Calculate air-flow using pressure drop

In summary: Thanks for any answer :)In summary, the flow velocity of air leaving an outlet decreases as the pressure difference between the inside and outside of the pipe increases.
  • #1
apekattenico
6
0
At what speed (flow/volume for that matter) does air leave an outlet when you know it holds x bars in the pipe?

(It goes from x to zero (or athmospheric) -- you know the pressure difference)

I assume you can neglect relative roughness, resistance, and so on as the pressure drop is over such a short distance they do not affect it. The only thing that matters is the size of the hole. Am I in the wrong here?

How would I calculate such a thing?

Thanks for any answer :)
 
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  • #2
How do you know the pressure drop? Just because it vents to atmosphere doesn't mean the gage pressure immediately goes to zero. I don't think we have enough information to give an easy answer.
 
  • #3
apekattenico said:
At what speed (flow/volume for that matter) does air leave an outlet when you know it holds x bars in the pipe?
I don't think that you (can) even know that. The upper limit for gas speed in a pipe is the local 'speed of sound'. Continuity suggests that speed goes to zero as flow area goes to infinity. But 'speed of sound' is very complicated and your question is intimate with the design of rocket motor's bells.

Don't think of pressure (force per area) but use mole rate, moles per time.

I was given a facility where I directed operations. One was to pressurize a large tank as quickly as possible. It was challenging to explain/convince the operators that once the adjustable pressure regulator was squealing there was no point increasing the desired pressure. The squeal was the shock wave of the gas at the local speed of sound.
 
  • #4
Brian_C said:
Just because it vents to atmosphere doesn't mean the gage pressure immediately goes to zero.
Actually, for all intents and purposes, it does.

apekattenico, you haven't given us much to go on, but if you have an orifice (such as a slightly opened valve where you can be certain virtually all of the pressure drop occurs, then you can simply apply Bernoulli's equation to calculate the flow veloctiy (up to a few hundred feet per second).
 
  • #5
Brian_C said:
How do you know the pressure drop? Just because it vents to atmosphere doesn't mean the gage pressure immediately goes to zero. I don't think we have enough information to give an easy answer.


Doug Huffman said:
I don't think that you (can) even know that. The upper limit for gas speed in a pipe is the local 'speed of sound'. Continuity suggests that speed goes to zero as flow area goes to infinity. But 'speed of sound' is very complicated and your question is intimate with the design of rocket motor's bells.

Don't think of pressure (force per area) but use mole rate, moles per time.

I was given a facility where I directed operations. One was to pressurize a large tank as quickly as possible. It was challenging to explain/convince the operators that once the adjustable pressure regulator was squealing there was no point increasing the desired pressure. The squeal was the shock wave of the gas at the local speed of sound.

Fair point, but I think you are over complicating things.

russ_watters said:
Actually, for all intents and purposes, it does.

apekattenico, you haven't given us much to go on, but if you have an orifice (such as a slightly opened valve where you can be certain virtually all of the pressure drop occurs, then you can simply apply Bernoulli's equation to calculate the flow veloctiy (up to a few hundred feet per second).


And I did so on purpose.

I might be in the wrong here, but allow me to explain how I think, and feel free to correct me.

The difference in pressure causes a net force, which drives the air forward.

Given a pressure difference it's critical to understand where you get it. p1 is inside the pipe at speed v1, while p2 is outside at speed v2 (which must be 0 as p2 equals zero. (this might very well be incorrect)

When I open a valve on the pipe there excists a pressure difference over the pipe which makes the air move forward.

I believe you might very well apply The Bernoulli equation (imagine two cross sections - one before, and one after)

Is my logic flawed, or can I go with this?

EDIT:
To give some context.
I'm interested in checking how much air-leakage costs. Yes, I can find data on this other places, but I want to know if these are rough estimates (just like my calculation will be), or they have done it in detail.

But then again I do know that this problem is highly nonlinear coupled problem (the gas is cooling, the pressure is changing, and the flow rate may or may not be choked flow, etc...).

Are there any easy ways out here?
 
Last edited:
  • #6
russ_watters said:
Actually, for all intents and purposes, it does.

apekattenico, you haven't given us much to go on, but if you have an orifice (such as a slightly opened valve where you can be certain virtually all of the pressure drop occurs, then you can simply apply Bernoulli's equation to calculate the flow veloctiy (up to a few hundred feet per second).

How can you apply Bernoulli's equation? He's talking about a pressure difference of several atmospheres. The density of the gas will change dramatically. Viscous forces could also have a significant impact depending on the flow geometry.
 
  • #7
For times like this, I always reference people to Milton Beychok's web site. He has some really good stuff for calculating the accidental release of gasses under pressure and I think this is applicable now. It's mostly compressible flow equations that should look familiar.

http://www.air-dispersion.com/usource.html
 
  • #8
Good link. Thanks FredGarvin.

Matt
 
  • #9
Brian_C said:
How can you apply Bernoulli's equation? He's talking about a pressure difference of several atmospheres. The density of the gas will change dramatically. Viscous forces could also have a significant impact depending on the flow geometry.
I'm not sure - he just said "x" bars and I don't know what that means. If it's 2 or 3, the simple form of Bernoulli's still works pretty well. If it's 5-10 you need compressibility and viscocity effects. If it's 10 or 20, you get choked flow. I guess this is where the "need more info..." bit comes in.
 
  • #10
It appears that American engineering language is not the OP's forte and that he knows more than we do about the question at hand.
Fair point, but I think you are over complicating things.
 

Related to Calculate air-flow using pressure drop

1. What is air-flow and why is it important to calculate?

Air-flow refers to the movement of air through a specific space or system. It is important to calculate because it allows us to determine the efficiency and performance of ventilation systems, as well as ensuring proper air circulation and quality in indoor environments.

2. How is air-flow measured?

Air-flow is typically measured in cubic feet per minute (CFM) using an anemometer, which measures the speed and volume of air moving through a specific area. Other methods include using a manometer to measure pressure differentials or a pitot tube to measure air velocity.

3. What is pressure drop and how does it relate to air-flow?

Pressure drop refers to the decrease in air pressure as it travels through a system or duct. It is important to consider in air-flow calculations because it can impact the overall efficiency and effectiveness of the ventilation system. A higher pressure drop means more resistance and potentially lower air-flow.

4. What factors affect air-flow and pressure drop calculations?

The size and shape of the space or system, the type and condition of the ductwork, the speed and volume of air being pushed through, and any obstructions or blockages in the air pathway can all affect air-flow and pressure drop calculations. Temperature, humidity, and altitude can also have an impact.

5. Can air-flow and pressure drop be calculated for any system?

Yes, air-flow and pressure drop can be calculated for any system that involves the movement of air. However, the accuracy of the calculations may vary depending on the complexity of the system and the availability of accurate measurements. It is important to use proper equations and consider all relevant factors when calculating air-flow and pressure drop.

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