# I Bernoulli Equation and Navier-Stokes

1. Oct 17, 2016

### joshmccraney

Hi PF!

I was reading about Bernoulli's equation for steady, inviscid, incompressible flow. Now it's my understanding this equation is derived from the Navier-Stokes (momentum balance); then these two equations are identical regarding information offered. However, while thinking about applications of Bernoulli I came to a problem I couldn't reason: suppose you have a tug boat that's pulling a cone underwater, and that the process is steady and the velocity of the boat (and the cone) is constant. Let the entrance of the cone have pressure $P_1$ and velocity $V_1$ and let pressure and velocity at exit be $P_2$ and $A_2$. Bernoulli's equation gives $$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$$. However, Navier-Stokes is expressed $$\partial_t\iiint_v \rho \vec{V} \, dv+ \iint_{\partial v} \rho \vec{V} (\vec{V} \cdot \vec{n}) \, dS = \sum \vec{F}\implies \\ \rho(V_2^2 A_2-V_1^2 A_1) = P_1A_1-P_2A_2+F_{boat}$$
But clearly the final expression is not the same, namely the force of the boat is present. Can someone explain why I am having a discrepancy?

2. Oct 17, 2016

What are you defining as $F_{boat}$?

3. Oct 17, 2016

### joshmccraney

$F_{boat}$ would be the tension force from the boat pulling the cone. Does that make sense or have I not described it clearly?

4. Oct 17, 2016

It makes some sense but you are now essentially double-counting the tension force due to the boat. You are using a control volume analysis and you have already zeroed out the time-varying terms that are in there, which assumes the flow is steady. That assumption carries with it the constant tension force, in effect. That tension force isn't acting on the fluid (at least directly) anyway. Its effect on the fluid is essentially distributed over the surface of the cone in your example, which means it is essentially translated into the pressure gradient through the cone.

5. Oct 17, 2016

### joshmccraney

Awesome, thanks for clearing this up! So to be clear, even if we were accelerating $F_{boat}$ would still be disregarded and the pressure forces would account for the tension force?

6. Oct 17, 2016

Well you'd have to take the acceleration into account somehow. The easiest way to do it would be to take it into account via the resulting changing inflow conditions to the cone. After all, those changes are captured in the $\partial_t\int\int\int(\rho \vec{V})dv$ term, which is, at the end of the day, a force (time rate of change of momentum). The boat accelerating simply makes that term nonzero.

7. Oct 17, 2016

### Staff: Mentor

NO. In my judgment, your macroscopic momentum balance is perfectly valid (as reckoned by an observer traveling with the velocity of the cone). However, the Bernoulli equation tells you something else. The microscopic Bernoulli equation is obtained by dotting the Navier Stokes Eqn. (microscopic momentum balance equation) with the fluid velocity. This is usually referred to as the mechanical energy balance equation. In addition to the usual Bernoulli terms, this also includes viscous dissipation terms. The viscous dissipation terms are then neglected to obtain the Bernoulli equation. This is all addressed in Bird, Stewart, and Lightfoot.

8. Oct 17, 2016

### joshmccraney

Chet, are you saying that $F_{boat}$ SHOULD BE considered even if we are traveling at a constant velocity?

9. Oct 17, 2016

### Staff: Mentor

Yes. F is the force that the cone exerts on the water passing through it, and it is also equal to minus the tension in the cable pulling the cone.

10. Oct 17, 2016

### joshmccraney

So in my post #1, it seems Bernoullis equation $P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$ and this force balance give different information. But how is this possible since Bernoulli is derived from Navier-Stokes?

11. Oct 17, 2016

I disagree. Consider a nozzle at the end of a hose. If you want to apply Bernoulli's equation through that nozzle, you don't have to consider the tension in the hose/connection that is holding the nozzle in place in order to calculate the flow. It does not enter into the analysis of the control volume at all.

12. Oct 17, 2016

### joshmccraney

(I'll wait on further comments until you two agree. I don't want to get in the way, and I've seen both of you in action: you both know what you're talking about. A showdown of giants)

13. Oct 17, 2016

### Staff: Mentor

They don't. You can combine them to determine F.
The control volume analysis determines the force that the nozzle is exerting on the fluid flowing through it. This is equal an opposite to the force that the fluid is exerting on the nozzle. And the latter is equal an opposite to the tensile force that the hose exerts on the nozzle (to hold it in place). For a complete analysis of how the Bernoulli equation is properly combined with the control volume macroscopic momentum balance to determine the force that the fluid exerts on the nozzle, see the following thread: https://www.physicsforums.com/threads/nozzle-reaction-forces.888983/

14. Oct 17, 2016

Right, so in applying Bernoulli's equation (or the Navier-Stokes equations) to a control volume, one need not account for the force exerted by the boat directly provided that the inlet flow field, steady or unsteady, is known (as is the case in the OP's example). Since that tension in the wire does directly equate to the force of the water on the object being dragged, you can of course then back out the tension required for a given flow condition, but that force needn't be considered in the actual analysis of the flow field given the assumptions about a known inflow.

If the inflow is not known a priori, then it's a different story. Trying to determine the flow field that results from a given tension force, for example, would be considerably more difficult. Then you would need to know the tension in the cable, but even then, you really only need to know it in order to convert that into a pressure distribution and you don't need to take it into account as an added body force on the flow.

15. Oct 17, 2016

### Staff: Mentor

Yes. I agree. This is different from the nozzle problem because we don't know how much fluid is entering the cone. So, a much more comprehensive flow analysis would be required that takes into account the flow diverted around the cone. As a first approximation, however, I might assume that the flow through the cone is some fraction of the boat velocity times the inlet area.

In any event, the original question was whether Josh's two original equations, as written, were correct renditions of the macroscopic momentum balance and Bernoulli's equation, and whether they were independent of one another. In my judgment, the answer to this is Yes.

16. Oct 18, 2016

### joshmccraney

Awesome, thank you both so much! The link really helped (well it was confusing but your derivation was clear!)

Why wouldn't the flow through the cone simply be boat velocity, rather than some fraction?

Last edited: Oct 18, 2016
17. Oct 18, 2016

### Staff: Mentor

Suppose that the cone is totally closed at the small end. Then no flow goes through, and it is just about the same as a solid being dragged through the water.

18. Oct 18, 2016