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.~..

Wiley International Edition'

Alan S. Foust Leonard A. Wenzel Curtis W. Clump Louis Maus

L. Bryce Andersen

Department of Chemical Engineering Lehigh University Bethlehem, Pennsylvania

Principles of •,

_Unit Operations
17, _SIMULTANEOUS HEAT AND . 18~ SIMULTANEOUS HEAT AND 19 SIMULTANEOUS HEAT AND .

MASS TRANSFER I: HUMIDIFICATION MASS TRANSFER II: DRYING

87

89 113 128 139 152 202 219 221

223 267 289 320

MASS1'itANSFER 111: EVAPORATION AND

352

CRYSTALLIZATION

20 . MOMEl'ITUM"n~ER

65 81 86

I: THE ENERGY BALANCE AND ITS

APPLICATIONS

21

MOMENTUM TRANSFER II: PUMPS AND COMPRESSORS

22

MOMENTUM TRANSFER III: PHASE SEPARATIONS BASED UPON FLUID MECHANICS PART III: NOTATION AND NOMENCLATURE

A

DIMENSIONS AND UNITS, DIMENSIONAL ANALYSIS, AND MODEL THEORY

B

DESCRIPTION OF PARTICULATE SOLIDS

C

EQUIPMENT DESIGN DATA

D

PHYSICAL DATA

391 421 449 510 515 525 540 550 569

l!.

chapter I

Unit Operations in Chemical Engineering oui)0QJ~~[ cent. ~!nce an equilibrium s~ives the greatest composition change. possible for, a given set of operating c,£nditions, it is also referred to as . an ideal, or theoretical"stage. --' The calculation of equipment requirements for industrial multistage operations usually involves the determination of the number of equilibrium stages followed by the application of stage efficiency to give the number of actual stages required. An equilibrium stage can be represented schematically:

illdustry

~... ethe transf~()r!llass f!Q.~l!e phase to another.

C'mally one comp!!ent of the phase ,will tra!!sfer ~ a ~...-r extent th~n another, i!t.e~ebI_~~~~ing _a_~_~arat!on C!l:]:c components of the E1!~t~r~. For example, crude ~_ ,-'ileum can be s~a_rated __ ~~~o several c()mponents by l&Ji$$ transfer between, a liquid and a vapor phas~... The i""i!t!nmical engineer is concerned'Wlththe dlst~!l..tiolJ. of ~nents between two"phas~~a..~,eq~llibriumaJld,,'!Vith ~ cue of transfer of the components from one phase JO - _.. .. .. ~

-

~

~r.

,.

7"!re rate 9f mass transfer must be considered in the

~ ~f'eqt1ii'mentwnereihei~oPhases' are .in con~

I

i.

mass

contact lmd Js irit~Clfa!1ged continuously the phases. This subject wIll be considered in a 1taEr >eetion. In many mass-transfer operations, equipDaJ[ :;$ designed to give discontinuous contact of phases • ;a ~ of stages. The initial calculations involved in a!iIJID~t design or in the evaluation of performance of ~ equipment~~ based on relatively simple ~c and equilibrium relationships. A ~e may be defined as a unit of equipment in .;~ :wo dissimilar phases are brought int~ intimate ~ with each other and then are mechanically iIII:IJ!IIIIIl!.med. During the contact various diffusing com~ '\ ~ of the mixture r~dis!ribute themselves between ... • ~_ The resultant two ·phases have approached ~ and therefore have' compositions different Innaa :iJC initial phases. By successive contact and ~n of dissimilar phases (a multistage operation) lirBnp ::amges in the compositions of the phases are ~ In an equilibrium 'SttJge the two phases are . . m:ud for a time sufficient to allow establishment ..rlknDodynamic equilibrium between the phases ~ Ute stage. At equilibrium nO' further net change "@py!""~ osition of the phases is possible for a given set 1II8DC1IJtS

.....-

.----.----.

-.-----.'

"'-J.;

---~

"~'----'~-'-

-

".

Equilibrium

stage

Two phases, Vin and Lin, are mixed and allowed to come to equilibrium. The phases are then mechanically separated and leave the stage as Vout and Lout, which are in equilibrium with one another. An everyday example of a single-stage mass-transfer operation is the vacuum coffee maker. Here, the hot water (Vin) and -th'!"grobhd coffee (lin) are contacted to distribute the soluble constituents of the coffee between the liquid and solid. If given sufficient time, the dissolved coffee would come to equilibrium with that in the grounds . The coffee solution (Vout) is then mechanically separated (by pouring, for example) from the grounds (Lout). In actual coffee making, the time of contact with the 11

12

PRINCIPLES OF UNIT OPERATIONS

In general, each stage of the equipment mixes the incoming two phases thoroughly so that the material car be transferred as rapidly as possible from one phas~ to the other. Each stage then must separate the n;su!· tant two phases as completely as possible and pass therr on to the adjacent stages. Some industrial equipmen ~ may consist of a single stage, but more often multistage units are employed with countercurrent flow of the twc phases. Multistage operation permits greater changes ir. the compositions of the two phases than can be accomplished in one stage. The names usually attached to the various mass-tllmsfer operations evolved before the similarities among the stage operations were fully understood. The primar: difference between the various stage operations is. the nature of the two phases involved in each operation In distillation a vapor phase contacts a liquid phase, and mass is transferred both from the liquid to the vapor anc

Plan view of top plate

Downcomer seal

L-phase flow (heavy phase)

-I

t t -""'o-:o:---P,

= vapor pressure of component a at the given

temperature " = total pressure

:::::l;uations indicate that the vapor evolved from a ,:m...,;:ure will be a mixture of the same components llIIliIUd. The vapor will normally be richer in the

21

Since Yb = 1 - Ya and

Xb

[1 -

=

1-

Xa,

Ya] X a] _ Pa _ -- [ -xa 1 - Ya Pb

'Xab

(3.6)

Equation 3.6 defines the relative volatility (otab ) of component a relative to b. For systems which do not fOlIOw Raoult's law, the relative volatility is defined as lXab = Ka/ Kb • The relative volatility is constant when either Henry's or Raoult's law holds. In other cases it varies with composition. Figure 3.1 shows vapor-liquid composition data for two systems where the relative volatility is constant and also for two systems where the relative volatility varies with composition. The systems shown in Figure 3.1 are at constant pressure, but the temperature varies with composition. The variation of temperatures is shown in Figures 3.2 and 3.3.

",il"

...;:.-

"', -:p.~~'i

r' •

SOLUTION. (a) Vapor-pressure data are found in tables of physical data (Appendix D): Vapor pressure of pentane at lOO°F = 830 mm Hg Vapor pressure of butane at 100°F = 2650 mm Hg Total pressure: P = 3 x 760 = 2280 mm Hg

For butane:

YB =

For pentane:

Yp =

PB

P

Pp

P

Raoult's law can be used to calculate the dew point and bubble point of ideal mixtures. A mixture does not boil at a single temperature for a constant total pressure, in contrast to the behavior of a pur(! liquid. The temperature at which a liquid mixture of a given composition begins to vaporize as the temperature is increased is called the "bubble point." Conversely, the temperature at which a vapor mixture first begins to condense on cooling is called the "dew point." For a pure liquid the bubble point and dew point are identical and equal to the boiling point, since a pure component vaporizes or condenses at one temperature.

