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Need a confirmation. Bernoulli.

  1. Jan 4, 2005 #1

    Clausius2

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    I'm not sure completely of this. Can you confirm if, assuming IDEAL (Bernoulli) flow there is a loss of stagnation(total) pressure in the geometry attached.

    I think there is a loss of stagnation pressure indeed.

    The flow enters with a stagnation pressure Po1, pass through an orifice at 2 and leaves at section 3 (which area is equal to section 1) in a atmosphere of pressure Pa.
     

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  3. Jan 4, 2005 #2
    The way i see it is that pressure will remain the same when the flow passes through, only the velocity of the flow will be higher when it passes through the orifice...I just used the continuity law : Av = constant where A is the area...

    regards
    marlon
     
  4. Jan 4, 2005 #3

    Clausius2

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    What pressure remains constant?? Static or total?

    Caution: take the downstream of the tube (3) as a chamber of homogenous pressure Pa. So that, once the jet is discharging at (3) the static pressure just at the exit of the orifice must be Pa in order to be fully expanded. That's part of the enunciated of the problem.
     
  5. Jan 4, 2005 #4
    Isn't P01 just equal to Pa ???

    Isn't this what the continuity law is about ???


    regards
    marlon
     
  6. Jan 4, 2005 #5

    Clausius2

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    Maybe my explanation is not enough clear, sorry Marlon and thanks by the way.

    Po1=total pressure at section1
    Pa=static pressure at section 3 and static pressure at section 2 because the boundary constraint of the jet in ideal flow just at the exit of the orifice must be P2=Pa in order to be fully expanded.

    Continuity: U1=U3 but here is the doubt:

    For example, when a pipe ends in a large chamber of pressure Pa, I always take the static pressure just at the outlet of the pipe to be Pa. It seems to be a boundary constraint for the flow (it makes sense because eventually the flow jet is surrounded by an atmosphere of pressure Pa).

    Thus P2=P3=Pa and Po1 isn't equal to Po3. That was my reasoning. I hope to obtain some opinion about this.

    Thanks.
     
  7. Jan 4, 2005 #6

    Doc Al

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    total pressure remains constant

    I don't understand why total pressure (static + dynamic = [itex]P + 1/2\rho v^2[/itex]) would change along the streamline. The static pressure in sections 1 & 3 will equal atmospheric pressure. What am I missing?
     
  8. Jan 4, 2005 #7

    arildno

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    But you CAN'T use Bernoulli in this case!!
    In particular, between 2&3, the energy loss due to viscous dissipation is simply too large.
    (I haven't gone through all the arguments presented as yet, though..)
     
  9. Jan 4, 2005 #8

    Clausius2

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    Right, I was hoping a confirmation rather than a explanation by myself. Anyway I must acknowledge your effort only by merely trying to understand my poor set up of the problem.

    See the geometry attached. There is a pipeline, whose fluid has a total pressure Po. The pipeline discharges into a reservoir of pressure Pa. Which is the lost of total pressure?

    [tex] P_o=P+1/2\rho U^2[/tex]

    Into the reservoir total pressure=Pa.

    So that the fluid has a loss of total pressure:

    [tex] \Delta P_o=1/2\rho U^2[/tex]

    Because just at the pipe outlet P=Pa. What do you think about that? The question here is that in sudden expansions the total pressure is not conserved. Why? See the figure and realise there is a jet flowing into the chamber. The unique boundary constraint a jet can feel while flowing into an atmosphere is a pressure condition---> P=Pa.

    Try to assimilate this to the downstream widening of the pipe of the original problem I posted.

    Thanks.
     

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  10. Jan 4, 2005 #9

    Clausius2

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    Yes, I can't use Bernoulli. BUT in my last post I used Bernoulli with a particular boundary constraint (pressure outlet). Moreover:

    i) Although I assume ideal flow, there is a loss enhanced by the static pressure boundary condition (P2=Pa). (I always employ the subindex "o" when writting stagnation or total pressures, if not it is a static pressure).

    ii) In my analysis: have you seen any experimental coefficient of pressure loss anywhere? Answer: You don't. Such experimental coefficient K measures viscous dissipation and turbulence:

    [tex] \Delta P_o=1/2 \rho K U^2[/tex] and it is the responsible of stagnation pressure loss.
     
