Q33E
Question
The mixing tank in Figure 7.18 initially holds 500 L of a brine solution with a salt concentration of 0.02 kg/L. For the first 10 min of operation, valve A is open, adding 12 L/min of brine containing a 0.04 kg/L salt concentration. After 10 min, valve B is switched in, adding a 0.06 kg/L concentration at 12 L/min. The exit valve C removes 12 L/min, thereby keeping the volume constant. Find the concentration of salt in the tank as a function of time.
Step-by-Step Solution
VerifiedThe concentration of salt in the tank as a function of time is\(C = 0.04 - 0.02{e^{ - \frac{3}{{125}}t}} + 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10)\)
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\((0 < t < 10)\)and after 10min\((t > 10):\)
\(\begin{array}{c}{\rm{\;ir\;}} = \left\{ {\begin{array}{*{20}{l}}{12 \cdot 0.04,0 < t < 10}\\{12 \cdot 0.06,t > 10}\end{array}} \right.\\ = \left\{ {\begin{array}{*{20}{l}}{0.48,0 < t < 10}\\{0.72,t > 10}\end{array}} \right.\\ = 0.48 + 0.72u(t - 10)\end{array}\)
Let
\(\begin{array}{c}{\rm{\;ir\;}} = f(t)\\ = 0.48 + 0.72u(t - 10)\end{array}\)
Therefore (1) becomes
\(m' + \frac{3}{{125}}m = 0.48 + 0.72u(t - 10)\)
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
\(m' + \frac{3}{{125}}m = 0.48 + 0.72u(t - 10),m(0) = 10\)
Applying Laplace transform and use its linearity we get
\(\begin{array}{c}\mathcal{L}\left\{ {m' + \frac{3}{{125}}m} \right\} = \mathcal{L}\{ 0.48 + 0.72u(t - 10)\} \\\mathcal{L}\left\{ {m'} \right\} + \frac{3}{{125}}\mathcal{L}\{ m\} = 0.48\mathcal{L}\{ 1\} + 0.72\mathcal{L}\{ u(t - 10)\} \end{array}\)
\(\begin{array}{c}(sM(s) - m(0)) + \frac{3}{{125}}M(s) = 0.48\frac{1}{s} + 0.72\frac{{{e^{ - 10s}}}}{s}\\\left( {s + \frac{3}{{125}}} \right)M(s) = \frac{{0.48}}{s} + \frac{{0.72{e^{ - 10s}}}}{s} + 10\\M(s) = \frac{{10s + 0.48}}{{s\left( {s + \frac{3}{{125}}} \right)}} + \frac{{0.72{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}\end{array}\)
Using partial fraction
\(\begin{array}{*{20}{c}}{\frac{{10s + 0.48}}{{s\left( {s + \frac{3}{{125}}} \right)}} = 10\left( {\frac{2}{s} - \frac{1}{{s + \frac{3}{{125}}}}} \right)}\\{\frac{{0.72}}{{s\left( {s + \frac{3}{{125}}} \right)}} = 30\left( {\frac{1}{s} - \frac{1}{{s + \frac{3}{{125}}}}} \right)}\end{array}\)
Now we get,
\(\begin{array}{c}m(t) = {\mathcal{L}^{ - 1}}\left\{ {\frac{{10s + 0.48}}{{s\left( {s + \frac{3}{{125}}} \right)}} + \frac{{0.72{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\}\\ = {\mathcal{L}^{ - 1}}\left\{ {\frac{{10s + 0.48}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\} + {\mathcal{L}^{ - 1}}\left\{ {\frac{{0.72{e^{ - 10s}}}}{{s\left( {s + \frac{3}{{125}}} \right)}}} \right\}\\ = 10{\mathcal{L}^{ - 1}}\left\{ {\frac{2}{s} - \frac{1}{{s + \frac{3}{{125}}}}} \right\} + 30{\mathcal{L}^{ - 1}}\left\{ {\frac{{{e^{ - 10s}}}}{s} - \frac{{{e^{ - 10s}}}}{{s + \frac{3}{{125}}}}} \right\}\\ = 10\left( {2 - {e^{ - \frac{3}{{125}}t}}} \right) + 30\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10)\end{array}\)
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
\(C = 0.04 - 0.02{e^{ - \frac{3}{{125}}t}} + 0.06\left( {1 - {e^{ - \frac{3}{{125}}(t - 10)}}} \right)u(t - 10)\)