study unit power plant water treatment, part 2

Study Unit

Power Plant WaterTreatment, Part 2By

Megan Brummett Assistant Professor of Power Plant TechnologyJohnson County Community CollegeOverland Park, Kansas

Copyright © 2003 by Education Direct, Inc.

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About the Author

Megan Brummett is the career

program facilitator and professor

for the power plant technology

program at Johnson County

Community College located in

Overland Park, Kansas. She devel-

oped the power plant technology

program offered at the college in

2001. Brummett writes textbooks,

college course curriculum, and

online training material. She has a B.S. in engineering from

Kansas State University and an MBA from Baker University.

Before embarking on a career in academia, she was part of

a design team that built power plants all over the globe.

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Water treatment practices and quality specifications depend

on the intended use for the water. Generally, the discussion

of water quality can be divided into two groups, water treat-

ment and wastewater treatment. Water treatment typically

refers to the treatment of safe water for public use, with

“drinking water” implied. Wastewater treatment refers to the

treatment of wastewater from either municipal or industrial

sources and limits treatment to the amount of contamina-

tion that can safely be discharged into the environment.

This study unit’s objectives fall under the heading of water

treatment, but water treatment for a power plant can be

even further divided by the water’s intended use. There are

three major categories of water usage in a power plant: cir-

culating, or cooling water, drinking, or service water, and

feed, or boiler water. The water quality requirements for

each are substantially different. The circulating or cooling

water needs the least amount of treatment to meet the

requirements of its intended use. Drinking, or service, water

in a power plant is typically equivalent to potable water

treatment in municipal water treatment plants. It requires

more treatment than cooling water but has less stringent

standards than boiler water. Before we study the process

used to treat the different water systems within a power

plant, we need to become familiar with the terminology and

methods for quantifying water quality. This study unit will

discuss the terminology associated with water quality and

the treatment process in power plants up to and including

the supply of service water.

When you complete this study unit, you’ll be able to • Distinguish between water quality characteristics and water

quality parameters

• Use the terminology associated with the water treatmentprocess

• Describe the first six steps in a power plant’s water treat-ment process

• Perform chemical calculations to determine the amount ofsludge produced by the front-end process

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WATER QUALITY CHARACTERISTICS 1

Physical Characteristics 2Chemical Characteristics 6Biological Characteristics 7

WATER QUALITY PARAMETERS 11

Parameters Typically Used in a Power Plant 12Turbidity 12Hardness 14Total Dissolved Solids 16Conductivity 17Cation Conductivity 19pH 20Silica Content 20Sodium Content of Steam and Water 21Dissolved Gases 22

FRONT-END WATER TREATMENT 25

Screening 25Discrete Settling 28Coagulation 29Flocculation 32Settling 32Clarifiers 34Filtration 35

CHEMICAL CALCULATIONS FOR WATER TREATMENT PLANTS 41

Calculating Sludge on a Dry Weight Basis 42Calculating Sludge on a Wet Weight Basis 48

SELF-CHECK ANSWERS 57

EXAMINATION 59

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1

WATER QUALITY CHARACTERISTICS

Water is known as the universal solvent because of its ability

to dissolve so many different materials. In its purest form,

H2O is highly corrosive and is sometimes called hungry

water. Due to its natural ability to dissolve minerals, it will

eat away at copper, iron, or lead in piping systems and will

naturally absorb minerals and other materials in the earth.

However, pure water is never found in nature. Water always

contains some level of contamination. The question is, what

level of contamination is safe or acceptable for the intended

use of the water? To define an acceptable level of contamina-

tion, we must first look at the characteristics by which we

define contaminants.

Note: Don’t worry if you notice redundancies between

terms like water quality characteristics and water quality

parameters, which are discussed in the next section. You

may also note some redundancy between the types of

characteristics. That’s because a physical characteristic

may also be a chemical characteristic, and vice versa. Just

remember that characteristics and parameters aren’t exclu-

sive. A characteristic can fit into two separate categories,

and a characteristic can also be a parameter (Figure 1).

There are three broad categories of water characteristics:

physical, biological, and chemical. As broad, general classifi-

cations, they will serve as a good starting point. Changes

in a characteristic in one category often will affect the

characteristics in another category. We’ll discuss these

secondary changes after we define the major characteristics.

Power Plant Water Treatment, Part 2

While a characteristic

is most often a distin-

guishing trait or quality,

a parameter is most

often a quantifiable

value used to describe

the extent to which a

characteristic applies to

an object or process.


Physical Characteristics

The physical characteristics of water include color, taste,

odor, suspended solids, turbidity, total suspended solids,

and total dissolved solids. Color, taste, and odor are terms

we’re all familiar with, and they play an important role in

the quality of drinking water. Would you drink a glass of

brown water that smelled of rotten eggs? What if a chemist

assured you that the water was perfectly safe to drink, and

that lab results showed it contained no harmful substances

in excess of safe limits? The chemist’s opinion wouldn’t

matter, because you would probably say, “There’s no way

I’m drinking that stuff!”

Now, let’s look at some less familiar characteristics of water

beyond color, taste, and odor.

Suspended solids (SS) are those solid particles that are held

or suspended in water. They include sand, grit, clay, organic

matter, and any other solid that wouldn’t dissolve. For exam-

ple, if you dropped a carrot in a tub of water, it would float

2

Water Qualities Characteristics Parameters

Physical

Color *

Taste *

Turbidity * *

Total suspended solids * *

Total dissolved solids * *

Chemical

Chemicals * *

Dissolved gases * *

Biological

Algae *

Bacteria * *

Viruses * *

Miscellaneous

Hardness *

Conductivity *

Cation conductivity *

pH level * *

Sodium content *

Silica content *

Temperature *

FIGURE 1—Classification systems often have redundancies between the characteristics of theobjects being classified andthe parameters by which theobjects are judged.

Power Plant Water Treatment, Part 2 3

(it certainly wouldn’t dissolve!). It would be a suspended

solid. But the term usually refers to very small particles.

Chemists don’t spend much time trying to remove carrots

from water, but this example should help you remember the

definition. Sand, like carrots, doesn’t dissolve into the water

but is suspended in it.

Tiny suspended solids cause turbidity, or cloudiness, in

water (Figure 2). If you fill a clear glass with water and stir

in some dirt from your backyard, you would have a glass of

muddy water. If you let it stand undisturbed for 15 minutes,

you would see that most of the dirt had settled to the bottom

of the glass, but there would still be some particles suspend-

ed in the water. Particles that stay suspended are typically

fine clay, organic, or fecal material. These tiny particles do

not weigh enough to settle to the bottom under the force of

gravity. Eventually, these particles clump together. These

suspended solids called colloids are the cause of turbidity

(Figure 3).

The term Total suspended solids (TSS) refers to the sum of all

the solids suspended in the water or solution. In our exam-

ple, it would be the total weight of the dirt divided by the

volume of the water. The technical definition of TSS is the

concentration of material that can be filtered out of a solu-

tion. It’s calculated by taking the dry weight of the material

collected on a filter and dividing it by the original volume of

the sample (Figure 4). The units used are milligrams per

liter (mg/l).

Floating Particles(Suspended Solids)

Settled Dirt

FIGURE 2—Turbidity is the cloudiness caused by tinyfloating particles in water.


_

_

FIGURE 3—Colloids are clumps oftiny particles whose negativecharges prevent them from settlingto the bottom.

FilteredDry Sample

Scale

1 Liter

Volume ofOriginal Sample

= 2 mg

1 Liter

FIGURE 4—Total suspendedsolids are calculated bydividing the filtered dryweight by the volume of theoriginal sample


To determine the amount of TSS in a solution, you would

measure out a quantity of the water to be tested. For simplic-

ity we’ll use one liter of the water to be tested. Put the filter

on the scale you’ll later use to weigh the dry sample and zero

the scale. Next, carefully measure one liter of water and run

it through the filter. Once all the solution has run through,

carefully remove the filter from the apparatus that held it

and place it an oven to dry at 217°F (103°C). Finally, weigh

the dry sample and record the weight. For our example, let’s

say the dry sample weighed 2 mg (milligrams). The TSS is

calculated by taking the dry weight of the filtered sample

divided by the original volume of the tested sample, in our

case one liter of water, so the concentration of TSS in our

example is 2 mg/l.