  11. Jan 4, 2005 #10

    arildno

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    All right, I'll check up on your conundrum after supper!
     
  12. Jan 4, 2005 #11

    dextercioby

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    Clausius,your notation is very decieving.It can miselead.Mybe it's traditional among engineers,but us physicists tend to take them very seriously and if one does not understand our notations,then it's bad,really bad...
    Hopefully mine will e very explicit.
    Let's assume that along the pipe the ideal fluid has a laminary stationary flow,which means that the equations of Euler admit the solution due to DANIEL ( :wink: ) Bernoulli ("Hydrodynamica",1738).
    Let's label the 3 portions of the pipe with "1","2" and "3".

    Then considering the gravity contribution as well,we can write
    [tex] P_{1}=P_{2}=P_{3} [/tex]
    ,which becomes
    [tex] P^{0}_{1}+\frac{\rho_{1}U_{1}^{2}}{2}= P^{0}_{2}+\frac{\rho_{2}U_{2}^{2}}{2}= P^{0}_{3}+\frac{\rho_{3}U_{1}^{3}}{2} [/tex]
    ,where gravity constribution is canceled/reduces.
    [itex]P_{i}^{0}[/itex] is the contribution to total pressure by the pressure field in the absence of movement (velocity field is zero and the gravity does not act) in the portion 'i' of the pipe.
    The fluid is assumed incompressible,therefore
    [tex] \rho_{1}=\rho_{2}=\rho_{3}=\rho;S_{1}U_{1}=S_{2}U_{2}=S_{3}U_{3} [/tex]

    I found the difference [itex] P_{3}^{0}-P_{2}^{0} [/itex] to be
    [tex] P_{3}^{0}-P_{2}^{0}=\frac{1}{2}\rho U_{2}^{2}(\frac{S_{3}^{2}-S_{2}^{2}}{S_{3}^{2}}) [/tex]

    Maybe that's not the loss (actually it's a gain,the difference is larger than 0),but at least it may give you ideas to correctly approach the problem.

    Daniel.
     
  13. Jan 4, 2005 #12

    Q_Goest

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    Yes, Bernoulli's is incomplete. Bernoulli's is a conservation of energy equation, and energy is not conserved for a fluid moving through a pipe. Regardless of there being an orifice there or not, there is energy loss as defined by various equations, such as Darcy Weisbach (sp?).

    Consider for example a very long pipe. There is a drop in pressure due to the viscous affect/turbulence in the pipe. An orifice can often be viewed as a very long section of pipe. So you're correct that at the outlet of the pipe, the pressure is Pa (assuming incompressible flow). In compressible flow, there could be a shock wave present, just as there could be in an orifice. But neglecting that, the flow is then determined by the dP and the total restriction.

    Note that a converging/diverging nozzle may actually provide for very little pressure drop (stagnation pressure drop), but for a flat plate or sharp edged orifice, there is significant loss of energy.
     
  14. Jan 4, 2005 #13

    Clausius2

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    Well, in the last posts I used the notation:
    [tex] P_{oi}[/tex] as stagnation or total pressure
    [tex] P[/tex] as static pressure.

    Your notation:
    [tex] P_i^0[/tex] is the static pressure
    [tex] P[/tex] is the stagnation or total pressure.

    You have started by P1=P2=P3 which is wrong. The last formula gives the jump of static pressure between 2 and 3, which must be compensated with the one between 1 and 2 to give zero. The flow is not as an ideal as we could think.
     
  15. Jan 4, 2005 #14

    Clausius2

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    Thanks for the aid, but here I was not trying to get oneself into turbulence and rugosity (Darcy). My considerations are only involving ¿¿ideal?? flow.
     
