answers 10 spring 2021 - university of manchester · 2021. 1. 1. · cfd answers 10 – 1 david...

CFD Answers 10 – 1 David Apsley

Answers 10 SPRING 2021

Q1.

(a)

RANS solver – a computer code which solves for average, not instantaneous, flow quantities.

(For theoretical work “average” refers to a probabilistic or “ensemble” average).

(b)

(i) Eddy-viscosity models: (deviatoric) stress mean rate of strain.

(ii) Non-linear eddy-viscosity models: Reynolds stresses modelled as non-linear functions of

the mean-velocity gradients.

(iii) Differential stress models: solve transport equations for all Reynolds stresses.

(iv) Large-eddy simulation: carry out a full time-dependent simulation of resolvable motions,

with modelling only of subgrid (i.e., unresolvable) scales.


Q2.

Dimensions:

[ρ] = ML−3

[μ𝑡] = ML−1T−1

[𝑘] = L2T−2

[ε] = L2T−3

[ω] = T−1

Since only μ𝑡 and ρ contain M as a dimension, the expression for μ𝑡 in terms of ρ, 𝑘 and ε must be of the form

μ𝑡 = constant × ρ𝑘𝑎ε𝑏

Equating powers of the fundamental dimensions:

L: −1 = −3 + 2𝑎 + 2𝑏T: −1 = −2𝑎 − 3𝑏

Solving gives 𝑎 = 2, 𝑏 = −1; i.e.

μ𝑡 = constant × ρ𝑘2

ε

Similarly, the expression for μ𝑡 in terms of ρ, 𝑘 and ω must be of the form

μ𝑡 = constant × ρ𝑘𝑐ω𝑑

Equating powers of the fundamental dimensions:

L: −1 = −3 + 2𝑐 T: −1 = −2𝑐 − 𝑑

Solving gives 𝑐 = 1, 𝑑 = −1; i.e.

μ𝑡 = constant × ρ𝑘

ω


Q3.

(a)

D𝑈1D𝑡

=∂𝑈1∂𝑡+ 𝑈1

∂𝑈1∂𝑥1

+ 𝑈2∂𝑈1∂𝑥2

+ 𝑈3∂𝑈1∂𝑥3

D𝑈

D𝑡=∂𝑈

∂𝑡+ 𝑈

∂𝑈

∂𝑥+ 𝑉

∂𝑈

∂𝑦+𝑊

∂𝑈

∂𝑧

Similarly, by expanding, or by inspection of the result above:

D𝑉

D𝑡=∂𝑉

∂𝑡+ 𝑈

∂𝑉

∂𝑥+ 𝑉

∂𝑉

∂𝑦+𝑊

∂𝑉

∂𝑧

(b) It is required to sum over all 𝑖 and all 𝑗. A good way is to separate parts 𝑖 = 𝑗 and 𝑖 ≠ 𝑗:

𝑃(𝑘) = − (𝑢2∂𝑈

∂𝑥+ 𝑣2

∂𝑉

∂𝑦+ 𝑤2

∂𝑊

∂𝑧)

−𝑣𝑤 (∂𝑉

∂𝑧+∂𝑊

∂𝑦) − 𝑤𝑢 (

∂𝑊

∂𝑥+∂𝑈

∂𝑧) − 𝑢𝑣 (

∂𝑈

∂𝑦+∂𝑉

∂𝑥)

(c) When 𝑖 = 𝑗 the two parts of the sum are the same, so

𝑃11 = −2𝑢1𝑢𝑘∂𝑈1∂𝑥𝑘

= −2 (𝑢1𝑢1∂𝑈1∂𝑥1

+ 𝑢1𝑢2∂𝑈1∂𝑥2

+ 𝑢1𝑢3∂𝑈1∂𝑥3

)

= −2 (𝑢𝑢∂𝑈

∂𝑥+ 𝑢𝑣

∂𝑈

∂𝑦+ 𝑢𝑤

∂𝑈

∂𝑤)

When 𝑖 ≠ 𝑗 there are, unfortunately, twice as many terms:

𝑃12 = −𝑢1𝑢𝑘∂𝑈2∂𝑥𝑘

− 𝑢2𝑢𝑘∂𝑈1∂𝑥𝑘

= − (𝑢1𝑢1∂𝑈2∂𝑥1

+ 𝑢1𝑢2∂𝑈2∂𝑥2

+ 𝑢1𝑢3∂𝑈2∂𝑥3

) − (𝑢2𝑢1∂𝑈1∂𝑥1

+ 𝑢2𝑢2∂𝑈1∂𝑥2

+ 𝑢2𝑢3∂𝑈1∂𝑥3

)

