Question

A tank contains 300 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 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.

Answer 1

`A(t)=300-290e^{-(t)/(60)}`

Explanation:

The volume of fluid in the tank = 300 Liters

Initial amount of Salt in the tank, A(0)=10 grams

Change in the Amount of Salt in the Tank

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

Rate In =(concentration of salt in inflow)(input rate of brine)

`R_(in)=(1(gram)/(L))( 5(L)/(min))=5(grams)/(min)`

Rate Out =(concentration of salt in outflow)(output rate of brine)

`R_(out)=((A(t))/(300))( 5(L)/(min))=(A(t))/(60)`

Therefore:

`(dA)/(dt)=5-(A(t))/(60)\\$Rearranging, we have:\\(dA)/(dt)+(A(t))/(60)=5`

We solve the resulting linear differential equation for A(t)

`\text{The integrating factor: } e^{\int (1)/(60)dt} =e^{(t)/(60)}\\$Multiplying by the integrating factor all through\\(dA)/(dt)e^{(t)/(60)}+(A)/(60)e^{(t)/(60)}=5e^{(t)/(60)}\\(Ae^{(t)/(60)})'=5e^{(t)/(60)}`

Taking the integral of both sides

`\int(Ae^{(t)/(60)})'=\int 5e^{(t)/(60)} dt\\Ae^{(t)/(60)}=5*60e^{(t)/(60)}+C, $(C a constant of integration)\\Ae^{(t)/(60)}=300e^{(t)/(60)}+C\\$Divide all through by e^{(t)/(60)}\\A(t)=300+Ce^{-(t)/(60)}`

Recall that when t=0, A(t)=10 grams (our initial condition)

`10=300+Ce^{-(0)/(60)}\\10-300=Ce^0\\C=-290`

Therefore, the number A(t) of grams of salt in the tank at time t is:

`A(t)=300-290e^{-(t)/(60)}`