CIR 113 Study Guide

8/2/2019 CIR 113 Study Guide

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Fokuiteit ng+1teir svr 'ese, ou-omgowli: ' : ,5: TFcr;uitY f [ng;\eer ir ig, Buii t nvircnn+ni & lT

Sclrool f tr 'c;r la' i ' ingr-)epar;rnert cf Chet'''ricaini;iiee;'!rr.9

[itt.lti]i ri'rr;i-:*{l 'h*lr i ir:,r j[ "1::ngLne*rinil i : i : lf': i ';.! i -i .l

C. \i.X I i:

ri i ' l vrl ; l ! i l i t i i \ r, , i l Ft ' l l 0!-i l 1i . , ,r iY! i t : i l t r O' Ft: ' f 0i l rt-!-i l-i .r.,ij---lj- ' : "- ! !:1

old


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_ TABLEOF CONTENTS

ORGANISATIONALOMPONENT

STUDY HiMES

TU I l - 1' t

Page1. GENERALRElvl iSENDEDUCIONAL PPROACI-1.. . . . .......... .....2. LECTURER.ENUES NDCO}ISIJL] 'INGOURS,.,,.-....3. STUDY ATERIALNDPURCHASES.-.-. . .,.,,...4. LEARNTNGACTT\./ triES...........5. RULESOFASSESSI\.IENT6, GENERAL, , . , , . . ,

B. STUDYCOMPONFNT1, MODLII.F SJECTiVES,.R'CULAIIOI,I I.JLTEARNING UT'JCi\4E$2. MCDULESTRUCTURr.. . . . . . . . . . . . . . . .

.34445

It5

.. . . . . . . . . . . . . 7,. . . , , . ' ' . '' t

. . . . . . )"t.

s_Eci-o-bl]Dimensirns. ni is, '' d n-: i i ica:v;r: ! i rr, .The mol unit, denrity, cti?)ei-r!.,it- l, i:r'r'llcri{fuii; elic rli:):isuraEnrpirrcalcrrnul?+....

EEC-.]ONI.I.1 . l rrtrcJucticno m.rter;al-. ! lanrDirSi i i r. .Ey or ,,1,, , : i . . .rcL\i r, t i : . . . ... .. . . . . : :rr5. lJaterial balan,reswiiiroiJt 'heliticei :iiirin ... . 2rt

D. AfiACHMdNTS


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. ;' , _; : r1, ; : r ;s- : :5r ,1 ' ', :t - - ,i ' '_ ; : - . \ ; __ ;, r., .::i*:. -


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3. STUDYMATERIALNDPURCHASESPrescribedext : Himmelblau [,i], Basic Principles nd calculationsn ChemicalEngineeringSeverrthdition).Every tudentmusthavea copyof hisbook.Studyguide: This studyguidewith attachmentss used n the developmentf thelearninorocess.t mustbe broughto all ectures nd utorial essions.Classbook/Tutorialook Every tudontmustalso havea classnotebooututorjalook(which an be a loose eaf ile). Thiswill be used or additional otesn he classand crtutorial foblerrls.4. LEARNING CTMTIES4.1 CoirtaEtire aRdijamingimel.ectureser r,eek 2Tutorial;essionser.reek: I c,f wohoursTest essign I of Dne ourThi.r ir.idule arriesa weightof B Q'?ditg hich ndicateshai 3 studenl n average':,/illrvork .J ours o,lsin lle Intendedkills4.? Lel*r-iresLroiur3 r,Jprcrr|ntedr'r he styleof siudent --ntredeanlinq. A shortdescriptlc'rlndr:(planationf the module ontent nd ssncepts regivendu ng lle lectures Prcblgmsrelati!1go ihe rnodule ontenl rregetand solvedby mean$ f classex.lmples.Studentsare rltvilied noi to take copiolrstotes dudng ectureg, ut rather o sFeoll he ;lmecc.]ocniidlingnJ a.tivclypadicipatingn classdiscussions Ffequentefe.encesJiii :tIrdn to llustiativexairpiesn he extbook.d :-r ir;r.jFil .:i'93iod3Tii. i:i,iule rnrk..r irrDvi5ionbr 12 tutcrial sssions. lt !! assunred hal eier/ slud;n! rYi:liir!vr-.iicrnpt:rd aii tire ti"rtorial roblems r)r : padicuiiu-.e55icn efort ll'!es.:;lei:luledat.r.i"lnarrFounueillrs! tests \1.illtequentlybe set duiinq tutcfial s\)ssiong. lf nt] ill:rsi test islvriilen, h4 teiii sci.iions will he l,lilixed or discussionof,itudents'problems.

ii. i(l il.i-i3(li A: i:\l:$fliunN'!-:;irngstirt ntafl{lwo semest,)i ests ,v!ll be rvritten'lriFngthe semester These ests will conttiiltJt'rB0%towards he sE|.nqdter 3rk. The .rther20% of the sefllesterntarkwillbe detennhled y tlierhlderifs perlnnrarcc jn lass ests. A semestermork of at lerist30% s .eq{ired f,li exam:-5frlier!ce

FinalmarkAt theendof the semester ne wohourexam s written The semestermarkand exammark achcontribute 0% o the inalmark. The exampaperwill consist f woseparatesections:SECTION on chapters , 2 and 3 andSECTIONl on chapters and 5 AiuUr-inirra of aoy" s requiredor each ection.e to pass he module final iark of atleast 0olos required NDa mark f at east 0% or every ection ftheexampaper'SlppleinentaryxamSupplementaryxamswillbe ewardedo students ith inalmarksof 40% o 49%AND osiud;ntswith lnalmarks bove 0%who donotmeet lle subminimumequirementThesubminimumequirementillalsoaopiyo hesupplgmeritaryxam.

6. GENERALProgrammableocketcalculators,/ith lFfia-nurrj?riialbiliq/3re not piinitied As ageniralguidelinet can be assutnedhat irccket altulatoGhat ate accef)tablen e.gMath andPhysics exacis,ma) alsobe uscclil Cll( 113


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1B. STUDYCOMPONENT

1, MODULE BJECTIVES,RTICULATIONNDLEARNINGUTCOMES1,1 Oan6ral blectivesThcAenerat bjectvL.s re as descibed nparagraphA1.Speolllc hJectives rc detailed ur eachstr.ldyheme.1,2 Proroquisiteno'{ledge ndskillsThotedreqoDrerecuisiles.1.3 Articulfltion,vith thermodulesn heprogrammelhe knowlodgeand lrki s devetoped n mis noaue arc an abso/uteequiament iotovotylhiv lhat follotvsn lateryp,ars f studyof lhe chemical ngineeingprogramflte,1 4 Crltlcaleamin'J utconEsI ht.tollour'ing ntentediate level autcames are aCdressed n this module, i.e. on condclicnttl ihls nxJttula he student wil sa(Ltfy he {ollowing out ames:. Enginoorit|! probletn nolvingTht! su@esslitl leatner will be able to tolve chan ical engi ''Eering problens su.h astttsftrrial balances clealivaly andscieotifrcally: includinll thc abilily to do a fo nal anal)/s!.tind tnodolline of the prablent and to communicate the saiution iogicaly to llE readei.. .ltlpticahbn of funJamentulanJ tciennic knowledgcih.1 lt!.nresstu! leanet wll be able to apply hislhe, knawledge oi concenlratbn, !.{err:;i':..ilt,1cif(: volu;iie, tDnp.eiurc, p!'essur


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. l

Just ike inSl) hecombinationf units g m/s2 scalled Newton,hecombinationb'ft/s2is called poundal.l poundal= l lbmfvsz

poundals r it can be

482,6Poundals - : : - lb . = 15lbr32,t'7r new dentity s created

The weightof the 15 Ib mass canconverteds ollowsWeighi

From heabove t should ; clear hat

Dinensionlessroups :Nussetiumbei lu = f ;irrrrsiJlLZ]

now be reporteC s 482,6orr,u t-.t' t I!.t'ls E Ib^Jl19?'6o, r; ri'.32.11

:,:,ffi'Y,::;,#f''fr:#Do tire ssif assessment test in Fiirr'rn1.. 23. NJi$ {hat rit !risv'.g to tiltr:e lrifalisessmeniesis appear n AppendixA in Hin..n'1p. S97)aaw-^ l . lL . " r ,h1r l le : . , i ' . , . I

Piandllnrmber Pp = !,:it .;,.tYzznnndbu ,qc g1l{cp46t- u f,:ra fuqttIflE [IeL__Ut{I-!Er'lsj.i.Y,aNcEili&:.:rjQ.]j--Ei:llii:BATliI-LAfl:r,eB!E:il-rr'i,,.

l_hi,J -:,) :,rir en.-N/ tii i Lo9.,I .j pr,jul-Jm! assirCiate,l ?,:ih.i ,r i r - : , . rer r : lL i l r i r r I i ! , . 1,448 \ l qn d { jo tai at trnpt : ) u) r , " ,n,1rol r , i l i rr ,15, ," i i . - . i ii i ' r l j r r . t r , \ ) i \ , ) f i i ,or : ,1,1. . ,1n1' l i re , . ' i lai r r r jh; tJ , i . ( i c , rir r ai , .J . , . , ! ; i , i11:

r.:,rNt!ta.)..?,V',-irats til!.) .^,.ci,,jhii ! |l.nsrioi ri l]?:roi r ; ': i. -. i: ; .-: ' , t . ,1l : 4U.d.: i t , j \ i , : .

