MT Unit 5 Multiphase Flow Prof R R Joshi

UNIT-5

MULTIPHASE FLOW

Prof Ratnadip R Joshi

Hydrodynamics in Porous Media

Darcy’s Law for saturated media– In 1856 Darcy hired to figure out the water supply to the

town’s central fountain. – Experimentally found that flux of water porous media could

be expressed as the product of the resistance to flow which characterized the media, and forces acting to “push” the fluid through the media.

– Q - The rate of flow (L3/T) as the volume of water passed through a column per unit time.

– hi - The fluid potential in the media at position i, measured in standing head equivalent. Under saturated conditions this is composed of gravitational potential (elevation), and static pressure potential (L: force per unit area divided by ρg).

– K - The hydraulic conductivity of the media. The proportionality between specific flux and imposed gradient for a given medium (L/T).

– L - The length of media through which flow passes (L).– A - The cross-sectional area of the column (L2).

MT Unit 5: Prof R R Joshi 3

Darcy’s LawDarcy then observed that the flow of water in a vertical column was well described by the equation

Darcy’s expression is written in a general form for isotropic media as

– q is the specific flux vector (L/T; volume of water per unit area per unit time),

– K is the saturated hydraulic conductivity tensor (second rank) of the media (L/T), and

– ∇H is the gradient in hydraulic head (dimensionless)


Q= K )H(HLA

0-1

q = -K ∇H

Aside on calculus ...What is this up-side-down triangle all about?

The “dell” operator: short hand for 3-d derivative

• The result of “operating” on a scalar function (like potential) with ∇is the slope of the function

• ∇F points directly towards the steepest direction of up hill with a length proportional to the slope of the hill.

• Later we’ll use ∇•F. The dot just tells us to take the dell and calculate the dot product of that and the function F (which needs to be a vector for this to make sense).

• “dell-dot-F” is the “divergence” of F. • If F were local flux (with magnitude and direction), ∇•F would be the

amount of water leaving the point x,y,z. This is a scalar result! ∇F takes a scalar function F and gives a vector slope

∇•F uses a vector function F and gives a scalar result.


∇ = ∂∂xi, ∂

∂y j, ∂∂z k

Now, about those parameters...Gradient in head is dimensionless, being length per length

Q = Aq Q has units volume per unit time Specific flux, q, has units of length per time, or velocity. For vertical flow: speed at which the height of a pond of fluid would drop

CAREFUL: q is not the velocity of particles of water– The specific flux is a vector (magnitude and direction). – Potential expressed as the height of a column of water, has units of

length.


LHH = H 01 −∇

About those vectors...

Is the right side of Darcy’s law indeed a vector? – h is a scalar, but ∇H is a vector– Since K is a tensor (yikes), K∇H is a vector

So all is well on the right hand side

Notes on K: – we could also obtain a vector on the right hand side by

selecting K to be a scalar, which is often done (i.e., assuming that conductivity is independent of direction).


q = -K∇H

A few words about the K tensor

– Kab relates gradients in potential in the b-direction to flux that results in the a-direction.

– In anisotropic media, gradients not aligned with bedding give flux not parallel with potential gradients. If the coordinate system is aligned with directions of anisotropy the "off diagonal” terms will be zero (i.e., Kab=0 where a≠b). If, in addition, these are all equal, then the tensor collapses to a scalar.

– The reason to use the tensor form is to capture the effects of anisotropy.


q = -

K xx K xy K xz

K yx K yy K yzK zx K zy K zz

∂h

∂x ∂h∂y

∂h∂z

= -

K xx

∂h∂x +K xy

∂h∂y + K xz

∂h∂z ; K yx

∂h∂x +K yy

∂h∂y + K yz

∂h∂z ; K zx

∂h∂x +K zy

∂h∂y + K zz

∂h∂z

flux in x-direction flux in y-direction flux in z-direction

Looking holisticallyCheck out the intuitively aspects of Darcy’s result. The rate of flow is:

– Directly related to the area of flow (e.g., put two columns in parallel and you get twice the flow);

– Inversely related to the length of flow (e.g., flow through twice the length with the same potential drop gives half the flux);

– Directly related to the potential energy drop across the system (e.g., double the energy expended to obtain twice the flow).

The expression is patently linear; all properties scale linearly with changes in system forces and dimensions.


Why is Darcy Linear?Because non-turbulent?

No.Far before turbulence, there will be large local accelerations: it is the lack of local acceleration which makes the relationship linear.