Total dissolved solids (TDS) differ from suspended solids

because they dissolve in water. The technical definition is the

amount of solids that can be removed by a fine filter. Table

salt, or sodium chloride (NaCl), readily dissolves in water but

is otherwise a solid. Sodium chloride is the chemical name

for common table salt. If you mix a teaspoon of salt into a

16-ounce glass of water, it dissolves. If you continue to add

salt, you’ll eventually reach a point where the salt won’t all

dissolve. The dissolved amount of salt would be considered

dissolved solids and the salt in the bottom of the glass would

be suspended solids.

This limit to the amount of a solid that will dissolve is called

its solubility. The solubility of a substance is defined as the

maximum amount of the substance that can dissolve into a

given amount of solvent (water, in our case). The solubility of

a substance is defined for a specific temperature and specific

amount of solvent. Sodium chloride’s solubility is defined as

39.12 grams of NaCl per 100 milliliters of water at 100°C

(212°F).

The solubility of a substance typically increases with temper-

ature. Tea and sugar are good examples of this phenomenon.

If you like sugar in your tea you’ve probably noticed that it’s

easier to dissolve sugar in hot tea than in iced tea (Figure 5).

Chemical Characteristics

Chemical characteristics of water are simply defined as the

chemical constituents found in the water. Some common

chemicals found in water include chlorides, phosphates,

nitrogen, pesticides, lead, copper, and sulfate (Figure 6).

Chemical characteristics are generally defined by analysis.

For example, when operators check the pH of the water, they

are actually checking the concentration of H+ ions, which is a

specific chemical constituent.

Dissolved gases are another type of chemical characteristic,

which are often addressed as contaminants in their own

right, especially in power plants. Dissolved gases are gases

dispersed throughout a body of water. Fish get their oxygen

from oxygen dissolved in the water. Although dissolved gases

are essential for marine life, they wreak havoc on power plant

equipment. The most common forms of dissolved gases pres-

ent in water are oxygen, carbon dioxide, and nitrogen.


Sugar Cube

Iced Tea

Sugar Cubes

Hot Tea

FIGURE 5—The solubility of a substance increases with temperature. More sugar dissolves in an equal amount ofhot tea compared to iced tea.

Chemicals can be

classified as organic

or inorganic. Organic

chemicals contain carbon

or are carbon based.

Inorganic chemicals

don’t contain carbon

or are not carbon based.


Biological Characteristics

Biological characteristics of water refer to the living matter

found in the water. This obviously includes marine life of

all sizes and shapes, both plant and animal. However, the

biological characteristics that chemists and plant operators

typically deal with are called microorganisms, and include

algae, bacteria, and viruses. Bacterial levels for drinking water

are controlled using standards applied to the following specific

bacteria: coliform, Giardia lamblia, and fecal streptococci

(Figure 7). When these particular harmful bacteria are held

to a specified minimum, it’s assumed that the rest, including

viruses, have also been adequately treated.

Chemicals

or or

pH Dissolved Gases

chlorides

phosphates

chemicals H +

H+

H+

O2

O2

CO2

CO2

(A)

Blue Water

CuSO4

(B)

FIGURE 6—Copper sulfate (CuSO4 ) dissolved in water is a chemical characteristic. It turns the water blue andthereby alters a physical characteristic, color. (A) The presence of chemicals and dissolved gases is also consid-ered to be a chemical characteristic.


In a power plant, the main issues with biological characteris-

tics are not the treatment of drinking water but the treatment

of cooling water. So-called cooling water is actually very warm,

because it picks up heat from the steam in the condenser.

Cooling towers, which are commonly used in power plants,

have a tendency to concentrate the contaminants in water.

This system provides ideal growing conditions for algae

and bacteria. A large number of microorganisms can form a

biomass or slime, which may be obviously visible (Figure 8).

Now that we have defined all three types of major water

characteristics, physical, chemical, and biological, let’s look

at how they affect each other. Imagine water that has high

chlorine content, which is a chemical characteristic. How

would water with a high chlorine content taste, compared to

water with a low chlorine content? Think of your last trip to a

public swimming pool. Did the water in the pool smell like

tap water? This example shows how a chemical characteristic

like chlorine content can affect physical characteristics like

taste and odor. Would water that was highly acidic affect the

biological properties of water? Dissolved solids, which are a

physical characteristic, can drastically change the pH of

water, which is a chemical characteristic. Although we are

categorizing the characteristics of water separately, they

don’t affect the water quality in isolation. A change in one

typically has an effect on the others.

8

FIGURE 7— Microorganism levels inwater are regulated using tests forcoliform, Giardia lamblia, and fecalstreptococci.


Slimy Biomass

FIGURE 8—A Slimy Biomass


Self-Check 1

At the end of each section of Power Plant Water Treatment, Part 2, you’ll be asked to pauseand check your understanding of what you’ve just read by completing a “Self-Check” exercise.Writing the answers to these questions will help you to review what you’ve studied so far.Please complete Self-Check 1 now.

Indicate whether the following statements are True or False.

_____ 1. Water in nature is always pure.

_____ 2. The acceptable level of contamination is determined by the intended use of thewater.

_____ 3. Water quality characteristics and parameters are always exclusive.

_____ 4. Alkalinity is a physical characteristic of water.

_____ 5. Suspended solids readily dissolve in water.

Choose the correct answer.

6. Turbidity is caused by

a. dissolved solids. c. excessive pH.b. colloids. d. dissolved gases.

7. If the dry weight of a 100 ml filtered sample is 3 mg, the TSS is

a. 3 mg/l. c. 30 mg/l.b. 6 mg/l. d. 60 mg/l.

8. Which of the following will dissolve the most NaCl?

a. 100 ml of water at 10° F c. 1 liter of water at 40° Fb. 100 ml of water at 80° F d. 1 liter of water at 75° F

9. Carbon dioxide in water is a

a. dissolved gas. c. suspended gas.b. dissolved solid. d. suspended solid.

Check your answers with those on page 57.


WATER QUALITY PARAMETERS

Water quality parameters form the basis for judging the qual-

ity of water. Parameters are typically specific, individual

water characteristics. They can be general (we’ll judge the

water based on its pH value) or specific (we want the boiler

water pH to be in a range from 9.5 to 10) (Figure 9). The

parameters used to judge water quality consumed by the

public are set and regulated by the Environmental Protection

Agency (EPA).

Boiler

pH 9.5 to 10

Steam Drum

FIGURE 9—A specific parameter might require boiler water to be in the pH range of 9.5 to 10.

To protect public health and safety, federal regulations establish the safe level of contaminationfor potable water. These contamination limits are called MCLs (maximum contamination levels) and are enforceable under the Safe Drinking Water Act (SDWA). These standardsapply to water systems that supply public water to at least 15 connections or to a minimumof 25 people. The regulations set limits for contamination by class and name. There are generally five classes of contamination under these regulations: organic chemicals, inorganicchemicals, disinfection by-products, radioactive chemicals, and microorganisms. The regulationscontain tables that list the names of the contaminants under the appropriate classification.The limits are frequently listed in units of milligrams per liter (mg/l). For specific limits, visitthe EPA Web site at http://www.epa.gov/safewater/mcl.html#1.

http://www.epa.gov/safewater/mcl.html#1.


The limits set by the EPA are important to power plants that

treat their own drinking water. EPA contamination limits

would generally apply to the service water treatment within

the plant. Drinking water that meets EPA standards is cleaner

than cooling water, but it still isn’t clean enough for such

power plant equipment as the boiler. Plant designers and

equipment manufacturers establish the parameters and limits

used to judge water quality in a power plant. Rarely, if ever,

will you find any two power plants in which water quality levels

are identical. This section won’t cover the specific limits of any

particular plant but rather the basic parameters commonly

used to judge the water quality in all power plants.

Parameters Typically Used in a PowerPlantThe water taken in by a water treatment plant is called

influent and enters the water treatment process at the intake

structure (Figure 10). These structures can contain various

devices that screen large objects from the influent. Influent

is often called raw water in power plant treatment processes.

The raw water may come from surface or groundwater and,

in some cases, may be a combination of the two. The raw

water contaminants will vary from source to source and

location to location. Contaminants are removed from the raw

water in stages, using a variety of different processes. The

quality of that water can be measured at any point required

in the treatment process, using one or more of the following

parameters: turbidity, hardness, total dissolved solids, con-

ductivity, cation conductivity, dissolved gases, pH, sodium

content, silica content, and temperature.

