Lin S.D. Water and Wastewater Calculations Manual

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Groundwater 225

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Chapter

Fundamental and Treatment Plant

Hydraulics

1 Deﬁnitions and Fluid Properties 228

1.1 Weight and mass 228

1.2 Speciﬁc weight 228

1.3 Pressure 229

1.4 Viscosity of water 231

1.5 Perfect gas 234

2 Water Flow in Pipes 236

2.1 Fluid pressure 236

2.2 Head 237

2.3 Pipeline systems 254

2.4 Distribution networks 258

2.5 Sludge ﬂow 265

2.6 Dividing-ﬂow manifolds and multiport diffusers 265

3 Pumps 265

3.1 Types of pump 265

3.2 Pump performance 265

3.3 Cost of pumping 270

4 Water Flow in Open Channels 272

4.1 Che’zy equation for uniform ﬂow 272

4.2 Manning equation for uniform ﬂow 273

4.3 Partially ﬁlled conduit 276

4.4 Self-cleansing velocity 281

4.5 Speciﬁc energy 284

4.6 Critical depth 285

4.7 Hydraulic jump 288

5 Flow Measurements 291

5.1 Velocity measurement in open channel 291

5.2 Velocity measurement in pipe ﬂow 291

5.3 Discharge measurement of water ﬂow in pipes 292

5.4 Discharge measurements 296

References 306

227

1 Deﬁnitions and Fluid Properties

1.1 Weight and mass

The weight (W ) of an object, in the International System of Units (SI),

is defined as the product of its mass (m, in grams, kilograms, etc.) and

the gravitational acceleration ( g ⫽ 9.81 m/s

on the earth’s surface) by

Newton’s second law of motion: F ⫽ ma. The weight is expressed as

W ⫽ mg (4.1a)

The unit of weight is kg ⋅ m/s

and is usually expressed as newton (N).

In the SI system, 1N is defined as the force needed to accelerate 1kg

of mass at a rate of 1m/s

. Therefore

1N ⫽ 1 kg ⋅ m/s

⫽ 1 ⫻ 10

g ⋅ 10

cm/s

⫽ 10

g ⋅ cm/s

⫽ 10

dyn

In the US customary units, mass is expressed in slugs. One slug is

defined as the mass of an object which needs one pound of force to accel-

erate to one ft/s

, i.e.

(4.1b)

Example: What is the value of the gravitational acceleration (g) in the US

customary units?

solution:

g ⫽ 9.81 m/s

⫽ 9.81 m/s

⫻ 3.28 ft/m

⫽ 32.174 ft/s

Note: This is commonly used as g ⫽ 32.2 ft/s

1.2 Speciﬁc weight

The specific weight (weight per unit volume) of a fluid such as water, ␥,

is defined by the product of the density (␳) and the gravitational accel-

eration (g), i.e.

␥ ⫽ ␳g (in kg/m

⋅ m/s

)

⫽ ␳g (in N/m

) (4.2)

m 5

Wslbd

g sft/s

sin slugsd

228 Chapter 4

Water at 4⬚C reaches its maximum density of 1000 kg/m

or 1.000 g/cm

The ratio of the specific weight of any liquid to that of water at 4⬚C is

called the specific gravity of that liquid.

Example: What is the unit weight of water at 4⬚C in terms of N/m

and

dyn/cm

solution:

Step 1. In N/m

, ␳ ⫽ 1000 kg/m

␥ ⫽ ␳g ⫽ 1000 kg/m

⫻ 9.81 m/s

⫽ 9810 N/m

Step 2. In dyn/cm

, ␳ ⫽ 1.000 g/cm

␥ ⫽ 1 g/cm

⫻ 981 cm/s

⫽ 981 (g ⋅ cm/s

)/cm

⫽ 981 dyn/cm

1.3 Pressure

Pressure (P) is the force (F ) applied to or distributed over a surface

area (A) as a measure of force per unit area:

(4.3a)

In the SI system, the unit of pressure can be expressed as barye, bar,

N/m

, or pascal. One barye equals one dyne per square centimeter

(dyn/cm

). The bar is one megabarye, 10

dynes per square centime-

ter (10

barye). One pascal equals one newton per square meter

(N/m

). In the US customary units, the unit of pressure is expressed

as pounds per square inch, lb/in

(psi), or lb/ft

, etc.

Pressure is also a measure of the height (h) of the column of mer-

cury, water, or other liquid which it supports. The pressure at the

liquid surface is atmospheric pressure (P

). The pressure at any point

in the liquid is the absolute pressure (P

) at that point. The absolute

pressure is measured with respect to zero pressure. Thus, P

can be

written as

(4.3b)

where ␥ ⫽ specific weight of water or other liquid.