= − (𝑢𝑢∂𝑉

∂𝑥+ 𝑢𝑣

∂𝑉

∂𝑦+ 𝑢𝑤

∂𝑉

∂𝑧) − (𝑣𝑢

∂𝑈

∂𝑥+ 𝑣𝑣

∂𝑈

∂𝑦+ 𝑣𝑤

∂𝑈

∂𝑧)

(d)

𝑀𝑖𝑖 = 𝑀11 +𝑀22 +𝑀33 = trace(M) (also written as {M} )

𝑀𝑖𝑗𝑀𝑗𝑖 = (M2)𝑖𝑖 (by definition of matrix multiplication)

= (M2)11 + (M2)22 + (M

2)33

= trace(M2) (also written as {M2} )

Similarly,

𝑀𝑖𝑗𝑀𝑗𝑘𝑀𝑘𝑖 = trace(M3)


Q4.

(a) By definition,

𝑃(𝑘) ≡ −𝑢𝑖𝑢𝑗∂𝑈𝑖∂𝑥𝑗

This is a sum over all 𝑖 and 𝑗. However, if ∂𝑈/ ∂𝑦 is the only non-zero mean-velocity gradient then there is only a non-zero contribution when 𝑖 = 1 and 𝑗 = 2. Hence, in simple shear,

𝑃(𝑘) → −𝑢1𝑢2∂𝑈1∂𝑥2

= −𝑢𝑣∂𝑈

∂𝑦

(b) By definition,

𝑃𝑖𝑗 ≡ −(𝑢𝑖𝑢𝑘∂𝑈𝑗

∂𝑥𝑘+ 𝑢𝑗𝑢𝑘

∂𝑈𝑖∂𝑥𝑘

)

This is a sum over all 𝑘. However, if ∂𝑈/ ∂𝑦 is the only non-zero mean-velocity gradient then there is only a non-zero contribution to the first term when 𝑗 = 1 and 𝑘 = 2 and to the second term when 𝑖 = 1 and 𝑘 = 2. Hence, in simple shear,

𝑃11 = −2𝑢𝑣∂𝑈

∂𝑦, 𝑃22 = 𝑃33 = 0

𝑃12 = −𝑣2∂𝑈

∂𝑦 , 𝑃23 = 𝑃31 = 0 (𝑃𝑗𝑖 = 𝑃𝑖𝑗 in any flow)

(c) By definition,

𝑆𝑖𝑗 =1

2 (∂𝑈𝑖∂𝑥𝑗

+∂𝑈𝑗

∂𝑥𝑖)

In simple shear,

(𝑆𝑖𝑗) =

(

0

1

2

∂𝑈

∂𝑦0

1

2

∂𝑈

∂𝑦0 0

0 0 0)

𝑆2 ≡ 2𝑆𝑖𝑗𝑆𝑖𝑗 = 2 × (sum of elements)2 = 2 × 2 × (

1

2

∂𝑈

∂𝑦)2

= (∂𝑈

∂𝑦)2

Hence,

𝑆 = |∂𝑈

∂𝑦|


(d) By definition,

Ω𝑖𝑗 =1


−∂𝑈𝑗

∂𝑥𝑖)

In simple shear,

(Ω𝑖𝑗) =

(

01

2

∂𝑈

∂𝑦0

−1

2

∂𝑈

∂𝑦0 0

0 0 0)

Ω2 = 2Ω𝑖𝑗Ω𝑖𝑗 = 2 × (sum of elements)2 = 2 × 2 × (

1

2

∂𝑈

∂𝑦)2

= (∂𝑈

∂𝑦)2

Hence,

Ω = |∂𝑈

∂𝑦|


Q5.

(a)

𝑃(𝑘) ≡ −𝑢𝑖𝑢𝑗∂𝑈𝑖∂𝑥𝑘

= −𝑢𝑖𝑢𝑗(𝑆𝑖𝑗 + Ω𝑖𝑗) (*)

where

𝑆𝑖𝑗 ≡1


+∂𝑈𝑖∂𝑥𝑗) = mean strain (symmetric)

Ω𝑖𝑗 ≡1


−∂𝑈𝑖∂𝑥𝑗) = mean vorticity (antisymmetric)

But 𝑢𝑖𝑢𝑗 is symmetric, whereas Ω𝑖𝑗 = −Ω𝑗𝑖 so that all Ω-related contributions to (*) will cancel

in pairs. Hence,

𝑃(𝑘) = −𝑢𝑖𝑢𝑗𝑆𝑖𝑗

(b) The anisotropy tensor is defined by

𝑎𝑖𝑗 =𝑢𝑖𝑢𝑗

𝑘−2

3δ𝑖𝑗

Hence,

𝑢𝑖𝑢𝑗 = 𝑘𝑎𝑖𝑗 +2

3𝑘δ𝑖𝑗

Then,

𝑃(𝑘) = −𝑢𝑖𝑢𝑗𝑆𝑖𝑗

= −(𝑘𝑎𝑖𝑗 +2

3𝑘δ𝑖𝑗)𝑆𝑖𝑗

But for the second term in the summation:

