# Model pressure loss vs time for air in a cylinder w/ hole

• bad throwaway name
In summary: I found an example product. A small 3 gal. tank/compressor supports .6CFM @ 90PSI and 1CFM at 40PSI...I think, (but do not KNOW for sure), that this means the tank can release 90PSI air for .6 of a minute and 40PSI air for a full minute without a compressor...In summary, according to the product, a small 3 gal. tank/compressor can support .6CFM @ 90PSI and 1CFM at 40PSI, meaning the tank can release 90PSI air for .6 of a minute and 40PSI air for a full minute without a compressor.
bad throwaway name
Related to an engineering project I will be working on.

The situation is that essentially I will have a closed cylinder (5.5 in x 26 in) with a small 15mm hole on the side (not on a circular face), which allows the air inside to be at equilibrium pressure/temperature with the outside. Then a substance in the cylinder undergoes a chemical reaction to rapidly expand into a gas, pressurizing it so that the pressure difference is about 1 atmosphere. I am interested in looking at the pressure over time inside this cylinder.

The method I have in mind is using a calculation similar to the method described in http://www.efunda.com/formulae/fluids/calc_orifice_flowmeter.cfm (which is to describe volume flowrate as ##Q = C_f A_{orifice} \sqrt{\frac{2\Delta P}{\rho}}##, then simply describe mass flowrate as ##\rho Q##) to model the mass loss for some given time step, then recalculate the internal pressure based on PV = nRT with a reduced mass. However, I just want to make sure of a couple things. As I expect pressure and mass inside the volume to change over time, obviously I can't use a fixed density- is it sufficient to simply divide mass by the volume of the cylinder and assume that is the density of the fluid flowing through the orifice?

Additionally, I am unsure of how to approach finding the ##\beta=\frac{D_0}{D_{inlet}}## value they define near the bottom, as the hole in this case is not on the a circular face of the cylinder.

In general, a lot of sources on Bernoulli's principle I find are meant specifically for flow along a pipeline, which is not the situation I have here

Last edited:
bad throwaway name said:
I've edited my post. No effusion, the initial pressure difference is about 1 atmosphere across a 15 mm hole, in a 26 inch tall/5.5 inch wide cylinder.
So, what do you get?

Chestermiller said:
So, what do you get?
Nothing yet, I'm not sure how to proceed. I want to see if my process is sound, and additionally if this equation I found can be applied to situations other than flow through a pipe.

bad throwaway name said:
Nothing yet, I'm not sure how to proceed. I want to see if my process is sound, and additionally if this equation I found can be applied to situations other than flow through a pipe.
Who said it applies to flow through a pipe?

Chestermiller said:
Who said it applies to flow through a pipe?
A lot of the stuff on the page is based on comparisons of area, which don't apply in my case (the hole is not on a circular face)

bad throwaway name said:
A lot of the stuff on the page is based on comparisons of area, which don't apply in my case (the hole is not on a circular face)
In your application, you are going from a large region where the flow velocity is very low (negligible), and is exiting an orifice into the atmosphere, where the fluid flow velocity is high. So the equation does apply to your situation.

I have a similar inquiry, did a search, and came to this posting. I honestly never have heard the term effusion and this seems to, at least in this example, imply a threshold between accidental, negligible release of air/fluid from the tank out versus purposeful [measurable] release...which doesn't make sense. (A formula is a formula, even if a variable is super-duper tiny, ey?) So here's my [similar] question:

I have a small hose (a "fish tank" vinyl hose 1/4" ID) that I will use to release air from a tank at a constant 4psi (a trickle, let's say, like the above example). There's a lot of ways to ask the question, but I am shopping for a tank/compressor that will, with a scientific certainty, support this release...constantly. What formula can I use to know, based obviously [somehow] on the size of the tank and it's max PSI (or the PSI I have in it that I'd purposely, for safety sake, never exceed about 90% of its max-PSI) to know, "How long in seconds will this tank/compressor support the constant release of 4 PSI air?"

I found an example product. A small 3 gal. tank/compressor supports .6CFM @ 90PSI and 1CFM at 40PSI...I think, (but do not KNOW for sure), that this means the tank can release 90PSI air for .6 of a minute and 40PSI air for a full minute without a compressor cut-in.

Aaron Luckette said:
3 gal. tank/compressor supports .6CFM @ 90PSI and 1CFM at 40PSI

This is the rating of the compressor. It has nothing to do with the size of the tank. Air compressors are rated in SCFM (Standard Cubic Feet per Minute).

And "4 psi" says nothing about the flow rate. If you do not have a pressure regulator, the flow rate will vary with tank pressure. If a regulator controls the pressure to, say, 4 PSI, then the flow will be controlled by the total restriction downstream from the regulator. You can measure the flow rate by inverting glass jar over the air outlet and timing how long it takes to displace the water in the jar.

