J. Szantyr – Lecture No. 3 – Fluid in equilibrium

J. Szantyr – Lecture No. 3 – Fluid in equilibrium

Internal forces – mutual interactions of the selected mass elements of the analysed region of fluid, forces having a surface character, forming pairs acting in the opposite directions, thus reducing to zero.

External forces – the result of action of masses which do not belong to the analysed region of fluid – they may be divided into mass forces and surface forces.

Mass forces act on every fluid element and they are proportional to Mass forces act on every fluid element and they are proportional to its mass.

τ ρ

τ

ρ

d

F d F

m F F

m

= ′

∆

∆ ′

∆ =

∆ ′

=

lim 1 lim 1

0 0

 

  s

F m



 m3

ρ kg density of fluid

Surface forces act on the surface surrounding the selected region of fluid and they are proportional to the area of this surface.

ds p d s

P p

s

= ′

∆

∆ ′

=

lim



 m2

P N unit surface force

In general the surface force depends on the orientation of the surface element, defined by the unit length normal vector n, thus it should be symbolized by:

P

The fluid remains in equilibrium under the action of the given external forces if the forces acting on an arbitrarily selected part of the fluid form the system of vectors equivalent to zero.

In the fluid in the state of equilibrium the pressure in an

arbitrary point has a constant value and it does not depend on the orientation of the surface element passing through this point.

Equilibrium conditions of the tetrahedron:

( ^, ) ⁰

cos =

− pdS p x dS

p

( ^, ) ⁰

cos =

− pdS p y dS

p

( ^, ) ⁰

cos =

− pdS p z dS

p ^{dS − pdS cos} ( ^{p , z} ) ^{= 0}

p

but we have:

^dS

^{= dS cos} ( ^{p , x} )

0

; 0

;

0 − = − =

=

− p p p p p

p

p = p

= p

= p

Conclusion: the hydrostatic state of stress in the fluid has the character of a scalar field.

Conditions of the fluid equilibrium Unit mass force:

( ^{x y z} )

F k

Z j

Y i

X

F = + + = , ,

Density:

( ^{x , y , z} )

ρ ρ =

Conditions of equilibrium of the fluid element:

element:

= 0

 

 





∂ + ∂

−

+ dx dydz

x p p

pdydz dxdydz

X ρ

= 0

 



 



∂ + ∂

−

+ dy dxdz

y p p

pdxdz dxdydz

Y ρ

= 0

 

 





∂ + ∂

−

+ dz dxdy

z p p

pdxdy dxdydz

Z ρ

hence we obtain:

x X p

∂

= ∂ ρ 1

y Y p

∂

= ∂ ρ

1

z Z p

∂

= ∂

ρ

what leads to the basic hydrostatic Euler equation:

gradp F

ρ

= 1

or in the differential form:

ρ

Ydy

Xdx + + =

ρ

if the mass force field has the potential U, such that:

ρ

dU = − dp

p = − ρ U + C

The integration constant may be determined from the given pressure and mass force field potential in the selected point in the fluid.

then we obtain:

gradU

F = −

For example in the gravitational field near the Earth we have X=Y=0

z g U

Z ∂

− ∂

=

−

= or: U = gz what gives: p = −

ρ

Conclusion: in the gravitational field of the Earth the surfaces of constant hydrostatic pressure (isobaric surfaces) are horizontal.

If by we denote the pressure on the fluid free surface at the elevation H we get:

pa

General conclusion: isobaric and equipotential surfaces are

perpendicular to the vector of mass forces (see example at the end)

elevation H we get:

C gH

p

= − ρ +

C ⁼ p

⁺ ρ gH

( ^{H z} )

g p

p =

+ ρ −

p = p

+ ρ gh

h=H-z – immersion of the point under free surface

p

Examples of application Connected vessels – at the level A-B we have:

1 1

gh p

p =

+ ρ ^p = ^p

+ ρ

^gh

czyli:

2 2 1

1

h ρ h

ρ =

1 2

1 ²

ρ

= ρ h

h

Hydrostatic measurement of pressure Hydrostatic measurement of pressure

p

p

= ^p

^{= p}

⁺ ρ ^{g ∆ h}

B

A

p

p = ^{p − p}

⁼ ρ ^{g ∆ h}

In this way we measure the overpressure (relative pressure) in the container, or the difference between the absolute pressure p and the atmospheric pressure.

