Brine containing 1.5lb/gal of salt enters a tank at the

oliviayychengwh 2022-01-03 Answered
Brine containing 1.5lb/gal of salt enters a tank at the rate of 2gal/min and the mixture well stirred leaves at the same rate. The tank initially contains 100 gal of brine with10 lbs of dissolved salt. When will the amount of salt in the tank be 20 lbs?

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Gerald Lopez
Answered 2022-01-04 Author has 1097 answers
\(\displaystyle{y}=\) amount of salt at time t
\(\displaystyle{r}_{{{i}}}={2}\)
\(\displaystyle{c}_{{{i}}}={1.5}\)
\(\displaystyle{r}_{{{o}}}={2}\)
\(\displaystyle{v}{\left({t}\right)}={v}_{{{0}}}+{\left({r}_{{{i}}}-{r}_{{{o}}}\right)}{t}\)
\(\displaystyle{v}{\left({t}\right)}={100}+{\left({2}-{2}\right)}{t}\)
\(\displaystyle{v}{\left({t}\right)}={100}+{0}{t}\)
\(\displaystyle{v}{\left({t}\right)}={100}\)
\(\displaystyle{c}_{{{o}}}={\frac{{{y}}}{{{v}{\left({t}\right)}}}}\)
\(\displaystyle{c}_{{{o}}}={\frac{{{y}}}{{{100}}}}\)
DE is given by
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={r}_{{{i}}}{c}_{{{i}}}-{r}_{{{o}}}{c}_{{{o}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={2}\cdot{1.5}-{2}\cdot{\frac{{{y}}}{{{100}}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={3}-{\frac{{{y}}}{{{50}}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\frac{{{y}}}{{{50}}}}+{3}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\left({\frac{{{y}}}{{{50}}}}-{3}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{\frac{{{1}}}{{{50}}}}{\left({y}-{150}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}=-{0.02}{\left({y}-{150}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left({y}-{150}\right)}}}}=-{0.02}{\left.{d}{t}\right.}\)
\(\displaystyle\int{\frac{{{\left.{d}{y}\right.}}}{{{\left({y}-{150}\right)}}}}=\int-{0.02}{\left.{d}{t}\right.}\)
\(\displaystyle{\ln{{\left({y}-{150}\right)}}}=-{0.02}{t}+{c}\)
\(\displaystyle{y}-{150}={e}^{{-{0.02}{t}+{c}}}\)
\(\displaystyle{y}-{150}={e}^{{-{0.02}{t}}}\cdot{e}^{{{c}}}\)
\(\displaystyle{y}-{150}={e}^{{-{0.02}{t}}}\cdot{C}\)
\(\displaystyle{y}-{150}={C}{e}^{{-{0.02}{t}}}\)
\(\displaystyle{y}={150}+{C}{e}^{{-{0.02}{t}}}\) (1)
The tank initially contains initially 10 lb of salt
\(\displaystyle{10}={150}+{C}{e}^{{-{0.02}\cdot{0}}}\)
\(\displaystyle{10}={150}+{C}\cdot{1}\)
\(\displaystyle{10}={150}+{C}\)
\(\displaystyle{C}={10}-{150}\)
\(\displaystyle{C}=-{140}\)
\(\displaystyle{y}={150}-{140}{e}^{{-{0.02}{t}}}\)
find t when \(\displaystyle{y}={20}\)
\(\displaystyle{20}={150}-{140}{e}^{{-{0.02}{t}}}\)
\(\displaystyle{140}{e}^{{-{0.02}{t}}}={150}-{20}\)
\(\displaystyle{140}{e}^{{-{0.02}{t}}}={130}\)
\(\displaystyle{e}^{{-{0.02}{t}}}={\frac{{{130}}}{{{140}}}}\)
\(\displaystyle{e}^{{-{0.02}{t}}}={\frac{{{13}}}{{{14}}}}\)
\(\displaystyle-{0.02}{t}={\ln{{\left({\frac{{{13}}}{{{14}}}}\right)}}}\)
\(\displaystyle{t}=-{\frac{{{\ln{{\left({\frac{{{13}}}{{{14}}}}\right)}}}}}{{{0.02}}}}\)
\(\displaystyle{t}={3.705398}\)
\(\displaystyle{t}\approx{3.71}\) minutes
the amount of salt in the tank be 20 lbs after 3.71 minutes
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Matthew Rodriguez
Answered 2022-01-05 Author has 2684 answers

