Ammonia Pulse in a Recirculating Fish Tank
A fish farm uses a 500 m³ holding tank. Fresh water enters at 20 m³/h and well-mixed water leaves at the same rate, so the volume stays constant. At time t = 0 the tank contains 5 kg of ammonia. For the first 6 hours a feeder releases ammonia into the tank at a constant 0.5 kg/h; after t = 6 hours the release stops. The farm manager models the amount A(t) kg of ammonia in the tank using a first-order differential equation, assuming the tank is perfectly mixed and that no ammonia is lost by any route other than the outflow. A regulatory threshold is 2 kg: if the model predicts the tank is above 2 kg for too long, the stock must be moved. The manager wants to know the maximum amount, when it occurs, and when the tank first falls below 2 kg after the release stops.
1. Set up and solve the differential equation for A(t) during the first 6 hours. Find the maximum amount of ammonia and the time at which it occurs.
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During 0 ≤ t ≤ 6, ammonia enters at 0.5 kg/h. The outflow concentration is A/500 kg/m³, so ammonia leaves at 20(A/500) = 0.04A kg/h. Hence
dA/dt = 0.5 − 0.04A, A(0) = 5.
This is linear. The equilibrium value is 0.5/0.04 = 12.5, so A(t) = 12.5 + (5 − 12.5)e^(−0.04t) = 12.5 − 7.5e^(−0.04t).
Since dA/dt = 0.5 − 0.04A and A < 12.5 throughout this interval, dA/dt > 0. Thus A is increasing for 0 ≤ t ≤ 6, and the maximum during the release occurs at t = 6:
A(6) = 12.5 − 7.5e^(−0.24) ≈ 12.5 − 5.900 = 6.60 kg.
2. After t = 6 the feeder stops. Solve for A(t) for t ≥ 6 and find the first time the tank falls below 2 kg.
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For t ≥ 6, dA/dt = −0.04A, so A(t) = A(6)e^(−0.04(t − 6)) ≈ 6.60e^(−0.04(t − 6)).
Set A = 2: 2 = 6.60e^(−0.04(t − 6)). Thus e^(−0.04(t − 6)) = 2/6.60 ≈ 0.3030, so −0.04(t − 6) = ln(0.3030) ≈ −1.194. Hence t − 6 ≈ 29.85 and t ≈ 35.9 hours.
So the model predicts the tank first falls below 2 kg at about 35.9 hours after the start, i.e. about 29.9 hours after the release stops.
3. The manager claims the model is conservative because ammonia also evaporates. Discuss whether evaporation would make the 2 kg threshold occur earlier or later, and identify one assumption that may fail in a large tank.
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Evaporation would add another loss term to the differential equation, for example −kA with k > 0. That makes dA/dt more negative for every positive A, so A(t) would decrease faster than in the model. Therefore the tank would fall below 2 kg earlier, not later; the original model is conservative in that sense.
One assumption that may fail in a large tank is perfect mixing. In reality ammonia may be concentrated near the feeder or near the outflow, so the outflow concentration may not equal A/500. Other assumptions that may fail are constant volume, no biological uptake of ammonia, and a uniform release rate from the feeder.