IsruDY BJEcrvESIl_ -| . uer :ne ano u5e'rnei mcr) ki lomcl ,pcLtni . ino ;an


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2.1 Thenulgils-i{},apotr 2.1{\tiimm..S43.47. sn-d5xampre.1.,*lr tl,1,rl rEil i! Uendrally accepted ss ! dimensionlessquantity. This is hawever 5trifl]yi i l r i .akirgot ifl ie. lf wesaythemol In?ss iwater is 18, henwe rfleax hat 18 gramsJater:s cne elrrnr nol water, 1.r:.lte Inol iSss of watet is:

lf a fluid lows n a duct (pipeline tc) at a votrFetric ftcw ratepf Q m3/8and the cross-sectional reaof the duct s A m', then he linear elocity s-r i l' :l t ; \ ru=a/A l ' " ' =rt , , lf t ta' )

This s requenily sedin he calculatidRoiReynotds umUlrs. " Jn'2.5 MolractionndMass ractlolt / /alilnL ftWtln . . - ,l ll' t lRI t | |Studyparagraph .5 a Himrnpp. 57-53andespeciallyExample .6-) l\p1.' ,'Averaqemol mass

* Dil{donoshn nilbt" 'liluru.2.4 Flow ateStudy aragraph.e nHimmp. 66 andalsoLinearvelocitv of fio\r,t

the-individualcl masses.- .5 . /^rir\.!v{:,/av - ,1,

with xi = molfractionof cornF,c ntili'rll, '.

Ncte he units n tiriscal.:0!3tiai,. i iv/occ)mpi-)undsand B aremix"d tc givem'ri r'.ri);!o;r'\

18 ]./gnrol or 13 b/lbnrcji;,ie lhat tlre rrcle unit is essentiallye mass unit.i,!/k! K, t,:irh carl alsobe g'ven ;r KJ/kmolK.

or '18kg/kmolCullsidei e.g. tl]e unil.jor hcat capa.ity,

: '. 'l - i\ . . , i i i iSfi,!,.".":r:,r"_.lgn oStrrr* tlin $I?ardgraph 22pp 4tr.51. Nete the cL)rlieptsdensity, specific volume,bull(.!. 'i.|lrr, }ttlf densiry, tqlitr voh'n1c ,ndsDlut:on.UUi!, e_ntjty s ?iv;rriableLhstd$peMs on ii e S(i (or reai density)and on tn.: paitrle ej1earrd por')sityof tne Inaferisl. .lt rrriiv happelr hai a lirterial witil an SG of 2,6 (in .)thei'dorcli r dLrnsity i 26C0 kg/rni) cdn be chopped nto ;rnall pieces. lf the line pieces a!'er',inv l.rcecir1 '/olijrne

of 1 rnr, t v/rllaot bc possiblc ! fit 2600!(g of maienal hrto hr: 1 ri",,,rlir:ne,br-lausr) he pafiicler will settlc irr he tontainer in $ucit a manner h3t opcnings,.,,iri:xi$t betv/een ire,'n lf it ir then pcssible o pack eg l3o0 kc of the particles rto a 1,llj v.jLrme,we say the bulk density of thq rnaterjal s 1800 ki/m'. lt ougltt o be clEar ltati l .!e Lr kdensiLyf i iquids s the rame as ttr: tnre r lc, 'stty"/ni l . ithe buikdeFsity i io l i . i-.ariictes..vrllrllw.iyrjbe ttrrrr ihan Lhe rle d+rrtiry of if|c sotrrt ^ /q ..f/at^?.:.1 $f-r i /) 'ShrJyp{aqrapii 2.3 in t{inrrnpi:. 5'l-56. t mt 'f/O4/h//n' 9.Sroyd"r,uorl

r! and a. hen f T cjenot:s ol AL :6l [6"/rhtw'Mlr)." =-': ' ' ' , ' l i '\ / , " r, l/ ! yib)kt:loi A i(g ,'t l;irol iJ

. i;,lol TLi' T

kntcl T

ke B.kroi B 'nol T .innoi .!,

= r-loi .l.tl4 .=kncl T_FJcnrple2Sl

MM oi tdij total -,'ixture

Calculatehe avelagemol mAssof a mixture onsistingof i4"1:' l2,2e9i' Or anC0f)% lil:('/olume asis).,Assuni5h'i dcai aii ar holds ruefifoimasses: lJ2 = 23, Ot = 9-; CC" = 44

X fi''4t mr'/*rL -, -,// a/uoyf,41/utt*' y'Ao/ z.. r/q/ Vr hY lablr /raz'

Calculatehe massor number f mol of a componentn a mixturef he ractionorpercentage)ompositionsgiven.Mass r molunits.Calculateveragemolmass.Realisehe mgortancef he units f SGDoconversionsrom SG o density ndviceversaUhderstandnduse he concrrptbulkdensityCalcllate hedensity f deal iquidmixturesKnow,explain nd use he difbrencebetween bsolute nd relativeelnperatlreConvedromone emperaturenit o anyotharKnow nduse herefeaenceoints f he our emperaturecalesDefineand use prcssure, tmosphericressure, arometric ressuie, tandardoressltre ndvacuumKnuw nduse hc differenceetwee4 bsoluteressurendgauge iessuteCon,/ertromgauge ressureoabsolute resaurendviceversaCcnvertromanysetof pressurenitso anyotherC;,lc,Jlatehenressr,fexerted ya certaiiheight f iquic l a certain ensit'j

*ficvn{w


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4(f)

(s)(h )

I xi MM)r

tr k l i r,- i i r. l . , i l ' rr,,vrri ;r.F);rfispcif |nilltrtr!: i'llis is always expressed on a ntass basiij. li ii is s.:td ihat;lsclul,on coniails 40 ppm solid, then ii i:o,ltains 4(,r rnass piirts oi sct;il irer rniilli)nine:r.s .ri.. ril tli] s.)lutioe. gcrnetimes tior qases) ii i-j exi\rcss{)cic:.!a n.ol b3siq.illirijri:;y !-!-n':;::.,.'ir.rlo{',1r,'ic[. f t-\i! {isii(rh/ed soii('j -j,:.i;r'e,)i ihi scj!!,*j ;r,]ntlc'.j{ i ' i . . - .( \12' i ' r: i i , . : r j- , ; , t : i ' l i . :. n.al. 'L) .

geglg: 1 kg water. lt contains ,24 k9 Na2So4.Mass olut ion 1.24 g.o )4iat kq Jarsoa/kgsolution= :::-- j 1]"19.1g l'lr/il ).'lk! sriutic.n.i1 4

ib ) Mass ract ion a2SOa = n,2a!1,24 ' 9. iVti,43ssractionH2O = il1,2,l = qSqE(c) kmol Na2SO4 = 0,24t142 - l,tigl'r : 1'i-3 'totalkr'ol HzO = th 1 - 5. i i : ;6 10'> l . 'r i j -:11q'? nr. l

rnol% Na2.soa= (1,690.x O-3i5, i)-5t l t )r '10i j . -2.95%mo l% Ha O = (5,556 l0-?lc,72i ( 0'?)10 0 = !i.A%

fuaefu-21.2On e grnolNaC! (M M = 58,5) s di '. isoir '.cii i rn' w.rtr, l Wil. t i : lf + concert i , l i iorlrlprlm?Basis: l m3wateri,lassv,/ater1000 and density1 kgr; ,= I0',r0 'l1 gmol . iacl = 58.5SN4ass olution = 1000 kg + 58,5g = .t-Cq0_q{i.iSIn 1 000058,5 solut icns 58,5 Ne,aii .e. n 1 occ ooc o solui;on, -_1!0. i ' i ' l '56,5) :J.rcr' I JOnli X '= 58.499d ao l

r.d)

tYttgnorW(MM)"" =

Since he analysis f the ga s s givenon a volume asis, nd he volume nalvsiss hogameas the mrl anaivsis if the ideatgas taw hotds rue;. it fc ows frllii-e-?-o'fiiiEtr-6ilar e0,14:0,26an.l0.60.' i .e. (MM)""= 6,1a12U, 0,26 32) + 0,60 44 ) = 38.642.6 Alalyce-sruXtlcetlpeleltselllbls-aldlllisures NR lllr I tu l .. rStudyparagraph .6 n Himmpp 59-61.=-ri:-_ -co/,/F/ab* h.ry "?zhruL.7 ConcentraiionStudvparagraph . in Hitnmpp. 52-67and atsoCoDcent.ations no[t']allyexpressedn rFrmsof a cerlalnquantirymass,mol olrglum.r) ota roatefleitlies('luteor dissolvedmaterial) er giveqquantity mass, nol )i volume) i ihesolr/irnl r rf he soltticrt. The ollovlingways o express oncentration regenerally|sed:l! : t t . j in lm.ru i ' . : i i . fo i i ik i . l he or af eajso ommoniy sed. I t \rl) rviirrspr)r..riritvotr,n:eofoi,.riun @qtt dfnllgl pr n* 4law flalhl

^gln' : i . , i t ' : . j / r ' , ,": kg/r-, etc.i ,r i i rr. , : .rs.. :oin olut icnof sai t inwater this aanfcr exan,pl- br 2,10 g sal fp.: i rnl 'c iil'; :llili i{)lrtion or 0 Z |\g salt per litre sol,.'ljon.etc. llass pet unt? (.rlun:e ijusi |'x)1 ,:coiiu icd vJjthdijn3iitr/.

(b) fu'lxss]i]fllDilvolume cf soivertik91i;;:; kg,t 31c but frol\' ii i$ kg sall peri:tl of v.;atet eir.:i ,;ct_ ig salt nsi;n:i r3h"'.1' i ,r, ,n

(ci M$iFii iilrij ,,t,:rili'L ,:,is.,i!r-rnt

ijit r: ,i .;,:i,.'t:; tt ti:

Normallty: This s he number f gramequivalentsf he dissolved atrlal or l lrrtof solution geq/Lor gelL ).(Self tudy ssignment: ind uthow henormality JaFolut igns calculated)Molality:his s he umbertsmots"rrn ot"t(/r!kn"r$.f{rt l{ kilr,(Note notper'10009 olut ion)gmol/ '1000solvent),Mass fraction, Mass percentage,Mol fractionand Mol percentage s already as-

;H *o;$ ('htitV,4ulolwfty*A solution n wateraontains ,24kg Na2SO4 er kg \^/ater.Express his concenftation s(a l ksNa2so4perkssorution. ftal4le 'vkl ,il| d'V/nApl:i l$:ilJ[il**" solpul,ht.s/4k wt/6'^(,rl* molmass rNarsons r42. ttlalth : /''ntl g"I

',*i13


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i.e. concentrations 58.4966 pmNg!g: Theconcentrations 58,5 /1000 = 5gl mg/LFor dilutb solutibdsn Water, he concentrationn ppm is the same as the concentrati onnmg/L.FurtherxamolesExample.7.3A solutiont methanol, thanolandwaterhas he ollowing nalysis

Ethanol= 30%; Methanot 45%: Watet = 25o/oVYhats he analysis n a rnolbasis?. . ' . iBas:s: 1q0kq ofthesolution.

