Question

Plateau Creek carries 5 m³/s of water with a selenium (Se) concentration of 0.0015 mg/L. A farmer starts withdrawing 1 m³/s of water from the creek for irrigation. During irrigation, the water picks up selenium from soil. One half of the irrigation water is lost to the atmosphere or infiltration to the groundwater and the other half is returned to Plateau Creek. The irrigation runoff contains 1 mg/L Se. Assume that selenium is a conservative substance (i.e., it does not degrade) and the stream does not receive selenium from any other source.

c. Fish are sensitive to concentrations of selenium above 0.04 mg/L. How much irrigation water can the farmer remove while still protecting the health of the fish?

Answer 1

First calculate the concentration in mol/L of Se:

C = 0.0015 mg/L * 1g/1000 mg * 1mol/79g = 1.9x10^-8 mol/L

Concentration the fish can take: 0.04 mg/L * 1g/1000mg * 1mol/79g = 5.1x10^-7 mol/L

According to the general dillution principle:

M₁V₁ = M₂V₂

Solve for V₂ (volume that can take the farmer)

V₂ = 1.9x10^-8 M * 5x10³ L / 5.1x10^-7 M

V₂ = 186.27 L

Converting this into m³ : 0.18627 m³

Finally the volume the farmer can remove would be:

V = 5-0.18627 = 4.8137 m³

Hope this helps