fluid mechanics for civil engineering, flow patterns in pipelines

FLUID MECHANICS FOR
CIVIL ENGINEERING
Chapter 4: Flow in Pipelines

SEQUENCE OF CHAPTER 4
Introduction
Objectives
4.1 Pipe Flow System
4.2 Types of Flow
4.2.1 Laminar Flow
4.2.2 Turbulent Flow
4.3 Energy Losses due to Friction
4.3.1 Friction Losses in Laminar Flow
4.3.2 Friction Loss in Turbulent Flow
4.4 Minor Losses
4.4.1 Losses due to Pipe Fittings
4.4.2 Sudden Enlargement
4.4.3 Sudden Contraction
4.5 Energy Added and Extracted
4.6 Pipe Flow Analysis
4.6.1 Simple Pipeline
4.6.2 Pipes in Series
4.6.3 Pipes in Parallel
4.6.4 Pipe Network
Summary

Introduction
 In considering the convenience and necessities in every day
life, it is truly amazing to note the role played by conduits in
transporting fluid.
 For example, the water in our homes is normally conveyed
through pressure pipelines, from the distribution system, so
that it will be available when and where we want it.
 Moreover, virtually all of this water leaves our homes as
dilute wastes through sewers, another type of conduits. Oil
is often transferred from their source by pressure pipelines
to refineries while gas is conveyed by pipelines into a
distribution network for supply.
 Thus, it can be seen that the fluid flow in conduits is of
immense practical significance in civil engineering.

Objectives
1. Differentiate between laminar and turbulent flows in
pipelines.
2. Describe the velocity profile for laminar and turbulent
flows.
3. Compute Reynolds number for flow in pipes.
4. Define the friction factor, and compute the friction losses in
pipelines.
5. Recognize the source of minor losses, and compute minor
losses in pipelines.
6. Analyze simple pipelines, pipelines in series, parallel, and
simple pipe networks.

4.1 Pipe Flow System
• This chapter introduces the fundamental theories of flow in
pipelines as well as basic design procedures.
• In this chapter, the pipeline system is defined as a closed conduit
with a circular cross-section with water flows (flowing full) inside
it.
• It is a closed system, the water is not in contact with air (i.e. no
free surface). Flow in a closed pipe results from a pressure
difference between inlet and outlet. The pressure is affected by
fluid properties and flow rate.
• The following diagram gives the geometrical properties for circular
pipe. In the diagram, D represents the diameter of pipe, R is the
pipe radius and L is the pipe length. The cross-sectional area of
the pipe can be calculated using A = πR2
.

 In hydraulic applications, energy values are often converted into
units of energy per unit weight of fluid, resulting in units of length.
 When using these length equivalents, the energy of the system is
expressed in terms of ‘head.’ The usual unit used is meter. The
energy at any point within a hydraulic system is often expressed in
three parts, i.e. the pressure head (P/ρg), elevation head (z), and
velocity head (V2
/2g) The sum of all these components is known as
the total head (H).
(4.1)
RR
D=2R
L
R
Pipe
Centerline
Figure 4.2: Schematic diagram of a circular pipe
g2
V
z
g
P
H
2
++
ρ
=

Figure 4.3:
Flow in a
uniform pipeline
 A line plotted of total head versus distance through a system is called the
total energy line (TEL).
 The TEL is also known as energy grade line (EGL).
 The sum of the elevation head and pressure head yields the hydraulic
grade line (HGL).
 In a uniform pipeline, the total shear stress (resistance to flow) is constant
along the pipe resulting in a uniform degradation of the total energy or
head along the pipeline.
 The total head loss along a specified length of pipeline is referred to head
loss due to friction and denoted as hf.