0,65221,4063

Example .7.5 Densitv f iouidmixtures

Similar o the method of calculatingMlil)avas MM)"u =it is frequentlyassumed n enor that :

p. r = Exipi = 0,4(1400) 0,6(2100)This s WRONG. Note he units

. N+100' 2100

1630 fromwhichollows: x=wk!

rx(MM)r

= [email protected]

.Y.!!!- - --I-i6--- = -.L = *,Densi^, 21CA g/ rl 2100-],/./ is - l0 or t , , , , , i - roo - "'

16.1 + -'l-- 1 63 .. 17' 2 00'iafttss _. ,tr i 100 -.

From it follolvs-: p,I

p,

No'vnote heunitsk, ,1t tN pi ) : Kg ntxrte

I

I

30452510 0 _ 3A!!

/, !;qi.iiilf den:ily i$30 kg/m3mustbe prepared y mixing laywith water. l'iow nucildayi-li;'. 2,1) nustbe added c 100kg cf water o achie,rehe requiiedSG?Assumemixings dealAagS. i00 kg wsterLei X kg otaybe rerlt.;ii'edith 100kg HzO.

kg/m3

\/'rl'-[rre f he day

YLlurne f thewilter

i.e. inalvolume

ind f indldsnsily

mt

A mixture f iquidsA'and B coitains40yo (i.e.xA= 0,4; xg = 0,5). TheSG of A is '1,4and he SG of B s 2,1. Assume hat he iquidsorman dealmixture ndcalculateheSG

kg Bmt B(0,652213,4474) O0= '18,92o/o40,794/"L0"23b

k{A ,ts A kg BLXio, ---_-' kg nirture m-A kg nixturcTheunits renot hesameand he \,voerms annotbe added!

gg[dg!: consideionekg rni]rture,.e.0,4kgA and0,6 kg tsVoturne = m.iss,/ciensity(04/1400)r43-Votume = mass/, lensity (0,6/2100) .Volume ixture = VolumeA+VclumeBlfideal rixiu.lanbeassumed)= (D.41400+ 0,5/2100)m"Massmixture = 1k g'Densitvmixture = Mass/yolurne- ---- I ...-_-- = 175C- 0,4/14000,6/ 100

-- =-L,--= h,iih x = rnaas raction)-\x, iA) _

t(.r,/p,) i.e.l- = -1'-.+ 5 r.-!.. P" P4_ pu P,

kC .1 --8,!_iq ,nixture." ." f ks14


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What bout hedensity fgasmixturesf x j = mol raction?trCToRTAL-X2.8 TemDerature

= !.'-!!!!tkg mixture I

HirnO 0t HitnnyL@(a) 1'C emperatureifferences hesameas 1 K emperatufeifference

(b) 1"C emperatureifferences he sameas 1,8'Fdifference

_ ^' A + mt Bkg mixture klkcK

i .e.The Celsius and fahrenheit temperature scales are relative and arbitrary. lt is oftennecessary o $/orkwith absolute emperatures Absolute scales have heirzeros at thelowest temperature v/hich can exist from consideratiol)of the laws of Thermodynamics.The absolute scale lvhich correspondswith the Fahrenheitscale in size of the unit is thaRank:nescale and that which has the same unit size as the Celsiusscale is the Kg[j!scale. The Kelvinscale s the acceptedscale or the Sl system. From he aboveequal unitsi:e follovJs

d" F = d' RA' C = A KFi.iilher, ronr l)e definitionof the Celsiusand Fahrenheitscales, he following alio existsbetdJeen nit size ol "F 3nd C :Between he ircez;ng poilri and the bciliirli pcint of lvater therc are 212 -32 - i80 unitson the FahrenheitsDaleand 100 uniison rhe Celsius soals; a ratio of.1_,:q.Coitversions.en thereforebe done with his factor and the neuessary ero reterencepointadjrjstmenis.It s clear hat ihe sarne ratiowill exist between'R and K.The atsolute zero on the Rankine scale is at -459,58'F (roundedoff to -460"F) and thoabsoluiezefo on ire KElvinscale s at -273,15"C (roundedoff to -273'C).Reacicirapier4 :limm. pp. $g-r5. Study.example4.1. Ignoreexamplea.2 and the sejie$g$s.neFtqucsiionNr)4.}l!{E: li, piofrertiessuch as heat capacity, kJikgx), hFat t.ansfer cceffiiierrt 0ry/n'?Kletc, all i'lnperatuit-< refer to a !ggpg3!r.9_!iffefer!rie of I K (of 1'C). li conversionof thi."type af unit l!:-,st e dr,ne, rl2 i&g.e oointadiugtugft !s necessary. Only he i!rllo!',ngi!lenlij.,/ ill ba usc|j: a1 C = A 1,8.Fsi'iiii, is gLir r:i,/ wiiite4 ,rs (,1- l,r"F cr K . '1,8'Ror .r\n 'C = I ,8'R oi K = 1,8'F..fxin'4.e._?glhe rreit capa.ity r,t w:itef i6 4,18i kJlkg"C. Wh:t is tha heat capaciiy n

&r ljl-l - Rru- o'4s36g(c ) 4. '187J/kg'C = a lat 1,r"c1t.8"Fl,055ll tb= lp BTU/tb"F

lJUroRrAL l2.9 PressutePressure,rst like emperature,an alsobe expressed n"an dbsclute r relative asis'Pressurei cieflned s &M per unit 3rea. Noie hat kg/munJicates massdistributionanonoia pressui-e-Read hopter Himm. P. 9'102.E&nllcJs.1

tlre pressure in kPa)exeriedby a 10 t c.inrnn of waterat 60'F cnly'due oThedensiiv fwaterat 60"F s 62,4 b./ft".l0 t 0 = -3-!i9 m

$/hat isqravitJ'lColumn eight

Density rr-. ,rs.]6q '_fl_T I - -1000s/m.J''i Ih . 10,3048,n-li.e.pressuref't;;"x I then P = 100013807)(-1,048)fF ;t ,, = #t]t---i

= -"cccoim'= too'o'" = to n'uit l

lf tfre orrnula e - p gh lis noi knowii. he pres,sute an be calculated rom iirst princi-ples,as ,'ollcwsLet he columncross sectionalarea be ,a (rn'z)

k.l/ (!t K (b) kJlrig'F (c) niU/lb"i-


11/26

i .e. olumnolume 3,048 (m')and hemassof water n he column mass= density voluhe)= 10003.o4BA)9 .' = r" ) -r t [n ' ') e{ frhe orcexertedy hismasss

^tttia)

F=ma | - , hdJ. uot'= 1oc03,0484)aozlrs { - .v l , ,n\ -.n{ \ r.i.e. he ressufe- rorrerarea {\inil ' qlf ' t .lA}/i.e. p = l{I1o't u.1rxoz 'v ,-1 U' - l*' '. A -, i . ; r=, . ) l \ i t tvI a' '= 3ooooa oaq'oo5 }.ililcalg.elles.s-ulqand Absolute eressurt | -'0-

Most pressure measuring devlces measure gauoe pressure. lf atmosphe c pressure salsobrought nto accountan absqlgtqlGqsllq is ohtai ed.Absotqteprsssure = Gaugepressure + Atrnosphericpressure

l he atmospherlcpressure s not necessarilyone atmosphere hut va es dependingonthe heiglrt above sea ievel and meteoiological conditions. On the Witwatersrand heaverageatmosphericpresiure is 83 kPa. In Pietoria l is 87 kPa. AtmosphcricplesoufF salso sr\nretimes alled baromettic lressrlre. The standaid atmospheie s deiii'ed as thetressLr-.equivglent o 761 mtn Hg in a standardgraviiatioll al ield or1 ?hr = 70 Umrn {g = i4.696 b,/inch'? '101325 /m'?= 101325 a " 1.013bar = 101,3kPa= 29,92 rrch -lg = :13,9'!t HzO

NOTE: AVacuum must be seen as i: negati\re auge preg9ure.In cases wher., confusior! s Possibie t is advisable o distinguishbe.tween l)sclute3rlogai,,rfleressufeas tollows. kPar. kPRg,psia,psig,etc.,L-)il,!pl? _i2-9.2An ob!;oletepres.il/regoLt.le hi)ws I ieaditq cf 1 tur/,,m2. Uvhat s ihe pressure?fhe coi:.,rrr)i gr is noi deiined. S nillr to ibr he f'Jllowing an be lrssumed

1r' t: l r = 11tiNi ' 9,807 i .Ji i i ,q ' g,BD7rr/sz 1p

1ks,tcm2s,eoz'l l*#l' l-#"] =*-o'"Example .9.3A Vacuum f 4 incheswater s measuredn a gaspipeline-f the at'nosphericressures730mm Hg,calculateheabsolute ressuren hepipelinen nchesHgand n N/m'.

4 nchesH,ol fi H,o l2g.s2nchesHF.vacuum -..ost7 t1oAbsolute ressure= Atmospheric Gauge = 28,739 0,294 = 2g{45 inchesHgor i""hetHg toj325N tn, | = 96330Nim,bsoluteressure= 28.445 ""-:::-- pg,g2 nchesgExample .9.4The gaugepressure n a gas cylindern Pr-'toriaatmpressure= 87 kPa) 5 300 kPane i/inAer is transpo(edo a campsite at ihe coast atm pressure'= 1010miliibar).Whatwill hegauge ressuren hecylinder e? Assume he emperaturetays onstantPretoria: Pabs 300+ 87 = 337kPaQe.e$: p*- = 1L19o* ,Y0911"1 1o1 Pai000 | u, IThe abroluteoressuren the cyiinder an not changewithoutchanglnghe nurrrdcr lmolesof gas n the cyhrrder,.s ai the coast. f tlle ternperatureemains onstarrtheabsolute rcssure il! e."naincns:ant nd

Paos 387kPaan d Ps = Pa"-Pam = 387-'101= 286 kPa2.'l0 DiffeEltlal !'^s!rEStudy i.nm aragiap'r.3pp 'i '14-i1S.

