03-Control of Primary Patriculates

8/9/2019 03-Control of Primary Patriculates

1/239

1

Control of

Primary Particulates

Hsin Chu

ProfessorDept. of Environmental Engineering

National Cheng Kung University


2/239

2

Many of the primary particles, e.g., asbestosan heavy metals, are more to!ic than mostseconary particles.

"he average engineer is more li#ely toencounter a primary particle control problemthan any other type of air pollution problem.


3/239

3

1. Wall Collection Devices

"he first three types of control evices $econsier%gravity settlers, cyclone separators,an electrostatic precipitators%all function by

riving the particles to a soli $all, $here theyahere to each other to form agglomeratesthat can be remove from the collectionevice an ispose of.


4/239

4

1.1 Gravity Settlers

"he mathematical analysis for gravity settlersis very easy& it $ill reappear in moifie formfor cyclones an electrostatic precipitators.

Ne!t slie '(ig. ).*+ gravity settler.


5/239


6/239

6

"o calculate the behavior of such a evice,engineers generally rely on one of t$omoels.

Either $e assume that the flui going throughis totally unmi!e 'bloc# flo$ or plug flo$moel+ or $e assume total mi!ing, either inthe entire evice or in the entire cross sectionperpenicular to flo$ 'bac#mi!e or mi!emoel+.


7/239

7

"he observe behavior of nature most often fallsbet$een these t$o simple cases.

(or either bloc# or mi!e flo$, the average hori-ontalgas velocity in the chamber is

avg

Qv

WH=


8/239

8

(or the bloc# flo$ moel, $e assume'*+ "he hori-ontal velocity of the gas in the chember

is eual to vavgevery$here in the chamber.

'/+ "he hori-ontal component of the velocity of the

particles in the gas is al$ays eual to vavg.'0+ "he vertical component of the velocity of theparticles is eual to their terminal settling velocityue to gravity, vt.

'1+ 2f a particle settles to the floor, it stays there anis not re%entraine.


9/239

9

Consier a particle that enters the chamber someistance h above the floor of the chamber.

"he length of time the gas parcel it entere $ith $illta#e to traverse the chamber in the flo$ irection is

avg

Lt

=


10/239

10

During that time the particle $ill settle by gravity aistance

2f this istance is greater than or eual to h, then it$ill reach the floor of the chamber an be capture.

vartical settling distance t tavg

Lt

= =


11/239

11

2f all the particles are of the same si-e 'an hencehave the same value of vt+, then there is some

istance above the floor 'at the inlet+ belo$ $hich allof the particles $ill be capture, an above $hich

none of them $ill be capture.

2f all the particles are istribute uniformly across theinlet of the chamber, the fractional collectionefficiency is

Fraction captured = for block flow (1)t

avg

LH

=


12/239

12

"o compute the efficiency%particle iameterrelationship, $e replace vtin E. '*+ $ith the gravity%

settling relations escribe in last chapter '3to#es4la$ euation+ an fin

2

for block flow (2)18

p

avg

LgD

Hv

=


13/239

13

No$ to consier the mi!e flo$ moel, $eassume that the gas flo$ is totally mi!e inthe - irection but not in the ! irection.

Mi!ing in the - irection leas to a ecreasein collection efficiency.


14/239

14

5e then consier a section of the settler $ith length!.

2n this section the fraction of the particles that reachthe floor $ill eual the vertical istance an averageparticle falls ue to gravity in passing through thesection, ivie by the height of the section, $hich$e may $rite as

Fraction collected=

tdt

H


15/239

15

"he change in concentration passing this section is

"he time the average particle ta#es to pass throughthis section is

(fraction collected) tc dt

dc cH

= = g

avg

dxdt

v

=


16/239

16

Combining these euations an rearranging, $e have

5hich $e may integrate from the inlet '! 6 7+ to theoutlet '! 6 8+, fining

t

avg

dcdx

c H

=

ln out t

in avg

c Lc H

=


17/239

17

9r

(inally $e can substitute for vtfrom 3to#es4 la$,

fining

( ) ( )1 1 expout t in avg

c L

c H

= =

( )2

18mixed flow1 exp (3)p

avg

LgD

H

=


18/239

18

Comparing this result $ith that for the bloc# flo$, E.'/+, $e see that E. '0+ can be re$ritten as

( )1 exp (4)mixed block flow =


19/239

19

Example 1

Compute the efficiency%iameter relation for agravity settler that has H 6 / m, 8 6 *7 m, an

vavg6 * m:s for both the bloc# an mi!e flo$moels, assuming 3to#es4 la$.

Solution:

Here $e can get the result using only onecomputation an then using ratios.


20/239

20

(irst $e compute the bloc# efficiency for a *%;particle, vi-.,

"he mi!e assumption leas to practically the sameresult, vi-.,

2 2 2 34

!

(1" )(#$81 % )(1" ) (2""" % )3$"3 1"

18 (18)(1$8 1" % % )(2 )(1 % )

p

block

avg

LgD m m s m kg m

H kg m s m m s

= = =

4 41 exp( 3$"3 1" ) 3$"2# 1"mixed

= =


21/239

21

"o fin the efficiencies for other particleiameters, $e observe that the bloc#efficiency is proportional to the particle

iameter suare, so $e ma#e up a table ofbloc# flo$ efficiencies by simple ratios to thevalue for *;, an then compute thecorresponing mi!e flo$ efficiencies as


22/239

22

Particle iameter, ; =bloc#'>+ =mi!e'>+

* 7.7070 7.7070

*7 0.70 /.)?

07 /@.0 /0.)A7 @B.7 A0.7

[email protected] *77.7 B0.7

?7 % ?B.7

*77 % )A.7

*/7 % )).7

"hese values are sho$n in (ig. )./ 'ne!t slie+.


23/239


24/239

24

(rom E. '/+, , the gravity settler $oul beuseful for collecting particles $ith iameters ofperhaps *77; 'fine san+ but not for particles of airpollution interest.

5e $oul increase the efficiency by ma#ing 8 larger'e!pensive+, by ma#ing H smaller '$hich issometimes one by subiviing the chamber $ithhori-ontal plates, $hich ma#es the cleanup much

more ifficult+, by lo$ering vavg'e!pensive+, or byincreasing g.

2

18

p

avg

LgD

H

=


25/239

25

1.2 Centrifugal Separators

2f a boy moves in a circular path $ith raius r anvelocity vcalong the path, then it has angular velocity

6 vc: r, an2

2

cm

Centrifugal force m r r

= =


26/239

26

Example 2

particle is traveling in a gas stream $ith velocity B7ft:s '*? m:s+ an raius * ft.5hat is the ratio of centrifugal force to the gravityforce acting on it

Solution:2 2

2

% (" % ) %1111$8 &

32$2 %

cm rCentrifugal force ft s ft

Gravity force mg ft s

= = =


27/239

27

(or further $or# $e $ill use a centrifugaleuivalent of 3to#es4 la$.

"o obtain the centrifugal euivalent, $e nee

only substitute the centrifugal force for thegravitational force.

Ne!t slie '(ig. ).0+"he terminal settling velocity in the raialirection vtan the velocity along the

circular pathvc.


28/239


29/239

29

No$ substituting the centrifugal acceleration for thegravitational one in the euation for vtan ropping

the fterm, $e fin

2 2

' (!)18

c pt centrifugal

Dr

=


30/239

30

Example 3

Fepeat the computation of the terminal settlingvelocity sho$n in E!ample * of last chapter for aparticle in a circular gas flo$ $ith velocity vc6 B7 ft:s

'*?./) m:s+ an raius * ft '7.071? m+. "he ensity ofthe flui can be ignore.

Solution:2 2 3

!

(18$2# % ) (1" ) (2""" % )"$""8 % "$8 %(18)(1$8 1" % % )("$3"48 )

t

m s m kg mm s cm skg m s m

= = =


31/239

31

"his ans$er is **/ times as large as thevalue foun in E!ample * of last chapter.

9ne may compute the particle Feynols

number here, fining that is about 7.7771B. Hence the assumption of a 3to#es4 la$ type

of rag seems reasonable.


32/239

32

"here are many types of cyclones, but themost successful is s#etche is (ig. ).1 'ne!t slie+.

cyclone consists of a vertical cylinricalboy, $ith a ust outlet at the conical bottom.


