WATER TREATMENT TECHNOLOGY (TAS 3010) LECTURE NOTES 3 - Water Chemistry

Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT

© SHAHRUL ISMAIL, DESc.University College of Science and Technology Malaysia

CHAPTER 3:Environmental Microbiology

CHAPTER 3 : CHAPTER 3 : CHAPTER 3 : CHAPTER 3 :

TAS 3101 : WATER TREATMENT TECHNOLOGY

Water ChemistryWater Chemistry

Water ChemistryWater Chemistry


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1) Introduction

2) Organic Compound

A) BOD

B) COD

C) Suspended Solid

3) Inorganic Compound

A) Metal

B) Non Metal


1 - Organic Compound

“All organic compound contain carbon in

combination with one or more elements”

Organic Compound Organic Compound


Organic Compound - Properties

• Combustible

• Have lower melting and boiling points

• Less soluble in water

• Very high molecular weight

• Mostly serve as a source of food for micro organisms


1 - Organic Compound

SOURCES????SOURCES????


Organic Compound - Sources

• Nature :

• Synthesis:

• Fermentation:


Organic Matter - Classification

• (a) biodegradable organics

• (b) non-biodegradable organics


Biodegradable Organics

• food to microorganism

• fast and easily oxidized by micro organism

• e.g. . .


Non-biodegradable Organics

• difficult and much more longer to biodegrade

• or toxic to micro organisms

• e.g. . .


NBO - Effects

• depletion of dissolved oxygen in the water– destroying aquatic life– damaging the ecosystem

• some organic can cause cancer

• trihalomethanes (THM – carcinogenic compound) are produced in water and wastewater treatment plants when natural organic compounds combine with chlorine added for disinfection purposes.


Laboratory Analysis

• Various parameters are used as a measure of the organic strength of wastewater:

• (a) BOD – Biochemical oxygen demand

• (b) COD – Chemical oxygen demand

• (c) TOC – Total organic carbon

• (d) VSS – volatile suspended solid


A- Biochemical Oxygen Demand

• Def: the quantity of oxygen utilized by a mixed population of micro organisms to biologically degrade the organic matter in the wastewater under aerobic condition.

• BOD is the most important parameter in water pollution control

• It is used a measure of organic pollution as a basis for estimating the oxygen

• Needed for biological processes, as and indicator of process performance

• BOD test


Test Method

• A water sample containing degradable organic matter is placed in a BOD bottle.

• If needed, add dilution water (known quantity). • Dilution water is prepared by adding phosphate buffer

(pH 7.2), magnesium sulphate, calcium chloride and ferric chloride into distilled water. Aerate the dilution water to saturate it with oxygen before use.

• Measure DO in the bottle after 15 minutes (DOi)• Closed the bottle and placed it in incubator for 5 days,at

temperature 20oC• After 5 days, measure DO in the bottle (DOt)


BOD - Calculation

BODt = DOi – DOt

p

Where:

• BODt=biological oxygen demand, mg/L

• DOi = initial DO of the diluted wastewater sample

about 15 min. After preparation, mg/L

DOt = final DO of the diluted wastewater sample

after incubation for t days, mg/L• P = dilution factor

= ml of wastewater sample

ml BOD bottle


BOD test - Dilution

• for a valid BOD test, the final DO should not be less than 1 mg/L. BOD test is invalid if DOt value near zero

• dilution can be decrease organic strength of the sample. By using dilution factor, the actual value can be obtained.


Dilution of Waste

• by direct pipetting into 300 mL BOD bottle


Volume of sample (mL) Range of BOD value (mg/L)

0.02 30,000 – 105,000

0.05 12,000 – 42,000

0.10 6,000 – 21,000

0.20 3,000 – 10,500

0.50 1,200 – 4,200

1.00 600 – 2,100

2.00 300 – 1,050

5.00 120 – 420

10.00 60 – 210

20.00 30 – 105

50.00 12 – 42

100.00 6 – 21

300.00 0 – 7


BOD Analysis

• In aerobic processes (O2 is present), heterotropic bacteria oxydize about 1/3 of the colloidal and dissolved organic matter to stable end products (CO2 + H2O) and convert the remaining 2/3 into new microbial cells that can be removed from the wastewater by settling.

