Question

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t. A(t) =

Answer 1

`A(t)=240-220e^{-(t)/(40)}`

Explanation:

A tank contains 240 liters of fluid in which 20 grams of salt is dissolved.

• Volume of the tank = 240 liters
• Initial Amount of Salt in the tank, A(0)=20 grams

Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 6 L/min

`R_(in)=`(concentration of salt in inflow)(input rate of fluid)

`R_(in)=(1(gram)/(liter))( 6(Liter)/(min))=6(gram)/(min)`

`R_(out)=`(concentration of salt in outflow)(output rate of fluid)

`R_(out)=((A(t))/(240))( 6(Liter)/(min))\\R_(out)=(A)/(40)`

Rate of change of the amount of salt in the tank:

`(dA)/(dt)=R_(in)-R_(out)`

`(dA)/(dt)=6-(A)/(40)`

We then solve the resulting differential equation by separation of variables.

`(dA)/(dt)+(A)/(40)=6\\$The integrating factor: e^{\int (1)/(40)dt} =e^{(t)/(40)}\\$Multiplying by the integrating factor all through\\(dA)/(dt)e^{(t)/(40)}+(A)/(40)e^{(t)/(40)}=6e^{(t)/(40)}\\(Ae^{(t)/(40)})'=6e^{(t)/(40)}`

Taking the integral of both sides

`\int(Ae^{(t)/(40)})'=\int 6e^{(t)/(40)} dt\\Ae^{(t)/(40)}=6*40e^{(t)/(40)}+C, $(C a constant of integration)\\Ae^{(t)/(40)}=240e^{(t)/(40)}+C\\$Divide all through by e^{(t)/(40)}\\A(t)=240+Ce^{-(t)/(40)}`

Recall that when t=0, A(t)=20 (our initial condition)

`20=240+Ce^{-(0)/(40)}\\20-240=C\\C=-220\\$Therefore, the number A(t) of grams of salt in the tank at time t\\A(t)=240-220e^{-(t)/(40)}`