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In some books I find that, among the assumptions for proving bernoulli equation using work-energy theorem there are:

  • stationary flow
  • perfect fluid ($\rho=$constant, $\eta=0$)
  • laminar flow

But if one looks to the derivation of bernoulli equation from Euler equation instead, the conditions are only

  • stationary flow
  • perfect fluid ($\rho=$constant, $\eta=0$)

And there is a further condition

  • $\mathrm{rot} \vec{v}=0$ (irrotational flow)

In order to say that the constant of Bernoulli equation does not depend on the streamline.

In this context can I say that laminar flow $\implies \mathrm{rot} \vec{v}=0$ (irrotational flow) so that the condition imposed in the first derivation automatically include the condition that in the second derivation guarantees that the constant of the equation does not depend on the streamline?

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Laminar flow occurs when the liquid layers move parallel. Otherwise stated, the motion of the particles of the fluid is very orderly with all particles moving in straight lines parallel to the pipe walls. The condition in which Bernoulli's principle has been derived is that the viscosity of the fluid should be negligible. Hence we can assume that the given fluid has got nearly zero viscosity. In such a case, the velocity vector will be steady and has no rotation at all and is a constant at every points.

In such a case, you may write laminar flow corresponds to irrotational flow. However, in reality the effects of viscous force are not that negligible. Hence no real flows are irrotational. Hence in general, if you cannot demand that a laminar flow should correspond to an irrotational flow. In reality, due to the effects of viscosity, the increase in both its dynamic pressure and kinetic energy does not occurs with a simultaneous decrease in (the sum of) its static pressure, potential energy and internal energy. There will be a drag or a delaying effect.

However, in deriving the Bernoulli's equation from the work energy principle, you should neglect that the liquid has got negligible viscosity and can assume that the fluid element has to be along the flow. So, in the present context, $rot\vec{v}=0$ corresponds to a stream line flow, as long as the assumptions holds good.

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    $\begingroup$ -1: because laminar flow does not imply $\operatorname{curl} \mathbf{u} = \mathbf{0}$. Example $\mathbf{u}=[u(y),0,0]^T$. The rotation is clearly not equal to $\mathbf{0}$. $\endgroup$
    – MrYouMath
    Mar 6, 2017 at 11:11
  • $\begingroup$ Why should that not be legible? For example the Couette-flow has such a form. $\endgroup$
    – MrYouMath
    Mar 6, 2017 at 18:03
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MrYouMath said it correctly.

Laminar flow doesn't imply $\Delta \times \vec{u} = 0$.

Best expample is laminar boundary layer. Assume in a flat-plate you have a boundary layer which is laminar with $u = (y/\delta)^2 $ (a parabolic velocity profile with boundary layer thickness $\delta$.)

$ \Delta \times \vec{u} = (2y/\delta^2)$ -> Vorticity exists in laminar flow !!!

So you can apply Bernouli's equation in laminar flow with zero vorticity.

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