\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={\left(\text{rate in}\right)}-{\left(\text{rate out}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={2}{\left({1}\cdot{5}\right)}-{\left({\frac{{{y}}}{{{100}}}}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{y}}}{{{100}}}}={3}-{\left({1}\right)}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}+{p}{\left({x}\right)}{y}={q}{\left({x}\right)}\)
Which is a first order linear ode.
\(\displaystyle{I}.{f}={e}^{{\int{p}{\left.{d}{x}\right.}}}={e}^{{\int{\frac{{{1}}}{{{100}}}}{\left.{d}{x}\right.}}}={e}^{{{\frac{{{x}}}{{{100}}}}}}\)
\(\displaystyle{y}\times{I}{F}=\int{q}\times{I}{F}.{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{y}.{e}^{{{\frac{{{x}}}{{{100}}}}}}=\int{3}\cdot{e}^{{{\frac{{{x}}}{{{100}}}}}}\cdot{\left.{d}{x}\right.}\)
\(\displaystyle\Rightarrow{y}\cdot{e}^{{{\frac{{{x}}}{{{100}}}}}}={300}{e}^{{{\frac{{{x}}}{{{100}}}}}}+{c}\)
\(\displaystyle\Rightarrow{y}={300}+{c}{e}^{{{\frac{{{x}}}{{{100}}}}}}\)
So \(\displaystyle{y}={300}+{c}{e}^{{\frac{{t}}{{100}}}}-{\left({2}\right)}\)
\(\displaystyle{y}{\left({0}\right)}={\frac{{{10}}}{{{100}}}}={\frac{{{1}}}{{{10}}}}\)
(2) \(\displaystyle\Rightarrow{300}+{c}{e}^{{{0}}}={\frac{{{1}}}{{{10}}}}\Rightarrow{C}={\frac{{{1}}}{{{10}}}}-{300}\)
\(\displaystyle\Rightarrow{C}={\frac{{{2.999}}}{{{10}}}}=-{299.9}\)
So (2) \(\Rightarrow 300-299.9 \cdot e^{-t/100}=100\)
\(\displaystyle\Rightarrow{299.9}\cdot{e}^{{-\frac{{t}}{{100}}}}={200}\)
\(\displaystyle\Rightarrow{e}^{{-\frac{{t}}{{100}}}}={0.66688}\)
\(\displaystyle\Rightarrow-{\frac{{{t}}}{{{100}}}}={\log{{\left({0.66688}\right)}}}\)
\(\displaystyle\Rightarrow{t}={100}{\log{{\left({0.66688}\right)}}}\)
\(\displaystyle{t}={100}\times{0.4051}\)
\(\displaystyle{t}={40.51}\)

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karton
Answered 2022-01-09 Author has 8454 answers

Let x(t)= The amount of salt in the tank at time t.
Rate of increase of x,
\(\frac{dx}{dt}=\text{Rate of salt in}-\text{Rate of salt out}\)
\((\text{Rate})in=3lb/gal \times 4gal/min\)
\(\begin{array}{}=12 lb/min \\(\text{Rate})out=2gal/min \times \frac{x(t)}{2t+20} lb/gal \\=\frac{2x}{2t+20} lb/min \\\frac{dx}{dt}=12-\frac{2x}{2t+20} \\\Rightarrow \frac{dx}{dt}=+\frac{2x}{2t+20}=12 \\\text{This is linear first order differential equation of the form} \\\frac{dx}{dt}+px=q \\p=\frac{2}{2t+20}\ q=12 \\\text{Integrating factor}\ (I.F)=e^{\int pdt} \\=e^{\int \frac{2}{2t+20}dt} \\\text{Put}\ 2t+20=z \\2dt=dz \\\therefore IF=e^{\int \frac{dz}{z}} \\=e^{\ln |z|} \\=e^{\ln |2t+20|} \\IF=2t+20 \\\text{The solution is} \\x \times I.F=\int q \times IF dt \\\Rightarrow x(2t+20)=\int 12(2t+20)dt \\\Rightarrow x(2t+20)=12 \int 2(t+10)dt \\=24 \int (t+10)dt \\=24 [\frac{t^{2}}{2}+10t]+c \\\Rightarrow x(t)=\frac{12t^{2}+240t+c}{2t+20} \\\text{Given},\ x(0)=0 \\0=\frac{12(0)^{2}+240 \times 0+c}{2 \times 0+20} \\\Rightarrow 0=\frac{c}{20} \\\text{After}\ t=5min \\x(5)=\frac{12(5)^{2}+240 \times 5+20}{(2 \times 5)+20} \\=\frac{(12 \times 25)+1200+20}{10+20} \\=\frac{300+1200+20}{300} \\=\frac{1520}{30} \\=50.67lb \\\Rightarrow c=0 \\x(t)=\frac{12t^{2}+240t}{2t+20} \\=\frac{\not{2}(6t^{2}+120t)}{\not{2}(t+10)} \\x(t)=\frac{6t^{2}+120t}{t+10} \\\text{After}\ t=5min \\x(5)=\frac{6(5)^{2}+120(5)}{5+10} \\=\frac{(6 \times 25)+600}{15}=\frac{150+600}{15} \\=\frac{750}{15} \\=50lb \end{array}\)
The amount of salt in the tank after 5 min is 50lb.

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