Dr. R. Nagarajan

Professor

Dept of Chemical Engineering

IIT Madras

Advanced Transport PhenomenaModule 4 - Lecture 16

Momentum Transport: Flow in Porous Media & Packed Beds

Volumes of interest may contain a solids fraction, ,

made up of:

Granular particles (sand, pebbles)

Wool (steel wool, fiberglass, etc.)

Gauzes, screens (woven metals)

Porous pellets (absorbent, catalyst support)

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

FLOW IN PORUS MEDIA AND PACKED BEDS

> 0.8:

Large void fraction

Flow about each object may be considered as “perturbed”

external flow

Each wetted object contributes drag

< 0.5:

View as internal flow through tortuous ducts between

particles

Used in following module

FLOW IN PORUS MEDIA AND PACKED BEDS

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,eff

Ad

P

,

4 / 4,

/ '''i eff

volume available for flow total volumed

wetted area total volume a

'''

,

. 1

6. 1

p

p

p eff

Aparticle surface area particlevolumea

particlevolume total volume V

d

FLOW IN PORUS MEDIA AND PACKED BEDS

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 d

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

bed eff

v d G dconst const

FLOW IN PORUS MEDIA AND PACKED BEDS

i iG v

Empty-duct (superficial) mass velocity

and

and

0 ,i

GG

0 ,

,Re . ,1

p effbed eff

G d

0 0 0/G v m A

FLOW IN PORUS MEDIA AND PACKED BEDS

fbed dimensionless momentum transport coefficient

Function of Re

For a single straight duct of short length, dimensionless

momentum-transfer coefficient

where2 2

41 12 2

eff

wf

d dPdz

CU U

| |zP p g z

FLOW IN PORUS MEDIA AND PACKED BEDS

Hence:

and

Correlates well with experimental data (next slide)

Ergun’s approximation:

3

,

20

1

/

p eff

bed

dPd

dzf

G

1501.75

Rebedbed

f

,

2

1/

2.12

i eff

bed

i

d dP dzf const

v

FLOW IN PORUS MEDIA AND PACKED BEDS

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

FLOW IN PORUS MEDIA AND PACKED BEDS

Laminar flow region:

Rebed < 10

fbed ≈ 150/Rebed

Darcy’s law:

Linear relationship between G0 and (-dP/dz)

Effective local permeability = G0 (-dP/dz)

>> intrinsic permeability of each particle

Fully-turbulent asymptote: (Burke-Plummer)

Rebed > 1000

fbed ≈ 1.75

FLOW IN PORUS MEDIA AND PACKED BEDS

Above equations are basis for most practical pressure-

drop calculations 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 flows

through isotropic fixed beds

Can be used to estimate inter-phase forces between a

dense cloud of droplets and host (carrier) fluid

FLOW IN PORUS MEDIA AND PACKED BEDS

For a combustion turbine materials-test program it is

desired to expose specimens to a sonic but atmospheric

pressure jet of combustion-heated air with a post-

combustion (stagnation) chamber temperature of 2000 K.

a.What should the pressure in the combustion chamber

(upstream of the nozzle) be?

b.What should the shape of the nozzle be?

c. How much air flow (g/s) must be handled if the exit jet must

be 2.5 cm. in diameter?

PROBLEM 1

d. What will the exit (jet) velocity be?

e. What will be the flow rate of axial momentum at the

nozzle exit?

f. By what factor will the gas density change in going

from the nozzle inlet to the nozzle outlet?

PROBLEM 1

a. Upstream (Chamber) Pressure

Isentropic gas flow from T0 = 2000 K to sonic speed @

1 atm. Assume

Therefore,

and

1.28

10.140

2

14.571, 1.14

1

SOLUTION 1

Now,

Also,

/ 120*

*

4.571200

p 11 Ma

p 2

pTherefore, 1 0.140 1 p 1.82 atm

1

0*

*

T 11.14 T 1754 K

T 2

SOLUTION 1

For a perfect gas EOS:

b. Nozzle shape: see figure

d. Exit (Jet) Velocity

Therefore

0 0 *

* * 0

p T 11.82 . 1.6

p T 1.14

( gas density ratio across nozzle ),

1/ 281/ 2

** *

1.28 0.83144 10 1754RTu a

M 28.97

4* *u a 8.03 10 cm / s

SOLUTION 1

c. Mass Flow Rate

But, for a perfect gas:* * *m u A

4 3**

*

4*

22 2*

*

4 4

1 28.97p M2.01 10 g / cm

RT 82.06 1754

u 8.03 10 cm / s Part d

and

dA 2.5 4.91 cm

4 4

Therefore m 2.01 10 8.03 10 cm / s 4.91

79.1 g / s

SOLUTION 1

e. Flow rate of Axial Momentum, at exit:

f.

