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

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.. 4.2 CALCULATIO

PART II

Water/Wastewater Operations: Math and Technical Aspects

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

a universal language Mathematical symbolshave the same meaning to people speakingmany different languages throughout the globe

The key to learning mathematics is to learn thelanguage — the symbols, definitions and terms,

of mathematics that allow you to understand theconcepts 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 theyreally looking for?” Writing down exactly what

is being looked for is always smart Sometimesthe answer has more than one unknown Forinstance, you may need to find X and then find Y

4 If the calculation calls for an equation, write itdown

5 Fill in the data in the equation and look to seewhat 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 solveproblems (for practical application or for taking licensureexaminations), it is essential to understand the language,equivalents, symbols, and terminology used

Because this handbook is designed for use in practicalon-the-job applications, equivalents, formulae, and symbolsare 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 treatmentplant, considerable mathematical data are collected Thedata, if properly evaluated, can provide useful informationfor trend analysis and indicate how well the plant or unitprocess is operating However, because there may be muchvariation in the data, it is often difficult to determine trends

in performance

Arithmetic average refers to a statistical calculationused 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 commonlyused measurement of average value

4

TABLE 4.1 Equivalents, Formulae, and Symbols

Formulae

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

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

For the primary influent flow, the following

composite-sampled solids concentrations were recorded for the week:

measure-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.

the central item when the data are arrayed bysize First, arrange all of the readings in eitherascending or descending order Then find themiddle value

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.

mg L SS 7

Day Chlorine Residual (mg/L)

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:

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:

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

A ¥ D = B ¥ C

If one of the four items is unknown, we solve the ratio

by dividing the two known items that are multipliedtogether by the known item that is multiplied by theunknown 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 H2O, 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

H2O as pounds of alum (or x) is to 12,000 gal We set this

CD

lb alum H

,

12, 000

1000

lb alum

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

wish to express as a percent by the total quantity

then multiply by 100

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 thehandbook using percent values require thepercent be converted to a decimal before use

first convert the percent to a decimal thenmultiply 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?

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?

94 5 10

y

y y

4.4.4 U NITS AND C ONVERSIONS

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 H2O 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 Thedifficulty arises when we must recall these formulas frommemory

Probably the easiest way to recall these importantformulas is to remember three basic steps for both Fahr-enheit and Celsius conversions:

On the Fahrenheit scale, the freezing point of water is 32°,and 0° on the Celsius scale The boiling point of water is212° on the Fahrenheit scale and 100° on the Celsius scale

TABLE 4.3 Common Conversions

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

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°:

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 temperatureconversion calculations is useful However, in practical(real world) operations, you may wish to use a temperatureconversion table

4.4.4.2 Milligrams per Liter (Parts per Million)

One of the most common terms for concentration is grams per liter (mg/L) For example, if a mass of 15 mg ofoxygen 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

milli-in terms of micrograms per liter (µg/L) For example, aconcentration of 0.005 mg/L is preferably written as itsequivalent, 5 µg/L Since 1000 µg = 1 mg, simply movethe decimal point three places to the right when convertingfrom mg/L to µL Move the decimal three places to theleft when converting from µg/L, to mg/L For example, aconcentration of 1250 µ/L is equivalent to 1.25 mg/L.One liter of water has a mass of 1 kg But 1 kg isequivalent to 1000 g or 1,000,000 mg Therefore, if wedissolve 1 mg of a substance in 1 L of H2O, we can saythat 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 watersitting 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 asthe expression µg/L is preferred over its equivalent of partsper billion (ppb) However, both types of units are stillused, and the waterworks/wastewater operator should befamiliar with both

4.5 MEASUREMENTS: AREAS AND VOLUMES

Water and wastewater operators are often required to culate surface areas and volumes

cal-Area is a calculation of the surface of an object Forexample, the length and the width of a water tank can bemeasured, but the surface area of the water in the tankmust be calculated An area is found by multiplying twolength measurements, so the result is a square measure-ment For example, when multiplying feet by feet, we getsquare feet, which is abbreviated ft2 Volume is the calcu-lation of the space inside a three-dimensional object, and

is calculated by multiplying three length measurements,

or an area by a length measurement The result is a cubic

measurement, such as cubic feet (abbreviated ft3)

4.5.1 A REA OF A R ECTANGLE

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

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 ft2

4.5.2 A REA OF A C IRCLE

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 r2, 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

4.5.3 A REA OF A C IRCULAR OR C YLINDRICAL T ANK

If you were supervising a work team assigned to paint awater or chemical storage tank, you would need to knowthe surface area of the walls of the tank To determine thetank’s surface area, visualize the cylindrical walls as arectangle wrapped around a circular base The area of arectangle is found by multiplying the length by the width;

in this case, the width of the rectangle is the height of thewall and the length of the rectangle is the distance aroundthe 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):

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

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,

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 ft2 Thus, the amount of paint

needed must cover 1570.8 ft2 + 314 ft2 = 1884.8 or 1885

ft2 If the tank floor should be painted, add another 314 ft2

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

to know the volume of the tank in gallons ratherthan 1 ft3 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 acylinder rising above it (see Figure 4.4) The volume of acircular tank is calculated by multiplying the surface areatimes the height of the tank walls

2

2

p p

4.5.4.3 Example Volume Problems

E XAMPLE 4.19: T ANK V OLUME (R ECTANGULAR )

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: T ANK V OLUME (C IRCULAR )

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 ft3 in

the operation to ensure the result in gallons

E XAMPLE 4.21: C YLINDRICAL T ANK V OLUME

Note: Channels are commonly used in water and

wastewater treatment operations Channelsare typically rectangular or trapezoidal inshape 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.)

3

3

50 12 8 4800

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.)

E XAMPLE 4.23: C HANNEL V OLUME (T RAPEZOIDAL )

where

B1 = distance across the bottom

B2 = 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:

B1 = 4 ft across the bottom

B2 = 10 ft across water surface

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:

E XAMPLE 4.26: P IT OR T RENCH V OLUME

Note: Pits and trenches are commonly used in water

and wastewater plant operations Thus, it isimportant 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.)

,

v ft( )3 =0785¥Diameter2 ¥L

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

.

=

282 7 48 2110

ft gal ft gal

.

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 :

E XAMPLE 4.27: T RENCH V OLUME

E XAMPLE 4.28: P OND V OLUMES

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.)

yd yd

3 1 67

200 0 83 1 67 277

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 recalcu-lationship 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 ft3 H2O

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/ft2 The area of

the bottom in square inches is:

1 ft2 = 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 referencepoint, the head would be one foot From this we can seethat 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 ships between pressure and head

relation-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 Amarble at a water depth of one foot would have0.433 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

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

2 2

ft2 .433

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

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 aboveprocess 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 touse Notice that the relationship between psi and feet isalmost two to one It takes slightly more than 2 ft to make

1 psi Therefore, when looking at a problem where thedata are in pressure and the result should be in feet, theanswer 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, wemust 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.)

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?

Step 1: Convert pressure to feet of water:

Step 2: Convert psi to psf:

55 psi 1 ft 0.433 psi 7 ft

Pressure psi( )=0 433 ¥Head ft( )

Head ft( )=2 31 ¥Pressure ( )psi

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 (ft3/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 ft3/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 orstream, we use the following equation:

Q = A ¥ Vwhere

Q = cubic foot per second (ft3/sec)

A = area in square feet (ft2)

V = velocity in feet per second (ft/sec)