10.10-Pumps and Systems of Pipes, fluid mech
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Pumps and Systems of Pipes
10.10 Pumps and Systems of Pipes
Previous sections have presented information for modeling flow in a single round pipe. This section extends this
information by describing how to model flow in a network of pipes and how to incorporate performance data
for a centrifugal pump. These topics are important because pumps and pipe networks are common.
Modeling a Centrifugal Pump
As shown in Fig. 10.17, a
centrifugal pump
is a machine that uses a rotating set of blades situated within a
housing to add energy to a flowing fluid. The amount of energy that is added is represented by the head of the
pump
h
p
, and the rate at which work is done on the flowing fluid is
.
Figure 10.17
A centrifugal pump drives flow with a rotating impellor.
To model a pump in a system, engineers commonly use a graphical solution involving the energy equation given
in Eq. (7.29) and data from the pump manufacturer. To illustrate this approach, consider flow of water in the
system of Fig. 10.18
a
. The energy equation applied from the reservoir water surface to the outlet stream is:
For a system with one size of pipe, this simplifies to
(10.9)
Hence, for any given discharge, a certain head
h
p
must be supplied to maintain that flow. Thus, construct a
headversusdischarge curve, as shown in Fig. 10.18
b
. Such a curve is called the
system curve.
Now, a given
centrifugal pump has a headversusdischarge curve that is characteristic of that pump at a given pump speed.
This curve is called a
pump curve
. A pump curve can be acquired from a pump manufacturer, or it can be
measured. A typical pump curve is shown in Fig. 10.18
b
.
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Figure 10.18
(a) Pump and pipe combination.
(b) Pump and system curves.
Figure 10.18
b
reveals that, as the discharge increases in a pipe, the head required for flow also increases.
However, the head that is produced by the pump decreases as the discharge increases. Consequently, the two
curves intersect, and the operating point is at the point of intersection—that point where the head produced by
the pump is just the amount needed to overcome the head loss in the pipe.
To incorporate performance data for a pump, use the energy equation to derive a system curve. Then acquire a
pump curve from a manufacturer or other source and plot the two curves together. The point of intersection
shows where the pump will operate. This process is illustrated in Example 10.9.
Interactive Application: Pumping System Operating Point
EXAMPLE 10.9 FIDIG A SYSTEM OPERATIG POIT
Problem Definition
Situation:
1. A pump has the headversusdischarge curve shown in Fig. 10.18b.
2. The friction factor is
f
= 0.015.
Find:
Discharge (m
3
/s) in the system.
Plan
1. Develop an equation for the system curve by applying the energy equation.
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2. Plot the given pump curve and the system curve on the same graph.
3. Find discharge
Q
by finding the intersection of the system and pump curve.
Sketch:
Q
(m
3
/s)
h
p
= 30 m + 127
Q
2
m
0
30
0.1
31.3
0.2
35.1
0.3
41.4
Solution
Energy equation
Here
K
e
= 0.5,
K
b
= 0.35, and
K
E
= 1.0. Hence
Now make a table of
Q
versus
h
p
(see below) to give values to produce a system curve that will be
plotted with the pump curve. When the system curve is plotted on the same graph as the pump curve,
it is seen (Fig. 10.19
b
) that the operating condition occurs at
Pipes in Parallel
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Consider a pipe that branches into two parallel pipes and then rejoins, as shown in Fig. 10.19. A problem
involving this configuration might be to determine the division of flow in each pipe, given the total flow rate.
No matter which pipe is involved, the pressure difference between the two junction points is the same. Also, the
elevation difference between the two junction points is the same. Because
h
L
= (
p
1
/γ +
z
1
) (
p
2
/γ +
z
2
), it follows
that
h
L
between the two junction points is the same in both of the pipes of the parallel pipe system. Thus,
Then
If
f
1
and
f
2
are known, the division of flow can be easily determined. However, some trialanderror analysis
may be required if
f
1
and
f
2
are in the range where they are functions of the Reynolds number.
Pipe etworks
The most common pipe networks are the water distribution systems for municipalities. These systems have one
or more sources (discharges of water into the system) and numerous loads: one for each household and
commercial establishment. For purposes of simplification, the loads are usually lumped throughout the system.
Figure 10.20 shows a simplified distribution system with two sources and seven loads.
Figure 10.19
Flow in parallel pipes.
Figure 10.20
Pipe network.
The engineer is often engaged to design the original system or to recommend an economical expansion to the
network. An expansion may involve additional housing or commercial developments, or it may be to handle
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increased loads within the existing area.
In the design of such a system, the engineer will have to estimate the future loads for the system and will need to
have sources (wells or direct pumping from streams or lakes) to satisfy the loads. Also, the layout of the pipe
network must be made (usually parallel to streets), and pipe sizes will have to be determined. The object of the
design is to arrive at a network of pipes that will deliver the design flow at the design pressure for minimum
cost. The cost will include first costs (materials and construction) as well as maintenance and operating costs.
The design process usually involves a number of iterations on pipe sizes and layouts before the optimum design
(minimum cost) is achieved.
So far as the fluid mechanics of the problem are concerned, the engineer must determine pressures throughout
the network for various conditions—that is, for various combinations of pipe sizes, sources, and loads. The
solution of a problem for a given layout and a given set of sources and loads requires that two conditions be
satisfied:
1. The continuity equation must be satisfied. That is, the flow into a junction of the network must equal the
flow out of the junction. This must be satisfied for all junctions.
2. The head loss between any two junctions must be the same regardless of the path in the series of pipes
taken to get from one junction point to the other. This requirement results because pressure must be
continuous throughout the network (pressure cannot have two values at a given point). This condition
leads to the conclusion that the algebraic sum of head losses around a given loop must be equal to zero.
Here the sign (positive or negative) for the head loss in a given pipe is given by the sense of the flow with
respect to the loop, that is, whether the flow has a clockwise or counterclockwise direction.
Only a few years ago, these solutions were made by trialanderror hand computation, but modern applications
using digital computers have made the older methods obsolete. Even with these advances, however, the engineer
charged with the design or analysis of such a system must understand the basic fluid mechanics of the system to
be able to interpret the results properly and to make good engineering decisions based on the results. Therefore,
an understanding of the original method of solution by Hardy Cross 17 may help you to gain this basic insight.
The Hardy Cross method is as follows.
The engineer first distributes the flow throughout the network so that loads at various nodes are satisfied. In the
process of distributing the flow through the pipes of the network, the engineer must be certain that continuity is
satisfied at all junctions (flow into a junction equals flow out of the junction), thus satisfying requirement 1. The
first guess at the flow distribution obviously will not satisfy requirement 2 regarding head loss; therefore,
corrections are applied. For each loop of the network, a discharge correction is applied to yield a zero net head
loss around the loop. For example, consider the isolated loop in Fig. 10.21. In this loop, the loss of head in the
clockwise direction will be given by
(10.10)
The loss of head for the loop in the counterclockwise direction is
(10.11)
For a solution, the clockwise and counterclockwise head losses have to be equal, or
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