TurbidityTurbidity, as you’ve learned, is caused by colloids in the

water, which make the water look cloudy. The visually per-

ceived cloudiness is caused by the scattering and absorption

of light rays by the suspended particles. Water treatment is

required to remove the particulate contamination that causes

turbidity. The most common form of turbidity treatment is a

chemical process called coagulation. We’ll discuss this

process in more detail later in this study unit.

12


River Influent

To Water Treatment

(A)

Bridge Crane

Access Hatch

Bridge Crane

Traveling Screens

(B)

Intake Structure

High-Water Level

FIGURE 10—The raw water enters the water treatment process at the intake structure.


Optical devices are used to measure turbidity. One of the

oldest and most basic turbidity devices is the candle

turbidimeter, which is still used in some areas to measure

highly turbid water. A candle turbidimeter uses a candle

flame and human visual perception to determine the turbidity

of a water sample, which is expressed in units of measurement

called the Jackson Turbidity Unit (JTU). A more sensitive

turbidimeter is the nephelometer, which can measure the

turbidity of water even when the sample appears clear to

the human eye. The nephelometer electronically measures

the amount of light transmitted through the water sample.

Results are expressed in Nephelometric Turbidity Units (NTUs),

which are different from JTUs. It has become common in

today’s digital age to simple say “turbidity meter” unless you

specifically mean a candle turbidimeter (Figure 11).

HardnessThe amount of hardness in water depends on its source. In

general, groundwater is harder than surface water. Water

hardness usually comes from contact with soils and rocks.

Hard water contains substantial concentrations of salts,

commonly salt compounds formed with calcium (Ca+2)

and magnesium (Mg+2). Other elements, such as sodium

(Na+), aluminum (Al+3), and iron (Fe+2), also form salts

that contribute to water hardness but do so much less

than magnesium and calcium.

Salt compounds formed by combinations of calcium,

magnesium, or sodium combined with bicarbonates (HCO3-),

sulfates (SO4-2), and chlorides (Cl-) readily dissociate in water.

When these compounds dissociate, they form positively

charged ions of calcium, magnesium, or sodium (Ca+2, Mg+2,

and Na+), and negatively charged ions of bicarbonate, sulfate,

and chloride (HCO3-, SO4

-2, and Cl-). Once they dissociate in

the water, chemical analysis can’t determine which positive

ion was originally combined with which negative ion. All the

calcium ions look and act alike, as do all the magnesium

and sodium ions. The same is true of the bicarbonate,

sulfate, and chloride ions. There’s no way to tell whether

the bicarbonate in a chemical analysis came from calcium

bicarbonate, magnesium bicarbonate, sodium bicarbonate,

14


FIGURE 11—This turbidime-ter has a digital readout thatdisplays values in units ofNTUs.


or any combination of these. So it’s standard chemical practice

to assume that all of the ions came from calcium carbonate,

CaCO3, and to express hardness in terms of equivalent

amounts of CaCO3. This is often called calcium hardness

(Figure 12). Hard water doesn’t cause health problems and

can be used for drinking, but it may harm steam plant

equipment by forming scale. Removing the maximum

possible amount of the water hardness—in a process called

softening—minimizes boiler problems. The two principal

processes for softening water are demineralization and

coagulation/flocculation.

Total Dissolved SolidsTotal solids (TS) is the sum of total dissolved solids and total

suspended solids. Power plant personnel often use the terms

“total solids” and “total dissolved solids” interchangeably

(Figure 13). Total dissolved solids is the commonly used

term in this industry, so we’ll use it here for consistency.

Total dissolved solids are calculated in the lab by filtering,

drying, and weighing a sample, a process called a gravimetric

analysis. In gravimetric analysis, a water sample is typically

filtered through a 0.45-micron membrane disk. The material

captured on the disk is then dried and weighed. The results

of the analysis are given in milligrams per liter (mg/l) or

parts per million (ppm) by weight. To see a sample of a simi-

lar calculation, refer back to the first section on suspended

solids. The difference between the lab method described

earlier and the one noted above is in the filtering medium.

16

Ca+2

Mg+2Fe+2

Al+3

Ca+2

Ca+2

Ca+2

Ca+2

Ca+2

Ca+2

Ca+2

Ca+2

Mg+2

Mg+2

Mg+2

Measured asEquivalent

CaCO3

FIGURE 12—Water hardnessis reported as equivalents ofcalcium carbonate (CaCO3)hardness.


Total dissolved solids concentrations are often estimated in

power plants using a conductivity meter. In some cases,

gravimetric analysis is only performed when the conductivity

meter or other source indicates there’s a problem.

Conductivity

Conductivity is a measure of how well a substance conducts

electricity. In low-turbidity water, conductivity tests are more

common than TDS tests for determining water quality. Water

and electricity are a dangerous mix, but it’s the contami-

nants in the water that conduct the electrical current, not

the water itself. Pure water contains few ions and is actually

a poor conductor of electricity. However, when compounds

dissolve in water, they dissociate into positive and negative

ions. Colloids also have an electrical charge, typically nega-

tive. All of these charged particles in the water are what

make it such a good conductor. Measuring water’s conductiv-

ity, therefore, gives you a good idea of the concentration of

ions or contaminants in the water (Figure 14). Conductivity

is expressed in units of microSiemens/cm (µS/cm) and is

measured with a conductivity meter and cell.

Total Suspended Solids

TotalSolids = +

Total Dissolved Solids

Na+

Cl_

_Cl

Na+

FIGURE 13—Total solids are equal to the total suspended solids plus the total dissolved solids.


Ions

Ions

Ions

Ions

Ions

Dirty Water

Battery

Pure Water

Battery

Lightbulb Not Lit

Lightbulb Lit

FIGURE 14—Conductivitycan be used to measure thepureness of water, since purewater doesn’t conduct elec-tricity and dirty water does.


Cation Conductivity

Cation conductivity is a conductivity test performed on

condensed steam samples to indicate the presence of carry-

over. Carryover occurs when water droplets are carried along

with the steam to the turbine. Water, and the impurities it

carries, can seriously damage turbine blades. Steam sent to

the turbine is superheated dry steam under normal operating

conditions. Conductivity tests performed on samples of

condensed steam provide a valuable indicator of steam purity

by measuring the presence or absence of dissolved solids.

Cations are positively charged ions, and anions are negatively

charged ions. A cation conductivity test is performed after

the sample to be tested has been sent through a cation

exchange resin. It’s a more precise indicator of the source of

the ions present in a sample. In Figure 15 the conductivity

of a condensate sample is tested before and after the sample

passes through a cation exchange resin.

Cation conductivity is typically used to indicate steam purity

in a power plant. The steam sample is condensed and sent

through a cation exchange resin to remove any boiler treat-

ment chemicals that were added to the system. This process

is typically used when ammonia or other amines are used for

feedwater pH control and oxygen scavenging. By eliminating

the treatment chemicals before testing, the cation conductivi-

ties run much lower and provide more accurate indications

of carryover.

pHpH measurements are taken to determine the amount of

acidity or alkalinity in the steam and water in the power

plant. The pH of water has an effect on many phases of water

treatment, including coagulation and water softening, which

will be described in the section on front-end treatment. The

pH of water also affects the scaling potential and corrosive-

ness of water. It can be determined by various means, such

as color indicators, pH paper, or pH meters. The pH meter is

used most often because it provides the most accurate meas-

urement. Conductivity tests and pH are often used together

to help determine the contaminants in the water.

Amines are a family of

compounds derived from

ammonia by replacing

hydrogen with hyro-

carbons.

Evaporation is a natural

process of purification.

Pure vapor or steam

carries no impurities

with it. When water

droplets are carried

along with the steam,

impurities will be car-

ried with the water.

Dry steam carries no

liquid water because its

temperature is above the

saturation temperature.

As you learned in earlier

studies, the saturation

point represents the

highest temperature (at

a given pressure) at

which liquid water can

be suspended in steam.


