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Incompressible flow in a funnel

  1. Apr 4, 2014 #1
    I am having a few issues reconciling Bernoulli's principle and the continuity equation for an incompressible flow in a horizontal funnel where there is significant difference in the area from the start to the end. More specifically, I want to work out exactly what the fluid is doing once it reaches the narrowest point.

    Opening of funnel: radius 10m, fluid flow at 1m/s
    Narrowest point: radius 1m, fluid flow x
    Fluid density is 1000kg/m3, fluid pressure ~200kPa and there is constant fluid flow into the funnel.

    If I use A1V1 = A2V2, I get a flow x of 100m/s

    However, if I use Bernoulli's principle of v2/2 + p/ρ = constant, and substitute in the values from the previous equation, I wind up with a really negative pressure.

    Now, I understand that the fluid would begin to spin in the funnel - so my guess is that 100m/s is not horizontal flow but incorporates the distance traveled while spinning. Furthermore, I would guess that the pressure cannot drop below 0. As such, I am arriving at an answer that the fluid would flow horizontally at ~20m/s, at close to 0 pressure, with any one fluid molecule that is located toward the edge of the funnel travelling at a speed of 100m/s.

    Helping me understand this all would be fantastic; but if not, the specific thing I am looking for is exactly what is happening to the fluid at the narrow point of funnel e.g. horizontal speed, rotational speed, velocity of the fluid at the edge of the funnel, fluid pressure.

    Thanks for any help you can provide.
     
  2. jcsd
  3. Apr 4, 2014 #2
    You left out the gravitational term from the Bernoulli equation. You don't know the height of the liquid level, but you do know that the pressure on the bottom is zero, assuming that the 200 kPa is gage pressure.

    Chet
     
  4. Apr 4, 2014 #3
    Thanks for helping out, but does gravity matter if I am talking about a horizontal funnel where the center of the narrow end is at the same height as the center of the wide end? For the elevation point would be 0, meaning that the gravity term cancels out.
     
  5. Apr 4, 2014 #4
    I don't know, but isn't a funnel usually oriented vertically. Anyway, that's how I interpreted the problem.

    Chet
     
  6. Apr 4, 2014 #5

    SteamKing

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    What if the OP isn't talking about a funnel, with all that funnel baggage, but a horizontal, converging pipe or duct?
     
  7. Apr 4, 2014 #6
    You would be right, my apologies for the misuse of the terminology. I honestly did not think that turning a funnel horizontal would change its name. I am talking about a converging pipe or duct with a 10m radius at the wide opening down to a 1m at the narrowest point.

    For the purpose of the hypothetical, the duct is immersed entirely within a large body of this moving fluid and the narrowest part of the pipe leads to another dimension - so we do not have to worry about the effects of the faster fluid inside the pipe slowing down when returning to the main body of fluid.
     
  8. Apr 5, 2014 #7
    If the funnel has a circular cross section then the area would be ##\pi*100 m^2## at the inlet and the flow would be ##\pi*100 m/s## at the outlet. To avoid complicating things I will assume a square cross section with sides of 10m and 1m at the inlet and outlet respectively and stick with your figures.
    At the inlet we have v2/2 + p/ρ = constant = 200.5 and at the outlet the principle predicts the pressure is (200.5-10000/2)*1000 = -4800 kPa approx. Yep, that's negative.

    You are correct that the pressure cannot drop below zero, but assuming a spinning flow in the nozzle does not solve the problem. At the inlet of the funnel there is 100,000 kgs of water entering every second, and so 100,000 kgs of water has to leave the nozzle every second (conservation of mass) so the exit velocity has to be 100 m/s horizontally. (I think conservation of momentum also demands this). A horizontal flow of only 20 m/s would only allow 20,000 kgs of water to leave the nozzle every second and any spinning motion does not change this fact. I think Bernoulli's equation beaks down here and what actually happens is that viscous forces restrict the flow (and presumably accelerating the water in the nozzle to 100 m/s causes a back pressure) so that it becomes impossible to force water into the funnel mouth at a rate of 100,000 kgs a second without increasing the inlet pressure significantly. (Note that increasing the pressure significantly at the inlet raises the value of the 'constant' so that the pressure at the outlet is no longer negative). Basically, such a narrow nozzle become a throttled valve or tap. Shutting a tap restricts the flow and reduces the total flow through the system.

    <EDIT> Actually, if we assume the outlet pressure is zero we can write this equality:

    ##v_{out}^2 +0 = v_{in}^2 + 200##

    Since ##v_{in} = v_{out}/100## we can also write:

    ##v_{out}^2 = (v_{out}/100)^2 + 200##

    ##v_{out} \approx 20 m/s## and ##v_{in} \approx 0.2 m/s## and this gives an indication of how the flow is restricted. However it should be noted that Bernoulli's equation is only valid when viscous forces are insignificant and it's not clear that is the case here.
     
    Last edited: Apr 5, 2014
  9. Apr 5, 2014 #8
    Yes, I agree with yuiop. If you are going to maintain that throughput rate with zero pressure at the outlet, then you are going to have to increase the inlet pressure. If we knew more about the details of the convergent geometry (such as the length of the converging section), we could probably make some additional approximations, and possibly include the effects of turbulence. But, we don't. Yikes!!! A 20 meter diameter pipe is a huge pipe. What would such a pipe be used for, and what kind of pump would handle that kind of flow rate?

    Chet
     
  10. Apr 5, 2014 #9
    Thank you very much for that definately cleared many things up for me. Now I can get to work on calculating the spin rate.

