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Question

In this exercise, a 1000-liter 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 R_{\text {in }}=80 L/min. The fluid mixes instantaneously and is pumped out at a specified rate Rout. R_{\text {out. }} Let y(t) denote the quantity of salt in the tank at time t.

Assume that Rout R_{\text {out }}=40 L/min.

(a) Set up and solve the differential equation for y(t)y(t).

(b) What is the salt concentration when the tank overflows?

Solution

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dydt\dfrac{dy}{dt} is the difference between the two rates of change:

dydt=RateinRateout\dfrac{dy}{dt}=Rate_{in}-Rate_{out}

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