜕 ! 𝜕𝜉! ! 𝜕! 𝜕𝜉! 𝜕𝜉! 𝜕 ! 𝜕𝜉! ! = +2 + ! 𝜕𝜉! 𝜕𝜉! 𝜕𝑥! 𝜕𝑥! 𝜕𝑥!! 𝜕𝜉!! 𝜕𝑥! 𝜕𝜉! 𝜕𝑥!

Therefore the Laplacian becomes 𝜕! ∇ = ! 𝜕𝜉! !

𝜕! + ! 𝜕𝜉!

𝜕𝜉! ! 𝜕𝜉! ! 𝜕! 𝜕𝜉! 𝜕𝜉! 𝜕𝜉! 𝜕𝜉! + +2 + 𝜕𝑥! 𝜕𝑥! 𝜕𝜉! 𝜕𝜉! 𝜕𝑥! 𝜕𝑥! 𝜕𝑥! 𝜕𝑥! 𝜕𝜉! ! 𝜕𝜉! ! + 𝜕𝑥! 𝜕𝑥!

Hence, in two-dimensions we are left with the following equations to be solved

29

𝜕 ! 𝑥! 𝜕 ! 𝑥! 𝛼 +𝛽 =𝑺 𝜕𝜉!! 𝜕𝜉!! 𝛼

𝜕 ! 𝑥! 𝜕 ! 𝑥! + 𝛽 =𝑺 𝜕𝜉!! 𝜕𝜉!!

i.e. 𝜕!𝜙 𝜕!𝜙 𝛼 ! +𝛽 ! =𝑺 𝜕𝜉 𝜕𝜂 where 𝜙 = 𝑥 or 𝑦.

Note, in the above 𝜕𝜉! ! 𝜕𝜉! ! 𝛼= + 𝜕𝑥! 𝜕𝑥! 𝜕𝜉! ! 𝜕𝜉! ! 𝛽= + 𝜕𝑥! 𝜕𝑥! 𝑆 = −2

𝜕𝜉! 𝜕𝜉! 𝜕𝜉! 𝜕𝜉! 𝜕 ! 𝑥! + 𝜕𝑥! 𝜕𝑥! 𝜕𝑥! 𝜕𝑥! 𝜕𝜉! 𝜕𝜉!

Note, for a complete set of useable equations α, β and the source term above needs the application of the chain-rule to convert the derivatives with respect to the physical coordinates into differentials with respect to the transformed coordinates.

30

3.2 The equations are restated below 𝜕!𝜉 𝜕!𝜉 + = 𝑆! (𝜉, 𝜂) 𝜕𝑥 ! 𝜕𝑦 ! !

!

𝜕 𝜂 𝜕 𝜂 + = 𝑆! (𝜉, 𝜂) 𝜕𝑥 ! 𝜕𝑦 !

Figure S.3.2.1 gives a schematic of the physical and computational planes.

Figure S.3.2.1 Schematic comparing grids in the physical and computational planes.

The above equations can be expressed generally as 𝜕!𝜙 𝜕!𝜙 + =𝑆 𝜕𝑥 ! 𝜕𝑦 !

In polar coordinates the above becomes 1 ∂ ⎛ ∂φ ⎞ 1 ∂ ⎛ 1 ∂φ ⎞ ⎜r ⎟ + ⎜ ⎟= S r ∂r ⎝ ∂r ⎠ r ∂θ ⎝ r ∂θ ⎠

31

(1)

For a uniform grid, θ = constant π, where if ξ ranges between zero and unity the constant = 2 π.

Hence, lets just focus on the radial mapping ( η = f(r) ). Therefore, we have 1 ∂ ⎛ ∂η ⎞ ⎜r ⎟ =S r ∂r ⎝ ∂r ⎠

For S = 0, the (easily derived) fundamental solution to the above is given by Equation (4.141) in the main text of the book having the form

η = c ln r Hence, uniform η values will not map back to uniform r distributions. Now lets consider the equation with the source term S = c2/η 1 ∂ ⎛ ∂η ⎞ c 2 ⎜r ⎟= r ∂r ⎝ ∂r ⎠ η

(2)

Note, the constant, c, having the dimensions of the inverse of length, has been added for dimensional consistency. The uniform mapping could be defined as

η=c r Hence, now let us see if this satisfies Equation (2). Using the above in this equation gives 1 ∂r c 2 = r ∂r η

i.e.

1 c = r η

i.e. η = c r satisfying Equation (2) and corresponding to a uniform mapping.

32

3.3 To generate the Figure 3.65a mesh, using basic transfinite interpolation, Equation (3.1) given in the text is sufficient. To define the geometry, as shown in Figure S3.3.1, two strings of points that define the upper (north - n) and lower (south - s) boundaries can be defined. These strings of points can be stored in the arrays xbn(i), xbs(i), ybn(i) and ybs(i). Then basic linear interpolation is all that is required. For a FORTRAN type procedure the following would be needed Do j = 0, nj Wf = η = j/nj x(i,j) = (1 - Wf)*xbs(i) +Wf*xbn(i) y(i,j) = (1 - Wf)*ybs(i) +Wf*ybn(i) end do

Or for an array ranging from 1 (FORTRAN default) Do j = 1, nj Wf = η = (j-1)/(nj-1) x(i,j) = (1 - Wf)*xbs(i) +Wf*xbn(i) y(i,j) = (1 - Wf)*ybs(i) +Wf*ybn(i) end do

In the above Wf corresponds to η in Equation (3.1). Also, with FORTRAN great care is needed to distinguish between and handle integer and real variables. Hence, the float command below is needed to convert the integers to default real values. Using this, for example

Wf = float(j)/float(nj)

33

xbn(i)

xbs(i) Figure S3.3.1. Simple transfinite interpolation based grid generation.

For laminar flow, to generate a non-uniform distribution of points, in a simple straight channel, Equation (3.13) in the text could be used. In FORTRAN the application of this equation would look as below where the L1 and L2 in the text correspond to L1 and L2 in the equations below. With FORTRAN L1 and L2 must be explicitly defined as real numbers.

******cosine grid do 10 j=1,nj y(j)=.5*(L2+L1)-.5*(L2-L1)*cos((j-.5)*pi/(nj)) 10

continue y(0)=y(1)-(y(1)-L1)*2 y(nj+1)=(L2-y(nj))*2+y(nj)

Note pi = π in the above. For a geometric expansion in a channel the code below could be used.

The code below is for a geometric expansion in a channel. Equation (3.15)

******geometric alpha=1.1 if(alpha.ne.1.)then dy(0)=(L2-L1)/(1.+2*(1.-alpha**((nj+1)/2-1))/(1-alpha)) 34

else dy(0)=(L2-L1)/float(nj) endif dy(1)=dy(0) do 50 j=2,jd-2 if(j.lt.(nj+1)/2)dy(j)=dy(j-1)*alpha if(j.eq.(nj+1)/2)dy(j)=dy(j-1) if(j.gt.(nj+1)/2)dy(j)=dy(j-1)/alpha 50

continue dy(nj)=dy(nj-1) if(n.eq.1)y(0)=L1-dy(0)/2. y(1)=dy(0)/2+L1 do 60 j=1,nj-1 y(j+1)=y(j)+dy(j)

60

continue y(jd)=y(nj)+dy(nj) endif

Note, in the above it is assumed that nj+1 is an even number. Also, the geometric and cosine grids are ideal for the grid generation needed for the questions 1-3 and 7-8 given in Chapter 4. Similar expressions can be written for the x direction. To adapt for the Figure 3.65 case the above y values should be scaled to range between zero and unity and used to replace Wf. Note, for the above the grid is stored in array locations that range between zero and nj or nj +1. For FORTRAN access to an array location zero must be explicitly defined. Alternatively, the range could be made to start from unity to some upper level.

To visualize the above mesh a simple initial expedient is to write the coordinates out in an order such that the mesh can be plotted as a single continuous line. ParaView is free flow visualization software.

This would read in a mesh written out in the plot3d format. – see Section 7.7 in Chapter 7. Please note, plot3d files can be quite daunting to read into ParaView and use of the native csv format is worth considering.

35

3.4 & 3.5 The vectors shown in Figure 3.71 can be defined as below VD1 = a1 i + b1 j where a1 = -(x(i+1,j+1) - x(i,j)) b1 = -(y(i+1,j+1) - y(i,j)) VD2 = a2 i + b2 j a2 = x(i+1,j) - x(i,j+1) b2 = y(i+1,j) - y(i,j+1) The area is equal to half the cross product of the above two vectors and hence Area =

1 (a1 b2 – a2 b1) 2

The above can be straightforwardly tested and applied to the mesh generated as part of Question 3.

Print the areas out for the Question 3 mesh to see that they match expectations and have the correct sign.

36

3.6 Looping over all the i and j subscripts For the west face (see Figure 3.78a) LIX(I,J) = Δy = = y(i,j+1) - y(i,j) LIY(I,J) = -Δx = -(x(i,j+1) - x(i,j)) For south face (see Figure 3.78b) LJY(i,j) = Δx = x(i+1,j) - x(i,j) LJX(i,j) = -Δy = -(y(i+1,j) - y(i,j)) Note, the above can evaluated just west and south faces the values for east and north will arise as a consequence. These values are simply the west and south values from adjacent cells in the x and y directions, respectively.

37

3.7 Basically we can test the edge areas per unit depth in the above by looking for conservation of mass flow. For mass conservation (in the steady state) the sum of the mass flows around the w, e, n and s faces of the control volume shown in Figure 3.78 should be zero. If we assume that the fluid has a constant density (say unity), then since the mass flow rate is the product of normal velocity times density times area we can just look for the conservation of velocity times area (volume flow rate).

However, if we assume the x and y velocities are unity we can just sum areas with an appropriate sign convention. Hence, looking at Figure 3.78 for u= v = 1 we have the following flows (LIXw + LIYw + LJYs + LJXs) – (LIXe + LIYe + LJYn + LJXn) = 0 Note, in the above there is a subtraction between date at the w and e faces and also the n and s. This is because, for example, at the e faces a flow for left to right is an outflow. The same flow at the w face is an inflow. For a rectangular control volume the above reduces to LIXw + LJYs – LIXe + LJYn = 0 with the obvious expectation of a summation to zero. It is important that the above edge projected areas and their signs carefully checked at this stage. Otherwise it could result in a lot of wasted time later. Every stage, when developing a computer code should be carefully checked.

