ME2134E - Chapter 8
ME2134E - Chapter 8
ME2134E - Chapter 8
1
8.1 Introduction
To Malaysia
Water pipelines
(Acknowledgement: skyscrapercity.com)
7
8.2 Laminar and Turbulent Flows
8
8.2.1 Laminar Flow
FLOW
Laminar flow
FLOW
V d
Vd Vd
Re d or
where =density of fluid, =dynamic
viscosity, and =kinematic viscosity.
As shown in Chapter 7, Reynolds number can be viewed as the
ratio of inertial forces to viscous forces acting on a fluid element.
11
At large Reynolds numbers, the inertial forces, which are
proportional to the fluid density and the square of the fluid
velocity, are large relative to the viscous forces, and thus the
viscous forces cannot prevent the random and rapid
fluctuations of the fluid (turbulent).
At small or moderate Reynolds numbers, the viscous forces
are large enough to suppress these fluctuations and to keep the
fluid “in line” (laminar).
Critical Reynolds number, Recr: The Reynolds number at
which the flow becomes turbulent. It was first observed by
Osborne Reynolds (a British Engineering Professor) to
determine whether the flow is Laminar or Turbulent in a
circular pipe. The value of the critical Reynolds number is
different for different geometries and flow conditions.
13
8.3 Entrance Region
8.3.1 Velocity Boundary Layer
No-slip condition: A fluid in
direct contact with a solid
“sticks” to the surface due to
viscous effects, and there is
no slip.
16
8.3.3 Entrance Length
d
A B C
Boundary Layer
Entrance Length (Le)
Le Vd
Re d
d
17
Laminar Flow:
Fully Developed
Velocity Profile Profile
d
A B C
Laminar Boundary Layer
Laminar Flow
Entrance Length (Le)
18
Turbulent Flow:
Laminar Turbulent
Boundary Layer Boundary Layer
Fully
Developed
Profile
A B C
Solution:
d2 4Q
Q= V V
4 d 2
Vd 4 Q
Re = Laminar or Turbulent?
d
20
(a) Q=0.001 m3/s
4(870)0.001
Re= =355 < 2300 Laminar
(0.104)(0.03)
From equation (8.2), we have
21
8.4 Energy Loss and Friction in A Circular Pipe
gsinf
p1 gf
Consider a steady Control
flow in a circular volume
pipe of constant d r=R w
p2= p1+p
radius R. The V1
pipe is inclined to
horizontal level at f f
w V2
an angle f and
the flow is driven L =x2-x1
by pressure or z1
gR2L z2= z1+z
gravity or both.
Arbitrary reference datum
1 2
By applying energy equation between sections 1 and 2, we get
p1 V12 p2 V22
z1 z2 h f (8.4)
g 2 g g 2 g
where hf = friction head loss, V = averaged velocity 22
Since there is no change for the section area, the average
velocity will have no change. Thus, the energy loss (or head
loss) is given by p2 p1 p
h f z2 z1 z (8.5)
g g
pR 2 g R 2 z w 2RL
p 2w L
or z hf (8.6)
g gR
24
Define the dimensionless parameter f (Darcy Friction Factor)
8w
f (8.7)
V 2
By combining Eqs. (8.6) and (8.7), the desired equation for pipe
head loss is
For simplicity, we have
LV2 used L to replace ∆L
hf f (8.8)
d 2g
(Darcy-Weisbach Equation)
f -dz
dx
R
p
v1
f
dz r p +dp
f
v2 x
dx
gr2dx
Consider fully developed laminar flow in a circular pipe. Applying
Newton’s second law to a control volume represented by dashed
red line, we have
pr 2 ( p dp)r 2 g r 2 dx sin 2rdx ṁ (v 2 -v1 ) 0 26
dpr 2 g r 2 dx sin 2rdx
dpr 2 g r 2 dz 2rdx
r d
p gz (8.9)
2 dx
The shear stress is related to the velocity gradient by
du
(8.10)
dr
Note that velocity gradient, du/dr, is negative i.e. as r
increases, u decreases.
Substituting Eq. (8.10) into Eq. (8.9) gives
du r d
p gz
dr 2 dx
27
du r d
p gz
dr 2 dx
Any assumption involved?
