Question

Consider two brine tanks connected as shown in the figure. The brine solution is pumped from tank 1 into tank 2 at a rate of 10 L/min10 L/min, and from tank 2 into tank 1 at a rate of 10 L/min10 L/min. Suppose there are 50 L50 L of brine in tank 1 and 25 L25 L of brine in tank 2. Let xx be the amount of salt, in kilograms, in tank 1 after tt minutes have elapsed, and let yy the amount of salt, in kilograms, in tank 2 after tt minutes have elapsed. Assume that each tank is mixed perfectly.

1. If x(0)=8 kgx(0)=8 kg and y(0)=10 kgy(0)=10 kg, find the amount of salt in each tank after ttminutes.

Tank 1 Tank 2 (3 points) Consider two brine tanks connected as shown in the figure. The brine solution is pumped from tank 1 into tank 2 at a rate of 10 L/min, and from tank 2 into tank 1 at a rate of 10 L/min. Suppose there are 50 L of brine in tank 1 and 25 L of brine in tank 2. Let x be the amount of salt, in kilograms, in tank 1 after t minutes have elapsed, and let y the amount of salt, in kilograms, in tank 2 after t minutes have elapsed. Assume that each tank is mixed perfectly a. If x(0) = 8 kg and y(0) = 10 kg, find the amount of salt in each tank after t minutes x(t) = kilograms y(t) kilograms b. As t o0, how much salt is in each tank? In the long run, tank 1 will have kilograms of salt. In the long run, tank 2 will have kilograms of salt. (Reread the question and think about why this answer makes sense.)

Answer 1

1oL/Min amount of Salt in kg = ()X Tank 1 SOL n 2SL Binly armount o y salt in kg y ct 201 Tank 2 in Tank 2 Tank mixed Peofecty each tankis co) 8 k9, Assume, hat and ind o1 xct) and yt) xo Ct and yo Scanned with Canrscanner d

dx dt 25S LJ amount of Salt Comning from tunk 2 amount of Salt gaing into Tank 2 dy at 25 amount nto tunk Salt aMount of Salt Coming goirg. ftan dx ouy o.22t C dt 22-04y dy dt 9n matix firSt dt 0-2 - -0-2 -o.4 O.2 det -0-2-)0-4-A)00 0-2 A = O.o8 to-2A to 4t--o O 2 K-h-O ho O-2-o.4 -0-2 eigenYedor for A0 0-2V, to 4 V2 0 +V,-2V2 o Y2 V O-2V, -o-UV220 eigenator prAio eigenyeuter tr d6 41 o-4 -0.6 0-2 O-2 ut2 0 O-2u1t 0.2U2 eirewvator Th-z'h Ohus te general Solution is 2 e (0X y (o) tzo at Finaly canied Withs. mscanne

2 Ct 10 2 3 & + 1o 12 12 - +20 HvP given by wr Thus Sol 2 79.0- e yet 39.0- 12 OR Xct) 6 t 4 e0.6 6 12 As t0 ecwith CamScanna