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Bernoulli equation questions

  1. Oct 10, 2013 #1
    Hey! So if the vorticity of a fluid = 0, it is in steady state laminar flow and friction is negligible (and viscosity too?), you can use the bernoulli equation between any two arbitrary points in the fluid, regardless if they are connected by a streamline.

    If the vorticity is non-zero, can you still use bernoulli, but only if you follow a streamline?

    And how does the viscosity of a fluid come into this? I believe the viscosity must be negligible for the Bernoulli-equation to be used. However, without viscosity you wouldn't even have laminar flow, so what's up with that?
    Last edited: Oct 10, 2013
  2. jcsd
  3. Oct 10, 2013 #2


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    Well, Bernoulli's equation says nothing about laminar versus turbulent flow because it applies only to inviscid flows, and the concept of laminar and turbulent flow are meaningless. Bernoulli's equation requires that the flow be steady, inviscid and incompressible to be valid and applies generally to flows along a streamline. If the flow is also irrotational, then the Bernoulli equation is no longer restricted to a streamline and applies to the whole flow.

    Of course, all fluids are viscous, so you might ask when you can ever use Bernoulli's equation. One consequence of boundary-layer theory is that outside of boundary layers, flows behave as if they were inviscid, so outside of the boundary layer, you can use Bernoulli''s equation to your heart's content. Inside the boundary layer, though, you can't, as viscosity is a major factor in those flows.
  4. Oct 11, 2013 #3
    So assuming there is high viscosity inside a fluid: the fluid wouldn't just get deformed, energy would also be lost?

    And could you give a small explanation as to why you can use bernoulli regardless of streamlines during irrational flows?
  5. Oct 11, 2013 #4


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    A fluid could be incredibly viscous but if there is no shear stress in the fluid, that high viscosity doesn't matter and the fluid will still behave as an inviscid fluid. That's why as long as you are outside of the boundary layer, for example on an airplane wing, you can still use Bernoulli's equation to get the pressure distribution.

    As for the streamlines question, how familiar are you with vector calculus? It would be hard to explain without familiarity with calculus. Basically, without starting too early in the process, the motion of an inviscid fluid is governed in the fullest sense by what is called Euler's equation, and one possible method of writing Euler's equation gives you
    [tex]\rho\left( \dfrac{\partial\vec{V}}{\partial t} + \vec{V}\cdot \nabla\vec{V} \right) = \rho \vec{f} - \nabla p.[/tex]
    Through an identity, this is also equal to
    [tex]\rho\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\nabla\times\vec{V} \right) = \rho \vec{f} - \nabla p,[/tex]
    where [itex]\vec{f}[/itex] is the body force (e.g. gravity). If you assume the flow is irrotational and thus [itex]\nabla \times \vec{V} = 0[/itex] and the velocity can therefore be described in terms of a potential function [itex]\vec{V} = \nabla \Phi[/itex], then Euler's equation simplifies to the following
    [tex]\rho\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \rho \vec{f} - \nabla p,[/tex]
    [tex]\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.[/tex]
    Now, body forces are conservative and also governed by a potential function, [itex]\vec{f} = \nabla\Phi_f[/itex], so
    [tex]\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} \right) = \nabla\Phi_f - \nabla \dfrac{p}{\rho}.[/tex] We can gather all the gradient terms together then and take the gradient of the entire expression
    [tex]\left( \nabla\dfrac{\partial\Phi}{\partial t} + \nabla\dfrac{V^2}{2} - \nabla\Phi_f + \nabla \dfrac{p}{\rho}\right) = 0.[/tex]
    [tex]\nabla\left( \dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} - \Phi_f + \dfrac{p}{\rho}\right) = 0.[/tex]
    From calculus, then, you know that this is the equivalent of saying
    [tex]\dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} - \Phi_f + \dfrac{p}{\rho} = \text{constant}.[/tex]
    If the gravity is the only body force, then [itex]\Phi_f[/itex] is simply the gravitational potential and is [itex]-gh[/itex], which gives you the a modified Bernoulli's equation that can handle unsteady flows,
    [tex]\dfrac{\partial\Phi}{\partial t} + \dfrac{V^2}{2} + \dfrac{p}{\rho} + gh= \text{constant}.[/tex]
    Or, in a steady-state flow, you get the classical Bernoulli's equation:
    [tex]\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh= \text{constant}.[/tex]
    This required the following assumptions to derive:
    • Inviscid flow
    • Irrotational flow
    • Incompressible flow
    • Steady flow (if you want the classic form)

