1.7 Turbulent Flow
Turbulent flow is characterized by the random
and chaotic motion of fluid molecules.
1.7.1 Time-smoothed variables
Figure 1.7-1 referred steady and unsteady
laminar and turbulent flows. The velocity
refers to that at a fixed position in the fluid.
A turbulent flow can still be considered
as steady if the time-smoothed velocity
remains constant. As illustrated in Fig. 1.7-2,
we assume the fluid velocity at a fixed point
in space over a given finite time interval t0
can be resolved into the time-smoothed velocity
and a fluctuation velocity term.
v  v  v'
Where
1
v
t0

t  t0
t
[1.7-1]
vdt
[1.7-2]
55
Fig. 1.7-2 Time-smoothed velocity in turbulent flow
The time interval to is small with respect to the time over which v varies but large
with respect to the time of turbulent fluctuations. From Eqs. [1.7-1] and [1.7-2],
it is obvious that
1
v 
t0
'

t t 0
t
v ' dt  0
[1.7-3]
Since the local pressure is affected by the velocity, we can write a similar
expression for the pressure at a fixed point in pace over the same time internal.
p  p  p'
[1.7-4]
The significance of turbulent can be illustrated by fluid flow through a circular
pipe. Form Eqs. [1.5-53] and [1.5-54], the expression for laminar flow in a circular
pipe is
and
vz
r
 1  ( )2
vmax
R
[1.7-5]
vav 1

vmax 2
[1.7-6]
56
In contrast for turbulent flow in a circular pipe, it has been proposed and verified
experimentally that
1
vz
r 7
 (1  )
[1.7-7]
vmax
R
And
vav
4
~
[1.7-8]
vmax 5
Eq. [1.7-7] is known as the Blasius one-seventh power law.
As shown in Fig. 1.7-3, the velocity distribution is significantly more uniform in
the case of turbulent flow, as a result of better mixing in the bulk fluid.
1.7-2 Time-smoothed governing equations
v  v  v'
 v  0
1
v
t0

t  t0
t
vdt
Substituting Eqs.[1.7-1] into Eq.[1.3-5] and then taking the time average
according to Eqs [1.7-2] and [1.7-3], we obtain
[1.7-9]
 v  0
which is the time-smoothed equation of continuity for incompressible fluids.
57
Similar, substituting Eqs. [1.7-1]and [1.7-4] into [1.5-6] and then taking the time
average according to Eqs.[1.7-2] and [1.7-3], we get the following time-smoothed
equation of motion
[1.7-10]
v  v  v
1
v
t0

t  t0
t
1
vdt 
t0

t  t0
t
1 t  t0
vdt   vdt  v  (v)  v  0
t0 t
  v    (v  (v))    v    v    v  0
 v' xv' x

     v' y v' x

 v'z v'x

v' xv' y
'
'
v yv y
v'z v' y
v' xv' z 

v ' y v ' z  [1.7-11]

' '
v zv z 

  , resulted from turbulent velocity,
is sometimes called the Reynolds
stresses.
Since the fluid in Ω exerts a stress ．n on its surroundings, the surroundings
can be considered to exert a stress –．n on the fluid in Ω. Note thatτis a tensor
(p.36), and ．n is a vector (p.46).
The vector ．n is the force exerts on its surrounding, which is due to the
viscous force and can be expressed by the velocity components.
58
1.7.3 Turbulent momentum flux
Several semiempirical relations have been proposed for the Reynolds
stresses ’, in order to solve Eq. [1.7-10] for velocity distribution in turbulent flow.
1.7.3.1 Eddy viscosity
dv z
Boussinesq proposed the following form  yz'   m '
dy
[1.7-12]
By analogy with Eq.[1.1-2], Newton’s law of visiosity. The coefficient m’ is a turbulent
or eddy viscosity and is position-dependent.
1.7.3.2 Prandtl’s mixing length
Prandtl proposed the concept of the mixing length based on the assumption
that eddies move about in a fluid like molecules do in a gas. The mixing length plays
a role somewhat similar to that of the mean free path in the gas kinetic theory.
 yz'   l 2
dv z dv z
dy dy
[1.7-13]
and from Eq.[1.7-12]
dv z
m  l
dy
'
2
[1.7-14]
Where l, called the mixing length, is a function of position, for example l= ay, where y
59
is the distance from the solid surface.
1.8 Momentum transfer correlations
1.8.1 External flow
1.8.1.1 Flow over a flat plate
Flow over a flat plate is laminar for local Reynolds number Rez < 2 x 105.
The following correlation can be used for laminar flow over a flat plate
0.664
C fz 
(Re z  2 105 )
Re z
where, as shown in Eq. [1.1-6]
and
Re z 
zv


[1.8-1]
C fz 
 v z
m
 yz
y 0
[1.8-2]
1
 v 2
2
[1.8-3]
From these equations the friction coefficient averaged over a distance L from
the leading edge of the plate is
[1.8-4]
1 L
C fL 
Substituting Eq. [1.8-1] into q. [1.8-4]
12
 1   
C fL  0.664    
 L   v 