33/239


34/239

34

During the outer spiral of the gas the particlesare riven to the $all by centrifugal force,$here they collect, attach to each other, an

form larger agglomerates that slie o$n the$all by gravity an collect in the ust hopperin the bottom.


35/239

35

"hus the outer heli! is euivalent to thegravity settler.

"he inlet stream has a height 5iin the raial

irection, so the ma!imum istance anyparticle must move to reach the $all is 5 i.

"he comparable istance in a gravity settler isH.


36/239

36

"he length of the flo$ path is NGD7, $here N

is the number of turns that the gas ma#estraversing the outer heli! of the cyclone,

before it enters the inner heli!, an Dois theouter iameter of the cyclone.

"his length of the flo$ path correspons to 8in the gravity settler.


37/239

37

Ma#ing these substitutions irectly into the gravitysettler euations, Es. '/+ an '0+, $e fin

an

" ()

tblock

i c

N D

W

=

"

1 exp ()tmixedi c

N D

W

=


38/239

38

2f $e then substitute the centrifugal 3to#es4 la$e!pression, E. 'A+, into these t$o euations, $e fin

an

value of N 6 A represents the e!perimental atabest.

2

(8)

#

c p

block

i

N D

W

=

2

1 exp (#)#

c p

mixed

i

N D

W

=


39/239

39

Example 4

Compute the efficiency%iameter relation for a cycloneseparator that has 5i6 7.A ft, vc6 B7 ft:s, an N 6 A, for

both the bloc# an mi!e flo$ assumptions, assuming

3to#es4 la$. Solution:

(or a *%; particle,2 2 2 3

2 4

( )(!)(" % )(1" ) (3$28 % ) (124$8 % )=

# (#)("$! )(1$8 1" ) $2 1" %( ) "$"232

c p

block

i

N D ft s m ft m lbm ft

W ft cp lbm ft s cp

=

=


40/239

40

"hen as $e i in E!ample *, $e can use this number,plus the fact that the particle iameter enters the euationto the secon po$er, to ma#e up the follo$ing table

Particle iameter, ; =bloc#

'>+ =mi!e

'>+

7.**/0

1AB.AA)

*7*A

7.7/0//.0/).07

/7.)

0@./A?./*77.7

%%

7.7/0//.07?.??

*?.)

0*.*11.*B0./)7./)).A


41/239

41

Comparing this result to that for gravitysettling chambers in E!ample *, $e see theform of the result is the same, but the

ma!imum particle si-e for $hich the evice iseffective is much smaller.

Ne!t $e introuce a ne$ term, the cutiameter, $hich is $iely use in escribing

particle collection evices.


42/239

42

2f $e consiere only spherical peas in acolaner 'a sheet metal ish $ith uniform,circular holes+, then the cut iameter $oul

be the iameter of the holes. (or peas larger than the cut iameter the

collection efficiency $oul be *77>, an forthose smaller it $oul be 7>.


43/239

43

(or all practical particle collection evices theseparation is not that sharp.

"here is no single iameter at $hich the

efficiency goes suenly from 7> to *77>. "he universal convention in the air pollution

literature is to efine cut iameter as theiameter of a particle for $hich the efficiencycurve has the value of A7>.


44/239

44

5e can substitute this efinition into E. '?+ ansolve for the cut iameter that goes $ith 3to#es4 la$,bloc# flo$ moel, fining 'Fosin%Fammler euation+

lthough one might logically e!pect that E. ')+, $ithits more realistic mi!e flo$ moel, $oul better

represent e!perimental ata, E. '?+ appeare in theliterature earlier.

1% 2

# (1")2

icut

c p

WDN

=


45/239

45

Example 5

Estimate the cut iameter for a cyclone $ith inlet$ith 7.A ft, vc6 B7 ft:s, an N 6A.

Solution:

"his e!ample sho$s that for a typical cyclone si-ean the most common cyclone velocity an gasviscosity, the cut iameter is about A;.

1% 2!

3

(#)("$! )(1$8 1" % % )4$3 1" ! &

2 (!)(" % )(2""" % )cut

ft kg m sD m

ft s kg m

= =


46/239

46

Comparing this calculation $ith that inE!ample 1 sho$s that the cut iameter $e$oul calculate by the mi!e moel is

some$hat larger, but not ramatically so. "he cyclone $or#s $ell on a gas stream

contains fe$ particles smaller than A;, e.g.,sa$ust from $oo shops an $heat grains

from pneumatic conveyers.


47/239

47

(rom E. '*7+, suppose $e $ish to apply acyclone for even smaller particles, thealternatives are to ma#e 5ismaller or vclarger.

Ienerally $e cannot alter the gas viscosity orthe particle ensity.

Ma#ing vclarger is generally too e!pensive

because, the pressure rop across a cycloneis generally proportional to the velocitysuare.


48/239

48

"o ma#e 5ismaller, $e must ma#e the $holecyclone smaller if $e are to #eep the sameratios of imensions.

Jut the inlet gas volumetric flo$ isproportional to 5isuare, so that a smallcyclone treats a small gas flo$.

ery small cyclones have been use to collect

small particles from very small gas flo$s forresearch an gas%sampling purposes, but theinustrial problem is to treat large gas flo$s.


49/239

49

3everal practical schemes have been $or#eout to place a large number 'up to severalthousan+ small cyclones in parallel, so that

they can treat a large gas flo$, capturingsmaller particles.

"he most common of these arrangements,calle a multicyclone '(ig. ).A, ne!t slie+.


50/239


51/239

51

2f the iniviual cyclone $ere one%half foot iniameter, the 5iin E. '*7+ $oul be about

7.*/A ft.

Fepeating E!ample A for a 5iof 7.*/A ft, $efin a preicte cut iameter of /.0 ;.


52/239

52

Ne!t slie '(ig. ).B+"he comparisons of the preictions of Es. '?+ an')+ an the follo$ing totally empirical ata%fittingeuation

2

2

( % ) (11)

1 ( % )

cut

cut

D D

D D=

+


53/239


54/239

54

"he lo$ collection efficiency, about ?7>, of atypical cyclone sho$s that it cannot meetmoern control stanars 'usually L )A>

reuire control efficiency+ for any particlegroup that has a substantial fraction smallerthan A ; in iameter.


55/239

55

Example 6

2n E!ample A ho$ long oes the gas spen in thehigh centrifugal force fiel near the $all $here aparticle has a goo chance of being capture

Solution:(ollo$ing the assumptions leaing to Es. 'B+ an'@+, ! 2

"$!2!

" %

o

c

N DL ftt s

ft s

= = = =g


56/239

56

"he istance the particle can move to$arthe $all is eual to the prouct of this timean vt, but vtis proportional to vcsuare, so

that to get better collection efficiency $e mustgo to lo$er an lo$er times in the cyclone.

Previously $e state that the typical velocityat a cyclone inlet is B7 ft:s an that this

velocity is selecte for pressure rop reasons.


57/239

57

(or a given cyclone one $ill generally fin that thepressure rop, for various conitions, can berepresente by an euation of the form

$here gis the gas ensity an viis the velocity at

the inlet to the cyclone.

2

(12)2

g i

in out P P P

= =


58/239

58

ll suen e!pansions have a K of *.7, allsuen contraction have a K of 7.A, etc.

Most cyclone separators have Ks of about ?.

2t is common in air%conitioning esign torefer to the uantity 'g v/ : /+ as a velocity

hea, so one coul say that most cycloneshave pressure rops of about ? velocityheas.


59/239

59

Example 7

cyclone has an inlet velocity of B7 ft:s an areporte pressure loss of ? velocity heas. 5hat isthe pressure loss in pressure units

Solution:

2 2 2 2

2

2

2

18("$"!)(" % ) ( % 32$2( ))( %144 $ )

2

"$23 % $

1'" % 1$1 "$23 $4 $ &

P ft s lbf s lbm ft ft in

lbf in

N m kPa psi in H !

=

=

= = = =


60/239

60

1.3 Electrostatic Precipitators (ESP)

"he electrostatic precipitator is li#e a gravitysettler or centrifugal separator, butelectrostatic force rives the particles to the

$all. 2t is effective on much smaller particles than

the previous t$o evices.


61/239

61

2n all three #in of evices, the viscous'3to#es4 la$+ resistance of the particle to beriven to the $all is proportional to the particle

iameter '(6 0G;Dv+. (or gravity an centrifugal separators, the

force that can be e!erte is proportional tothe mass of the particle, $hich, for constant

ensity, is proportional to the iameter cube.