38

3.8 This involves the solution of Equation (3.6) for φ = x and y with two different sources S. Note, these are coupled equations. Tyically, grid spacings in the computational plane would be assumed to be unity i.e. Δξ = Δη = 1. Since the computational plane is a square, with uniform grid spacings finite differences are the best solution approach for Equation (3.6). The form of these is shown in Section 4.2. Section 4.2.4 shows how to express a convection-diffusion equation in a matrix form. Setting u = v = 1 and S to desired values, based again on finite differences will allow Equation (3.6) to be discretized. Appendix C gives a TDMA solver to enable solution of the matrix system. See also Section 4.6 to use this. Alternatively a simple, Jacobi or Gauss-Siedel solution procedure can be used as a starting point. Again, see Section 4.6. The boundary conditions are the values used the array xb(i) (see Figure S3.1) and yb(i). Similar coordinate definitions can be used at the west and east boundaries.

Directly applying the finite difference equations given in Section 4.2 (equations (4.4 and (4.5)), Equation (3.8) becomes 𝛼=

𝑥!,!!! − 𝑥!,!!! ! 𝑦!,!!! − 𝑦!,!!! ! + 2 · ∆𝜂 2 · ∆𝜂

Equation (3.9) becomes 𝛽=

𝑥!!!,! − 𝑥!!!,! ! 𝑦!!!,! − 𝑦!!!,! ! + 2 · ∆𝜉 2 · ∆𝜉

The source term takes the following form

𝑆𝑥 = 2

𝑥!!!,! − 𝑥!!!,! 𝑥!,!!! − 𝑥!,!!! 𝑦!!!,! − 𝑦!!!,! 𝑦!,!!! − 𝑦!,!!! ∙ + ∙ 2 · ∆𝜉 2 · ∆𝜂 2 · ∆𝜉 2 · ∆𝜂 ·

𝑥!!!,!!! − 𝑥!!!,!!! − 𝑥!!!,!!! + 𝑥!!!,!!!

4 · ∆𝜉 · ∆𝜂

Equation (3.6) in its discrete form then looks like 𝛼

𝑥!!!,! − 2 ∙ 𝑥!,! + 𝑥!!!,! 𝑥!,!!! − 2 ∙ 𝑥!,! + 𝑥!,!!! +𝛽 = S! ! ∆𝜉 ∆𝜂 !

The above can be expressed in the matrix form for ‘x’ below where φ = x, 𝐴! 𝜙! + 𝐴! 𝜙! + 𝐴! 𝜙! + 𝐴! 𝜙! + 𝐴! 𝜙! = 𝑆!

39

where 𝐴! = 2(α + β) 𝐴! = −α 𝐴! = - α 𝐴! = - β 𝐴! = -β and 𝑆! = 𝛾 𝑥!!!,!!! − 𝑥!!!,!!! − 𝑥!!!,!!! + 𝑥!!!,!!! where 𝛾=

𝑥!!!,! − 𝑥!!!,! 𝑥!,!!! − 𝑥!,!!! 𝑦!!!,! − 𝑦!!!,! 𝑦!,!!! − 𝑦!,!!! ∙ + ∙ 2 · ∆𝜉 2 · ∆𝜂 2 · ∆𝜉 2 · ∆𝜂

The above can either be solved using simple Jacobi or Gauss-Siedel, for the latter iterating an equation of the form (4.154). Alternatively, TDMA can be used. A sample TDMA solver (subroutine tdmauy) is supplied as an appendix to the main book. Note, in this simultaneous equation solver fj = Sx sp = AE am = AW ac = AP Also, if the boundary conditions are absorbed into the equation coefficients n = nnode – 2 where nnode is the total number of nodes. If the boundary conditions are not absorbed into the equation coefficients n = nnode.

Note the subroutine tdpauz is a periodic TDMA algorithm for use with the Chapter 4 problems. With this one can imagine that the mesh nodes are wrapped around in a circle. However, there is no nodal

40

overlap. The boundary node is expecting to pick up data from the final node at the other end of the mesh.

If the equation system is unstable, the under-relaxation Equation (4.168) can be used. This can either be applied directly after the variable φP has been updated. Alternatively, Equation (4.168) can be made an integral part of the discretized equation system. If the equation system for x and y is slow to converge over-relaxation can be used in Equation (4.168). With this α

φ

> 1. Hence, rather than

variable changes being damped they are magnified. A similar discretized equation needs to be formulated for φ = y. Note it is a sensible idea to use the algebraically generated grid as a starting point for solving the above equation system. For an example of a working program please see Appendix A (Note, the code in Appendix A expects a data file – geom.dat - that has the upper (n) and lower boundaries (s) defined as a string of points).

Figure S3.8.1. Bi-linearly interpolated and elliptically generated grids for the supersonic ramp case (blue line is the elliptically generated grid and the red line the bi-linearly interpolated).

Figure S3.8.2. Bi-linearly interpolated and elliptically generated grids for the ‘bump case’ (blue line is the elliptically generated grid and the red line the bi-linearly interpolated).

41

Figure S3.8.1 contrasts the bi-linearly interpolated grid (see Figure 3.65) with an elliptically generated grid. The blue line is the elliptically generated grid and the red line the bi-linearly interpolated. Figure S3.8.2 makes a similar comparison for the ‘bump case’ also considered in the text. The smoother nature of the elliptically generated grid is evident.

3.9 Equations having a form akin to (3.52) can be used to generate finer grid spacings in areas where there are steep gradients. In practice this extension can prove challenging. Clearly, the relevant problems in Chapter 4 must be completed before this problem is tackled.

42

CHAPTER 4

4.1 The governing equation can be expressed as

∂φ ∂φ +u = 0 ∂t ∂x when discretized for u > 0 with first order upwinding, this becomes

φpn+1 − φpn Δt

+u

φpn − φwn

=0

Δx

(1)

We can express the Taylor series about xi, tn as below

φ

n+1 p

n

n

n

p

p

p

n

n

n

p

p

p

⎛ ∂φ ⎞ (Δt )2 ⎛ ∂2φ ⎞ (Δt )3 ⎛ ∂3φ ⎞ = φ + Δt ⎜ ⎟ + + + ...

2 ⎜⎝ ∂t 2 ⎟⎠ 6 ⎜⎝ ∂t 3 ⎟⎠ ⎝ ∂t ⎠ n p

⎛ ∂φ ⎞ (Δx)2 ⎛ ∂2φ ⎞ (Δx)3 ⎛ ∂3φ ⎞ φ = φ − Δx ⎜ ⎟ + − + ...

2 ⎜⎝ ∂x 2 ⎟⎠ 6 ⎜⎝ ∂x 3 ⎟⎠ ⎝ ∂x ⎠ n w

n p

These expressions can then be substituted into the discretized equation to give n

n

p

p

n

n

n

n

⎛ ∂φ ⎞ ⎛ ∂φ ⎞ Δt ⎛ ∂2φ ⎞ (Δt )2 ⎛ ∂3φ ⎞ Δx ⎛ ∂2φ ⎞ (Δx)2 ⎛ ∂3φ ⎞ + u = − − + u − + ... ⎜⎝ ∂t ⎟⎠ ⎜⎝ ∂x ⎟⎠ 2 ⎜⎝ ∂t 2 ⎟⎠ 6 ⎜⎝ ∂t 3 ⎟⎠ 2 ⎜⎝ ∂x 2 ⎟⎠ 6 ⎜⎝ ∂x 3 ⎟⎠ p

p

p

(2)

p

Hence, in the above, the left hand side is the original differential equation and the right hand side the modified equation terms. Now we need to make the right hand side temporal derivatives spatial derivatives. To do this we differentiate Equation (2) with respect to time as below

∂2 φ ∂2 φ Δt ∂3φ (Δt )2 ∂4 φ Δx ∂3φ (Δx)2 ∂4 φ + u = − − + u − u + ... ∂x ∂t 2 ∂t 3 6 ∂t 4 2 ∂x 2 ∂t 6 ∂x 3 ∂t ∂t 2

(3)

We also need to differentiate Equation (2) with respect to x and multiple it by u 2 3 2 4 ∂2 φ Δt ∂3φ (Δt )2 ∂4 φ 2 ∂ φ 2 Δx ∂ φ 2 (Δx) ∂ φ u +u = −u −u +u −u + ... ∂t ∂x 2 ∂t 2 ∂x 6 ∂t 3 ∂x 2 ∂x 3 6 ∂x 4 ∂x 2

(4)

we then subtract Eqaution (4) from Equation (3) ignopring higher order terms to give the equation below 2 3 ⎤ Δx ⎡ ∂3φ ⎤ ∂2 φ Δt ⎡ ∂3φ ∂3 φ 2 ∂ φ 2 ∂ φ = u + − + u + O(Δt ) + u − u + O(Δx) ⎢ ⎥ ⎢ ⎥ 2 ∂t 2 ∂x 2 2 ⎣ ∂t 3 ∂t 2 ∂x ∂x 3 ⎦ 2 ⎣ ∂x ∂t ⎦

43

(5)

We then differentiate Equation (5) with respect to x and also t as below 3 ∂3 φ 2∂ φ = u + O[Δt ,Δx] ∂t 2 ∂x ∂x 3

(6)

3 ∂3 φ 2 ∂ φ =u + O[Δt ,Δx] ∂t 3 ∂x 2 ∂t

(7)

We also differentiate Equation (5) with respect to x

∂3 φ ∂3 φ = −u + O[Δt ,Δx] ∂x 2 ∂t ∂x 3

(8)

From equations (6) and (7) it can be seen that 3 ∂3 φ 3 ∂ φ = −u + O[Δt ,Δx] ∂t 3 ∂x 3

(9)

Since u=dx/dt or dt =dx/u this could have been concluded more simply and also identities (8) and (9).

Substituting equations (6), (8), (9) into Equation (5) gives 2 ∂2 φ ∂3 φ 2∂ φ 2 = u + u uΔt − Δx + O[Δt ,Δx] ∂t 2 ∂x 2 ∂x 3

(

)

(10)

Substituting equations (9) and (10) into Equation (2) gives the following modified equation

∂φ ∂φ Δt ⎡ ∂2φ ∂3φ ⎤ 3 (Δt )2 ∂3φ Δx ∂2φ (Δx)2 ∂3φ + u = −u2 ⎢ 2 + uΔt − Δx + u + u − u + ... ⎥ ∂t ∂x 2 ⎣ ∂x 6 ∂x 3 2 ∂x 2 6 ∂x 3 ∂x 3 ⎦

(

)

(11)

This can be written more compactly using the CFL number

C =u

Δt Δx

to give:

(

)

∂φ ∂φ Δx ∂2 φ (Δx)2 ∂3 φ 2 +u = u 1− C + u 3C − 2C −1 + ... ∂t ∂x 2 6 ∂x 2 ∂x 3

(

)

(12)

For the backward Euler (see Gresho et al. (1984)) the unsteady discretization contribution to the false diffusion term just changes sign hence giving greater stability. Gresho et al. (1984) give more compact version of the above analysis process (equations (18)-(29) in the text of Gresho et al). This approach allows more focus to be placed on the temporal integration scheme. This is more useful for the continuing part of this problem. The above demonstrates the broad process for the spatial derivatives.