Consider the Navier Stokes Equation for fluid flow. The x-component of flow in a velocity field with velocities u, v, and w in the x, y, and z (vertical) directions, may be written


∂u∂t + u

∂u∂x + v

∂u∂y + w

∂u∂z =

-1ρ

∂P∂x - g

∂z∂x +

µρ ∇2u

Creeping flowNow impose the conditions needed for which Darcy’s Law

– “Creeping flow”; acceleration (du/dx) terms small compared to the viscous and gravitational terms

– Similarly changes in velocity with time are small

so N-S is:

– Linear in gradient of hydraulic potential on left, proportional to velocity and viscosity on right (same as Darcy).

– Proof of Darcy’s Law? No! Shows that the creeping flow assumption is sufficient to obtain correct form.


0zu

yu

xu =

∂∂=

∂∂=

∂∂

0tu ≈

∂∂ u gzPx

2∇=+∂∂

µρ

∂u∂t + u

∂u∂x + v

∂u∂y + w

∂u∂z =

-1ρ

∂P∂x - g

∂z∂x +

µρ ∇2u

Capillary tube model for flowWidely used model for flow through porous media is a group of cylindrical capillary tubes (e.g.,. Green and Ampt, 1911 and many more).

– Let’s derive the equation for steady flow through a capillary of radius ro

– Consider forces on cylindrical control volume shownΣ F = 0


V

0

∆ s

r

ro

s

C y l i n d ri c al C o n t ro l V o lu m e

Momentum Transport: Flow in Porous Media & Packed BedsFlow in Porous Media

And Packed Beds

Volumes of interest may contain a solids fraction, φ, made up of:

Granular particles (sand, pebbles)

Wool (steel wool, fiberglass, etc.)

Gauzes, screens (woven metals)

Porous pellets (absorbent, catalyst support)

Void fraction, ε = 1 – φ Usually, packing geometry is random


FLOW IN PORUS MEDIA AND PACKED BEDS

ε > 0.8:Large void fractionFlow about each object may be considered as

“perturbed” external flowEach wetted object contributes drag

ε < 0.5:View as internal flow through tortuous ducts between

particlesUsed in following module



In case of flow through a straight duct of noncircular cross-section, effective duct diameter

For flow through a packed bed, effective average interstitial (duct) diameter

where


4 ,effAd

P≡

( )( ),

4 / 4 ,/ '''i eff

volume available for flow total volumed

wetted area total volume aε

≡ =

( )

( )

'''

,

. 1

6 . 1

p

p

p eff

Aparticle surface area particlevolumeaparticlevolume total volume V

d

ε

ε

≡ = −

= −


Effective diameter of each particle in bed

Therefore:

Appropriate Reynolds number for internal flow

where interstitial mass velocity


, 6 pp eff

p

Vd

A

≡

, ,2 . . ,3 1i eff p effd dε

ε = −

, ,,Re . . ,i i eff i i eff

bed eff

v d G dconst const

ρµ µ

= =


i iG vρ= =

Empty-duct (superficial) mass velocity

and

and


0 ,iGGε

=

( )0 ,

,Re . ,1

p effbed eff

G dµ ε

≡−

0 0 0/G v m Aρ= =


fbed dimensionless momentum transport coefficient

Function of Re For a single straight duct of short length, dimensionless

momentum-transfer coefficient

where


2 2

41 12 2

eff

wf

d dPdzC

U U

τ

ρ ρ

− ≡ =

| |zP p g zρ≡ +


Hence:

and

Correlates well with experimental data (next slide)Ergun’s approximation:


3

,

20

1/

p eff

bed

dPddz

fG

εε

ρ

− − ≡

150 1.75Rebed

bed

f ≅ +

( ),

2

1 /2. 1

2

i eff

bed

i

d dP dzf const

vρ

−≡


Experimentally determined dependence of fixed-bed friction factor fbed onthe bed Reynolds’ number(adapted from Ergun (1952))



Laminar flow region:Rebed < 10 fbed ≈ 150/Rebed

Darcy’s law:Linear relationship between G0 and (-dP/dz)

Effective local permeability = G0 ν/ (-dP/dz)

>> intrinsic permeability of each particle

Fully-turbulent asymptote: (Burke-Plummer)Rebed > 1000 fbed ≈ 1.75



Above equations are basis for most practical pressure-dropcalculations in quasi-1D packed ducts

Can be used to estimate incipient fluidization velocity(by equating –∆p to bed weight per unit area)

Can be generalized to handle multidimensional flowsthrough isotropic fixed beds

Can be used to estimate inter-phase forces betweena dense cloud of droplets and host (carrier) fluid



