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Air-Flow Between Two Different Pressures

  1. Jan 12, 2005 #1
    In an engineering class, we are working in teams to build hovercrafts from scratch. My team is trying to derive all equations with our knowledge instead of looking them up at hovercraft hobbyists' websites, etc. We're only in high school, so our knowledge of physics, especially fluid physics, is limited. We'd appreciate any sort of help.

    There is a volume, V, of air in a chamber with pressure, P, which is higher than the standard 14.7 psi. There is a hole in this chamber that leads to a second chamber; the hole's area is A. The second chamber's pressure, p, is at standard (14.7 psi). What is the instantaenous rate of airflow from the first chamber into the second chamber, in terms of the variables V, P, A, and p? (Feel free to create any other necessary variables.)

    I say "instantaneous" because I realize that as more air flows from one chamber to another, the pressure decreases in the first chamber, and the air flow must reduce in speed. I imagine that the air fllow's velocity must decrease at a constant rate, but I'm not sure.

    To summarize, I need an equation that gives me the air flow (in units such as CFM) dependent on the pressure of the starting chamber, P, and/or the pressure of the ending chamber, p. Other variables such as A (area of the hole) and V (volume of air) are simply constants, and may be ignored.
  2. jcsd
  3. Jan 12, 2005 #2


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    What you need is a "basic" orifice calculation. There are a couple of things to consider however. I put "basic" in quotes because I am not sure how in depth you want to get since you are in the high school stage.

    The simplified basic flow equation through an orifice is:

    [tex] q = YCA \sqrt{(2g(144)\Delta P)/\rho} [/tex]

    q = flow rate in [tex]ft^3/sec[/tex]
    Y = Net Expansion Factor for compressible flow
    C = Orifice flow coefficient (use an approximate value of .6 for a sharp edge orifice). This is a corrected form of the discharge coef.
    A = Cross sectional area in [tex] ft^2[/tex]
    g = acceleration due to gravity in [tex]ft/sec^2[/tex]
    [tex]\Delta P[/tex] = Pressure difference in [tex]Lb_{f}/in^2[/tex]
    [tex]\rho[/tex]= Density of the fluid in [tex]Lb_{m}/ft^3[/tex]

    If the pressure in your hovercraft is enough, then compressibility will have an effect on the flow, hence the need for the net expansion factor Y in the above equation. If you choose to neglect it or assume non-compressible flow, set it = 1.

    The other thing to look out for is that if the [tex]\Delta P[/tex] across the orifice is about a 2:1 ratio you will have choked flow. That means that despite lowering the downstream pressure, the velocity will not increase through the orifice. The mass flow can increase by increasing the upstream pressure, but the velocity will not change.

    I hope this helps.
    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3
    Thank you very much for all your help! I am still a little confused though.

    First, the Net Expansion Factor (variable Y). I have no idea what this refers to, but I can infer that some fluids have different compressive qualities than others. You seem to suggest that this factor is not important at low pressures ("If the pressure in your hovercraft is enough, then compressibility will have an effect on the flow, hence the need for the net expansion factor Y in the above equation.") The pressure should not exceed 15.2 psi. I don't think that this is necessarily high, so I choose to ignore the Net Expansion Factor. Please let me know if this is inappropriate.

    Orifice Flow? The hole is made in vinyl-coated nylon. There is no tube between the two chambers, except technically, the thickness of the vinyl-coated nylon. I assume that I can consider this "sharp", and I'll use the value of .6 that you provided for me. Again, let me know if I am making a mistake.

    The cross-sectional area is the area of the whole. I don't know the acceleration due to gravity in terms of ft/sec^2 but I can find out easily. The pressure difference is, for example, 14.7 psi subtracted from 15.2 psi. And I'm sure the density of air is in some reference table.

    Thanks again for the explanation.
  5. Jan 12, 2005 #4


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    Using a value of 1 for Y in your case is a good assumption.

