WSO Water Treatment Grade 1: Operator Math 1, Ch. 2

let's start with a few math Basics and

terms that are necessary when working

with water operations and

calculations this material will be

reviewed for some viewers so we'll move

through it fairly

quickly for example poers notation is

really just a shorthand way to show

multiplication 4 * 4 can be expressed as

4 the 2 power or 4 sared 4 to the 3

power or 4 cubed is 4 * 4 * 4

scientific notation is a method by which

any number can be expressed as a term

multiplied by a power of 10 examples of

numbers written in scientific notation

are 5.4 * 10 to the^ of 1 and 1.2 * 10

cubed the numbers can be taken out of

scientific notation by performing the

indicated multiplication such as 1.2 *

10 * 10 * 10 which equals

1,200 an easier way to take a number out

of scientific notation is by moving the

decimal point the number of places

indicated by the exponent a positive

exponent indicates a decimal point move

to the right as we just saw a negative

exponent indicates a decimal point move

to the

left dimensional analysis is a tool used

to determine whether a math problem has

been set up correctly by working only

with the dimensions or units of measure

and not the numbers

themselves to use dimension analysis you

must know how to express a horizontal

fraction as a vertical fraction how to

divide by a fraction and how to divide

out or cancel terms in the numerator and

denominator of a fraction when a problem

involves division by a fraction the rule

is to invert the numerator and

denominator of the fraction being

divided and then multiply for every term

canel in the numerator of a fraction a

similar term must be cancelled in the

denominator dimensional analysis is used

to determine if a particular formula

uses multiplication or Division and if

the equation will provide you an answer

in the desired

unit let's try dimensional analysis in

the following problem the flow rate in a

water line is 2.3 cubic feet per second

what is the flow rate expressed as

gallons per

minute now let's check the math setup by

converting the problem to a vertical

formula and by cancelling similar terms

you can see that the answer will

ultimately be in gallons per minute

getting the desired units indicates that

you've probably set up the problem

correctly another method for checking

your work before you get too far into a

calculation is by rounding and

estimating if you round the numbers and

estimate you can do a quick calculation

to make sure your specific or detailed

answer makes sense you'd be surprised

how many mistakes occur simply when

multiplying dividing or punching the

wrong button on a

calculator rounding means replacing the

final digits of a number with zeros thus

expressing the number as 10 hundreds

thousands or tenths hundreds and so on

when rounding whole numbers if the digit

is less than five replace all digits to

the right of the rounding place with

zeros if the digit to the right of that

place is five or greater increase the

digit in the rounding place by one and

replace all digits to the right of the

increase with zeros

when estimating the answer to a problem

round each number in the calculation so

that only one digit remains with the

rest of the digits in the number either

changed to zeros or dropped in

accordance with the rounding rules you

can add the number of zeros back to your

answer after your quick math to get an

appropriate estimation in this way the

estimation can often be done in your

head or at least with a minimum of

computing with a pencil or

calculator in treatment plan operations

we use equation for calculations such as

the detention time of a tank flow rate

in a channel filter loading rate or

chlorine dosage but to make these

calculations we need to know the values

for all but one of the terms of the

equation used for example let's look at

flow rate we know that the equation is

flow rate equals area time velocity to

solve this puzzle we need to know two of

the

values when solving unknown values most

of the equations we use in water

operations involve multiplication and

division from time to time we also

encounter equations with addition and

subtraction in the case of

multiplication and division you must

rearrange the terms so that X the

unknown value is by itself A good rule

to remember is that anything done to

both sides of the equation does not

change the

equation move x to the top or numerator

of the fraction if it's not already

there move other terms away from X to

the other side of the equal sign so that

X stands alone to move a term from one

side of the equation to the other move

from the top of one side to the bottom

of the other or from the bottom of one

side to the top of the other side this

is called the rearrangement

rule now let's look at equations

involving addition and subtraction such

as might be used when calculating

chlorine dosages the rule here is to

rearrange the terms so that X is

positive and by itself however since

there isn't a numerator or denominator

you can't crisscross terms instead you

move it from one side of the equation to

the other as you do you must change the

sign of the term if positive to negative

if negative to positive if there is no

sign in front of the number assume it is

positive once you have ensure that the

unknown value is a positive and by

itself you can solve the equation

here's how this type of equation is used

at a treatment facility chlorine dosage

equals chlorine demand plus chlorine

residual if you know that chlorine

dosage is 7 mg per liter and chlorine

residual is2 mg per liter then we can

determine that chlorine demand the

unknown value must be 6.8 mg per liter a

ratio is the relationship between two

numbers and may be written using a colon

or as a fraction

when the relationship between two

numbers in a ratio is the same as

between two other numbers in another

ratio they are said to be in proportion

a method used to determine if the ratios

are proportionate is to cross

multiply if the answers are the same

when numbers diagonally across from each

other are multiplied then the ratios are

proportionate sometimes only three of

the four numbers in a proportion are

known it is possible to determine the

unknown value by solving the rest of the

equation as shown

here in practical application problems

the ratios will not be arranged as in

the previous example you must know how

to set up the proportion before you can

solve the unknown value since a

proportion establishes a simple relation

between two ratios if any number in one

of the ratios changes then another must

also change to maintain

equality in many treatment plan

operations a change in one operating

condition produces a proportionate

change in some other condition for

example changing the speed of a metering

pump changes the quantity metered or

changing the amount of chemical added to

a certain volume of water changes the

strength of the solution if you

establish a relation or ratio for a

particular condition then when one of

the quantities changes you can calculate

what the new value of the other quantity

will be to set up a proportion first

analyze the problem to decide what is

the unknown then decide whether you

would expect the unknown value to be

larger or smaller than the known value

of the same

unit write all values including the

unknown factor in fraction form grouping

values having the same units such as

pounds and pounds time with time also be

sure to put the smaller value in the

numerator or top of each fraction and

the larger value in the denominator you

will need to follow the these three

rules first if the unknown is expected

to be smaller than the known value put

an X in the numerator of the first

fraction and put the known value of the

same unit in the

denominator the second rule is that if

the unknown is expected to be larger

than the known value put an X in the

denominator of the first fraction and

put the known value of the same unit in

the

numerator and finally make the two

remaining values of the problem into the

second fraction with the smaller value

in the numerator and the larger in the

denominator let's put these rules into

an actual water treatment problem if 5

pounds of chemical are mixed with 2,000

gallons of water to obtain a desired

solution how many pounds of chemical

would be mixed with 10,000 gallons of

water to obtain a solution of the same

concentration the unknown is the number

of pounds of chemical you would expect

the more water used the more chemical

would be needed so the unknown number of

pounds or X in our equation will be the

denominator and the formula would look

like this it's then possible to solve

the unknown value with the answer of 25

lb here's another problem and you can

pause the video to work on it if You'

like a pump working at 86% efficiency

can fill a certain tank in 49 minutes

how long will it take the pump to fill

the tank if the pump's efficiency is

only

59% the unknown Val Val is the time it

takes the pump to fill the tank at a

reduced efficiency and we would expect

the unknown number of minutes to be

larger than the known number of 49

minutes the known values of the other

unit are 59% and 86% so we arrang these

numbers into a fraction by which to

multiply 49 minutes the answer should be

larger than 49 minutes so the fraction

should be larger than 1 86 over 59 we

can now determine X by by multiplying

886 * 49 and then dividing it by 59 the

answer is 71 minutes