1 Consider a large tank holding 100 L of pure water into whi

1. Consider a large tank holding 100 L of pure water into which a brine solution of salt begins to flow at a constant rate of 6 L/min. The solution inside the tank is kept well stirred and is flowing out at a rate of 6 L/min. The concentration of salt in the brine entering the tank is 0.1kg/L.

a. Carefully set up the differential equation for the amount of salt in the tank.

b. Do a six step phase line analysis for the differential equation of part (a).

c. Using five step separation of variables method, solve the differential equation from part (a). Include exact input/output from the TI. d. Find when the concentration of salt in the tank will reach 0.05 kg/L.

2. Change the conditions of problem 1, so that the outflow is 5 L/min, while all the other conditions remain the same.

a. Carefully set up the differential equation for the amount of salt in the tank. b. Explain why a phase line analysis is inappropriate for the differential equation from part (a). c. Use deSolve to solve the equation from part (a). Include input/output. d. Determine the concentration of salt c at time t.

Solution

Let S(t) be the amount (in kg) of salt in the tank at time t > 0 (t in minutes). Then

dS/dt = (rate in) - (rate out)

rate in = (brine concentration, in kg/L)(inflow rate, in L/min) = (0.1)(6) = 0.6 kg/min

rate out = (concentration in the tank, in kg/L)(outflow rate, in L/min)

For the first problem, the outflow rate equals the inflow rate, and so the tank always contains 1000L of solution. For this case, the concentration in the tank is S(t)/1000 and the outflow rate is 6 L/min.

Thus, the DE for the first problem is

dS/dt = 0.6 - 0.006S
S(0) = 0

You should have no trouble solving this first-order linear DE. The solution is

S(t) = 100(1 - e^(-0.006t))

For the second problem, the amount of solution in the tank is a function of t. There is a net increase of (6 - 5) = 1 L/min, so the amount of solution in the tank at time t is (1000 + t) L. Therefore, the concentration in the tank at time t is S(t)/(1000 + t)

Thus, the DE for this problem is

dS/dt = 0.6 - S/(1000 + t)
S(0) = 0