PIPE FLOW 3/16/2019
January - April, 2019
A. W. Jayawardena
1
Introduction
 Flow in pipes is important for many reasons, for example,
 Study of turbulence in pipes have led to a better
understanding of the turbulence in general, e.g. boundary
layer
 The need to determine head losses in pipe systems connected
via pumps and turbines
 Boundary shear (stress) is important in gas flow
 Heat and fluid transfer in engineering is always via ducts.
 Laminar flow exists for Reynolds Number up to about 2300
(Re = VDν ; where D is the hydraulic diameter which is 4
times the hydraulic radius). For engineering applications,
Re < 2000 is considered laminar
Re > 2000 is considered turbulent

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Momentum equation for fully
developed flow
 Fully developed flow refers to the condition when 0, p
, u have attained their maximum values. The
momentum flux for a fully developed flow is

 D2
4
x
V V
where  is the momentum correction factor, given by

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2
u
 dA
V2A
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Momentum equation for fully
developed flow....
 Considering a control volume inside a pipe, we may
write
pD 2
4
 0Ddx
( p  dp)
D 2
4
Fig. 1: Forces acting on a fluid element in pipe flow
πD2
πD2
πD2 d

p
  p  dp 
 τ 0 π Ddx 
ρβV 2  dx
4 dx
4
4
which, for fully developed incompressible flow,
simplifies to


dp 4τ 0
dβ
dV
d
 ρV 

 ρV 2
 β ρV
 βV
dx
D
dx
dx
dx

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dp Δp 4τ 0


dx
L JanuaryD- April, 2019
A. W. Jayawardena
4
Momentum equation for fully
developed flow....
d
 ρV   0
dx
 Because,
not depend on x.
βV
dβ
dV

0
,
0
; dx
dx
and 0 does
 i.e. the pressure decreases with increasing x and
the pressure drop is used in overcoming wall shear
stress.
 For developing flow, at the entrance of a pipe  =
1 and approaches  = 4/3 for fully developed
laminar flow in circular pipes. For turbulent flow,
  1.03
 Therefore, for developing flow, the momentum
4τ
dβ
equation becomes  dp

 ρV
dx
D
dx
0
because
3/16/2019
2
D2V1January
= D2V-2April,
2019V1 A.
= W.
V2 Jayawardena
; therefore,
dV
0
dx
and
d
 ρV   0 .
dx
5
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Momentum equation for fully developed
flow....
 For compressible flow,
dβ
0
dx
but V varies with x
because of density variations. Therefore, 0 and dp vary
dx
with x, giving,
dp 4τ 0
dV


 βρV
dx
D
dx
0.5
0
t0
1.0
t
r0
V
r
u max
Velocity profile
and
shear
in laminar flow
January
- April,
2019 A.distribution
W. Jayawardena
3/16/2019
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Fully developed flow in pipes
 The pressure head per unit length, , can be obtained
using dimensional analysis.
Δp
 f ( D, k ,V ,  ,  )
L
where D is the diameter of the pipe, k is the pipe
roughness, V is the average velocity,  and  are the
density and viscosity of fluid.
 There are six (n = 6) variables with three (m = 3)
fundamental dimensions, M, L and T.
Therefore,
according to Buckingham’s π theorem there will be at
least one set of (n - m) independent dimensionless groups
in a dimensional analysis.
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Fully developed flow in pipes….
 Each of these will consist of m (= 3) variables in common,
and are called repeating variables. The rules for selection
of repeating variables are
The repeating variables must include among them, all
fundamental quantities
Dependent variables should not be used as repeating
variables
 For a fluid system, the most significant groups will result
if the repeating variables are chosen so that one is a
geometrical characteristic, one is a fluid property and
one is a flow characteristic
 After the repeating variables are chosen, each one of the
remaining variables is included in these groups.
 In this case, the repeating variables are D, ρ and V (average
velocity). Therefore,
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 p 

L


 1  D a1  b1V c1 
 2  D a 2  b 2V c 2 k
 3  D a3  b3V c3 
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Fully developed flow in pipes….
 Substituting the fundamental quantities M, L and T
for each, it can be shown that

(Twice the friction factor, f)


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(Relative roughness of a pipe)
3 
μ
VDρ
(
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)
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Fully developed flow in pipes….
 Thus, we can write
Δp 4f ρV 2

L
D 2
 This form of the head loss equation is called the Darcy -
Weisbach equation. It can also be written as
 ∆p = ℎ𝑓 ρg⇒

Δp
 hf
ρg
2
LV
 4f  
 D  2g
4fV 2

L
2Dg
hf
 Note: The relationship that π1 is twice the friction factor f
is based on the values of f given in the English system. In
the US, the value of f is 4 times that defined by the above
relationships. With their values of friction factor, the
Darcy-Weisbach equation becomes
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Fully developed flow in pipes….
f 'V 2

