Question

The average sewage from a city is 90 million litres per day and the average 5-day Biochemical Oxygen Demand (BOD₅) is 300 mg/l. Average standard BOD₅ of the domestic sewage is 0.08 kg/day/person. The population equivalent of the city is

• A. 337500
• B. 216000
• C. 270000
• D. 168750

Hint:

Find the total BOD load (concentration times flow, unit-converted to kg/day), then divide by the standard per-person BOD load.

Answer 1

The Correct Option is A

Step 1: Recall a direct shortcut used in environmental engineering.
If flow is expressed in million litres per day (MLD) and concentration in mg/l, the load in kg/day is simply the product of the two numbers, with no extra conversion constant needed. Let's confirm why that works, then use it.

Step 2: Derive the shortcut.
1 million litres (ML) $= 10^6$ litres $= 10^3$ m$^3$. A concentration of $C$ mg/l means $C$ grams dissolved in every 1000 litres, that is, $C$ grams per m$^3$. So the mass present in $Q$ million litres per day is
\[ \text{mass} = C \ (\text{g/m}^3) \times (Q \times 10^3) \ (\text{m}^3/\text{day}) = C \times Q \times 10^3 \ \text{g/day} \]
Dividing by 1000 to convert grams to kilograms cancels the $10^3$:
\[ \text{Load (kg/day)} = C \times Q \]
So with $C$ in mg/l and $Q$ in MLD, the load in kg/day is just $C \times Q$.

Step 3: Apply the shortcut.
Here $C = 300$ mg/l and $Q = 90$ MLD, so
\[ \text{Load} = 300 \times 90 = 27{,}000 \ \text{kg/day} \]
This matches the load found by first-principle unit conversion, confirming the shortcut.

Step 4: Convert the BOD load into an equivalent population.
Each person is assumed to contribute a standard $0.08$ kg/day of $BOD_5$ through domestic sewage. Dividing the city's total load by this per-person figure gives the number of standard people that would generate the same load:
\[ P.E. = \frac{27{,}000}{0.08} \]
Multiply top and bottom by 100 to clear the decimal:
\[ P.E. = \frac{2{,}700{,}000}{8} = 337{,}500 \]

Final Answer:
\[ \boxed{P.E. = 337{,}500 \text{, option (A)}} \]