MIXTURE PROBLEMS

Question

Determine how much salt is in the tank when it is full if a 30-gallon tank initially contains 15 gallons of salt water containing 5 pounds of salt and suppose salt water containing 1 pound of salt per gallon is pumped into the top of the tank at the rate of 2 gallons per minute, while a well-mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute?

Answer 1

Rate in, r_i=2g/m

Rate out, r_o=1g/m

r_i-r_o= 2-1 = 1 g/m

To fill the tank, need to add 15 gallons, which will take 15 minutes or t = 15

Let A(t) be the amount of salt in the tank at any time. A(0) = 5 lbs which was given.

Let c_i be the density in, which will be 1 lb/15gallons, since 15 gallons is needed to fill the tank.

c_i = 1

Let c₀ the density of the well mixed solution, which is changing in time because 1 gallon/minute is being

pumped in.

c_o=A(t)/(15 + (1)t) = A/(15+t)

dA/dt = Amount_in-Amount_out = c_ir_i-c_or_o = 2 - A/(15+t)

dA/dt + A/(15+t)=2

The integrating factor is μ = e^{∫1/(15+t)dt}=e^ln(15+t)=15+t

μdA/dt + μA/(15+t)=2μ

(15+t))dA/dt + A = 2(15+t)

d/dt((15+t)A) =2(15+t)

Integrating both sides

(15+t)A(t) = 30t+t²+C

A(t) = (30t+t²+C)/(15+t)

A(0) = C/15 = 5 --> C=75

The tank will be full after 15 minutes

A(15) = (30(15)+15²+75)/450 = 750/30 = 25 pounds of salt