Question

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flows into the tank at a rate of Rin=80 L/minutes (min). The fluid mixes instantaneously and is pumped out at a specified rate Rout. Let y(t) denote the quantity of salt in the tank at time t. Assume that Rout=40L/min. (a) Set up and solve the differential equation for y(t). (b) What is the salt concentration when the tank overflows?

Answer 1

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$, where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$, and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain.  $\frac{dy}{dt}=4000-\frac{2y}{25}$, then

$\frac{dy}{dt}+\frac{2y}{25}=4000$, and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$, therefore  $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$, we get   $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$, after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$, therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$, to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$, so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$, due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$, so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$, is the salt concentration when the tank overflows