Turbulence Prof. E.G.

Tulapurkara
Chapter-3

Chapter 3

Lecture 18

Equations for turbulent flows - 3

Topics

3.6. Momentum integral equation for turbulent boundary layer

3.7 Reynolds Average equations for compressible flow
3.7.1 Averaging procedures
3.7.2 Reynolds form of continuity equation for compressible flow
3.7.3 Reynolds form of momentum equations
3.7.4 Reynolds form of energy equation

3.6. Momentum integral equation for turbulent boundary layer

In Appendix C, section C.10, the momentum integral equation for the laminar boundary
layer is derived by integrating the boundary layer equations across the boundary layer.
In a similar manner, the momentum integral equation for turbulent case can be obtained
by integrating, across the boundary layer, the boundary layer equations for turbulent
flow. The momentum integral equation is used (a) as a check on numerical calculations
of thin shear flow and (b) in approximate methods called ‘integral methods’ to predict
the boundary layers.
The momentum integral equation for a two-dimensional incompressible flow is derived
below.

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

Fig.3.1 Boundary layer profile

The boundary layer equations in this case are:

U  V
 =0 (3.66)
x y

U U 1 dp 2U 
U +V =-   (u' v ' ) (3.67)
x y ρ dx y 2
y

The boundary conditions are (a) y=0, U=0; (b) y= δ, U=Ue(x)

1 dp dUe
Noting that = -Ue and integrating the momentum equation (Eq.3.67) gives:
ρ dx dx

δ 
 U U dUe    U 
0 U
 x
V
y
-Ue  dy = 
dx 

0 y  y
- u' v'  dy

(3.68)

δ= boundary layer thickness

Now, the r.h.s. of Eq.(3.68) equals :

  U 0 U
0 -   = - , as (a) u' v' is zero at y=0 and (b) u'v' and are both zero at
  y 0 ρ y
y=δ.
y
 U 
From continuity, V = 0  x  dy

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

Hence, Eq. (3.68) becomes:

δ
 U U y
U dUe  
0  U  0  x -U  dy = - 0 (3.69)
 x y
e
dx  ρ

Integrating by parts yields:

δ
U

y
U  
δ
U U
0 
 y
0 x
dy  dy = Ue

0 x
dy   U
x
dy (3.70)
 

Thus, Eq.(3.67) can be written as:

δ  U U dUe  0
0  2U
 y
-Ue
x
-Ue  dy = -
dx  ρ
(3.71)

δ δ
 dUe 
Or  [ U (Ue -U)dy   (Ue -U)dy = 0 (3.72)
0 x dx 0 ρ

Now, it is known that the displacement thickness (δ1 or δ* ) and momentum thickness

(δ2 or θ) are defined as:

δ
 U
δ1 or δ*    1-  dy (3.73a)
0 Ue 

δ
U U (3.73b)
δ2 or θ =  1-  dy
0 Ue  Ue 

Substituting from Eqs. (3.73a) and (3.73b) in Eq.(3.72) gives:

0 d 2 dUe
= (Ueθ) +δ* Ue (3.74)
ρ dx dx

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

0 δ*
Further, Cf = and H = (3.75)
1 2
ρUe θ
2

Hence, Eq.(3.74) can be written as :

Cf dθ θ dU
= + (H+ 2) e (3.76)
2 dx Ue dx

Equation (3.76) is the momentum integral equation for two-dimensional turbulent

boundary layer.
Remarks:
i) The momentum integral equation for laminar flow was derived by Von Karman in 1921
and is known as Von Karman momentum integral equation.
ii) Reference 3.2, chapter 3 gives the integral equation for two dimensional
compressible flow as:

dθ θ dU C
+ (H + 2 - M2e ) e = f (3.77)
dx Ue dx 2

iii) References 3.3, chapter 2 can be referred to for mean energy integral equation .
iv) It may be pointed out that equations (3.76) and (3.77) are same for both laminar and
turbulent flows. However, relationships between H ,θ and Cf are needed to solve these
equations. These relationships are different for laminar and turbulent cases.
v) The boundary layer equations can also be used for wakes, jets and shear layers.
Hence, the momentum integral equation can also be used in these flows which in
general are called thin shear flows.
vi) Alternate form of the momentum integral equation
Following chapter 5 of Ref.3.5 an alternate form of the momentum integral equation is
derived below.
For convenience Eq.(3.72) is reproduced below.
 
 0
  dU
0 x U Ue  U  dy  dxe  Ue 
 U dy 

(3.72)
0

The following may be pointed out.