5 gh 1 P

P 5

Fundamental and Treatment Plant Hydraulics 229

In engineering work, gage pressure is more commonly used. The gage

pressure scale is designed on the basis that atmospheric pressure (P

)

is zero. Therefore, the gage pressure becomes:

P ⫽ ␥h (4.3c)

and pressure head

(4.3d)

According to Pascal’s law, the pressure exerted at any point on a con-

fined liquid is transmitted undiminished in all directions.

Example: Under normal conditions, the atmospheric pressure at sea level

is approximately 760-mm height of mercury. Convert it in terms of m of water,

N/m

, pascals, bars, and psi,

solution:

Step 1. Solve for m of water

Let ␥

and ␥

be specific weights of water and mercury, respectively, and h

and h

be column heights of water and mercury, respectively; then

approximately

␥

⫽ 1.00, ␥

⫽ 13.5936

From Eq. (4.3c):

Step 2. Solve for N/m

Since ␥

⫽ 9810 N/m

P ⫽ ␥

⫽ 9810 N/m

⫻ 10.33 m ⫽ 101,337 N/m

⫽ 1.013 ⫻ 10

N/m

Step 3. Solve for pascals

Since 1 pascal (Pa) ⫽ 1 N/m

, from Step 2:

P ⫽ 1.013 ⫻ 10

N/m

⫽ 1.013 ⫻ 10

Note: 1 atm approximately equals 10

N/m

or 10

Pa.

5 10.33 m sof waterd

13.5936

3 760 mm 5 10,331 mm

P 5 g

5 g

h 5

230 Chapter 4

Step 4. Solve for bars

Since approximately ␥

⫽ 981 dyn/cm

P ⫽ ␥

⫽ 981 dyn/cm

⫻ 1033 cm

⫽ 1,013,400 dyn/cm

⫽ 1.013 bars

Note: 1 atm pressure is approximately equivalent to 1 bar.

Step 5. Solve for lb/in

(psi)

From Step 1,

⫽ 10.33 m ⫽ 10.33 m ⫻ 3.28 ft/m

⫽ 33.88 ft of water

P ⫽ 33.88 ft ⫻ 1(lb/in

)/2.31 ft

⫽ 14.7 lb/in

1.4 Viscosity of water

All liquids possess a definite resistance to change of their forms, and

many solids show a gradual yielding to force (shear stress) tending to

change their forms. Newton’s law of viscosity states that, for a given rate

of angular deformation of liquid, the shear stress is directly propor-

tional to the viscosity. It can be expressed as

(4.4a)

where ␶ ⫽ shear stress

m ⫽ proportionality factor, viscosity

du/dy ⫽ velocity gradient

u ⫽ angular velocity

y ⫽ depth of fluid

The angular velocity and shear stress change with y.

The viscosity can be expressed as absolute (dynamic) viscosity or kine-

matic viscosity. From Eq. (4.4a), it may be rewritten as

(4.4b)

The viscosity m is frequently referred to as absolute viscosity, or mass

per unit length and time. In the SI system, the absolute viscosity m is

␮ 5

du/dy

t 5 ␮

5 1.013 3 10

dyn/cm

1 bar

1 3 10

dyn/cm

Fundamental and Treatment Plant Hydraulics 231

expressed as poise or in dyne ⋅ second per square centimeter. One poise

is defined as the tangent shear force (␶) per unit area (dyn/cm

) required

to maintain unit difference in velocity (1 cm/s) between two parallel

planes separated by 1 cm of fluid. It can be written as

Also

1 poise ⫽ 100 centipoise

The usual measure of absolute viscosity of fluid and gas is the cen-

tipoise. The viscosity of fluids is temperature dependent. The viscosity

of water at room temperature (20⬚C) is one centipoise. The viscosity of

air at 20⬚C is approximately 0.018 centipoise.

The British system unit of viscosity is 1 lb ⋅ s/ft

or 1 slug/ft ⋅ s.