δ𝑖𝑗𝑆𝑖𝑗 = 𝑆𝑖𝑖 = 𝑆11 + 𝑆22 + 𝑆33 =∂𝑈

∂𝑥+∂𝑉

∂𝑦+∂𝑊

∂𝑧 = 0 (incompressibility)

Hence, in incompressible flow,

𝑃(𝑘) = −𝑘𝑎𝑖𝑗𝑆𝑖𝑗

(c) The eddy-viscosity hypothesis gives

−𝑢𝑖𝑢𝑗 = ν𝑡𝑆𝑖𝑗 −2

3𝑘δ𝑖𝑗

−𝑘𝑎𝑖𝑗 = ν𝑡𝑆𝑖𝑗

Hence, in incompressible flow,


𝑃(𝑘) = ν𝑡𝑆𝑖𝑗𝑆𝑖𝑗 = ν𝑡 × (sum of (elements of 𝑆𝑖𝑗 squared))

≥ 0


Q6.

(a)

ϕ = 𝑘𝑚ε𝑛 ⇒ ε = ϕ1/𝑛𝑘−𝑚/𝑛

Hence,

ν𝑡 = 𝐶μ𝑘2

ε = 𝐶μ𝑘

2+𝑚/𝑛ϕ−1/𝑛

(b) For ANY differential of ϕ,

dϕ = d(𝑘𝑚ε𝑛)

= 𝑚𝑘𝑚−1ε𝑛d𝑘 + 𝑘𝑚𝑛ε𝑛−1dε

Hence,

dϕ

ϕ= 𝑚

d𝑘

𝑘+ 𝑛

dε

ε

In particular,

1

ϕ

Dϕ

D𝑡=𝑚

𝑘

D𝑘

D𝑡+𝑛

ε

Dε

D𝑡 and

1

ϕ∇ϕ =

𝑚

𝑘∇𝑘 +

𝑛

ε∇ε

From the first of these,

𝑛

ε

Dε

D𝑡=1

ϕ

Dϕ

D𝑡−𝑚

𝑘

D𝑘

D𝑡

=1

ϕ{∇ • (Γ(ϕ)∇ϕ) + (𝐶ϕ1𝑃

(𝑘) − 𝐶ϕ2ε)ϕ

𝑘+ 𝑆(ϕ)} −

𝑚

𝑘{∇ • (Γ(𝑘)∇𝑘) + 𝑃(𝑘) − ε}

=1

ϕ∇ • (Γ(ϕ)∇ϕ) −

𝑚

𝑘∇ • (Γ(𝑘)∇𝑘) + {(𝐶ϕ1 −𝑚)𝑃

(𝑘) − (𝐶ϕ2 −𝑚)ε}1

𝑘+𝑆(ϕ)

ϕ

(*)

The first two terms on the RHS of (*) give: 1

ϕ∇ • (Γ(ϕ)∇ϕ) −

𝑚

𝑘∇ • (Γ(𝑘)∇𝑘)

=1

ϕ∇ • [Γ(ϕ)(

𝑚

𝑘∇𝑘 +

𝑛

ε∇ε)ϕ] −

𝑚

𝑘∇ • (Γ(𝑘)∇𝑘)

= ∇ • [Γ(ϕ)(𝑚

𝑘∇𝑘 +

𝑛

ε∇ε)] + Γ(ϕ)(

𝑚

𝑘∇𝑘 +

𝑛

ε∇ε) •

∇ϕ

ϕ−𝑚

𝑘∇ • (Γ(𝑘)∇𝑘)

=𝑚

𝑘∇ • (Γ(ϕ)∇𝑘) − Γ(ϕ)

𝑚

𝑘2(∇𝑘)2 +

𝑛

ε∇ • (Γ(ϕ)∇ε) − Γ(ϕ)

𝑛

ε2(∇ε)2 + Γ(ϕ)(

𝑚

𝑘∇𝑘 +

𝑛

ε∇ε)2 −

𝑚

𝑘∇ • (Γ(𝑘)∇𝑘)

=𝑚

𝑘∇ • [(Γ(ϕ) − Γ(𝑘))∇𝑘] +

𝑛

ε∇ • (Γ(ϕ)∇ε) + Γ(ϕ) [(𝑚

∇𝑘

𝑘+ 𝑛

∇ε

ε)2 −𝑚(

∇𝑘

𝑘)2 − 𝑛(

∇ε

ε)2]

