Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

Please explain this clearly to understand.

Pressure act at the centre of the rectangle because of the centre of gravity point so H =h/2,
Pressure= ρ* gH,
W =ρ*g,
w=W/h.
Answer=> Pressure = wh2/2.

The integration of p=[wH] with limit 0 to H because water pressure increases for 0 level to H depth. This way we can calculate the total pressure on the wall ie wH^2/2.

P = F/A.
P = (m.a)/(A).
P = (m.a.h)/(A.h).
P = (m.a.h)/(V^3).
P = p.a.h As p = (m)/(V^3).
P = pgh As a = g.
P = w.h As w = pg.

Now for vertical wall immersed in a fluid.
P = w.A.x.
P = w.A.H/2 x = H/2 for rectangular object.
Now pressure per unit length is given by;
P = (w.A.H)/(2L).
P = (w.A)/(2) As H = L.
P = (w.H^2)/(2) As A = H^2.

Where,
P, pressure.
m, mass.
a, acceleration.
g, gravitational acceleration.
w, specific weight of liquid.
h, height.
p, density of liquid.
w, specific weight of liquid.
x, Depth of the center of gravity of the immersed surface from the liquid surface.

Total pressure = Rho*g*A*H/2.

But pressure/length = Rho*g*(H*l)*H/(2*l) (because A=H* length).

= Rho*g*(H^2)/2 = w*g*(H^2)/2.

It should be total force per unit length instead of pressure.

Total pressure, p = row*g*A*h.

H = Distance from free surface to CG of the body.

Here, h = H/2.

Load will be maximum at bottom and is equal to WH.

Minimum at top = 0.

So distribution is triangular, total load is area of triangle.

= 1/2*WH*H.

Consider the fluid pressure exerted on the vertical wall w (pressure per unit height) wall height H as UDL. The location where the pressure acts is, w*H*(H/2).

For vertical plane force = density*g*h^-
w = ROW*G,H^- = h/2.
f = wh/2.

Please verify anyone.