CVE 372
HYDROMECHANICS
FLOW IN CLOSED CONDUITS
II
Dr. Bertuğ Akıntuğ
Department of Civil Engineering
Middle East Technical University
Northern Cyprus Campus
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Overview
2.2 Fully Developed Flow in Closed Conduits
2.2.1 Derivation of Darcy-Weisbach Equation
2.2.2 Laminar Flow in Pipes
2.2.3 Turbulent Flow in Pipes
2.2.4 Moody Chart
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
„ Consider a steady fully developed flow in a prismatic pipe
V1=V2=V
A1=A2=A
α1=α2=1
β1=β2=1
Assumptions:
- Fully developed flow (uniform)
- Circular tube (pipe)
- Steady Flow
CVE 372 Hydromechanics
- Incompressible fluid
- Constant diameter
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
(a) Relationship between wall shear stress
and head loss
„ Continuity Equation
Q = V1 A1 = V2 A2 = VA = constant
„
Momentum Equation
0, since β1= β2
p1 A1 − p2 A2 + W sin θ − F f = ρQ(β 2V2 − β1V1 )
where
W sin θ = γAL sin θ
= γA( z1 − z 2 )
CVE 372 Hydromechanics
and
F f = τ w PL
P: wetted perimeter
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
„ Momentum Equation gives
p1 A1 − p2 A2 + γA( z1 − z 2 ) − τ w PL = 0
τ w LP τ w L
=
=
z1 + − z 2 −
γ
γ
γA
γRH
p1
p2
where RH is hydraulic radius
A πD 2 4 D R
=
RH = =
P
πD 4 2
W sin θ = γAL sin θ
= γA( z1 − z 2 )
CVE 372 Hydromechanics
P: wetted perimeter
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
„ Momentum Equation gives
τ wL
z1 + − z 2 −
=
γ
γ γRH
p1
„
p2
Energy Equation gives
z1 +
p1
γ
− z2 −
p2
γ
= hL
τ wL
hf =
γRH
2τ L
hf = w
γR
4τ w L
hf =
γD
Note that the above equation is applicable for both laminar and
turbulent flows and for open channel flows as well.
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
(b) Relationship between wall shear stress
and velocity
„ Dimensional Analysis
τ w = F (V , D, ρ , µ , ε )
 shear stress 
τw
= f′⇒ 
π1 =

2
ρV
 dynamic pressure 
ρVD VD
=
⇒ Reynolds Number
π2 =
µ
ν
ε
π 3 = ⇒ Relative Roughness
D
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π 1 = φ (π 2 , π 3 )
 ρVD ε 
τw
= φ 
, 
2
ρV
 µ D
τ w = f ′ρV 2
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.1.1 Derivation of Darcy-Weisbach Equation
(c) Relationship between head loss and velocity
τw
ε 

f′=
= function Re, 
2
ρV
D

τ w = f ′ρV 2
′ V 2L
4τ w L 4 f ρ
=
hf =
γD
γD
let 8f'=f and g=γ/ρ
L V2
ε

⇒ f = function Re, 
hf = f
D
D 2g

CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
„
Local acceleration is zero. Convective acceleration is zero as
well as a result of fully developed flow
Motion of a cylindrical fluid element within a pipe.
Assumptions:
- Fully developed flow (uniform)
- Circular tube (pipe)
- Steady Flow
CVE 372 Hydromechanics
- Incompressible fluid
- Constant diameter
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Local acceleration is zero. Convective acceleration is zero as well as a
result of fully developed flow
Free-body diagram of a cylinder of fluid
Momentum Equation
p1πr 2 − ( p1 − ∆p )πr 2 − τ 2πrl = 0
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∆p 2τ
=
l
r
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Shear stress distribution within the fluid in a pipe
(laminar or turbulent flow) and typical velocity profiles.
2τ w r
τ=
D
CVE 372 Hydromechanics
4lτ w
∆p =
D
Small shear stress can produce large
pressure difference if l/D is large
(relatively long pipe)
Applicable to both laminar and turbulent flow in pipes11/54
2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
For laminar flow of a Newtonian fluid, the shear stress is
simply proportional to the velocity gradient (Section 1.6)
τ = µ (du/dr). For our pipe flow,
∆p
du
τ=
r = −µ
2l
dr
The negative sign is included to give τ > with du/dr < 0 ( the
velocity decreases from the pipe centerline to the pipe wall)
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
 ∆p 
du
r
= −
dr
 2 µl 
∆p 2
)r + C1
integration yields u ( r ) = −(
4 µl
We can evaluate the integration constant using the no slip
condition at the pipe wall so that u = 0 at r=D/2 and
C1=(∆pD2/(16µl)). Hence
Centerline (max) velocity
Velocity:
Vc=(∆pD2/(16µl)).
2
  2r  2 
 ∆pD 2   2r  
 1 −   = Vc 1 −   
u (r ) = 
 D 
 16µl   D  


