Handbook of Water and Wastewater Treatment Plant Operations - Chapter 4 pps

PART II

Water/Wastewater Operations:
Math and Technical Aspects

© 2003 by CRC Press LLC

Water and Wastewater
Math Operations

To operate a waterworks and/or a wastewater treatment
plant, and to pass the examination for an operator’s license,
you must know how to perform certain mathematical oper-
ations. However, do not panic, as Price points out, “Those
who have difficulty in math often do not lack the ability
for mathematical calculation, they merely have not
learned, or have not been taught, the ‘language of math.”

1

4.1 INTRODUCTION

Without the ability to perform mathematical calculations,
operators would have difficulty in properly operating water
and wastewater systems. In reality, most of the calculations
operators need to perform are not difficult. Generally, math
ability through basic algebra is all that is needed. Experi-
ence has shown that skill with math operations used in
water and wastewater system operations is an acquired skill
that is enhanced and strengthened with practice.

Note:

Keep in mind that mathematics is a language —
a universal language. Mathematical symbols
have the same meaning to people speaking
many different languages throughout the globe.
The key to learning mathematics is to learn the
language — the symbols, definitions and terms,
of mathematics that allow you to understand the
concepts necessary to perform the operations.
In this chapter, we assume the reader is well grounded
in basic math principles. We do not cover basic operations
such as addition, subtraction, multiplication, and division.
However, we do include, for review purposes, a few basic
math calculations in the Chapter Review Questions/Prob-
lems at the end of the chapter.

4.2 CALCULATION STEPS

As with all math operations, many methods can be suc-
cessfully used to solve water and wastewater system prob-
lems. We provide one of the standard methods of problem
solving in the following:
1. If appropriate, make a drawing of the informa-
tion in the problem.
2. Place the given data on the drawing.
3. Determine what the question is. This is the first
thing you should determine as you begin to
solve the problem, along with, “What are they

really looking for?” Writing down exactly what
is being looked for is always smart. Sometimes
the answer has more than one unknown. For
instance, you may need to find X and then find Y.
4. If the calculation calls for an equation, write it
down.
5. Fill in the data in the equation and look to see
what is missing.
6. Rearrange or transpose the equation, if necessary.
7. If available, use a calculator.
8. Always write down the answer.
9. Check any solution obtained.

4.3 TABLE OF EQUIVALENTS, FORMULAE,
AND SYMBOLS

In order to work mathematical operations to solve
problems (for practical application or for taking licensure
examinations), it is essential to understand the language,
equivalents, symbols, and terminology used.
Because this handbook is designed for use in practical
on-the-job applications, equivalents, formulae, and symbols
are included, as a ready reference, in Table 4.1.

4.4 TYPICAL WATER AND WASTEWATER
MATH OPERATIONS
4.4.1 A

RITHMETIC

A

VERAGE

(

OR

A

RITHMETIC

M

EAN

)

AND

M

EDIAN

During the day-to-day operation of a wastewater treatment
plant, considerable mathematical data are collected. The
data, if properly evaluated, can provide useful information
for trend analysis and indicate how well the plant or unit
process is operating. However, because there may be much
variation in the data, it is often difficult to determine trends

in performance.
Arithmetic average refers to a statistical calculation
used to describe a series of numbers such as test results.
By calculating an average, a group of data is represented
by a single number. This number may be considered typical
of the group. The arithmetic mean is the most commonly
used measurement of average value.
4

© 2003 by CRC Press LLC

TABLE 4.1
Equivalents, Formulae, and Symbols

Equivalents

12 in. = 1 ft
36 in. = 1 yd
144 in.

2

= 1 ft

2

9 ft

2

= 1 yd

2

43,560 ft

2

= 1 ac
1 ft

3

= 1728 in.

3

1 ft

3

H

2

0= 7.48 gal
1 ft

3

H

2

0= 62.4 lb
1 gal of H

2

0= 8.34 lb
1 L = 1.000 mL
1 g = 1.000 mg
1 MGD (million gal[MG]/d) = 694 gal/min, 1.545 ft

3

/sec
average BOD/capita/day = .17 lb
average SS/capita/day = .20
average daily flow = assume 100 gal/capita/day

Symbols

A= Area
V= Velocity
t= Time
SVI = Sludge Volume Index
v= Volume
#= Pounds
eff = Effluent

W= Width
D= Depth
L= Length
H= Height
Q= Flow
C= Circumference
r= Radius

p

= pi (3.14)
WA S= Waste activated sludge
RAS = Return activated sludge
MLSS = Mixed liquor suspended solids
MLVSS = Mixed liquor volatile suspended solids

Formulae

SVI =

¥

100
Q = A

¥

V
Detention time = v/Q
v = L

¥

W

¥

D
area = W

¥

L
Circular area =

p



¥

Diameter

2

C =

p

d

Hydraulic loading rate = Q/A
Sludge age =
Mean cell residence time =
Organic loading rate =

Source:

From Spellman, F.R.,

Standard Handbook for Wastewater Operators,

Vol. 1–3,
Technomic Publ., Lancaster, PA, 1999.
v
Concentration
# MLSS in aeration tank
# SS in primary eff d
# SS in secondary system aeration tank sec. clarifier
# WAS d + SS in eff d
+
()
# BOD d
v

© 2003 by CRC Press LLC

Note:

When evaluating information based on aver-
ages, remember that the average reflects the

general nature of the group and does not nec-
essarily reflect any one element of that group.
Arithmetic average is calculated by dividing the sum
of all of the available data points (test results) by the
number of test results:
(4.1)

E

XAMPLE

4.1

Problem:

Effluent biochemical oxygen demand (BOD) test results
for the treatment plant during the month of September are
shown below:
What is the average effluent BOD for the month of
September?

Solution:

E

XAMPLE

4.2

Problem:

For the primary influent flow, the following composite-
sampled solids concentrations were recorded for the week:

Solution:

E

XAMPLE

4.3

Problem:

A waterworks operator takes a chlorine residual measure-
ment every day. We show part of the operating log in
Table 4.2.
Find the mean.

Solution:

Add up the seven chlorine residual readings: 0.9 + 1.0 +
1.2 + 1.3 + 1.4 + 1.1 + 0.9 = 7.8. Next, divide by the
number of measurements (in this case 7): 7.8

∏

7 = 1.11.
The mean chlorine residual for the week was 1.11 mg/L.

Definition:

The median is defined as the value of
the central item when the data are arrayed by
size. First, arrange all of the readings in either
ascending or descending order. Then find the
middle value.

E

XAMPLE

4.4

Problem:

In our chlorine residual example, what is the median?

Solution:

Arrange the values in ascending order:
0.9, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4
The middle value is the fourth one — 1.1. Therefore, the
median chlorine residual is 1.1 mg/L. (Usually, the median
will be a different value than the mean.)
If the data contain an even number of values, you must
add one more step, since no middle value is present. You
must find the two values in the middle, and then find the
mean of those two values.

Test 1 20 mg/L
Test 2 31 mg/L
Test 3 22 mg/L
Test 4 15 mg/L
Monday 300 mg/L SS
Tuesday 312 mg/L SS
Wednesday 315 mg/L SS
Thursday 320 mg/L SS
Friday 311 mg/L SS
Saturday 320 mg/L SS
Sunday 310 mg/L SS

Total 2188 mg/L SS
Test 1 Test 2 Test 3 Test N

++++
()
L
Number of Tests Performed N
Average =
+++
=
20 mg L 31 mg L 22 mg L 15 mg L
22 mg L
4
Average SS
Sum
Number of M

of All Measurements

easurements Used
mg L SS
7
mg L SS
=
=
=
2188
312 6.

TABLE 4.2
Daily Chlorine Residual Results

Day Chlorine Residual (mg/L)

Monday 0.9
Tuesday 1.0
Wednesday 1.2
Thursday 1.3
Friday 1.4
Saturday 1.1
Sunday 0.9

Source:

From Spellman, F.R.,

Standard
Handbook for Wastewater Operators,

Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.

© 2003 by CRC Press LLC

E

XAMPLE

4.5

Problem:

A water system has four wells with the following capac-
ities: 115, 100, 125, and 90 gal/min. What are the mean
and the median pumping capacities?

Solution:

The mean is:
To find the median, arrange the values in order:
90 gal/min, 100 gal/min, 115 gal/min, 125 gal/min
With four values, there is no single middle value, so we
must take the mean of the two middle values:

Note:

At times, determining what the original num-
bers were like is difficult (if not impossible)
when dealing only with averages.

E

XAMPLE

4.6

Problem:

A water system has four storage tanks. Three of them have
a capacity of 100,000 gal each, while the fourth has a
capacity of 1 million gallons (MG). What is the mean
capacity of the storage tanks?

