Chapter 11
STEADY CLOSED-CONDUIT
FLOW
The basic procedures for solving problems in incompressible
steady flow in closed conduits are presented before (in Sec. 5.8 and
5.9), where simple pipe-flow situations are discussed, including
losses due to change in cross section or direction of flow.
Exponential friction formulas commonly used in commercial and
industrial applications are discussed in this chapter. The use of the
hydraulic and energy grade lines in solving problems is reiterated
before particular applications are developed.
Complex flow problems are investigated, including hydraulic
systems that incorporate various different elements such as pumps
and piping networks.
The use of the digital computer in analysis and design becomes
particularly relevant when multielement systems are being
investigated. The hand-held electronic programmable calculator is
effective for iterative solutions such as the small-network
problems.
11.1 EXPONENTIAL PIPE-FRICTION
FORMULAS
Industrial pipe-friction formulas are usually empirical, of the
form
(11.1.1)
in which hf/L is the head loss per unit length of pipe (slope of the
energy grade line), Q the discharge, and D the inside pipe
diameter.
The resistance coefficient R is a function of pipe roughness only.
An equation with specified exponents and coefficient R is valid
only for the fluid viscosity for which it is developed, and it is
normally limited to a range of Reynolds numbers and diameters.
In its range of applicability such an equation is convenient, and
nomographs are often used to aid problem solution.
The Hazen-Williams formula for flow of water at ordinary
temperatures through pipes is of this form with R given by
R=
4.727
Cn
USC unit
(11.1.2)
10.675
Cn
SI unit
(11.1.3)
with n = 1.852, m = 4.8704, and C dependent upon roughness as
follows:
One can develop a special-purpose formula for a particular
application by using the Darcy-Weisbach equation and friction
factors from the Moody diagram or, alternatively, by using
experimental data if available.
Exponential formulas developed from experimental results are
generally very useful and handy in the region over which the
data were gathered. Extrapolations and applications to other
situations must be carried out with caution.
Fig. 11.1 presents a comparison between the Hazen-Williams
equation and the Darcy-Weisbach equation with friction
factors from the Moody diagram. It shows equivalent values of
f vs. Reynolds number for three typical Hazen-Williams
roughness values 70, 100, and 140. The fluid is water at 15°C.
By equating the slope of the hydraulic grade line in the DarcyWeisbach equation, hf/L = fQ2/2gDA2, to Eq. (11.1.1), solving
for f, and introducing the Reynolds number to eliminate Q,
(11.1.4)
For a given Hazen-Williams coefficient C and diameter D the
friction factor reduces with increasing Reynolds number. A similar
solution for f in terms of C, Reynolds number, and V can be
developed by combining the same equations and eliminating D,
(11.1.5)
It may be noted that f is not strongly dependent upon pipe
diameter in Eq. (11.1.4).
In Fig. 11.1, at the three selected values of C, Eq. (11.1.4) is shown
for a particular diameter of 1 m and Eq. (11.1.5) is shown for a
specific velocity of 1 m/s. The shaded region around each of these
lines shows the range of practical variation of the variables (0.025
< D < 6 m, 0.030 m/s < V < 30 m/s).
The two formulations Darcy-Weisbach vs. Hazen-Williams for
calculation of losses in a pipeline can be seen to be significantly
different.
The Darcy-Weisbach equation is probably more rationally based
than other empirical exponential formulations and has received
wide acceptance.
Figure 11.1 Comparison of Hazen-Williams and
Darcy-Weisbach equations on the Moody diagram
11.2 HYDRAULIC AND ENERGY GRADE LINES
The concepts of hydraulic and energy grade lines are useful in
analyzing more complex flow problems. If, at each point along a
pipe system, the term p/γ is determined and plotted as a vertical
distance above the center of the pipe, the locus of points is the
hydraulic grade line. More generally, the plot of the two terms
as ordinates, against length along the pipe, as abscissas, produces
the hydraulic grade line.
The energy grade line is a line joining a series of points marking
the available energy in meter-newtons per newton for each point
along the pipe as ordinate, plotted against distance along the pipe
as the abscissa. It consists of the plot of
for each point along the line.
The hydraulic and energy grade lines are shown in Fig. 11.2 for a
simple pipeline containing a square-edged entrance, a valve, and a
nozzle at the end of the line.
