Question

15. Consider a two tank system pictured below. Suppose tank A contains 100 gallons of water in which 120 pounds of salt are dissolved initially. Suppose tank B has 100 gallons of water in which zero pounds of salt are dissolved initially. Liquid is pumped into and out of the tanks as indicated in the figure; the mixture exchanged between the two tanks and the liquid pumped out of tank B are assumed to be well mixed. How many pounds of salt are in tank B after 25 minutes? Round your final answer to four decimal places. Hint, work slowly, the numbers are nice. 6 gal/min 2 gal/min pure water Tank A Tank B 8 gal/min 6 gal/min

Answer 1

Solution:

Let x be the amount of salt in tank A and v in tank B

According to the diagram

100 100 dt

4.xy 一(1) dt _ _50 + 5。2(0) = 120

工 -. 2 * 100 100 dt

dly .Ty dt 50 50

taking laplace transform of equation (1) and (2) we get

sX(s) - r 50 50

4X(s) 4Y(s) => X(s) ( s-50 sy (s) - y(0)-1X0)_4Ys) 4Y (s) 504) (s) ( s 50

eliminating Y(s) from (3) and (4) we get,

$=>X(s)\left (s+\frac{4}{50} \right )=-\frac{X(s)}{4}\left (s-\frac{4}{50} \right )+120$

$=>X(s)\left (s+\frac{4}{50} \right )+\frac{X(s)}{4}\left (s-\frac{4}{50} \right )=120$

$=>X(s)\left (\frac{5s}{4}+\frac{3}{50} \right )=120$

120 => X(s)-

=> X(s) = 96

$\text{Now taking the inverse laplace transform we get, }$

$=>x(t)=96e^{-\frac{6t}{125}}$

$\text{Now, }$

$=>X(s)\left (s-\frac{4}{50} \right )=-\frac{4Y(s)}{50}$

$=>\left ( \frac{96}{s+\frac{6}{125} } \right )\left (s-\frac{4}{50} \right )=-\frac{4Y(s)}{50}$

$=>\left ( \frac{24}{s+\frac{6}{125} } \right )\left (s-\frac{4}{50} \right )=-\frac{Y(s)}{50}$

$=>Y(s)=- \frac{1200\left (s-\frac{4}{50} \right )}{s+\frac{6}{125} }$

$=>Y(s)=-1200+\frac{19200}{125 s + 6}$

1200 + 19200

$\text{Now taking the inverse laplace transform we get, }$

$=>y(t)=-1200 \delta\left(t\right) + \frac{768}{5} e^{- \frac{6 t}{125}}$