Solving the tank salt model after ten minutes

Key takeaway: This follow-up problem uses the same mixing equation to compute a specific value, then checks the model against a concentration threshold.

Part (b): Salt after 10 minutes

From part (a), the model is

$\frac{dS}{dt}=3-\frac{2S}{100+t}.$

To find $S$ after 10 minutes, we solve the differential equation. Rewrite it in linear form:

$\frac{dS}{dt}+\frac{2}{100+t}S=3.$

The integrating factor is

$\mu(t)=e^{\int \frac{2}{100+t}\,dt}=e^{2\ln(100+t)}=(100+t)^2.$

Multiply through by $(100+t)^2$:

$(100+t)^2\frac{dS}{dt}+2(100+t)S=3(100+t)^2.$

So

$\frac{d}{dt}\left[(100+t)^2S\right]=3(100+t)^2.$

Integrate:

$(100+t)^2S=\int 3(100+t)^2\,dt = (100+t)^3 + C.$

Hence

$S=100+t+\frac{C}{(100+t)^2}.$

Use the initial condition $S(0)=0$ because the tank initially contains pure water:

$0=100+\frac{C}{100^2}$

$C=-1{,}000{,}000.$

So

$S=100+t-\frac{1{,}000{,}000}{(100+t)^2}.$

At $t=10$:

$S(10)=110-\frac{1{,}000{,}000}{110^2}$

$S(10)=110-\frac{1{,}000{,}000}{12{,}100}\approx 27.36.$

So there are about 27.4 g of salt after 10 minutes.

Part (c): When concentration reaches 0.9 g/L

Concentration is

$\frac{S}{100+t}.$

Using the formula for $S$:

$\frac{S}{100+t}=1-\frac{1{,}000{,}000}{(100+t)^3}.$

Set this equal to 0.9:

$1-\frac{1{,}000{,}000}{(100+t)^3}=0.9$

$\frac{1{,}000{,}000}{(100+t)^3}=0.1$

$(100+t)^3=10{,}000{,}000.$

Taking the cube root:

$100+t\approx 215.44$

so

$t\approx 115.44.$

To the nearest minute, the valve should be closed after 115 minutes.

Part (d): Evaluate the model

The model is useful because it captures dilution, inflow, and outflow with reasonable assumptions. However, it assumes perfect instantaneous mixing and constant flow rates. In a real tank, mixing may take time, flow rates may vary, and the density of the solution may change. So the model is mathematically sound, but it is still an approximation of the plant process.

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Pitfalls the pros know 👇 A common mistake is to treat the differential equation as if it could be solved by simple inspection without checking the initial condition. That often leads to the wrong constant of integration. Another error is to compute concentration from the mass of salt alone, forgetting to divide by the changing volume. In part (c), some students solve for $S=0.9$, but the question asks for concentration, so the correct equation is $S/(100+t)=0.9$.

What if the problem changes? If the initial tank contained 40 litres of salt-free water instead of 100 litres, the same integrating-factor method would still work, but every occurrence of 100+t would become 40+t. If the inflow concentration were 0.5 g/L, the right-hand side would become 1.5 instead of 3. If the question asked for the time when concentration reached 1 g/L, you would set $S/(100+t)=1$ and solve the resulting cubic in the same way.

Tags: integrating factor, mixed concentration, initial value problem