Fixed bed and fluidized bed


Why fixed (or fluidized) bed? Expensive Catalyst enzyme (immobilized) Large Surface area

Used in reaction/adsorption/ elution (for example)

Ref: BSL, McCabe & Smith

Fixed bed


Filled with particles Usually not spherical

To increase surface area To increase void fraction

To decrease pressure drop

For analytical calculation, assume all particles are identical Usable, because final formula can be modified by a constant factor

(determined by experiment)

Fixed bed


What are important parameters? (For example, for adsorption of a protein from a

broth) rate of adsorption (faster is better) saturation concentration (more is better)

From the product requirement (eg X kg per day), density and product concentration in broth ==> volumetric flow rate

Fixed bed


Sphericity Volume of particle = Vp

Surface Area of particle = Ap

Surface Area of sphere of same volume (Vs =Vp) = As

Sphericity = As/Ap

May be around 0.3 for particles used in packed beds lower sphericity ==> larger surface area

Assume quick adsorption (rate of adsorption is high) Calculate the surface area of particles needed for

operation

As, Vs

Ap,Vp

Sphericity <=> specific surface area <=> average particle diameter

Fixed bed


Specific surface area = Ap /Vp

Minimal value for sphere Some books use S to denote area (instead of A) Assume all the particles are identical

==> all particles have exactly same specific surface area

Tarus saddlePall Ring

Rings (Raschig,etc)

Fixed bed


What is the pressure drop we need, to force the fluid through the column?

(i.e. what should be the pump spec) We know the volumetric flow rate (from adsorption equations, productivity

requirements etc) We know the area per particle (we assume all particles are identical). And the

total area for adsorption (or reaction in case of catalytic reactor). Hence we can calculate how many particles are needed Given a particle type (eg Raschig ring) , the approximate void fraction is also

known (based on experimental results)

Fixed bed


What is void fraction? Volume of reactor = VR

Number of particles = Np

Volume of one particle = Vp

Volume of all the particles = Vp * Np = VALL-PARTICLES

R ALL PARTICLES

R

V VV

ε −−=

VOIDS

R

VVoid fractionV

ε= =

R P P

R

V V NV

ε −=

( )1RP

P

VN

Vε−

=

Knowing void fraction, we can find the reactor volume needed Alternatively, if we know the reactor volume and void

fraction and the Vp, we can find the number of particles

Fixed bed


To find void fraction experimentally Prepare the adsorption column (or reactor....) and fill it

with particles Fill it with water Drain and measure the quantity of water (= void volume) Calculate void fraction

Fixed bed


Since we know Vp, Np, ε, we can find VR

Choose a diameter and calculate the length (i.e. Height) of the column (for now) In normal usage, both the terms ‘height’ and ‘length’ may be used

interchangeably (to mean the same thing) Adsorption rate, equilibrium and other parameters will also influence the

determination of height & diameter

To calculate the pressure drop Note: columns with large dia and shorter length (height) will have lower

pressure drop What can be the disadvantage(s) of such design ? (tutorial)

Fixed bed


To calculate the pressure drop You want to write it in terms of known quantities Length of column, void fraction, diameter of particles, flow rate of

fluid, viscosity and density Obtain equations for two regimes separately (turbulent and laminar) Consider laminar flow

Pressure drop increases with velocity viscosity inversely proportional to radius

Actually, not all the reactor area is available for flow. Particles block most of the area. Flow path is not really like a simple tube

Hence, use hydraulic radius

Fixed bed - pressure drop calculation (Laminar flow)


To calculate the pressure drop, use Force balance

Force P Area= ∆2

Area where flow occurs = 4Dπ ε

2

4DForce P π ε∴ = ∆

Resistance : due to Shear Find Contact Area Find shear stress

Contact areaForce τ∴ =

Until now, we haven’t said anything about laminar flow. So the above equations are valid for both laminar and turbulent flows

Fixed bed - pressure drop calculation (Laminar Flow)


Find contact area

Wetted Area= ppN A ( )=1

pR

p

VA

Vε− ( )= 1 p

Rp

AV

Vε−

To calculate the shear stress, FOR LAMINAR FLOW

max 42 avgVVR R

µµτ = =

r R

dVdr

τ µ=

= −

8 avgVD

µ=

2

max 21 rV VR

= −

max 2 avgV V=

Here V refers to velocity for flow in a tube

However, flow is through bed, NOT a simple tube



Find effective diameter (i.e. Use Hydraulic radius), to substitute in the formula Also relate the velocity between particles to some quantity we know

