Question

How do you determine how much salt is in the tank when it is full if a 30-gallon tank initially contains 15 gallons of salt water containing 5 pounds of salt and suppose salt water containing 1 pound of salt per gallon is pumped into the top of the tank at the rate of 2 gallons per minute, while a well-mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute?

Answer 1

There will be 25 pounds of salt.

Let `Q` be the amount of salt in the tank, in pounds and let `t` be the time elapsed, in minutes.

The rate at which salt is entering the tank is

`(2\frac{\mbox{gal}}{min})\cdot (1 \frac{\mbox{pound}}{\mbox{gal}}) = 2 \frac{\mbox{pounds}}{\mbox{min}}`

The rate at which salt is exiting the tank is

`(1\frac{\mbox{gal}}{\mbox{min}})\cdot (\frac{Q\mbox{ pounds}}{t+15\mbox{ gal}}) = \frac{Q}{t+15} \frac{\mbox{pounds}}{\mbox{min}}`

This leads to the differential equation `\frac{dQ}{dt}=2-\frac{Q}{t+15}` and the initial condition is `Q(0)=5`

To solve this differential equation, write it as

`\frac{dQ}{dt}+\frac{1}{t+15}Q=2`

Then multiply both sides of this equation by the integrating factor

`\mu(x)=e^{\int \frac{1}{t+15}dt}=e^{\ln(t+15)}=t+15`

to get

`(t+15)\frac{dQ}{dt}+Q=2(t+15)=2t+30`

or (by the Product Rule)

`\frac{d}{dt}((t+15)Q)=2t+30`

Now integrate both sides to get

`(t+15)Q=t^{2}+30t+C`

so that

`Q=\frac{t^{2}+30t+C}{t+15}`

The initial condition implies that

`5=Q(0)=\frac{C}{15}`

so that

`C=75` and `Q=\frac{t^{2}+30t+75}{t+15}`

The tank will be full when `t=15` minutes.

The final answer is therefore:

`Q(15)=\frac{225+450+75}{30}=\frac{750}{30}=25` pounds of salt.

Go to the following video for more help in setting it up:

Mixing Problem at 20:45

Go to the following video for an integrating factor example:

Integrating Factor at 27:32