Flow in Systems with Multiple Pipes

Flow in Systems with Multiple Pipes

In many systems of interest, pipes are connected in series, parallel, or complex networks. Thus far, we have analyzed single-pipe systems by writing and solving equations that describe fundamental relationships among the key variables. Often, we can specify enough parameters so that only one variable remains unknown (hL, V or Q, or D), and use the Moody diagram or an equivalent equation to solve for that unknown. In multi-pipe systems, we typically do not know as much about the flow conditions applicable to the individual pipes (e.g., we might know the flow rates entering and leaving, but not in individual pipes in the middle of a system), and so we need to write more equations relating the unknown variables and solve those equations simultaneously. The additional equations generally fall into two categories:

Expressions of continuity at junctions, stating that all the flow that enters a junction must leave it.

Expressions relying on the fact that the total head has a single value at each point in the system, so that the headloss between any two points must be the same regardless of the path the fluid follows between these locations.

The application of these equations is most easily seen by considering some example systems.

______________________ Example 1. Characteristics of the system shown schematically below are summarized in the table following the schematic. (Note that, to simplify the math, friction factors are assumed for each pipe, rather than being calculated from the flow conditions in the pipe.) Water is discharged to the atmosphere at point D. The system geometry is such that the change in velocity head is negligible compared to the change in piezometric head between any two points of interest, so the velocity heads can be ignored in an energy analysis. Compute (a) the flow in each pipe and (b) the pressures at points A and B.

E

2

A 1

Water

B

3

C

4 D

C:\AData\CLASNOTE\345\Spring 2012\Pipes_networks_Spr2012.docx; saved 4/29/2012 4:47:00 PM

Point Elevation (ft) Pipe Length (ft) Diameter (in)

E

150

1

1200

8

A

120

2

800

6

B

70

3

900

4

C

60

4

500

6

D

30

Friction factor 0.035 0.025 0.045 0.025

Solution. (a) The general principles that we will apply are that the headloss must be identical through the two pipes in parallel (#2 and #3), and that the flow into the junctions at points B and C must equal the flow out of them. Furthermore, of course, the total headloss between points E and D must satisfy the energy equation between those points. The application of these principles is as follows.

The areas of the pipes can be computed as D2/4, from which:

A1 0.349 ft2 A2 A4 0.196 ft2

A3 0.0873 ft2

The headloss between points B and C must be the same regardless of whether the water flows through pipe 2 or pipe 3. Writing the Darcy-Weisbach (D-W) equation for flow through these pipes, we find:

hL,2

f2

l2 D2

V22 2g

f3

l3 D3

V32 2g

hL,3

0.025 800 ft V22 0.045 900 ft V32

0.5 ft 2g

0.33 ft 2g

V2 1.74V3 The corresponding relationships among the flow rates are:

Q2 1.74 Q3

A2

A3

Q2

1.74

A2 A3

Q3

1.74

0.196 0.087

ft 2 ft 2

Q3

3.92Q3

The principle that the flow into the junctions must equal the flows out of them is, essentially, just a statement of the principle of continuity. Formally, we can express this idea at point B as:

Q1 Q2 Q3 3.92Q3 Q3 4.92Q3 A1V1 4.92 A3V3

2

V1

4.92

A3 A1

V3

4.92

0.0873 ft2 0.349 ft2

V3

1.23V3

The corresponding calculations at point C are:

4.92Q3 Q4

V4

4.92

A3 A4

V3

4.92

0.0873 ft2 0.196 ft2

V3

2.19V3

We now know all the velocities in terms of V3 and can write the energy equation between points E and D with V3 as the only unknown. We can write this energy equation using either the path through pipes 1, 2, and 4, or that through pipes 1, 3, and 4; choosing the latter option, keeping in mind that we are ignoring the change in velocity head between points E and D, and noting that the pressure is zero at both points, we find:

pE

zE

pD

zD hL,1 hL,3 hL,4

150

ft

30

ft

f1

l1 D1

1.23V3 2

2g

f3

l3 D3

V32 2g

f4

l4 D4

2.19V3 2

2g

120 ft 0.035 1200 ft 1.231V3 2 0.045 900 ft V32 0.025 500 ft 2.19V3 2

0.667 ft 2 32.2 ft/s2

0.333 ft 2g

0.50 ft 2g

V32

22.9

ft 2 s2

V3

4.79

ft s

Substituting this result back into the expressions for the relative velocities and the flow rates yields:

V1

1.23

4.79

ft s

5.90

ft s

Q1 A1V1

0.349 ft2

5.90

ft s

2.06

ft3 s

V2

1.74

4.79

ft s

8.35

ft s

Q2 A2V2

0.196 ft2

8.35

ft s

1.64

ft3 s

V3

4.79

ft s

Q3 A3V3

0.0873 ft2

4.79

ft s

0.42

ft3 s

3

V4

2.19

4.79

ft s

10.5

ft s

Q4 A4V4

0.196 ft2

10.5

ft s

2.06

ft3 s

(b) To determine the pressures at points B and C, we utilize the energy equation. For point B, we write the equation between points E and B, as follows:

pE

zE

pB

zB

hL,1

150 ft pB 70 ft 0.035 1200 ft 5.90 ft/s2

0.667 ft 2 32.2 ft/s2

pB 45.96 ft

pB

45.96

ft

62.4

lb ft3

1 ft2

144

in 2

19.9

psi

Similarly, the pressure at point C can be computed by writing the energy equation between points B and C:

pB

zB

pC

zC

hL,2

45.96 ft 70 ft pc 60 ft 0.025 800 ft 8.35 ft/s2

0.50 ft 2 32.2 ft/s2

pC 12.65 ft

pB

12.65

ft

62.4

lb ft3

1 ft2

144

in

2

5.48

psi

Representing Simple Pipe Networks as a Smaller Number of Equivalent Pipes

The systems analyzed above each have one point at which all the flow enters the system and another at which all the flow exits, allowing the flow direction in each pipe to be inferred unambiguously. In such cases, if we are just interested in the relationship between total flow and total headloss, it is sometimes convenient to simplify the analysis by representing the whole group of pipes as a single, hydraulically equivalent pipe.

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For such a representation, the `friction intensity' parameter for each pipe must be known, and it must be independent of the flow conditions for the range of flow conditions of interest. If the D-W equation is used to relate hL to V or Q, the friction intensity factor is f, which has a constant value in a given pipe for fully turbulent flow, but not for laminar or transitional flow (even though we assumed otherwise in Example 1). If the Hazen-Williams (H-W) equation is used, the friction intensity factor is CHW, which is assumed to be known for a given pipe.

To demonstrate how we can identify the characteristics of an equivalent pipe for a given real pipe network, we consider again the system analyzed in Example 1. This time, however, we will use the H-W equation and assume that the CHW values are 110 in pipes 1 and 2, and 90 in pipes 3 and 4. In addition, we will specify that the diameter of the equivalent pipe is 8 inches, and that it has a CHW of 100; the only parameter that must be calculated is the equivalent pipe length.

E

2

A

1

Water

B

3

C

4 D

We begin by representing just pipes 2 and 3 by a section of the equivalent pipe. Both these pipes link points B and C, so the headloss through the two pipes must be identical. Expressing this equality using the H-W equation for headloss in BG units, we find:

4.73Q21.85

l2 D 4.87

2

1 C1.85

HW ,2

4.73Q31.85

l3 D 4.87

3

1 C1.85

HW ,3

Q2 Q3

1.85

D2 D3

4.87

l3 l2

CHW ,2 CHW ,3

1.85

Q2 Q3

D2 D3

4.87

1.85

l3 l2

1

1.85

CHW CHW

,2 ,3

1.85 1.85

D2 D3

2.54

l3 l2

0.54

CHW CHW

,2 ,3

Substituting given values yields:

Q2 Q3

0.5 ft 0.33 ft

2.54 900 800

ft ft

0.54 110 90

3.74

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A pipe that is hydraulically equivalent to the combination of pipes 2 and 3 must carry a total flow rate of Q2 + Q3 at the same headloss as actually occurs in each of those pipes in the real system. The flow in the equivalent pipe would therefore be:

Qeq,23

Q2

Q3

Q2 Q3

1 Q3

3.74 1Q3

4.74Q3

Again using the H-W equation to equate headlosses, this time between pipe 3 and the equivalent pipe, we find:

4.73Qe1q.8,523

leq , 2 3 D 4.87

eq , 2 3

1 C1.85

HW ,eq,23

4.73Q31.85

l3 D 4.87

3

1 C1.85

HW ,3

leq,23 l3

Q3 Qeq,23

1.85

Deq,23 D3

4.87

CHW ,eq,23 CHW ,3

1.85

1 4.74

1.85

8 4

in in

4.87

100 90

1.85

2.00

leq,23 2.00l3 2.00900 ft 1800 ft

We now know that the parallel pipes in the middle of the network can be represented by a single, 8-in diameter pipe that is 1800 ft long and has a CHW of 100. We can use that information to represent the whole network as three pipes in series: Pipe 1, the equivalent pipe to pipes 2 and 3, and pipe 4. An pipe that is equivalent to the whole network must have the same total headloss as the real system, when the total flow is Q2 +Q3, or Qtot. We can represent this equality as:

hL,eq,tot hL,1 hL,eq,23 hL,4

4.73Qe1q.8,t5ot

leq,tot D 4.87

eq ,tot

1 C1.85

HW ,eq,tot

Q11.85

l1 D 4.87

1

1 Q l 1.85 C D HW ,1

1.85 eq , 2 3

eq , 2 3

4.87 eq , 2 3

1 C1.85

HW ,eq,23

Q1.85 4

l4 D 4.87

4

1

C1.85 HW ,4

Since all the Q's are identical (equal to Qtot), this equality can be simplified to:

leq ,tot

1

l1

1 leq,23

1 l4 1

D C 4.87 1.85 eq,tot HW ,eq,tot

D C D C D C 4.87 1.85

1

HW ,1

4.87

1.85

eq,23 HW ,eq,23

4.87 1.85

4

HW ,4

All the terms in this equation except leq,tot are known, so we can substitute and solve for that value:

leq ,tot

0.667 ft 4.87

1

100 1.85

1200 ft

0.667 ft 4.87

1

1101.85

2000 ft

0.667 ft 4.87

1

1001.85

500 ft

0.5 ft 4.87

1

90 1.85

6

leq,tot 5270 ft

Thus, we conclude that the network is hydraulically equivalent to a single, 8-in diameter, 5270-ft long pipe, with CHW of 100. Note that these calculations did not utilize any specific value of Qtot, so the results are valid no matter what the total flow rate through the system is.

