Chapter 2:
Analysis of Steady Flow in Pipelines
Learning outcomes
• To apply energy equation in pipes
• To analyze of flow and piping systems
including pipe in series and pipes in parallel by
applying energy equations
• To analyze pipe systems consist numerous
pipes connected in a complex manner with
general entry and withdrawal points by using
Quantity Balance Method
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Content
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Introduction
Incompressible flow through pipes
Flow through pipes in series
Flow through pipes in parallel
Flow through branching pipes
Pipe networks
Quantity Balance Method
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Introduction
• This chapter concerned with the analysis of flow in
pipes and piping systems including pipe in series and
pipes in parallel.
• Dealing with flow in circular pipes flowing full under
steady conditions and flowing under gravity.
• Analyses problems involving pipelines, e.g. flow
between reservoirs, water flow from a reservoir
discharging to atmosphere, from higher to lower
elevation reservoir or vice versa.
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Introduction
• Normally uses energy and continuity equation
to solve problems.
• Method of solving:
– Number of equations needed is the same as the
number of unknowns to be solved
– Successive corrections until the continuity
equation is satisfied (trial & error)
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Incompressible flow through pipes
• Incompressible : density is constant throughout
• For a circular cross-section conduit flowing
full:
– Head loss due to friction by Darcy Weisbach
(2.1)
4 fLv 2
hf 
2 gd
– Pressure loss by Darcy Weisbach
4 fLv 2
p 
2d
(2.2)
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Incompressible flow through pipes
• But
v
Q
d 2
4
• Therefore (2.1) & (2.2),
2
flQ
hf 
3.03d 5
3.24 flQ
p 
5
d
(2.3)
2
(2.4)
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Example 2.1(Douglas, 2006)
• Water discharges from reservoir A through a 100 mm
diameter pipe, 15 m long, which rises to the highest
point at B, 1.5 m above the free surface of the
reservoir and discharges direct to the atmosphere at
C, 4 m below the free surface at A. The length of pipe
L1 from A to B is 5 m and the length of pipe L2 from
B to C is 10 m. Both the entrance and exit of the pipe
are sharp and the value of f is 0.08. Calculate,
– The mean velocity of the water leaving the pipe at point C
– The pressure at point B
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B
2m
A
L1
L2
4m
C
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Flow through pipes in series
• Pipes in series are pipes of different diameters
are connected end-to-end to form pipeline to
form a pipeline.
• The total loss of energy, or pressure loss, over
the whole pipeline will be the sum of the
losses for each pipe together with any
separation losses that might occur at the
junctions.
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Example 2.2 (Douglas, 2006)
• Two reservoirs A and B have a difference level of 9 m
and are connected by a pipeline 200 mm in diameter
over the first part AC, which is 15 m long, and then
250 mm diameter for CB, the remaining 45 m length.
The entrance to and exit from the pipes are sharp and
the change of section at C is sudden. The friction
coefficient f is 0.01 for both pipes. List the losses of
head that occur and calculate the system flow rate.
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Apr 2010
• Two pipes are connected in series and is used to connect
two reservoirs as shown in Figure Q2(b). Both pipes are 500
m long and have diameters of 0.7 m and 0.5 m respectively.
The discharge from reservoir A to reservoir B is 0.8 m3/s
and the elevation of water surface in reservoir A is 85 m.
Using the appropriate equations and assuming sharp exit
and entry as well as existence of sudden contraction at the
junction, determine :
– velocity for each pipe
– head losses due to friction, separation and fittings
– elevation of water surface in reservoir B
(Cc = 0.674, μ = 1.14 x 10-3 Pa.s and k = 0.15 mm)
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Incompressible flow through pipes in parallel
• When two reservoirs are connected by two or
more pipes in parallel, the fluid can flow from one
to the other by a number of alternative routes.
• Difference in head h available to produce flow
will be the same for each pipe.
• Thus pipe can be considered separately,
independent of other parallel pipes.
• Steady flow energy equation can be applied for
flow by each route.
• The total volume rate of flow will be the sum of
volume rates of flow in each pipe.
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Example 2.3 (Douglas, 2006)
• Two sharp-ended pipes of diameter d1 = 50 mm,
and d2 = 100 mm, each of length L = 100 m, are
connected in parallel between two reservoirs
which have a difference of level h = 10 m, as in
figure. If the Darcy coefficient f = 0.008 for each
pipe, calculate,
a. the rate of flow for each pipe,
b. diameter D of a single pipe 100 m long that
would give the same flow if it was substituted for
the original two pipes.
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Incompressible flow through branching pipes
• Branching pipes - consist of one or more pipes
that separate into two or more pipes
• The basic principles that must be satisfied are:
• Continuity - at any junction the total mass flow
rate towards the junction must be equal the total
mass flow rate away from it
• There can be only one head value at any point
• The energy equation must be satisfied for each
pipe
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• Figure 2.4 shows a simple branching pipe
system where there are three tanks connected
by three pipes that join at D
• Actual direction of flow depends on
– Tank pressures and elevation
– Diameters, length and kinds of pipe
• The system must satisfy two equation i.e.
Darcy Weisbach and continuity equation
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• The inflow and outflow at the junction D must
be equal (continuity equation)
• The flow must be from the highest tank to the
lowest tank.
• Therefore, continuity equation can be
Q1  Q2  Q3 or Q1  Q2  Q3
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• If the HGL at junction D is above the elevation
of the intermediate tank B – the water is
flowing into tank B.
• If the HGL at junction D is below tank B – the
water is flowing out from tank B.
• General problem: to find the flow rate in each
pipe
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Example 2.4 (Douglas, 2006)
• Water flows from reservoir A through a pipe of
diameter d1 = 120 mm and length L1 = 120 m to
junction D, from which a pipe of diameter d2 = 75
mm and length L2 = 60 m leads to reservoir B in
which the water level is 16 m below that in reservoir
A. A third pipe of diameter d3 = 60 mm and length L3
= 40 m, leads from D to reservoir C, in which the
water level is 24 m below that in reservoir A. Taking f
= 0.01 for all the pipes and neglecting all losses other
than those due to friction, determine the discharge in
each pipe.
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Pipe networks
• A pipe network is a set
of pipes which are
interconnected so that
the flow from a given
input or to a given outlet
may come through
several different routes.
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• An alternative approach is used to analyse the
systems by means of successive approximations,
assuming values for the flow in each pipe, or the
heads at each junction, and checking whether the
values chosen satisfy the requirements:
– The loss of head between any two junctions must
be the same for all routes between these junctions
Loss of
head in
pipe bc
+
Loss of = Loss of
head in
head in
pipe de
pipe bd
+
Loss of
head in
pipe ce
– The inflow to each junction must equal the outflow
from that junction
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Quantity balance method
• Can be used when the heads at various points in a
pipe network are known.
• Estimate the head at each junction and Q is calculated
for each pipe from the difference of head h between
the junctions at each end of pipe and its resistance
coefficient K by,
2
h  KQ
• If the inflow does not equal to outflow for each
junction, the original estimated head must be
corrected.
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• If the head at b has been overestimated by δh
relative to the head at a, c, and d, the values of
Qab and Qcb will be too small and the value of
Qbd will be too great.
• Differentiating h  KQ 2
Q   Q nh h