Solution:

The mean capacity of the storage tanks is:

Note:

Notice that no tank in Example 4.6 has a
capacity anywhere close to the mean. The
median capacity requires us to take the mean
of the two middle values; since they are both
100,000 gal, the median is 100,000 gal.
Although three of the tanks have the same
capacity as the median, these data offer no
indication that one of these tanks holds
1 million gal — information that could be
important for the operator to know.

4.4.2 R

ATIO

Recall that a ratio is the comparison of two numbers by
division or an indicated division. A ratio is usually stated
in the form A is to B as C is to D, and is written as two
fractions that are equal to each other:
(4.2)
We solve ratio problems by cross-multiplying; we
multiply the left numerator (A) by the right denominator
(D) and say that A is equal to the left denominator (B)
times the right numerator (C):
A

¥

D = B

¥

C
AD = BC (4.3)
If one of the four items is unknown, we solve the ratio
by dividing the two known items that are multiplied
together by the known item that is multiplied by the
unknown. This is best shown by a couple of examples:

E

XAMPLE

4.7

Problem:

If we need 4 lb of alum to treat 1000 gal of H

2

O, how
many pounds of alum will we need to treat 12,000 gal-
lons?

Solution:

We state this as a ratio: 4 lb of alum is to 1000 gallons of
H

2

O as pounds of alum (or

x

) is to 12,000 gal. We set this
up this way:
Cross-multiplying:

E

XAMPLE

4.8

Problem:

If 10 gal of fuel oil costs $5.25, how much does 18 gal
cost?
1 gal min + 1 gal min 25 gal min 0 gal min
1 gal min
15 00 1 9
4
07 5
++
=
.
100 115
2
07 5
gpm gpm
1 gpm
+
= .
1111
4
325 000
00, 000 + 00, 000 + 00, 000 + , 000,000
gal= ,

A
B
C
D
=
4
22

1000 gal O

12, 000 gal O
lb alum
H
lb alum
H
=
x
1000 4 12 000
4
48
¥=¥
=
¥
=
x
x
x
,
12, 000
1000

lb alum

© 2003 by CRC Press LLC

4.4.3 P

ERCENT

Percent (like fractions) is another way of expressing a part
of a whole. The term percent means per hundred, so a
percentage is the number out of 100. For example, 22%
means 22 out of 100, or if we divide 22 by 100, we get
the decimal 0.22:

4.4.3.1 Practical Percentage Calculations

Percentage is often designated by the symbol %. Thus,
10% means 10 percent, 10/100, or 0.10. These equivalents
may be written in the reverse order: 0.10 = 10/100 = 10%.
In water and wastewater treatment, percent is frequently
used to express plant performance and for control of
sludge treatment processes.

Note:

To determine percent divide the quantity you
wish to express as a percent by the total quantity
then multiply by 100.
Percent = (4.4)

E

XAMPLE

4.9

Problem:

The plant operator removes 6000 gal of sludge from the
settling tank. The sludge contains 320 gal of solids. What
is the percent of solids in the sludge?

Solution:

E

XAMPLE

4.10

Problem:

Sludge contains 5.3% solids. What is the concentration
of solids in decimal percent?

Solution:

Note:

Unless otherwise noted all calculations in the

handbook using percent values require the
percent be converted to a decimal before use.

Note:

To determine what quantity a percent equals
first convert the percent to a decimal then
multiply by the total quantity.
Quantity = Total

¥

Decimal Percent (4.5)

E

XAMPLE

4.11

Problem:

Sludge drawn from the settling tank is 8% solids. If 2400
gal of sludge are withdrawn, how many gallons of solids
are removed?

Solution:

E

XAMPLE

4.12

Problem:

Calcium hypochlorite (HTH) contains 65% available
chlorine. What is the decimal equivalent of 65%?

Solution:

Since 65% means 65 per hundred, divide 65 by 100
(65/100), which is 0.65.

E

XAMPLE

4.13

Problem:

If a 50-ft high water tank has 32 ft of water in it, how full
is the tank in terms of the percentage of its capacity?

Solution:

Thus, the tank is 64% full.
11
10 18 5 25

18 5 25
10
94 5
10
945
0 al
$5.25
8 al
$
gg
y
y
y
y
y
=
¥= ¥
=
¥
=
=
$.
$.
.
$.
22
22
022%.==
100
Quantity 100¥

Total
Percent
gal
=¥
=
320
100
53

6000 gal
.%
Decimal Percent
100
==
53
0 053
.%
.
gallons gal gal=¥ =
8
2400 192
%
100

32
50
064
064 100 64

ft
ft
decimal equivalent=
¥=
()
.
.%

© 2003 by CRC Press LLC
4.4.4 UNITS AND CONVERSIONS
Most of the calculations made in the water and wastewater
operations involve using units. While the number tells us
how many, the units tell us what we have. Examples of
units include: inches, feet, square feet, cubic feet, gallons,
pounds, milliliters, milligrams per liter, pounds per square
inch, miles per hour, and so on.
Conversions are a process of changing the units of a
number to make the number usable in a specific instance.
Multiplying or dividing into another number to change the
units of the number accomplishes conversions. Common
conversions in water and wastewater operations are:
1. Gallons per minute to cubic feet per second
2. Million gallons to acre-feet
3. Cubic foot per second to acre-feet
4. Cubic foot per second of water to weight
5. Cubic foot of H
2
O to gallons
6. Gallons of water to weight

7. Gallons per minute to million gallons per day
8. Pounds to feet of head (the measure of the pres-
sure of water expressed as height of water in
feet — 1 psi = 2.31 feet of head)
In many instances, the conversion factor cannot be
derived — it must be known. Therefore, we use tables
such as the one below (Table 4.3) to determine the com-
mon conversions.
Note: Conversion factors are used to change measure-
ments or calculated values from one unit of
measure to another. In making the conversion
from one unit to another, you must know two
things: (1) the exact number that relates the two
units, and (2) whether to multiply or divide by
that number
Most operators memorize some standard conversions.
This happens because of using the conversions, not
because of attempting to memorize them.
4.4.4.1 Temperature Conversions
An example of a type of conversion typical in water and
wastewater operations is provided in this section on tem-
perature conversions.
Note: Operators should keep in mind that temperature
conversions are only a small part of the many
conversions that must be made in real world
systems operations.
Most water and wastewater operators are familiar with
the formulas used for Fahrenheit and Celsius temperature
conversions:
(4.6)

(4.7)
These conversions are not difficult to perform. The
difficulty arises when we must recall these formulas from
memory.
Probably the easiest way to recall these important
formulas is to remember three basic steps for both Fahr-
enheit and Celsius conversions:
1. Add 40°.
2. Multiply by the appropriate fraction (5/9 or 9/5).
3. Subtract 40°.
Obviously, the only variable in this method is the
choice of 5/9 or 9/5 in the multiplication step. To make
the proper choice, you must be familiar with two scales.
On the Fahrenheit scale, the freezing point of water is 32°,
and 0° on the Celsius scale. The boiling point of water is
212° on the Fahrenheit scale and 100° on the Celsius scale.
TABLE 4.3
Common Conversions
Linear Measurements
1 in. = 2.54 cm
1 ft = 30.5 cm
1 m = 100 cm = 3.281 ft = 39.4 in.
1 ac = 43,560 ft
2
1 yd = 3 ft
Volume
1 gal = 3.78 L
1 ft
3
= 7.48 gal

1 L = 1000 mL
1 ac-ft = 43,560 ft
3
1 gal = 32 cups
1 lb = 16 oz dry wt.
Weight
1 ft
3
of water = 62.4 lb
1 gal = 8.34 lb
1 lb = 453.6 g
1 kg = 1000 g = 2.2 lb
1% = 10,000 mg/L
Pressure
1 ft of head = 0.433 psi
1 psi = 2.31 ft of head
Flow
1 ft
3
/sec = 448 gal/min
1 gal/min = 1440 gal/d
Source: From Spellman, F.R., Standard
Handbook for Wastewater Operators, Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.
∞ = ∞ - ∞
()
CF59 32
∞ = ∞ + ∞
()
FC59 32

© 2003 by CRC Press LLC
What does this mean? Why is it important?
Note, for example, that at the same temperature,
higher numbers are associated with the Fahrenheit scale
and lower numbers with the Celsius scale. This important
relationship helps you decide whether to multiply by 5/9
or 9/5. Let us look at a few conversion problems to see
how the three-step process works.
E
XAMPLE 4.14
Problem:
Convert 220°F to Celsius.
Solution:
Using the three-step process, we proceed as follows:
Step 1: Add 40°F:
220°F + 40°F = 260°F
Step 2: 260°F must be multiplied by either 5/9 or 9/5. Since
the conversion is to the Celsius scale, you will be moving
to number smaller than 260. Through reason and observa-
tion, obviously we see that multiplying 260 by 9/5 would
almost be the same as multiplying by 2, which would
double 260, rather than make it smaller. On the other hand,
multiplying by 5/9 is about the same as multiplying by 1/2,
which would cut 260 in half. Because we wish to move to
a smaller number, we should multiply by 5/9:
5/9 ¥ 260°F = 144.4°C
Step 3: Now subtract 40°:
144.4°C ¥ 40°C = 104.4°C
Therefore, 220°F = 104.4°C
EXAMPLE 4.15

