Mathematics of profiles of water diffusers and nozzles

Discussion in 'Physics & Math' started by Peter Dow, Aug 4, 2017.

  1. Peter Dow Registered Senior Member

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    I derived this system of differential equations this week as I was researching possible profiles for water diffusers & nozzles, used when joining pipes with different bores.

    Mathematics: Solve this system of differential equations.

    \( x' = y^{-2} \)

    \( y' = - \sqrt {(t+1)^2- y^{-4}} \)​

    \(x'\) and \(y'\) are derivatives with respect to \(t\).​

    I have obtained a numerical solution (which was non-trivial because of the numerical instability of the Euler method with this system of differential equations) but I am curious to know "does an analytical solution exist?", which would be more efficient and convenient to use.

    Derivation of the system of differential equations

    Water is accelerated in a nozzle or a pipe of reducing width, which is rotationally symmetrical about the X-axis, with the bore, the inner diameter and the inner radius proportional to a function \(y(x)\) .

    Neglecting viscosity and considering averages for simplicity, the velocity of the water in the X-axis is inversely proportional to the bore's area of cross section and to \(y^2\).

    \(x' = y^{-2}\)

    The average velocity r', of a radial element, a thin slice of water "pie", is composed of the vector addition of the velocity along the X-axis x' and in the radial direction y', which are related in magnitude by Pythagoras,

    \( r'^2 = x'^2 + y'^2 \)

    So

    \( y' = \sqrt{r'^2 - x'^2} \)

    Assume an acceleration \( r' = at + u \), at a time t, with acceleration a and initial velocity u, but for simplicity here, both a and u are assumed to be 1.

    \( r' = t + 1 \)

    Therefore

    \( y' = - \sqrt{ (t+1)^2 - y^{-4}} \)

    choosing the negative root corresponding to an radially inward y' when water accelerates in a nozzle.

    Numerical Solution

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    The numerical instability was managed by writing a computer program which could calculate in t-increments corresponding to the square root of linear increments in \(t^2\).

    \( t_{i+1} = t_i + \Delta t_i \) where \( \Delta t_i = \min(h_1,\sqrt{t_i^2+h_2}-t_i) \) and \(h_1 \), \( h_2 \) are step size constants.

    Approximate solution for \( (x,y,t) = (0,1,0)\)

    With the initial conditions \( t=0, x=0, y=1 \) then \( x' = r' = 1 \) and \( y'=0 \).

    Assuming that then \( y' << x' \) for all \( t>0 \) then approximately

    \( x' = t + 1 \)

    Integrating with respect to t and substituting for t (or simplifying this equation for linear acceleration, \( x'^2 - u^2 = 2ax \) with \(a=1\) and \(u=1\) ) gives

    \( x'^2 - 1 = 2x \)

    Substituting for \( x' = \sqrt{2x+1} \) in the system equation \( x' = y^{-2} \) and rearranging for y gives

    \( y = (2x+1)^{-0.25} \)

    As this graph shows, this is a good approximate solution for these starting conditions.

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    Note on numerical instability
    The graph also shows how numerical stability interrupted the numerical solution part plotted using the free on-line Two Dimensional Differential Equation Solver and Grapher V 1.0. Investigate the numerical instability by selecting the option "System of first order DEs: x' = f(x, y, t), y' = g(x, y, t)" and typing for x', y' -
    x' = y^(-2)
    y' = -1*sqrt((t+1)^2-y^(-4))
    with initial values (x=0, y \(\ge\)1)
     
    Last edited: Aug 4, 2017
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