Silica ContentSilica is one of the most common substances in the Earth’s

crust, and is found in many types of rock and soil. As water

runs over and through the earth, it absorbs silica. Both

ground- and surface waters contain silica, usually as silicon

dioxide (SiO2). Silica, which can be detected with an analyzer

like the one shown in Figure 16, has a crystal structure and

is the major constituent of glass and sand. When heated,

silica melts and becomes a molten glass. The glass adheres

to the boiler tubes and is called scale. Silica scale is hard to

remove from power plant equipment; it typically requires

sandblasting, a corrosive process that shortens the life of the

equipment. Silica is detrimental to steam plant equipment

even in minute concentrations. Silica deposits on boiler tubes

20

FIGURE 15—These gauges monitor the condensate flow rate through two conductivity test devices. The flowgauge on the right monitors the flow of untreated condensate as it moves toward a conductivity tester. After passing through the resin, condensate is subjected to a cation conductivity analysis. The gauge on the left monitors the flow rate through the cation conductivity tester.


reduce heat transfer rates and can lead to boiler tube failure.

Carryover of silica from the boiler to the turbine will cause

severe turbine blade damage. Plant startups are delayed

when silica levels are high.

Sodium Content of Steam and WaterSodium salts and sodium compounds of silica are common

in groundwater. Sodium contamination causes significant

problems in power plants, especially in the turbine. Sodium,

FIGURE 16—Silica SamplingEquipment Used in a PowerPlant


as either sodium chloride (NaCl) or sodium hydroxide

(NaOH), causes corrosion at high temperatures and high

pressures. Steam turbines are especially susceptible to this

type of corrosion, which leads to corrosion stress cracking.

So it’s common practice to measure the amount of sodium

present in both condensate and steam systems. Since the

condensate is relatively pure and the cooling water is signifi-

cantly more contaminated than the condensate, a quick

check of sodium levels is a good indication of leaks in the

condenser tubes (Figure 17). Sodium levels in steam are an

indication of steam purity or quality. Sodium can only be

present in the steam if it’s contaminated with water droplets.

Carryover—the contamination of water in the steam—is the

most common problem associated with turbine damage.

Dissolved Gases

Water contains many dissolved gases. The most common are

oxygen (O2), nitrogen (N2), and carbon dioxide (CO2). Water

will absorb any gas it contacts and may contain other gases

such as ammonia (NH3) from treatment processes used in the

plant. Generally, which dissolved gases are found in power

plant water will depend on the water source and treatment

processes employed.

22

CoolingWater

Cooling WaterInto Condenser

Tubes

HotwellCondensate

Condenser

Steam In

FIGURE 17—Condenser tubes carry cooling water, which condenses the steam back into water. Unless a condens-er tube leaks, the cooling water and steam never touch, because the cooling water is carried in tubing and thesteam flows over the tubing.

The condenser’s bottom

contains a reservoir,

known as a hotwell,

where condensate

collects.


Dissolved gases corrode most metals. Oxygen in the water

causes oxygen pitting. This is concentrated corrosion in a

small area, creating a small hole or pit (Figure 18). Oxygen

pitting commonly occurs in the boiler and economizer tubes

during normal operation and in the superheaters and

reheaters during periods when the plant is shut down. This

type of corrosion seriously weakens the structural integrity of

the tubes and will eventually lead to a tube rupture. As you

learned from studying acid rain in an earlier study unit, car-

bon dioxide mixed in water forms a weak acid called carbonic

acid. Carbonic acid reduces the pH of water, which acceler-

ates the corrosion rate of metal piping.

Dissolved gases are removed from the power plant’s water

supply by air ejectors, degasifiers, and deaerators. These

pieces of equipment also remove any other dissolved gases

that may be present. Chemical additives are often used in

addition to mechanical treatments to remove dissolved

oxygen, since that is by far the biggest culprit among the

dissolved gases. Chemical additives used to remove dissolved

oxygen are called oxygen scavengers. Two of the most

common oxygen scavengers are hydrazine (N2H4) and sodium

sulfite (Na2SO3). Dissolved oxygen concentration in power

plants is usually limited to 5–7 parts per billion (ppb) after

treatment.

FIGURE 18—Oxygen pitting isthe result of concentratedlocal corrosion. (Photo sup-

plied by Ondeo Valco

Company)


Self-Check 2


_____ 1. Water quality parameters are always general.

_____ 2. The acronym “ppb” stands for phosphate-phosphate-boil.

Answer the following questions.

3. The ______ regulates the safety of public drinking water.

4. Who establishes the quality criteria for the boiler water in a power plant?

__________________________________________________________________________

5. Explain when and why cation conductivity is used.

__________________________________________________________________________

Complete the following sentences with the correct answer.

6. The term MCLs stand for

a. motor controlled logics. c. maximum contamination levels.b. major colloids listed. d. master chemical laboratories.

7. Drinking water can be characterized as

a. cleaner than boiler water. c. cleaner than cooling water.b. cleaner than condensate. d. cleaner than steam.

8. The units mg/l means

a. megagrams per liter. c. mecagrams per liter.b. milligrams per liter. d. minigrams per liter.

9. Pure water

a. exists naturally. c. won’t conduct electricity.b. contains large quantities of ions. d. has a pH of 11.

10. The most common turbidity treatment is

a. boiling. c. oxygen scavenging.b. coagulation. d. cooling.



FRONT-END WATER TREATMENT

Water treatment processes can generally be divided into seven

basic stages: screening, settling, coagulation, flocculation,

settling, filtering, and advanced treatment (Figure 19). This

section will discuss in detail the first six stages, which we call

front-end treatment. The seventh stage, advanced treatment, is

called secondary treatment, and will be covered later.

First, we’ll discuss the basics of each component of a simpli-

fied water treatment plant shown in Figure 19. Then we’ll take

a more specific look at common practices in power plants.

ScreeningThe first stage in the water treatment process is to screen the

water at the intake source, which can be a river, lake, or even

the ocean. Screens over the intake pipes or ducts prevent

large pieces of debris from entering the system and damaging

the equipment. Typically, large spaced bars are placed direct-

ly over the intake structure. These are called bar screens or

bar racks (Figure 20). Once the water passes through the bar

racks, it passes through a smaller set of screens. This second

set, shown in Figure 21, can be stationary or traveling. These

screens block smaller debris, such as twigs, leaves, and trash,

from being sucked into the pump. In addition to the screens,

many intake structures have trash rakes (Figure 22), which

remove the garbage collected on the screens for disposal. If

the debris and trash aren’t periodically removed from the

screens, they will block the flow of intake water. Sprays and

washes are often used alone or with trash rakes to wash

debris from the screens.

Intake andScreening Settling

Cooling Water

CoagulationFlocculation Settling

Filtration

DrinkingWater

Makeup Water

AdvancedTreatment

FIGURE 19—An Overview of Water Treatment Process


FIGURE 20—Bar racks screen large debris from the intake water.

Water Into Screen

Water Passed Through Screen

FIGURE 21—A Stationary Screen


A common problem at intake structures in fresh water is

clogging by zebra mussels, freshwater animals that look like

clams (Figure 23). They multiply rapidly and are considered

a pest because they clump together and attach themselves

to various plant structures. If unchecked, they can block the

water flow or completely clog an intake structure. Zebra

mussels aren’t easily removed from screens with washing

alone, and many plants have found that treating the wash

water with chlorine helps.

Trash Rake

Intake Water

Screen

Cable

FIGURE 22—A Cable-Operated Bar Screen


Discrete Settling

The second stage in the water treatment process is settling.

This occurs when large contaminants like sand or gravel are

allowed to fall to the bottom of the tank. This type of settling

is called discrete particle settling, which means an individual

particle falls by gravity under its own weight (Figure 24).

High-velocity water can carry a significant amount of sand,

28

FIGURE 23—Zebra mussels have become the bane of dam and power plant operators because they are commonand difficult to remove. (Photo provided by the U.S. Geological Survey, Upper Midwest Environmental Sciences

Center)


gravel, and/or dirt. The settling stage simply allows the water

to slow down in a large tank so that the large particles of sand

or dirt can settle out on their own without any additional

treatment. The discrete settling stage is bypassed in some

plants. Plants that don’t use a discrete settling phase often

use more chemical treatment in their coagulation process.

Settling occurs again in the fifth stage of the process and

will be discussed in more detail later.

