Recently I learned a new word for what I'm looking for -- rotameter -- a device that uses a weight in a conical bore to measure fluid flow.

How can I make an uncalibrated flow meter for water flows of a few liters per hour

I've found many for gasses, but not many for liquids, and none for the range I'm looking for.

To first order approximation do I need to do more than compare the viscosity of air to that of water to use an air calibrated one for water?

The viscosities differ by about a factor of 50. Would a 5 liter/minute gas rotometer be a 6 liter per hour water rotameter.

Caveats: In a gas meter the bead may be plastic, which has no buoyancy effect in air, but a substantial one in water. Indeed the plastic might float, and I'd have to use the thing upside down. Plastic with a specific gravity of .9 would make the device 10 times more sensitive.

Depending on what's dissolved in the water, the viscosity can change somewhat. I will be using more or less pure tapwater with about 600 ppm of assorted hardwater crud.

  • $\begingroup$ would you object to replacing the possibly-plastic bead with a metal one to improve the response in water? $\endgroup$
    – fred_dot_u
    May 1, 2018 at 21:23
  • $\begingroup$ How about something like this? omega.com/pptst/FL2000.html $\endgroup$
    – Mark
    May 2, 2018 at 12:30
  • $\begingroup$ @Mark Exactly what I want at 10 times the price. Thanks for the tip, though I'll keep hunting. I don't need accuracy at all. Just precision. $\endgroup$ May 3, 2018 at 3:32

1 Answer 1


There are water ones that exist, what flow range are you attempting to use? A very low flow rate (in the Liter / Hour range) would be difficult to size, as the uncertainty would vary substantially for the thing to be readable. This would be the main reason why they would not be offered in low flow rates - the uncertainty. What good is a measuring device if you can't read it?

The uncertainty part works as this: In order to have consistent drag coefficient of a spherical float, the Reynolds number needs to be pretty near constant, and in the range of 10,000. Otherwise the drag coefficient isn't constant, and you have high uncertainty:

enter image description here

To have a Reynolds number around 10,000, at say 6L/hr with water we'd need a cross sectional area of the ball to be around 0.029 mm^2. Then there are the tolerances that would be associated with ensuring the tube linearly expands, and the ball stays within the proper cross sectional area. All of this then would vary drastically if the tube expands too drastically as that would cause compressible flow effects to emerge and the temperature of the water would change.

In short, use a positive displacement meter or variable area meter for low flowrates. There is a reason why there are different flowmeters for different applications.

  • $\begingroup$ I'm confused. Variable area flow meter and rotameter seem to be used synonymously. $\endgroup$ May 3, 2018 at 3:26
  • $\begingroup$ A rotameter is a kind of variable area flow meter, but there are other kinds of variable area flow meters. In particular, by taking a paddle that is pinned off-center in a circular housing, the open area for the flow will vary as the plate rotates. A torsional spring acting on the plate resisting the rotation about the pin will allow for a balance of forces, forming a device who's rotation will be related to the flow. Since a plate's drag coefficient is significantly more linear across different Reynolds numbers, and springs rates are more controllable, this is better for low flow rates. $\endgroup$
    – Mark
    May 3, 2018 at 17:18
  • $\begingroup$ Overall, an internet search for "flowmeter pros and cons" of a good source will list different flow meters and their different applications. One such source is flowmeters.co.uk/… which, while in Imperial units, still rates the different applications. They list a thermal flowmeter as the best for low flow rates, but it has modest cost and isn't very linear. On the other hand, they list ultrasonic as handling most flowrates at reasonable cost. $\endgroup$
    – Mark
    May 3, 2018 at 17:34
  • $\begingroup$ Good article -- thanks. I don't understand where you got the area of the ball down to .029 mm. Nor the requirement for a Re of 10000. My limited understanding of Re is that you use a 'chararistic lengh' usually an average length such as the waterline length of a ship, or the chord lengh of an airplane wing. 1 liter per hour is about 5 cm/s in a 5 mm diameter tube. If you have a conical bore that goes from 5 mm to 10 mm, then the annular area around a 5 mm diameter 1 cm long tapered slider goes from 0 to 50 mm2 while the flow velocity drops from 5 cm/s to 1 cm/s $\endgroup$ May 4, 2018 at 20:22
  • $\begingroup$ The characteristic length for a small object (such as a sphere) is sqrt(A). If you take velocity as volumetric flow rate / area then the Re = Q/(vsqrt(A)), with Q as flow rate and v as kinematic viscosity of water. This can be solved for A in terms of the Reynolds number, A = (Q/(vRe))^2. $\endgroup$
    – Mark
    May 5, 2018 at 0:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.