First published in internet mathematics forums

Posted: Friday 4th August 2017

## 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,

$r'^2 = x'^2 + y'^2$

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

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.

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)