Chemical Engineering Thermodynamics II

Transcription

1 Chemcal Engneerng hermdynamcs II (CHE 303 Curse Ntes).K. Nguyen Chemcal and Materals Engneerng Cal ly mna (Wnter 009)

2 Cntents Chapter : Intrductn. Basc Defntns -. rperty -.3 Unts -3.4 ressure -4.5 emperature -6.6 Energy Balance -7 Example.6-: Gas n a pstn-cylnder system -8 Example.6-: Heat ransfer thrugh a tube -0 Chapter : hermdynamc rperty Relatnshps. ype f hermdynamc rpertes - Example.-: Electrlyss f water -3 Example.-: Vltage f a hydrgen fuel cell -4. Fundamental rperty Relatns -5 Example.-: Fndng the saturatn pressure -8.3 Equatns f State -.3- he Vral Equatn f State - Example.3-: Estmate the tank pressure -.3- he Van de Walls Equatn f State -3 Example.3-: Expansn wrk wth Van de Walls EOS Save-Redlck-Kwng (SRK) Equatn -7 Example.3-3: Gas ressure wth SRK equatn -7 Example.3-4: Vlume calculatn wth SRK equatn -8.4 rpertes Evaluatns - Example.4-: Exhaust temperature f a turbne - Example.4-: Change n temperature wth respect t pressure -4 Example.4-3: Estmatn f thermdynamc prperty -6 Example.4-4: Heat requred t heat a gas -7 Chapter 3: hase Equlbra 3. hase and ure Substance 3-3. hase Behavr 3-4 Example 3.-: Specfc vlume frm data Intrductn t hase Equlbrum ure Speces hase Equlbrum Gbbs Free Energy as a Crtern fr Chemcal Equlbrum he Chemcal tental Vapr Lqud hase Equlbrum 3-6 Example 3.4-: Hrsepwer f a cmpressr 3-8 Example 3.4-: w-level refrgeratn system he Clapeyrn Equatn 3-3 Example 3.4-3: rple pnt estmatn Refrgeratn 3-30

3 Example 3.5-: A vapr cmpressn refrgeratn cycle 3-3 Example 3.5-: A refrgeratn cycle usng refrgerant 34a 3-3 Example 3.5-3: A refrgeratn cycle usng refrgerant 34a 3-34 Example 3.5-4: A refrgeratn cycle wth nndeal cmpressr artal Mlar rpertes 3-36 Example 3.6-: artal mlar vlume 3-37 Example 3.6-: Enthalpy f mxng 3-38 Example 3.6-3: Enthalpy-cncentratn chart 3-39 Chapter 4: rncple f hase Equlbrum II 4. he hase Rule 4- Example 4.-: he degree f freedm 4- Example 4.-: Degree f freedm fr vapr-lqud equlbrum 4-4. he Fugacty 4-3 Example 4.-: Fugacty f lqud water 4-5 Example 4.-: Fugacty usng Van der Waals EOS 4-6 Example 4.-3: Fugacty f steam 4-0 Example 4.-4: Change n Gbbs energy Fugacty f Speces n a Gas Mxture 4-5 Example 4.3-: Fugacty usng eng-rbnsn EOS Fugacty n the Lqud hase 4-9:0 Chapter 5: Appled hase Equlbrum 5. Vapr-Lqud Equlbrum fr Ideal Systems 5- Example 5.-: Bubble pnt temperature calculatn 5-3 Example 5.-: Bubble pnt pressure calculatn 5-5 Example 5.-3: Dew pnt temperature calculatn 5-6 Example 5.-4: Dew pnt pressure calculatn 5-8 Example 5.-5: xy dagram fr benzene-tluene mxture 5-9 Example 5.-6: Dew pnt pressure usng K-values 5-5. Isthermal Flash Calculatn Usng K-values 5-5 Example 5.-: Isthermal flash wth V/F specfed 5-8 Example 5.-: Isthermal flash wth specfed Vapr-Lqud Equlbrum wth Nn-deal Lqud 5-3 Example 5.3-: Bubble pnt pressure wth Van Laar mdel 5-5 Example 5.3-: Bubble pnt temperature wth Wlsn mdel 5-6 Example 5.3-3: Dew pnt temperature wth Wlsn mdel 5-8 Example 5.3-4: xy dagram wth Wlsn mdel Fttng Actvty Ceffcent Mdels wth VLE Data 5-35 Example 5.4-: Evaluatn f Margules and Van Laar parameters 5-37 Example 5.4-: Evaluatn f Wlsn parameters Azetrpes 5-45 Example 5.5-: Evaluatn f Wlsn parameters Estmatn f Actvty Ceffcents 5-5 Example 5.6-: Acetne mle fractn n a system wth ar 5-5 Example 5.6-: Slublty f ethane n n-heptanl hase Behavr n artally Mscble Systems 5-55

4 Example 5.7-: Lqud mxture n tw separate phases 5-56 Example 5.7-: Lqud mxture n a sngle phase 5-56 Example 5.7-3: LLE fr a bnary mxture Vapr-Lqud-Lqud Equlbrum: VLLE 5-63 Example 5.8-: Cmpstn and pressure fr a VLLE system 5-64 Example 5.8-: VLLE calculatns 5-66 Example 5.8-3: VLLE estmatn 5-68 Example 5.8-4: VLLE fr water and hydrcarbn he hermdynamcs f Osmss 5-73 Example 5.9-: Estmatn f the VC mlecular weght 5-76 Example 5.9-: Dsscatn f NaCl mlecules Dstrbutn f a Slute Between Lqud hases a Slublty f a Sld n a Lqud hase 5-79 Example 5.0-: Slublty f a Drug a Dstrbutn f a Slute Between Lqud hases 5-84 Example 5.0-: Drug Extractn frm the Aqueus hase 5-87 Example 5.0-3: urfcatn f an Antbtc c Sngle-Stage Equlbrum Extractn 5-89 Example 5.0-4: Drug Extractn frm the Aqueus Stream 5-90:9 Chapter 6: Chemcal Equlbrum 6. Intrductn 6- Example 6.-: Extent f reactn and fractnal cnversn 6- Example 6.-: me t reach 90% cnversn n a batch reactr Chemcal Reactn and Gbbs Energy he Cndtn f Equlbrum fr a Chemcal Reactn Calculatn f Equlbrum Cnstant frm Data 6- Example 6.4-: Equlbrum cnstant fr methanl reactn 6- Example 6.4-: Equlbrum cnstant fr ntrgen xde Varatn f Equlbrum Cnstant wth emperature 6-4 Example 6.5-: Methanl reactn at 60 C 6-7 Example 6.5-: yrlyss f methanl Hmgeneus Gas hase Reactn 6-0 Example 6.6-: Decmpstn f Ethane 6- Example 6.6-: artal pressure f mnatmc hydrgen Hetergeneus Reactn 6-30 Example 6.7-: Actvty f water 6-3 Example 6.7-: Equlbrum cnversn fr smerzatn 6-3 Example 6.7-3: Dsscatn f CaCO 3 (s) hermdynamcs f ack Cementatn Equlbrum n Electrchemcal Systems 6-37 Example 6.9-: Cpper etchng Cmplex Chemcal Equlbrum 6-4 Example 6.0-: Number f ndependent reactns 6-4 Example 6.0-: Number f ndependent reactns 6-43

5 Appendx Appendx A: Slvng Algebrac Equatns A. he Newtn-Raphsn Methd A- Example A.-: Newtn-Raphsn methd fr a rt n [, ] A- A. Newtn s Methd fr Systems f Nnlnear Algebrac Equatns A-3 Example A.-: Newtn methd fr 3 nnlnear equatns A-4 Slvng set f nnlnear equatns wth Excel A-6 Appendx B: Curve Fttng B. Nnlnear Curve Fttng B- Example B.-: Ft the functn (t; ε, h) ε( e ht ) t the data B-4 Appendx C: rcess Smulatr (rvsn) Appendx D: revus Exams Quz D- Quz D-3 Quz 3 D-7 Quz 4 D-9 Quz 5 D- Answer t Quzzes v

6 Chapter Intrductn. Basc Defntns hermdynamcs s the scence that seeks t predct the amunt f energy needed t brng abut a change f state f a system frm ne equlbrum state t anther. Whle thermdynamcs tells us nthng abut the mechansms f energy transfer, rates f change, and tme asscated wth a system changng frm ne equlbrum state t anther, t s stll the lynch-pn that allw us t answer these questns. Defntn f 'heat': Heat s energy n transt slely as a result f a temperature dfference. Defntn f 'wrk': Wrk s energy exchange between system and surrundngs due t any phenmenn except a temperature dfference. Defntn f 'temperature': emperature s a measure f the mean knetc energy f mlecules. Abslute zer (0 K) s a state f cmplete mtnless f mlecules. 'Rate': 'Rate' mples an element f speed, hw fast an event happens, and tme. 'System': In thermdynamcs, the unverse can be dvded nt tw parts. One part s the system, the ther part s the rest f the unverse called the surrundngs. System can be classfed as () slated system where n mass r energy s transferred acrss the system bundares, () clsed system (system) where nly energy s transferred acrss the system bundares, r (3) pen system (cntrl vlume) where mass and energy can be transferred acrss the system bundares. A system s any desgnated regn f a cntnuum f fxed mass. he bundares f a system may be defrmable but they always enclse the same mass. Surrundngs Bundary System ` Fgure. Schematc dagram f the "unverse", shwng a system and the surrundngs. -

7 'Cntrl vlume': A 'cntrl vlume' s als any desgnated regn f a cntnuum except that t may permt matter t crss ts bundares. If the bundares f a cntrl vlume are such that matter may nt enter r leave the cntrl vlume, the cntrl vlume s dentcal t a system. In these respects, a 'system' s a subset f a 'cntrl vlume'. 'Equlbrum': 'Equlbrum' means that there are n spatal dfferences n the varables that descrbe the cndtn f the system, als called the 'state' f a system, such as ts pressure, temperature, vlume, and mass (,, V, m), and that any changes whch ccur d s nfntesmally slwly. he laws f thermdynamcs are applcable nly t equlbrum states whch means that the state des nt really change sgnfcantly wth tme, dfferences n varables between the state f a system and ts surrundngs are f nfntesmal magntude and that wthn the system tself there are n spatal varatns f the varables that determne ts state. Usng thermdynamcs, we can predct the amunt f energy needed t change a system frm an equlbrum state t anther. Fr example t wll take abut 75 kj t change kg f ar at 5 C and atm t 00 C and atm. It wll take much mre energy, abut 57 kj, t change kg f water at 00 C and atm t water vapr (steam) at the same temperature and pressure. State 75 kj Ar, atm State Ar, atm requred 00 C 5 C kg kg Water atm 00 C Steam atm State kg 57 kj 00 C requred kg State. rperty Fgure. Energy requred changng ar r water frm state t state. A prperty s a macrscpc characterstc f a system such as pressure, temperature, vlume, and mass. At a gven state each prperty has a defnte value ndependent f hw the system arrved at that state. he prpertes f ar n state shwn n Fgure. are: pressure at atm, temperature at 5 C, and mass f kg. A prperty can be classfed as extensve r ntensve. An extensve prperty depends n the sze f the system whle an ntensve prperty s ndependent n the sze f the system. Cnsder systems () and () shwn n Fgure. bth at 00 C and atm cntanng and 5 kg f steam, respectvely. -

8 System () 00 C atm kg System () 00 C atm 5 kg Fgure. Example f ntensve and extensve prpertes. emperature, pressure, and specfc vlume f bth systems are ntensve prpertes. tal mass and ttal vlume f each system are extensve prperty. At 00 C and atm, the specfc vlume v f each system s.674 m 3 /kg. he mass f system () s m kg and that f system () s m 5 kg. he ttal vlume f system () s V m v ( kg)(.674 m 3 /kg) m 3. he ttal vlume f system () s V m v (5 kg)(.674 m 3 /kg) 8.37 m 3. An ntensve prperty mght be btaned frm an extensve prpery by dvdng the extensve prperty by the mass f the system..3 Unts he SI unts (Système Internatnal d'untès, translated Internal System f Unts) are used n ths text. It happens that seven prmary quanttes are needed t cmpletely descrbe all natural phenmena. he decsn as t whch quanttes are prmary s arbtrary. he unts f the prmary quanttes and ther symbls are lsted n able.3- and are defned arbtrarly as fllws: Meter: the length f the trajectry traveled by lght n a vacuum per /99,79,458 s, Klgram: the mass f the platnum cylnder depsted at the Internatnal Offce fr Weghts and Measures, Sèvres, France, Secnd: 9,9,63,770 tmes the perd f radatn n energy level transtns n the fne spectral structure f 33 Cs, Kelvn: /73.6 f the trple pnt temperature f water wth naturally ccurrng amunts f H and O stpes, Amperes: the current whch, n passng thrugh tw parallel nfnte cnductng wres f neglgble crss sectn, separated by m and n vacuum, nduces a frce (per unt length) f 0-7 N/m, Mle: the amunt f a matter cntanng the number f partcles equal t the number f atms n 0.0 kg f the pure stpe C, Candela: the amunt f perpendcular lght (lumnsty) f / m f the surface f an abslute black bdy at the meltng temperature f platnum and a pressure f 0,35 a. -3

