Question

A tank contains 60 kg of salt and 1000 L of water. A solution of a concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate. Let y be the number of kg of salt in the tank after t minutes. Write the differential equation for this situation

Answer 1

Answer:

$\dfrac{dy}{dt}=0.27-0.009y(t), y(0)=60kg$

Step-by-step explanation:

Volume of water in the tank = 1000 L

Let y(t) denote the amount of salt in the tank at any time t.

Initially, the tank contains 60 kg of salt, therefore:

y(0)=60 kg

Rate In

A solution of concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min.

$R_{in}$ =(concentration of salt in inflow)(input rate of solution)

$=(0.03\frac{kg}{liter})( 9\frac{liter}{min})=0.27\frac{kg}{min}$

Rate Out

The solution is mixed and drains from the tank at the same rate.

Concentration, $C(t)=\dfrac{Amount}{Volume} =\dfrac{y(t)}{1000}$

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

$=\dfrac{y(t)}{1000}* 9\dfrac{liter}{min}=0.009y(t)\dfrac{kg}{min}$

Therefore, the differential equation for the amount of Salt in the Tank  at any time t:

$\dfrac{dy}{dt}=R_{in}-R_{out}\\\\\dfrac{dy}{dt}=0.27-0.009y(t), y(0)=60kg$