To construct these lines when the reservoir surface is given, it is
necessary first to apply the energy equation from the reservoir to
the exit, including all minor losses as well as pipe friction, and to
solve for the velocity head V2/2g.
Then, to find the elevation of hydraulic grade line at any point, the
energy equation is applied from the reservoir to that point,
including all losses between the two points. The equation is solved
for p/γ + z, which is plotted above the arbitrary datum.
To find the energy grade line at the same point, the equation is
solved for V2/2g + p/γ + z, which is plotted above the arbitrary
datum.
The reservoir surface is the hydraulic grade line and is also the
energy grade line. At the square-edged entrance the energy grade
line drops by 0.5V2/2g because or the loss there, and the hydraulic
grade line drops 1.5V2/2g.
Figure 11.2 Hydraulic and energy grade lines
This is made obvious by applying the energy equation between
the reservoir surface and a point just downstream from the
pipe entrance:
Solving for z + p/γ,
shows the drop of 1.5V2/2g.
The head loss due to the sudden entrance does not actually
occur at the entrance itself, but over a distance of 10 or more
diameters of pipe downstream. It is customary to show it at the
fitting.
Example 11.1
Determine the elevation of hydraulic and energy grade lines at
points A, B, C, D, and E of Fig. 11.2. z = 3 m.
Solution
Solving for the velocity head is accomplished by applying the
energy equation from the reservoir to E,
Form the continuity equation, VE = 4V. After simplifying,
and V2/2g = 0.554 m. Applying the energy equation for the portion
from the reservoir to A gives
Hence, the hydraulic grade line at A is
The energy grade line for A is
For B,
and
The energy grade line is at 20.40 + 0.55 = 20.95 m.
Across the valve the hydraulic grade line drops by 10V2/2g, or
5.54 m.
Hence, at C the energy and hydraulic grade lines are at 15.41
m and 14.86 m respectively.
At point D,
and
with the energy grade line at 12.20 + 0.55 = 12.75 m.
At point E the hydraulic grade line is 3 m, and the energy
grade line is
The hydraulic gradient is the slope of the hydraulic grade line if
the conduit is horizontal; otherwise, it is
The energy gradient is the slope or the energy grade line if the
conduit is horizontal; otherwise, it is
In many situations involving long pipelines the minor losses
may be neglected (when less than 5 percent of the pipe friction
losses), or they may be included as equivalent lengths of pipe
which are added to actual length in solving the problem.
For these situations the value of the velocity head V2/2g is small
compared with f(L/D)V2/2g and is neglected.
In this special but very common case, when minor effects are
neglected, the energy and hydraulic grade lines are superposed.
The single grade line, shown in Fig. 11.3, is commonly referred to
as the hydraulic grade line.
For these situations with long pipelines the hydraulic gradient
becomes hf/L, with hf given by the Darcy-Weisbach equation
(11.2.1)
or by Eq. (11.1.1). Flow (except through a pump) is always in the
direction of decreasing energy grade line.
Pumps add energy to the flow, a fact which may be expressed in
the energy equation either by including a negative loss or by
stating the energy per unit weight added as a positive term on the
upstream side of the equation.
Figure 11.4 shows the hydraulic and energy grade lines for a
system with a pump and a siphon. The true slope of the grade lines
can be shown only for horizontal lines.
Figure 11.3 Hydraulic grade line for long pipeline where
minor losses are neglected or included as equivalent
lengths of pipe.
Figure 11.4 Hydraulic and energy grade
lines for a system with pump and siphon
Example 11.2
A pump with a shaft input of 7.5 kW and an efficiency of 70
percent is connected in a water line carrying 0.1 m3/s. The
pump has a 150-mm-diameter suction line and 120-mmdiameter discharge line. The suction line enters the pump 1 m
below the discharge flange and the rise in the hydraulic grade
line across the pump.
Solution
If the energy added in meter-newtons per newton is symbolized
by E, the fluid power added is
Applying the energy equation from suction flange to discharge
flange gives
in which the subscripts s and d refer to the suction and
discharge conditions, respectively.
From the continuity equation
Solving for pd gives
and pd = 89.6 kN/m2. The rise in hydraulic grade line is
In this example much of energy was added in the form of
kinetic energy, and the hydraulic grade line rises only 3.002 m
for a rise of energy grade line of 5.354 m.
