Mechanical Engineering - Hydraulics and Fluid Mechanics - Discussion

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*H*A.
Total pressure/lenth = (rho*g*H*A)/l.
H = h/2.
Then,
= (rho*g*h*A) / 2l (A= h*l)
= (rho*g*h*h*l) / 2l.
= (rho*g*h^2) / 2.
= (w*h^2) / 2 ( specific weight (w) = rho*g).

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.

For finding pressure per unit of length,
P = W*H.
Where we are finding pressure effect on unit length,
So, H = Integral H dh.
H = H^2/2.
So,
P = w * H.
P = wH^2/2.

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.

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).

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.

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.

P = Pressure;

P/h = Total pressure per unit length = integral((w*h)dh).

= (w*h^2)/2.

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

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

Here, h = H/2.