# Few Basic Fluid Flow Questions

1. Aug 29, 2011

### Red_CCF

Hi

1. I was wondering why Bernoulli’s equation does not work for unsteady flows (where there is acceleration). If there was a horizontal pipe initially at steady flow, if the pressure at one end suddenly increases to cause a constant acceleration for the entire fluid body in the pipe (say from t = 0 to t = 5), why can’t I analyze the properties of flow using bernoulli’s equation at an instant of time (e.g. t = 2)?

2. I was wondering if there is a relationship between viscosity (internal friction) and pipe friction for flow in a pipe? i.e. can there exist pipe friction for inviscid flows?

3. This may be related to the above question but, what is the relationship between viscosity and the ability for a fluid to stick to a solid surface (like when fluid is placed between two plates and the plates are sheared, a velocity gradient is produced due to fluids sticking to the plates)?

Any help is appreciated

2. Aug 29, 2011

Bernoulli's equation neglects the time-dependent terms that would be needed to apply it to an unsteady flow. There is an unsteady analog to Bernoulli's equation that essentially includes a derivative of the potential function.

Viscosity is representative of the physical mechanism that leads to pipe friction affecting the flow. In other words, if it weren't for the property that we call viscosity (which is actually more of a proportionality coefficient than anything else), fluid would not feel any effect from the walls of a pipe or whatever other solid objects may be in the flow. There cannot be pipe friction in an inviscid flow. That is why Bernoulli's equation requires correction factors in the form of head losses to correct for friction and other viscous effects.

Viscosity, as I said before, is effectively the means by which shear is transmitted through a fluid. It is because of a fluid's viscosity that the flow tends to follow along (although lag behind a bit) with an accelerating plate and why a boundary layer forms along solid/fluid boundaries.

3. Aug 29, 2011

### Red_CCF

Hi, thanks very much for the response

I understand this from a mathematical point of view but I was wondering if there is a physical explanation on why the equation does not hold for unsteady flows.

Thanks very much

4. Aug 30, 2011

Math and physics are one in the same. If you neglect the math associated with unsteady flow, you are neglecting to capture the proper physics associated with unsteady flow.

5. Aug 30, 2011

### Red_CCF

Hi

I want to try to understand this situation from both math and physics perspectives. I have a lot of trouble “visualizing” the invalidity of Bernoulli’s equation for unsteady flow; is there a good way for me to understand physically why this does not hold for an unsteady flow (i.e. constant acceleration of fluids in a horizontal pipe)?

Thanks

6. Aug 31, 2011

### Travis_King

Do you see a time variable in bernoulli's equation? Do you see any consideration of viscosity or boudary layer thickness? Bernoulli's equation is a simplification of the fluid flow phenomena based on inviscidity. Because the math involved in properly accounting for viscosity is much more extensive, it makes sense to, as bonehead said, just use bernoulli and correct accordingly.

Unsteady flow is on another level, mathematically. Bernoulli's equation is valid only insofar as you neglect viscous effects, consider a flow constant and laminar, etc. There are corrections/estimations you can use to groom your results, but true analysis of unsteady flow is way above bernoulli's equation.

7. Aug 31, 2011

### RandomGuy88

In the derivation of Bernoulli's equation the flow is assumed to be steady and that is why the equation is only valid in steady flows. As Boneh3ad mentioned there is an unsteady form of Bernoulli's equation. Here is a derivation of the unsteady equation.

You can see that there are two differences between the steady and unsteady equation. The unsteady equation has the additional term that boneh3ad mentioned which is the time derivative of the velocity potential, and the left hand side is no longer equal to a constant. It is now a function of time.

Last edited by a moderator: Apr 26, 2017
8. Aug 31, 2011

The problem is that the physics is in the math. When you derive an equation, each term in the equation represents something physical. The equations aren't arbitrary. For Bernoulli's equation, the mathematical terms representing unsteady and viscous phenomena were assumed to be zero (or never accounted for) in the derivation of the equation, therefore the equation cannot properly capture the physics.

9. Sep 18, 2011

### Denys.Ca

Could you help me visualize why is the Pressure drops when the Velocity increases.

10. Oct 7, 2011

### Dever

I'll throw my picture out there to the wolves (be gentle). Physically, without the use of equations, it may be helpful to visualize what is actually occurring to create the pressure in the first place. Pressure in fluids is just the sum of the change in momentum of all of the particles that contact the surface. Additionally, those particles contact the particles around them, which is why the pressure is equivalent theoretically inside the fluid (disregarding gravity and other external factors besides the boundary). Now let's create flow. This means there's a pressure differential from one part of a container to the other. If the pressure, say, to the particle's right is less than the pressure to the left, then statistically it will move right after many collisions (I find this also to be a good visual for other forms of potential such as thermal and chemical). That statistical probability of movement to the right means the kinetic energy of the particle is no longer divided between three dimensions evenly, which is why you get net flow. Or, in the case of flow increased from smaller flow, it is less evenly divided than it was before. This means the component of velocity in the direction of the wall (as well as neighbors orthogonal to flow direction) is reduced. If the velocity in the direction of the wall is reduced, that means the pressure is also reduced.

11. Oct 16, 2011

### antonima

I have a related question- how does momentum travel in a closed container? IE a full pipe, closed at both ends
For every macroscopic flow there will have to be an equal and opposite flow because the volume of the pipe is constant. The only mechanism left for momentum transport is then compression waves - is this true?

12. Oct 16, 2011

### Dever

Related to the general topic, that would be my interpretation. There are a few ways to make that happen, such as a movable end to the enclosed container (think drum) or similarly an acoustic disturbance on a relatively rigid container, a temperature change, etc., but essentially you'd have a standing wave until their were enough collisions for entropy to take over.

However, relative to my previous post, which starts from microscopic phenomena and is observed as a macroscopic one; That is the point of pressure - that equal and opposite as an average quantity of momenta will result in having equivalent pressure on all surfaces in the container (after equilibrium, and excluding gravity). In that case, you only see momentum when you move within the mean free path of the individual particles.