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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Integration of u(r) over the pipe area gives volume flowrate
R
Q = ∫ udA = ∫ u (r )2πrdr =
0
πR 2Vc
2
by definition, the average velocity is the V=Q/A = Q/πR2
π R 2Vc Vc ∆pD 2
V=
= =
⇒
2
2π R
2
32 µl
πD 4 ∆p
Q=
128µl
Above equation is referred to as Poiseuille’s Law.
Recall that all of these results are restricted to laminar flow in
a horizontal pipe.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
„
„
„
„
„
Flowrate:
πD 4 ∆p
Q=
128µl
Proportional to the pressure drop
Inversely proportional to viscosity
Inversely proportional to the pipe length
Proportional to the pipe diameter to the fourth power
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
„
For Non-horizontal pipe, the adjustment is required by
replacing ∆p by (∆p-γl sinθ) where θ is the angle between
the pipe and the horizontal.
Free-body diagram of a fluid cylinder for flow in a nonhorizontal pipe.
∆p − γl sin θ 2τ
(∆p − γl sin θ ) D 2
π (∆p − γl sin θ ) D 4
=
V=
Q=
32 µl
128µl
l
r
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Example 3: An oil with a viscosity of µ = 0.40 N.s/m2 and
density ρ = 900 kg/m3 flow in a pipe of diameter D = 0.020 m.
a) What pressure drop, p1-p2, is needed to produce a flowrate of
Q = 2.0 x 10-5 m3/s if the pipe is horizontal with x1=0 and x2=10 m?
b) How steep a hill, θ, must the pipe be on if the oil is to flow
through the pipe at the same rate as in part (a), but with p1 = p2?
c) For the conditions of part (b), if p1 = 200 kPa, what is the
pressure at section x3 = 5 m, where x is measures along the pipe?
Solved in the class room
CVE 372 Hydromechanics
17/54
2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
From dimensional analysis:
„ Assume that ∆p is a function of
∆p = f (V , l , D, µ )
# of variables = 5 and # of dimensions = 3 (M,L,T)
According to the result of dimensional analysis
(CVE 371), this flow can be described in terms of 5-3=2 dimensionless
Group. One such represents
D∆p
l
= φ( )
µV
D
A further assumption: the pressure drop is directly proportional to the
pipe length
4
D∆p Cl
∆p CµV
π
C
pD
(
/
4
)
∆
=
⇒
=
⇒ Q = AV =
2
D
µV
l
D
µl
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
The same functional form as theory implies
(π / 4C )∆pD 4
Q = AV =
µl
Recall average velocity was found to be
D 2 ∆p
V =
32µl
We can divide both sides by the dynamic pressure
(recall from Chapter 3)
1
ρV 2
2
 µ  l  64  l 
32 µlV / D 2
  =
=
= 64
 