9 able.3- he seven prmary quanttes and ther unts n SI rmary quantty Length Mass me emperature Electrc current Amunt f matter Amunt f lght Unt Meter (m) Klgram (kg) Secnd (s) Kelvn (K) Ampere (A) Mle (ml) Candela (cd) Several f the derved quanttes wth unts are lsted n able.3-. A derved unt s a quantty expressed n terms f a prduct and/r qutent f tw r mre prmary unts. able.3- he derved quanttes and ther unts n SI Derved quantty C p, specfc heat capacty E, energy F, frce k, thermal cnductvty p, pressure q, heat transfer rate q", heat flux q, heat generatn rate per unt vlume μ, vscsty ρ, densty Unt J/kg K J N m, jule N kg m/s, newtn W/m K a N/m, pascal W J/s kg m /s 3, watt W/m J/s m W/m 3 N/m kg/s m kg/m 3.4 ressure Any frce actng n a surface cnssts f a cmpnent perpendcular t the surface and a cmpnent parallel t the surface. hese tw cmpnents are called nrmal frce and shear frce as shwn n Fgure.4-. ressure s defned as a nrmal frce per unt area n whch the frce acts. he SI pressure unt, N/m, s called a pascal (a). ressure at any pnt s a flud s the same n any drectn. Nrmal frce Nrmal stress Nrmal frce/a F n Area ( A) F s Shear frce Fgure.4- Nrmal and parallel cmpnents f a frce n a surface. -4

10 Cnsder a hle n the wall f a tank r a ppe as shwn n Fgure.4-. he flud pressure p may be defned as the rat F/A, where F s the mnmum frce that wuld have t exerted n a frctnless plug n the hle t keep the flud frm emergng A(m ) A(m ) F(N) Flud F(N) (N/m ) (N/m ) Flud flwng thrugh a ppe Fgure.4- Flud pressure n a tank and a ppe. he pressure at a gven pstn measured relatve t abslute zer pressure r abslute vacuum s called the abslute pressure. Mst pressure-measurng devces are calbrated t read zer n the atmsphere as shwn n Fgure.4-3. hese pressure gages ndcate the dfference between the abslute pressure and the lcal atmspherc pressure. ressures belw atmspherc pressure are called vacuum pressures and are measured by vacuum gages that ndcate the dfference between the atmspherc pressure and the abslute pressure. Abslute, gage, and vacuum pressures are all pstve quanttes and are related t each ther by gage abs atm vac atm abs atm vac ka 0 atm gage atm abs abs atm abs 0 abs 0 Fgure.4-3 Abslute, gage, and vacuum pressures. w cmmn pressure unts are the bar and standard atmsphere: bar 0 5 a 0. Mpa 00 ka atm 0,35 a 0.35 ka.035 bar ps R. M. Felder and R. W. Russeau, Elementary rncples f Chemcal rcesses, Wley, 000, p

11 .5 emperature When tw bjects are brught nt cntact and slated frm the surrundng, energy tends t mve spntaneusly frm ne t the ther. he bject that gves up energy s at a hgher temperature, and the bject that receves energy s at a lwer temperature. We wuld be able t bserve that the electrcal resstance f the warmer bject decreases wth tme, and that f the clder blck ncreases wth tme; eventually there wuld be n change n the electrcal resstances f these bjects. he tw bjects are then n thermal equlburm. hey are at the same emperature. We culd then defne temperature as a measure f the tendency f an bject t spntaneusly gve up energy t ts surrundngs. Any bject wth at least ne measurable prperty that changes as ts temperature changes can be used as a thermmeter. Mst thermmeters perate n the prncple f thermal expansn: Materals tend t ccupy mre vlume at a gven pressure when they are at a hgher temperature. A mercury thermmeter s just a cnvenent devce fr measurng the vlume f a fxed amunt f mercury. defne actual unts fr temperature we arbtrary assgn 0 t the freezng pnt and 00 t the blng r steam pnt f water. We then mark these tw pnts n ur mercury thermmeter, measure ff a hundred equally spaced ntervals n between, and declare that ths thermmeter nw measures temperature n the Celsus (r centgrade) scale, by defntn. he Kelvn scale s an abslute temperature scale that measures temperature frm abslute zer nstead f frm the freezng pnt f water. he relatnshp between these tw temperature scales s gven by ( C) (K) 73.5 (.6-) By defntn, the Rankn scale, s related t the Kelvn scale by a factr f.8: ( R).8(K) (.6-) A degree f the same sze as that n the Rankne scale s used n the Fahrenhet scale but the zer pnt s shfted accrdng t the relatn ( F) ( R) (.6-3) Frm equatns (.6-), (.6-), and (.6-3) the Fahrenhet scale can be related t the Celsus scale by ( F).8( C) + 3 Many f the equatns f thermdynamcs are crrect nly when yu measure temperature n the abslute scale, Kelvn r Rankne. here wll be n prblem n usng the Celcus and Fahrenhet scales when the dfference between tw temperatures s needed. -6

12 .6 Energy Balance Energy balance s the crnerstne f heat transfer analyss. he frst law f thermdynamcs s the cnservatn f energy, whch states that energy s nether created nr destryed. he frst law can be wrtten fr a system as δq δw + de where δq heat transfer between the system and the surrundngs δw wrk exchanged between the system and the surrundngs de accumulated energy f the system he frst law pstulates the exstence f a "functn f state" called the accumulated energy such that fr an adabatc system (δq 0) the wrk utput s balanced by a reductn n the accumulated energy: de - δw Whle δq and δw are nt themselves a "functn f state", the dfference δq - δw s a functn f state. A quantty s a functn f state when the dfference n ts values between tw states nly depends n the ntal and fnal states and nt n the paths cnnectng these tw states. he accumulated energy E s a state functn s that, [ E (,,...) E(,,...)] a [ E (,,...) E(,,...)] b he dfferental f E s an exact dfferental fr whch the ntegral frm state t state s smply the dfference E,,...) E(,,...). ( de E E(,,...) (,,...) E Heat and wrk are path functns and the dfferentals f heat and wrk, δq and δw, respectvely, are nnexact dfferentals s that δ Q Q and δ W W. he Q fllwng example wll shw that wrk s path dependent. W -7

13 Example.6- A gas s cntaned wthn a cylnder and pstn system shwn. Assumng a 'smple' system (expansn and cmpressn wrk nly), calculate the wrk dne by the system n transfrmng frm state, V t state, V. ath a ath b Slutn V V. Cmpute the wrk usng path a wth cnstant vlume fllwed by cnstant pressure δ W, V, V dv, V, V dv +, V, V dv 0 + (V V ) W a. Cmpute the wrk usng path b wth cnstant pressure fllwed by cnstant vlume δ W, V, V dv, V, V dv +, V, V dv (V V ) + 0 W b Clearly W a W b It shuld be nted that a cnstant pressure prcess makes δw a functn f state. dv d(v) Vd δ W d (V ) Vd d(v ) s a functn f state whle Vd s nt a functn f state. Fr cnstant δ W d (V ) V V V r (V V ) A functn f state s ne whse ntegral f a dfferental f tself recvers the rgnal functn, fr example du U ; d ; d (V ) V -8

14 Frst law as a rate equatn Apply the frst law t the system shwn ver tme nterval Δt ΔQ ΔW + ΔE (.6-) Dvde the abve equatn by Δt ΔQ Δt ΔW Δt + ΔE Δt (.6-) We are departng frm the classcal thermdynamc vew that deals wth equlbrum because tme s nt a relevant parameter fr equlbrum systems. ake the lmt f Eq. (.6-) as Δt 0 lmt Δt 0 ΔQ Δt lmt Δt 0 ΔW Δt + lmt Δt 0 ΔE Δt δq δt δw δ t de + dt (.6-3) q W & de + dt (.6-4) where q net heat nput plus heat generated (W) q q n q ut + q gen (.6-5) W & net wrk utput (W) de accumulated energy change (W) dt A cntrl vlume must be defned t apply Eqs. (.6-4) and (.6-5). -9

15 Example Saturated steam at C cndenses n the utsde f a 5-m lng, 4-cm-dameter thn hrzntal cpper tube by clng lqud water that enters the tube at 5 C at an average velcty f 3 m/s and leaves at 45 C. Lqud water densty s 997 kg/m 3, c p f lqud water s 4.8 kj/kg C. (a) Determne the rate f heat transfer t water. (b) If the rate f heat transfer t water s 00 kw, determne the rate f cndensatn f steam Slutn Cndensng steam Q e (a) he rate f heat transfer t water s gven by Q & m& c p ( e ) In ths equatn, the mass flw rate f water s gven by m& ρv vel A tube he heat transfer rate s then m& (997 kg/m 3 )(3 m/s)(π 0.0 m ) kg/s Q & ( kg/s)(4.8 kj/kg C)(45 5) C 34. kw (b) If the rate f heat transfer t water s 00 kw, determne the rate f cndensatn f steam. We need the enthalpy fr saturated lqud and saturated vapr Specfc emp ressure Enthalpy Qualty hase C Ma kj/kg Saturated Vapr Saturated Lqud he rate f heat transfer t water can als be determned frm Q & m& steam m& steam (h g h f ) m& steam Q& h h 00 kj/s kg/s ( ) kj/kg g f -0

16 Chapter hermdynamc Relatns. ypes f hermdynamc rpertes he thermdynamc state f a system can be characterzed by ts prpertes that can be classfed as measured, fundamental, r derved prpertes. We want t develp relatnshps t relate the changes n the fundamental and derved prpertes n terms f the measured prpertes that are drectly accessble frm labratry measurements. Sme f the measured prpertes are, v,, cmpstn, c p, and c v. he small letters are used t dente specfc quanttes fr example v s specfc vlume. he fundamental prpertes are nternal energy u and entrpy s. hese prpertes arrve frm the frst and secnd law f thermdynamcs. he frst law states that energy s cnserved, and the secnd law states that entrpy f the unverse always ncreases. he derved prpertes are defned t facltate the energy balance f systems n whch the cmbnatn f nternal energy and ther prpertes ften ccurs. In pen systems, the mass that crsses the bundary between the surrundngs and the system always cntrbutes t tw terms n the energy balance: nternal energy and flw wrk (v). Fr cnvenent we can defne an enthalpy (h) as h u + v (.-a) In terms f the ttal enthalpy H, we have H U + V (.-b) We can then make an enthalpy balance fr an pen system n whch the flw wrk s ncluded n the enthalpy term. Fgure.- shws a randrp created frm the surrundng super saturated vapr n the atmsphere. Nt nly the energy U f the randrp s needed but als sme addtnal energy, equal t V, s requred t push the atmsphere ut f the way t make rm fr the drp. A randrp wth vlume V and nternal energy U Fgure.- An energy f U + V s requred t create a randrp. Enthalpy s the ttal energy we wuld need, t create a system ut f nthng and put t n an envrnment wth cnstant pressure. Or, f we culd cmpletely annhlate a system, H s -

17 the energy we culd recver: the system s energy plus the wrk dne by the cllapsng atmsphere. Hwever, we usually are nt nterested n the ttal energy needed r the ttal energy that can be recvered frm a system. We wll be mre nterested n the wrk nvlved n a system. Fr sthermal surrundngs, the system can extract heat frm the surrundngs fr free, s the wrk requred t create the system frm nthng s equal t the nternal energy mnus the heat receved. And f we annhlate the system, we generally cannt recver all ts energy as wrk snce we have t dspse f ts entrpy by dumpng sme heat nt the surrundngs. herefre t s mre cnvenent t defne the Helmhltz free energy, A, fr an envrnment at cnstant temperature A U S (.-) A s the energy that must be prvded as wrk f we create the system ut f nthng. he heat extracted frm the surrundngs s ΔS (S f S ) S f where S f s the system fnal entrpy and S the system zer ntal entrpy. If we annhlate a system wth ntal entrpy S, A s the amunt f recvered wrk, snce we have t dump sme heat, equal t S, nt the envrnment t get rd f the system s entrpy. Equatn (.-) ncludes all wrk, even the wrk dne by the system s surrundngs. If the system s n an sthermal and sbarc envrnment, t s mre cnvenent t use the Gbbs free energy G U S + V (.-3) Gbbs free energy s the wrk requred t create a system frm nthng n an envrnment wth cnstant and cnstant temperature. We usually are mre nterested n the change n states f a system rather than ts creatn r annhlatn. We then want t lk at the changes n A and G. he change n A at cnstant temperature s gven by ΔA ΔU ΔS Q + W ΔS (.-4) In ths expressn Q s the heat added and W s the wrk dne n the system. If the prcess s reversble then Q ΔS and the change n A s precsely equal t the wrk dne n the system. If the prcess s rreversble then Q < ΔS and ΔA < W, the change n A s less than the wrk dne n the system. Fr an envrnment wth cnstant and cnstant temperature, the change n G s gven by Fr any prcess we have ΔG ΔU ΔS + ΔV Q + W ΔS + ΔV (.-5) Q ΔS 0 (equal sgn fr reversble prcesses) (.-6) -

