43
2)
•
CAmo UNIVERSITY !caculty of Engineering
Chemical Engineering Department
Degree: B.Sc. May 2005
Subject: Mass Transfer Operations Time allowed: 3h.
Attempt All Questions
An aqueous feed solution of 1000 Kg/hr containing 23.5 wt% acetone and 76.5 wt% water is being extracted in a countercurrent multistage extraction system using pure methyl isobutyl ketone (MIK) solvent. The o utlet water raflinate will contain 2 .5 wt% acetone. The equilibrium data can be found in the table below.
a) Calculate the minimum solvent that can be used. b) Using a solvent flow rate of 1.5 times the minimum, calculate the number of theoretical stages
Composit ion Data Acetone Distribution Data wt% (wt%
MlK Acetone Water Waler Phase MlK Phase 98.0 0 93.2 4.6 77.3 ! 8.95 71.0 24 .4 65.5 28 .9 54.7 37.6 46.2 43.2 12.4 42 .7 5.01 30.9 3.23 20.9 2. 12 3 .73 2.20 0
-a-Draw neat skctches ill ustrating :i· Bollman bucket type extractor.
2.00 2.33 3,86 4.66 5,53 7.82 10.7 45.0 64.2 75.8 94 .2 97.8
ii - Hildebrandt screw-conveyor extractor.
2.5 5.5 7.5 10.0 12.5 15.5 17.5 20.0 22.5 25 .0 26. 0
The table above gives thc composition of streams sampled from thc two stage solid liquid extraction battery.1t may be asswned that the underflow from the first and the second stage retain the same mass of so lution per unit mass of dry inso lub le solids. Calculate:-
• ,-.. ,,-· .. 11 1-• <V-
The operating solvent/feed ratio. The mass of extract produced per unit mass of feed . The compos ition of the underflow from the first stage (XI)' The overflow Murphree efficiencies of the tow stages.
4.5 10.0 13.5 17.5 21.3 25 .5 28,2 31.2 34 .0 36.5 37.5
(3)
(4)
(5)
A 5O.0-L tank contains :10 air-carbon tetrachloride gas milL:lure al an absolute preSSure of 1 aim, a temperature of 34°e. and a relative saHl nll ion o f 30%. Activated carbon is added to the tank to remove the CC'" from the gas by adsorption and the tank is then sealed. lne vol Uine of added activated carbon may be assumed negligible in comparison to the lank volume. (a) Calculate Pco. at the mOmen t the la nk is sC3/cd, assuming ideal gas behavior and neglecting
adsorption Ihal occurs prior to sealing. (b) Calcula te the 10lai pressure in the lank and the partial pressure of carbon tetrachloride OI l a point
when half of the CCl. initially in the lank has been adsorbed.
Langmuir adsorption isotherm for carbon tetrachloride on activated ca rbOIl at )4°C. is
X' (8 Co. adsOrbed ) = g carbon
O.0762pca, I + O.096pco..
where Pea.. is the parti:1I pressure (mm Hg) of carbon lel rachloride in Ihe gas con lac ling Ihe carbQn.
(c) How much aCllv tl lerJ carbon musl be added 10 Ihe lank. 10 reduce the mole frac liOIl of ca~ in Ihe gas 10o.OOI?
Why would Ihe actual amolJnt placed in the tank be large r th;1I1 Ihe calc ula lCll value?
From Ihe Antoine cqu ation . Ihe vapor prcssure of 'Moon tetrachloridc at }4°C is
Pco, '" 169 I1IIlI lig. Con~qucnl ly.
Give reasons for :
·In Ihe clay treatment ofperrolcum - lubricant fraction the adsorbent - oil mixture Illlly be pumped
I I
tl () r h 'I through a tubular fu rnace 10 be heated to as much as 120 to I SO" C • and lor very cavy 01 s even to 300 to 380~ C . • In waler cooling lowers it is perfectly possible to cool water to a value less than the entering air drybulb temperature . but water cannot cool below the air wet - bulb temperature .
. In batch adsorption if the adsorption factor is sufficiently large. the solid - phase mass transfer resistance can be neglected . - Air fed to a COntact sulfuric acid plant must be gried . .,.. When very large flow rates are treated· 500 m '/ s of flue gas !Tom a power statlOll for example a fluid ized bed adsorber or a fi xed bed adsorber is beller to usc and why?
A hydrogen stream a t 300 K and a tmospheric pressure has a dew point of 275 K. It is to be fu rther humidifi ed by add ing to it (through a nozzle) saturated steam at 240 kN/ m
1 at the rate o f I kg steam to 30 kg o f hydrogen feed. What wi ll be the
temperature and humidity of the res ul tan t stream ?
At 275 K the vapour pressure of water == 0-72 kN /m1 (fro m tab les) and the hydrogen is saturated .
At 240 kN/m1 pressure. lal e nt heat is 2185 kJlkg.
. s team --is sa ~u!ated at 400 K at which temperature the
Taking the mean specifi c heat of hyd rogen as 14·6 kJlkg K,
The lalent heat of water at 275 K is 2490 kJ/kg and laking the specific hent o f
water vapo ur as 2·01 kJ/kg K,
Water has a vapo ur pressure o f I · 11 3 kN/m1 at 28 1 K at wh ic h the latent heat is 2477 kJ lkg. .'
(6) . _ , ':Three fixed-bed adsorbers containing 10,000 lb of gra nules of activated carbon (PI> "" 30 [b/ft J )
each arc to be used to treat 2S0 gplll of water contai ning 4.6 mg/ L of 1.2- dichlorocl hane (D) to reduce
Ihe CQnccntralion to less than 0.001 mg/L. Each carbon bed has a height equal to twicc Ihe diameter. Two beds are to be placed in series so that when bed I (the Ic:Ld bed) becomes ~atura t cd wit h D at the feed concentration. tll .. t bed is removed. Bcd 2 (Ihe tr3iling bed). which is pa rtially saturated at this I>oint, depend ing upon the width o f the MTZ, becomes Ihe lead bed. and previously idlc bed 3 wkes the p13ce of bec.l 2. While bed I is off-line, its spent carbon is remMed and replaced with fresh carbon. The spent ca rbon is incinerated. The equil ibrium adsorption isot herm for 0 is give n by q = 8 cQ
·n . where q is in mg/g and (: is in mgfL. Once the cycle is estab lished, how often must the ca rbon in a bed be replaced? What is the maximum width o f thc MTZ that will allow s:lw ratcd loading or the lead bed?
8000 , 0000 '000
'000
." 3000 • " , ., • , • 2000 ! ~ " ~ • , ,
> " , • • >000 • , 9 < -~ .00 " " , • , •
" 600 , , '" , ,
< '" " , ~
300
• • " '""
Tem pe .. """. ·c
", ... ) PsychrOmetric chart for air-water vapor. I sid atm abs. in Sf units.
\ ,' . . '. :: ~ ,-',- ~' ·;·"'·' ", ' .... · ... ·,·. ·,,· .. · ~ .. ·,·.·,· .. ~_ .. ~~=a .... =~ ___ ~ .. ~ ____ ~ ___ ,~'''' ____ ~ ______ •. , ..... ,.~.
2S00 2000
, "", • " •
" ,
• '000 • , • " • ."" " • " • • ''''
, t • • • ,
. ", " • • • I. "
< -, 300 " • " , • • 2" " , -~ ," , < " '" , .. -
''" • • • •
om ,
0,02
0,01
,., ~) p.ychrometoc d,art for air_w.o.ltl Vllpor. 1 sId aim abs. in Engli.h enpncaing units.
,
A. I - l
APPEN DI X A. I
Fundamental Constants and Conversion Factors
Cas La", Constant R
N~IfWn.:1II V <IIwe U""J
1.9812 gcaVgmo!' K • 19872 hlu/lb mol ·o R 82.0S7 em'· alm/s mo!' K 8314.34 l/kg mol · K • • 82.057 II 10-' mJ·atm!kgmol - K, 831 4,34 kg _ml/, I , )'; g mol· K 10.73 1 ft' ·lbJin.I · lb mol·OR 0.7302 fl ' · lun,llb mol · oR 1545.3 fl 'lb,f1b mol- "R 8314.34 m' · Pajkg mol- K
A.I-2 Voilimt Ind Ocnlity
I I mol ideal p,:11 O·c, 760 mm Hg .. 22.41 40 liters .. 2241 4 em' I Ib mol ideal gu II O"C, 160 mm Hg ~ 359.05 fl' , I kg mol ideal gas II O"C. 760 mm Hg .. 22.414 m' Densi ty of d ry air al O·C. 760 mm Hg "" ] .2929 &/1iler
~ 0.080711 lb..lft ' Molecular weight of air " 28.97 Ibmflb mol " 28.97 ~g mol I gjcm l .. 62,43Ib..lft l .. lOOOkg/m ' I gJem' ... 8.l45Ib.lU.S. &al Ilb../fI' .. 16.018Hg/m '
A.l-J Length
1m .. 2.540em IOOem .. I m{mcler)
I , :
,
I micron _IO--m .. IO··c,n .. 10-' mm ... 1 Jim (micrometer) I ;. (angstrom) .. 10 - ,0 m = 10- ' Jim I mile .. 5280 (I 1m .. 3.2808 ft .. 39.37 in.