Integrating above equation gives
r2 d
u p gz C1 (8.11)
4 dx
R2 d
C1 p gz (8.12)
4 dx
28
After substituting Eq. (8.12) into Eq. (8.11), the velocity
distribution is given by
r2 d R2 d
u p gz p gz
4 dx 4 dx
1 d 2 2
u p gz R r (8.13)
4 dx
which is a parabolic distribution as shown below. Equation
(8.13) is called the Hagen-Poiseuille Flow.
u
R
r
FLOW
R2 d
umax p gz (8.14)
4 dx
30
To find volume flow rate (Q) [m3/s],
we consider flow through a circular ring
dr
R r
1 d 2 2
R
Q p gz R r 2rdr
0
4 dx
R 4 d
Q
dx p gz (8.15)
8
31
Comparing Eqs. (8.14) and (8.15), it can be seen that
R 4 d 1 2 R d
2
Q
8 dx p gz 2 R 4 dx p gz
umax
1 2
Q R umax
2
Q umax
But V (average velocity ) (8.16)
A 2
32
8.5.3 Shear Stress, Friction Factor and Head Loss
Shear stress (), friction factor (f) and head loss (h) can be
determined as follows
1 d 2 2
du u
4 dx
p gz
R r (8.13)
w
dr r R
R d 2 R 2 d 2umax
w p gz p gz (8.17)
2 dx R 4 dx R
From Eq. (8.7)
8 w 64 and umax=2V
fla min ar (8.18)
V 2
Re d
L V 2 128LQ (8.19)
h f ,la min ar fla min ar
d 2g gd 4
33
From Eq. (8.8)
8.5.4 Pressure Drop for a Laminar Flow in a Horizontal Pipe
Datum 34
Equation (8.20) becomes
R 4 p z
Q
L g
8 L
8LQ
p (8.21)
R 4
for laminar flow only
p 128LQ
h f z
Introducing R=d/2 g gd 4
128LQ
p (8.22)
d 4
35
Example 8.2
Oil of specific gravity 0.9 and kinematic viscosity 330×10-6
m2/s is pumped over a distance of 1.5 km through a 75 mm
diameter tube at a rate of 25×103 kg/hr. Determine the shear
stress at the wall and the head loss through the pipe.
Solution: Given:
ν=330×10-6 m2/s
SG (oil)=0.9
L=1.5 km
d=75 mm = 0.075m
ṁ 25 103 kg/hr
Calculation of shear stress and head Vd
Re
loss is related to Reynolds number 36
d 2 25 103
ṁ V kg/s
4 3600
1 4 25 103
V 1.746 m/s
0.9 1000 (0.075) 3600
2
Vd 1.746 0.075
Re 396.94 2300 Laminar
330 10 -6
37
From the definition of friction factor in Eq. (8.7):
8 w
f
V 2
f 0.1612
w V
2
(900)(1.746) 2 55.28 N / m 2
8 8
Head loss:
38
8.5.5 Friction in Non-circular Pipes
• Most of the pipes or conduits used in engineering applications
are circular in cross-section.
• On some occasions, we also use rectangular ducts and cross
sections of other geometry.
• We can modify many of the equations that we have derived
earlier for circular cross-sections to noncircular sections by
using the concept of Hydraulic Diameter.
(Acknowledgement: itctubeco.com)
(Acknowledgement: Rigidtools.org)
H
W
4WH 4WH
DH 2H
2 H W 2(W ) 40
If W=H (i.e. Square)
64 64
For laminar flow, f (from eqn (8.18))
Re d VDH
will be changed a little bit 41
Laminar Flow in
Noncircular Pipes
The friction factors f
given in the table are for
fully developed laminar
flow in pipes of various
cross sections. The
Reynolds number for flow
in these pipes is based on
the hydraulic diameter Dh
= 4Ac /p, where Ac is the
cross-sectional area of the
pipe and p is its wetted
perimeter.
From: Fluid Mechanics:
Fundamental and
Applications, 3rd Edition
by Cengel and Cimbala 42
8.6 Turbulent Flow Through A Smooth Pipe
Laminar
(PARABOLIC)
d V
Turbulent
umax
r
R u
u r
1 R r u * (8.23)
n B
u*
where =0.4 and B=5.0. =kinematic viscosity of fluid
w
u* is called the friction velocity.
Note: u[r] means that u is a function of r. 45
Relationship between average velocity (V) and maximum
velocity (umax) for a turbulent flow in a pipe can be given by
V 1
(8.24)
umax 1 1.33 f
Friction Factor:
The relationship between friction factor and Reynolds number
for a smooth turbulent pipe flow is given as (semi-empirical)
1
f
2.0log Re d
f 0.8 (8.25)
46
Some numerical values of equation (8.25) are listed below
1
f
2.0log Re d
f 0.8
0.02
1
f
2.0log Re d
f 0.8
0.01
p LV2
h f z f
g d 2g
1
Introducing Q d 2V
4
3 1
p 0.241L d 4.75Q1.75 for turbulent flow
4 4 (8.29)
128Ld 4Q
p For Laminar flow (8.22)
P1
Q P2 2d Pipe 2
27
Turbulent flow
Limit of laminar
sublayer
53
8.7.1 Type of Rough Pipe
Roughness elements
protrude into the main flow
Limit of laminar sublayer
u *
70 (8.31)
55
Transitional Rough Walls
u *
70 rough
57
1 1
V f 1000 1.62 0.0204 6.53 Pa
2
8 8
w 6.53
u* 0.0808 m/s
1000
Thus, the reference Reynolds number is:
u *
0.046 10 -3
0.0808 3.7 5
10 -6
58
8.7.2 Velocity Profile in a Rough Pipe
Hydraulically Smooth
For hydraulically smooth pipe, the roughness elements are
submerged inside laminar sublayer. We may use the
velocity profile for a smooth pipe, i.e.
