From Eq.(4) we obtain τρw = 2νU δ
0
. Substituting this formula into the defini-
tion of cf in Eq. (7.14), we obtain
.  2νU0 2 4
c f ≡ τw 1
2
ρU02 = 2
= . (7)
δ U0 Re

Similarly, from the definition of Cf in Eq. (7.15), we have
.  9
Cf ≡ τw 1
2
ρU¯ 2 = . (8)
Re0
Then Eqs.(7.17 - 7.20), which are rewritten at following, can be very easily
obtained from Eqs.(4-8):
τw δ
U0 = = 32 U¯ , (9)
2ρν

Re = 43 Re0 , (10)

4 16
cf = = , (11)
Re0 3Re
and
9 12
Cf = = . (12)
Re0 Re
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2
 Turbulent Flows
Stephen B. Pope
Cambridge University Press (2000)

Solution to Exercise 7.2
Prepared by: Renfeng Cao Date: 3/25/03


The friction velocity is defined by
q
uτ = τw /ρ. (1)

From this definition we can get

τw /ρ = u2τ . (2)

Substituting Eq. (2) into the definition of cf in Eq. (7.14), we obtain
. 
cf = u2τ 1
2
U02 = 2(uτ /U0 )2 . (3)

For laminar flow, from Eq. (7.19) in Ex. (7.1), we obtain
s s
uτ cf 2 8
r
= = = . (4)
U0 2 Re0 3Re

For the upper limit of laminar flow, i.e., Re=1,350, we have
uτ
= 0.044. (5)
U0

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1
 Turbulent Flows
Stephen B. Pope
Cambridge University Press (2000)

Solution to Exercise 7.3
Prepared by: Zhuyin Ren Date: 3/27/03


According to Re = 2δ U¯ /ν, we get
Re ν
U¯ = = 105 × 1.14 × 10−6 /(2 × 2 × 10−2 ) = 2.85m/s. (1)
2δ
q
τw τw
According to Cf = 1 ¯2
ρU
and uτ = ρ
, we get
2

 12 !1
uτ Cf 4.4 × 10−3 2

= = = 4.69 × 10−2 . (2)
U¯ 2 2
uτ δ
According to Reτ = ν
, we get

uτ 2δ U¯ uτ
Reτ = ¯ = ¯ Re = 4.69 × 10−2 × 105 /2 = 2345. (3)
2U ν 2U
And
ν 1
δν /δ = = = 4.26 × 10−4 . (4)
uτ δ Reτ
The viscous wall region corresponds to y + < 50, so
50δ
T he thickness of the viscous region = 50δν = = 0.426mm.
Reτ
The viscous sublayer corresponds to y + < 5, so
5δ
T he thickness of the viscous sublayer = 5δν = = 0.0426mm.
Reτ
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1
 Turbulent Flows
Stephen B. Pope
Cambridge University Press (2000)

Solution to Exercise 7.4
Prepared by: Zhuyin Ren Date: 3/27/03


Equation (5.139) shows the turbulent kinetic energy evolves by
¯
Dk
¯ + ∇ · T0 = P − ε, (1)
Dt
where
1
Ti0 ≡ hui uj uj i + hui p0 i/ρ − 2νhuj sij i. (2)
2

Equation (5.164) shows that the turbulent kinetic energy equation can alter-
natively be written as

Dk ∂ h1 0
i
+ hu u u
i j j i + hu i p i/ρ = ν∇2 k + P − ε˜. (3)
Dt ∂xi 2
For fully developed channel flow, it is statistically stationary and statistically
one-dimensional, with velocity statistics depending only on y. So Eq. 1 can
be reexpressed as
!
d 1 ∂v 1 ∂(uj uj )
hvu.ui + hvp0 i/ρ − νhuj + i = P − ε, (4)
dy 2 ∂xj 2 ∂y

With divergence-free ( ∂u
∂xj
j
= 0) and k = 12 uj uj , we get
!
d 1 d
hvu.ui + hvp0 i/ρ − ν (k + hv 2 i) = P − ε. (5)
dy 2 dy

Equation 3 can be reexpressed as

d 1 d2 k
 
hvu.ui + hvp0 i/ρ = ν 2 + P − ε˜, (6)
dy 2 dy

1
i.e. !
d 1 dk
hvu.ui + hvp0 i/ρ − ν = P − ε˜. (7)
dy 2 dy
Comapring Eq. 5 with Eq. 7, we get

d2 hv 2 i
ε = ε˜ + ν . (8)
dy 2

Using Eq.(5.161) (see Exercise 5.25) for fully developed channel flow, we can
also get the above relation between ε and ε˜.

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2
 Turbulent Flows
Stephen B. Pope
Cambridge University Press (2000)

Solution to Exercise 7.5
Prepared by: Michael Stumpf Date: 12/08/17


The expansions near the wall are

u = b1 y + O(y 2 ),
v = c2 y 2 + O(y 3 ),
w = b3 y + O(y 2 ).

a) In channel flow, hV i = hW i = 0 and statistics only vary in y direction.
...