44

Following the procedure of Gresho et al. we can write, for example, a convection-diffusion operator where represents diffusivity.

∂φ = ∇ i ⎡⎣ Γ i ∇ − u φ ⎤⎦ ∂t

(

)

(A)

Note, for the current problem Γ = 0 and ∇ ∙ 𝑢 = 0. We can define the spatial operator as

(

A ≡ ∇i Γ i∇−u

)

(B)

This allows us to write

∂φ = Aφ ∂t

(C)

Using Taylor’s series, and having 𝜙 ! at 𝑡 ! , the exact solution 𝜙(𝑡 !!! ) at 𝑡 !!! = 𝑡 ! + Δ𝑡 is n

n

φ(t

n+1

∂φ Δt 2 ∂2φ ) = φ + Δt + + O(Δt 3 ) 2 ∂t 2 ∂t n

(D)

Using Equation (B) and the above, ∂ ⎛ ∂φ ⎞ ∂ ∂ = Aφ = A φ = A2φ ⎜ ⎟ ∂t ⎝ ∂t ⎠ ∂t ∂t

( )

()

we can write

φ(t

n+1

( )

(Δt )2 ⎡ 2 n ⎤ ) = φ + Δt Aφ + A φ ⎥ + O(Δt )3 ⎢ 2 ⎣ ⎦

( )

n

n

(E)

The approximate solution of Equation (C) with the forward Euler method is

( )

φ n+1 = φ n + Δt A!φ where 𝐴

n

(F)

is an approximate operator. Equating 𝜙 !!! to 𝜙(𝑡 !!! ) allows us to estimate the

opproximate operator to 𝑂(Δ𝑡 ! ) as

( )

Δt 2 A! = A + A 2 Note

( )

A2φ ≡ A Aφ

and

45

(G)

⎡⎛ ⎛ ∂Γ ⎞ ∂A ∂u ⎞ ⎤ φ = ∇ i ⎢⎜ ⎜ ⎟ i ∇ − ⎟ φ ⎥ ∂t ∂t ⎠ ⎥⎦ ⎢⎣⎝ ⎝ ∂t ⎠ Assuming Γ is invariant with time and focusing on just parts of 𝐴 that give us the expected dissipative behaviour we have

Δt A!φ = Aφ + ∇ i ⎡⎣u∇ i uφ ⎤⎦ + O(Δt ) 2

( )

Or since ∇ ∙ 𝑢 = 0

Δt A!φ = Aφ + ∇ i uui ∇φ + O(Δt ) 2

(

)

Hence, we finally have

⎡⎛ ⎤ Δt ⎞ A!φ " ∇ i ⎢⎜ Γ + τ ⎟ i ∇φ − uφ ⎥ 2 ⎠ ⎣⎝ ⎦ where 𝜏!" = 𝑢! 𝑢! and hence ⎛ ∂φ Δt ⎞ + ui ∇φ = ∇ ⎜ Γ − τ ⎟ i ∇φ ∂t 2 ⎠ ⎝

(H)

Therefore the above shows the anti-diffusive term that tends to be associated with the use of many explicit temporal schemes and is consistent with the initial derivation.

If we repeat the above process for a backwards difference in time then Equation (D) becomes

φ(t ) = φ n

n+1

∂φ − Δt ∂t

n+1

Δt 2 ∂2φ + 2 ∂t 2

n+1

− O(Δt 3 )

Also, Equation (E) becomes

( )

φ(t n ) = φ n+1 − Δt Aφ

n+1

+

( )

(Δt )2 ⎡ 2 n+1 ⎤ A φ ⎥ − O(Δt )3 ⎢ 2 ⎣ ⎦

Equation (F) becomes

( )

φ n+1 = φ n + Δt A!φ Equation (G) becomes

( )

Δt 2 A! = A − A 2

46

n+1

Hence, there is now a change of sign. Carrying this form of expression forwards with the change of sign to the subsequent terms results in Equation (H) changing as follows ⎛ ∂φ Δt ⎞ + ui ∇φ = ∇ i ⎜ Γ + τ ⎟ i ∇ ∂t 2 ⎠ ⎝

i.e. there is now substantial additional diffusion which is associated with many implicit temporal schemes.

Next we consider the Crank-Nicolson scheme. Again using Taylor’s series we can write the following

⎛ ∂φ ⎞ φ n = φ n+1 − Δt ⎜ ⎟ ⎝ ∂t ⎠

n+1

(Δt )2 ⎛ ∂2φ ⎞ + 2 ⎜⎝ ∂t 2 ⎟⎠

φ

+ O(Δt )3

n

n

n+1

n+1

⎛ ∂φ ⎞ (Δt )2 ⎛ ∂2φ ⎞ = φ + Δt ⎜ ⎟ + + O(Δt )3 ⎜ 2⎟ 2 ⎝ ∂t ⎠ ⎝ ∂t ⎠ n

From the above it follows that Δt ⎡⎛ ∂φ ⎞ ⎛ ∂φ ⎞ = φ + ⎢⎜ ⎟ + ⎜ ⎟ 2 ⎢⎝ ∂t ⎠ ⎝ ∂t ⎠ ⎣ n

φ

n+1

n+1

n

⎤ (Δt )2 ⎡⎛ ∂2φ ⎞ n ⎛ ∂2φ ⎞ n+1 ⎤ ⎢ ⎥+ − ⎜ 2 ⎟ ⎥ + O(Δt )3 ⎜ 2 ⎟ 4 ⎢⎝ ∂t ⎠ ⎝ ∂t ⎠ ⎥ ⎥ ⎦ ⎣ ⎦

This is the equivlanet of Equation (D) above. Equation (E) becomes

φ

n+1

( ) ( )

n+1 n n+1 Δt ⎡ (Δt )2 ⎡ 2 n ⎤ + O(Δt )3 ⎤ 2 =φ + Aφ + Aφ + A φ − A φ ⎢ ⎥⎦ ⎢ ⎥ ⎦ 2 ⎣ 4 ⎣ n

( ) ( )

The approximate solution of Equation (C) with the second order Crank-Nicolson can be expressed as below (the equivalent of Equation (F) above)

φ n+1 = φ n +

(

)

Δt A φ n + φ n+1 + O(Δt )2 2

The equivalent of Equation (G) is

( )

Δt A! = A + ⎡⎢ A2φ 4 ⎣

explicit

( )

− A2φ

implicit

⎤ ⎥⎦

Hence, we now have both the diffusive and anti-diffusive elements found above in the same scheme.

47

4.2 Discussion on how this problem can be tackled is provided in the main text of the book. Note, for use of the TDMA solver please see discussion in the document relating to Solution 8 in Chapter 3. This case is a useful step towards solving Problem 7. Figure S.4.7.1 gives the program flow chart for Problem 7. It would be useful here if this program structure was initiated at this stage. Hence, for this problem the program structure would look like that shown in Figure S.4.2.1. Appendix B gives the form of a FORTRAN subroutine for solving a basic convection-diffusion transport equation (modifications for solving the transport of the staggered momentum variables are also given in Appendix B).

Read data Generate grid Initialize flow

Main solution loop

Setup u – momentum coefficients (v = 0, p = constant = 0) Simultaneous eqn. solver – u for n sweeps

No

Yes

Write out data

Figure S.4.2.1 Program flow chart for Couette flow solution.

To start with it would be good to start with by initializing the flow with the exact analytical solution (see Chapter 2). Also, as can be seen from the analytical solution the only active terms for this case should be the vertical diffusion terms i.e. ∂2v/∂y2 = 0. Hence, to start with the AE and AW coefficients of the discretized equation could be set to zero and all the convective terms in all directions ignored i.e. in essence this is a simple conduction of heat problem that will allow a more complex solver to be built up in careful stages.

48

4.3 As noted in the question a large part of this problem is tackled through addressing Question 2 above.

However, the key aspect is the inclusion of the temporal discretizations discussed in Section 4.8.

Cleary spatial and temporal schemes that are dissipative will rapidly round the edges of the cylinder and will cause its height to drop and for their to be a spread in the cylinder’s diameter.

4.4 The code development process is outlined in the text. Please also the solution to Question 7 and the accompanying text in Chapter 7. This will help with interpreting your results and the performance of your code.

4.5 Low Storage Runge-Kutta Scheme There are a range of options for implementing the low storage Runge-Kutta scheme and when the smoothing and other subroutines are called. However, Figure S.4.5.1 gives a potential modified (from Figure 4.41) program flow chart for the Runge-Kutta scheme. Values of Wf for different values of n are given in Table 4.8 in the main text body. A wide range of options can be found it the literature. Also, the number of stages can be experimented with. This scheme, when working should allow the use of much larger time steps and much lower values of smoothing. However, the 4 stages have a clear increase in computational cost.

49

Set basic variables u, v, p, h0 Update boundary conditions Feed in relevant Wf

Do n = 1 -> 4 Wf = ?

Evaluate fluxes

Main solution loop

Sum fluxes (Call, x4)

Continue as normal

Update solution (x 4) (ΔΦ = Δt Σ Flux/Area) Scatter cell centered ΔΦ to corners (x 4) Make primary solution variable update Φinew = Φiold + ΔΦi

Figure S.4.5.1. Potential modified program flow chart for the Runge-Kutta scheme.

Deferred (smoothing) Correction There are different ways of implementing the above approach.

In classic CFD form it would

involve say making a stable 1st order estimate of the solution and then adding on a correction based on a high order spatial discretization to yield stability and accuracy. A crude method proposed by Denton (2007) is outlined. With this we: ● Make more accurate the basic estimate [?]I = (1-ε) [?]unsmoothed + ε [?]smooth ● By adding the à correction = [?]unsmoothed - [?]smooth ● Hence we have [?]I = (1-ε) [?]unsmoothed + ε ([?]smooth + ‘correction’) 50

● The correction is introduced slowly (initial ‘correction’ = 0) ‘correction’ = 0.99 ‘correction’ + 0.01 correction Hence, the unsmoothed variable, the one we gain prior to smoothing is the accurate estimate. The smoothed update variable is the lower order and the difference is the higher order correction that can be added to the lower order. Initially the correction = 0 and so the solution migrates slowly to the deferred correction scheme.

