A bankrupt chemical firm has been taken over by new management. On the property they found a 20,000-m3 brine pond containing 25,

Question

2 months ago

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A bankrupt chemical firm has been taken over by new management. On the property they found a 20,000-m3 brine pond containing 25,

000 mg · L−1 of salt. The new owners propose to flush the pond into their dis- charge pipe leading to the Atlantic Ocean, which has a salt concentration above 30,000 mg · L−1 .
What flow rate of fresh water (in m3 · s−1 ) must they use to reduce the salt concentration in the pond to 500 mg · L−1 within one year?

Engineering

1 answer:

choli [298] · Answer 1 · 2026-03-22T10:25:25+03:00

Response:

Flow-rate = 0.0025 m^3/s

Clarification:

We must assume that the inflow of pure water to the pond matches the outflow of brine, meaning the pond volume remains constant at 20,000 m^3. Based on this, we can compute the water inflow or outflow rate (which are equal) by establishing a separable differential equation dQ/dt, where Q represents the salt in milligrams (mg) over time t. To solve our issue, we set up a differential equation as follows:

dQ/dt = -(Q/20,000)*r where r is the flow rate in m^3/s

This equation shows that the rate at which salt departs the pond (salt lost over time) equals the concentration (salt amount per unit volume) multiplied by the outflow rate of the liquid from the tank; the negative is included because it describes the rate of salt leaving the pond.

By rearranging the equation, we achieve dQ/Q = -(r/20000) dt then integrating both sides ∫dQ/Q = -∫(r/20000) dt and arriving at ln(Q) = -(r/20000)*t + C where C is the constant (initial value) determined from the integrals. Note that the integral of dQ/Q yields ln(Q) and r/20000 is a constant, thus the integral of dt is t.

To find C, we evaluate the integrated equation at t = 0, hence, ln(Q) = C, given that at t=0 we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q equals this concentration (mg/m^3) multiplied by the liquid amount (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. The final equation is ln(Q) = -(r/20000)*t + 26.9378, now the only remaining task is to determine the constant flow rate (r) necessary to lower the pond's salt concentration to 500 mg/L = 500000 mg/m^3 in a year (equating to 31536000 seconds). Therefore, we aim for Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, hence, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s approximately equal to 0.0025 m^3/s.

Note:

ln() indicates the natural logarithm
The volume in the tank remains constant since the inflow matches the outflow rate
While solving the differential equation, we computed the outflow rate but were asked for the inflow rate, and since they are equivalent, we could solve the dilemma
Throughout the solution process, units were consistently converted to m^3 and seconds because we required the answer in m^3/s.