A large tank initially holds 1200 L of a brine solution in which 300 kg of salt

Question

A large tank initially holds 1200 L of a brine solution in which 300 kg of salt is dissolved. For the first 10 min of operation, valve A is open, adding 6L/min of brine containing a 4 kg/L salt concentration. After 10 min, valve B is switched in, adding 2 kg/L concentration at 6 L/min. The exit valve removes 6 L/min, thereby keeping the volume constant.

A.) Set up an IVP(initial value problem) using the step function
b.) Solve it to find the mass of salt in the tank at t minutes after the process begins,

A picture of the two tanks would be lovely

Explanation / Answer

If Q(t) gives the amount of the salt dissolved in the liquid in the tank at any time t, we have

Rate of change of Q(t) = dQ/dt = Rate at which Q(t) enters the tank - Rate at which Q(t) exits the tank

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of salt in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of salt in liquid exiting)

Concentration of exiting solution at time t is = Amount of salt in the tank at any time t / Volume of solution at any time t

Since, entry and exit flow rates are same (both are 6 L/min), Voume of tank at any time t remains constant = 1200 L

So, for t <= 10, Q'(t) = 6 *4 - 6*(Q/1200), Q(0) = 300 kg

and for t >= 10, Q'(t) = 6 *2 - 6*(Q/1200), Q(10) = calculated from above DE

Solving first DE, for t <= 10, Multiplying both the sides by e^(6t/1200) we get

Q' * e^(6t/1200) + (6Q/1200) * e^(6t/1200) = 24*e^(6t/1200)

(Q * e^(6t/1200))' = 24*e^(6t/1200)

Integrating both the sides, Q * e^(6t/1200) = (24*1200/6)*e^(6t/1200) + c1

At t = 0, Q = 300, so 300 = (24*1200/6) + c1

c1 = -4500

So, Q * e^(6t/1200) = (24*1200/6)*e^(6t/1200) - 4500

or Q * e^(6t/1200) = 4800*e^(6t/1200) - 4500

or Q = 4800 - 4500*e^(-6t/1200)

thus at t = 10, Q = 519.467 kg

Using this to solve second DE, for t >= 10, Q'(t) = 6 *2 - 6*(Q/1200), Q(10) = 519.467 kg

Q * e^(6t/1200) = (12*1200/6)*e^(6t/1200) + c2

At t= 10, Q = 519.467

so c2 = -1976.949

so, Q * e^(6t/1200) = (12*1200/6)*e^(6t/1200) - 1976.949

or Q = 2400 - 1976.949*e^(-6t/1200)

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