Differential Equations: Salt tank; inflow/outflow DE — MI 2025 H2 Math Prelim Paper 1
What this question tests
Question
A tank initially contains 100 litres of pure water. A brine solution, with a salt concentration of 0.2 kg per litre, is pumped into the tank continuously at a constant rate of 5 litres per minute. At the same time, a well-stirred mixture of the tank's contents is continuously drained from the tank at the same rate of 5 litres per minute. At time \(t\) minutes, the mass of salt in the tank is denoted by \(Q\) kilograms.
- Explain briefly why the volume of the mixture in the tank remains constant.
- By considering the salt concentration of the brine solution, show that salt enters the tank at a constant rate of 1 kg per minute.
- Hence, by also considering the mass of salt leaving the tank, show that the rate of change of the mass of salt in the tank can be expressed as \[ \frac{\mathrm{d}Q}{\mathrm{d}t} = 1 - \frac{Q}{20}. \]
- Solve the differential equation found in part (iii) to find \(Q\) in terms of \(t\).
- Determine the time taken for the salt concentration in the tank to reach 0.1 kg per litre.
- What happens to \(Q\) for large values of \(t\)?
Show full worked solution▾
(i) At any point in time, the inflow rate is equal to the outflow rate (both 5 litres per minute). Therefore, the volume of the mixture in the tank remains constant.
(ii) Salt inflow rate \(= \) salt concentration of brine \(\times\) brine inflow rate \(= 0.2\times 5 = 1\) kg/min (shown).
(iii) Mass of salt in tank at time \(t\) is \(Q\) kg.
Volume of mixture \(= 100\) litres.
Salt concentration in tank at time \(t\) \(= \dfrac{Q}{100}\) kg/litre.
Salt outflow rate \(= \dfrac{Q}{100}\times 5 = \dfrac{Q}{20}\).
\[ \frac{\mathrm{d}Q}{\mathrm{d}t} = \text{salt inflow rate} - \text{salt outflow rate} = 1 - \frac{Q}{20} \quad\text{(shown)} \](iv) \[\begin{aligned} \frac{\mathrm{d}Q}{\mathrm{d}t} &= 1-\frac{Q}{20} = \frac{20-Q}{20}\\ \int\frac{1}{20-Q}\;\mathrm{d}Q &= \int\frac{1}{20}\;\mathrm{d}t\\ -\ln|20-Q| &= \frac{t}{20}+c\\ \ln|20-Q| &= -\frac{t}{20}-c\\ |20-Q| &= \mathrm{e}^{-t/20-c}\\ 20-Q &= A\,\mathrm{e}^{-t/20} \end{aligned}\]
When \(t = 0\), \(Q = 0\) (initially pure water): \[ 20-0 = A\,\mathrm{e}^{0} \;\Rightarrow\; A = 20. \]
\(\therefore\; Q = 20-20\,\mathrm{e}^{-t/20}\).
\(Q = 20-20\,\mathrm{e}^{-t/20}\)(v) Salt concentration in tank \(= 0.1\) kg/litre.
Mass of salt \(= 0.1\times 100 = 10\) kg. Sub \(Q = 10\): \[\begin{aligned} 10 &= 20-20\,\mathrm{e}^{-t/20}\\ \mathrm{e}^{-t/20} &= \frac{1}{2}\\ -\frac{t}{20} &= \ln\frac{1}{2}\\ t &= -20\ln\frac{1}{2} = 20\ln 2 \approx 13.9\text{ (3 s.f.)} \end{aligned}\]
\(t = 20\ln 2 \approx 13.9\) min
(vi) \(Q = 20-20\,\mathrm{e}^{-t/20}\).
When \(t\to\infty\), \(\mathrm{e}^{-t/20}\to 0\), so \(Q\to 20\).
\(Q\) increases and approaches 20.