Question

Each of the tanks shown in Figure 3.2.9 contains a brine solution. Assume that Tank 1 initially contains 30 gallons (gal) of water and 55 ounces (oz) of salt, and Tank 2 initially contains 20 gal of water and 26 oz of salt. Water containing 1 oz/gal of salt flows into Tank 1 at a rate of 1.5 gal/min, and the well-stirred solution flows from Tank 1 to Tank 2 at a rate of 3 gal/min. Additionally, water containing 3 oz/gal of salt flows into Tank 2 at a rate of 1 gal/min (from the outside The well-stirred solution in Tank 2 drains out at a rate of 4 gal/min, of which some flows back into Tank 1 at a rate of 1.5 gal/min, while the remainder leaves the system. Note that the volume of solution in each tank remains constant since the total rates of flow in and out of each tank are the same: 3 gal/min in Tank 1 and 4 gal/min in Tank 2. (a) Denoting the amount of salt in Tank 1 and Tank 2 by 21(t) and 020). respectively, use the principle of mass bal- ance (see Example 1 of Section 2.3) to show that de - -0.10 +00750, +1.5, dt dQ2 d00.12,-0.202 +3, dt Qi (0) = 55, Q2(0) = 26. (b) Write the initial value problem (i) using matrix notation. (c) Find the equilibrium values @f and 0% of the system (d) Use a computer to draw component plots of the initial value problem (i), and the equilibrium solutions, over the time interval 0s1350. (e) Draw a phase portrait for the system centered at the crit- ical point.

Answer 1

Given that: - rightarrow Tank1 contains 30 gallons (gal) of water 55 ounces of salt rightarrow Tank 2 contains 20 gal of water 26 o^F of salt rightarrow Q_1 (t) = Amount of salt in tan 1 at time T. rightarrow Q_2 (t) = Amount of salt in tank 2 at time T. rightarrow calculate the system of differential equations for Fun Q_1 t and Q_2 t \r\n rightarrow The rate of change of salt Q (t) equals to the rate of which salt is (a)Q_i (t) = 1.5 + 1.5 times Q_2/20 - 3 times Q_5/30 Q (t) = 1.5 + Q_2/13.3 - Q_1/10 rightarrow initial conditions are Q_2 (0) = 55 rightarrow are the differential equation tank2 Q_2 (t) = 3 times 1 + 1 times Q_1/30 - 4 times Q_2/20 Q^1_2 (t) = 3t - Q_1/30 - Q_2/5 (b) rightarrow The equilibrium occurs at Q_1 (t) = 0 Q^1_2 (t) = 0 \r\n rightarrow Find equilibrium