Lin S.D. Water and Wastewater Calculations Manual

Подождите немного. Документ загружается.

Pipelines in series

Example: As shown in Fig. 4.5a, two concrete pipes are connected in series

between two reservoirs A and B. The diameters of the upstream and down-

stream pipes are 0.6 m (D

⫽ 2 ft) and 0.45 m (D

⫽ 1.5 ft); their lengths are

300 m (1000 ft) and 150 m (500 ft), respectively. The flow rate of 15⬚C water

Fundamental and Treatment Plant Hydraulics 255

Figure 4.5 Pipeline system for two reservoirs.

from reservoir A to reservoir B is 0.4 m

/s. Also given are coefficient of

entrance K

⫽ 0.5, coefficient of contraction K

⫽ 0.13, and coefficient of dis-

charge K

⫽ 1.0. Determine the elevation of the water surface of reservoir B

when the elevation of the water surface of reservoir A is 100 m.

solution:

Step 1. Find the relative roughness e/D, from Fig. 4.1 for a circular concrete

pipe, assume e ⫽ 0.003 ft ⫽ 0.0009 m

then

e/D

⫽ 0.0009/0.60 ⫽ 0.0015

e/D

⫽ 0.0009/0.45 ⫽ 0.0020

Step 2. Determine velocities (V ) and Reynolds number (R)

From Table 4.1a, ␯ ⫽ 1.131 ⫻ 10

⫺6

/s for T ⫽ 15⬚C

Step 3. Find f value from Moody chart (Fig. 4.1) corresponding to R and e/D

values

⫽ 0.022

⫽ 0.024

Step 4. Determine total energy loss for water flow from A to B

5 f

5 0.024

150

0.45

5 8

5 K

5 0.13

5 f

5 0.022 3

300

0.6

5 11

5 K

5 0.5

2.51 3 0.45

1.131 3 10

5 9.99 3 10

1.41 m/s 3 0.6 m

1.131 3 10

5 7.48 3 10

0.4

s0.45d

5 2.52 m/s

0.4 m

/ s

s0.6 md

5 1.41 m/s

256 Chapter 4

Step 5. Calculate total energy head H

Step 6. Calculate the elevation of the surface of reservoir B

Ele. ⫽ 100 m ⫺ H ⫽ 100 m ⫺ 4.12 m

⫽ 95.88 m

Pipelines in parallel

Example: Circular pipelines 1, 2, and 3, each 1000 ft (300 m) long and of 6,

8, and 12 in (0.3 m) diameter, respectively, carry water from reservoir A and

join pipeline 4 to reservoir B. Pipeline 4 is 2 ft (0.6 m) in diameter and 500 ft

(150 m) long. Determine the percentages of flow passing pipelines 1, 2, and

3 using the Hazen–Williams equation. The C value for pipeline 4 is 110 and,

for the other three, 100.

5 4.12 smd

5 11.5 3

s1.41d

2 3 9.81

1 9.13

s2.52d

2 3 9.81

5 s0.5 1 11d

1 s0.13 1 8 1 1d

H 5 h

1 h

5 K

5 1 3

Fundamental and Treatment Plant Hydraulics 257

solution:

Step 1. Calculate velocity in pipelines 1, 2, and 3

Since no elevation and flow are given, any slope of the energy line is OK.

Assuming the head loss in the three smaller pipelines is 10 ft per 1000 ft of

pipeline:

S ⫽ 10/1000 ⫽ 0.01

For a circular pipe, the hydraulic radius is

then, from the Hazen–Williams equation, Eq. (4.20a)

Step 2. Determine the total flow Q

Step 3. Determine percentage of flow in pipes 1, 2, and 3

For pipe 1 ⫽ (Q

) ⫻ 100% ⫽ 0.58 ⫻ 100/5.42

⫽ 10.7%

pipe 2 ⫽ 1.24 ⫻ 100%/5.42 ⫽ 22.9%

pipe 3 ⫽ 3.60 ⫻ 100%/5.42 ⫽ 66.4%

2.4 Distribution networks

Most waterline or sewer distribution networks are complexes of looping

and branching pipelines (Fig. 4.6). The solution methods described for

the analysis of pipelines in series, parallel, and branched systems are

5 5.42 sft

/sd

5 Q

1 Q

5 1.58 1 1.24 1 3.60

5 3.60 sft

/sd

5 A

5 0.785s1d

3 4.58

5 1.24 sft

/sd

5 A

5 0.785s0.667d

3 3.54

5 0.58 sft

/sd

5 A

5 0.785s0.5d

3 2.96

5 10.96 a

0.63

5 4.58 sft/sd

5 10.96 a

0.667

0.63

5 3.54 sft/sd

5 2.96 sft/sd

5 10.96 3 a

0.5

0.63

5 1.318 CR

0.63

0.54

5 1.318 3 100 3 a

0.5

0.63

s0.01d

0.54

R 5

258 Chapter 4

not suitable for the more complex cases of networks. A trial and error

procedure should be used. The most widely used is the loop method

originally proposed by Hardy Cross in 1936. Another method, the nodal

method, was proposed by Cornish in 1939 (Chadwick and Morfett, 1986).