=𝑚

𝑘∇ • [(Γ(ϕ) − Γ(𝑘))∇𝑘] +

𝑛

ε∇ • (Γ(ϕ)∇ε) + Γ(ϕ) [𝑚(𝑚 − 1)(

∇𝑘

𝑘)2 + 𝑛(𝑛 − 1)(

∇ε

ε)2 + 2𝑚𝑛(

∇𝑘

𝑘) • (

∇ε

ε)]

Substituting into (*) and multiplying by ε/𝑛 gives


Dε

D𝑡= ∇ • (Γ(ϕ)∇ε) + {(

𝐶ϕ1 −𝑚

𝑛)𝑃(𝑘) − (

𝐶ϕ2 −𝑚

𝑛)ε}ε

𝑘

+ε

𝑛{𝑆(ϕ)

ϕ+𝑚

𝑘∇ • [(Γ(ϕ) − Γ(𝑘))∇𝑘] + Γ(ϕ) [𝑚(𝑚 − 1)(

∇𝑘

𝑘)2 + 𝑛(𝑛 − 1)(

∇ε

ε)2 + 2𝑚𝑛(

∇𝑘

𝑘) • (

∇ε

ε)]}

This is of the form:

Dε

D𝑡= ∇ • (Γ(ε)∇ε) + {𝐶ε1𝑃

(𝑘) − 𝐶ε2ε}ε

𝑘+ 𝑆ε

where 𝑆ε is of the form given in the question and

Γ(ε) = Γ(ϕ)

𝐶ε1 =𝐶ϕ1 −𝑚

𝑛

𝐶ε2 =𝐶ϕ2 −𝑚

𝑛


Q7.

In simple shear flow,

(𝑠𝑖𝑗) =

(

0

1

2σ 0

1

2σ 0 0

0 0 0)

=

1

2σ(0 1 01 0 00 0 0

)

(ω𝑖𝑗) =

(

01

2σ 0

−1

2σ 0 0

0 0 0)

=

1

2σ(

0 1 0−1 0 00 0 0

)

Hence,

s =σ

2(0 1 01 0 00 0 0

), 𝛚 =σ

2(0 1 0−1 0 00 0 0

)

s2 =σ2

4(1 0 00 1 00 0 0

), {s2} =σ2

2, s

2 −1

3{s2}I =

σ2

12(1 0 00 1 00 0 −2

)

ωs = σ2

4(1 0 00 −1 00 0 0

), sω = σ2

4(−1 0 00 1 00 0 0

), ωs− sω=σ2

2(1 0 00 −1 00 0 0

)

ω2 =σ2

4(−1 0 00 −1 00 0 0

), {ω2} = −σ2

2, ω

2 −1

3{ω2}I = −

σ2

12(1 0 00 1 00 0 −2

)

Substituting these in the constitutive relationship,

a = (𝑢𝑖𝑢𝑗

𝑘−2

3δ𝑖𝑗) = −𝐶μσ(

0 1 01 0 00 0 0

)

+β1σ2

12(1 0 00 1 00 0 −2

) + β2σ2

2(1 0 00 −1 00 0 0

) − β3σ2

12(1 0 00 1 00 0 −2

)

Extracting components:

𝑢2

𝑘=2

3+ (β1 + 6β2 − β3)

σ2

12

𝑣2

𝑘=2

3+ (β1 − 6β2 − β3)

σ2

12

𝑤2

𝑘=2

3− (β1 − β3)

σ2

6


𝑢𝑣

𝑘= −𝐶μσ

𝑣𝑤

𝑘=𝑤𝑢

𝑘= 0


Q8.

(a) In plane strain,

s = σ(1 0 00 −1 00 0 0

)

ω = 𝟎

Hence,

s2 = σ2 (1 0 00 1 00 0 0

), {s2} = 2σ2, s2 −1

3{s2}I =

σ2

3(1 0 00 1 00 0 −2

)

ωs− sω = ω2 −1

3{ω2}I = 𝟎



𝑘−2

3δ𝑖𝑗) = −2𝐶μσ(

1 0 00 −1 00 0 0

) + β1σ2

3(1 0 00 1 00 0 −2

)


𝑢2

𝑘=2

3− 2𝐶μσ +

β13σ2

𝑣2

𝑘=2

3+ 2𝐶μσ +

β13σ2

𝑤2

𝑘=2

3−2β13σ2

𝑢𝑣

𝑘=𝑣𝑤

𝑘=𝑤𝑢

𝑘= 0

(b) In axisymmetric strain,

s =σ

2(2 0 00 −1 00 0 −1

)

