Answer:
12.84 mg/L
Explanation:
We are given;
Volume of lake; V = 1.1 x 10^(6) m³
decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s
Factory rate: Q_f = 4.3 m³/s
Factory concentration: C_f = 100 mg/L
Stream rate: Q_s = 34 m³/s
Stream Concentration: C_s = 2.3 mg/L
Now, to find the steady state concentration of pollutant in the lake, we will use the formula;
(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)
Where C_L is the steady state concentration of pollutant in the lake.
Thus, making C_L the subject, we have;
C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)
Plugging in the relevant values gives;
C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))
C_L = 12.84 mg/L