CoagulationStage three in the water treatment process is coagulation, or

the addition of chemicals to increase the settling of particles

(Figure 25). Colloids are typically negatively charged clusters

of small organic matter. These particles have like charges,

which repel each other. This tends to evenly disperse the

particles in the water and keep them suspended—even if

they grow large enough to settle slowly on their own. By

adding chemicals with a positive charge, or chemicals that

dissociate to positive ions, the negative particles will be

High-Energy Water CarryingSand and Small Debris

Particles Fall Under Their Own Weight in Calm Water

Low-VelocityWater Out

FIGURE 24—A Discrete Settling Tank


attracted to the opposite charges and will clump together.

The mixing of the positive and negative will create an overall

neutral environment, and the particles, now larger and no

longer repelling each other, will rapidly settle to the bottom.

The process of coagulation is a bit like making popcorn balls,

where the sticky caramel or molasses helps the loose bits of

popcorn stick together in clumps.

Coagulants are typically an aluminum salt or an iron salt.

The two most common chemicals used as coagulants in water

treatment plants are alum—its chemical name is aluminum

sulfate, Al2(SO4)3—and ferrous sulfate (2FeSO4). Alum is by

far the most common chemical coagulant.

The coagulation process requires more than just the addition

of chemicals to work properly. You need rapid mixing initially

to get chemicals evenly dispersed throughout the fluid

(Figure 26A). This is frequently accomplished by turbulent

(a jet stream) flow through an injecting system as the water

enters the tank, or with rapid mixers, which use fast-moving

blades to stir the water. But if the mixing were to continue

at high speed throughout the process, the particles would be

torn apart by the agitation and would not clump or have the

opportunity to settle. So the next phase of coagulation is slow

mechanical stirring. The slow stirring increases the number of

collisions between particles and promotes further clumping

(Figure 26B). The larger the particles grow, the faster they

will settle.

30

Chemical In

Al2(SO )4 3or

2 FeSO4or

NH Al2(SO )4 34

FIGURE 25—Chemicals areadded to water to promotecoagulation.


For the coagulation process to be as efficient as possible,

an appropriate quantity of chemical is required to promote

proper clumping. The contaminants in the water and the pH

both affect the dosage of chemicals, so the proper dosage is

determined based on the analysis of the water to be treated.

It’s sometimes necessary to add additional chemicals, like

lime or soda ash, to adjust the pH and aid in the coagulation

process. Chemists must perform frequent tests to ensure the

dosage levels are correct. The amount of contaminants in the

raw or intake water can vary greatly with the seasons and

Fast

or

Jet Stream Turbulence

(A)

Slow

(B)

FIGURE 26—(A) Coagulation begins with rapid mixing, then continues with slow mechanical stirring (B).


cycles of nature. For example, a heavy rain upstream of

the intake area can rapidly change the constituents of the

incoming water.

Flocculation

Flocculation is the actual agglomeration, or grouping together

of particles suspended in the water (Figure 27). The chemicals

added during the coagulation process promote clumping.

The particles grow and settle and they are called “floc” or

“chemical floc.” To distinguish between the two processes,

remember that coagulation is the adding of chemicals that

cause flocculation. These two processes are closely related

and are often referred to as “coagulation/flocculation.”

Settling

Settling is the process whereby particles fall to the bottom of

the tank or equipment. As they settle out of the water, an

accumulation, called a sludge blanket, builds. Scrapers on

the bottom of the tank remove the sludge blanket. The scrap-

ers move the sludge toward the blowdown piping, typically

located in the center of a circular tank (Figure 28).

Settling in a clarifier happens through four different processes,

which include discrete settling, floc settling, hindered settling,

and compression settling (Figure 29). Although the four types

of settling all occur in the same tank, the discrete and floc

settling are more likely to occur near the top of the tank, and

the hindered and compression settling will occur near the

bottom as particles build up and become congested. The

four types happen at different speeds.

32

Chemical Floc

FIGURE 27—The actual clumping oragglomeration of particles caused bycoagulation is called flocculation.


In discrete settling, a single particle falls at normal velocity.

In flocculation settling, the agglomerated particles fall at an

increased velocity; clumped particles are heavier, so they set-

tle faster. In hindered settling, the velocity is decreased or

limited due to particle collisions. Particles in this zone settle

more slowly, because as they frequently collide, they bounce

back. Think of what happens in a pinball game. As the ball

hits posts or flippers, it frequently changes direction and it

slows down. But if it happens to find a straight path, it zips

toward the exit. This is the same principle that’s used to slow

the settling of particles in the hindered zone, which lies

above the sludge blanket. The collisions here differ from the

collisions in flocculation. In flocculation, the particles clump

and grow larger so they settle faster; collisions that create

clumping increase settling. In the hindered zone, however,

the collisions cause bounce-back, not clumping. Slowest of

the four is compression settling, which occurs near the

FIGURE 28— Sludge Scrapers on the Bottom of a Clarifier


bottom of the tank and is extremely slow. As particles come

into contact with the particles in the sludge blanket, they

stop or settle very slowly. They can only settle further by

compressing the particles underneath them.

Clarifiers

Clarification is typically the combination of processes three

(coagulation), four (flocculation), and five (settling). These

processes can be combined in a single piece of equipment

called a clarifier. The two most common types are horizontal

clarifiers, shown in Figure 30, and up-flow clarifiers.

A horizontal flow clarifier is basically a large rectangular tank

divided into two sections. Coagulation and chemical disper-

sion take place in the first, and flocculation and settling take

place in the second. The sludge accumulates on the bottom,

and the clear or clarified water flows over a weir-type exit.

34

CompressionHindered

Discrete

Flocculation

Chemical Added

Need to Remove Sludge WhenCompression Settling Occurs

FIGURE 29—The varioustypes of settling include discrete, flocculation, hindered, and compression.


In cross section, up-flow clarifiers are shaped like cones.

Seen from the top, they display circular rings, or chambers.

The dirty water enters at the top in the center of the clarifier.

The chemicals are added and mixing is done in the center

(inner) chamber. This inner chamber is called the mixing

chamber or detention zone. Clean water overflows and is

collected near the top through a system of perforated pipes.

Filtration

Filtration is the second most common process in water

treatment and is required for public water supplies. Small

particles in the water—the ones not removed by the previous

processes—are now removed as the water moves through a

bed of filter media (Figure 31). Filter beds typically consist of

sand, gravel, coal, garnet, or some combination of the four.

Circular Clarifier

Horizontal Clarifier

FIGURE 30—Horizontal and Up-Flow Clarifiers


Filters can remove only a certain amount of solid materials

before the openings or pores in the filter media become

clogged with removed particles. Once the filter becomes

clogged it must be cleaned. The process of cleaning filters is

called backwashing. Backwashing involves reversing the flow

of water through the filter to partially lift or suspend the bed

(sand and other media), allowing the trapped particles to be

washed away by the backwash water (Figure 32).

36

Under Drains

Filtering Medium

Water In Gravity Filter

Fill with Media

Water Out

FIGURE 31—A Gravity-Fed Filter

Backwash In

Media Lifted

Backwash Trough

Dirty Water Out

FIGURE 32—Backwashing is used to clean filter beds.


There are two basic types of filters: gravity bed filters and

pressurized bed filters. Gravity bed filters are more common

for large-scale water treatment plants. In gravity systems, the

filters are either slow filters or rapid filters, depending on the

velocity of the flow through the filter.

The simplest gravity filter consists of a single bed of sand

(Figure 33). This type of filter has a uniform pore or opening

size throughout the filter bed, so the top layers of the filter

tend to get clogged before the bottom ones. Single-bed filters

need frequent backwashing and don’t make full use of the

lower or bottom portions of the filter bed.

Multimedia filters use a combination of filter media (sand,

gravel, coal, or garnet). These filters use heavier (more dense)

and smaller material on the bottom of the filter with a lighter

(less dense) and larger media on the top of the bed (Figure 34).

The density difference is an important component in multi-

media filters. The density difference allows the smaller media

material to stay on the bottom of the filter during backwash-

ing, which allows larger contaminant particles to be removed

at the top of the filter and smaller contaminant particles to

be removed at the bottom. This reduces the frequency of

Sand

DP

DifferentialPressure Sensor

Underdrain Chamber

Effluent(Clean Water)

Water Surface Above Sand

Influent(Dirty Water)

FIGURE 33—A Single-BedFilter


backwashing and makes the filtering process more efficient.

Since the filter media’s smaller particles are heavier or

denser than the larger media particles, they settle faster.