• The overall biological conversion proceeds sequentially, with oxidation of carbonaceous material as the first step (known as nitrification oxygen demand):

-------------------------------------------

• Under continuing aerobic conditions, autotrophic bacteria then convert the nitrogen in organic compounds to nitrates (known as nitrification oxygen demand):

-------------------------------------------


BOD Analysis


Ultimate BOD

• The ultimate BOD (L0) is defined as the maximum BOD exerted by the waste. The carbonaceous oxygen demand curve can be expressed mathematically as:

BODt = y at any time t, (day) and Lo is the ultimate BOD

K= K’/2.303• Where

BODt = biochemical oxygen demand at time t, mg/L

L0 = ultimate BOD, mg/L

K = reaction rate constants, day-1

KdtL

dL

KtKtt eL

L 10 KtKtt eL

L 10

Kto

tKot LeLL 10'

)101( KtoLy


BOD K-rate

Time (day) BODt (mg/L) [time/BODt]1/3

1 X [1/X]1/3

2 Y [2/Y]1/3

3 Z [3/Z]1/3

• from the experiment results of BOD for various values of t, calculate [time/BODt]1/3 for each day

• plot [t/BODt]1/3 versus t

• determine the intercept (A) and slope (B) from the plot• calculate K = 2.61 (B/A)


• K (base 10)

• K (base e)

• K = k/2.3

• Simple compounds such as sugars and starches are easily utilized by micro organisms have high k rate

• More complex materials such as phenols and cellulose are difficult to assimilate have

low k value

BOD constant, k, per day


Water type K, per day (base 10)

Tap water 0.04

Surface water 0.04 – 0.1

Raw sewage 0.15 – 0.30

Well-treated sewage 0.05 – 0.10

Typical values of K for various water


Temperature• Most biological processes speed up as temperature

increases and slow down as the temperature drops. The rate utilization is effected by temperature.

• The relationship for the change in the reaction rate constant (K) with temperature is expressed as:

KT = K20 x Ө(T-20) Where • KT = reaction rate constant at temperature T,

per day• K20 = reaction rate constant at 20°C, per day • Ө = temperature coefficient = 1.047• T = temperature of biological reaction, 0°C


--------------=----------------------------

Where

• TL0 = ultimate BOD at temperature T, mg/L

• 20L0 = temperature BOD at 20OC, mg/L

Ultimate BOD (L0)


Example:

The following data were obtained from an experiment to determine the BOD rate

constant.

T = 30oC, As = 100mL (total amount of water samples used in the experiment)

Time (days) DO (mg/L)

0 7.4

1 5.5

2 4.5

3 3.7

4 2.5

5 2.1

Question:

• calculate values of BOD3

• determine the BOD rate constant, K30

• calculate value of BOD5 at 20 OC


Definition

• the quantity of oxygen needed to chemically oxidize the organic compound in sample, converted to carbon dioxide and water

• commonly used to define the strength of industrial wastewaters

B) Chemical oxygen demand


• Add measured quantities of potassium dichromate, sulfuric acid reagent containing silver sulphate, and a measured volume of sample into a flask,

• The mixture is refluxed (vaporized and condensed) for two hours. The oxidation or organic matter converts dichromate to trivalent chromium,

• The mixture is titrated with ferrous ammonium sulphate (FAS) to measure the axcess dichromate remaining in sample.

• A blank sample of distilled water is carried through the same COD testing procedures as the wastewater sample.

• COD is calculated from the following equation:

Where • COD = chemical oxygen demand, mg/L• A = amount of ferrous ammonium sulphate titrant added to blank, mL• B = amount of titrant added to sample, mL• A = volume of sample, mL• 8000 = multiplier to express COD in mg/L of oxygen

COD - Test procedure


• COD > BOD

• COD ~ ultimate BOD

• COD/BOD ~ 2, biodegradable organic

• COD >> BOD, non-biodegradable organic

Relation between COD and BOD:


COD - Effects

• Diseases

• Aesthetic

• disturb human activity


COD - Laboratory analysis

• Diagram of laboratory procedure to determine total solid and total volatile solids concentration of a water or wastewater sample.


Diagram of laboratory procedure to determine the suspended solid and volatile suspended solids concentrations of a water or wastewater sample.


Suspended solid concentrations

1. weigh a filter paper on an electrical balance2. place the filter paper on the filter apparatus3. apply vacuum and filter 100 mL (or a larger volume if total

suspended matter is low) well mixed sample4. dry the filter paper in an oven at 1030C to 1050C for at least 1 hour5. after 1 hour, cool the filter appear in a desiccator and weigh.6. repeat the drying cycle until a constant weigh is attained or until

weight loss is less than 0.5 mg.mg/L suspended solids = [(A – B) x 1000]/mL samplewhere A = weigh of filter paper + suspended matter

B = weigh of filter paper

Total suspended = suspended solids + dissolved solids


2 - Inorganic Compound

• When placed in water, inorganic compounds dissociate into electrically charged atoms referred to as ions

• All atoms linked in ionic bond

• Can be classified into two:

Metal (e.g. Pb2+, Hg+2, Cu +2) non-metal (e.g. Si+4, Cl_, NO3-)


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WATER TREATMENT TECHNOLOGY (TAS 3010) LECTURE NOTES 3 - Water Chemistry

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