Figure S.4.5.2 shows the difference between standard and deferred correction schemes for the ‘bump case’. The yellow lines are the deferred correction scheme and black lines are for the standard scheme. As can be seen the deferred correction has dramatically increased the accuracy.

Figure S.4.5.2. Difference between standard and deferred correction schemes for the ‘bump case’ (yellow lines are the deferred correction scheme and black lines are the standard scheme).

Residual Averaging In the solutions we have smoothed solved for variables, Φ, for stability. Essentially we are solving ΔΦ/Δt=Residual(=Σflux) At convergence Residual = 0. Hence, we can smooth R – like we did for Φ and not change answer.

This is residual averaging. It gives more stability allowing the use of a bigger Δt. Here, for convenience, it is possible smooth ΔΦ (ΔΦ α R) as the solution progresses.

4th Order Smoothing In the basic program we have been explicitly adding as smoothing 51

Smoothing = ∇ 2 ⎡⎣?⎤⎦ =

∂ 2 ⎡⎣?⎤⎦ ∂ 2 ⎡⎣?⎤⎦ + ∂x2 ∂y2

For greater accuracy we can use

4

Smoothing = ∇ ⎡⎣?⎤⎦ =

∂ 4 ⎡⎣?⎤⎦ ∂ 4 ⎡⎣?⎤⎦ + ∂x4 ∂y 4

As well is the geometric and Taylor series based methods for generating finite difference expressions outlined in the main body of the text polynomials can also be used to generate expressions (as noted in the text). The 4th order smoother hence can be practically coded by fitting a cubic curve through I-2, I-1, I+1 and I+2 – missing out I. We then form a uniform grid write an expression where the target value is [ ]i. The smoothing would do nothing if the numerical solution is exact for a cubic variable distribution. This makes the approach higher order. Hence, to do this the smoothing value at a node I,J can be expressed an average of the values from cubic fits through the values at nodes I-2, I-1, I+1, I+2 and also through J-2, J-1, J+1 and J+2. Therefore, the following can be written

At the walls forwards and backwards difference based expressions can be used

Please see also the solution to Question 7 and the accompanying text in Chapter 7. This will help with interpreting your results and the performance of your code.

52

4.6 In relation to the smoothing discussed above, commercial CFD codes generally have

Smoothing = ε1f1 ⎡⎣?⎤⎦ ∇ 2 ⎡⎣?⎤⎦ + ε2f2 ⎡⎣?⎤⎦ ∇ 4 ⎡⎣?⎤⎦

The 2nd order smoother is usually activated around shocks through f1[]. nd

Different weightings

th

between the 2 order and 4 order smoothing components can be tried the weighting on the 2nd order component being lower. A further refinement is to vary the smoothing factors with the second derivatives of the flow variables so that high smoothing is only applied when the second derivatives are large hence migrating to the use of a shock switch.

Please see also the solution to Question 7 and the accompanying text in Chapter 7. This will help with interpreting your results and the performance of your code.

4.7 The code development process is outlined in the text. Figure S.4.7.1 gives a program flow chart, storage locations and potential grid configurations for this task. Appendix C gives samples FORTRAN subroutine structures for evaluation the pressure correction equation coefficients and performing the velocity and pressure corrections. This problem links to Problem 2 above.

53

Read data Generate grid Initialize flow

Setup u or w- momentum coefficients

Grid structure and storage locations

Simultaneous eqn. solver – u or w for n sweeps Setup v or u-momentum coefficients

Main solution loop

Simultaneous eqn. solver – v or u for n sweeps

No

Update differential boundary conditions Yes

Write out data

for derivative conditions not in their own direction

Setup p’ – eqn. coefficients Simultaneous eqn solver– p’ Correct pressure & velocity ϕ = ϕ* + ϕ’

Figure S.4.7.1. Program flow chart, storage locations and potential grid configuration for incompressible, pressure based, Navier-Stokes flow solver.

4.8 The code development process is outlined in the text.

54

CHAPTER 5

5.1 The discussion in Section 5.2.4.1 is useful. Studying the paper of Nee and Kovasznay (1969) is also recommended. However, much of this paper focuses on solving equations in the days when CFD was in its infancy. The information in Section 5.2.3.1 on mixing length formulations is also important.

Ideally, mixing length expressions suitable for the inner and outer parts of the boundary layer are needed. The inner layer mixing length itself needs a multiple characterization. For channel flow cases, outlined in the problems in Chapter 4, the flow can be made fully developed, through the use of periodic boundary conditions or just more manually feeding outflow data back into the inflow periodically. This will mean the convection and diffusion terms in the transport equation will become inactive. Hence, the model is basically of the mixing length type. A long channel could be used to allow the flow to develop.

Initially, to quickly glimpse the impact of the turbulence model,

comparison could be made with the 1/7th power law profile for turbulent flow i.e. u/ Ucl = (y/δ)1/7 where δ is the boundary layer thickness. For fully developed flow this is equal to the channel half height. Note, also for fully developed flow Ucl is the channel centerline velocity, otherwise being the boundary layer edge. The profile should be seen to migrate from the laminar Blasius profile which can be approximated as u/ Ucl = sin(πy/(2δ)) . The first key test would be to understand how your model matched the standard ‘law of the wall’ velocity profile outlined in Section 5.11.5.

Alternatively, Equation (5.19) could be utilized and comparisons made with Equation (5.20). The critical parameter to plot the law of the wall is the wall shear stress. The local value can be gained by a simple first order one-sided difference. Since the wall velocity is zero, the wall shear stress is simply the velocity at the first off-wall grid node multiplied by the dynamic viscosity divided by the distance that the first off-wall node is from the wall. Equation (4.23) gives an alternative 2nd order backward difference equation for evaluating the wall shear stress. If you are using periodic boundary conditions to generate fully developed flow then a simple force balance can be made based on the axial pressure gradient treating the whole channel as a control volume (Δp × channel height = 2 × τw × channel length).

Figure S.5.1.1 gives a modified program flow chart when solving an additional

transport equation for turbulence. The expectation is that the code developed as part of Problem 7, Chapter 4 will be used and modified in accordance with Figure S.5.1.1.

55

Setup u or w- momentum coefficients Simultaneous eqn. solver – u or w for n sweeps etc

Simultaneous eqn solver– p’ Correct pressure & velocity ϕ = ϕ* + ϕ’ Setup turbulence eqn coefficients Simultaneous eqn. solver – µt for n sweeps

Figure S.5.1.1. Modified program flow chart when solving an additional transport equation for turbulence.

5.2 The k and ε equations can be expressed as below !! !! !! !!

!

= 𝒫 − 𝜀 + !!

!

!"

𝜈 + !!

!

!

!

= ! 𝐶!! 𝒫 − 𝐶!! 𝜀 + !!

!

! !!

, 𝑈𝜀! = − !!!

!"

=> 𝑈 !" = −𝜀

,

!!

! !!

𝑈 !" = − !!!

Divide: 𝑑𝑘 −𝜀 𝑘 = ! !! = 𝑑𝑥 − !! 𝐶!! 𝜀 !

=>

!"

!

!!

!!

!

=! !

!

=> ln 𝑘 = ! ln 𝜀 !!

56

!

!"

!

!!!

𝜈 + !!

In the free stream the above equations reduce to. 𝑈𝑘! = −𝜀

(1)

! !!!

(2)

=> 𝑘 !!! = 𝜀

(1)

Now consider !"# !"

=>

= − 𝜀 = −𝑘 !!!

!"

!"

! !!!

=−!

Integrate: ! !!!! !! !!!! !!

=> =>

!

!

= − ! + 𝐶𝑜𝑛𝑠𝑡. !

! !!!

= ! 𝐶!! − 1 + 𝐶𝑜𝑛𝑠𝑡.

!

!

!

= ! 𝐶!! − 1 + 𝐶𝑜𝑛𝑠𝑡.

@ 𝑥 = 0 , 𝐶𝑜𝑛𝑠𝑡. = =>

! !

!

= ! 𝐶!! − 1 +

! ! !"#

! ! !"#

Continuing: 𝑘 !!! = 𝜀

=> 𝜀!"# = 𝑘!"#

𝑘 𝑥 𝑘 = 𝐶!! − 1 + 𝜀 𝑈 𝜀 !"#

Replace with 𝑘 !!!

=>

! ! !!!

!

!

= ! 𝐶!! − 1 + ! |!"# !

!

=> 𝑘 !!!! !! = ! 𝐶!! − 1 + ! |!"# !

!

=> 𝑘 = ! 𝐶!! − 1 + ! |!"# !!

!!! !! 𝑘 𝑥𝜀 𝑘= 1+ | 𝐶 −1 𝜀 !"# 𝑈 𝑘 !"# !!

=

𝑘!"#

!! !!! !!

𝑘!"# !!!

= 𝑘!"#

!!!!!

!! !!! !!

!! !!! !!

!! !!! !!

𝑥𝜀 1+ | 𝐶 −1 𝑈 𝑘 !"# !! 𝑥𝜀 1+ | 𝐶 −1 𝑈 𝑘 !"# !!

57

!! !!! !!

!! !!! !!

𝑥𝜀 𝑘 = 𝑘!"# 1 + | 𝐶 −1 𝑈 𝑘 !"# !!

!! !!! !!

(1)

˙.˙ 𝑘 !!! = 𝜀 Take Eq.(1) => Rower by 𝐶!! : 𝑘 !!!! = 𝑘!"#

𝑥𝜀 1+ | 𝐶 −1 𝑈 𝑘 !"# !!

!!!

!!

=> 𝜀 = 𝜀!"# 1 + ! ! |!"# 𝐶!! − 1

!!!! !!! !!

!!!! !!! !!

For both the k-ω and k-ω SST equations the process is similar. The governing equations are as below D𝑘 𝜕 = 𝒫 − 𝛽 ∗ 𝜔𝑘 + D𝑡 𝜕𝑥!

D𝜔 𝛾 𝜕 = 𝒫 − 𝛽𝜔! + D𝑡 𝜇! 𝜕𝑥!

𝜈+

𝜈! 𝜕𝑘 𝜎! 𝜕𝑥!

