Momentum and Mass transport and transfer: an introduction...

BackgroundMomentum transport Mechanisms

Moving fluid : Mass and Momentum transport equationsInternal flows, external flows and boundary layers

Momentum transfer at interfaces : the friction factorEnergy balance in a flow : Bernoulli’s theorem

Momentum and Mass transport and transfer:an introduction to fluid mechanics

Yannick Hallez

LGC-UPS

24 fevrier 2012

Yannick Hallez Momentum and Mass transport and transfer: an introduction to fluid mechanics




What is a fluid ?Microscopic or macroscopic approach ?Compressible or incompressible (static) fluidCompressible or incompressible flowLaminar or turbulent flow

Plan

1 Background

2 Momentum transport Mechanisms

3 Moving fluid : Mass and Momentum transport equations

4 Internal flows, external flows and boundary layers

5 Momentum transfer at interfaces : the friction factor

6 Energy balance in a flow : Bernoulli’s theorem






1 BackgroundWhat is a fluid ?Microscopic or macroscopic approach ?Compressible or incompressible (static) fluidCompressible or incompressible flowLaminar or turbulent flow











The microscopic point of view (physics/chemistry)

A solid is ...

... made of a set of atoms fixed on a net,apart from vibrations due thermal agita-tion. The atoms are held together by strongbonds. The structure can be ordered (crys-tal) or disordered.






The microscopic point of view (physics/chemistry)

A fluid is ... not a solid.

It is a set of atoms or molecules not strongly lin-ked together, even if they can interact at distance.They can rearrange. Some examples are : liquid,gas, plasma, molten salt, supercritical fluid... Inan ideal fluid, interactions are neglected and onlyshocks between molecules or atoms are considered(perfect gas for instance). The average distancetravelled by molecules or atoms between shocksis called the ”mean free path“. It is of the orderof the size of the molecules in liquids, and muchlarger in gases.






The macroscopic point of view (mechanics)

Response to shear stress

The material is characterised by its responseto an external shear stress. A solid will onlydeform, but atoms will remain at their po-sition relative to the other atoms. A fluidwill ”flow“, i.e. the atoms or molecules willrearrange (see MFM-p153). Another way toexpress the same idea : a liquid can alwaysfit inside a container. However, this mecha-nical distinction can be vague in some cases(a glacier for instance).






The microscopic approach

Method

Any flow problem could be solved in principle by computing the tra-jectory of every atom or molecule individually thanks to Newton’slaw. This is what ”molecular dynamics“ simulations do. Averagesover a large number of particles are then needed to recover macro-scopic quantities such as temperature, fluid velocity, pressure... Thisis however not a reallistic method fot usual systems. Example : toperform a simulation of 1g of water (molecular mass : 18 g/mol)we need to store in memory at least three values for each particleposition, stored with at least 4 bytes. Hence in the end we needapproximately a 400 billion TeraBytes RAM ! (Today’s largest su-percomputers have around 0.3 TeraByte RAM.)






The macroscopic approach

Method

The presence of individual molecules is igno-red and the fluid is considered as a ”conti-nuous media“. It is then possible to definequantities continuously in space. Thesequantities are called fields. Example : velo-city field, temperature field, pressure field,concentration field. In particular, we consi-der it is possible to define values forthese fields at any point in space. Scalar field in a numerical simulation

of a turbulent flow by Brethouwer &Nieuwstadt. Sc = 25.






The macroscopic approach

Limits of validity

If a system is considered on a large scale, there are many particleseverywhere and defining the value of a field at any point is possible.Imagine we do an average on a very small zone near the ”point“.This zone would still contain many particles and the average wouldbe meaningful.If a system is considered at a very small scale, the discrete natureof the problem cannot be ignored. At a given point we may be invacuum or in a particle, and local averages cannot be defined.Large or small scale is characterized bu the Knudsen number.






The Knudsen number

Definition of the Knudsen number

Kn =l

L,

where l is a characteristic scale of the molecular motion (mean freepath) and L is the characteristic length of the system.






Fluid compressibility

In a static situation (no flow), a pressure variation applied to a fluidelement leads to a change of its volume :

δVV

∣∣∣∣S

= −χSδP,

where we suppose the compression is adiabatic and χS is called theisentropic compressibility coefficient (or just compressibility). Anyfluid is compressible, but the model of incompressible fluid is so-metimes used and assumes that no finite pressure variation couldchange the volume of a fluid element. Example : air in a bicyclepump is compressible, whereas water in a bicycle pump could beconsidered incompressible with good accuracy.






Compressibility and speed of sound

Compressibility is responsible of the propagation of sound waves.Hence the incompressible fluid model is not adapted to treat acous-tics problems, water hammer problems...Speed of sound is given by a = 1√

ρχS. In an incompressible fluid,

since χS → 0, a → ∞ (any pressure variation on the boundary ofthe problem is transmitted insantaneously to the whole fluid !).

χS (Pa−1) a (m/s)

air 10−5 330water 5 10−10 1400






Compressible or incompressible flow

The Mach number

The Mach number is defined as the ratio of the fluid velocity to thevelocity of sound :

M =V

a

Fluid flows are said to be incompressible if M ≤ 0.2 and compressibleif M > 0.2. Hence, if M ≤ 0.2 the flow behaves as if the fluidwere an incompressible fluid. At higher Mach numbers the flow isinfluenced by compressibility effects. Examples : for air (STP), aflow can be considered incompressible if V ≤ 60 − 70m/s (200 −250 km/h). For water a flow is incompressible up to V ' 280m/s(1000 km/h). In the rest of the course, we focus in incompressibleflows.






Flow classification with Mach number






Laminar or turbulent flow

Turbulent flow : a difficult definition and problem

”I am an old man now, and when I die and go to heaven there aretwo matters on which I hope for enlightenment. One is quantumelectrodynamics, and the other is the turbulent motion of fluids.And about the former I am rather optimistic.”

Sir Horace Lamb (famous mathematician and physicist)

But we can try a definition anyway...







Turbulent flow

Flow in which any function of the movement has fluctuationsintrinsic to the movement and covering a continuous range ofspatial and temporal scales.A characteristic of turbulence is the very high sentitivity of the flowto initial conditions (butterfly effect). Even if equations of motion areknown, the impossibility to impose initial conditions exactly makesthe full solution often unknown. It has however known statistics.

