Lin S.D. Water and Wastewater Calculations Manual

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the gas. For a perfect gas, changing from pressure P

and volume V

and V

at constant temperature, the following law exists:

⫽ P

(4.6)

According to the Boyle–Mariotte law for perfect gases, the product of

pressure P and volume V is constant in an isothermal process.

PV ⫽ nRT (4.7)

where P ⫽ pressure, pascal (P

) or lb/ft

V ⫽ volume, m

or ft

n ⫽ number of moles

R ⫽ gas constant

T ⫽ absolute temperature, K ⫽⬚C ⫹ 273, or ⬚R ⫽ ⬚F ⫹ 459.6

␳ ⫽ density, kg/m

or slug/ft

On a mole (M) basis, a pound mole (or kg mole) is the number of

pounds (or kg) mass of gas equal to its molecular weight. The product

MR is called the universal gas constant and depends on the units used.

It can be

The gas constant R is determined as

(4.8a)

or, in slug units

(4.8b)

For SI units

(4.8c)

Example: For carbon dioxide with a molecular weight of 44 at a pressure of

12.0 psia (pounds per square inch absolute) and at a temperature of 70⬚F

(21⬚C), compute R and its density.

R 5

8312

n N/kg

R 5

1545 3 32.2

lb/slug

R 5

1545

lb/lb

MR 5 1545 ft

lb/lb

P 5

RT 5 rRT

Fundamental and Treatment Plant Hydraulics 235

solution:

Step 1. Determine R from Eq. (4.8b)

Step 2. Determine density ␳

Based on empirical generation at constant pressure in a gaseous

system (perfect gas), when the temperature varies, the volume of gas

will vary approximately in the same proportion. If the volume is exactly

one mole of gas at 0⬚C (273.15 K) and at atmosphere pressure, then,

for ideal gases, these are the so-called standard temperature and

pressure, STP. The volume of ideal gas at STP can be calculated from

Eq. (4.7) (with R ⫽ 0.08206 L ⋅ atm/K ⋅ mol):

Thus, according to Avogadro’s hypothesis and the ideal-gas equation,

one mole of any gas will occupy 22.4l L at STP.

2 Water Flow in Pipes

2.1 Fluid pressure

Water and wastewater professionals frequently encounter some funda-

mentals of hydraulics, such as pressure, static head, pump head, veloc-

ity of flow, and discharge rate. The total force acting on a certain entire

space, commonly expressed as the force acting on unit area, is called

intensity of pressure, or simply pressure, p. The US customary of units

generally uses the pound per square inch (psi) for unit pressure. This

quantity is also rather loosely referred to simply as pounds pressure; i.e.

“20 pounds pressure” means 20 psi of pressure. The International

System of Units uses the kg/cm

(pascal) or g/cm

To be technically correct, the pressure is so many pounds or kilo-

grams more than that exerted by the atmosphere (760 mm of mercury).

However, the atmospheric pressure is ignored in most cases, since it is

applied to everything and acts uniformly in all directions.

V 5

nRT

s1 moleds0.08206 L

atm/mol

Kds273.15 Kd

1 atm

5 22.41 L

r 5

12 lb/in

144 in

/ft

s1130 ft

lb/slug

8Rds460 1 708Rd

5 0.00289 slug/ft

R 5

1545 3 32.2

lb/slug

1545 3 32.2

5 1130 ft

lb/slug

236 Chapter 4

2.2 Head

The term head is frequently used, such as in energy head, velocity head,

pressure head, elevation head, friction head, pump head, and loss of head

(head loss). All heads can be expressed in the dimension of length, i.e.

ft ⫻ lb/lb ⫽ ft, or m ⫻ kg/kg ⫽ m, etc.

The pump head equals the ft ⋅ lb (m ⋅ kg) of energy put into each

pound (kg) of water passing through the pump. This will be discussed

later in the section on pumps.

Pressure drop causes loss of head and may be due to change of veloc-

ity, change of elevation, or friction loss. Hydraulic head loss may occur

at lateral entrances and is caused by hydraulic components such as

valves, bends, control points, sharp-crested weirs, and orifices. These

types of head loss have been extensively discussed in textbooks and

handbooks of hydraulics. Bend losses and head losses due to dividing

and combining flows are discussed in detail by James M. Montgomery,

Consulting Engineers, Inc. (1985).

Velocity head. The kinetic energy (KE) of water with mass m is its

capacity to do work by reason of its velocity V and mass and is

expressed as .

