There are some things I encountered, studying the Bernouilly equation, that I don't understand. I was studying in the following book: http://www.unimasr.net/ums/upload/files/2012/Sep/UniMasr.com_919e27ecea47b46d74dd7e268097b653.pdf. At page 72-73 they derive the Bernouilli equation for the first time, from energy considerations. They state that it can be applied if the flow is steady, incompressible, inviscid, when there is no change in internal energy and no heat transfer is done (p.72 at the bottom). I understand this derivation, my problems arise when they derive the equation again at page 110-111, this time from the Navier-Stokes equation.

I don't quite understand the derivation yet, but I am more confused about the outcome. It seems that the Bernouilli equation can now be applied under the four conditions they state (inviscid flow, steady flow, incompressible flow and the equation applies along a streamline) while it seems that there are no limitations to the internal energy of the flow and the heat transfer done. If I understand the last paragraph right, they state that the requirement that the flow is inviscid already comprises the claim that there is no change in internal energy ("the constant internal energy assumption and the inviscid flow assumption must be equivalent, as the other assumptions were the same", I don't really understand this reasoning, because the "inviscid flow assumption" was also already made in the previous derivation).

Furthermore, if I follow the reasoning (so if I assume that inviscid flow indeed implies that there is no change in internal energy or if I assume that the Bernouilli equation can also be applied, without the assumption of "no change in internal energy") then example 4 at page 74 seems strange to me. It shows that in this situation the Bernouilli equation can't be applied directly (because headloss has to be included in the equation). However I think that we can easily repeat the derivation, assuming that the flow is inviscid (and assuming that the other conditions are satisfied), but this example shows that the internal energy increases, so inviscid flow can't imply "no change in internal energy". And the example also shows that the Bernouilli equation can't be applied, because head loss has to be considered (so this example seems to contradict the result at page 110-111).

I hope someone can explain where my understanding is lacking, because this really confuses me about the conditions under which the Bernouilli equation can be applied.

It would also be helpful, if someone could explain the derivation at p.110: what I don't understand is the definition of "streamline coordinates". Do they just take a random point on the streamline, where they place a cartesian coordinate system?

  • $\begingroup$ viscosity (basically internal fluid friction) results in the "dissipation" of kinetic energy in the form of heat into the fluid, resulting in a rise in the internal energy of the fluid. that defines the statement: the constant internal energy assumption and the inviscid flow assumption must be equivalent. $\endgroup$ – aditya kp Aug 17 '13 at 2:11

Yes, they look at how the point (on the streamline) moves as a result of the defined quantities (pressure, density, velocity, etc).

Since Bernoulli's equation results from an equation of conservation of energy, you are assuming no loss of energy, which means no friction, which for fluids means no viscosity, which means inviscid flow.
So the equation is $v^2/2+gy+P/\rho = const$, which has no change in internal energy (i.e. $U_1=U_2$).

Example on pg74, however, has a change in internal energy, but that's because here the fluid is not inviscid.
OK was going to leave this for you to puzzle through, but just for completeness: because of the immediate area change of the pipe, with the constant steady pressure $P_1>0$, there is turbulent flow (high Reynolds number, depicted in the picture by the swirls), corresponding to viscosity and hence not inviscid.

  • $\begingroup$ Thank you. I'm still confused though with the fact that if we f.i. image a pipeline which broadens, then we see that the speed of the fluid after the enlargement has decreased, so because the kinetic energy has decreased the internal energy of the fluid increases to satisfy conservation of energy, no matter the viscosity of the fluid. This is derived more rigorously in the book I linked at page 74. In this derivation they don't use the viscosity of the fluid. So it seems to me that even for inviscid fluid the energy will increase, which would mean that the two statements are not equivalent. $\endgroup$ – Rayman Aug 17 '13 at 8:07
  • $\begingroup$ Huh? By definition, "inviscid flow" is flow without viscosity. Nothing more, nothing less. $\endgroup$ – Chris Gerig Aug 17 '13 at 19:35
  • $\begingroup$ Yes, I know that ;). My problem is that "flow without viscosity", doesn't imply "flow with constant internal energy", which is demonstrated in the example I mentioned, and that the Bernouilli equation is not valid for incompressible, inviscid, steady flow (example at p.74 shows this), which is in contradiction with the fact that the Bernouilli equation is derived for this kind of flow. $\endgroup$ – Rayman Aug 17 '13 at 20:57

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