A large tank is filled with 500 liters of pure water. At time t = 0, an inflow v
ID: 2982575 • Letter: A
Question
A large tank is filled with 500 liters of pure water. At time t = 0, an inflow valve is opened and a brine solution with a concentration of 500 grams of salt per liter flows into the tank at 5 liters/minute. At the same time (t = 0), an outflow valve is opened and the thoroughly mixed solution flows out of the tank at 5 liters/minute. Derive the ODE that describes the x(t), mass of salt in the tank, for all times t 0. Solve this ODE together with the appropriate initial condition. Find the function that gives the concentration of salt in the tank for all times t > 0. Graph and interpret both the mass and concentration of salt. In particular, what can you say about the concentration in the tank as t rightarrow oo? Now imagine that the experiment of part (1) is repeated, except that at t = 0, the mixing tank springs a leak and the thoroughly mixed solution also flows out through the leak at a rate of .5 liter/minute. Derive the ODE that describes the mass of salt in the tank for all times t 0. Solve this ODE together with the appropriate initial condition. Find the function that gives the concentration of salt in the tank for all times t 0. Graph and interpret both the mass and concentration functions. At what time does the mass function have a maximum? When does the tank become empty and what is the concentration of the solution in the tank just as the tank becomes empty?Explanation / Answer
V = 500 gallons
A-in = 10 pounds/min
5 gal/min is pumped out from the tank. 5 gal is 1% of the tank's volume
A-out = 1% x S
(Where A-in is the rate of salt being pumped into the tank, and A-out for the rate of that being pumped out of the tank)
Always use A-in and A-out if you are confused on how to write down the first order differential ;) .
dA/dt = A-in - A-out = 10 - (A/100)
Solve the equation by separating variables.
dA / {10 - (A/100)} = dt
100 dA / (1000-A) = dt
integrate both sides
-100 ln|1000-A| = t
ln|1000-A| = -t/100
1000 - A = e^(-t/100)
A = 1000 - e^(-t/100)
b). c(t) = A(t) : V = {1000 - e^(-t/100)} : 500
The algebra should look much easier when done on paper I guess, but the main idea is to state the rate at which salt goes in (A-in) and the rate at which it goes out (A-out).
Then, the rate at which the mass of the salt changes can be stated as A-in - A-out.
Finally, use separating variable method to solve the first order differential equation.
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