# Question about dynamic pressure vs pressure on plate

1. Jul 11, 2011

### kenmcelvain

Dynamic pressure, from Bernoulli's principle, for a moving fluid with a given density is
$$P = \frac{1}{2}\rho {v^2}$$
On the other hand, we can compute the force on a plate hit by that fluid from the rate of momentum transfer. Assuming that the fluid exits sideways after impact, we have
$$F = \frac{{dmv}}{{dt}} = \frac{{(Avdt)\rho v}}{{dt}} = A\rho {v^2}$$
Dividing by the area, we have the pressure
$$P = \rho {v^2}$$
I have seen a number of other postings that seem to suggest that the pressure on the plate should be equal to the dynamic pressure of the fluid before it hits the plate. I tend to trust the rate of momentum transfer. Could someone clarify why dynamic pressure is supposed to be pressure on the plate (or tell me that I am right).

Thanks!

2. Jul 12, 2011

### Andrew Mason

The pressure in a pipe carrying a fluid is the sum of the static and dynamic pressures. This is the pressure on the walls of the pipe carrying the fluid - the pressure in the direction perpendicular to the direction of flow. Who is saying that this is the same as the pressure of the fluid striking a vertical plate - eg. a fire hose spraying a wall?

The product rule for differentiation is:

$$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = v\frac{dm}{dt} + m\frac{dv}{dt}$$

The liquid does not stop instantaneously so I think you have to take the m(dv/dt) term into account. It is complicated. Does the area over which the change in velocity occurs remain the area of the pipe/hose? Does it not spread out?

AM

Last edited: Jul 13, 2011
3. Jul 13, 2011

### Studiot

You are right to consider the momentum balance equation correct.

Bernouilli's equation is an energy balance and includes all the energy in the moving fluid.

The fluid is not brought to rest at any point so possesses energy.

Unless the plate moves the pressuure force exerted by the fluid does no work so no energy is tansferred so the application of Bernoulli does not help.

Momentum is a vector and you have only stated the momentum balance in the x direction.
Here the momentum is reduced to zero and it is this which produces the pressure on the plate.

Consideration of momentum balance in the y direction (since the fluid is now moving in that direction) would yield the new pressures in the fluid. However this requires further information about the fluid stream whjich spreads sideways and also about the shear forces that now come into play between the fluid and the plate.

4. Jul 13, 2011

### kenmcelvain

5. Jul 13, 2011

### cjl

In an inviscid, incompressible flow, Bernoulli always applies, whether the object is moving or not. That having been said, the assumptions made in a problem can make a large difference in the results. For example, in the case of a jet impinging on a plate (where the plate is much larger than the jet, such that all the fluid exits sideways), a momentum balance is by far the easiest way to solve the problem. To correctly apply bernoulli, you need to know details of the flow that are somewhat challenging to compute, while a simple momentum balance gives the answer in one straightforward step.

Basically, the reason you get two different answers is because you start from two separate assumptions about how the flow behaves. If you simply have a plate submerged in a flow, you can't assume that a region of fluid the size of the plate is completely stopped by the plate, since the flow will actually flow around the plate. The only way to completely stagnate the flow is if the plate is much larger than the fluid jet.

In addition, the pressure distribution on the plate will never be even - in the case of a plate fully submerged in a flow (perpendicular to the flow), it will have a relatively smooth gradient from the center of the plate (which will be at approximately the stagnation pressure) to the edges (which will be at approximately the ambient pressure). In the case of a jet, there will be a large high pressure region where the jet impinges on the plate, surrounded by relatively negligible pressure where the fluid is simply escaping to all sides.

In general, for fluids problems, there are a number of ways to get the solution, however some methods are much easier than others. In addition, you have to be careful to have the same assumptions and boundary conditions in each case, as if you aren't careful, methods can appear to disagree (as seen here) when in reality, it was just a lack of care with the setting up of the problem.

6. Jul 13, 2011

### rcgldr

On a related subject, a pitot tube stops a portion of the flow of air (at speeds where compression of the air is a factor), and the pressure inside the pitot tube is normally termed "impact pressure". For incompressable flows (which air approximates at speeds below Mach .3), the term dynamic pressure and a simpler formula is used. (Note some use the term dynamic pressure for compressable flows as well).

Pitot_Calculation_from_impact_pressure

http://en.wikipedia.org/wiki/Impact_pressure

http://en.wikipedia.org/wiki/Dynamic_pressure

http://en.wikipedia.org/wiki/Stagnation_pressure

Last edited: Jul 13, 2011
7. Jul 13, 2011

### cjl

I've usually heard that called dynamic pressure rather than impact pressure (actually, that's the first time I've heard the term "impact pressure", and I've taken quite a few courses in fluid dynamics). However, it is true that a pitot is specifically designed to capture as close to the true stagnation pressure as possible.

8. Jul 13, 2011

### rcgldr

I was updating my previous post. Dynamic pressure is normally used for incompressable flows, and impact pressure for compressable flows, but some use the term dynamic pressure for compressable flows. My guess is the main reason for the term impact pressure is that it makes is clear the more complicated formula is being used.

Still this doesn't solve the conflict beween dynamic pressure and momentum formula from the original question. I recall some post or web site that explained this, but can't find it now.

Last edited: Jul 13, 2011