Answer to Question #122451 in Differential Equations for JSE

Question

1. A tank contains 200 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 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.

A(t) = ____ g

2. A tank contains 420 liters of fluid in which 30 grams of salt is dissolved. Pure water 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) = ___ g

Answer 1

(1)

initial amount of salt = 20 gram

rate of pumping in = 5L/min

Let A be amount of salt at any time t,

then

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

R is rate of flow.

$R_{in} = \frac{1g}{1L} \frac{5L}{1 min} = 5 g/min$

$R_{out} = \frac{A}{200}*{5} = \frac{A}{40} g/min$

Hence equation of state will be,

$\frac{dA}{dt} = 5-\frac{A}{40}$

solving equation,

$\frac{dA}{dt} + \frac{1}{40}A =5$

$e^{\frac{1}{40}t}A = \int 5e^{\frac{1}{40}t}dt$

$e^{\frac{1}{40}t}A = 200e^{\frac{1}{40}t} + C$

$A = 200 + Ce^{-\frac{1}{40}t}$

applying condition, A=20 at t=0

$20 = 200 + C \implies C = -180$

Hence desired equation of the state is given by,

$A = 200 - 180e^{-\frac{1}{40}t}$

(2) For this part, since we are pumping pure water so $R_{in} = 0$

similarly, making the equation of the state,

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

$R_{out} = \frac{A}{420}*6 = \frac{A}{70} g/min$

Then

$\frac{dA}{dt} = -\frac{A}{70}$

solving equation,

$\int \frac{dA} {A} = - \frac{1}{70} \int dt$

$ln A = -\frac{1}{70}t + C$

Applying condition that at t=0, A = 30

$ln(30) = C$

then

$lnA = -\frac{1}{70}t +ln(30)$

$A = 30e^{-\frac{1}{70}t}$