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Linearising a Mixing Tank

  1. Sep 9, 2010 #1

    danago

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    I have a problem to solve where i am required to linearise a nonlinear system. Basically, the system involves a mixing tank of height H(t) and cross section area A with two input streams with flow rates Q1(t) and Q2(t) with concentrations C1 and C2 of some species (concentrations are constant). There is one outlet stream with flow rate Q(t) and concentration C(t). We are assuming that the tank is always perfectly mixed and that the outlet flow rate is proportional to [tex]\sqrt{H(t)}[/tex].

    So what i need to do is express the concentration of the outlet stream as some nominal steady-state value plus an additional term for deviations from this nominal value

    i.e. C(t) = C0 + c(t)

    Where c(t) is a linear approximation.


    My first thought is to write a material balance on the species in the fluid:

    i.e. [tex]\frac{d}{{dt}}\left( {CAH(t)} \right) = {C_1}{Q_1} + {C_2}{Q_2} - CQ[/tex]

    It wouldn't be too hard to solve this DE for C(t), but im not sure where to go from there. I guess i could then differentiate C(t) with respect to each variable to linearise it, but this seems a little tedious so im not sure if there is something else i should be doing?

    Any input is greatly appreciated.

    Thanks,
    Dan.
     
  2. jcsd
  3. Sep 10, 2010 #2

    CEL

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    Expand your nonlinear equation in a Taylor series and take the first two terms.
     
  4. Sep 11, 2010 #3

    danago

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    Are you referring to the analytical solution of the differential equation? If i solve the DE (i used the integrating factor method), i get:

    [tex]C(t) = \frac{{\int {({C_1}{Q_1} + {C_2}{Q_2}){e^{\int {\frac{k}{{A\sqrt H }}dt} }}dt + \alpha } }}{{AH{e^{\int {\frac{k}{{A\sqrt H }}dt} }}}}[/tex]

    where alpha is a constant of integration.

    Im not sure if this is correct, and even if it is, it seems a little difficult to linearise, especially since i dont know the exact form of the flow rate functions Q1 and Q2

    Thanks for the reply by the way :)
     
  5. Sep 11, 2010 #4

    CEL

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    Do you need an analytical solution, or a numerical solution will do?

    If you can use a numerical solution, you can start from

    [tex]\frac{d}{dt}C(t) = f(C, t)[/tex]

    and

    [tex]C(t_0) = C_0[/tex]

    and calculate

    [tex]C(T) = C_0 + \frac{d}{dt}C(t_0) T[/tex]

    then calculate [tex]C(2T) [/tex] from [tex]C(T) [/tex] and so on.

    If you need an analytical solution, you can make

    [tex]f(C, t) = f(C_0, t_0) + \frac{df}{dC}(C_0, t_0) \right( C - C_0 \left )[/tex]
    and integrate.
     
  6. Sep 11, 2010 #5

    danago

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    I think i understand what you are saying; using the product rule of differentiation with the species material balance gives me:

    [tex]C\frac{{d(AH)}}{{dt}} + AH\frac{{dC}}{{dt}} = {C_1}{Q_1} + {C_2}{Q_2} - CQ[/tex]

    I can then express the derivative of AH as a sum of flow rates into and out of the tank:

    [tex]C({Q_1} + {Q_2} - Q) + AH\frac{{dC}}{{dt}} = {C_1}{Q_1} + {C_2}{Q_2} - CQ[/tex]

    Rearranging this and substituting the relevant values for t=0 (denoted by a zero in the subscript) gives:

    [tex]{\left. {\frac{{dC}}{{dt}}} \right|_{t = 0}} = \frac{{{C_{1}}{Q_{10}} + {C_{2}}{Q_{20}} - {C_0}({Q_{10}} + {Q_{20}})}}{{A{H_0}}}[/tex]

    I can then use this to give me the linearised form:

    [tex]C(t) = {C_0} + {\left. {\frac{{dC}}{{dt}}} \right|_{t = 0}}t[/tex]

    Have i taken the right approach with this?
     
  7. Sep 11, 2010 #6

    CEL

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    Yes, it is correct.
     
  8. Sep 11, 2010 #7

    danago

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    Alright, thanks very much :smile: Your help is much appreciated!
     
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