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Water rocket and water velocity

  1. Jan 28, 2015 #1
    I found a few questions on water rockets but am not able to understand the equations. The guide I am using has this for the exit velocity of the water:
    V = sqrt( ( 2 * pressure * nozzle_area ) / water_density )
    I plug in the values for my water rocket to get
    V = sqrt( ( 2 * 17.24newtons/cm2 * 3.80 cm2 ) / 1 gram/cm3 )
    After some simplification:
    V = sqrt( 131 newtons * cm2 / 1 gram )
    That does not look anything like a velocity. How do I get the terms to work our correctly?
     
  2. jcsd
  3. Jan 29, 2015 #2
    Firstly, I'd suggest convert all units to SI units. It'll be easier.
    Secondly, Newton can be further broken down into two basic units. Force = mass x acceleration
    Try it
     
  4. Jan 29, 2015 #3

    Orodruin

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    You are correct, the expression does not have correct dimensions to be a velocity (it has dimension length^2/time), although you did a minor mistake in the units that does not make it better either.

    Are you certain the area is supposed to be there?
     
  5. Jan 29, 2015 #4
    Re siddharth: The newton is the SI unit for force. The nozzle area is 3.8 square cm. Converting that to 0.000308 meters2 does not help much. But OK, I’ll do that.
    Re Orodruin: No, I am not sure at all. That is what the document says. It can be found with a search for “water rocket predictors.pdf” It is the simplest equation I could find for the water flow . Reasonably accurate is good for me. If you have one I should use instead, Please show it and describe each of the terms.


    ( 2 * pressure * nozzle_area ) / water_density
    filling in my values

    ( 2 * 17 newtons * 0.000308 meters2 ) / ( 1,000 kgrams/m3 )

    The denominator is a fraction so invert and multiply, and consolidate the constants

    ( 0.010472 newtons meters2 * meters3 ) / 1,000 kgrams

    Consolidate the meters to get

    0.000 10472 newtons meters5 / kgrams

    F = MA
    or
    newtons = kgrams * ( meters / second2)

    yielding

    ( 0.000,10472 kgrams * meters * meters5 ) / ( kgrams * sec2 )
    yielding

    0.000 10472 meters6 / sec2

    It becomes obvious to the casual observer that I have totally blown this. Please advise.

    FYI, the critical values are: Pressure = about 17 newtons/ square cm and a nozzle area of 3.80 square cm. It is a water rocket and I need to find the flow rate of the water. The velocity combined with the area of the nozzle will yield the flow rate.
    Thank you for your time.
     
    Last edited: Jan 29, 2015
  6. Jan 29, 2015 #5
    Get the units right first, then concern yourself with the values.I agree with Orodruin. You get length^2/time as the final units. Check the initial formula again.
     
  7. Jan 29, 2015 #6

    Orodruin

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    Pressure is not measured in Newton in SI units, it is measured in Pascal = N/m^2 ...

    "It can be found with a search for" is not a good reference. Even if it does appear we have no way of verifying we actually found the document you refer to. If you want to refer to it, you need to provide a link.
     
  8. Jan 30, 2015 #7
    Point taken. The document is here: http://www.schulphysik.de/strutz/predictors.pdf
    On the second page, right side, is the equation for flow rate. Is that a valid equation? Do you recommend something different?

    I am still working the equation and the units are not working out. I will continue for a while then post my results.

    Edit: I suspect I am miss-using the equation from this page, or just flat not understanding what that author is trying to tell me. As I understand the equation, the velocity of the water exiting the bottle is the square root of:
    ( 2 * pressure * nozzle_area ) / water_density

    Ignoring the constants for now, and just trying to work out the units I start with

    ( pascal * m^^2 ) / ( kilograms / m^^3 )
    invert the denominator and multiply to get

    (pascal * m^^2 * m^^3 ) / kilograms
    which simplifies to
    pascal * m^^5 / kilograms

    Pascal must be expanded to what is represents, newtons / m^^2 so the equation becomes

    ( newtons * m^^5 ) / ( kilograms * m^^2)

    The meters simplifies resulting in
    ( newtons * m^^3 ) / kilograms

    This does not look like a velocity. Velocity is composed of ( distance / time ). If I use F = MA re-arranged to M = F/A and replace kilograms then I introduce seconds^^2. That will not work because velocity is seconds^^1, not ^^2.

    Please reveal my fundamental error to me.
     
    Last edited: Jan 30, 2015
  9. Jan 31, 2015 #8

    Orodruin

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    The equation is wrong. As the dimensions are not correct, the equation simply cannot be correct. I would not trust a source making this error.
     
  10. Jan 31, 2015 #9

    Nathanael

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    Sorry for being pedantic :s but this may be my only opportunity to ever correct you. :) It's length^(3/2)/time
     
  11. Jan 31, 2015 #10

    Orodruin

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    The OP has made an error in the simplification of units. The dimensional analysis goes:
    pressure x area / density = force / density = (mass x length / time^2)/(mass / length^3) = length^4 / time^2
    Taking the square root gives length^2 / time.
     
  12. Jan 31, 2015 #11
    It's m^2/sec
     
  13. Jan 31, 2015 #12

    Nathanael

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    I just looked at the final equation in the OP :H:headbang: sorry hehe
     
  14. Jan 31, 2015 #13
    I guess today is not the day you get the better of him :-p
     
  15. Jan 31, 2015 #14
    Well then. I will shift my reference. I will start here: http://tbpmindset.org/modules/SECME/Rockets/Rockets_Calculations_Manual.pdf
    and search for a few more. If you have a known good reference, please post.

    My goal is to enter instantaneous formulas into an Excel spread sheet and iterate my way through the flight, maybe 100 steps per second, more if needed.

    I thank each of you for your time.
     
    Last edited: Jan 31, 2015
  16. Jan 31, 2015 #15
    The document referenced in my previous post provided a formula for mass flow per seconds. After working it for a while the result was 6.98 kg per seconds. With a water load of 0.5 kg, the water rocket is out of water at about the 0.07 second mark. That agrees with our rather imprecise observation of “whooose and the water is gone.” It also agrees with an earlier calculation I made that the rocket would be something less than 2 meters high when out of water. Unless someone asks I won't show the calculations because I am sure everyone reading this can already do the math.

    Thanks for confirming that the first equation was bogus (or maybe I was unable to understand the author's intent).
    Again, thanks for your time.

    Is there a way to mark this answered?
     
  17. Jan 31, 2015 #16
    Did your iterations adjust the pressure value as the water flows out of the rocket?
     
  18. Feb 1, 2015 #17
    I am going rather slow, spending too much time on Quora. But yes, and I will add in a phrase to consider temperature drop and the resultant further pressure drop. I am doing this because I am a volunteer at an elementary school STEM class, one hour a week. We launched a water rocket a few times and the kids all thought that was just great. One of them noticed the fog remaining in the bottle and I was impressed at his taking notice. I am putting my lesson plans on my personal web site and that takes me a rather long time.
    Thank you for asking / reminding me about that.
     
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