Superheated vapor

2650 x B = 2280 x B = 1.16xB

830

1

Xp = 2280 xp - 0.363xp

220

Since only butane and pentane are present in ~he liquid and vapor, x B + xp = I, and YB + Yp = I. lIIr~.,,~re now four equations and four unknowns. Solving,

LL.

I!:'

YB

+ Yp = I = 1.16xB + 0.363x p I = I.16xB + 0.363(1 -

0_210 ~

E

~

~

x B ), etc.

~ 200 I-

x B = 0.80 YB = 0.93

= 0.20

xp

Yp

= 0.07 190

(b) Experimental1,lflues of K at 3 atm and 100°F are for butane K B = 1.15 and for pentane K p = 0.36 (Appendix .

D).

~.

Therefore,

YB = 1.15xB

xB

= 0.81

Yp = 0.36x p YB

and as before,

180

Subcooled'liquid

= 0.93

Xp = 0.19

Yp = 0.07 The small difference between the values in parts (a) and (b) is within the precision with which the charts for vapor pressure and K can be read. Therefore, mixtures of Dutane and pentane follow Raoult's law at 3 atm pressure. In general, deviations from Raoult·s law are greater at higher pressures.

Xa •

mole fraction benzene in the liquid

Ya • mole fraction benzene in the vapor

Figure 3.2. Temperature-composition diagram for

liquid-vapor equilibrium of benzene and toluene at 1 atm .

.f

I

i

.~i__ -._~

.

~~-k-}~

.' • .""e'

PHASE RELATIONSHIPS - - - I~n of the phase rule to the vaporization of ".ICIIIIIIIIDponlent liquids shows that there can be one.,. liIII_adent variable. Thus, if the over-all pressure has ~. the temperature of the vaporization is set. a other hand, in the vaporization of a two-comIp!Il1llllISI! liquid, there are two independent variables. In Ili!I!IInnJB1,,,~ after the pressure is specified either the tempera11tIIIUi1I1I!: . . the composition may be varied independently. !ilmnuI::m: the liquid composition will vary as vaporization iP"""'"""'s, the temperature must then vary. A boiling ~ results, extending between the bubble point and _ dew point. ~

3.3. Calculate the dew point of a gaseous containing 20 mole percent benzene, 30 mole percent IIIIDIIIiIIICIIII:.. and 50 mole percent o-xylene at I atm total pressure.

1I1t1111m!!t:ItIII!IIl:

0.2

At this moderate pressure for these similar Raoult's law will be sufficiently accurate. The \~ ::RSSUres of the three components depend upon ,~. which is to be determined. A trial-and-error ., ~ is necessary, since the vapor pressures cannot be '!iiU ! as simple mathematical functions of temperature. 'lilllIDDI..=noN.

~

(a)

toluene:

Ethanol-water (8).

~

152

;.- tienzene:

1.0

0.4 %a' Y a , mole fraction ethanol

~:11IIIIIIIQIIlIIn

~

xB=PB

~ 140

228 xT =-

E ~

8.

PT

%a' Ya, mole fraction acetone

•

"1111:

!lie

correct temperature x B

+ x T + Xx

=

(b)

1;

Figure 3.3. Temperature-composition diagram for mixtures forming an azeotrope at 1 atm.

152 228 380 -+-+-=1 PB PT Px

The check at 263°P is sufficiently close to consider this the dew point of the mixture. . .. At the dew point, the first liquid formed has a compOSItIOn such that the pressure exerted by each component of t~ liquid is equal to the partial pressure of the .~o~pone?t ~n. the vapor. The composition of the first eqUIlibrIUm hquld formed as condensation begins at 263°P is

- oflow In distillation reflux at both ends of a cascade is very solvent concentration, part of which is extract product frequently used to obtain distillate (D) and bottoms (B) (D) and part extract reflux (Lo). The solvent separator of high purity. The V-phase reflux (Vs) is produced by adding heat (qs) in the reboiler (S) to part of the liquid usually is a distillation column. Extract reflux is particularly beneficial for ternary liquid systems with two from the cascade (LN ) and vaporizing it. Similarly, pairs of partially miscible components, such as the L-phase reflux is produced by removing heat (-qd in the condenser (C) from part of the vapor product (VI) diethylene glycol-styrene-eth .bemene .system._~igUre 3.9). In such systems it . 'tble t,?..:..o~ain "Ii~r .. n and condensing it to give liquid reflux (£0). Use of reflux into an extract conta' )legliglble amO'tm~Jr~ \ in distillation permits recovery of very high purity extracted raffinate co ~t,r ~~d a.~~_,.,;e~@~"',ihIdiij"" products, providing no azeotrope occurs. \ ..,A ( < . · 1 r. llllvcrsi:y ') ~ \ ..... 'L;\!':O;{E, ):p.. t

\~">'i'" ',-I

I{,~ oJ-!-:"'/ . . . . . . . . . '>~'-Is .,.....,.-.. '-'~ - .';/} J~~! t" '." ," " .... ~:,~.:::() I~ \ \~: >'"'

Sl----PRINCIPLES OF UNIT OPERATIONS

•

J

COUNTERCURRENT MULTISTAGE OPERATIONS WITH REFLUX----53

•

.

Figure 6.1. Ethylene purification by distillation. The six columns pictured separate ethylene from other light hydrocarbons. The pure ethylene product is used in the manufacture of polyethylene plastic. The first column on the right (No.1) is an absorption tower with reflux at the lower end which absorbs ethylene and less volatile materials with a heavier hydrocarbon solvent, separating them from methane and hydrogen. The sixth column from the right (No.6) is a distillation column with reflux which removes propane and lighter components from the solvent. The second column separates ethane and ethylene from propane and propylene by distillation. The fourth and fifth columns separate ethylene from ethane. The two columns are in reality one distillation cascade, split into two sections to reduce the height of the unit. The third column. separates the ethylene product from any methane which was not removed in the absorber. Physical data for the columns are tabulated below. Column 1 2 3 4

•

Il:

Absorber De-ethanizer Demethanizer Ethylene fractionator, lower section 5 Ethylene fractionator, upper section 6 Depropanizer

Diameter, ft

Height, ft

No. of Actual Stages

3t 3,4t 3 4

75 77 73 196

30 30 28 80

4

83

40

4

67

26

Columns of this type may employ either bubble-cap or perforated plates. The operating pressures in this application vary between 280 and 500 psig. The temperatures in the columns range between -45°F at the top of the demethanizer and 300°F at the base of the depropanizer. The dark-painted columns operate above atmospheric temperature, and the light painted columns operate below. The unit requires a substantial quantity of refrigeration for the reboilers and condensers on the columns operating below atmospheric temperature. Some of the large quantity of auxiliary equipment necessary to operate the columns is also pictured. (Designed fot E. I. duPont de Nemours ,and Co. by The Lummus Company.)

