Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Suppose in Problem 33 valve B is initially opened for10 min and then valve A is switched in for 10 min. Finally, valve B is switched back in. Find the concentration of salt in the tank as a function of time.

Step-by-Step Solution

Answer

The concentration of salt in the tank as a function of time is

\(C = 0.06 - 0.04{e^{ - \frac{3}{{125}}t}} - 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10) + 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 20)\).

Step 1: find the mass of salt

Let\(m(t)\) be the mass of salt in the tank at time t.

Then the rate of change of the mass will be

\(m' = {\rm{ir}} - {\rm{\;or\;}} \to {\rm{(1)}}\)

where "ir" is input rate and "or" output rate.

Since the mixing tank holds 500L and the exit valve removes 12L/min, we have that output rate is:

\(\begin{array}{c}{\rm{\;or\;}} = \frac{m}{{500}} \cdot 12\\ = \frac{{3m}}{{125}}\end{array}\)

Since in input rate we have valve A and valve B, we need to broke input rate in two intervals, for the first 10 min\((0 < t < 10)\), between 10min and 20min\((10 < t < 20)\)and after \((t > 20):\)

\(\begin{array}{c}{\rm{\;ir\;}} = \left\{ {\begin{array}{*{20}{c}}{12 \cdot 0.06,0 < t < 10}\\{12 \cdot 0.04,10 < t < 20}\\{12 \cdot 0.06,t > 20}\end{array}} \right.\\ = \left\{ {\begin{array}{*{20}{c}}{0.72,0 < t < 10}\\{0.48,10 < t\left\langle {20,0.72,t} \right\rangle 20}\end{array}} \right.\\ = 0.72 - 0.48u(t - 10) + 0.48u(t - 20)\end{array}\)

 

Let

\(\begin{array}{c}{\rm{\;ir\;}} = f(t)\\ = 0.72 - 0.48u(t - 10) + 0.48u(t - 20)\end{array}\)

Therefore (1) becomes

\(m' + \frac{3}{{125}}m = 0.72 - 0.48u(t - 10) + 0.48u(t - 20)\)

and using initial condition, which is mass of the salt in the tank at time\(t = 0\)

\(\begin{array}{c}m(0) = 500 \cdot 0.02\\ = 10\end{array}\)

We get the IVP

\(m' + \frac{3}{{125}}m = 0.72 - 0.48u(t - 10) + 0.48u(t - 20),m(0) = 10\)

Step 2: Apply Laplace transform

Applying Laplace transform and use its linearity we get

\(\begin{array}{c}\mathcal{L}\left\{ {m' + \frac{3}{{125}}m} \right\} = \mathcal{L}\{ 0.72 - 0.48u(t - 10) + 0.48u(t - 20)\} \\\mathcal{L}\left\{ {m'} \right\} + \frac{3}{{125}}\mathcal{L}\{ m\}  = 0.72\mathcal{L}\{ 1\}  - 0.48\mathcal{L}\{ u(t - 10)\}  + 0.48\mathcal{L}\{ u(t - 20)\} \end{array}\)\(\begin{array}{*{20}{c}}{\underline{\phantom{xx}}}\\{}\end{array}\)

\(\begin{array}{c}(sM(s) - m(0)) + \frac{3}{{125}}M(s) = 0.72\frac{1}{s} - 0.48\frac{{{e^{ - 10s}}}}{s} + 0.48\frac{{{e^{ - 20s}}}}{s}\\\left( {s + \frac{3}{{125}}} \right)M(s) = \frac{{0.48}}{s} - \frac{{0.48{e^{ - 10s}}}}{s}\frac{{0.48{e^{ - 20s}}}}{s} + 10\\M(s) = \frac{{10s + 0.72}}{{s\left( {s + \frac{3}{{125}}} \right)}} - \frac{{0.48{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}} + \frac{{0.48{e^{ - 20s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}\end{array}\)

 

Using partial fraction

\(\begin{array}{*{20}{c}}{\frac{{10s + 0.72}}{{s\left( {s + \frac{3}{{125}}} \right)}} = 10\left( {\frac{3}{s} - \frac{2}{{s + \frac{3}{{125}}}}} \right)}\\{\frac{{0.48}}{{s\left( {s + \frac{3}{{125}}} \right)}} = 30\left( {\frac{1}{s} - \frac{1}{{s + \frac{3}{{125}}}}} \right)}\end{array}\)

Now we get,

\(\begin{array}{c}m(t) = {\mathcal{L}^{ - 1}}\left\{ {\frac{{10s + 0.72}}{{s\left( {s + \frac{3}{{125}}} \right)}} - \frac{{0.48{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}} + \frac{{0.48{e^{ - 20s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\}\\ = {\mathcal{L}^{ - 1}}\left\{ {\frac{{10s + 0.72}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\} - {\mathcal{L}^{ - 1}}\left\{ {\frac{{0.48{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{{0.48{e^{ - 20s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\}\\ = 10{\mathcal{L}^{ - 1}}\left\{ {\frac{3}{s} - \frac{2}{{s + \frac{3}{{125}}}}} \right\} - 30{\mathcal{L}^{ - 1}}\left\{ {\frac{{{e^{ - 10s}}}}{s} - \frac{{{e^{ - 10s}}}}{{s + \frac{3}{{125}}}}} \right\} + 30{\mathcal{L}^{ - 1}}\left\{ {\frac{{{e^{ - 20s}}}}{s} - \frac{{{e^{ - 20s}}}}{{s + \frac{3}{{125}}}}} \right\}\\ = 10\left( {3 - 2{e^{ - \frac{3}{{125}}t}}} \right) - 30\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10) + 30\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 20)\end{array}\)\(\begin{array}{*{20}{c}}{\underline{\phantom{xx}}}\\{}\end{array}\)

 

To find a concentration (C) we need to divide m(t) by 500L which is the volume of the solution in the tank, so we get 

\(C = 0.06 - 0.04{e^{ - \frac{3}{{125}}t}} - 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10) + 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 20)\)