# Mathematics of profiles of water diffusers and nozzles

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

1. ### Peter DowRegistered 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

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.

with initial values (x=0, y $\ge$1)