Compressed Air Velocity

  • Thread starter BradP
  • Start date
  • #1
38
0
Suppose I have a straight, frictionless duct, with consistent cross-sectional area, and air is flowing straight through and exiting into a free stream. A fan at the beginning of the duct draws stagnant air at atmospheric pressure, and then the air exits the end of the duct into another room also at atmospheric pressure. The air immediately after the fan is at 20 psi, due only to the fan itself.

From that initial pressure, is there any way to determine the velocity when the stream of air is further down the duct and only at 10 psi?

edit: I realize many people will think that I need to know the velocity at the 20 psi point, but notice that I have outlined a determinate system. Given a duct of a certain cross-sectional area, there is only a certain fan power (or speed) that will produce an initial 20 p.s.i. The air stream will have a definite velocity profile.
 
Last edited:

Answers and Replies

  • #2
44
1
If you know the velocity at one point, you should be able to get the velocity everywhere with conservation of mass. You will need to know the temperature and pressure everywhere to compute the air density. I might be misunderstanding the issue though.
 
  • #3
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
The typical way of doing this is to use the Darcy Weisbach equation for fluid flow. In this case, since it's air with a fairly significant density change along the length, you would want to seprate into sections so the change in density was no more than about 20%-40%, the less density change, the more accurate will be the pressure drop calculated for each section.

You can read more from the Pipe-Flo Pro manual I posted here:
https://www.physicsforums.com/showthread.php?t=179830&
 
  • #4
38
0
Hm, okay. Appreciate it!
 
  • #5
44
1
Brad, you mentioned that the duct was frictionless. The Darcy-Weisbach equation isn't going to help you if that is the case.

I'm not sure if that was just a bad assumption on your part or if you were interested in the theoretical case of a frictionless duct, but I thought it was worth questioning you what you meant by that.
 
  • #6
russ_watters
Mentor
20,584
7,252
Agreed, MrMatt - I was thinking it was a very odd and impractical question.
 
  • #7
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
Brad, you mentioned that the duct was frictionless. The Darcy-Weisbach equation isn't going to help you if that is the case.

I'm not sure if that was just a bad assumption on your part or if you were interested in the theoretical case of a frictionless duct, but I thought it was worth questioning you what you meant by that.
Thanks Matt, I missed that. As you say, the DW equation doesn't apply.

Brad, there's no irreversible pressure loss due to flow through a frictionless duct, but a frictionless duct is a hypothetical situation. It doesn't exist in the real world.
 
  • #8
38
0
I don't remember why I said frictionless. I do want to account for friction, but I just don't care about a specific value right now. I will have one eventually, but right now I am just trying to get the concept straight.

I was reviewing the Darcy-Weisbach, and I can enter everything into it and get the pressure drop. But how will this give me the velocity at any point? I do not know the initial velocity within this pressurized stream of air. I just know that it draws stagnant air from a room at atmospheric pressure.
 
  • #9
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
Knowing the flow rate gives you the velocity. So if you know the (nominal) velocity at some point, and knowing density, you have the flow rate. Flow rate through a duct without branches is constant (conservation of mass). With flow rate and density (denisity is a function of pressure and temperature) at every point, you can calculate velocity at every point.
 
  • #10
38
0
Knowing the flow rate gives you the velocity. So if you know the (nominal) velocity at some point, and knowing density, you have the flow rate. Flow rate through a duct without branches is constant (conservation of mass). With flow rate and density (denisity is a function of pressure and temperature) at every point, you can calculate velocity at every point.

Yeah, if I knew flow rate velocity would be simple. I don't know it. All I will know is:

- friction factor (using an estimated velocity)
- initial pressure
- duct length
- duct diameter

And it is assumed the air is room temperature. This is a determinate system. If you set up a pressure gauge in the duct wall and increase the fan speed until 20 psi is reached, the air is going to have a definite velocity. But how do you calculate it?
 
  • #11
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
Ok, I think I got ya now. You have pressure at two locations and the duct between these two locations is a known geometry (length, diameter, etc...). From duct geometry you can determine the restriction and from the pressure you can then determine the flow rate. Note that the DW equation can be written in a couple of different ways. The way it's shown in the attachment (Pipe-flo pro - see link above) in equation 1, you have dP, rho, f, L, and D (g is a constant so you have that too). The only variable you don't have is velocity so you can solve for that.

Note that f requires some iteration because it has to be pulled from the Moody diagram or an equation that is a function of Reynolds number and Reynolds number is a function of velocity also. The way around this is to 'guess' at velocity, calculate f, then see if the equation 1 solves for the same velocity. Normally this is done on a computer so you don't need to do the iterations yourself.
 
  • #12
38
0
I only know the pressure at one location: immediately after the fan. I just know that at some point further down the duct, it will be at some lower pressure. I don't know the exact location where it will reach 10 psi. I just know that it is pulled from a room where it is roughly stagnant and atmospheric pressure. I can't use Bernoulli because I don't know how much energy the fan ads to the stream of air as it passes through it.
 
  • #13
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
I only know the pressure at one location: immediately after the fan.
Actually, don't you have the pressure at 2 locations? You have 20 psi downstream of the fan and 0 psig at the duct outlets. Once you have that, you can use the DW equation on the known geometry of the duct and determine mass flow rate.

If the duct branches off, you still can do it. You simply chop the entire duct up into smaller bits and do each separately. Breaking up a network of ducts like this is fairly easy using a computer program but can be a bear doing by hand because of all the iterations.