Barometer – measurement of the atmospheric pressure.

gh p

= ρ

Pascal theorem

Increase of pressure in an arbitrary point of the homogenous incompressible fluid in the state of

equilibrium, placed in the potential mass force field, results in the same increase of pressure in any other results in the same increase of pressure in any other point of the fluid.

( ^{U U} )

p

p −

= ρ

−

( ^{p p} ) ( ^{U U} )

p

p + δ −

+ δ

= ρ

−

0

= 0

− p p δ δ

p

p δ

δ =

Example No.1: Determination of the inclination of the liquid free surface in the container moving in a straight constantly accelerated motion in an arbitrary direction.

a

a = a

projections of the unit mass force:

= 0 X

β cos a

Y = −

β

sin a

g Z = − −

liquid equilibrium equations:

1 0

∂ =

∂ x p ρ

ρ ^cosβ

1 a

y

p = −

∂

∂

ρ ^sin β

1 g a

z

p = − −

∂

∂

or:

^{dp = −} ( ^a β ^{dy + a} β ^{dz + gdz} )

ρ ^{cos sin}

after integration with ρ=const we obtain:

1 1

sin

cos z C

g y a

g g a

p  +



 











 +

+

−

=

ρ β β

the constant is determined for the pressure at the free surface in the point M1:



  

( )



 



  +







 +

+ +

= z t

g y a

g g a p

C_{1 a} ρ cosβ sin β 1

after substituting we get:

g t g a

p

p _a 







 + +

=

ρ

β

the equation of the isobaric (constant pressure) surfaces has the form:

C y

g

z a +

+

−

=

β β sin cos

it describes a family of planes inclined at an angle α such that:

β α β

sin cos

+

−

=

g tg a

g

but the angle of inclination of the resultant mass force φ is:

β α β

ϕ g ctg

a Y

tg Z =

+

=

= cos

sin

Conclusion: the resultant mass force is perpendicular to the isobaric surfaces.

Example No. 2: Determination of the relation describing the pressure distribution in a tank rotating with constant angular velocity ω. The tank is filled with liquid having density ρ, and the ambient pressure is equal p.

In the cylindrical system of co-ordinates the basic hydrostatic equation has the form:

dz q

d r q dr

dp q

⋅ +

⋅

⋅ +

⋅

= q dr q r dϑ q dz dp

z

r ⋅ + ⋅ ⋅ + ⋅

= ϑ

ρ ^ϑ

where the respective terms are equal to:

r q

= ω

⋅

= 0 q

g

q

= −

After substitution we get:

^{dp = ρ ⋅} ( ^ω

^{⋅ r ⋅ dr − g ⋅ dz} )

Integration leads to:

p = ρ ⋅ ω

⋅ r

− ρ ⋅ g ⋅ z + C 2

For the liquid surface point at the tank axis we have:

= 0

r z = z

^p = ^p

Hence the integration constant is equal to:

C = p + ρ ⋅ g ⋅ z

Hence the integration constant is equal to:

C = p

+ ρ ⋅ g ⋅ z

Finally, the pressure distribution in the liquid is described by equation:

(

)

0 g z z 2 r

p

p = − ⋅ ⋅ − + ρ ⋅ω ⋅ ρ

Example No. 3: Three pistons of areas A1=0,6 m**2, A2=0,8 m**2, A3 0,4 m**2, respectively loaded with forces P1=1 kN, P2=2 kN i P3=3 kN, act on water having density ρ=1000

kg/m**3. Determine at which elevations h1 i h2 the system of pistons remains in equilibrium.

Pressure under piston 2 is:

2 2 1

1 1

A g P

A h

P + ⋅

ρ

Pressure under piston 3 is:

3 3 2

2 2

A g P

A h

P + ⋅

ρ

The elevations are determined from the above equations:

[ ]

P

h P_{2 1} 1 0.085

1 =

⋅ ⋅







 −

=

ρ ^h

^P

^P

^{1 = 0 . 51} [ ] ^m

⋅ ⋅







 −

= ρ

[ ]

g A

h A 0.085

1 2

1 =

⋅ ⋅



 −

=

ρ [ ] ^m

g A

h A 0 . 51

2 3

2

=

⋅ ⋅

 

  −

= ρ

Correctness of the solution may be checked using the equilibrium equation referring to pistons 1 and 3:

( ) ( )

4 . 0 51 3000 .

0 085 . 0 81 . 9 6 1000

. 0 1000

3 3 2

1 1

1 + ⋅ ⋅ + = → + ⋅ ⋅ + = → =

A h P

h A g

P ρ