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August 4th, 2017, 05:53 AM  #1 
Newbie Joined: Aug 2017 From: Aberdeen, Scotland Posts: 1 Thanks: 0  Mathematics of profiles of water diffusers and nozzles
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 nontrivial 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 Xaxis, 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 Xaxis 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 Xaxis \(x'\) 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 a radially inward \(y'\) when water accelerates in a nozzle. Numerical Solution See (attached) graph "Diffusers and Nozzles Profiles". The numerical instability was managed by writing a computer program which could calculate in tincrements 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 the (attached) graph shows, this is a good approximate solution for these starting conditions. Note on numerical instability The graph also shows how numerical stability interrupted the numerical solution part plotted using the free online 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)^2y^(4)) with initial values (x=0, y \(\ge\)1) Last edited by skipjack; August 4th, 2017 at 06:41 AM. 

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