How do I work out air and water speed from a measured static force? This is a real question - not homework - I am NOT a student...
How fast is the water going through my 140mm diameter propeller when my boat is pulling 12kg against the tree it is tied to? Water is 1013kg/m3 density.  These are real numbers I actually measured.
I'm really struggling with the units here - especially gravity/newtons/kg !
To attempt to get a handle on whether or not proposed online formulas are right.  Another example: I hooked up an air-propeller to a scale. It pushes at 100 grams force from it's 26 cm diameter propeller and I measured 4.8 m/s wind at the fastest place in the outgoing prop wash. Air is 1.22kg/m3 here today.
Every time I find a formula that seems "ball park" right for the air, the water answer is totally ridiculous - so I obviously can't find the right formula...
Help?!?!
 A: Turbulence means that you can't really get a good answer to this question the way you've asked it. There won't be a single speed for the fluid leaving the propeller; there'll be a distribution of speeds which vary in space and time as the propeller moves, some of which variation will average out as you move away.
A "toy model" we can use is to imagine the propeller (with radius $r$) is enclosed in a cylinder with area $A=\pi r^2$ and some length $L$ which is bigger than the length scale of any turbulent eddies.  Then we can pretend all of the fluid enters and exits the tube with the same speed $v$.  The mass in the tube is $m = \rho A L$, and the amount of time it takes to replace the fluid in the tube is $\Delta t = L/v$.
The force exerted by the propeller on the fluid is equal to the rate of change of the fluid's momentum.  The fluid outside the cylinder starts at rest, with no momentum, so we have
\begin{align}
F = \frac{\Delta p}{\Delta t}
&= \frac{mv}{\Delta t}
\\
&= \frac{\rho A L \cdot v}{L / v}
\\
F &= \rho A v^2
\end{align}
If you're measuring forces using scales which are calibrated for determining masses in Earth's gravity, you should use the unit conversion $1\,\text{newton} = 100\,\text{gram}$.  (By using $g=10\rm\,m/s^2$ instead of The Mythical Nine Point Eight, you're introducing a 2% error, which is tiny compared to the effects of turbulence.)  You see your $26\,\rm cm$ air fan producing one newton of thrust, which ought to correspond to an average air velocity of $3.9\rm\,m/s$.  You identified a maximum airflow speed about 25% more than this predicted average, which is ... pretty good agreement, I'd say.  That tells you something interesting about how much variation there is in the velocity field.
In addition to turbulence meaning there isn't just one velocity to measure, there are some other messy things that make this kind of computation harder in a real-world system:

*

*Viscosity means that fluid from some ring outside the area $A$ of the propeller is getting caught up in the flow.  Water is more viscous than air, so the difference between actual area and effective area is bigger for a water propeller than for an air propeller.  An engineering reference book might tell you that, for one propeller design, you should use $A_\text{effective}=1.23\pi r^2$, while for some other more efficient propeller design you should use $A_\text{effective}=1.34\pi r^2$.  Computing those fudge factors is very hard, and people only really believe measurements.


*Compressibility means you can't quite assume the fluid density $\rho$ is constant, which changes the entire argument in a subtle way.  Your air-propeller experiment doesn't seem to be operating in a regime where compressibility matters.  Water is incompressible to a good approximation — unless your propeller introduces bubbles, so that your actual working fluid is a much more compressible air-water mixture.  Engineers also have reference books about this.