ω = 𝟎

Hence,

s2 =σ2

4(4 0 00 1 00 0 1

), {s2} =6

4σ2, s2 −

1

3{s2}I =

σ2

4(2 0 00 −1 00 0 −1

)


ωs− sω = ω2 −1

3{ω2}I = 𝟎



𝑘−2

3δ𝑖𝑗) = −𝐶μσ(

2 0 00 −1 00 0 −1

) + β1σ2

4(2 0 00 −1 00 0 −1

)


𝑢2

𝑘=2

3− 2𝐶μσ +

β12σ2

𝑣2

𝑘=𝑤2

𝑘=2

3+ 𝐶μσ −

β14σ2

𝑢𝑣

𝑘=𝑣𝑤

𝑘=𝑤𝑢

𝑘= 0


Q9.

(a)

𝑃𝑖𝑗 is the rate of production of from the mean flow.

Φ𝑖𝑗 is the rate of redistribution of energy amongst components by pressure forces.

ε𝑖𝑗 is the rate of dissipation by viscous action.

(b)

The purpose of is to try to maintain the correct level of anisotropy near a boundary (the

rest of Φ𝑖𝑗 is trying to isotropise the turbulence).

In an equilibrium turbulent boundary layer 𝑓 = 1.

(c)

Production

𝑃11 = −2𝑢𝑣̅̅̅̅𝜕𝑈

𝜕𝑦 = 2𝑃(𝑘), 𝑃22 = 𝑃33 = 0

𝑃12 = −2𝑣2̅̅ ̅𝜕𝑈

𝜕𝑦 =

𝑣2̅̅ ̅

𝑢𝑣̅̅̅̅𝑃(𝑘), 𝑃23 = 𝑃31 = 0

where

𝑃(𝑘) = −𝑢𝑣̅̅̅̅𝜕𝑈

𝜕𝑦

Dissipation

ε11 = ε22 = ε33 =2

3ε

ε23 = ε31 = ε12 = 0

Pressure-strain (return-to-isotropy part)

Φ11(1) = −𝐶1ε (

𝑢2̅̅ ̅

𝑘−2

3), Φ22

(1) = −𝐶1ε (𝑣2̅̅ ̅

𝑘−2

3), Φ33

(1) = −𝐶1ε (𝑤2̅̅ ̅̅

𝑘−2

3)

Φ23(1) = −𝐶1ε (

𝑣𝑤̅̅ ̅̅

𝑘), Φ31

(1) = −𝐶1ε (𝑤𝑢̅̅ ̅̅

𝑘), Φ12

(1) = −𝐶1ε (𝑢𝑣̅̅̅̅

𝑘)

Pressure-strain (rapid part)

Φ11(2) = −

4

3𝐶2𝑃

(𝑘), Φ22(2) = Φ33

(2) =2

3𝐶2𝑃

(𝑘)

Φ23(2) = Φ31

(2) = 0, Φ12(2) = −𝐶2

𝑣2̅̅ ̅

𝑢𝑣̅̅̅̅ 𝑃(𝑘)

jiuu

)(Φ wij


Pressure-strain (wall-reflection)

Φ11(𝑤) = Φ̃22, Φ22

(𝑤) = −2Φ̃22, Φ33(𝑤) = Φ̃22

Φ23(𝑤) = 0, Φ12

(𝑤) = −3

2Φ̃12, Φ31

(𝑤) = 0

where

Φ̃22 = 𝐶1(𝑤)ε

𝑣2̅̅ ̅

𝑘+2

3𝐶2(𝑤)𝐶2𝑃

(𝑘), Φ̃12 = 𝐶1(𝑤)ε

𝑢𝑣̅̅̅̅

𝑘− 𝐶2

(𝑤)𝐶2𝑣2̅̅ ̅

𝑢𝑣̅̅̅̅𝑃(𝑘)

(d)

Setting 𝑃22 +Φ22 − 𝜀22 = 0:

0 − 𝐶1ε (𝑣2̅̅ ̅

𝑘−2

3) +

2

3𝐶2𝑃

(𝑘) − 2 [𝐶1(𝑤)ε

𝑣2̅̅ ̅

𝑘+2


(𝑘)] −2

3ε = 0

Dividing through by 𝑃(𝑘) (= ε):

−(𝐶1 + 2𝐶1(𝑤))

𝑣2̅̅ ̅

𝑘+2

3𝐶1 +

2

3𝐶2 −

4

3𝐶2(𝑤)𝐶2 −

2

3= 0

2

3(−1 + 𝐶1 + 𝐶2 − 2𝐶2

(𝑤)𝐶2) = (𝐶1 + 2𝐶1(𝑤))