Problem:
Convert 22°C to Fahrenheit.
Step 1: Add 40°F:
22°F + 40°F = 62°F
Step 2: Because we are converting from Celsius to Fahr-
enheit, we are moving from a smaller to larger number,
and should use 9/5 in the multiplication:
9/5 ¥ 62°F = 112°F
Step 3: Subtract 40°:
112°F – 40°F = 72°F
Thus, 22°C = 72°F
Obviously, knowing how to make these temperature
conversion calculations is useful. However, in practical
(real world) operations, you may wish to use a temperature
conversion table.
4.4.4.2 Milligrams per Liter (Parts per Million)
One of the most common terms for concentration is milli-
grams per liter (mg/L). For example, if a mass of 15 mg of
oxygen is dissolved in a volume of 1 L of water, the con-
centration of that solution is expressed simply as 15 mg/L.
Very dilute solutions are more conveniently expressed
in terms of micrograms per liter (µg/L). For example, a
concentration of 0.005 mg/L is preferably written as its
equivalent, 5 µg/L. Since 1000 µg = 1 mg, simply move
the decimal point three places to the right when converting
from mg/L to µL. Move the decimal three places to the
left when converting from µg/L, to mg/L. For example, a
concentration of 1250 µ/L is equivalent to 1.25 mg/L.
One liter of water has a mass of 1 kg. But 1 kg is
equivalent to 1000 g or 1,000,000 mg. Therefore, if we

dissolve 1 mg of a substance in 1 L of H
2
O, we can say
that there is 1 mg of solute per 1 million mg of water, or
in other words, 1 part per million (ppm).
Note: For comparative purposes, we like to say that
1 ppm is analogous to a full shot glass of water
sitting in the bottom of a full standard swim-
ming pool.
Neglecting the small change in the density of water
as substances are dissolved in it, we can say that, in gen-
eral, a concentration of 1 mg/L is equivalent to 1 ppm.
Conversions are very simple; for example, a concentration
of 18.5 mg/L is identical to 18.5 ppm.
The expression mg/L is preferred over ppm, just as
the expression µg/L is preferred over its equivalent of parts
per billion (ppb). However, both types of units are still
used, and the waterworks/wastewater operator should be
familiar with both.
4.5 MEASUREMENTS: AREAS
AND VOLUMES
Water and wastewater operators are often required to cal-
culate surface areas and volumes.
Area is a calculation of the surface of an object. For
example, the length and the width of a water tank can be
measured, but the surface area of the water in the tank
must be calculated. An area is found by multiplying two
length measurements, so the result is a square measure-
ment. For example, when multiplying feet by feet, we get
square feet, which is abbreviated ft

2
. Volume is the calcu-
lation of the space inside a three-dimensional object, and
is calculated by multiplying three length measurements,
© 2003 by CRC Press LLC
or an area by a length measurement. The result is a cubic
measurement, such as cubic feet (abbreviated ft
3
).
4.5.1 AREA OF A RECTANGLE
The area of square or rectangular figures (such as the one
shown in Figure 4.1) is calculated by multiplying the
measurements of the sides.
A = L ¥ W (4.8)
To determine the area of the rectangle shown in
Figure 4.1, we proceed as follows:
A = L ¥ W
A = 12 ft ¥ 8 ft
A = 96 ft
2
4.5.2 AREA OF A CIRCLE
The diameter of a circle is the distance across the circle
through its center, and is shown in calculations and on
drawings by the letter D (see Figure 4.2). Half of the
diameter — the distance from the center to the outside
edge — is called the radius (r). The distance around the
outside of the circle is called the circumference (C).
In calculating the area of a circle, the radius must be
multiplied by itself (or the diameter by itself); this process
is called squaring, and is indicated by the superscript

number 2 following the item to be squared. For example,
the radius squared is written as r
2
, which indicates to
multiply the radius by the radius.
When making calculations involving circular objects,
a special number is required — referred to by the Greek
letter pi (pronounced pie); the symbol for pi is p. Pi always
has the value 3.1416.
The area of a circle is equal to the radius squared times
the number pi.
A = p ¥ r
2
(4.9)
E
XAMPLE 4.16
Problem:
Find the area of the circle shown in Figure 4.2.
Solution:
A = p ¥ r
2
A = p ¥ 12.5 ft ¥ 12.5 ft
A = 3.1416 ¥ 12.5 ft ¥ 12.5 ft
A = 490.9 ft
2
At times, finding the diameter of a circular object is
necessary, under circumstances that allow you to measure
only the circumference (e.g., a pump shaft). The diameter
and the circumference are related by the constant p:
C = p ¥ Diameter or (4.10)

4.5.3 AREA OF A CIRCULAR OR CYLINDRICAL TANK
If you were supervising a work team assigned to paint a
water or chemical storage tank, you would need to know
the surface area of the walls of the tank. To determine the
tank’s surface area, visualize the cylindrical walls as a
rectangle wrapped around a circular base. The area of a
rectangle is found by multiplying the length by the width;
in this case, the width of the rectangle is the height of the
wall and the length of the rectangle is the distance around
the circle, the circumference.
The area of the sidewalls of a circular tank is found
by multiplying the circumference of the base (C = p ¥
Diameter) times the height of the wall (H):
A = p ¥ Diameter ¥ H (4.11)
FIGURE 4.1 Rectangular shape showing calculation of sur-
face area. (From Spellman, F.R., Spellman’s Standard Hand-
book for Wastewater Operators, Vol. 1–3, Technomic Publ.,
Lancaster, PA, 1999.)
L
W 8 ft
12 ft
FIGURE 4.2 Circular shape showing diameter and radius.
(From Spellman, F.R., Spellman’s Standard Handbook for
Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster,
PA, 1999.)
C = ?
r = 12.5 ft
D = 25 ft
C
p

© 2003 by CRC Press LLC

For a tank with Diameter = 20 ft and H = 25 ft:
To determine the amount of paint needed, remember
to add the surface area of the top of the tank, which in
this case we will say is 314 ft

2

. Thus, the amount of paint
needed must cover 1570.8 ft

2

+ 314 ft

2

= 1884.8 or 1885
ft

2

. If the tank floor should be painted, add another 314 ft

2

.

4.5.4 V

OLUME

C

ALCULATIONS

4.5.4.1 Volume of Rectangular Tank

The volume of a rectangular object (such as a settling tank
like the one shown in Figure 4.3) is calculated by multi-
plying together the length, the width, and the depth. To
calculate the volume, you must remember that the length
times the width is the surface area, which is then multi-
plied by the depth.
v = L

¥

W

¥

D (4.12)

E

XAMPLE

4.17

Problem:

Using the dimensions given in Figure 4.3, determine the
volume.

Solution:

Note:

For many calculations involving water, we need
to know the volume of the tank in gallons rather
than 1 ft

3

contains 7.48 gal.

4.5.4.2 Volume of a Circular or Cylindrical Tank

A circular tank consists of a circular floor surface with a
cylinder rising above it (see Figure 4.4). The volume of a
circular tank is calculated by multiplying the surface area
times the height of the tank walls.

E

XAMPLE

4.18

Problem:

If a tank is 20 ft in diameter and 25 ft deep, how many
gallons of water will it hold?

Solution:

In this type of problem, calculate the surface area first,
multiply by the height and then convert to gallons.

