Two tanks are interconnected. Tank A contains 50 grams of salt in 20 liters of w

Question

Two tanks are interconnected. Tank A contains 50 grams of salt in 20 liters of water, and Tank B contains 80 grams of salt in 40 liters of water.

A solution of 1 gram/L flows into Tank A at a rate of 9 L/min, while a solution of 5 grams/L flows into Tank B at a rate of 6 L/min. The tanks are well mixed.

The tanks are connected, so 12 L/min flows from Tank A to Tank B, while 3 L/min flows from Tank B to Tank A. An additional 15 L/min drains from Tank B.

Letting xx represent the grams of salt in Tank A, and yy 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:

Inital Volume of Tank A = 20L
Initial Volume of Tank B = 40L

First, you have to note the inflow and outflow of water for each tank:
Tank A: In= 12 L/min (9 L/min + 3 L/min) Out= 12 L/min
Tank B: In= 18 L/min (6 L/min + 12 L/min)Out= 18 L/min ---->(3L/min + 15L/min)

Now let x(t) = amount of salt, in grams, in Tank A as function of time per min
andy(t) = amount of salt, in grams, in Tank B as function of time per min

where at time t = 0;

==> x(0) = 50 grams of salt and y(0) = 80 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 of 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 in Tank A)(concentration of salt in flow) + (Incoming flow from Tank B)[x(t) / (Volume of water in Tank B)] - (Outgoing flow from Tank A)[y(t)/(Volume of water in Tank A)]

x'(t) = (1g/L)(9L/min) + (3L/min)[x(t)/40L] - (12L/min)[y(t)/20L]

x'(t) = 9 + (3/40)x(t) - (12/20)y(t) g/min
x'(t) = 9 + (3/40)x(t) - (24/40)y(t) g/min = dx/dt


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 Tank B)[y(t)/(Volume of water in Tank B)]

y'(t) = (5g/L)(6L/min) + (12L/min)[x(t)/20L] - (18L/min)[y(t)/40L]
y'(t) = 30 + (12/20)x(t) - (18/40)y(t) g/min
y'(t) = 30 + (12/20)x(t) - (9/20)y(t) g/min

Therefore p' and q' is:
x'(t) = 9 + (3/40)x(t) - (24/40)y(t) g/min = dx/dt for Tank 1
y'(t) = 30 + (12/20)x(t) - (9/20)y(t) g/min = dy/dt for Tank 2

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