L
0
L
0
C fz dz
12
  
2
z dz  1.328 

 Lv  
1
[1.8-5]
60
Hence
where
C fL 
1.328
(Re L  2 105 )
Re L
Re L 
Lv 


L  v
m
[1.8-6]
[1.8-7]
For turbulent flow over a flat plate the following empirical correlation
has been suggest
C fz  0.0592 Re z 1 5 (5 105  Re z  107 )
[1.8-8]
C fL  0.074 Re L 1 5 (5 105  Re L  107 )
[1.8-9]
and from Eq. [1.8-4]
61
1.8.1.2 Flow normal to a cylinder
For incompressible flow normal to a cylinder of diameter D, the drag coefficient
is a function of Reynolds number, as shown in Fig. 1.8-1(a)
CD  CD  ReD 
[1.8-10]
Where, according to Eq. [1.1-37],
CD 
FD
Af   v 2 2 
and
Re D 
Dv


D  v
m
Fig. 1.8-1
62
1.8.1.3 Flow past a sphere
The creeping incompressible flow around a sphere of diameter D is
CD 
24
( Re D  1)
Re D
[1.8-13]
where CD and ReD are defined in Eqs. [1.8-11] and [1.8-12], respectively.
The following empirical correlations can be used for higher ReD:
CD 
18.5
(2  Re D  5 102 )
35
Re D
[1.8-14]
and
CD  0.44 (5 102  Re D  2 105 )
[1.8-15]
The experimental data of Cd as
a function of ReD are shown in
Fig. 1.8-1b.
63
1.8.2 Internal flow
1.8.2.1 Flow through a circular tube
For fully developed laminar flow in a circular tube of inner diameter D:
f 
64
Re D
(ReD < 2100)
[1.8-16]
where the friction factor is defined by
 ( p   gh0 )  ( pL   ghL )  D
f  0

2

v
2
av

L
[1.8-17]
For fully developed turbulent flow in pipes, the experimental results of the
friction factor f called the Moody diagram, as shown in Fig. 1.8-2.
For fully developed incompressible turbulent flow in a smooth tube
f  0.316 Re D
1 / 4
f  0.184 Re D
1 / 5
(ReD < 2x104)
(ReD > 2x104)
When the fluid velocity is known, the Reynolds number can be calculated
and the friction factor can be found from Figure 1.8-2. From the pressure
drop po-pL along a pipe of length L can be found from Eq. 1.8-17.
64
To find the velocity from a known pressure drop, one approach is described below.
For a horizontal pipe, Eq. 1.8-17 can be written in the form of
1
2
f Re D 
 D 2( p0  pL ) D
m
L
[1.8-20]
If a plot was prepared to show the relationship between log(ReD) and log(ReDf1/2),
then the velocity can be found from ReDf1/2.
For turbulent flow in smooth pipes a log(ReD)/log(ReDf1/2) plot is as shown in
Fig. 1.8-3. The relationship happens to be close to a linear one in this particular case.
log ReD  0.380  1.115log(ReD
Where ReD
f)
[1.8-21]
f can be obtained from Eq. [1.8-20]
65
66
Example 1.8-1 Flow through a cooling tube
log Re D  0.38  1.115(Re D
f)
Given: Tube length Height
Pressure, D, L, m, , smooth wall
Find: f, Q
1
2
f Re D 
 D 2( p0  pL ) D
m
L
log Re D  0.38  1.115(Re D
[1.8-20]
f)
[1.8-21]
67
1.9 Overall Mechanical Energy Balance
Let us consider the flow of a fluid through a pipe and a turbine, as shown in
Fig. 1.9-1a. Let the fluid in the pipe and the turbine be the control volume Ω.
The control surfaces at the inlet A1 and the outlet A2 are chosen to be perpendicular
to the pipe walls.
 rate of mechanical 
 rate of mechanical


(1)  
(2) 
 energy accumulati on 
 enrgy in by convection 
 rate of mechanical energy 
 rate of work done by 

(3)  
(4)
 out by convection

 system on surroundin gs 
[1.9-1]
68
1.9 Overall Mechanical Energy Balance
Let us consider the flow of a fluid through a pipe and a turbine, as shown in
Fig. 1.9-1a. Let the fluid in the pipe and the turbine be the control volume Ω.
The control surfaces at the inlet A1 and the outlet A2 are chosen to be perpendicular
to the pipe walls.
 rate of mechanical 
 rate of mechanical


(1)  
(2) 
 energy accumulati on 
 enrgy in by convection 
 rate of mechanical energy 
 rate of work done by 

(3)  
(4)
 out by convection

 system on surroundin gs 
Let us consider the case of steady-state flow of an incompressible fluid. At the
steady state, the rate of mechanical energy change in Ω is zero (term 1).
The mass flow rate of fluid entering Ω through a differential area dA1 at the
entrance is ρvdA1. The kinetic and potential energy per unit mass of the fluid are
v2/2 and gz, respectively.
As such, the kinetic and potential energy of the fluid enter through the inlet A1
and leave the outlet A2, respectively, at the rate of
69
1 2
A1 ( 2v  gz ) vdA1
(term 2)
1 2
A2 ( 2v  gz ) vdA2
(term3)

The rate of the pressure work done by the fluid on the system is
pvdA1. As
A1
such, the system can be considered to do work on the surrounds at the rate of

pvdA1 . At the outlet, the work done by the system is
pvdA2 .