FIGURE 4.3

Rectangular settling tank illustrating calcula-
tion of volume. (From Spellman, F.R.,

Spellman’s Standard
Handbook for Wastewater Operators,

Vol. 1–3, Technomic
Publ., Lancaster, PA, 1999.)
ADH
Aftft
Aftft
Aft
=¥ ¥
=¥ ¥
=¥¥
=
p

p 20 25
3 1416 20 25
1570 8
2



.
.
vL W D
vA D
vftftft
vft
=¥ ¥
=¥
=¥¥
=
12 6 6
432
3


12 ft
L
D
6 ft
W
6 ft

FIGURE 4.4

Circular or cylindrical water tank. (From Spell-
man, F.R.,

Spellman’s Standard Handbook for Wastewater
Operators,

Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)
H = 25 ft
r = 10 ft
r Diameter ft ft
Ar
Aftft
Aft
= ∏ = ∏ =
=¥
=¥ ¥
=
220 210
10 10
314
2
2
p
p
vA H
vft ft
vftgal ft gal
=¥
=¥

=¥ =
314 25
7850 7 5 58 875
2
33


© 2003 by CRC Press LLC
4.5.4.3 Example Volume Problems
E
XAMPLE 4.19: TANK VOLUME (RECTANGULAR)
Problem:
Calculate the volume of the rectangular tank shown in
Figure 4.5.
Note: If a drawing is not supplied with the problem,
draw a rough picture or diagram. Make sure
you label or identify the parts of the diagram
from the information given in the question (see
Figure 4.5).
Solution:
E
XAMPLE 4.20: TANK VOLUME (CIRCULAR)
Problem:
The diameter of a tank is 70 ft. When the water depth is
30 ft, what is the volume of wastewater in the tank in
gallons?
Note: Draw a diagram similar to Figure 4.6.
Solution:
Note: Remember, the solution requires the result
in gallons; thus, we must include 7.48 ft

3
in
the operation to ensure the result in gallons.
E
XAMPLE 4.21: CYLINDRICAL TANK VOLUME
Problem:
A cylindrical tank is 12 ft in diameter and 24 ft in height.
What is the approximate capacity in liters?
Note: Draw a rough diagram of the tank and label
dimensions.
Solution:
Note: Remember, the problem asked for the tank
capacity in liters.
Convert 2713 ft
3
to liters:
E
XAMPLE 4.22: CHANNEL VOLUME CALCULATIONS
Note: Channels are commonly used in water and
wastewater treatment operations. Channels
are typically rectangular or trapezoidal in
shape. For rectangular channels, use:
v = L ¥ W ¥ D
Problem:
Determine the volume of wastewater (in ft
3
) in the section
of rectangular channel shown in Figure 4.7 when the
wastewater is 5 ft deep.
FIGURE 4.5 Rectangular tank. (From Spellman, F.R., Spell-

man’s Standard Handbook for Wastewater Operators, Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.)
50 ft
8 ft
12 ft
vft L W D
ft ft ft
ft
3
3
50 12 8
4800
()
=¥ ¥
=¥¥
=
v gal Diameter D gal ft
ft ft ft gal ft
gal
()
()
=¥ ¥¥
= ¥¥¥¥
=
0 785 7 48
0 785 70 70 30 7 48
863 155
23
3

,
FIGURE 4.6 Circular tank. (From Spellman, F.R., Spell-
man’s Standard Handbook for Wastewater Operators, Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.)
70 ft
30 ft
v
d
H
ft
ft
ft
k()



tan
=
¥
¥
=
¥
¥
=
()
p
p
2
2

3
4
12
4
24
2713
=¥ ¥
=
2713 7 48 3 785
76 810
3

,
gal ft L gal
l
© 2003 by CRC Press LLC
Solution:
E
XAMPLE 4.23: CHANNEL VOLUME (TRAPEZOIDAL)
where
B
1
= distance across the bottom
B
2
= distance across water surface
L = channel length
D = depth of water and wastewater
Problem:
Determine the volume of wastewater (in gallons) in a

section of trapezoidal channel when the wastewater depth
is 5 ft.
Given:
B
1
= 4 ft across the bottom
B
2
= 10 ft across water surface
L = 1000 ft
Solution:
E
XAMPLE 4.24: VOLUME OF CIRCULAR PIPELINE
Problem:
What is the capacity in gallons of wastewater of a 10-in.
diameter, 1800-ft section of pipeline (see Figure 4.8)?
Note: Convert 10 in. to feet (10 in./12 in./ft = .833 ft).
Solution:
EXAMPLE 4.25: PIPE VOLUME
Problem:
Approximately how many gallons of wastewater would
800 ft of 8-in pipe hold?
Note: Convert 8 in. to feet (8 in./12 in./ft = .67 ft).
Solution:
Convert: 282 ft
3
converted to gallons:
EXAMPLE 4.26: PIT OR TRENCH VOLUME
Note: Pits and trenches are commonly used in water
and wastewater plant operations. Thus, it is

important to be able to determine their vol-
umes. The calculation used in determining pit
or trench volume is similar to tank and channel
FIGURE 4.7 Open channel. (From Spellman, F.R., Spell-
man’s Standard Handbook for Wastewater Operators, Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.)
1000 ft
5 ft
4 ft
vft L W D ft ft ft
ft
3
3
600 6 5
18 000
()
=¥ ¥= ¥ ¥
= ,
vft
BB
DL
3
12
2
()
()
=
+
¥
v gal

BB
DL gal ft
ft ft
ft ft gal ft
gal
()
()
()
=
+
¥¥
=
+
¥¥
=¥¥ ¥
=
12
3
3
2
748
410
2
5 1000 7 48
751000 7 48
261 800
.
.
.
,

vft Diameter L
32
0785
()
=¥ ¥
FIGURE 4.8 Circular pipe. (From Spellman, F.R., Spell-
man’s Standard Handbook for Wastewater Operators, Vol.
1–3, Technomic Publ., Lancaster, PA, 1999.)
1800 ft
10 in.
v gal Diameter L gal ft
ft gal ft
gal
()
=¥ ¥¥
= ¥¥¥ ¥
=
0785 7 48
0 785 833 833 1800 7 48
7334
23
3
.
.
v
Diameter
L
ft
ft
=

¥
¥
=
¥
¥
=
p
p
2
2
3
4
067
4
800
282
.
=¥
=
282 7 48
2110
33
ft gal ft
gal
.
© 2003 by CRC Press LLC
volume calculations with one difference —
the volume is often expressed as cubic yards
rather than cubic feet or gallons.
In calculating cubic yards, typically two approaches are

used:
1. Calculate the cubic feet volume, then convert to cubic
yard volume.
2. Express all dimensions in yards so that the resulting
volume calculated will be cubic yards.
Problem:
A trench is to be excavated 3 ft wide, 5 ft deep, and 800 ft
long. What is the cubic yards volume of the trench?
Note: Remember, draw a diagram similar to the
one shown in Figure 4.9.
Solution:
Now convert ft
3
volume to yd
3
:
EXAMPLE 4.27: TRENCH VOLUME
Problem:
What is the cubic yard volume of a trench 600 ft long,
2.5 ft wide, and 4 ft deep?
Solution:
Convert dimensions in ft to yds before beginning the
volume calculation:
EXAMPLE 4.28: POND VOLUMES
Ponds and/or oxidation ditches are commonly used in
wastewater treatment operations. To determine the volume
of a pond (or ditch) it is necessary to determine if all four
sides slope or if just two sides slope. This is important
because the means used to determine volume would vary
depending on the number of sloping sides.

If only two of the sides slope and the ends are vertical,
we calculate the volume using the equation:
However, when all sides slope as shown in Figure 4.10,
the equation we use must include average length and
average width dimensions. The equation:
Problem:
A pond is 6 ft deep with side slopes of 2:1 (2 ft horizontal:
1 ft vertical). Using the data supplied in Figure 4.10,
calculate the volume of the pond in cubic feet.
FIGURE 4.9 Trench. (From Spellman, F.R., Spellman’s
Standard Handbook for Wastewater Operators, Vol. 1–3,
Technomic Publ., Lancaster, PA, 1999.)
vft L W D
yd
ft
ft yd
3
3
3
3
27
()
=¥ ¥
=

yds yds yds yd¥¥ =
()
3
vft L W D
ft ft ft

ft
3
3
800 3 5
12 000
()
=¥ ¥
=¥¥
= ,
=
=
12, 000 ft
27 ft
3
3
yd
yd
3
3
444
800 ft
3 ft
5 ft
L
ft
ft yd
yd
W
ft
ft yd

yd
D
ft
ft yd
yd
vyd L W D
yd yd yd
yd
==
==
==
=¥ ¥
=¥ ¥
=
()
600
3
200
25
3
083
5
3
167
200 0 83 1 67
277
3
3

.