lib .... 453.59 g .. 0.45359 kg lib .... 16 oz .. 1000 grains I kg .. 1000 g .. 2.2046 lb. lion (shore) .. 2000 lb. I Ion (long) _ 2240 lb", I Ion (metric) .. 1000kg
A.I-S St.nd3 rd Accc!ru,;on of GfI,ily
9 .. 9.80665 mis ' g ... 980.66Scm/s l g .. 32.174(1/5' f1, (gfllvilalional conversion facl(.o(j .. 32.1740 lb .. · (I jib, . Sl
.. 9so.665 s..' em/&:' Sl
A.I-6 Volunle
1 L (liler)" 1000 tm' I ill.l .. 16JS7 tm' 1 [.' .. 28.317 L (liler) I fl' .. 0.028317 m' 1 hi .. 7.48] U.S. gal 1 m' .. 264.17 U.S. gal
A.] ·7 Forn
I m l .. 1000 L (li ler)
IU.S.gal.4qt I U.S. gal .. 3.7854 L (lilu) I U.S. 8~1 • 3785.4 tm l
J B,ili5h gIl .. 1.20094 U.S_ gal ] m) .. 35.3]) rll
I g 'em/s' (dYIl)" 10- 1 kg- rn/s 1 .. 10- 1 N (newton) I 8 ' em/s' .. 7.2330 x ]0-' Ib"' · rV,l (poundll ) I kg ·rn/Sl .. I N(newIOn) ] Ib, .. 4.4482 N Ig cm/s l ... 2.2481 '" 10-6Ib,
A.I·8 Pr~lIlt
1 b~r .. I x 10' Pa (pascal) .. I x 10' N/rn' I p,;a .. Ilb,lin. ' I p~ia .. 2.0)60 in. IIg at O'C I [l.ia '" 2.311 fl H,D al 70"F
J Jl5ia .. 51.715 mm Hg ~t O°C II' ... .. 13.5955 glern'} I aim" I ~ .69(, psi~ .. 1.01325 '" 10' N/ m' __ 1.01 J2S In, I aim" 760mm IlgllO"C "" 1.01325 x 10' PI I aIm .. 29.921 in IIgltO· C I aIm., .\.1.90fl / 1, 0111 4· C
I psi. = 6.&9476 x IO~ g,lem ' I' 1 psi. = 6.&9476 " 10' dyn/cm' 1 dyn/cm' .. 2,0886 x 10-) lb,/ft' I psin :- 6.89476 x 103 N/m 1 - 6.89476 x 10' Pa llb,lfl' .. 4.7880 x 10· dyn/cm' .. 47.880 N"/rn' I mm Hg {ooq .. \.333224 )! 101 N/ m· .. 0.1))3224 ~Pa
A.I-9 PO"'er
o
I hp .. 0.74570 kW I hp .. 550 f, ' I~fs I hp .. 0 ,7068 btu/s
I .... an (W) _ 14 340 eal/ min I btufh .. 0.29lO7 W l"'lu) 1 1/, (joule/5) .. I W
A. I-IO Hut, Entru. Work
IJ_IN m _ lkg-m'/s' Il::g m'/s' .. I 1 (joule) .. 10' g -cm ' /$' (eT"::) I btu _1055.06J .. 1.05506kJ 1 btu .. 252.16 cal (thennochemical) I keai (thermoehemical) '"' 1000 c,a1 ~ 4.1840 kJ I cal (thermochemic,al) .. 4.18401 I eal (IT) .. 4 1868 J I btu - 251 .996clIl (IT) I btu -778.11ft·lbr Ihp · h_0.74SH W ·h Ihp · h-2s.u S btu I ft Ib,- _ I 3SS&2 J I ft· I~llb . .. 2 9890 l/1<g
A.I-II Th~rmal Conductivi lY
I btufh · ft -~F .. 4 1J6S X 10 - ' eal/s -em ' °c 1 btu/h oft .O F • 1.73073 W/m' K
A. I-12 Heat-Tnnsfer Coefficient
1 btufh -ft" OF _ 1.3571 x 10 - · ~ I/s em: ··C I htufh · fl' ,· F .. 5,678l x 10- ' W/cm" oC 1 btu{h-ft" oF .. S.678lW/m' K 1 kcaljh m" oF_0.2048btu/h Ttl OF
A. I·13 Vi'\Cosi ly
Icl' ~ 10-'g,lem·s(poise) IeI' '" 2_4191Ib,.lft-h 1 cl' = 6.71\17 x IO-'lb,.lft S lcp ~ 10 ' lPJ s_IO - ' kg/m s_JO - 1N .. 1m' I ep ~ 2_0886 " 10 - Sib, ' s/fl'
•
II'a-s" t N·,;!m 1 .. I kg/m· _, · IO()()~p '" l).b71\17Ibm'f\ ·s
A"" A .I
T ,
I
1
i
I I •
i
I , ,
I 1
,
,
I
-
A.I_14 Dlffll'litity
1 eml/I ~ l87S rt'/h 1 eml/s" 10-' mIll 1 m1/h ... 10_764 ftl/h
I ml/I" 1.875 )! 10' rt 'fh 1 centistoke .. JO- ' em' /s
A.I-IS Mass Flux and Molar Flux
I g/s' em' .. 7.3734 )! 10· !b..fh · ft' Igmolh-em'-73734)( IO'lbmolfh re Igmol/s'em' _ IOkgmol/s·ml .. l)( 1O~,mol/s ' m' Ilbmoljh·ft'_1 .3562)( 10- 1 ",mo\ls m'
A. I-16 Ilut Flux and lin t Flo ..
I btu/h·ft ' = 3.1S46W/m' I btu,lh '" 0_29307 W I nlfh .. 1.1622)( 10- 1 W
A.I-17 lIu. Capaci.y ud Enthalpy
I blu/lb. ·"F "" 4.1868 U/I<g - K I btu/lb.· OF '" 1.000 callt· "C I blu/lb .. .. 2326.0J/I<& 1 rl ' Ibr/lb .. '" 2.9890 J/I<g ! ea! (ITVg· · C .. 4,!&6S1::1/l::g ' K I ~ cal/il mol. 4.1840 " 10] kJ/kg mol
A.I·18 1Io1llss-Tnnsftr Codficj~Qt
I l , em/ s _IO·lmjs 1 1,;, rt/h .. 8.4668 x 10-' mls 11.;.&mol/, cm' · molfrae .. l0 kgmol/ s·m' molrrlC II,; , g mol/ s'em" mol frae "" I )( 10' g molfs m' · mol frac lk. lbmolfh · ft' · molfrac ", l.J562 " 10-.' \:.gmol/s · m'·mol frlc 14. alb molfh · fll . mol frae .. 4449 x 10-' kg mol/s· m" mol frlc I ~ .. kg moV$ ' ml ' ltm .. 0.98692 x 10- ] kgmol/s' m" PI I 1,; .. .. kg molls ' mi . atm .. 0.98692 )( 10-' kg moljs- m" Pa
ApPMd',f A .I
• ,
Subject: Degree: Date:
Part One
CAIRO UNIVERSITY Faculty of Engineering
Chemical Engineering Department
Mass Transfer Operations B.Sc. May 2006
Attempt All Questions
Time allowed: 3h.
) ,. Water at 40°C enters 8. coo ling lower at a rate of 200 kg/s o The water is cooled to 25°C in t]
coo ling tower by the air which enters the tower at 1 atm, 20°C, 60 per cent relative humidi ty al
le(!vcs saturated at 30°C. Neglec ting the power input to the fan, dClennine (a) the volume fl c rate of the air entering the cool ing lower, and (b) ule required mass flow rate of the make-! water.
b- Determine the req ui red height of adsorlxnt in an adsorption column th:ll trcalS a dcgrcaser-vcntil ;'l.l ion stream con!..3.lII inated with trichloroethylene (TeE). Design and oper. a t ing data arc as follows:
Volumetric fl ow .-ate of con(.amin:llcd air: 10,000 std ft'/min
(4.72 m'/s), stand:ll"d cond itions being 60" r and 1 aun
O perat ing tCOlpe l'·:-Iturc: 70" F (294 K)
Opc r.l(ing pressure: 20 psia (138 kPa)
Adsor bclll: activated caroon
Bulk dcnsity P, of activated carbon: 36 lb/ rt J (576 kg/lll')
Working capacity of acti va ted carbon: 28 lb TeE pe r 100 Ib carbon
In let conce ntration ofTCE: 2000 pp m (b y volume)
Molecula r weight ofTCE: 131.5
The adsorption column is a vertical qlinder with an inside diameter of 6 fl (1.8 m) and a he ight of 15 ft (4.5711\). Il operates on the followi ng cycle: 4 h In the adsorption mode. 2 h fo r hcaling and desor bing. I h for cooling. I h for standby. A II iden tical colum n treatS the contam ina ted gas while the fi nlone is not in the adsorption mode. T he system is requ ired to reco .... er 99.5 JXrcent of the TeE by weight.
