# Separable differential equation modeling salt dissolved in water

I have no idea what to do for this. The equation is supposed to model a solution having salty water dumped into a tank that leaks the solution.

? A tank contains 1000L of pure water. Brine that contains .05 kg of salt per liter of water enters the tank at a ra te of 5L/min. Brine that contains .04 kg of salt per liter of water enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. From Stewart Calculus 7e

I am supposed to find the salt at t minutes and at one hour.

I try to simplify this problem by making it .07kg per lit of water at 5L/m and from here I have no idea how to make an equation for this.

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Could you copy the problem exactly? –  Pedro Tamaroff Jun 19 '12 at 23:17
A tank contains 1000L of pure water. Brine that contains .05 kg of salt per liter of water enters the tank at a ra te of 5L/min. Brine that contains .04 kg of salt per liter of water enters the tank at a rate of 10L/min. Te solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. –  user138246 Jun 19 '12 at 23:18
Please copy this into the question (with proper attribution) –  robjohn Jun 19 '12 at 23:36

Let $s(t)$ be the amount of salt (in kg) in the tank at time $t$. $s(0)=0$ since the tank starts with pure water.

The amount of salt leaving the tank would be the density of salt in the tank times the flow leaving the tank: $\frac{s}{1000}\frac{\text{kg}}{\text{L}}\times15\frac{\text{L}}{\text{min}}$

The amount of salt entering the tank from a particular source would be the density of salt entering times the flow entering the tank: $.05\frac{\text{kg}}{\text{L}}\times5\frac{\text{L}}{\text{min}}+.04\frac{\text{kg}}{\text{L}}\times10\frac{\text{L}}{\text{min}}$

Taking the combination of sources and drains, we get $$\frac{\mathrm{d}s}{\mathrm{d}t}=\left(.05\times5+.04\times10-\frac{s}{1000}\times15\right)\frac{\text{kg}}{\text{min}}$$

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This is a kind of standard mixing problem that basic differential equations can be used to solve. It's not an entirely realistic model, since you will need to assume that when salty water goes in the tank, that it instantly homogenizes throughout the tank rather than introducing a slowly growing salty cloud. But anyway, this is how it's done:

Introduce a quantity $A(t)$ that changes over time - $A(t)$ is the amount of salt (in kg) that is in the tank at time $t$.

There are only two factors that cause $A$ to change. $A$ is motivated to increase over time because of the salty input. $A$ is motivated to decrease over time because of the mixture leaking out. All together, \begin{align*} \frac{dA}{dt} & = \mbox{rate in} - \mbox{rate out}\\ \end{align*} Now, the units everywhere in that equation are kg/min. This helps you translate "rate in" and "rate out".

The "rate in" needs to be the rate of the volume of brine going in (in L/min), multiplied by the concentration of that brine (in kg/L). Both of these are constant as time changes, so "rate in" is constant.

The "rate out" needs to be the rate of the volume of liquid coming out (in L/min) multiplied by the concentration of that liquid (in kg/L). This time however, the concentration is changing over time. In fact, it is $\frac{A(t)}{V(t)}$ where $V(t)$ is the volume of liquid in the tank at time $t$. $V(t)$ is a linear function. If the volume of liquid flowing in per minute equals that which is flowing out, then $V(t)$ is just a constant function.

This should all leave you with a differential equation like $$\frac{dA}{dt}=c_1+\frac{c_2A}{c_3+c_4t}$$

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Let $A(t)$ denote the amount of salt in the tank at time $t$. Note that the amount of brine in the tank remains constant, since it gains 15 L/min and loses 15 L/min.

The rate of salt in is a constant .05 kg/L * 5 L/min + .04 kg/L * 10 L/min = .65 kg/min.

The rate of salt out is derived by the realization that since 15/1000 of the water leaves per minute, 15/1000 of the salt must leave as well (this is where the well-mixedness comes in). Thus, the rate out is $\frac{15}{1000} A(t)$

So

$$\frac{dA}{dt} = .65 - \frac{15}{1000} A(t)$$

We also need an initial condition: we start with pure water, and therefore no salt. $A(0) = 0$.

This is a separable equation. It can be rearranged as follows (first clear the fractions; they're ugly):

$$1000 \frac{dA}{dt} = 650 - 15A$$ $$1000 \frac{dA}{650-15A} = dt$$

Now, integrate and go!

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