.
.

vft
BB
DL
3
12
2
()
()
=
+
¥
v gal
BB
DL gal ft
()
()
=
+
¥¥
12
3
2
748.
© 2003 by CRC Press LLC
Solution:
4.6 FORCE, PRESSURE, AND HEAD
Force, pressure, and head are important parameters in

water and wastewater operations. Before we study calcu-
lations involving the relationship between force, pressure,
and head, we must first define these terms.
1. Force — the push exerted by water on any
confining surface. Force can be expressed in
pounds, tons, grams, or kilograms.
2. Pressure — the force per unit area. The most
common way of expressing pressure is in
pounds per square inch (psi).
3. Head — the vertical distance or height of water
above a reference point. Head is usually
expressed in feet. In the case of water, head and
pressure are related.
Figure 4.11 illustrates these terms. A cubical container
measuring one foot on each side can hold one cubic foot
of water. A basic fact of science states that 1 ft
3
H
2
O
weights 62.4 lb. The force acting on the bottom of the
container would be 62.4 lb. The pressure acting on the
bottom of the container would be 62.4 lb/ft
2
. The area of
the bottom in square inches is:
1 ft
2
= 12 in. ¥ 12 in. = 144 in.
2

(4.13)
Therefore the pressure in pounds per square inch (psi) is:
(4.14)
If we use the bottom of the container as our reference
point, the head would be one foot. From this we can see
that one foot of head is equal to 0.433 psi.
Note: In water and wastewater operations, 0.433 psi
is an important parameter.
Figure 4.12 illustrates some other important relation-
ships between pressure and head.
Note: Force acts in a particular direction. Water in a
tank exerts force down on the bottom and out
of the sides. Pressure acts in all directions. A
marble at a water depth of one foot would have
0.433 psi of pressure acting inward on all sides.
Key Point: 1 ft of head = 0.433 psi.
FIGURE 4.10 Pond. (From Spellman, F.R., Spellman’s Standard Handbook for Wastewater Operators, Vol. 1–3, Technomic
Publ., Lancaster, PA, 1999.)
320 ft 350 ft
Pond Bottom
600 ft
650 ft
v
LL WW
D
ft ft ft ft
ft
ft ft ft
ft
=

+
¥
+
¥
=
+
¥
+
¥
¥¥
()( )
()()
12 1 2
3
22
600 650
2
320 350
2
6
625 335 6
1 256 250


=
=, ,
FIGURE 4.11 One cubic foot of water weighs 62.4 lbs. (From
Spellman, F.R., Spellman’s Standard Handbook for Wastewater
Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)
62.4 lbs

of
water
1 ft
1 ft
1 ft
62.4 lb ft
1 ft
62.4 lb ft
144 in.
0 psi
2
2
2
2
==
ft
2
433.
© 2003 by CRC Press LLC
The parameter, 1 ft of head = 0.433 psi, is valuable
and should be committed to memory. You should also
know the relationship between pressure and feet of head —
in other words, how many feet of head 1-psi represents.
This is determined by dividing 1 by 0.433.
What we are saying here is that if a pressure gauge were
reading 12 psi, the height of the water necessary to represent
this pressure would be 12 psi ¥ 2.31 ft/psi = 27.7 ft.
Note: Again, the key points: 1 ft = 0.433 psi, and
1 psi = 2.31 ft
Having two conversion methods for the same thing is

often confusing. Thus, memorizing one and staying with
it is best. The most accurate conversion is: 1 ft =
0.433 psi — the standard conversion used throughout this
handbook.
E
XAMPLE 4.29
Problem:
Convert 50 psi to ft of head.
Solution:
Problem:
Convert 50 ft to psi.
Solution:
As the above examples demonstrate, when attempting
to convert psi to ft, we divide by 0.433; when attempting
to convert feet to psi, we multiply by 0.433. The above
process can be most helpful in clearing up the confusion
on whether to multiply or divide. Another way, however,
may be more beneficial and easier for many operators to
use. Notice that the relationship between psi and feet is
almost two to one. It takes slightly more than 2 ft to make
1 psi. Therefore, when looking at a problem where the
data are in pressure and the result should be in feet, the
answer will be at least twice as large as the starting num-
ber. For example, if the pressure were 25 psi, we intu-
itively know that the head is over 50 ft. Therefore, we
must divide by 0.433 to obtain the correct answer.
E
XAMPLE 4.30
Problem:
Convert a pressure of 55 psi to ft of head.

FIGURE 4.12 Shows the relationship between pressure and head. (From Spellman, F.R., Spellman’s Standard Handbook for
Wastewater Operators, Vol. 1–3, Technomic Publ., Lancaster, PA, 1999.)
1 ft
0.433 lb of water
1 sq. in. AREA
1 ft water = 0.433 psi
1 psi = 2.31 ft water
1 sq. in. AREA
2.31 ft
1 lb of water
feet of head
1 ft
0.433 psi
ft psi==231.
50 psi 1 ft
0.433 psi
ft
1
115 5¥=.
50 psi 0.433 psi
1 ft
2 .7
psi
1
1¥=
© 2003 by CRC Press LLC
Solution:
EXAMPLE 4.31
Problem:
Convert 14 psi to ft of head.

Solution:
E
XAMPLE 4.32
Problem:
Between the top of a reservoir and the watering point, the
elevation is 115 ft. What will the static pressure be at the
watering point?
Solution:
Using the preceding information, we can develop the
following equations for calculating pressure and head.
E
XAMPLE 4.33
Problem:
Find the pressure (psi) in a 12-ft deep tank at a point 15 ft
below the water surface.
Solution:
E
XAMPLE 4.34
Problem:
A pressure gauge at the bottom of a tank reads 12.2 psi.
How deep is the water in the tank?
Solution:
E
XAMPLE 4.35
Problem:
What is the pressure (static pressure) 4 mi beneath the
ocean surface?
Solution:
Change mi to ft, then to psi.
EXAMPLE 4.36

Problem:
A 50-ft diameter cylindrical tank contains 2.0 MG H
2
O.
What is the water depth? At what pressure would a gauge
at the bottom read in psi?
Step 1: Change MG to ft
3
:
Step 2: Using volume, solve for depth.
EXAMPLE 4.37
Problem:
The pressure in a pipe is 70 psi. What is the pressure in
feet of water? What is the pressure in lb/ft
2
?
Solution:
Step 1: Convert pressure to feet of water:
Step 2: Convert psi to psf:
55 psi 1 ft
0.433 psi
7 ft
1
12¥=
14 psi 1 ft
0.433 psi
32.3 ft
1
¥=
115 ft 0.433 psi

1 ft

psi
1
49 8¥=.
Pressure psi
()
=¥
()
0 433. Head ft
Head ft psi
()
=¥
()
231. Pressure
Pressure
rounded
psi ft
psi
()
()
=¥
=
0 433 5
217
.
.
Head ft psi
ft
()

()
=¥
=
231 122
28 2

. rounded
5 380 4 21 120
143
,,
,
ft mi ft
rounded
¥=
=
()
21,120 ft
2.31 ft psi
9 psi
2 000 000
748
267 380
3
,,
.
,
gal
ft=
v Diameter D
ft D

Dft
=¥ ¥
=¥¥
=
0 785
267 380 0 785 150
15 1
2
32
.
,.
.
70 2 31 161 7psi ft psi¥= ft of water
70 144 10 080
22 2
psi in ft lb ft¥=.,
© 2003 by CRC Press LLC
EXAMPLE 4.38
Problem:
The pressure in a pipeline is 6,476 lb/ft
2
. What is the head
on the pipe?
4.7 FLOW
Flow is expressed in many different terms (English System
of measurements). The most common flow terms are:
1. Gallons per minute (gal/min)
2. Cubic foot per second (ft
3
/sec)

3. Gallons per day (gal/d)
4. Million gallons per day (MGD)
In converting flow rates, the most common flow conver-
sions are: 1 ft
3
/sec = 448 gal/min and 1 gal/min = 1440 gal/d.
To convert gal/d to MGD, divide the gal/d by
1,000,000. For instance, convert 150,000 gallons to MGD:
In some instances, flow is given in MGD, but needed
in gal/min. To make the conversion (MGD to gal/min)
requires two steps.
1. Convert the gal/d by multiplying by 1,000,000
2. Convert to gal/min by dividing by the number
of minutes in a day (1440 min/d)
E
XAMPLE 4.39
Problem:
Convert 0.135 MGD to gal/min
Solution:
Step 1: First, convert the flow in MGD to gal/d:
Step 2: Now convert to gal/min by dividing by the number
of minutes in a day (24 hrs per day X 60 min per hour) =
1440 min/day:
In determining flow through a pipeline, channel or
stream, we use the following equation:
Q = A ¥ V
where
Q = cubic foot per second (ft
3
/sec)

A = area in square feet (ft
2
)
V = velocity in feet per second (ft/sec)
E
XAMPLE 4.40
Problem:
Find the flow in ft
3
/sec in an 8-in. line, if the velocity is
3 ft/sec.
Solution:
Step 1: Determine the cross-sectional area of the line in
square feet. Start by converting the diameter of the pipe
to inches.
Step 2: The diameter is 8 in.; therefore, the radius is 4 in.
4 in. is 4/12 of a foot or 0.33 ft.
Step 3: Find the area in square feet:
A = p ¥ r2
A = p ¥ 0.33 ft
2
A = p ¥ 0.109 ft
2
A = 0.342 ft
2
Step 4: Use the equation, Q = A ¥ V:
EXAMPLE 4.41
Problem:
Find the flow in gal/min when the total flow for the day
is 75,000 gal/d.

