29103278 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 :
TAS 3101 : WATER TREATMENT TECHNOLOGY
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 CompoundOrganic 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 20o
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
= 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
Kdt
L
dL
=
− KtKtt
e
L
L −−
== 10 KtKtt
e
L
L −−
== 10
Kt
o
tK
ot LeLL −−
== 10'
)101( Kt
oLy −
−=

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

• 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 20O
C, mg/L
Ultimate BOD (L0)

Example:
The following data were obtained from an experiment to determine the BOD rate
constant.
T = 30o
C, 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 O
C

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 balance
2. place the filter paper on the filter apparatus
3. apply vacuum and filter 100 mL (or a larger volume if total
suspended matter is low) well mixed sample
4. dry the filter paper in an oven at 1030
C to 1050
C for at least 1 hour
5. 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 sample
where 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-
)

THANK YOUR FORTHANK YOUR FOR
YOUR ATTENTION…..YOUR ATTENTION…..

29103278 water-treatment-technology-tas-3010-lecture-notes-3-water-chemistry

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