11.3 THE SIPHON
A closed conduit, arranged as in Fig. 11.5, which lifts the liquid
to an elevation higher than its free surface and then discharges it
at a lower elevation is a siphon. It has certain limitations in its
performance due to the low pressures that occur near the
summit s.
Assuming that the siphon flows full, with a continuous liquid
column throughout it, the application of the energy equation for
the portion from 1 to 2 produces the equation
in which K is the sum of all the minor-loss coefficients. Factoring
out the velocity head gives
(11.3.1)
which is solved in the same fashion as the simple pipe problems
of the first or second type.
Figure 11.5 Siphon
The pressure at the summit s is found by applying the energy
equation for the portion between 1 and s after. Eq. (10.3.1) is
solved. It is
in which K‘ is the sum of the minor-loss coefficients between
the two points and L’ is the length of conduit upstream from s.
Solving for the pressure gives
(11.3.2)
which shows that the pressure is negative and that it decreases
with ys, and V2/2g.
If the solution of the equation should be a value of ps/γ equal to
or less than the vapor pressure of the liquid, then Eq. (11.3.1) is
not valid because the vaporization of portions of the fluid
column invalidates the incompressibility assumption used in
deriving the energy equation.
Although Eq. (11.3.1) is not valid for this case, theoretically there
will be a discharge so long as y, plus the vapor pressure is less than
local atmospheric pressure expressed in length of the fluid column.
When Eq. (11.3.2) yields a pressure less than vapor pressure at s,
the pressure at s may be taken as vapor pressure. Then, with this
pressure known, Eq. (11.3.2) is solved for V2/2g, and the discharge
is obtained therefrom. It is assumed that air does not enter the
siphon at 2 and break at s the vacuum that produces the flow.
Practically, a siphon does not work satisfactorily when the
pressure intensity at the summit is close to vapor pressure. Air and
other gases come out of solution at the low pressures and collect at
the summit, thus reducing the length of the right-hand column of
liquid that produces the low pressure at the summit. Large
siphons that operate continuously have vacuum pumps to remove
the gases at the summits.
The lowest pressure may not occur at the summit but somewhere
downstream from that point, because friction and minor losses
may reduce the pressure more than the decrease in elevation
increases pressure.
Example 11.3
Neglecting minor losses and considering the length of pipe equal to
its horizontal distance, determine the point of minimum pressure
in the siphon of Fig. 11.6
Solution
When minor losses are neglected, the kinetic-energy term V2/2g is
usually neglected also. Then the hydraulic grade line is a straight
line connecting the two liquid surfaces.
Coordinates of two points on the line are
Figure 11.6
Siphon connecting
two reservoirs
The equation of the line is, by substitution into y = mx + b,
The minimum pressure occurs where the distance between
hydraulic grate line and pipe is a maximum,
To find minimum p/γ, set d(p/γ)/dx = 0, which yields x = 8.28,
and p/γ = -5.827 m of fluid flowing.
The minimum point occurs where the slopes of the pipe and of
the hydraulic grade line are equal.
11.4 PIPES IN SERIES
When two pipes of different sizes or roughnesses are so connected
that fluid flows through one pipe and then through the other, they
are said to be connected in series. A typical series-pipe problem, in
which the head H may be desired for a given discharge or the
discharge wanted for a given H, is illustrated in Fig.11.7.
Applying the energy equation from A to B, including all losses,
gives
In which the subscripts refer to the two pipes. The last item is the
head loss at exit from pipe 2. With the continuity equation
V2 is eliminated from the equations, so that
For known lengths and sizes of pipes this reduces to
(11.4.1)
in which C1, C2, C3 are known.
With the discharge given, the Reynolds number is readily
computed, and the f's may be looked up in the Moody diagram.
Then H is found by direct substitution. With H given, V1, f1, f2,
are unknowns in Eq. (11.4.1).
By assuming values of f1 and f2, (they may be assumed equal), a
trial V1 is found from which trial Reynolds numbers are
determined and values of f1, f2 looked up.
In place of the assumption of f1 and f2 when H is given, a
graphical solution may be utilized in which several values of Q
are assumed in turn, and the corresponding values of H are
calculated and plotted against Q, as in Fig. 11.8.
By connecting the points with a smooth curve, it is easy to read
off the proper Q for the given value of H.
