Lin S.D. Water and Wastewater Calculations Manual

Подождите немного. Документ загружается.

Step 4. Similar calculations for the 2-MGD pump

Step 5. For the 4-MGD pump

Step 6. For the 3 MGD: using the 1- and 2-MGD capacity pumps

Step 7. For the 5 MGD: using the 1- and 4-MGD capacity pumps

Step 8. For the 6 MGD: using the 2- and 4-MGD capacity pumps

Step 9. For the 7 MGD: using all 3 pumps simultaneously

Hazen–Williams equation. Due to the difficulty of using the Darcy–

Weisback equation for pipe flow, engineers continue to make use of an expo-

nential equation with empirical methods for determining friction losses in

pipe flows. Among these the empirical formula of the Hazen–Williams

equation is most widely used to express flow relations in pressure conduits,

while the Manning equation is used for flow relations in free-flow conduits

and in pipes partially full. The Hazen-Williams equation, originally devel-

oped for the British measurement system, has the following form:

(4.20a)V 5 1.318 CR

0.63

0.54

5 s48.2 1 48.8 1 51.2d ft 5 148.2 ft

5 s0.2 1 0.8 1 3.2d ft 5 4.2 ft

5 s48.8 1 51.2d ft 5 100.0 ft

5 s0.8 1 3.2d ft 5 4.0 ft

5 s48.2 1 51.2d ft 5 99.4 ft

5 s0.2 1 3.2d ft 5 3.4 ft

5 s48.2 1 48.8d ft 5 97 ft

5 s0.2 1 0.8d ft 5 1.0 ft

5 48 ft 1 3.2 ft 5 51.2 ft

5 0.2 ft 3 4

5 3.2 ft

V 5 0.493 ft/s 3 4 5 1.972 ft

5 48 ft 1 0.8 ft 5 48.8 ft

5 0.2 ft 3 2

5 0.8 ft

V 5 0.493 ft/s 3 2 5 0.986 ft/s

5 48 ft 1 0.2 ft 5 48.2 ft

5 0.20 ft

Fundamental and Treatment Plant Hydraulics 245

where V ⫽ average velocity of pipe flow, ft/s

C ⫽ coefficient of roughness (see Table 4.3)

R ⫽ hydraulic radius, ft

S ⫽ slope of the energy gradient line or head loss per unit

length of the pipe (S ⫽ h

/L)

The hydraulic radius R is defined as the water cross-sectional area A

divided by the wetted perimeter P. For a circular pressure pipe, if D is

the diameter of the pipe, then R is

(4.21)

The Hazen–Williams equation was developed for water flow in large pipes

(D ⱖ 2 in, 5 cm) with a moderate range of velocity (V ⱕ 10 ft/s or 3 m/s).

The coefficient of roughness C values range from 140 for very smooth

(new), straight pipe to 90 or 80 for old, unlined tuberculated pipe. The

value of 100 is used for the average conditioned pipe. It is not a function

of the flow condition (i.e. Reynolds number R). The major limitation of this

equation is that the viscosity and temperature are considered. A nomo-

graph (Fig. 4.2) can be used to solve the Hazen-Williams equation.

R 5

246 Chapter 4

TABLE 4.3 Hazen–Williams Coefﬁcient of Roughness C for

Various Types of Pipe

Pipe material C value

Brass 130–140

Brick sewer 100

Cast iron

tar coated 130

new, unlined 130

cement lined 130–150

uncertain 60–110

Cement–asbestos 140

Concrete 130–140

Copper 130–140

Fire hose (rubber lined) 135

Galvanized iron 120

Glass 140

Lead 130–140

Plastic 140–150

Steel

coal tar enamel lined 145–150

corrugated 60

new unlined 140–150

riveted 110

Tin 130

Vitrified clay 110–140

Wood stave 110–120

SOURCES: Perry (1967), Hwang (1981), and Benefield et al. (1984)

Fundamental and Treatment Plant Hydraulics 247

Figure 4.2 Nomograph for Hazen–Williams formula.

248 Chapter 4

In the SI system, the Hazen–Williams equation is written as

(4.20b)

The units are R in m, S in m/1000 m, V in m/s, and D in m, and discharge

Q in m

/s (from chart).