For a pipe with mean flow velocity V (m/s or ft/s) and pipe cross-

sectional area A (cm

or sq. in), the total mass of water flowing through

the cross section in unit time is m ⫽ ␳VA , where ␳ is the fluid density.

Thus the total kinetic energy for a pipe flow is

(4.9)

The total weight of fluid W ⫽ mg ⫽ ␳AV g, where g is the gravitational

acceleration. It is commonly expressed in terms of energy in a unit

weight of fluid. The kinetic energy in unit weight of fluid is

(4.10)

This is the so-called velocity head, i.e. the height of the fluid column.

Example 1: Twenty-two pounds of water are moving at a velocity of 2 ft/s.

What are the kinetic energy and velocity head?

solution:

Step 1.

5 44

5 44 slug

ft/s ssince 1 slug 5 1 lb

ft/s

KE 5

3 22 lb 3 s2 ft/sd

rAV

rgAV

2 g

KE 5

srVAdV

rAV

Fundamental and Treatment Plant Hydraulics 237

Step 2.

Example 2: Ten kilograms of water are moving with a velocity of 0.61 m/s.

What are the kinetic energy and velocity head?

solution:

Step 1.

Step 2.

Note: Examples 1 and 2 are essentially the same.

Pressure head. The pressure energy (PE) is a measure of work done by

the pressure force on the fluid mass and is expressed as

PE ⫽ pAV (4.11)

where P ⫽ pressure at a cross section

A ⫽ pipe cross-sectional area, cm

or in

V ⫽ mean velocity

The pressure head (h) is the pressure energy in unit weight of fluid. The

pressure is expressed in terms of the height of the fluid column h. The

pressure head is

(4.12)

where ␥ is the weight per unit volume of fluid or its specific weight in

N/m

or lb/ft

. The general expression of unit pressure is

P ⫽ ␥h (4.13)

and the pressure head is h

⫽ p/␥ in ft or m.

h 5

pAV

rgAV

pressure

wt.

5 0.062 ft

5 0.019 m 3 3.28 ft/m

5 0.019

s0.61 m/s

2 3 9.81 m/s

5 1.86 N

m, since 1 N 5 1 kg

m/s

5 1.86 kg

5 1.86 m

m/s

KE 5

s10 kgds0.61 m/sd

s2 ft/sd

2 3 32.2 ft/s

5 0.062 ft

238 Chapter 4

Example: At the bottom of a water storage tank, the pressure is 31.2 lb/in

What is the pressure head?

solution:

Step 1. Convert the pressure to lb/ft

Step 2. Determine h

; for water

Elevation head. The elevation energy (EE) of a fluid mass is simply the

weight multiplied by the height above a reference plane. It is the work

to raise this mass W to elevation h and can be written as

(4.13a)

The elevation head, z, is EE divided by the total weight W of fluid:

(4.13b)

Example: (1) Ten kilograms and (2) 1 lb of water are at 50 ft above the

earth’s surface. What are their elevation heads?

solution:

Step 1.

For 10 kg

Step 2.

For 1 lb

Both have 50 ft of head.

1 lb 3 50 ft

1 lb

5 50 ft

10 kg 3 50 ft

10 kg

5 50 ft

Elevation head 5

5 z

EE 5 W z

31.2 3 144 lb/ft

62.4 lb/ft

5 72 ft

g 5 62.4 lb/ft

P 5 31.2 lb/in

5 31.2 lb/in

3 144 in

/ft

5 31.2 3 144 lb/ft

Fundamental and Treatment Plant Hydraulics 239

Bernoulli equation. The total energy (H ) at a particular section of water

in a pipe is the algebraic sum of the kinetic head, pressure head, and

elevation head. For sections 1 and 2, they can be expressed as

(4.14)

and

(4.15)

If water is flowing from section 1 to section 2, friction loss (h

) is the major

loss. The energy relationship between the two sections can be expressed

as the Bernoulli equation

(4.16)

where h

is the friction head. This is also called the continuity equation.

Example: A nozzle of 12-cm diameter is located near the bottom of a storage

water tank. The water surface is 3.05 m (10 ft) above the nozzle. Determine

(a) the velocity of effluent from the nozzle and (b) the discharge.

solution:

Step 1. (a) Determine V

Let point 1 be at the water surface and point 2 at the center of the nozzle. By

the Bernoulli equation

Since and

Step 2. (b) Solve for flow

Q 5 A

5 pr

5 3.14 s0.06 md

3 7.73 m/s

5 0.087 m

5 7.73 m/s

5 22gH

5 22 3 9.806 3 3.05

5 z

2 z

5 H 5 3.05 m

5 0P

5 P

5 0,

1 z

5 h

1 z

240 Chapter 4

Friction head. Friction head (h

) equals the loss of energy by each unit

weight of water or other liquid through friction in the length of a pipe,

in which the energy is converted into heat. The values of friction heads

are usually obtained from the manufacturer’s tables.