L
2Dg
hf
where f’ = 4f
 The friction factor f is a function of relative roughness and Reynolds
Number.
Δp D
4τ
τ0 
 From Eq.  dp  Δp  , 0
dx
L
2
Δp
4f
ρV
 Since 
, τ0
L
D 2
L 4
D

4f ρV 2 D fV 2

D 2 4
2
 The shear velocity is defined as u* 
0
= V

f
2
 [For circular pipes, the characteristic length is D; for non-circular pipes
it is the hydraulic diameter (not radius!) Dh = 4A
; hydraulic diameter is
P
the same as the geometrical diameter]
 When using US equations, appropriate adjustments should be made
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Fully developed flow in pipes….
f 'V 2

L
2Dg
hf
where f’ = 4f
 The friction factor f is a function of relative roughness and Reynolds
Number.
Δp D
4τ
τ0 
 From Eq.  dp  Δp  , 0
dx
L
2
Δp
4f
ρV
 Since 
, τ0
L
D 2
L 4
D

4f ρV 2 D fV 2

D 2 4
2
 The shear velocity is defined asu* 
0

= V
f
2
 [For circular pipes, the characteristic length is D; for non-circular pipes
4A
it is the hydraulic diameter (not radius!) Type equation here.Dh
= ;
P
hydraulic diameter is the same as the geometrical diameter]
 When using US equations, appropriate adjustments should be made
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Critical Reynolds Number
 The transition from laminar to turbulent flow takes place
when the Reynolds number is around 2000. The critical
velocity for this Re is
2000
ucrit 
 For water, ν
=1.15x10-6
m2/s;
d
therefore, for a pipe of 25 mm
diameter, ucrit =92 mm/s.
 This is a very small velocity when compared to the
velocities that are normally encountered. If the velocity is
to be greater, a smaller diameter would be needed.
However, a diameter smaller than 25 mm is not of any
practical value.
 For oil, ν is 200 times that of water; therefore, for a pipe of
25 mm diameter, ucrit =18.4 m/s.
 This means that the transition takes place when the
velocity is around 18.4 m/s which is a very high velocity.
Therefore, all oil flows can be considered as laminar.
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Laminar fully developed flow
Rounded entrance
from tank
Fully developed
parabolic profile
A
Boundary layer
L'
Unestablished flow
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B
Established flow
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Laminar fully developed flow....
Laminar boundary layer
Turbulent
boundary
layer
Uniform flow
y
r0
r
Fully developed
turbulence
U
U
Viscous sublayer
Xc
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Laminar fully developed flow....
 For circular pipes, (because y = R-r; dy = -dr)
u
du
du
 

 
y
dy
dr
Therefore,
du
τ

dr
μ
dp Δp 4τ
(Because,    0
dx
L
D

Δp r
L 2μ
)
 Integration gives (boundary conditions: u = 0 at r = R)
u
1 Δp 2

R  r2 
4μ L
which is parabolic. The discharge through the cross
section is
 p R
 p R
R 4 p
2
2
Q   2rudr 
rudr 
r ( R  r )dr 


2 L 0
2 L 0
8 L
0
R
 This is usually known as Hagan-Poiseuille formula.
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Laminar fully developed flow....
 From this velocity profile it can be seen that the
1 Δp 2
R
4μ L
maximum velocity is
and the average velocity
is half the maximum velocity. QA   udA
A
 It can also be shown that the friction factor f is

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f

16μ
16

VDρ Re D
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Turbulent fully developed flow
 In turbulent flow the velocity profile is flatter than that for
laminar flow. The higher the Re, the flatter the velocity
profile. This is because of the radial component of the
velocity. This means steep velocity gradients at the
boundary and larger boundary shear stress. (for the same
Re)
 Turbulent flow is not as simple as laminar flow. Using
statistical representation of turbulence, Taylor simplified
the solution of the Navier-Stokes equations by assuming
the turbulence to be homogeneous and isotropic.
 As a result, several velocity profiles have been proposed;
e.g. logarithmic velocity profile, universal velocity profile
etc.
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V
Laminar flow
R<2000
Smooth pipe
R=10 7, f=0.012
Rough pipe
R=10 7, f=0.04
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 For example,
u*
u
 5.75log
y  5.5
u*

u max  u
τ0
ρ

u mas  u 1  R 
 n 
u*
κ  y
 ……………
 …………….
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Flow regimes in viscous flow
 In viscous flow, 3 flow regimes can be identified:
 Hydraulically smooth region - roughness elements
submerged within the sub-layer
0
u* k
5
ν
 Transitional region - roughness elements partly
submerged within the sub-layer and partly projecting
outside the sub-layer
5
u* k
 70
ν
 Hydraulically rough region - all roughness elements
project outside the sub-layer
u* k
 70
ν
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Empirical equations for friction
factor for smooth and rough pipes
 There are many empirical equations. e.g. Blasius,
Moody, Nikuradse, Colebrook etc.
 Moody equation for smooth pipes is of the form
3
 
k 10 6  
 
f  0.001375 1   20,000 
D Re  
 

1
 Colebrook equation for smooth pipes is of the
form
1
Re
 1.8log ( )
6.9
f
 Colebrook equation for rough pipes is of the form
1
  2 log
f
 2.51
k 


