Section VIII Water Treatment- Introduction Dissolved Oxygen (DO): - The source of D.O in water is photosynthesis and aeration - It is one of important parameters to measure the water quality. - It gives pleasure taste to water - As the temp D.O - If the D.O concentration decreases to less than 4mg/l all fish die - If the D.O concentration is less than 2 mg/l all organism dies and the water is called septic water - Best D.O concentration is between 8-10 ppm. Optimum is 9ppm. - The maximum naturally accrued is 14 mg/l.
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Dissolved Oxygen (DO) - Philadelphia University · Biochemical Oxygen demand (BOD) : - is the quantity of oxygen that is used by microorganism to stabilize the wastewater, Usually
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Section VIII Water Treatment- Introduction
Dissolved Oxygen (DO):
- The source of D.O in water is photosynthesis and aeration
- It is one of important parameters to measure the water quality.
- It gives pleasure taste to water
- As the temp D.O
- If the D.O concentration decreases to less than 4mg/l all fish die
- If the D.O concentration is less than 2 mg/l all organism dies and the water is
called septic water
- Best D.O concentration is between 8-10 ppm. Optimum is 9ppm.
- The maximum naturally accrued is 14 mg/l.
Section VIII Water Treatment- Introduction
Biochemical Oxygen demand (BOD) :
- is the quantity of oxygen that is used by microorganism to stabilize the wastewater,
Usually measured after 5 days.
- a BOD test can be used to measure waste loadings to treatment plants, plant
efficiency and the effects of a discharge on a receiving stream, and to control the
plant process.
- It is indicator for the required aeration amount.
- The main equation describes the process is:
DO + organic matter CO2 + biological growth
- Drinking water usually has a BOD of less than 1 mg/L
- Ordinary domestic sewage may have a BOD of 200 mg/L.
-Any effluent to be discharged into natural bodies of water should have BOD less than
30 mg/L.
Test Summary:
1- The sample is filled in an airtight bottle and incubated at 20 oC for 5 days.
2- The dissolved oxygen (DO) content of the sample is determined before and
after five days of incubation at 20°C
3- and the BOD is calculated from the difference between initial and final DO.
The initial DO is determined shortly after the dilution is made; all oxygen uptake
occurring after this measurement is included in the BOD measurement
Section VIII Water Treatment- Introduction
Calculations:
BOD5 mg/l = (Initial DO - DO5) x Dilution Factor
Dilution Factor =Bottle Volume (300 ml)
𝑠𝑎𝑚𝑝𝑙𝑒 𝑉𝑜𝑙𝑢𝑚𝑒
BOD At any time:
BODt = L(1-10-kt
)…. Or BODt = L(1-exp-kt
)….
Where:
BODt : BOD at any time
L: ultimate bio-oxygen demand
k: oxygen decay constant, [day-1
]
t: time, [days]
However: The BOD reaction rate constant (K) is dependent on the following:
1. The nature of the waste:
- degradability of the organic matter. For example , Simple sugar
and starches are rapidly degraded
2. The ability of the organisms in the system to utilize the waste
3. The temperature
- the water temperature may vary from place to place for the same river; hence, the
BOD rate constant is adjusted to the temperature of receiving water using
following relationship:
KT = K20 θ (T-20)
Where
T = temperature of interest, oC
KT = BOD rate constant at the temperature of interest, day-1
K20 = BOD rate constant determined at 20oC, day
-1
θ = temperature coefficient. This has a value of 1.056 in general and 1.047 for higher
temperature greater than 20oC
Section VIII Water Treatment- Introduction
BOD incubator
Example:
Determine ultimate BOD for a wastewater having 5 day BOD at 20oC as 160 mg/L.
Assume reaction rate constant as 0.23 per day (base exp).
Solution
BOD5 = Lo ( 1 – exp-k.t
)
160 = Lo (1 – exp-5 x 0.23
)
Therefore, Lo = 234.1 mg/L
Example:
A BOD test is done by pipiting 5 ml of waste water into 300 ml testing bottle. If the
initial DO was 8.4 mg/l and the DO after 5-days of incubation at 20 oC was 3.7mg/l,
calculate the BOD and estimate the 20-days BOD value assuming the reaction decay