 Referring to the above figure, the Bernoulli equation
can be written from section 1 to section 2 as;
(4.2)
f
2
2
2
2
2
1
1
1 h
g2
V
z
g
P
g2
V
z
g
P
+
α
++
ρ
=
α
++
ρ

4.2 Types of Flow
• The physical nature of fluid flow can be categorized into three types, i.e.
laminar, transition and turbulent flow. It has been mentioned earlier that
Reynolds Number (Re) can be used to characterize these flow.
(4.3)
where ρ = density
µ = dynamic viscosity
ν = kinematic viscosity (ν = µ/ρ)
V = mean velocity
D = pipe diameter
In general, flow in commercial pipes have been found to conform to the
following condition:
Laminar Flow: Re <2000
Transitional Flow : 2000 < Re <4000
Turbulent Flow : Re >4000
ν
=
µ
ρ
=
VDVD
Re

4.2.1 Laminar flow
 Viscous shears dominate in this type of flow and the
fluid appears to be moving in discreet layers. The
shear stress is governed by Newton’s law of viscosity
(equation 1.1):
(1.1)
 In general the shear stress is almost impossible to
measure. But for laminar flow it is possible to
calculate the theoretical value for a given velocity,
fluid and the appropriate geometrical shape.
dy
du
µ=τ

Pressure loss during a laminar flow in a pipe
 Up to this point we only considered ideal fluid
where there is no losses due to friction or any
other factors.
 In reality, because fluids are viscous, energy is lost
by flowing fluids due to friction which must be
taken into account.
 The effect of friction shows itself as a pressure (or
head) loss. In a pipe with a real fluid flowing, the
shear stress at the wall retard the flow.
 The shear stress will vary with velocity of flow and
hence with Re. Many experiments have been done
with various fluids measuring the pressure loss at
various Reynolds numbers.
 Figure below shows a typical velocity distribution
in a pipe flow. It can be seen the velocity
increases from zero at the wall to a maximum in
the mainstream of the flow.
Figure 4.4:
A typical velocity
distribution in a
pipe flow

 In laminar flow the paths of individual particles of fluid do
not cross, so the flow may be considered as a series of
concentric cylinders sliding over each other – rather like the
cylinders of a collapsible pocket telescope.
 Lets consider a cylinder of fluid with a length L, radius r,
flowing steadily in the center of pipe.
Figure 4.5: Cylindrical of fluid flowing steadily in a pipe

 The fluid is in equilibrium, shearing forces equal the pressure forces.
 Shearing force = Pressure force
(4.5)
Taking the direction of measurement r (measured from the center of
pipe), rather than the use of y (measured from the pipe wall), the
above equation can be written as;
(4.6)
Equating (4.5) with (6.6) will give:
2
r
L
P
rPPArL2 2
∆
=τ
π∆=∆=πτ
dr
dur
µ−=τ
µ
∆
−=
µ−=
∆
2
r
L
P
dr
du
dr
du
2
r
L
P

 In an integral form this gives an expression for velocity, with the
values of r = 0 (at the pipe center) to r = R (at the pipe wall)
(4.7)
where ∆P = change in pressure
L = length of pipe
R = pipe radius
r = distance measured from the center of pipe
The maximum velocity is at the center of the pipe, i.e. when r = 0.
It can be shown that the mean velocity is half the maximum velocity, i.e. V=umax/2
∫
µ
∆
−= =
=
Rr
0r rdr
2
1
L
P
u
( )
L
P
4
rR
u
22
r
∆
µ
−
−=
L
P
4
R
u
2
max
∆
µ
−=

Figure 6.6: Shear stress and velocity distribution in pipe for laminar flow
The discharge may be found using the Hagen-Poiseuille equation,
which is given by the following;
(4.8)
The Hagen-Poiseuille expresses the discharge Q in terms of the
pressure gradient , diameter of pipe, and viscosity of the
fluid.
Pressure drop throughout the length of pipe can then be calculated
by
(4.9)
µ
π∆
=
128
D
L
P
Q
4