18 he wrk term W cnssts f the wrk dne by the envrnment, ΔV, and any ther wrk dne n the system. W ΔV + W ther (.-7) Substtutng equatns (.-6) and (.-7) nt equatn (.-5) we btan ΔG W ther at cnstant, (.-8) Example Determne the electrcal wrk requred t prduce ne mle f hydrgen n the electrlyss f lqud water at 98 K and atm. he chemcal reactn s H O(l) H (g) + 0.5O (g) Data (at 98 K and atm): ΔH 86 kj fr ths reactn, S HO 70 J/ K, S H 3 J/ K, and S O 05 J/ K. Slutn G H S At cnstant we have ΔG ΔH ΔS he change n system entrpy s gven by he change n G s then ΔS S H + 0.5S O S HO (05) J/ K ΔG 86 kj (98 K)(63.5 J/ K) 37 kj hs s the amunt f energy n terms f electrcal wrk requred t prduce ne mle f hydrgen by electrlyss. If we burn ne mle f hydrgen, the amunt f heat we wuld get s 86 kj. If we can cmbne ne mle f hydrgen and half a mle f xygen n a fuel cell t prduce water we can extract 37 kj f electrcal wrk. he dfference ΔH ΔG ΔS 49 kj s the waste heat that must be expel by the fuel cell t get rd f the excess entrpy that was n the gases. herefre the maxmum effcency, ε fuel cell, f the fuel cell s ε fuel cell 37/ hs effcency s hgher than the 40% effcency f electrcal pwer plants. -3

19 Example In a hydrgen fuel cell shwn n Fgure.-, hydrgen and xygen gas pass thrugh prus electrdes and react t frm water. Electrns are released at the ande (negatve electrde) and depsted at the cathde (pstve electrde). he verall reactn s H (g) + 0.5O (g) H O(l) Calculate the vltage f the cell. Data (at 98 K and atm): ΔG 37 kj fr ths reactn. Fgure.- A hydrgen fuel cell Slutn In a hydrgen fuel cell, the steps f the chemcal reactn are Ande Reactn: H > H + + 4e - Cathde Reactn: 0.5O + H + + e - > H O w electrns are pushed thrugh the crcut each tme the full reactn ccurs. he electrcal wrk prduced per electrn s 37 kj/( ) J.3 ev (Nte: ev J) Snce vlt s the vltage needed t gve each electrn ev f energy, s the fuel cell has a vltage f.3 V. Schreder, D. V., An Intrductn t hermal hyscs, Addsn Wesley Lngman, 000-4

20 . Fundamental rperty Relatns he frst law fr a clsed system undergng reversble prcess wth nly V wrk s gven by du δq rev + δw rev Frm the secnd law, δq rev ds, the frst law can be wrtten as du ds pdv (.-) We can wrte the change n nternal energy n terms f tw ndependent varables S and V, that s, U U(S, V): U du S V U ds + V Cmparng equatns (.-) and (.-) we have S dv (.-) U S V U and p V S (.-3) he fur functns U, H, A, and G are cllectvely called thermdynamc ptentals. he expressn represented by U U(S, V) results n the partal dervatves f the thermdynamc ptentals crrespndng t thermdynamc prpertes and p as defned n Eq. (.-3). Whle U can be a functn f any tw prpertes, n ther grupng f ndependent prpertes x and y, U U(x, y), allws us t wrte partal dervatves n terms f thermdynamc prpertes as gven by Eq. (.-3). he three prpertes {U, S, V} frm a fundamental grupng. Smlarly, the change n enthalpy (H U + pv) s gven by dh du + d(pv) ds pdv + pdv +Vdp dh ds + Vdp (.-4) he change n enthalpy n terms f tw ndependent varables S and p s H dh S ds + H p S dp (.-5) Cmparng equatns (.-4) and (.-5) we have H S and V H p S (.-6) -5

21 We have tw mre relatns fr the change n Helmhltz free energy, da, and the change n Gbbs free energy, dg, da Sd pdv (.-7) dg Sd + Vdp (.-8) he fundamental grupngs {A,, V} and {G,, p} result n S A V and p A V (.-9) S G and V G p (.-0) he grupng can be btaned frm the fllwng dagram where there prpertes at a crner frm a fundamental grupng, fr example {A,, V} and {G,, p}. G p H S Great physcsts Have Study Under Very Able eachers A V U hermdynamc prpertes can then be btaned frm the partal dervatve f the thermdynamcs ptentals. emperature can be btaned frm the dervatve f H r U. In bth cases, the varable ppste n the dagram wll be n the denmnatr f the dervatve and the remanng varable belng t the grupng wll be the varable keepng cnstant. G p H G p H S S A V U A V U Usng ths rule, we have U S V H and S Smlarly, p can be expressed n terms f the partal dervatve f thermdynamc ptentals usng the fllwng dagram. Nte the mnus sgn snce V s n the negatve drectn f. -6

22 G p H G p H S S A V U A V U U p V S and p A V Smlarly, we have G p H G p H S S A V U A V U S A V and S G and G p H G p H S S A V U A V U V H p S and V G p When we study phase equlbrum, the Gbbs free energy wll be use frequently t btan the relatns between the measured prpertes. -7

23 Example Fr the Van Der Waals stherm shwn n the fllwng fgure, shw that the saturatn pressure can be determned by lcatng the hrzntal, tw-phase segment f the stherm s that tw equal areas are enclsed between t and the Van de Waals curve. B A A B V Slutn G H G H S S A V U A V U Frm the grupng {G,, }, we have G G(, ), therefre dg G G d + d Sd + Vd Alng an stherm f the equatn f state, d 0, therefre ΔG Vd At the saturatn pressure ΔG G V G L 0, we have A ΔG G V G L Vd V Vd B Vd Vd V A V A Snce area () Vd, and area () +, the saturatn V Vd V B Vd Vd A V pressure can be determned by lcatng the hrzntal, tw-phase segment f the stherm s that tw equal areas are enclsed between t and the Van Der Waals curve. V B V B Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999-8

24 Maxwell Relatns he fundamental prperty relatns are wrtten s that the left-hand sde s the dfferental f a state functn as fllw du ds dv dh ds + Vd da Sd dv dg Sd + Vd (8.3-a) (8.3-b) (8.3-c) (8.3-d) he exact dfferental f a functn z z(x, y) s wrtten as dz z x y dx + z dy Mdx + Ndy (8.3-) y x In ths expressn, we have M z x y and N z y x A useful prperty f exact dfferental s that the rder f partal dfferentatn des nt matter. hat s z y x y x z x y x y (8.3-) M y x N x y (8.3-3) Applyng equatn (8.3-3) t equatn (8.3-a) we get V S S V (8.3-4a) hs relatn s called the Maxell relatn. Smlarly, frm the ther three fundamental prperty relatnshps we have S V S, S V V S, and V (8.3-4b,c,d) -9

25 he Maxwell relatns can be used t calculate entrpy frm V data. he dervatve relatns f equatns (8.3-a) (8.3-d) then enable us t calculate U, H, A, and G. he Maxwell relatns can als be btaned frm the dagram G H S A V U V S S V he sgn f the dervatve s btaned frm multplcatn f the ndvdual sgn f each varable n a grup. Fr VS the sgn s ( )( )(+) +, fr SV the sgn s (+)(+)( ). Smlarly, fr the ther three Maxwell relatns we have G H S A V U S V S G H S A V U S V V G H S A V U S V -0

26 Chapter.3 Equatns f State In the calculatns f energy, enthalpy, and entrpy f a substance we need an accurate representatn f the relatnshp amng pressure, vlume, and temperature. Besdes the tabular and graphcal presentatns f the p-v- relatnshp, analytcal frmulatns, called equatn f state, cnsttute anther way f expressng the p-v- relatnshp. he equatns f state are cnvenent fr perfrmng the mathematcal peratns requred t calculate u, h, s, and ther thermdynamc prpertes. We wll dscuss the vral, Van de Walls, and Save- Redlck-Kwng (SRK) equatn f states..3- he Vral Equatn f State he vral equatn f state can be derved frm the prncple f statcal mechancs t relate the p-v- behavr f a gas t the frces between mlecules. A vral equatn f state expresses the quantty pv as a pwer seres n the nverse f mlar vlume v. R Z pv R + B( ) v C ( ) + v D ( ) + 3 v + (.3-) In ths equatn, B, C, and D are called vral ceffcent and are functns f temperature. Fr a truncated vral equatn wth tw terms we have pv R + B( ) v (.3-) In ths equatn, B() can be estmated frm the fllwng equatns: B() R c p (B 0 + ωb ) (.3-3) c 0.4 B R 0.7, B R In equatn (.3-3), ω s the tzer acentrc factr, whch s a parameter reflectng the gemetry and plarty f a mlecule. In the lmtng case where there are n nteractns between the mlecules, all the vral ceffcents are equal t zer. Eq. (.3-) becmes Z pv R (.3-4) Eq. (.3-4) s the deal gas equatn f state. Example.3- wll llustrate the use f the vral equatn f state wth ne vral ceffcent. -

27 Example A three-lter tank cntans tw gram-mles f ntrgen at 50.8 C. Estmate the tank pressure usng the deal gas equatn f state and then usng the vral equatn f state truncated after the secnd term. akng the secnd estmate t be crrect, calculate the percentage errr that results frm the use f the deal gas equatn at the system cndtns. Data fr ntrgen: c 6. K, p c 33.5 atm, and ω Slutn Frm the deal gas law, v 3.0 L/ ml.5 L/ml, K p deal R v ( ) L atm/ml K (.4 K).50 L/ml atm Frm the truncated vral equatn, pv R + B( ) v R.4/ B R B R 0.7 B B() R c p (B 0 + ωb ) c B() ( ) L atm/ml K (6. K) [ (0.04)( )] 0. L/ml 33.5 atm p R v 0..5 ( ).50 L/ml L atm/ml K (.4 K) (0.953) p 6.96 atm Errr n usng deal gas law ε pdeal p % p Felder R. M., Russeau R. W., Elementary rncples f Chemcal rcesses, Wley, 005, pg. 0 -

28 .3- he Van de Walls Equatn f State We wll use the Van de Walls equatn f state t llustrate the evaluatn f thermdynamc prpertes. Bth the Van de Walls and the SRK equatns f state have tw adjustable cnstants but the Van de Walls equatn s smpler. he Van de Walls equatn f state s R v b a v (.3-5) In ths equatn, the cnstant b accunts fr the fnte vlume ccuped by the mlecules and a the term accunts fr the attractve frces between mlecules. v Fgure.3- Istherms frm the Van der Waals equatn. he Van der Waals parameters a and b can be determned frm the crtcal prpertes snce there s an nflectn pnt at the crtcal stherm as shwn n Fgure.3-. At the crtcal pnt we have v c v c 0 (.3-6) he stherm passng thrugh the crtcal pnt s gven by Rc v b a v he frst and secnd dervatves f wth respect t v are gven by v c c ( ) v c R b a + 3 vc 0 (.3-6a) -3

29 v c c ( ) 3 v c R b 6a 4 vc 0 (.3-6b) We can slve the tw equatns (.3-6a) and (.3-6b) fr the tw unknwns a and b. Multplyng equatn (.3-6a) by and equatn (.3-6b) by ( ν c b) and add them tgether we get 4a v 3 c 6a 4 vc ( ν c b) 0 (.3-7) 4aν c 6aν c + 6ab 0 ν c 3b (.3-8) Substtutng b ν c /3 nt equatn (.3-6a) and slvng fr a gves 9 a ν c R c 8 At the crtcal pnt we have c Rc v b a v c c (.3-9) We can use equatn (.3-9) t slve fr ν c n terms f crtcal temperature and crtcal 9 pressure. Substtutng a ν c R c and b ν c /3 nt equatn (.3-9) we btan 8 c 3 R v 9ν R c c c c 8vc 3 R v c c R 8v c c Slvng fr ν c n terms f c and c we have Hence R ν c 3 c 8c 9 7 a ν c R c 8 64 ( Rc ) c Usng R 8.34 J/(ml K) m 3 bar/(ml K) and fr prpane, c K, c 4.46 bar, we have 7 a 64 ( Rc ) c 7 64 ( ) m 6 bar/ml