𝜈+

𝜈! 𝜕𝜔 1 𝜕𝑘 𝜕𝜔 + 2 1 − 𝐹! 𝜎! 𝜕𝑥! 𝜎!! 𝜔 𝜕𝑥! 𝜕𝑥!

In the free stream these will reduce to D𝑘 = −𝛽 ∗ 𝜔𝑘 D𝑡 D𝜔 = −𝛽𝜔!

D𝑡

The above equations would then be solved along the previous lines. However, since the k-ω and k-ε equation sets are related the equations for the latter can be converted.

Note 𝜔≡

𝜀 𝐶! 𝑘

Hence,

58

𝑘 𝑘 𝑥 = + 𝐶!! − 1 𝜀 𝜀 !" 𝑈 can be written as 1 1 𝑥 = + 𝐶! 𝐶!! − 1 𝜔 𝜔!" 𝑈 Since 𝛽 is equivalent to 𝐶! 𝐶!! − 1 (with 𝐶! = 𝛽 ∗ = 0.09) 𝜔 = 𝜔!"

𝜔!" 𝛽𝑥 !!

1+ 𝑈

59

5.3 Basically, for this question boundary layer thickness and skin friction coefficient expressions are needed. Chapman (1979) was the first to carry out this type of procedure. Later Chapman’s estimates where reconfirmed by Choi and Moin (2012) using more realistic outer boundary layer cost estimates and empirical data. Spalart et al (1998) present a slightly altered approach that emphasizes the modeling of a boundary layer developing from zero thickness and the implications of this to the downstream grid, unless a complex grid design is used (see Figure 3.59). However, largely in these texts the procedures are the same. Further complications arise with regards to grid design and cost as the geometric complexity increases. For example, it is worth reflecting on the grid design when two flat surfaces are orthogonal to each other – this being a canonical form of a junction flow. There are a very wide range of alternatives, but Chapman (1979) used the equations below !!/!

𝐶! = 0.045𝑅𝑒! !

𝜃 = !" 𝛿

(1) (2)

where !

𝐶! = ! ! ! !

!!

(3)

is the skin friction coefficient. Note 𝜏! is the wall shear stress. The Reynolds number is defined as !"

𝑅𝑒! = !

(4)

In the above 𝛿 is the boudary layer thickness, 𝑈 is the free stream velocity and 𝜈 the kinematic viscosity. Note 𝜃 is the momentum thickness. Equation (1) is based on measurements for a round pipe. Equation (2) is based on the 1/7th paper law velocity distribution.

Since 𝐶! =

2𝑑𝜃 𝑑𝑥

(5)

for a developing boundary layer 𝛿 !!/! = 0.37𝑅𝑒! 𝑥 And

60

(6)

!!/!

(7)

𝐶! = 0.0577𝑅𝑒!

!"

where the Reynolds number is now defined as 𝑅𝑒! = ! where x is the distance from say the leading edge of a flat plate. Using the above expressions Chapman showed that for a wall modeled LES the !/!

!/!

number of grid points scaled as 𝑅𝑒!! and that for a wall resolved LES there was a 𝑅𝑒!! scaling.

The equations above are based on more modest Reynolds number (𝑅𝑒! ≤ 10! ) data and the momentum thickness is not that accurate. Hence, Choi and Moin (2012) use correlations that are more accurate at high Reynolds numbers (10! ≤ 𝑅𝑒! ≤ 10! ). These equations are given below 𝛿 !!/! = 0.16𝑅𝑒! 𝑥 !!/!

𝑐! = 0.027𝑅𝑒!

(8) (9)

The date in Tables 5.1 and 5.2, respectively, can be applied quite directly once the above equations for boundary layer thickness and skin friction are available. The latter provides the wall shear stress, thus allowing the grid spacings in wall units to be converted to direct physical spacings. The use of either the equations of Chapman or Choi and Moin etc. will not make a substantial difference. Note Chapman’s estimates assume the use of hanging nodes (to coarsen the grid as the wall is moved away from). Such toplogy assumptions with respect to the underlying grid etc. will have the greater impact on your estimates along with how much emphasis you place in resolving potentially laminar activity at a leading edge.

61

5.4 4) For eddy resolving simulations, near walls, Table 5.2 gives typical grid aspect ratios. These can be used to explore potential variations between the filters given in Table 5.13. The Figure S5.4.1 based on Tyacke et al. (2016) gives grid spacing ratios for meshes typically used when making eddy resolving simulations for jets. See figures 3.49 and 3.50 for examples of meshes. Figure S5.4.1 indicates that the axial, x, to radial, r, grid aspect ratio can reach 100. However, as can be inferred from Tyacke et al. and is perhaps not surprising the tangential grid resolution at large diameters can be hard to keep to acceptable levels. Hence, the circumferential to radial grid spacing ratio can be much higher than 100 (see Tyacke et al.) potentially tending to being an order of magnitude higher.

Figure S5.4.1 Grid spacing ratios for meshes typically used when making eddy resolving simulations for jets.

62

CHAPTER 6

6.1 To integrate weightless particles Equation (6.32) needs to be solved. Equation (6.36) gives a discrete form for solution. The process is relatively simple. The more complex aspect, is efficiently searching to find say the notional cell that a particle is in and interpolating velocity, based on surround values to give the velocity local to the particle. Ideally, the velocity interpolation needs to have an order consistent with the underlying discretization. You can generate a mesh and either initialize the nodes with an arbitrary velocity field or, if you have a working CFD code take one of your existing simulations. To start with, the coordinates of the points defining the particle trajectory can be simply plotted using a graph-plotting package. The line will be a streamline.

To integrate particles that have weight, as well as the above, Equation (6.33) must be solved.

Equation (6.39) gives the dicretized form of the equation that needs to be typed into your computer program. As a first test, you could check you code to see that it gives the expected results relative to the analytical solution for a particle dropping in a stationary fluid. You could explore the impact of some of the different temporal integration schemes described in Chapter 4.

6.2 The follows the same procedure as the first part of Question (1) above. However, this time a cylindrical grid is generated and the equations solved in this system. Intrinsically the process is the same.

63

6.3 Appendix B gives the discretized equation coefficients when using the finite-difference/finitevolume procedure outlined in Section 4.2.4 of Chapter 4. These coefficients are for 2nd order central differences with a convection-diffusion transport equation. The 2nd order central differences are only suitable for low velocities with problems where there is natural diffusion. Here we wish to, ideally, switch the diffusion off. The Appendix B discretization does not have the derivatives with respect to the tangential direction. The extra terms relating to the tangential derivatives are given below.

1 ∂(Fr ) ∂(Fz ) 1 ∂(Fθ ) + + = Sϕ r ∂r ∂z r ∂θ Fθ = ρ vφ − Γθ

1 ∂φ r ∂θ

We now introduce the front (f) and back (b) stations. These are made use of below Ff =

1 ⎡ 1 ∂φ ⎤ ⎢ρ f v f φ f − Γθ f ⎥ rP Δθ ⎢⎣ rP ∂θ f ⎥⎦

Fb =

1 ⎡ 1 ∂φ ⎤ ⎢ρb vb φb − Γθ b ⎥ rP Δθ ⎣ rP ∂θ b ⎦

The convection coefficients take the form below

Cf =

ρ f vf rp Δθ

Cb =

ρ b vb rp Δθ

The diffusion coefficients have the form

Df =

Db =

Γθf 2 P

r δθ f Δθ

Γθb 2 P

r δθ b Δθ

AF = D f −

Cf 2

64

AB = Db +

Cb 2

The final discretized equation then looks as below, AP,φ φ P = AN ,φ φ N + AS ,φ φ S + AF ,φ φ F + AB,φ φ B

A more stable coefficient formulation is given below and would be best used,

where the construct

⎡ AN = ⎢ ⎣

Dn ,

Cn 2

⎤ Cn ⎥ − 2 ⎦

⎡ AS = ⎢ ⎣

Ds ,

Cs 2

⎤ Cs ⎥ + 2 ⎦

⎡ AF = ⎢ ⎣⎢

Df ,

Cf

⎤ Cf ⎥ − 2 ⎦⎥

⎡ AB = ⎢ ⎣

Db ,

2

Cb 2

⎤ Cb ⎥ + 2 ⎦

[ A, B ] means that the maximum value of A and B is taken. Use of the above is

especially important if the diffusion is fully omitted – this being a critical aspect if we are to model just the convection of a tracer. The coefficients above correspond to what is known as the HYBRID convective scheme (see Spalding (1972))). This uses central differences when the cell Peclet number (a type of cell Reynolds number = C/D) is less than two. For values greater than two first order upwinding is used. This scheme will be stable but it is too diffusive and, especially for the current purpose, more advanced high order schemes are needed. Another crude option would be to stay with the 2nd order central differences and use the diffusion terms in the above to provide a low level of smoothing. For formulations of convective terms, in the above framework, for higher order scheme interpolations see Tyacke (2009).

6.4 Figure 6.58 shows the expected traits. The left hand images are based on particle tracking. The right hand images involve particle tracking. Some discussion contrasting these approaches is given in the text, the discussion being centered around Figure 6.58.

65

CHAPTER 7

7.1 Derive the three eigenvalues λ1 ≤ λ2 ≤ λ3 of the symmetric matrix S 2 + Ω 2 According to the definition, the tensor S 2 + Ω 2 can be written as Sik Skj + Ωik Ω kj =

1 1 (ui /k + uk /i ) (uk /j + u j /k ) + (ui /k − uk /i ) (uk /j − u j /k ) 4 4

1 ⎡( ui /k uk /j + ui /k u j /k + uk /i uk /j + uk /i u j /k ) + ( ui /k uk /j − ui /k u j /k − uk /i uk /j + uk /i u j /k )⎤ ⎦ 4⎣ 1 = ( ui /k uk /j + uk /i u j /k ) 2 =

(1)

Here “/” indicates partial differential operator; for example, ui /k can be also written as

∂ui . ∂xk

In order to get the eigenvalues of tensor S 2 + Ω 2 , we need to diagonalize the matrix. According to the definition of eigenvalues, the elements of the diagonal matrix are the required eigenvalues. This is because if λi is one of the elements of the diagonal matrix Λ = diag [λ1 , λ2 ,L , λi ,L , λn ], then the equation Λ − λi I = 0 is satisfied. Here we can diagonalize this matrix S 2 + Ω 2 by postmultiplying an orthogonal matrix Ω (which is formed by eigenvectors) and premultiplying its transpose ΩT , and this can be expressed in a tensor form as

(

)

ΩT ⋅ S 2 + Ω 2 ⋅ Ω = Ωnm Sik Skj + Ωik Ωkj Ωmnδ jmδij

(

)

= ⎡⎣Ωnj Sik Skj + Ωik Ωkj Ω jn ⎤⎦δij = Ωni ( Sik Ski + Ωik Ωki ) Ωin

(

)

= Ωni Sik SkiΩin + ΩniΩik ΩkiΩin == ( Sik Ski + Ωik Ωki ) ΩniΩin

(2)

= ΩT ⋅ ( Sik Ski + Ωik Ωki ) ⋅ Ω = ΩT ⋅ ⎡⎣ui/k uk /i ⎤⎦ ⋅ Ω ⎡ u u ⎢ 1/k k /1 = ΩT ⋅ ⎢ u2/k uk /2 ⎢ ⎢ ⎣

⎤ ⎥ ⎥⋅ Ω ⎥ u3/k uk /3 ⎥ ⎦

It should be noticed that in a tensor form, the post-multiplication can be rewritten in a way that the last index of pre-dot-product matrix and the first index of post-dot-product matrix should be the same (e.g. A ⋅ B = Aij Bmnδ jm = Aim Bmn ). Finally, the eigenvalues are λi = ui /k uk / i

(3)

(Notice: i is not a dummy index, so there is no summation for index i, and thus we add a bar over the second one). Alternatively, see Cucitore et al. (1999) and discussion around Equation (7).

