How to Model Salt Dissolving in a Well-Mixed Tank

A salt-and-water tank problem is modeled by tracking how quickly salt enters and leaves. The key is to write a rate balance for the amount of salt, not for the concentration alone. Under the usual assumptions—constant tank volume and a well-mixed solution—the resulting differential equation is separable.

Set up the salt balance

Let $S(t)$ be the amount of dissolved salt in the tank at time $t$, measured, for example, in pounds. Suppose the tank holds a constant volume $V$ gallons. Brine flows in at $q$ gallons per minute with salt concentration $c_{\mathrm{in}}$ pounds per gallon, and the same volume flows out at the same rate.

The rate of salt entering is

$$q c_{\mathrm{in}}.$$

Because the tank is well mixed, the outflow has the same concentration as the tank, $S(t)/V$. Therefore, the rate of salt leaving is

$$q\frac{S(t)}{V}.$$

Subtracting the outflow rate from the inflow rate gives the differential equation

$$\frac{dS}{dt}=q c_{\mathrm{in}}-\frac{q}{V}S =\frac{q}{V}\left(Vc_{\mathrm{in}}-S\right).$$

This is the conservation-of-salt equation: the change in salt equals salt in minus salt out. Its terms have units of pounds per minute.

Separate variables and solve

For $S\ne Vc_{\mathrm{in}}$, separate the variables:

$$\frac{dS}{Vc_{\mathrm{in}}-S}=\frac{q}{V}\,dt.$$

Integrating both sides gives

$$-\ln\left|Vc_{\mathrm{in}}-S\right|=\frac{q}{V}t+C.$$

Equivalently, the difference between the salt amount and its equilibrium value decays exponentially. Applying the initial condition $S(0)=S_0$ yields

$$S(t)=Vc_{\mathrm{in}}+\left(S_0-Vc_{\mathrm{in}}\right)e^{-qt/V}.$$

Here, $Vc_{\mathrm{in}}$ is the equilibrium salt amount: at that amount, the tank concentration equals the incoming concentration, so salt enters and leaves at the same rate. The time constant is $V/q$ minutes. After one time constant, the difference from equilibrium has been multiplied by $e^{-1}$.

Worked example

A 50-gallon tank initially contains 20 pounds of salt dissolved in water. Brine containing 2 pounds of salt per gallon flows in at 3 gallons per minute, while the well-mixed solution leaves at 3 gallons per minute. Find the salt amount after 25 minutes.

The volume is constant, and the salt inflow rate is

$$(3\text{ gal/min})(2\text{ lb/gal})=6\text{ lb/min}.$$

The tank concentration is $S/50$ pounds per gallon, so salt leaves at

$$(3\text{ gal/min})\frac{S}{50\text{ gal}}=0.06S\text{ lb/min}.$$

Thus the initial-value problem is

$$\frac{dS}{dt}=6-0.06S,\qquad S(0)=20.$$

The equilibrium amount is $6/0.06=100$ pounds. Substituting $V=50$, $q=3$, $c_{\mathrm{in}}=2$, and $S_0=20$ into the solution gives

$$S(t)=100+(20-100)e^{-3t/50}=100-80e^{-3t/50}.$$

At $t=25$ minutes,

$$S(25)=100-80e^{-1.5}\approx82.15\text{ lb}.$$

The amount is moving toward 100 pounds, not away from it. The exponent is dimensionless because $qt/V$ has units $(\text{gal/min})(\text{min})/\text{gal}$.

Special case: fresh water flows in

If the incoming liquid contains no salt, then $c_{\mathrm{in}}=0$. The model becomes

$$\frac{dS}{dt}=-\frac{q}{V}S, \qquad S(t)=S_0e^{-qt/V}.$$

This is exponential washout: the salt amount approaches zero as fresh water replaces the solution. The tank concentration is $S(t)/V$ when the volume stays constant.

Common mistakes

• Using concentration where the equation needs an amount. The variable $S$ is salt in pounds; the tank concentration is $S/V$ pounds per gallon.
• Forgetting the concentration of the outflow. Well mixed means the outflow concentration matches the tank concentration at that time, not the inlet concentration.
• Ignoring the volume assumption. Equal inflow and outflow rates keep $V$ constant. If the rates differ, then $V$ changes and the outflow term becomes $q_{\mathrm{out}}S/V(t)$; the simple constant-volume solution no longer applies.
• Leaving out the initial condition. The differential equation describes a family of solutions. The value $S(0)=S_0$ determines which one models the tank.
• Treating the equilibrium as the only possible solution. If $S_0=Vc_{\mathrm{in}}$, the amount stays at equilibrium. Otherwise, it approaches that value exponentially.