62/239

62

"hus the ratio of riving force to resistingforce is proportional to 'iametercube:iameter+ or to iameter suare.

s the iameter ecreases, this ratio fallsrapily.


63/239

63

2n E3Ps the force moving the particle to$arthe $all is electrostatic.

"his force is practically proportional to the

particle iameter suare, an thus the ratioof riving force to resisting force isproportional to 'iameter suare:iameter+or to the iameter.


64/239

64

"he basic iea of all E3Ps is to give theparticles an electrostatic charge an then putthem in an electrostatic fiel that rives them

to a collecting $all. "his is an inherently t$o%step process.

2n one type of E3P, calle a t$o%stageprecipitator, charging an collecting are

carrie out in separate parts of the E3P 'usein builing air conitioners+.


65/239

65

(or most inustrial applications, the t$oseparate steps are carrie out simultaneouslyin the same part of the E3P.

"he charging function is one much moreuic#ly than the collecting function, an thesi-e of the E3P is largely etermine by thecollecting function.


66/239

66

Ne!t slie '(ig. ).@+ $ire%an%plate E3P $ith t$o plates.


67/239


68/239

68

"he gas passes bet$een the plates, $hichare electrically groune 'i.e., voltage 6 7+.

Jet$een the plats are ro$s of $ires, hel at a

voltage of typically %17,777 volts.


69/239

69

"he po$er is obtaine by transformingorinary alternating current to a high voltagean then rectifying it to irect current through

some #in of soli%state rectifier. "his combination of charge $ires an

groune plates prouces both the freeelectrons to charge the particles an the fiel

to rive them against the plates.


70/239

70

9n the plates the particles lose their chargean ahere to each other an the plate,forming a ca#e.

3oli ca#es are remove by rapping theplates at regular time intervals $ith amechanical or electromagnetic rapper thatstri#es a vertical or hori-ontal blo$ on the

ege of the plate.


71/239

71

(or liui roplets the plate is often replaceby a circular pipe $ith the $ire o$n itscenter.

3ome E3Ps 'mostly the circular pipe variety+have a film of $ater flo$ing o$n thecollecting surface, to carry the collecteparticles to the bottom $ithout rapping.


72/239

72

"here are many types of E3Ps.

(ig. ).? 'ne!t slie+ sho$s one of the mostcommon in current use.


73/239


74/239

74

Each point in space has some electricalpotential .

2f the electrical potential changes from place

to place, then there is an electrical fiel, E 6O : O!, in that space.

"he units of E are :m.


75/239

75

2n a typical $ire%an%plate precipitator, theistance from the $ire to the plate is about 1to B in., or 7.* to 7.*A m.

5ith a voltage ifference of 17 # an 7.* mspacing, one $oul assume a fiel strength of17 #:7.* m 6 177 #:m.

"his is inee the fiel strength near the

plate.


76/239

76

"he surface area of the $ires is much lo$erthan that of the plate& thus, by conservation ofcharge, the riving potential near the $ires

must be much larger. "ypically it is A to *7 M:m.


77/239

77

5hen a stray electron from any of a variety ofsources encounters this strong a fiel, it isaccelerate rapily an attains a high

velocity. 2f it then collies $ith a gas molecule, it has

enough energy to #noc# one or moreelectrons loose, thus ioni-ing the gas

molecule.


78/239

78

"hese electrons are li#e$ise accelerate bythe fiel an #noc# more electrons loose, untilthere are enough free electrons to form asteay corona ischarge.

"he positive ions forme in the coronamigrate to the $ire an are ischarge.

"he electrons migrate a$ay from the $ire,

to$ar the plate.


79/239

79

9nce they get far enough a$ay from the $irefor the fiel strength to be too lo$ toaccelerate them fast enough to ioni-e gasmolecules, the visible corona ceases anthey simply flo$ as free electrons.

s the electrons flo$ to$ar the plate, theyencounter particles an can be capture by

them, thus charging the particles.


80/239

80

(or particle larger than about 7.*A ;, theominant charging mechanism is fielcharging.

s the particles become more highly charge,they ben the paths of the electrons a$ayfrom them.


81/239

81

"hus the charge gro$s $ith time, reaching a steaystate value of

here is the charge on the particle, an is theielectric constant of the particle Qa imensionlessnumber that is *.7 for a vacuum, *.777B for air, an 1to ? for typical soli particles.

"he permittivity of free space ois a imensionless

constant $hose value in the 32 system of units is ?.?A! *7%*/C:' R m+. D is the particle iameter, an Eois the local fiel

strength.

23 (13)2

o o" D #

= +


82/239

82

Example 8 *%; iameter particle of a material $ith a ielectric

constant of B has reache its euilibrium charge in anE3P at a place $here the fiel strength is 077 #:m.

Ho$ many electronic charges has it Solution:

( ) ( ) ( )2

12

1#1

3 8$8! 1" % % 1" 3"" %

2

1$"2 1"1$88 1" 3"" &

" C $ m m k$ m

electronsC electrons

C

= +

= =


83/239

83

2f the particle is smaller than about 7.*A ;,then $e $oul ma#e a serious errorcomputing its charge by E. '*0+.

5e must consier the aitional charge thatis acuire by iffusion charging, resultefrom particle%electron collisions cause by theranom motion superimpose on that motion

by electron%gas molecule collisions, $hichma#e the electron behave li#e a gas molecule$ith a Jolt-mann velocity istribution.


84/239

84

"he electrostatic force on a particle is

Here Epis the local electric fiel strength causing theforce.

2f $e substitute for from E. '*0+, $e fin

"he t$o subscripts on the Es remin us that onerepresents the fiel strength at the time of charging,the other the instantaneous 'local+ fiel strength.

p% "#=

23 (14)2

o o p% D # #

= +


85/239

85

(or all practical purposes $e use an averageE& an in the rest of this chapter $e $ill useEo6 Ep6 E.

2f the particle4s resistance to being riven tothe $all by electrostatic forces is given by the3to#es rag force, $e can set the resistanceforce eual to the electrostatic force in E.

'*1+ an solve for the resulting velocity,fining


86/239

86

"his velocity is calle the rift velocity in the E3P

literature, an is given the symbol .

( )2 2 (1!)

o

t

D #

+= =


87/239

87

Example 9 Calculate the rift velocity for the particle in E!ample

?.

Solution: 12 ! 2

! 2

(1" )(8$8! 1" % % )(3 1" % ) ( % 8) ( %( ))

(1$8 1" % % )( %( ))

"$"33 % 3$3 % "$1"# % &

m C $ m $ m N m C $

kg m s N s kg m

m s cm s ft s

=

= = =


88/239

88

E. '*A+ sho$s that the rift velocity isproportional to the suare of E, $hich isappro!imately eual to the $ire voltageivie by the $ire%to%plate istance.

2f $e coul raise the voltage or lo$er the $ire%to%plate istance, $e shoul be able toachieve unlimite rift velocities.

"he limitation here is spar#ing.


89/239

89

9ccasionally an ioni-e conuction path $illbe forme bet$een the $ire an the plate&this ioni-e path is then a goo conuctoran forms a continuous staning spar#.

"he po$er supply to the $ire must sense thissuen increase in current an stop the flo$into it to prevent a burnout of the transformer.


90/239

90

Normally the current is shut off for a fractionof a secon, the lighting stro#e ens, anthen the fiel is reestablishe.

s one raises the values of E, the freuencyof spar#s increases.

"hese spar#s are energetic events thatisrupt the ca#e on the plate, thus reucing

the collection efficiency, so a large number ofspar#s are ba.


91/239

91

Most E3P control systems are set for aboutA7 to *77 spar#s per minute.

(urthermore, it is common practice to

subivie the po$er supply of a largeprecipitator into many subsupplies so thateach part of the precipitator can operate atthe optimum voltage for its local conitions.


92/239

92

5hen $e compare the rift velocity here $iththe terminal settling velocity compute for thesame particle in a cyclone separator inE!ample 0, $e see that this is only about fivetimes as fast.

5hy then is an E3P so much more effectivethan a cyclone for fine particle collection


93/239

93

"o obtain a high rift velocity in a cyclone, onemust use a high gas velocity, thus, the resiencetime of the particle in a cyclone is very short.