Eufo-RrAL6 |

= 28.739nchesHg

= q4l!! inchesHg

730 n He otm12g.g2nches sAtmptessure tov n ngl


12/26

2.1'l BuovancvndFlotation'Thewell-knownrchimedes rincipleanbe stated

lf a body s wholly r partiallyubmergedn a liquid, napparentossof welght fthebodywill be observed. heapparent eightoss s equalo hewelght flhaliquid isplaced y he body.In he caseof a floating ody, uchas a piece fwoodor a ship, he apparent eight lthe loating ody s zero,whichmeans hat he loating odyhas ostall ts weight, nthiscase he otalweight f he body sequal o heweight f he water isplaced. hhmeans hat he heavierhe loating ody, he more iquidt mustdisplaceo enablet ofloat,which xplains hyempty hips loatshigher"n hewater han ully aden nes.lf the immersed odydoes not float, he weight oss s not total, but is still easllydetermined y meansof a scale. lf the bocjy s fully mmersed,he volume f waterdisplaceds equal o hatof he body tself, nd he apparent eight oss anbe used ocalculatehedensity f hebody, r tsvolume.

II

Since he wholebodywas mmersed, his s also he volurneof the body .e.Volumof Body = O.0O17 3 = 1.7 i+Er

Thedensityof th body s calculatd rom :Density = lvlass^/olurn= 4.0/0.0017 = 2350kg/m3

SG of he body = Densityof body/Density f warf = 2350/1000= 2.35Note hat if the body floats, he mass of the body s equal to the mass of theliquiddisplaced,.e. otal "weightoss'.Example .11.2A glasshydrometer s madeby puttingead shot nto a glass bulb and addingastem as shown on the sketch. The volumeof the bulb is 50 am' and he stemhas an outsidediameterof 12mm. The otal massofthe hydrometers 65 g. lt isfound o float in a liquid as detailed n the sketch. What is the density of thel iquid?

il

An irfegular haped bject asa massof 4.0 kg and t is necessaryo find tsvolumeand density. fhe object s immersedn water SG= 1.0)and s then ound o havea'weight- foniy2.3kg.Weight f body = 4.0x 9.8 'kS m/sj= N].WeEht'oibody n vater = 2.3x 9.8 kgx m/s' = N]

Solution:Apparent eight oss = 4.0 9.8 2.3 9.8= '16.68

'ihis apparcntweight oss s known s the buoyantorceexerted y the iquid l! ihebody. The body s pushed p or buoyed y he iquid nd he body llereiore eerni c'be lightof'han t reallys.Accordingo the ArchimedesrincjFle,he buolrantorce s equal o the vJeiglrtf thew3terwhich r displaced,.eWeight fv/ater ispiased 16,68Nmass fwaterdi.iplaced !l

S-e!de!The mass of the hydrometer s 65 g and it floats. This nEans that he mass ofthe liquiddisplaced s also 65 g and if one can find th volunreof the liquiddisplaced,t is possible o calculate he density.The bulbdisplaces50 cm3of liquid. The volumeof liquiddispl?ed y the stem,is equal o tt volurnof the stm belo\ivi'le huid surfacc, . r/4 d' x depth.Stemvolumebelows urt'a


13/26

EMPIRICALORMULAESTUDY BJECTIVES

. Applicationf theconceptf dimensionalonsistencyo determineheunlts lany erm n a function.. Conversionf anyempiricalormulavalidor a certain et of units)o other nlttwithoutosing hevalidity fthe ormula.

(Thiswork s notcoveredn Himm.)Formulae erived rom basicprinciplesre valid, egardless f the system f untis nuse.Take orexample Speed = Distanceltimeu = s/tlf s is measured n kilometersand t in hours, u will be calculatedkm/h. Confirmationihat it is a basic ormula can be done by doing dimensioncomparision

Theequations dimensionallyorreci nd f it s^for xample equiredo calculate infvs, t willonlybe equiredo set d in t. D in t'ls andg In vs' in heequationconsider tso heequation u = l'08kCo PWhich olds rue,on conditionhat u jn vs, k in BTU/ft "F, Co n BTU/lb"F ndpin b/ft3 reset n heequation.

t orr t a,1Lhs = [fys] Rhs = l-:j-Y- '"' ':, - '', IL f rhF Bru tb h)

Theequations notdimensionallyorrect,n otherwords t is valid lly if u is n t/s'kin BTU/fth"F,tc.lf suchan equation ustbeused, aremustbe aken hat hecorrect nits reused orwhich heequation asdesigned.Sucha formula s knownas an 9!0piIgal&Itrgla.lf it snecessaryo calculatewith his ormula)u in m/s, t will benecessaryo write llthedata(k,Cpand )in unfsofthe Brit ish ystem; alculate (i n ys)i and henconverthe inalanswer o units f m/s.Example .1

, = X, asdiscussedbove.

: ILength/timels rs a distance r ength : Ilengthl i.e. q - Ilenath/timeltt ]s ime i ltinrelIn otherwofds he netdimensionn he efthand ide s thesameas henetdimensionon he gh thand ide.It s,however, otalways ositnple: Dnsideror example

Ltdt{ - Dl tgtrv'ifh u = speed (e.9. m/s)d = diameter e.9.m) -D - diffusivity(e.9. m'ls)g = gravitational cceleration (e.g mls'?)lhe dimcnsioFal correctness of the equation can be tested by ensuring hat the netdimension n the Lhs ; to haton he Rhs.

l.hs : la ,,,' 'o - .''o rulL.t I

tocalculate inm/hif k = 0, 1WmK Cp = s,2kJlksK p = l20okg/rn3k = 0, 1WmK "r I ar z l:ooolo.:o+a,,rl


14/26

t_e. u 1,08(0,0578)1,24(14,9)

6,72 , 1o1

= 6.72x104 fus

I 0,304813600l = 0.736 /hsl .f i h I

Examole .3Theheat apacity Cp) f sulphursgiven y

Cp = 3,63+ 0,640 with Cp in cal/gmol and T in KNOIEi In the eq Cp = 3,63 + 0,640 T, T is a real temperatureand !9t atemperaturedifference. In the units of q (caygmolK) the K denotes6temperatureifierencen K(or n otherwords C).Rewritehis ormulao he orm

Cp = a+btto calculate pdirectlyn cal/gmolF f in "F s set nto he new ormula.Let be he emperaturen'F, i .e.

c" = 3.63 0.640 '-32 * ztt ] --fL 6161in pL 1,8 ) gnotKCq = 165,92 0,35556 with Cp n cal/gmol ; t in"F

Note hat he unitsof Cpare stillnot ght.thisstage.Testai 32"F (273K)

Cp = 363+0640i2731or Cp = 156.97 0,3555632 ) =

It is usually dvisableo test he ormuia t

Conversion f an emDiricalomulalf calculationsf this ype mustbe done egularly ith he same ormulat will be lmcconsuming,o ccnvertall the dataand hen againconverthe finalanswer. n suchinstancest is ha[dy o rewriteheempiricalormula Drdirectapplicationn hesystemof units n which t s required.lf thismustbe done, he constantn heequation,in he above ase heconstant ,08)is seenas f it has unitsand hat heunitsof the constant re such hat heequationsdirnensionallyorrect.Only he constants thenconvertedo unitswhich onespondothese equiredor he newcalculation.Example .2

. 1.08t . . .The otmulau = :r::-: is vdlid n conditionhat k, Cpand p are n British nits ndco Pit can henbe used o calculate in us. Rewritehe ormulaor direct pDlicationithunits i the Sl system ndcalculate in m/h f k = 0,'1WmK, Cp= 5,2kJ/kgK ndp = 1,2glaftf

178,35cal/gmol17e.3q cal,'gmol

u = 12756LCo P rlFor appl;cation n this formula, k = 0,1 W/mK, Cp = 52C0Jrh! K, Ii +. ,, = -r-2f:q0,.ll = 2.04 1oa nts52002c0)

= 2.04 104 4116!0,!= o.73rj n/n,t l h

[t y kgK m.--= -- -|n rrrK J kg

Te cori 'ecihe uni isoi C,: [C p =166,97 + 0,35556 ]TIle units of the 166,97afe cal/gmol K

j66.e7 -:!l)=:l -- s?.7e callsmotFgnoi K)1,8"tiThe !n;ts ot ihe 0,35556are callgmol K'F

0,3555e --"9:l-+i = o,1s/-/iar/Emorr"Fgnol K "Fl\,8"Fi. aatti. --- wtt|t I tn -rgrn,tl" F

= 1200 g/,r3

i ,e, Cp = 92,76+ 0,1975

( e,nrrlo,r' ") "


15/26

Test gain t32'FCp = 92,76+ 0,1975(32) 99,08 al/gmolF

= 99,08(1,8) lf_gJ5 cat/smolAlsostudyHimmelblau xample .2,p. 93 and he self assessmentuestion o.4 Onp. 95.t@iAilE -dfrr,// /4/u? ltt'nrt!,,!,ttu

sEcTtoN l IEM?I}ANN!

sll,f coNcEltT.4ttoitut Sargta?m50!ofroN

ReadHimmelblauhapters aitd7. Concentraten he oltowingopics . O\. system closed ndopen) . 137. t O"). Steady taleandunsteadytatepp.138143. n \ - /n ) ChernicaleaLtionspp.14$150). ^\ \Naieltsre:,rysrare \ilt\Ii.] + GENERA.I-FD OUT + CONSUMED \

. Batclr.andgjni-batchrocessespp151-155).' ' S$ategy or problemsolving pti 168-175and pp. 182-18tt).. Ohoosinq busls pp.78-81).-lh. lhrti! il!esti!.,]' method of chocsing a basis for the calculation s q!it-. vdl;abl,.StuiJyg\surpl.:s :1.13.3 and note he serious errof in exarnple3.1 . fhe chcseir bs-cisis 1 (ic,: oi a--C, con$istingot 1 knrolof Ce and 1 kmoi of O.iJtie that 1 krnol of Na2CO3 lrnsistsof 2 kmci I'ia, 1 kmol C and 3 kiroj C) 1,5 [,nair.t).