66

7.2 1 2

Proof: Q = − ( λ1 + λ2 + λ3 ) According to the definition, the trace of a matrix is the summation of all the eigenvalues and also the summation of all diagonal elements. Therefore, λ1 + λ2 + λ3 = tr ( Sik Skj + Ωik Ω kj ) = ui /k uk / i ,

(4)

(Notice: i and k are dummy indices respectively, which indicates that the item should be added for i and k separately according to Einstein summation convention.) where λ1 , λ2 , λ3 are the eigenvalues.

Since the second invariant Q can be also written as Q=

1 2 2 Ω − S 2

(

)

(5)

where S 2 = tr ( Sik Skj ) = 1 (ui /k + uk /i )2 , Ω 2 = tr (Ωik Ωkj ) = 1 (ui /k − uk /i )2 . Then 4

Q=

4

1 1 2 2 (ui /k − uk /i ) − (ui /k + uk /i ) = − ui /k uk /i .

8 2

)

(6)

1 ( λ1 + λ2 + λ3 ) .

2

(7)

(

therefore, Q=−

Alternatively, see Cucitore et al. (1999) and discussion around Equation (5).

67

7.3 The eikonal equation can be expressed as

𝜕𝜙 ! =1 𝜕𝑥!

If we define the pseudo velocity 𝑢! =

𝜕𝜙 𝜕𝑥!

Then the above yields a form of convective transport equation with a source term of unity

𝑢!

𝜕𝜙 =1 𝜕𝑥!

68

7.4 Consider 3 solutions φ1 , φ 2 and φ3 on grids of Δx 1 , Δx 2 and Δx 3 .

Using Equation (7.54) with

φ1 and φ2 gives

φ = φ1 + cΔx1n = φ2 + c Δx2n

(1)

φ1 − φ 2 = c (Δx2n − Δx1n )

(2)

φ1 − φ 2 Δx2n − Δx1n

(3)

or after rearranging

Hence we can write

c=

Considering solution φ 3 , with Equations (7.54) ( φ = φ1 + cΔx n ) and (3) gives

φ = φ3 +

(φ1 − φ 2 )Δx3n Δx2n − Δx1n

(4)

Equating (4) to (7.54) involving φ2 gives

φ3 +

(φ1 − φ 2 ) Δx3n = φ 2 + cΔx2n n n Δx2 − Δx1

(5)

Replacing c in Equation (5) with Equation (3) gives

φ3 +

(φ1 − φ 2 ) (φ1 − φ 2 ) Δx3n = φ 2 + Δx2n n n n n (Δx2 − Δx1 ) (Δx2 − Δx1 )

φ3 − φ 2 =

(φ1 − φ 2 ) (Δx2n − Δx3n ) n n (Δx2 − Δx1 )

(6)

(7)

Equation (7) can be rearranged as:

φ1 − φ 2 Δx1n − Δx2n = φ 2 − φ3 Δx2n − Δx3n

(8)

69

7.5 λ=

Δx1 Δx 2 = Δx 2 Δx3

(1)

We can write

Δx1 = λΔx 2 , Δx3 = λ−1 Δx 2

(2)

Re-stating Equation (8) in question 7.4 above

φ1 − φ 2 Δx1n − Δx2n = φ 2 − φ 3 Δx2n − Δx3n

(3)

We eliminate Δx 1 and Δx 3 from Equation (3) by using Equation (2)

φ1 − φ 2 Δx2n (λn − 1) = φ 2 − φ 3 Δx2n (1 − λ−n )

(4)

Multiplying the right hand side of Equation (4) top and bottom by λn gives

φ1 − φ 2 λn (λn − 1) = = λn n φ 2 − φ3 (λ − 1)

(5)

Taking the natural log of Equation (5) gives

⎡ φ − φ2 ⎤ n = 1n ⎢ 1 ⎥ ⎣φ 2 − φ3 ⎦ 1n λ

(6)

7.6 The

code

for

this

format

is

given

in

Section

http://www.grc.nasa.gov/WWW/wind/valid/plot3d.html.

70

7.7.

It

can

also

be

found

at

7.7 The bump, results for which are shown in Figure 4.4, is a useful case for this exercise. As shown in Figure 4.4 a lack of general flow field symmetry, evident from the Mach number contours, arises from false diffusion. The loss in total pressure, also shown in Figure 4.4 is evident. This again arises from false diffusion. Figure S7.7.1 contrasts Mach number and entropy contours. The entropy contours show a region of generation and the appearance of downstream convection. The all takes place in the regions where the velocity gradients are steep.

(a)

(b) Figure S7.7.1 Contrasting Mach number and entropy contours for flow over a bump computed with Euler solver (see Question 4, Chapter 4) with a high level of numerical diffusion: (a) Mach number contours and (b) entropy contours.

Figure S7.7.2 contrasts Mach number (Frame (a)) and entropy (Frame (b)) contours for flow over the bump computed with an Euler solver. This time there is a low level of numerical diffusion error. As can be seen there is now more symmetry and less entropy generation. The extensions to be attempted as part of questions 5 and 6 in Chapter 4 should help improve symmetry. Your results should tend towards those shown in Figure S.7.7.2.

71

If the false diffusion is low enough and central differences used, then dispersion error will become evident. This can be seen in Figure S7.7.3, which contours a function of entropy. Note, this plot was generated using ParaView.

(a)

(b) Figure S7.7.2 Contrasting Mach number and entropy contours for flow over a bump computed with an Euler solver with a low level of numerical diffusion error: (a) Mach number contours and (b) entropy contours.

Figure S7.7.3. Entropy related contours for subsonic flow over a bump showing the dispersion error.

72

Figure 3.65, in the main book text, shows Mach number contours for the flow over the supersonic wedge. Figure 3.66, again in the main book text, shows contours of total pressure. Figure S7.7.3 shows entropy contours for this case.

Figure S7.7.4. Entropy contours for supersonic flow over a wedge.

73

7.8 Generate a finer grid (See Figure S8.8.1, Frame (b)) with double the number of grid points in both directions. Directly apply the equation derived in Question (5). Note, the order is unlikely to be that close to 2nd order. Figure C8.8.1 shows a useful way to plot the performance of your scheme. The x axis gives the log(Δx) - Δx being the grid spacing. The y axis :gives the log(rms error in %). Since error of a scheme is - rms error = (Δx)n, taking log on both sides means we have log(rms error) = n log(Δx) n being slope of the line = order of scheme.

(a)

(b) Figure S.8.8.1. Coarse and fine meshes to test numerical order of spatial scheme: (a) coarse mesh and (b) finer mesh.

74

Figure S8.8.1. Plot showing the order of a numerical scheme.

Acknowledgements Solutions for questions 2.6-2.8 where provided Dr A. P. Morse and 7.1 and 7.2 by Yunfei Ma. Also, I am grateful for Cen Zhang and Andrea Massi for helping with the preparation of this document.

Thanks are also due to Hardeep Kalsi for drawing some of the figures above. Also, I am grateful to Callum Mantell for providing the mesh generation code on the Appendix. The TDMA solver was kindly provided by Iftekhar Naqavi. I am grateful to the various students who have generated some of the different solution plots shown here.

75

REFERENCES

G. A. BRES, J. W. NICHOLS, S. K. LELE AND F. E. HAM. 2012. Towards Best Practices for Jet Noise Predictions with Unstructured Large Eddy Simulations. AIAA Journal, 2012-2965, 1–21.

D. CHAPMAN. 1979. Computational aerodynamics, development and outlook. AIAA Journal, 17, 1293–1313.

H. CHOI AND MOIN. 2012. Grid-point requirements for large eddy simulation: Chapman's estimates revisited. Physics of Fluids, 24:011702.

R. CUCITORE, M. QUADRIO AND A. BARON. 1999. On the effectiveness and limitations of local criteria for the identification of a vortex. European Journal of Mechanics B/Fluids, 18, 261282.

J. DENTON, 2007. Euler extensions, 4A2 Lecture Course Notes, Cambridge University.

P. M. GRESHO, S. T. CHAN, R. L. LEE AND C. D. UPSON. 1984. A modified finite element method for solving the time-dependent, incompressible Navier-Stokes equations. Part 1: Theory.

International Journal for Numerical Methods in Fluids, 4, 557-598.

S. MENDEZ, M. SHOEYBI, A. SHARMA, F.E. HAM, S. K. LELE AND P. MOIN. 2012. LargeEddy Simulations of Perfectly Expanded Supersonic Jets Using an Unstructured Solver. AIAA Journal, 50 (5), 1103–1118.

V. W. NEE AND L. S. KOVASZNAY. 1969. Simple phenomenological theory of turbulent shear flows. Physics of Fluids (1958-1988), 12, 473-484.

R. H. PLETCHER, J. C. TANNEHILL, D. A. ANDERSON. 2013. Solution's Manual Computational Fluid Mechanics and Heat Transfer, Taylor & Francis Inc. R. H. PLETCHER, J. C. TANNEHILL, A. A. ANDERSON. 2013. Computational fluid mechanics and heat transfer, CRC Press.