Laminar flow

Flow that is not turbulent. Example : no fluctuations (MFM-202),or well-defined fluctuations (no chaos, MFM-636).







Common traps

Do not mix up unsteady and turbulent. A flow can beunsteady and laminar.

Do not assume a flow is turbulent because there is a vortexsomewhere. Laminar flows may have vortices also.

Refs : MFM-785, MFM-229 (several)





AdvectionViscous momentum diffusionMomentum transport (or rather source) due to external forcing

Plan

1 Background











Introduction

Quantity of interest

As far as momentum is concerned the quantity of interest is themomentum per unit of volume : ρu, where ρ is the fluid density andu is the fluid velocity. Momentum can be expressed in kg/s/m2 inSI units. More generally it has the dimension ML−2T −1.

Momentum flux

A momentum flux as the dimension of momentum times velocity,and is thus expressed in kg/m/s2 or Pa in SI units. More generallyit has the dimension ML−1T −2.






Mechanisms : an overview

Momentum source or sink

Pressure gradients (pumps, meteo)

Gravity/Buoyancy (rivers, free convection)

Magnetic fields (convection in the sun)

Viscous dissipation (wall friction)

Electrochemical potential gradients (ion transport)

Momentum transport mechanisms

Advection

Diffusion






1 Background

2 Momentum transport MechanismsAdvectionViscous momentum diffusionMomentum transport (or rather source) due to externalforcing




6 Energy balance in a flow : Bernoulli’s theoremYannick Hallez Momentum and Mass transport and transfer: an introduction to fluid mechanics





Transport by advection

Definition

Consider a blob of high momentum fluid(high density or high velocity or both) in-side stream with density ρ and velocity fieldu. The advective momentum flux densityfor the ith momentum component ρui (i =1, 2, 3) is Fi = ρuiu It means that thehigh momentum spot is transported froma point to another with a velocity u. It isa more complex phenomenon than heat ormass transport since momentum is transpor-ted by a velocity field that is inside its defi-nition.







Definition

The flux density of the ith component of momentum is a vectorFi = ρuiu. It is possible to define the full momentum flux with asecond order tensor F = ρu⊗ u where ⊗ is the dyadic product.

F = ρu⊗ u = ρ

u1u1 u1u2 u1u3

u2u1 u2u2 u2u3

u3u1 u3u2 u3u3

or in index notation

Fij = ρuiuj







Advective flux and inertia

Imagine the whole fluid is at rest (u = 0)apart from a fluid particle released with mo-mentum ρu 6= 0. The momentum advec-tive flux will ”create“ a zone with non-zeromomentum downstream of the fluid particleinitial position after some time and withoutany external forcing. This is a manifestationof the inertia force.







Characteristic advective time scale

If the movement of fluid takes place on a length scale L and witha velocity scale U , it is possible to define a characteristic advectivetime scale by

Ta =L

U(1)






1 Background










Newton-Stokes law

From a simple experiment

Consider some viscous fluid between two parallel plates separated bya distance h. The bottom plate is fixed while the top plate is pulledwith a constant and controlled force F . Obviously the force requiredto pull the plate depends on the surface area A of the plate. Thuswe are interested in the force per unit surface area F/A.






Newton-Stokes law

From a simple experiment

The result of the experiment is that at steady state the force perunit surface area is proportional to U/h. Hence it is possible to write

F

A= µ

U

h,

where µ is called the dynamic viscosity of the fluid. It is expressed inkg/m/s or Po (Poises) in SI units. In this experiment the velocityprofile is linear, so that U/h is actually ∂u

∂y , the gradient in thetransverse direction of the x component of the velocity field.






Newton-Stokes law

Definition of a stress

A stress is a force per unit surface area. It can be a normal force, ornormal stress, like the pressure. It can also be a tangential force, ortangential stress, like F/A in the present experiment. A tangentialstress is called a shear stress. It is often denoted τij where i isthe index of the force direction and j is the index of the surfacenormal direction. In the previous experiment F/A was actually τ12

(force exerted in the x direction and applied to surfaces with normaldirection y). The experimental result an thus be written τ12 = µ∂u1∂x2

.






Newton-Stokes law

Definition of the viscous stress tensor

Generalisation : the strain rate tensor (symmetric part of the ve-locity gradient tensor ∇u = ∂ui

∂xj) is

S =1

2

(∇u +∇uT

)or Sij =

1

2

[∂ui∂xj

+∂uj∂xi

]and for an incompressible fluid, the viscous stress tensor is

τ = 2µS = µ(∇u +∇uT

)or τij = 2µSij = µ

[∂ui∂xj

+∂uj∂xi

]This is the Newton-Stokes law.






Diffusive momentum flux

A stress has the dimension of a momentum flux...

It is also possible to define the diffusive momentum flux (due toviscous effects)

F = −τ = −2µS = −µ(∇u +∇uT

)Analogy with heat and mass transfer

Since µ = ρν where ν is the kinematic viscosity, note that thisexpression (Newton-Stokes law F = −ν

(∇(ρu) +∇(ρu)T

)) is for-

mally similar to Fourier’s law for heat transfer q = −D∇(ρcpT ) =−λ∇T and Fick’s law for mass transfer j = −D∇C. This is whyPrandtl and Schmidt numbers compare ν and D.






Diffusive momentum flux






Non-Newtonian fluids

Limitation of Newton-Stokes’ law

In the previous experiment, thesimple proportionality law betweenthe shear stress and the strainrate (gradient of velocity) is va-lid only for the so-called Newto-nian fluids. Several other fluid be-haviours exist and can be describedin a τ(∂u∂y ) plot.

0.0 0.2 0.4 0.6 0.8 1.0uy

0.0

0.5

1.0

1.5

2.0

2.5

τ

Newtonian

Shear thinning

Shear thickening

Bingham plastic






Non-Newtonian fluids

Examples of Newtonian and non-newtonian fluids

Newtonian : water, air, most common fluids

Shear thinning : shampoo, ketchup, paint, lava, blood, nailpolish

Shear thickening (or dilatant) : cornstarch, silly putty, clay,wet sand, fluids used in traction control, body armors

Bingham plastic : mayonaise, toothpaste, slurries, someyogurts






Back to Newtonian fluids

Viscosity of a Newtonian fluid

In this course, only Newtonian fluids are considered. Hence the dy-namic viscosity µ is a constant. It is possible to define the kinematicviscosity ν whose dimension is L2T −1 (m2/s in SI units). Becauseof the dimension of ν, the obvious diffusive time scale is

Td = L2/ν

Examples

ρ kg/m3 µ (kgm−1s−1) ν (m2/s)

air 1.29 1.85 10−5 14.3 10−6

water 1000 10−3 10−6

oil ' 1000 ' 1 ' 10−3






Viscosity and temperature

The dynamic viscosity represents the difficulty to move a fluidwhen a velocity gradient is applied, thus ...