30 Example One mle f prpane gas s t be expanded frm 0.00 m 3 t m 3 whle n cntact wth a heatng bath at 00 C. he expansn s nt reversble. he heat extract frm the bath s 600 J. Determne the wrk fr the expansn usng the Van der Waals equatn f state. R v b a v Slutn Frm the frst law we have Δu q + w. Snce the amunt f heat transfer q s gven (600 J/ml), we nly need t evaluate Δu t fnd w. We wrte u as a functn f the ndependent varables and v snce the values f these tw varables are gven n the prblem u u(, v) du u d + v u v dv Snce the prcess ccurs at cnstant, d 0, we have du u v dv We need t express u v n terms f measurable prpertes by usng the dagram G H G H S S A V U A V U u du ds + s v u dv ds dv v s herefre s v du v dv s v dv 3 Kresky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg. -5

31 G H S A V U S V V s Usng the Maxwell relatn we have, the change n nternal energy can nw v v be expressed slely n terms f measurable prpertes: du v dv In terms f mlar quanttes du v d v Frm the Van der Waals equatn: R v b a v v R v b v a v Hence Δu m s m v d v m m a v a d v v m m Usng R 8.34 J/(ml K) m 3 bar/(ml K) and fr prpane, c K, c 4.46 bar, we have 7 a 64 ( Rc ) c 7 64 ( ) m 6 bar/ml herefre Δu m 3 bar/ml -6

32 Δu 93 J/ml he requred wrk fr the expansn s w Δu q J/ml.3-3 Save-Redlck-Kwng (SRK) Equatn he Save-Redlck-Kwng (SRK) equatn belngs t a class f cubc equatns f state because, when expanded, they yeld thrd-degree equatns fr the specfc vlume. he SRK equatn f state s p R v b αa v v ( + b) (.3-0) In ths equatn, the parameter a, b, and α are emprcal functns f the crtcal temperature and pressure, the tzer acentrc factr, and the system temperature. he fllwng crrelatns can be used t estmate these parameters: ( Rc a ) p R b p c c c m ω 0.56ω α + m( ) R Example A gas cylnder wth a vlume f.50 m 3 cntans.00 kml f carbn dxde at 300 K. Use the SRK equatn f state t estmate the gas pressure n atm. Data fr carbn dxde: c 304. K, p c 7.9 atm, and ω Slutn R 300/ R c ( L atm/ml K)(304. K) 4.96 L atm/ml ( Rc a ) p c ( 4.96 L atm/ml ) 7.9 atm L atm/ml 4 Felder R. M., Russeau R. W., Elementary rncples f Chemcal rcesses, Wley, 005, pg. 03-7

33 R b p c c L atm/ml 7.9 atm L/ml m ω 0.56ω α + m( ) R ( ).05 p R v b αa v v ( + b) L atm/ml K (300 K) p ( ) p 9.38 atm ( ) L/ml ( ) ( ).05 (3.654 L atm/ml ).50 L/ml ( ) L/ml Example A stream f prpane at temperature 43 K and pressure p(atm) flws at a rate f 00.0 kml/hr. Use the SRK equatn f state t estmate the vlumetrc flw rate V & flw f the stream fr p 0.7 atm, 7 atm, and 70 atm. In each case, calculate the percentage dfferences between the predctns f the SRK equatn and the deal gas equatn f state. Data fr prpane: c K, p c 4.0 atm, and ω Slutn We frst calculate a, b, and α frm the fllwng expressns: ( Rc a ) p c Rc, b p c m ω 0.56ω, α + m( ) he SRK equatn s wrtten n the frm R f( v ) p R v b + αa v v ( + b) 0 v s then calculated usng Newtn s methd: v v f ( v ) f '( v ) v d v, where 5 Felder R. M., Russeau R. W., Elementary rncples f Chemcal rcesses, Wley, 005, pg. 04-8

34 R f ( v ) ( ) v b α a( v + b) ( + b) v v he ntal value fr v s btaned frm deal gas law: v deal R p. he teratn prcess stps when v /d v s less than he percentage dfference between v SRK and v deal s v deal v 00% v Once v s knwn fr a gven p, the vlumetrc flw rate crrespndng t a mlar flw rate f 00.0 kml/hr s btaned as V & flw (m ml /hr) v (L/ml) kml m L (00 kml/hr) 00 v (L/ml) he calculatns are perfrmed usng the fllwng Matlab prgram: % Example.4-3 c369.9; % K pc4.0; % atm w0.5; % acentrc factr Rg0.0806; % L*atm/(ml*K) 43; % K pnput('p(atm) '); r/c; a0.4747*(rg*c)^/pc; b *(rg*c)/pc; m *w *w^; alfa(+m*(-r^0.5))^; vdealrg*/p;vvdeal; fr :0; fp-rg*/(v-b)+alfa*a/(v*(v+b)); dfrg*/(v-b)^-alfa*a*(*v+b)/(v*(v+b))^; dvf/df; vv-dv; f abs(dv/v)<e-4, break, end end D(vdeal-v)/v*00; Flwrate00*v; fprntf('vdeal %6.f, v(l/ml) %6.f, ercentage Dfference %6.3f\n',vdeal,v,D) fprntf('flw rate (m3/hr) %6.f\n', Flwrate) -9

35 >> ex3d4d5 p(atm).7 vdeal , v(l/ml) , ercentage Dfference 0.37 Flw rate (m3/hr) >> ex3d4d5 p(atm) 7 vdeal 4.959, v(l/ml) 4.775, ercentage Dfference 3.86 Flw rate (m3/hr) >> ex3d4d5 p(atm) 70 vdeal 0.496, v(l/ml) 0.89, ercentage Dfference 7.57 Flw rate (m3/hr) 8.9 he SRK equatn f state (and every ther equatn f state) s tself an apprxmatn. At 43 K and 70 atm, the actual value fr v s L/ml. he percentage errr n the SRK estmate ( v 0.89 L/ml) s %, and that n the deal gas estmate ( v 0.50 L/ml) s 9%. -0

36 Chapter.4 rpertes Evaluatns hs sectn wll shw the use f equatn f state and prperty relatns t evaluate thermdynamc prpertes n several examples. Example rpane at 350 C and 600 cm 3 /ml s expanded n an sentrpc turbne. he exhaust pressure s atmspherc. What s the exhaust temperature? v behavr has been ft t the Van der Waals equatn wth a [(atm cm 6 )/ml ] and b 9 [cm 3 /ml]. Slve ths usng as v as the ndependent varables, that s s s(, v). he Van der Waals equatn s gven as R v b a v he deal gas heat capacty fr prpane s R c p Slutn Snce prpane s expanded n an sentrpc turbne, we can cnstruct a path such that the sum f Δs s zer., v Δs Ideal gas v Δs Δs 0, v Chsng and v as the ndependent varables the change n entrpy s gven by ds s d + v s v dv Q Snce s s v Q v c v 6 Kresky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg. 45 -

37 Usng Maxwell relatn as shwn n the fllwng dagram, we have s v v G H S A V U S V V herefre ds c v d + dv v Frm the Van der Waals equatn: R v b va v v R b Fr an sentrpc prcess, we have c Δs d + v v v R dv 0 v b c v c p R R 0.3ln( / ) ( ) ( v b ) + ln 0 (E-) v b In terms f mlar quanttes 0.3ln( / ) ( ) ( v b ) + ln v b 0 (E-) Snce R 8.06 cm 3 R atm/ml K, v

38 Substtutng the fllwng numercal values: 350 C 63.5 K, v 600 cm 3 /ml, atm, and b 9 [cm 3 /ml] nt equatn (E-), we have 0.3ln( /63.5) ( 63.5) ( 63.5 ) ln (E-) We can use the fllwng Matlab statements t slve the nnlnear equatn >> funnlne('0.3*lg(x/63.5) e-3*(x -63.5) - 4.4e-6*(x^ ^)+lg((8.06*x-9)/509)'); >> fslve(fun,500,ptmset('dsplay','ff')) >> he temperature f the exhaust frm the turbne s K. -3

39 Example Develp a general relatnshp fr the change n temperature wth respect t pressure at cnstant entrpy s (a) Evaluate the expressn fr an deal gas. (b) Frm the result n part (a), shw that fr an deal gas wth cnstant c, an sentrpc expansn frm state and state yelds equatn. v k cnst (c) Evaluate the expressn fr a gas that beys the Van der Walls equatn f state. Slutn (a) Evaluate the expressn fr an deal gas he cyclc rule can be emplyed t gve s s s Substtutng the relatn s c and s v yelds s v c Fr an deal gas: v R R v herefre, s c v c R (b) Shw that fr an deal gas wth cnstant c, an sentrpc expansn frm state and state yelds equatn. v k cnst Separatn f varables prvdes c R 7 Kresky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg. 45-4

40 Integratn prvdes c ln ln R hs expressn can be rewrtten as c R he deal gas law s nw emplyed c R v v v v c R c R where k c c c R c c R v If we rase bth sdes f the equatn by a pwer f k, we fnd k k v v. v k cnst (c) Evaluate the expressn fr a gas that beys the Van der Walls equatn f state. In art (a), we fund s v c Usng the dervatve nversn rule, we fnd fr the van der Waals equatn v b R v a ( ) ( ) 3 b v a Rv b v Rv v 3 ( herefre, ) ( ) 3 b v a Rv b v Rv c s 3-5

41 Example Yur cmpany has just develped a new refrgeratn prcess. hs prcess uses a secret gas called Gas A. Yu are tld that yu need t cme up wth thermdynamc prperty data fr ths gas. he fllwng data have already been btaned fr the superheated vapr: 0 bar bar [ C] v [m 3 /kg] s [kj/kg K] v [m 3 /kg] s [kj/kg K] ? As accurately as yu can, cme up wth a value fr s n the table abve. Clearly ndcate yur apprach and state any assumptns that yu make. D nt assume deal gas behavr. Slutn In rder t slve ths prblem we need t relate the change n entrpy frm 0 t bar t the change n mlar vlume (fr whch we have cmplete data). Frst, we can rewrte the change n entrpy as bar s Δs s s d 0 bar Applyng a Maxwell relatn, we can relate the abve equatn t the change n mlar vlume: As 0 bar: bar bar s v s s + d s + d 0 bar 0 bar At bar: v v Δv Δ Δv Δ m kg K 3 m kg K ntegrate the abve entrpy equatn, we need an expressn that relates pressure. hus, we wll ft a lne t the data. We btan v t 8 Kresky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg. 45-6

42 v 3 0 m kg K a 3 4 m kg K Nw ntegrate the equatn t fnd the entrpy: s. 0 s [( a 0 4 kj kj ) ] d a kg K kg K Example Yu need t desgn a heater t preheat a gas flwng nt a chemcal reactr. he nlet temperature s 7 C and the nlet pressure s 50 bar. Yu desre t heat the gas t 7 C and 50 bar. Yu are prvded wth an equatn f state fr the gas. v a Z + R 0. 5, wth a [ K / /bar] and wth deal gas heat capacty data: c where s n [ K] R As accurately as yu can, calculate, n [J/ml], the amunt f heat requred. Slutn Energy balance: h h q Because the gas s nt deal under these cndtns, we have t create a hypthetcal path that cnnects the ntal and fnal states thrugh three steps. One hypthetcal path s shwn belw: [bar] 50, q Δh, Δh step step 3 Δh 3 0 Δh step deal gas [K] 9 Kresky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg. 45-7

43 Chsng and as the ndependent prpertes: h dh h d + d Usng Equatn h v + v yelds dh c v d + he gven EOS can be rewrtten as + v d v R + a / akng the dervatve gves: Hence v R + 0.5aR 0.5 Fr step Fr step Fr step 3: 0.5 dh cd + ( 0.5aR )d J ( 0.5aR ) d 0.5aR 5 Δh ml 50 bar J ( ) K 3.58 Δh R d 300 K J ( ar ) d 0.5aR bar 0.5 Δh 3 ml 0 ml Fnally summng up the three terms, we get, q Δh + Δh + Δ h J ml -8

