Q: A large tank initially contains 100 gal of brine in which 10lb of salt is sissolved. Starting at t=0, pure water flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and the well-stirred mixture simulaneously flows out at the slower rate of 2 gal/min.(adsbygoogle = window.adsbygoogle || []).push({});

a) How much salt is in the tank at the end of 15 min and what is the concentration at that time?

A: Initial condition x = 10, t=0

[tex]IN = (0 lb/gal)(5 gal/min) = 0 lb/min[/tex]

[tex]OUT =(\frac{x}{100+3t} lb/gal)(2 gal/min) = \frac {2x}{100 + 3t}lb/min [/tex]

then

[tex]\frac{dx}{dt} = 0 - \frac{2x}{100+3t}[/tex]

[tex]\frac{dx}{dt} + \frac{2x}{100+3t} = 0[/tex]

[tex]IF = (3t + 100) ^\frac{2}{3}[/tex]

so [tex]\frac{d(3t+100)^\frac{2}{3}x}{dt} = dt[/tex]

[tex](3t + 100)^\frac{2}{3}x = c[/tex]

[tex]x = \frac{c}{(3t + 100)^\frac{2}{3}}[/tex]

Substituting initial condition x = 10, t=0,

[tex]10 =\frac{C}{(3t +100)^\frac{2}{3}}[/tex]

[tex]10(100)^\frac{2}{3}= C[/tex]

C = 215.443

Hence the amount of salt in time t is

[tex]x = \frac{215.443}{(3t + 100)^\frac{2}{3}}[/tex]

Solving problem a) using this equation, I got

t = 15

[tex]x =\frac{215.44}{(3*15 + 100)^\frac{2}{3}}= 7.8lb.[/tex]

But the textbook answer was 0.054lb. I understand that since there was only 10lb in 100 gal to start with, it doesn't make sence it still contains 7.8lb. of salt after 15min., but I don't know where I made a mistake.

Will somebody tell me where I should revise?

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# Mixture Problem (differential equations)

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