I •
I
, •
I
I
(2) a-Give reasons for.
· 1n a tray (shelf) dlyer , part o f the drying alr may be recycled . • For efficient operation of a flXed bed adsorber the run must be stopped just before
the breakpoint. · For amorphous, fib rous or gellike o rganic solids, the dryi.ng curves show ollly
very shOlt constanl - rate periods, eoding at higb values of cri tical moisture conte nt.
b- Pure-component hqUid-phase adsorption data are used with the extended Langnlwr Isotherm 10 predict a binary. solute dam pomt Use the foHo .... ing mixture data to oomin the best fit to an e:<lcmded Langmuir-Freundlich IsOthern of the form
(qO) ik iC}lll, qi = 1 -1-. '" k 1/"
I L..,; --je) J
/
Data for binary-mixture ad~orp t lOn on activated carbon (1000 m2lg) at 25°C for acetone ( I) and proplOrltIrile (2) are as foHows:
Solutiun Concentration, mollL L oadi ng, mmoJ/g
c, C2 'I, q2
- -, Ie -).:L . - .) 7,46E - 5 0.0192 0.0199 6 1410 - 5 7.71 E - 5 0.0191 0.0198 LOnE - 4 1.35E - 4 0.0308 0.0320 1.12E - 4 1.46E - 4 0.0307 0.0319 3.0JE - 4 732E - 3 0.0378 0.263 J.17 E - 4 2.34E -- 3 0.0378 0.264 325E - 4 3.89E - 4 0.0644 0.0672 1.426 - J 1.58E -
, 0.161 0.169 .,
IA2E ,
1.61 E - 3 0.161 0.1 69 J
1.431:: , 1.60E - 3 0.161 0.169 - J
2,OlJE -,
3.84E - 4 0.250 0.0390 ., 2. 176
, 3.85 E 4 0.251 (10392 .' ;~" 991:,. ,
5.24E 3 0.291 11.307 - .' -'.06E 3 5.31 E , 0.288 0.305 - ., 7<11E - 3 2AZE - , 0.237 0.900 -"7 ,,'IF ,
'.471' , 0.236 0.896 -' -. - . -.., 79 r ,
7.51)[ ,
0.802 0.251 - , -..: !)f) I , 3.44 [ , 11.71, I' X22 - -1 i!.'" 1
, ~ 1"11- , D.7 17 n.:-<.H .'- -
-- -- -"" ---,- - ---~- ... -
(J)
4) ,.
b·
A tram of fOllI 55-gallon carullslers of 3CtlVllt~d carbon is to be used 10 reduce the ni troglycenne (NC) content of ~oo gph of wastewater from 2,000 Rpm by weight to less than 1 ppnl Each cannis ter has a diameter of 1 ft and holds ZOO lb acttvated carbon (Pb = 32 Ih/f! ) Each canmster is equipped with a I1quid-f1ow distributor to promote plug flow through the bed of carbon. The eff1uenl from the first cannister;s monitored so thaI when II 1 ppm threshold of NG IS reached, that canmster is rem oved from the train and a fl"eslt (:UllliSlt.o IS added to the end of the tram The spent carbon IS mlxed wIth coal for use as a fuel In II CO<1.1-fired power plant at the process si te USing Ihe follOWing pilol-pl:lIH d;U!l, estlm ate how rn :IllY c:lrl il is tcrs are nceded each month and the monlhly cll nni Slcr' COS! at $iOO per canruster
Pilol-IJI:ulI dala:
Tests \.".'lth the same 55-gallon cannIster to be used in the commerCIal process, water flow rate '" 10 gpm. NC conlent In feed = 1,020 ppm by weight
Ureakllll'ough co rrel:lI ion:
In ':: 3.90 L - 2.05, where ' )) .. tIme, II , at breakthrough of the I ppm threshold and L '" bed dellill of carbon III fect
DI':l\V nen l sketches for the fOllowing: - Troy or shelf dlYCr. - Mullislage nuidized bed adsorber with regenerarion . - PrOcess flow diagram of:1 typical spray drying system.
An air-conditlotung system t<lkes in atmospheric air at 12°C and 25 per cent relative hurrudtty a a s teady rate o f 50 mJ/ m.i n and conditions it to 22°C and 50 per cent re lat ive humjdi ty. Thl atmospheriC air is first heated to 20°C in the heal ing section and then humid ified by the injecti Ol of hot steam In the humidifying sec tion. Assuming that the entire p rocess takes place at . pressure of 100 kPa , determine (a) Ihe rate of hea t supply in the heating sec tion and (b) th' requ ired mass flow rate of the steam in the humidifying section.
PA RT II 1- a- Draw neat sketches 01': i- Bollman Extractor. ii-Fixed Bed Leaching Extractor.
b- Seeds containing 25% by we ight of oil are extracted in a countercurrent plant, and 90% of oil is to be recovered in a solution containing 50% of oil. [t has been founcl experimentally that the amount of solution removed in the unclerflow in assoc iation with every kilogram of insoluble matter is given by the equation: k=0.7+ 0.5y,+ 3y/ Where y, is the concentration of the overflow solution (weight fracti on of solute). If the seeds are e,tracted with fresh solvent, how many ideal stages are required?
2-a-" 'rite short notes on:-i- Advantage and limitation of extract with reflux. IL- T he relation between the required number of stages and reflux rat io in li quid extraction battcry. iii- Effect of temperature on solub ility curves.
b-!\n aqueous solution containing 40% acid is to be extracted in a countercurrent cascade wi th extract retlux. The extraction will be conducted with pure solvent. The solvent leaving the solvent recovery unit is water free and contains lOO kgs sol vent / kg acid. The Gna! raffinatc and extract products are solvent free and should have 5% acid, 95% water and 90% acid and 10% water respectively.
Dcterminc:-i- The composition of the extract phase leaving thc battery. ii- The maximum ac id content, on so lvent free basis obtained in the extract phase from the eX1T3ction battery ifno extract reflux were to be used, iil- The mini mum refl ux ratio that may be used [or the extract enriching section, j\'. The mmimum number of ideal stages. Data:-The ex tract branch of solubility curve on the mass ra tio diagram IS straight li ne tllrough ( Y, =0 ,N =29) and ( Yo. = I, t\ = 5) , the solvent is completel y insol uble in the [afftna te liquid. The equilibrium Curve on sol vent free basts is:-
0 0.06 O. I I o. 18 0.26 0.36 O~ 0.48 0.62 0.79 1.0 .
0 0.2 0.3 I 0.4 . 0.5 0.6 0.63 0.7 0.8 1 09 1.0 1
bject: !gree: .te:
art One
CA ll<U UNIVERS.ITY •
Faculty of Engineering C hemical Engineering Department
Mass Transfer Operations B.Sc. May 2006
Atte mpt All Questions
Time allowed: 3h.
Determine tht: r equ ired height of adsorbelll in an adsoq)lion column that treats :1 deg,'casc r-venti lation strea m contaminated with trichloroethyle ne (T eE). Design and operat ing data a re as follows:
Volumetric Aow roue of contaminated air: 10,000 std fll / min
(4.72 m ) /s), Slandard condi tions be ing 60 · F and I aun
Oper.u ing te mperature: 70· F (294 K)
Opera ting pl'essure: 20 psia (138 kPa)
Adsorbent: act ivated carbon
Bulk densi ty P. of activated ca rbon: 36 1b/ ft3 (576 kg/ lIl l)
Working cap.."I.ci l)· of activated ca rbon : 28 1b TeE per 100 Ib carbon
Inlet concentration of TeE: 2000 ppm (by volume)
Molecula r weight ofTCE: 131.5
The adsorption column is a vertical cylinder wit h an inside: diameter or 6 rt (1.8 m) and a height or t 5 rt (4 .57 m). It o~ra tes on the rollowing cycle: 4 h in the adsorption mode. 2 h ror heating and deso rbing. 1 h ror cool ing. I. h ror standby. An identical column treats the contaminated gas while the hrst one is not in theadsorplion mode. T he system is req uired to recover 99.5 percent or the TeE by weight.
Ai r at I atm. 35°C. and 70 per cent relative humidity enters an <IIr-conditioner at a rate of 9 .5 m)/rni and leaves as sanlrated air al 15°C. Part of the moisture in the air which condenses dur ing II process is removed at 15°C. Detenlllne the rate of heat ;md moisture removal from the air.