If you only need air at 4 PSI, a 90 PSI compressor will waste a lot of energy and make noise. Those small compressors are designed for light intermittent use. They make aquarium compressors designed to deliver air at low pressure and run continuously. Those are far more efficient, last longer, and run more quietly.

I actually bought a small fish tank compressor for testing. Although I've not found any information on its pressure specs, it doesn't deliver anywhere near the pressure I need. What I want - literally - is air coming out of a hose that is strong enough to move the reeds on a harmonica. I found this article (https://saintsreport.com/threads/how-hard-can-a-human-blow.67835/) and this person notes a 2.4PSI as hard as he could expel air...so I'm sure even as I say above 4 PSI is WAY too much...but I want to know...for sure...that what I buy will do what I need...but to my surprise, there doesn't seem to be a formula. My thought is def. along the lines of a \$30 tank from Harbor Freight (kinda thing).

Please note that there is a very good chance this flow is compressible. Your situation is also not an orifice plate, so I am not sure how that link you provided would even apply. At any rate, you will almost certainly need to use compressible flow relations for a large portion of the problem. Depending on the gases and pressure involved, it may or may not be initially choked at the exit as well.

Instead of guessing the necessary pressure, run a test. Play a harmonica by blowing through a piece of tubing. Put a Tee fitting in the tubing and connect to a water manometer. Measure the pressure needed to play the harmonica in inches (or cm) of water. No need to convert to PSI. Then connect the fish tank compressor to the water manometer (without the harmonica). If the fish tank compressor will produce enough pressure without the harmonica, but the pressure drops too low when the harmonica is connected, then the fish tank compressor has enough pressure, but not enough flow. In that case, get a larger fish tank compressor or add a second one in parallel.

Water manometers are an excellent way to measure pressure. Cheap, easy to use, never need calibration, and 100% accurate.

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boneh3ad said:
Please note that there is a very good chance this flow is compressible. Your situation is also not an orifice plate, so I am not sure how that link you provided would even apply. At any rate, you will almost certainly need to use compressible flow relations for a large portion of the problem. Depending on the gases and pressure involved, it may or may not be initially choked at the exit as well.

This started out simple. I simply want to attach an air hose to a surface (glue it down to a table) that is blowing out air into a harmonica. As I move the harmonica back and forth, it will make different pitches, like normal. My first Q was, I wonder what pressure I will need, so I that up (basically) and found that link. I don't see how it does not apply. When you blow out for a harmonica, you're not even close to blowing as hard as you can (like the poster there notes)...so my PSI requirements must be less, even far less. My take from that post: 4 PSI would be overkill, so that's a good # to go with.

BTW: My plan is to definitely have a regulator on the discharge, but, again, I don't really care about the exact #s, as long as the harmonica makes noise.

Well you originally suggested a 1 atm pressure difference. A 4 psi difference would almost definitely not be choked, but may still be compressible.

Oh...my topic title...that's the specs for a random small compressor/tank I saw at Harbor Freight. The original Q stands (I think unanswered):

Does that spec mean that the tank can release 90PSI air for .6 of a minute and 40PSI air for a full minute without a compressor cut-in?

## 1. What is the purpose of studying model pressure loss vs time for air in a cylinder with a hole?

The purpose of studying this model is to understand the behavior of air inside a cylinder with a hole and how it affects the pressure over time. This information can be applied to various industrial and engineering applications, such as designing pneumatic systems or predicting air leakage in compressed air tanks.

## 2. How is the pressure loss affected by the size of the hole in the cylinder?

The pressure loss is directly influenced by the size of the hole in the cylinder. A larger hole will result in a faster pressure drop, while a smaller hole will have a slower pressure drop. This is because a larger hole allows more air to escape, reducing the pressure in the cylinder at a faster rate.

## 3. What factors can affect the accuracy of the model?

The accuracy of the model can be affected by several factors, including the temperature and humidity of the air, the shape and size of the cylinder, and the smoothness of the hole's edges. Additionally, external factors such as air flow and turbulence can also impact the accuracy of the model.

## 4. How does the pressure loss change over time in this model?

The pressure loss in this model follows an exponential decay pattern, meaning that it decreases rapidly at first and then levels off over time. This is due to the decreasing pressure difference between the inside and outside of the cylinder as the air escapes through the hole.

## 5. How can this model be applied in real-world situations?

This model can be applied in various real-world situations, such as predicting the pressure drop in pneumatic systems, calculating the air leakage rate in compressed air tanks, and designing ventilation systems. It can also be used to study the effects of different hole sizes and shapes on pressure loss in different types of cylinders.

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