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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?

User MaxiWheat
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Answer:

a)
y(t)=50000-49990e^{(-2t)/(25)}

b)
31690.7 g/L

Explanation:

By definition, we have that the change rate of salt in the tank is
(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(L)/(min)*50(g)/(L)=4000(g)/(min), and


R_(o)=40(L)/(min)*(y)/(500) (g)/(L)=(2y)/(25)(g)/(min)

So we obtain.
(dy)/(dt)=4000-(2y)/(25), then


(dy)/(dt)+(2y)/(25)=4000, and using the integrating factor
e^{\int {(2)/(25)} \, dt=e^{(2t)/(25), therefore
((dy )/(dt)+(2y)/(25)}=4000)e^{(2t)/(25), we get
(d)/(dt)(y*e^{(2t)/(25)})= 4000 e^{(2t)/(25), after integrating both sides
y*e^{(2t)/(25)}= 50000 e^{(2t)/(25)}+C, therefore
y(t)= 50000 +Ce^{(-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^{(-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^{(-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
(500L)/(40L/min) =(25)/(2) min=t, so we can evualuate the expression of a)
y(25/2)=50000-49990e^{(-2)/(25)(25)/(2)}=50000-49990e^(-1)=31690.7, is the salt concentration when the tank overflows

User Illishar
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