mixing prob: A storage tank contains 2000 gallons of gasolin

[thebenji]

Problem: A storage tank contains 2000 gallons of gasoline that initially has 100 pounds of an additive dissolved in it. In preparation for winter, gasoline containing 2 pounds of this additive per gallon is pumped into the tank at a rate of 40 gallons per minute. The well-mixed solution is pumped out at a rate of 45 gallons per minute. How much of the additive is in the tank 20 minutes after the pumping process begins?

[I believe my setup is incorrect.]

dy/dx = (rate in)-(rate out)

y(0)=100

rate in: (2/2000)*40 = 1/25
rate out: (y/2000)*40 = y/50

dy/dx = (1/25)-(y/50)

dy/dx = (2/50)-(-y)

-Int[dy/y] = Int[dx/25]

-ln(y) = x/25 + C
C=-ln(100)

y = e^[ln(100)-(x/25)]

y(20) = e^[ln(100)-(4/5)]

Thanks in advance!

 

[galactus]

\(\displaystyle \L\\\text{rate in}=(2\frac{lb}{gal})(40\frac{gal}{min})=80\frac{lb}{min}\)

\(\displaystyle \L\\\text{rate out}=(\frac{y(t)}{2000}\frac{lb}{gal})(45\frac{gal}{min})=\frac{9y(t)}{400}\frac{lb}{min}\)

\(\displaystyle \L\\\frac{dy}{dt}=80-\frac{9y(t)}{400}\)

\(\displaystyle \L\\\frac{dy}{dt}+\frac{9y(t)}{400}=80\)

Integrating factor:

\(\displaystyle \L\\e^{\int\frac{9}{400}dt}=e^{\frac{9t}{400}}\)

Initial condition => y(0)=100

\(\displaystyle \L\\\frac{d}{dt}[e^{\frac{9t}{400}}y]=80e^{\frac{9t}{400}}\)

Integrating gives:

\(\displaystyle \L\\e^{\frac{9t}{400}}y=\frac{32000}{9}e^{\frac{9t}{400}}+C\)

\(\displaystyle y(t)=\frac{32000}{9}+Ce^{\frac{-9t}{400}}\)

Using initial condition we find \(\displaystyle \L\\C=\frac{-31100}{9}\)

\(\displaystyle \H\\y(t)=\frac{32000}{9}-\frac{31100}{9}e^{\frac{-9t}{400}}\)

\(\displaystyle \L\\y(20)=\frac{32000}{9}-\frac{31100}{9}e^{\frac{-9(20)}{400}}=1352.2 \;\ \text{lbs}\)

Check it out. Easy to go astray.