Question

The flow system shown in the figure is activated at time t = 0. Let Qi(t) denote the amount of solute present in the ith 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 0.2 kg/L, and the concentration of solute in the inflow to Tank 2 (from a source other than Tank 1) is 0 kg/L. Assume each tank is mixed perfectly. Tank 1 Tank 2 a. Set up a system of first-order differential equations that models this situation. Qí (3) Qź b. If Qı(0) = 20 kg and Q2(0) = 10 kg, find the amount of solute in each tank after t minutes. Qı(t) = kg Q2(t) = kg Instructions

Answer 1

Given, Tonkl Tank 2. Now a) Q2 9:- Inflow -outflow - (0:2)60)+ (10) Q1 (10) 1000 1000 (10) 1000 Q2 100 t 2 and. Q2 Q2 = Inflow outflow Q 1000 (10)_02 (10)- 1000 1ooo Q2 loo 5o. 1 -1 so loo 2 - o 100 50 Assume, -/ /oo 1/100 so A= and F(t)= - [ ] chasa steric equation (A-XII = 0. 7 so 100 → -X so loa

By solving the determinant, we get , and x 1 Too These & values are eigen values Find Eigen vectors, POS x = -3 log solve matrix equation, {A - [ 28 ] K = 0 =) 100 loo O -loo 1 Too kitkz=0. choose k,=-1, K2=1 NO Eigen vector for x = 3 is 100 and 6 the matrix equation {A - [ 100 2 } } x = 0 Too Solve the I loo 100 IN] -D Too loo -k, tkz = 0. choose ki=1 and k₂=1 loo Eigen vector for x = -1

The solution of homogeneous ean is, 100 9, c. [1] + c[i] e Now determine particular solution, Bp=[] o loo [a so gt- sli G ENICI li So. 9, + 1 b, +2=0 - ① 9,-Eb; - = 9, = 26, -0 1 Too so Now substitute a in - - - 2 (2 6,) + 1 b, +2=0 200 6, = 3 400 of 3. 400 3. 400 particular solution is ap- 200 The general solution is Q = Qct Qp yoo I mot ca + [Jean 200 -> Q(t):

Now From above, we p-3 di (t) = -c, e + Cze. can write Toot + 400 -It 200 'le 100 + c₂e 100 + Given &, (o)= and Q2(0)=10 8, (o)=-ci + C a + 400 20=-catcht 400 3 - - C., &C2 = 340 2 and 3 Q2 (0) = c + C₂ + 200 10=citet 200 & 170 = cth - we get by solving ② g ②, 85 C = 3 C -85

The amount of solute at each tanke at timet, Q, (t)= -85 -3 100 -It -85e 100 400 3 e + 3 Q₂ (t) = 85 pot 3 C -It - 85 e 100 200 3, 400 3 when tto, 8(t)= 200 3 i. Tankl will have 400 - 3 kg of solute Tanks will have 200 kg of Solute.