# Bernoulli's Principle on Unsteady State and Compressible Fluid

1. Feb 12, 2008

Do any one know about any web site were the Bernoulli equation is developed from the Newton Second law and without the assumption of steady state and incompressible fluid?
With differential calculus, please.

My textbook just give out the equations for unsteady state and incompressible, but I want a really good explanation for a real understanding.

thanks

2. Feb 12, 2008

### arildno

Derive it for yourself.

Start out with the equation of motion (i.e, the Euler equations for the inviscid fluid), take the dot product with the velocity vector and integrate.

3. Feb 12, 2008

Should I put preassure as time dependant too for unsteady state?

P(s,n,t)
$$dP$$=$$\partial P$$/$$\partial s$$ $$ds$$ + $$\partial P$$ / $$\partial n$$ $$dn$$ + $$\partial P$$/$$\partial t$$ $$dt$$

Last edited: Feb 12, 2008
4. Feb 12, 2008

### FredGarvin

Bernoulli equation, by definition, assumes those conditions. Therefore you can't use Bernoulli for compressible, non steady flows. You want to look at the Navier-Stokes equations.

5. Feb 12, 2008

Bernoulli's Equation have to many assumptions, so on the textbook they explain how this equation could be use for unsteady state but the just give out the equations without a good explanation.
I just like to really understand how this equation is developed to use it properly.

6. Feb 12, 2008

Here is Bernoulli's Equation on unsteady state:

$$\rho*\frac{v^{2}}{2} + \rho \int \frac{\partial V}{\partial t} ds+ P +\gamma z = C$$

7. Feb 12, 2008

### Hootenanny

Staff Emeritus
You want a proof of the Navier-Stokes equations? How much mathematics have you done? Have you met tensors for example? Are you familiar with the convective derivative?

Last edited: Feb 12, 2008
8. Feb 12, 2008

Nope, no tensors and no convective derivative.

I don't really know if I'm trying to proof Navier-Stokes Equations. The fluid mechanics book state the bernoulli's equation for non-steady state but with no mathematical proof and I'm looking for the proof.

Is tensor a grad course? I know is really useful but in my years in college any course that I had taken talk about how to deal with tensor.

9. Feb 12, 2008

### John Creighto

The derivation of Navier Stokes equations is on Wikipedia.

http://en.wikipedia.org/wiki/Navier-Stokes_equations
http://en.wikipedia.org/wiki/Navier-Stokes_equations/Derivation

You do not need to understand tensors follow it. The Navier Stokes equations are a statement about conservation of momentum.

http://en.wikipedia.org/wiki/Reynolds_transport_theorem#Momentum_Formulation

Bernoulli’s equation is a statement about the conservation of energy:

http://en.wikipedia.org/wiki/Reynolds_transport_theorem#Energy_Formulation
http://en.wikipedia.org/wiki/Bernoulli's_equation

Thus Navier Stokes equations are not a generalization of Bernoulli’s equation.

Wikipedia does have a form of Bernoulli’s equation for compressible flows which is based on adiabatic compression:
http://en.wikipedia.org/wiki/Bernoulli's_equation#Compressible_flow_in_fluid_dynamics

10. Feb 13, 2008

### arildno

Let us clear up a few things first:

1. Bernoulli's equation is first and foremost a "first integral" of the equations of motion along a "streamline".
A streamline at a given instant is a curve in space whose tangent at a particular point equals the field velocity at that point (and instant).

2. In the "steady" flow, since the velocity field is locally independent of the time variable, the streamlines will coincide with particle trajectories, and since integrating the particle's equations of motion along its trajectory over a time interval gives us the energy balance, the standard Bernoulli equation does, indeed, COINCIDE with energy conservation.

3. Note, however, that the underlying DEFINITIONS are different:
For Bernoulli(-like) quantities, we integrate SPATIALLY, whereas for energy(-like) quantities, we integrate TEMPORALLY.

