Mixing Tank Alarm: Salt Concentration During Filling and After Outlet Adjustment
A 500 L mixing tank initially contains 250 L of solution with 100 kg of dissolved salt. Brine containing 0.20 kg/L of salt enters at 8.0 L/min. The tank is well mixed, and solution leaves through an overflow line at 6.0 L/min. The tank has a high-level alarm at 500 L; the operator must know when the alarm will sound and how much salt will be present at that moment. At the alarm, a control valve can increase the outlet flow to 8.0 L/min while the inlet remains at 8.0 L/min; the tank then stays at 500 L. The plant's downstream process cannot tolerate a salt concentration above 0.21 kg/L for more than 10 minutes after the alarm, because the elevated concentration can upset a crystallizer. The operator's log currently says 'concentration will approach 0.20 kg/L from below, so no overshoot is possible.' The engineer suspects this is wrong during the filling phase and must check both the alarm-time concentration and the post-alarm transient.
1. Derive the volume V(t) during filling and find the time of the high-level alarm. State the assumptions needed.
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Assume constant density, perfect mixing, and negligible volume contributed by dissolved salt. The net volume rate is 8.0 - 6.0 = 2.0 L/min, so V(t) = 250 + 2.0t litres. The alarm sounds when V = 500: 250 + 2t = 500, so t = 125 min. The tank reaches 500 L at 125 minutes after the start.
2. Set up and solve the salt mass balance during filling. Find S(125) and the concentration at alarm. Is the operator's log statement correct?
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Let S(t) be kg of salt. Salt enters at 0.20(8.0) = 1.6 kg/min. The outlet concentration is S/V and the outlet flow is 6.0 L/min, so dS/dt = 1.6 - 6S/(250 + 2t) = 1.6 - 3S/(125 + t). This is linear. The integrating factor is exp(∫ 3/(125+t) dt) = (125+t)^3. Then d/dt[S(125+t)^3] = 1.6(125+t)^3, so S(125+t)^3 = 0.4(125+t)^4 + C. Using S(0) = 100 gives 100(125)^3 = 0.4(125)^4 + C, hence C = 97,656,250. Therefore S(t) = 0.4(125+t) + 97,656,250/(125+t)^3. At t = 125 min, 125+t = 250, so S = 0.4(250) + 97,656,250/250^3 = 100 + 6.25 = 106.25 kg. The concentration is 106.25/500 = 0.2125 kg/L. The operator's log is wrong: the concentration is above 0.20 kg/L at the alarm and approaches 0.20 from above, not from below. The reason is that the tank starts at 100/250 = 0.40 kg/L and is diluted by feed that is lower in salt but not low enough to bring the concentration below 0.20 before the tank fills.
3. After the alarm the outlet is increased to 8.0 L/min. Solve the post-alarm salt balance and find when the concentration first falls below 0.21 kg/L. Does the plant meet its 10-minute requirement? Suggest one simple operational change that would help.
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After the alarm the volume is constant at 500 L and inlet = outlet = 8.0 L/min. Let τ = t - 125. Then dS/dτ = 1.6 - 8S/500 = 1.6 - 0.016S, with S(0) = 106.25. The equilibrium is 1.6/0.016 = 100 kg, so S(τ) = 100 + (106.25 - 100)e^{-0.016τ} = 100 + 6.25e^{-0.016τ}. The concentration is C(τ) = 0.2 + 0.0125e^{-0.016τ}. Set C = 0.21: 0.0125e^{-0.016τ} = 0.01, so e^{-0.016τ} = 0.8 and τ = -ln(0.8)/0.016 ≈ 13.95 min. The concentration is still about 0.2107 kg/L at 10 min, so the plant does not meet the 10-minute requirement; it needs about 14 minutes to fall below 0.21 kg/L. A simple operational change is to start equalizing the outlet flow before the tank reaches 500 L, or switch to a lower-salt feed earlier, so the tank does not enter the alarm period with concentration well above 0.21 kg/L. Waiting until the alarm to adjust the outlet only stabilizes the volume; it does not remove the existing excess salt quickly enough.