To find hydraulic radius ( and hence effective dia)

RFlowvolume Vε=

Wetted Area= ppN A ( )=1

pR

p

VA

Vε−

4HFlow AreaD

ContactPerimeter=

Hydraulic diameter*4

*Flow Area Column Height

ContactPerimeter Column Height=

4 Flowvolumewetted area

=



( )

4

1H

p

p

DA

V

ε

ε=

−

8 avg

H

VDµ

τ∴ =( )8 1

4

pavg

p

AV Vµ ε

ε

− =

( )2 1 pavg

p

AV Vµ ε

ε

− =

Vavg is average velocity of fluid “in the bed”, between particles

Normally, volumetric flow rate is easier to find



Can we relate volumetric flow rate to Vavg ? Use a new term “Superficial velocity” (V0)

0Volumetric flowrateV

Column Area= 0 2

4

QVDπ

=

I.e. Velocity in an ‘empty’ column, that will provide the same volumetric flow rate

Can we relate average velocity and superficial velocity?

0avg

VVε

=



( )2 1gp

pav

AV Vµ ετ

ε

− =

( )0

2

2 1 p

p

AVVµ ε

ε

− =

2

4DForce P π ε= ∆

( )( )

02

2

2 1 1

4

p

p pR

p

AV V ADP VV

µ επ ε ε

ε

− ∆ = −

Force balance: Substitute for τ etc.

Contact areaForce τ=

2

4RDV Lπ

= Volume of reactor (say, height of bed = L)

( )2

02

2

2

2

4

1

4

2

p

p

AV

PV

D D Lµ ε

εε

π π − ∆ =



Pressure drop

( )2

20 2

2

2

4

1

4

2

p

p

AV VP LD Dπ

εε

επ

µ − ∆ =

( )2

20

3

2 1

p

p

ALV VP

µ ε

ε

− ∆ =

Specific surface area vs “average diameter”

p

p

AV

Define “average Dia” of particle as6

pp

p

DAV

=

Some books (BSL) use Dp



Pressure drop

( )2

20

3

62 1 pDLV

Pµ ε

ε

−∆ =

( )202 3

72 1

p

LVD

µ εε

−=

However, using hydraulic radius etc are only approximations

Experimental data shows, we need to multiply the pressure requirement by ~ 2 (exactly 100/48)

( )2

20

3

25

6

1

p

p

ALV VP

µ ε

ε

− ∆ =

In terms of specific surface area

( )202 3

1

150

p

LVP

Dµ ε

ε−

∆ =

In terms of average particle diameter

Fixed bed - pressure drop calculation (Turbulent Flow)


Pressure drop and shear stress equations2

4DForce P π ε= ∆ Contact areaForce τ=

Only the expression for shear stress changes

f

Re

For high turbulence (high Re),

2=constant1

2 avg

fV

τρ

=

21=constant 2 avgVτ ρ∴

202= VKτ ρ

ε

0avg

VVε

= However



We have already developed an expression for contact area

Wetted Area= ppN A ( )= 1 pR

p

AV

Vε−( )=

1 p

R

p

VA

Vε−

( )2

02 1 p

Rp

AVK VV

ρ εε

= −

2

Contact area4DForce P π ε τ= ∆ =

Hence, force balance

2

4RDV Lπ

=

Volume of reactor (say, height of bed = L)

( )2

30 1 p

p

AVP K LV

ρ εε

∆ = −



( )2

03 1 6

p

VP LD

K ρ εε

∆ = −

In terms of average particle diameter

( )2

03 1 p

p

AVP K LV

ρ εε

∆ = −

In terms of specific surface area

Value of K based on experiments ~ 7/24

What if turbulence is not high? Use the combination of laminar + turbulent pressure drops: valid for all regimes!

( )20

Laminar 2 3

150 1

p

LVP

Dµ ε

ε−

∆ =( )2

03

1

74Turbulent

p

LVP

Dρ ε

ε −

+ ∆ =

( ) ( )2 20 02 3 3

150 1 7 1

4totalp p

LV LVP

D Dµ ε ρ ε

ε ε− −

∆ = +Ergun Equation for packed

bed

Fixed bed - pressure drop calculation (Laminar OR Turbulent Flow)


If velocity is very low, turbulent part of pressure drop is negligible

If velocity is very high, laminar part is negligible

( ) ( )2 20 02 3 3

150 1 7 1

4totalp p

LV LVP

D Dµ ε ρ ε

ε ε− −

∆ = +Ergun Equation for packed

bed

( )0 20

2 2

2

2 1724

12

p

p

avg

AV V V

fV

µ ερ

ε ε

ρ

− +

=

Some texts provide equation for friction factor

212 avg

fV

τρ

= laminar21

2

turbulent

avg

fV

τ τρ+

=



( )0

2

2

20

2

2

0

2 1

1

2

p

p

AV

K

fV

VVρ

ε ε

ρ

µ

ε

ε −

+ =

( )

0

4 17

12

p

p

V

AV

ρ

µ ε − = +

For pressure drop, we multiplied the laminar part by 2 (based on data) . For the turbulent part, the constant was based on data anyway.