...µ decreases with increasing T in a liquid

Momentum can be transfered only if particles can move and in aliquid they are quite packed so that it is difficult to trigger a mo-vement (high viscosity). If T ↗, particles are more agitated andmovement is facilitated : momentum transfer can occur more easily.

...µ increases with increasing T in a gas

In a gas momentum transfer requires shocks between particles, whichare more frequent if the temperature is higher. Hence at high T itis more difficult to move the fluid : viscosity increases.






How to measure viscosity ?

Measure of viscosity is performed with a rheometer

Plane and plane

Cone and plane

Couette’s rheometer

Pipe or capillary






Comparing momentum advection and diffusion

The Reynolds number

This is the ratio of diffusive to advective time scales

Re =TdTa

=L2/ν

L/U=UL

ν

It is also the ratio of the advective flux to the diffusive flux

Re =FaFd

=ρU2

µU/L=UL

ν

If the advective flux is much larger than the diffusive flux, Re � 1and the flow is dominated by inertia effects. If the diffusive fluxis much larger than the advective flux, Re � 1 and the flow isdominated by viscous effects.






1 Background










Momentum sources

Volume and surface forcing

Surface forcing

pressure (normal stress)shear stress (tangential stress)

Volume forcing

gravityelectromagnetism...





Total mass conservationMomentum balance : the Navier-Stokes equations

Plan

1 Background











1 Background


3 Moving fluid : Mass and Momentum transport equationsTotal mass conservationMomentum balance : the Navier-Stokes equations









From mass conservation to the continuity equation

The continuity equation

Consider the total mass transport problem as seen before. Whatplays the role of concentration (mass per unit volume) here is thedensity of the fluid. Now we express the advective flux ja = ρu andthere is no net diffusive mass flux across a given surface. Hence themass transport equation is

∂ρ

∂t+ div(ρu) = 0

It is called the continuity equation. For an incompressible fluid, ρis independant of pressure, and if T is constant ρ is also constantand the equation becomes

div(u) = ∇.u = 0Yannick Hallez Momentum and Mass transport and transfer: an introduction to fluid mechanics





From mass conservation to the continuity equation

The continuity equation : no diffusion term

Imagine a fluid at rest with a zone with higher density. Any macro-scopic displacement of the particles from the high density zone toanother neighbouring zone whould imply an average movement ofthe fluid at some time, and thus a non-zero fluid velocity. (Remindthe fluid velocity is by definition the average displacement of a setof particles around a given point per unit of time). So any diffusionof atoms or molecules from a dense to a less dense zone is takeninto account in the fluid velocity u and the only mass flux is ρu.






1 Background


3 Moving fluid : Mass and Momentum transport equationsTotal mass conservationMomentum balance : the Navier-Stokes equations









From momentum balance to Navier-Stokes equations

It is possible to write a momentum balance on ρu in a volume Vdelimited by a closed surface S. The total momentum inside thevolume V is ∫

VρudV

and so the total momentum accumulation is

∂

∂t

∫VρudV.

There are advective and diffusive momentum fluxes going in and outof the volume through the surface S. The net flux (advective anddiffusive) is ∮

S

[ρu⊗ u− µ

(∇u +∇uT

)]dS







As usual, we express that the accumulation plus divergence of fluxesthrough S is equal to the source terms, among them the momentumsource due to pressure gradients

∂

∂t

∫VρudV+

∮S

[ρu⊗ u− µ

(∇u +∇uT

)]dS =

∮S−PndS+

∫VsdV

Then using the divergence theorem and the fact that Pn = P In,where I is the identity matrix we get∫V

∂ρu

∂t+div

[ρu⊗ u− µ

(∇u +∇uT

)]dV =

∫V−div(P I)+sdV







Since the previous result is true for any volume V , we necessarilyhave

∂ρu

∂t+ div

[ρu⊗ u− µ

(∇u +∇uT

)]= −div(P I) + s

It is possible to simplify this expression for constant density ρ andusing the continuity equation for incompressible fluids ∇.u = 0

ρ

[∂u

∂t+ (∇u)u

]= −∇P + µ∆u + s

This is the Navier-Stokes equations for an incompressible fluidwith constant density and constant viscosity.







The NS equation must be supplemented by boundary conditions.Some of them are

No-slip condition on a solid wall : u = 0

Slip condition : ∂u∂n = 0 (model for liquid-gas interface)

Continuity of normal velocity at a liquid-gas interface :uin = uon

Continuity of normal and tangential stress at a liquid-gasinterface, ...






NS equation as a momentum balance or Newton’s secondlaw

ρ

[∂u

∂t+ (∇u)u

]= −∇P + µ∆u + s

The NS equation can be seen as a momentum balance as derivedhere but it is also the expression of Newton’s second law(force=mass × acceleration) on a fluid element. The left hand sideis the mass (per unit volume) times acceleration for a deformablesystem and the right hand side is made with the external forcesapplied to the fluid element. Pressure and shear stress are surfaceforces and s is then any volume force applied to the fluid. Forexample s = ρg for gravity. The term ρ(∇u)u can also be seen asan inertial force.






Non-dimensional form of the NS equations

We introduce the non-dimensional (star) variables with

u = u∗U , t = t∗T , x = x∗L, P = P ∗P0,

where U , T , L and P0 are respectively velocity, time, length andpressure scales. Replacing this in the NS equations leads to (sourceterm is dropped for simplicity)

ρ

[∂u∗

∂t∗U

T+ (∇∗u∗)u∗U

2

L

]= −∇∗P ∗P0

L+ µ∆∗u∗

U

L2

Multiplying everything by L/(ρU2) gives(L

TU

)∂u∗

∂t∗+ (∇∗u∗)u∗ = −

(P0

ρU2

)∇∗P ∗ +

( ν

LU

)∆∗u∗






Non-dimensional form of the NS equations

We recognized the Strouhal number St = TU/L, the Reynoldsnumber Re = UL/ν and we introduced the Euler number Eu =P0/(ρU

2).

