Laminar & turbulent Flow
• Shear stress on fluid
• Newton’s law of viscosity
du

dy
• Shear stress,, in fluid is proportional to the
velocity gradient - the rate of change of velocity
across the fluid path.
du

dy
• Viscosity, µ is the constant of proportionality
Laminar and turbulent flow in pipes
• Flow can be either
 Laminar - low velocity
 Turbulent – high velocity
 A small transitional zone between
Phenomenon was first investigated in the 1880s by
• Osbourne Reynolds in an experiment which has
become a classic in fluid mechanics.
ud

ud ud
Reynold' s Number Re or N R 




 

Kinematic viscosity 
dynamic viscosity

3
(m/s)(m)(k g/m )
Re 
kg/(m.s)
Unitless, or non-dimensional number

high  : Fluids tend to flow as laminar


low  : Fluids tend to flow as turbulent
Inertial force on fluid element
Re 
Viscous force on fluid element
F  ma Newton’s 2nd law
F  A shear stress over
fluid surface
;
Re
is different for different Conduits,
Non-circular x-sections, open channels etc.
The flow in round pipes is
f (  ,  , D (pipe  ),  (mean velocity ))
Laminar flow: Re < 2000
Transitional flow:2000 < Re < 4000
Turbulent flow: Re > 4000
• Pipe flow nearly always turbulent
Example
Given the following: Water flowing,
T  25o C
D  150 mm
u  3.6 m/s
Take  at 25 to be 1258 kg/m and   0.96 Ns/m
o
3
Find the flow regime.
Soln.
ud 1258  3.6  0.15
Re 

 708

0.96
Implies Laminar
2
Example
Given that:
• Pipe diameter: 0.5m
• Crude oil:
Kinematic viscosity = 0.0000232 m²/s
• Water:
Dynamic viscosity μ = 8.90 × 10-4 Ns/m2
What are the velocities when
Turbulent flow would be expected to start?
ud ud
Re 



• Crude oil:
u0.5
4000 
5
2.23  10
u  0.1784m / s
• Water:
1000u0.5
4000 
4
8.90  10
u  0.007m / s
Pressure loss due to friction in a pipe
Consider fluid flowing in a pipe
L
1
p
2
The pressure at 1
(upstream) is
higher than the
pressure at 2.
If a manometer is attached the pressure (head)
difference due to the energy lost by the fluid
overcoming the shear stress is seen
• Consider a cylindrical element of
incompressible fluid flowing in a pipe
1
w
2
Area A
Direction of flow
Pressure
p
1
w
2
Pressure (p-p)
The driving force (due to pressure)
(F = Pressure x Area)
Driving force = Pressure force at 1 - pressure force at 2
Driving Force  pA  ( p  p) A  pA  p
d
4
2
1
w
2
Area A
Direction of flow
Pressure

p
1
w
2
Pressure (p-p)
Retarding force (due to shear stress at wall)
Retarding force = shear stress x area over which it acts
Retarding Force   w  area of pipe wall   wdL
1
w
2
Area A
Direction of flow
Pressure p

1
w
Pressure (p-p)
2
Flow is in equilibrium
Driving force = Retarding force
p
d
4
2
  wdL
p 
 w 4L
d
Pressure loss in terms of Shear Stress at wall
log pL
p  u
p  u
1.7 to 2.0
log u
p  u
1.7 to 2.0
p 
log pL
p  u
 w 4L
d
log u
Pressure loss in laminar flow
• In laminar flow it is possible to do theoretical
analysis
 Fluid particles move in straight lines
Consider a cylinder of fluid element, length L, radius r,
flowing steadily in the centre of a pipe.
L
• In equilibrium, the shear stress on the cylinder
equal the pressure force.
 2r L  pA  pr
p r

L 2
Remember this
2
p 
• By Newton’s law of viscosity we have
du

dy
Where y is the distance
from the wall
 w 4L
d
y
Measuring from the pipe centre, we change the sign
and replace y with r distance from the centre, giving
du
  
dr
Hence
p r
du
 
L 2
dr
or
du
p r

dr
L 2
In an integral form we have
p 1
u
rdr  c

L 2
Integrating gives the velocity at a point distance r
from the centre
p r
ur  
c
L 4
2
At r = R (the pipe wall) uR = 0
p R
c
L 4
2
Hence
Hence an expression for velocity at a point r from
the pipe centre when the flow is laminar
p 1
2
2
ur 
(R  r )
L 4
This is a parabolic profile (of the form y = ax2 + b )
Velocity profile in a pipe
Flow through the element (dr )  dQ  area  velocity
Hence
p ( R  r )
dQ  2rr 
L
4
2
Integrating for the limits
2
r  0
r  R

2 ( p) R 2
2
Q
( R  r )rdr

4L 0
 (p)d
Q
128L
d
R
2
4
Equation for laminar flow in a pipe
This expresses the discharge

p

p
gradient (

),
x
L
Q
in term of the pressure
diameter of the pipe and
the viscosity of the fluid.
• The mean velocity is determined as
Q
 ( p)d 4
( p)d
um 