s. !,1it lilAl.lAL4UsEF-W[FrolrrEuG,4:8EAA]10-N\\t0(,./' t

i,td0\|3:rad Hir rl (,itapter J 1!. 196-209 nd exairpies 8. 1-95. \) , _ \4t5. q.l-.ili.lliqeilo.Ian.Urtr]I|S 0\\Sl-\prr?i s,ii!L.iiiiyc.urjii;r ,,r r1!r'-ganrcaits n Waterare shown

jlIt

The curves or IQSO4, 'lac and NarSO4.1oHrOhow hat f saturated oluiions fthese altsarecooled, rystsls f the saltswill be ormed o that he saltconcentiationin he inalcooled saturated) othei iquor,will be ower han n heoriginal atuGiedsolution. The crystalyieldcan be determined y meansof simple nassbalar]occalculations.Exa4le_5.1.!,The solubility f NaCl in wateral 1Co'C s 28,5olo,t 0'C it is 26.i% f/l:.ii ri ihepercentageield f NeC/ r.ystalfa saturatedolution t 100'(j scooler:1il ij'd

Saturaiedsoiution1AO'C

,M{,5 9!$S: 100kg eatr,rateciulution t 100'Ctj3ct in = L,1,285ito0)Z8 l kSLel x kg NaCl cryi.tais e ui Ed i... (100-x) g saturcted olutiorrt C'CNaCl-r,alan.e 2f i.5=.y.0.263(,00-x) .e . r = Z-99kr. l/ roD\% \'ietd -:::r i I ri,l : iq.r_%i- !5-) )

E&qtde-!1.e5OO g of a soluiion, sirtlifaierdwlth l'lalico3, at 60"C, is (.oled to iO"C i'ti ti r:r hecrystal ic ldinkgt SJiLjb!: i tval aareas:Jl lovr: t t _,t1* Cqltalqlclt trypatx 0l,uhrrtoh :'i6 ;;;;/,/ ;,'r/ ij ol/rrul

/ t'/

* (q\r."\q\c,hc. 0,


16/26

io,\,iqbilih l+- q\ wh;r.!r ol (Temperatufe"C ) M 40 3o 20 _.10

8,16ra mNaHCO3 er 10 0gramH, O '16,4 12,7 11,1 9,6

x r,f flrl rl ' \ rJSe!!e!

X Basis 500ka saturatedolution t 60"C.

I 16J I7o NaHCO: = | '' ;* _: 100 = i{_qg_%\ i o !!r + ruu./f jifl 1oo = 7s36%i rud,r)

-....* 30o/"Na2CO3olution

Se t x = kg Na2CO310HrOBasis: 100 q 10% olutionMass-raction azC03n Nazc03 10H20= 106/(106 180)= o.szooNa2cO3 alance: ,3706 + 0 1(100)= 0'3 (100+ x)- x = 283.3!9 of Na2CO30H2O-Erampre.1.4-. lOrfin!, ,t ;fuqlaA milksupnliercan buy 'lull creammilk"(whichconiains6% butter et) for R3-i0 psrlitre. tle can also buy "no fat milk" whichcontains0,6%butter at) for oniyR1-3U erlirre. He wants o marketa 'lcw fat milk" whichmust contain2% b'ltter 3t) :y nixingthe t\,vo avvmaterialsand packaging he product n plasticcontoin+rs Al! thesepercentages re on a volumebasisThe costcf nlixingnnd packagings R0-80psr iiire n'lk. VJ-h3t ilouldhis seli;ngpiiccbe ii he warrts o rltake130o/orofit,basedon ille cotnbined r:ri of tx.w nr-tteriali.{pacl(agrng.Basir: 1 L of ihe 2% niilk. Set x = i of full ctesirl iiiik'B-ut terfat 'alance:,06x +'J,0C5(1x) '= 0,C2(- i )=. x = !359_1, of 6% milk e.qzlLL of ir,6?i, riikcost = 0,259 3,3ir) 0,741 1,80) 0,80 = B2aq9'r9r-!=Sell ' lrg fice = ?.30 2,99)= E9.9-t

E&EIA!--rJ

NaHCO:

6o'c saturatedl 16'4 NaHc-o-a- . t0ogH,Ol_0_!,Satua!ed: % NaHCO3 =Lei x kg crystals e orroed:NaHC03 alance0,1409500) = x+ (500-x) C,07536)i.e. x = 35,14 ks,gtudyxample.6Himm.p.209., r a. t I^.1, , "Ex.2111pe_Er=i!> Lelort lriola,j l.qW


17/26

5.3 Combinationsf EouiDmentMater ia lba|ancesarenoton|ydoneonindiv idualp iecsofapparatus,butmoregenerallyn combinationsf equipmenthich ormprocesses'Think or exampleaboutprocesses.hich ncludeneutralisation'rystallisationndorving i tf," "wit"r" or a piocesswhich onsists f an absorber' distillation

olumn'condensertc. or evena matenal alance ver a wholeproductionlante g an oilrefinery.Theprinciplestay hesame.5,4 Tie elements r ie comDoundsA tie element r tie compounds gne whichpassesunchangedromone stream oanother.of tenmorethanonet ieelementwi llbepresent , inwhichcasetheonepresent inthe;reatest uantity illgive hebest esultsn calculationsTie elements re iseful ven f themissesandcompositionsrenotknown ecausetenables ne o place woslreams n he samebasls'Example .4.1Streams , B andC are mixed o produce treamD stream'\ is a l0% solution iilio. i"'Jt"t"i ""J "tieama is a 3o%solution f HNo3 n water' Str-eam

is pure;'#t: tfi;t'l;;;

-t"-"" no" t"t" of streanr if it is requiredhat-slreamD must;;;" dilirlq il a"/"HNc.r rhe flcw ateof stream mustbe 160 g/h

TotalmassbalanceA+B+C = D16 0+ 71.1+C = 266,7.. C = 35.6kq/h +glggK: Waterbalanceo,9A + 0.78 + C = Water in0, s 160)+ 0,7(71,1) 35,6 = 2291s/h , /l!Andwater inD 0.86 266,7) 229.4kslht / ^l t: .. t t -t.-r..hl,>,t C//rth^ f4q Ul,, 'Examote.4.2- l/t ql 4 flrJv'r,r-,Benzene nd Toluene re separatedn a distillationolumn a yielda distillate l 95%Ben@and a bottoms roduct f 3% Benzene. he eedcontain$5%BenzEnendthe@|,/tates 100kilomouh.The eflux atio s 3 (i.e.L"/Don hediagrsms 3 on a molbasis) nd t canbeassumedthat Vr = Vr.What re he low ates f all hestreams , B,V, Loand LN n kilomolih?

, / t , .z*.i1- ------,-, tlc,lq)il*'-4dr//n atufiltV/ I u, ,I', 4,nptrutd,vtteAllpercentagesreon a molbasls.Nochemicaleactionakesplace*i^Lal

/ I " ( ; l l^f r \ ,//. /lookmouh l.. t*u' 1/xF = 0,65 | |

/ / l \-/t . ___r_.-1 )^ \ ", - l l h-l----- .1-1+-- -r.-.----Fl i l I ' I i / L" Yr_ rn-oes.-;-1-l ?lI i , -)4rL^ t*i-;l* 1-t,:,-.1-Qda' . r .1 l ' / | \ ,6% HzSOr8%HNO336'/oH2O

Eagig: 160kg/hstreamH,SO. inD= 0.1(160) = lg lqh an d hat is6%otD.. . O,OOO tO > D = 266.7kuihllNOgn D = O,0S266'7)= A-L3 ko/hAnd hat s 30%of stream

0,39 a 21,33 + A-:lLLLg4l

Solution Eeqjq 100 mol/h f F

an/ndvt, t->3>3{t -c e2uq/auti .e. Lo - 3D


18/26

NarCO3 n =

Total HzO n = 1gg + 1 lteOl tgl_* - / l t * -9 ' l=0461e- x = s78ks-zaottldolt-

Example .4.4The otlowingwo equirementsustbe met at he same ime)by a solution f NaC{andCUSOan water.. Thesolutionmust ontain 0,0% issolvedolids. Thesolution lustcontain ,0% molbasis)Nacl

What nasses aClandCUSO4 ustbe dissolvedn 100kg watefEegig: 100kg H2O x kg NaCland y kg CUSO4

* (roo)g

kmot: oo/18 5,55 mot rO;;h'kmcl _L kmel6ugg.3o7ool ids: - : :+ - : = 0, 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . 1).r +Jr'+ luuMo lo/ oNact: - t/ 58'5 -.- . -,55-ti58,5+,/159,5 0,0s . . . . . . . . . . . . .. . . . . 2)From1)and 2) :j-.17 ,!g_!gi!.qg! y_=_5,27 gclQor-


19/26

ATTACHMENT: TUTORIALPROBLEMSuseowLv te colvenstoruactcins twt't Water lows n a 4 inch pipe at a velocityof 0,6m/s p = 19/cm3andP = 1cP.

TUT 1Do he ollowing onversions:(a) Flow ateof 30 m3perday o cm3/s(p) . Flo\,\/ateof 2 cubicmilesper^yearo us(c) Density 152,3bJff to kg/m'(d) Density f 1,3ozft3 q kg/m3(e) KineticeneKiyof 50 t-lbro J(0 Kinetic nergy f400kJ o ftlbr(S) Kinetic nergy f400 kJ o t-poundals'(h) Viscosity f 20 cP centipoise)o bmfth -(i) Viscosity f 1,5cP o Sl units0) 40 t-lbr/ho W(k) 2,02million bl/dayo t3/s(l) Density f 1 kg/L o b/lmp. alland b/ft3(m) 182hp c BTU/h nd t.lb/s(n) 800kg/m3o Poundal"

*o, Converthegiven ata o Sl units ndcalculatehe Reynoldsunlber'Conved he gjven data to Britishunits and calculatehe Reynoldsnumber.A light ear s hedistancehat ight ravElsn oneyear'Thevelocily f ight s3. 0 10"m/ sThedistance e,tweenlanet andearths 8,4 ight ears,The nhabitantst plarlet recentiy aidus a visit How onqdid t take hemto kavel the distance f their spaceshipmaintained n average peed ,l3,8x 1oB bmft?Give heanswernhouls.