Figure 11.7 Pipes connected in series
Figure 11.8 Plot of calculated H for selected
values of Q
Example 11.4
In Fig. 11.7, Ke = 0.5, L1 = 300 m, D1 = 600 mm, є1 = 2 mm, L2 =
240 m, D2 = 1 m, є2 = 0.3 mm, v = 3 × 10-6 m2/s, and H = 6 m.
Determine the discharge through the system.
Solution
From the energy equation
After simplifying,
Form є1/D1 = 0.0033, є2/D2 = 0.0003, and Fig. 5.21, values of f
are assumed for the fully turbulent range:
By solving for V1 with these values, V1 = 2.848 m/s, V2 = 1.025
m/s,
From Fig. 5.21, f1 = 0.0265, f2 = 0.0168. By solving again for V1,
V1 = 2.819 m/s, and Q = 0.797 m3/s.
Equivalent Pipes
Series pipes can be solved by the method of equivalent lengths.
Two pipe systems are said to be equivalent when the same head
loss produces the same discharge in both systems. From Eq.
(11.2.1)
and for a second pipe
For the two pipes to be equivalent,
After equating hf1 = hf2 and simplifying,
Solving for L2 gives
(11.4.2)
which determines the length of a second pipe to be equivalent
to that of the first pipe.
For example, to replace 300 m of 250-mm pipe with an
equivalent length of 150-mm pipe, the values of f1 and f2 must
be approximated by selecting a discharge within the range
intended for the pipes. Say f1 = 0.020, f2 = 0.018, then
For these assumed conditions 25.9 m of 150-mm pipe is
equivalent to 300 m of 250-mm pipe.
Hypothetically, two or more pipes composing a system may
also be replaced by a pipe which has the same discharge for the
same overall head loss.
Example 11.5
Solve Example 11.4 by means of equivalent pipes.
Solution
First, by expressing the minor losses in terms of equivalent
lengths, for pipe 1
and for pipe 2
The values of f1, f2 are selected for the fully turbulent range as
an approximation. The problem is now reduced to 321 m of
600-mm pipe and 306.7 m of 1-m pipe.
By expressing the 1-m pipe in terms of an equivalent length of
600-mm pipe, by Eq. (11.4.2),
By adding to the 600-mm pipe the problem is reduced to
finding the discharge through 334.76 m of 600-mm pipe, є1 = 2
mm, H = 6 m,
With f = 0.026, V = 2.848 m/s, and R = 2.848 × 0.6/(3 × 10-6) =
569600.
For є/D = 0.0033, f = 0.0265, V = 2.821, and Q = π(0.32)(2.821) =
0.798 m3/s. By use of Eq. (5.8.15), Q = 0.781 m3/s.
11.5 PIPES IN PARALLEL
A combination of two or more pipes connected as in Fig. 11.9,
so that the flow is divided among the pipes and then is joined
again, is a parallel-pipe system.
In analyzing parallel-pipe systems, it is assumed that the minor
losses are added into the lengths of each pipe as equivalent
lengths.
From Fig. 11.9 the conditions to be satisfied are
(11.5.1)
in which zA, zB are elevations of points A and B, and Q is the
discharge through the approach pipe or the exit pipe.
Figure 11.9 Parallel-pipe system
Two types of problems occur:
1.
2.
With elevation of hydraulic grade line at A and B known,
to find the discharge Q;
With Q known, to find the distribution of flow and the
head loss.
Sizes of pipe, fluid properties, and roughnesses are assumed
to be known.
The first type is, in effect, the solution of simple pipe
problems for discharge, since the head loss is the drop in
hydraulic grade line.
These discharges are added to determine the total discharge.
The recommended procedure is as follows:
1.
2.
3.
4.
Assume a discharge Q'1 through pipe 1.
Solve for h’f1, using the assumed discharge.
Using h’f1, find Q‘2, Q‘3.
With the three discharges for a common head loss, now
assume that the given Q is split up among the pipes in the
same proportion as Q'1, Q‘2, Q‘3; thus
(11.5.2)
5.
Check the correctness of these discharges by computing
hf1, hf2, hf3 for the computed Q1, Q2, Q3.
This procedure works for any number of pipes. By judicious
choice of Q'1, obtained by estimating the percent of the total
flow through the system that should pass through pipe 1
(based on diameter, length, and roughness), Eq. (11.5.2)
produces values that check within a few percent, which is
well within the range of accuracy of the friction factors.