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• Therefore, the estimated head at b is reduced
by δh, the flows in ab and cb will be increased
to Qab + δQab and Qcb + δQcb.
• If the flows are correct, inflow and outflow at
b will be balanced:
Flow in ab  Flow in cb  Flow in bd
Qab  Qab   Qcb  Qcb   Qbd  Qbd 
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• If hAD, hBD and hDC , are
the assumed losses in
pipe ab, cb and bd, used
to calculate Qab, Qcb and
Qbd,
Qab
Qab 
h
nhab
Qcb
Qcb 
h
nhcb
Qbd
Qbd 
h
nhbd
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• Substituting,
Flow in ab  Flow in cb  Flow in bd

 
 

Q
Q
Q
 Qab  ab h    Qcb  cb h    Qbd  bd h 
nhab  
nhcb  
nhbd 

Q  Qcd  Qbd 
h   ab
Qab Qcb Qbd


nhab nhcb nhbd
Q

h  
Q
 nh
• Sign convention: flow towards junction b both Q and
h are posititve
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Q  Q
AD
 QBD  QDC  Algebraicsum of theflows
towardsthe junction
• The process is repeated until ΣQ has been
reduced to negligible quantity (approaching
zero)
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Example 2.5 (Douglas, 2006)
• A reservoir A with its surface 60 m above the
datum supplies water to a junction D through a
300 mm diameter pipe, 1500 m long. From the
junction, a 250 mm diameter pipe, 800 m long,
feeds reservoir B, in which the surface level is
30 m above datum, while a 200 mm diameter
pipe, 400 m long, feeds reservoir C, in which
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Example 2.6 (Douglas, 2006)
• Water flows in the parallel pipe system as shown in
figure for which the following data are available:
Pipe
Diameter (m)
Length (m)
f
AaB
0.10
300
0.0060
AbB
0.15
250
0.0050
AcB
0.20
500
0.0050
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• The supply pipe to point A is of 0.30 m
diameter and the mean velocity of water in it is
3 m/s. If the elevation of point A is 100 m and
the elevation of point B is 30 m above datum,
calculate the pressure at point B if that at A is
200 kPa. What is the discharge in each pipe?
Neglect all minor losses.
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Review of past semesters’
questions
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Oct 2010
• Water flows from reservoir A through pipe 1 to
junction J which in turn flows to reservoir B
through pipe 2 and reservoir C through pipe 3
as shown in figure. Using analytical method
and neglecting all losses other than those due
to friction, determine the flow rate and velocity
for each pipe. Assume that the friction factor, f
is 0.01 for all pipes.
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