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

(a) Since the integrand in Eq.(3.72) vanishes both at y = 0 and y =  , the differentiation
with respect to ‘x’ can be brought outside the integral sign.
(b) Further, the integrals in Eq.(3.72) vanish for y>  hence, the upper limit of the
integrals can be changed from  to .

Thus Eq.(3.72) can be rewritten as:

 
0
d
dx 0
U 
Ue  U dy  
dUe
dx  Ue 
 U dy 

(3.77a)
0

The following equations are obtained by multiplying Eq.(3.73a) and Eq.(3.73b) by

suitable factors.


Ue*   Ue  U dy  (3.77b)
0


Ue    U Ue  U dy
2
 (3.77c)
0

Consequently, Eq.(3.77a) can be rewritten as:

dUe 0
d 2
dx

U e   * Ue dx


(3.77d)

 dUe 
It may be pointed out that the term  Ue  denotes the acceleration (or deceleration)
 dx 
of the free stream. It also indicates presence of the free stream pressure gradient. As a
simple case, consider a boundary layer developing under a uniform external stream. In

 dUe 
this case Ue is constant or   is zero and Eq.(3.77d) reduces to:
 dx 

d 
   02 (3.77e)
dx U e

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Turbulence Prof. E.G. Tulapurkara
Chapter-3


Equation (3.77e) indicates that the rate of loss of momentum U e is equal to  0 the
2

shear stress at the wall. This is as it should be

Coles and Hirst (Ref.3.6, p.53) use Eq.(3.77d) to apply a cross-check on the
experimental data on turbulent boundary layers. The expression they use is slightly
different in form and can be derived as follows. Consider that a turbulent boundary layer
is developing on a wall. Let ‘ x 0 ’ be the initial station where the boundary layer is already

turbulent. At x 0 , Ue  Ue0 and   0 . Let ‘x’ be a downstream station. Further, let

Ue , , * and  w be the external velocity, momentum thickness, displacement

thickness and wall shear stress respectively at ‘x’. Note  0 in Eq. (3.77d) is replaced

with  w to avoid confusion.

Integrating Eq.(3.77d) between x 0 and x gives:


x x
dUe
U e  U e00    Ue
2 2
dx   w dx
*
(3.77f)
x0
dx x0


Dividing by U
2
e0 0 yields:

U2e * Ue dUe w
x x

U2e00
 1  U2e0 0 dx
x0
dx   U2e0 0 dx
x0

 U2e 
d 2 
U2e 1 *  U e0  w
x x
Or 2  1  dx   dx
U e00 2 x0 0 dx x
x0 0
U 2
e0 0

w
Noting that w  U
2
 where U  = friction velocity = , yields:

 U2e 
d  2 
U2e  1 *  U e0  U2   x 
x x
 1  dx  x U2e0 d  0  (3.77h)
U2e0 0 2 x0 0 dx 0

Use of this equation is explained in example 4.5.

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

Samuel and Joubert (Ref 3.7) express Eq.(3.77h) in the following form.

U2e  Ue * dUe 1  Cf Ue0

x x
1  2 dx   dx  RNS (3.77i)
U2e0 0 x0
U e0 0 dx 2 x0 0
w
Where  Cf U  and RNS is the extra term due to inclusion Reynolds stress
1 2
U e0
e0

2
term in the momentum equation (Eq.3.65a). It may be pointed out that for an attached
boundary layer, the term ‘RNS’ in Eq.(3.77i) is negligible. In Ref.3.7, it is found that this
term is only 1.8% of the other terms on the R.H.S of Eq.(3.65a).

3.7 Reynolds Average equations for compressible flow

The equations of motion for a compressible viscous fluid are derived in Appendix ‘C’
(Eq. C.38, C.50 and C.65). They are reproduced below :

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

ρ 

 t  xj
ρ Uj  = 0 (3.78)

   p 
ρUi   ρUi Uj  = -  ij (3.79)
t x j  xj  xj

  p 
ρH  ρUj H   Ui  ij  qj  (3.80)
t  xj  t  xj

where, stress tensor ij and heat flux vector qi and total enthalpy H, are given by:

2  Uk   U  Uj 
 ij = - μδij μ i 
  x j  xi 
(3.81)
3  xk  

T
qj = -  (3.82)
x j

1
H = h+ UU
i i (3.83)
2

ij is Kronecker delta, having the value 1 for i=j and 0 for ij. The energy equation in
terms of enthalpy, h is:

  p p U q
ρh  ρhUj  =  Uj  ij i  j (3.84)
t x j t x j x j x j

Remark:

While using tensor notation the normal stresses σ,xx , σ,yy and σ,zz are denoted as 11,
22, and 33.