    The question was purely hypothetical, so no need to worry about pumps and so forth. Thanks again
     
  11. Apr 5, 2014 #10

    boneh3ad

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    You guys are completely missing the actual problem here. You can't just arbitrarily choose numbers and say it's (1) incompressible or (2) even physically realizable. There are a few things here that lead to problems.

    First, the OP specifies the upstream pressure as 200 kPa but not whether that is gauge or absolute. That is an important distinction.

    Regardless of which interpretation, the first problem is that there simply isn't enough pressure difference between the inlet and outlet to accelerate the fluid that much. If it is absolute and the greatest possible pressure difference you can achieve is 200 kPa (assuming you could somehow get zero pressure on the outlet side, which is itself impossible), and 200 kPa is simply not enough pressure difference to support that sort of flow acceleration. If it is a gauge pressure, then the maximum pressure difference is 301.325 kPa and you run into the same problem. Or, the pressure could be a differential pressure, in which case the inlet and outlet ambient pressures are exactly 200 kPa apart and we run again into the same problem.

    Second, 10,000 m/s is several times the speed of sound even in water. That means you really can't assuming the situation is incompressible, even in water. That doesn't mean that you would suddenly see large density changes in water, of course, but you would see the flow moving much faster than pressure disturbances can travel, so you won't be able to make all of the incompressible assumptions.

    Last, viscosity certainly could be important, but unless you have a fully-developed duct flow, it shouldn't matter all that much, and with such large ducts here, that shouldn't be the case. The effects of viscosity would be small here. Even in relatively small ducts, Bernoulli's equation is usually a "pretty good" approximation and will underpredict the required pressure differential rather than overpredict it, so accounting for viscosity here would not tend to solve the problem.

    The flow would not necessarily spin here.
     
  12. Apr 5, 2014 #11
    The point is for the OP is that velocities and pressures cannot be picked just out of a hat. Trying to solve with Bernoulli will just give nonsense. Attempting to justify the nonsense with derivations of the flow, such as a hypothetical rotation, does not solve anything, if the original assumptions were incorrect to start with.
     
  13. Apr 5, 2014 #12
    Your welcome ;)

    It might help to note that Bernoulli's equation is basically a statement of energy conservation. The expanded equation for a vertical funnel that includes head pressure is:

    ##\frac{v^2}{2}+gh+\frac{p}{\rho} = Constant##

    Multiplying through by the mass of the volume element gives:

    ##\frac{1}{2} mv^2+mgh+pV = Constant##

    which is kinetic energy plus gravitational potential energy plus energy due to pressure.

    For an incompressible fluid (Constant density ##\rho##) the mass and volume (V) of the element are constant.

    Now if you have some sort of vanes in the nozzle that induce rotation in the fluid, you will need to include a term that takes account of the angular kinetic energy of the fluid element, that will look something like this:

    ##\frac{1}{2} mv_{in}^2+mgh_{in}+p_{in}V +\frac{1}{4}mr_{in}^2\omega_{in}^2 = \frac{1}{2} mv_{out}^2+mgh_{out}+p_{out}V +\frac{1}{4} mr_{out}^2 \omega_{out}^2 ##

    where ##\omega## is the angular velocity of the fluid. For a horizontal reducing duct with no rotation at the inlet (and dividing through by m) this simplifies to:

    ##\frac{v_{in}^2}{2} +\frac{p_{in}}{\rho} = \frac{v_{out}^2}{2} +\frac{p_{out}}{\rho} +\frac{r_{out}^2 \omega_{out}^2 }{4} ##

    Also note that even with rotation, the relation:

    ##A_{in}v_{in} = A_{out}v_{out}##

    still has to hold, due to conservation of mass. It follows that ##v_{in} = v_{out}*A_{out}/A_{in} = v_{out}*r_{out}^2/r_{in}^2##.

    Substituting this into the equation above it and rearranging gives:

    ## p_{in} =\rho \left( \frac{v_{out}^2}{2} +\frac{p_{out}}{\rho} +\frac{r_{out}^2 \omega_{out}^2 }{4} - \frac{v_{out}^2 r_{out}^4}{2 r_{in}^4}\right)##

    so if you know the linear and angular velocity and pressure (p>=0) of the outlet flow, you can calculate the required pressure at the inlet, (with the usual assumptions).
     
    Last edited: Apr 5, 2014
  14. Apr 5, 2014 #13
    Thought I'd let you know that the numbers weren't picked out of a hat. There is a real world example of what I am talking about. I just phrased it as a hypothetical. Thanks to those that helped
     
  15. Apr 5, 2014 #14
    If it's a real world example, how come the parameters are un-physical? Did you really measure a flow of 1 m/s at the inlet with a pressure of 200 kPa? It would be interesting to know what the real world example is and what was actually measured.
     
  16. Apr 5, 2014 #15
    Just curious to know where in the real world is a 20m diameter pressurized pipe located?
     
  17. Apr 5, 2014 #16

    etudiant

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    The OPs problem description sounds like a hydroelectric generating facility, as I can't think of any other place where water is ducted through 10 m radius pipes.
    If that is so, there should be a recognition that the flows will accelerate under gravity and that the funneling to a 1/100th the input surface area implies a huge speedup from the intake flow.
     
  18. Apr 5, 2014 #17

    boneh3ad

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    Well the problem with that assumption is that he already said it was laid down horizontally, so it doesn't sound like it is a hydro plant. I agree the numbers are unphysical and, if they really are based on some real-world application, must have been either incorrectly measured or incorrectly interpreted.
     
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