J

4*

6 2

J mu 79.1 g / s 8.03 10 cm / s

6.35 10 g.cm / s

0

*

1.6 calculated above

SOLUTION 1

Consider a dilute acetylene-air mixture at atmospheric pressure

and 300 K with a C2H2 mass fraction of 4.5%. If the heat of

reaction of acetylene is 11.52 kcal/gm C2H2, estimate:

a.The speed with which a Chapman- Jouguet dectonation would

propagate through this mixture (km/s);

b.The stagnation pressure (atm) immediately behind such a

detonation wave;

c. The corresponding Chapman –Jouguet deflagration speed

(km/s)

PROBLEM 2

SOLUTION 2

2 2 1C H

32 2

2 2

2

0.045

g C H 11.52 kcal 10 calq 0.045

g mix g C H 1 kcal

calq 5.184 10

g

1

M 28.97

1.35

R 1.987

T 300 K

2

1

2 2

1 MqH .

2 RT

28.97 5.184 101.35 1H .

2 1.35 1.987 300

H 7.675or

1 2

1/ 2 1/ 2

1 CJ det on

1 1 1

1

Ma 1 H H 5.7156 5.716

kmu a .Ma 0.3472 5.716

s

u 1.98 km / s

2Calculation of T

2 2

effect of '' heat addit"

2 22 1

2 0 2 0 1 1p p p p

since u a

RT uq qT T T T

2c M c 2c c

SOLUTION 2

therefore

Since

we calculate:

21 1

2

p

1T u q

2TR / M

12c

1

2

p

u 1.98 km / s

q 5.184 10 cal / g

c 0.263 cal / g.K

2T 3453 K

SOLUTION 2

This fixes since:

or, equivalently:

But

therefore

2 1/ 2

21 1 2 2 2

RTG .u u .

M

1/ 2

1 1 12 1 2 1

2 2

a Ma T. Ma

a T

1 1 3 32

1

p M 1 28.971.177 10 g / cm

RT 82.06 300

3 32 1.983 10 g / cm

SOLUTION 2

0

Therefore

And, since Ma 2 =1(Chapman-Jouguet Condition)

conclusion: Detonation Speed=1.98 km/s ((Ma)1 =5.7). Stagnation Pressure:

2 12

RTp 19.39 atm;

M

/ 1

0 2 2.

1p p

2

19.39 1.863 36.1 atm

0 2p 36.1 atm;

SOLUTION 2

Consider the steady axisymmetric flow of hot air in a

straight circular tube of radius aw and cross sectional area

A

Conditions (at exit):

p=1 atm (uniform)

T= 1500 K (uniform)

aw= 5 mm

PROBLEM 3

Suppose it has been observed that the axial-velocity profile

is, in this case, well described by the simple

equation:

where U=103 cm/s. Using this observation, the conditions

above, answer the following questions:

a. If the molecular mean-free-path in air is approximately

given by the equation:

zv ( r,z( exit ))

2

zw

rv 2U 1

a

PROBLEM 3

1.2T 1

l( air ) 0.065 ( in m ),300 p

Estimate the prevailing mean free path l and the ratio of l to

the duct diameter- i.e., relevant Knudsen number for the gas

flow:

What conclusions can you now draw concerning the

validity of the continuum approach in this case?

b. Calculate the convective mass flow rate (expressed in

g/s) through the entire exit section. For this purpose assume

the approximate validity of the “perfect “ gas law, viz,:

m

PROBLEM 3

w

lKn

2a

3pMg / cm

RT

Here p is the pressure (expressed in atm), M is the

molecular weight (g/g-mole)(28.97 for air), R=82.06 (univ

gas const), and T is the absolute temperature (expressed

in kelvins). Also note that for this axisymmetric flow a

convenient area element is the annular ring sketched

below

( where is the unit vector in the z- direction).

zd 2 r.drA e

ze

PROBLEM 3

c. Also calculate the average gas velocity

at the exit section and the corresponding Reynolds’

number;

avgU m / A

avg avg avg w

avg

w

U U U 2aRe

U

2a

PROBLEM 3

d. Calculate the convective axial momentum flow rate

(expressed in g. cm/s2) through the exit section. Is your result

equivalent to Why or why not?

e. Calculate the convective kinetic-energy flow rate (expressed in g.

cm2/s3). Is your result equivalent to Why or

why not?