9n the other han, the gas velocity oes not

enter E. '*A+, $e can ma#e the precipitatorlarge enough that the particle spens a longtime in it an has a high probability of capture.

"ypical moern E3Ps have gas velocity of * to /

m:s, an the gas spens from 0 to *7 seconsin them.


94/239

94

2f $e no$ consier the section bet$een thero$ of $ires an one plate on (ig. ).@, $esee its collecting area is

6 8 h

"he volumetric flo$ through the section isS 6 Hhvavg


95/239

95

Ma#ing these substitutions in Es. '*+ an '1+, $efin

an

E. '*@+ is the Deutsch%nerson euation, the most

$iely use simple euation for esign, analysis, ancomparison of E3Ps.

(1)block&

Q

=

1 exp (1)mixed&

Q

=


96/239

96

Example 10 Compute the efficiency%iameter relation for an E3P

that has particles $ith a ielectric constant of B an':S+ 6 7./ min:ft. $e $ill use only the mi!e flo$

euation. Solution:

Using the results of E!ample ) $e #no$ that a *%;iameter particle $ill have a rift velocity of 7.*7) ft:s,

an that the rift velocity $ill be linearly proportionalto the particle iameter.


97/239

97

"hus for a *%; particle $e may compute

5e can then ma#e up a table using the rift velocity

proportional to the particle iameter

( )"min1 exp ("$1"# % )("$2 min% ) ="$3sft s ft =

Particle iameter, ; = '>+

7.*7.A*0A

*/[email protected])?.7)).?


98/239

98

"his e!ample sho$s that this fairly typicalprecipitator has a cut iameter of about 7.A ;,one%tenth of that of a typical cyclone.

E. '*@+ suggests that if $e pass a particle%laen gas stream through variousprecipitators, all of the ata for this stream $illform a straight line on a plot of log p vs. :S,

$here p 6 *%=. (ig. ).) 'ne!t slie+ is such a plot.


99/239


100/239

100

Example 11 (rom (ig. ).), estimate the value of for coal

containing *> sulfur.

Solution:

(rom the figure at )).A> efficiency $e rea that for*> sulfur coal,

(rom E. '*@+

2

3

31""$31

1""" %

& ft min

Q ft min ft = =

ln ln "$""!1 1$"# "$28 8$ &

% "$31 % min

p ft ft cm

& Q min ft s s= = = = =


101/239

101

"he ifferent lines for ifferent coal sulfurcontents on (ig. ).) are cause by sulfur4sinirect effect on fly ash resistivity 'iscusselater+.

E3Ps $or# $ell $ith meium%resistivelysolis, but poorly $ith lo$%resistivity or high%resistivity solis.

5e can see $hy by referring to (ig. ).*7'ne!t slie+.


102/239


103/239

103

(or the case of a lo$%resistivity soli, e.g., carbonblac#, the material forms a ca#e that is a gooconuctor of electricity.

"he voltage graient in the ca#e is small.

9n reaching the plate the particles areischarge an hence there is very littleelectrostatic force holing the collecte particlesto the plate.

"he collecte particles o not ahere an areeasily re%entraine& the overall collection is poor.


104/239

104

(ig. ).*7 also sho$s particles of very highresistivity, e.g., elemental sulfur, on the plate.

Here most of the voltage graient occurs

through the ca#e, causing at least t$oproblems.

(irst, the voltage graient near the $ire hasno$ fallen so much that it cannot prouce a

goo corona ischarge.


105/239

105

3econ, the voltage graient insie the ca#eis so high that in the gas spaces bet$een theparticles stray electrons $ill be accelerate tohigh velocities an $ill #noc# electrons off of

gas molecules an form a bac# corona insiethe ca#e.

"his bac# corona is a violently energeticconversion of electrostatic energy to thermal

energy that causes minor gas e!plosions,$hich blo$ the ca#e off the plate an ma#e itimpossible to collect the particles.


106/239

106

"he practical resistivity range is grater than*7@an less than / ! *7*7ohm T cm.

2f the resistivity of the particles is too lo$,

complete combustion may be the onlysolution.

2f the resistivity is too high, there are somepossibilities.


107/239

107

"he resistivity of many coal ashes is too high at077o( for goo collection, but satisfactorily lo$ atB77o(.

"hus Hot%sie E3Ps are use in some coal%fire

po$er plants. 2f one coul conense on its surface a hygroscopic,

conucting material, the ash resistivity $oul bereuce.

3ome of the sulfur in coal is converte in thefurnace to 390, $hich collects on the ash, absorbs

$ater, an ma#es the ash more conuctive.


108/239

108

nother approach to the ash resistivityproblem is to separate the charging ancollecting functions.

2f the particles are charge in a separatecharger, one can use a higher voltage an not$orry much about the resulting spar#s,because they o not pass through the ca#e

an isrupt it.


109/239

109

"able )./ 'ne!t slie+ sho$s somerepresentative values of 'effective riftvelocity+ for inustrial precipitators.


110/239


111/239

111

Example 12 9ut E3P has a measure efficiency of )7>.

5e $ish to upgrae it to ))>. Jy ho$ much must $eincrease the collecting area

Solution:Using E. '*@+

1 "$1 exp

1 "$"1 exp

existing

existing existing

newnew new

&p

Q

&p

Q

= = =

= = =


112/239

112

Unfortunately life is harer than that.2n E. '*A+ $e foun that the rift velocity isproportional to the particle iameter.

"hus, as the percentage efficiency by $eight

increases, the remaining particles become smalleran smaller an harer an harer to collect.

%ln "$1 "$!ln "$"1 %

t*us' 2 &

existing existing

new new

new

existing

& Q && Q &

&

&

= = =

=


113/239

113

"o ta#e this phenomenon into account, someesigners use a moifie Deutsch%nersoneuation $ith the form

5here # is an arbitrary e!ponent, typically about 7.A.

1 exp ( % ) (18)k

p & Q = =


114/239

114

Example 13 Fe$or# E!ample */ using E. '*?+ instea of E.

'*@+, ta#ing # 6 7.A.

Solution:

Here for the e!isting unit ':S+ 6 '%ln p+*:# 6 '%ln 7.*+/6 A.07.(or the upgrae precipitator $e nee ':S+ 6 '%ln7.7*+/6 /*./7.

3o the ne$ value of ':S+%assuming constant %is'/*./7:A.07+ 6 1.7 times the ol value of ':S+.


115/239

115

Example 14 precipitator consists of t$o ientical

sections in parallel, each hanling one%half of

the gas. 2t is currently operating at )A>efficiency. 5e no$ hol the total gas flo$constant, but malistribute the flo$ so thatt$o%thir of the gas goes through one of the

sections, an one%thir through the other.5hat is the preicte overall collectionefficiency


116/239

116

Solution:(or the e!isting situation 'using E. '*@++

(or the ne$ situation

"$"! exp

ln "$"! 2$##!

&p

Q

&Q

= =

= =

1

2

1% 2exp( 2$##!) "$1"!

2 % 3

1% 2exp( 2$##!) "$"111

1% 3

p

p

= = = =


117/239

117

"his e!ample sho$s mathematically ho$malistribution egraes precipitatorperformance.

1 1

2 2

1 2

2 ("$1"!) "$""3

1("$"111) "$""4

3

( ) "$"4

"$"4+ 1 #2$, &

Q p Q Q

Q p Q Q

Q Q p Q

p p

= =

= =

+ == = =


118/239

118

"he typical pressure rop of a E3P is 7.* to7.A in. H/9, much lo$er than that in a cyclone.

"he pressure rop in the ucts leaing to an

from the precipitator is generally more than inthe E3P itself.

"he collection reuirements of E3Ps havebeen pushe from the )7V)A> range typical

in *)BA to the )).A>%plus range no$commonly specifie.

2. Diviing Collection Devices


119/239

119

2. Diviing Collection Devices

(ilters an scrubbers o not rive theparticles to a $all, but rather ivie the flo$into smaller parts $here they can collect theparticles.

2.1 Surface Filters

surface filter is a membrane 'sheet steel,cloth, $ire mesh, or filter paper+ $ith holes

smaller than the imensions of the particlesto be retaine.


120/239

120

Ho$ever, one only nees to poner themechanical problem of rilling holes of 7.*%;iameter or of $eaving a fabric $ith threasseparate by 7.*; to see that such filters arenot easy to prouce.