1

St

∂u∗

∂t∗+ (∇∗u∗)u∗ = −Eu∇∗P ∗ +

1

Re∆∗u∗

You can forget about the Euler number in most applications, thesame way you can forget about the Strouhal number. Usually wecan drop terms from the equations according to the values of thenon-dimensional numbers but there are very good reasons toalways keep the pressure term in the NS equations forincompressible flows (but this is a long long story...). Do car gameMFM on similarity.






Large-Re limit of the NS equations (Euler’s equation)

If Re� 1, inertia dominates over viscosity and the time scale is theadvective one T = L/U so that St = 1 and the pressure scale isthe dynamic pressure P0 = ρU2 so that Eu = 1. Then the equationbecomes

∂u∗

∂t∗+ (∇∗u∗)u∗ = −∇∗P ∗

This is Euler’s equation in non-dimensional form, valid in the Re� 1limit. Area of application : aeronautics, meteorology...In dimensional form it reads (with source term)

ρ

[∂u

∂t+ (∇u)u

]= −∇P + s






Low-Re limit of the NS equations (Stokes’ equation)

If Re� 1, viscosity dominates over inertia and the time scale is thediffusive one T = L2/ν so that St = Re and the pressure scale isthe viscous stress scale P0 = µU/L so that Eu = 1/Re. Then theequation becomes

∂u∗

∂t∗= −∇∗P ∗ + ∆∗u∗

This is Stokes’ equation in non-dimensional form, valid in the Stokesflow (or creeping flow) limit Re� 1. Area of applications : microflui-dics (technology, biology), viscous flows, glacier flows, lava flows...In dimensional form it reads (with source term)

∂u

∂t= −1

ρ∇P + ν∆u +

s

ρ






Low-Re limit of the NS equations (Stokes’ equation)

Note : The Stokes’ equation is linear, and therefore if u is asolution, −u is also a solution. In other words, time has nodirection, or any flow is reversible if the external forcing is reversed.(See G.I. Taylor’s experiment MFM-232).






About laminar and turbulent flows

Laminar flows are encountered at low Reynolds numbers andturbulent flows at high Reynolds numbers. But ”high“ and ”low“has nothing to do with Re < 1 or Re > 1. The value Re = 1discriminates between creeping and inertial flows, but there areinertial (Re > 1) flows that are also laminar (”low Re“) (see wakebehind a cylinder MFM-131). Transition to turbulence in a pipe isusually around Re = 2000 (Reynolds apparatus MFM-731,MFM-686) and transition to turbulence on a flat plate is aroundRe = 105.





Poiseuille flow in a cylindrical pipeExercisesPressure gradient/Flow rate relationFlow in a porous mediumDrag and lift forces exerted on an objectFlow on a flat plate : boundary layers

Plan

1 Background











1 Background



4 Internal flows, external flows and boundary layersPoiseuille flow in a cylindrical pipeExercisesPressure gradient/Flow rate relationFlow in a porous mediumDrag and lift forces exerted on an objectFlow on a flat plate : boundary layers








Example of solution of the NS equations

We consider a steady laminar flow in a cylindrical pipe of radiusR. We use cylindrical coordinates (r, θ, x) where x is aligned withthe pipe axis. The velocity vector is u = (ur, uθ, ux). We supposethe flow is parallel, i.e. ur = 0, uθ = 0 and ux 6= 0. For simplicityreasons, we now denote ux = u and so u = (0, 0, u). In this context,the NA equations simplify greatly and can be solved. We recall themhere :

ρ

[∂u

∂t+ (∇u)u

]= −∇P + µ∆u + ρg

The time derivative is 0 because the flow is steady.







Let us write the equation in the direction er. Since ur = 0 andgr = 0 we find

∂P

∂r= 0

which means that P does not depend on r. Similarly, if we write theequation in the direction eθ we find

∂P

∂θ= 0

(P does not depend on θ), which was obvious for symmetry reasons.The continuity equation for an incompressible fluid div(u) = 0becomes 0 + 0 + ∂u

∂x = 0. So u does not depend on x. An forsymmetry reasons, u does not depend on θ.







Now we write the equation for velocity component u (equation pro-jected on ex).

ρ∇u.u = −∂P∂x

+ µ∆u− g

Now recall that only derivatives against r are non-zero, so that ∆u =1r∂∂r

(r ∂u∂r

)and ∇u = (∂u∂r , 0, 0). Since u = (0, 0, u), ∇u.u = 0. So

now the equation simplifies to

∂

∂r

(r∂u

∂r

)= r

1

µ

[∂P

∂x+ ρg

]Since P does not vary with r, we integrate once to obtain

r∂u

∂r=

1

2µ

[∂P

∂x+ ρg

]r2 + a







and a second time to obtain

u(r) =1

4µ

[∂P

∂x+ ρg

]r2 + a ln(r) + b

where a and b are two constants to be fixed with boundary condi-tions. On the axis of the tube, for symmetry reasons, ∂u

∂r

∣∣r=0

= 0 =1µ

[∂P∂x + ρg

]× 0 + a ln(0) which is possible only if a = 0. On the

wall at r = R the no-slip condition tells that u(r = R) = 0 =1

2µ

[∂P∂x + ρg

]R2 + b = 0 so that b = − 1

2µ

[∂P∂x + ρg

]R2 and the

final solution reads

u(r) = −R2

4µ

[∂P

∂x+ ρg

] [1− r2

R2

]Yannick Hallez Momentum and Mass transport and transfer: an introduction to fluid mechanics






The solution

u(r) = −R2

4µ

[∂P

∂x+ ρg

] [1− r2

R2

]is the well-known solution for the Poiseuille laminar flow in acylindrical pipe. The term R2

4µ is positive and[1− r2

R2

]> 0. So if

g = 0 (a horizontal pipe), the flow direction is the opposite of thepressure gradient direction : the flow is from high pressure zones tolow pressure zones. Note that ∂P

∂x is typically generated by a pump.