II

-

. ;\

~ .Ii

I

, ,;

54

I ~

PRINCIPLES OF UNIT OPERATIONS

In distillation (Figure 6.2), heat (qs) is added at the still (S) to vaporize part (L S - 1) of the liquid flow (LN ) from the bottom stage. The vapor reflux (Vs) formed in the still rises through the stages, coming to equilibrium with the liquid dow,nflow at each stage. Ultimately the _.xapQr from the top stage (VI) is condensed in the .(OJ,>>,,. ~ \/1 _,' >_>'C ! 1....-

•

F

ff\

./

Figure 6.2. Countercurrent stage operation with reflux.

total condenser (C); part is removed as distillate product (D) and the remainder (Lo) is the liquid reflux. In

flow in the enriching section in Figure 6.2 is

binary distillation the succession of stages above the feed enrich the vapor in the more volatile component, and the stages below the feed strip the more volatile component from the liquid downflow. The reflux rates of streams Lo and Vs , as well as the number of equilibrium stages, determine the degree of separation of the two components present in a given feed. In extraction there is a net flow of solvent carrying the extracted solute from stage N to stage 1. Similarly, in distillation there is a net flow of heat which might be thought of as "carrying" the more volatile component from stage N to stage 1. Just as the solvent distribution between extract and raffinate must be considered in extraction, so must the heat contents (enthalpies) of the liquid and vapor be considered in distillation. The reflux may constitute a large fraction ofthe material reaching the ends of the cascade. The products would then be but a small fraction of the phases circulating internally within the cascade. Reflux may be employed in other stage operations where it pr6ves beneficial. For example, certain separations in gas absorption and adsorption are improved by using reflux.

VI = Ln -

ST~CALCULATIONS WITH REFLUX

In a typical design calculation the quantity-and composition of the feed and products may be specified. If the reflux ratio (Lol D) has been chosen, the required number of equilibrium stages for the separation may be calculated. Another situation often encountered involves prediction of the operation of existing equipment in the separation of a new system. Usually the construction of the cascade allows the variation of reflux ratio over a limited range. Then the possible product compositions for a given feed may be calculated at the allowable reflux

~ = Lo -

Vn+l = -(D

+ VO>

where n is any stage in the enriching section. the net flow in the stripping section is

Li

~ l..~'"

= ~ -

Vs = Lm -

Vm+l

= (B -

(6.1) ~

Similarly,

I

VS+l) (6.2)

where m is any stage in the stripping section. The introduction of feed changes the net flow between stages F-l and F. An over-all material balance around the cascade in Figure 6.2 gives F=.vo

/'

+ D + 1J -

-

VS+l

(6.3)

Combining Equations 6.1 and 6.2 wiUt 6.3 gives

F = -~

+ Li

or

Li = ~ + F

(6.4)

Equation 6.4 relates the net flows in the two sections of the cascade. The addition of the feed increases the net flow by an amount equal to F. The feed may consist of a mixture of L-phase and V-phase, such as liquid and vapor in distillation. In this situation the feed will split between the L-phase andY-phase when it isinlroduced .j into the cascade, but Equations 6.3 and 6.4 are still -, correct. It should be emphasized that the magnitude of the internal circulation (L and V) bears no dit:ect'relation to the quantities of F, D, and B. It is therefore impossible to predict the valu~s of L and V solely from the values of feed and products, m>coritrastto ~ and ho may be substituted into Equation 6.13 to evaluate h.:l' or h.:l may be located graphically using the equation as a guide. For a total condenser, ho = hn and xn = YI' A similar analysis of the stripping section will yield the following relationships:

~=LN- Vs=Lm- Vm+1=B h = hB _ QSB =

a where QSB = qs

B

Xa =xB Vs

Ii o Figure 6.7.

Distillation calculations with vapor and liquid reflux.

Combining this equation with Equation 6.9 gives !1h:::,.

= LohO -

V1H1 = -D(h n - Qcn)

Solving for the enthalpy coordinate of the delta point gives· h - -D(hJ) - QeD) - h Q (6.11) ,l !1 - J) CJ) net flow of heat total net flow of mass The x-coordinate of the delta point is obtained from !1x,l = Loxo - VI!h = - DxJ) x,l =

-D

T

Xl)

=

XJ)

(6.12)

The two coordinates (h,l, x.:l) of the delta point, as defined by Equations 6.11 and 6.12 can be used to locate the delta point when distillate quantity and composition and condenser duty are known. If the reflux ratio is known, h.;,. may be determined from it, analogous to the derivation of Equation 6.5 for extraction. A material balance around the condenser gives

0::::

hN - ha Hs-hN

net flow of heat total net flow of mass .....

(6.14) (6.1~

.

(6.16){ (6.17) (6.18)

The direction of the net flows in the distillation column should be considered. Distillation columns are built vertically with the condenser at the top and the reboiler at the bottom. Therefore, net flow defined as positive ... to the right in Figure 6.2 will be positive in a downward direction in a vertical distillation column. Equation 6.6 shows that the total net flow of mass in the .enriching section is upwards, but Equation.6. 7 shows that the total net flow in the stripping section is downward. Since heat is added at the reboiler and removed at the condenser, the net flow of heat is upward through both the stripping and enriching sections. Therefore, h.:l is always positive and ha is always negative.. The use of the equations -derived previously for graphical calculations is shown in Figure 6.7. For the illustration, the composition and thermal condition of-'* the feed.Jl~ttoms, and distillate are assumed to be known. In addition,either the reboiler or condenser duty is known. The feed is shown as a subcooled liquid (zF' hF)' The distillate is a liquid at its bubble point (xn, hn ), and the bottoms is a liquid at its bubble poin~ (x B , hB)' The reflux is also a liquid at its bubble point (xo, ho)' Since h.:l = hn - QeD and X.:l = Xo = x n ' the delta point for the enriching section can be determined from the condenser duty. The delta point for the stripping section is located by the line x.:lzFextended and the vertical line XB = x;\. Stages may be stepped off from either end with use of the appropriate delta poirtt. In Figure 6.7 the stages are stepped off from

COUNTEllCURRENT MULTISTAGE OPERATIONS WITH REFLUX----S9

the top of the column downward, by alternate use of the equilibrium data and the delta point. Since the column has a total condenser Yl = Xo = XD and the .,.\ point (YI' HI) can be located. This point represents the vapor leaving the first stage. The liquid leaving the first stage is in equilibrium with this vapor. Therefore, its composition (xJ is determined from the equilibrium curve, and its enthalpy is determined from its composition and the saturated liquid enthalpy curve. To locate (Y2' HJ a line is drawn from (xl> hJ to (xa, ha>. From (Y2' H 2) the equilibrium relation is applied to find {~, hJ. This stepping-off of stages is continued using (xa, ha) until ZF is reached. Then the construction switches to the delta point for the stripping section (x~, h~). The calculation is completed when the liquid composition equals or exceeds the bottoms composition xB • The calculation of stages across the feed stage as outlined is not rigorous, considering the enthalpy and pha&e characteristics of the feed. However, it is a .. sufficiently accurate approximation and is generally ~-> used. About 5.4 stages are required in the enriching section and about 3.4 are required in the stripping section. The steps in this calculation are identical to those out-

B

/ /

il/

1/

lJf/ I

I

/ /

/

~htl~~~~==============

-qc

v2

l'I

D

1

C

Lo

F".-e 6.8.

hA

.1 Ll

Partial condenser equivalent to one equilibrium stage.

iii! iii!iI'

;;;;;;;iI"U"P:;'I!!!!!""W!!!

tulj'liiiliiiHW""!!