 ∆
=
L
P
dx
dP
4
R
LQ8
P
π
µ
=∆

4.2.2 Turbulent flow
 This is the most commonly occurring flow in engineering
practice in which fluid particles move erratically causing
instantaneous fluctuations in the velocity components.
 These fluctuations cause additional shear stresses. In this
type of flow both viscous and turbulent shear stresses exists.
 Thus, the shear stress in turbulent flow is a combination of
laminar and turbulent shear stresses, and can be written as:
where µ = dynamic viscosity
ε = eddy viscosity which is not a fluid property but
depends upon turbulence condition of flow.
( )
dy
dU
turbulentarminla ε+µ=τ+τ=τ

 The velocity at any point in the cross-section will be proportional
to the one-seventh power of the distance from the wall, which
can be expressed as:
(4.10)
where Uy is the velocity at a distance y from the wall, UCL is the velocity at
the centerline of pipe, and R is the radius of pipe. This equation is known
as the Prandtl one-seventh law.
Figure 6.7 below shows the velocity profile for turbulent flow in a pipe.
The shape of the profile is said to be logarithmic.
7/1
CL
y
R
y
U
U






=
Figure 4.7: Velocity profile for turbulent flow

For smooth pipe:
(4.11a)
 For rough pipe:
(4.11b)
In the above equations, U represents the velocity at a distance y
from the pipe wall, U* is the shear velocity =
y is the distance form the pipe wall, k is the surface roughness and
ν is the kinematic viscosity of the fluid.
5.5
yU
log75.5
U
U *
10
*
+





ν
=
5.8
k
y
log75.5
U
U
10
*
+





=
ρτ

Example 4.1
Glycerin (ρ = 1258 kg/m3, µ=9.60 x10-1
N.s/m2
) flows with a
velocity of 3.6 m/s in a 150-mm diameter pipe. Determine
whether the flow is laminar or turbulent.
Solution:
Since Re=708, which is less than 2000, the flow is laminar.
µ
ρ
=
VD
Re
708
10x60.9
15.0x6.3x1258
Re
1
==
−

4.3 Energy Losses due to friction
 When a liquid flows through a pipeline, shear stresses
develop between the liquid and the pipe wall.
 This shear stress is a result of friction, and its magnitude is
dependent upon the properties of the fluid, the speed at
which it is moving, the internal roughness of the pipe, the
length and diameter of pipe.
 Friction loss, also known as major loss, is a primary cause
of energy loss in a pipeline system.

4.3.1 Friction Losses in Laminar flow
 In laminar flow, the fluid seems to flow as several layers, one on another.
Because of the viscosity of the fluid, a shear stress is created between
the layers of fluid.
 Energy is lost from the fluid to overcome the frictional forces produced by
the shear stress.
 Energy loss is usually represented by the drop of pressure in the direction
of flow.
 Therefore, the frictional head loss, hf, can be written in terms of pressure
drop along the pipeline, as follows:
(4.12)
Substituting the Hagen-Poiseuille equation and applying the continuity
equation, Q = VA, to the above resulted into the following expression:
(4.13)
g
P
hf
ρ
∆
=
g2
V
D
L
Re
64
gD
LV32
h
2
2f =
ρ
µ
=

4.3.2 Friction Losses in Turbulent flow
 In turbulent flow, the friction head loss can be calculated by considering
the pressure losses along the pipelines.
 In a horizontal pipe of diameter D carrying a steady flow there will be a
pressure drop in a length L of the pipe.
 Equating the frictional resistance to the difference in pressure forces, and
manipulating resulted into the following expression:
(4.14)
This equation is known as Darcy-Weisbach (D-W) equation, in which λ is
the friction factor. It should be noted that λ is dimensionless, and the
value is not constant
g2
V
D
L
h
2
f λ=