This maintains the filter media order, or layering, during

the backwashing process.

In real-world applications, filtration is a combination of three

or four processes occurring simultaneously inside the filter

(Figure 35). Straining removal occurs when a suspended solid

that is larger than the opening in the filter media is trapped

in the opening between two media pieces. Interception removal

occurs when the flow through the filter media brings a parti-

cle in contact with a piece of the media and the particles

adhere or stick together. With interception, the removed

particle size has nothing to do with its removal. It was

removed because it got stuck to a piece of the filter—the way

gum gets stuck to the bottom of your shoe. Sedimentation

removal occurs when a particle settles onto a piece of the

media. Again, this has nothing to do with the particle’s size

but rather where it lands. If a small particle, small enough to

fit through the openings in the media, lands directly on top

of a grain of sand or gravel instead of in an opening between

the pieces, then it is removed by sedimentation rather than

38

Fine Media(Silica Sand)

DP

DifferentialPressure Sensor

Underdrain Chamber

Effluent(Clean Water)

Water Surface Above Sand

Influent(Dirty Water)

Coarse Media(Anthrafilt)Ê

FIGURE 34—A multimediagravity filter has several layers of filter material.


straining. Finally, when coagulation/flocculation is a part

of the water treatment process, the flocculation process

can continue inside the filter and allow particles to grow

large enough so they are removed by one of the other three

methods.

Straining Interception Sedimentation

FIGURE 35—A filter systemworks by straining, intercep-tion, sedimentation, and flocremoval.


Self-Check 3


_____ 1. Bar racks are used to hold glassware in the lab.

_____ 2. Discrete settling is aided by chemicals.

_____ 3. Coagulation is a test for the presence of gases.

Answer the following questions.

4. List four types of settling.

__________________________________________________________________________

5. List the first six steps in the water treatment process.

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

__________________________________________________________________________

Complete the following sentences with the correct answer.

6. The first stage in the water treatment process is

a. screening. c. scavenging.b. settling. d. sulfur dosing.

7. Gravity filters are cleaned by

a. vacuuming. c. simple replacement.b. shaking. d. backwashing.

(Continued)


CHEMICAL CALCULATIONS FORWATER TREATMENT PLANTS

Water treatment plants often need to calculate how much

coagulant to use to remove suspended solids properly, and

how much sludge or waste is being produced. Sludge pro-

duced through the water treatment process is often hauled

off for further treatment or sent to a sludge pond (Figure 36).

This section will use the concepts you have learned earlier

and build on them to show how those calculations are

performed.

Self-Check 3

8. Straining removes particles that

a. get stuck to the media regardless of the particle size.b. are larger than the opening in the filter media.c. settle on the media rather than landing in an opening in the filter media.d. are smaller than the openings in the filter.

9. Which of the following coagulants would you expect to see in most water treatment systems?

a. Al2(SO4)3 c. NH4Al2(SO4)3b. FeSO4 d. Al2(CO3)4

10. At the bottom of the clarifier, you would expect to find

a. clear water. c. sludge.b. large fish. d. ice.



Calculating Sludge on a Dry WeightBasis

If a water treatment plant is processing 7 Mgal/day (million

gallons a day) and is using 15 mg/l of alum with the addition

of lime, as calcium hydroxide, to enhance the coagulation

process, how many pounds of aluminum hydroxide sludge

are produced each day on a dry weight basis? The following

equation represents the reactions:

This may look like a tough question, but you can solve it by

applying the concepts we have just learned. We’ll break it

down into the following steps:

Step 1: Balance the equation.

42

FIGURE 36—An Onsite Sludge Pond

432342 CaSOAl(OH)Ca(OH))(SOAl +→+

Alum Calcium Aluminum Calcium

Hydroxide Hydroxide Sulfate


Step 2: Calculate the molecular weights of each product and

each reactant.

Step 3: Check to see that you’ve calculated the molecular

weights correctly.

Step 4: Determine which product or reactant has a known

concentration.

Step 5: Convert all molecular weights to a fractional equiva-

lent of the known concentration’s molecular weight.

Step 6: Calculate the concentration of the desired

(unknown) molecule.

Step 7: Calculate appropriate unit conversions.

Now that we know what the steps are, let’s apply them in

order.

Step 1: Balance the equation.

ElementNo. of Atoms of

ReactantsNo. of Atoms of

Products

Al 2 1

S 3 1

O 14 7

Ca 1 1

H 2 3



Products

Al 2 2

S 3 1

O 14 10

Ca 1 1

H 2 6

432342 CaSOAl(OH)Ca(OH))(SOAl +→+ 2




Balance Al

Balance S

Balance Ca

Our equation is balanced.

Step 2: Calculate the molecular weights of each product and

each reactant.

Look up the atomic weight of each element.

44



Products

Al 2 2

S 3 3

O 14 18

Ca 1 3

H 2 6



Products

Al 2 2

S 3 3

O 18 18

Ca 3 3

H 6 6

432342 CaSOAl(OH)Ca(OH))(SOAl 32 +→+










Calculate each molecular weight.

Alum 2(27) + 3(32) +12(16) = 342

Calcium Hydroxide 3{(40) + 2(16) + 2(1)} = 222

Aluminum Hydroxide 2{27 + 3(16) + 3(1)} = 156

Calcium Sulfate 3{40 + 32 +4(16)} = 408

Step 3: Check to see that you have calculated the molecular

weights correctly.

If we calculated the molecular weights correctly, the right

side of the equation should equal the left side of the

equation.

342 + 222 = 156 + 408

Since the two sides are equal, we calculated the molecule

weights correctly: 564 = 564.

Step 4: Determine which product or reactant has a known

concentration.

The problem stated that the plant is using 15 mg/l of alum,

so that is our known concentration. Concentrations are

given in units of mass per volume or weight per volume, so

milligrams per liter is a common unit.

Step 5: Convert all molecular weights to a fractional equiva-

lent of the known concentration’s molecular weight.

Element Atomic weight

Al 27

S 32

O 16

Ca 40

H 1







You will recall that atomic weights are relational: An atomic

weight is a comparison of how much an element weighs when

compared to carbon. We can use this relational property

when solving problems. We will divide each molecular weight

by the molecular weight of our known (in this example our

known is alum). This gives us a fractional representation of

how the different molecules are related to each other.

In Step 2 we calculated:

Alum 2(27) + 3(32) +12(16) = 342



Calcium Sulfate 3{40 + 32 +4(16)} = 408

Now we will convert each weight to a fractional equivalent of

the molecular weight of alum. It’s easy; you just divide each

by the molecular weight of alum, which is 342.

Alum 342/342 = 1

Calcium Hydroxide 222/342 = .649

Aluminum Hydroxide 156/342 = .456

Calcium Sulfate 408/342 = 1.193

These fractions can now be used to determine the concentra-

tions of each of the other terms. For every 1 mg/l of alum,

you will have .649 mg/l of calcium hydroxide, .456 mg/l of

aluminum hydroxide, and 1.193 mg/l of calcium sulfate. In

other words, 1 mg/l of alum reacts with .649 mg/l of calcium

hydroxide to produce .456 mg/l of aluminum hydroxide and

1.193 mg/l of calcium sulfate. Keep this principle in mind;

we will use it in the next step.

Step 6: Calculate the concentration of the desired

(unknown) molecule.

The unknown is the quantity of substance we are looking for,

in this case, how many pounds of aluminum hydroxide

sludge are produced each day. Ultimately we will calculate

how many pounds per day of Al(OH)3 sludge are produced,

46



(1 mg/l) Al2 (SO4)3 + (.649 mg/l) Ca(OH)2 → (.456 mg/l) Al(OH)3 + (1.193 mg/l) CaSO4


but in this step we want to know what the concentration is

for Al(OH)3. In Step 5 we discovered how to calculate the con-

centration of each part of the chemical equation if we had 1

mg/l of the known. Now we need to go back to the question

and see what the actual concentration of the known is and

multiply by that concentration. The known was 15 mg/l of

alum. We know that for every 1 mg/l of alum we added, the

process produces 0.456 mg/l of Al(OH)3. Now all we have to

do is multiply 15 times 0.456 and we obtain the concentra-

tion of Al(OH)3 for this problem.

15 mg/l × .456 = 6.84 mg/l of aluminum hydroxide

Step 7: Calculate appropriate unit conversions.

