Water flow rate vs pressure for a very small nozzle (0.1mm)

[user37]

TL;DR

This problem is too big for my brain:

What pump pressure (psi) do I need to push one quart of water per hour through a nozzle with an orifice of .1 mm? Point one. Not 1mm. One quart in an hour ( like a humidifier) assume friction losses to be negligible.

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[berkeman]

Welcome to PF.

Is this a question for schoolwork? Or is it a question for a personal project of yours? What links have you found so far in your Google searching?

 

[user37]

Personal project
I saw something about Bernoulli equation but don’t know if it applies. At 47 I’m a few years out of school

 

[jack action]

This is essentially the same problem as evaluating the flow through a hole at the bottom of a bucket. The water pressure at the bottom of the bucket is equivalent to the pump pressure you are looking for.

All you need is the Bernoulli equation. Can you figure it out with these two problems:

​

 

[russ_watters]

Jack is right that as stated you can just plug and chug with Bernoulli's equation, but if you want this to come anywhere close to real world you can't assume it is lossless.

 

[user37]

I can try to use the suggested equations. I’m not concerned about minor losses as long as it doesn’t exceed 15% of my target.

 

[gmax137]

If I did the arithmetic right, the velocity in the 0.1 mm nozzle is ~33.5 meter/second. I don't think losses can reasonably be neglected.

 

[erobz]

Unfortunately, I highly doubt we can apply Bernoulli's (or anything remotely close to it) and get a reasonable answer. I believe this nozzle belongs to the domain of Microfluidics.

https://books.rsc.org/books/edited-volume/1076/chapter/901085/Governing-Equations-in-Microfluidics

Which is unfortunately advanced material. It's possible there is an equation in there that could describe what you are after Section 1.5.3. Poiseuille Flow, Eq. 1.63a.

 

[user37]

Well there’s always the brute force approach. I can buy a 100psi 12v pump and a variety of nozzles, give it 12v power, and see how long it takes to empty a quart. I know that the smaller the orifice, the stronger the pump required for a fine mist. The point of all this was to avoid buying 4 pumps and 6 nozzles and spending considerable time on trial and error but that may be all I’ve got.

 

[erobz]

Well there’s always the brute force approach. I can buy a 100psi 12v pump and a variety of nozzles, give it 12v power, and see how long it takes to empty a quart. I know that the smaller the orifice, the stronger the pump required for a fine mist. The point of all this was to avoid buying 4 pumps and 6 nozzles and spending considerable time on trial and error but that may be all I’ve got.

Like I said eq 1.63 a in the linked paper seems like its what you are after, or equivalently:

$$ \Delta P = \frac{8 \mu L}{\pi r^4}Q $$

$\mu$ is the viscosity of the fluid
$L$ is the length of the channel
$r$ is the channel radius
$Q$ is the volumetric flowrate

I would give that a shot first.

just need to look up the viscosity, and tell us the length.

 

[berkeman]

Well there’s always the brute force approach. I can buy a 100psi 12v pump and a variety of nozzles, give it 12v power, and see how long it takes to empty a quart. I know that the smaller the orifice, the stronger the pump required for a fine mist. The point of all this was to avoid buying 4 pumps and 6 nozzles and spending considerable time on trial and error but that may be all I’ve got.

No need to buy much, do you ride bicycles? Do you already have a bicycle floor pump? If not, do you have any neighbors or friends who have one? You could probably prototype this pretty easily with a 2 quart container that you fit a bicycle tire valve to and try a few different drain hole sizes...

https://sc04.alicdn.com/kf/H3468d0b884a5494bb9fc09df355e25316.jpg

 

[russ_watters]

Well there’s always the brute force approach. I can buy a 100psi 12v pump and a variety of nozzles, give it 12v power, and see how long it takes to empty a quart. I know that the smaller the orifice, the stronger the pump required for a fine mist. The point of all this was to avoid buying 4 pumps and 6 nozzles and spending considerable time on trial and error but that may be all I’ve got.

The data sheets for the nozzles should give you the performance so you don't need to guess, test or calculate.

 

[erobz]

The data sheets for the nozzles should give you the performance so you don't need to guess, test or calculate.

Thats an even better approach!

 

[user37]

Lots of ideas thank you. The math route isn’t so good for me. Once I got to pre-algebra I was lost. And my dad was a math teacher d’oh! I have to find another way.

 

[berkeman]

I have to find another way.

Sweat equity. Pump it up!

