Relative equilibrium of Relative equilibrium of horizontally moving liquidhorizontally moving liquid

• Liquid in vessel moving in horizontal direction with acceleration (see Fig.3.12) is acted by gravity and inertial forces, characterised by accelerations ax and g and also

by surface forces.

g

X

Z

po

u

axFig. 3.12 Relative

equilibrium of a liquid moving with

acceleration in horizontal direction

If axis Z is directed upward and axis X – in vessel motion direction, in expression of (3.12) ax=-ax, ay=0 and az=-g.

From it follows such equation of equipotential surface 

dU=-axdx-gdz=0,

or -xax-zg=const.

x=0 when y=0, from what follows const=0. Thus equipotential surface equation receives such expression:

.xg

az x

• It is equation of inclined plane. Free surface of liquid represent one of equipotential surfaces and has shape of

plane, inclined by angle (see Fig. 3.12).

The main law of hydrostatics (3.9) for this case receives shape

or

const.1

-x pzgxa

,1

gdz-dxa x dp

g

axarctang

Constant of integration may be received from condition:

p=po when x=z=0.

From there follows and

  

Solution the equation with respect to p gives result: 

(3.14) Received formula suits for computation of pressure of

liquid acted simultaneously by gravity and inertial forces.

.11

0pp-zgxax

0

1-const p

.pp o

x

g

azg x

Relative equilibrium of Relative equilibrium of rotating liquidrotating liquid

• Liquid in rotating vessel is acted by centrifugal and gravity forces (Fig. 3.13). Centrifugal force may be characterised by centrifugal acceleration ac=2r,

components of which along axis x and y are ax=2x and

ay=2y. The liquid is acted also by gravity force, which

is characterised by gravity acceleration along axis z, i.e. az=-g.

Equation of equipotential surface (3.13) in this case obtains expression: 2xdx+2ydy—gdz=0 solution of which leads to: 

const.2

1

2

1 2222 gzyx

X

Y

Z

zo

po

X

Fig. 3.13 Relative

equilibrium of fluid rotated

around vertical axis

For point on axis on free surface of the liquid x=y=0 and z=zo (Fig. 3.13) and const=-gzo. Now equipotential surface

obtains shape: 

or 

It is equation of parabolic. Free surface of the liquid being one of equipotential surfaces has shape of concave meniscus with the lowest point in the middle and the highest in periphery.

ogzgzyx

2

)( 222

.

2

222

0 g

yxzz

• Applying indicated acceleration components to the main law of hydrostatics leads to: 

dpgdzydyxdx

122

or

where for x=y=0, z=zo and p=po .0

p

gzconst

const,2

1

2

1 2222

pgzyx (3.15)

• Using this constant in (3.15) and solving it with respect to p leads to: 

(3.16) 

where and 2 r2 = u2. Here r

is radius or distance from revolution axis, u is linear velocity of revolution.

222 ryx

,

2

222

00

g

yxzzgpp