P.R. SPALART, W.-H. JOU, M. STRELETS, AND S.R. ALLMARAS. 1998. Comments on the feasibility of LES for wings, and on a hybrid RANS/LES approach. In Advances in DNS/LES., edited by C. Liu and Z. Liu, Greyden Press, Colombus.

D. B. SPALDING. 1972. A novel finite difference formulation for differential expressions involving both first and second derivatives. International Journal for Numerical Methods in Engineering, 4, 551-561.

P. G. TUCKER. 2001. Computation of unsteady internal flows, Springer, ISBN 978-1-4615-1439-8.

J. F. THOMPSON, F. C. THAMESA, C. W. MASTINA. 1974. Automatic numerical generation of body-fitted curvilinear coordinate system for field containing any number of arbitrary twodimensional bodies. Journal of Computational Physics, 15, 3, 299–319.

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J. TYACKE et al. 2016. Predictive LES for Jet Aeroacoustics - Current Approach and Industrial Application. Proceedings of ASME Turbo Expo 2016 GT2016, 13-17, Seoul, South Korea, GT2016-56673 J. TYACKE. 2009. Low Reynolds number heat transfer prediction employing Large Eddy Simulation for electronics geometries. Civil and Computational Engineering Centre, Swansea University.

77

APPENDIX A Elliptic and Bi-linear Interpolation Mesh Generator implicit none C

Variables taken from the Common Block integer i_max,j_max parameter (i_max=440, j_max=101) character*80 title integer ni,nj real xlow(i_max),ylow(i_max),xhigh(i_max),yhigh(i_max) real x(i_max,j_max), y(i_max,j_max)

C

Variables for Smoothing integer j real xold(i_max,j_max),yold(i_max,j_max),aflowx,aflowy,aflowin real delta_eta,delta_xi,dx,dy,xdiff,ydiff real alpha1,alpha2,alpha,beta1,beta2,beta,gam,a,b,c,Ap,Aw,Ae,An,As real Sx,Sy,tol real lenx,leny,lenin,gradhigh(i_max),gradlow(i_max)

C

Variables for Interpolation integer scaling real xhighold(i_max),yhighold(i_max),xlowold(i_max),ylowold(i_max)

C

Variables used in both (Separately) integer i,count

C

Variables use integer K real eta,XS(3,I_MAX,J_MAX,2) real y_lin(i_max,j_max),x_lin(i_max,j_max) real interp_lin real interp_cubic

C C C C

################################################################### Read Data ################################################################### Read in the boundaries OPEN(UNIT=1,FILE='geom.dat') READ(1,*) TITLE, ni, nj DO 100 i = 1,ni 78

READ(1,*) xlow(i), ylow(i), xhigh(i), yhigh(i) 100 CONTINUE scaling = 2

C C C

################################################################### New grid interpolating code ###################################################################

C

Store old boundaries if(scaling.ne.1) then xhighold = xhigh yhighold = yhigh xlowold = xlow ylowold = ylow

C

Copy over orginal values into their new locations do i = ni,1,-1 xhigh(scaling*i-(scaling-1)) = xhighold(i) yhigh(scaling*i-(scaling-1)) = yhighold(i) xlow(scaling*i-(scaling-1)) = xlowold(i) ylow(scaling*i-(scaling-1)) = ylowold(i) enddo count = 1

C

C

!variable used to keep track of which gap we are in

First gap is filled using linear interpolation (For function see end of program) do i = 2,scaling if(MOD(i-1,scaling).ne.0.0) then xhigh(i) = interp_lin(xhighold(count),xhighold(count+1),i &,scaling) yhigh(i) = interp_lin(yhighold(count),yhighold(count+1),i &,scaling) xlow(i) = interp_lin(xlowold(count),xlowold(count+1),i &,scaling) ylow(i) = interp_lin(ylowold(count),ylowold(count+1),i &,scaling) else count = count +1 endif enddo Subsequent gaps are filled using a cubic fit (Catmull-Rom spline) do i = scaling+1,scaling*(ni-2)+1 if(MOD(i-1,scaling).ne.0.0) then 79

xhigh(i) = interp_cubic(xhighold(count-1),xhighold(count) &,xhighold(count+1),xhighold(count+2),i,scaling) yhigh(i) = interp_cubic(yhighold(count-1),yhighold(count) &,yhighold(count+1),yhighold(count+2),i,scaling) xlow(i) = interp_cubic(xlowold(count-1),xlowold(count) &,xlowold(count+1),xlowold(count+2),i,scaling) ylow(i) = interp_cubic(ylowold(count-1),ylowold(count) &,ylowold(count+1),ylowold(count+2),i,scaling) else count = count+1 endif enddo C

Last gap is filled using linear interpolation do i = scaling*(ni-2)+2,scaling*(ni-1)+1 if(MOD(i-1,scaling).ne.0.0) then xhigh(i) = interp_lin(xhighold(count),xhighold(count+1),i &,scaling) yhigh(i) = interp_lin(yhighold(count),yhighold(count+1),i &,scaling) xlow(i) = interp_lin(xlowold(count),xlowold(count+1),i &,scaling) ylow(i) = interp_lin(ylowold(count),ylowold(count+1),i &,scaling) else count = count +1 endif enddo ni = scaling*(ni-1)+1 nj = scaling*(nj-1)+1 write(6,*) 'ni =',ni,'nj =',nj endif

C C C

################################################################### Generate Grid ###################################################################

C

Generate the linear grid DO 200 i = 1,ni DO 200 j = 1,nj eta = float(j-1)/float(nj-1) 80

x(i,j) = (1-eta)*xlow(i) + eta*xhigh(i) y(i,j) = (1-eta)*ylow(i) + eta*yhigh(i) 200 continue x_lin = x y_lin = y

C C C

################################################################### New grid smoothing code ###################################################################

C

Copy orginal algebraic grid into 'old' values xold = x yold = y

C

Calculate computational step sizes aflowx = xhigh(1)-xlow(1) aflowy = yhigh(1)-ylow(1) aflowin = sqrt(aflowx*aflowx + aflowy*aflowy) !total channel 'height' delta_eta = aflowin/(float(nj-1))

!average cell 'height'

lenx = xlow(2)-xlow(1) leny = ylow(2)-ylow(1) lenin = sqrt(lenx*lenx + leny*leny) delta_xi = (lenin) C

Conditions to start loop dx = 1.0 dy = 1.0

C

Assign tolerance value - total motion allowed at convergence tol = 0.00000001

C

Nested loop to determine elliptical grid xdiff=0.0; ydiff=0.0; count=0;

!value to store total x motion !value to store total y motion !value to store number of steps to convergence 81

do while (dy.gt.tol.or.dx.gt.tol) do i = 2,ni-1 do j = 2,nj-1 alpha1=(xold(i,j+1)-x(i,j-1))/(2*delta_eta) alpha2=(yold(i,j+1)-y(i,j-1))/(2*delta_eta) alpha=alpha1*alpha1+alpha2*alpha2 beta1=(xold(i+1,j)-x(i-1,j))/(2*delta_xi) beta2=(yold(i+1,j)-y(i-1,j))/(2*delta_xi) beta=beta1*beta1+beta2*beta2 gam=alpha1*beta1+alpha2*beta2 a=alpha/(delta_xi*delta_xi) b=beta/(delta_eta*delta_eta) c=-1.0*gam/(2.0*delta_xi*delta_eta) Ap=(2.0*a+2.0*b) Aw = a Ae = a An = b As = b Sx=c*(xold(i+1,j+1)-xold(i+1,j-1)-xold(i-1,j+1)+x(i &-1,j-1)) x(i,j)=(1/Ap)*(Ae*xold(i+1,j)+Aw*x(i-1,j)+An*xold(i, &j+1)+As*x(i,j-1)+Sx) xdiff=x(i,j)-xold(i,j)+xdiff Sy=c*(yold(i+1,j+1)-yold(i+1,j-1)-yold(i-1,j+1)+y(i &-1,j-1)) y(i,j)=(1/Ap)*(Ae*yold(i+1,j)+Aw*y(i-1,j)+An*yold(i, &j+1)+As*y(i,j-1)+Sy) ydiff=y(i,j)-yold(i,j)+ydiff enddo enddo dx=xdiff dy=ydiff xdiff=0.0 ydiff=0.0 xold=x yold=y count=count+1 enddo write(6,*) 'The solution took',count,'iterations to converge'

C C C

################################################################### Output file ################################################################### 82

C

Output File DO I=1,NI DO J=1,NJ XS(1,I,J,1) = X_lin(I,J) XS(2,I,J,1) = Y_lin(I,J) XS(3,I,J,1) = 0.0 XS(1,I,J,2) = X_lin(I,J) XS(2,I,J,2) = Y_lin(I,J) XS(3,I,J,2) = 0.1 ENDDO ENDDO OPEN(UNIT=61,FILE='euler_linear.csv') WRITE(61,*) 'X,Y,Z' DO K=1,2 DO J=1,NJ DO I=1,NI WRITE(61,"(8(F13.6,A1,1X))") XS(1,I,J,K),',', & XS(2,I,J,K),',', & XS(3,I,J,K) ENDDO ENDDO ENDDO CLOSE(61)

DO I=1,NI DO J=1,NJ XS(1,I,J,1) = X(I,J) XS(2,I,J,1) = Y(I,J) XS(3,I,J,1) = 0.0 XS(1,I,J,2) = X(I,J) XS(2,I,J,2) = Y(I,J) XS(3,I,J,2) = 0.1 ENDDO ENDDO OPEN(UNIT=61,FILE='euler.csv') WRITE(61,*) 'X,Y,Z' DO K=1,2 DO J=1,NJ DO I=1,NI 83

WRITE(61,"(8(F13.6,A1,1X))") XS(1,I,J,K),',', & XS(2,I,J,K),',', & XS(3,I,J,K) ENDDO ENDDO ENDDO CLOSE(61) RETURN END

C C C

################################################################### Functions for grid interpolation ################################################################### real function interp_lin(y0,y1,i,scaling) integer i,scaling real y0,y1 interp_lin = (float(MOD(i-1,scaling))/float(scaling)) &*y1+(1.0-(float(MOD(i-1,scaling))/float(scaling)))*y0 return end real function interp_cubic(y0,y1,y2,y3,i,scaling) integer i,scaling real y0,y1,y2,y3,a0,a1,a2,a3,mu,mu2 mu = (float(MOD(i-1,scaling))/float(scaling)) mu2 = mu*mu a0 = -0.5*y0 + 1.5*y1 - 1.5*y2 + 0.5*y3 a1 = y0 - 2.5*y1 + 2*y2 - 0.5*y3 a2 = -0.5*y0 + 0.5*y2 a3 = y1 interp_cubic = a0*mu*mu2+a1*mu2+a2*mu+a3 return end

84

APPENDIX B Basic finite difference / finite volume transport equation subroutine for a variable vPHI. This aligns with Section 4.2.4 in the main book text. However, the discretization here is for a cylindrical polar coordinate system. If all radial values in the program are set to unity then the code will correspond to that for a Cartesian system.