Figure 6.9. Partial reboiler equivalent to one equilibrium stage.

lined in Illustration 6.1 for extraction. The analogy between the two operations should be carefully studied. Figures 6.2 and 6.6 show a total condenser in which all the vapor (VI) is condensed and then split into distillate (D) and reflux (4). In a partial condenser, on the other hand, only the reflux is condensed and the distillate is drawn off as a vapor. Since the liquid reflux and vapor distillate are in contact, they may approach equilibrium and give as much as one additional equilibrium stage, as shown in Figure 6.8. In actual practice, the liquid and vapor may not have sufficient contact to equilibrate, so that a partial condenser functions as less than one equilibrium stage. Figures 6.2 and 6.7 show the bottoms wi~hdrawn before the reboiler, so that the stream entering the reboiler is totally vaporized. In most distillation towers the bottoms are withdrawn from the still, as shown in Figure 6.9. This results in a reboiler that way be equivalent to as much as one equilibrium stage. The reboiler is frequently a steam-heated heat exchanger. Many types of reboiler construction are used in industry. The design of reboilers is based upon heat-tn.nsfer considerations covered in later sections of this book. '

60

PRINCIPLES OF UNIT OPERATIONS OPERA~G

Xa

Figure 6.10. Solution to Illustration 6.2.

Illustration (6.2) = LN - Vs = Lm - Vm+l = (B - VS+l) (6.19) 3. =L· - v.+1 The subscript i refers to any stage in the intermediate section of the column between stage I and stage F-I. The net flow at the left end of the cascade is~. Mter the intermediate stream (I) is added, the net flow becomes (6.20)

a

. .

Figure 6.14.

Graphical determination of the delta points for liquid extraction with an intermediate product stream.

J!1

l

COUNTERCURRENT MULTISTAGE OPERATIONS WITH REFLUX

."i'

added to give ~; i.e., x a will lie on the straight line X.lZ~:- The intersection~f this line with XBYS+l is Xa' The sum L may be con~dered as a "fictitious feed." The sections of the cascade between I and I and between F and N behave as if there were a single feed L = I + F at some point between I and F. The net flows in these two sections may be determined by combining I and F into the fictitious feed (L) and treating the cascade as the usual case of one feed. Since the stream I was a ;>roduct in the illustration of Figure 6.13, the numerical value of I is negative and z~ lies beyond ZF on the line zIZF' as in the usual case of subtraction. The concept of the "fictitious feed" is abandoned in order to determine xX, which is an expression of the net flow from I to F. The third delta point is located by inference from Equations 6.20 and 6.21 at the intersection of the lines XaZI and xazF: The stages may now be stepped off using the appropriate delta point in w.ach section of the cascade. Intermediate streams in distillation and other operations are calculated in a similar manner. An "intermediate stream" in distillation may consist of only 'the addition or removal of heat. PROBLEMS

.

6.2. Using the same rates and compositions of solvent and feed as in l11ustration 6.1, determine the maximum possible concentration of styrene in the extract product of a simple countercurrent cascade. 6.3. Locate the delta points of Illustration 6.1 on a ternary diagram. Attempt to step off the required number of stages on this diagram. 6.4. Prove that LN'/F' = XDZp/XDXN . 6.5. Derive Equation 6.5b. JA... Derive Equations 6.14 through 6.18. ~A flow of 100 Ib moles/hr of a vapor containing 0.40 mole fraction NH. and 0.6 mole fraction H 20 is to be enriched in ammonia in a distillation tower which consi;ts of an enriching §'ctjgD and a total condenser. The feed into the bottom of the column is a saturated vapor (at its dew point) at 100 psia, the operating pressure. The distillate is withdrawn from a total condenser and ~ has a composition of 0.90 mole fraction NH.. Part of the liquid condensed in the total condenser is withdrawn as the distillate product and part is returned to the column as I'eflux. 85 per cent of the ammoni!l ~h,ar.geQJmI8t be recovered in the d,il>tillate. (a) Calculate the number of equilibrium stages required and the reflux ratio (Lo/D). (b) What is the condenser duty (heat removed per hour from the condenser) ? (c) After the column has been built and is operating, a change in the process changes the feed to 0.2 plole fr,action NH3 at its dew point. Can the existing column of part (a) ~ u~d? The reflux ratio can be adjusted for this new fe.ed, but the number of theoretical stages is fixed. The distillate composition and percentage recovery of NHs must be unchanged . 6.8. A liquid ethanol-water feed is to be stripped of ethanol in distillation column consisting of a striEpj!!g.~ection and a reboiler. .J

~-

.

The liquid will be fed to the top plate and vapor reflux will be furnished by a reboiler which takes liquid from the bottom plate and totally vaporizes it. The bottoms product is also withdrawn from the bottom plate. (a) Calculate the number of equilibrium stages required for the following conditions: Feed:

100 lb moles/hr, 0.2 mole fraction ethanol, 0.8 mole fraction water at 100°F and 1 atm

Vapor product: 0.5 mole fraction ethanol Bottom product: (1.03 mole fraction ethanol (b) What is the reboiler duty (Btu/hr added at reboiler)? (c) Can a vapor product of O. 7 mole fraction ethanol be obtained

with a stripping unit like this? (You may use more stages and higher reflux ratio but use the same feed and bottoms composition.)

Ex.n. V{W72lb moles/hr of an ammonia-water mixture at 100 psia and 70°F, containing 25 mole percent ammonia, is to be fractionated in a disti\lation tower into an overhead product of 95 per cent ammonia and a bottoms of 4 per cent ammonia. The overhead product is withdrawn as a vapor in equilibrium with the reflux from a partial condenser; and the bottoms is withdrawn from the reboiler. The reboiler duty is 700,000 Btu/hr. (a) What is the condenser duty (Btu/hr)? (b) What is the reflux nitio (Lo/D)? • (c) How many equilibrium stages are required? (d) What is the net flow of heat in the column? (e) What is the minimum reflux ratio for this separation? (f) What' he minimum number of stages required at total refl? 0 ....~.::;,1.,. 100 lb moles/hr of an aqueous solution containing 20 mole

6.1. Evaluate the total and component net flows in Illustration 6.1.

·63

.