u 1 yu *
ln 5.0 (8.33)
u * 0.4
Completely Rough y Rr
u 1 y
ln 8.5 (8.34)
u * 0.4
59
Transitional Rough
The velocity profile is also given by
u 1 y
ln B (8.35)
u * 0.4
where B is given from the experimental data as
shown below
u *
ln
Acknowledgment: F.M. White:
Fluid Mechanics, 3rd Edition 60
8.7.3 Moody Diagram for Friction Factor
C
A
LAMINAR
FLOW
D
1
f
2.0log Red f 0.8
SMOOTH
PIPES
B
F
61
Moody Diagram
A
COMPLETELY ROUGH PIPE
Smooth pipe
1
2.0 log Red f 0.8
f
Eqn (8.25) B
For all the curves on the right hand side of AB (red curve), f
versus Reynolds number relationship becomes horizontal, indicating
that friction factor is independent of the Reynolds number. This
region is identified as a fully rough flow, and is described by
1 d
2.0log (8.36)
f 3.7 62
A
COMPLETELY ROUGH PIPE
1 d
E 2.0log
f 3.7
Eqn (8.36)
Smooth pipe
1
f
2.0 log Re d f 0.8
Eqn (8.25)
B
F
1 d
E 2.0log
f 3.7
Eqn (8.36)
Smooth pipe
1
f
2.0 log Re d f 0.8
Eqn (8.25)
B
F
1.8log (8.38)
f Re d 3.7
which varies less than 2% from equation
1 d 2.51
2.0log (8.37)
3.7 Re f 64
f d
COMPLETELY ROUGH PIPE
0.0125
0.0185
0.0128
65
COMPLETELY ROUGH PIPE
Material (mm)
Riveted steel 0.9-9.0
Concrete 0.3-3.0
Cast iron 0.26
Galvanised iron 0.15
Asphalted cast iron 0.12
Commercial steel or wrought 0.046
iron
Drawn tubing 0.0015
Glass “Smooth”
From: F.M. White, Fluid Mechanics, 3rd Edition
67
8.8 Minor Losses in Pipes
Fittings
Inlet Bends
Valve
Outlet
Sudden Sudden
enlargement contraction
68
A minor loss is expressed in terms of loss coefficient K,
defined by
V2 (8.39)
hL K
2g
It is often the practice to express a loss coefficient as an
equivalent length (Leq ) of the pipe.
V2 Leq V 2
K f
2g D 2g
D
Therefore Leq K (8.40)
f
69
8.8.1 Loss in Sudden Enlargement
(1) (2)
F p’ E
B
A1v1=A2v2
C D
p1 p2
A1 A2
v1 v2
z1 z1
v 2 v 2 v1 v12 v 22
hL
g 2 2g
v1 v 2
2g 71
v1 v 2
2 (1) (2)
F p’ E
hL A
2g
A1
v2 v1 B
A2 A1v1=A2v2
A1v1=A2v2 C D
A2 p1 p2
v1 v2 A1 A2
A1 v1 v2
z1 z1
2
v A1
2
hL 1
1
(8.41)
2g A 2
2
v A2
2
hL 2
1 (8.42)
2g A1
The above equation was first obtained by J.C. Borda (1733-
99) and L. Carnot (1753-1823) and is frequently known as
the Borda-Carnot head loss. 72
Exit Loss
A1 A2
2
Large tank
v A1
2
h L 1
1
2g A 2
A2>> A1
A1 A2
A1 A2 Equivalent
v12
hL (exit loss)
A1 v12 2g
0 hL
A2 73
2g
8.8.2 Loss in Sudden Contraction
Vena Contracta
(Area=Ac)
V1 V2 d2
d1
D
C
Area (A2)
Area (A1) (Section 2)
(Section 1)
V22 (8.43)
hL K sc
2g
75
V22
hL K sc (8.43)
2g
Because of the complexity of the flow, the loss coefficient Ksc is
obtained experimentally. Representative values are shown below.
Note:
Square-edge Protrusions
(flush) V V
K=0.5
(a) (b)
t/d=0.02
Bell-mouthed
V
r
Rounded
0.04 Edge
78
8.8.4 Gradual Expansion Energy loss can be considerably
reduced if the pipe transition is
more gradual.
