kuhse4461a

2021-12-30

Brine containing 3lb/gal of salt enters a tank at the rate of 4gal/min and the solution well-stirred leaves at 2gal/min. The tank initially contains 20 gal of water. What is the amount of salt in the tank after 5 minutes?

Pansdorfp6

Beginner2021-12-31Added 27 answers

Solution:

At a rate, salt enters the tube.

$R=\left(4\text{}ga\frac{l}{min}\right)\times \left(3\text{}l\frac{b}{g}al\right)$

$R=12$ lbs/min

Due to the tenk's 2 gal/min loss of liquid

$=4\text{}ga\frac{l}{min}-2ga\frac{l}{min}$

$=2\text{}ga\frac{l}{min}$

after to time: (minutes). The number of gallons of Brine in the tenk

$=20+2t$

So after 5 min

$=20+2\times 5$

$=30$ gallons

and 2 gallons $=.3454lbs$

So amount of salt $=30\times 8.3454$

$=250.362$

Heather Fulton

Beginner2022-01-01Added 31 answers

Given that the initial amount of salt is zero( because of pure water)

Rate in 4gal/min

Rate out 2gal/min initial volume is 20 gal

Brine containing 3lb/gal of salt

The inteving rate 4gal/min

So intering concentration

$=3\times 4$

$=12$

Volume$=20gal$

Let the Amount of Brine is y(+)

Then rate change

$y}^{1}=rat{e}_{\mathrm{ln}}-rat{e}_{out$

$=12-\frac{2\times y}{20+(4-2)t}$

${y}^{1}=12-\frac{y}{10+t},y\left(0\right)=0$

Salving${y}^{\prime}+\frac{1}{10+t}\text{}y=12$

$IF={e}^{\int \frac{1}{lot\text{}t}dt}={e}^{\mathrm{ln}(10+t)}=10+t$

$y\times (10+t)=\int 12\times (10+t)dt+c=6{(10+t)}^{2}+c$

$y(+)=6(10+t)+\frac{c}{10+t}$

By$y\left(0\right)=0\Rightarrow 0=60+\frac{c}{10}\Rightarrow c=-600$

$y\left(t\right)=6(10+t)-\frac{600}{10+t}$

The amount after 5 min

$y\left(5\right)=6\times (10+5)-\frac{600}{10+5}=90-\frac{600}{15}$

$y\left(5\right)=90-40=50lb$

Rate in 4gal/min

Rate out 2gal/min initial volume is 20 gal

Brine containing 3lb/gal of salt

The inteving rate 4gal/min

So intering concentration

Volume

Let the Amount of Brine is y(+)

Then rate change

Salving

By

The amount after 5 min

karton

Expert2022-01-09Added 613 answers

xleb123

Skilled2023-05-13Added 181 answers

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