Brine Tank After Inlet Shutdown: Why Concentration Stops Rising
A 100 L mixing tank initially contains pure water. Brine containing 0.20 kg of salt per litre enters at 5 L/min, and the well-stirred mixture leaves at the same rate, so the volume remains 100 L. After 10 minutes a pump fault closes the inlet, but the outlet keeps running at 5 L/min. The operator wants the salt concentration to reach 0.15 kg/L and believes that because water continues to leave while salt stays behind, the concentration will rise to that target before the tank empties. The tank would empty 20 minutes after the inlet closes if nothing changes. A supervisor wants the actual concentration at 10 minutes, the concentration behaviour after the inlet closes, and a clear decision on whether the target can be met without restarting the inlet or adding salt. Assume perfect mixing and constant density.
1. Find the amount Q(t) of salt in the tank and the concentration at t = 10 minutes while the inlet is open.
Read the worked answer
Let Q(t) be the mass of salt in kg at time t minutes. Inflow adds 5 × 0.20 = 1 kg/min. Outflow removes 5 × Q/100 = 0.05Q kg/min. Therefore
dQ/dt = 1 − 0.05Q, Q(0) = 0.
This is separable: dQ/(20 − Q) = 0.05 dt, giving Q(t) = 20(1 − e^(−0.05t)). At t = 10,
Q(10) = 20(1 − e^(−0.5)) ≈ 20(1 − 0.6065) = 7.869 kg.
The concentration is Q/100 ≈ 0.0787 kg/L.
2. After the inlet closes, set up and solve the differential equation for Q(t), and find the concentration as a function of time. Does it reach 0.15 kg/L before the tank empties?
Read the worked answer
After t = 10 the inlet is closed, so no salt enters. The volume is
V(t) = 100 − 5(t − 10) = 150 − 5t, 10 ≤ t ≤ 30.
The outlet removes salt at rate 5 × Q/V = 5Q/(150 − 5t) = Q/(30 − t). Thus
dQ/dt = −Q/(30 − t).
Separate: dQ/Q = −dt/(30 − t), so ln Q = ln(30 − t) + constant, hence Q = C(30 − t). At t = 10, Q = 7.869 and 30 − t = 20, so C = 0.3935. Therefore
Q(t) = 0.3935(30 − t).
The concentration is
Q/V = 0.3935(30 − t) / [5(30 − t)] = 0.0787 kg/L,
which is constant. It does not rise; it stays at the value reached at t = 10 and never reaches 0.15 kg/L before the tank empties at t = 30. Physically, salt leaves in the same proportion as water, so removing water alone cannot enrich the solution.
3. Recommend one practical action, with a quantitative justification.
Read the worked answer
Restart the inlet immediately. With the inlet running, Q obeys dQ/dt = 1 − 0.05Q starting from Q = 7.869 kg at t = 10. The target is Q = 0.15 × 100 = 15 kg. The time needed after restart is
(1/0.05) ln[(20 − 7.869)/(20 − 15)] = 20 ln(12.131/5) ≈ 20 × 0.886 = 17.7 min.
So the target would be reached about 27.7 min after the start. Alternatively, at t = 10 add 15 − 7.869 ≈ 7.13 kg of salt and mix; the concentration becomes 0.15 kg/L immediately and stays there while the inlet remains closed. Restarting the inlet is preferable because it is continuous and avoids a manual salt slug.