V1 A1 A2 V2
V1 V2
2
9
hL K L
2g
KL A2/A1=2.25
2
(8.44)
A2/A1=4 A1 V12
K L 1
A2 2 g
6o
79
Some Nominal Loss Coefficients ( K) (turbulent Flow)
Type of fitting K
2
A1 A2 A1
1
Sudden enlargement A 2
0.2
90o mitre bend (with vanes)
=30o 0.02
a) Pipes connected in
Series
b) Pipes connected in
Parallel
c) Branched Pipes
81
8.9.1 Pipes in Series
3
1 2
Q1 Q2 Q3
The total head loss is the sum of the total losses in each of the
individual pipes and fittings
H L H L1 H L 2 H L 3 (8.46)
82
In terms of friction and minor losses in each pipe, equation
(8.46) may be re-written as
Friction loss
Minor losses
V12 f1 L1 V22 f 2 L2
HL K1,i K 2,i
2 g d1 i 2g d2 i
(8.47)
V32 f3 L3
K 3,i
2 g d3 i
83
8.9.2 Pipes in Parallel
1
Q1 (B)
(A) 2
Q Q2 Q
Q3 3
1
Q1
(A) 2 (B)
Q Q2 Q
Q3 3
Therefore, all the fluid in Each Pipe suffers the same loss of
head HL
H L1 H L 2 H L 3 (8.49) 85
8.9.3 Branched Pipes
Tank 1
Tank 2
1
2
Q1
z1 Q2 z2
3 J
Q3
Tank 3
z3 HJ
Arbitrary datum
86
The basic principles to solve the problem:
87
Tank 1
Tank 2
1
2
Q1
z1 Q2 z2
3 J
Q3
Tank 3
z3 HJ
Arbitrary datum
a) Tank 1=H1,
b) Tank 2 =H2,
c) Tank 3=H3,
d) Location J = HJ
88
Under steady condition
f1 L1 V12 f3 L3 V32
H1 H 3 H L1 H L 3
1d 2 g d 3 2 g
89
P1 P2
Tank 1 If P1 = P2 = P3 = Patmosphere
Tank 2
1 2 Assuming that Tank 1, Tank 2
Q1
z1 Q2 and Tank 3 are so large that
P3 J z2
3
Tank 3
Q3 V1=V2=V3 0
z3 HJ
Arbitrary datum If gauge pressure is used
throughout, it can be shown that
In general,
P1 V12
H1 z1 H1 z1
g 2g
P2 V22
H2 z2 H2 z2
g 2g
P3 V32
H3 z3 H3 z3
g 2g
PJ VJ2 VJ2
PJg
HJ zJ HJ zJ
g 2g g 2g
90
Example 8.4
Water at 20oC flows through a 40 mm diameter wrought
iron horizontal pipe with a flow rate of 0.003 m3/s. If the
pipe is 1000 m long, and the roughness element of the
pipe (ε) is 0.046 mm, determine the pressure drop over
the length of the pipe.
Solution:
Q 0.003
V 2.39m / s
A 0.04 4
2
VD 2.39 0.04
Re= = 9.6 10 4
ν 10-6
0.046
Also 0.00115
D 40
91
From the Moody
diagram: f=0.023 0.023
1 6.9 d 1.11
1.8log
f d 3.7
Re
93
Solution:
h f z1 z2 z
94
From Darcy-Weisbach friction Eq. (8.8)
LV2
hf f z (8.8)
d 2g
The flow rate Q is given by
d2
Q V
4
4Q V2 8Q 2
V 2 4
d2 2g d g
Substituting above expression into Eq. (8.8) gives
L 8Q 2
z f a Q 2
d 2d 4 g
8 fL
where a 2 4
d gd 95
Example 8.6
Water flows from reservoir A through a 100 m long pipe of
diameter 120 mm to a branch point D where it is diverted to
reservoirs B and C in separate pipes as shown in the figure below.
Assuming that f =0.02 for all the pipes and neglecting all losses
other than those due to friction, determine the elevation of the
reservoir B. The flow from the reservoir A is 0.02 m3/s.
96
The velocity of water
between A and D is
QA 20 103
V
A 2
(0.12)
4
1.768 m / s
.
The head loss in friction between A and D is
Gauge pressure
fLA VA2 0.02 100 1.7682 for tanks (zero)
h fA 2.655m
DA 2 g 2 9.81 0.12 is used, but
gauge pressure
H D z A h fA 50 2.655 47.365 at D is not zero.
97
The head loss in friction
between D and C is
H D - zC 47.365 - 25 22.365
Therefore,
3.78 10-3
VB 0.855 m/s
2
(0.075)
4
Head loss in friction between D and B is
0.02 60 0.8552
h fB 0.596 m/s
2 9.81 0.075