Two tanks are interconnected. Tank A contains 70 grams of salt in 60 liters of w

Question

Two tanks are interconnected. Tank A contains 70 grams of salt in 60 liters of water, and Tank B contains 60 grams of salt in 30 liters of water. A solution of 4 gram/L flows into Tank A at a rate of 7 L/min, while a solution of 1 grams/L flows into Tank B at a rate of 6 L/min. The tanks are well mixed. The tanks are connected, so 10 L/min flows from Tank A to Tank B, while 3 L/min flows from Tank B to Tank A. An additional 13 L/min drains from Tank B. Letting x represent the grams of salt in Tank A, and y represent the grams of salt in Tank B, set up the system of differential equations for these two tanks. dx/dt = dy/dt = x (0) =, y (0) =

Explanation / Answer

Solution:

Given that

Inital Volume of Tank A = 60L
Initial Volume of Tank B = 30L

First, you have to note the inflow and outflow of water for each tank:
Tank A: In =7 L/min Out= 10 L/min
Tank B: In = 6 L/min Out= 16 L/min ---->(3L/min + 13L/min)

Now let x(t) = amount of salt, in grams, in Tank A as function of time per min
and y(t) = amount of salt, in grams, in Tank B as function of time per min
where at time = 0 -->

x(0) = 70 grams of salt
y(0) = 60 grams of salt

Thus the derivative is:

x'(t) = the rate of change with respect to time of the amount of salt in Tank A

y'(t) = the rate of change with respect to time of the amount of salt in Tank B

x'(t) = (Incoming flow in Tank A)(concentration fo salt in flow)
+ (Incoming flow from Tank B)(Concentration of salt in Tank B)
- (Outgoing flow from Tank A)(concentration in Tank A)

x'(t) = (Incoming flow)(concentration of salt in flow)

+ (Incoming flow from TankB)[x(t)/(Volume of water in Tank B)]
- (Outgoing flow from Tank A)[q(t)/(Volume of water in Tank A)]

x'(t) = (4g/L)(7L/min) + (3L/min)[x(t)/30L] - (10L/min)[y(t)/60L]

x'(t) = 28 + (1/10)x(t) - (1/6)y(t) g/min

Similerly

y'(t) = (Incoming flow)(Concentration of salt in flow)

+ (incoming flow from Tank A)[x(t)/(Volume of water in Tank A)]
- (Outgoing flow from TankB)[y(t)/(Volume of water in Tank B)]

y'(t) = (1grm/L)(6L/min) + (10L/min)[x(t)/60L] - (16L/min)[y(t)/30L]

y'(t) = 6 + (1/6)x(t) - (8/15)y(t) g/min

Therefore x'(t) and y'(t) is:
x'(t) = dx/dt = 28 + (1/10)x(t) - (1/6)y(t) grams/min
y'(t) = dy/dt = 6 + (1/6)x(t) - (8/15)y(t) grams/min

x(0) = 70 grams

y(0) = 60 grams

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