𝑣2̅̅ ̅

𝑘

𝑣2̅̅ ̅

𝑘=2

3(−1 + 𝐶1 + 𝐶2 − 2𝐶2

(𝑤)𝐶2

𝐶1 + 2𝐶1(𝑤)

)

Setting 𝑃11 +Φ11 − 𝜀11 = 0:

2𝑃(𝑘) − 𝐶1ε (𝑢2̅̅ ̅

𝑘−2

3) −

4

3𝐶2𝑃

(𝑘) + 𝐶1(𝑤)ε

𝑣2̅̅ ̅

𝑘+ 2


(𝑘) −2

3ε = 0

Dividing through by 𝑃(𝑘) (= ε):

2 − 𝐶1 (𝑢2̅̅ ̅

𝑘−2

3) −

4

3𝐶2 + 𝐶1

(𝑤) 𝑣2̅̅ ̅

𝑘+2

3𝐶2(𝑤)𝐶2 −

2

3= 0

4

3+2

3𝐶1 −

4

3𝐶2 + 𝐶1

(𝑤) 𝑣2̅̅ ̅

𝑘+2

3𝐶2(𝑤)𝐶2 = 𝐶1

𝑢2̅̅ ̅

𝑘

𝑢2̅̅ ̅

𝑘=2

3(2 + 𝐶1 − 2𝐶2 + 𝐶2

(𝑤)𝐶2𝐶1

) +𝐶1(𝑤)

𝐶1

𝑣2̅̅ ̅

𝑘

The other results follow similarly.


Q10.

Expand each velocity component in a Taylor series about 𝑦 = 0:

𝑢 = 𝑎1 + 𝑏1𝑦 + 𝑐1𝑦2 +⋯

𝑣 = 𝑎2 + 𝑏2𝑦 + 𝑐2𝑦2 +⋯

𝑤 = 𝑎3 + 𝑏3𝑦 + 𝑐3𝑦2 +⋯

where the 𝑎𝑖, 𝑏𝑖, etc. are functions of 𝑥, 𝑧 and 𝑡.

Non-slip condition:

𝑢 = 𝑣 = 𝑤 = 0 on 𝑦 = 0

𝑎1 = 𝑎2 = 𝑎3 = 0

Continuity:

∂𝑢

∂𝑥+∂𝑣

∂𝑦+∂𝑤

∂𝑧= 0

∂𝑣

∂𝑦= −

∂𝑢

∂𝑥−∂𝑤

∂𝑧 = 0

𝑏2 = 0

Hence, to leading order:

𝑢 = 𝑏1𝑦 +⋯

𝑣 = 𝑐2𝑦2 +⋯

𝑤 = 𝑏3𝑦 +⋯ (*)

Hence,

𝑢2 = 𝑏22𝑦2 + O(𝑦3)

𝑣2 = 𝑐22𝑦4 + O(𝑦5)

𝑤2 = 𝑏32𝑦2 + O(𝑦3)

𝑢𝑣 = 𝑏1𝑐2𝑦3 + O(𝑦4)

𝑘 ≡1

2(𝑢2 + 𝑣2 + 𝑤2) =

1

2(𝑏12 + 𝑏3

2)𝑦2 + 𝑂(𝑦3)

By definition,

ν𝑡 =−𝑢𝑣

∂𝑈/ ∂𝑦

But

τ𝑤 ≡ ρ𝑢τ2 = μ

∂𝑈

∂𝑦− ρ𝑢𝑣

𝑢τ2 = ν

∂𝑈

∂𝑦+ O(𝑦3)


∂𝑈

∂𝑦=𝑢τ2

ν+ O(𝑦3)

Hence,

ν𝑡 = −𝑏1𝑐2ν

𝑢τ2𝑦3 + O(𝑦4)


Q11.

The 𝑘 equation is

D𝑘

D𝑡=∂

∂𝑥𝑗{(ν +

ν𝑡σ(𝑘)

)∂𝑘

∂𝑥𝑗} + 𝑃(𝑘) − ε

As 𝑦 → 0 this reduces to

0 =∂

∂𝑦(ν∂𝑘

∂𝑦) − ε

ε = ν∂2𝑘

∂𝑦2 (*)

But, from Q12,

𝑘 = 𝑘0𝑦2 + O(𝑦3)

whence,

∂2𝑘

∂𝑦2= 2𝑘0 + O(𝑦)

= 2𝑘

𝑦2+ O(𝑦)

Hence, from (*),

ε =2ν𝑘

𝑦2+ O(𝑦) (𝑦 → 0)


Q12.

(a) Assuming constant shear stress (ρ𝑢02) the mean-velocity profile can be derived from the

viscosity law in viscous and turbulent regions, with the former providing a boundary condition

for the latter at 𝑦 = 𝑦𝜈.

