Pressure in stationary andPressure in stationary and moving fluidg

Lab-On-Chip: Lecture 2Lab On Chip: Lecture 2

Lecture plan

• what is pressure and how it’s distributed in a s p essu e a d o s d s bu edstatic fluid

ater press re in engineering problems• water pressure in engineering problems• buoyancy and archimedes law; stability of y y y

floating bodies• fluid kinematics 2nd Newton law for fluid• fluid kinematics. 2nd Newton law for fluid

particles. • Bernoulli equation and its application

Fluid Statics

• No shearing stressg• No relative movement between adjacent fluid particles, i.e. static or j pmoving as a single block• Main question: How pressure is q pdistributed through the fluid

Pressure at a point

δ δ δNewton’s second law:

Question: How pressure depends on the orientation of a plane in fluid?

sin2

cos

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

x y z x y zF

δ δ δδ δ δ δ θ ρ

δ δ δ δ δ δδ δ δ δ θ γ ρ

= − =∑

∑ cos2 2z z s zy yF p x y p x s aδ δ δ δ θ γ ρ= − − =∑

iδ δ θ δ δ θcos sin

2y s y

y s z syp p pa

δ δ θ δ δ θδ

= =

− =

( )2

, , 0, ,

z s z

y s z s

zp p pa g

if x y z p p p p

δρ

δ δ δ

− = +

→ = =

• Pascal’s law: pressure doesn’t depend on the orientation of

y s z sf y p p p p

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

Basic equation for pressure fieldQuestion: What is the pressure distribution in liquid in absence shearing stress variation from point to point

• Forces acting on a fluid element:

– Surface forces (due to pressure)

– Body forces (due to weight)

Surface forces:

( ) ( )2 2y

p y p yF p x z p x zy yδ δδ δ δ δ δ∂ ∂

= − − +∂ ∂

Surface forces:

ypF y x zypF y x z

δ δ δ δ

δ δ δ δ

∂= −

∂∂

= −x

z

F y x zxpF y x zz

δ δ δ δ

δ δ δ δ

= −∂∂

= −∂

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

x y zδ δ δ δ∂ ∂ ∂

= − + +∂ ∂ ∂

Resulting surface force in vector form:

Fi j k px y z y x z

δδ δ δ

∂ ∂ ∂∇ = + + = −∇

∂ ∂ ∂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 maδ δ δ− =

p y x z g y x z k g y x z aδ δ δ ρ δ δ δ ρ δ δ δ−∇ − = −

p gk gaρ ρ−∇ − = −General equation of motion for a fluid without shearing stressesshearing 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 staticsSame pressure –much higher force!

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

• Pressure depends on the depth in the solution

e.g. in hydraulic lifts

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

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

i t tinto account:

nRTp RTρ= =p RTV

dp gpgd RT

ρ

ρ= − = − ~8 km

/1 22 1 0

( )assuming exp ; ( ) h H

dz RTg z zT const p p p h p e

RT−⎡ ⎤−

= ⇒ = =⎢ ⎥⎣ ⎦0RT⎣ ⎦

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

p h pγ +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. Ai t b th id f th ll d ill l• Air pressure acts on both sides of the wall and will cancel.

h2R 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 wall2H H( )2

0 2RA

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= =∫

Centroid (first moment of the area)

• Momentum of force acting on the o e u o o ce ac g o ewall

( )3H HF y gh ydA g H y y bdy g bρ ρ ρ= = − ⋅ =∫ ∫ ( )

0 63

R RA

R

F y gh ydA g H y y bdy g b

y H

ρ ρ ρ

=

∫ ∫

2dA∫• Generally,2

AR

c

y dAy

y A=∫

Pressure force on a curved surface

Buoyant force: Archimedes principleh b d i t t ll ti ll b d fl id• 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( )BF F F W

F F g h h Aρ= − −− = −

[ ]2 1 2 1( ) ( )BF g h h A g h h A Vρ ρ= − − − −

BF gVρ=B t f i ti th t idBuoyant force is acting on the centroid of displaced volume

Stability of immersed bodies

• Totally immersed bodyo a y e sed body

Stability of immersed bodies• Floating body

Elementary fluid dynamics:Elementary fluid dynamics: Bernoulli equationq

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

conductivity

F= ma

Net pressure force + Net gravity forceNet pressure force + Net gravity force

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

vsvdv ∂∂∂Acceleration along the vstsdt

as ∂=

∂∂==

v2

cce e at o a o g t estreamline:

Rvan =Centrifugal acceleration:

Along the streamline

vsvVmaF ss ∂∂

==∑ ρδδ VspVgFWF psss δθδρδδδ∂∂

−−=+=∑ )sin(s∂ s∂

∂∂ vsv

spg

∂∂

=∂∂

−− ρθρ )sin(

Balancing ball

Pressure variation along the streamlineC id i i id i ibl t d fl l th• 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 t i t B AA to point B. Assume: 3

0 31 aV Vx

⎛ ⎞= +⎜ ⎟

⎝ ⎠

p vvs s

ρ∂ ∂= −

∂ ∂

Equation of motion:

s s∂ ∂

3 323 1v a av v⎛ ⎞∂

= − +⎜ ⎟0 3 4

3 2 30

3 1

3 1

v vs x x

a vp aρ

= − +⎜ ⎟∂ ⎝ ⎠⎛ ⎞∂

= +⎜ ⎟4 3

63 2 3 320

1

3 11a

x x x

a v a a ap dx vρ ρ−

= +⎜ ⎟∂ ⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞Δ + +⎜ ⎟⎜ ⎟ ⎜ ⎟∫ 0

04 3 312

p dx vx x x xρ ρ

−∞

Δ = + = − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠∫

Raindrop shape

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

Bernoulli equationv

sv

spg

∂∂

=∂∂

−− ρθρ )sin(Integrating

dzds

, n=const along streamlinedpds

212

dz dp dvgds ds ds

ρ ρ− − =

We find 0)(21 2 =++ gdzvddp ρρ Along a streamline

t++ 21Along a streamlineAssuming constgzvp =++ ρρ 2

2Along a streamline

B lli ti

gincompressible flow:

Bernoulli equation

Example: Bicycle• Let’s consider coordinate system fixed to the bike.

Now Bernoulli equation can be applied to

21012 2

vpp ρ=−

Pressure variation normal to streamlinePressure variation normal to streamline

RvVmaF nn

2

ρδδ ==∑ VpVgFWF pnnn δθδρδδδ∂∂

−−=+=∑ )cos(Rnn ρ∑ np ∂∑

Rv

np

dndzg

2

ρρ =∂∂

−−

constgzdnvp ++ ∫ ρρ2

Across streamlinesconstgzdnR

p =++ ∫ ρρ

compare1

Across streamlines

constgzvp =++ ρρ 2

21

Along a streamline

Free vortex

Example: pressure variation normal to streamlineL t’ id 2 t f ti ith th l it• Let’s consider 2 types of vortices with the velocity distribution as below:

lid b d free vortexsolid bodyrotation

free vortex

Rvp

ddzg

2

ρρ =∂∂

−−Rndn ∂

,∂ ∂= −

∂ ∂as

22

1p V C rρ ρ∂= =

2213

Cp Vρ ρ∂= =,

n r∂ ∂ 1r rρ

∂ 3r r rρ

∂

21 1 1p C pρ⎛ ⎞

= +⎜ ⎟( )2 2 21p C r r pρ= + 2 02 202

p C pr r

ρ= − +⎜ ⎟⎝ ⎠

( )1 0 02p C r r pρ= − +

1 constgzvp =++ ρρ 2

21

Static, Stagnation, Dynamic and TotalPressure

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

constgzvp =++ ρρ 21 gp ρρ2

d i

hydrostatic pressure,

static pressure,point (3)

dynamic pressure,

point (3)

21 vpp ρ+12 2vpp ρ+=

Stagnation pressureVelocity 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 p"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 waterAs 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 p p g p (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 = 222111 ρρ

Determine the flow rate to keep the height constant

22 11 gzvpgzvp ρρρρ ++=++ 222111 22gzvpgzvp ρρρρ ++=++

vAvAQ == 2211 vAvAQ ==

Venturi channel

Measuring flow rate in pipes

222

211 2

121 vpvp ρρ +=+

2211 vAvAQ ==

Restriction on use of Bernoulli equation

• Incompressible flowco p ess b e o• Steady flow• Application of Bernoulli equation across the

stream line is possible only in irrotational flowp y• Energy should be conserved along the

streamline (inviscid flow + no active devices)streamline (inviscid flow + no active devices).

Probelms• 2.24 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. The initial differential reading is 30cm y gas shown.

• 2.39 An open tank contains gasoline ρ=700kg/cm at a depth of 4m The gate is 4m high and 2m wide Water is slowlyof 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• 3.29 The circular stream of water from a

faucet is observed to taper from a ( f )diameter 20 mm (at the faucet) down to

10 mm in a distance of 50 cm. Determine the flow rateDetermine the flow rate.

• Water flows through a pipe contraction as shown g p pbelow. Calculate flowrate as a function of smaller pipe diameter for both manometer configuration