3.7.1 Averaging procedures

(I) Conventional averaging :

As mentioned earlier, in conventional averaging procedure, a time averaged quality f is
defined as :
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Turbulence Prof. E.G. Tulapurkara
Chapter-3

t0 +T
1
f =
T 
t0
f dt (3.85)

In the conventional Reynolds decomposition, the instantaneous value is written as sum

of the time average plus the fluctuation about the time average, i.e.
U = U+u ; V = V + v  ; W = W + w  ; ρ = ρ+ρ ;

p = p+p'; h = h+h'; T = T+ θ'; H = H+H'; (3.86)

The fluctuations in fluid properties such as viscosity, thermal conductivity and specific
heat are usually small in a turbulent flow and are generally neglected.

(II) Mass-weighted averaging

For the treatment of compressible flows, it is convenient to take the average of the
product of the instantaneous value of density and the flow variable. This average is
called ‘mass-weighted average’ or ‘Favre average’. This average is denoted by a ~
over the quantity. i.e.

ρU ρV ρW ρh ρT ρH
U= ;V= ;W= ;h= ;T= ;H= (3.87)
ρ ρ ρ ρ ρ ρ

It is noted that, only the velocity components and thermal variables are mass-averaged.
 and p are averaged as before. The instantaneous values of mass-averaged variables
are given by :
U = U+u ,V = V + v , W = W + w ,h = h+h, T = T +   , H = H+H (3.88)

It must be pointed out that the time averages of u", v" etc. i.e. u", v" etc. are not zero
unless  is zero.

But, ρ f  = 0 (3.89)

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

ρU ρ u ρu
Hence, u = U - U = U - = U- U- = - (3.90)
ρ ρ ρ

ρv
Similarly, v =  and so on.
ρ

3.7.2 Reynolds form of continuity equation for compressible flow

I) With conventional averaging

Substituting for ρ = ρ+ρ and Ui = Ui +ui in Eq. (3.78) gives:


t
 ρ +ρ 

x j
 ρ +ρ U +u  = 0
j j

Expanding and taking time averages yields:

 ρ  ρ 
 
 t  t  xj
ρUj 

 
 xj
ρ Uj 

 xj
ρuj 

 xj
ρ uj  =0 (3.92)

Use of Eq. (3.9) simplifies Eq. (3.92) as:

ρ 

 t  xj

ρUj  ρuj = 0  (3.93)

II) With mass-weighted averaging

To obtain the continuity equation when mass-weighted averages are used, substitute
ρ = ρ  ρ and Ui = Ui + ui in Equation (3.78). On taking time average the following
equation is obtained :

ρ ρ    
 
t t x j
ρUj + 
x j

ρ Uj 
x j
ρuj 
x j

ρ uj = 0  (3.94)

The second and the fourth terms in Equation (3.94) are obviously equal to zero. The
last two terms can be combined as:

 xj
 
ρuj 

 xj
 
ρuj =

 xj
 
ρuj , which is equal to zero by Eq.(3.89).

Hence, Eq. (3.94) simplifies to :

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

ρ 

 t  xj

ρ Uj = 0  (3.95)

This is the continuity equation with mass-weighted averages. It is more compact than
Eq. (3.93).
For an incompressible flow  is zero and both conventional and mass-averaged
continuity equations take the form:
 Uj
=0 (3.96)
 xj

3.7.3 Reynolds form of momentum equations

I) With conventional averaging:

Here, also one proceeds in the same manner as above. However, for the sake of clarity
only the X-momentum equation, i.e. Eq.(C.50), is considered. Substitute
ρ = ρ +ρ ; U = U+u , V = V + ν ; W = W + w  and p = p +p in Eq.(C.50).This gives:

 

t 
 x
  
 ρ +ρ U+u  ρ+ρ U+u U+u  p+p   xx 

   
   

y 
   
ρ+ρ U+u  V +u  xy  
 z   
ρ+ρ U+u W + w   zx  = 0
    (3.97)

When time averages are taken, some terms in Eq.(3.97), which are linear in fluctuating
quantities, become zero. Some others when grouped together are found to be zero
through the use of continuity equation. The resulting equation is:
    p
t

