Question

(1 point) Consider the two tank apparatus shown in the figure. Each tank has capacity 750 liters and initially contains 150 liters of fresh water. At time t = 0, the well-stirred mixing process begins. Suppose that the concentration of brine flowing into Tank 1 via the top tube is 0.75 kilograms per liter, and that the flow rates are r = 13 = 3 liters per minute, and r2 = 14 = 10 liters per minute. (a) Determine the volume of solution in each tank as a function of time, t, in minutes V:1) = liters V2(1) = liters (b) Determine the time interval of interest. (The process stops when a tank is full or empty.) Stopping time is minutes (c) Lot (t) and 0.(t) denote the amounts of salt (in kilograms) in the tanks at time in minutes). Derive the initial value problem with Q.(t) and 02(1) as dependent variables describing the moding process. Enter Q (1) as 01(t) and Q (1) as 02(t). kg/min (0,()) = 020 - kg/min with Q.(0) = kg and Q,(0) =

Answer 1

a) volume of solution in terms of time 't Vi(t) = Vo + (n + 74-72) t. 7 Vilt) = 150 + ( 3+10-10 t Vict) = 150+ 3t Also, V2(t) = Vot (r2- 74Y3)t xicotrie 12+ = 150 + (10-10-3)t V2c4l = 150-3t I b) Tank 1 is fall when, Vict) = 150+3t = 750 3 t = 600 t = 200 minutes = Tank 2 is empty when, 150-3t=0 » 3t = 150 te 50 minutes

Stopping time is so minutes C the rates Vi=13 =3 we have given 82 = 84 =10 further we have into Tank 1 via given the top tha brine flows tube at rate of 0.75. Thule, we can write system of diff. equn as shown Inflow rate - out flow rate | 7 d (Qi(t)) = 0.75V, + Q2Ct) 84 - QiCt).r2 V2ct) VICE at (10) - (1 0.75 (3) + Q2L) 150-3t QICL) 150+3t 3 d (Q () = 2.25 + 10 Q24+) 150-3t - 10Qice) 150+34