1
1
 ρVD  D  Re  D 
ρV 2
ρV 2
2
2
∆p
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
This is often written as
l ρV 2
∆p = f
D 2
where the dimensionless quantity
 D  2 

f = ∆p 
2 
 l  ρV 
Darcy friction factor
For a fully developed laminar flow the friction factor is
Alternate expression as a dimensional wall shear stress
CVE 372 Hydromechanics
f =
64
Re
8τ w
f =
ρV 2
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Energy Consideration
Recall from Chapter 5 (in the absence of energy sources)
p2
V 22
V 21
+ α1
+ z1 =
+ α2
+ z2 + hL
2g
2g
γ
γ
p1
where alpha values (always >=1) compensate for the fact that
velocity profile across the pipe is not uniform (Chapter 5).
α1 = α 2
α1V12 / 2 g = α 2V22 / 2 g
For fully developed flow velocity profile is constant
Hence
CVE 372 Hydromechanics
 p1
  p2

 + z1  −  + z 2  = hL
γ
  γ

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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
Remember
p1πr 2 − ( p1 − ∆p )πr 2 − τ 2πrl = 0
∆p 2τ
=
l
r
(
p1
γ
+ z1 ) − (
p2
γ
2τl
hL = h f =
γr
+ z 2 ) = hL
Using the shear stress at the pipe wall, we have
4lτ w
hf =
γD
CVE 372 Hydromechanics
No minor loss
Friction loss is proportional to the shear stress,
which is proportional to viscosity.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
In summary
Velocity:
max. velocity
2
 ∆pD 2   2r  
 1 −  
u (r ) = 
 16 µl   D  
Vc = umax
  2r  2 
or u (r ) = Vc 1 −   
 D 


Average Velocity:
Vc ∆pD
V= =
2
32 µl
2
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α1V12 / 2 g = α 2V22 / 2 g
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
In summary
Wall Shear Stress:
∆pD
τw =
4l
or
8µV
τw =
D
Shear Stress:
du
τ = −µ
dr
CVE 372 Hydromechanics
or
2r
τ =τw
D
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.2 Laminar Flow in Pipes
In summary
Flow Rate:
πD 4 ∆p
Q = VA =
128µl
Friction Loss:
4lτ w
hf =
γD
Friction Factor:
CVE 372 Hydromechanics
64
f =
Re
32lµV
hf =
γD 2
f =
l V2
hf = f
D 2g
8τ w
ρV 2
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Example 4: The flow rate Q of corn syrup through the horizontal pipe shown in
the figure is to be monitored by measuring the pressure difference between
sections (1) and (2). The variations of the syrup’s viscosity and density with
temperature are given in the following table
a) Determine the wall shear stress and
the pressure drop ∆p=p1-p2 for Q=14 lt/s
for T=40°C.
b) For the condition of part (a), determine
the net pressure force and the net shear
force on the fluid within the pipe between
the section (1) and (2).
T (°C)
ρ (kg/m3)
µ (N.s/m2)
15
1067
1.9152
25
1062
0.9097
40
1057
0.1819
50
1051
0.0211
60
1046
0.0044
70
1041
0.0011
2m
10 cm
CVE 372 Hydromechanics
Solved in the class room
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Transition from Laminar to Turbulent Flow
CVE 372 Hydromechanics
We slowly increase the flowrate in a long section of a pipe.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Transition from Laminar to Turbulent Flow
u(t): instantaneous velocity in the x-direction
u’(t): fluctuating part of u(t)
The time-averaged, ū, and fluctuating, ú, description of a parameter for tubular flow.
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Shear Stress
We can write the velocity vector as
u = u + u'
Fluctuation (time average of fluctuations is zero)
Average velocity:
Fluctuations are equally
distributed on either side of
the average. However, the
square of fluctuation is
always greater than zero.
1
(u ' ) =
T
2
1
u=
T
t o +T
∫ u ( x, y, z, t )dt
to
t o +T
2
(
u
−
u
)
dt > 0
∫
to
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
Turbulent Shear Stress
The structure and characteristics of turbulence may vary
from one flow situation to another. A measure of turbulence
is called ‘turbulence intensity’
1