Example 1: How much in British units is 1 poise absolute viscosity?

solution:

On the other hand:

1 lb ⋅ s/ft

⫽ 479 poise

⫽ 47.9 N ⋅ s/m

5 0.00209 lb

s>ft

5 0.02248 lb

s>10.758 ft

5 0.1 N 3

1 lb

4.448 N

s/m

s3.28 ftd

1 poise 5 0.1 N

s/m

5 0.1 N

s/m

5 1 dyn

s/cm

1 N

dyn

1 m

1 poise 5 1 dyn

s/cm

5 1 g/cm

5 1 dyn

s/cm

cm/s

1 dyn

1 poise 5 1 dyn

s/cm

232 Chapter 4

The kinematic viscosity ␯ is the ratio of viscosity (absolute) to mass

density ␳. It can be written as

␯⫽m/␳ (4.5)

The dimension of ␯ is length squared per unit time. In the SI system, the

unit of kinematic viscosity is the stoke. One stoke is defined as

1 cm

/s. However, the standard measure is the centistoke (⫽10

⫺2

stoke or

⫺2

/s). Kinematic viscosity offers many applications, such as the

Reynolds number R ⫽ VD/␯. For water, the absolute viscosity and kine-

matic viscosity are essentially the same, especially when the tempera-

ture is less than 10⬚C. The properties of water in SI units and British units

are respectively given in Tables 4.1a and 4.1b.

Fundamental and Treatment Plant Hydraulics 233

TABLE 4.1b Physical Properties of Water—British Units

Absolute Kinematic

Specific viscosity viscosity Surface Vapor

Temperature Density ␳, weight m ⫻ 10

␯ ⫻ 10

tension pressure

T, ⬚F slug/ft

␥, lb/ft

lb ⋅ s/ft

/s ␴, lb/ft P

, psia

32 1940 62.42 3.746 1.931 0.00518 0.09

40 1938 62.43 3.229 1.664 0.00514 0.12

50 1936 62.41 2.735 1.410 0.00509 0.18

60 1934 62.37 2.359 1.217 0.00504 0.26

70 1931 62.30 2.050 1.059 0.00498 0.36

80 1927 62.22 1.799 0.930 0.00492 0.51

90 1923 62.11 1.595 0.826 0.00486 0.70

100 1918 62.00 1.424 0.739 0.00480 0.95

120 1908 61.71 1.168 0.609

140 1896 61.38 0.981 0.514

160 1890 61.00 0.838 0.442

180 1883 60.58 0.726 0.385

200 1868 60.12 0.637 0.341

212 1860 59.83 0.593 0.319

SOURCES: Benefield et al. (1984) and Metcalf & Eddy, Inc. (1972)

TABLE 4.1a Physical Properties of Water—Sl Units

Specific Absolute Kinematic Surface Vapor

Temperature Specific weight viscosity ␮, viscosity tension pressure

T, °C gravity ␥, N/m

N ⋅ s/m

␯, m

/s ␴, N/m

, N/m

0 0.9999 9805 0.00179 1.795 ⫻ 10

⫺6

0.0756 608

4 1.0000 9806 0.00157 1.568 ⫻ 10

⫺6

0.0750 809

10 0.9997 9804 0.00131 1.310 ⫻ 10

⫺6

0.0743 1226

15 0.9990 9798 0.00113 1.131 ⫻ 10

⫺6

0.0735 1762

21 0.9980 9787 0.00098 0.984 ⫻ 10

⫺6

0.0727 2504

27 0.9966 9774 0.00086 0.864 ⫻ 10

⫺6

0.0718 3495

38 0.9931 9739 0.00068 0.687 ⫻ 10

⫺6

0.0700 6512

93 0.9630 9444 0.00030 0.371 ⫻ 10

⫺6

0.0601 79,002

SOURCE: Brater et al. (1996)

Example 2: At 21⬚C, water has an absolute viscosity of 0.00982 poise and a

specific gravity of 0.998. Compute (a) the absolute (N ⋅ s/m

) and kinematic

viscosity in SI units and (b) the same quantities in British units.

solution:

Step 1.

For (a), at 21⬚C (69.8⬚F)

Step 2.

For (b), from Step 1 and Example 1

1.5 Perfect gas

A perfect gas is a gas that satisfies the perfect gas laws, such as the

Boyle–Mariotte law and the Charles–Gay-Lussac law. It has internal

energy as a function of temperature only. It also has specific heats with

values independent of temperature. The normal volume of a perfect gas

is 22.4136 liters/mole (commonly quoted as 22.4 L/mol).

Boyle’s law states that at a constant temperature, the volume of a

given quantity of any gas varies inversely as the pressure applied to

n 5 1.059 3 10

n 5 0.984 3 10

/s 3 a

3.28 ft

1 m

5 2.05 3 10

slug/ft

m 5 0.000982 N

s/m

1 slug/ft

47.9 N

s/m

5 0.984 3 10

n 5

␮

0.000982 kg/m

0.998 3 1000 kg/m

5 0.000982 kg/m

5 0.000982, N

s/m

1 kg

m/s

1 N

5 0.000982 N

s/m

2 

␮ 5 0.00982 poise 5 0.00982 poise 3

0.1 N

s/m

1 poise

234 Chapter 4