Variable Flow Problem

If you are going to have 4 dispensing point with flow meters the defiantly recommend a reservoir with level switches the pump just keep it full
gravity supplies the dispensing points each point has a pulse flow meter and valve
gravity will be constant no matter how many valves are open. The flow through each would be constant
with the pump as valves open and close the pressure and flow on the system will vary
the meters will be far less accurate
 
These are all great ideas! I have looked at the pulse option, I think that would be better with a constant flow rate. The kicker i forgot to mention is that I am eventually going to have 4+ of these puppies connected in parallel. And when one shuts off the others are going to get a major boost in flow rate.


Not quite sure what you mean here; [whoops, never mind, I get it - when you shut off one five gallon sink the others get the total flow.]
 
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If you are going to have 4 dispensing point with flow meters the defiantly recommend a reservoir with level switches the pump just keep it full
gravity supplies the dispensing points each point has a pulse flow meter and valve
gravity will be constant no matter how many valves are open. The flow through each would be constant
with the pump as valves open and close the pressure and flow on the system will vary
the meters will be far less accurate


yes, this definitely a reasonable and feasible way to go: re-create the Cornell hydraulics lab (without the leaky weigh tank). Oh, and if the level was kept more or less constant then valves run by a timers would also work, and no need for four FD-Qs.


Not exactly KISS any more but then neither are four (somethings) in parallel.


You don't even need level control: oversize the pump so the reservoir is always overflowing and providing constant head (just like Beebe Lake;)).
 
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It should not matter that much for the flow meters providing you do not exceed the flow rate for the flowmeter, after all, that's what they are "FLOW METERS".
 
FYI: I gave my email and phone number to the Keyence site to download and read the FD-Q manual, and I got a call from a pleasant fellow named Conrad. He realized he was not going to sell me anything but was nice enough to listen to a description of the problem of this thread. He said the FD-Qs could do the job alone i.e. without a PLC; also the FD-X is a newer product with better accuracy.
 
These are all great ideas! I have looked at the pulse option, I think that would be better with a constant flow rate....


I don't think so; I think the flow meter integrates flow over time and outputs a pulse for every accumulated fixed volume regardless of rate. If that volume is 5gal, then maybe that pulse can switch in the next bucket.


Of course if this is hand sanitizer, the viscosity could be high; would that affect the calibration of the flow meter? Do you know the Reynolds number range for possible flow rates?
 
Your diaphragm pump is a fixed displacement device. Barring leaks or diaphragm ruptures every stroke should give you a fixed volume of fluid. Count the strokes, multiply by the displacement and you have the volume dispensed. You would really only use the flow meter for confirmation that the pump is moving fluid. That is assuming you aren't using all four dispensing points simultaneously.

Just a side note: you indicate you are pumping into an open container. That would qualify as free discharge, so your pump should see minimal pressure. I suspect the 40-60 psi is a pump rating, not an actual pressure reading. If that is not the case, you have some more explaining to do.
 
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I just made some quick numbers. Assuming hand-sanitizer has the following properties inside the 1/2" pipe


  • 0.0127m = D = pipe ID
  • 890kg/m**3 = ρ = fluid density
  • 8000mPa s = 8 kg m**-1 s**-1 = μ = fluid viscosity
Then Reynolds number Re = DVρ/μ = 1.4 V, where V is speed of the fluid in the pipe (mean speed, IIRC). Say we fill a 5gal pail in one minute, the volume flow would be


5gal x 231 in**3 gal**-1 x 0.0254**3 m**3 in**-3 / 60s



= 0.000315 m3/s



The cross-sectional area of the pipe is π D**2 / 4 = π 0.0127**2 / 4 = 0.000127 m**2


So the mean speed is 2.5m/s.


That puts Re at 3.5, ouch. Even if we filled the bucket in a second Re would only go as high as 210, so it's going to be laminar flow. Maybe I missed a few powers of 10?



I'm not sure my suggestion to do this using an analog of the Beebe Lake gravity feed system is going to work unless the driving head is somewhere near the ceiling or maybe the third floor, eh?



Also, I wonder if the Keyence flowmeter calibration assumes laminar flow (velocity profile is a parabola (1-(r/R)**2) across the pipe), IIRC) or turbulent flow (velocity profile approximated by (1-(r/R)**7) across the pipe), the latter being more or less plug flow.


Oh dear, I just calculated the pressure drop per meter of pipe using Darcy-Weisbach: 641220 Pa/m or 93psi/m. So we will use some of that 60psi, depending on the length of the pipe.


19psi for 20cm (0.2m ~ 8") or 9.3psi for 10cm (4"). At 890kg/m**3 that would be over 40' or 20' of head.


That pressure drop is inversely proportional to V, I think (doing a bunch of stuff in my head here), so increasing the 5gal fill time to two minutes reduces the pressure drop to 23psi/m, so only 10' or 5' of head for 8" and 4" drain pipes into the buckets. Also at those short lengths (L/D ~ 16), Darcy-Weisbach may not be valid.


I think Beebe Lake is a pipe dream ;)
 
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Oh dear, I just calculated the pressure drop per meter of pipe using Darcy-Weisbach: 641220 Pa/m or 93psi/m. ...


Whoops, I just went back and found I had divided instead of multiplying by one of the 2.5m/s values. Also, I had been using the nominal ID 0.5", not the actual ID, 0.602". So the actual pressure drop at 5gal/min is around 2 MPa/m or 273psi/m. Holy cow!


Can that be right? If so, it's going to take a bit more than a minute to fill that pail.


Assuming laminar flow, and at this point that's a given, the Darcy-Weisbach (cf. its Wiki) reduces to
f507c812765bbf75e32b4a4d321a26da2110a3c7

where
Dc is circular pipe diameter, 0.602" x 0.0254 m/"
μ
is the dynamic viscosity of the fluid = 8 Pa·s
Q is the volumetric flow rate (m3/s) = (qUs USgal/min x 231 * .0254**3 / 60)
which in our case reduces to
deltaP/L = (1.88E6, Pa-min/m) / (T, min)
deltaP/L = (273, psi-min/m) / (T, min)
where T is the time it takes to fill a 5gal pail in minutes.

That Dc**-4 is really killing you here (with 3/8" pipe its another factor of 2.6 more); can you use a bigger pipe?
 
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if I remember correctly
Those carrousel bottle fillers are filled by a timer from a reservoir the reservoir is filled from a bulk tank
They are accurate enough for retail sales
 
if I remember correctly
Those carrousel bottle fillers are filled by a timer from a reservoir the reservoir is filled from a bulk tank
They are accurate enough for retail sales


deltaP/L = (273, psi-min/m) / (T, min)
To fill a 5gal pail in 5 minutes through an 8" (0.2m) long, 1/2" tubing would require 10.8psi. At 890kg m**-3 density, that reservoir level would have to be over 28ft above wherever the tubing exits the reservoir.

10.9psi = (273 psi-min/m) * 0.2m / 5min

28.3ft = 10.9 lb in**-2 / (2.2046 lb/kg x 890 kg m**-3 x 0.0254**3 m**3 in**-3 x 12 in ft**-1)

Again, this assumes

  • The sanitizer is a Newtonian fluid
  • Darcy-Weisbach applies to a short 8" 1/2"-ID pipe
 

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