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Bernoulli's Theorem

  1. Oct 6, 2011 #1
    Two of the assumptions in Brenoulli's Theorem say

    1. The fluid should be non-viscous
    2. The velocity of fluid should be less than its critical velocity

    Now how can these two co-exist. We know that

    Vc = (N.n)/(D.p)

    Vc= Critical Velocity
    N= Raynolds No.
    n= viscosity coef.
    D= Dia of pipe
    p= dencitu of fluid

    if n=0 the Vc=0 i.e. fluid is static.

    If there is no flow then what are we studying here.
    Please explain anybody.
  2. jcsd
  3. Oct 6, 2011 #2


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    Isn't what they mean by the "critical velocity" in this case actually the speed of sound?
  4. Oct 6, 2011 #3
    No. Its the velocity till which the flow of a perticular fluid is laminar.
  5. Oct 6, 2011 #4


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    There is no general equation for which a fluid is laminar in a given situation. There is no magic velocity where laminar flow stops.
  6. Oct 6, 2011 #5


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    I take it your Reynolds No. is actually supposed to be a "critical Reynolds No.," which is around 2000 for pipe flow.

    If so, the assumptions are really saying:

    1. the fluid flow can be approximated as an inviscid flow (viscous dissipation is not big), yet

    2. the flow is not turbulent

    Here I assume the non-turbulent part is included because you are going to use (or derive) the form of Bernoulli for an irrotational flow (Also, dissipation is going to become important at some point if you have turbulence).

    I know they're funny if you think about the Reynolds Number carefully, but well there you have it. Most physics are done in a "magic region" of parameter space.
    Last edited: Oct 6, 2011
  7. Oct 6, 2011 #6
    The four assumptions for bernoulli's are as follows:
    1) inviscid/frictionless - in other words no forces on the fluid due to viscosity
    2) applied along a streamline
    3) steady state
    4) incompressible

    The other assumption you refer to has do do with turbulence. When we have turbulence assumptions 2 and 3 break down.

    Remember these are assumptions, no fluid is truly without friction, but that doesn't mean inviscid isn't a good assumption in certain flows. i.e. away from walls and/or boundary layers.
  8. Oct 6, 2011 #7


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    Inviscid flows are indeterminant. Since there's no viscosity, there's no interaction between stream lines, so all of the net mass flow of an inviscid fluid could be constrained to a very narrow stream line in an otherwise non-moving inviscid fluid, such as the high speed flow exiting a narrow diameter section of pipe into a larger diameter section of pipe.

    Incompressable fluids also presents problems. Pressure times volume is no longer a form of potential energy, because no energy is transferred when pressure is changed (you have force but zero distance). The speed of sound in an incompressable fluid is infinite.

    Idealized Bernoulli relies on an idealized fluid with magical properties where the fluid is inviscid and non-inviscid at the same time depending on when it's convenient to fit the model.
  9. Oct 7, 2011 #8
    Thanks Guys.. :-)
  10. Oct 7, 2011 #9
    Not true. The internal energy has a PV term even for incompressible fluids.

    Just consider water going through a nozzle. It enters the nozzle with high pressure and low velocity and exits with lower pressure and higher velocity. Where do you think the kinetic energy came from? A nozzle swaps pressure energy for kinetic energy. Indeed, that is precisely the meaning of Bernoulli's equation: P/ρ is the pressure energy per unit mass, and V2/2 is the kinetic energy per unit mass. The mechanical energy of a fluid is constant along a streamline in steady flow, provided the flow is inviscid.

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