The flow system shown in the figure is activated at time t= 0. Let Q_i (t) denot

Question

The flow system shown in the figure is activated at time t= 0. Let Q_i (t) denote the amount of solute present in the i th tank at time t. Assume that all the flow rates are a constant 10 L/min. it follows that the volume of solution in each tank remains constant; assume this volume to be 1000 L. The concentration of solute in the inflow to Tank 1 (from a source other than Tank 2) is 0.5 kg/L, and the concentration of solute in the flow to Tank 2 (from a source other than Tank 1) is 0 kg/L. Assume each tank is mixed perfectly. a. Set up a system of first-order differential equations that models this situation. [Q_1' Q_2] = [-1/50 01 01 -0.02] [Q_1, Q_2] + [5 0] If Q_!(0) = 30 kg and Q_2 (0)=10 kg, find the amount of solute in each tank after t minutes. Q_1 (t)= kg Q_2(t) = kg As t rightarrow infinity , how much solute is in each tank? In the long run, Tank 1 will have 1000/3 kg of solute. In the long run, Tank 2 will have 500/3 kg of solute. (Reread the question and think about why this answer makes sense.)

Explanation / Answer

Q1' = -Q1 / 50 + Q2/100 +5;

Q2' = Q1/100 - Q2/50;

We get;

Q1/100 = Q2'+Q2/50;

Q1' = -2(Q2'+Q2/50) + Q2/100 +5;

Q1' = -2Q2' - 3Q2/100 + 5;

Also;

Q2'' = Q1'/100 - Q2'/50;

We get; 100Q2 '' +2Q2' = Q1';

100Q2 '' +2Q2' = -2Q2' - 3Q2/100 + 5;

Or;

100Q2''+4Q2' + 3Q2/100 =5;

Solution for above equation is; Q2 = 500/3 + Ae^(-t/100) +Be^(-3t/100);

Now given Q2(0)=10; and Q2'(0)=0;

We get Q2= 5/3(47e^(-3t/100)-141e^(-t/100)+100);

Now we have Q2; putting in equation 1 we get Q1;

Q1 = 100Q2'+2Q2;

Q1 = 1000/3 - 235/3 e^(-3t/100) - 235 e^(-t/100);

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