Two large tanks are shown below with the corresponding intake and output of salt

Question

Two large tanks are shown below with the corresponding intake and output of saltwater: Tank A is initially contains initially 50 gallons of salt water (but it is not full and has large capacity) with of 0.25 lb/gal. Tank B is also a high capacity tank that has an initial volume of 40 gallons and it contains only pure water. Pure water is pumped into tank A at a rate of 4 gal/min. Mixture quickly mix the liquid in tank A which then sows out of A and into B at the rate of 2 gal/min. Finally, the mixture in tank B is released at the same rate of 2 gal/min [see figure]. Write the appropriate differential equations with the corresponding initial conditions for x(t) and y(t). Then solve the system to obtain the correct expressions for x(t) and y(t).

Explanation / Answer

a)

It is given that x(t) and y(t) are the amount of salt in lb in tank A and tank B respectively at time t.

Thus the concentrations in tank A and tank B at time t are x/50 and y/40 respectively (amount/vol).

Now the rate of change in concentration of salt in tank A is:

dx/dt = - [net rate of outflow of salt from tank A]

          = - [flow rate of water]*[initial conc.of salt]

          = - [4x/50 - 2*x/50]

Now for tank B, rate of change in concentration of salt is :

dy/dt   =    [inflow rate salt from tank A] – [outflow rate of salt from tank B]

          =    2x/50 – 2y/40

Finally we have,

dx/dt + x/25 = 0,

dy/dt + y/20 – x/25 = 0

these are the appropriate differential equations with the corresponding initial conditions for x(t) and y(t).

b)

Now to find expressions for x(t) and y(t), proceed likewise:

dx/dt = -x/25 where x(t=0) = 12.5                 …(1)

dy/dt = x/25 – y/20 where y(t=0) = 0            …(2)

integrating (1), we get dx/dt = -x/25 => dx/x = -1/25 dt     => x = C’e-t/25

now with the initial condition of (1), we get C’ = 12.5, so we have     x = 12.5*e-t/25

now putting this value of x in (2), we get

dy/dt + y/20 = 0.5*e-t/25

we will solve this diff equation by using integrating factor method,

I.F = et/20

Upon integration, we get >   y* et/20 = 50*t1.01/101   =>        y = (e-t/20 * 50*t1.01) / 101

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