(3)
(4)
Equ ilibrium isotherm data for adsorption of glucose from an aqueous solution to act i va ted alumina are as fo ll ows (H3):
c(glem') 0.0040 0.0087 0.0 19 0.027 0.094 0. 195
q (g sol u(e/g alum;na) 0.076 0.053 0.075 0.082 0.123 0. 129
De te rmine the isotherm tha t fi ts the data a nd give the consta nts of the equatio n using the give n unit s.
Using molecu lar sieves, water vapor was removed from nitrogen gas in a packed bed (C3) al 28.3°C. The column height was 0.268 m, wi th the bulk density of the solid bed being equal to 712.8 kglm 3 . The in itial wat er concc ntnllion in the sol id was 0.0 1 kg watcrlkg solid and the mass velocit y of the nitrogen gas was 4052 kg/Ill!' h.
'I he initial waler concentratio n in the gas was (.' " == 926 x 10 .6 kg wa ter/kg nit rogen. The bn:nkthrough data arc as follows.
, (h) 0 9 9.2 9.6 10 10.4
c (kg H 20 /kg N2 x 106) <0.6 0.6 2.6 21 9 1 235
I (h) 10.8 11.25 11.5 12.0 12.5 12.8
c (kg H, O/kg N ,x 10") 4 18 630 717 855 906 926
A value of ell' " := 0 .02 is de si red at the break point. Do as fo llows. (a) Determine the break-point lime. the frac tion o f total capacil y used up to
the break poin t . the length of the unused bcd. and the satura tion loading capac ity of the so lid.
(b ) For a proposed column length H T = 0.40 m, calcul ate the break-poi nt time and frac tion o f 10 lal capac it y used .
""" , .... , .. ... -• lOOO -• - • • , - - • - -- • • • -• • • • , - • - • • - - <C." •
• • • • ~ - l'" • • • • • • • • 600 < , , ..
• • ... • 2 ,.. 1 • • • '" I
• .. , .. '"
,,. Tempt •• I ...... 'c
'"
HOG " " , .. ~ .l " '"'' , ., " - , .. • , "'" • • • • • • -• " • • "" -- • " • ,
• • I~_' • • , " - "
;; , , -• -- • ,; < ,. , • ' 00 • • , - • • -•
, , .. • , "
• • - • • • • • , ; - "" " • - • , • • , • • '" • ", • , • , • • - • • '00 - "
, - l • " • ,
" •
"
TeMpt""'" 'F
'"
Conversion Factors
Are" (A) I ft ' ~ 0.0929 m' I in ,l "'" 645. 16 mm2
Density (p) Ilbm / ft' - 16.018 kg/m'
•
I slug/ft ' - 515.379 kg / m' Energy (E, H, U, P.E. , K.E., Q, W)
I Btu ~ 1.015 51 kJ I Btu - 778.169 ft-Ibf I IT cal ~ 4. 1868 J
Eo tropy (5) I Btu/ R - 1.8992 kJ / K
Force (F) I Ibf ~ 4.4482 N I dyne = [ X 10-' N
Heat (Q) see £Il~rgy Heal flow rate (Q)
I Btu / h ~ 0.2931 W I Btu/s - 1.0551 kW -Heat transr~r coefficient (h() 1 Btu/hr-ft 2-oF"'" 5.678 W I m2-K
Length (Ll I ft - 0.3048 m 1 in . = 2.54 cm = 0.0254 m I m = 39.37 in. t mi = 1.6093 km = 5280 rt
Mass (m) I kg - 2.20462 Ibm I sl ug - 32. 1739 Ibm 1 tonne = 1000 kg 1 shan ton ... 2000 Ibm
Power ( ~V) I Btu/h - 0 .2931 W t Blu /s = 1.055 [ kW I hp=745 .7 W I hp - 550 ft-Ibf/s I hp "" 254 .. L5 Btu / hr I kW - 1.341 hp
Pressure (p) I psi, - 6.8948 kP,
I bar - 100 kPa I in. Hg = 0.4912 psia I mm Hg - 0.1333 kPa I atm - 101.325 kPa ~ 14.696 psia I aIm "'" 760 mm Hg "'" 29.92 in. Hg
Specific energy (q, W, e, h, II, p.e., k.e.) I Btu I lbm - 2.326 1 kJ Ikg
Specific entropy (5) f Btu/lbm-R - 4.1868 kJ Ikg-K I Btu/lbm-R ~ 778. 169 fi-Ib f/ lbm-R
Specific heat (cp , c .. ) see Specific entropJ' Specifi c vo lume (v)
I rt l /lbm=O.062418 mJ jkg Specific work (IV) see Specific energy Temperature (T)
nR) - 1. 8 nK) nCJ - 5/9 (nF) - 32) nF) - 915 nC) + 32 nC)-nK)-273.15 nF) - nR) - 459.67
Thermal conductivity (J.) I Btu / hr·ft-F~ 1.731 W / m-K
Thermal diffusivity (a) I ft' /sec ~ 0.0929 m' l s I rt 2/scc = 2.581 X 10-' m~ / s
Velocity (v) I fl /sec - 0.3043 m /s ! mph = 0.44703 m/s
Viscosity. d::-"n.:lmic (u) I Jbr'l/rt-s = 1.488 N-s/m" I centipoise = 0.00 I N-s / m!
V iscosity, kinematic (v) 1 ft lfsec = 0.0929 m!/s 1 ft 2/hr = 2.581 X 10- 1 mIls
Volume (V) 1 m1 = 35.3147 ftJ = 1000 liters I U.S. gal = 3.7853 li ters 1ft' - 7.481 U.S. gal 1 U.S. barrel = 42 gal
Work ( J J") sec Energy
Chemical Engineering Department Final Exam
Part Two:
1- a-Write short notes on:-
Faculty of Engineering Mass Transfer Operations
( c/ G, /vJ) May2008
i- Contacting and separation steps in leaching. ii- Thennodynamic equilibrium in leaching.
b- Draw neat sketches illustrating, Hildbrandt extractor.
. c- Meal containing 55 mass % oil is to be extracted at a rate of 4000 Iblhr, using 6000 Iblhr of n-hexane, containing 5 mass % oil, as a solvent. A counter current multi-stage extraction system which is equivalent to tow ideal stages is to be employed. The meal will retain I lb of solution! pound of oil -free meal. The overflows will be a mixture of solution and fine meal particles with an estimated ratio of 0.05 Ib of solids! pound of solution. Detennine the percent recovery of oil obtained under the above conditions.
2- a-Write short notes on:-i-Properties of good solvent in liquid extraction. ii- Sketch the relation between number of stages and reflux ratio, and optimum reflux ratio. iii- assumptions used in extract with reflux.
b- Draw neat sketches for:i- mixer settler cascades. ii- packed tower extractor.
c-An aqueous solution containing 45% acid is to be extracted in a counterCWTent cascade with extract with reflux. Pure solvent is to be used at the exhausting section end. The top product of the solvent recovery distillation column is water- free and is to contain 10% acid. The bottom product of this column and the fmal raffmate stream are solvent -free and contain 90% and 10'/. acid respectively. Detennine:-i- the minimum reflux ratio. ii- the minimum number of theoretical stages. Data:- the locus of the extract phase composition is a straight line through (XA ={) , Xs= 0.75) and (XA = 0.7 ,Xs = 0.3 ). The locus of the raffmate is given by Xs = O. The equilibrium relation is Y A = XA .
Cairo University Faculty of Engineering Chemical Engineering Department Mass Transfer Time Allowed 3 hr ~?-; LJ J.-...- 1101.j 2008 ATTEMPT ALL QUESTIO NS,·
l-a-A solid silica gel is used for adsorption of water vapor from humid air 25° C and I .
i - when 1 room at atm and hunl;< 1 Dmm Hg for long time, what will be the final concentration of water in gel and in air? ii - When lkg of dry gel is placed for long time with flow of stream of air 25°C , 1 atm and humid ity 10 mm Hg, what is final concentration of water in ge l? iii - how can you produce completely dry air? b- Choose proper equipment, give reasons & draw neat sketcbes for : i -column fo r batch fractiona l disti llation. ii - plate su itable for use for distillation with high tu m -down ratio. iii - economic industrial evaporation.
2- A feed solution of 1000 kg Ihr at 50°C containing 50 kg Mg SOJ 100kg water is cooled to 20°C where Mg S04.7 H20 crystals are removed. The solubi lity at 20cC IS
32 kg Mg S04/ 100 kg water. The average heat capac ity of feed solution is 3 KJI Kg K. The heat ofcrystalJization is 1400 KJ 1 Kgmol MgS0 4.7}hO.
Calculate:-i-yield and cooling load assuming 1 % evaporation (from feed) Mg=24, S=32, latent heat of evaporation of water = 580 Kcal /Kg of water. ii- Draw net sketch fo r proper crystallizer for this object ? iii - How can you prevent caking of product? iv- What is the proper cooling medium for thi s process?
