Salt Tank With a Mid-Run Brine Switch
A water-treatment tank holds 200 L of solution with 40 kg of dissolved salt. At t = 0, brine containing 0.3 kg/L of salt enters at 5 L/min and the well-stirred mixture leaves at the same rate, so the volume stays 200 L. The operator wants to know when the salt content will reach 55 kg, because a sensor alarms above that level. After 30 minutes, the inflow is switched to a weaker brine of 0.1 kg/L while flow rates stay the same. A trainee claims that because the inflow is weaker, the salt level will immediately start falling, and that the tank will be back to 40 kg within 10 minutes. The plant needs a correct prediction and a check on the trainee's claim.
1. Set up and solve the differential equation for S(t), the kg of salt at time t during the first 30 minutes. Find S(30) and the time at which 55 kg would be reached if the first brine continued.
Read the worked answer
Salt enters at 5 × 0.3 = 1.5 kg/min. The outflow concentration is S/200, so salt leaves at 5(S/200) = S/40 kg/min. Thus dS/dt = 1.5 − S/40, S(0) = 40.
Rewrite as S' + S/40 = 1.5. The integrating factor is e^(t/40), so d/dt [S e^(t/40)] = 1.5 e^(t/40). Integrating gives S e^(t/40) = 60 e^(t/40) + C, hence S = 60 + C e^(−t/40). The initial condition gives 40 = 60 + C, so C = −20 and S(t) = 60 − 20e^(−t/40).
At t = 30, e^(−0.75) ≈ 0.47237, so S(30) = 60 − 20(0.47237) ≈ 60 − 9.45 = 50.55 kg.
If the first brine continued, 55 = 60 − 20e^(−t/40) gives e^(−t/40) = 0.25, so t = 40 ln 4 ≈ 55.45 min. That is beyond 30 min, so the switch occurs before the first-brine formula would reach 55 kg.
2. For t ≥ 30, solve the new differential equation with S(30) = 50.55 kg. Find S(60) and the long-run salt level. Does the trainee's claim hold?
Read the worked answer
With 0.1 kg/L brine, salt enters at 5 × 0.1 = 0.5 kg/min. The equation is dS/dt = 0.5 − S/40, S(30) = 50.55.
The equilibrium is S = 20. The solution is S(t) = 20 + (50.55 − 20)e^(−(t−30)/40) = 20 + 30.55e^(−(t−30)/40).
At t = 60, e^(−0.75) ≈ 0.47237, so S(60) = 20 + 30.55(0.47237) ≈ 20 + 14.43 = 34.43 kg. As t → ∞, S → 20 kg.
The trainee is partly right: at t = 30, dS/dt = 0.5 − 50.55/40 ≈ 0.5 − 1.264 = −0.764 kg/min, so the salt level does begin to fall immediately. But it falls continuously, not instantly. After 10 more minutes, at t = 40, S = 20 + 30.55e^(−0.25) ≈ 20 + 23.78 = 43.78 kg, still above 40 kg. The tank reaches 40 kg when 20 + 30.55e^(−(t−30)/40) = 40, giving e^(−(t−30)/40) = 20/30.55 ≈ 0.6547, so t − 30 ≈ 16.94 min, or t ≈ 46.94 min. The claim of returning to 40 kg within 10 minutes is false.
3. The sensor alarms above 55 kg. Using both phases, determine whether the alarm ever sounds, and find the maximum salt level.
Read the worked answer
During the first phase, S(t) = 60 − 20e^(−t/40) is increasing because S' = 1.5 − S/40 > 0 while S < 60. Its maximum on 0 ≤ t ≤ 30 is at t = 30, where S ≈ 50.55 kg.
After the switch, S(t) = 20 + 30.55e^(−(t−30)/40). This is decreasing because the exponential decreases and the derivative is negative for S > 20. Its maximum for t ≥ 30 is therefore the starting value, 50.55 kg.
The overall maximum salt content is about 50.55 kg, which is below 55 kg. The alarm never sounds during this run.
A careless analysis would extend the first-phase formula past 30 minutes and predict 55 kg at t ≈ 55.45 min, but that ignores the switch. The correct piecewise model shows the peak occurs at the switch and is only 50.55 kg, so the operator should not expect an alarm unless the first brine is kept longer or the flow rates change.