"hey are much too e!pensive an fragile foruse as high%volume inustrial air cleaners.


121/239

121

lthough inustrial air filths rarely have holessmaller than the smallest particles capture,they often act as if they i.

"he reason is that, as fine particles are

caught on the sies of the holes of a filter,they ten to brige over the holes an ma#ethem smaller.

"hus as the amount of collecte particles

increases, the ca#e of collecte materialbecomes the filter.


122/239


123/239


124/239

124

2f $e follo$ the gas stream in (ig. ).*/ frompoint * to point 0 $e see that the flo$ ishori-ontal an has a small change in velocitybecause the pressure rop, causing the gas

to e!pan, an because the gas is leavingbehin its containe particles.


125/239

125

2n most inustrial filters, both for gases an liuis,the flo$ velocity in the iniviual pores is so lo$ thatthe flo$ is laminar.

"herefore, $e may use the $ell%#no$n relations for

laminar flo$ of a flui in a porous meium, $hichinicate

Here, # is the permeability, a property of the be.

(1#)sQ p k

& x

= =


126/239

126

(or a steay flui flo$ through a filter ca#e supporteby a filter meium, there are t$o resistances to flo$in series, but the flo$ rate is the same through eachof them.

5e fin

3olving for P/, $e get

2 31 2

cake filter

-=s

PP P k k

x x

=

2 1 3

cake filter

=s sx x

P P Pk k

= +


127/239

127

"hen solving for vs

"his euation escribes the instantaneous flo$ ratethrough a filter& it is analogous to 9hm4s la$ for t$oresistors in series.

"he W!:# terms are calle the ca#e resistance an

the cloth resistance.

1 3( ) (2").( % ) ( % ) /

s

cake filter filter

p p Qv

x k x k &

= =

+


128/239

128

"he resistance of the filter meium is usuallyassume to be a constant that is inepenent oftime, so 'W!:#+filteris replace $ith a constant X.

2f the filter ca#e is uniform, then its resistance is

proportional to its thic#ness 1

1

cake

cake

cake

mass of cakex

area

volume of gas mass of solids removed

area volume of gas

=

=


129/239

129

Customarily $e efine

Here 5 is the volume of ca#e per volume of gas

processe, $hich correspons to a collectionefficiency, =, of *.7.

(or most surface filters =6*.7, so the = is normallyroppe in the euation. "hus

Here is the volume of gas cleane.

1

cake

mass of solids removed volume of cakeW

volume of gas volume of gas processed

= =

( ) and (21)cakecake sd x$x W W& dt

= =


130/239

130

3ubstituting E. '/*+ for the ca#e thic#ness in E. '/7+, $efin

(or most inustrial gas filtrations the filter is supplie by acentrifugal blo$er at practically constant pressure, so 'P*%P0+ is a constant, an E. '//+ may be rearrange an

integrate to

1 3(- - )0 1 d= = = (22) dt .( %k /

sW

2

1 3( ) (23)2

$ W $

P P t& k &

+ =


131/239

131

(or many filtrations the resistance X of thefilter meium is negligible compare $ith theca#e resistance, so the secon term of E.'/0+ may be roppe.

"he t$o most $iely use esigns ofinustrial surface filters are sho$n in (igs.).*0 an ).*1 'ne!t an secon slies+.


132/239


133/239


134/239

134

(or the baghouse in (ig. ).*0 there must be some$ay of removing the ca#e of particles thataccumulates on the filters.

Normally this is not one uring gas%cleaning

operations. $ea# flo$ of gas in the reverse irection may also

be ae to help isloge the ca#e, thus eflating thebags.

9ften metal rings are se$n into filter bags at regularintervals so that they $ill only partly collapse $henthe flo$ in reverse.


135/239

135

Jecause it cannot filter gas $hile it is beingcleane, a sha#e%eflate baghouse cannot serveas the sole pollution control evice for a source thatprouces a continuous flo$ of irty gas.

"ypically, for a ma


136/239

136

"he other $iely use baghouse esign,calle a pulse%


137/239

137

"he bags are cleane by intermittent


138/239

138

Example 15 "he sha#e%eflate baghouse on a po$er

station has si! compartments, each $ith **/bags that are ? in. in iameter an // ft long,for an active area of 1B ft/per bag.

"he gas being cleane has a flo$ rate of?B,/17 ft0:min.

"he pressure rop through a freshly cleanebaghouse is estimate to be 7.A in. H/9.


139/239

139

"he bags are operate until the pressure ropis 0 in. H/9, at $hich time they are ta#en out

of service an cleane.

"he cleaning freuency is once per hour.

"he incoming gas has a particle loaing of *0grains:ft0.


140/239

140

"he collection efficiency is ))>, an the filterca#e is estimate to be A7> solis, $ith thebalance being vois.

Estimate ho$ thic# the ca#e is $hen the gagsare ta#en out of service for cleaning. 5hat isthe permeability, #, of the ca#e


141/239

141

Solution:"he average velocity coming to the filter surface

"he A is use here because one of the si!compartment is al$ays out of service for cleaning.

vsis commonly referre to as the air%to%cloth ratio,

face velocity, or superficial velocity.

3

2

824" %3$3! 1$"2

(!)(112)(4 )

s

Q ft min ft m

& ft min min

= = = =


142/239

142

2f the filter remains in service for * hour beforecleaning an vsis constant, the * suare foot of bag

$ill collect the follo$ing mass of particles

( )3

2 2

"13 3$3! "$## (1 )

"""

"$3# 1$8"

s

m gr lbm ft minc t '

& ft gr min '

lbm kg

ft m

= =

= =


143/239

143

"he thic#ness of the ca#e collecte in * hour is

"a#ing

2

3 3 3

3

% "$3# %

(2 % )("$!)(2$4 % % )

!$# 1" "$"1 $ 1$8

m & lbm ft

g cm lbm cm ft g

ft in mm

=

= = =

"' we can solve 5$ (2") for k'filter

x

k

= =

! 2

2

2 212 2 13 2

(3$3! % min)("$"1 %12 $)("$"18 )(2$"# 1" % % )( % " )

( ) (3 $ )(!$2"2 % % $ )

$# 1" $4" 1"

s x ft ft in cp lbf s ft cp min skp in H ! lbf ft in H !

ft m

= =

= =


144/239

144

Compare this $ith values foun in groun $ater flo$

"he calculate permeability of this material is roughlythe same as that of a highly permeable sanstone.

12 2

11 2($# 1" ) "$!

1$" 1"

darcyk ft darcies

ft

= =


145/239

145

"his calculation sho$s that the collecte isabout *.? mm thic#.

2f the cleaning $ere perfect, this $oul be theca#e thic#ness.

Ho$ever, it is har to clean the bagscompletely, an in po$er plant operation it iscommon for the average ca#e thic#ness on

the bags to be up to *7 times this amount.


146/239

146

9ne of the avantages of the pulse%


147/239


148/239

148

2f the superficial velocity increases, theefficiency falls& for a superficial velocity of0.0A m:min the outlet concentration is about/7 percent of the inlet concentration.

"he particles that pass through such a filtero not pass through the ca#e but throughpinholes, $hich are regions $here the ca#e

i not establish properly '(ig. ).*B, ne!tslie+.


149/239


150/239

150

"he pinholes are apparently about *77 ; iniameter, much too large for a single particleto bloc# because there are rarely *77%;particles in the streams being treate.

5hen the superficial velocity is high, morepinholes form.


151/239

151

Example 16 Estimate the velocity through a pinhole in a

filter $ith a pressure rop of 0 in. of $ater.

ssuming that this is the pressure ropcorresponing to the curve for 7.0) m:min on(ig. ).*A, that the steay%state penetration atthat velocity is 7.77*, an that the pinholes

have a iameter of *77 ;, estimate ho$ manypinholes per unit area there are in the ca#e.