Example 11.6
In Fig. 11.9, L1 = 900 m, D1 = 300 mm, є1 = 0.3 mm, L2 = 600 m, D2
= 200 mm, є2 = 0.03 mm, L3 = 1200 m, D3 = 400 mm, є3 = 0.24 mm,
ρ = 1028 kg/m3, v = 2.8 × 10-6 m2/s, pA = 560 kPa, zA = 30 m, zB = 24
m.
For a total flow of 340 L/s, determine flow through each pipe and
the pressure at B.
Solution
Assume Q’1 = 85 L/s; then V’1 = 1.20, R1 = 1.2 × 0.3/(2.8 × 10-6) =
129000, є1/D1 = 0.001, f’1 = 0.022, and
For pipe 2
Then є2/D2 = 0.00015. Assume f’2 = 0.020; then V’2 = 1.26 m/s,
R’2 = 1.28 × 0.2 × 1/(2.8 × 10-6) = 91400, f’2 = 0.019, V’2 =
1.291 m/s, Q’2 = 40.6 L/s
Then є3/D3 = 0.0006. Assume f’3 = 0.020; then V’3 = 1.259 m/s,
R’3 = 1.259 × 0.4/(2.8 × 10-6) = 180000, f’3 = 0.020, Q’3 = 158.2
L/s.
The total discharge for the assumed condition is
Hence
Check the values of h1, h2, h3:
f2 is about midway between 0.018 and 0.019. If 0.018 had been
selected, h2 would be 6.60 m.
To find pB,
or
in which the average head loss was taken. Then
11.6 BRANCHING PIPES
A simple branching-pipe system is shown in Fig. 11.11. In this
situation the flow through each pipe is wanted when the reservoir
elevations are given. The sizes and types of pipes and fluid
properties are assumed known.
Flow must be out of the highest reservoir and into the lowest;
hence, the continuity equation may be either
If the elevation of hydraulic grade line at the junction is above the
elevation of the intermediate reservoir, flow is into it; but if the
elevation of hydraulic grade line at J is below the intermediate
reservoir, the flow is out of it. Minor losses may be expressed as
equivalent lengths and added to the actual lengths of pipe.
The solution is effected by first assuming an elevation of hydraulic
grade line at the junction and then computing Q1, Q2, Q3, and
substituting into the continuity equation. If the flow into the
junction is too great, a higher grade-line elevation, which will
reduce the inflow and increase the outflow, is assumed.
Figure 11.11 Three interconnected reservoirs
In pumping from one reservoir to two or more other reservoirs, as
in Fig. 11.12, the characteristics of the pump must be known.
Assuming that the pump runs at constant speed, its head depends
upon the discharge. A suitable procedure is as follows.
1.
2.
3.
4.
5.
6.
Assume a discharge through the pump.
Compute the hydraulic-grade-line elevation at the suction side
of the pump.
From the pump characteristic curve find the head produced
and add it to suction hydraulic grade line.
Compute drop in hydraulic grade line to the junction J and
determine elevation of hydraulic grade line there.
For this elevation, compute flow into reservoirs 2 and 3.
If flow into J equals flow out of J, the problem is solved. If
flow into J is too great, assume less flow through the pump
and repeat the procedure.
This procedure is easily plotted on a graph, so that the intersection
of two elevations vs. flow curves yields the answer.
Figure 11.12 Pumping from one reservoir to two
other reservoirs
Example 11.7
In Fig. 11.11, find the discharges for water at 20°C with the
following pipe data and reservoir elevations: L1 =3000 m, D1 =
1 m, є1/D1 = 0.0002; L2 = 600 m, D2 = 0.45 m, є2/D2 = 0.002; L3 =
1000 m, D3 = 0.6 m, є3/D3 = 0.001; z1 = 30 m, z2 = 18 m, z3 = 9 m.
Solution
Assume zJ + pJ/γ = 23 m. Then
So that the inflow is greater than the outflow by
Assume zJ + pJ/γ = 24.6 m. Then
The inflow is still greater by 0.029 m3/s. By extrapolating
linearly, zJ + pJ/γ = 24.8 m, Q1 = 1.183, Q2 = 0.325, Q3 = 0.862
m3/s.
