Determine ultimate BOD for sewage having 5-day BOD at 20°C as 180 ppm. Assume the de-oxygenation constant as 0.8 per day.
Biochemical Oxygen Demand (BOD) is a measure of the amount of dissolved oxygen required by aerobic biological microorganisms to break down organic material present in a water sample at certain temperature over a specific time period.
The question asks to find the ultimate BOD ($L_0$) of a sewage sample. The ultimate BOD represents the total amount of oxygen required for the complete decomposition of the organic matter.
We are given the 5-day BOD ($BOD_5$) at 20°C, which is the amount of oxygen consumed over 5 days at 20°C. We are also given the de-oxygenation constant ($k$). The relationship between BOD at any time $t$ ($BOD_t$) and the ultimate BOD ($L_0$) is given by the formula, assuming a first-order reaction rate.
The formula relating $BOD_t$, $L_0$, and the de-oxygenation constant is typically given as:
$\text{BOD}_t = L_0 (1 - e^{-kt})$ (using base-e constant $k$)
or
$\text{BOD}_t = L_0 (1 - 10^{-k't})$ (using base-10 constant $k'$)
Often, in practical problems, the given constant is the base-10 constant ($k'$) unless specified otherwise, especially when its value is relatively high like 0.8 per day. Let's assume the given de-oxygenation constant of 0.8 per day is the base-10 constant $k'$.
We can calculate the ultimate BOD ($L_0$) using the given 5-day BOD ($BOD_5$), time ($t$), and the de-oxygenation constant ($k'$).
$\text{BOD}_t = L_0 (1 - 10^{-k't})$
We are given $\text{BOD}_5 = 180$ ppm and $t=5$ days, $k'=0.8$ per day.
$180 = L_0 (1 - 10^{-(0.8 \times 5)})$
First, calculate the term in the exponent:
$0.8 \times 5 = 4$
So, the equation becomes:
$180 = L_0 (1 - 10^{-4})$
Calculate $10^{-4}$:
$10^{-4} = 0.0001$
Substitute this back into the equation:
$180 = L_0 (1 - 0.0001)$
$180 = L_0 (0.9999)$
Now, solve for $L_0$:
$L_0 = \frac{180}{0.9999}$
Calculating the value:
$L_0 \approx 180.018$ ppm
The calculated ultimate BOD is approximately 180.018 ppm. Looking at the options, 180 ppm is the closest value, which suggests either a slight rounding or that the 5-day BOD value itself was rounded in the problem statement leading to this result.
| Term | Meaning |
|---|---|
| BOD | Biochemical Oxygen Demand - amount of oxygen consumed by microorganisms |
| Ultimate BOD ($L_0$) | Total oxygen required for complete decomposition of organic matter |
| 5-day BOD ($BOD_5$) | Oxygen consumed in 5 days at a standard temperature (usually 20°C) |
| De-oxygenation Constant ($k$ or $k'$) | Rate at which the BOD reaction proceeds |
Based on the calculation using the de-oxygenation constant as a base-10 value (0.8 per day), the ultimate BOD is approximately 180 ppm, which matches one of the provided options.
Conventionally, the biochemical oxygen demand (BOD) is measured for _______ days.
Consider the following statements:
1. Ammonia nitrogen is a measure of nitrogen present as ammonium hydroxide and ammonium salts. It will progressively decrease as sewage gets treated.
2. Organic nitrogen is the total nitrogenous matter in sewages excepting that present as ammonia nitrogen, nitrites and nitrates. It becomes ammonia in anaerobic decomposition and nitrites or nitrates in aerobic decomposition.
Which of the above statements is/are correct?Identify the correct relation from the following:
Where BOD, COD, and TOD are Biochemical, Chemical, and Theoretical oxygen demand, respectively.
What is the correct range of \(\frac{{BO{D_u}}}{{COD}}\) for a waste water to be considered as fully biodegradable?
(where BODu = ultimate Bio-chemical oxygen demand and COD = chemical oxygen demand)
What does Chemical Oxygen Demand (COD) indicate?