*

(b)

(lfuc\"W,kt3i The d.insity af solutionA is 9,0 lb/lnf, tDlio!]. Whai is ihe'iolu.ne in litr?r of

10,2tJ b of the solution?fhe rJensity f ienzene at 60't- i$ 0,8/9 :rict!"r3.ihe df,ilsifynI \,,/dlefi 6l'j :--.',;62,4 b/ft3. Calculate SG5c.r.60"r.ol binz..'ne, TUT 22.1 30 000bblof oil s spilt n h esea. Accordingo a newspaperrticlehis ed oan oil slick2 miles cng,halfa milevJide nd25 mm hick. ls thenewspaperfepott onecn

2.2 Do he ollolviogroblemsn Himm.pp.33-35),. 1. 1 1. 2 :1.8 1.13 1.14

2.3 Whar s hekinetic nergylr t-poundalsnd n ftib of an 8 b masswhich asa velocity f 60 mil.es erhour?Show ow heunits reobteined.2.4 A certainmass. asa potential nergy f 700 tlh relaiiveo a surfEce hich s4 mbelowthemass.What s hemass n kg?

D0 he ollowing roblemsnHimn.1.2C nP,351.27 nP.3a

3.:! Dt ti're irllc,\,1rig r!)bl.nls hr Hirnnl.2.-12 .7.1,'.3'i p. i4 - i\ir,ie n flimr': liall : US tnii'-rrl

* ' , \llcvr ,-ni:n!,ii.--sot liquid { ($'f = 2, ) Ir';:ii be i(id5'J il a 100 itresel ii'.iL':il''(5i-j = 1,:2.)o g!!c :i:.: f!nal jens:ir'-:. ii0 li:iii:'? iii:r l;.jri'is rrlr'i:'-ic,':lii ll:'ini:rlngir deai hg Fx l i tr !-- , i') f -l ( ?;i i y i l ls:i ci l !t; i i i i ixioclvcixi Y) r!i r : i.niYtur,iiThe butk dens;ty ot grElel rs i ll0lJ ':giitri. Tlrc ;; oi iil.' ltDne is 2,t' li '., .tliiit: oontainer s idi 4i iJravol,lc,vt 'ila'r'.,l:.tl.g i v,rai6r arl be po'rI.rt ilii? liieco;r '?;r ict?A S iife co tainer i iiiler,!vith ri"ia: sthei4$. Iho iI.) ot ihe lll'jtai i5 I l. :i !"now pcasible tc t-,oijr15 iiiie ti !",'nt3rirto tflc ccrlitln.r. vvfiri :s th. tuji'densiti oi Lh ni:t?l r!'h,.ree?

>i ,u


20/26

4.7 1Okg ot a 12'/" MgC/2 solutionSG = 1,1) s mixedwith20 kg of I 28%MgCl2 solution SG = 1,3). The mixing s ideal. No ternperaturehanootakes lace.Calculateof he lnalmixture(a) Thedensity n b./ft3(b) Themass ractionMgCl,(c) kg MgCl2 Per gwater(d) Molarity(e) MolalityTh emo lmass f Mgcl, is95,2.To neutralize 0kg of a 75%solution J H2SOan water equkes ovr' anylitresot :13,8N NaOH olution?2,4 N Ca(OH)2olution?2.4N Al(OH)3 olution?6,2 , Ba(OH), olution?'1,8V! l(Oll)3 olution?

5.1 The boiling point of ethanol b 173,1?-'Fand the frepljng poiiri is -16.rr.6'ir.Whatafe hese emperature-"n "R, C andK?5.2 The thennalconductivity f steel r: 26,2 B"i'U/ft.h."F.Wh3l i5 il!.' ri:.rriir::!corrdLrctivityn Sl units'/5.3 Tt'e heat qapocily of rl,ater 5 '1 :al/g'U

kJ/kg K, B u/lb"F an'J n callgmol"F?lvhet ia ilic noej cepicily cl!v3i1-'r'rl

5.4 Do prDblem .4 in Himnron D.97:5.5 The value of the Stefair-Soiijrmgnn onsiriri is 0,121 x 108 il fulit2h'Pjl. vv;rjiis the !.'umericl!value n Sl-uritst55 The 20113 ergion o! the GRIM ternlrr iatrr ie ( ,ale vi l i l trevl :20L1 L r

temperitrire of -50'C and 80"G ?t a temperaLxri: f ./tl)';:1. !i it.-. ii:lllv1 irlrshowsa t.irnpciatlre f 6r ic, what s the cn per ' :r tu.c|' i, ;

Do he ourself-assessmentroblemsn Himm .59.Doproblem.13n Himm. .71.

4. 1

4. 2

rurlDo he ollowing roblemsn Himm.2.38p.742.33p.744 kg Na2SOa, kg Na2CO3nd 5 kg NaC/ are dissol'/e dn water. Enoughwater s addedo bringhe inalmasstd120kg. \ryhats themassaciion andmole raction f each ompoundn hesolution?A nitricacidsolution n watei contains 6% HNO3 n a massbasis. The,density of the solution s 1,345g/cm3. Calculate le concentratronf thesolution xpresseds:(a) Mol% HNO3(c) g l.lNob/litreolution(e) Molarjty120 kg CH4,40 kg H2and 30 kg N: are stoEd n a cylinder.What s theaveragemolmassofthegasmixture?

.A solution f NaCl in watercontaini20% NaCf. 15 kg Na2SO4.10H2Ocrystals nd 30 kg CuSOa.Sl-i2Orystals re added o 120kg of ihis soluiion.Themixtures heatedo dissolve ltthe rystals.Calculate(a) Thep.rcentagemass)tlacl, I'la?SOt rd CuSOl n he inal oiution.{r)) Themasspercentage aandO, n he inalsolrrtion.A solution in watercontains mixture f saiis (Na2CO3nd NazSOt anCanalyses s ollows

Water = 70o/ol Salts = ,?0%j'he sallsare present n lte mol ratio l.la2SOy'NarCO3 1.6.Calculate he o/oNazSOa nd % Na2CO3n the solution.

4. 3

(b) kg HNO3/kg 2O(d) kg HNOy'm3olution(0 Normality

, ., 4.t)


21/26

6. 2 Do he ollowing roblemsn Himrnelblau5.22on 0.123.5.26on p. 125.Thepressurn hegawe is6,6psig.5.30on p.126. TheSGof BenzrEs 0,88.5.34onp.128 TheSG of Mercurys 13,6.6.3. A system f twocylindricalanksA andB is llusbated.

Do he ollowing onversions:(a) 10,732 sia.ft3lb ol R to cm3 tm/gmol.K(b) A pressuref 47 t H2Oabs)o psigat sea evel(c) A vacuum f 640mmHg opsiaat sea evelThepressure auge n a compressedir cylinder eads 00psi. What s theabsolute ressuren he cylindern kPa?Atrnosphericressures 700mmHg.A Manometers used.o measure vacuum f 500mm BenzeneSG = 0,88)on a reactor. What s the pressuren the reactor n Paa f the barometricpressures 660mmHg?

5.8

5.10 A "Handigas" j/linderhowsa gaugepressure f 300 kPa n Pretoria. f thecylinder s transpor'{edo Dur ban he gaugereadsonly 285 kPa. Did anappreciable mount of gas leak from the cylinderduring rarisportation?,Atnosphericressuren Pretorias 87 kPa.5.11 Convert12,44 l cm/s t "C to the generally cceptedSl units or thermai. conducti\./ity.5.12 Ttte dealgasconstants reportedo havea value

R = 8,956 1041b.ft2llbmol"cCcnveil liis o ft3 psi)/lbmolR

TankA: Dia- 2 mTankB Dia=1m

tl . l mtAB

The wo anksareconnected y means f a pipeas shown-The anks ontainwatertoa depth f 1.1m.Oil SG= 0,72) s nowpumped n op of the water n ankA. The oil dcsnotmixwith he vater. The oil willcausewater o be transbned ton aank tank

tII1, 5mI,5mI

TUT 6.61 For the configuraiion hown in the

sketch,calculatehegaugc ress-ureatA and thc absolute ressureai d. Valve C i! shrt. Atmos-phsiic prcssures 98,5kPa andthedenslty f heoil s 50 b/ft3.

--*-l -

What volume (litr.e5) f oil can be purnped

6.4 A tank with a manometersystem E shown.The tank containsair, oil and water as indi-cated (The density of the air is negligihlecompad to that ofthe liquids).The manometer contains liquid [,'las Indi-catcd. The space above the manorneterliq'rid n the long eg, is filledwiih oil. Intheshort eg, this space s fill ed with water.The prss,Jre gaug shows a reading ofi30 kPag in the air space above the oil ,tevet.th? icllowing nform;rtign efF {o dimen-sions$hown cn th? sketch :a : .1,5 n b - 0,3 nr h= ?.72mmSG of oil = o,8 SG of waier - 1,00SG of the mancmekr liquid M) = 2,5x: waterd?pth y=orldepth x+y:4,0m

into tank A before icnk 3 ',fi:l

45.

oit -+ lta/z

X bt fr'll')'^l rtl'"


22/26

Vvhats hedepth f water n he ank?Howwill hemanometereadingh)be nfluenpedf hepressurebovethe oil n he ank s ncreasedrom130kPag o 150kPag?TUT 7

In t\^,o different te)ds the following equations are given for thethermodynamicelationship.(d . [* ] - -- ! lo ' l\d P .c . \df )

Wih V the molar olume;c the molar eatcapacityiT{h6temperaturei heDressurend 4 dimensionlessonstant,Testbothequationso determine hich ne s correct.