Manning equation. Another popular empirical equation is the Manning

equation, developed by the Irish engineer Robert Manning in 1889, and

employed extensively for open channel flows. It is also commonly used

for pipe free-flow. The Manning equation is

(4.22a)

where V ⫽ average velocity of flow, ft/s

n ⫽ Manning’s coefficient of roughness (see Table 4.4)

R ⫽ hydraulic radius (ft) (same as in the Hazen–Williams

equation)

S ⫽ slope of the hydraulic gradient, ft/ft

V 5

1.486

2/3

1/2

V 5 0.85 C R

0.63

0.54

TABLE 4.4 Manning’s Roughness Coefﬁcient n for Various

Types of Pipe

Type of pipe n

Brick

open channel 0.014–0.017

pipe with cement mortar 0.012–0.017

brass copper or glass pipe 0.009–0.013

Cast iron

pipe uncoated 0.013

tuberculated 0.015–0.035

Cement mortar surface 0.011–0.015

Concrete

open channel 0.013–0.022

pipe 0.010–0.015

Common clay drain tile 0.011–0.017

Fiberglass 0.013

Galvanized iron 0.012–0.017

Gravel open channel 0.014–0.033

Plastic pipe (smooth) 0.011–0.015

Rock open channel 0.035–0.045

Steel pipe 0.011

Vitrified clay

pipes 0.011–0.015

liner plates 0.017–0.017

Wood, laminated 0.015–0.017

Wood stave 0.010–0.013

Wrought iron 0.012–0.017

SOURCES: Perry (1967), Hwang (1981), and ASCE & WEF (1992)

Fundamental and Treatment Plant Hydraulics 249

This equation can be easily solved with the nomograph shown in

Fig. 4.3. Figure 4.4 is used for partially filled circular pipes with vary-

ing Manning’s n and depth.

The Manning equation for SI units is

(4.22b)

where V ⫽ mean velocity, m/s

n ⫽ same as above

R ⫽ hydraulic radius, m

S ⫽ slope of the hydraulic gradient, m/m

Example 1: Assume the sewer line grade gives a sewer velocity of 0.6 m/s

with a half full sewer flow. The slope of the line is 0.0081. What is the diam-

eter of the uncoated cast iron sewer line?

solution:

Step 1. Using the Manning equation to solve R

From Table 4.4, n ⫽ 0.013

Step 2. Compute diameter of sewer line using Eq. (4.21)

Example 2: Determine the energy loss over 1600 ft in a new 24-in cast iron

pipe when the water temperature is 60⬚F and the flow rate is 6.25 cfs using

(1) the Darcy–Weisback equation, (2) the Hazen–Williams equation, and

(3) the Manning equation.

5 4 in

5 0.102 smd

D 5 4R 5 4 3 0.0254

R 5 0.0254

5 0.0864

5 0.6 3 0.013/0.0081

1/2

2/3

5 V n/S

1/2

2/3

5 Vn/S

1/2

V 5

2/3

1/2

V 5

2/3

1/2

250 Chapter 4

Figure 4.3 Nomograph for Manning formula.

Fundamental and Treatment Plant Hydraulics 251

Figure 4.4 Hydraulic elements graph for circular sewers (Metcalf and Eddy, Inc.,

Wastewater Engineering: Collection and Pumping of Wastewater, Copyright, 1990, McGraw-

Hill, New York, reproduced with permission of McGraw-Hill).

solution:

Step 1. By the Darcy–Weisback equation

From Table 4.2, e ⫽ 0.0102 in

(a) Determine e/D, D ⫽ 24 in ⫽ 2 ft

(b) Compute velocity V, r ⫽ 1 ft

5 1.99 ft/s

V 5

6.25 ft

ps1 ftd

0.0102

5 0.000425

From Table 4.1b, ␯⫽1.217 ⫻ 10

⫺5

/s for 60⬚F

(d) Determine f value from the Moody diagram (Fig. 4.1)