3-a-Write short notes on: i- Chil ton-colburn analogy. ii- Overall driving fo rce for mass transfer operation benveen two phases. iii- Diffe rence between equimoler counter diffusion and diffusion in one direction. iv-Penetration theory for convective mass transfer.
b-Discuss: i- Special types of plates are used for vacuum fractional distillation. ii- Desalination by reverse osmosis requires operation at high pressure (50-80atm.). iii- Plate efficiency is usually higher than point efficiency. iv- Asymmetric membranes are wide ly used for membrane separation process.
4-a-Complete the following statements:-i- We obtain ------- at in finite number of stages, zero driving fo rce, and min imum amount of solvent.
ii- In the case of absorption with chemical reaction the rate is affected positive ly with ------- in temperature. iii-Kremser method is an approximated method to calculate ------ . iv-Absorption process is usually fo llowed by -------- . v- --------- is the most important factor in choosing good absorption so lvent.
b-Draw neat sketches and write short notes on:i-U-tube absorber and spray column. ii-Effect of temperature and pressure on equilibri um relations in absorption. iii-Absorption with tow solvents.
c-Ammonia is to be absorbed from an a ir mixture by contacting it with water at 20 °c . The entering gas composition is 10 mole% ammonia and the exit gas is to conta in I mole% ammonia. Flow of entering gas is 10 molesls at 200 C at normal atmospheric pressure. i- Determine the minimum entering water flow rate needed to accomplish this separation. ii-Determine the number of equilibrium stages required and the exit liquid composition at 1.25 of the minimum ente ring water fl ow rate. iii- Suppose that all the liquid leaving the absorption column in ( ii) is fed to the top of a stripping tower operating at 200 C. Pure ai r at 200 C is fed into the bonom at a rate of 9 moles/s. Determine the exit gas and liquid compositions fo r these conditions, assuming that the stripper is equivalent to the same number of equ il ibrium stages as the absorbe r in (ii). E T b' I ' 20 ' C . Ii II UI I [Ium re atlOns at tS are as a ows. Kmole of NH,I 0.021 I<.tno&.ofwatcr
...I!:I'...of NH, mmf!gl 12
5-a-Writc short notes on:i- Steam disti ll ation. ii- Molecular distillation.
0.031 0.042 0.053
18.2 24.9 31.7
iii-Compare between partial and total condensers. iv- Using o f pipe still heater. v- Linde double column.
0.079 0.106 0.159
50 69.6 114
b-A mixture of 30 mo le% ethanol, 70 mole% water is to be fracti onated at I atm into a dist illate of 82 mole% ethanol and a bottoms of 4 mole% ethanol. There will be a total condenser at the top of the tower, but, instead of the reboiter , saturated steam will be directly fed at the bonom of the tower to supply necessary heat. The feed is 40 mo le% vapor and 60 mole% liquid. Operating reflux ratio is 1.5 the minimum value. i- Draw a flow diagram for the tower, and labe l the enteri ng and leaving streams. ii- Derive an equation for the operating line in the stripping section. iii- Determine the number of the equilibrium stages required. iv- Determine the amounts of steam, liquid and vapor in the bonom section. v-Determine the minimum steam rate(per mole of bottoms) which would permit this separation. The equilibrium re lation between the mole fraction of ethanol in vapor and liquid is
• ven below Where x and are mole fractions)
~- 0 0.32 0.44 0.57 0.61 0.65 0.7 0.82 0.9
0 0.05 0.1 0.3 0.4 0.5 0.6 0.8 0.9 x
Cairo University Chemical Engineering Department Final Exam
Mass Transfer Operations
Part Two: Question I:
Attempt aJl Questions:
I-Complete:
Faculty of Engineering May 2008
Time allowed 1.5 hr
a- In leaching the solution resulting from separation step is termed ............. , and solid left over is termed ..................... .
b- In leaching we obtain the ideal stage when .......................... .
c- One of the applications in leaching is ................................ .
d- The factors affecting the efficiency of leaching tank are ....................... ", .................... ", ............................... .
2- Draw neat sketches illustrating a basket-type oil seed extractor.
3- 70%of the oil present in seeds containing 39.1 % oil is to be extracted with pure hexane in an extractor equivalent to two ideal stages. The underflow retains 1 kg solution per 1 kg of solids, and the over·flow from the second stage is solids-free, while the rich extract overflowing from the first stage entrai"s 1 kg of solids per 19 kg of liquid.
Determine:
i-The required solvent to feed ratio.
ii-The corn positions of the overflow and underflow streams leaving each stage.
Question II:
I-Draw neat sketches illustrating: i- A counter current liquid. extraction piece of equipment. ii- The effect of temperature on the terna.·y solubility diagram of three immiscibility liquids. iii- Inte'·polation of tie line data in liquid extraction.
2-Nicotine (C) in a water (A) solution containing I % is to be extracted with kerosene (B) at 20°C. Water and kerosene are essentially insoluble.
i) Determine the % extraction of nicotine if 100 Ib of seed solution is extr·acted once with 150 Ib solvent each. Equilibrium data:
X=lb Nicotine 0 0.001011 0.00246 0.00502 0.00751 0.00998 0.020 Ib Water +
Y=lb Nicotine 0 0.000807 0.001 961 0.00465 0.00686 0.00913 0.0087 Ib kerosene I
ii) 100 Ib per hou.· of Nicotine (C) - water (A) solution containing 1 % Nicotine is to be counter-currently extracted with ke.·osene at 20°C to reduce the Nicotine content to 0.1 %.
a) Determine the minimum kerosene rate. b) Determine the no. of theoretical stages required if 150 Ib of
kerosene is used per hour.
3- For the extr·action system shown in the figu re: i) Complete the table. ii) For stage (1) calculate the maximum concentration obtainable
. th ttl III e ex rac lase. Lo "3 V, L, "2 I "2 Vf
Ko/hr 1500 + %A 40 40 9 %S 0 100 %B
L,
Equilibrium data:
• YA=(4/3) XA • Extract locus: straight line between (0,0.9) & (0.85,0.05) • Raffinate locus: straight line between (0,0.05) & (0.85,0.05)
v, v.' -1
Lo
V I
L,
v, V3
') -L,
(I) "
Degree: SSe May 2008
Part One
CAIRO UNIVERSITY Faculty of Engineering
Chemical Engineering Department
Subject: Mass Tnmsfer Operations Time Allowed: 3h
Attempt All Questions
Wet solids pass through a continuous dryer. Hot dry aI r enters the dryer at a rate of 400 kg/min and picks up the walcr lha! evaporates from the solids. Humid air leaves the dryer at sOGe containing 2.44 wl % wate r vapor and passes through a condenser in which it is cooled to 100e. The pressure is constant at I aIm throughout the system. (a) At what ratc (kg/mi n) is waler evaporating in Ihc dryer? (b) Use thc psych rometric chart to estimate the wet-bulb temperature, rc[alive humidi ty. dew point,
and specifi c en thalpy of the air leaving the dryer. (c) Use the psychrometric chart to estimate the absolu te humidity and speci fic enthalpy of the air
leaving the condenser. (d) Use the results of parts (b) and (c) to calculate the rate of condensat ion of wate r (kg/min) and
the ra te at which heal must be transferred from the condenser (k W). (e) If the dl)'er operates adiabatically, what can you conclude about the tempe rat ure of the entering
air'! Briefl y explain your reasomng. What additIonal information would you need to calcu late this temperature?
b The separalJ\,n \)f propane and propylene is accomplIshed by distillation, but :u the expense of more than 100 trays and a reflux rJtio of greak'r th,",n 10. Consequently, the use of adsorption has been tnvest igJI~J l!1 a number of studies. Jarvelin and Fai r [{lid. Eng. Chew. Rtseolt!J . .\:!, 2201-2207 (1993)J measured adsorption. eqUilibrium data al ~5 C for three different zeolite molecular sieves (ZMSs) and activatcI! ':Irbon. The data were fitted to the Langmuir isotherm wllh the foll,"\"ing results:
Adsorbent Sorbate q. K
ZMS4A C, 0.226 9.770 • cr 2.092 95.096
ZMS5A C, 1.919 100.223 Cr 2.436 147.260
ZMS 13X C, 2.130 55.412 Cr 2.680 100.00J
Activated carbOIl C, 4.239 58.458
or 4.889 34.915
where q and q,. are in mmoVg and p is in bar.
(a) Which component is most strongly adsorbed by each of the adsorbents? (b) Which adsorbent has the greatest adsorption capac· ity? (c) Which adsorl:"ltnt has {he greatest selectivity? (d) Based on equilibrium con~ider ations. which adsorbent is best for the separation '!
K q .. ,P q == -
I f K I'
I{ ,\ ::=; - ((fA)", 1<. A 1','"'_ l ..L KAfiA + KnfJO
(4n L,. Ktl/'ll
{f1,.)/II Ki P, </1 ~ - , .......
J f L K, PI ;
(2) •
b
The selection of a batch or continuous dryer is determined largely by the condition of the feed. the temperawre-sensitivity of the dried material , and the form of the dried product. What types of batch and continuous dryers would be Suil .. ble for the following cases:
(a) A tempe rature-insensitive paste that must be maintained in slab
form.