Darcy–Weisback equation. The Darcy–Weisback formula can be calcu-

lated from the friction head:

(4.17)

where h

⫽ head of friction loss, cm or ft

f ⫽ friction factor, dimensionless

L ⫽ length of pipeline, cm or ft

D ⫽ diameter of pipe, cm or ft

⫽ velocity head, cm or ft

V ⫽ average velocity of flow, cm/s or ft/s

Example 1: Rewrite the Darcy–Weisback formula for h

in terms of flow

rate Q instead of velocity V.

solution:

Step 1. Convert V to Q

In a circular pipe,

The volumetric flow rate may be expressed in terms of velocity and

area (A) as

Q ⫽ VA

then

Step 2. Substitute in the formula

(4.18)

5 f a

16Q

2gp

5 16Q

V 5

5 4Q/pD

A 5 psD/2d

5 pD

/4 5 0.785 D

5 f a

Fundamental and Treatment Plant Hydraulics 241

The head of loss due to friction may be determined in three ways:

1. The friction factor f depends upon the velocity of flow and the diam-

eter of the pipe. The value of f may be obtained from a table

in many hydraulic textbooks and handbooks, or from the Moody

chart (Fig. 4.1) of Reynolds number R versus f for various grades and

sizes of pipe R ⫽ DV/␯, where ␯ is the kinematic viscosity of the fluid.

2. The friction loss of head h

per 1000 ft of pipe may be determined

from the Hazen-Williams formula for pipe flow. A constant C for a

particular pipe, diameter of pipe, and either the velocity or the

quantity of flow should be known. A nomograph chart is most com-

monly used for solution by the Hazen-Williams formula.

3. Another empirical formula, the Manning equation, is also a popu-

lar formula for determining head loss due to free flow.

The friction factor f is a function of Reynolds number R and the rel-

ative roughness of the pipe wall e/D. The value of e (Table 4.2), the

roughness of the pipe wall (equivalent roughness), is usually deter-

mined from experiment.

For laminar pipe flow (R < 2000) f is independent of surface rough-

ness of the pipe. The f value can be determined from

f ⫽ 64/R (4.19)

When R > 2000 then the relative roughness will affect the f value. The

e/D values can be found from the manufacturer or any textbook. The

relationship between f, R, and e/D are summarized in graphical expres-

sion as the Moody diagram (Fig. 4.1).

242 Chapter 4

TABLE 4.2 Values of Equivalent Roughness

e for

New Commercial Pipes

Type of pipe e, in

Asphalted cast iron 0.0048

Cast iron 0.0102

Concrete 0.01–0.1

Drawn tubing 0.00006

Galvanized iron 0.006

PVC 0.00084

Riveted steel 0.04–0.4

Steel and wrought iron 0.0018

Wood stave 0.007

0.04

SOURCE: Benefield et al. (1984)

243

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

Example 2: A pumping station has three pumps with 1 MGD, 2 MGD, and

4 MGD capacities. Each pumps water from a river at an elevation of 588 ft above

sea level to a reservoir at an elevation of 636 ft, through a cast iron pipe of

24 in. diameter and 2600 ft long. The Reynolds number is 1600. Calculate the

total effective head supplied by each pump and any combinations of pumping.

solution:

Step 1. Write Bernoulli’s equation

Let stations 1 and 2 be river pumping site and discharge site at reservoir,

respective; h

is effective head applied by pump in feet.

Since V

⫽ V

is not given (usually negative), assumed zero

⫽ zero, to atmosphere

then z

⫹ h

⫽ z

⫽ h

⫽ (z

⫺ z

) ⫹ h

⫽ (636 ⫺ 588) ⫹ h

⫽ 48 ⫹ h

Step 2. Compute flow velocity (V ) for 1 MGD pump

Step 3. Find f, h

and h

From Eqs. (4.19) and (4.17),

5 0.04a

2600

s0.493d

2 3 32.2

5 f a

f 5 64/R 5 64/1600 5 0.04

5 0.493 ft/s

V 5

1.547 cfs

3.14 ft

5 1.547 ft

/s scfsd

gal 3 0.1337 ft

/gal

1 day 3 24 3 60 3 60 s/d

Q 5 1 MGD

A 5 pr

5 3.14s1 ftd

5 3.14 ft

1 z

1 h

1 z

1 h

244 Chapter 4