f ⫽ 0.018

(e) Compute loss of head h

Step 2. By the Hazen–Williams equation

(a) Find C

From Table 4.3, for new unlined cast pipe

C ⫽ 130

(b) Solve S

Eq. (4.21): R ⫽ D/4 ⫽ 2/4 ⫽ 0.5

Eq. (4.20a): V ⫽ 1.318CR

0.63

0.54

Rearranging

⫽ S ⫻ L ⫽ 0.000586 ft/ft ⫻ 1600 ft

⫽ 0.94 ft

Step 3. By the Manning equation

(a) Find n

From Table 4.4, for unlined cast iron pipe

n ⫽ 0.013

S 5 c

1.318CR

0.63

1.852

5 c

1.99

1.318 3 130 3 s0.5d

0.63

1.852

5 0.000586 ft/ft

5 0.885 sftd

5 f a

5 0.018a

1600

s1.99d

2 3 32.2

5 3.27 3 10

R 5

1.99 ft/s 3 2 ft

1.217 3 10

252 Chapter 4

(b) Determine slope of the energy line S

Rearranging

Example 3: The drainage area of the watershed to a storm sewer is 10,000

acres (4047 ha). Thirty-eight percent of the drainage area has a maximum

runoff of 0.6 ft

/s ⭈ acre (0.0069 m

/s ⋅ ha), and the rest of the area has a

runoff of 0.4 ft

/s ⭈ acre (0.0046 m

/s. ha). Determine the size of pipe needed

to carry the storm flow with a grade of 0.12% and n ⫽ 0.011.

solution:

Step 1. Calculate the total runoff Q

Q ⫽ 0.6 cfs/acre ⫻ 0.38 ⫻ 10,000 acre ⫹ 0.4 ft

/(s ⋅ acre) ⫻ 0.62 ⫻ 10,000 acre

⫽ 4760 ft

Step 2. Determine diameter D of the pipe by the Manning formula

Minor head losses due to hydraulic devices such as sharp-crested orifices,

weirs, valves, bend, construction, enlargement, discharge, branching, etc.

have been extensively covered in textbooks and handbooks elsewhere.

Basically they can be calculated from empirical formulas.

D 5 20.78 sftd

3265 5 D

8/3

4760 5 0.785D

1.486

0.011

2/3

s0.0012d

1/2

Q 5 AV 5 A

1.486

2/3

1/2

R 5 D/4

A 5

5 0.785D

5 S 3 L 5 0.000764 ft/ft 3 1600 ft

5 1.22 ft

S 5 a

1.486R

2/3

5 c

1.99 3 0.013

1.486 3 s0.5d

2/3

5 0.000764 sft/ftd

From Eq. s4.22ad, V 5

1.486

2/3

1/2

From Eq. s4.21d, R 5 D/4 5 2/4 5 0.5

Fundamental and Treatment Plant Hydraulics 253

2.3 Pipeline systems

There are three major types of compound piping system: pipes con-

nected in series, in parallel, and branching. Pipes in parallel occur in a

pipe network when two or more paths are available for water flowing

between two points. Pipe branching occurs when water can flow to or

from a junction of three or more pipes from independent outlets or

sources. On some occasions, two or more sizes of pipes may be connected

in series between two reservoirs. Figure 4.5 illustrates all three cases.

The flow through a pipeline consisting of three or more pipes connected

in various ways can be analyzed on the basis of head loss concept with

two basic conditions. First, at a junction, water flows in and out should

be the same. Second, all pipes meeting at the junction have the same water

pressure at the junction. The general procedures are to apply the Bernoulli

energy equation to determine the energy line and hydraulic gradient

mainly from friction head loss equations. In summary, using Fig. 4.5:

1. For pipelines in series

Energy: H ⫽ h

⫹ h

Continuity: Q ⫽ Q

⫽ Q

Method: (a) Assume h

(b) Calculate Q

and Q

⫽ Q

, the solution is correct

(d) If Q

⫽ Q

, repeat step (a)

2. For pipelines in parallel

Energy: H ⫽ h

⫽ h

Continuity: Q ⫽ Q

⫹ Q

Method: Parallel problem can be solved directly for Q

and Q

3. For branched pipelines

Energy: h

⫽ z

⫺ H

⫽ z

⫺ H

⫽ H

⫺ z

where H

is the energy head at junction J.

Continuity: Q

⫽ Q

⫹ Q

Method: (a) Assume H

(b) Calculate Q

, Q

, and Q

⫹ Q

⫽ Q

, the solution is correct

(d) If Q

⫹ Q

⫽ Q

, repeat step (a)

254 Chapter 4