 Blasius (1913) was the first to propose an accurate empirical
relation for the friction factor in turbulent flow in smooth
pipes, namely
(4.15)
Thus, the calculation of losses in turbulent pipe flow is
dependent on the use of empirical results and the most
common reference source is the Moody diagram. A Moody
diagram is a logarithmic plot of λ vs. Re for a range of k/D.
A typical Moody diagram is shown in Figure 4.9.
λ = 0.316/Re0.25

Table 4.1: Typical pipe roughness
(Reference: White, 1999)
Material Roughness, k (mm)
Glass smooth
Brass, new 0.002
Concrete
Smoothed
Rough
0.04
2.0
Iron
Cast, new
Galvanised, new
Wrought, new
0.26
0.15
0.046
Steel
Commercial, new
Riveted
0.046
3

4.4 Minor Losses
• In addition to head loss due to friction, there are always
other head losses due to pipe expansions and contractions,
bends, valves, and other pipe fittings. These losses are
usually known as minor losses (hLm).
• In case of a long pipeline, the minor losses maybe negligible
compared to the friction losses, however, in the case of short
pipelines, their contribution may be significant.

6.4.1 Losses due to pipe fittings
where hLm= minor loss
K = minor loss coefficient
V = mean flow velocity
g2
V
Kh
2
Lm = (6.16)
Table 4.2: Typical K values
Type K
Exit (pipe to tank) 1.0
Entrance (tank to pipe) 0.5
90° elbow 0.9
45° elbow 0.4
T-junction 1.8
Gate valve 0.25 - 25

4.4.2 Sudden Enlargement
 As fluid flows from a smaller pipe into a larger pipe through
sudden enlargement, its velocity abruptly decreases; causing
turbulence that generates an energy loss.
 The amount of turbulence, and therefore the amount of energy, is
dependent on the ratio of the sizes of the two pipes.
 The minor loss (hLm)is calculated from;
(4.16a)
where is KE is the coefficient of expansion, and the values depends on the
ratio of the pipe diameters (Da/Db) as shown below.
g2
V
Kh
2
a
ELm =
Da
/Db 0.0 0.2 0.4 0.6 0.8
K 1.00 0.87 0.70 0.41 0.15
Table 4.3: Values of KE vs. Da/Db

Figure 4.9: Flow at Sudden Enlargement

4.4.3 Sudden Contraction
The energy loss due to a sudden contraction can be calculated using
the following;
(4.16b)
The KC is the coefficient of contraction and the values depends on
the ratio of the pipe diameter (Db/Da) as shown below.
g2
V
Kh
2
b
CLm =
Db
/Da 0.0 0.2 0.4 0.6 0.8 1.0
K 0.5 0.49 0.42 0.27 0.20 0.0
Table 4.4: Values of KC vs. Db/Da
Figure 4.10: Flow at sudden contraction

Example 4.2
 Water at 10°C is flowing at a rate of 0.03 m3/s through a pipe. The pipe
has 150-mm diameter, 500 m long, and the surface roughness is estimated
at 0.06 mm. Find the head loss and the pressure drop throughout the
length of the pipe.
Solution:
 From Table 1.3 (for water): ρ = 1000 kg/m3
and µ =1.30x10-3
N.s/m2
V = Q/A and A=πR2
A = π(0.15/2)2
= 0.01767 m2
V = Q/A =0.03/.0.01767 =1.7 m/s
Re = (1000x1.7x0.15)/(1.30x10-3
) = 1.96x105 > 2000 → turbulent flow
To find λ, use Moody Diagram with Re and relative roughness (k/D).
k/D = 0.06x10-3
/0.15 = 4x10-4
From Moody diagram, λ ≈ 0.018
The head loss may be computed using the Darcy-Weisbach equation.
The pressure drop along the pipe can be calculated using the relationship:
ΔP=ρghf = 1000 x 9.81 x 8.84
ΔP = 8.67 x 104 Pa
.m84.8
81.9x2x15.0
7.1x500
x018.0
g2
V
D
L
h
22
f ==λ=