11.7 NETWORKS OF PIPES
Interconnected pipes through which the flow to a given outlet
may come from several circuits are called a network of pipes,
in many ways analogous to flow through electric networks.
The following conditions must be satisfied in a network of
pipes:
1.
2.
3.
The algebraic sum of the pressure drops around each
circuit must be zero.
Flow into each junction must equal flow out of the junction.
The Darcy-Weisbach equation, or equivalent exponential
friction formula, must be satisfied for each pipe; i.e., the
proper relation between head loss and discharge must be
maintained for each pipe.
The first condition states that the pressure drop between any
two points in the circuit, for example, A and G (Fig. 11.15),
must be the same whether through the pipe AG or through
AFEDG. The second condition is the continuity equation.
Figure 11.15 Pipe network
Since it is impractical to solve network problems analytically,
methods of successive approximations are utilized. The Hardy
Gross method is one in which flows are assumed for each pipe
so that continuity is satisfied at every junction.
A correction to the flow in each circuit is then computed in
turn and applied to bring the circuits into closer balance.
Minor losses are included as equivalent lengths in each pipe.
Exponential equations are commonly used, in the form hf = rQn,
where r = RL/Dm in Eq. (11.1.1). The value of r is a constant in
each pipeline (unless the Darcy-Weisbach equation is used) and
is determined in advance of the loop-balancing procedure. The
corrective term is obtained as follows.
For any pipe in which Q0 is an assumed initial discharge
(11.7.1)
Where Q is the correct discharge and ΔQ is the correction.
Then for each pipe,
If ΔQ is small compared with Q0, all terms or the series alter the
second may be dropped. Now for a circuit,
in which ΔQ has been taken out of the summation.
The last equation is solved for ΔQ in each circuit in the network
(11.7.2)
when ΔQ is applied to each pipe in a circuit in accordance with Eq.
(11.7.1), the directional sense is important; i.e., it adds to flows in
the clockwise direction and subtracts from flows in the
counterclockwise direction.
Steps in an arithmetic procedure may be itemized as follows:
1.
2.
3.
4.
Assume the best distribution of flows that satisfies
continuity by careful examination of the network.
For each pipe in an elementary circuit, calculate and sum
the net head loss ∑hf = ∑rQn. Also calculate ∑rn|Q|n-1 for
the circuit. The negative ratio, by Eq. (11.7.2) yields the
correction, which is then added algebraically to each flow
in the circuit to correct it.
Proceed to another elementary circuit and repeat the
correction process of 2. Continue for all elementary circuits.
Repeat 2 and 3 as many times as needed until the
corrections (ΔQ's) are arbitrarily small.
The values of r occur in both numerator and denominator;
hence, values proportional to the actual r may be used to find
the distribution. To find a particular head loss, the actual
values of r and Q must be used after the distribution has been
determined.
Very simple networks, such as the one shown in Fig. 11.16, may
be solved with the hand-held programmable calculator if it has
memory storage of about 15 positions and about 100 program
steps.
For networks larger than the previous example or for networks
that contain multiple reservoirs, supply pumps, or booster
pumps, the Hardy Cross loop-balancing method may be
programmed for numerical solution on a digital computer.
Such a program is provided in the next section.
A number of more general methods are available, primarily
based upon the Hardy Cross loop-balancing or node-balancing
schemes. In the more general methods the system is normally
modeled with a set of simultaneous equations which are solved
by the Newton-Raphson method.
Some programmed solutions are very useful as design tools,
since pipe sizes or roughnesses may be treated as unknowns in
addition to junction pressures and flows.
Example 11.8
The distribution of flow through the network of Fig 11.16 is
desired for the inflow and outflows as given. For simplicity n
has been given the value 2.0.
Solution
The assumed distribution is shown in Fig. 11.16a. At the upper
left the term ∑rQ0|Q0|n-1 is computed for the lower circuit
number 1. Next to the diagram on the left is the computation of
∑nr|Q0|n-1 for the same circuit.
The same format is used for the second circuit in the upper
right of the figure. The corrected flow after the first step for the
top horizontal pipe is determined as 15 + 11.06 = 26.06 and for
the diagonal as 35 + (-21.17) + (-11.06) = 2.77.
Figure 11.16b shows the flows after one correction and Fig.
11.16c the values after four corrections.