44 Chapter 3 3. hase and ure Substance rncples f hase Equlbrum A phase s a quantty f matter characterzed by bth unfrm physcal structure and unfrm chemcal cmpstn. A phase can be sld, lqud, vapr r gas. he atms n a sld phase are fxed relatve t ther atms n the sld. hey are hwever can vbrate abut ths fxed pstn. Lke a sld, mlecules wthn the lqud phase are n clse prxmty t ne anther due t ntermlecular frces. Hwever, the mlecules n a lqud are nt fxed relatve t ther mlecules n the lqud. hey are cnstantly n mtn, free t mve relatve t ne anther. Mre than ne lqud phase can cexst such as l and water. hey are cnsdered separate lqud phases snce they are nt mscble. Smlarly, slds can cexst n dfferent phases. Fr example, sld carbn can exst n the damnd phase r the graphte phase. Gas mlecules mve randmly t fll the entre vlume f the cntaner n whch they are hused. he mlecules cntnuusly change drectn as they clde wth each ther and bunce ff the cntaner surface creatng the measurable pressure p. A substance that has a fxed chemcal cmpstn thrughut the system s called a pure substance. Water, hydrgen, ntrgen, and carbn mnxde, fr example, are all pure substance. A pure substance can als be a mxture f varus chemcal elements r cmpunds as lng as the mxture s hmgeneus. Ar, a mxture f several cmpunds, s ften cnsdered t be a pure substance because t has a unfrm chemcal cmpstn. A mxture f tw r mre phases f a pure substance s stll a pure substance as lng as the chemcal cmpstn f all phases s the same. A mxture f ce and lqud water, fr example, s a pure substance because bth phases have the same chemcal cmpstn. he state f a system at equlbrum s defned by the values f ts thermdynamc prpertes. System f pure substances r smple cmpressble systems are defned by tw ntensve prpertes such as temperature and pressure. Fr any hmgeneus, pure substance, all thermdynamc functns r prpertes may be expressed n terms f any tw gven ndependent prpertes. v f(, p) r v v(, p) u u(, p) r u u(, v) s s(, p) r s s(, v) he functnal relatns fr these prpertes must be develped frm expermental data and are dependent n the partcular chemcal mlecules makng up the system. Only expansn r cmpressn wrks gven by pdv are permssble n smple cmpressble systems underges quasequlbrum prcesses. Cengel Y. A., Bles M. A., hermdynamcs: An Engneerng Apprach, Mc-Graw Hll, 998, pg

45 ressure can be expressed as a functn f temperature and specfc vlume: p p(, v). he plt f p p(, v) s a surface called p-v- surface. Fgure 3. shws the p-v- surface f a substance such as water that expands n freezng. Fgure 3.- p-v- surface and prjectns fr a substance that expands n freezng. (a) 3-D vew (b) p- dagram (c) p-v dagram. he lcatn f a pnt n the p-v- surface gves the values f pressure, specfc vlume, and temperature at equlbrum. he regns n the p-v- surface labeled sld, lqud, and vapr are sngle-phase regns. he state f a sngle phase s determned by any tw f the prpertes: pressure, temperature, and specfc vlume. he tw-phase regns where tw phases exst n equlbrum separate the sngle-phase regns. he tw-phase regns are: lqud-vapr, sld-lqud, and sld-vapr. emperature and pressure are dependent wthn the tw-phase regns. Once the temperature s specfed, the pressure s determned and vce versa. he states wthn the tw-phase regns can be fxed by specfc vlume and ether temperature r pressure. he prjectn f the p-v- surface nt the p- plane s knwn as the phase dagram as shwn n Fgure 3. (b). he tw-phase regns f the p-v- surface reduce t lnes n the phase dagram. A pnt n any f these lnes can represent any tw-phase mxture at that partcular temperature and pressure. he trple lne f the p-v- surface prjects nt a pnt n the phase dagram called the trple pnt. hree phases cexst n the trple lne r the trple pnt. Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, p

46 he cnstant temperature lnes f the p-v dagram are called the stherms. Fr any specfed temperature less than the crtcal temperature, the pressure remans cnstant wthn the twphase regn even thugh specfc vlume changes. In the sngle-phase lqud and vapr regns the pressure decreases at fxed temperature as specfc vlume ncreases. Fr temperature greater than r equal t the crtcal temperature, there s n passage acrss the tw-phase lqud-vapr regn. Fgure 3.- -v dagram fr water (t scale). Fgure 3.- s a -v dagram fr water. Fr pressure greater than r equal t the crtcal pressure, temperature ncreases cntnusly at fxed pressure as the specfc vlume ncreases and there s n passage acrss the tw-phase lqud-vapr regn. he sbarc curve marked 50 Ma n Fgure 3.- shws ths behavr. Fr pressure less than the crtcal value, there s a tw-phase regn where the temperature remans cnstant at a fxed pressure as the twphase regn s traversed. he sbarc curve wth values f 0 Ma r less n Fgure 3.- shws the cnstant temperature durng the phase change. At 00 C, the saturated vlumes f lqud and vapr water are.0434 cm 3 /g and,673.6 cm 3 /g, respectvely. he qualty f steam s the mass fractn f water vapr n a mxture f lqud and vapr water. he specfc vlume f 00 C steam wth a qualty f 0.65 s gven by v ( 0.65)v L v V (0.35)(.0434) + (0.65)(,673.6) 088. cm 3 /g 3-3

47 3. hase Behavr We wll cnsder a phase change f kg f lqud water cntaned wthn a pstn-cycnder assembly as shwn n Fgure 3.-a. he water s at 0 C and.04 bar (r atm) as ndcated by pnt () n Fgure 3.-. Fgure 3.- hase change at cnstant pressure fr water 3 Fgure 3.- Sketch f -v dagram fr water 4 As the water s heated at cnstant pressure, the temperature ncreases wth a slght ncrease n specfc vlume untl the system reaches pnt (f). hs s the saturated lqud state crrespndng t.04 bar. he saturatn temperature fr water at.04 bar s 00 C. he lqud states alng the lne segment -f are called subcled r cmpressed lqud states. When the system s at the saturated lqud state (pnt f n Fgure 3.-) any addtnal heat wll cause the lqud t evaprate at cnstant pressure as shwn n Fgure 3.-b. When a mxture f lqud and vapr exsts n equlbrum, the lqud phase s a saturated lqud and the vapr phase s a saturated vapr. Lqud water cntnues t evaprate wth addtnal heat untl t becmes all saturated vapr at pnt (g). Any further heatng wll cause an ncrease n bth temperature and specfc vlume and the saturated vapr becmes superheated vapr dented by pnt (s) n Fgure 3.-. Fr a tw-phase lqud-vapr mxture, the qualty x s defned as the mass fractn f vapr n the mxture x m m vapr vapr + m lqud (3.-) 3 Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, p Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, p

48 he prpertes f pure subtances can be btaned frm tables such as able 3.- fr saturated prpertes f water and able 3.- fr prpertes f superheated water vapr. hermdynamc prpertes can als be btaned frm sftwares, whch are mre cnvenent snce nterplatns fr values nt drectly avalable n the tables are nt necessary. able 3.- hermdynamc prpertes f saturated water 5. Specfc Vlume m 3 /kg Internal Energy kj/kg Enthalpy kj/kg Entrpy kj/kg K emp. C ress. bar Sat. Lqud v f 0 3 Sat. Vapr v g Sat. Lqud u f Sat. Vapr u g Sat. Lqud h f Evap. h fg Sat. Vapr h g Sat. Lqud s f Sat. Vapr s g emp. C Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, p

49 able 3.- hermdynamc prpertes f superheated water vapr 6. C v m 3 /kg u kj/kg h kj/kg s kj/kg K v m 3 /kg u kj/kg h kj/kg s kj/kg K p 0.06 bar Ma ( sat 36.6 C) p 0.35 bar Ma ( sat 7.69 C) Sat p 0.70 bar 0.07 Ma ( sat C) p.0 bar 0.0 Ma ( sat C) Sat Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, p

50 Example (a) Determne the specfc vlume fr saturated water vapr and nternal energy fr saturated lqud at 4.4 C frm the fllwng data emp. C ress. bar Specfc Vlume m 3 /kg Sat. Lqud v f 0 3 Sat. Vapr v g Internal Energy kj/kg Sat. Lqud u f Sat. Vapr u g emp. C (b) Determne the specfc vlume and the nternal energy fr superheated water vapr at 90 C and 0.0 bar frm the fllwng data C v m 3 /kg u kj/kg h kj/kg s kj/kg K p 0.06 bar Ma ( sat 36.6 C) v m 3 /kg u kj/kg h kj/kg s kj/kg K p 0.35 bar Ma ( sat 7.69 C) Slutn (a) Determne the specfc vlume fr saturated water vapr at 4.4 C. y y y x x x x Usng lnear nterplatn we have y y x x y x y x v 57.3 g v g (0.4)( ) m 3 /kg 3-7

51 We can als use Matlab nterplatn cmmand nterp. >> vgnterp([4 5],[ ],4.4) vg Determne the nternal energy fr saturated lqud at 4.4 C ul u l (0.4)( ) 9.9 kj/kg We can als use Matlab nterplatn cmmand nterp. >> ulnterp([4 5],[ ],4.4) ul rpertes f water can als be btaned frm the Steam4 prgram wrtten by.k. Nguyen. Yu shuld nte that the values frm the sftware mght be dfferent than the values btaned frm lnear nterplatn snce the sftware des nt assume lnear relatn between the varables. he values calculated frm sftware are n general mre accurate. Fgure 3.-3 rptertes f saturated lqud and vapr water. 3-8

52 he Catt prgram (Cmputer-Aded hermdynamc able ) prduced by Snntag cntans prpertes f ther cmpunds besde water. Fgure 3.-4 rptertes f saturated lqud and vapr water frm Catt. (b) Determne the specfc vlume fr superheated water vapr at 90 C and 0.0 bar C v m 3 /kg v m 3 /kg 0.06 bar 0.35 bar We frst need t determne v (90 C, 0.06 bar) and v (90 C, 0.35 bar) v v (0)( )/ m 3 /kg v v (0)( )/ m 3 /kg We nw determne v(90 C, 0.0 bar) frm v (90 C, 0.06 bar) and v (90 C, 0.35 bar) v

53 v (0.4)( )/ m 3 /kg We can als use the fllwng Matlab cmmands. >> vnterp([80 0],[ ],90); >> vnterp([80 0],[ ],90); >> vnterp([.06.35],[v v],.) v Determne the nternal energy fr superheated water vapr at 90 C and 0.0 bar C u kj/kg u kj/kg 0.06 bar 0.35 bar Use the fllwng Matlab cmmands, we btan u 500. kj/kg >> unterp([80 0],[ ],90); >> unterp([80 0],[ ],90); >> unterp([.06.35],[u u],.) u.500e+003 he nternal energy fr superheated water vapr at 90 C and 0.0 bar can als be btaned drectly frm thermdynamc prperty sftware. 3-0

54 Chapter 3 rncples f hase Equlbrum 3.3 Intrductn t hase Equlbrum A system s sad t be n equlbrum when ts prpertes are unfrm ver the system and are ndependent wth tme. 70 C 60 C t 0 70 C t 5 mn 45 C 40 C 30 C 30 C A B C Fgure 3.3 Example f equlbrum and nn-equlbrum systems. System A shwn n Fgure 3.3 s nt at equlbrum snce ts temperature s a functn f bth pstn and tme. It s an unsteady state system. System B s nt at equlbrum snce ts temperature s a functn f pstn. It s a steady state system. System C s at equlbrum snce ts temperature s unfrm and ndependent f tme. We can use thermdynamcs t frm relatnshps between the states f a system that underges certan prcesses. Frm the frst and secnd law we can calculate the wrk r heat requred fr the change f states r the value f an unknwn prperty f a gven state. We can als use thermdynamcs t determne the cmpstn f a mxture when t reaches equlbrum between cexstng phases wth r wthut chemcal reactns. he crtern fr equlbrum can be based n the energy r entrpy f the system. Fr a spntaneus prcess, the system s at equlbrum f ts energy s a mnmum r ts entrpy s at a maxmum. At equlbrum we have ΔG, 0 (3.3-a) Equatn (3.3-a) says that n a clsed system at cnstant temperature and pressure, spntaneus prcesses prduce negatve changes n Gbbs free energy, and at equlbrum the Gbbs free energy s at ts mnmum value. G H S Great hyscsts Have Study Under Very Able eachers A V U Frm the abve dagram, we can see that equatn (3.3-) nvlves three varables n a grup. We can use varables belng t dfferent grups fr the equlbrum crtern. 3-

55 ΔH,S 0, ΔU S,V 0, r ΔA V, 0. (3.3-b,c,d) he crtern based n entrpy s gven as ΔS U,V 0, r ΔS H, 0. (3.3-a,b) Equatn (3.3-a) says that n a clsed system at cnstant nternal energy and vlume, spntaneus prcesses prduce pstve changes n entrpy, and at equlbrum the entrpy s at ts maxmum value. 3.4 ure Speces hase Equlbrum 3.4- Gbbs Free Energy as a Crtern fr Chemcal Equlbrum Cnsder a clsed system f a pure cmpnent n mechancal and thermal equlbrum and, therefre, at cnstant and. he system can absrb energy Q at cnstant. herefre the surrundngs wll have a reductn n the entrpy gven by ΔS sur Q (3.4-) Frm the secnd law we have ΔS sur + ΔS sys 0 (3.4-) In ths expressn, the equal sgn apples fr reversble prcesses and the unequal sgn apples fr rreversble prcesses. Cmbnng equatns (3.4-) and (3.4-) gves ΔS sys Q We wll nw drp the subscrpt sys fr the entrpy f a system (ΔS ΔS sys ), the abve equatn becmes ΔS Q (3.4-3) he frst law fr a clsed system wth nly V wrk s gven by Q ΔU W ΔU + ΔV Substtutng Q ΔU + ΔV nt equatn (3.4-3) gves ΔU + ΔV ΔS 0 (3.4-4) he Gbbs free energy s related t ther prpertes by the fllwng equatn G H S U + V S 3-

