Relative Equilibrium( Linear Acceleration)

Relative Equilibrium( Linear Acceleration)
The components of forces on the element(dxdz) at (x & z)plan are as shown in the figure:
??F_x =((-?p)/?x) dxdz (1)
??F_z =((-?p)/?z- ?) dxdz (2)
Mass of element:
dM=?/g dxdz (3)
By applying Newton s second law of motion:
(F=Ma) we can get:
((-?p )/?x)dxdz=(?/g dxdz)ax (4)
((-?p )/?z- ?)dxdz=(?/g dxdz)az (5)
Simplifying Eqs. (4)& (5):
((-?p )/?x)=(?/g )ax
((-?p )/?z)=(?/g )(az+g) (7)
These two equations show the change in pressure at accelerated fluid.
The differential equation of pressure is:
By substituting eqs.( 6& 7) in eq.( 8):
dp= - ?/g (ax)dx-?/g ( az+g)dz (9)
On a line with constant pressure (dp=0) its equation will be:
dz/dx=-(ax/(g+az)) in x& z plan.
This equation is the slope of constant pressure line.
There is three special cases of accelerated fluid:
Vertical acceleration
There is no change in pressure in x direction( ?p/?x=0 )so Eq.(7) will be:
dp/dz=-?((az+g)/g) (10)
When (az = constant) the pressure gradient from liquid surface to the bottom will be linear as if it is static as shown in the Fig. for (az> 0 &