(b) A temperature- insensitive paste that can be extruded.
(c) A temperature-i nsensitive slurry.
(d) A thin liqu id from which flakes al'e to be produced.
(e) Pieces of lumber.
(f) Pieces of pollery. (g) Temperature-insensitive inorganic c ry~la!s where particle size is to be maintained and only surface moisture jl> to be removed.
(h) OrJnge j uice to prOO lICc a powder.
Air al a flow rate of [2.000 scfm (60"F, 1 aIm) and con· taining 0.5 mol% ethyl acetate (EA) and no water vapor i!; [0 be treated with activated carbon (e) (Pt>:; 30 lb/ft) with an equiva·
• lent panicle diameter of 0.011 ft in a lixed-bed adsor!:.er to remove the elhyl acetate, which will be subsequently $tripped from the car· bon by steam a1 230~ F. Based on the foll owing dat a. determine the diameter and height of the carbon bed, assuming ad~orplion at 100 F and I aIm and 0 (une·lO-breaklhrough of 8 h with a superficial gas veloci ty of 60 f!lmin. If lhe bed hcight-to-diameter is unr~asonab!e. what change in design ';'asis would you suggest?
Adsorption isotherm data (IOrrr) jQr EA:
0.0002 0.0005 0.0010
q. Ib ENlb C
0. 125 O. !64 0.195
0.0020
0.0050 0.0 I 00
(/' Ib ENiu C
0,227 0.270 0.304
Breakthrough data at 10000F lind I aIm for EA in air at a gas superficial velocity of 60 fVm;n in a 2-ft dry bed:
Mole Fraction Mole Fraction EA in Eftlucnl Time, Mm EA in Erfluclll Time, Min
0.00005 60 0.00100 95 0,000 10 66 0.00250 [20 0.00025 75 0.00475 [60 0,00050 84
(3) a
b
Give reasons for :
-In the clay trearment ofpctroleum - lubricant fraction the adsorbent - oil mixture may be pumped through a tubular furnace to be he-'u ed to as much as 120~o l Scf C • and for very heavy oils even to 300 to 380'" C . - In water cooling towers it is perfectly possib le to cool water to a value less than the entering air dry bulb temperature • but water cannot cool below the air wet - bulb temperature .
- In batch adsorption if the adsorption factor is sufficiently large , the solid - phase mass transfer resistance can be neglected . - Air fed to a contact sulfuric acid plant must be dried . . , .,.. When very large flow rates are treated - 500 rri"/ s of fl ue gas from a power station fo r example a flu idiz.ed bed adsorber or a fixed bed adsorber is better to use and why?
An insolub le granu lar material wet with water is be ing dried in a pan 0.457 X 0.457 Tn and 25.4 mm deep. The mate rial is 25.4 mm deep in th e meta l pan, which has a metal boltom o f thickness 1.,\/ = 0.6 10 mm having a thermal con duct ivity k,\/ = 43.3 \V!m · K, The thermal co nductivity o f the solid can be assum ed as ks =. 0.865 W /m· K. Heat transfer is by convection from an air st ream fiowing p;lralle llo the to p drying su rface and the bottom metal surface at a veloci ty o f 6. 1 r.1 ls and having a te mperature of 65.6°C and hu midi ty H = 0.010 kg H 20 / kg dry air. Th e lor surfa ce also receives direct radi ation fro m s team- heated pipes whose surface te mperature Tf{ = 93.3°C, T he emissivity of the so lid is £; == 0.92. Es tim ate the rale or drying fo r th e cons tant -rate period.
he = 62.45 W /ml . K. -
As ~ 2424 X !OJ J/ kg.
(4)
•
Adsorpt ion with activated carbon, made from bituminous coal, of soluble organ ic compounds (SOCs) to purify surface and ground water is a proven technology, as discussed by Stenzel [Chern. Ellg. Prog .. 89 (4) , 36-4~ (1993)J . 'Ille less-soluble organic compounds. such as chlorinated organ ic solvents and aromatic sol ventS, are the more strongly adsorbed. Water containi ng 3.3 mg/L of trichloroethylene (TeE) is to be treated wi th ac tivated carbon to obta in an effluent wi th only 0.01 mg TCEIL. At 25~C, adsorpt ion equi librium data for TeE on activated carbon are correlated with the following Freundl ich equation:
•
where
q = mg TCElg carbon and c = mg TCEIL so[ulion
The TeE is to be removed by slurry adsorption using a powdered form of the ac tivated caroon, with an average part icle diameter of 1.5 min . In the absence of any laboratory d;l \a on mass- transfer rates. assume thaI Ihe ra te of adsorption for the small particles is contro1!ed by external mass transfer with a Shcrwood nUlJlber of 30. Particle surface area is 5 m"!kg. The molecu lar diffusivi ty of TeE in low concentrations in water :H 25°C may be determined from the Wil ke- Ch:lng equat ion.
(a) Determine the mi nimum amount of adsorbent needed.
(b) For operation in lhe batch mode with twice the min illlum amount of lIdsorbl!nl, determi ne the lime to reduce the TeE conteru to the desired value.
(c) For operation in the continuous mode using twice the minimuln amount of adsorbem, detemline lhe required res idence time.
(d) For operat ion in the semicontinuous mode at a feed rate of 50 gpm and fo r a liquid residence time equal 10 1.5 times that corn· puted in pan. (c), de tennine the amount of activated carbon to give a reasonable vol% sol ids in the tank a lld a run time of not less !l1an 10 limes the liquid residence time.
k, = 5 X 10-5 rnls .
(I)
(2)
Degree: BSc May 2008
Part One
CAIRO UNIVERSITY Faculty of Engineering
Chemical Engineering Department
Subject: Mass Transfer Operations Time AJ lowed: 311
A ttempt All Questions
Small amounts of VOCS in water can be removed by adsorption. Generally, two or more VOCs are present. An. aqueous stream continuiog smalJ amounts o~ acetone (1) and propionitrile (2) is to be treated with activated carbon. Sing1e-solute equilibrium data a· .. ailable from Radke and
• . ' 0 ." . . _, -Prausnitz . have been fitted to the Freundlich and langmuir isothenns, with the average deviations indicated, for solute concentrations up to 50 mmol/L: .. ' '. -" .
, .;, ~ .• , ~ _,0. , ' .. Acetone in Waler (2S0 C): . ". . Absolute Average
Deviation of q. %
ql '" O.t4lctm O.l9Oc.
ql'" 1 + O.l46cl
(1)
(2)
Propionitrile io water (2S'C):
where:
•
(3)
(4)
,,.; 'j
14.2
27.3
10.2
26.2 I ~--: ' I
ql =- amount of solute adsorbed, mmol/g _ CI = solute concentration in aqueous solution, mmol/L"
. ' - • I
Use ~.~~e, single-solute r~s~lts wi~~ an extended L:mgmuir-~ isotherm to predict the equilib~um 'adsorption iIi a bina!y-solute ~qu,~6us system cO'!tainmg ~o and 34.4 mmollL, respecti':.ely,
" .. , of,ace.tone -and, propionitrile: af 'z.s:C',with the, s:lme adsorbent. Compare tlJe results 'with Itll e~ '", _" .~· ' .'. · l !, '~", "" " ), ' foUowing experimental values from Radke and Prausnitz ': ' ", .. " . Jl .:;t' , \. ..... ',- " ... r, 'if
, ql - 0.715 mmol/g. Q2'" 0.822 mmol/g,--: ; "7 >.: an~ qio<aJ - 1.537 mmollg , .
A counterflow in~uced--draft cooling tower operates with inlet and exit water temperatures! 'of 105 and 85°F when the air haS dry-bulb and : wet~bulb
. temperatures, respectively, of 90 and 76°P. 111e tower has 4 ft of stacked plastic fiU, and the flow rates are Gy= 2,000 J~, f12 and GL = 2,200 lblb· ft2. (a). Detennine the number of transfer units, the heigh,t of a transfer unit based on lfle overaU- gas~phase driving force, and the temperature approach. (b) If the cooling load remains the same but
the air temperature drops to 70°F with a wet-bulb temperature of 60°F, predict the water temperature and the temperature approiJ.ch,
(3)
(4)
A waste stream of alcohol vapor in air from a process was adsorbed by activated carbon particles in a packed bed having a diameter of 4 em and length of 14 em containing 79.2 g of carbon. The inlet gas s tream having a concentration Co of 600 ppm and a density of 0.00115 glcm J entered the bed at a How rate of754 cm J/s. Data in Table 1. give the concentrations of the breakthrough curve. The break·point concentration is set at clc o = 0.01. Do as follows.
(a) Determine the break·point time, the fraction of total capacity used up to the break point , and the length of the unused bed. Also determine the saturation loading capacity of the carbon.