56 he change n Gbbs free energy s then ΔG ΔU + ΔV + VΔ ΔS SΔ At cnstant and, Δ Δ 0, the abve equatn becmes ΔG ΔU + ΔV ΔS 0 (3.4-5) he abve equatn s essentally equatn (3.3-a), ΔG, 0. herefre n a clsed system at cnstant temperature and pressure, spntaneus prcesses prduce negatve changes n Gbbs free energy, and at equlbrum the Gbbs free energy s at ts mnmum value. It shuld be nted that the specfcatn f tw ntensve prpertes fxes the values f all ther state prpertes nly n the unfrm equlbrum state f a sngle-cmpnent, sngle-phase system. hus, the equlbrum crtern specfed by Eq. (3.3-a) r (3.4-5) can be used fr dentfyng the fnal equlbrum state n an slated system that s ntally nnunfrm, r n whch several phases r cmpnents are present he Chemcal tental G H S Great hyscsts Have Study Under Very Able eachers A V U Cnsder the fundamental grupng {U, S, V}, we have U U(S, V) Fr a system wth C cmpnent, the nternal energy U als depends n the number f mles, n, f each cmpnent. Let C 3, we have U U(S, V, n, n 3, n 3 ) he dfferental change n U s gven by du U S V, n, n, n3 ds + U V S, n, n, n3 dv + U n S, V, n, n3 dn + U n S, V, n, n3 dn + U n 3 S, V, n, n dn 3 In general, we have U U(S, V, n,,n,, n C ) (3.4-6) and U du S V, n U ds + V S, n C U dv + n 3-3 S, V, n j dn (3.4-7)

57 he subscrpt j dentes a cmpnent dfferent than. he frst tw ceffcent f Eq. (3.4-7) U refer t cndtns f cnstant cmpstn. hs refers t the same cndtns as and S V U. V S G H G H S S A V U A V U herefre U S V, n U and V S, n We can smplfy Eq. (3.4-7) by defnng the chemcal ptental μ : μ U n S, V, n j Equatn (3.4-7) nw becmes C du ds dv + μ dn (3.4-8) Addng d(v) t each sde f ths equatn, we btan C dh du + d(v) Vd + ds + μ dn (3.4-9) Subtractng d(s) frm each sde f Eq. (3.4-8) yelds C da du d(s) Sd dv + μ dn (3.4-0) Smlarly, subtractng d(s) frm each sde f Eq. (3.4-9) yelds C dg dh d(s) Vd Sd + μ dn (3.4-) 3-4

58 Equatns (3.4-8)-(3.4-) are applcable t clsed systems where the change n the number f mles f ne r mre cmpnents s caused by a chemcal reactn. hey are als applcable t pen systems where the change s frm a transfer f matter. G H S Great hyscsts Have Study Under Very Able eachers A V U We defned the chemcal ptental as μ U n S, V, n j Frm equatns (3.4-9)-(3.4-), we can als defne the chemcal ptental n terms f H, A, and G μ U n S, V, n j H n, S, n j S n, V, n j G n,, n j (3.4-) hese defntns nvlve the 3 prpertes at each crner f the dagram. hey can be cnsdered as alternatve ways f determnng a thermdynamc prperty. he last term n equatn (3.4-) s the mre useful defntn snce a prcess f cnstant temperature and pressure s expermentally the mst cnvenent. Fr a sngle cmpnent system G Ng, and μ G n,, n j ( Ng) N, g A physcal meanng f the chemcal ptental may be btaned when Eq. (3.4-8) s appled t a prcess. C du ds dv + μ dn (3.4-8) Frm the frst law, we have du δq + δw. herefre C δq + δw ds dv + μ dn (3.4-3) Snce ths equatn apples nly t changes between equlbrum states, δq ds. Equatn (3.4-3) becmes 3-5

59 C δw dv + μ dn (3.4-4) Eq. (3.4-4) s the wrk nteractn f a system. It cntans a v wrk and a wrk nvlved a change n mle numbers due t chemcal reactn r mass transfer Vapr Lqud hase Equlbrum Cnsder an slated system f a sngle cmpnent shwn n Fgure he equlbrum system cnssts f a vapr phase (I) and a lqud phase (II). Fr the entre system, the ttal mles, nternal energy, vlume, and entrpy, all f whch are extensve varables, are the sums f these quanttes n the tw phases, that s, N N I + N II U U I + U II V V I + V II and S S I + S II Vapr phase I Lqud phase II Fgure 3.4- An slated equlbrum system. We nw cnsder the nternal energy t be a functn f vlume, entrpy, and mle number. U I U I (S I, V I, N I ) We can calculate the change n the nternal energy fr the vapr phase I due t changes n S I, V I, and N I. du I U S I I I I V, N ds I U + V I I I I S, N dv I + U N I I S I, V I dn I (3.4-5) he partal dervatves can be wrtten n terms f thermdynamc prpertes. du I I ds I I dv I + g I dn I (3.4-6) Slvng the abve equatn fr ds I yelds 3-6

60 ds I I du I I + I dv I g I I dn I (3.4-7a) Smlarly, fr the lqud phase we have ds II II du II II + II dv II II g II dn II (3.4-7b) he entrpy change f the entre system s ds ds I + ds II du I + II du II I + I dv I II + II dv II I g I dn I II g II I dn II Snce the system s slated, the ttal number f mles, ttal nternal energy, and ttal vlume are cnstant. herefre dn dn I + dn II 0 dn I dn II du du I + du II 0 du I du II dv dv I + dv II 0 dv I dv II ds I II du I + I I II II dv I g I I g II II dn I (3.4-8) At equlbrum, the value f S s at a maxmum. Hence ds 0. Snce U I, V I, and N I are ndependent, ds can be equal t zer nly f the ceffcents f du I, dv I, and dn I must all equal t zer. I II I II g I I I I g II II II II I II g I g II herefre, the equlbrum cndtn fr the tw phases system shwn n Fgure 3.4- s satsfed f bth phases have the same temperature, the same pressure, and the same chemcal ptental r Gbbs free energy. 3-7

61 ressure, psa - 40 F Example Calculate the hrsepwer fr cmpressng 5,000 lbs/hr f ethylene frm 00 psa, 40 F t 00 psa. he adabatc effcency f the cmpressr s 75%. Include yur calculatns and Mller chart. Slutn he Mller chart (ressure-enthalpy dagram) fr ethylene can be btaned frm the text by Darby, R., Chemcal Engneerng Flud Mechancs, Marcel Dekker, 00, p s.64 Btu/lb F Enthalpy, Btu/lb he enthalpy f ethylene at 00 psa and 40 F can be lcated frm the dagram t gve h 044 Btu/lb Fr sentrpc cmpressn, Δs 0, we fllw the curve where s s a cnstant untl t reaches the lne where 00 psa t lcate a value fr the enthalpy h 064 Btu/lb herefre (h h ) sentrpc Btu/lb. he actual ncrease n enthalpy f the gas s btaned frm the adabatc effcency (h h ) actual (h h ) sentrpc 6.7 Btu/lb 0.75 he hresepwer suppled by the cmpressr t cmpress 5000 lb/hr f ethylene s wer ,500 Btu/hr wer (33,500 Btu/hr)( hp/(btu/hr)) 5.5 hp 7 Dr. ang's Ntes 3-8

62 Chapter 3 Example Desgn a refrgeratn system, whch supples tw levels f ammna refrgerant at 0 C and at 0 C t tw exchangers requrng dutes f 850 and,500 kw, respectvely. Draw a prcess flw dagram (FD) shwng majr equpment, flw rates n kg/s f refrgerant, duty f each heat exchanger, duty, and hrsepwer f each cmpressr. Als, shw temperatures and pressures f all streams. here are n nter-clers between cmpressrs. he cmpressed gas frm the last stage cmpressr can be cled t saturated lqud at 30 C. Use 75% adabatc effcency fr all cmpressrs. Slutn C, 3, 4 C, 5 3 E3, 3 F V F V E E Fgure 3.4- A tw level refrgeratn system. A desgn f a tw level refrgeratn system s shwn n Fgure Saturated lqud ammna at s partally vaprzed thrugh a let-dwn valve V t and. art f the lqud separated by the flash drum F s used t prvde the frst level refrgeratn thrugh the heat exchanger E. he remanng lqud passes thrugh a secnd valve where ts temperature and pressure s reduced t and. he lqud stream frm the secnd flash drum F s used fr the secnd level refrgeratn thrugh the heat exchanger E where t 3-9

63 becmes saturated vapr. hs stream s cmbned wth the saturated vapr frm the secnd flash drum and cmpressed back t pressure. Istherm at Crtcal pnt Saturated lqud, 3, Saturated vapr, 3,deal Isentrpc lne, 3, Enthalpy Fgure ressure-enthalpy dagram. he prblem can be slved ether wth a ressure-enthalpy dagram f ammna as shwn n Fgure r wth a prgram slvng an equatn f state t btan thermdynamc prpertes f a pure substance gven any tw ther prpertes (fr example, temperature and pressure r saturatn temperature). We wll use the cmputer prgram CA t btan thermdynamc values lsted n able Fr ths example the pressure drp acrss the heat exchangers and acrss the flash drums wll be neglected. Saturated lqud ammna at 30 C s a defned state at 3.67 Ma wth enthalpy equal t 3.4 kj/kg as shwn n the frst lne f the prperty table Smlarly, the saturated prpertes f ammna at 0 C and 0 C can be btaned. he mass flw rated f saturated lqud ammna, m&, requred t remve,500 kw n heat exchanger E s gven by L, m& L, 500kW h V, h L, In ths equatn, h V, and h L, are the enthalpy f saturated vapr and lqud ammna at 0 C, respectvely. herefre 3-0

64 m& L, 500kW h V, h L, kg/s able 3.4- hermdynamc prpertes f ammna. Specfc Specfc Specfc ype emp ressure Vlume Enthalpy Entrpy Qualty hase C Ma m3/kg kj/kg kj/kg/k Ammna Saturated Lqud Ammna Saturated Vapr 3 Ammna Saturated Lqud 4 Ammna Saturated Vapr 5 Ammna Saturated Lqud 6 Ammna Saturated Vapr 7 Ammna Superheated Vapr 8 Ammna Superheated Vapr 9 Ammna Superheated Vapr 0 Ammna Superheated Vapr Ammna Superheated Vapr m& L, s als the lqud flw rate leavng flash drum F. he vapr flw rate, m& V,, leavng the secnd flash drum F can be determned frm the mass and energy balance arund valve V and flash drum F. m& m& L, + m& V, In ths equatn m& s the saturated lqud ammna flw rate at and enterng valve V. he energy balance s gven by ( m& L, + m& V, )h L, m& L, h L, + m& V, h V, Slvng fr m& V, we have m& V, m& L, h h L, V, h h L, L, kg/s he lqud flw rate thrugh valve V s then kg/s m& he mass flw rated f saturated lqud ammna, exchanger E s gven by m& E,, requred t remve 850 kw n heat 3-

65 m& E, 850kW h V, h L, In ths equatn, h V, and h L, are the enthalpy f saturated vapr and lqud ammna at 0 C, respectvely. herefre m& E, 850kW h V, h L, kg/s he lqud flw rate, m& L,, leavng flash drum F s the sum f the lqud flw rate thrugh the heat exchanger E and the lqud flw rate thrugh the valve V m& L, m& E, + m& kg/s Smlarly, the vapr flw rate, m& V,, leavng the frst flash drum F can be determned frm the mass and energy balance arund valve V and flash drum F. m& V, m& L, h h h V, L, h kg/s he ttal mass flw rate m& f ammna requred fr the system s then m& m& L, + m& V, kg/s he vapr flw rate m &,vap frm bth the exchanger E and the flash drum F s gven by m &,vap m& E, + m& V, kg/s We nw determne the hrsepwer f the secnd cmpressr. he saturated vapr leavng the secnd exchanger E and the flash drum F must be cmpressed frm Ma t 0.65 Ma. Fr sentrpc wrk, the gas enterng and extng the cmpressr wll have the same entrpy at kj/kg K. he enthalpy f the gas leavng the secnd cmpressr C due t sentrpc cmpressn s 530 kj/kg. Fr 75% adabatc effcency, the actual enthalpy f the extng gas s gven by h 3,actual 43 + (530 43)/ kj/kg he hrsepwer requred f the secnd cmpressr C s H.0764 (530 43)/ kw 368 hp he actual temperature 3 f the extng gas s btaned frm 0.65 Ma and h 3,actual 563 kj/kg. We have the gas leavng the secnd cmpressr at C that must be mxed wth the saturated gas leavng the frst heat exchanger E and the frst flash drum F at 0 C. he enthalpy f the gas enterng the frst cmpressr C s then 3-