If the pipe is vertical, there are several possibilities according to thevalues of ∂P

∂x and ρg.

if ∂P∂x < −ρg (large pressure at the bottom), ∂P

∂x + ρg < 0 andu > 0 : the flow is rising against gravity.

if −ρg < ∂P∂x < 0, ∂P

∂x + ρg > 0 and u < 0 : the flow goesdownwards even if there is a larger pressure at the bottomthan at the top. The pressure gradient imposed by the pumpis not sufficient to win against gravity.

if ∂P∂x > 0, the pump pushes at the top, in the same direction

as gravity, and the flow goes downwards.






1 Background











Characteristics of a Poiseuille flow

Characteristics of a Poiseuille flow

Compute the maximum velocity in the pipe.

Compute the flowrate Q in m3/s. (hint : u is a volume fluxdensity, and Q is the volume flux)






Solution of the exercise

Since 0 ≤ r2/R2 ≤ 1, the maximum velocity is obtain for r = 0 :

umax = −R2

4µ

[∂P∂x + g

].

The flowrate (or volume flowrate) is

Q =

∫SudS =

∫ 2π

θ=0

∫ R

r=0u(r)rdrdθ

Q = −R2

4µ

[∂P

∂x+ g

] ∫ 2π

θ=0

∫ R

r=0

[1− r2

R2

]rdrdθ

Q = −R2

4µ

[∂P

∂x+ g

]2π

[r2

2− r4

4R2

]r=Rr=0

= −R2

4µ

[∂P

∂x+ g

]2πR2

4






Solution of the exercise

Hence we find the flow rate :

Q = −R4π

8µ

[∂P

∂x+ ρg

]and we can define the average velocity as U = Q/S = Q/(πR2) so

U = −R2

8µ

[∂P

∂x+ ρg

]Note that we can absorb ρg in the pressure and define an extendedpressure P ′ = P + ρgx so that ∂P

∂x + ρg = ∂P ′

∂x without loss ofgenerality.






Laminar parallel flow between two infinite plates separatedby a distance h

Laminar parallel flow between two infinite plates separated by adistance h

Recompute the solution in this geometry, the maximum velocity, theflowrate and the average velocity. Compare it to the pipe flow.






1 Background











Pressure gradient/Flowrate relation in internal flows

For a straight cylindrical pipe, we have computed the Poiseuille so-lution for a laminar flow, and deduced the flow rate

Q = −R4π

8µ

∂P ′

∂x

Now we drop the prime for clarity. We can consider ∂P∂x = constant =∆PL where L is the pipe length. Then, looking only at the norm, the

flow rate is

Q =R4π

8µL∆P

where the term in brackets is called the hydraulic conductance.






Pressure gradient/Flow rate relation in internal flows

Hydraulic conductance

For a system with an internal flow, the flow rate through the systemcan be related to the pressure drop in the system by the hydraulicconductance C

Q = C∆P

The hydraulic conductance depends on the geometry of the system(straight cylindrial pipe, straight square pipe, bend, orifice, expan-sion, contraction...), of the walls roughness, of the Reynolds number,etc.







Hydraulic resistance

The hydraulic resistance is R = 1/C, so that

Q =∆P

R

Electric analogy : ∆P is a difference of potential around the elementand Q is the electric current.Analogy with heat and mass transfer : Q is like the mass or heatfluxes J and Q. It is a volume flux (m3/s). ∆P is the driving force,like a difference of concentration or temperature for mass and heattransfer.






Energy loss

Energy loss by viscous dissipation

If we impose the flow rate, the difference of pressure between bothsides of the system corresponds to a loss of energy for the flow.Some energy has been dissipated in the system by friction betweenthe fluid and the walls and by friction within the fluid. The energyof the flow is actually converted into heat. The hydraulic resistanceis directly representative of this energy dissipation.If we impose a pressure difference between both ends of the system(with a pump for instance), a small hydraulic resistance R leads to alarge flow rate, and a large hydraulic resistance to a small flow rate.






Flow with imposed pressure or flow rate

Imposed pressure

If pressure is imposed, for example by a pump or a water tower,varying the hydraulic resistance of a system will directly affect theflow rate without danger for the network (e.g. when you open orclose a valve).

Imposed flow rate

If flow rate is imposed, for example by a volumetric pump, the pres-sure will increase if a valve is progressively closed (if R increases),and can potentially increase to infinity if the valve is closed. As aconsequence, the system will be damaged if care is not taken whena valve is closed.






1 Background











Characterisation of a porous medium

Porosity

ε =empty volume

total volume

Ex : for a random identical spheres pileup, ε = 0.32.

Compacity

C = 1− ε =solid volume

total volume






Characterisation of a porous medium

Specific surface area

As =total surface area

total volumeor As =

total surface area

volume of solidor

As =total surface area

total mass

Example for one sphere : As = 4πR2

4/3πR3 = 3/R

Example for N spheres piled up with a porosity ε :As = N.4πR2/Vtotand Vtot = empty volume + solid volume = εVtot + N 4

3πR3 so

Vtot = N4πR3

3(1−ε) and As = 3R(1− ε)

Note that the specific surface area can also be given by unit ofmass instead of unit of volume.






Fluid flow through a porous medium

Darcy’s law

The volume flux (or flow rate, or discharge rate) per unit surfacearea in a porous medium is given by Darcy’s law :

q = −kpµ∇P

where kp is the porous medium permeability (in m2) and µ is thedynamic viscosity of the fluid. The permeability only depends on thesolid and not on the flow. If we consider a section S in a porousmedium of length L and a constant pressure gradient, it simplifiesto :

Q =kpS

µL∆P






Fluid flow through a porous medium

Superficial velocity

The superficial velocity is defined as us = Q/S where S is the crosssection, including solid and fluid phase. This velocity is therefore notphysical and lower than the real velocity through the pore, but it iseasily defined. It is thus often used in the porous media world.

Interstitial velocity

The interstitial velocity is the average velocity inside the pores ui =Q/(εS) where εS is the fluid cross section. This velocity is thereforecloser to physics than the superficial velocity but is more difficultto define precisely since ε has to be known. Hence the superficialvelocity is encountered more often.






Evaluation of the permeability kp

Experimentally

It is possible to measure the permeability of a porous medium witha mesure of electric conductivity through the porous medium.

Analytically

A precise analytical computation of permeability would involve sta-tistics on the pore size and distribution, orientation, connections, etcand is almost impossible. It is however possible to do it with simple”models” of porous medium...







