Question

Environmental engineers and scientists are becoming concerned about pharmaceuticals in the environment. The moodstabilizing drug carbamazepine is discharged into a small lake (through a river feeding the lake) at an influent concentration of 11 mg/L. The lake has a surface area of 1.1 acres and a mean depth of 6 feet. The flow of the rivers feeding and draining the lake is 3000 cubic feet per second (cfs). The drug is degraded naturally in the lake with a first-order rate constant of 15hr−1 and generate with zero order rate constant of 21(mg/L)∗hr−1. What is the drug concentration ( mg/L ) in a river draining the lake at steady state? (1 acres = 43,560ft2 )

Answer 1

To determine the drug concentration in the river draining the lake at steady state, we need to consider the inflow of the drug, the degradation of the drug, and the outflow of the drug.

First, let's calculate the volume of the lake:

Volume of the lake = Surface Area of the lake * Mean Depth

= 1.1 acres * 43,560 ft^2/acre * 6 ft

= 285,912 ft^3

Since the flow of the rivers feeding and draining the lake is given as 3000 cubic feet per second (cfs), we can assume that the volume of water entering and leaving the lake per unit time is 3000 ft^3/s.

Next, let's calculate the inflow rate of the drug:

Inflow Rate = Influent Concentration * Inflow Volume Rate

= 11 mg/L * 3000 ft^3/s

= 33,000 mg/s

Now, let's consider the degradation of the drug in the lake. The degradation rate can be calculated using the first-order rate constant:

Degradation Rate = Drug Concentration in the lake * First-Order Rate Constant

= Drug Concentration in the lake * 15 hr^(-1)

Since we are interested in the steady state, the degradation rate should be equal to the generation rate:

Degradation Rate = Generation Rate

Drug Concentration in the lake * 15 hr^(-1) = 21 (mg/L) * hr^(-1)

Solving for the Drug Concentration in the lake:

Drug Concentration in the lake = 21 (mg/L) * hr^(-1) / 15 hr^(-1)

= 1.4 mg/L

Finally, let's calculate the drug concentration in the river draining the lake at steady state:

Outflow Rate = Drug Concentration in the lake * Outflow Volume Rate

= 1.4 mg/L * 3000 ft^3/s

= 4200 mg/s

Since the outflow volume rate is the same as the inflow volume rate, the drug concentration in the river draining the lake at steady state is:

Drug Concentration in the river = Outflow Rate / Outflow Volume Rate

= 4200 mg/s / 3000 ft^3/s

= 1.4 mg/L

Therefore, the drug concentration in the river draining the lake at steady state is 1.4 mg/L.