For analysis flows in a network of pipes, the following conditions

must be satisfied:

1. The algebraic sum of the pressure drops around each circuit must

be zero.

2. Continuity must be satisfied at all junctions; i.e. inflow equals out-

flow at each junction.

3. Energy loss must be the same for all paths of water.

Using the continuity equation at a node, this gives

(4.23)

where m is the number of pipes joined at the node. The sign conven-

tionally used for flow into a joint is positive, and outflow is negative.

Applying the energy equation to a loop, we get

(4.24)

where n is the number of pipes in a loop. The sign for flow (q

) and head

loss (h

) is conventionally positive when clockwise.

Since friction loss is a function of flow

(4.25)h

5 fsq

⌺

i51

5 0

⌺

i51

5 0

Fundamental and Treatment Plant Hydraulics 259

Figure 4.6 Two types of pipe network.

Equations (4.23) and (4.24) will generate a set of simultaneous non-

linear equations. An iterative solution is needed.

Method of equivalent pipes. In a complex pipe system, small loops within

the system are replaced by single hydraulically equivalent pipes. This

method can also be used for determination of diameter and length of a

replacement pipe; i.e. one that will produce the same head loss as the

old one.

Example 1: Assume Q

⫽ 0.042 m

/s (1.5 ft

/s), Q

⫽ 0.028 m

/s (1.0 ft

/s),

and a Hazen–Williams coefficient C ⫽ 100. Determine an equivalent pipe for

the network of the following figure.

260 Chapter 4

solution:

Step 1. For line BCE, Q ⫽ 1 ft

/s. From the Hazen–Williams chart (Fig. 4.2)

with C ⫽ 100, to obtain loss of head in ft per 1000 ft, then compute the total

loss of head for the pipe length by proportion.

(a) Pipe BC, 4000 ft, 14 in ␾

(b) Pipe CE, 3000 ft, 10 in ␾

BCE

⫽ 1.76 ft ⫹ 7.20 ft ⫽ 8.96 ft

(d) Equivalent length of 10-in pipe:

Step 2. For line BDE, Q ⫽ 1.5 ⫺ 1.0 ⫽ 0.5 ft

/s or 0.014 m

1000 ft 3

8.96

2.4

5 3733 ft

2.40

1000

3 3000 ft 5 7.20 ft

0.44

1000

3 4000 ft 5 1.76 ft

(a) Pipe BD, 2500 ft, 12 in ␾

(b) Pipe DE, 5000 ft, 8 in ␾

BDE

⫽ h

⫹ h

⫽ 9.725 ft

(d) Equivalent length of 8-in pipe:

Step 3. Determine equivalent line BE from results of Steps 1 and 2, assum-

ing h

⫽ 8.96 ft

(a) Line BCE, 3733 ft, 10 in ␾

(b) Line BDE, 5403 ft, 8 in ␾

From Fig. 4.2 Q

BDE

⫽ 0.475 ft

/s ⫽ 0.013 m

⫽ 1 ft

/s ⫹ 0.475 ft

/s ⫽ 1.475 ft

/s ⫽ 0.0417 m

(d) Using Q ⫽ 1.475 ft

/s, select 12-in pipe to find equivalent length

From chart

S ⫽ 1.95

Since assumed total head loss H

⫽ 8.96

Answer: Equivalent length of 4600 ft of 12-in pipe.

