Assignment #2 - | University of Toronto Institute for...

AER1310: TURBULENCE MODELLING

Assignment #2

1. The Reynolds-averaged form of the boundary layer equations for steady incompressible floware

∂U

∂x+

∂V

∂y= 0 ,

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x+

∂

∂y

(ν

∂U

∂y− u′v′

),

where ρ is the density (a constant), U and V are the x and y components of the mean velocity,u′ and v′ are the fluctuating components of velocity, x and y are the flow aligned and normalcoordinate directions, and P is the Reynolds averaged mean pressure. The parameter ν = µ/ρis the kinematic viscosity (a constant), µ is the dynamic viscosity, and λxy = −ρu′v′ is theReynolds shear stress.

(a) For boundary layer flow over a flat plate, ∂P/∂x = 0. Introducing the Boussinesqapproximation for the Reynolds shear stress, show that, in the log layer of a flat plateboundary layer, the x-direction momentum equation can be simplified and re-expressedas (

1 + µ+T

) dU+

dy+= 1 ,

where U+ = U/uτ , y+ = uτy/ν, µ+T = µT /µ, and µT is the so-called “eddy viscosity”.

The friction velocity, uτ , is given by

u2τ =

τw

ρ= ν

dU

dy

∣∣∣∣y=0

,

where τw = τxy(y = 0) is the wall shear stress.

(b) Show that if Prandtl’s mixing length theory is used to prescribe the eddy viscosity, thenthe non-dimensional eddy viscosity, µ+

T , can be expressed as

µ+T =

(`+mix

)2∣∣∣∣∣dU+

dy+

∣∣∣∣∣ ,

where `+mix = `mixuτ/ν and `mix is the mixing length.

(c) Using the mixing length model, the ordinary differential equation above for the non-dimensional velocity component aligned with the plate, U+, can be re-expressed as

[1 + (`+

mix)2dU+

dy+

]dU+

dy+= 1 .

Show that for `+mix < y+ and y+ << 1, the series expansion

dU+

dy+≈ 1− (`+

mix)2 + 2(`+mix)4 + · · · ,

provides an estimate for dU+/dy+ very near the wall (y+ = 0).

Assignment #2 — of 5

(d) Using a standard numerical scheme, such as the fourth-order Runge-Kutta method,integrate the non-linear ordinary differential equation

[1 + (`+

mix)2dU+

dy+

]dU+

dy+= 1 ,

from y+ = 0 to y+ = 500 and solve for U+. Perform the calculation using two differentexpressions for the mixing length: first try the simple approximation for the mixinggiven by

`mix = κy ,

where κ = 0.41 is the von Karman constant; and then try Van Driest’s expression forthe mixing length given by

`mix = κy[1− exp(−y+/A+

◦ )]

,

where A+◦ = 26. HINTS: Solve the preceding quadratic equation for dU+/dy+ to arriveat a first-order ordinary differential equation for U+ that can be directly integrated anduse the series expansion determined in part (c) for dU+/dy+ near y+ = 0 to avoidnumerical difficulties and ensure accurate results.

(e) In the log layer, experimental evidence has shown that

U+ =1κ

ln y+ + C .

Using the results of part (d), calculate the limiting value of C as y+ →∞ by examiningthe value of

C = U+ − 1κ

ln y+ ,

for y+ ∈ {250, 300, 350, 400, 450, 500}. Determine values for C for both mixing lengthmodels considered in part (d).

2. The Reynolds-averaged form of the Navier-Stokes equations for steady incompressible two-dimensional axisymmetric mean flow are

∂U

∂x+

1r

∂

∂r(rV ) = 0 ,

U∂U

∂x+ V

∂U

∂r= −1

ρ

∂P

∂x+

1ρ

[∂

∂x(τxx + λxx) +

1r

∂

∂r(rτxr + rλxr)

],

U∂V

∂x+ V

∂V

∂r= −1

ρ

∂P

∂r+

1ρ

[∂

∂x(τxr + λxr) +

∂

∂r(τ rr + λrr) +

(τ rr + λrr)− (τ θθ + λθθ)r

],

where ρ is the density (a constant), U and V are the axial and radial components of the meanvelocity, u′ and v′ are the fluctuating components of velocity, x and r are axial and radialcoordinates, and P is the Reynolds averaged mean pressure. The laminar viscous stresses canbe related to the flow strain rates as follows

τxx = 2µ∂U

∂x, τrr = 2µ

∂V

∂r, τθθ = 2µ

V

r, τxr = µ

[∂U

∂r+

∂V

∂x

],


��

��

��

��

r = R

r

x

y

U = U(r)

Figure 1: Geometry for fully developed pipe flow.

where ν = µ/ρ is the kinematic viscosity (a constant), and µ is the dynamic viscosity. Thevariables λxx, λrr, λθθ, and λxr are the Reynolds stresses given by

λxx = −ρu′2 , λrr = −ρv′2 , λθθ = −ρw′2 , λxr = −ρu′v′ .

