water treatment technology (tas 3010) lecture notes 3c - water chemistry extended - organic and...
DESCRIPTION
WATER TREATMENT TECHNOLOGY (TAS 3010) LECTURE NOTESUNIVERSITY MALAYSIA TERENGGANU 2009Disclaimer: I don't own this file. If you believe you do, and you don't want it to be published here, please let me know. I will remove it immediatelyTRANSCRIPT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Menu
1) Introduction
2) Organic Compound
A) BOD
B) COD
C) Suspended Solid
3) Inorganic Compound
A) Metal
B) Non Metal
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
SOURCES???SOURCES???
Organic Compound
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Organic Compound - Lists
Antigens Polysaccharides, carbohydrates and sugars Enzymes Hormones Lipids and fatty acids Nucleic acids Proteins, peptides and amino acids Vitamins
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Organic Matter – Classification
• (a) biodegradable organics
• (b) non-biodegradable organics
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Biodegradable Organics
• food to microorganism
• fast and easily oxidized by micro organism
• Examples :
- starch, fat protein, alcohol, human and animal waste
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Non-biodegradable Organics
• difficult and much more longer to biodegrade
• or toxic to micro organisms
• Examples :
- pvc, pesticide, industrial waste, cellulose, phenol, lignic acid.
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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.
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
BOD Test
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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.
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
BOD Analysis
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
• 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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
• 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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
• COD > BOD
• COD ~ ultimate BOD
• COD/BOD ~ 2, biodegradable organic
• COD >> BOD, non-biodegradable organic
Relation between COD and BOD:
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
COD - Effects
• Diseases
• Aesthetic
• disturb human activity
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
TSS - Laboratory analysis
• Diagram of laboratory procedure to determine total solid and total volatile solids concentration of a water or wastewater sample.
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Diagram of laboratory procedure to determine the suspended solid and volatile suspended solids concentrations of a water or wastewater sample.
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
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-)
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
Inorganic Compound – Con’d
• Any compound that doesn't contain carbon (except for carbon dioxide, carbon monoxide, and carbonates).
Izan Jaafar, Engineering Science, FST, UMTIzan Jaafar, Engineering Science, FST, UMT
THANK YOUR FOR THANK YOUR FOR
YOUR ATTENTION…..YOUR ATTENTION…..
THANK YOUR FOR THANK YOUR FOR
YOUR ATTENTION…..YOUR ATTENTION…..