Modelling salt concentration with a mixing differential equation

Key takeaway: This mixing problem uses conservation of salt and volume change to build a first-order differential equation.

Key modelling principle

Let $S$ be the grams of salt in the tank after $t$ minutes.

The tank starts with 100 litres of water, and salt water enters at 3 litres per minute. Since the outflow is 2 litres per minute, the volume increases by 1 litre per minute.

So after $t$ minutes, the volume in the tank is

$100+t\text{ litres}.$

Salt entering and leaving

Salt enters with concentration 1 gram per litre, at 3 litres per minute. Therefore the rate of salt entering is

$3\text{ g/min}.$

The salt leaving depends on the concentration in the tank. Since the tank is well mixed, the concentration is

$\frac{S}{100+t}\text{ g/L}.$

The outflow rate is 2 litres per minute, so the salt leaving per minute is

$2\cdot \frac{S}{100+t}=\frac{2S}{100+t}.$

Form the differential equation

Rate of change = rate in - rate out:

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

This is exactly the required model.

Why the volume matters

A mixing problem is not just about salt amount. The changing volume affects concentration, and concentration determines how much salt leaves each minute. If the volume were constant, the differential equation would be much simpler. Here, the rising volume is the reason the denominator is $100+t$, not just 100.

Final answer

The model is

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

This equation expresses conservation of salt in the tank over time.

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Pitfalls the pros know 👇 A common mistake is to use the inflow rate 3 litres per minute as if it were the salt rate itself without checking the concentration. In this problem that happens to be correct only because each litre of incoming solution contains exactly 1 gram of salt. Another error is to forget that the tank volume changes over time. If you write the outflow concentration using a fixed 100 litres, the model becomes wrong immediately. The denominator must match the actual volume at time $t$.

What if the problem changes? If the inflow were still 3 litres per minute but each litre contained 2 grams of salt, then the input term would be 6 instead of 3, giving $\frac{dS}{dt}=6-\frac{2S}{100+t}$. If the outflow rate changed to 4 litres per minute, the volume would increase by only 1 litre per minute net, so the same volume formula $100+t$ would remain, but the leaving term would become $\frac{4S}{100+t}$. The modelling structure stays the same; only the coefficients change.

Tags: mixing model, concentration function, first-order differential equation