f. If, in a addition to the axial component of the velocity vz, the air in

the duct also has a swirl component how would this

influence your previous estimates ( of mass flow rate,

momentum flow rate, kinetic energy flow rate)? Briefly discuss.

avgmU ?

avg

2m U / 2 ?

v r,z

PROBLEM 3

g. If the local shear stress is given by the following

degenerative form of Newton’s law:

at what radial location does maximize? Calculate the maximum

value of and express your result in dyne cm -2 and Newton m-

2. Calculate the skin-friction coefficient, c f (dimensionless), at

the duct exit. At what radius does take on its minimum

value? Can- be regarded as the radial diffusion flux of axial

momentum? Why or why not? Does the rate at which work is

done by

PROBLEM 3

zrz

v

r

rz

rzrz

rz rz

the stress maximize at either of the two locations

found above? Why or why not?

h. Characterize this flow in terms of flow descriptors and

defend your choices.

rz

PROBLEM 3

well into the continuum regime

1.2

1.2

w w

6 2w

4

T 1a. l( air ) 0.065 . m

300 p

1500 1l( air ) 0.065 . m 0.448 m

300 1

d 2a 2 0.5 cm 1.00 cm

Kn l / d 0.448 10 m / 1.00 10 m

Kn 0.448 10 ( )

SOLUTION 3

b.

w w

43

2a a

z0 0w

22 4 3w

1 28.97 gpM / RT 2.35 10

82.06 1500 cm

rm v ( r ).2 r dr 2U 1 2 r dr

a

m U a 2.35 10 10 0.5 cm

therefore m 0.185 g / s

SOLUTION 3

c.

d.

laminar range

3z ,avg

4 3

w4

2

v m / A U 10 cm / s

2.35 10 10 1.0UdRe

5.40 10

4.36 10

w w

22

a a

z z0 0w

1 22 2 2 2 2w w0

rJ v v 2 r dr 2U 1 2 r dr

a

4J 4 U a 1 2 d U a

3

SOLUTION 3

6( where 540 10 poise )

But since

e.

2w

32

m U a we find :

4 4 g.cmJ mU 0.185 10 246.5

3 3 s

Note that m / A J / m

w

w

2a

zz0

32

a

0

vK v 2 r dr

2

2U 1 r / a.2 r dr

2

SOLUTION 3

Note that:

that is,

f. would not influence

Discuss.

1/ 2m J 2K

A m m

v ( r,z ) 0 zm,J ,K if v were

313 2 2w 0

2 6

5 2 3

4 U a . 1 2 d

2m U / 2 2 0.185 10 / 2

K 1.85 10 g cm / s

1/ 24U U 2 U

3

2

w2U 1 r / a

SOLUTION 3

g.

Thus

rz rz www

4 U0 at r 0 and

a

4 3

w 2

wf

2

4 5.40 10 10 dyne4.32 0.452 Pa

0.5 cm

16C

1 ReU2

SOLUTION 3

therefore

h. Descriptors:

Continuum Laminar

“Incompressible” Quasi-one-dimensional

Newtonian (viscous) Internal

Steady Single-Phase

2f 2

16C 3.67 10

4.359 10

SOLUTION 3

Estimate the drag force (Newtons) per meter of length for

each of the following long objects of transverse dimension 5

cm if placed in a heated air stream with the following

properties.

a. A circular cylinder.

b. A thin “plate” perpendicular to the stream (i.e., at 90o

incidence).

c. A thin plate aligned with the stream (i.e., at 0o incidence).

T 1200 K , p 1 atm, and U 10 m / s.

PROBLEM 4

In each case qualitatively discuss how the drag is apportioned

between “form” (pressure difference) drag and “ friction” drag

For Part ( c), is the application of laminar boundary-layer

theory likely to be valid? (Briefly discuss your reasoning) If so,

what would be the estimated BL thickness, , at the trailing

edge of the plate, i.e., at x=L? Suppose two such adjacent

plates were separated by a distance much greater than

--- would they strongly “interact” with respect to momentum

transfer?

( L )

PROBLEM 4

d. Justify the use of an incompressible Newtonian fluid

CD(Re, shape) curve to solve Part (a) ( involving the gas

air) by showing that is small enough under these

conditions to neglect .

