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Fluid Mechanics: a nozzle that pulls nozzle forward

  1. Sep 19, 2012 #1
    This is not a homework problem, but an example that was demonstrated that I am confused about. Sorry if it should've been in the other forum nevertheless.

    I understand the mathematics of the derivation and the result; but I'm having difficulty reconciling this against common experience with compressed-air hose, garden hose, images of firefighters pushing on an actively discharging hose, etc. I've simplified the example and skipped a few trivial algebraic steps to what is most relevant about my confusion.

    Hose of diameter Dh = 0.10m
    Nozzle diameter Dn = 0.04m
    p(water) = 1000kg/m^3

    Taking point 1 to be inside the hose, where the hose diameter is Dh, and point 2 to be just outside the nozzle where pressure is atmospheric, both points at the same z:

    m1 = m2 = m = 20 kg/s
    u1 = 2.55 m/s
    u2 = 15.9 m/s

    P1 = P2 + p/2*(u2^2 - u1^2) = 218 kPa

    Drawing a rectangular control volume around the nozzle (see attached), assuming steady-flow (no accumulation of momentum), and assuming that F reqd to hold nozzle stationary is in the same direction as the discharge flow:

    Force balance:
    P1Ah + F + mu1 = P2Ah + mu2

    F = (P2-P1)Ah + m(u2-u1) = -917 N + 259 N = -658 N

    The negative force indicates that the force required is against the direction of flow; that is, one must pull on the hose to keep it stationary while it discharges. But how can this be? I've never heard of a nozzle that pulls itself forward in the same direction as the discharge flow. The only self-feeding nozzles I've seen/heard of are ones that have jets pointing backwards (sewer cleaning, etc.) which in of themselves prove my intuition about reality to be true and this example to be flawed somehow.

    What am I missing?

    Attached Files:

  2. jcsd
  3. Sep 20, 2012 #2


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    No, this is correct. Imagine that you just placed a funnel in the way of the flow, which way would the funnel be pushed? Well, along the flow, of course. That's basically what you have there.

    The reason you get a recoil from an actual hose is that you need to take into account the pressure acting on the walls of the hose. Presumably, the hose has a bend, a back wall, a piston, or some other surface on which pressure in the fluid is acting on the opposite side from the nozzle. If you add that pressure in, now, because the pressure at the terminal edge of the nozzle is lower due to Bernoulli effect, you end up with a net recoil force.
  4. Sep 20, 2012 #3
    I see. If I were to allow an open reservoir to drain vertically into the same nozzle, now pointing towards gravity, I would have to pull the nozzle/funnel upwards against the direction of flow to keep it stationary. Now there's a gravitational effect, I'm aware, but because the system is open to atmospheric on either end, the concept doesn't fundamentally change.

    Interesting. So the recoil force that we experience is not a function of the nozzle itself, but is a result of Newton's 3rd law from elsewhere in the hose-nozzle system; and because water is modeled nicely by continuum mechanics, that force is transmitted through the water and experienced at the nozzle. Which also explains why valves should be opened/closed gradually (the water hammer effect.)

    Thanks for clearing that up!
  5. Sep 20, 2012 #4
    A couple of errors seems to me. I think for a flexible canvas-style fireman's hose, it's ok to assume the net force is owing to pressure differential at the nozzle end. But - in your last expression, how do you finish up with P1 and F opposed, when they have the same sign in the png? Vice versa for P2. I'd check the earlier force balance expression. Also, you finish up having both P1 and P2 acting over the same area Ah. How about introducing an An, and having the appropriate pressure operate over it. Things will probably add up 'normal' then.
  6. Sep 20, 2012 #5

    I assumed a direction for F in my sketch. If I'd placed F pointing left on the control volume, then yes, my calculation would have it result as a numerically positive quantity but the interpretation remains the same: to hold the nozzle stationary, one exerts a force on the nozzle against the direction of fluid flow.

    As far as the area of the nozzle: that was used earlier for the linear velocity of fluid discharging. In my control volume as drawn, I must consider the entire area of that side of the CV, therefore the use of Ah on both sides of the force balance. If I'd drawn the CV to be identically shaped as the nozzle, then I would use An as you suggest, but I'd also have to use the x-components of the pressure*area force on the angled sides of the CV. The y-components on the top and bottom of the nozzle would cancel.