We’re almost there! We know that we have 6.84 mg/l of alu-

minum hydroxide, that the plant processes 7 Mgal/day, and

we want to know how many pounds of aluminum hydroxide

sludge are produced a day. All that is left to do is multiply

the two quantities, while carefully applying the appropriate

conversion factors.

Conversion factors:

8.34 lb/gal

1,000,000 gal/Mgal

1000 g/l

1000 mg/g

Set up the equation so all the units cancel with the exception

of lb/day.

Cancel all the units that will cancel and reduce the equation.

gal

lb

Mgal

gal

day

Mgal

g

l

mg

g

l

mg 8.341,000,0007

100010006.84 ×××××

lb/daylb

day399

8.3476.84 =××

gal

lb

Mgal

gal

day

Mgal

g

l

mg

g

l

mg 8.341,000,0007

100010006.84 ×××××


Notice that to convert mg/l to lb/day, if the flow rate is given

in Mgal/day, that the following conversion factor can be used:

mg/l × Mgal/day × 8.34 = lb/day.

Calculating Sludge on a Wet WeightBasis

Let’s try another problem using the same common coagulant,

but with a different question. If a water treatment plant is

processing 10 Mgal/day (million gallons a day) and is using

12 mg/l of alum with the addition of lime, as calcium

hydroxide, to enhance the coagulation process, how many

pounds of aluminum hydroxide sludge are produced each

day if the wet sludge is 90 percent aluminum hydroxide by

weight?

The following equation represents the reactions:

To solve this problem we will repeat what we did last time,

and at the end we will add one more step. First, balance the

equation. Once the equation is balanced, calculate the molec-

ular weights and verify that you’ve balanced, the equation

properly. If you have trouble following any of the steps, look

back at the previous problem. This one is identical, so all the

details won’t be repeated. When you are sure you have a

properly balanced equation and properly calculated molecular

weights, you are ready to convert all your molecular weights

to fraction equivalents based on your known value’s molecular

weight. The known value is the one that was stated as a con-

centration in the original problem; in this case it’s 12 mg/l

of alum. Once you’ve calculated the fraction percent, multiply

the fraction for your unknown by the concentration of your

known. This will give you the concentration of aluminum

hydroxide in mg/l. Take the concentration of aluminum

48





hydroxide times the flow rate in Mgal/day and multiply by

the 8.34 conversion factor to get the lbs per day on a dry

weight basis. You should end up with something like this:

Calculated molecular weights:

Alum 2(27) + 3(32) +12(16) = 342



Calcium Sulfate 3{40 + 32 +4(16)} = 408

Fractional relationships with known:

Alum 342/342 = 1

Calcium Hydroxide 222/342 = 0.649

Aluminum Hydroxide 156/342 = 0.456

Calcium Sulfate 408/342 = 1.193

It’s given in the problem that the plant is using 12 mg/l

of alum.

12 mg/l × .456 = 5.47 mg/l of aluminum hydroxide

5.47 (mg/l) × 10 (Mgal/day) × 8.34 = 456 lb/day

You will remember from the previous example that to

convert mg/l to lb/day if the flow rate is given in

Mgal/day, we could use the following conversion factor:

mg/l × Mgal/day × 8.34 = lb/day. This is a common

conversion factor used in treatment plants, and it’s easy to

remember because 8.34 is conveniently the weight of water

(in lbs) per gallon. It’s not magic that it works out this way;

there are simply one million gallons to an Mgal, obviously.

What may not be so obvious is that there are one million

mg in a liter. These two conversions cancel each other out,

leaving only the conversion for lbs of water per gallon.

Here is another wrinkle to consider: What does it mean that

the sludge is 90 percent Aluminum Hydroxide by weight?

The sludge doesn’t come out of the bottom of the tank

perfectly dry, which is the way we calculated it. It’s actually

When qualifying the

amount of a substance

found in a solution of

weak concentration,

you can approximate

the value of mg/l as

being equal to parts per

million.



Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3CaSO4


wet and holds water that was removed with the sludge. The

actual weight of the sludge removed from the bottom of the

clarifier is the weight of the sludge plus the weight of the

water that is removed with it. If the sludge is 90 percent

aluminum hydroxide by weight, that means 10 percent of

the weight is from water. We calculated the weight of the 90

percent aluminum hydroxide, and if we want a more accurate

calculation we must add in the weight of the water as well.

The equation that represents this would be written as

Total weight (.9) = Dry weight. We calculated the dry weight,

so all you have to do is rearrange the equation to look like

this and plug in the values.

Now let’s try another problem similar to the last one, but this

time let’s use a different coagulant. If a water treatment plant

is processing 8 Mgal/day (million gallons a day) and is using

17 mg/l of ferric chloride with the addition of lime, as calci-

um hydroxide, to enhance the coagulation process, how

many pounds of ferric hydroxide sludge are produced each

day if the sludge is 70 percent ferric hydroxide by weight?

Assume the sludge is purely ferric hydroxide and water.

The following equation represents the reactions:

The first step to the solution is to balance the equation.

50

2323 CaClFe(OH)Ca(OH)FeCl +↓→+

Ferric Calcium Ferric Calcium

Chloride Hydroxide Hydroxide Chloride



.90

weightDryweightTotal =

456 lb

dayTotal weight = ___________ = 507 lb

.90 day

2FeCl3 + 3Ca(OH)2 → 2Fe(OH)3 ↓ + 3CaCl2


Next you calculate the molecular weights and check that

both sides equal each other.

Ferric Chloride = 325

Calcium Hydroxide = 222

Ferric Hydroxide = 214

Calcium Chloride = 333

325 + 222 = 547 = 214 + 333. Once you are sure you have

balanced the equation correctly you can proceed to calculate

the fractional weights in comparison to the known.

Ferric Chloride 325/325 = 1

Calcium Hydroxide 222/325 = 0.68

Ferric Hydroxide 214/325 = 0.66

Calcium Chloride 333/325 = 1.02

It is given in the problem that the plant is using 17 mg/l of

ferric chloride. We will produce .66 mg/l of ferric hydroxide

for every mg/l of ferric chloride.

17 mg/l × .66 = 11.2 mg/l of ferric hydroxide

This means that for every liter of water treated by the plant,

we will have 11.2 milligrams of ferric hydroxide precipitate.

Since the plant is processing 8 million gallons of water a day,

we convert to determine how many lbs of ferric hydroxide

sludge the plant produces on a dry weight basis.

The sludge pumped out of the bottom of the clarifier each day

contains 747 pounds of ferric hydroxide, but the ferric hydrox-

ide only represents 70 percent of the sludge. We want to know

how many total pounds are hauled away, so we perform the

following calculation to determine the total weight of sludge.

Although we have calcu-

lated all the fractional

relationships in each

problem, we didn’t really

need to. We have only

used the fractional pro-

portions for the known

and the unknown. If we

were performing multi-

ple calculations for the

same equation, this

information for all the

compounds represented

in the equation would

probably come in

handy. Since we are

actually only using two

of them in each calcula-

tion of this type, you

can save some time by

just calculating the

unknown. The known is

always 1. The unknown

divided by the known is

the information required

to solve the problem.

.70

weightDryweightTotal =

11.2 mg

� 8 Mgal

� 8.34 = 747lb/dayl day

747 lb

dayTotal weight = __________ = 1067 lb

.70 day


The plant disposes of 1,067 pounds of sludge a day, slightly

more than half a ton. The sludge is 70 percent ferric hydrox-

ide and 30 percent water. Can you figure out how many

pounds of sludge the plant produces a year if it operates

every day of the year? There are 365 days in a year, so multi-

ply 1,067 by 365, then divide by 2,000 (remember that 2000

lbs = 1 ton), and you will find that the plant produces almost

195 tons of sludge a year.

52

If you see an equation like this for ferrous sulfate

you need to notice that there are 7 water molecules included in the formula for ferrous sul-fate (2FeSO4 × 7H2O). This means that for every molecule of ferrous sulfate there are 7 mol-ecules of water. To calculate the molecular weight of ferrous sulfate you’ll need to calculatethe molecular weight for 1 molecule of FeSO4 plus 7 molecules of H2O, then multiply thisnumber by 2.