Viscous Sublayer (𝑦 < 𝑦𝜈)

τ𝑤ρ= ν

∂𝑈

∂𝑦, 𝑈(0) = 0

This integrates immediately to give

𝑈 =τ𝑤ρν𝑦

or, in a non-dimensional form,

𝑈

𝑢0=τ𝑤

ρ𝑢02

𝑢0𝑦

ν

Turbulent Region (𝑦 > 𝑦𝜈)

τ𝑤ρ= [ν + κ𝑢0(𝑦 − 𝑦ν)]

∂𝑈

∂𝑦, 𝑈(𝑦ν) =

τ𝑤ρν𝑦

Rearrange:

∂𝑈

∂𝑦=

τ𝑤/ρ

ν + κ𝑢0(𝑦 − 𝑦ν)

Integrate, using the velocity at 𝑦 = 𝑦𝜈 as a lower limit:

[𝑈]𝑦ν𝑦=τ𝑤/ρ

κ𝑢0[ln{ν + κ𝑢0(𝑦 − 𝑦ν)}]𝑦𝜈

𝑦

𝑈 −τ𝑤/ρ

ν𝑦ν =

τ𝑤/ρ

κ𝑢0lnν + κ𝑢0(𝑦 − 𝑦ν)

ν

𝑈

𝑢0=τ𝑤

ρ𝑢02 {𝑢0𝑦νν+𝑢0κln [1 +

κ𝑢0(𝑦 − 𝑦ν)

ν]}

Hence, the complete mean velocity profile is:

𝑈

𝑢0=τ𝑤

ρ𝑢02 × {

𝑦+ 𝑦 ≤ 𝑦ν

𝑦ν+ +

1

κln[1 + κ(𝑦+ − 𝑦ν

+)] 𝑦 > 𝑦ν


(b) The cell-averaged rate of production of turbulent kinetic energy is

𝑃av(𝑘)≡1

Δ∫ 𝑃(𝑘) d𝑦Δ

0

=1

Δ∫ ν𝑡 (

∂𝑈

∂𝑦)2

d𝑦Δ

𝑦ν

=1

Δ∫ κ𝑢0(𝑦 − 𝑦ν) ×

(τ𝑤/ρ)2

{ν + κ𝑢0(𝑦 − 𝑦ν)}2d𝑦

Δ

𝑦ν

To simplify the integral, change variables to

𝑌 =κ𝑢0(𝑦 − 𝑦ν)

ν, d𝑦 =

ν

κ𝑢0d𝑌

Then

𝑃av(𝑘)=1

Δ∫ ν𝑌 ×

(τ𝑤/ρ)2

{ν + ν𝑌}2ν

κ𝑢0d𝑌

κ𝑢0(Δ−𝑦ν)ν

0

=(τ𝑤/ρ)

2

κ𝑢0Δ∫

𝑌

(1 + 𝑌)2 d𝑌

κ(Δ+−𝑦ν+)

0

=(τ𝑤/ρ)

2

κ𝑢0Δ∫ {

1

1 + 𝑌−

1

(1 + 𝑌)2} d𝑌

κ(Δ+−𝑦ν+)

0

=(τ𝑤/ρ)

2

κ𝑢0Δ[ln(1 + 𝑌) +

1

1 + 𝑌]0

κ(Δ+−𝑦ν+)

=(τ𝑤/ρ)

2

κ𝑢0Δ{ln[1 + κ(Δ+ − 𝑦ν

+)] +1

1 + κ(Δ+ − 𝑦ν+)− 1}

=(τ𝑤/ρ)

2

κ𝑢0Δ{ln[1 + κ(Δ+ − 𝑦ν

+)] −κ(Δ+ − 𝑦ν

+)

1 + κ(Δ+ − 𝑦ν+)}

(b) The cell-averaged dissipation rate is

εav ≡1

Δ∫ ε d𝑦Δ

0

=1

Δ(∫ ε𝑤 d𝑦

𝑦ε

0

+∫𝑢03

κ(𝑦 − 𝑦𝑑) d𝑦

Δ

𝑦ε

) , where ε𝑤 =𝑢03

κ(𝑦ε − 𝑦𝑑)

=𝑢03

κΔ(

𝑦ε𝑦ε − 𝑦𝑑

+∫d𝑦

𝑦 − 𝑦𝑑

Δ

𝑦ε

)

=𝑢03

κΔ[𝑦ε

𝑦ε − 𝑦𝑑+ ln (

Δ − 𝑦𝑑𝑦ε − 𝑦𝑑

)]


Q13.