 

[Dullard]

This will get you in the ballpark:

your orifice has a Cv of approx 0.00035. That's an approximate empirical value which eliminates all of the precise math based on imprecise assumptions. Measure with a micrometer, mark with chalk, cut with an axe...

GPM = Cv / SQRT(1/dP)

where dP is the pressure difference across the orifice (supply pressure in your case)

I get something like 140 PSIG

 

[Baluncore]

Record the water temperature at the start of the experiment.

The viscosity of water will vary, typically by ±25%, depending on the temperature of the water.

 

[hutchphd]

Here's an online calculator: https://www.tlv.com/global/US/calculator/water-flow-rate-through-orifice.html
I consulted and the oracle said 100psi will get you 0.7 liters/hr

 

[user37]

Thank you everyone. I didn’t want to come here and just use people to get an answer in a selfish transient kind of way but hopefully to learn something. 100 psi keeps the parts affordable and .7 liter/hr would be ideal.

 

[hutchphd]

Thank you everyone. I didn’t want to come here and just use people to get an answer in a selfish transient kind of way but hopefully to learn something. 100 psi keeps the parts affordable and .7 liter/hr would be ideal.

Just to be crystal clear: I do not know nor can I vouch for the accuracy of the online orifice oracle I mentioned . The result seems within the range dictated by the "sniff test" and other estimates presented......so trust but verify (Why am I quoting Ronny Reagun??)

 

[Chestermiller]

Like I said eq 1.63 a in the linked paper seems like its what you are after, or equivalently:

$$ \Delta P = \frac{8 \mu L}{\pi r^4}Q $$

$\mu$ is the viscosity of the fluid
$L$ is the length of the channel
$r$ is the channel radius
$Q$ is the volumetric flowrate

I would give that a shot first.

just need to look up the viscosity, and tell us the length.

If you calculate the Reynolds number, you will find that, for a 0.1 mm tube, the Reynolds number is 3300, indicating turbulent flow, and a higher pressure drop than this Hagen-Poiseulle equation.

 

[Chestermiller]

What is the cross sectional schematic diagram of the nozzle? If the final opening is D=0.1 mm over a length L, from fluid flow correlation of friction factor vs Reynolds number, I get $$\Delta P=\frac{4L}{D}(0.81\ psi)$$

 

[erobz]

If you calculate the Reynolds number, you will find that, for a 0.1 mm tube, the Reynolds number is 3300, indicating turbulent flow, and a higher pressure drop than this Hagen-Poiseulle equation.

Yeah, did end up reading in the paper that is valid under low Reynolds number. Since the OP didn't seem interested in that route, I didn't bother withdrawing it.

 

[jrmichler]

What pump pressure (psi) do I need to push one quart of water per hour through a nozzle with an orifice of .1 mm? Point one. Not 1mm.

Another way to get an answer is to look at similar nozzles with published pressures and flow rates. Spraying Systems Company lists a nozzle with zero degree spray angle that has 0.008" (0.2 mm) equivalent orifice diameter in their Flat Spray Nozzles catalog (https://www.spray.com/-/media/dam/industrial/usa/sales-material/catalog/cat75hyd_us_flat-spray_c.pdf). Since that nozzle is twice the diameter in the OP, the flow rate will be four times higher. Four times 1 quart per hour equals 1 gallon per hour which equals 0.017 gallon per minute. Their Size 0009 model QVVA nozzle is rated 0.014 GPM at 100 PSI and 0.020 GPM at 175 PSI. This agrees well with the calculations in Posts #16 and #18.

 

[jack action]

Using Bernoulli as in post #4, you get:
$$\Delta P = \frac{1}{2} \rho v^2 = \frac{1}{2} \rho \left(\frac{4Q}{\pi D^2}\right)^2$$
Where:

• $Q$ is the desired flow rate - 1 quart per hour (2.63x10^-7 m³/s)
• $D$ is the orifice diameter - 0.1 mm (0.0001 m)
• $\rho$ is the density of water (100 kg/m^3)

This gives a pressure $\Delta P$ of 560663 Pa or 81 psi - assuming no losses. This is most likely the value you would have if your orifice is shaped like a venturi nozzle.

If you have a simple sharp orifice, you can assume a coefficient of contraction of 0.611 ($= \frac{\text{actual area}}{\text{orifice area}}$) which would increase the necessary pressure to 218 psi.

This is pretty much the range you should expect between the best and worst scenarios.

Reference:

• https://en.wikipedia.org/wiki/Vena_contracta
• https://www.engineeringtoolbox.com/orifice-nozzle-venturi-d_590.html