Question

At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon of brine is allowed to enter the tank at a rate of 55 gal/min and the mixed solution is drained from the tank at the same rate.a. How much salt is in the tank at an arbitrary time t? b. How much salt is in the tank after 25 min?

Answer 1

Answer:

a) $s(t) =400- 385 e^{-\frac{1}{10} t}$

b) $s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

$Rate change = In-Out$

And we can name the rate of change as $\frac{ds}{dt}=rate change$

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is $\frac{s}{50gal}$

And now we can find the expression for the amount of salt after time t like this:

$\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}$

And we have this differential equation:

$\frac{dS}{dt} +\frac{1}{10} s = 40$

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

$\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}$

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

$e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt$

$e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C$

Now we can divide both sides by $e^{\frac{1}{25} t}$ and we got:

$s(t) =400 + C e^{-\frac{1}{10} t}$

Now we can apply the initial condition in order to solve for the constant C like this:

$15 = 400+C$

$C=-385$

And then our function would be given by:

$s(t) =400- 385 e^{-\frac{1}{10} t}$

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

$s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$