*********************************************************************** subroutine coeffPHI(u,v,vPHI,ap,an,as,ae,aw,su,dz,z,dr,r,id,jd) ************************************************************************ implicit real (a-h,o-z) dimension u(0:id,0:jd),v(0:id,0:jd),vPHI(0:id,0:jd) dimension ap(0:id,0:jd),an(0:id,0:jd),as(0:id,0:jd) dimension ae(0:id,0:jd),aw(0:id,0:jd),su(0:id,0:jd) dimension z(0:id),dz(0:id),dr(0:jd),r(0:jd) c c

Data ************************ dc = diffusion coefficient. Kinematic viscosity for momentum. dc = 1.0 do 20 j=1,jd-1 do 10 i=1,id-1

******Set radial coordinates rn=r(j)+dr(j)/2. rs=r(j)-dr(j-1)/2. rp=r(j) ******Set grid spacings dze=dz(i) dzw=dz(i-1) dzp=(dz(i)+dz(i-1))/2. if(i.eq.1)dzw=dz(i-1)/2. if(i.eq.id-1).le.0)dze=dz(i)/2. drn=dr(j) drs=dr(j-1) drp=(dr(j)+dr(j-1))/2. if(j.eq.1)drs=dr(j-1)/2. if(j.eq.jd-1)drn=dr(j)/2. c c

Interpolate velocities************************************* Note based on a staggered grid

c c

Note u is axial velocity and v is the radial i.e. non-standard naming convention 85

vn=v(i,j+1,k) vs=v(i,j,k) ue=u(i+1,j,k) uw=u(i,j,k) c

Set convection coefficients******************************* cn=vn*rn/rp/drp cs=vs*rs/rp/drp ce=ue/dzp cw=uw/dzp

c

Set diffusion coefficients******************************** dn=dc*rn/rp/drn/drp ds=dc*rs/rp/drs/drp de=dc/dze/dzp dw=dc/dzw/dzp

c c c c c c c c c c c c c c c c c c

The commented out text below would allow the diffusion terms to have a 2nd order discretisation near walls. if(i.eq.1)then de=dc/dze/dzp+dc*dzw/(dze+dzw)/dze/dzp dw=dc*(dze+dzw)/dzw/dze/dzp-dc*dzw/(dze+dzw)/dze/dzp endif if(i.eq.id-1)then dw=dc/dzw/dzp+tce*dze/(dzw+dze)/dzw/dzp de=tce*(dzw+dze)/dze/dzw/dzp-tce*dze/(dzw+dze)/dzw/dzp endif if(j.eq.1)then dn=rn*dc/drn/drp/rp+rs*dc*drs/(drn+drs)/drn/drp/rp ds=rs*(dc*(drn+drs)/drs/drn/drp-dc*drs/(drn+drs)/drn/drp)/rp endif if(j.eq.jd-1)then ds=rs*dc/drs/drp/rp+rn*dc*drn/(drs+drn)/drs/drp/rp dn=rn*(dc*(drs+drn)/drn/drs/drp-dc*drn/(drs+drn)/drs/drp)/rp endif

c

Use mass conservation error to enhance diagonal dominance cmp=abs(cn-cs+ce-cw+cf-cb)

c

******First Order Upwind*********** awq=dw+dmax1(cw,0.d0) aeq=de+dmax1(-ce,0.d0) anq=dn+dmax1(-cn,0.d0) asq=ds+dmax1(cs,0.d0) abq=db+dmax1(cb,0.d0) afq=df+dmax1(-cf,0.d0) ***********************************

c

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c c c c c c c c

*******Std Central difference******* awq=dw+0.5*cw aeq=de-0.5*ce asq=ds+0.5*cs anq=dn-0.5*cn abq=db+0.5*cb afq=df-0.5*cf ***********************************

c c c c c c c c c c c

***************hybrid************** First order unwinding for cell Reynolds number > 2. Other wise 2nd order central difference anq=(dmax1(dabs(0.5*cn),dn)-0.5*cn) asq=(dmax1(dabs(0.5*cs),ds)+0.5*cs) aeq=(dmax1(dabs(0.5*ce),de)-0.5*ce) awq=(dmax1(dabs(0.5*cw),dw)+0.5*cw) afq=(dmax1(dabs(0.5*cf),df)-0.5*cf) abq=(dmax1(dabs(0.5*cb),db)+0.5*cb) *********************************** an(i,j,k)=anq as(i,j,k)=asq ae(i,j,k)=aeq aw(i,j,k)=awq ap(i,j,k)=an(i,j,k)+as(i,j,k)+ae(i,j,k)+aw(i,j,k) 1 +cmp su(i,j,k)= ADD SOURCE TERMS HERE 1 + cmp*vPHI(i,j,k) endif

10 20

continue continue end

******************************************************************************************************** ******************************************************************************************************** c Below are the modifications to CoeffPHI when using staggered grids and solving for the c u (x - direction) momentum variable. do 10 i=2,id-1 dze=(dz(i)+dz(i-1))/2 dzw=(dz(i-1)+dz(i-2))/2 87

dzp=dz(i-1) vn=(v(i,j+1)+v(i-1,j+1))/2 vs=(v(i,j)+v(i-1,j))/2 uw=(u(i-1,j)*dz(i-1)+u(i,j)*dz(i-2))/(dz(i-1)+dz(i-2)) ue=(u(i+1,j)*dz(i-1)+u(i,j)*dz(i))/(dz(i-1)+dz(i))

su(i,j)= cmp*u(i,j)+(p(i-1,j)-p(i,j))/dzp c c

rf is an under relaxation parameter For SIMPLE - rf=1 du(i,j)=1. / ( ap(i,j) - rf*(an(i,j)+as(i,j)+ae(i,j)+aw(i,j)) )/dzp ******************************************************************************************************* ******************************************************************************************************** ******************************************************************************************************** c Below are the modifications to CoeffPHI when using staggered grids and solving for the c v (y - direction) momentum variable. rn=r(j) rs=r(j-1) rp=(rn+rs)/2. drn=(dr(j)+dr(j-1))/2. drs=(dr(j-1)+dr(j-2))/2. drp=dr(j-1) vn=(v(i,j)*dr(j)+v(i,j+1)*dr(j-1))/(dr(j-1)+dr(j)) vs=(v(i,j)*dr(j-2)+v(i,j-1)*dr(j-1))/(dr(j-1)+dr(j-2)) ue=(u(i+1,j-1)+u(i+1,j))/2 uw=(u(i,j-1)+u(i,j))/2 su(i,j)= cmp*v(i,j)+(p(i,j-1)-p(i,j))/drp c c

rf is an under relaxation parameter For SIMPLE - rf=1 dv(i,j)=1. / ( ap(i,j) - rf*(an(i,j)+as(i,j)+ae(i,j)+aw(i,j)) )/drp *******************************************************************************************************

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APPENDIX C Code for Evaluation of the Pressure Field The code below evaluates the coefficients for the pressure correction equation. ************************************************************************ subroutine coeffpr(u,v,du,dv,ap,an,as,ae,aw,su,dz,dr,r,id,jd) ************************************************************************ implicit real (a-h,o-z) dimension u(0:id,0:jd),v(0:id,0:jd) dimension du(0:id,0:jd),dv(0:id,0:jd) dimension ap(0:id,0:jd),an(0:id,0:jd),as(0:id,0:jd) dimension ae(0:id,0:jd),aw(0:id,0:jd),su(0:id,0:jd) dimension dz(0:id),dr(0:jd),r(0:jd) do 20 j=1,jd-1 rn=r(j)+dr(j)/2. rs=r(j)-dr(j-1)/2. rp=r(j) drp=(dr(j)+dr(j-1))/2. do 10 i=1,id-1 dzp=(dz(i)+dz(i-1))/2. vn=v(i,j+1) vs=v(i,j) ue=u(i+1,j) uw=u(i,j) c c c

Note it is expected that these have been evaluated elsewhere. In Their simplest form they are 1/Ap where Ap is the respective momentum equation coefficient. dvn=dv(i,j+1) dvs=dv(i,j) due=du(i+1,j) duw=du(i,j) an(i,j)=(dvn*rn/rp/drp) as(i,j)=(dvs*rs/rp/drp) ae(i,j)=(due/dzp) aw(i,j)=(duw/dzp) if(j.eq.1)as(i,j)=0.0 if(j.eq.jd-1)an(i,j)=0.0 if(i.eq.1)aw(i,j)=0.0 if(i.eq.id-1)ae(i,j)=0.

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ap(i,j)=an(i,j)+as(i,j)+ae(i,j)+aw(i,j) c

Continuity residual su(i,j)=(vs*rs-vn*rn)/rp/drp +(uw-ue)/dzp 10 continue 20 continue end The code below corrects the velocity and pressure fields. ****************************************************************** subroutine prc3(u,v,r,dr,dz,du,dv,p,pp,id,jd,rf) ************************************************************************ implicit real (a-h,o-z) dimension u(0:id,0:jd),v(0:id,0:jd) dimension du(0:id,0:jd),dv(0:id,0:jd) dimension p(0:id,0:jd),pp(0:id,0:jd) dimension r(0:jd),dr(0:jd),dz(0:id)

c For staggered grids, note the loop limits of 2 below do 20 j=2,jd-1 do 10 i=2,id-1 u(i,j)=u(i,j) + (pp(i-1,j)-pp(i,j))*du(i,j) v(i,j)=v(i,j) + (pp(i,j-1)-pp(i,j))*dv(i,j) endif 10 continue 20 continue do 50 j=1,jd-1 do 40 i=1,id-1 c

rf is a tuneable relaxation factor. rf = 0.3 p(i,j)=p(i,j)+pp(i,j)*rf

40 50

continue continue

do 642 j=1,jd-1 p(id,j)=p(id-1,j) p(0,j)=p(1,j) 642 continue

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do 643 i=1,id-1 p(i,jd)=p(i,jd-1) p(i,0)=p(i,1) 643 continue do 330 j=0,jd do 320 i=0,id pp(i,j)=0.0 320 continue 330 continue end

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