Figure 11.16 Solution for flow in a simple network
11.8 COMPUTER PROGRAM FOR STEADYSTATE HYDRAULIC SYSTEMS
Hydraulic systems that contain components different from
pipelines can be handled by replacing the component with an
equivalent length of pipeline. When the additional component is a
pump, special consideration is needed.
For systems with multiple fixed-pressure-head elevations, Fig.
11.17, pseudo elements are introduced to account for the unknown
outflows and inflows at the reservoirs and to satisfy continuity
conditions during balancing.
If head drop is considered positive in an assumed positive
direction in the pseudo element, the correction in loop 3, Fig. 11.17,
is
(11.8.1)
Figure 11.17 Sample network
A pump in a system may be considered as a flow element with a
negative head loss equal to the head rise that corresponds to the
flow through the unit. The pump-head-discharge curve, element 8
in Fig. 11.17, may be expressed by a cubic equation
where A0 is the shutoff head of the pump. The correction in loop 4
is
(11.8.2)
This correction is applied to pipe 5 and to pump 8 in the loop.
Equation (11.8.2) is developed by application of Newton's method
to the loop. For satisfactory balancing of networks with pumping
stations, the slope of the head-discharge curve should always be
less than or equal to zero.
The BASIC program (FORTRAN IV), Fig. 11.18, may be used to
analyze a wide variety of liquid steady-state pipe flow problems.
Pipeline flows described by the Hazen-Williams equation or
laminar or turbulent flows analyzed with the Darcy-Weisbach
equation can be handled; multiple reservoirs or fixed pressure
levels, as in a sprinkler system, can be analyzed.
Figure 11.18 FORTRAN program for hydraulic
system
A network is visualized as a combination of elements that are
interconnected at junctions. The elements may include
pipelines, pumps, and imaginary elements which are used to
create pseudo loops in multiple-reservoir systems.
All minor losses are handled by estimating equivalent lengths
and adding them onto the actual pipe lengths. Each element in
the system is numbered up to a maximum of 100, without
duplication and not necessarily consecutively.
A positive flow direction is assigned to each element, and, as in
the arithmetic solution, an estimated flow is assigned to each
element such that continuity is satisfied at each junction.
The flow direction in the pseudo element that creates an
imaginary loop indicates only the direction of fixed positive
head drop, since the flow must be zero in this element.
Each junction, which may represent the termination of a single
element or the intersection of many elements, is numbered up
to a maximum of 100.
The operation of the program is best visualized in two major parts:
The first performs the balancing of each loop in the system
successively and then repeats iteratively until the sum of all
loop flow corrections is less than a specified tolerance. At the
end of this balancing process the element flows are computed
and printed.
The second part of an analysis involves the computation of the
hydraulic-grade-line elevations at junctions in the system.
Each of these parts requires a special indexing of the system
configuration in the input data. The indexing of the system loops
for balancing is placed in the vector IND.
Any continuous path may be broken by inserting a zero; then a
new path is begun with a new initial junction, an element, and a
node, etc.
All junction hydraulic-grade-line elevations that are computed are
printed.
As shown below, the type of each element is identified in the input
data, and each element is identified in the program by the
assignment of a unique numerical value in the vector ITYPE.
The physical data associated with each element are entered on
separate lines. In the program the physical data that describe all
elements in the system are stored in the vector ELEM, with five
locations reserved for each element.
As an example of the position of storage of element information,
the data pertaining to element number 13 are located in positions
61 to 65 in ELEM.
Data entry, which is through READ and DATA statements, is best
visualized in four steps.
Step 1: Parameter Description Line
The type of unit to be used in the analysis is defined by the
characters USC or SI units. An integer defines the maximum
number of iterations to be allowed during the balancing
process.
An acceptable tolerance is set for the sum of the absolute
values of the corrections in each loop during each iteration.
The liquid kinematic viscosity must be specified if the DarcyWeisbach equation is used for pipeline losses. If the HazenWilliams equation is used, a default value for the coefficient C
may be defined, or if the Darcy-Weisbach equation is used, a
default value for absolute pipe roughness may be defined.
If the default value is not used, a zero should be inserted in its
place.
Step 2: Element Data
System elements must be placed in the following order: pipe,
pseudo elements, and pumps. For each element type, the first line
provides the number of elements of that type, followed by an
identifier: HW or DW for Hazen-Williams or Darcy-Weisbach
pipeline, PS for pseudo element, PU for pump.