 Re f
3.71D 

 Von Karman equation for rough pipes 3/16/2019
1
 2 log
f
 3.71D 


 k 
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Empirical equations for friction factor
for smooth and rough pipes
 It is important to note that
 For laminar flow, f depends on Re only and is not
affected by k.
 For hydraulically smooth pipes, f for turbulent flow
also depends on Re only.
k
 For hydraulically rough pipes, f depends only on D
and not on Re.(von Karman Eq.)
k
 For smooth-rough transition zone, f depends on D
and Re.
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Moody diagram: (Moody, 1944)
 In this diagram, a single line on the left hand side
corresponds to laminar flow (Re < 2000) where f varies
according to f  16Re
or, f  Re64
 For turbulent flow, the curves are based on Nikuradse’s
experiments.
 It is also important to note that the friction factors used
in some text books (US origin) are four times those used
in some other text books (British). This difference can
be seen in the two Moody diagrams shown below. Care
should therefore be taken when using these values in the
appropriate head loss equation. It is important to check
the formula used in Moody diagram. If the straight line
part (for laminar flow) is represented by f = 64/Re, then
the equation for head loss is hf = fLV2/2gD; if the straight
line part is represented by f = 16/Re, then the equation
2/2gD.
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2019 4flV
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for head loss should
Moody diagram (British system)
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Moody diagram (US system)
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Typical pipe flow problems
 Typical pipe flow problems can be categorised
into 3 types:
 Type I: Given Q, L, D, , k
Find hf
 Type II: Given hf,, L, D, , k Find Q
 Type III: Given hf,, L, Q , k Find D
• In all cases, it is necessary to use
 Darcy-Weisbach equation
 Continuity equation
 Moody diagram
• Very often, an iterative approach is needed.
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Common problems
Water supply distribution (from WSD):
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Common problems
Gravity flow from one reservoir to the other (whenever possible):
H
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Common problems
Gravity flow from one reservoir to the other (whenever possible):
H
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Common problems
rvice reservoirs:
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 For a sudden contraction, it can be shown (Ref. p 300,
Fluid Mechanics by Streeter, Wylie, and Bedford) that
2

1
2 V 
 hL  ( - 1 )

C
2g
c


where Cc is the coefficient of contraction.
 Entrance losses - In calculating the pressure drop, it
is customary to use fully developed friction factors and
add a correction factor hL to account for entrance
effects.
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Entrance conditions
h'e
hf
EL
V2/ 2g
HGL
P/ v
V
B
D
C
A
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Submerged discharge losses
EL
V2/ 2g
HGL
(c)
(a)
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(b)
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Minor losses in pipes
 Losses at entrance, exit, bends, contractions,
expansions and pipe fittings etc. are called minor
losses. They are usually expressed as a velocity head in
the form
2

V 
 hL  k L

2g 

 where kL is loss coefficient which needs to be
experimentally determined, except for sudden
expansions.
 For a sudden expansion from A1 to A2, the loss
coefficient is
A
k  (1  )
1
L
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2
A2
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Minor losses
Device losses of elbows and tees.
Elbow
45o :
90o :
KL = 0.4
KL = 0.75
Tee
line flow:
KL = 0.4
branch flow: KL = 1.5
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Minor losses
Device losses of valves:
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Minor losses
Device losses of valves:
Globe valve
KL = 10.0 (fully open)
KL = 12.5 (½ open)
Gate valve
fully open KL = 0.2
¾ open
0.9
½ open
4.5
¼ open
20.0
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Example 1: Determine the size of galvanized steel pipe needed to carry water a distance of
180 m at 85l/s with a head loss of 9 m.
hf  9 
4 f (180) 0.085 2
1
( 2 )
D
D / 4 2(9.81)
 D 5  0.0478 f
0.085
D
9.49 x10 4