Head on pipe
Pressure
=
=¥
=¥
=
()
ft of pressure
Weight H
lb ft lb ft H
Hftrounded
6476 62 4
104
23
.
150, 000 gal d

1 000 000
0 150
,,
.= MGD
0 135 1 000 000 135 000.,,, MGD ¥=gal d
135, 000 gal d

r 94 gal min
1 440
93 8
, min
.
d

o=
QA V
Qft ft
Qft
=¥
=¥
=
30342
103
2
3
sec .
. sec
© 2003 by CRC Press LLC
Solution:
E
XAMPLE 4.42
Problem:
Find the flow in gal/min when the flow is 0.45 ft
3
/sec.
Solution:
4.7.1 FLOW CALCULATIONS
In water and wastewater treatment, one of the major con-
cerns of the operator is not only to maintain flow, but also
to measure it. Normally, flow measurements are deter-
mined by metering devices. These devices measure water
flow at a particular moment (instantaneous flow) or over
a specified time (total flow). Instantaneous flow can also
be determined mathematically. In this section, we discuss

how to mathematically determine instantaneous and aver-
age flow rates and how to make flow conversions.
4.7.1.1 Instantaneous Flow Rates
In determining instantaneous flows rates through channels,
tanks and pipelines, we can use Q = A ¥ V.
Note: It is important to remember that when using an
equation such as Q = A ¥ V the units on the
left side of the equation must match those units
on the right side of the equation (A and V) with
respect to volume (cubic feet or gallons) and
time (seconds, minutes, hours, or days).
E
XAMPLE 4.43
Problem:
A 4-ft wide channel has water flowing to a depth of 2 ft.
If the velocity through the channel is 2 ft/sec, what are
the cubic feet per second (ft
3
/sec) flow rate through the
channel?
Solution:
4.7.1.1.1 Instantaneous Flow into and out of
a Rectangular Tank
One of the primary flow measurements water and waste-
water operators are commonly required to calculate is flow
through a tank. This measurement can be determined
using the Q = A ¥ V equation. For example, if the dis-
charge valve to a tank were closed, the water level would
begin to rise. If you time how fast the water rises, this
would give you an indication of the velocity of flow into

the tank. This information can be plugged into Q = V ¥ A
to determine the flow rate through the tank. Let us look
at an example.
E
XAMPLE 4.44
Problem:
A tank is 8 ft wide and 12 ft long. With the discharge
valve closed, the influent to the tank causes the water level
to rise 1.5 ft in 1 min. What is the gal/min flow into the
tank?
Solution:
Step 1: First, calculate the ft
3
/min flow rate:
Step 2: Convert ft
3
/min flow rate to gal/min flow rate:
How do we compute flow rate from a tank when the
influent valve is closed and the discharge pump remains
on, lowering the wastewater level in the tank?
First, we time the rate of this drop in wastewater level
so that the velocity of flow from the tank can be calculated.
Then we use the Q = A ¥ V equation to determine the
flow rate out of the tank, as illustrated in Example 4.45.
E
XAMPLE 4.45
Problem:
A tank is 9 ft wide and 11 ft long. The influent valve to
the tank is closed and the water level drops 2.5 ft in 2
min. What is the gal/min flow from the tank?

Drop rate = 2.5 ft/2 min = 1.25 ft/min
75, 000 gal d

gal min
1 440
52
, min d
=
0.45 ft 448 gal d
1 ft
gal min
3
3
sec
sec1
202¥=
Qft A Vft
ft ft ft
ft
3
3
422
16
sec sec
sec
sec
()
()
=¥
=¥¥

=
Qft A Vft
ft ft ft
ft
3
3
81215
144
min min
. min
min
()
()
()
=¥
=¥ ¥
=
144 7 48 1077
33
ft gal ft galmin . min¥=
© 2003 by CRC Press LLC
Step 1: Calculate the ft
3
/min flow rate:
Step 2: Convert ft
3
/min flow rate to gal/min flow rate:
4.7.1.1.2 Flow Rate into a Cylindrical Tank
We can use the same basic method to determine the flow
rate when the tank is cylindrical in shape, as shown in

Example 4.46.
E
XAMPLE 4.46
Problem:
The discharge valve to a 25-ft diameter cylindrical
tank is closed. If the water rises at a rate of 12 inches in
4 minutes, what is the gal/min flow into the tank?
Solution:
Step 1: Calculate the ft
3
/min flow into the tank:
Step 2: Then convert ft
3
/min flow rate to gal/min flow rate:
4.7.1.2 Flow through a Full Pipeline
Flow through pipelines is of considerable interest to water
distribution operators and wastewater collection workers.
The flow rate can be calculated using the Q = A ¥ V. The
cross-sectional area of a round pipe is a circle, so the area,
A, is represented by 0.785 ¥ diameter
2
.
Note: To avoid errors in terms, it is prudent to express
pipe diameters as feet.
E
XAMPLE 4.47
Problem:
The flow through an 8-in. diameter pipeline is moving at
a velocity of 4 ft/sec. What is the ft
3

/sec flow rate through
the full pipeline?
Convert 8 in. to feet: 8 in./12 in. = 0.67 ft
4.7.2 VELOCITY CALCULATIONS
To determine the velocity of flow in a channel or pipeline
we use the Q = A ¥ V equation. However, to use the
equation correctly we must transpose it. We simply write
into the equation the information given and then transpose
for the unknown (V in this case), as illustrated in Example
4.48 for channels and 4.49 for pipelines.
E
XAMPLE 4.48
Problem:
A channel has a rectangular cross section. The channel is
5 ft wide with wastewater flowing to a depth of 2 ft. If
the flow rate through the channel is 8500 gal/min, what
is the velocity of the wastewater in the channel (ft/sec)?
Solution
Convert gal/min to ft
3
/sec:
EXAMPLE 4.49
Problem:
A full 8-in. diameter pipe delivers 250 gal/min. What is
the velocity of flow in the pipeline (ft/sec)?
Solution:
Convert: 8 in/12 in to feet = 0.67 ft
Qft A Vft
ft ft ft
ft

3
3
912125
124
min min
. min
min
()
()
=¥
=¥ ¥
=
124 7 48 928
33
ft gal ft galmin . min¥=
Rise in ft
ft
ft
==
=
=
12 1
14
025
.
min
. min
Qft A Vft
ft ft ft
ft

3
3
0 785 25 25 0 25
123
min min
min
min
()
()
=¥
= ¥¥¥
=
125 7 48 920
33
ft gal ft galmin . min¥=
Qft A Vft
ft ft ft
ft
3
3
0 785 0 67 0 67 4
14
sec sec
. sec
. sec
()
()
=¥
= ¥¥¥
=

8500 gal min
7.48 gal 60 sec
1 ft
3
¥
= 89. sec
Qfts A Vft
cfs ft ft ft
3
18 9 5 2
()
()
=¥
=¥ ¥
sec
. secx
V ( ft sec)
18.9
5 2
1. ft secx =
¥
= 89
© 2003 by CRC Press LLC
Convert gal/min to ft
3
/sec flow:
4.7.3 AVERAGE FLOW RATE CALCULATIONS
Flow rates in water and wastewater systems vary consid-
erably during the course of a day, week, month, or year.
Therefore, when computing flow rates for trend analysis

or for other purposes, an average flow rate is used to
determine the typical flow rate. Example 4.50 illustrates
one way to calculate an average flow rate.
E
XAMPLE 4.50
Problem:
The following flows were recorded for the week:
What was the average daily flow rate for the week?
Solution:
4.7.4 FLOW CONVERSION CALCULATIONS
One of the tasks involving calculations that the wastewater
operator is typically called on to perform involves con-
verting one expression of flow to another. The ability to
do this is also a necessity for those preparing for licensure
examinations.
Probably the easiest way in which to accomplish flow
conversions is to employ the box method illustrated in
Figure 4.13.
When using the box method it is important to remember
that moving from a smaller box to a larger box requires
multiplication by the factor indicated. Moving from a larger
box to a smaller box requires division by the factor indicated.
From Figure 4.13 it should be obvious that memoriz-
ing the 9 boxes and the units in each box is not that
difficult. The values of 60, 1440, 7.48, and 8.34 are not
that difficult to remember either — it is a matter of remem-
bering the exact placement of the units and the values.
Once this is accomplished, you have obtained a powerful
tool that will enable you to make flow conversions in a
relatively easy manner.