    Setting up the problem as you suggest would (and should) have the same result as mine.
  7. Sep 20, 2012 #6
    I regret jumping in on this one. My earlier assumption amounted to that the hose was not taught, but in practice that probably doesn't apply, at least not after some brief adjusting to equilibrium. So for taught hose there really aught to be just the differential momentum rate terms applying. Pressure terms all cancel me thinks, including any frictional contributions.
  8. Sep 20, 2012 #7
  9. Sep 20, 2012 #8

    Don't be so quick to throw in the towel.

    I worked through the example you linked, very similar (nearly identical, actually, but different values) to what I posed. I used my approach from above and calculated the same magnitude but opposite direction result as was provided. We both arrived at ~ -181 N, however my assumed direction of F_R was opposite the author's (that is, I assumed it to be in the same direction as discharge.) Which means that Dr. C's conclusion is concurrent with intuition and the fireman is pushing against the recoiling hose; whereas my fireman pulls.

    So now I'm confused all over again because K^2's justification for my "pulling nozzle" made sense too.
  10. Sep 21, 2012 #9
    Thanks for encouragement - I won't.
    It initially seemed like a simple problem, but the more I think about it, the more I doubt not only everything I have ventured before, but also the 'official' explanation. We assume minimal pressure change owing to difference of elevation between pump and nozzle - the hose is basically horizontal. Pressure is generated at the pump end, and supposing the hose is fully taut, then with the nozzle initially closed, there is a complete internal balance between a positive and uniform fluid pressure all along the hose/nozzle, and counteracting tension in the hose/nozzle wall. Now unmuzzle the nozzle, and suppose the pump maintains pressure P1 at it's end. In reality there will be a steady pressure drop from pump end to nozzle end owing just to friction of flow against the hose wall, but we can assume that will be fairly small so let's neglect it. Therefore, right up to the nozzle part, an even tension exists in the hose wall, exactly countering pump force P1Ah.
    Now let's ask what exactly the fireman has to resist against. It is just the hose - inclusive of nozzle, not the flowing fluid in the hose (neglecting gravitational weight). Sole role of fluid in force balance is in exerting a varying pressure on the solid structure - hose/nozzle. It's that solid structure the fireman has to grasp and press against to maintain static equilibrium. A tensile force -P1Ah in the hose wall, only partially balanced by a reduced net pressure force on the nozzle pointing forward.

    What is the reason for and value of that reduced nozzle force? Well you can apply the Bernoulli equation, but it seems a vexed approach because the assumption of uniform back pressure P2Ah is highly dubious. There is a continual change in flow velocity and acceleration all along the conical region, and a proper analysis likely calls for computer simulation. But there is imo a simpler way out. The idea behind a good nozzle design is to minimize any 'fanning' of the flow past the nozzle end. In other words, we assume the axial flow rate is minimally reduced from inside to outside. Which means net reaction force, the difference between nozzle capped and nozzle open situation, is just the total rate-of-change of momentum owing to axial component of fluid discharge. And that is just flow rate in hose dp/dt = ρvAh. it represents a reduced pressure on nozzle, so as expected the fireman must push forward to counteract the otherwise greater hose wall tensile force acting back towards the pump. Maybe we need actual test rig results as final check, but this third attempt seems right. May have it wrong yet again though. :uhh:
  11. Sep 22, 2012 #10
    Oops - screwed up that last bit in #9. Had just given a mass flow rate which is not even dimensionally correct for force. Assume it's safe to say the reduction in nozzle force is the negative of the dynamic pressure force in the outlet stream once stabilized to ambient pressure - as that is the force a plate held against the stream must absorb, assuming no stream rebound. And that value should be fairly close to just the 1/2ρAnv22 applying at outlet point of nozzle throat. Depends then on whether stream is converging or diverging at point where contact is lost with nozzle, and details of nozzle design will come into that. Ha - just trying to salvage some self-dignity! Fall-back position is, just trust that piece linked to in #7. :redface:
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