ρU+ρ u 
x

ρUU+Uρ u 
y

ρUV +Uρ v' 
z
 
ρUW +Uρ w = -
x
  
    U 2  Uk      U  V  
 μ  2 -   Uρ u  ρu u  ρ u u  μ     V ρ u  ρ u v   ρ u v  
 x    x 3  x k    y    y  x  

   U  W  
 μ     W ρ u  ρu w   ρ u w   (3.98)
 z    z  x  

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

The complete Reynolds momentum equation (all three components) can be written, in
tensor notation, as:

t

ρUi  ρ'u'i  
 xj  
ρ Ui Uj  Ui ρ'u' j = -
p 

 xi  x j

 ij - Uj ρ'u'i - ρu'i u' j - ρ'u'i u' j  (3.99)

  Ui  Uj  2  Uk 
where ,  ij = μ     δij  (3.100)
  x j  xi  3  xk 

Proceeding in a similar manner the Reynolds averaged momentum equation in mass-

weighted variables can be deduced. It appears as :
  p 
t
 
ρUi 
 xj

ρUi Uj = 
 xi  x j

 ij  ρui vj  (3.101)

Where,
  U  Uj  2  U     Ui  u j  2  u 
 ij = μ  i
   δij k
  μ     δij k
 (3.102)
  
  x j  xi  3  xk     x j  xi  3  xk 

Remarks:
(i) It is seen that the momentum equation, in mass-weighted variables i.e. Eq.(3.101), is
simpler in form than Eq.(3.98) which is with conventional averaging. However, the
expression for  ij in Eq.(3.102) is more complicated than that in Eq.(3.100). However, in

practice the viscous terms involving double primed fluctuations in Eq.(3.102) are likely
to be small and hence, negligible.

(ii) For incompressible flow, the Reynolds averaged momentum equation is same in
both mass-averaged and conventional average forms and can be written as:


t
 
ρUi 

x j

ρUi Uj = 
p 


xi x j

 ij  ρui uj  (3.103)

 Ui Uj 
where,  ij  μ    (3.104)
 x j xi 
 

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Turbulence Prof. E.G. Tulapurkara
Chapter-3

3.7.4 Reynolds form of energy equation

Substituting H = H+H , ρ = ρ+ρ , Ui +ui , T = T+θ , and p = p+p , in Eq.(3.80) and

taking time averages, gives on rearrangement the following Reynolds averaged form of
energy equation:

   T  p
t

ρ H+ρ H 
 xj 

 ρ Uj H + ρ uj H  ρuj H  ρ uj H  U j ρ H    
 xj  t


  2 
  
  Uk    Uj  Ui  2
Ui  - μδi j  +μUi  +  - μδij ui   uk /  xk  +μ ui   uj /  xi  +ui   ui /  x j 
 xj   3  xk    x i  x j  3
(3.105)
The same equation in terms of temperature becomes

  p p  p'    T 
t

Cp ρT + Cp ρ'θ' +
 xj
 
ρ Cp T Uj = +Uj
t  xj

+uj + 
 x j  x j   x j
-ρCp θ' u' j - Cp ρ'θ'u' j  +  (3.1 0 6)


where,

 =  ij
 Ui
 xj
=  ij
 Ui
 xj 
+  ij   ui /  x j   (3.107)

The energy equation with mass-weighted averaging is:

  ~   ~ ~ T  p   ~ 
 ρH    ρUj H  ρu" j H"   =   Ui  ij  ui  ij  (3.108)
 t    x j   xj   t  xj 
 

  ~
   ~ ~
  p ~  p   p 
Or   ρCp T    ρCp TUj  = +Uj +u j 
 t    xj  t  xj  x j 
 ~

  T  θ"
    Cp ρθ"uj    (3.109)
 xj   xj  xj 
 
~
 Ui  ui
Where,  =   (3.110)
ij  x j ij  x
j
Remark:
For incompressible flow the Reynolds averaged energy equation is:

 ρH   T  p     Uj  Ui    uj  u  
+  ρUj H+ρuj H -   = +  μUi  +  +μ  ui
  xi  x j    xi
+ui i   3.111
 t  xj   xj   t  xj      x j  
 

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 13

Turbulence Prof. E.G. Tulapurkara
Chapter-3

  p p   T 
Or
t
 
ρc T 
xj

ρc TU j = 
t
+ Uj +  
 xj  x j   xj
- ρc θ'uj   
 3.112 


Dept. of Aerospace Engg., Indian Institute of Technology, Madras 14