(a) For fully developed pipe flow (see figure above), ∂P/∂x = constant, ∂P/∂r = 0, ∂U/∂x =0. In this case, show that the only non-trivial equation is the axial momentum equationwhich can be re-expressed as

µdU

dr− ρu′v′ = −ρu2

τ

r

R,

where R is the pipe radius and uτ is the friction velocity given by

u2τ =

τw

ρ= −ν

dU

dr

∣∣∣∣r=R

= − R

2ρ

dP

dx.

(b) As in problem #1, introduce the Boussinesq approximation for the Reynolds shear stress.Also let y = R − r be the distance from the pipe wall. Then, by defining the followingnon-dimensional variables:

U+ =U

uτ, y+ =

uτy

ν, µ+

T =µT

µ,

where µT is the eddy viscosity, show that the axial momentum equation can be re-writtenas (

1 + µ+T

) dU+

dy+= 1− y+

R+,

where R+ = uτR/ν.(c) Using Prandtl’s mixing length model, show that the ordinary differential equation of

part (b) for U+ can be written as[1 + (`+

mix)2dU+

dy+

]dU+

dy+= 1− y+

R+.

Also show that dU+/dy+ very near the wall (y+ = 0) can then be approximated by theseries expansion

dU+

dy+≈

(1− y+

R+

)−

(1− y+

R+

)2

(`+mix)2 + 2

(1− y+

R+

)3

(`+mix)4 + · · · .

Note the similarities between these pipe flow results and the results for the boundarylayer flow over a flat plate of problem #1.


(d) Modify the numerical scheme you developed in part (d) of problem #1 to integrate thepreceding non-linear ordinary differential equation for U+ from y+ = 0 to y+ = R+

subject to the boundary condition that U+(y+ = 0) = 0. Use the following two-layermixing length model:

`mix =

{κy [1− exp(−y+/A+◦ )] for y ≤ ym ,0.09R for ym < y ≤ R ,

where κ = 0.41, A+◦ = 26, and ym is the minimum value of y for which the inner andouter layer values of the mixing length are equal. Plot the velocity and Reynolds shearstress profiles for the pipe flow for a Reynolds number of ReD = 40, 000, where theReynolds number based on pipe diameter is given by

ReD =2UavgR

ν= 2U+

avgR+ ,

and Uavg is the average mean velocity for the pipe cross section and given by

Uavg =1

πR2

∫ 2π

0

∫ R

0U(r)rdrdθ =

2R

∫ R

0U(y)

(1− y

R

)dy .

Compare your results to the experimental data of Laufer given in the table below.

(y/R, U/U(y = R))-data pairs obtained by Laufer for ReD = 40, 000.

0.010, 0.333 0.590, 0.9310.095, 0.696 0.690, 0.9610.210, 0.789 0.800, 0.9750.280, 0.833 0.900, 0.9990.390, 0.868 1.000, 1.0000.490, 0.902

Also compare the value of the predicted skin friction coefficient given by

cf =2

(U+avg)2

,

to the empirical result1√cf

= 4 log10

(2ReD

√cf

)− 1.6 ,

for ReD = 40, 000. HINTS: U+avg can be determined as part of the integration procedure

for U+(y+). You will also need to apply an iterative method, such as the bisectionmethod, to the overall solution procedure so as to determine the values of U+

avg and R+

that correspond to desired Reynolds number, ReD = 40, 000.

3. Repeat part (d) of problem #2, this time using the Baldwin-Lomax two-layer algebraic eddyviscosity model to specify the mixing length for the turbulent pipe flow. In the inner layerthe eddy viscosity is given by

νTi = `2mix

∣∣∣∣dU

dy

∣∣∣∣ ,

where `mix = κy[1 − exp(−y+/A+◦ )] and κ = 0.40 and A+◦ = 26. In the outer layer, eddyviscosity is prescribed in the Baldwin-Lomax model by the following approximation:

νTo = αCcpFwakeFKleb ,


where α = 0.0168, Ccp = 1.6,

FKleb =

[1 + 5.5

(y

ymax/CKleb

)6]−1

,

Fwake = min

(ymaxFmax, Cwk

ymaxU2dif

Fmax

),

Udif = U(y = R), CKleb = 0.30, and Cwk = 1. The value of Fmax is given by

Fmax =1κ

maxy

(`mix

dU

dy

),

and ymax is the value of y where `mix(dU/dy) is maximum. Note that your numerical so-lution procedure of problem #2 must be further modified to deal with non-linear nature ofthe Baldwin-Lomax model. Adopt a simple fixed-point iterative approach, with damping,whereby successively improved solutions for U+ are obtained using values for ymax, Fmax,and Udif based on the previous solution for U+. You could start by calculating values forymax, Fmax, and Udif using the solution from problem #2 and then allowing for them tochange when a new solution for U+ has been obtained. Compare your results to the pipeflow data of Laufer and the results obtained in problem #2. (HOPEFULLY) HELPFULHINTS: Apply damping to ymax and Fmax such that the n + 1 estimates of ymax and Fmax

in terms of the nth estimates are given by

yn+1max = yn

max + ω (ynewmax − yn

max)

Fn+1max = Fn

max + ω (Fnewmax − Fn

max)

where ynewmax and Fnew

max are the values of ymax and Fmax obtained following the evaluation ofthe updated solution for U+ and ω is a relaxation or damping parameter (0 < ω < 1). Asuggested value of the damping parameter, ω, is 1/2.


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