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August 4th, 2017, 04:53 AM   #1
Joined: Aug 2017
From: Aberdeen, Scotland

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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 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,

\[ r'^2 = x'^2 + y'^2 \]


\[ y' = \sqrt{r'^2 - x'^2} \]

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.

\[ r' = t + 1 \]


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

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 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 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 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)
Attached Images
File Type: jpg diffuser_nozzle_profiles.jpg (85.4 KB, 2 views)
File Type: jpg 2x+1y04-0.25_C_n.jpg (85.9 KB, 2 views)

Last edited by skipjack; August 4th, 2017 at 05:41 AM.
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