    An orifice is exactly what you have...a hole. Granted, most of us are used to a \n orifice that is used as a restriction in a pipe of some kind, but the idea is still the same. I realize that your situation is hardly text book in this regard, but you are going to have to make some simplifying judgements or else this situation could get complicated very quickly. Especially since you are dealing with a flexible geometry.

    The value for g is 32.2 ft/sec^2. You can calculate the density by using the ideal gas law.

    Good luck.
  6. Jan 12, 2005 #5
    Thanks again. But if you could help me just a little further, it would be great.

    The equation made sense and I have all the numerical values. However, I think there is a flaw in an assumption that I made previous to all this. I thought I might as well ask you...

    Imagine a long tube (about 18 feet in length). It is wrapped into a rectangular shape (6 feet by 3 feet). On top of this, the frame of the hovercraft is mounted (the frame is also 6 x 3 to fit on the tube). This tube must be inflated (lifting the 200 lb frame). The area of the surface of the tube that will be in contact with the frame is 612 square inches. Therefore, all the pressure from the frame's weight of 200 lbs will be concentrated in 612 square inches.

    You might be wondering how we got the area of 612 square inches for a complex meeting between the curved edge of a fabric and a frame. Please ignore this factor and pretend that the 612" value is correct; it is still under consideration, and likely to change, but not important. I need to understand the concept, not the numbers.

    1) My assumption was that a pressure of 200/612 is necessary to sustain the weight without collapsing (deflating), and more pressure is necessary to allow inflation. Is 0.3268 psi (which is 200/612) the correct critical pressure?

    Into this tube, air is pumped in at 820 CFM (through two leaf-blowers). Along the bottom of the tube, toward the inside of the rectangle, there are 36 (another arbitrary number) small holes that allow air to escape into the chamber enclosed by the rectangular tube. Earlier in the day, the pressurized chamber I was talking about was the rectangular tube; air was escaping through the 36 holes into the center chamber. The cross-sectional area of the "orifice" was simply the sum of the area of the 36 holes.

    Side-note: If it helps you to understand, this is the path that the air takes: from the leaf-blower into the rectangular tube; from the rectangular tube to the inner rectangular area; from this inner area to the outside (when enough pressure builds up in the inner area, air will try to escape by going under the rectangular tube, thus lifting the hovercraft and "creating" a frictionless surface.)

    2) My second assumption is this: if air enters a chamber at a certain velocity, it will exit at the same velocity as long as the pressure is kept the same throughout. If the pressure is higher inside the chamber, then the air will exit at a higher velocity. Is this assumption correct?

    3) The third assumption: I can keep "q" (from that equation you gave me) lower than the input of 820 CFM, by controlling the cross-sectional area of the 36 holes (as a total). Keeping this value lower would mean that the tube is inflating. Is this assumption correct?

    4) When plugging into the equation that you gave me, I used .3268 psi (remember that 200/612 value?) for the pressure difference. At all times, the 200lb frame will cause this pressure. Am I correct in deciding to use .3268 psi, or is it something more complex than this (like a pressure that continuously changes?)

    In the end, all I need to do is theoretically solve dimensions and parameters so that the tube will inflate and raise the hovercraft.

    Thank you for all your time. Our team appreciates all your advice.
  7. Jan 12, 2005 #6
    Sorry, I think I'm doing this completely wrong. I realize that there is much more going on here than I thought, and I conceptualized the entire thing wrong. The input air raises pressure, which in turn forces air out. Two rates need to be combined, and I did not account for any of this.

    Anyway, if you wish to answer the questions above, that'd be great for me to learn from, but I have to redo the entire thing, so you don't have to take the time to explain or answer anything.

    Thank you for all of your time.
  8. Jan 13, 2005 #7


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    I'll give some quick answers here since you're reconsidering:

    1. That pressure would (assuming your area calculation is correct) a net force exactly equal to the weight of the craft frame.

