Question

A \(2 \text{ m} \times 2 \text{ m}\) tank of \(3 \text{ m}\) height has inflow, outflow, and stirring mechanisms. Initially, the tank was half-filled with fresh water. At \(t = 0\), an inflow of a salt solution of concentration \(5 \text{ g/m}^3\) at the rate of \(2 \text{ litres/s}\) and an outflow of the well stirred mixture at the rate of \(1 \text{ litre/s}\) are initiated. This process can be modelled using the following differential equation: \[ \frac{dm}{dt} + \frac{m}{6000 + t} = 0.01 \] where \(m\) is the mass (grams) of the salt at time \(t\) (seconds). The mass of the salt in the tank at 75\% of its capacity is \_\_\_\_ grams (rounded off to 2 decimal places).

Hint:

Differential equations can model how quantities change over time. Solving them often involves integrating factors and assumptions about initial conditions.

Answer 1

Step 1: Define the volume of the tank at 75\% capacity. The tank's total volume is: \[ \text{Volume} = 2 \text{ m} \times 2 \text{ m} \times 3 \text{ m} = 12 \text{ m}^3 \] At 75\% capacity: \[ \text{Volume at 75\%} = 0.75 \times 12 = 9 \text{ m}^3 \] Step 2: Calculate the rate of salt mass change. Salt inflow rate: \[ \text{Salt mass inflow rate} = 2 \text{ l/s} \times 5 \text{ g/m}^3 = 10 \text{ g/s} \] Salt outflow rate, assuming concentration \(c = \frac{m}{V}\): \[ \text{Salt mass outflow rate} = 1 \text{ l/s} \times c = \frac{m}{9000} \text{ g/s} \] The differential equation becomes: \[ \frac{dm}{dt} = 10 \text{ g/s} - \frac{m}{9000} \text{ g/s} \] Step 3: Solve the differential equation using an integrating factor. Integrating factor, \(\mu(t)\): \[ \mu(t) = e^{\int \frac{dt}{9000}} \] \[ \mu(t) = e^{\frac{t}{9000}} \] Multiplying through by the integrating factor: \[ \frac{d}{dt} (m e^{\frac{t}{9000}}) = 10 e^{\frac{t}{9000}} \] Integrating both sides: \[ m e^{\frac{t}{9000}} = 9000 \times 10 \times (1 - e^{-\frac{t}{9000}}) + C \] At \(t = 0\), \(m = 0\) (initially fresh water), so \(C = 0\). \[ m = 9000 \times 10 \times (1 - e^{-\frac{t}{9000}}) \] Step 4: Calculate the mass of salt at 75\% capacity. Substituting \(t = \frac{9}{3} \times 3600 \) (time to fill to 75\% capacity): \[ m \approx 90000 \times (1 - e^{-\frac{1}{3}}) \] \[ m \approx 90000 \times 0.2835 \approx 25515 \text{ g} \] Rounding to two decimal places: \[ m \approx 25.52 \text{ g} \]