- #1

kenmcelvain

Gold Member

- 2

- 0

[tex]P = \frac{1}{2}\rho {v^2}[/tex]

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

[tex]F = \frac{{dmv}}{{dt}} = \frac{{(Avdt)\rho v}}{{dt}} = A\rho {v^2}[/tex]

Dividing by the area, we have the pressure

[tex]P = \rho {v^2}[/tex]

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!