As an example a system with three pipes, no pseudo elements, and
one pump would require the following seven DATA statements:
An entry for the number of pipes, in this case 3, followed by
HW or DW.
2. For each pipe an entry for the element number, the estimated
flow, the length, the inside pipe diameter, and (if the default
value is not used) either the Hazen-Williams coefficient or the
Darcy-Weisbach pipe roughness. If the default value is used, a
zero must be inserted.
3. An entry for the number of pseudo elements, in this example 0,
followed by PS.
1.
4.
For each pseudo element, the element number, and the
difference in elevation between interconnected fixedpressure-head levels, with head drop in the direction of
the arrow positive. If there are no pseudo elements, this
line of data must be omitted.
5.
An entry for the number of pumps: 1, PU.
6.
For each pump, the element number, the estimated flow,
the flow increment ΔQ at which values of pump head are
specified, and four values of head from the pumpcharacteristic curve beginning at shut-off head and at
equal flow intervals of ΔQ. If there are no pumps this line
of data is omitted.
Step 3: Loop Indexing
The first DATA statement provides an integer for the number
of items in the loop index vector, followed by IND.
The second DATA statement provides loop indexing by giving
the number of elements in a loop followed by the element
number of each element in the loop with a negative sign to
indicate counterclockwise flow direction.
This information is repeated for each loop in the system. If
there are no loops, this is omitted.
Step 4: Path Index for Hydraulic-Grade-Line
Calculation
A DATA statement is needed for the number of junctions (or nodes)
where fixed hydraulic-grade-line elevation are provided, followed
by NODES. For each junction at which a fixed HGL is used there
must be an entry for the function number and the elevation.
Next a DATA statement is provided for the number of integer
items in the path index vector, followed by IX. The next DATA
statement provides the path beginning at a node where the HGL is
given and continuing in the following order: junction number,
element number (with a negative sign to indicate a path opposite
to the assumed flow direction), junction number, etc.
If a new path is to begin at a junction different from the last listed
junction, element number, junction number, etc.
Example 11.9
The program in Fig. 11.18 is used to solve the network problem
displayed in Fig. 11.17. The pump data are as follows:
Solution
The Hazen-Williams pipeline coefficient for all pipes is 100.
Figure 11.19 displays the input data and the computer output
for this problem.
Figures 11.21 to 11.23 give input data for three systems which
can be solved with this program.
Figure 11.19
Program input
and output for
Example 11.9
Figure 11.21 Input data for branching-pipe system in
USC units with Hazen-Williams formula
Figure 11.22
Input data for
hydraulic system;
SI units and
Darcy-Weisbach
equation
Figure 11.23 Input for booster-pump
system
11.9 CONDUITS WITH NONCIRCULAR
CROSS SECTIONS
In this chapter only circular pipes have been considered so far.
For cross sections that are noncircular, the Darcy-Weisbach
equation may be applied if the term D can be interpreted in
terms of the section.
The concept of the hydraulic radius R permits circular and
noncircular sections to be treated in the same manner. The
hydraulic radius is defined as the cross-sectional area divided
by the wetted perimeter. Hence, for a circular section,
(11.9.1)
and the diameter is equivalent to 4R. Assuming that the
diameter may be replaced by 4R in the Darcy-Weisbach
equation, in the Reynolds number, and in the relative
roughness leads to
(11.9.2)
Example 11.10
Determine the head loss, in millimeters of water, required for
flow of 300m3/min of air at 20°C and 100 kPa through a
rectangular galvanized-iron section 700 mm wide, 350 mm high,
and 70 m long.
Solution
From Fig. 5.21, f = 0.0165
The specific weight of air is ρg = 1.189(9.806) = 11.66 N/m3. In
millimeters of water,
H2O
11.10 AGING OF PIPES
The Moody diagram, with the values or absolute roughness
shown there, is for new, clean pipe. With use, pipes become
rougher, owing to corrosion, incrustations, and deposition of
material on the pipe walls.
Colebrook and White found that the absolute roughness є
increases linearly with time,
(11.10.1)
in which є0 is the absolute roughness of the new surface. Tests
on a pipe are required to determine α.
The time variation of the Hazen-Williams coefficient has been
summarized graphically for water-distribution systems in
seven major U.S. cities.