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

Answer:

$A(t)=240-220e^{-\frac{t}{40}}$

Step-by-step 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)

$R_{in}=(1\frac{gram}{liter})( 6\frac{Liter}{min})=6\frac{gram}{min}$

$R_{out}=$(concentration of salt in outflow)(output rate of fluid)

$R_{out}=(\frac{A(t)}{240})( 6\frac{Liter}{min})\\R_{out}=\frac{A}{40}$

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

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{40}$

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

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

Taking the integral of both sides

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

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

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