Example 2: The sewer pipeline system below shows pipe diameter sizes (␾),

grades, pipe section numbers, and flow direction. Assume there is no surcharge

L 5

8.96

1.95

3 1000 < 4600 sftd

S 5

8.96

5403

5 1.66/1000

BCD

5 1.0 ft

S 5

8.96

3733

5 2.4/1000 5 24%

1000 ft 3

9.725

1.80

5 5403 ft

1.80

1000

3 5000 ft 5 9.0 ft

0.29

1000

3 2500 ft 5 0.725 ft

Fundamental and Treatment Plant Hydraulics 261

solution:

Step 1. Compute cross-sectional areas and hydraulic radius

Section Diameter, D (ft) A (ft) R ⫽ D/4 (ft)

1 1.167 1.068 0.292

2 0.667 0.349 0.167

3 0.833 0.545 0.208

4 1.0 0.785 0.250

BC 1.5 1.766 0.375

CD 1.667 2.180 0.417

Step 2. Compute flows with Manning formula

5 3.82 sft

/sd

5 Q

1 Q

5 3.22 1 0.60

5 0.60 sft

/sd

5 0.349 3 114.3s0.167d

2/3

s0.0025d

1/2

5 3.22 sft

/sd

5 1.068 3

1.486

0.013

s0.292d

2/3

s0.0036d

1/2

Q 5 A

1.486

2/3

1/2

262 Chapter 4

and full flow in each of sections 1, 2, 3, and 4. All pipes are fiberglass with

n ⫽ 0.013. A, B, C, and D represent inspection holes. Determine

(a) the flow rate and minimum commercial pipe size for section AB;

(b) the discharge, sewage depth, and velocity in section BC; and

Step 3. Compute D

Use 18-in commercially available pipe for answer (a).

Step 4. Determine Q

, Q

⫽ 0.545 ⫻ 114.3 ⫻ (0.208)

2/3

(0.0049)

1/2

⫽ 1.53 (ft

/s)

⫽ Q

⫹ Q

⫽ 3.82 ft

/s ⫹ 1.53 ft

⫽ 5.35 ft

At section BC: velocity for full flow

⫽ 114.3 ⫻ (0.375)

2/3

(0.0036)

1/2

⫽ 3.57 (ft/s)

⫽ AV

⫽ 1.766 ⫻ 3.57

⫽ 6.30 (ft

/s)

Percentage of actual flow

Step 5. Determine sewage depth d and V for section BC from Fig. 4.4

We obtain d/D ⫽ 0.70

Then

d ⫽ 0.70 ⫻ 18 in ⫽ 12.6 in

⫽ 1.05 ft

and

V/V

⫽ 1.13

Then

V ⫽ 1.13 ⫻ V

⫽ 1.13 ⫻ 3.57 ft/s

⫽ 4.03 ft/s

Step 6. Determine slope S of section CD

⫽ 0.785 ⫻ 114.3 ⫻ (0.25)

2/3

(0.0025)

1/2

⫽ 1.78 (ft

/s)

5.35 ft

6.30 ft

5 0.85

5 15.9 in

D 5 1.33 ft

8/3

5 2.146

5 3.82 5 0.785D

3 114.3a

2/3

3 s0.0025d

1/2

Fundamental and Treatment Plant Hydraulics 263

⫽ Q

⫹ Q

⫽ 1.78 ft

/s ⫹ 5.35 ft

⫽ 7.13 ft

Section CD if flowing full:

7.13 ⫽ 2.18 ⫻ 114.3(0.417)

2/3

1/2

⫽ 0.05127

S ⫽ 0.00263

Hardy–Cross method. The Hardy–Cross method (1936) of network analy-

sis is a loop method which eliminates the head losses from Eqs. (4.24)

and (4.25) and generates a set of discharge equations. The basics of the

method are:

1. Assume a value for Q

for each pipe to satisfy ⌺Q

⫽ 0.

2. Compute friction losses h

from Q

; find S from Hazen–Williams

equation.

3. If the solution is correct.

4. If apply a correction factor ⌬Q to all Q

, then repeat step 1.

5. A reasonable value of ⌬Q is given by Chadwick and Morfett (1986)

(4.26)

This trial and error procedure solved by digital computer program is

available in many textbooks on hydraulics (Hwang 1981, Streeter and

Wylie 1975).

The nodal method. The basic concept of the nodal method consists of

the elimination of discharges from Eqs. (4.23) and (4.25) to generate

a set of head loss equations. This method may be used for loops or

branches when the external heads are known and the heads within

the networks are needed. The procedure of the nodal method is as

follows:

1. Assume values of the head loss H

at each junction.

2. Compute Q

from H

3. If ⌺Q

⫽ 0, the solution is correct.

4. If ⌺Q

⫽ 0, adjust a correction factor ⌬H to H

, then repeat step 2.

5. The head correction factor is

(4.27)⌬H 5

⌺

⌬Q 5

⌺

2 0,

⌺h

5 0,

264 Chapter 4