(u ' ) 2  T
=
u


u
u
dt
(
−
)
∫t

o

u
t o +T
2
2
As this parameter increases, the fluctuations of the velocity
increases. Well designed wind tunnels have typical values of
0.01 or smaller.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Shear Stress
The random velocity components result in momentum transfer in
turbulent flow resulting in an additional term in the shear stress
expression:
du
τ =µ
− ρ u ' v' = τ lam + τ turb
dy
If the flow is laminar, then fluctuations vanish and we recover the
viscosity expression for Newtonian fluids. The second term is called the
turbulent shear stress and it is always positive. Hence the shear stress in
turbulent flow is always greater than shear stress in laminar flow.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Shear Stress
τlam is
dominant
τturb is
dominant
Structure of turbulent flow in a pipe. (a) Shear stress. (b) Average velocity.
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Shear Stress
„
„
Typically turbulent shear stress is 100 to 1000 times
greater than the shear stress in the laminar region, while
the converse is true in the viscous sublayer.
Note that an accurate model of turbulent flow requires the
knowledge of Reynolds stresses which require the
knowledge of velocity fluctuations which can not be solved
for most turbulent flow problems.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Velocity Profile
„
Fully developed turbulent flow in a pipe can be broken into
three regions which are characterized by their distances
from the wall:
„
„
„
CVE 372 Hydromechanics
Viscous sublayer: The viscous shear stress is dominant
compared with the turbulent (or Reynolds) stress and the
random (eddying) nature of flow is absent. In this layer fluid
viscosity is important parameter.
Overlap region: Transition region
Outer turbulent layer: The Reynolds stress is dominant, and
there is considerable mixing and randomness of the flow. In
this layer density is important parameter.
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Velocity Profile
„
In the viscous sublayer:
u
yu ∗
=
*
u
ν
where
y = R−r
τw
u =
ρ
∗
friction velocity
u = the time av. x component of the velocity
y = δs: thickness of the viscous sublayer
R
y
CVE 372 Hydromechanics
r
δs =
5ν
Law of the wall, valid
very near the smooth wall
for 0 ≤ yu ∗ /ν ≤ 5
u*
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Velocity Profile
„
In the overlap region:
 yu ∗ 
u
 + 5.0
= 2.5 ln
*
u
 ν 
where
y = R−r
R
y
r
u = the time av. x component of the velocity
τw
∗
u =
ρ
CVE 372 Hydromechanics
friction velocity
Law of the wall, valid
very near the smooth wall
for 0 ≤ yu ∗ /ν ≤ 5
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Velocity Profile
„
In the turbulent region:
R
Vc − u
= 2.5 ln 
*
u
 y
or
1/ n
u  r
= 1 − 
Vc  R 
CVE 372 Hydromechanics
Figure 8.17
Exponent, n, for power-law velocity profiles.
Power-Law velocity profile
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
Turbulent Velocity Profile
Typical laminar flow
and turbulent flow
velocity profiles.
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
Example 5: Water at 20°C (ρ=998 kg/m3 and ν=1.004 x 10-6 m3/s) flows
through a horizontal pipe of 0.1 m diameter with a flowrate of Q=4x10-2 m3/s
and a pressure gradient of 2.59 kPa/m.
a) Determine the approximate thickness of the viscous sublayer.
b) Determine the approximate centerline velocity, Vc.
c) Determine the ratio of the turbulent to laminar shear stress, τturb/ τlam at a
point midway between the centerline and the pipe wall (r=0.025 m).
Solved in the class room
CVE 372 Hydromechanics
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
„
„
Most turbulent pipe flow analyses are based on
experimental data and semi-empirical formulas which are
expressed in dimensionless forms.
We need to determine the head loss. For convenience,
we will consider two types of energy losses; minor (local)
and major (friction) losses
hL = hLmajor + hLminor
Note that major and minor losses do not necessarily reflect the magnitude
of the energy losses
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
Consider the pipe flow again. Pressure drop is
a function of a number of physical and
geometrical parameters:
∆p = F (V , D, l , ε , µ , ρ )
This time we included a parameter which is
a measure of the roughness of the pipe
wall (unit is length)
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
There are seven parameters and three
reference dimensions.
∆p
~ ρVD l ε
=φ (
, , )
2
1 / 2 ρV
µ D D
Reynolds number
Dynamic pressure
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
Let’s assume that pressure drop is proportional to the length of the pipe:
∆p
ε
l
= φ (Re, )
2
1 / 2 ρV
D
D
Recall that
f =
ε
∆pD
is
the
friction
factor.
Then
we
have
f
=
φ
(Re,
)
2
D
lρV / 2
The energy equation for steady, incompressible flow is given by
p1
γ
CVE 372 Hydromechanics
+
α1V 21
2g
+ z1 =
p2
γ
+
α 2V 2 2
2g
+ z 2 + hL
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
For constant diameter, horizontal pipe with fully
developed flow (alphas are equal)
hLmajor
l V2
= f
D 2g
This is called Darcy-Weisbach equation
For nonhorizontal pipes
p1 − p2 = γ ( z1 − z 2 ) + γhLmajor
CVE 372 Hydromechanics
l V2
= γ ( z1 − z 2 ) + f
D 2g
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed
Conduits
2.2.3 Turbulent Flow in Pipes
The Moody Chart.
For laminar flow:
f=64/Re
For turbulent flow:
ε / D
1
2.51
= −2 log
+
 3.7 Re f
f