  16. Jan 4, 2005 #15

    Q_Goest

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    The "ideal" flow you're refering to is not real though. There is no conservation of energy. In the first geometry you posted (first post), if you assume a fluid which somehow manages to conserve energy, then you can apply Bernoulli's and P1 = P3 (I'm assuming cross sectional areas are equal) and of course you could use Bernoulli's to determine pressure on the other geometry you've posted. In the case of the second geometry, note that Bernoulli's would also predict no drop it total pressure.

    When you say Pa, and "pipe outlet":
    . . . that seems to indicate you're thinking there's atmospheric pressure on the pipe outlet, and it implies you're trying to find a pressure drop across this orifice. Pressure drop meaning total pressure or stagnation pressure. The only pressure drop is from real, viscious affects. There are standard equations for pressure drop across an orifice. They relate the orifice geometry, fluid properties, and other variables - to flow and pressure drop. These equations give you the pressure drop due to viscous affects. Bernoulli's never predicts a drop in total pressure (ie: the sum of velocity, static, and head pressure).
     
  17. Jan 4, 2005 #16

    rcgldr

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    Bernoulli's logic for flow in pipes of varying diameters:

    The amount of mass flowing past any cross section of the pipe must be constant, otherwise mass will be accumulating. Assuming that this flow has settled into a constant state, no mass is accumulating anywhere. This requires the flow in the smaller pipe to be moving faster, or more compressed, or a combination.

    Assuming that the flow is faster in the smaller part of the pipe, then there must be a pressure differential to cause the accleration into the narrower section pipe and deceleration to slow the flow down after it leaves the narrower section of pipe. Based on this logic, the pressure in the narrower section of pipe is lower than the wider sections.

    If the fluid or gas is compressible / expandable, the pressure differential is even more, because the gas expands under reduced pressure, requiring even faster flow to maintiain a constant rate of mass flow.

    One factor being ignored here is temperature, which resists the expansion of gas, and therefore reduces the increase of speed somewhat, since the temperature is reduced along with any reduction in pressure.

    The temperature effect is real, it's not uncommon to see frost accumulate on the surfaces of venturis.

    Getting back to reality, because of viscosity, friction, and turbulence, a pressure differential force between the input and output ends of a pipe must exist in order to maintain a flow, or you can turn the pipe vertical and let gravity supply the force to draw a gas or fluid downwards in pipe (assuming that buoyancy issues are eliminated).
     
  18. Jan 5, 2005 #17

    Clausius2

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    Thanks for the comment, but the case was incompressible for simplicity. See below this post. I'm going to clarify this. It was my fault.
     
  19. Jan 5, 2005 #18

    Clausius2

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    Thanks QGoest. You're comment is very near of the truth. I'm going to clarify this because I have thinking of it while sleeping.

    -First assumption: incompressible flow
    -Second assumption: high Reynolds number: Re>>1 but small enough to remain laminar flow.
    -Third assumption: I thinking of the original geometry, an orifice inside a duct, the cross area of the orifice is A2 and the cross area of the pipe (recovered symmetrically after the orifice) is A1=A3.
    -Fourth assumption: the problem data is a total pressure Po1 at stage 1, and static pressure P3 at 3 where the flow is fully developed.
    -Fifth assumption: the total pressure loss between stages 1 and 2 is much smaller than the total pressure loss between stages 2 and 3. It can be demonstrated so. Therefore, I will consider losses only at the sudden expansion.

    There are 3 main approach to this problem:

    1) Considering entirely ideal flow. Thus, DocAl, Marlon and Daniel are right. Total pressure is conserved: Po1=Po2=Po3 as Bernoulli states. There is no losses of energy. Of course, this behavior is very far of reality, and forecasts a larger mass flow (unreal). There is no possibility of recirculations, the flow remains straight.