152/239

152

Solution:"he flo$ through the pinhole is best escribe by Jernoulli4seuation, from $hich $e fin the average velocity

Here the 'area Tvelocity+ of the pinholes must be 7.77* times the'area 6velocity+ of the rest of the ca#e. Hence

1% 21% 2

2

3 2

2

2 2(3 ) 24#"$1

(1$2" % )

21$! %

pin'oles

P in H ! Pa kgC

kg m in H ! Pa m s

m s

= = =

2

2

"$""1 "$""1 ("$3# % )3$" 1"

(21$! % )(" % )

pin'oles s

cake pin'oles

& m min m

& m s s min m

= = =


153/239

153

Each pinhole has an area of 6'*77 *7%Bm+/'G:1+ 6 @.?A *7%)m/, so there mustbe

"he calculate velocity through a pinhole is '/*.B !B7+ :7.0) 6 0077 times the velocity through the ca#e.

2

# 2

3$" 1"

38 pin*oles%m &$8! 1" m

=

2.2 Dept! "ilter


154/239

154

(ilters o not form a coherent ca#e on thesurface, but instea collect particlesthroughout the entire filter boy are calleepth filters.

"he e!amples $ith $hich the stuent isprobably familiar are the filters on filter%tippecigarettes an the lint filters on many home

furnaces.


155/239

155

2n both of these a mass of ranomly orientefibers 'not $oven to form a single surface+collects particles as the gas passes throughit.

2n (ig. ).*@ 'ne!t slie+, $e see a particle%laen gas flo$ing to$ar a target, $hich $emay thin# of as a cylinrical fiber in a filter.


156/239


157/239

157

"o etermine $hether a particle bumps intothe target or flo$s aroun it, $e can computethe relative velocity bet$een particle an gasusing the appropriate euivalents of 3to#es4

la$. "hat tas# $as first unerta#en by 8angmuir

an Jlogett, an (ig. ).*? 'ne!t slie+

conveniently summari-es the mathematicalsolutions for the small particles of interest inair pollution $or#.


158/239


159/239

159

"o see ho$ they obtaine (ig. ).*?, consiera single particle in a turning part of the gasstream as sho$n in (ig. ).*) 'ne!t slie+.


160/239


161/239

161

2n (ig. ).*), the appropriate velocity to use in 3to#es4la$ is not the overall stream velocity but rather theifference in y%irecte velocity bet$een the particlean the gas stream.

Ienerally, the gas stream $ill have a larger velocityin this irection than the particle, so

3 ( )y drag y gas y particle% D =


162/239

162

"he resisting 'inertial+ force of the particle is

2f there are no electrostatic, magnetic, or other forcesacting, then these t$o forces are eual an opposite,so

3

y particle

y inertial

d% ma D

dt

= =

2

18 ( )y particle y gas y particled

dt D

=


163/239

163

Fearrange,

Wt is the time available for the y%irecte forces tomove the particle aroun the garget, $hich must beproportional to the time it ta#es the main gas flo$ togo past the target.

2

18 (24)

( )

y particle

y gas y particle

d t

D

=


164/239

164

"his time 6 Db:v, $here Db is the iameter of thebarrier, so $e may substitute in E. '/1+, fining

Nsis the separation number, $hich appears on the

hori-ontal a!is in (ig. ).*?.

2t is eual to the iameter of the barrier ivie bythe sto#es stopping istance.

2

12

18 1 (2!)

( )

y y particle b

y

y gas y particle s

d D

D N

=


165/239

165

3ome authors call Nsthe impactionparameter or inertial parameter.

2n E. '/A+ $e can see that if the integral onthe left is a large number 'a lo$ value of Ns+,

then there is plenty of time an force for theflo$ to move the particle aroun the targetan the target efficiency '=t+ $ill be lo$.


166/239

166

Example 17 single, cylinrical fiber *7 ; in iameter is

place perpenicular to a gas stream that ismoving at * m:s.

"he gas stream contains particles that are * ;in iameter an the particle concentration is *mg:m0.

5hat is the rate of collection of particles onthe fiber


167/239

167

Solution:2f all the particles that start moving irectly to$ar thefiber hit it, then the collection rate $oul be eual tothe volumetric flo$ rate approaching the fiber timesthe concentration of particles

2f $e catch them all, $e $ill collect *7%?g:s for every

meter of fiber length.

! 3 3

8

maximum possible rate = (1 % )(1" )(1" % )

1" % %

bD c m s m g m

g m s

=

=


168/239

168

"he actual amount caught $ill be this number timesthe target efficiency.

"he separation number is

(rom (ig. ).*?, $e see that for cyliners this value ofNscorrespons to a target efficiency of about 7.1/,

so $e $oul e!pect to collect about 7.1/ *7%?g:m:s.

3 2

! !

(2""" % )(1" ) (1 % )

"$1(18)(1$8 1" % % )(1" )s

kg m m m s

N kg m s m

= =


169/239

169

Example 18 filter consists of a ro$ of parallel fibers

across a flo$, as escribe in E!ample *@,$ith the center%to%center spacing of the fibers

eual to five fiber iameters. 5hat collection efficiency $ill the filter have

for the particlesssume that the fibers are far enough apart

that each one behaves as if it $ere in aninfinite flui, uninfluence by the other fibers.


170/239

170

Solution:(rom E!ample *@, =t6 1/> for a single fiber.

2f the fibers are space five fiber iametersapart, then the open area is ?7>, an the

bloc#e area is /7>. 3o,

Collection efficiency 6 'target efficiency+ 'percentage bloc#e+6 1/>/7> 6 ?.1>


171/239

171

Example 19 filter consists of *77 ro$s of parallel fibers

as escribe in E!ample *?, arrange inseries.

"hey are space far enough apart that theflo$ fiel becomes completely uniformbet$een one ro$ an the ne!t.

5hat is the collection efficiency of the entirefilter.


172/239

172

Solution:=overall6 * % poverall6 * % 'piniviual+n

6 * % '*%7.7?1+*776 7.)))?

"hese three e!amples sho$, in ieali-eform, $hat goes on $ithin epth filters.

Most such filters o not have an orerly arrayof parallel fibers& the filter meium consists of

a tangle


173/239

173

2n epth filters that iffusion leas to smallparticle collection in aition to that computeabove by impaction.

(rielaner evelope a theoretical euation,

$ith constants etermine by e!periment, forthe case of iffusion collection of particlesfrom a gas steam flo$ing past a cyliner

uner circumstances $here impaction $asnegligible.


174/239

174

Most of the publishe ata coul be represente by

$here all the terms are as efine previously, Diis

the iffusivity an Y is the #inematic viscosity.

"he first term on the right is for iffusion collection,$here as the secon is for collection by noninertialcontact 'interception+

2%3 2 1% 2

1% 2 3% 21% 1% 2 1% 2

3 (2)it

b b

D D

D D

= +


175/239

175

Example 20 Fepeat E!ample *@ for particles having a

iameter of 7.* ;. "a#e into accountimpactions, iffusion, an interception.

Solution:2n this case Nsis '7.*+/6 7.7* times theprevious value, or 7.77B/, for $hich, from (ig.).*? $e can rea =t Z 7.

(rom (ig. ?.*, $e can rea that the iffusivityis about B *7%Bcm/:s 6 B *7%*7m/:s.


176/239

176

3o

"he iffusion term is '7.77?B : 7.777/A+ 6 01.1 timesthe interception term.

"able ).0 'ne!t slie+ sho$s the effect of changes invelocity an particle iameter on the collectionmechanisms.

1" 2 2%3 2 1% 2

! 2 ! 1% 2 1% 2 ! 2 1% 2 ! 3% 2

( 1" % ) 3(1" ) (1 % )

(1$4# 1" % )(1" ) (1 % ) (1$4# 1" % ) (1" )

"$""8 "$"""2! "$""88 "$#, &

t

m s m m s

m s m m s m s m

= +

= + =


177/239


178/239

178

"here is some particle si-e at $hich there is aminimum collection efficiency.

"ypically, this si-e is in the range 7.* to * ;,$hich is the si-e most li#ely to be eposite

in the human lung.

2.3 "ilter #eia


179/239

179

(or sha#e%eflate baghouses, the filter bagsare mae of tightly $oven fibers 'surfacefilter+, much li#e those in a pair of


180/239

180

(ilter (abrics are mae of cotton, $ool, glassfibers, an a variety of synthetic fibers.

Cotton an $ool cannot be use above *?7an /77o(, respectively, $ithout rapi

eterioration, $hereas glass can be use toA77o( 'an short%time e!cursions to AA7o(+.


181/239

181

2n aition the fibers must be resistant toacis or al#alis if these are present in the gasstream or the particles as $ell as to fle!ing$ear cause by the repeate cleaning.

