Question

A tank with capacity of 300 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Let Q(t) lb be the amount of salt in the tank, V(t) gal be the volume of water in the tank. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Q(t)

Answer 1

The amounts of salt in the tank at any time prior to the instant when the solution begins to overflow is.

`Q(t)=-\frac {4* 10^6}{(200+t)^2}+200+t`

Explanation:

Let Q(t) be the amount of salt in the tank at any time
`t`.

Then, its time of change,
`Q'(t)` by (Balance law).

Since three gallons of salt water runs in the tank per minute, containing
`1` lb of salt, the salt inflow rate is

`3\cdot 1=3`

The amount of water in the tank at any time
`t` is,

`V(t)=200+(3-2)t=200+t,`

Now, the outflow is
`2` gal of the solution in a minute. That is
`\frac 2{200+t}` of the total solution content in the tank, hence
`\frac 2{200+t}` of the salt salt content
`Q(t)`, that is
`\frac {2Q(t)}{200+t}`,

Initially, the tank contains
`100` lb of salt,

Therefore, we obtain the initial condition
`Q(0)=100`

Thus, the model is

`Q'(t)=3-\frac {2Q(t)}{200+t}, \;\;Q(0)=100`

`\Rightarrow Q'(t)+\frac {2}{200+t}Q(t)=3, \;\;Q(0)=100`

here,
`p(t)=\frac 2{200+t}\;\; \text{and}\;\; q(t)=3` Linear ODE

So, the integrating factor is

`e^(\int pdt)=e^{2\int (dt)/(200+t)=e^(\ln(200+t)^2)=(200+t)^2`

and the general solution is

`Q(t)(200+t)^2=\int q(t)(200+t)^2dt+c`

`\Rightarrow Q(t)=\frac 1{(200+t)^2}\int 3(200+t)^2dt+c`

`=\frac c{(200+t)^2}+200+t`

using initial condition and find the value of constant c.

`100=Q(0)=\frac c{(200+0)^2}+200+0\Rightarrow -100=\frac c{200^2}`

`\Rightarrow c=-4000000=-4* 10^6`

`\Rightarrow Q(t)=-\frac {4* 10^6}{(200+t)^2}+200+t`

Hence, is the amount of salt in the tank at any moment t.