7.2 Reynolds umberRe) is adimensionlessroupwhlch s used ocharacteraseflo\4/.t iscalculatedrcm

;ir" '/ ''' 'p = density f hesolutionn kgln3Thisequation anbe used o calculatehe heat ransfer oefficient whichenables s o determineheheat low rom,o)1zl : v ia ' . 4toq-/ - ln t . Q = Heatf lowinwA = hA(Al A = Heat ransfer rea n m200aC.b1A aT = TemperatureifieienceA stirred eactors ull of oil andheat ransferakesplaceo tireoil. ahe lirjerdiameters 600 mmand t rotates t20 rpm. Calculatehe hest llw i.r he otl fthe heat ransfer f ea s 30 ft2,and AT = 100"C. Data or the ,rit ir ai

k = 0,097BTU/ft "F;P = 0,8glcms

(?)[?]""'=w,l!\L)""'with h = heat ransfercoefficientn Wm'KD = tankdiameterk = thermal onductiviiyn Wmklp = heat apacityn J/kgKp = heat apacityn J/kgKI = viscosityi = JLi"iJr ottr.r""1ir",aan- (41-)N = rateof rotation f hestirrern 2!!!l^ ZP4l']=

followsp = 10Poise;cp - 0, 8BTU/lb'l-';

'ttilttt)iKqf t ,' . knv ,,1',rlql r,'lt,'ttnltt'l lK'

. ' / .

D ,' ri ro;

same.1 , f t. ( dr\. T (dY)|(D l 4 l - l = =l-; l\df ) ' r:. \at ,l

cP .(i )(i )

- uDou = flow ateD = diameterp = censityp :: viscoslty

Calculateheheat ransfer lefficient nd he heat low.7.4 Doprcblem .26 n Himrrt. .38and on'rert he orrnuleor dliarl :irpti!iii,:r rst .

-r:u.l.-e8. 1 -ai1?heat rpaciiy fethdnolsgive,r y

cp - a + bt vJith p ngTu/lemol R:l : enrpiriealonsiant= 14,0485b . empii ica!onstant 0,0215315t iir F {Y.B: t is nota ernpeialurelitterenoe)

R{.:v]ljiehis iquaii,tnor Cirect pplic.liior,) (;1, ,t oil,rv,.:Cp = A r' BT $ith C, in Jigrncl K a d T in K,q anrl B are r,ev/gntniricalD,)nst.tnts.

(a) Shc\y hatwithsuitable nitcholceRewillbe dinlensionl ess.(',r) Ina crtain ipeline ith nside ianleter 50mm he io$Jateoi 1/aiers4,948 107 m3/h.Thedensity fwater s 1 g/cmiarrdhevisccsitys 1

Converthed,rta c S.l.units ndcalcuiBteheReynoldsumber.Convail he data tc f.iii3h units 3nd calculatehe ReynoldsnurnbciThe lrllowiirgelationshipfCjtrglglgdCg:jl2llg is valid ci heai ranslern aslined anl( eactor


23/26

Theheatcapacity f a certain.compoundsgivenn kJ/kgK by% = 22,8t + 0,044221 O,0oOO1487t'?ittr in "C

Rewritehisequationo calculateheheatcapacity irectlyn BTU/lb F withTin R.ce = A + BT + CT? in BTU/lbFwithT in"R

The ollowing qua$onor heat ransfer n the shellside of a con denser asobtained mpirically

- _ ^. " lk 'P-g^l\ jDp6r )with h = heat ransfer oefficientn BTU/h f "Fk = thermalcondu-ctivityn BTU/h t "Fp = densityn b/ffg = accelerationue o gravityn Uh2,l = latent eatof evaporationn Bl-U/lbD = tubediametefn t' -l = viscosityn bffth'aT = temBeratureifferencen"F

i = j-,*^t", oJ ntuCalculatehe valueof the heat.transferoefficientn W/m':K ot a condenssiwirich ontains 00 ubesof 38 rnmdiameteff he hermal onduciivirys C,075BTU/h t 'F and the iemperatuie ifferences 20"C. The otherphysicalpropertiesreas ollows

p = 500kg/r.r3;p " 0,'14.P) = 274kJlkg

S.4 'ih,Jfollov,/ingmpirical or.elations kno\,,/nor the determlnaiionf .l heattransfermeffiDieniT I 0'56r , ., o.zgat i l ILp D

h 'j heat transferccefficientn BTL,/h t'?'Fk . mass trdnlier coefficient n b/hI - vistosity in lb/ft hp .. d.nsity in lb/rFu = didusivity n felh ' '

Rewritehisequationo calculate directlyn Wm'ZKf all he eih.?r arhblcsare known n units f heSl system.

lUrl9.1 A solution f NaCl in watercontains 0% NaC1. 45 kg of a salt maxture(containinga2SO4.10H2orystals nd CUSO.!.5F|2Orystals)s added o 120kg of the solution.The mixtures heated o dissolve ll the crystals nd hemixturehencontains ,00o/oa2SOa.

CalculateThemassNa2SO1.1oH:Ond CuSO+.5HzOn he originalsal tdixtrrreThemass raction t each ompoundn he inalsulution.Themass raction u ands in he inal olution.

9.2 100 kg CUSO. is dissolved n 200 kg water and he solltlrn is cooled o yiekl60 kg CuSOa.sH20 rystals.Calculate(a) The composition fthe remainingmother iquor(b) The crystaiyield h kg crystal$/kg riginal olution(c) lf the ciystals are renloved rorn he molh4r iqllci by iiiration, t rs D:ltl'jthat the noist crystalscontain 14,7;nl,Jth(\riquor. What is t!' r!r55scimorher iquorthat pas!1eCllrough he tiltet?

9.3 10% NallCOg soiuticn s fed to an e'/np$rator?t a !at4 of 1{.)'1 g/nrir in,!concentrated soirrilon iDnr ilrc evnpcratDi pa-Aie. into a crysialii?itr :rlli ;,-.cooled to '10'G. H.!'}i rnucll w.rier nusl be evaiofateil cer mintrti 19 3i.!r'r1'rihat /{)% ol the I'laHCU3 /ill cryst,iliiic fiom Fe s,llution. Lt 10't l':e : ri,ilr;liLof N,1HCO3s 8,'15 Nal1OO3/10t)Hro.

9.4 To niee.t r.rlaiir spe0iii';:riiriits !n,inui?dti..ur llixes t',T giue ial,v tjiidilai-, tr)produce i Inixturewiiirlr coiriaifrs16'10 r'nier.Cosipri'.:,,.

(a)(b )(c)

Ri-00/kgR2-eo/kgWhat is the cost Fi.Ebl ilrc ftix.)d glue?

9.5 The solubilltyof anhv,lr,)ui r1r1!io4n wntgi at 2li'i I ls 6?.9 9 ivrr:;Or"1.ir ..ll llH2O. What mass ci i',*r.tor.a,F!2'J-riust .: diss.'i't.td r 1u.l kg ol "r riot L4 ,,ii.ii.]a saturatedsoluiion a'i2ii'C?

il3v,/ naie alRav'/ lraierial i ' i 'r?:?-%


24/26

A gascylinder ontains kmolCaHs as. What s the massYoHz n the CrHegas?.What mass of CzHrgas must be pumped nto he cylindero giveamixture ontaining0o/oarbonmass asis).WhatmassC2Hagas illgive84%e?

TUT 1010.1 The solubility f Ba(NO3), t 90'C is 30,6 9/100g HzOand at 20'C it is8,69/100g HrO. 1200kg h(NO3), is dissolvedn waterat 90'C. 10olo orewater han equireds used. Howmuchwater s used? The solutions cooledto 20'C. What s hemassBa(NO3),rystalsorrned?10.2 A manulacturerf cheese roduces cheese r'hich,n the undried ondit;on,contains 3,7o./ooisture. esells hisproduct t Rl I,00perkg. lle aisodriesilris cheese o yielda productwhichcontains nl y 10yomoisture.The dryingcoet s R95,00 erkg water e moved.Whatmuct he selling rice f hedriedproduct e f hewishes c maintainhesameprofit?b' ->on',/ol /c/t q /*t-//z /r+Vt, '[, lh //a6d -/' t' '//l a*'10.3 A sol ution in watei contains 25o/. Na(jt anC 5ol" NarSOa. 100 kg/h cf thissolution $ fed to an evapoBtor crystallizerwhere 30 i(g/h water s evapgrated.The evaporation akes place lnder v3cuurn at 50"C. Thg crystalswhich areJormeC a mixturc of NaC/ and NsrSO.), are separated iom the mother iquorand th'rn contain 10% mother nuor. I'h-- compcsitign of the mcther iquor !s0?i, Ne:SOa and 24% NaAt. The ,vet crysbls are iried to remove all ihewnier. CaicLrlale ne prodlction r3tl ;rld lne crmpcsition of the ci.iedcrystelprodrct.ll.4 Ihe ;() of anhydrouli sodium sulphate NalSC)") s 2,70. The SG of sodiur'l(hicirde (NaCi) is.i.f,L$. The bulk density !f Na2SL)..1'H?Ocrystals is13?o g/m'. h.ja/, brrA ;aige tank contains 8460 it.es nf q/10ol6 olution of NaC in !,,ater/ Ti,r SGoi the solutioD s 1.C7. Crvstals of Na,SO".10 H,O m'Jstbe dissolvecin thirscl:'lion ic increise tnAfci,u' n ccrien@fapprc xim=relylo, QLlla) lne onty.igihoo avallable or nt3suiing the cqslals. s a bucker' Tlrevolcrne f th

tL,:.::et .i?5 l:ircs.lioJ, ffaily bLrckel.i i l.ja2SCr.10H2G',rystrls rust b9 dissol\'sd'1

'10.5 The uedF to a distillationolurnn as he ollowing omposition,ivenn orderof ascendingoiling oints

Ethylene, rHaEthane,CaHoPropylene, rHoPropatts, gHo

n-Pentane,CsH12

mgl %2, 43, 05, 015,025,0

15,0The (iistillerte does nct contirin componetltwith rcilinq ointhiqherii;n isobutane,nd hebotfoms roductE does |.lr)t ,Jntain compo-nent with oilingpointov;ef hanpropane. heisolJrtane oncenirati(liln ihe distiiletes 5%ani tlrc pfopane .'nceniuiioi]n theboitarnss0,8%. All hes(;pcircntagesrEJn a inolbasis.calcul3te ii tlrebasis f 1C0 molfll i the feed

lh..i rir{rl '1,cctnpositionof D and iJ.'l-!",iiflclv r3teJr f D irnli B.