    2. This is a tough one to answer. In your case I am leaning toward saying no. Since you didn't mention the diameter of your 18 ft long tube, I am making the assumption that it's diameter is quite a bit larger than that of the leaf blowers' nozzles. If the amount of enlargement is enough, you would probably experience a kind of "bellows" effect which would help to quiet the flow down. If you think of just a straight pipe, then the assumption you made is true as long as the diameter of the pipe does not change. It might make a bit more sense if you think of it in terms of mass flow rate instead of velocity. That way you can consider the conservation of mass, i.e. what goes in is going to come out.

    3. That is a good assumption assuming you have no leakage anywhere else.

    4. That is also a good starting point.

    In your last post I think you came to a good realization: You are going to have to bleed off a certain amount of air depending on the load of your craft. It is going to be a balance between the constant incoming air source and (what ideally should be a variable) the bled off air source to maintain your pressure that is doing the supporting.

    Keep the info coming.
  9. Jan 13, 2005 #8


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    And one more note, the pressure rise of .32 psi over atmospheric is possibly attainable with a leaf blower but the CFM will be dramatically lower.

    If you watch the show Mythbusters on the Discovery channel you might remember the show where they measured the static pressure that could be created by a leaf blower in inches of water using some clear plastic tubing and food dye in the water. I believe they ended up with less pressure than you're calculations but were using an electric leaf blower and they had choked off the output completely to reduce the CFM to zero.

  10. Jan 13, 2005 #9
    Thank you FredGarvin for all your help. It definitely makes us more confident about our calculations. Although I revisited all the concepts, the confirmation of my assumptions help. And thanks for the input, Cliff_J. I have not seen the show, but I'll take that into consideration.

    Here is the new theory, if you'd like to check over and inspect it for any flaws in logic...

    Necessary Final Pressure

    My latest method of calculation works backwards through the steps. At the end of inflation, I need 315 square inches of contact between the tube and the frame. The frame is 200 lbs. Therefore, 200/315 is the psi necessary to lift the weight. This is .6349 psi. During inflation, more of the tube will be in contact with the weight, and so the necessary psi is lower than .6349. Therefore, I conclude that maintaining a pressure of .6349 will take care of all the various pressures during inflation.

    Rate Of Escape

    Next, I found the equation for the rate at which air would escape using the information you gave me. Plugging in the following values...

    Y= 1
    C= .6
    g= 32.0873 (ft^2)/s
    p= .0750 lb/(ft^3)

    ...and leaving q, A, and P as variables, I got...

    q = 210.6347 * A * sqrt(P)

    This "q" is the air's output rate in terms of area and pressure.

    Relative View

    Leaving that for a second, next I reconceptualized the inflation/deflation situation in my mind. Instead of having air go in and out of the tube at the same time, I made the situation relative. I pretended that the tube was closed (no air can exit at all). To compensate, the air's exit rate is subtracted from the air's input rate. The input rate is 13.6667 CFS. The exit rate was found in the previous section as [q = 210.6347 * A * sqrt(P)].

    Relative CFS= 13.6667 - q
    Relative CFS= 13.6667 - [210.6347 * A * sqrt(P)]

    The relative CFS, when the gauge pressure reaches the maximum of .6349 psi (calculated further above), must become zero (inflation must stop). Therefore...

    0= 13.6667 - [210.6347 * A * sqrt(.6349)]
    A= 14.7155 square inches

    Applying The Answer

    Now that I know that 14.7155 square inches of cross-sectional area is needed for the air to exit at a rate such that inflation will cease when full... I can divide this value by 32, since I am making 32 holes in the tube for the air to exit out of. This gives the area per hole, and then, the diameter of each hole.


    If the area exceeds 14.7155, the tube will not inflate fully. If the area is smaller, the tube will stretch (our material, vinyl-coated nylon, can stretch by 12%). If the area is too small, the tube will break under pressure.


    1) Turn on leaf-blowers to input air at 13.6667 CFS.
    2) Tube inflates and lifts the 200 lb frame.
    3) When the tube is at full capacity, the output CFS = input CFS.
    4) The tube stays as it is, holding the frame up.

    Once again, thank you for all the help. You have been a valuable source of reference for our team.
  11. Jan 13, 2005 #10


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    You've put a good bit of thought into this.