Example 4.3
 Determine the energy loss that will occur as 0.06 m3
/s water flows
from a 40-mm pipe diameter into a 100-mm pipe diameter through
a sudden expansion.
Solution:
 The head loss through a sudden enlargement is given by;
Da/Db = 40/100 = 0.4
From Table 6.3: K = 0.70
Thus, the head loss is
g2
V
Kh
2
a
m =
s/m58.3
)2/04.0(
0045.0
A
Q
V
2
a
a =
π
==
m47.0
81.9x2
58.3
x70.0h
2
Lm ==

4.5 Energy added and extracted
• There are many occasions when energy needs to be added to
a hydraulic system to overcome elevation differences,
friction losses and minor losses.
• A pump is a common device to which mechanical energy is
applied and transferred to the water as total head of the
pump.
• The head added is called pump head (Hp), and is a function
of flow rate through the pump.
• On the other hand, fluid motor or turbines are common
examples of devices that extract energy from a fluid, and
the head extracted is called head of turbine (Ht), deliver it
in a form of work.

 Denoting the head loss due to friction and minor losses as HL, and
the external energy added/extracted by HE, then the Bernoulli
equation may be rewritten as
(4.17)
HE = Hp (positive for pump) when the head is added to the fluid, or
HE = Ht (negative for turbine) when the head is extracted from the
fluid. Note the similarity of this equation with equation (3.12). It
is often necessary to convert the total power (P) of a pump or
turbine to HE or vice versa. Recall from Chapter 3, the
relationship between P and HE is given by the following
P = ρgQHE (3.8)
In a pump HE = HP, the value is positive since power is added to the
fluid. In a turbine, HE = Ht is negative and power is extracted
from the flow.
21L
2
2
2
2
E
2
1
1
1
H
g2
V
z
g
P
H
g2
V
z
g
P
−+++
ρ
=±++
ρ

 The term efficiency is used to denote the ratio of the power delivered by
the pump to the fluid to the power supplied to the pump.
 Because of energy loses due to mechanical friction in pump components,
fluid friction in the pump, and excessive fluid turbulence in the pump,
not all of the input power is delivered to the fluid. Therefore, the
efficiency of a pump can be written as;
 The value of ηp is less than 1.0.
 Similarly, energy losses in a turbine are produced by mechanical and fluid
friction. Therefore, not all the power delivered to the motor is
ultimately converted to power output from the device. The efficiency of
a turbine is defined as;
 Here again, the value of ηt is less than 1.0.
in
p
in
p
P
gQH
P
P
pumptopliedsupPower
fluidtodeliveredPower ρ
===η
t
o
t
o
t
gQH
P
P
P
fluidbypliedsupPower
motorfromoutputPower
ρ
===η

Example 4.4
 Calculate the head added by the pump when the water system
shown below carries a discharge of 0.27 m3
/s. If the efficiency
of the pump is 80%, calculate the power input required by the
pump to maintain the flow.

Solution:
Applying Bernoulli equation between section 1 and 2
(1)
P1 = P2 = Patm = 0 (atm) and V1=V2≈ 0
Thus equation (1) reduces to:
(2)
∑HL1-2 = hf + hentrance + hbend + hexit
∑+++
ρ
=+++
ρ
−21L
2
2
2
2
p
2
1
1
1 H
g2
V
z
g
P
H
g2
V
z
g
P
∑+−= −21L12p HzzH
g2
V
4.39
14.05.0
4.0
1000
x015.0
g2
V
H
2
2
21L
=






+++=∑ −
81.9x2
V
4.39200230H
2
p +−=

The velocity can be calculated using the continuity
equation:
Thus, the head added by the pump: Hp = 39.3 m
Pin = 130 117 Watt ≈ 130 kW.
( )
s/m15.2
2/4.0
27.0
A
Q
V
2
=
π
==
in
p
p
P
gQHρ
==η
8.0
3.39x27.0x81.9x1000gQH
P
p
p
in =
η
ρ
=

4.6 Pipe Flow Analysis
• Pipeline system used in water distribution, industrial
application and in many engineering systems may
range from simple arrangement to extremely complex
one.
• Problems regarding pipelines are usually tackled by the
use of continuity and energy equations.
• The head loss due to friction is usually calculated using
the D-W equation while the minor losses are computed
using equations 4.16, 4.16(a) and 4.16(b) depending on
the appropriate conditions.

