Differential equations problem. 1. Two mixing tanks are arranged as shown in the

Question

Differential equations problem.

1. Two mixing tanks are arranged as shown in the diagram. Each tank holds 50 gallons of fluid. Initially, tank A contains 10 pounds of salt and tank B contains 0 pounds of salt. Suppose pure water is pumped into tank A at 3 gal/min, which is then pumped to tank B at 5 gal/min. The mixture in B is then pumped back into A at 2 gal/min while the mixture in B is also pumped out of the system at 3 gal/min. As a result of the input and outputs to the system, the volume of each tank remains constant 2. 50 ss Let A(): pounds of salt in tank A at time t (in minutes), while B(t) = pounds of salt in tank B at time t (in minutes). (a) Set up a system of differential equations to model the two-tank system. Include the initial conditions. (b) Solve the system of equations to find A(t) and B(t) (c) Graph the functions A(t) and B(t) over a reasonable time interval. Label both vertical and horizontal scale. Staple the graph to the assignment. (d) Estimate the time when both tanks have the same content of salt. (e) How much time until tank B has less than 0.1 pounds of salt?

Explanation / Answer

Given that

Inital Volume of Tank A = 50gal

Initial Volume of Tank B = 50gal

First, you have to note the inflow and outflow of water for each tank:
Tank A: In =7 gal/min Out= 5 gal/min
Tank B: In = 2 gal/min Out= 3 gal /min ---->(2gal/min + 3gal/min)

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

A(0) = 10 pounds of salt
B(0) = 0 pounds of salt

Thus the derivative is:

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

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

A'(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)

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

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

A'(t) = (0lb/gal)(3gal/min) + (2gal/min)[B(t)/50gal] - (5gal/min)[A(t)/50gal]

A'(t) = (1/25)B(t) - (1/10)A(t) pounds/min

Similerly

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

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

B'(t) = (A(t)lb/gal)(5gal/min) - (5gal/min)[B(t)/50L]

B'(t) = 5A(t) lb/min + ) - (1/10)B(t) lb/min

(a)

Therefore A'(t) and B'(t) are the differntial equations to model the two tank system include initial conditon

A'(t) = ((1/25)B(t) - (1/10)A(t)) lb/min

B'(t) = (5A(t) - (1/10)B(t)) lb/min

A(0) = 10 lb

B(0) = 0 lb

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