Note that the DW equation has essentially 3 variables: dP, mass flow rate and fluid system restriction. If you know 2, you can find the third. When doing fans or similar air movers, you may only have the system restriction but it seems in your case you also have dP so you can calculate mass flow rate directly. But the point is that fans have a flow "curve" in which mass flow rate is a function of dP across the fan. That's a pretty common situation, so you then have no ability to simply plug 2 of the variables into the DW equation and come out with the 3'rd. You need to determine how the duct or pipe system you have will react with that particular fan. Note that in this case you have a known fluid system restriction but not dP or mass flow rate which are dependent on the fan's curve so you have 2 unknowns and only 1 equation. The fan's curve then becomes your 'second equation' so to speak.

To solve this set of 2 equations and 2 unknowns you can graph the two flow curves and where they intersect is called the "operating point". So the first thing you do is guess a mass flow through your system and determine the dP from the DW equation. Do this a number of times and you get a 'curve' that shows how mass flow rate increases as dP increases. This is the green/gray line in the figure below that starts at 0,0 and moves up and to the right. The curve for the fan is the red line and they intersect at A, the operating point.
Ref: http://www.epa.gov/apti/bces/module5/fans/principle/principle.htm#operate [Broken]

[PLAIN]http://www.epa.gov/apti/bces/module5/fans/principle/images/fig07.gif [Broken]

But as mentioned above, I think you already have dP (20 psi downstream of the fan and 0 psig at the duct outlet). If you can be sure of that information, just calculate mass flow from the DW equation.
 
Last edited by a moderator:
  • #14
38
0
Actually, don't you have the pressure at 2 locations? You have 20 psi downstream of the fan and 0 psig at the duct outlets. Once you have that, you can use the DW equation on the known geometry of the duct and determine mass flow rate.

If the duct branches off, you still can do it. You simply chop the entire duct up into smaller bits and do each separately. Breaking up a network of ducts like this is fairly easy using a computer program but can be a bear doing by hand because of all the iterations.

Note that the DW equation has essentially 3 variables: dP, mass flow rate and fluid system restriction. If you know 2, you can find the third. When doing fans or similar air movers, you may only have the system restriction but it seems in your case you also have dP so you can calculate mass flow rate directly. But the point is that fans have a flow "curve" in which mass flow rate is a function of dP across the fan. That's a pretty common situation, so you then have no ability to simply plug 2 of the variables into the DW equation and come out with the 3'rd. You need to determine how the duct or pipe system you have will react with that particular fan. Note that in this case you have a known fluid system restriction but not dP or mass flow rate which are dependent on the fan's curve so you have 2 unknowns and only 1 equation. The fan's curve then becomes your 'second equation' so to speak.

To solve this set of 2 equations and 2 unknowns you can graph the two flow curves and where they intersect is called the "operating point". So the first thing you do is guess a mass flow through your system and determine the dP from the DW equation. Do this a number of times and you get a 'curve' that shows how mass flow rate increases as dP increases. This is the green/gray line in the figure below that starts at 0,0 and moves up and to the right. The curve for the fan is the red line and they intersect at A, the operating point.
Ref: http://www.epa.gov/apti/bces/module5/fans/principle/principle.htm#operate [Broken]

[PLAIN]http://www.epa.gov/apti/bces/module5/fans/principle/images/fig07.gif [Broken]

But as mentioned above, I think you already have dP (20 psi downstream of the fan and 0 psig at the duct outlet). If you can be sure of that information, just calculate mass flow from the DW equation.

Thanks for that thorough explanation. Essentially, what is throwing me off was whether the pressure was really zero at the duct outlet. It is going to have a significant velocity for some distance after exiting, so I was thinking it must have a different pressure than the stagnant air. Because of the Bernoulli principle. Is that right? Now that I think about it, that is why I said "frictionless" duct at first -- I wanted to know what is happening to the pressure in that region after it exits. I suppose that since it eventually reaches zero m/s outside of the duct, and there is no duct friction acting on it after it exits, I can just use the same equations.
 
Last edited by a moderator:
  • #15
Q_Goest
Science Advisor
Homework Helper
Gold Member
2,974
39
Thanks for that thorough explanation. Essentially, what is throwing me off was whether the pressure was really zero at the duct outlet. It is going to have a significant velocity for some distance after exiting, so I was thinking it must have a different pressure than the stagnant air. Because of the Bernoulli principle. Is that right?
Yes, if you want to be perfectly precise (and that's generally a good idea for students) then you should take into account the fact the air has some velocity at the plane of the duct outlet. The amount of kinetic energy the air has because of that velocity isn't substantial and can generally be neglected in real situations. In real situations, both the head of air in an AC system and the kinetic energy due to velocity are very small and are often neglected. However, you can still account for them by applying Bernoulli's equation. Read through the section entitled "System Fluid Pressure" on page 14 of Pipe Flo Pro. It explains how to incorporate frictional losses of a system into the Bernoulli equation.

For the case of a duct outlet, you can imagine a frictionless diffuser on the outlet. This frictionless diffuser does 2 things. It has no frictional losses (no need to apply the DW equation, only Bernoulli's) and it increases in area from the size of the duct to infinity such that the velocity of the air drops to zero in this diffuser. So as velocity decreases, static pressure increases along the diffuser. Applying Bernoulli's you now find that the velocity of the air at the outlet is converted completely to stagnation pressure, so the static pressure at the plane of the duct outlet is lower than atmospheric pressure by that amount. The actual duct obviously doesn't have this hypothetical frictionless diffuser on it, but it can be used to predict a bit more accurately how the static pressure at the plane of the duct outlet will be affected by velocity.
 

Related Threads on Compressed Air Velocity

  • Last Post
Replies
20
Views
9K
  • Last Post
Replies
4
Views
970
  • Last Post
Replies
7
Views
886
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
4K
Replies
3
Views
618
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
5
Views
34K
  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
6
Views
1K
Top