We consider that the complex porous medium of cross section Sand length L is replaced by a set of N capillary tubes of radiusR with the same porosity ε and the same specific surface area As.Then we are going to express the flow rate Q as a function of theseparameters, and kp will be found by identification. The porosity isε = (NπR2L)/(SL) = NπR2/S. The specific surface area is a =(Nπ2πRL)/(SL − Nπ2πRL) = (N2πR)/(S − N2πR). Knowingε and a, we want N and R. This is a 2 × 2 system whose solutionis R = 2ε/(a(1 − ε)) and N = a2S(1 − ε)2/(4πε). After replacingin Poiseuille’s solution for the flow rate and multiplying by N tubes,we find the total flow rate

Q =Sε3

2a2(1− ε)2µL∆P







When compared to Darcy’s law Q =SkpµL ∆P , we finally find

kp =ε3

2a2(1− ε)2

Actually, the hypothesis of straight capillaries is too strong sincethere is connectivity between the pores and the streamlines are notstraight, so the constant 2 in kp is not recovered experimentally. thisconstant is lose to 25/6. For an array of spheres of diameter d, thespecific surface area is a = n4πR2/(n4/3πR3) = 3/R = 6/d, andso we obtain the well known Kozeny-Carman semi-empirical relation

kp =d2ε3

150(1− ε)2







The previous relation holds for a laminar flow in the porous medium,which is encountered for

Re =qd

ν=usuperficiald

ν< 10

Examples : consider a pileup of spheres where ε = 0.32. We want aflow of water (µ = 0.001Pa.s) with flow rate 1mL/hour througha porous medium of section 1 cm2 and length 1m. For two spherediameters d, the permeability and pressure drop are

kp ∆P

d = 100µm 4.72 10−12m2 0.0059 bard = 1µm 4.72 10−16m2 59 bar






Conclusion on flows through porous media

Warning !

Do not mix up porosity and permeability. Porosity only measures thevoid fraction, and permeability tells about the difficulty for a flow topass through the porous medium. Two porous media with the sameporosity can have very different permeabilities.

Method

KozenyCarman

Darcy






External flows in membrane engineering ?

Why external flows

By “external flows”, we mean flows around objects, or boundarylayer flows. Of course the objects or boundary layers will often beinside a device, but as long as the system boundaries are far awayfrom the object or other boundaries, the flow can be considered“external”. Example : we want to compute the hydrodynamic forcesexerted on a 1µm colloid arriving at a membrane in a tube with a1 cm2 cross section. From the point of view of the particle, this isan external flow.






1 Background











Drag and lift forces

Definitions

The drag force D is the hydrodynamic force exerted in the directionof the relative flow. The lift force L is the hydrodynamic forceexerted in the direction perpendicular to the relative flow. They arenot related the object velocity direction alone.






Drag and lift forces

Encountered in ...

Aeronautics : we want a minimum drag and a maximum lift. Liftmust be directed upwards.Car industry : we want a minimum drag (and a maximum lift forrace cars). Lift must be directed downwards.Sport (football, tennis...) : lift can be upwards (slice, backspin),downwards (topspin), lateral (Roberto Carlos)






Ideas on the drag and lift forces

Note that the amplitude of the drag and lift forces depend on therelative flow velocity, i.e. the flow velocity as seen by the object :urel = ufluid − vobject. New York / Paris takes much less time thanParis / New York since the polar jet stream goes from west to eastwith velocities around 200 km/h.The drag and lift forces are computed by integration of the pressureand viscous stresses on the whole surface of the object.In order to create lift, the object must be asymmetric or rotating,which is rare in membrane engineering applications, so from nowwe consider L = 0.






Drag force expressions at low and high Reynolds numbers

For a Stokes flow (Re < 1) around a sphere (model for colloidalparticles) the drag is

D = 6πµRu

where u is the relative velocity. For a high Reynolds number flow itis

D = CD.1

2ρu2S

where S is the frontal area and CD is the drag coefficient. For highReynolds numbers it is constant, and close to 0.45 for a sphere.The drag coefficient depends on the geometry of the object. Wewill come back to this later.







Drag dependance on velocity

It is important to remark that the drag force grows linearly withspeed at low Reynolds numbers, and grows quadratically withspeed at high Reynolds numbers.

Exercise

Compare the drag force exerted on a car at 110 km/h and130 km/h.

Compare the drag force exerted on a car at 130 km/h and140 km/h.







Exercise solution

D130D110

= 1302

1102' 1.4 (from 4 to 5.6L/100km !)

D140D130

= 1402

1302' 1.16 (from 5.6 to 6.5L/100km !)






1 Background











Boundary layer

Definition

If a flow is imposed on a flat plate, right at the surface the flowvelocity is u = 0 because of the no-slip condition, and far fromthe plate u = U∞. The fluid velocity then varies in a zone calledthe boundary layer. One of the definitions of the boundary layerthickness is “the distance of the plate at which u = 0.99U∞”.






Boundary layer thickness

In a boundary layer, there is no obvious length scale representativeof the global flow. The Reynolds number is thus defined asRex = U∞x

ν where x is the distance from the upstream plate edge.This kind of flow is laminar up to Rex ' 105. Hence, a highReynolds number boundary layer is first laminar, and then forsufficiently large x, it becomes turbulent. Note that for a boundarylayer to exist, there must be an inertia dominated “outer” flow(outside the boundary layer) and an inner flow influenced byviscous effects. So we always have Rex > 1 in a boundary layer.Remark that the outer-inertial/inner-viscous+inertial description isalso a definition of the boundary layer : the zone close to a wallwhere viscous effects must be taken into account.MFM-special-wing, MFM-BL






Boundary layer thickness : Blasius solution

In a laminar boundary layer (Rex < 105)

δ

x= 4.91Re−1/2

x

In a turbulent boundary layer (Rex > 105)

δ

x= 0.38Re−1/5

x

Remark 1 : in a laminar BL, δ ∼√x and in a turbulent one δ ∼ x4/5,

so the boundary layer thickness always grows with x.Remark 2 : the BL thickness always decreases with U∞.






Boundary layer thickness : example

Boundary layer on a car

Compute the boundary layer thickness on the roof of a car. Velocityis 100 km/h, roof length is 2m.