-

percent ethanol is to be fractionated at 1 atm to produce a distiHal"e of 80 mole percent ethanol and a bottoms of 2 mole percent ethanol. The total condenser has a duty of 1,000,000 Btu/hr. The bottoms are withdrawn from t~ bottom plate. The feed is a liquid at 140°F. (a) How many equilibrium stages are required? (b) What is the reboiler duty? (c) J;>lot the number of stages required for this separation as a function of reflux ratio. The plot should extend from the minimum reflux to total reflux. (d) How many stages would be required if the feed were half liquid and half vapor? 6.11. 2000 Ib/hr of a cottonseed oil-oleic acid solution containing 30 per cent acid is t6 be extracted with propane in a continuous countercurrent cascade at 98.5°C using extract reflux. The extract product is to contain 85 per cent acid, and it is to have a negligible propane content. The raffinate product should contain 3 per cent acid. The extract reflux ratio is to be 3.5. The fresh solvent and the recovered solvent are pure propane. (a) Calculate the number of equilibrium stages required on the triangular diagram. (b) Calculate the number of equilibrium stages requi~ on solvent-free coordinates. (c) Determine the minimum reflux ratio and the minimum number of stages. 6.12. A gaseous mixture of propane and propylene can be separated by fractional adsorption using silica gel at 1 atm pressure. The silica gel will be fed into the top of the column and will pass countercurrent to the gas. At the bottom of the column all the adsorbed gas will be stripped from the silica gel. Part of the gas will be withdrawn as product, and part will be fed back into the column as reflux. The feed contains 60 mole percent propane, and the product streams are to contain 90 per cent and 5 per cent propane. (a) What is the minimum reflux ratio?

64 ----PRINCIPLES

OF UNIT OPERAnONS

(b) How many equilibrium stages are required at twice the minimum reflux ratio? (e) How much silica gel per pound mole of feed is required at twice the minimum reflux ratio? ..........,~50 Ib moles/hr of an aqueous ethanol' solution containing 23 mole percent ethanol is to be fractionated at I atm in a distillation column equipped with a total condenser and a reboiler from which the bottoms are withdrawn. The distillate is to have a composition of 82 mole percent ethanol and the bottoms 3 mole percent ethanol. The feed is at its bubble point. (a) At a reflux ratio (Lo/ D) of 3, how many equilibrium stages are required? (b) Plot the number of stages required for this separation as a function of the reflux ratio from minimum reflux to total reflux. (e) Plot the reboiler duty ,as a function of reflux ratio from minimum reflux to total reflux. (d) How many stages would be required if the reflux ratio (Lo/ D) is 3 but if the reflux is subcooled 30°F below its bubble point? 6.14. A mixture is to be fractionated into a distillate, a bottoms, and an intermediate stream between the feed and the bottoms. (a) Draw a flow sheet for the fractionating column, and label streams. (b) Sketch a typical enthalpy-concentration diagram and show how the location of the required delta points is determined. Assume that Lo/D is set and that the quantities, compositions, and enth I ies of F, D, B, and I are known. 6.1 An aqueous solution of ammonia containing 27 mole peren ammonia is to be fractionated at 100 psi a in a tower equivalent to five equilibrium stages. It is desired to recover 95 per cent of the ammonia in a distillate which has a concentration of 98 mole percent ammonia. The feed solution is 20°F below its bubble point. What reflux ratio (Lo/ D) should be used? _I ~DiphenYlhexane is to be separated from docosane by extraction with furfural in a simple countercurrent cascade at 45°C. There are two sources of furfural solvent, one pure and one containing a small amount-of diphenylhexane and docosane. Since it is desired to process a maximum quantity of feed, both solvents must be used. It will be desirable to feed the impure solvent separately from the pure solvent. Mass Mass Mass Fraction Mass Flow Fraction Fraction DiphenylRate,lb/hr Furfural Docosane hexane Stream

Entering raffinate Entering solvent I Entering solvent 2 Exit extract Exit raffinate

5000 3000

0.00 1.00 0.92

0.70 0 0.05

0.30 0 0.Q3 0.15 0.01

(a) How much entering raffinate !!an be processed with the available solvent? (b) Determine the mass of the f,ldt extract and exit raffinate. (e) Determine the number o(.;quilibrium stages required. (d) At what stage should the impure solvent be fed? (Data for this s stem are in Chapter 5, Protilem,5.l3.) PD 6.1 In Problem 6.10 how many equilibrium stages are required if t e iquid entrainment amounts to 10 mole percent of the vapor leaving each stage? 6.18. Derive Equations 6.20 and 6.21. !' 6.19. The oil from halibut livers is to be extracted using ethyl ether as a solvent in a simple countercurrent cascade. There are available two sources of halibut livers: 200 Ib/hr of livers of 10 per cent oil, and 90 per cent insoluble; 300 lb/hr of livers of 35 per cent oil and 65 per cent insoluble. The entering ethyl ether solvent contains 2 per cent halibut-liver oil. The final extract is to contain 75 per cent oil and 25 per cent ether,! and 95 per cent of the oil charged in the livers is to be recovered in the extract. . (a) Calculate the number of equilibrium stages required when each liver source is fed separately into the cascade at the best location. Underflow datu are in Fig. 3.14. (b) Calculate the number of equilibrium stages required when the two sources of livers are mixed and fed together into the cascade. Use the same solvent rate and recovery as in part (a). 6.20. A mixture of 100 Ib moles/hr of 40 mole percent a and \ 60 mole percent b is to be separated into a distillate of composition 90 mole percent a and a bottoms of 5 mole percent a. Instead of using a conventional reboiler or condenser, the following proposal has been made. ' Each theoretical plate in the enriching section will have a cooling coil which will remove 100,000 Btu/hr from the liquid on the plate. Each theoretical plate in the stripping section will have a heating coil which will supply 100,000 Btu/hr to the liquid on the plate. (a) Calculate the number of equilibrium stages required to give • the desired separation.

Relative volatility: Enthalpy:

O(a_b

H

= 3.0 = 9000 -

5000x; .h

=

1000x

where x = mole fraction a (b) Is there any advantage to this arrangement over the conventional condenser and reboiler? 6.21. Is reflux useful in leaching? Explain. 6.22. Although Figure 6.2 implies a raffinate reflux when applied to liquid extraction, raffinate reflux is never used in practice because it does not improve the separation. Using an Appropriate extraction diagram, show that raffinate reflux does not reduce the number of stages required for a given separation, either with or without extract reflux. 6.23. Compare the eirect of using a total rebeiler (Fig. 6.2) in place of a partial reboiler (Fig. 6.9) on the number of equilibrium stages required for a given separation. What conclusion can be drawn from this co_mparison?

chapter 7

Special.Cases in Stage Operations: Simplified Calculation Methods The methods presented in the previous chapters can be applied to the calculation of multistage equipment when the data for the system of interest are available. In many cases more simplified graphical and analytical \ methods may be ~sed. A number of such simplified methods and their limitations will be considered in this chapter. The methods are in many cases more rapid than those considered previously, and often they require ~~ less physical data. However, the methods are based upon certain simplifying assumptions, which the system under consideration must follow. Simplified graphical procedures will be developed for simple countercurrent flow and for countercurrent flow with reflux. All the concepts covered in the preceding chapters will be I redeveloped using the simplified graphical procedures. In addition, certain analytical procedures will be considered.

