Hydrostatics and Stevin theorem 4

[dianad]

Hi,

In hydrostatics everyone learned that P=F/A
p=Pressure
F=force
A=area
From this relation we can determine for a cilinder(for example):

P=(m x g)/A = (mv x V x g)/A = (mv x A x h x g)/A = mv x h x g
m=mass (kg)
mv=specific heigth (kg/m3)
V=volume
h=height
g=gravity acceleration

So, from this late relation: P=mv.h.g , wich corresponds to STEVIN theorem.

My problem is that with Stein theorem, says that Pressure only depends on the depth and always is presented the image that i've attached.
But if we use for both images the relation P=F/A, instead of the Stevin relation, to determine the pressure at the bottom, we have the following:

m=10kg (both)
A1=2 m2
A2=4 m2

P1=10kg * 10m/s2 / 2m2 = 50 Pa
P2=10kg * 10m/s2 / 4mw = 20 Pa

If we used the relation of Stein, we would have the same pressure for both.

What is your comment about this?

Thanks

 

[BigInch]

dianad,

If you still have problems with the pump and pipe configuration, post the question in the pump engineering forum, OK?

"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[dianad]

Thanks BigInck