Solution

The velocity is u∞ = 27.8m/s. The kinematic viscosity of air is ν =1.5 10−5m2/s. So in the laminar regime, δ(x) = 4.91xRe−1−2

x =√xνu∞

= 3.6 10−3√x. This solution holds for Rex < 105, or

x < 105ν/u∞ = 0.054m. So it is valid at x = 5 cm whereδ(5 cm) = 0.8mm. In the turbulent regime (x > 5.4 cm), δ(x) =

0.38xRe−1/5x = 0.0212x4/5. So δ(2m) = 3.7 cm.






Thermal boundary layer

Definition

If a flow with a bulk temperature T∞ is imposed on a flat plate whosetemperature is Tw, the fluid temperature varies in a zone called thethermal boundary layer. The thermal boundary layer thickness δT is“the distance off the plate where T = 0.99T∞”.






Thermal boundary layer

Thermal boundary layer thickness

In a laminar boundary layer (Rex < 105), and for Pr = ν/D > 0.6

δT = Pr−1/3δ = xRe−1/2x Pr−1/3

and if Pr = ν/D � 1

δT = Pr−1/2δ = xRe−1/2x Pr−1/2

If Pr = 1, the thermal and dynamic boundary layers have the samethickness. If Pr > 1 (D small, non conducting fluid) the thermalboundary layer thickness is smaller than the dynamic one. If Pr < 1(D large, very conducting fluid) the thermal boundary layer thicknessis larger than the dynamic one.






Thermal boundary layer : Pr number effect






Thermal boundary layer thickness effect on heat transfer

The heat flux density at the wall is

q = −λ ∂T∂y

∣∣∣∣w

= h∆T

and in the film model, − ∂T∂y

∣∣∣w' ∆T/δT , so injecting this in the

previous relation, we get λ/δT ' h. The order of magnitude of theNusselt number at position x is then (for Pr > 0.6)

Nux =hx

λ' x

δT=

x

Pr−1/34.91xRe−1/2x

∼ Pr1/3Re1/2x

In the heat/mass transfer course we said, for a laminar boundary

layer and Pr > 0.1 : Nux = 0.332Pr1/3Re1/2x .






Thermal boundary layer : Pr and Re number effect

High Pr (ex : water) or Re numbers⇒ small BL thickness⇒ high heat flux

Low Pr (ex : air) or Re numbers⇒ large BL thickness⇒ low heat flux






Concentration boundary layer

Definition

If a flow with a bulk solute concentration C∞ is imposed on a flatplate where concentration is Cw, the concentration in the fluid variesin a zone called the concentration boundary layer. Its thickness δCis “the distance off the plate where C = 0.99C∞”.






Concentration boundary layer

Concentration boundary layer thickness

In a laminar boundary layer (Rex < 105), and for Sc = ν/D > 0.6

δC = Sc−1/3δ = xRe−1/2x Sc−1/3

and if Sc = ν/D � 1

δC = Sc−1/2δ = xRe−1/2x Sc−1/2

If Pr = 1, the concentration and dynamic boundary layers havethe same thickness. If Sc > 1 (D small) the concentration boun-dary layer thickness is smaller than the dynamic one. If Sc < 1 (Dlarge) the concentration boundary layer thickness is larger than thedynamic one.






Concentration boundary layer : Sc number effect






Concentration boundary layer thickness effect on masstransfer

The mass flux density at the wall is

j = −D ∂C

∂y

∣∣∣∣w

= k∆C

and in the film model, − ∂C∂y

∣∣∣w' ∆C/δC , so injecting this in the

previous relation, we get D/δC ' k. The order of magnitude of theSherwood number at position x is then (for Sc > 0.6)

Shx =kx

D' x

δC=

x

Sc−1/34.91xRe−1/2x

∼ Sc1/3Re1/2x

In the heat/mass transfer course we said, for a laminar boundary

layer and Sc > 0.1 : Shx = 0.332Sc1/3Re1/2x .






Concentration boundary layer : Sc and Re number effect

High Sc (ex : water) or Re numbers⇒ small BL thickness⇒ high mass flux

Low Sc (ex : air) or Re numbers⇒ large BL thickness⇒ low mass flux





Momentum transfer at interfaces and hydrodynamic forceFlow in a pipeFlow around an object

Plan

1 Background











1 Background




5 Momentum transfer at interfaces : the friction factorMomentum transfer at interfaces and hydrodynamic forceFlow in a pipeFlow around an object






Momentum transfer at an interface

Momentum transfer at an interface is a force (or rather a stress) !

We are interested in the momentum flux at an interface. We haveshown that a momentum flux (diffusive or advective) could also beinterpreted as a stress. Hence momentum transfer at an interfacemeans stress transfer at an interface, or simply, force applied by thefluid on the solid (or vice-versa) by unit surface area.






Momentum transfer at an interface

Definition of the friction factor f

The total force exerted by a fluid with velocity scale U and densityρ on a solid with surface area S can be written

F = f.1

2ρU2.S

where f is called the friction factor. Here 12ρU

2 has the dimensionof a pressure, or stress, so that f is non-dimensional. The frictionfactor depends a priori of the solid geometry, roughness and of theReynolds number of the flow.






1 Background










Friction factor in a pipe

Consider a pipe section of length L, diameter d, with a flow withaverage velocity U . The upstream position of this section is A andthe downstream position is B. At steady state, no net force should beexerted on the fluid between A and B (since there is no acceleration).On face A, the upstream fluid exerts a force PAπd

2/4 on the fluidbetween A and B and on face B the downstream fluid exerts a force−PBπd2/4 on the fluid between A and B. On the side walls, theforce is F = −f 1

2ρU2.πdL. The minus sign is present since the flow

is in the x direction and so the force exerted by the pipe on the flowis in the −x direction. The force balance then reads

PAπd2/4− PBπd2/4− f 1

2ρU2.πdL = 0







Then we get the pressure loss due to friction on the lateral wallsbetween points A and B

PA − PB = ∆P =(4f)L

d.1

2ρU2

The factor 4f is sometimes denoted λ and is called the Darcy fric-tion factor or Darcy-Weisbach friction factor or Moody frictionfactor, whereas f is called the Fanning friction factor. Quite oftenif water is used, this result is written as a head loss between pointsA and B

HA −HB = ∆H =(4f)L

d.1

2

U2

g

where H = Pρwaterg

is called the hydraulic head.







