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Math Help - Mixing Problem

  1. #1
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    Mixing Problem

    A cistern (water tank) contains 200 litres of pure water. Brine (salty water) that contains 0.02 kg/L of salt is added at a rate of 3 L/min. Brine from a second source with 0.04 kg/L of salt is added at a rate of 1 L/min. Assume that the cistern is instantaneously well-mixed. The cistern is drained at a rate
    of 4 L/min. If Q(t) is the amount of salt (in kg) at time t (in min), how much salt is in the cistern after t minutes?
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  2. #2
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    Quote Originally Posted by calculuskid1 View Post
    A cistern (water tank) contains 200 litres of pure water. Brine (salty water) that contains 0.02 kg/L of salt is added at a rate of 3 L/min. Brine from a second source with 0.04 kg/L of salt is added at a rate of 1 L/min. Assume that the cistern is instantaneously well-mixed. The cistern is drained at a rate
    of 4 L/min. If Q(t) is the amount of salt (in kg) at time t (in min), how much salt is in the cistern after t minutes?

    See this thread

    CB
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  3. #3
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    In our problem, the level of water remains the same, as we have 3\frac{\mbox{L}}{\mbox{min}} and 1\frac{\mbox{L}}{\mbox{min}} flowing in and 4\frac{\mbox{L}}{\mbox{min}} flowing out.

    If we call the salt level S(t), then we have

    S(0)=0\,\mbox{kg}.

    To find \frac{dS}{dt}, we subtract the rate of salt flowing out from the rate flowing in:

    \frac{dS}{dt}=\frac{dS_{\mbox{\scriptsize{in}}}}{d  t}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}.

    The rate flowing in is

    <br />
\begin{aligned}<br />
\frac{dS_{\scriptsize{\mbox{in}}}}{dt}&=0.02\frac{  \mbox{kg}}{\mbox{L}}\cdot 3\frac{\mbox{L}}{\mbox{min}}+0.04\frac{\mbox{kg}}{  \mbox{L}}\cdot 1\frac{\mbox{L}}{\mbox{min}} \\<br />
&=0.06\frac{\mbox{kg}}{\mbox{min}}+0.04\frac{\mbox  {kg}}{\mbox{min}}\\<br />
&=0.1\frac{\mbox{kg}}{\mbox{min}}.<br />
\end{aligned}<br />

    Since the fluid is well-mixed, the rate flowing out will be in proportion to 4\,\mbox{L} as the current salt level is to the entire resevoir. Therefore,

    <br />
\begin{aligned}<br />
\frac{dS_{\scriptsize{\mbox{out}}}}{dt}&=\frac{S}{  200\,\mbox{L}}\cdot 4\frac{\mbox{L}}{\mbox{min}} \\<br />
&=\frac{S}{50}\,\mbox{min}^{-1}.\\<br />
\end{aligned}<br />

    For \frac{dS}{dt}, we obtain

    <br />
\begin{aligned}<br />
\frac{dS}{dt}&=\frac{dS_{\mbox{\scriptsize{in}}}}{  dt}-\frac{dS_{\mbox{\scriptsize{out}}}}{dt}\\<br />
&=0.1\frac{\mbox{kg}}{\mbox{min}}-\frac{S}{50}\,\mbox{min}^{-1}.<br />
\end{aligned}<br />

    Now, all that remains is to solve the differential equation using the initial conditions given.
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  4. #4
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    from the above, we get the equation : ds/dt = 0.1 - s/50

    thus, ds/dt + 1/50 s = 0.1 ==> this is linear first degree equation .....(1)

    find the integration factor, miu(t) = exp(integration of 1/50 dt)
    = exp(1/50 t)

    miu(t) x (1) : exp(1/50 t) x (ds/dt + 1/50 s) = 0.1 exp(1/50 t)

    d/dt[exp(1/50 t).s] = 0.1 exp(1/50 t)

    d[exp(1/50 t).s] = integration of [0.1 exp(1/50 t)] dt

    = 0.1 integration of [exp(1/50 t)] dt

    exp(1/50 t).s = 0.1/(1/50) exp(1/50 t) + c

    thus s(t) = 5 + c exp-(1/50 t) ...(2)


    applying initial condition : s(0) = 0 kg in (2) :

    0 = 5 + c exp-(1/50 (0))
    c= -5 in (2)


    thus, the solution : s(t) = 5 - 5 exp-(1/50 t)
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