66 h kj/kg At these cndtns ( 0.65 Ma, h kj/kg) C and entrpy s kj/kg K, the gas wll be cmpressed t.67 Ma. he enthalpy f the gas leavng the frst cmpressr C due t sentrpc cmpressn s 68 kj/kg. Fr 75% adabatc effcency, the actual enthalpy f the extng gas s gven by h 5,actual ( )/ kj/kg he hrsepwer requred f the frst cmpressr C s H ( )/ kw 534 hp he gas leavng the frst cmpressr C at 3.67 Ma, h 5,actual 66. kj/kg, ts temperature 5 s 00.8 C. he gas wll be cndensed t saturated lqud at 30 C (.67 Ma). he duty f the heat exchanger E3 s gven by Q ( ) 40 kw he Clapeyrn Equatn emperature and pressure are nt ndependent fr a pure speces that exsts n tw phases at equlbrum. We nw wsh t derve an expressn relatng the pressure at whch tw phases can cexst t the temperature f the system. At equlbrum between α and β phases, where α and β can represent the vapr, lqud, r sld phases, we have α β, α β, and g α g β (3.4-9) Hence, the dfferental changes n Gbbs free energy f each phase must be equal dg α dg β g α d + g α d g β d + g β d (3.4-0) Equatn (3.4-0) shws the dsplacement f the system frm ts rgnal equlbrum state descrbed by Eqs. (3.4-9) t anther equlbrum state. hs change wll mve the system alng ne f the cexstence lnes shwn n Fgure t a new and where + d and + d 3-3

67 g α hase equlbrum (cexstence lne) β hase β s mre stable hase α s mre stable Fgure Surfaces f Gbbs surfaces fr tw phases α and β. Applyng the fundamental prperty relatn fr g t each phase yelds s α d + v α d s β d + v β d (3.4-0) Rearrangement gves d d s v α α s v β β (3.4-) We have g α g β h α s α h β s β Slvng fr the dfference n entrpy yelds s α s β h h α β Substtutng the expressn fr entrpy nt equatn (3.4-) gves α β d h h α β d ( v v ) (3.4-) Equatn (3.4-), called the Clapeyrn equatn, s a general relatnshp amng pressure, temperature, vlume change, and enthalpy change fr a sngle cmpnent, tw-phase system at equlbrum. hs equatn can be appled t any knd f phase equlbrum ncludng sldvapr and sld-lqud equlbra by substtutng the alternatve sublmatn r fusn prpertes nt the equatn. 3-4

68 Fr sld-lqud equlbrum, we are usually nterested n the change f freezng pnt wth pressure d/d. We can nvert Eq. (3.4-) t btan l s d ( v v ) d h l s h (3.4-3) Fr mst substances d/d s small and pstve; hwever fr water d/d s negatve snce ce s less dense than lqud water. hs unusual behavr fr water means that ce wll melt when pressure s appled t t sthermally as shwn n Fgure Due t ths phenmenn, yu can skate n ce. Lqud Lqud lqud+vapr sld+lqud lqud+vapr sld+lqud Sld Vapr Sld Vapr Water Mst substance Fr vapr-lqud equlbrum, we have Fgure behavr f real flud. d sat v l h h v l d ( v v ) (3.4-4) When appled t equlbrum nvlvng a vapr phase, the pressure s referred t as the vapr r saturatn pressure and wll be dented by sat. In the regn far frm the crtcal pnt, v l << v v, the vlume f the lqud s neglgble cmpared t the vlume f the vapr. If the vapr beys deal gas law, Eq. (3.4-4) becmes d sat sat Δh d R vap (3.4-5) In ths equatn, Δh vap h v h l s the enthalpy f vaprzatn f the substance at temperature. Equatn (3.4-5) can be rewrtten n the frm d ln sat Δh vap R d (3.4-6) Integratng Eq.(3.4-6) between state and state yelds 3-5

69 sat ln Δh vap sat R (3.4-7) We can als ntegrate Eq. (3.4-6) usng a cnstant f ntegratn C ln sat Δh C vap R (3.4-8) Equatns (3.4-7) and (3.4-8) are ften called the Clausus-Clapeyrn equatn. hey can be used t crrelate, nterplate, and extraplate f vapr data r t estmate latent heats frm vapr pressure data. Equatn (3.4-8) suggests that a plt f ln sat vs. (/) shuld be a straght lne ver the regn where Δh vap s a cnstant. Actually, t has been fund that Eq. (3.4-8) s vald ver a surprsngly wde range f temperature due t a cancellatn f errrs ntrduced by the varus assumptns. Saturatn pressure data are cmmnly crrelated n terms f the Antne equatn ln sat B A C + (3.4-9) he Antne equatn can be regarded as an emprcal versn f the Clausus-Clapeyrn equatn. Values fr the Antne cnstants, A, B, and C can be fund fr varus substances n Kretsky s text 8. here are several mre cmplex crrelatns 9 avalable fr saturatn data that are accurate ver a wder range f temperature. Fr example, the water vapr pressure can be btaned accurately frm the fllwng equatn 0 sat 8 5 crt exp τ 0 ( cr ) F ( ) (3.4-30) In ths equatn, dentes saturatn temperature ( C), crt crtcal pressure (0.88 bars), cr crtcal temperature ( C), τ 000/( K), and F s gven by F F F 3 F 4 F 5 F 6 F 7 F Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg Red R. C., raunts J. M., lng B. E., he rpertes f Gases & Lquds, McGraw-Hll, Keenan J. H., Steam ables, Wley, pg

70 Chapter 3 Example he heat f fusn f ce s 80 cal/g at 0 C and atm, and the rat f the specfc vlume f water t that f ce s.000:.09. he saturated vapr pressure and the heat f vaprzatn f water at 0 C are 4.58 mmhg and 600 cal/g, respectvely. Estmate the trple pnt usng these data. Slutn Lqud sld+lqud Sld A lqud+vapr Vapr 4.58 mmhg B rple pnt 0C Fgure E behavr f water. he crdnates f pnt A and B are (0 C, atm) and (0 C, 4.58 mmhg), respectvely. he cexstence lne thrugh A wll ntersect the cexstence lne thrugh B at the trple pnt. Fr equlbrum between the sld and lqud phase we have dg s dg l s g s g d + l g d l g d + Applyng the fundamental prperty relatn fr g t each phase yelds Rearrangement gves s s d + v s d s l d + v l d d d s v l l s s s v d 6 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

71 We have g s g l h s s s h l s l Slvng fr the dfference n entrpy yelds s l s s h l h s Substtutng the expressn fr entrpy nt d/d yelds l s d h h l s d ( v v ) d 80 d ( 0.09)(73.5) 3.85 cal/cm 3 K We need t change the unt frm cal/cm 3 K t mmhg/ K. cal 4.84 J 4.84 kg m /s kg cm /s cal/cm kg/s cm kg/s m mmhg kg/s m kg/s m mmhg herefre d ( 3.85)( )( ) mmhg/ K d he equatn fr the straght lne thrugh A(0 C, 760 mmhg) wth slpe d/d mmhg/ K s gven by mmhg/ K (E-) Smlarly fr equlbrum between the sld and vapr phase we have v s d h h v s d ( v v ) Neglectng the sld vlume cmpared t the vlume f the vapr and usng deal gas law we btan 3-8

72 d d ( h v s h ) R (4.58)(600)( 8) mmhg/ K (.987)(73.5 ) he equatn fr the straght lne thrugh B(0 C, 4.58 mmhg) wth slpe d/d mmhg/ K s gven by mmhg/ K (E-) At the trple pnt t and t, Equatns (E-) and (E-) becme, respectvely t t (E-3) t t (E-4) Slvng the tw lnear equatns (E-3) and (E-4) gves t C t mmhg 3-9

73 3.5 Refrgeratn he mst cmmn refrgeratn cycle s the vapr cmpressn cycle shwn n Fgure In step 4, heat s remved at the temperature L frm the system beng refrgerated by the evapratn f a lqud under the pressure L. In step, saturated vapr at L s cmpress sentrpcally t H where t becmes superheated vapr. In step 3, heat Q H s transferred t the surrundng by cndensatn at H. In step 3 4, the cycle s clsed by thrttlng the lqud t the lwer pressure L. H 3 Q H Cndenser hrttle Lqud valve Cmpressr 3 Evapratr Isenthalp 4 Vapr W Q 4 L a b c s Fgure 3.5- A vapr-cmpressn refrgeratn cycle and ts s dagram. he heat transfer between the system and the surrundngs can be btaned frm the s dagram. Snce Q ds, the heat effect s the area under the curve representng the path. In Fgure 3.5- the heat Q H transferred frm the refrgeratr t the hgh temperature envrnment s the area -3-a-c, whch s negatve. he heat Q L remved frm the lw temperature system s the area 4--c-b and s pstve. Fr the cyclc prcess ΔU Q L + W Q H 0 W Q H Q L (3.5-) he effcency f the refrgeratn cycle, called the ceffcent f perfrmance CO, s gven by L CO W Q L (3.5-) he wrk requred fr the refrgeratn cycle can be btaned frm the s dagram W Q H Q L (Area -3-a-c) (Area 4--c-b) W Q H Q L (Area --3-4) + (Area 3-4-b-a) he refrgeratn cycle shwn n Fgure 3.5- s a sem-reversble cycle snce all steps except thrttlng are reversble. 3-30

74 Example A vapr cmpressn refrgeratn prcess usng NH 3 as the wrkng flud s t perate between 0 and 80 F. Determne the ceffcent f perfrmance fr the sem-reversble peratn. Slutn H h 4 h h Evapratr 4 Q L W 80 F 0 F Lqud 3 Isenthalp 4 a b c L Vapr s We wll frst lcate the fur states f the refrgeratn cycle: State 3: Saturated lqud at 80 F, h 3 h Btu/lb. State : Saturated vapr at 0 F, h 66.8 Btu/lb. State 4: Lqud and vapr mxture at h Btu/lb and 0 C State : Superheated vapr at s s.95 Btu/lb R, H 53. psa Makng energy balance arund the evapratr yelds 0 h 4 h + Q L Q L h h Btu/lb Makng energy balance arund the cmpressr yelds 0 h h + W W h h Btu/lb he ceffcent f perfrmance s then Q CO L 485. W Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

75 Example Refrgerant 34a s the wrkng flud n an deal vapr-cmpressn refrgeratn cycle that cmmuncates thermally wth a cld regn at 0 C and a warm regn at 6 C. Saturated vapr enters the cmpressr at 0 C and saturated lqud leaves the cndenser at 6 C. he mass flw rate f the refrgerant s 0.08 kg/s. Determne (a) the cmpressr pwer, n kw, (b) the refrgeratn capacty, n tns, (c) the ceffcent f perfrmance, and (d) the ceffcent f perfrmance f a Carnt refrgeratn cycle peratng between warm and cld regns at 6 and 0 C, respectvely. Slutn H 3 Q H Cndenser Expansn valve Cmpressr Evapratr 4 W Q L Lqud 3 4 a b Isenthalp 6 C 0C c L Vapr s Specfc Internal Specfc Specfc ype emp ressure Vlume Energy Enthalpy Entrpy Qualty hase C Ma m3/kg kj/kg kj/kg kj/kg/k R-34a Saturated Vapr 3 R-34a Saturated Lqud R-34a Superheated Vapr (a) he cmpressr wrk s gve by W & c m & (h h ) (0.08 kg/s)( ) kj/kg.408 kw (b) he refrgeratn capacty, n tns, s Q & L m & (h h 4 ) (0.08 kg/s)( ) kj/kg 3.0 kw Q & tn L (3.0 kj/s)(60 s/mn) kj/mn 3.7 tn (c) the ceffcent f perfrmance s 3 Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, pg