"ypical bag service life is 0 to A years.

2.$ Scru%%ers for Particulate

Control


182/239

182

3crubbers effectively ivie the flo$ ofparticle%laen gas by sening many smallrops through it.

Most fine particles $ill ahere to a liui rop

if they contact it.


183/239

183

Particles A7 ; an larger are easily collectein cyclones. 2f our problem is to collect a set of 7.A ;

particles, cyclones $ill not $or# at all.

Ho$ever, if $e $ere to introuce a largenumber of A7 ; iameter rops of a liui'normally $ater+ into the gas stream to collectthe fine particles, then $e $oul pass the

stream through a cheap, simple cyclone ancollect the rops an the fine particles stuc#on them.


184/239

184

complete scrubber has several parts, ass#etche in (ig. )./7 'ne!t slie+.


185/239

2.$.1 Collection of Particles in a &ainstorm


186/239

186

(ig. )./* 'ne!t slie+ sho$s the geometry for$hich $e $ill ma#e a material balance on theparticles an on the rops.


187/239


188/239

188

5e consier a space $ith imensions W!, Wy,W-.

"he concentration of particles in the gas inthis space in c.

No$ $e let one spherical rop of $ater ofiameter DDpass through this space.

Ho$ much of the particulate matter in the

space $ill be transferre to the rop


189/239

189

5e can see that the volume of space s$ept out bythe rop is the cylinrical hole sho$n in (ig. )./*,$hose volume is

"he total mass of particles that $as originally in thats$ept volume is that volume times the concentrationc.

2

4swept by one drop D$ D (

=


190/239

190

"he fraction of these that $ill be collecte by the ropis the target efficiency =t, $hich $e can etermine

from (ig. ).*? or its euivalent.

3o the mass of particles transferre from the gas tothe rop is

( )

2

7 t

mass transferred swept target= concentration

to one drop volume efficienc

9

= 7:;c


191/239

191

Ne!t $e consier a region of space 'still W!Wy W-+that is large $ith respect to the si-e of any oneiniviual rainrop through $hich a large number ofrainrops are falling at a steay rate ND, e!presse

as rops:time.

(rom a material balance on the particles in thespace, $e can say that

22

( )( o % )

( )

( % 4)( ) (2)

4

D t D DD t

dc mass transferred to eac' drop number f drops time

dt volume of t'e region

D (c N ND c

x y ( x y

=

= =


192/239

192

5e multiply top an bottom of E. '/@+ by the volumeof a single spherical rop an simplify to

"he final term in parentheses in E. '/?+ representsthe volume of rain that fell per unit time ivie by thehori-ontal area through $hich they fell 'rainfall rate+.

3( % )1$! (28)t D D

D

cdc N D

dt D x y

=


193/239

193

(or the rest of this chapter the total liui volumetricflo$ rate going to a scrubber $ill have the symbol S8,

so the rightmost term is S8:, $here is the

hori-ontal pro


194/239

194

Example 21 rainstorm is epositing 7.* in.:h, all in the

form of spherical rops * mm in iameter.

"he air through $hich the rops are fallingcontains 0 ; iameter particles at an initialconcentration *77 ;g:m0.

5hat $ill the concentration be after one

hour


195/239

195

Solution:3olving E. '/)+ for c, $e fin

(rom (ig. ?.@ $e can rea the terminal settlingvelocity of a *%mm iameter rop of $ater in still air isabout *1 ft:s 6 1./ m:s, so $e can compute Nsfrom

E. '/A+ as

1$! t Lo

D

Q tc c exp

D &

=

2 3 2

! 3(2""" % )(3 1" ) (4$2 % ) "$23

18 (18)(1$8 1" % % )(1" )

P)

D

D kg m m m sND kg m s m

= = =


196/239

196

(rom (ig. ).*? $e can rea =tZ 7./0, so

"his e!ample sho$s that the result epens on thetotal amount of rain that fell, S8Wt:, $hich is 7.*

inch in this case, not on the time or rainfall rateseparately.

&3 3 3

(1$! "$23)("$1 $% )(1 )1"" 43

1" 3#$3 $

g in ' ' m gc exp

m m in m

= = g


197/239

197

2f the e!ample ha as#e for the collectionefficiency for particles of * ; iameter, $e$oul have calculate an N3one%ninth as

large, an from (ig. ).*? $e $oul have

compute an =tof -ero. "his calculation suggests that a rainstorm

oes not clean the air $ell.

2.$.2 Collection of Particles in Crossflo'

Counterflo' an Coflo' Scru%%ers


198/239

198

1! "rossflo# Scru$$ers (ig. ).// 'ne!t slie+

3chematic of a crossflo$ scrubber.


199/239


200/239


201/239

201

3ubstituting E. '07+ into E. '/)+, $e fin

"his euation says that the smaller the rop an thetaller the scrubber, the more efficient it $ill be inremoving particles.

1$! (31)t L

o D G

c Qln p ln (

c D Q

= = g g


202/239

202

Ho$ever, a small rop has a much lo$ervertical velocity, so it $ill be carrie along inthe flo$ irection by the gas an not becollecte in the scrubber.

(or this reason, this type of scrubber is not$iely use.


203/239

203

2! "ounterflo# Scru$$er Ne!t slie '(ig. )./0+

3chematic of a counterflo$ scrubber


204/239


205/239

205

2f the rop is at its terminal settling velocityvtrelative to the gas that surrouns it, but

that gas is moving up$ar $ith velocity vI

6 SI : W!Wy, then the velocity of the rop

relative to the fi!e coorinates of thescrubber is vD%fi!e6 vtQ vI.

5 # i t i l b l


206/239

206

5e rema#e our previous material balance as

"o compute the uantity 'volume s$ept byrops:time+ $e must compute the instantaneous

number of rops per unit volume.

( ) ( ) ( )

( )

mass of particles transferred

to drops per unit time swept%time particle conc$ target efficienc

per unit volume

gas volumetric flow rate

volume

=

= ( )c*ange in particle conc$ (32)

"h li i fl i t th t i S thi i t


207/239

207

"he liui flo$ into the system is S8, an this consists

of NDrops:time, each of volume 'G:B+DD0.

"he average time each such rop spens in thescrubber is the vertical istance ivie by thevertical velocity relative to fi!e coorinates, or

( )t G

(average time

=

3 t ti th b f i th t i


208/239

208

3o at any time the number of rops in the system is

rops present at any time6

"he volume of gas that these rops s$eep out perunit time is their number times their cross%sectionalarea times the velocity at $hich they move relative tothe gas, $hich is vt.

(33)( )

D

t G

N (

3


209/239

209

3o,

5e substitute E. '01+ into E. '0/+, fining.

( )

2

2

3

4

% 4

1$! (34)

D Dt

t G

D tL

D t G

tL

D t G

volume swept N ( D

time

DQ (

D

Q (D

=

=

=

( ) ( )1$! t

L t G

D t G

Q ( c Q cD

=

2f $e no$ let the scrubber height be infinitesimally


210/239

210

2f $e no$ let the scrubber height be infinitesimallysmall, so that W- an Wc become - an c, $e canseparate the variables an integrate, fining.

1$!

ln ln 1$! (3!)

t tL

D G t G

t tL

o D G t G

dc Qd(

c D Qc Q

p (c D Q

=

= =

g g g

g g


211/239

211

(rom E. '0A+, $e can get *77> efficiency'c:co6 7+ if $e let vt6 vI.

Physically, that means that if the up$arvelocity of the gas euals the terminal settling

velocity of the liui, then the iniviual rop$ill stan still in the scrubber.


212/239

212

Ho$ever, if $e continue to put liui into thescrubber an no liui leaves, $e $ill fill thescrubber $ith liui.

2t $ill become flooe an $ill cease to

operate as a scrubber. "here are some important applications $here

they are use for aci gases removal, but

they o not play a ma


213/239

213

3! "o%flo# Scru$$ers Clearly $e nee a geometrical arrangement

in $hich $e can get very small rops to moveat high velocities relative to the gas being

scrubbe, to get a high Nsan high =t, $ithoutblo$ing the rops out the sie or top of thescrubber.

the solution to this problem is the Co%flo$

scrubber, sho$n schematically in (ig. )./1'ne!t slie+.


214/239


215/239

215

"he liui enters at right angles to the gas flo$. ery high gas velocities can be use in this type

of scrubber, as much as 177 ft:s '*// m:s+.

