Question

A tank contains 70 gallons of pure water. A brine solution with 2 lbs/gal of salt enters at 2.25 gal/min and the well - stirred mixture leaves at the same rate. Find the amount of salt in the tank at any time and the time at which the brine leaving the tank will contain 15 lbs of salt.

First order differential equations
ANSWERED
asked 2021-01-19
A tank contains 70 gallons of pure water. A brine solution with 2 lbs/gal of salt enters at 2.25 gal/min and the well - stirred mixture leaves at the same rate. Find the amount of salt in the tank at any time and the time at which the brine leaving the tank will contain 15 lbs of salt.

Answers (1)

2021-01-20

Form a differential equation that solves the amount of salt at any time t as follows
\(\displaystyle\frac{{{d}{A}}}{{\left.{d}{t}\right.}}=\) rate in - Rate out
\(\displaystyle={2}\cdot{2.25}-\frac{{{A}{\left({t}\right)}}}{{70}}\cdot{2.25}\)
\(\displaystyle=\frac{4.5}{{{9}{A}{\left({t}\right)}}}/{280}\)
\(\displaystyle=\frac{{{1260}-{9}{A}}}{{280}}\)
\(\displaystyle\frac{{{d}{A}}}{{\left.{d}{t}\right.}}=\frac{{{1260}-{9}{A}}}{{280}}\), A(0)= 0
\(\displaystyle\frac{{{d}{A}}}{{{1260}-{9}{A}}}=\frac{{\left.{d}{t}\right.}}{{280}}\)
\(\displaystyle\int\frac{{{d}{A}}}{{{1260}-{9}{A}}}=\int\frac{{\left.{d}{t}\right.}}{{280}}\)
\(\displaystyle-\frac{1}{{9}} \ln{{\left({1260}-{9}{A}\right)}}=\frac{t}{{280}}+{C}\)
A(0)= 0
\(\displaystyle\Rightarrow{C}=-\frac{1}{{9}} \ln{{\left({1260}\right)}}\)
\(\displaystyle\Rightarrow \ln{{\left({1260}-{9}{A}\right)}}=-{9}\frac{t}{{280}}+ \ln{{\left({1260}\right)}}\)
\(\displaystyle\Rightarrow{1260}-{9}{A}={e}^{{-{9}\frac{t}{{280}}+ \ln{{\left({1260}\right)}}}}\)
\(\displaystyle\Rightarrow{1260}-{9}{A}={1260}{e}^{{-{9}\frac{t}{{280}}}}\)
\(\displaystyle\Rightarrow{140}-{A}={140}{e}^{{-{9}\frac{t}{{280}}}}\)
\(\displaystyle\Rightarrow{A}{\left({t}\right)}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)
Obtain the time at which the brine leaving the tank will contain 15 lbs of salt as follows
\(\displaystyle{A}{\left({t}\right)}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)
\(\displaystyle{15}={140}-{140}{e}^{{-{9}\frac{t}{{280}}}}\)
\(\displaystyle{140}{e}^{{-{9}\frac{t}{{280}}}}={140}-{15}\)
\(\displaystyle{e}^{{-{9}\frac{t}{{280}}}}=\frac{125}{{140}}\)
\(\displaystyle-{9}\frac{t}{{280}}= \ln{{\left(\frac{125}{{140}}\right)}}\)
\(\displaystyle{t}=-\frac{280}{{9}} \ln{{\left(\frac{125}{{140}}\right)}}\)
\(t= 3.53 min\)

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