water treatment technology (tas 3010) lecture notes 3c - water chemistry extended - organic and...

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 3c : CHAPTER 3c : CHAPTER 3c : CHAPTER 3c :

TAS 3101 : WATER TREATMENT TECHNOLOGY

Water Chemistry : Water Chemistry :

Organic and Inorganic Organic and Inorganic CompoundCompound

Water Chemistry : Water Chemistry :

Organic and Inorganic Organic and Inorganic CompoundCompound


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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

“Naturally occurring (animal or plant-produced

or synthetic) substances containing mainly

carbon, hydrogen, nitrogen, and oxygen”

All organic compound contain carbon in

combination with one or more elements”

(except carbon dioxide, carbon monoxide, and

carbonates)

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


SOURCES???SOURCES???

Organic Compound


Organic Compound – Sources

• Nature : • fibre, vegetable oils, animal oils and fats, cellulose,

starch• Synthesis:

• (Formation of a compound from simpler compounds or elements)

• compounds produced by manufacturing process e.g. DDT, polyvinyl chloride

• Fermentation:• Any of a group of chemical reactions induced by

living or nonliving ferments that split complex organic compounds into relatively simple substances.

• alcohols, acetone, glycerol, antibiotic, acids


Organic Compound - Lists

Antigens Polysaccharides, carbohydrates and sugars Enzymes Hormones Lipids and fatty acids Nucleic acids Proteins, peptides and amino acids Vitamins


Organic Matter – Classification

• (a) biodegradable organics

• (b) non-biodegradable organics


Biodegradable Organics

• food to microorganism

• fast and easily oxidized by micro organism

• Examples :

- starch, fat protein, alcohol, human and animal waste


Non-biodegradable Organics

• difficult and much more longer to biodegrade

• or toxic to micro organisms

• Examples :

- pvc, pesticide, industrial waste, cellulose, phenol, lignic acid.


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


A- Biochemical Oxygen Demand

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

• Expressed in milligrams of oxygen required per liter of wastewater (mg/L).

• Made up of two (2) parts :

– 1) Carbonaceous oxygen demand (CBOD)

– 2) Nitrogeous oxygen demand (NBOD)


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 20°C

• 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/fraction

= volume of sample/volume of sample plus dilution water

= s

s + w


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 (0)

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

• BOD5 ??– Total amount of oxygen consumed by microorganisms

during the first five days of biodegradation


BOD Test


Dilution of Waste

• A standard BOD bottles holds 300 mL, so P is just the volume of sample divided by 300 mL

• by direct pipetting into 300 mL BOD bottle


A 10 mL sample of water mixed with enough water to fill a 300 mL bottle has an initial DO of 9.0 mg/L. To assure an accurate test, it is desirable to have an at least a 2.0 mg/L drop in DO during the five-day run and the final DO should be at least 2.0 mg/L. For what range of BOD5 would this dilution produce the desired results?

Dilution - Problems


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

Volume and Range


BOD Analysis - a

• In aerobic processes (O2 is present), hetherotropic

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.


CBOD

• The overall biological conversion proceeds sequentially, with

oxidation of carbonaceous material as the first step (known

as carbonaceous oxygen demand):

Organic matter + O2 CO2 + H2O + new cells


NBOD

• Under continuing aerobic conditions, autotrophic

bacteria then convert the nitrogen in organic

compounds to nitrates (known as nitrification oxygen

demand):

Organic – Nitrogen Ammonia-Nitrogen


BOD Analysis


Ultimate BOD (L0)

• The ultimate BOD (L0) is defined as the maximum BOD exerted by the waste – Ultimate carbonecous oxygen demand.

• Sum of the amount of oxygen already consumed by the waste in the first t days (BODt), plus the amount of oxygen remaining to be consumed after time t. That is;

L0 = BODt + Lt


Ultimate BOD (L0)

• The carbonaceous oxygen demand curve can be expressed mathematically as:

BODt = L0 (1 – e-kt)

• Where

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

L0 = ultimate BOD, mg/L

k = reaction rate constants, day-1


Ultimate BOD (L0) - Problems

The dilution factor P for an unseeded mixture of sample and water is 0.03. The DO of the mixture is initially 9.0 mg/L and after 5 days, it has dropped to 3.0 mg/L. The reaction rate constant k has been found to be 0.22 day -1.

Find :

a) Five day-BOD of the sample

b) Ultimate carbonaceous BOD


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)


BOD K-rate ??? • BOD rate constant k• “Factor indicates rate of biodegradation of wastes”• As k increase, the rate of DO increase• Reaction rate factors:

– Nature of the waste– Available microorganisms capability– Temperature

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


• K (base 10) , Lθ = BOD

1 – 10-kt

• K (base e) Lθ = BOD

1 – e-kt

• 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 θ(T-20)

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


TL0 = 20L0 [1 + 0.02 (T – 20)]

Where

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

• 20L0 = temperature BOD at 20°C, mg/L

Ultimate BOD (L0)


Example:

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

T = 30°C, s = 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°C


Example:• calculate values of BOD5 at 20°C

Solution :

1) Use formula;K20 = KT ---------------(from K30) 1.047 (T-20)

2) TL0 = 20L0 [1 + 0.02 (T – 20)]

3) Ultimate BOD @ 30°C

L30 = BOD BOD = BOD3 value 1 – 10-kt

4) values of BOD5 at 20°C = Ultimate BOD @ 30°C

L20 = BOD BOD = BOD5 value 1 – 10-kt


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:

(mL blank – mL sample titrant) x (mol of FAS)8000

COD = ___________________________________ mL sample

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


TSS - 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-)


Inorganic Compound – Con’d

• Any compound that doesn't contain carbon (except for carbon dioxide, carbon monoxide, and carbonates).


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