TABLE I. Breaklhrough Concen lralion
Time, h cleo Tim e , h cleo 0 0 5.5 0.658
3 0 6.0 0.903 3.5 0.002 6.2 0.933 4 0.030 6.5 0.975 4.5 0.155 6.8 0.993 5 0.3%
(b) If the break·point time required for a new column is 6.0 h, what is the new total length of the column required?
Cold air at 20"F. 76(J mm Hg pressure, and 70% relative hu midity is conditioned by being passed through a bank of heating coil s., then through a water spray. and finally through a second set of heating coils. In passing th rough the fi rst coil bank, the air is healed to 75°F. TIle temperature of the water supplied to the spray chamber is adjusted to the wet·bulb temperat ure of the air ad mitted to the chamber, so that the humidify ing unit may be assumed to opera te adiabatically. It is required that the air emerging from the conditioning unit be at 70"F and 35% rela tive humidi ty .
. (3) Calcula te the te mperature of the water supplied to the spray chamber and the relative humidity
and dry·bulb temperature of the air leaving the spray chamber. (b) Calculate the mass of wate r eva porated (Ibm) per cubic foot of air fed to the co nd itioning unit. (c) Calcu late the required heat tnli1sfer rates (Btulftl en teri ng air) in each of the hea ting coi l b~ n ks. (d) Sketch a psychrometric chart and show the path followed by the air in each of the three steps
of this process.
" 8000 , ..., >000 ...
. , " "'" • 0 • • " • ''''' . ~ • • ~ -, • " • " • - ' 000 " " • • < -• l'"
< • • • • , ... , ~ "" • < '00 • , , "" • " • ;.
001
T~nlp<u.u", . ·C
(.,
C.j Psy<:hrOmetric charI for air· ... al<:r ""pot". I "d 11m abs, in SI 'unill.
''''' " '000
• · - " , "" " •• •
" .; < " '000 •
" > •
" • '00 "
" • -• • • " • • • " ""
, • "
, " • •
" ,
t • • "
, '00 ,
< " • " • • " -• • , )00 < - • • " ,
" • • < • • ~
• " ;; '00 -•
• " • • , < i '" , •
• " • ,
< ~ • <
• - '00 • , .' " 1 ". • • " • 00)
;.
" "
(tj Psychromclric cha.rt lor .ir .... ater Vllpar, 1 sId atm ;obi., in En~sb cl'lciDocrinr; units.
Conversion Factors Area (A )
I ft' = 0.0929 m' I in.2 = 645. 16 mrn 2
Density (p) I lbm /ft' - 16.018 kg/m'
•
I slug/ ft' - 515.379 kg/ m' Ene rgy (E, H, U, P.E., K.E., Q, W)
I Btu - 1.01551 kI I Btu - 778. 169 ft- Ibf I IT cal = 4.1868 J
Entropy (S) I Btu / R = 1.8992 kI / K
Force (F) I Ibf - 4.4482 N I dyne - 1 X 10-' N
Heat (Q) see £n~rgy Heat flow rate (Q)
I Btu/h = 0.293 1 W I Btu /s - 1.0551 kW
Heat transfer coefficient (h, ) ! Btu/hr-ft2-oF= 5.678 W/ m2_K
Length (L) I ft - 0.3048 m ( in. = 2.54 cm = 0.0254 m I m - )9.37 in . \ mi = 1.6093 km "" 5280 ft
Mass (m) I kg ~ 2.20462 Ibm I sl ug - 32.1739 Ibm 1 tonne"'" 1000 kg 1 short to n "" 2000 Ibm
•
Power (H-") I Btu / h ~ 0.293' W I Btuls - 1.0551 kW 1 hp'"" 745.7 W I hp ~ 550 ft-Ibf/s I hp == 2544.5 Btu / hr I kW - 1.341 hp
Pressure (PJ I psia = 6.8948 kPa
I bar-\OOkPa 1 in. Hg = 0.49 t 2 psia I mm Hg=0. 1333kPa I atm =- 101.325 kPa = 14.696 psia 1 atm - 760 mm Hg""" 29.92 in. Hg
Specific energy (q, W, e, h, U, p.e., k.e.) I Btu / Ibm = 2.326 1 kI / kg
Specific entropy (s) l Btu/lbm-R - 4. 1868 kI / kg-K I Btu /lbm-R = 778. 169 ft -Ib f/ lbm-R
Specific heat (cp , cp ) see Specific entropy Specific volume (tI)
1 ft' / Ibm - 0.062 428 m' / kg Specific work (1<,.') see Specific energy Temperature (T)
TTR) - 1.8 TTK ) TTC) - 5/9 (TTF) - 32) TT F ) - 9/5 11CJ + 32 11CJ - 11K) - 273. 15 TTF) - 11R) - 459.6 7
Thennal conductivity (A) I Btu / hr-ft-F -1.73 1 W/ m- K
Thennal diffusivity ta) I ft' /sec = 0.0929 m' /s I fV /sec =< 2.581 X 10 -5 m! /s
Velocity (v) 1 rt /sec = 0.3048 m/s I mph - 0.44703 m /s
Viscosity, dynamic (,u) I lbr'l / rt-s = 1.488 N·s / m! 1 centipoise == 0.00 1 N-s/ml
Viscosity, kinematic (v ) I ftljsec-O.0929m!/s I ftz / hr=2.58\ X 1O - 'm2/s
Volume (V) 1 ml >:ll )5.3147 ftl = 1000 liters I U.S. gal == 3.7853 liters ! ft J -7.481 U.S. gal I U.S. barrel "'" 42 gal
Wo rk (W) sec Energ.v
Degree: B.Se. June 2007
PART ONE
(1) a- Give reasons for:
CAIRO UNIVERSITY Faculty of Engineering
Citemic.i I(nglneermg Department
Subject: Mass Transfer Operations Time Allowed: 311
Attempt All Questions
- The shape and lime of appearance oflhe breakthrough curve greally influence the method of operating a fixed bed adsorber.
- Control of ecological problems is as much a part of the design engineer's responsibility as is the prediction of the themlal and mechanical perfonnancec f the water cooling towers,
- Js ;t possjbJe to JeJ]]OYC water from air by spraying water jnto tlJe air? Explain your answer.
b oo A stream of air al 31"C and 50% relative humidity flo wing at a ratc of 1250 kg/h is to be cooled (0
1 5~C and dehumidified in a spray lowe r. The air is saturated as it emerges [rom the to ..... er. LIquid wa te r leaves the towe r at 12ee; some is withdrawn, and th e rest is cooled and recirculated . No heat is transrerred between the towe r and its surroundings.. Calculate the rate (kglh) at which water must be withdrawn (rom the recirculation loop and the heat duty o n the cooler (kW).
(2) a- For the systems other than air-water explain how to calculate the change in conditions of the liquid and the gas in a hwnidificatioo tower lIsing the method developed by Lewis and While.
.
b- A I m ) volume o f a mixture o f air and acetone vapour is O\t a temperature o f 303 K and a lo tal pressure of 100 kN/m2. If the relative saluratio n of the air by acetone is 40 per cent, /lo w muc h ac tivated carbon must be added to the space in o rde r to reduce the value to 5 per cent at 303 K? .
If 1.6 kg carbon is added, what is re lative saturation of the equilibrium mixture assu ming the temperature to
be unchanged? The vapour pressure o f acetone at 303 K is 31.9 k.N/m2 and the adsorpt io n equilib rium data (or acetone on
carbon at 303 K are :
Part ial pressure aceto ne x 10- 2 (N/m2)
Xr (kg acelone/kg carbon)
(3) a- Give reasons for.
o 5 o 0.14
10 0.19
30 0.27
50 0.3 !
90 0.35
- There is a criticnl minimum bed heigh1 below which tile operation or a fixed-berl adsorber is considered unsuccessfuL
. in watcr-cooiing towers the water can be cooled below the bulk inlet air temperature but the liquid cannot cool below the wet-bulb temperature of tbe air.
- The precise way in which adsorption and regeneration are achieved depends on the phases involved and the type of fluid-solid contacting employed.
b- An adult takes ro ughly 12 breaths each minute, inha ling approximately 500 mL with each breath. Oxygen and carbon dio xide are exchanged in the lungs. The amount of nit roge n exhaled eq uals the amount inhaled , and the mole fractio n o f nitrogen in the exhaled air is 0.75. TIle exhaled air is saturated with water vapor at body tempe rature, 37"C. Estimate the increase in the ra te of wate r loss (gfday) when a perso n breat hing air at 23°C and a rela tive humidi ty of 50% enters an airplane in which the tempera ture is also 23°C but the relative humidity is 10%.
(4) 9- Explain in brief the bask: components ora: fixed-bed adsorption unit [or gas deliydi'iit!Qii .
What process variabJes 1hat must be considered in design ing these components? What are the necessary information needed to design the required fixed-bed adsorber1
- Explain in brief how to a apply any of the available models for breakthrough curve prediction in designing the fixed-bed adsorber.