    A couple of items come to mind:

    1. The leaf blower is most likely not going to keep a constant output as the pressure inside increases. That number you have is undoubtedly with the blower exhausting to atmospheric pressure. If it is now exhausting to a higher pressure (you are backpressuring that pump). Without a pump curve from the manufacturer, there's no real way to tell what it will do until you try it. It may not have much effect at all. I do think that if you have any problems with this project, the capacity of the leaf blowers will be the number 1 culprit.

    2. As far as the outflow is concerned, you are creating that with fixed diameter holes, which is fine...in theory. I would make the suggestion that, if you can, put a valve in your system near the geometric center of lift so that you have a hole that is "adjustable." That way you can get close with your hand calcs and have some kind of adjustment to overcome inaccuracies and error that are always present. That way if you need a little more or less out-flow, you can simply turn the valve open or closed.
  12. Jan 13, 2005 #11
    Yes, I definitely forgot about the pumping power of the leaf-blower. The website does not show specs on the "pump curve." However, other information is available at http://www.sears.com/sr/javasr/prod...pid=07179734000&tab=spe&bidsite=CRAFT#tablink. Do you think the information there can be used to develop our own pump curve? Would it even be worth it, trying to figure it out? Because...

    Our idea was to have more holes than theoretically necessary, and then to patch them up one by one (through plastic-welding) until the desired inflation was achieved. So the adjustability is planned. Perhaps this adjustability will make the "pump curve" calculation negligible? Would .6349 psi cause such a great difference in input CFM that we must consider the pump curve?

    Also, something reminded me. I am not very familiar with the term, horsepower. According to my friends, a certain horsepower is needed to push the air through the tubes and out of the holes. So is there truly another factor that I must calculate dealing with horsepower? If so, how would I go about doing this?

    On a side note, out of curiosity, I assume your real name is Fred Garvin, right? May I ask for what engineering/science related profession you might be or have been involved in, if any? It may be possible that our team will have to reference you in our final presentation. If you'd like to keep that private, it is fine. Thank you.
  13. Jan 14, 2005 #12
    I think you may be underestimating the amount of area available to do the lifting. Although the ring has to lift at first by itself, this is only true until the air begins to fill the space inside the ring (the whole area under the 6 x 3 body). Once this begins the entire area works in your favor. To over come the initial lift you could manually lift the unit or set it up on blocks that allow the initial fill to occur downwards pressing the bladder to the ground and sealing the airspace in the center to help with the full lifting.

    I believe this is the case because there are examples of these lifting hundreds of pounds with a single leaf blower and as Cliff said, they don't have that much external static pressure available. However, spread out over 2592 square inches, they don't need as much static pressure to lift the weight.

    Just my opinion.

    Please let us know how you make out with this project. It sounds like fun.
  14. Jan 14, 2005 #13
    Yes, I did know that the middle would be in our favor. About a week ago, I planned on figuring out the dynamics of just the tube, and then incorporating the middle. But upon your reminding me of this, I realize that I may not be able to do separate calculations on the middle and just subtract the results from the tube calculations.

    Once again, I'll have to reconceptualize the entire situation. Now with three chambers (the third being the outside environment). My main concern right now is the cross-sectional area between the second and third "chamber." Between the first and second, holes were to be punctured in the fabric. Between the second and third, air was to escape from under the tube due to pressure building in the middle. We are going to drive the hovercraft on asphalt, not water. I dont believe asphalt is air-tight like water is. There are many holes due to the rough surface.

    Obviously, adding the middle section to the situation leads to many other problems to consider. I will be back when I figure out how to visualize this whole thing in understandable terms. Thanks to all for the considerable help you have provided!

    -Andrew Cheong & Team

    Edit: P.S. Would the pressure building up in the middle section exert force on the tube from the inside, and therefore and raise the pressure inside the tube? Any suggestions/comments on any topic related to any part of this thread from any person would be of great help, so please post anything that is on your mind as you read this. Thank you.
    Last edited: Jan 14, 2005
  15. Jan 14, 2005 #14
    Yes, but more due to the greater force required to push in the air in through the holes against the rising pressure. As the pressure rises the flow will reduce. If the pressure available still exceeds the lift required, the craft will rise and flow should stabilize by slipping under the skirt.
  16. Jan 14, 2005 #15
    Thanks Artman, I'll think about that.