For unsteady flows, then, a "Bernoulli" procedure does NOT give an energy conservation picture, but rather
A picture of the INSTANTANTENOUS spatial distribution of dynamical quantities like pressure along curves that do NOT represent the trajectories of any particles whatsoever.
(Energy is not, in general, conserved along a strip of the streamline, the local time derivative term tells us how much is lost or gained on it)

To develop the maths sufficient to derive it, let $\gamma$ be a variable along some streamline, and let the family of streamlines in the field at particular instant t be given as:
$$\vec{S}(\gamma;t,\vec{x}_{0}),\vec{S}(0;t,\vec{x}_{0})=\equiv\vec{x}_{0}$$
That the streamline tangent is everywhere parallell to the velocity field vector $\vec{v}(\vec{x},t)$ is given by the defining equation:
$$\frac{\partial\vec{S}}{\partial\gamma}=\vec{v}{\vec{S},t)$$

Varying $\gamma$ in $\vec{S}$ moves us therefore along the same streamline, varying $\vec{x}_{0}$ shifts us to a neighbouring streamline at the same instant t, whereas the quantity $\frac{\partial\vec{S}}{\partial{t}}$ tells us how the streamlines changes over time.

Now, for our general derivation in the inviscid case, using constant density for convenience, we introduce index notation for clarity.

We call the streamline $x_{i}(\gamma,t), i=1,2,3$, and our equations of motion reads:
$$v_{i,t}+v_{j}v_{i,j}=-\frac{1}{\rho}p_{,i}-g_{i}$$,
where indices after comma represent the variable we differentiate with respect to, and g_1=g_2=0.

Now, we take the dot product with $\frac{\partial{x}_{i}}{\partial\gamma}=x_{i,\gamma}(=v_{i})$

and gain (by making a legit switch of variable name):
$$v_{i,t}x_{i,\gamma}+(\frac{1}{2}v_{j}^{2})_{i}x_{i,\gamma}+\frac{1}{\rho}p_{,i}x_{i,\gamma}+gx_{3,\gamma}=0$$

Integrating the result between, say, $\gamma_{0}$ to $\gamma_{1}$ yields the already noted result in a previous post.
The remaining integral of the time-derivative of the velocity is usually difficult to estimate.

Last edited: Feb 13, 2008
11. Feb 13, 2008

### John Creighto

if S is a vector shouldn't v be scaller?

clarity? What does i mean?

12. Feb 14, 2008

### arildno

I don't get your question.
The velocity field is a vector, and so of course, is the position in space.

Furthermore, we have, for index notation the three spatial variables $x_{1},x_{2},x_{3}$ rather than x,y and z.

13. Feb 14, 2008

### FredGarvin

That still assumes non compressible flow.

14. Feb 14, 2008

### arildno

....................................

15. Feb 14, 2008

Yes. Let's forget the compressibility and focus on unsteady state.
How do I get to that equation that I post?

I have been working on it but I just get to:

$$\rho V \frac{\partial V}{\partial s} + \rho \frac{\partial V}{\partial t} + \frac{\partial P}{\partial s} + \gamma \frac{\partial z}{\partial s}=0$$

But don't know how to get to the equation posted before.

By the way, the only knowlege on math I have is Calculus and Ordinary Differential Equations.

Last edited: Feb 14, 2008
16. Feb 14, 2008

:grumpy:

17. Feb 14, 2008

### arildno

I'll take a simple example for you, in 2-D.

Let $$\vec{v}(x,y,t)=xt\vec{i}+y\vec{j}$$

Then, the equations for streamlines reads:
$$\frac{\partial{x}}{\partial\gamma}=xt,\frac{\partial{y}}{\partial\gamma}=y$$

This yields the family of streamlines:
$$\vec{S}=x_{0}e^{\gamma{t}}\vec{i}}+y_{0}e^{\gamma}\vec{j}$$

We have:
$$\frac{\partial\vec{v}}{\partial{t}}=x\vec{i}=\gamma{x_{0}}e^{\gamma{t}}\vec{i}}$$

The integral we are to evaluate is therefore:
$$\int_{\gamma_{0}}^{\gamma_{1}}t\gamma{x}_{0}^{2}e^{2\gamma{t}}d\gamma=\frac{\gamma_{1}x_{0}^{2}e^{2\gamma_{1}{t}}}{2}-\frac{x_{0}^{2}}{4t}e^{2\gamma_{1}t}-\frac{\gamma_{0}x_{0}^{2}e^{2\gamma_{0}{t}}}{2}+\frac{x_{0}^{2}}{4t}e^{2\gamma_{0}t}$$

18. Feb 14, 2008

Thanks arildno for helping with the partial of velocity, but I still having problems on going
from this:
To this

Maybe is simple but I don't know how to deal with these partials.

Thanks