Power Washer velocity reduction 200 mm away from tip Help with a little pressure washer test I need to perform.
I have a nozzle that has a 1mm diameter tip.  
15 degree spread in the water stream
Q flow rate 1 m^3/min
Pressure 200 bar
I need to figure out the force on the area that it is hitting  (assume) perpendicular to the water stream 200mm away.
I am having trouble with calculating the water velocity drop after the water leaves the nozzle tip.
Any help would be appreciated
.  
 A: Assuming no viscosity, we compute the velocity at the exit of the nozzle from the pressure and the density of the liquid (using Bernoulli's equation):
$$\frac12 \rho v^2 = \Delta p$$
For water at 200 bar, that gives a velocity of approximately
$$v = \sqrt{\frac{2\Delta p}{\rho}}=200 m/s$$
And the flow rate would be $Q = v\cdot A = 200 \cdot \frac{\pi}{4} \cdot 10^{-6}\approx 12.5 \rm{~\ell/s}$
It is quite hard to compute the drag on the water as it exits - experience says there is significant air entrained as the water jet expands, and this will transfer momentum to the air. We can easily estimate an upper limit on the force - because that is simply the momentum per unit time that needs to be stopped, and is
$$F = Q\cdot v\cdot \rho=v^2\cdot A\cdot \rho=2A\Delta p$$
As the jet expands, less liquid will hit a unit area - but it's still the same mass flow. So the pressure (force per unit area) decreases, but the total force (pressure times area) will change much less.
But without much more detail of the setup it will be VERY hard to give you an accurate number. Experimentation is your friend.
But IF you want to try to estimate the effect of drag on the water, I suppose we could do the following (no guarantees that I didn't make a mistake in the math... check this!)
Assume the water forms spherical drops of the diameter of the nozzle. Assume again that surface tension will keep them intact. I know this to be a poor assumption, but I'm looking for magnitudes here.
The drag on a sphere is given by
$$F_D = \frac12 \rho v^2 A C_D$$
Where we can use $C_C=0.5$ for spheres in a turbulent regime with $\mathrm{Re}\approx 10^4-10^6$.  Plugging in numbers $\rho = 1.2~\rm{kg/m^3}$, $A=\frac{\pi}{4} d^2 = 7.8e{-7}~\rm{m^2}$, $v=200~\rm{m/s}$, we find the initial force
$$F=9.4\cdot 10^{-3}~\rm{N}$$
And the Reynolds number is about 180,000 so the assumption about $C_D$ is valid.
Then we can write the equation of motion for the drop:
$$m\ddot{x} = -c\dot{x}^2$$
The solution to this equation is given here. Writing $\frac{m}{cv_0}=\tau$, the expression for velocity with time becomes
$$v(t) = \frac{v_0}{1+t~/~\tau}\\
x(t) = v_0 \tau \log(1+t~/~\tau)$$
and we get $\tau=11~ms$. With a velocity of 200 m/s that means the velocity of the air will follow approximately this curve:

As you can see, for a 1 mm drop there is some deceleration; assuming the drop atomizes down to 1/10th of the initial size, and it slows down considerably more (basically, $\tau$ scales with radius).
The code I used to generate this plot:
rho_air=1.2;
Cd = 0.5;
rho = 1e3; % water
A=pi*r^2;
V=4/3*pi*r^3;
m = V*rho;
v0=200;
C = 0.5*rho_air*A*Cd;
tau = m/(C*v0);
d=0.2; 
t=linspace(0,0.002,1000);

figure
for r=[0.5e-3, 0.05e-3]
    xt = 1000*v0*tau.*log(1+t/tau);
    vt = v0./(1+t/tau);
    ti = find(xt<1000*d);
    plot(xt(ti), vt(ti))
    hold on
end
title 'velocity vs position'
xlabel 'position (mm)'
ylabel 'velocity (m/s)'
legend('1 mm','0.1 mm')

