The Islamic University of Gaza
Faculty of Engineering
Civil Engineering Department
Environmental Engineering
(ECIV 4324)
Instructor: Dr. Abdelmajid Nassar
Lect. 10-11
Water Quality: Definition, Characteristics, and
Perspectives
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ORGANICS
•Biodegradable Organics
Biodegradable materials consists organics that can be utilized for
food by naturally occurring microorganisms within a reasonable
length of time.
Source of organics
• Organics include fats, proteins, alcohols, acids, aldehydes, and
esters.
• Organics are the end product of the initial microbial decomposition
of plant or animal tissue.
• Result from domestic or industrial wastewater discharge.
Microbial utilization of dissolved organics can be accompanied by
oxidation (addition of oxygen to elements of the organic molecule)
or by reduction (addition of hydrogen).
Biochemical Oxygen Demand - BOD, g/m3
Definition:
Mass of dissolved oxygen consumed in definite time by the
aerobic microbial utilization of organics substances.
Measurement (Standard Test):
• 300 mL BOD bottle is used and placed in an air-tight container
at 20oC for 5 days.
• Light must excluded from the bottle to prevent algal growth
that may produce oxygen in the bottle.
• Dilution of the sample with BOD–free , oxygen saturated
water is necessary to measure BOD values greater than just a
few milligrams per litre.
Determination of BOD
The BOD of a diluted sample is calculated by:
BOD (mg/L) = (DI - DF)/P
• Where:
– DI = initial DO in the diluted sample (mg/L)
– DF = DO in the diluted sample after incubation (mg/L)
– P = decimal fraction of wastewater in total volume
= mL wastewater/300mL
BOD Test Range
Assume an initial DO 9 (mg/L) in the mixture, with a minimum
of 2 and a maximum of 7 mg/L of O2 being consumed.
Example
• The BOD of a wastewater is suspected to range from 50 - 200
mg/L.
• Three dilutions are prepared to cover this range. The procedure
is the same in each case.
1. First the sample is placed in the standard BOD bottle and is then
diluted to 300 mL with organic- free, oxygen-saturated water.
2. The initial dissolved oxygen is determined, and
3. The bottles tightly stoppered and placed at 20oC for 5 days,
After which the dissolved oxygen is again determined.
wastewater DOI DO5 O2 used P BOD5
mL mg/L mg/L mg/L mg/L
5 9.2 6.9 2.3 0.017 138
10 9.1 4.4 4.7 0.033 141
20 8.9 1.5 7.4 (> max.) 0.067 111
The average BOD of the sample = (138+141)/2 = 140 mg/L
BOD for any time period can be determined as:
• The rate at which organics are utilized by microorganisms is
assumed to be a first- order reaction,
• The rate at which organics utilized is proportional to the amount
available. dL
t
= −k .Lt
dt
where:
• Lt is the oxygen equivalent of the organics at time t (mg/L),
• k is the reaction constant (d-1)
L t
dLt Lt − kt
∫L Lt = −k ∫0 dt ⇒ ln Lo = −kt ⇒ Lt = Lo e
0
• Lo is the total oxygen equivalent of the organics at time 0 (mg/L),
• Lt is the oxygen remaining at time t (mg/L),
٨
The amount of oxygen used in the consumption of organics =
BOD , can be found from the Lt value as the following:
If
• Lo is the total oxygen equivalent of the total mass organics at
time 0 (mg/L),
• Lt is the oxygen remaining at time t (mg/L),
− kt
Yt = Lo − Lt = Lo − Lo e −kt
⇒ Yt = Lo (1 − e )
Yt represents the BODt of the water
Note:
BOD (ultimate) = the initial oxygen equivalent of the water Lo
K constant
• The value of k determines the speed of the BOD reaction.
• K values range from (0.1-0.5) d-1 depending on the nature of the
organic molecules.
• K value for any given organic compound is temperature-
dependent, because microorganisms are more active at higher
temperatures. (k increase with increasing temperatures).
kT = k 20θ (T − 20 )
θ = 1.047
Example
The BOD5 of a wastewater is determined to be 150 mg/L at
20oC.
The k value is known to be 0.23 per day.
What would the BOD8 be if the test were run at 15oC?
Solution:
BODt = Lo (1 − e − kt ) ⇒ BOD5 = Lo (1 − e − kt )
− 0.23×5 150
150 = Lo (1 − e ) ⇒ Lo = − 0.23×5
= 220mg / L
1 −o e
kT = k 20θ (T − 20 ) ⇒ k15 = k 20 ×1.047 (15− 20 ) = 0.18
BODt = Lo (1 − o e − kt )
BOD8 = 220(1 − o e −0.18×8 ) = 168mg / L
Example
200 mL of a river water was collected. 2-
mL of river water diluted to 1 L, aerated
and seeded. The dissolved oxygen
content was 7.8 mg/L initially. After 5
days, the dissolved oxygen content had
dropped to 5.9 mg/L. After 20 days, the
dissolved oxygen content had dropped to
5.3 mg/L. What is the ultimate BOD?
Solution
BOD5 = 7.8 mg/L – 5.9 mg/L = 950 mg/L
2 mL/1000 mL
BOD20 = 7.8 mg/L – 5.3 mg/L = 1250 mg/L
2 mL/1000 mL
BOD5 = (1-e-k(5 days))
BOD20 (1-e-k(20 days))
950 = (1-e-k(5 days))
1250 (1-e-k(20 days))
0.76 = (1-e-k(5 days))
(1-e-k(20 days))
0.76 = (1-e-k(5 days))
(1-e-k(20 days))
0.76 – 0.76 e-k(20) = 1 – e-k(5)
e-k(5) – 0.76 e-k(20) = 1-0.76 = 0.24
We just graph the left side as a function of k and
look to see where it equals 0.24 (or you can use
solver on your calculator)
k e^-5k - 0.76e^20k
0 0.24
0.025 0.421534
0.6
0.05 0.499212
0.075 0.51771
0.5
0.1 0.503676
0.4 0.125 0.472877
0.15 0.434528
0.3 0.175 0.393912
0.2 0.35396
0.2
0.225 0.31621
0.25 0.281384
0.1
0.275 0.249734
0 0.3 0.221246
0 0.1 0.2 0.3 0.4 0.5 0.325 0.195769
0.35 0.173081
0.375 0.152935
0.4 0.13508
The k value is…
0.28 day-1
BOD5 = BOD (1-e-k(5 days))
950 mg/L = BOD (1 – e-(0.28)(5))
BOD = 1261 mg/L
Notes
•The water body is considered to be very clean if its BOD5 at
20oC is less than 1 mg/litre (i.e. ppm).
• The water body is considered poor if its BOD5 at 20oC is more
than 5 mg/litre.
• The BOD5 estimate however excludes complex organics such
as cellulose, and proteins, which cannot be readily biodegraded
by bacteria.
