A 2 m x 2 m tank of 3 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 g/m$^3$ at the rate of 2 litre/s and an outflow of the well stirred mixture at the rate of 1 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 in the tank at $t$ (seconds). The mass of the salt (in grams) in the tank at 75% of its capacity is ____________ (rounded off to 2 decimal places).
Tank Volume and Initial State
The tank has dimensions 2m x 2m x 3m, giving a total volume of $2 \times 2 \times 3 = 12$ m$^3$.
Since 1 m$^3$ = 1000 litres, the total tank capacity is 12000 litres.
Initially, the tank is half-filled with fresh water, meaning the initial volume is $0.5 \times 12000 = 6000$ litres.
As the water is fresh, the initial mass of salt is $m(0) = 0$ grams.
Differential Equation and Salt Inflow Rate
The inflow and outflow rates are given as:
- Inflow rate: $Q_{in} = 2$ litre/s
- Outflow rate: $Q_{out} = 1$ litre/s
The volume of liquid in the tank at any time $t$ (in seconds) is $V(t) = V_{initial} + (Q_{in} - Q_{out})t$.
Substituting the values: $V(t) = 6000 \text{ litres} + (2 - 1) \times t = 6000 + t$ litres.
The differential equation provided is: $\frac{dm}{dt} + \frac{m}{6000 + t} = 0.01$.
We verify the terms in the differential equation:
- Salt Inflow Rate: The concentration of the incoming salt solution is $C_{in} = 5$ g/m$^3$. To match the volume units (litres), we convert this to g/litre: $C_{in} = 5 \text{ g} / 1000 \text{ litres} = 0.005$ g/litre.
The rate at which salt enters the tank is $C_{in} \times Q_{in} = 0.005 \text{ g/litre} \times 2 \text{ litre/s} = 0.01$ g/s. This matches the constant term $0.01$ on the right side of the DE. - Salt Outflow Rate: Assuming the mixture is well-stirred, the concentration of salt in the tank at time $t$ is $C(t) = \frac{m(t)}{V(t)} = \frac{m}{6000+t}$ g/litre.
The rate at which salt leaves the tank is $C(t) \times Q_{out} = \frac{m}{6000+t} \text{ g/litre} \times 1 \text{ litre/s} = \frac{m}{6000+t}$ g/s. This matches the term $\frac{m}{6000+t}$ on the left side of the DE.
Solving the First-Order ODE for Salt Mass
The given differential equation is a first-order linear ordinary differential equation of the form $\frac{dm}{dt} + P(t)m = Q(t)$, where $P(t) = \frac{1}{6000+t}$ and $Q(t) = 0.01$.
To solve this, we find the integrating factor, $I(t) = e^{\int P(t) dt}$.
Calculate the integral of $P(t)$:
$ \int P(t) dt = \int \frac{1}{6000+t} dt = \ln(6000+t) $The integrating factor is $I(t) = e^{\ln(6000+t)} = 6000+t$.
Multiply the differential equation by the integrating factor:
The left side is the derivative of the product $(I(t)m)$:
$ \frac{d}{dt}[(6000+t)m] = 60 + 0.01t $Integrate both sides with respect to $t$:
$ (6000+t)m = \int (60 + 0.01t) dt $ $ (6000+t)m = 60t + \frac{0.01t^2}{2} + C $ $ (6000+t)m = 60t + 0.005t^2 + C $Use the initial condition $m(0) = 0$ to find the constant of integration $C$:
$ (6000+0) \times 0 = 60(0) + 0.005(0)^2 + C $ $ 0 = 0 + 0 + C \implies C = 0 $Therefore, the equation for the mass of salt at time $t$ is:
$ (6000+t)m(t) = 60t + 0.005t^2 $ $ m(t) = \frac{60t + 0.005t^2}{6000+t} $Time to Reach 75% Tank Capacity
The tank reaches 75% of its capacity when the volume is $0.75 \times 12000 = 9000$ litres.
We find the time $t$ when $V(t) = 9000$ litres:
Salt Mass at Target Capacity
Substitute $t = 3000$ seconds into the derived formula for $m(t)$ to find the mass of salt:
$ m(3000) = \frac{60(3000) + 0.005(3000)^2}{6000+3000} $ $ m(3000) = \frac{180000 + 0.005(9000000)}{9000} $ $ m(3000) = \frac{180000 + 45000}{9000} $ $ m(3000) = \frac{225000}{9000} $ $ m(3000) = 25 $The mass of salt in the tank when it is at 75% capacity is 25 grams.
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