Chapter 11STEADY 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 FORMULASIndustrial 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

with n = 1.852, m = 4.8704, and C dependent upon roughness as follows:R = USC unit

SI unit(11.1.2)

(11.1.3)

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 15C. By equating the slope of the hydraulic grade line in the Darcy-Weisbach 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 LINESThe 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/ 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/ + 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/ + 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/,

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.

SolutionSolving 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-mm-diameter 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.

SolutionIf 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 SIPHONA 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/ 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

SolutionWhen 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 areFigure 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/, set d(p/)/dx = 0, which yields x = 8.28, and p/ = -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 SERIESWhen 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 seriesFigure 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, 1 = 2 mm, L2 = 240 m, D2 = 1 m, 2 = 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 1/D1 = 0.0033, 2/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. SolutionFirst, 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, 1 = 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 /D = 0.0033, f = 0.0265, V = 2.821, and Q = (0.32)(2.821) = 0.798 m3/s. By use of Eq. (5.8.15), Q = 0.781 m3/s.

11.5 PIPES IN PARALLELA 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:

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:

Assume a discharge Q'1 through pipe 1.Solve for hf1, using the assumed discharge.Using hf1, find Q2, Q3.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, Q2, Q3; thus

(11.5.2)

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, 1 = 0.3 mm, L2 = 600 m, D2 = 200 mm, 2 = 0.03 mm, L3 = 1200 m, D3 = 400 mm, 3 = 0.24 mm, = 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.

SolutionAssume Q1 = 85 L/s; then V1 = 1.20, R1 = 1.2 0.3/(2.8 10-6) = 129000, 1/D1 = 0.001, f1 = 0.022, and

For pipe 2

Then 2/D2 = 0.00015. Assume f2 = 0.020; then V2 = 1.26 m/s, R2 = 1.28 0.2 1/(2.8 10-6) = 91400, f2 = 0.019, V2 = 1.291 m/s, Q2 = 40.6 L/s

Then 3/D3 = 0.0006. Assume f3 = 0.020; then V3 = 1.259 m/s, R3 = 1.259 0.4/(2.8 10-6) = 180000, f3 = 0.020, Q3 = 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 PIPESA 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.

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 20C with the following pipe data and reservoir elevations: L1 =3000 m, D1 = 1 m, 1/D1 = 0.0002; L2 = 600 m, D2 = 0.45 m, 2/D2 = 0.002; L3 = 1000 m, D3 = 0.6 m, 3/D3 = 0.001; z1 = 30 m, z2 = 18 m, z3 = 9 m.

SolutionAssume zJ + pJ/ = 23 m. Then

So that the inflow is greater than the outflow by

Assume zJ + pJ/ = 24.6 m. Then

The inflow is still greater by 0.029 m3/s. By extrapolating linearly, zJ + pJ/ = 24.8 m, Q1 = 1.183, Q2 = 0.325, Q3 = 0.862 m3/s.

11.7 NETWORKS OF PIPESInterconnected 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:

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 Q is the correction.

Then for each pipe,

If Q is small compared with Q0, all terms or the series alter the second may be dropped. Now for a circuit,

in which Q has been taken out of the summation.

The last equation is solved for Q in each circuit in the network

(11.7.2)

when Q 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:

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 (Q'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.

SolutionThe 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 STEADY-STATE HYDRAULIC SYSTEMSHydraulic 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 Darcy-Weisbach equation is used for pipeline losses. If the Hazen-Williams 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.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.An entry for the number of pseudo elements, in this example 0, followed by PS.

For each pseudo element, the element number, and the difference in elevation between interconnected fixed-pressure-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.

An entry for the number of pumps: 1, PU.

For each pump, the element number, the estimated flow, the flow increment Q at which values of pump head are specified, and four values of head from the pump-characteristic curve beginning at shut-off head and at equal flow intervals of Q. 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:

SolutionThe 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 SECTIONSIn 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 20C 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 g = 1.189(9.806) = 11.66 N/m3. In millimeters of water,H2O

11.10 AGING OF PIPESThe 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 increases linearly with time, (11.10.1)

in which 0 is the absolute roughness of the new surface. Tests on a pipe are required to determine . The time variation of the Hazen-Williams coefficient has been summarized graphically for water-distribution systems in seven major U.S. cities.