pressure in stationary and moving...

Pressure in stationary and moving fluid

Lab-On-Chip: Lecture 2

Fluid Statics

• No shearing stress•.no relative movement between adjacent fluid particles, i.e. static or moving as a single block

Pressure at a point

• Pascal’s law: pressure doesn’t depend on the orientation of plate (i.e. a scalar number) as long as there are no shearing stresses

sin2

cos2 2

y y s y

z z s z

x y zF p x z p x s a

x y z x y zF p x y p x s a

δ δ δδ δ δ δ θ ρ

δ δ δ δ δ δδ δ δ δ θ γ ρ

= − =

= − − =

∑

∑

Newton’s second law:

cos sin

2

( )2

, , 0, ,

y s y

z s z

y s z s

y s z syp p pa

zp p pa g

if x y z p p p p

δ δ θ δ δ θδ

δρ

δ δ δ

= =

− =

− = +

→ = =

Question: How pressure depends on the orientation of the plane ?

Basic equation for pressure field

• Forces acting on a fluid element::

– Surface forces (due to pressure)

– Body forces (due to weight)

Question: What is the pressure distribution in liquid in absence shearing stress variation from point to point

( ) ( )2 2y

y

x

z

p y p yF p x z p x zy y

pF y x zypF y x zxpF y x zz

δ δδ δ δ δ δ

δ δ δ δ

δ δ δ δ

δ δ δ δ

∂ ∂= − − +

∂ ∂∂

= −∂∂

= −∂∂

= −∂

Surface forces:

Basic equation for pressure field( )p p pF i j k y x z

x y zδ δ δ δ∂ ∂ ∂

= − + +∂ ∂ ∂

Fi j k px y z y x z

δδ δ δ

∂ ∂ ∂∇ = + + = −∇

∂ ∂ ∂

Resulting surface force in vector form:

If we define a gradient as:

The weight of element is: Wk g y x z kδ ρ δ δ δ− = −

Newton’s second law: F Wk ma

p y x z g y x z k g y x z a

δ δ δ

δ δ δ ρ δ δ δ ρ δ δ δ

− =

−∇ − = −

p gk gaρ ρ−∇ − = −General equation of motion for a fluid w/o shearing stresses

Pressure variation in a fluid at rest

• At rest a=0 0p gkρ−∇ − =

0 0p p p gx y z

ρ∂ ∂ ∂= = = −

∂ ∂ ∂

• Incompressible fluid

1 2p p ghρ= +

Fluid statics

• Pressure depends on the depth in the solution not on the lateral coordinate

Fluid equilibrium Transmission of fluid pressure, e.g. in hydraulic lifts

Same pressure –much higher force!

Compressible fluid• Example: let’s check pressure variation in the air (in

atmosphere) due to compressibility:

– Much lighter than water, 1.225 kg/m3 against 1000kg/m3 for water

– Pressure variation for small height variation are negligible– For large height variation compressibility should be taken

into account:

/1 22 1 0

0

( )assuming exp ; ( ) h H

nRTp RTV

dp gpgdz RT

g z zT const p p p h p eRT

ρ

ρ

−

= =

= − = −

⎡ ⎤−= ⇒ = =⎢ ⎥

⎣ ⎦

~8 km

Measurement of pressure• Pressures can be designated as absolute or gage (gauge) pressures

atm vaporp h pγ= +very small!

Hydrostatic force on a plane surface• For fluid in rest, there are no shearing stresses present and

the force must be perpendicular to the surface. • Air pressure acts on both sides of the wall and will cancel.

2R avhF p A g bhρ= =Force acting on a side wall in rectangular container:

Example: Pressure force and moment acting on aquarium walls

• Force acting on the wall

( )2

0 2

H

RA

HF gh dA g H y bdy g bρ ρ ρ= = − ⋅ =∫ ∫

Centroid (first moment of the area)

• Generally: sin sinR cA

F g ydA g y Aρ θ ρ θ= =∫

• Momentum of force acting on the wall

( )3

0 63

H

R RA

R

HF y gh ydA g H y y bdy g b

y H

ρ ρ ρ= = − ⋅ =

=

∫ ∫

• Generally,2

AR

c

y dAy

y A=∫

Pressure force on a curved surface

Buoyant force: Archimedes principle• when a body is totally or partially submerged a fluid

force acting on a body is called buoyant force

[ ]

2 1

2 1 2 1

2 1 2 1

( )( ) ( )

B

B

F F F WF F g h h AF g h h A g h h A V

ρρ ρ

= − −− = −

= − − − −

BF gVρ=Buoyant force is acting on the centroidof displaced volume

Stability of immersed bodies

• Totally immersed body

Stability of immersed bodies• Floating body

Elementary fluid dynamics: Bernoulli equation

Bernuolli equation – ”the most used and most abused equation in fluid mechanics”