...it

...

Solving Equation 7.2 for Y..+1 gives L.. Y..+I = - y x .. ..+1

In certain cases of stage operations the equilibrium x-y diagram may be used alone for a simplified calculation of the number of equilibrium stages required for a given separation. A material balance around any stage n and the left end of the simple countercurrent cascade in Figure 5.2 gives Total material balance:

+ V ..+1 =

L ..

+ VI

(7.3)

..+1

Equation 7.3 relates the compositions of a V-phase (y ..+1) and an L-phase (x .. ) flowing past each other between stages. This was precisely the purpose of the delta point defined in an earlier chapter. If the line represented by Equation 7.3 is plotted on X-Y coordinates it will be the locus of all possible values of (x m Yn+1)' It is usually referred to as the "operating line." If the composition (x ..) ofthe L-phase is known for any point in the cascade, the composition (Yn+l) of the V-phase flowing in the opposite direction at the same point can be determined from the plot of Equation 7.3. If in addition the equilibrium curve is plotted on the same x-y diagram, it is possible to calculate from stage to stage graphically. Equation 7.3 will yield a straight operating line if L and V are constant through the cascade. The terms Lo. VI' x o, and Yl have unique values, and therefore, if V is constant, the second term of Equation 7.3 is a constant. If in addition L is constant, the equation has the form of the standard slope-intercept equation of a straight line. The subscripts on L and V may be dropped, since the terms are constant.

GRAPIDCAL CALCULATIONS ON THE EQUILmRIUM DIAGRAM

Lo

+ V1YIy- Loxo

(7.3a)

The slope of the operating line is L/ V, and the y-intercept is the last term of Equation 7.3a. The line representing 7.3a is easily plotted if one point and the slope are known. If L and V are not constant, the line will not ~straight and more detailed calculations are required to. plot it.

(7.1)

Component balance: (7.2) 65

66---PRINCIPLES OF UNIT OPERATIONS

For most purposes the simplified method is more convenient only if Equation 7.3 describes a straight line. For this reason, it is of interest to examine several stage operations to determine under what circumstances Land V are constant. In distillation if the heat effects in the column do not change the molar flow rate of liquid or vapor from stage to stage, Equation 7.3 will describe a straight line. The liquid and vapor flows in a binary distillation column are influenced by four factors: 1. The molar heat of vaporization of mixtures. 2. The heat of mixing in the vapor and liquid. 3. The increase in sensible heat with increase in temperature through the column. 4. Heat losses from the column walls. If the molar heats of vaporization of. all mixtures are constant and if the other factors are negligible in comparison, for every mole of any liquid vaporized one mole of vapor will be condensed, and L and V will be constant through the cascade. Constant molar heats of vaporization give parallel vapor and liquid lines on the enthalpymolar composition diagram. Under this condition it can be shown geometrically that L and V do not vary. It is possible to have a straight operating line even if all four factors vary and are influential, as long as the net effect gives constant liquid and vapor flow. Many binary systems give sufficiently constant molar flow rates to permit use of the simplified calculation methods. When applied to distillation, the simplified procedures are often referred to as the McCabe-Thiele method (2). When the two solvent components (b and c) in liquidliquid extraction are completely immiscible at the concentrations of solute (a) under consideration, data can be reported as concentration (Ya) of a in phase V in equilibrium with concentration (xa) of a in phase L. Since components band c are mutually insoluble, a ternary diagram is unnecessary. Equilibrium data may be presented on x-y diagrams if the two solvent phases are mutually insoluble. However, the total mass flow rate ofthe raffinate (L) will decrease as the solute is transferred to the extract phase (V), and the extract phase will increase in total mass. This will not give a straight line with Equation 7.3. The mass of each solvent will not vary from stage to stage, so that Equation 7.3 will be a straight line.if the symbols are redefined in terms of mass ratios instead of mass fractions, as follows: L' Y,,+1 = V'X"

+

V' Y1

-

V'

L' Xo

Y,,+1 = mass of solute (a) per unit mass solvent (c)

X" = mass of solute (a) per unit mass of unextracted raffinate component (b) The extraction equilibrium data must be replotted on this basis on X-Y coordinates. The equations derived subsequently in the dimensions of Equation 7.3 may also be written in the dimensions of Equation 7.3b. In many cases of gas absorption, a single gas is absorbed from another nonabsorbed gas into a relatively nonvolatile liquid. An example is the absorption of ammonia from air-ammonia mixtures by water. Equation 7.3b can be used, where now L' = moles of liquid absorbent (b) V' = moles of the nonabsorbed constituent of the gas (c) Y,,+1 = mole ratio: moles of solute (a) in gas per mole of nonabsorbed gas (c) X" = mole ratio: moles of solute (a) in liquid per mole of liquid absorbent (b)

Where a gas is adsorbed by a solid from a nonadsorbed gas, Equation 7.3b may be used. An example of this would be the adsorption of water vapor by silica gel from humid air. Equilibrium data must be expressed in the appropriate units. GRAPHICAL CALCULATIONS FOR SIMPLE COUNTERCURRENT OPERATIONS

The graphical determination of the number of equilibrium stages involves alternate use of -the operating line and the equilibrium curve. An example of stage calculations for simple countercurrent operation is shown in Figure 7.1. The operating line may be plotted either by knowing all four of the compositions at both ends of

Equilibrium curve ~~-------~~

Ya

Multistage cascade

(7.3b)

where L' = mass of unextracted raffinate component (b), not including solute (a) V' = mass of extract solvent (c), not including solute (a)

Figure 7.1. Calculations on the equilibrium diagram for simple oountercurrent flow with transfer from the L-phase to the V-phase.

SPECIAL CASES IN STAGE OPERATIONS: SIMPLIFIED CALCULATION METHODS-----9

yl~-------------------~

~~------------------~~

(b) Transfer from L-phase to V-phase.

(a) Transfer from L-phase to V-phase.

Recovery specified; Yl to be determined.

Yl

specified;

XN

to be determined.

YN+l~--------------~

YN+l~---------~

~ ~--------------~----~ (c) Transfer from V-phase to L-phase. Recovery specified; x N to be determined.

Figure 7.2.

Limiting values of the ratio of phase flow rates.

the cascade or by knowing three compositions and the slope (L/ V) of the operating line. Equation 7.3a can be rearranged to give (7.3c)

or

(d) Transfer from V-phase to L-phase. X N specified; Yl to be determined.

