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Can we apply the inviscid Bernoulli equation

$$ \frac{P}{\rho} + \frac{1}{2}V^2 + gz = constant$$

along the center line of a fully developed pipe flow?

I think we can since the shear stress along the center line is zero ($\frac{\partial u}{\partial y}=0$ at the center ), thus resulting in zero shear stress on material elements at the center. There is therefore an inviscid streamline along the center line of the pipe flow.

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  • $\begingroup$ After writing my answer - thinking more about what you are asking - you probably want to say something more like - in the neighborhood of the center line. For a viscous fluid there is still a change in the velocity between streamlines at the center - but the gradient is minimum there. $\endgroup$ – docscience Dec 19 '18 at 21:19
  • $\begingroup$ An ideal gas model is often used to describe the flow of compressible gas in channels of variable cross section, for example, in rocket nozzles. $\endgroup$ – Alex Trounev Dec 19 '18 at 22:56
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    $\begingroup$ Why don't you look at the derivation of Bernoulli from the Euler equation, and see how it changes when you add the viscosity term? $\endgroup$ – mike stone Dec 19 '18 at 23:03
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We know from experience as well as from the solution to the Naiver Stokes equations for fully developed laminar flow of a viscous fluid in a horizontal pipe of constant diameter that the pressure drop is given by the Hagen-Poiseulle equation, and is thus not equal to zero. The Bernoulli equation predicts zero pressure drop for this situation. The Naiver Stokes equations also indicate that the pressure does not vary within the cross section, except for a tiny hydrostatic gradient. Thus, the pressure is essentially only a function of axial position. For turbulent flow, the axial pressure gradient is even larger than for laminar flow. None of this agrees with the Bernoulli equation, even along the centerline.

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One way to think about this - Bernoulli just considers the transformation of energy; energy that's potentially recoverable. The energy moves between potential (pressure) and kinetic (velocity) - but since it doesn't consider viscosity, no energy loss due to heat generation. Bernoulli's equation is often called the 'Energy Equation' in fluid dynamics. Without viscous losses, energy is conserved along stream lines in laminar flow including the streamline going down the center of the pipe.

If you want to throw viscosity back in you need an energy loss term. Energy losses due to heat generation can cross stream lines. So I don't know that you can do what you want to do if you insist viscosity is present.

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