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I Derivation of non-dimensional Navier Stoke equation

  1. Jun 17, 2017 #1
    Take the first three terms of Navier Stoke equation:
    $$\rho \cdot \left ( v_{x}\cdot \frac{\partial \vec{v}}{\partial x} + v_{y}\cdot \frac{\partial \vec{v}}{\partial y} + v_{z}\cdot \frac{\partial \vec{v}}{\partial z}\right )$$

    Define the length ##v## of the velocity vector field:
    $$\vec{v}=v\cdot \vec{v}^{0}$$
    where ##\vec{v}^{0}## is the unit vector field with same direction as ##\vec{v}##. Thus it is:$$v_{x}=v\cdot v_{x}^{0}$$$$v_{y}=v\cdot v_{y}^{0}$$$$v_{z}=v\cdot v_{z}^{0}$$

    Now both terms ##v\cdot \vec{v}^{0}## can change with ##x## coordinate, so their derivation is:
    $$\frac{\partial (v\cdot \vec{v}^{0})}{\partial x}=\vec{v}^{0}\cdot \frac{\partial v}{\partial x}+v\cdot \frac{\partial \vec{v}^{0}}{\partial x}$$

    When I substitute these derivations into the NS equation (1. equation), I get something like this:
    $$\rho \cdot\vec{v}^{0}\cdot v\cdot \left ( \vec{v}^{0}\cdot \left ( \frac{\partial v}{\partial x} , \frac{\partial v}{\partial y} , \frac{\partial v}{\partial z}\right ) \right )+\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )$$

    But according to the literature it should be just the second term:
    $$\rho \cdot v^{2}\left ( \vec{v}^{0}\cdot \left ( \frac{\partial \vec{v}^{0}}{\partial x} , \frac{\partial \vec{v}^{0}}{\partial y} , \frac{\partial \vec{v}^{0}}{\partial z}\right ) \right )=\rho \cdot v^{2}\cdot \left ( \vec{v}^{0}\cdot \left ( \mathrm{div} \vec{v}^{0} \right ) \right )$$

    What is wrong about this derivation?
     
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  3. Jun 17, 2017 #2

    Orodruin

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    I don't think you are understanding it right. If the point is to get rid of the dimensions then ##v## should be some constant reference speed, not the speed of the fluid at each point.

    Edit: I also suggest you stop using ##\cdot## when you don't mean scalar product, it makes your post difficult to read.
     
  4. Jun 17, 2017 #3
    So what is the definition of the unit velocity vector?

    Edit: If ##v## is the average constant speed, than the length of ##\vec{v}^0## is not generally unity.
     
    Last edited: Jun 17, 2017
  5. Jun 17, 2017 #4

    Orodruin

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    Indeed. It is just a dimensionless velocity. Note that ##v## can be any fixed non-zero speed. You would typically fix it to some speed that appears in your problem, such as the speed of the flow at infinity.
     
  6. Jun 21, 2017 #5
    The non-dimensional Navier Stokes equation is:
    $$\frac{L\cdot f}{v}\frac{\partial \vec{v}^*}{\partial \vec{\tau}^*}+ v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}=\frac{g\cdot L}{v^2}\vec{g}^*+\frac{p^0}{\rho \cdot v^2} \bigtriangledown^*p^*+\frac{\eta }{L\cdot v\cdot \rho } \bigtriangledown^{*2}\vec{v}^*$$
    I understand that by setting physical parameters (##v,\rho,L...##) I can achieve in my scaled flow e.g. identical ratio of friction force to net force as in the original flow.

    What I don't understand is how can I determine which terms in this equation are negligible. I know the physical dimensional parameters (##v,\rho,L...##) but I generally don't know the non-dimensional vectors (##v_x^*\frac{\partial \vec{v}^*}{\partial x^*}+ v_y^*\frac{\partial \vec{v}^*}{\partial y^*}+v_z^*\frac{\partial \vec{v}^*}{\partial z^*}##, ##\bigtriangledown^{*2}\vec{v}^*##...).
    So how can I know which values are standing behind every term in the equation?
     
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