21Ma

2 air

/

PROBLEM 4

Momentum Transfer to (Drag on) Immersed Objects

Drag/meter of axial length=? for objects of transverse

dimension 5 cm. in U =10 m/s, air @ 1 atm., 1200 K.

a. Cylinder in Cross flow

SOLUTION 4

4

43

2

4.7 10 poise

1 28.97pM g2.94 10

RT 82.06 1200 cm

v 1.60 cm / s

3

w2

3

D graph2

10 cm / s 5 cmUdRe

v 1.60 cm / s

3.13 10

Drag / LengthC 0.95

1U Frontal Area / Length

2

SOLUTION 4

SOLUTION 4

Frontal area/meter=5 cm x 100 cm = 5 x 102 cm2 /m

Therefore Drag/Length

22 4 3 22

1 1 dyneU 2.94 10 10 1.47 10

2 2 cm

2 2 2

4

4 5

1.47 10 0.95 5 10 cm

dyne6.99 10

meter

dyne N6.99 10 . 10

m dyne

0.699 N / m.

SOLUTION 4

Most of this drag is due to the p() distribution- that is, “ form” drag.

b. Plate Normal to flow: check literature

c. Plate Aligned with flow:

SOLUTION 4

In this case

and

Since ReL<106 (approx.) we expect flow in the momentum

defect Boundary Layer to be laminar. Then

where

f 1/ 2

L

1.328C ( Blasius )

Re

1/ 21/ 2 3 1LRe 3.13 10 5.59 10

L

ULRe

v

3LRe 3.13 10

SOLUTION 4

and

But total wetted area/meter=(2)(5x102)=103 cm2/m. Therefore

or

2f 1

2

2

1.328C 2.37 10

5.59 10Drag ( both sides )

therefore dimensionless1

U total wetted area2

2.37 10

SOLUTION 4

2

2 2 3 52

dyne cm NDrag 2.37 10 1.47 10 10 10

cm m dyne

2Drag 3.49 10 newtons

This drag is entirely due to - i.e., it is “friction drag”

2plates could be

brought to within

1cm apart w/o

interference

1/ 2 1L

5 55LL 0.45 cm therefore

Re 5.59 10

w x

SOLUTION 4

Reconsider the sonic jet test facility specified in problem1

from the viewpoint of turbulent jet momentum exchange

(mixing) with the surrounding atmosphere, and the

entrainment of that atmosphere

a. Calculate the Reynolds’ number at the

nozzle exit and compare it to the value above

which such jets almost certainly lead to turbulent mixing

with the surrounding atmosphere.

PROBLEM 5

j j jRe Ud / v43 10

b. Estimate the appropriate value of by assuming that the relevant

density is about the arithmetic mean between . How much

larger is the effective turbulent momentum diffusivity, , than the

intrinsic momentum diffusivity of the jet fluid ?

c. Estimate the downstream distance at which the time-averaged

velocity (axial momentum per unit mass) along the jet centerline

will be reduced to 10% of the initial jet velocity (axial

momentum per unit mass) as a result of momentum diffusion.

Compare this to the result that would have been obtained had the

jet

PROBLEM 5

tv

j and

tv j j jv /

z

jU

remained laminar (with kinematic viscosity ).

d. At this location what is the approximate ratio between the

entrained (laboratory air) mass flow and the “primary”

(combustion-heated air) jet?

j

PROBLEM 5

Momentum Transfer: Turbulent Round Jet

SOLUTION 5

Momentum Transfer: Turbulent Round Jet

is determined by . Using

we obtain:

4jet

6 2

cmJ mU 79.125 8.028 10

s

6.35 10 gm.cm / s

1/ 2J / j / 2

4 36.9 10 g / cm

SOLUTION 5

t

Therefore

that is,

Where does drop to Uj/10. For a turbulent jet:

1/ 2 5 3 2t

2j

J / 0.96 10 v 1.55 10 cm / s

cf . v 2.96 cm / s la min ar ( fluid ) momentum diffusivity

2t jv / v 5.2 10 zv 0,z

j

1/ 2

z

U / 10

3 1 J 1v 0,z . .

8 0.0161 z

SOLUTION 5

We find: z=88.6 cm =0.89 m

(This would have been 0.46 km if there had been no

turbulent enhancement in momentum diffusion.)

where

entrained primary jetm / m at this location?

SOLUTION 5

4 3

1/ 2 5 2

6.9 10 g / cm

J / 0.96 10 cm / s

1/ 2

entrained

Jm 8 . 0.0161 z ( turbulent case ),

therefore

so that at

cf.

Therefore

23

t

cmv 1.55 10

s

3entrained

2j

z 88.6 cm

m 4.05 10 g / s

m 0.792 10 g / s

jentrainedz

jet exit

Um51 at location where v 0,z

m 10

SOLUTION 5

Exercises:

1. Calculate the time-averaged profile at this

location.

2. Can you estimate the centerline, time averaged

temperature and CO2(g) concentration at this point?

(Itemize and discuss the underlying assumptions.)

SOLUTION 5

zv r,z