4.8 DETENTION TIME
Detention time (DT) is the length of time water is retained
in a vessel or basin, or the period from the time the water
enters a settling basin until it flows out the other end. To
calculate the detention period of a basin, the volume of
the basin must be first obtained. Using a basin 70 ft long,
25 ft wide and 12 ft deep, the volume would be:
Monday 8.2 MGD
Tuesday 8.0 MGD
Wednesday 7.3 MGD
Thursday 7.6 MGD
Friday 8.2 MGD
Saturday 8.9 MGD
Sunday 7.7 MGD
250 gal min
7.48 gal ft 60 sec min
ft
3
3
¥
= 056. sec
056 0785 0 67 0 67
3
. sec . . . secft ft ft ft= ¥¥¥x
V
0.56 ft
0.785 0.67 0.67
1. ft sec
3
=

¥¥
=
sec
6
Average Daily Flow
Total of All Sample Flows
Number of Days
MGD
d
MGD
=
=
=
55 9
7
80
.
.
FIGURE 4.13 Flow conversions using the box method. The
factors shown in the diagram have the following units associ-
ated with them: 60 sec/min, 1440 min/day, 7.48 gal./ft
3
, and
8.34 lbs/gal. (Adapted from Price, J.K., Applied Math for
Wastewater Plant Operators, Technomic Publ., Lancaster, PA,
1991, p. 32. With permission.)
60 1440
cfs cfm cfd
7.48 7.48 7.48
60 1440

gps gpm gpd
8.34 8.34 8.34
60 1440
lbs/sec lbs/min lbs/day
cfs = cubic feet per second gps = gallons per second
cfm = cubic feet per minute gpm = gallons per minute
cfd = cubic feet per day gpd = gallons per day
vL W D
vft ft ft
vft
=¥ ¥
=¥¥
=
70 25 12
21 000
3
,
Gallons V gal ft
Gallons gal
=¥
=¥=
748
21 000 7 48 157 080
3
.
,.,
© 2003 by CRC Press LLC

If we assume that, the plant filters 300 gal/min,
157,080 ÷ 300 = 524 min (rounded), or roughly 9 h of

detention time. Stated another way, the detention time is
the length of time theoretically required for the coagulated
water to flow through the basin.
If chlorine were added to the water as it entered the
basin, the chlorine contact time would be 9 h. That is, to
determine the contact time (CT) used to determine the
effectiveness of chlorine, we must calculate detention
time.

Key point:

True detention time is the “T” portion of
the CT value.

Note:

Detention time is also important when evaluat-
ing the sedimentation and flocculation basins
of a water treatment plant.
Detention time is expressed in units of time. The most
common are: seconds, minutes, hours, and days.
The simplest way to calculate detention time is to
divide the volume of the container by the flow rate into
the container. The theoretical detention time of a container
is the same as the amount of time it would take to fill the
container if it were empty.
For volume, the most common units used are gallons.
However, on occasion, cubic feet may also be used.
Time units will be in whatever units are used to
express the flow. For example, if the flow is in gal/min,

the detention time will be in days. If the detention time is
in the wrong time units, simply convert to the appropriate
units.

E

XAMPLE

4.51

Problem:

The reservoir for the community is 110,000 gal. The well
will produce 60 gal/min. What is the detention time in the
reservoir in hours?

Solution:

E

XAMPLE

4.52

Problem:

Find the detention time in a 55,000-gal reservoir if the
flow rate is 75 gal/min.

Solution:

E

XAMPLE

4.53

Problem:

If the fuel consumption to the boiler is 30 gal/d, how many
days will the 1000-gal tank last?

Solution:

4.8.1 H

YDRAULIC

D

ETENTION

T

IME

The term detention time or hydraulic detention time (HDT)
refers to the average length of time (theoretical time) a
drop of water, wastewater, or suspended particles remains
in a tank or channel. It is calculated by dividing the water

and wastewater in the tank by the flow rate through the
tank. The units of flow rate used in the calculation are
dependent on whether the detention time is to be calcu-
lated in seconds, minutes, hours or days. Detention time
is used in conjunction with various treatment processes,
including sedimentation and coagulation-flocculation.
Generally, in practice, detention time is associated
with the amount of time required for a tank to empty. The
range of detention time varies with the process. For exam-
ple, in a tank used for sedimentation, detention time is
commonly measured in minutes.
The calculation methods used to determine detention
time are illustrated in the following sections.

4.8.1.1 Detention Time in Days

The general hydraulic detention time calculation is:
(4.15)
This general formula is then modified based upon the
information provided or available and the normal range of
detention times for the unit being evaluated. Equation 4.28
shows another form of the general equation:
(4.16)
DT
h
==
=
110, 000 gal
60 gal min

or
1834 min
60 min h

1834
30 6
min
.
DT
h
==
=
55, 000 gal
75 gal min

or
634 min
60 min h

734
12
min
Days d==
1000 gal
30 gal d
33 3.
HDT =
Tank Volume
Flow Rate
HDT d

Tank Volume ft 7.48 gal ft
Flow gal d
33
()
=
()
¥
()

© 2003 by CRC Press LLC
EXAMPLE 4.54
Problem:
An anaerobic digester has a volume of 2,200,000 gal.
What is the detention time in days when the influent flow
rate is 0.06 MGD?
Solution:
4.8.1.2 Detention Time in Hours
(4.17)
EXAMPLE 4.55
Problem:
A settling tank has a volume of 40,000 ft.
3
What is the
detention time in hours when the flow is 4.35 MGD?
Solution:
4.8.1.3 Detention Time in Minutes
E
XAMPLE 4.56
Problem:
A grit channel has a volume of 1240 ft.

3
What is the
detention time in minutes when the flow rate is 4.1 MGD?
Solution:
Note: The tank volume and the flow rate must be
in the same dimensions before calculating
the hydraulic detention time.
4.9 CHEMICAL DOSAGE CALCULATIONS
Chemicals are used extensively in water and wastewater
treatment plant operations. Water and wastewater treat-
ment plant operators add chemicals to various unit pro-
cesses for slime-growth control, corrosion control, odor
control, grease removal, BOD reduction, pH control,
sludge-bulking control, ammonia oxidation, bacterial
reduction, fluoridation, and other reasons.
In order to apply any chemical dose correctly it is
important to be able to make certain dosage calculations.
One of the most frequently used calculations in water and
wastewater mathematics is the conversion of milligrams
per liter (mg/L) concentration to pounds per day (lb/d) or
pounds (lb) dosage or loading. The general types of mg/L
to lb/d or lb calculations are for chemical dosage, BOD,
chemical oxygen demand, or SS loading/removal, pounds
of solids under aeration and WAS pumping rate. These
calculations are usually made using either of the following
equation 4.31 or 4.32.
(4.19)
(4.20)
Note: If mg/L concentration represents a concentra-
tion in a flow, then MGD flow is used as the

second factor. However, if the concentration
pertains to a tank or pipeline volume, then MG
volume is used as the second factor.
4.9.1 CHLORINE DOSAGE
Chlorine is a powerful oxidizer commonly used in water
treatment for purification and in wastewater treatment for
disinfection, odor control, bulking control, and other
applications. When chlorine is added to a unit process, we
want to ensure that a measured amount is added.
In describing the amount of chemical added or
required two ways are used: (1) mg/L, and (2) lb/d.
In the conversion from mg/L (or ppm) concentration
to lbs/day, we use equation 4.33.
(4.21)
Note: In previous years it was normal practice to use
the expression parts per million (ppm) as an
expression of concentration, since 1 mg/L =
1 ppm. However, current practice is to use mg/L
as the preferred expression of concentration.
DT d

0.06 MGD 1, 000,000 gal MG
d
()
=
¥
=
2 200 000
37
,, gal

HDT h
Tank Volume ft 7.48 gal ft h d
Flow gal d
33
()
=
()
¥¥
()

24
DT h
40, 000 ft 7.48 gal ft h d
4.35 MGD 1,000, 000 gal MG
h
33
()
=
¥¥
¥
=
24
17.
HDT min
Tank Volume ft 7.48 gal ft min d
Flow gal d
33
()
=
()

¥¥
()