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



for large values, f is independent of 45/54
Re
2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
2.2.3 Turbulent Flow in Pipes
Smooth Pipe and Hydraulically Smooth Flow
 2.51
1
= −2 log
 Re f
f





Colebrook – White Transition Flow
ε / D
1
2.51

= −2 log
+

f
 3.7 Re f




Rough Pipe Hydraulically Rough Flow
1
ε / D
= −2 log

f
 3. 7 
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed
Conduits
2.2.4 Moody Diagram
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed
Conduits
2.2.4 Moody Diagram
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in
Closed Conduits
2.2.4 Moody Diagram
Loss coefficient for a sudden expansion.
Conservation of mass
A1V1 = A3V3
Conservation of momentum
p1 A3 − p3 A3 = ρA3V3 (V3 − V1 )
Note that:
Conservation of energy
V 21 p3 V 2 3
+
=
+
+ hL
γ 2g γ
2g
p1
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2
K L = hL /(V1 / 2 g )
A1 2
K L = (1 − )
A2
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Example 6:
A soft drink with the properties of 10°C water is
sucked through a 4 mm diameter, 0.25 m long
straw at a rate of 4 cm3/s. Is the flow at the outlet
of the straw laminar? Is it fully developed?
Solved in the class room
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Example 7:
0.5 m
D=10 mm
Entrance K=0.5
Given Q=3.6 lt/min
L=2 m
a) Determine the state of the flow and velocity in the pipe.
b) Draw E.G.L. and H.G.L. and determine the kinematics viscosity.
Solved in the class room
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Exercise 1:
Water at 20°C flows in a 15 cm diameter pipe with a
flowrate of 60 lt/s.
a) Determine the centerline velocity. (Ans: 4.06 m/s)
b) What is the approximate velocity at a distance 5 cm
away from the wall? (Ans: 3.86 m/s)
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Exercise 2:
For a smooth pipe of diameter 75 mm, the head
loss for a distance of 150 m is 21 m. When the
flowrate is 8.5 lt/s. Is the flow laminar or
turbulent? (Ans: laminar)
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2. FLOW IN CLOSED CONDUITS
Fully Developed Flow in Closed Conduits
Exercise 3:
The pressure heads measured in a 2 cm diameter circular pipe are
p1/γ = 22 m and p2/γ = 21.5 m. The distance between two measuring
points is 1500 m. Taking γ=9810 N/m3, ρ= 1000 kg/m3, ν=1x10-6 m2/s
determine:
a) State of the flow
b) The equation of velocity and shear stress profiles
c) Maximum velocity
d) Velocity and shear stress at r=5 mm and y=4 mm.
e) Discharge
ANS: (a) laminar, (b) u=0.0816-816r2, τ=1.635r, (c)=0.0816 m/s,
(d) u(r=5mm)=0.0612 m/s, τ(r=5mm)=0.008175 N/m2
u(y=4mm)=0.0522 m/s, τ(y=4mm)=0.00981 N/m2,
(e) 1.282 x 10-5 m3/s
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