Re =

D 2 / 4 1.14x10 6
D
VD
For galvanized steel, k = 0.15 mm
Try f = 0.006. Then D = 0.1956 and Re = 4.85 x 106;
k
 0.00077.
D
k
, Moody diagram gives f = 0.00475.
D
k
Substituting back, D = 0.1867 and Re = 5.08 x 105;
 0.00080.
D
For these values, f = 0.0048 which is close enough to the previous value. Therefore, use a
diameter of 200 mm.
For these values of Re and
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Fig. xx: Energy grade line (EGL) and Hydraulic grade line (HGL)
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Example 2: Calculate the steady rate at which water will flow through a cast iron pipe 100
mm diameter and 1200 m long under a head of 50 m.
For cast iron, k = 0.25 mm; therefore
k
 0.0025
D
Assume f = 0.0065 (guessed). Then,
4 fL V 2 4x 0.0065x1200 V 2
50 =

D 2g
0.10
19.62
Re 
 V  1.733m / s.
1.733x0.10
6

0
.
15
x
10
1.14 x10 -6
For this pair of Re and
Therefore, Q 
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D 2
4
k
, Moody diagram gives a f value of 0.0065 which is close enough.
D
V  14l / s.
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Example 3: Oil of specific gravity 0.83 and dynamic viscosity 0.08 kg/m.s passes through a
circular pipe of 12 mm diameter with a mean velocity of 2.3 m/s. Determine the
(i)
(ii)
(iii)
(iv)
Reynolds Number
Maximum velocity
Volumetric flow rate
Pressure gradient along the pipe
VD (0.83x1000)(2.3)(12 x10 3 )
Re 

 286.35

0.08
This value of Re is small enough to have laminar flow.
Maximum velocity = 2V = 4.6 m/s
Flow rate Q = Area x Average velocity =
 (0.012) 2
4
x2.3  2.601x10 4 m3/s.
8 x 2.601x10 4 x0.08
8Q
 40886 Pa / m
=
 (0.006) 4
R 4
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Example 4
 Derive an expression for the power required to pump a volumetric flow rate of Q through
a horizontal pipe-line of constant diameter D, length L and friction factor f assuming
that the only losses in the pipe-line are due to friction only
 For the purpose of project calculations the total cost of moving a fluid over a distance by
pipe-line at a steady flow rate Q can be broken down into two parts. First, the
manufacture, laying and maintenance including interest charges are represented by the
cost C1, which is proportional to D3. The second item C2 depends solely on the energy
required to pump the fluid.
 A preliminary design study for a particular project showed that the total cost was a
minimum for D = 600 mm. If fuel prices are increased by 150%, and assuming only C2 is
affected, make a revised estimate of the optimum pipe diameter. (Example 7.4,
Mechanics of Fluids by B. Massey and J. Ward-Smith, 8th edition)

•
•
gh f Q
p 4 fLV 2
4 fLV 2
P
; hf 

 p 

g
2 gD
2D
Power required, P:
From continuity,
V
Q
D 2 / 4
Eliminating V and substituting for ∆p,
32 fLQ 3
P
 2D 5
 From the expression for power, C2 is proportional to D-5
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Example 4
 Total cost C= C1 + C2 = aD3 + bD-5
 C is a minimum when
 i.e. when
D8 
dC
 3aD 2  5bD 6  0
dD
5b
5b
, or , D  8
3a
3a
 The diameter of the original pipe is given by
 The diameter of the new pipe, D2 is given by
D2  8
 Therefore,
D1  8
5b1
3a
5b2 8 5(2.5b1 )

3a
3a
D2  8 2.5 D1  8 2.5(600)  673mm
3/16/2019
January - April, 2019
A. W. Jayawardena
51
Calculation Example with Minor Losses
Finding pipe size for a given discharge (previous example)
 A smooth-walled pipeline 3500 m long is to connect two
reservoirs whose free water surfaces are 7 m different in
elevation. If a flow of water (under gravity) at 0.03m3/s is
required, what pipe diameter should be used?
 Answer considering only pipe friction  d = 0.228 m
Now include the following device losses:
 Entry is abrupt
 Outlet is a conical diffuser
 Two globe valves
 Five 60 bends
3/16/2019
January - April, 2019
A. W. Jayawardena
52
Calculation Example with Minor Losses
Solution
 d = 0.228 m, without minor losses
 Use the data for minor losses from handbook
 In the form of equivalent pipe length:
–
–
–
–




3/16/2019
an abrupt entry is equivalent to
a diffuser outlet is equivalent to
a globe valve is equivalent to
a 60 bend is equivalent to
25 diameters of pipe
6 diameters of pipe
75 diameters of pipe
22 diameters of pipe
The total additional length of pipe equivalent to all fittings is:
25 + 6 + (275) + (522) = 291 diameters
8 flQ 2
5
d  2
or a effective length of: 2910.228 = 66 m
 gh f
Thus total (equivalent) length of pipeline is : 3500+66 = 3566 m
With this length, new d5 = old d5  3566/3500
 new d = 0.229 m (Minor losses!)
January - April, 2019
A. W. Jayawardena
53