Answer to Question #122452 in Differential Equations for JSE

Question

1. A large tank is filled to capacity with 1000 L of pure water. Brine containing 0.25 kg of salt per liter is pumped into the tank at a rate of 10 L/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of kilograms of salt in the tank at time t.

A(t) = ___ kg

2. A tank contains 250 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.

A(t) = ___

Answer 1

The process is described by differential equation $dA=\frac{mt-A}{V}Mdt$ where $M$ -is the brine of salt rate $(L/min)$ , $m$ - is salt rate $(kg/min)$ , $V$ - is initial volume of water in a tank $(L)$ .The solution to this equation is $A(t)=m(t-\frac{V}{M}(1-e^{-\frac{Mt}{V}})+Ce^{-\frac{Mt}{V}})$ then

1. As $A(0)=0$ hence $C=0$ then $A(t)=2.5(t-100(1-e^{-0.01t}))kg$
2. As $A(0)=10g$ hence $C=2min$ then $A(t)=5(t-50(1-e^{-0.02t})+2e^{-0.02t})g$