(7.3d)

Even though the mathematical expression for the operating line is valid for any values of Yn+l and Xno it has physical meaning only for compositions which actually occur in the cascade. That is, the actual operating line extends from the point (xo, Yl) at one end of the cascade to the point (xN' YN+l) at the other end, as shown on Figure 7.1. The equilibrium curve can be plotted in the same range of x and y, as shown in Figure 7.1. The procedure for

stepping off stages may begin at either end of the cascade. For example, if the calculation is begun with the composition (Yl) of the V-phase leaving the cascade, the composition (Xl) of the L-phase leaving the first stage is determined by drawing a horizontal at Yl' on the equilibrium diagram. The horizontal intersects the equilibrium curve at Xl' as shown. Next, the composition Y2 must be determined from Xl by use of the equation for the operating line. This equation written specifically for the flow between stages I and 2 is '!12 =

L

-yXI +

VYI - Lxo V

(7.3a)

Since the operating line is a plot of the general form of this equation, the value of Y2 is determined by the intersection of a vertical through Xl with the operating line, as shown. Now x 2 is determined by the intersection of a horizontal at Y2 with the equilibrium curve, and this

68

PRINCIPLES OF UNIT OPERATIONS

0.06,----,---,--.,.---,--,----,----,---,---,--,----,---,

'OJ G)

(5

E .0

.:::::. G)

c

G)

N

C

G)

.0

en

G)

(5

E :!2

J

~

..

'-

0.6 X a , Ib moles benzene/lb mole absorbent oil

Figure 7.3. Solution to Illustration 7.1.

stepwise calculation is continued until (x N, YN+1) is reached. As shown in Figure 7.1 slightly over four stages are required. The location of the operating line below the equilibrium curve indicates that mass transfer is from the L-phase to the V-phase. On the other hand, an operating line above the equilibrium curve shows that mass transfer is from the V-phase to the L-phase. Operating Variables. Limiting values of the L/V ratio may be determined for several cases. For transfer from the L-phase to the V-phase, the minimum V/L ratio, which gives the desired separation with an infinite number of stages, can be determined as shown in Figure 7.2. For a specified recovery from the L-phase, the point (xN, YN+1) is fixed. As the quantity of V-phase is reduced, the slope (L/ V) of the operating line through (xN, YN+1) increases until the operating line first touches the equilibrium curve, as shown in Figure 7.20. An attempt to step off stages shows that an infinite number of stages is required to change the composition across the point at which the operating line and equilibrium curve intersect. The term "pinch" applies here to the

pinched-in stages at the point of intersection. If the concentration of the V-phase leaving the cascade is specified, the point (xo, Yl) is fixed, and the pinch occurs as shown in Figure 7.2h. However, the recovery is not specified; that is, X N is not fixed. In this case the operating line through (xo, Yl) which first intersects the equilibrium curve gives the limiting yalue of V/L, which is a maximum value. Limiting ratios of L/ V can)e determined in a similar .. manner for the case where mass transfer is from the V-phase to the L-phase, such as in gas absorption. For transfer from the V-phase to the L-phase, the operating line will be above the equilibrium curve. When the recovery of component 0 is specified, the value of Yl .is set since YN+1 is known. Since the entering L-phase composition (xo) is usually known, the point (xo, 'ill) is fixed and the limiting L/ V is located as shown in Figure 7.2c. In this case the limiting L/V is a minimum. On the other hand, when Xo and YN+1 are known instead of the recovery, XN may be specified. Then the point (x N, YN+1) is fixed and the limiting L/V is determined as shown in Figure 7.2d. This limiting ratio is a maximum. t

,

SPECIAL CASES IN STAGE OPERATIONS:

Illustration '1.1. A benzene-air mixture is to be scrubbed _". in a simple countercurrent absorption tower using a non'1\' volatile hydrocarbon oil as solvent. The inlet gas contains 5 per cent benzene and the entering gas flow is 600 lb moles/hr. Solubility of benzene in oil follows Raoult's law. The tower operates isothermally at 80°F. The average molecular weight of the oil is 200 and the tower pressure is 1 atm. (a) What is the minimum oil rate (lb/hr) needed to recover 90 per cent of the entering benzene? (b) How many theoretical stages are required if the oil rate is 1.5 times the minimum ?

Since the recovery is specified and the mass transfer is from the V-phase to the L-phase, the limiting L/ V ratio will be determined as shown in Figure 7.2c. To assure a straight operating line, mole ratios will be used. This necessitates calculating the equilibrium curve for mole-ratio coordinates. SOLUTION.

Pa

Ya = P

Xa

69

SIMPLIFIED CALCULATION METHODS

The stages may be stepped off from either end. About 5.3 stages are required for a liquid rate of 15,600 lb/hr, as shown on Figure 7.3. Use of mole fraction coordinates with the assumption of constant L/ V in the problem would result in a substantial error. Although the gas phase is dilute, the liquid phase is not. The equilibrium curve shows appreciable curvature. COUNTERCURRENT FLOW WITH REFLUX

The introduetion of reflux (Figure 6.2) results in two net flows and two operating lines. A material balance from the solute-rich end to include any stage n in the enriching section gives Ln

Y n+1 = - V xn n+l

+ DXD V+n+lVcYc

(7.4)

When L and V are constant, the subscripts may be dropped.

for benzene

At 80°F, Pa .= 103 mm Hg; P = 760 mm Hg.

(7.40)

103 Ya = 760 xa = O.136.1:a Ya

Since Ya = 1 + Y

a

A material balance between any stage m in the stripping section and the solute-poor end of the cascade gives

Xa and xa = 1 + Xa

Ya Xa - - =0.136-1 + Ya 1 + Xa Y

a

Xa

1

0

I

0.

005

Ym =

1

0.01

I

0.02

1 0.03

1 0.04

I

0.05

V = 600 lb moles/hr ii

Xo = 0.0,

0.05

YN+1

= 0.95 = 0.0526

For 90 per cent recovery, the benzene leaving in the gas will be (0.IOXO.05)(600) = 3.0 lb moles. Therefore, Y I = 3.0/570 = 0.00526 Ib mole benzene/lb mole air, since VI' = V.~+1 = (0.95X600) = 570 lb moles air. The conditions at the upper end of the tower are set and can be plotted: Xo = 0, Y1 = 0.00526. The minimum liquid rate occurs when the operating line through (Xo• Y1.) first touches the equilibrium curve, as shown. The slope of this line is 0.0526 - 0.00526 --0-.-:520--"--,0,----

=

0.091

Therefore, the minimum oil rate is (0.091)(570)(200) = 10,390 Ib/hr

=

I.5(L'/V')mln = (1.5)(0.091)

= 0.137,

L' = (0.137)(570)(200) = 15,600 lb/hr ~

BXB - VS+1YS+1

V

L

The actual operating line has a slope of 0.137, and it is determined that X N = 0.345.

Ym =

r

Xm- 1 -

(7.5)

m

The values ofL and V may change at the feed.

