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Beez
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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.
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 sense 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?
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 sense 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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