For engineering problems, the value of f or λ = 4f is needed :

Laminar flow, Re < 2000 : λ = 64/Re

Turbulent flow, smooth pipe, Blasius formulae,3.104 < Re < 106 : λ = 0.316Re−1/4

Turbulent flow, rough pipe, Colebrook formulae, ε is the

roughness length scale : 1√λ

= −2 log10

(ε/dH3.7 + 2.51

Re√λ

)dH is the hydraulic diameter computed as dH = 4Ap where A is thecross sectional area of the pipe filled with fluid and p is the wettedperimeter of the cross section. For a cylindrical pipe, dH = d. For asquare pipe, dH is the square side length.






Friction factor in a pipe : Moody diagram






1 Background










Friction coefficient on an object in a flow

Consider an object of frontal area S in a flow with average velocity U .The drag force exerted by the flow on the object is F = CD.

12ρU

2.S,where CD is called the drag coefficient. It is the same as the Fanningfriction factor.For a sphere :

If Re < 1 : CD = 24/Re

If 2 < Re < 500 : CD = 18.5Re−3/5

If 500 < Re < 2.105 : CD ' 0.44






Friction coefficient on an object in a flow

But the value of the drag coefficient depends strongly on the geo-metry of the object. At large Reynolds numbers, here are some orderof magnitude (the definition for planes is slightly different actually)

Objet CDPlane 0.02− 0.03Car 0.25 at bestSphere 0.5Cyclist 0.7






Drag coefficient for simple geometries





Bernoulli’s theoremHead losses

Plan

1 Background











1 Background





6 Energy balance in a flow : Bernoulli’s theoremBernoulli’s theoremHead losses






Bernoulli’s first theorem

Bernoulli’s first theorem, or principle

If a flow is

steady,

incompressible,

irrotational,

then the hydraulic H head is the same everywhere in the flow

H ≡ P

ρg+U2

2g+ z = constant






Bernoulli’s second theorem

Bernoulli’s second theorem, or principle

If a flow is

steady,

incompressible,

inviscid,

then the hydraulic head H is constant along a streamline

H ≡ P

ρg+U2

2g+ z = constant






Bernoulli’s second theorem

What does it mean ?

It states that the total mechanical energy is conserved along astreamline. Indeed, let’s multiply it by ρg and we get

P +1

2ρU2 + ρgz = constant

where 12ρU

2 is a kinetic energy per unit volume, ρgz is a potentialenergy per unit volume and P can be seen as an “elastic” energyper unit volume. (P ×L2 is a force, and an energy is a force times alength, so P ×L3 is an energy and P is an energy per unit volume).






Bernoulli’s generalized theorem


If a flow is

steady,

incompressible,

then between points A and B along a streamline

∆HAB =

(PAρg

+U2A

2g+ zA

)−(PBρg

+U2B

2g+ zB

)= ξAB > 0

where ξAB is called the head loss between A and B.







Explanation

If the fluid is no more inviscid, then energy dissipation occurs bet-ween points A and B because of viscous friction. Remember thatρgHA is the total mechanical energy at point A and ρgHB is thetotal mechanical energy at point B. Then ρg(HA −HB) = ρgξABis the total mechanical energy loss between points A and B. Thisenergy loss is due to viscous dissipation which converts mechanicalenergy into heat (or thermal energy). From this point of view Ber-noulli’s generalized theorem is the first theorem of thermodynamics.







Bernoulli’s theorem as first law of thermodynamics

Bernoulli’s theorem is

∆P +1

2ρ∆(U2) + ρg∆z = ρgξAB

and the first law of thermodynamics can be expressed as

∆U + ∆Ek + ∆Ep = W +Q

where ∆U is the change in internal energy (∆P ), ∆Ek is the changein kinetic energy (1

2ρ∆U2), ∆Ep is the gange in potential energyρg∆z, W is the amount of work due to external forces and Q is theamount of heat transmitted to the system (source or sink).







Bernoulli’s theorem as first law of thermodynamics

Here ρgξAB is a sink of energy for our system and is therefore in Q.But what about W ? It is possible to include an external mechanicalforcing in Bernoulli’s theorem to take into account a pump betweenA and B. If the power of a pump (in Watts) is denoted P, it trans-mits to the fluid a work P during one second, and the volume of fluidreceiving this work is the volume flowrate Q. Hence, the fluid receivesa work W = P/Q per unit volume. Since we write ρgHupstream =ρgHdownstream + ρgξ, and since the pump is a source of energy forthe fluid, the term will appear on the left hand side (since the downs-tream head will be the upstream head plus the pump’s additionalhead) : ρgHupstream + P/Q = ρgHdownstream + ρgξ







The new Bernoulli’s theorem with both viscous dissipation and amechanical energy source between A and B reads :

Bernoulli’s generalized theorem (know it by heart !)

If the flow between an upstream point A and a downstream pointB along a streamline, is steady and incompressible, then

∆P

ρg+

∆(U2)

2g+ ∆z = ξAB −

PρgQ

where ξAB > 0 and PρgQ > 0 and where for any function X,

∆X = XA −XB = Xupstream −Xdownstream.






1 Background





6 Energy balance in a flow : Bernoulli’s theoremBernoulli’s theoremHead losses






Examples of head losses in classical devices

They are provided under ther form ξAB = ksU2

2g , where ks is the

head loss coefficient and U is the maximum average velocity. For apipe it is a linear head loss coefficient and for any other geometry itis a singular head loss coefficient. Examples :

Straight pipe ks =λLdH

where λ = 4f is Moody’s friction factor and dHis the hydraulic diameter

Contraction ks = 0.45(1 − β) where β is the small section to largesection ratio

Expansion ks =(

1β− 1

)2

Rounded 90 bend ks = 0.6

Sharp 90 bend ks = 1.6

Open sharp 45 bend ks = 0.3

there are hundreds of them, often determined experimentally.Check in textbooks when needed.






The End

To know more about fluid mechanics (by increasing mathematicalcomplexity) :

Multimedia Fluid Mechanics DVD, Homsy et al.

Physical Hydrodynamics, Guyon, Hulin & Petit.

Mecanique des fluides, Chassaing

An Introduction to Fluid Dynamics, Batchelor

Advanced Transport Phenomena : Fluid Mechanics andConvective Transport Processes, Leal

Theoretical Hydrodynamics, Milne-Thomson


Momentum and Mass transport and transfer: an introduction...

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