- What energy saving recommendations you would propose to optimize Ole perfonuance ot the fixed-bed adsorption lUli! selected?
b- TIle fo ll owing equilibrium data ', have been obtained for the adsorption of nitrogen dioxide, NOl •
0 11 silica ge l a t ZS'C and 1 aIm:
PN~ (mm Hg) 0 2 4 6 S 10 12 . -X'(kg N021l00 kg silica gel) 0 0.4 0.9 1.65 2.GO 3.65 4E5 J
(a) Confirm Ihat these data are reasonably co rre laled by the frcundlich iso the rm
X' = KF~O~
and determine the \'alue~ o f KF <lnd f3 thaI provide the best co rrela tion. ,
(b) TIle ocisorption col umn s!J own in the fi gure below has an intern al diameter (I f 10.0 em and a bed hcig!J! of 1.00 m. Th~ bltd o f silico gel has a bulk density of 0.75 k£.lL 111e adsorbcr is to remove NO I [rom a stream co nt llining 1.0 11101e% N02 ,!I1d the ltalance air that ente rs the adsorber at ROO kg/h. The pressu re and tempera ture arc ma intained <.1 1 a ttn and 25°C. Pas: experience with this system has s!Jown l!Jal a plo t of the partial pressure ratio « Ptlo: ) o",,",I(pN~ ),.1 • .] versus time produces a breakthrough cun'c with thc fo llowing appeorance.
Inle! eas 6 kg/min' 1.0.01% N0 2Ig)
Sil ica ,. '-..
F .. 760 mm He T. Z5'C
,
ADSORPTION COLUMN
Oullet
LO
(PttOl )"", .. ,
{Pml/ );" ..
ll , e3~lh'ovg~ c~,~
---Breakthrough time
---::--0-/~-Tim;! (m,n)
Using the iso the rm derived in pan (a ), de termine the time (in min) requ ired fo r break through of the N02.
(c) Silica gel in the colum n can be regefl erated (i.e., adsorbed N02 can be removed so thatlhe silica gel column c~ n be reused) by elev~ting Ihe bed tempera ture Ilnd/or pu rging the bed wilh clean air. Suppose such a regeneration process requi res 1.5 hours 10 accomplish. Process shutdowns can be avoided by insta lling several silica gel columns ill paralle l, using one to carry out the purification while the o thers are being regenef~ted. Wlut is Ihe minimum number of columns required to achieve conl inuous process ope ral ion?
Part Two:
1- a-Write ShOlt notes on:-
i- Contacting and separation steps in leaching, ii- Thermodynamic equilibrium in leaching,
,
b- Draw neat sketches illustrating, Hildbrandt extractor.
c- Meal containing 55 mass % oil is to be extracted at a r~te of 4000 Ib/hr, using 6000 lblhr of n-hexane, contai ning 5 mass % oi l, as a solvent. A counter current mult i-stage extraction system which is equivalent to tow ideal stages is to be employed, The meal wil l retai n 1 lb of so lut ionl pou nd of o il - free mea l. The overflows wi ll be a mixture of solution and fine meal particles with an estimated rat io of 0,05 Ib of solidsl pound of solution, Determine the percent recovery of oil obtained under the above conditions,
2- a-Write short notes on:-i-Properties o f good solvent in liquid extraction, ii - Sketch the relat ion between number of stages and reflux ratio, and optimum reflux ratio. iii- assumptions llsed in extract with refl ux.
b- Draw neat sketches for :i- mixer settler cascades. ii- packed tower extractor.
c-An aqueous solution containing 450/0 acid is to be extracted in a countcrcurrcnt cascade with extract with reflux, Pure solvent is to be used at the exhausting section end, The top product of the so lvent recovery distill at ion column is water- free and is to contain 10% acid, The bottom product of til is column and tl,e fi nal raffinate stream are solvent -free and contain 90% and 10% acid respectively, Determi ne: -i- the minimum refl ux rat io. ii - the mini mum number of theoretical stages, Data:- the locus of the extract phase composition is a straight line through (XA =0 , Xs= 0,75) and ( XA = 0,7 , Xs = 0,3 ), The loclls of the raffinate is given by Xs = 0, The equ ili bri um relation is Y A = XA,
Conversion Factors
Arc. (A) I ft' - 0.0929 m' 1 in .l- 645.16 mm1
Densily (p) I Ibm / fl ' - 16.018 kg/m'
•
I slug/ fl ' - 515.379 kg/ ln ' Energy (E, fI, U, P.E., K.E., Q, IV)
I BIU - 1.0 155 1 kJ I BIU - 778. 169 fI-lbr IIT czl - 4.1868 J
Entropy (S) I Blu/ R - 1.8992 kJ I K
Force (F) I Ibr - 4.4482 N I dyne - I X 10- ' N
Heat (Q) see En,!rgy Heat Dow ra le (Q)
I Blu l h - 0.293 I W I Blu/, - 1.0551 kW _
Heat transfer coefficient (lit) I Blu/ hr-ft'-· F - 5.678 W / m' -K
Lenglh (L) I fI - 0.3048 m I in. - 2.54 em - 0.0254 m I m - 39 .31 in. I mi - 1.6093 km - 5280 fI
Mass (101) I kg - 2.10462 Ibm I slug- 32.1739 Ibm I tonne - t 000 kg I short ton - 2000 Ibm
Power (J~) I BIU / h - 0.293 I W I Blu ls - 1.055 1 kW I hp -745.7 W I hp - 550 fI -lbrls I hp - 2544.5 Blu/ hr I kW - 1.341 hp
Pressure (p ) I psi. - 6.8948 kP.
I bar - 100 kPa I ill. Hg - 0.49 12 psi. I mm Hg - 0.1J33 kPa I aim - 101.325 kPa - 14.696 psi. I aim - 760 mm Hg - 29.92 in. Hg
Specific energy (q, IV, e, h, U, pc., k.c.) I Blu/ ibm - 23261 kJ Ik&
Specific en:lfJPY (s) f Btu / lbm ·l<- - 4.1868 kJ/kg-K I Blu/lbm-R - 778.169 fI- lbr/lbm-R
Specific heat (c". c.) see Specific entrop),' Specific volume: (tI)
I fI'/lbm - 0.062428m'/ kg Speciftc "," ork (w) see Specific energy Temperature (T
TIR ) - I 8 71KJ TIC! - 519 (T1F) - j!)
TIF) - 915 TIC) + 32 TIC) - TIK) - 273.15 TIF) - TIR) - 459.67
Themlal conducthil}' (1) I Blu / hr-fl-F - I 7) I W I m -K
Thcmlai diffusl ... ity (0') I ft l/sec - u.O<'~9 mJ/s I ft J lsec - 2.58 I X 10 ., m1/li
Velocit y (v) I fl/5« - 0.3048 m/s I mph - 0.44703 m/s
Viscosity. dynam ic (P) I Ibr,'/ft-, - 1.488 N-slm' I centipoise - 0.001 N_s / ml
VISCOSity. kinemat ic (v) I fI '/= - U.On9 m' /s I n1/hr - 2.581 X 10- \ mf/s
Volume (q I m1 _ 35.3147 1'1' - 1000 li ters I U.S. gal - 3.7853 I,lers I fI' -7.481 U.s gal I U.S. bam:l- 42 g.1
Work (H 1 see Ellers."
• i 1.4 IS BOOO
7 -6000 SOOO
I.J 4000 • , .. -.- 3000 • -.-" • u " - 1.2 ~ u
• ~ 2000 Jt. ~ - -.- -- • 8. -u -~
~ • ~ - > ~ u • - - ~ - ~ u ~
• 1.1 ... - 1000 .' - • • -• - 9 -u ~ -E • ~
\00 c·
~ ~ u , , - ~ • " . 0 - ~ • • • > - 600 ". ~ .-
u 1.0 - ~ .- c .- SOO E -u C 0 • , ~ ~ 400 •
" " • -- , ".- ] 300 ·
0.9 D • • • 200 ,.
0.8
0.7
o 10 so 60 70 80 90 100 110 120 130
Temperatu re,o(
Psychrome tric chart [or air-water vapor, I sId atm abs, In SJ units. •
2500
" '000
~ J IJ ISOO -. ,
• "
, - ~ , . , -- • 1000 ~ - ~ D - -• ~
, - ~ 800 -. , - - " D • - ~
~ ~ D - - 600 , - '0 ~ ~ -- = v D --~ ~ ~ -" v 400 ~
" ~ " -, > • v
, - ~ - -=
• 300 e -• "
• • , • e ~ " , ~ D '-- -0 • - 200 -> 17 ~ ~ -:9 e -. ,
, • :2 ISO e , II, • E ~ '-
, , - ~
- IS • 'DO ,- -, ~
14 0 .04 D • .. , . . '- •
IJ 0.0]
, , 0.02 -
(bl
Psychrometric chart for air-water vapor, I std a1m abs, in Engl.ish engineering units.