0 0.038 0.0785 O. Hi8 0.272 0.395 0.539

At (L'/V!)actual

y-Xm-l-

m

These values are plotted on Figure 7.3.

!

Lm-l

BXB - VS +1YS+1

V

Therefore, (7.5a)

where L and V are the constant values of L-phase and V-phase flow in the stripping section. In distillation calculations V S+1 = 0 and Vc = 0; and Equations 7.4a and 7.5a become

L Yn+ l

or

D

=-X

V

n

L

= L

I

(7.4b)

+-xD V

D

+ D xn + L + D x D

(7.4c)

(7.5b)

or

(7.5c)

Inspection of the equation for the enriching-section operating line (Equation 7.4b) shows that it is a straight line when L and V are constant. The slope of the line is L/ Vand its intercept at x = 0 is Y = DXDJV. Furthermore, the operating line intersects the diagonal (x = y) at xn = Yn+1 = x D • Similarly, for the stripping section (Equation 7.5b), the operating line has a slope L/v and a y-intercept at -BxB/V. It intersects the diagonal at

J ]1

-,1\

~, I~


tf'liA,n,nl in air or to see a humidity, diagram for the lIihe~ne-'~Nn system (10).

In humidification calculations, it is frequently necessary know the volume and specific heat of the' gas-\(apor For~~ humid, volume is defined as of I lb of dry gaSji1u5 itsConrarnea- ; _ _ _----'--"-" lb

VII = (1

+ Y)

...,

X

359

X

T

~ 492P

, (17.7)

to those conditions where the perfect-gas law is Here; VII is the molal humid volume. Similarly, humid heat is defined as the heat capacity of Qf-dry gas plus that of its associated vapor. (17.8)

y. vapor-gas mixture is not generally Y'lltHet;lal name. However it frequently appears By analogy to Equation 17.8, (17.9)

where H is the enthalpy of I lb mole of dry gas pJus of its contained vapor, and Hb and H(J are"the 1ftbIal enthalpies of components b and a respectively. These enthalpies must, of course, be computed relative ~he-­ enthalpy of some arbitrary base condition at which H = O. The base conditions cannot be identi.::al for b and a since different components are being considered. However, they may be at the same temperatures, pressures, and phase conditions, though this is in no way necessary. For the air-water system, it is common practice to take the enthalpy to be zero for liquid water at the t.!:i.Q1e.... }Joint The base for the enthalpy of air is often taken as the dry gas at 32°F, I atm pressure. These bases will be used here. On these bases, the enthalpy of moist air may be related to the appropriate latent and specific heats.

= cb(T - To) + Y[Ao + ciT - To») (17.10) To = the base temperature, here,32°F for !Joth H

where

components

Ao = the latent heat of evaporatioJL:;f ll,(a~w

.;.t.,.:

the base temperature, Btu/lb mole and the effect of pressure on the enthalpy of the liquid is ignored. In this equation, both C b and C have been a taken as constant. If this assumption is not tenable, : these specific heats must be the integrated mean values applicable between T and To. Other terms used to describe vapor-gas mixtures are :. the dew point, wet-bUlb temperature, and adia,baticsaturation temperature. The dew point is th~._ ~hich condensation begins when the press~~2r temperature is changed Over a mixture of fixed composition. Usually the temperature is lowert!d at constant pressure, so that the dew-point tel1)perature is obtained. The wet-bulb tem eiiiiureis the skad -state temperature atta.! wet-bUlb t ermometer ex osed to a rapid moving stream of the va or- as mixture~ The bulb of the wet- u t ermometer is coated with the-' same liquid that forms the vapor in the vapor-gas mixture. The wet-bulb temperature is related t9 the humidity of the gas phase as wiJI be shown below. This relation allows the use of the, wet-bulb temperature coupled with the temperature itself as a measure of humidity. The adiabatic-saturaiion temperature is t~emperature ' that the' vapor-gas mixture would_~h ~irjJ, saturated through an adiabatic procesS-- Many of the ' proces~s discussed in this chapter and in Chapter 18 occur approximately adiabatically, so that the adiabaticsaturation temperature is a particularly useful and important quanti~y.The relation between this temperature and the gas-phase htfl'ffldity will also be shown below.

wet.e-:

For convenience these definitions are collected in Table 17.1. The saturation locus and partial-saturation curves on the temperature-concentration diagram are

.-

SIMULTANEOUS HEAT AND MASS TRANSFER I:

shown schematically on Figure 17.10 and quantitatively

~--ppendix D-14, the humidity chart for the air-water system. Adiabatic-Saturation Process. The relation between the adiabatic-saturation temperature and the gas-phase humidity can be developed by considering the process by which the gas can be saturated. A general humidification process is shown in Figure 17.6. Here the subscripts 1 and 2 refer to the bottom and top of the column respectively; subscripts L and V refer to liquid and vapor phases rcespectively; £ and V are the molal rates of flow of liquid. and vapor respectively; and V' is the molal flow rate of noncondensable gas, lb moles/hr. A material balance around the tower gives '

~

V'

(

.

Figure 17.6. General humidification process.

For this general process to be specialized and to ~ made adIabatic, several restrictions must be imposed. First, no heat can be transferred to the tower, or = 0 Second, the Ii uid stream will be this wi at steady state T' = T TJ As the process continu t e lquid temperature will become constant, and thus no net sensible heat will be brought to the tower or taken

L:'

,,,and an enthalpy balance gives \., q L H - ~HLl = V'(HV2

+

2

u

H Vl)

-

Table 17.1.

v/~

2. Molal humidity V

I.

Units

\ /''I/) / . Symbol Iv.' ......)

;>

olb vapor/lb noncondensable gas //~ moles

vapor/mole noncondensable gas

vapor content of fl gas

./

3. Relative saturation or relative humidity

atm/atm, or mole fraction/ mole fraction, often expressed as per cent

ratio of partial pressure of vapor to partial pressure of vapor at saturation

4. Per cent saturation or per cent humidity

ratio of concentration of vapor to the concentration of vapor at saturation with concentrations expressed as mole ratios

5. Molal humid volume

volume of lib mole of dry gas plus its associated vapor

mole ratio/mole ratio, expressed as per cent

cu ft/lb mole of dry gas

heat required to raise the temperature of 1 lb mole of dry gas plus its associated vapor 1°F

~ 6. Molal humid heat

7. Adiabatic-saturation temperature

temperature that would be attainediJ-theglls weresaturatecW n an adiabatic process

Btu/lb mole of dry gas, OF

OF or OR

OF or OR 8 .. Wet-bulb temperature

~

9.

.

DEFINITIONS OF HUMIDITY TERMS

vapor content of a gas

1. Humidity

-.;tf/'"

HI-I: #J.J. -"""1- II

Meaning

Term

"-

(l7~li)

~

Dew_po~_ature

steady-state temperature attained by a wet-bulb thermometer under - standardized conditions OF or OR temperature at which vapor begins to condense when the gas phase is cooled at constant pressure

.,.::L r

)JtA_J :.

N~-~'~~

')