    I just have one simple question. 212 square inches of the 200 pound weight will be in contact with the tube. 2380 square inches of the 200 pound weight will be in contact with the middle. I want to calculate the psi in the tube and the psi in the middle. I know that the weight has to be divided up, but I'm not sure how.

    Do I just take the ratio between 212 and 2380, and apply that to the 200 pound weight for the respective pressure calculations? If so, is that only for a uniform solid of 200 pounds (like a block with evenly distributed weight throughout)?

    Our frame is not a uniform solid. Imagine the frame of a rectangular prism (1" thick rectangular tubes make up the frame). The bottom is covered by sheet metal (which will be in contact with the middle 2380 square inches). The one-inch thick perimeter will be in contact with the tube (making up 212 square inches). Along the sides, there are trusses to provide the frame with support. My plan, if the non-uniformness matters, is to take the weight on the side (trusses, etc.), and divide that by 212 square inches for the tube pressure. And then I want to take all of the weight in the middle (sheet-metal, pilot, leaf-blowers, etc) and divide that by 2380 square inches for the middle pressure. I will continue my calculations in this way until I see some reply and tell me otherwise.

    For a future post, I will try to export an image of the CAD drawing that our team has made.

    -Andrew & Team
  17. Jan 14, 2005 #16


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    I wouldn't worry too much about that right now...unles you have a lot of extra time on your hands. You have bigger fish to fry.

    Power is defined as work per unit of time. You have pressure and volumetric flow rate. If you assume 100% efficiency on your pump (which is definitely not the case) the it would be defined as:

    [tex]\wp = (psi*GPM)/1714[/tex]

    No it isn't. Sorry. It's a throwback to an old SNL skit.

    I believe if you click on my name you can see the public info I have posted about my profession. If you must reference something, reference the board in general. Don't specify just me.

    BTW...Good info presented by Artman.
  18. Jan 14, 2005 #17
    Thanks FredGarvin. Ditto for your info. :smile: This is a cool project.
  19. Jan 15, 2005 #18
    There is something that I really don't understand. And I think it's the cause of most of my problems.

    Inflation. What is the exact process that occurs during inflation? The tube is crumpled up into a seemingly random shape while deflated. Air enters the tube. And then what? Does the tube begin lifting the 200 pounds when the pressure exceeds the amount needed to match the 200 pounds? If the pressure exceeds, why wouldn't that just force air out of the holes, thus not lifting the weight? Maybe some of the air escapes but the pressure is still enough to lift 200 pounds? Well supposing that the tube did lift. Then the volume is higher, so pressure decreases dramatically. Ahhh!

    I have no idea what is going on. It seems like calculations during inflation are realted to continuous rates... and... calculus?

    Is there any way to simplify this situation?
  20. Jan 15, 2005 #19
    Actually, you and your group seem to be doing very well at working out these design problems.

    Yes, I think it would be calculus because of the changing volumes and pressures.

    One way to simplify is by creating design points and determining the status at that point in time. (Graph those points and you can start to see your system curve.)
  21. Jan 16, 2005 #20
    Thank you for the compliment. Our group would not be this far without your help. I was actually trying the graph idea, but I couldn't get anywhere. It seemed this is dependent on that, and that is dependent on this. And it looped.

    For this entire post, pretend that there is no inner chamber; when the air exits out of the tube, it is always exiting into a 14.7 psi atmosphere. Later, the inflation of the inner chamber will be incorporated. For now, pretend the inner chamber is the atmosphere.