"he liui enters $ith -ero or negligible velocity

in the ! irection so that at the inlet the relativevelocity may be as high as 177 ft:s.

"his may be *77 times the ma!imum tolerablerelative velocity in a crossflo$ or counterflo$

scrubber.

t the inlet the rops have -ero velocity but by the time


216/239

216

t the inlet the rops have -ero velocity, but by the timethey reach the outlet they normally $ill have come to thesame velocity as the gas.

2f $e $rite the material balance for a ifferential length !of the scrubber, $e can say that the average time it ta#esa rop to pass that section is

5here vD%fi!e 6 rop velocity referre to fi!ecoorinates

vI 6 local gas velocityvFel 6 velocity of the rops relative to the

gas 6 vIQ vD%fi!e

(3)D fixed G +el

dx dxaverage time

= =

"hen $e can $rite out the analog of E '01+ using


217/239

217

"hen $e can $rite out the analog of E. '01+, usingvFelin place of vt an E. '0B+ in place of E. '00+.

5e fin

( )2

4

1$!

1$! (3)

D D+el

G +el

+elL

D G +el

+elLt

D G G +el

volume swept N x Dd

time

Q dxD

and

dc Q dxc D Q

=

=

=

g g g


218/239

218

(or the previous e!amples $e coul integratethe euations corresponing to E. '0@+because all the terms on the right $ereinepenent of istance or time.

(or a Co%flo$ scrubber they are not. (ig. )./A 'ne!t slie+ sho$s ho$ some of

these variables change for a simple, constantcross%sectional area scrubber.


219/239


220/239

220

2f these changes $ith istance $here all that$as involve, $e coul presumably use therag force relationships previously introuceto calculate vFelas a function of istance,calculate =

tat every point, an then integrate

E. '0@+ to fin the preicte performance ofany such scrubber.

lthough the simple geometry sho$n in (ig.

)./1 is occasionally use, for large flo$s theventuri esign sho$n in (ig. )./B 'ne!t slie+is much more economical.


221/239

(or the venturi sho$n in (ig ) /B the throat


222/239

222

(or the venturi sho$n in (ig. )./B the throatcross%sectional area is about one%fifth that at theinlet or outlet, so the velocity there must be aboutfive times the velocity at the inlet or outlet.

"o achieve high velocities in a gas flo$ $e musthave a rop in pressure& for steay, hori-ontalfrictionless flo$ $e #no$ that

( )2 2

2 1 1 22P P

=


223/239

223

venturi co%flo$ scrubber seems the mosteconomical $ay to get a high velocity forrapi liui brea#up into rops an highcollection efficiency $ith minimum fan po$er.


224/239

224

Example 22 2n a venturi scrubber the throat velocity is *//

m:s. "he particles to be collecte have iameters

of * ;, an the roplet iameter is *77 ;. 5e are feeing *7%0m0of liui per m0of gas

to the scrubber 'S8: SI6*7%0+.

t a point $here vFelis 7.) vI, $hat is the rateof ecrease in particle concentration in thegas phase

Solution:


225/239

225

Solution:"he appropriate velocity to use in Nsis vFel.

(rom (ig. ).*?, $e see that =t6 7.)/.

"hen from E. '0@+,

2n this part of the scrubber, the concentration of particlesin the gas is ecreasing by */.1> for every millimeter ofgas travel in the flo$ irection.

2 3 2

! 4

(2""" % )(1" ) ("$# 122 % )$8

18 (18)(1$8 1" % % )(1" )

P +el)

D

D kg m m m sN

D kg m s m

= = =

( )( )34"$#% 1$! 124 "$124

"$#2 1"1" "$#

G

G G

dc c

dx m m mm

= = =

Example 23


226/239

226

Example 23

Ho$ rapily is vFelchanging for the rop in E!ample

//

Solution:

"he particle Feynols number for the roplet is

2n any flui mechanics te!t $e can loo# up thecorresponing rag coefficient for a spherical roplet,fining that it is about 7.@.

4 3

!

(1" )(1$2" % )("$# 122 % )32

1$8 1" % %

D +elP

D m kg m m s+

k g m s

= = =

"hen $e can compute the acceleration of the rop as


227/239

227

"hen $e can compute the acceleration of the rop as

2 2

3

2 3 2

4 3

4 2

( % 4) ( % 2)

( % )

(1$!)("$)(1$2" % )(1"$ % ) 1$! 2 2(1" )(1""" % )

$2 1" %

3"" times t*e acceleration of gravit &

+el D d air +elD

D D

+eld air

D D

d D C%a

dt m D

kg m m sC D m kg m

m s

= = =

= =

= =


228/239

228

Calvert mae several simplifications an thus$as able to perform the integration of E. '0@+numerically& the results of that integration, fora typical venturi scrubber, are summari-e in

(ig. )./@ 'ne!t slie+. "he figure4s use $ill be illustrate later.


229/239

2.$.3 Pressure Drop in Scru%%ers


230/239

230

enturi scrubbers have higher pressure ropsan higher efficiencies than the crossflo$ ancounterflo$ scrubbers.

"he po$er cost of the fan that rives the

contaminate gas through a venturi scrubberoften is much more important than thepurchase cost of the scrubber.


231/239

231

Example 24 typical venturi scrubber has a throat area of

7.A m/, a throat velocity of *77 m:s, an apressure rop of *77 cm of $ater 6 )?7B

N:m/. 2f $e have a *77> efficiency motor an

blo$er, $hat is the po$er reuire to forcethe gas through this venturi

Solution:


232/239

232

2f the fan an scrubber operate ?@B7 h:yr an theelectricity costs A cents:#5h, the annual po$er cost$ill bePo$er cost 6 '/1A #5+'?@B7 h:yr+ '[7.7A:#5h+ 6 [ *7@,077:yr

( )2 2 31""

"$! #8"1"

24! 328

G

m N kW sPower Q p m

s m N m

kW 'p

= =

= =

g

g


233/239

233

venturi as sho$n in (ig. )./1, for steayflo$ an negligible $all friction, $e have'letting * stan for inlet conitions an /stan for outlet conitions+

'P*%P/+'W!Wy+6SII'vI/%vI*+\S88'v8/%v8*+ '0?+ 2n most cases the gas velocity changes very

little from inlet to outlet, although it may havea very high value at the throat, so the first

term on the right is negligible.

Normally the liui inlet velocity v8*is negligible, an


234/239

234

y y g g

the liui outlet velocity v8/eual to the gas velocity.

3o E. '0?+ normally is simplifie to

2

1 2 2 (3#)L L G L L G LG L

G

Q Q Q Qp p

x y & Q

= = =

Example 25


235/239

235

(or a scrubber using $ater as the scrubbing liui,estimate the pressure for vI6 SI:W!Wy 6 *77 m:s,

an S8:SI6 7.77*.

Solution:

(rom E. '0)+

( ) ( )( )2

2 3

1 2

4 2 4

2

1"" % 1""" % "$""1

1" % 1" "$1 1"2 &

N sP P m s kg m

kg m

N m Pa atm cm H !

=

= = =

g

g

Unfortunately, the very properties that cause


236/239

236

y, y p p$et scrubbers to have high pressure ropsare the same ones that ma#e them efficientparticle collectors the rapi acceleration of

liui by the fast%moving gas prouces botheffects.

(ig. )./? 'ne!t slie+ sho$s test results ta#enby 8apple an Kamac# in $hich they use

one #in of particle, a talc, in a variety ofscrubber esigns.


237/239


238/239

238

Example 26 5e $ish to treat a gas stream to remove

most of the particles.

5e conclue that if $e have a cut iameter of

7.A ; $e $ill have mae a satisfactoryparticle removal.

2f $e use S8:SI6 7.77*, $hat gas velocity at

the throat $ill $e nee, an $hat $ill thee!pecte pressure rop be

Solution:


239/239

(rom (ig. )./@, using the results in E!ample 1 of lastchapter for a 7.A ; particle, $e can compute aCunningham correction factor of *./1, so theaeroynamic cut iameter is

( (i ) /@ it i l th t thi l i

( )1% 2 1% 2

3 3"$! 2 1$24 "$#pca

gm gmD

cm cm = =

03-Control of Primary Patriculates

Documents

Transcript of 03-Control of Primary Patriculates