1440
(4.18)
DT min
1240 ft 7.48 gal ft min d
4,1000,000 gal d
minutes
33
()
=
¥¥
=
1440
326.
mg L MGD lb gal lb d¥¥= flow 8 34.
mg L MGD lb gal lb¥¥= volume 8 34.
mg L MGD lb d¥¥=834.
© 2003 by CRC Press LLC
EXAMPLE 4.57
Problem:
Determine the chlorinator setting (lb/d) needed to treat a
flow of 8 MGD with a chlorine dose of 6 mg/L.
Solution:
E
XAMPLE 4.58
Problem:
What should the chlorinator setting be (lb/d) to treat a
flow of 3 MGD if the chlorine demand is 12 mg/L and a

chlorine residual of 2 mg/L is desired?
Note: The chlorine demand is the amount of chlorine
used in reacting with various components of the
wastewater, such as harmful organisms and
other organic and inorganic substances. When
the chlorine demand has been satisfied, these
reactions stop.
Solution:
In order to find the unknown value (lb/d), we must first
determine chlorine dose.
Then we can make the mg/L to lb/d calculation:
4.9.2 HYPOCHLORITE DOSAGE
At many wastewater facilities sodium hypochlorite or cal-
cium hypochlorite are used instead of chlorine. The reasons
for substituting hypochlorite for chlorine vary. However,
with the passage of stricter hazardous chemicals regulations
under the Occupational Safety and Health Administration
and the U.S. Environmental Protection Association many
facilities are deciding to substitute the hazardous chemical
chlorine with nonhazardous hypochlorite. Obviously, the
potential liability involved with using deadly chlorine is
also a factor involved in the decision to substitute it with
a less toxic chemical substance.
For whatever reason the wastewater treatment plant
decides to substitute chlorine for hypochlorite, there are
differences between the two chemicals of which the waste-
water operator needs to be aware.
Chlorine is a hazardous material. Chlorine gas is used
in wastewater treatment applications at 100% available
chlorine. This is an important consideration to keep in

mind when making or setting chlorine feed rates. For
example, if the chlorine demand and residual requires
100-lb/d chlorine, the chlorinator setting would be just
that — 100-lb/24 h.
Hypochlorite is less hazardous than chlorine; it is sim-
ilar to strong bleach and comes in two forms: dry HTH
and liquid sodium hypochlorite. HTH contains about 65%
available chlorine; sodium hypochlorite contains about
12 to 15% available chlorine (in industrial strengths).
Note: Because either type of hypochlorite is not 100%
pure chlorine, more lb/day must be fed into the
system to obtain the same amount of chlorine
for disinfection. This is an important economic
consideration for those facilities thinking about
substituting hypochlorite for chlorine. Some
studies indicate that such a substitution can
increase operating costs, overall, by up to 3 times
the cost of using chlorine.
To calculate the lb/d hypochlorite required a two-step
calculation is used:
Step 1: Calculate the lb/d chlorine required using the mg/L
to lb/d equation:
(4.22)
Step 2: Calculate the lb/d hypochlorite required:
(4.23)
E
XAMPLE 4.59
Problem:
A total chlorine dosage of 10 mg/L is required to treat a
particular wastewater. If the flow is 1.4 MGD and the

hypochlorite has 65% available chlorine how many lb/d
of hypochlorite will be required?
mg L MGD lb d
mg L MGD lb gal lb d
lb d
¥¥=
¥¥ =
=
834
68834
400
.
.
mg L MGD lb d¥¥=834.
Cl
mg L mg L
mg L
Dose mg L Cl Demand mg L
Cl Residual mg L
() ()
()
=+
=+
=

12 2
14
12 3 8 34 300mg L MGD lb gal lb d¥¥ =.
mg L MGD lb d¥¥=834.
Cl lb d

% Available
lb d
()
¥=
()
100 Hypochlorite
© 2003 by CRC Press LLC
Solution:
Step 1: Calculate the lb/day chlorine required using the
mg/L to lb/d equation:
Step 2: Calculate the lb/d hypochlorite required. Since
only 65% of the hypochlorite is chlorine, more than 117
lb/d will be required:
EXAMPLE 4.60
Problem:
A wastewater flow of 840,000 gal/d requires a chlorine
dose of 20 mg/L. If sodium hypochlorite (15% available
chlorine) is to be used, how many lbs/day of sodium
hypochlorite are required? How many gal/day of sodium
hypochlorite is this?
Solution:
Step 1: Calculate the lb/day chlorine required:
Step 2: Calculate the lb/day sodium hypochlorite:
Step 3: Calculate the gal/d sodium hypochlorite:
EXAMPLE 4.61
Problem:
How many pounds of chlorine gas are necessary to
5,000,000 gal of wastewater at a dosage of 2 mg/L?
Solution:
Step 1: Calculate the pounds of chlorine required:

Step 2: Substitute the numbers into the equation:
4.10 PERCENT REMOVAL
Percent removal is used throughout the wastewater treat-
ment process to express or evaluate the performance of
the plant and individual treatment unit processes. The
results can be used to determine if the plant is performing
as expected or in troubleshooting unit operations by com-
paring the results with those listed in the plant’s operations
and maintenance manual. It can be used with either con-
centration or quantities (see Equations 4.24 and 4.25).
For concentrations use:
(4.24)
For quantities use:
Note: The calculation used for determining the per-
formance (percent removal) for a digester is
different from that used for performance (percent
removal) for other processes such as some pro-
cess residuals or biosolids treatment processes.
Ensure the right formula is selected.
E
XAMPLE 4.62
Problem:
The plant influent contains 259 mg/L BOD and the plant
effluent contains 17 mg/L BOD. What is the percentage
of BOD removal?
Solution:
4.11 POPULATION EQUIVALENT OR UNIT
LOADING FACTOR
When it is impossible to conduct a wastewater characteriza-
tion study and other data are unavailable, population equiv-

alent (PE) or unit per capita loading factors are used to
mg L MGD lb d
mg L MGD lb gal lb d
¥¥=
¥¥ =
834
10 1 4 8 34 117
.

117 lb d Cl
65% Available Cl
180 lb d hypochlorite¥=100
mg L MGD lb d mg L
MGD lb gal
lb d Cl
¥¥= ¥
¥
=
834 20
084 834
140
.
. .
140 lb d Cl
15% Available Cl
lb d hypochlorite¥=100 933
933 lb d
8.34 lb gal
1 gal d sodium hypochlorite= 12
v gal Cl

lb Cl
10 8 34
6
()
()
=¥
=
concentration mg L .
510 2 83483
6
¥¥¥=gal mg L lb Cl.
%

emoval
Influent Conc.
Influent Conc.
R
Eff Conc
=
-
[]
¥ 100
%
.
emoval
Influent Quantity uantity
Influent Quantity
R
Eff Q
=

-
[]
¥ 100
(4.25)
%.% emoval
259 mg L mg L
259 mg L
R =
-¥
=
[]
17 100
93 4
© 2003 by CRC Press LLC
estimate the total waste loadings to be treated. If the BOD
contribution of a discharger is known, the loading placed
upon the wastewater treatment system in terms of equivalent
number of people can be determined. The BOD contribution
of a person is normally assumed to be 0.17 lb BOD/d.
(4.26)
E
XAMPLE 4.63
Problem:
A new industry wishes to connect to the city’s collection
system. The industrial discharge will contain an average
BOD concentration of 349 mg/L and the average daily
flow will be 50,000 gal/d. What is the population equiv-
alent of the industrial discharge?
Solution:
Step 1: Convert flow rate to MGD:

Step 2: Calculate the population equivalent:
4.12 SPECIFIC GRAVITY
Specific gravity (sp gr) is the ratio of the density of a
substance to that of a standard material under standard
conditions of temperature and pressure. The standard
material for gases is air, and for liquids and solids, it is
water. Specific gravity can be used to calculate the weight
of a gallon of liquid chemical.
(4.27)
EXAMPLE 4.64
Problem:
The label states of the chemical states that the contents
of the bottle have a specific gravity of 1.4515. What is
the weight of 1 gal of solution?
Solution:
4.13 PERCENT VOLATILE MATTER
REDUCTION IN SLUDGE
The calculation used to determine percent volatile matter
(VM) reduction is complicated because of the changes
occurring during sludge digestion.
(4.28)
E
XAMPLE 4.65
Problem:
Using the digester data provided below, determine the
percent volatile matter reduction for the digester.
Data:
Raw Sludge Volatile Matter = 72%
Digested Sludge Volatile Matter = 51%
4.14 HORSEPOWER

In water and wastewater treatment operations, horsepower
(hp) is a common expression for power. One horsepower
is equal to 33,000 foot pounds (ft-lb) of work/min. This
value is determined, for example, for selecting a pump or
combination of pumps to ensure adequate pumping capac-
ity. Pumping capacity depends upon the flow rate desired
and the feet of head against which the pump must pump
(also known as effective height).
Calculations of horsepower are made in conjunction
with many treatment plant operations. The basic concept
from which the horsepower calculation is derived is the
concept of work.
Work involves the operation of a force (lb) over a
specific distance (ft). The amount of work accomplished
is measured in foot-pounds:
ft ¥ lb = ft-lb (4.29)
P
BOD Contribution lb d
E people

0.17 lb BOD d person
()
=
()
Flow ==
50, 000 gal d
gal MG
MGD
1 000 000
0 050

,,
.
P
MGD
E people
49 mg L lb mg L MG
0.17 lb BOD d person
people d
()
=
¥¥
=



30050 8 34
856
Chemical lb gal
Chemicals sp gr
()
=
()
¥HO lbgal
2

Weight lb gal
()
=¥
=
1 4515 8 34

12 1

.
lb gal
lb
%

%%
%%%
VM reduction
VM VM
VM VM VM
in
out
in in
out
()
=
-
[]
¥
-¥
()
[]
100
%
.
.
%VM reduction
.51

.72 .51
()
[]
()
[]
=
-¥
-¥
=
072 0 100
072 0 0
59
© 2003 by CRC Press LLC