    As I understand, this is what happens as the air enters the tube:

    The Three Factors Of Pressure (j, k, m, r, q, v)

    1) Let j(t) be the function for the rate at which pressure rises because of the air that flows in at a rate of r(t).
    2) Let k(t) be the function for the rate at which pressure lowers because of the air that flows out at a rate of q(t).
    3) Let m(t) be the function for the rate at which pressure lowers because of the increasing of the tube's volume at a rate of v(t)

    The Instantanous Pressure Available (f, F)

    Let f(t) be j(t) + k(t) + m(t). j(t) will be positive since pressure is rising. k(t) and m(t) will be negative, since pressure is lowering. The integration of f(t) in respect to dt would give the instantaneous pressure at t. Let F(t) be the integration of f(t) in respect to dt.

    The Instantaneous Pressure Required (p)

    Let p(t) be the pressure required to lift the weight. p(t) is 200 pounds divided by how much surface area is in contact with the tube. p(t) changes as the volume of the tube changes. It is hard to explain, so I'll ask Brad (the frame's main CAD designer) to export an image eventually. At peak inflation, 212 square inches will be in contact. At complete theoretical deflation ("theoretical" because in the real world, the tube is sure to fold and twist and leave gaps), 612 square inches will be in contact. As the tube fills with air, less surface area is available, and thus, more pressure is required to lift.

    Acceleration Of Weight Due To Net Instantaneous Pressure

    Whenever F(t) [instantaneous pressure] is greater than p(t) [instantaneous pressure required to lift], a force is available going upward. Thus, the 200 pound object experience upward acceleration related to p(t) subtracted from F(t). From this, an acceleration of the object (and thus, the height of the tube) can be found. With plenty more contemplating and bottles of aspirin, v(t) [mentioned before as the rate at which the volume of the tube increases] is concluded to be dependent on F(t) - p(t). However, m(t), one of the three parts of f(t), is dependent on v(t). This is the first loop that confuses me. Maybe the loop doesn't matter, and something can be done. But I don't know how.

    First Summary

    So far, I have defined all I know about the following functions:

    f(t), F(t), m(t), p(t), v(t)

    What is left?

    j(t), k(t), r(t), q(t)

    Function Detail: j(t)

    j(t), to repeat, is the function for the rate at which pressure rises because of the air that flows in at a rate of r(t). The pressure is the instantaneous volume of air (at STP) inside the tube... divided by the instantaneous volume of the tube.

    The integration of r(t) gives R(t), the instantaneous volume of air at (STP) that has entered the tube. The integration of v(t), the rate at which the volume of the tube changes, is V(t), the instantaneous volume of the tube. The integration of q(t), the rate at which the air leaves the tube, is Q(t), the instantaneous amount that the air has left the tube. Q(t) subtracted from R(t) gives the amount of air at STP that is contained within V(t). Therefore...

    j(t) = {14.7 * [R(t) - Q(t)] / V(t)} - 14.7

    The 14.7's have to do with ratios and absolute pressures. j(t) is the rate of gauge pressure.

    Function Detail: k(t)

    I just realized that j(t) takes k(t) into consideration. To repeat, k(t) is the function for the rate at which pressure lowers because of the air that flows out at a rate of q(t). Well, I think j(t) takes care of that. If anyone disagrees, please let me know.

    Function Detail: m(t)

    I've already discussed this above. It is part of the loop. It seems to be dependent on itself, in some twisted way. This is the part I need most help with.

    Function Detail: r(t)

    This has to do with the pump curve. For now, I'd like to keep it constant. But as pressure increases inside the tube, the harder it is for the leaf-blower to pump air inside the tube. And thus, the rate decreases.

    Function Detail: q(t)

    This is determined by the very first equation that FredGarvin told me. However, the delta-pressure in that equation is dependent on F(t). This is the second loop that I see.

    Second Summary

    I think I've outlined everything in my head. I could go into more variables such as "h" for the height of the tube, etc. But I'm already having trouble just remembering the function names.

    Can someone please check through this and confirm that my logic is correct? Are there any flaws? Anything I'm missing?

    As always, every comment is appreciated. Thank you.

    PS - I am not proof-reading this. It is 1:15AM and I am at the point of delirium. If something really doesn't make sense, let me know, and I'll look into it tomorrow.
    Last edited: Jan 16, 2005
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