4.6.2 Pipes in Series
 When two or more pipes of
different diameters or
roughness are connected in
such a way that the fluid
follows a single flow path
throughout the system, the
system represents a series
pipeline.
 In a series pipeline the total
energy loss is the sum of the
individual minor losses and
all pipe friction losses.
Figure 6.11: Pipelines in series

 Referring to Figure 4.11, the Bernoulli equation can be written
between points 1 and 2 as follows;
(4.18)
where P/ρg = pressure head
z = elevation head
V2
/2g = velocity head
HL1-2 = total energy lost between point 1 and 2
Realizing that P1=P2=Patm, and V1=V2, then equation (4.14) reduces to
z1-z2 = HL1-2
Or we can say that the different of reservoir water level is equivalent to the
total head losses in the system.
The total head losses are a combination of the all the friction losses and the
sum of the individual minor losses.
HL1-2 = hfa + hfb + hentrance + hvalve + hexpansion + hexit.
Since the same discharge passes through all the pipes, the continuity
equation can be written as;
Q1 = Q2
21L
2
2
2
2
2
1
1
1
H
g2
V
z
g
P
g2
V
z
g
P
−+++
ρ
=++
ρ

4.6.3 Pipes in Parallel
• A combination of two or
more pipes connected
between two points so
that the discharge
divides at the first
junction and rejoins at
the next is known as
pipes in parallel. Here
the head loss between
the two junctions is the
same for all pipes.
Figure 4.12:
Pipelines in parallel

 Applying the continuity equation to the system;
Q1 = Qa + Qb = Q2 (4.19)
 The energy equation between point 1 and 2 can be written
as;
 The head losses throughout the system are given by;
HL1-2=hLa = hLb (4.20)
 Equations (4.19) and (4.20) are the governing relationships
for parallel pipe line systems. The system automatically
adjusts the flow in each branch until the total system flow
satisfies these equations.
L
2
2
2
2
2
1
1
1
H
g2
V
z
g
P
g2
V
z
g
P
+++
ρ
=++
ρ

4.6.4 Pipe Network
 A water distribution system consists of complex interconnected pipes, service
reservoirs and/or pumps, which deliver water from the treatment plant to the
consumer.
 Water demand is highly variable, whereas supply is normally constant. Thus,
the distribution system must include storage elements, and must be capable
of flexible operation.
 Pipe network analysis involves the determination of the pipe flow rates and
pressure heads at the outflows points of the network. The flow rate and
pressure heads must satisfy the continuity and energy equations.
 The earliest systematic method of network analysis (Hardy-Cross Method) is
known as the head balance or closed loop method. This method is applicable
to system in which pipes form closed loops. The outflows from the system are
generally assumed to occur at the nodes junction.
 For a given pipe system with known outflows, the Hardy-Cross method is an
iterative procedure based on initially iterated flows in the pipes. At each
junction these flows must satisfy the continuity criterion, i.e. the algebraic
sum of the flow rates in the pipe meeting at a junction, together with any
external flows is zero.