Analysis of Complex Pipe Networks: Multiple Inlets/Outlets in a Single Loop

The analysis techniques used above are satisfactory if the pipe system is simple enough that the flow direction in each pipe is known unambiguously. In more complex systems, pipes might be combined in interconnected loops in ways that make it difficult to determine even the direction of flow in any given pipe. The fundamental relationships that we have used up to this point ? the energy and continuity equations and the relationships between flow and headloss in any given pipe ? still apply in such a system, but the sheer number of equations that need to be satisfied to determine the complete flow conditions is daunting. The conditions in such systems are usually solved with specialized computer programs designed specifically for that purpose. However, before such programs were widely available, manual and spreadsheet techniques were developed for analyzing the systems. These techniques provide a bridge between very simple problems like those analyzed above and the massive ones that can be solved only with special software. These techniques, as well as more sophisticated ones, allow us to answer such questions as:

* For a given set of flow rates, what will be the head loss in each pipe in the network? * Will additional head have to be supplied by pumps in order to obtain the desired flows? * How much will the flow rates change in various parts of the system if a new pipe is

installed, connecting two previously unconnected points, or to replace an older, smaller pipe? * How much will the pressure at a consumer's tap drop if the fire hydrant outside his or her home is in use?

Solving the System of Equations

The methods used historically to solve pipe network problems are "relaxation" techniques, in which a few of the variables are estimated, the others are computed, and the original estimates are revised based on algebraic manipulations of the computed variables. The approach bears many similarities to the iterative technique used to determine the flow rate through a pipe connecting two reservoirs at known elevations, but rather than iterating on the velocity in a single pipe, the velocities in all the pipes in specified loops are adjusted simultaneously. As is the case for individual pipes, the numerical solution depends to some extent on which equation is used to relate flow to headloss.

With respect to pipe network analysis, the traditional approach is known as the Hardy Cross method. This method is applicable if all the pipe sizes (lengths and diameters) are fixed, and either the headlosses between the inlets and outlets are known but the flows are not, or the flows at each inflow and outflow point are known, but the headlosses are not. This latter case is explored next.

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The procedure involves making a guess as to the flow rate in each pipe, taking care to make guesses in such a way that the total flow into any junction equals the total flow out of that junction. Then the headloss in each pipe is calculated, based on the assumed flows and the selected flow vs. headloss relationship. Next, the system is checked to see if the headloss around each loop is zero. Since the initial flows were guessed, this will probably not be the case. The flow rates are then adjusted in such a way that continuity (mass in equals mass out) is still satisfied at each junction, but the headloss around each loop is closer to zero. This process is repeated until the adjustments are satisfactorily small. The detailed procedure is as follows.

1. Define a set of independent pipe loops in such a way that every pipe in the network is part of at least one loop, and no loop can be represented as a sum or difference of other loops. The easiest way to do this is to choose all of the smallest possible loops in the network.

2. Arbitrarily choose values of Q in each pipe, such that continuity is satisfied at each pipe junction (sometimes called nodes). Use a sign convention such that Q in a given pipe is designated to be positive if the (assumed) direction of flow is clockwise in the loop under consideration. This convention means that the same flow in a given pipe might be considered positive when analyzing one loop, and negative when analyzing the adjacent loop.

3. Compute the headloss in each pipe, using the same sign convention for headloss as for flow, so that hL in each pipe has the same sign as Q, when analyzing any given loop.

4. Compute the headloss around each loop. If the headloss around every loop is zero, then all the pipe flow equations are satisfied, and the problem is solved. Presumably, this will not be the case when the initial, arbitrary guesses of Q are used.

5. Change the flow in each pipe in a given loop by Q. By changing the flow rates in all the pipes in a loop by the same amount, we assure that the increase or decrease in the flow into a junction is balanced by the exact same increase or decrease in the flow out, so that we guarantee that the continuity equation is still satisfied. The trick is to make a good guess for what Q should be, so that the headloss around the loop gets closer to zero after each adjustment. To achieve this, we assume that we can choose a value of Q that is exactly what is needed to make the headloss zero, and then see how this value of Q is expected to be related to other system parameters. The relationship is derived as follows.

Recall that, if the term characterizing the frictional "intensity" in a pipe is assumed to be constant, the D-W, H-W, and Manning equations can all be written in the form:

hf KQn

(1)

where K is a constant that includes geometric variables and the frictional intensity term. This intensity term is f in the D-W equation, CHW in the H-W equation, and nM in the Manning equation. The value of the exponent n is 2.0, 1.85, and 2.0 in those three equations, respectively. If the frictional intensity term is not constant (as in the D-W equation), we can nevertheless

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