(a) Equation (*) can be written conveniently as

𝑈 =𝑢τκln (𝐸

𝑢τ𝑦

ν), where 𝐸 =

eκ𝐵

1 + 𝑐𝑘𝑠+

The total volumetric flow rate in the pipe is

𝑄 ≡ 𝑈avπ𝑅2 = ∫ 𝑈. 2π𝑟 d𝑟

𝑅

0

or, writing 𝑟 = 𝑅 − 𝑦, d𝑟 = −d𝑦 and switching the integral limits:

𝑈avπ𝑅2 = 2π

𝑢τκ∫ ln (𝐸

𝑢τ𝑦

ν) (𝑅 − 𝑦)d𝑦

𝑅

0

Substituting the boundary-layer coordinate,

η =𝑦

𝑅 or 𝑦 = 𝑅η, d𝑦 = 𝑅dη

then

𝑈av𝑢τ=2

κ∫ ln (𝐸

𝑢τ𝑅

νη) (1 − η) dη

1

0

=2

κ{[ln(𝐸

𝑢τ𝑅

νη)(η −

η2

2)]0

1

−∫ (1 −1

2η) dη

1

0

}

=2

κ{1

2ln(𝐸

𝑢τ𝑅

ν) − [η−

1

4η2]

0

1

}

=1

κ{ln(𝐸

𝑢τ𝑅

ν) −

3

2}

Hence

𝑈av𝑢τ=1

κln (

𝐸𝑒−3/2

2×𝑢τ𝑈𝑎𝑣

×𝑈𝑎𝑣𝐷

ν)

But

𝑢τ𝑈av

= √𝑐𝑓

2

Hence

√2

𝑐𝑓=1

κln (

𝐸e−3/2

2× Re√

𝑐𝑓

2)

or

1

√𝑐𝑓=

1

κ√2ln (

eκ𝐵−3/2

2√2

Re√𝑐𝑓

1 + 𝑐𝑘𝑠+)


But

𝑘𝑠+ =

𝑢τ𝑘𝑠ν=𝑢τ𝑈𝑎𝑣

×𝑘𝑠𝐷×𝑈𝑎𝑣𝐷

ν

= √𝑐𝑓

2×𝑘𝑠𝐷× Re

Hence

1

√𝑐𝑓=

1

κ√2ln (

eκ𝐵−3/2

2√2×

Re√𝑐𝑓

1 + 𝑐√𝑐𝑓/2Re(𝑘𝑠/𝐷))

or, inverting the argument of the logarithm and rearranging:

1

√𝑐𝑓=−1

κ√2ln (

2√2e3/2−κ𝐵

Re√𝑐𝑓+ 2e3/2−κ𝐵𝑐

𝑘𝑠𝐷)

(b) To compare with the Colebrook-White formula change the base of logarithms:

1

√𝑐𝑓=

−1

κ√2log10elog10 (

2√2e3/2−κ𝐵

Re√𝑐𝑓+ 2e3/2−κ𝐵𝑐

𝑘𝑠𝐷)

Comparing with the Colebrook-White formula:

1

√𝑐𝑓= −4.0log10 (

1.26

Re√𝑐𝑓+𝑘𝑠3.7𝐷

)

gives

1

κ√2log10e= 4.0 or κ =

1

4.0√2log10e= 0.4070

2√2e3/2−κ𝐵 = 1.26 or 𝐵 =1

κ(3

2− ln(

1.26

2√2)) = 5.672

2e3/2−κ𝐵𝑐 =1

3.7 or 𝑐 =

√2

3.7 × 1.26= 0.3033

Answer: 𝜅 = 0.41, 𝐵 = 5.7, 𝑐 = 0.30

(c) Rewrite equation (*) as

𝑈+ =1

κln𝑦+ + 𝐵 −

1

κln(1 + 𝑐𝑘𝑠

+)

Hence,

𝑈+ =1

κln𝑦+ + �̃�(𝑘𝑠

+)

where


�̃�(𝑘𝑠+) = 𝐵 −

1

κln(1 + 𝑐𝑘𝑠

+)

(d) The last can be further rearranged as

�̃�(𝑘𝑠+) = 𝐵 −

1

κln 𝑐 −

1

κln(𝑘𝑠

+ +1

𝑐)

The mean-velocity profile is then

𝑈+ =1

κln

𝑦+

𝑘𝑠+ +

1𝑐

+ 𝐵 −1

κln 𝑐

In the limit as 𝑘𝑠+ becomes very large, this asymptotes to

𝑈+ =1

κln𝑦

𝑘𝑠+ 𝐵𝑘

where

𝐵𝑘 = 𝐵 −1

κln 𝑐 = 8.60

Answer: 𝐵𝑘 = 8.6.

answers 10 spring 2021 - university of manchester · 2021. 1. 1. · cfd answers 10 – 1 david...

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