A1
A2
The rate of the shaft work done by the fluid in the system on the surroundings
through the turbine (work done by the surroundings on the system) is Ws.
The rate of the viscous work done by the fluid in the system on the pipe walls,
that is , the friction loss, is Wv.
Substituting all the above terms into [1.9-1], we obtain
1 3
A1 2  v dA1  A1  gzvdA1  A1 pvdA1
1 3
   v dA2    gzvdA2   pvdA2  Ws  Wv
A2 2
A2
A2
[1.9-3]
70
Assume the inlet and the outlet are in the same height. However, the velocity
v can still vary significantly. For steady-state flow of an incompressible fluid
(vavA)1 = (vavA)2 =vavA
The average velocity vav is defined as follows
v av 
 vdA
[1.9-4]
A
[1.9-5]
A
For an incompressible fluid, Eq.[1.9-3] can be rewritten as follows
3
1



1
v
2
(  v av )1 (v av A)1   
dA    gz1 (v av A)1  p1 (v av A)1

[1.9-6]
A
 A  v av 

2

1
3
1



1
v
2
 (  v av ) 2 (v av A) 2   
 dA    gz2 (v av A) 2  p2 (v av A) 2  ws  wv
A

2
A  v av 

2
Dividing by the volume flow rate Q(=vavA), the equation becomes
b
b
2
2

v


gz

p


v
av 
1
1
av    gz2  p2  ws  wv


2
1
2
2
[1.9-7]
where b is the correction factor and is defined as
[1.9-8]
3
1
v
b   ( ) dA
A A v av
The shaft work and viscous work per unit volume of the fluid are defined71
as ws = Ws/Q [1.9-9] and wv=Wv/Q [1.9-10], respectively.
vz
r 2
v
1

1

(
) ) and [1.7-6] ( av 
By substituting Eqs.[1.7-5] (
) into
vmax
R
vmax 2
Eq.[1.9-8], that for laminar flow through a circular pipe, we obtain b = 2.
1
vav
4
vz
r 7

(
1

)
~
Similarity, By substituting Eqs.[1.7-7] (
) and [1.7-8] (
)
vmax
R
vmax 5
into Eq.[1.9-8], that for turbulent flow through a circular pipe, we obtain b = 1.
A special case of Eq.[1.9-7] is the flow of an inviscid fluid (wv=0) with an
essentially uniform velocity (b=1) and in the absence of any shaft work (ws=0).
We obtain the Bernoulli equation.
1
1
2
 v1   gz1  p1   v 2 2   gz2  p2
2
2
[1.9-13]
The friction loss wv can be considered to consist of two parts that associated
with pipes wvp and that associated with fittings and valves wvf.
The friction loss due to pipes is defined as
1
2 L
wvp  [( po  gho )  ( pL  ghL )]i  ( v av
f )i
2
D
i
i
72
According to the definition of friction factor f in Eq.[1.1-38]
 ( p0   gh0 )  ( pL   ghL )  D
f 

2

v
2
av

L
[1.1-38]
Define the loss coefficient K: (for pipe fitting or valve)
[1.9-17]
K
pin  pout
v av 2 / 2
Energy loss associated with fittings and valves is denoted as wvf and is
defined as
1 2
wvf   ( pin  pout )i   ( v av K )i
2
i
i
[1.9-18]
Substituting Eq.[1.9-16] and [1.9-18] into Eq. [1.9-7], we obtain the following
overall mechanical-energy balance for turbulent flow (b~1) in pipe lines:
1
(  v av 2 )1   gz1  p1 
2
1
L 
1
1

(  v av 2 ) 2   gz2  p2  ws     v av 2 f      v av 2 K  [1.9-19]
2
D i i  2
i
i 2
73
The friction factor f can be obtained from Fig.1.8-2 or from Eqs.[1.8-18] and
[1.8-19]. The loss coefficient K has been provided for various fittings and valves.
Two equations are given for the sudden expansion and contraction.
2

Ain 
pin  pout
K  1 


Aout 
(  v av 2 2)in

(sudden
[1.9-20]
expansion)
 A 
pin  pout
K  0.45 1  out  
Ain  (  v av 2 2)out

(sudden
[1.9-21]
contraction)
74
Example 1.9-1 Flow through a siphon
75
Example 1.9-2 Friction loss in a sudden expansion
76
Example 1.9-3 Pressure force acting on a nozzle
77
Example 1.9-4 Power requirement for pipe-line flow
78