Question

actual application of hydrostatic law in field work?

Answer 1

if u like it then reward me with brainley
A Hydrostatics Law state that rate of increase of pressure in a vertically downward direction in fluid/liquid is equal to weight density of the liquid. To prove that

Consider a three dimensional fluid element at rest along with the forces acting on it as shown in fig.



Weight of the element = w(dx∗dy∗dz)w(dx∗dy∗dz)

Px=pressure acting on the left face along x direction.

Px+∂p∂xdxPx+∂p∂xdx= pressure acting on the right face …..(as per Taylor expansion)

For equilibrium of the element; 

∑Fx=0Px(dy×dz)=[Px+∂px∂x.dx].dy.dz∑Fx=0Px(dy×dz)=[Px+∂px∂x.dx].dy.dz

Divide above equation throughout by ‘dy×dz′‘dy×dz′we get

px=px+∂px∂x.dx∴∂px∂x.dx=0But  dx≠0∴∂px∂x=0…………….(1)px=px+∂px∂x.dx∴∂px∂x.dx=0But  dx≠0∴∂px∂x=0…………….(1)

Similarly for ‘z’ direction we can write;

∴∂pz∂z=0……..............(2)∴∂pz∂z=0……..............(2)

From (1) and (2) it can be concluded that there is no change in pressure in x and z direction at the same level or height. Hence ‘pressure at the same level in the fluid at rest remains constant’. Now for equilibrium in y-direction;

∑Fy=0∑Fy=0

py.(dx×dz)=[py+∂pydy.dy].dx.dz+w.(dx×dy×dz)py.(dx×dz)=[py+∂pydy.dy].dx.dz+w.(dx×dy×dz)

Divide above equation throughout by ‘dx×dz′‘dx×dz′we get ;

Therefore , py=[py+∂pydy.dy]+w.dy∂pydy.dy+w.dy=0∂pydy+w=0py=[py+∂pydy.dy]+w.dy∂pydy.dy+w.dy=0∂pydy+w=0

Substitute w=ρ.g in above equation we get ;

 ∂pydy+ρ.g=0∂pydy=−ρ.g………………….(3)∂pydy+ρ.g=0∂pydy=−ρ.g………………….(3)

Since py=f(y)py=f(y) hence using perfect differentiation in relation (3)

Separating variables we get;

dp=−ρ.g.dydp=−ρ.g.dy

integrating above equation we get;

p=−ρ.g.y+c……………………………..(4)p=−ρ.g.y+c……………………………..(4)

equation (4) is known as basic hydrostatic equation or hydrostatic law.