10e Sdi t i s we: ihrci t i , r i r crud. t t l rJY. r , :x; r l J hD 1)( rd. :1! l l ' ' l j i i ' ' l . i4 i r !1; i , ' : ii ir 'r3t ic . i : i ( i t j l i ! , : , i , rdr ( ; i l : t , : r?! ) . P ; r t L, i the l r i t di5, ia. l ' , ' : : : ri th? ??f : i : i , r . : j i , :mi l lu ic i : n, l4 ' . r , j : i r i ;eo'1r?ie. I i i . r ,J i ! l ra! i r t ion ccn: i ' r i : i r i l j ,i l ; l t i . . : ! i i : , r - . r ' iwila;-) i:: (1i:tn llT.lt tlil irip ind tl.41!sed' vt;l3ll y.,,irr--,..:-iir.;ifi:ili1 1r'r ri i -. jDr:) ijr,r'.:irri if ihr r:rLt,J; .J;l 4itaiii:i .l% r):lii an.i llrl i..;rr(i ir!;il *rii:- ;ir,)r3it, ...,:riti;r ihe si!lt )unt;jjri'-nf!c!1 ,'l i 'lil "degrrlieC" rdl'l

10 r' !'. lri:,(ii,,'.- l lJ6..iil]r cry,)trl: d.J f i:r,la),i poi?lEr c*r'iai[i i/iii lir,r-]i ) , ' , ,qr,J Et-ria. :f 400 It ui\ire n:r)il'ire i3 chaigt'l to a re;r.:mr. t''iriri i' ii:: , : '-l.it2(;i:)ir !ll ilt4 ie:i ,tor?l-iic .;i,y'rii;rl ',i-(il,!: iri lirxded iito ll!-r."ei,l.rr by riieir'r:: o; 1 ,,rl..i i; I i , rv. lur i i i r i j1, .1! i l rei i . i ! . lw oi i iNy! - j , ( 'k : i5 fui l . f ! : i ) ! . r1: ir r : . i i : - , ; : . i : , . , r ' r i , 'L,ihe 4f0 iiaj rlrd,q:"/ '!-l'r?lial ilr l.l.zf:C3 is .1,:,i 1,rii :r::r SJ.ir)i ir;.'i 'l ':)r ;: ..Tir! i:i-iik'iqn-riir -f;o,) tiyst?l n'lixtirre s I Jirr) k(a,"n

(a,(b)

*fodk*/1.Auqrrq[ar(. riII. i. I

rn ,fqni:#,,lE3:ll;;" ' fee( ^9fl^i,il *lpttYII


25/26

TemperatureC 20 30 40 60 80 10 0Solubility(q Na,CO" er100q H,O) 21,5 48.5 46.4 45,8

TU T1111.1 The able hows olubilityataorsodiumarbonaGnwater.

Inierpolateinearlyf required.(a) What mass of J\larCO3.1o2O must be dissolved n 100kg ofwaterto give a saturatedolution t 67'C?

\^/hat s the normality f this solution? he densitvoi the soluiion s1,32 gll and tsviscositys 98 cP.(b) A mixtureof Na2CO3.102Ocrystalsand mother iquor at 2C"C) savailable t a flow rate of 100 kg/h. The mixtur cnsists l 600/0NarCO3.102O rystals nd40% saiuratedoluticn f Na2CO3n v,/atei.

The mixtures passedhrough separatoro separatehe motheriquorfronr he crystals.The separatedmother iquoicontains o crysials uilir? crystal roducts wetwith mother iquor, tillconiaining 5%mcil]elliquor, Thiswet mixture s passed hrougha driei and aii the lvatei(irciudinghewaterol crystallisation)s evapoBted.i Drawand abet block lowdiagram f heprocss.? At wh3t aie s hedryNa2CO3roduced?

11.5 Theheatcapacity f water apourHrO)sgivenby= 8,22+ 1,5x 101-l 1,34 1OT2 withCp n cal/gmol andT nK

Calculatehe heat apacity f water apour t 250'F n units f kJlkgC.a)(b )

(c )Write he ormula sTo calculate pdirectlyr kJ/kg c if in C is set nto henew ormula.Use he newfcrmulao calculate,! n kJ/kg"C l250'F.

ihe Sior-irce flliraia v,,hir)h cl9"9 lirr'ru'-lili:i: irlli:-, ii iiieil io,l!;ehii:ii::r i!! ::!lvaDorJtor. Tiraj |)iod!!{ itrir'trr': ,,.lvtn lir.,'c'Jai),rr'licl45 .l rr;tr-lillljc :jiii ii;lrar 94,3").( a) Drawand Bbd;rbiock0aw.riftldi"morllidafio{."e5s.

ijalculate tirr! r;i:)j.r fii:i;ar'_ii: j li.l2Par. in tiF ciihidi:rte i;tlrii:l; .,:NaHaPCr.'.tHr(r ,:lnd also iirr naii-. frt(:tiotr i'loil:r-'i-r.r irr iii{':i,it'Jj]sol l4icr :s at : r. l a, : f l a t' ,j

Ii{: :i 3-.j!t.i :r': :.:lr ltrr ijr',v rqrl:rii!;, i.:r ri.tltrjlalr: ile ctrnpiEle r?niriliirxib::ance ii. iirii ;r{,i }i,,..

cp = a+bt+cf

'i1.2 T!1r ailowqg bble shows lensrty ata for liquidwater

(:) 'lhe 3e of ethyl a!.ohoJ?r 20'C ii 0,7893. What is the densityoi eih\'iiti.-oitrl at 20'C ir lbJt'?{b) Thi. lii:io r/5D.i f a certain oil s r,eported s 0,7281. Whai is the cinsityci th,,' : i l ii k3/L)at 6!"F?

i ' .- i i ) , prcblem'11.10r l l i r r | i 'n..335.

1!.4 ln vJhat atio (rnass )asis) must Nac,ii (soliC)ani NarSOacrystalsLe rnixei iflhe mix.roprr)ilct n'rrrsi ontain60% Na (nrassbasis).

wil,--ft4wt\

---7 e,rat4Z TfArsalo

11,6 The following olubifiryata ate avrilable or sodiumdihydrogen hosphate(NaHzPOr)nwater.frernperat'rrecT- 60 -T 70 -I- -Bt--T_- 90 T 1o0 - |Fa,6i|it-----L(g! 00stqater) _l-__1_4i1_L_ .eQj!r.-i __2w_3_)__453__l 246,4 --iInterpolate./Extrapolai,rnearlyi requif.tl.A rawmatgdals avarlable,orrsisiiricti a rnixtlreol puresodium ilrl/drog,:fiphcsphate i-hyCnte rylrtalsNiliilir, 1.2ll2o) and sllica ijio2). A sampic ithe mixturewas analysed nd th{-'l .rrjoiatoryepotted l i l t : i cont3irts'14,0570i.

l4 tA soiutio0 ot sodiuii, .jlliydrogen plo:rih:ie (Nn!-!?FrOi )c !',/aief s Dicllr.j,r:idfrom this raw mateij?lihy {:iss')l}in0 i!,] itydr-nted:i:/sjai3n wafcr'at 58'il. -i1.}!lpgrcnr r ore $'atr)i than iireorFtii:3llvr:iiujlll is l!;e.i. iN:lie ihii ii:x !ilC?doe:rn.rt dirsoive). 1l,s SiO, i|r iil{:rl iii.ni}ri oif and iemoved lli :'r lil.,i Dri'i::,Thefi l tercakec.nh;nionlyts7%Sli ' j r- l i i . .:! ja i . i i rc,.) i) :: inOti i i ia ij - .-SldJ| lahln

i D.l

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1'1.7 Benzene nd olueneorm dealmixtures.TheSG of benzenes 0,879and heSGoitoluene s 0,866. A mixture f benzeneand toluene 40oloenzene)s in a large.container. glass phereloats n hecon-tainer.What nrass f solid lasspelletsmust beput insideheglass phereo ensure hatexactlyhalf thesphere iameters underthe iquid?Thedjameter f heglass pheres 110mm;irsmass s 182 . TheSG of heglass el-lets s 2,25and the bulkdensity fthepel-lets s 51.S b/'ft3.The olume fa spheres grven yV = 1o f where r is he.adi us fthesphere.

11.8 Strawbeniesontain15%solidsand 85%water. To makestrawbetryam,ci,Jshed trawberriesndsugarare r'lixed n a 45:55 atioand he mixtur+sh.raiedo evaporate aterulltil he residuc ontains ne hirdwaier. Draw ndlebet biock lolvdiagramor hisp|ciess.(:;rlljulaiehe ccinpleiemassbalanceor he pioduction f 10C g strawberryjarn

I 1.9 A slliry is a nrl\1i]re i a s olid anda liquid. 4 s lurry can he conien-traied i|l a thickencr, where lhesolidsaie allowed o seltle, yieid^i,rg a concentraiedunderflowandar, orerflow (fhe oveifio'riwill ol.iainonlysrnail udnti t iesl solids)!n iFj proces! iltustrated,_F c.!i-i:rins 100k9 Sio' pe i m" slurry(ih.J riurry is a ,11ixlJre,,i ancrno waier). F is chargcd ai araie of 200 i(g SiC?pet hour Ihsthi'-;iener auses 0% cf ,hewatel

The low ateofthe eed n m'slurry/hour.The low ateofthewater n he eed n m3water/hThe low ateoftheSiO2n heunderiown kg Si02/hTheconcentrationf SiO2n heoverflown kgsio2perm' of slrrry'NqIE: In hisSiOz/HzOlurryt canbeassumedhevolume fthe sllrryequalshevolume f the SiO, plus he volume f the water ldealmixing). 'he SG of silica s2.642.

(a)(b )(c)(d)

i:rirariledo be rcccvered n the undedlcv!. Tl:e underilowconi?lns4C0kl] SiO,p.lr Ir" slurry. Calculate:hr follow;ng.

X &Ly {on'1ifqlqloilt;k

CIR 113 Study Guide

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