Similarly...( )

0

10048

4 17

12

p

p

AV

Vf

ρ

µ ε − = +

( )

0

25 17

3 12

p

p

AV

Vρ

µ ε − = +



( )

0

25 17

3 12

p

p

AV

fV

µ ε

ρ

−

= +

Multiply by 3 on both sides (why?)

( )

0

25 17

3 1

6

2p

VDµ ε

ρ

− = +

( )0

150 1 734p

fD Vµ ε

ρ −

= +

( )0

150 1 734pD V

fεµ

ρ −

= +

Packed bed friction factor = 3 f

( )150 13 1

R.75

e ppf f

ε −= = +

Eqn in McCabe and Smith

Reynolds number for packed bed

Example


Adsorption of Cephalosporin (antibiotic) Particles are made of anionic resin(perhaps resin coatings on ceramic

particles) void fraction 0.3, specific surface area = 50 m2/m3(assumed) column dia 4 cm, length 1 m feed concentration 2 mg/liter (not necessary to calculate pressure drop, but

needed for finding out volume of reactor, which, in this case, is given). Superficial velocity about 2 m / hr

Viscosity = 0.002 Pa-s (assumed) What is the pressure drop needed to operate this column?

Fixed Bed


What is the criteria for Laminar flow? Modified Reynolds Number Turbulent flow:- Inertial loss vs turbulent loss

Loss due to expansion and contraction Packing uniformity

In theory, the bed has a uniform filling and a constant void fraction Practically, near the walls, the void fraction is more

Edge Center Edge

ε

0.2

0.4

0.8 Ergun Eqn commonly

used, however, other empirical correlations are also used

e.g. Chilton Colburn eqn

Re Ren

A Bf C= + +

( )1p oD V ρ

µ ε−

Fixed Bed


Sphericity vs Void Fraction

φ

ε0 1

1

~0.4

Fixed Bed


Alternate method to arrive at Ergun equation (or similar correlations) Use Dimensional analysis

dependent variableP∆ −

( subscript, means fluid density or )fwithoutρ ρ

, , , , , , (i.e. sphericity)p o columnD L V Dµ ε φ

2

2 ( , , , )p p o p

o column

D D V DP fV L D

ρε φ

ρ µ∆

=

Fluidized bed


When the fluid (moving from bottom of the column to the top) velocity is increased, the particles begin to ‘move’ at (and above) a certain velocity.

At fluidization, Weight of the particles == pressure drop (area) Remember to include buoyancy

( )( )2

14 s f RDP Vπ ρ ρ ε∆ = − −

( )( )2

14s fD Lπρ ρ ε= − −

Fluidized bed: Operation


Empirical correlation for porosity

n

t

VV

ε=

Types of fluidization: Aggregate fluidization vs Particulate fluidization

Larger particles, large density difference (ρSOLID - ρFLUID) ==> Aggregate fluidization (slugging, bubbles, etc)

==> Typically gas fluidization Even with liquids, lead particles tend to undergo

aggregate fluidization Archimedes number 3

2f pg D

Arρ ρ

µ∆

=



Porosity increases Bed height increases Fluidization can be sustained until terminal velocity is reached If the bed has a variety of particles (usually same material, but

different sizes) calculate the terminal velocity for the smallest particle

Range of operability = R Minimum fluidization velocity = incipient velocity (min range) Maximum fluidization velocity = terminal velocity (max range) Other parameters may limit the actual range further

e.g. Column may not withstand the pressure, may not be tall enough etc

R = Vt/VOM

Theoretically R can range from 8.4 to 74



Range of operation depends on Ar

Ar

100 104108

R

0

80

40



Criteria for aggregate fluidization Semi empirical

0.5

2 0.6 ( )

0.3 ( )

p

s

Dfor liquid

for gas

ρρ µ

∆ >

>

Particulate fluidization Typically for low Ar numbers More homogenous mixture

Documents

MT Unit 5 Multiphase Flow Prof R R Joshi