2
2
128L d
32L
d
(
)
4
4
2
hf
,
32Lu
p 
2
d
But
p  gh f
Writing pressure loss in terms of head loss
32Lu
hf 
gd 2
Hagan–Poiseuille equation
Example
Pressure loss in Turbulent Flow

Consider the forces on the element of fluid
flowing down the slope (open channel)
p1 A  p2 A  LP  W sin  0
W  gAL,
sin   z L
( p1  p2 )  gz
P
  0
L
A
Pressure loss in Turbulent Flow

The first pressure loss term is the piezometric
head, p*, loss per unit length,
dp
P

dx
A

dp dx
R
A
 Hydraulic mean depth (Hydraulic radius), m m 
P

dp
 m
Writing p  gh f
dx
•
Gives shear stress in terms of head loss
 m
gh f
L
Introduction of Friction factor

To make use of this equation we introduce the
friction factor, f
 f

u
2
2
Equating and rearranging gives

fLu 2
hf 
2 gm
For a circular pipe,
A d 2 4 d
m 

P
d
4
2

Giving
4 fL u
hf 
d 2g
Darcy-Weisbach Equation and the Friction factor
 This is the Darcy-Weisbach equation
4 fL u 2
hf 
d 2g

Gives head loss due to friction in a circular pipe

Often referred to as the Darcy equation

In terms of Q
Q  Au 
d
4
2
u
4Q
u 2
d
64 fL Q 2
hf  2 5
 d 2g
Darcy-Weisbach Equation and the Friction factor
In metric terms, g = 9.81m2/s, so
2
fLQ
hf 
5
3.03d
or
fLQ
hf 
5
3d
2
Darcy-Weisbach Equation and the Friction factor
This equation describes
 Head-loss due to friction


In terms of velocity u
In terms of Discharge Q
2
4 fL u
hf 
d 2g
64 fL Q 2
hf  2 5
 d 2g
 And friction factor, f
 The value of f is crucial to calculation of hf
 How do we find f?
 The f described here is that common in UK
(in text books and practice)
4 fL u 2
hf 
d 2g
 In US (and some text book) famerican = 4f,
fL u 2
hf 
d 2g

To try and avoid confusion this is
sometime written as ,
hf 
LQ 2
3d
5
 BE CAREFULL !!!
 When using any book,
look at the equation for hf
Example
 Two reservoirs have a height difference 15 m.
 They are connected by a pipeline 350 mm in
diameter and 1000 m long with a friction factor f of
0.005.
 What is the flow in the pipe? (ignore all local losses)
1
2
Z1
Z2
Datum
Soln.

Write the general energy equation for the system
ignoring all minor losses
2
1
P1
2
2
v
P2
v
 z1 
 hA  hR  h f 
 z2 

2g

2g
z1  z 2  h f
fLQ
0.005  1000  Q
2
15 

 317.33Q
5
5
3d
3  0.35
2
2
Q  0.0473  0.217 m / s
3
What
Whatis
isf fdependent
dependent on?
on?
4 fL u 2
hf 
d 2g
What is the value of f ?
• The friction factor depends on many physical
things
 ud k k

f   
, , ,  
  d d

'
• For laminar flow theoretical expression can be
derived
• For Turbulent flow, it is complex
f
in Laminar Flow
• We have the Hagen-Poiseuille equation
 Head loss in laminar flow
• We also have the Darcy equation
•
Equate the two equations
32Lu
hf 
2
gd
4 fL u 2
hf 
d 2g
32Lu 4 fL u
hf 

2
gd
d 2g

f  16
ud
16
f 
Re
2
Laminar flow example
• Calculate the head loss due to friction in a circular
pipe of 50 mm diameter, length 800m, carrying water
(μ = 1.14 ×10-3 Ns/m2) at a rate of 5 litres/min.
• Use both Hagen-Poiseuille and Darcy equations.
A
d 2
4
 0 .00196 m 2
5
Q
 0.8 10 4 m 3 / s
1000  60
u  Q / A  0.04 m/s
Check Re
ud 1000  0.04  0.05
Re 

 1754  2000

0.00114
32Lu 32  1.14  10 3  800  0.04

Hagan - Poiseuille h f 
2
1000  9.81 0.052
gd
 0.045 m
Darcy
16
f 
 0.00912
Re
4 fLu 2 4  0.00912  800  0.04 2
hf 

 0.047 m
2 gd
2  g  0.05
Smooth / Rough pipes in Turbulent Flow
y
r


k
u
Smooth Pipe
Wall
y
r

Wall
Rough Pipe
k
u
Smooth / Rough pipes in Turbulent Flow
• Let k be the average height of projection from the
surface of a boundary.
• Classification based on boundary characteristics
 If the value of k is large, then the boundary is
called rough boundary
 If the value of k is less, then boundary is
known as smooth boundary.
• Classification based flow and fluid characteristics
 Turbulent flow along a boundary is divided into two
zones
 First zone. Thin layer of fluid in the immediate
neighbourhood of the boundary where viscous
shear stress predominates, and shear stress due
to turbulence is negligible. This zone is known
as laminar sub-layer.
Height of this layer denoted by
 .
 The second zone of flow, where shear stress due
to turbulence are large as compared to viscous
stress is known as Turbulent zone.
  k
  k
If k   ,
• Outside the laminar sub-layer the flow is turbulent.
• Eddies of various size present in turbulent flow try to
penetrate the laminar sub-layer and reach the
roughness projection of the boundary.
• Due to the thickness of the laminar sub-layer, the
eddies are unable to reach the roughness projection
of the boundary
• Hence the boundary behaves as a smooth boundary.
• This type of boundary is called hydrodynamically
smooth boundary
 If the Re of flow is increased the