    So then what happens if you have a rotational flow and [itex]\nabla \times \vec{V} \neq 0[/itex]? We can start again with
    [tex]\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\nabla\times\vec{V} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.[/tex]
    Here, we can't assume a potential function exists since the flow field is not irrotational. What we can do is use the concept of a streamline (the direction of the velocity vector, [itex]\vec{V}[/itex]). To make things easier, let's introduce the vorticity, [itex]\nabla\times\vec{V} = \vec{\zeta}[/itex] so
    [tex]\left( \dfrac{\partial\vec{V}}{\partial t} + \nabla\dfrac{V^2}{2} - \vec{V}\times\vec{\zeta} \right) = \vec{f} - \nabla \dfrac{p}{\rho}.[/tex]
    We can then assume that the entire thing only applies along a streamline, which is the equivalent of taking the dot product with the velocity vector [itex]\vec{V}[/itex]. Looking at that vorticity term in the streamline direction
    [tex]\vec{V}\cdot\left(\vec{V}\times\vec{\zeta}\right) = \vec{\zeta}\cdot\left(\vec{V}\times\vec{V}\right) = 0.[/tex]
    So, along a streamline, we then have, analogous to before,
    [tex]\left[\dfrac{\partial\vec{V}}{\partial t} + \nabla\left(\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh \right)\right]_{s} = 0[/tex]
    where the [itex]s[/itex] subscript means the whole expression was taken in the streamline direction. If you then assume a steady flow once again, you get
    [tex]\left(\dfrac{V^2}{2} + \dfrac{p}{\rho} + gh \right)_{s} = \text{constant}.[/tex]
    This holds for any streamline, but the value of the constant will change from one streamline to another. The assumptions (and thus limitations) here were slightly different, and were
    • Inviscid flow
    • Incompressible flow
    • Steady flow (a hard requirement this time)
    Last edited: Oct 11, 2013
  6. Oct 16, 2013 #5
    Thanks for the proof! :) I enjoyed reading it (and yes, I have taken a vector calc course), but now I have stumbled yet again with the bernoulli equation.. Might I ask some questions?

    1) With "inviscid flow", you mean flow where you do not lose energy thru friction, right?

    Because in my textbook inviscid flow is defined as flow where the Reynold's number is very high (I think..), so viscous forces are dominated by inertial ones. Then the flow is turbulent (High Re number => more turbulence) and thus chaotic and unsteady, so you definitely cannot use bernoulli on those.

    BTW, what exactly do they mean by "inertial forces" - do they mean the fluid's "resistivity" to acceleration on account of its momentum?

    2) A bonus question about reynold's number if you can bother: Have I understood it correctly?

    "Most flow-problems can be cooked down to a solution of navier stokes equation which has 4 parameters: Density, scale, viscosity and velocity. If the solution could be rearranged so it depends on only 1 parameter (Reynold's number), modelling would be greatly simplified because then you only need to vary one parameter to properly understand how the flow behaves."

    So basically, all the real-life flow-problems can be simplified and then solved by a function with only Reynold's number as sole input?
    Last edited: Oct 16, 2013
  7. Oct 16, 2013 #6


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    Alright, so let's start with the Navier-Stokes equatiuons for an incompressible, viscous flow with no body forces (usually a good assumption, though this works fine when including them as well). We can start with the standard equations, using [itex]*[/itex] to denote dimensional quantities.
    [tex]\rho^*\left(\dfrac{\partial \vec{v}^*}{\partial t^*} + \vec{v}^* \cdot \nabla \vec{v}^*\right) = -\nabla p^* + \mu^* \nabla^2\vec{v}^*.[/tex]
    One thing that is often useful is nondimensionalizing the equations. We can do this by referencing the length variables to a reference length scale, [itex]L^*[/itex] (e.g. airfoil chord, pipe diameter, etc.); the velocity variables to the free-stream velocity, [itex]V^*[/itex]; pressure to twice the dynamic pressure, [itex]\rho^* V^{*2}[/itex]; and time to [itex]L^*/V^*[/itex]. Carrying this out, we are left with
    [tex]\dfrac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} = -\nabla p + \dfrac{\mu^*}{\rho^*V^*L^*} \nabla^2\vec{v},[/tex]
    [tex]\dfrac{\partial \vec{v}}{\partial t} + \vec{v} \cdot \nabla \vec{v} = -\nabla p + \dfrac{1}{Re} \nabla^2\vec{v}.[/tex]

