A large mixing tank initially contains 500 gallons water in which 40 pounds of s

Question

A large mixing tank initially contains 500 gallons water in which 40 pounds of salt have been dissolved. Another brine solution is pumped into the tank at the rate of 4 gallons per minute, and the resulting mixture is pumped out at the same rate. The concentration of the incoming brine solution is 2 pounds of salt per gallon. If A(t) represents the amount of salt in the tank at time t, find the differential equation for A. a. Solve the differential equation. b. How many pounds of salt are in the tank after 125 minutes? c. After a long time, how many pounds of salt will be in the tank?

Explanation / Answer

a)initial amount of water =500 gallons

initial amount of salt A(0)= 40 lbs

amount of salt input = 4*2 =8 pounds per minute

amount of salt output = 4*(A(t)/500)= (A(t)/125)

net rate=amount of salt input -amount of salt output

A'(t)=8- (A(t)/125)

or simply

dA/dt =8-(A/125)

dA +(A/125)dt=8dt

integrating factor =e(1/125)dt

integrating factor =e(t/125)

multiply on both sides by e(t/125)

dAe(t/125) +(A/125)e(t/125)dt=8e(t/125) dt

(Ae(t/125))'=8e(t/125) dt

integrate on both sides

(Ae(t/125))'=8e(t/125) dt

Ae(t/125)=1000e(t/125) +c

A=1000+ce(-t/125)

A(0)= 40

1000+ce(-0/125)=40

1000+c=40

c=-960

so amount of salt A(t)=1000 -960e(-t/125)

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b)

after 125 minutes, t=125

amount of salt ,A(125)=1000 -960e(-125/125)

amount of salt ,A(125)=646.84 pounds

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c)after long time

amount of salt=lim[t->][1000 -960e(-t/125)]

amount of salt=[1000 -960*0]

amount of salt=1000 -0

amount of salt=1000 pounds

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