will decrease.
 If the
becomes much smaller than the average
height k, the boundary will act as rough boundary.
 Because the roughness projection are above the
laminar sub-layer and the eddies present in the
turbulent zone come in contact with the roughness
projection a lot of energy will be lost.
 Such boundary is called hydrodynamically rough
boundary
From Nikuradse’s experiment
k
 0.25

k
 6.0

k
0.25 
6

the boundary is called a smooth boundary
the boundary is rough
the boundary is in transition
Blasius equation
 In 1913 Blasius examined a lot of experimental
measurements
 Found 2 distinct friction effects
 Smooth pipes and Rough pipes
0.079
f 
0.25
Re
Valid for Re < 100 000
Blasius equation
Nikuradse’s Experiment

Nikuradse made great progress in 1930’s

Artificially roughened pipes with sand of known
size, k
16
0.07
0.06
0.05
Relative roughness
Re
Transition
turbulence
A
Rough turbulence
C
0.04
B
0.03
ro / k
15
30.5
F
G 60
E
120
252
0.02
507
0.079
Blasius
f 
equation
Re0.25
0.01
10 3
2
4 5 104
2
D
5
5 10 2
5
Increasing grain size
f 
The following points must be noted from the curve
•
The line AB is common to all the pipes having
different relative roughness. This line indicates
laminar flow. The friction factor in this range is
given as
16
f 
Re
Valid for
•
Re  2000
As the value Re increases beyond 2000, the flow
passes through a transition stage represented by the
curve BC.
 The flow becomes turbulent at point C.
 Transition stage occurs in the range of Re
between 2000 to 4000.
•
•
The curve CD represents smooth turbulent flow
For turbulent flow, it is observed that the higher the
ratio ro/k (the smoother the pipe), the greater is the
tendency of the pipe to follow the line CD.
 Meaning the pipe with the roughness surface
causes the earliest breakaway from the line CD.
•
There is a Transition stage between the smooth
turbulent flow to the rough turbulent flow.
 For example, the relative roughness ro/k = 60,
the transition stage is represented by the curve
EF.
 In the transition stage, f depends on both
Re and ro/k.
•
After the flow is completely established as rough
turbulent, the curves become horizontal.


For example, the relative roughness ro/k = 60,
it is represented by FG.
In this stage, f depends only on the relative
roughness and is independent of Re.
Moody Diagram
Example
B pB  550 kPa
+
+
A
Given Benzene (S. G. = 0.86)
3
Q  0.11m / min  1.8  10 m / s
3
3
  4.2 10 Ns / m
4
2
Ignore all minor losses
Solution
pA
2
2
uA
pB
uB
 zA 
 hA  hR  hL 
 zB 

2g

2g
QA  QB ;
u A AA  uB AB ;
AA  AB
pA  pB   ( zB  z A  hL )
hL  h f
First find Re
1.83  10 3
[( 0.86)(1000)](
)( 0.050)

3
ud
 (Q / A)d
1.93  10
Re 




4.2  10  4
Re  9.54  10
4
Implies turbulent
Hence use Darcy equation:
( L) ( u2 )
hf  f
D 2g
We need
 D
for Moody
Re
pA  550 10  0.86  9.81 (21  3.83)
3
 550209.5 N/m
2
Velocity of Flow
 Chezy’s Formula
u  C RS
C - Chezy’s coefficient
•
Chezy from Darcy-Weisbach equation
2 g hL
u 
d( )
f
L
( L) ( u2 )
hf  f
d 2g
2
For a circular pipe
Hence
R
d
4
Hydraulic Radius
8 g d hL
8g
u 
( )
( R )( S )
f 4 L
f
2
A
R - Hydraulic radius R 
P
hL
S
L
A - Area of flow
Slope
8g
u
RS  C RS
f
8g
C
f
The value of
C = 55 – 75
P - Wetted perimeter
Manning’s Formula (Manning’s Velocity)
 Manning proposed that,
1 2 / 3 1/ 2
u R S
n
n - Manning’s coefficient. It depends on the type of material.
Relation between Manning and Chezy
 From Chezy and Mannings equations
1/ 2
CR S
1/ 2
1 2 / 3 1/ 2
 R S
n
1 1/ 6
C R
n
 Hazen Williams’ formula
Valid for diameter  5cm  1.9m, V  3m/s,
T = 16oC
u  0.85C h R
0.63
S
0.54
Hazen Williams
Ch  Hazen Williams cofficient , (dimension less)
R  Hydraulic Radius, m
hL
S  Ratio ( ) energy loss/length of conduit (m/m)
L
As pipe smoothness increases () C h also increases ()