Assumptions:• steady flow: each fluid particle that passes through a given

point will follow a the same path • inviscid liquid (no viscosity, therefore no thermal

conductivity

F= maF= ma

Net pressure force + Net gravity force

Streamlines: the lines that are tangent to velocity vector through the flow field

vsv

ts

sv

dtdvas ∂

∂=

∂∂

∂∂

==

Rvan

2

=

Acceleration along the streamline:

Centrifugal acceleration:

Streamlines

vsvVmaF ss ∂∂

==∑ ρδδ VspVgFWF psss δθδρδδδ∂∂

−−=+=∑ )sin(

vsv

spg

∂∂

=∂∂

−− ρθρ )sin(

Along the streamline

Balancing ball

Pressure variation along the streamline• Consider inviscid, incompressible, steady flow along the

horizontal streamline A-B in front of a sphere of radius a. Determine pressure variation along the streamline from point A to point B. Assume: 3

0 31 aV Vx

⎛ ⎞= +⎜ ⎟

⎝ ⎠

p vvs s

ρ∂ ∂= −

∂ ∂

Equation of motion:

3 320 3 4

3 2 30

4 3

63 2 3 32004 3 3

3 1

3 1

3 112

a

v a av vs x x

a vp ax x x

a v a a ap dx vx x x x

ρ

ρ ρ−

−∞

⎛ ⎞∂= − +⎜ ⎟∂ ⎝ ⎠

⎛ ⎞∂= +⎜ ⎟∂ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞∆ = + = − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠∫

Raindrop shape

The actual shape of a raindrop is a result of balance between the surface tension and the air pressure

vsv

spg

∂∂

=∂∂

−− ρθρ )sin(Integrating

dzds

We find 0)(21 2 =++ gdzvddp ρρ

constgzvp =++ ρρ 2

21

Along a streamline

Along a streamline

Bernoulli equation

Bernoulli equation

, n=const along streamlinedpds

212

dz dp dvgds ds ds

ρ ρ− − =

Assuming incompressible flow:

Example: Bicycle

2012 2

1 vpp ρ=−

• Let’s consider coordinate system fixed to the bike.Now Bernoulli equation can be applied to

Pressure variation normal to streamline

RvVmaF nn

2

ρδδ ==∑ VnpVgFWF pnnn δθδρδδδ∂∂

−−=+=∑ )cos(

Rv

np

dndzg

2

ρρ =∂∂

−−

constgzdnRvp =++ ∫ ρρ

2

compare

constgzvp =++ ρρ 2

21

Across streamlines

Along a streamline

Free vortex

Example: pressure variation normal to streamline

Rv

np

dndzg

2

ρρ =∂∂

−−

• Let’s consider 2 types of vortices with the velocity distribution as below:

,n r∂ ∂= −

∂ ∂as

22

1p V C rr r

ρ ρ∂= =

∂

2213

Cp Vr r r

ρ ρ∂= =

∂

solid bodyrotation

free vortex

22 02 2

0

1 1 12

p C pr r

ρ⎛ ⎞

= − +⎜ ⎟⎝ ⎠

( )2 2 21 0 0

12

p C r r pρ= − +

constgzvp =++ ρρ 2

21

Static, Stagnation, Dynamic and TotalPressure

212 2

1 vpp ρ+=

Stagnation pressure

• each term in Bernoulli equation has dimensions of pressure and can be interpreted as some sort of pressure

constgzvp =++ ρρ 2

21

static pressure,point (3)

dynamic pressure,

hydrostatic pressure,

Velocity can be determined from stagnation pressure:

On any body in a flowing fluid there is a stagnation point. Some of the fluid flows "over" and some "under" the body. The dividing line (the stagnation streamline) terminates at the stagnation point on the body.As indicated by the dye filaments in the water flowing past a streamlined object, the velocity decreases as the fluid approaches the stagnation point. The pressure at the stagnation point (the stagnation pressure) is that pressure obtained when a flowing fluid is decelerated to zero speed by a frictionless process

Pitot-static tube

Steady flow into and out of a tank.

222111 vAvA ρρ =

22

2212

11 21

21 gzvpgzvp ρρρρ ++=++

2211 vAvAQ ==

Determine the flow rate to keep the height constant

Venturi channel

Measuring flow rate in pipes

222

211 2

121 vpvp ρρ +=+

2211 vAvAQ ==

Probelms• 2.43 Pipe A contains gasoline (SG=0.7), pipe B contains oil

(SG=0.9). Determine new differential reading of pressure in A decreased by 25 kPa.

• 2.61 An open tank contains gasoline r=700kg/cm at a depth of 4m. The gate is 4m high and 2m wide. Water is slowly added to the empty side of the tank. At what depth h the gate will open.

Problems• 3.71 calculate h. Assume water inviscid and

incompressible.