Fe = 56S =32O = 16

H = 1

2FeSO4 × 7H2O, m.w. = 556

OHCOCaSOFe(OH)O)Ca(HCO0HFeSO 224322324 13422.5272 +++↓→+× +

Ferric Calcium Ferric Calcium Carbon

Sulfate Bicarbonate Hydroxide Sulfate Dioxide

[ ]16)(1)(274(16)32562 ++++

[ ] 5561261522 =+


Self-Check 4

Complete the following statements with the correct answer.

1. If a water treatment plant is processing 11 Mgal/day (million gallons a day) and is using7 mg/l of ferric chloride with the addition of lime, as calcium hydroxide, to enhance thecoagulation process, how many pounds of ferric hydroxide sludge are produced each dayif the sludge is 70 percent ferric hydroxide by weight? Assume the sludge is purely ferrichydroxide and water.

a. 424 lbs/day c. 400 lbs/dayb. 605 lbs/day d. 72.5 lbs/day

2. If a water treatment plant is processing 9 Mgal/day (million gallons a day) and is using 6mg/l of alum with the addition of lime, as calcium hydroxide, to enhance the coagulationprocess, how many pounds of aluminum hydroxide sludge are produced each day on adry weight basis?

a. 100 lbs/day c. 62 lbs/dayb. 205 lbs/day d. 190 lbs/day

3. If a water treatment plant is processing 13 Mgal/day (million gallons a day) and is using15 mg/l of ferrous sulfate to remove calcium bicarbonate, how many pounds of ferrichydroxide sludge are produced each day on a dry weight basis?


(Continued)




OHCOCaSOFe(OH)O)Ca(HCOOHFeSO 224322324 13422.5272 +++↓→+× +







Self-Check 4


4. If a treatment plant needs to remove 3 mg/l of magnesium sulfate MgSO4 hardness,how much lime as Ca(OH)2 should the operator add?

a. 1 mg/l c. 3 mg/lb. 2 mg/l d. 4 mg/l

5. If a water treatment plant is processing 25 Mgal/day (million gallons a day) and is using 4 mg/l of ferric chloride with the addition of lime, as calcium hydroxide, to enhance thecoagulation process, about how many pounds of ferric hydroxide sludge are producedeach day on a dry weight basis?


6. If a water treatment plant is processing 50 Mgal/day (million gallons a day) and is using 5 mg/l of alum with the addition of lime, as calcium hydroxide, to enhance the coagula-tion process, about how many pounds of aluminum hydroxide sludge are produced eachday if the sludge is 80 percent aluminum hydroxide by weight? Assume the sludge ispurely aluminum hydroxide and water.

a. 1,107 lbs/day c. 1,188 lbs/dayb. 1,205 lbs/day d. 1,090 lbs/day

(Continued)







4224 CaSOMg(OH)Ca(OH)MgSO +↓→+


Self-Check 4


7. If a water treatment plant is processing 10 Mgal/day (million gallons a day) and is using3 mg/l of ferrous sulfate to remove calcium bicarbonate, about how many pounds of ferric hydroxide sludge are produced each day on a dry weight basis?


8. If a water treatment plant is processing 9 Mgal/day (million gallons a day) and is using 6 mg/l of ferric chloride with the addition of lime, as calcium hydroxide, to enhance thecoagulation process, about how many pounds of ferric hydroxide sludge are producedeach day on a dry weight basis?






OHCOCaSOFe(OH)O)Ca(HCO0HFeSO 224322324 13422.5272 +++↓→+× +




NOTES

57

Self-Check 1

1. False

2. True

3. False

4. False

5. False

6. b

7. c

8. d

9. a

Self-Check 2

1. False

2. False

3. EPA or Environmental Protection Agency

4. Equipment manufacturers and design engineers

5. Cation conductivity is used to check the purity of steam

for carryover.

6. c

7. c

8. b

9. c

10. b

An

sw

er

sA

ns

we

rs

Self-Check Answers58

Self-Check 3

1. False

2. False

3. False

4. Discrete, floc, hindered, and compression

5. Screening, settling, coagulation, flocculation, settling,

and filtration

6. a

7. d

8. b

9. a

10. c

Self-Check 4

1. b

2. b

3. b

4. b

5. c

6. c

7. c

8. b

59

925 Oak Street

Scranton, Pennsylvania 18515-0001


When you feel confident that you have mastered the material in this study unit, complete the following examination. Then submitonly your answers to the school for grading, using one of theexamination answer options described in your “Test Materials”envelope. Send your answers for this examination as soon as youcomplete it. Do not wait until another examination is ready.

Questions 1–20: Select the one best answer to each question.

1. Chemicals used to create floc are called

A. flocculants. C. coagulants.B. conglomerates. D. distillers.

2. If a water treatment plant is processing 25 Mgal/day (million gallons a day) and is using 3 mg/l of ferric chloride with theaddition of lime, as calcium hydroxide, to enhance the coagula-tion process, how many pounds of ferric hydroxide sludge areproduced each day on a dry weight basis?

A. 309 lbs/day C. 547 lbs/dayB. 413 lbs/day D. 572 lbs/day

EXAMINATION NUMBER:

78600900Whichever method you use in submitting your exam

answers to the school, you must use the number above.

For the quickest test results, go to http://www.takeexamsonline.com

2CaCl3Fe(OH)2Ca(OH)3FeCl +↓→+



Ex

am

ina

tion

Ex

am

ina

tion

http://www.takeexamsonline.com

Examination60

3. Turbidity is an indication of the amount of

A. dissolved solids. C. acidity.B. dissolved gases. D. suspended solids.

4. A common chemical used to remove dissolved oxygen is sodium

A. chloride. C. sulfite.B. carbonate. D. hydroxide.

5. Which of the following will dissolve the most CaCO3?

A. 100 ml of water at 40° F C. 1 liter of water at 40° FB. 100 ml of water at 75° F D. 1 liter of water at 75° F

6. Water hardness is expressed in terms of

A. Mg(OH)2. C. Ca(OH)2.B. CaCO3. D. Al2(SO4)3.

7. The coagulation process requires

A. rapid dispersion of chemicals. C. stagnant still water conditions.B. continuous high-speed agitation. D. frequent screening.

8. Cation conductivity is used in power plants to check for ______ purity.

A. influent C. steam B. circulating water D. drinking water

9. The slowest type of settling is

A. discrete. C. hindered.B. floc. D. compression.

10. Sodium in a steam plant’s condensate is an indication of

A. carryover. C. boiler tube failure.B. condenser tube leaks. D. normal conditions.

11. Which process is commonly used to remove dissolved gases?

A. Deaeration C. FlocculationB. Distillation D. Filtration

12. Hindered settling occurs

A. at the top of the tank. C. in the sludge blanket.B. below the sludge blanket. D. above the sludge blanket.

Examination 61

13. If a water treatment plant is processing 11 Mgal/day (million gallons a day) and is using 7mg/l of alum with the addition of lime, as calcium hydroxide, to enhance the coagulationprocess, how many pounds of aluminum hydroxide sludge are produced each day if thesludge is 85 percent aluminum hydroxide by weight? Assume the sludge is purely alu-minum hydroxide and water.

A. 146 lbs/day C. 345 lbs/dayB. 292 lbs/day D. 460 lbs/day

14. Nitrogen (N2) in water is a

A. dissolved gas. C. suspended gas.B. dissolved solid. D. suspended solid.

15. If the dry weight of a 1000 ml filtered sample is 6 mg, the TSS is

A. 3 mg/l. C. 30 mg/l.B. 6 mg/l. D. 60 mg/l.

16. Sodium levels in the steam indicate

A. high-quality steam. C. circulating water failure.B. carryover. D. excessive blowdown.

17. Biological contaminants have the largest effect on the power plant’s

A. drinking water. C. feed water.B. service water. D. cooling water.

18. The flocculation process could be represented by

A. making popcorn balls. C. fueling your car.B. mowing the lawn. D. pruning a tree.

19. Clarifiers include which three processes?

A. Screening, coagulation, filtrationB. Flocculation, settling, filtrationC. Coagulation, flocculation, settlingD. Coagulation, flocculation, screening

20. Which of the following dissolved gases creates the most problems in power plants?

A. Carbon dioxide C. Oxygen B. Nitrogen D. Chlorine

4CaSO3Al(OH)2Ca(OH)3)4(SO2Al +→+



study unit power plant water treatment, part 2

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