• Assigning clockwise flows and their associated head losses are
positive, the procedure is as follows:
 Assume values of Q to satisfy ∑Q = 0.
 Calculate HL from Q using HL = K1
Q2
.
 If ∑HL = 0, then the solution is correct.
 If ∑HL ≠ 0, then apply a correction factor, ∆Q, to all Q and repeat from
step (2).
 For practical purposes, the calculation is usually terminated when ∑HL
< 0.01 m or ∆Q < 1 L/s.
 A reasonably efficient value of ∆Q for rapid convergence is given by;
(4.21)∑
∑−=∆
Q
H
2
H
Q
L
L

Example 4.5
• A pipe 6-cm in diameter, 1000m long and with λ = 0.018 is
connected in parallel between two points M and N with
another pipe 8-cm in diameter, 800-m long and having λ =
0.020. A total discharge of 20 L/s enters the parallel pipe
through division at A and rejoins at B. Estimate the
discharge in each of the pipe.

Solution:
Continuity: Q = Q1 + Q2
(1)
Pipes in parallel: hf1 = hf2
Substitute (2) into (1)
0.8165V2 + 1.778 V2 = 7.074
2
2
22
1
2
V)08.0(
4
V)06.0(
4
02.0
π
+
π
=
074.7V778.1V 21 =+
)2(V8165.0V
V
08.0
800x020.0
V
06.0
1000x018.0
gD2
VL
gD2
VL
21
2
2
2
1
2
2
22
2
1
2
11
1
=
=
λ=λ

Q2 = 0.0137 m3/s
From (2):
V1 = 0.8165 V2 = 0.8165x2.73 = 2.23 m/s
Q1 = 0.0063 m3/s
Recheck the answer:
Q1+ Q2 = Q
0.0063 + 0.0137 = 0.020
(same as given Q ⇒ OK!)
73.2x)08.0(
4
VAQ 2
222
π
==

Example 4.6
• For the square loop shown, find the discharge in all the
pipes. All pipes are 1 km long and 300 mm in diameter, with
a friction factor of 0.0163. Assume that minor losses can be
neglected.

Solution:
 Assume values of Q to satisfy continuity equations all at nodes.
 The head loss is calculated using; HL = K1
Q2
 HL = hf + hLm
 But minor losses can be neglected: ⇒ hLm = 0
 Thus HL = hf
 Head loss can be calculated using the Darcy-Weisbach equation
g2
V
D
L
h
2
f λ=

First trial
Since ∑HL
> 0.01 m, then correction has to be applied.
554'
'
554
3.0
4
77.277.2
81.923.0
1000
0163.0
2
2
2
2
2
2
2
2
2
2
=∴
=
=






==
=
==
K
QKH
QH
x
Q
x
A
Q
H
x
V
xxH
g
V
D
L
hH
L
L
L
L
fL
π
λ
Pipe Q (L/s) HL
(m) HL
/Q
AB 60 2.0 0.033
BC 40 0.886 0.0222
CD 0 0 0
AD -40 -0.886 0.0222
Σ 2.00 0.0774

Second trial
Since ∑HL
≈ 0.01 m, then it is OK.
Thus, the discharge in each pipe is as follows (to the nearest integer).
s/L92.12
0774.0x2
2
Q
H2
H
Q
L
L −=−=
∑
∑
−=∆
Pipe Q (L/s) HL
(m) HL
/Q
AB 47.08 1.23 0.0261
BC 27.08 0.407 0.015
CD -12.92 -0.092 0.007
AD -52.92 -1.555 0.0294
Σ -0.0107 0.07775
Pipe Discharge (L/s)
AB 47
BC 27
CD -13
AD -53

Summary
 Overall, chapter 4 has introduced basic design
procedures of a flow in pipelines and more elaboration
on the type of flows.
 A detail calculation of pipe flow analysis which consists
of simple pipeline, pipe in series, pipe in parallel and
pipe network were discussed.
 The pipe network used the Hardy Cross method and at
the end, the discharge of water of each pipe can be
known.

fluid mechanics for civil engineering, flow patterns in pipelines

Recommended

Recommended

More Related Content

More from Usman Shah

More from Usman Shah (20)

Recently uploaded

Recently uploaded (20)

fluid mechanics for civil engineering, flow patterns in pipelines