    So, going back to your questions:

    If you look at the final equation I wrote, you can see that the final term is the only term involving viscosity and when the Reynolds number is large, this term gets very small. In other words, when the Reynolds number is very large, that term can effectively be neglected and the equations become the Euler equations for inviscid flow. This is only valid in regions where [itex]\nabla^2\vec{v}[/itex] is not large. For sufficiently large [itex]\nabla^2\vec{v}[/itex], that term is no longer effectively zero since this term may very well be large enough to counteract the [itex]1/Re[/itex] term. The only region where this is true is the boundary layer. Essentially, this means that outside of the boundary layer, even real, viscous flows can be treated as inviscid.

    Also, just to note, Reynolds number does not mean that a flow is or isn't turbulent. In general, a higher Reynolds number flow is more likely to transition to turbulence eventually, but it varies greatly on a case by case basis. You can even put a different airfoil into two different wind tunnels at the same chord Reynolds number and get totally different transition behavior. Just be careful here, because there is a lot of funky information floating around about laminar-turbulent transition and Reynolds number, and the truth is that about the only thing we knows about the relationship is that higher Reynolds number means greater likelihood of transition.

    When you hear people describing the Reynolds number as the ratio of inertial forces to viscous forces, it really just a means of describing whether the flow is dominated by the bulk motion of the fluid (inertial) or the interactions between different fluid particles (viscosity).

    If you go back to my case of nondimensionalizing the equations, this is, in some sense, what has occurred there. The equations still have explicit dependence on space and time, but they are independent of scale, so you could take two airfoils that have identical profiles but different scales and the only difference in the solution would be the Reynolds number. The same thing goes for identical shapes with different free-stream velocity, viscosity or density (or altogether different fluids). They all scale the same way. This is why you can make a scale model of something (e.g. a car or a plane or a bridge) and place it in a wind tunnel and get data that still corresponds perfectly well with the flow over the full-size object. All you have to do is hold the Reynolds number constant. As an example of this, you can take a small toy car and place it in a wind tunnel and get exactly the same fluid response as it's full-size version by appropriately increasing the speed of the flow.
  8. Oct 17, 2013 #7


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    Regarding the general applicability of Bernoulli's equation, my professor used to say:

    "Bernoulli's equation is just brilliant, because its actual realm of applicability is much greater than the realm it ought to have been restricted to."
  9. Oct 17, 2013 #8
    So to sum up: Bernoulli is applicable only when the shear-stress/friction of a fluid-flow can be neglected (it's "inviscid"). So, in theory, you can apply Bernoulli on any non-turbulent stream, even if the viscosity is huge (as long as the velocity distribution is uniform and thus shear-stress is equal to 0)?

    OK, thanks. I really must thank you for the effort you put into this bonehead :)

    However, if you have time to spare, can I ask you for further help? I'm getting murdered by potential theory: https://www.physicsforums.com/showthread.php?t=717145

    Thanks for the quote, arildno. Yeh the Bernoulli is a tenacious little thing
    Last edited: Oct 17, 2013
  10. Oct 17, 2013 #9


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    And steady, and incompressible...

    No. The concept of turbulence is irrelevant since we are talking about inviscid flows. If a flow is such that you can apply Bernoulli's equation, then the concept of laminar versus turbulent flows are in essence undefined. A laminar boundary layer is equally intractable for Bernoulli's equation as a turbulent one. You can apply Bernoulli's equation outside the boundary layer, however.
  11. Oct 17, 2013 #10
    Yes, of course, but we were talking about the viscous effects.

    I understand. What I was thinking was that since Reynold's number is often high in bernoulli problems, I should point out that bernoulli stops working when the flow becomes turbulent.

    Anyway I think I understand this stuff now. thanks
  12. Oct 17, 2013 #11


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    And I am trying to say that these are disjoint concepts and this statement is not true. If the flow is turbulent, then it is viscous anyway and therefore you can't use Bernoulli's equation.

    I know what you are trying to say, but that also doesn't make sense. A high Reynolds number does not necessarily mean turbulent flow will ensue. Take the flow over an airfoil in flight conditions, for example. You will see Reynolds numbers well over 1 x 106 and still have laminar flow, yes Bernoulli's equation is still equally invalid inside the boundary layer and equally valid outside the boundary layer.
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