    2) Considering [tex] A_2/A_3[/tex] is of the order 1, and non ideal flow. In fact there are losses. Here I link with Arildno and his explanation about lift loss and vortex implications. Recirculations will be formed due to the violent turn of the flow at the intake and sudden expansion at the outlet. Such recirculations are various orders of magnitude smaller than the characteristic lenght [tex] L=\sqrt{A_1}[/tex], and have a characteristic lenght [tex] l [/tex], where [tex] L>>l[/tex]. So that, the Reynolds number based on L is much larger than the Reynolds number based on l: [tex] Re_L>>Re_l[/tex]. Therefore, viscous effects cannot be neglected at recirculation zones. Total Pressure loss will occur (Arildno, this could explain too the loss of lift when boundary layer separation and vortex effects on it, do you realise of that?).

    The outflow at stage 2 will be jet-shaped. Maybe its section is stretched. But I don't think the pressure outlet condition (P2=P3) will be satisfied in these conditions. I haven't found any evidence of that. Moreover, the velocity [tex] U_3\approx U_2[/tex] (I mean, they are not the same because of continuity, but are of the same order) and surely there will be a small gradient of static pressure between 2 and 3. It can be calculated the mass flow is smaller than in the case 1, and the coefficient of pressure loss can be calculated via integral equations: [tex] K=(1-A_2/A_3)^2[/tex]

    3)Considering [tex] A_2<<<<A_3[/tex] so that K-->1 and the total pressure loss is the kinetic energy itself based on the velocity of the orifice, as I predicted in one of my lasts posts, when I was explaining the case of a pipe discharging into a large reservoir. Yes Qgoest, Bernoulli can't be employed, in part because there is none streamline which ends at the center of the reservoir, where the fluid is at rest, without passing through a recirculation zone (where the viscous effects cannot be neglected).

    Moreover, I have seen analitycally that as [tex] A_3\rightarrow \infty[/tex] then the static pressures [tex] P_2 \rightarrow P_3[/tex] so that I recover the result that the total pressure loss through the pipe and orifice is [tex] \Delta P_o=1/2 \rho U_2^2[/tex]

    All of this is curious, isn't it?. :smile:

    A final question: in which case the mass flow is the largest?, remaining the data problem constant (except the quotient A_2/A_3)
     
    Last edited: Jan 5, 2005
  20. Jan 5, 2005 #19

    Q_Goest

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    Clausius, I think it's worth also pointing out that Bernoulli's equation would predict zero pressure drop for an infinitely long pipe with an infinitely high flow rate. (ie: total pressure drop)

    (PS: We must have posted at the same time. I'll have to go back and see what you wrote.)
     
  21. Jan 5, 2005 #20

    Q_Goest

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    Hi Clausius. You have a pretty good handle on this, but I think there's some significant points worth mentioning.

    - Bernoulli's is applicable to high or low flow rates. It is applicable to laminar or turbulent flow. It does not distingquish. Real, frictional pressure drop as predicted by various other equations exists regardless of the pressure drop being laminar or turbulent. The pressure drop is slightly higher in proportion for turbulent flow, but still exists for laminar flow.

    - Bernoulli's is the ONLY equation that predicts complete pressure recovery (ie: no pressure drop) after this type of restriction. If a fluid is flowing through this restriction in real life, there will actually be a pressure drop.

    - The restriction coefficient, K, as given by Crane paper #410, for a sudden, sharp edged contraction is 0.5. For a sudden expansion it is 1.0. In other words, the amount of pressure loss (ie: not calculated by Bernoulli's equation) for the expansion is roughly twice the pressure loss for the sudden contraction. The contraction can not be ignored. Regardless, the geometry is of a flat plate orifice, which is usually handled as a single restriction as opposed to two separate restrictions. It is acceptable to do it either way.

    Yes, this is exactly correct.

    You talked a bit in your last post about how the (real) pressure recovers after such a restriction. Of course, Bernoulli's doesn't really say anything about that, but it is an important feature of the flow. Pressure taps used on orifice meters for example, must be placed far enough upstream and downstream of the orifice to ensure the flow is fully developed across the cross section of the pipe. Bernoulli's really doesn't make that distinction. For Bernoulli's equation, we assume the velocity is dependant on the continuity equation, regardless of location to the restriction.

    My appologies if any of this sounds like lecturing, I don't mean to be standing on a soap box talking down. You seem to understand this stuff pretty well.
     
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