76 Q& L β W& c (d) he ceffcent f perfrmance f a Carnt refrgeratn cycle peratng between warm and cld regns at 6 and 0 C, respectvely β Q& W& L c Q& Q& Q& L H L L H L s 3 Q H Cndenser Lqud H Expansn valve Cmpressr Evapratr 4 W Q L H L 3 4 Isenthalp Fgure 3.5- A vapr-cmpressn refrgeratn wth rreversbltes. L s Vapr Fgure 3.5- shws several features exsted n actual vapr cmpressn systems. he heat transfers between the refrgerant and the warm and cld regns are nt accmplshed reversbly: the refrgerant temperature n the evapratr s less than the lw reservr temperature, L, and the refrgerant temperature n the cndenser s greater than the hgh reservr temperature, H. he cmpressr wll nt have 00% effcency s that the flud leavng the cmpressr wll be at state (), whch s at hgher entrpy than the sentrpc cmpressn state (s). he ceffcent f perfrmance decreases as the average temperature f the refrgerant n the evapratr decreases and as the average temperature f the refrgerant n the cndenser ncreases. If the thermal effcency f the cmpressr s knwn, the enthapy h at state () can be determned frm the fllwng expressn: ( W& ) c / m& s η c ( W& c / m& ) h h h h s Due t frctns, there wll be pressure drps as the refrgerant flws thrugh the evapratr, cndenser, and ppng cnnectng the varus cmpnents. he pressure drps are gnred n subsequent calcualtns fr smplcty. 3-33

77 Example Refrgerant 34a s the wrkng flud n an deal vapr-cmpressn refrgeratn cycle that cmmuncates thermally wth a cld regn at 0 C. Saturated vapr enters the cmpressr at 0 C and saturated lqud leaves the cndenser at 9 bar. he mass flw rate f the refrgerant s 0.08 kg/s. Determne (a) the cmpressr pwer, n kw, (b) the refrgeratn capacty, n tns, (c) the ceffcent f perfrmance. Slutn H 3 Q H Cndenser Expansn valve Evapratr 4 Q L Cmpressr W Lqud 3 4 a b Isenthalp 9 bar -0 C c L Vapr s Specfc Internal Specfc Specfc ype emp ressure Vlume Energy Enthalpy Entrpy Qualty hase C Ma m3/kg kj/kg kj/kg kj/kg/k R-34a Saturated Vapr R-34a Superheated Vapr 3 R-34a Saturated Lqud (a) he cmpressr wrk s gve by W & c m & (h h ) (0.08 kg/s)( ) kj/kg.44 kw (b) he refrgeratn capacty, n tns, s Q & L m & (h h 4 ) (0.08 kg/s)( ) kj/kg.43 kw Q & tn L (.43 kj/s)(60 s/mn) kj/mn 3.5 tn (c) he ceffcent f perfrmance s Q& L β.43 W& c 4 Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, pg

78 Example Refrgerant 34a s the wrkng flud n an deal vapr-cmpressn refrgeratn cycle that cmmuncates thermally wth a cld regn at 0 C. Saturated vapr enters the cmpressr at 0 C and lqud leaves the cndenser at 9 bar and 30 C. he cmpressr has an effcency f 80%. he mass flw rate f the refrgerant s 0.08 kg/s. Determne (a) the cmpressr pwer, n kw, (b) the refrgeratn capacty, n tns, (c) the ceffcent f perfrmance. Slutn Q H Cndenser Expansn valve Cmpressr Evapratr 4 W Q L Lqud 30 C -0 C 3 4 Isenthalp s H 9 bar L s Vapr Specfc Specfc Specfc ype emp ressure Vlume Enthalpy Entrpy Qualty hase C Ma m3/kg kj/kg kj/kg/k R-34a Saturated Vapr s R-34a Superheated Vapr 3 R-34a Cmpressed Lqud 4 R-34a Lqud Vapr Mxture (a) he cmpressr wrk s gve by W & hs h m & c (h h ) m& η (0.08 kg/s) kj/kg 3. kw (b) he refrgeratn capacty, n tns, s Q & L m & (h h 4 ) (0.08 kg/s)( ) kj/kg.08 kw Q & tn L (.08 kj/s)(60 s/mn) kj/mn 3.44 tn (c) the ceffcent f perfrmance s Q& L β.08 W& c 5 Mran, M. J. and Shapr H. N., Fundamentals f Engneerng hermdynamcs, Wley, 008, pg

79 3.6 artal Mlar rpertes If we mx 50 cm 3 f ethanl wth 0 cm 3 f water at 5 C we wll btan 67 cm 3 f the resultng slutn 6. he dfference n the slutn vlume and the sum f the pure vlumes s due t the nature f the unlke ethanl-water nteractns and the fact that they are dfferent frm the water-water r ethanl-ethanl pure speces nteractns. When a speces becmes part f a mxture, ts prpertes change; hwever t stll cntrbutes t the prpertes f the mxture, snce the ttal slutn prpertes f the mxture depend n the amunt present f each speces and ts resultant nteractns. We can defne a partal mlar prperty t accunt fr the cntrbutn f a speces t the mxture prperty. We wll frst cnsder the vlume f the mxture V, an extensve prperty, whch at a specfed temperature and pressure wll depend n the number f mles f each speces n the mxture. he partal mlar vlume f speces, V, n a mxture s defned as V V n,, n j (3.6-) We use the ntatn n j t specfy that we are hldng the number f mles f all speces except speces cnstant when we take the partal dervatve wth respect t n. A partal mlar prperty s always defned at cnstant temperature and pressure, tw f the crtera fr phase equlbrum. In general, the slutn prperty K s a functn f,, and the number f mles f m dfferent speces: K K(,, n, n,..., n,..., n m ) (3.6-) he dfferental f K can then be wrtten as the sum f partal dervatves f each f these ndependent varables, as fllws: K dk, n K d +, n m K d + n,, n j dn (3.6-3) If we multply the number f mles n a system by an arbtrary amunt α at a gven and, the extensve prperty K shuld als ncreased by that amunt: αk K(,, αn, αn,..., αn,..., αn m ) (3.6-4) Accrdng t Euler s therem, the abve equatn s a frst-rder, hmgeneus functn f n. Dfferentatng Eq. (3.6-4) wth respect t α yelds ( αk) α, K K n ( n ) K + n α ( n,, n ) +... α j,, n j 6 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

80 K + n ( n ) K n m α ( n,, n j m ) α,, n j (3.6-5) Equatn (3.6-5) must be vald fr any value f α, s at α, we get m K K n n,, n j (3.6-6) he ttal slutn vlume s then gven by m V V n n,, n j m n V (3.6-7) Example We need,000 cm 3 f an antfreeze slutn cnsstng f 30-ml% methanl n water. What vlumes f pure methanl and f pure water at 5 C must be mxed t frm the,000 cm 3 antfreeze, als at 5 C? artal mlar vlumes fr methanl and water n a 30-ml% methanl slutn and ther pure-speces mlar vlumes, bth at 5 C, are: Methanl (): V cm 3 ml - v cm 3 ml - Water (): V cm 3 ml - v cm 3 ml - Slutn he mlar slutn vlume at 30-ml% methanl s gven as v m nv m V n n x V x V + x V v (0.3)(38.63) + (0.7)(7.765) 4.05 cm 3 ml - he ttal number f mle requred s V 000 n ml v Vlume f pure methanl s V x n v (0.3)(83.46)(40.77),07 cm 3 Vlume f pure methanl s V x n v (0.7)(83.46)(8.068),053 cm 3 7 Smth J. M., Van Ness H. C., and Abbtt M. M., Intrductn t Chemcal Engneerng hermdynamcs, Mc Graw Hll, 00, pg

81 Example A 0 wt% chlrfrm slutn s btaned by mxng chlrfrm, CHCl 3, wth acetne, C 3 H 6 O, n an nsulated mxer at steady state. Fr every gram f slutn, 9.83 J s remved t keep the system temperature cnstant at 4 C. Determne the enthalpy f mxng fr ths slutn. Slutn n, h n, h (n +n ), h Q Chlrfrm (), CHCl 3 Mw 9.39 Acetne (), C 3 H 6 O Mw Fr 00 g f slutn, we have n 0/ ml CHCl 3 n 80/ ml C 3 H 6 O herefre n n + n ml A frst law ver the system gves n h + n h (n + n )h Q 0 Q/(n + n ) h (x h + x h ) Δh mx Q/n h (x h + x h ) x 0.675/ x.3774/ he heat f mxng s equal t the heat remved per mle f slutn Δh mx Q/n q q ˆ (x Mw + x Mw ) Δh mx (9.83)( ) J/ml 3-38

82 Chapter 3 Example A 5.0 wt% H SO 4 slutn at 60 F s t be cncentrated t 45 wt% by evapratn f water. he cncentrated slutn and water vapr leavng the evapratr at 80 F and atm. Calculate the rate at whch heat must be suppled t the evapratr t prcess 500 lb/hr f the feed slutn. he enthalpy-cncentratn chart fr H SO 4 -H O s gven n Fgure E- where the reference states pure lqud water at 3 F and pure lqud H SO 4 at 77 F. 5 Wt % HSO 4 slutn at 60 F 45 wt% HSO 4 slutn at 80 F Fgure E- Enthalpy-cncentratn chart fr H SO 4 -H O. 8 Felder R. M., Russeau R. W., Elementary rncples f Chemcal rcesses, Wley,

83 5 wt% slutn 500 lb/hr h F Evapratr Q water vapr m,h v 45 wt% slutn m, h Slutn Makng an acd balance arund the evapratr yelds (0.05)(500) (0.45)(m ) m lb/hr he rate f water vapr leavng the evapratr s then m lb/hr Enthalpy f the 5 wt% feed slutn at 60 F s btaned frm the chart h F 0 Btu/lb Smlarly, the enthalpy f 45 wt% slutn at 80 F frm the chart s h 30 Btu/lb Yu shuld nte that the slutn enthalpy culd nt be btaned frm the pure prpertes. If t were the value wuld be n the straght lne cnnectng the enthalpes f pure water and pure acd as shwn n Fgure E-. he enthalpy f water vapr, h v, at 80 F, atm can be apprxmated by the enthalpy f saturated water vapr at 80 F usng steam table. h v 38 Btu/lb Makng an energy balance ver the evapratr gves Q h v h 500h F Q (444.44)(38) + (55.56)( 30) (500)(0) lb/hr 3-40

84 Chapter 4 rncple f hase Equlbrum II 4. he hase Rule We want t determne the number f ntensve varables that must be specfed t defne a system cnsstng f m nn-reactng cmpnents dstrbuted amng π phases. descrbe each phase we need m ndependent cmpstns,, and t a ttal f m+ ntensve varables. he ttal number f varables requred t specfy π phases s π(m+). Fr phase equlbrum we have the fllwng set f equatns: α β π α β π α μ α μ β μ β μ π μ π μ M α β π μ m μ m μ m Each rw n the set f equatns abve cntans (π ) ndependent equatns. hus there are a ttal f (π )(m+) ndependent equatns between the varables. he number f varables we can ndependently pck (the s-called degree f freedm, F) s btaned by subtractng the ttal π(m+) varables we need t specfy by the (π )(m+) ndependent equatns. F π(m+) (π )(m+) m π + We can ndependently specfy (m π + ) t cmplete defne a system wth m cmpnents and π phases. Example Determne the degree f freedm fr the fllwng system: x x + 3x 3 0 x + x + 3x 3 5 Slutn here are three varables and tw ndependent equatns. Hence, the system has ne degree f freedm. 4-

85 Example Determne the degree f freedm fr the fllwng vapr lqud equlbrum system wth three speces A, B, and C. (a) A, B, and C are present n bth gas and ne sngle lqud phase. (b) A, B, and C are present n gas phase but nly A and B are present n the sngle lqud phase. (c) A, B, and C are present n bth gas and lqud phase (I) but nly A and B are present n the lqud phase (II). Slutn (a) A, B, and C are present n bth gas and ne sngle lqud phase. Number f ntensve varables t specfy the system s π(m+) (3 + ) 8 Fr phase equlbrum we have the fllwng set f ndependent equatns: L G, L G L μ A G μ A L μ B he degree f freedm s then G μ B L μ C,, and G μ C F # f varables # ndependent equatns (b) A, B, and C are present n gas phase but nly A and B are present n the sngle lqud phase. Number f ntensve varables t specfy the system: Gas phase: + 4 Lqud phase: + 3 Fr phase equlbrum we have the fllwng set f ndependent equatns: L G, L G L μ A G μ A L μ B, and he degree f freedm s then G μ B F # f varables # ndependent equatns

86 (c) A, B, and C are present n bth gas and lqud phase (I) but nly A and B are present n the lqud phase (II). Number f ntensve varables t specfy the system: Gas phase: + 4 Lqud phase (I): + 4 Lqud phase (II): + 3 Fr phase equlbrum we have the fllwng set f ndependent equatns: L G, L G, L G, and L G L μ A L μ A G μ A he degree f freedm s then L μ B L μ B G μ B L μ C,, and G μ C F # f varables # ndependent equatns he Fugacty Fugacty s a derved thermdynamc prperty that has the unts f pressure. Fugacty s an abstract cncept defned t facltate certan calculatns nvlved wth phase and chemcal equlbrum. Fr a pure substance at gven and, we have dμ vd sd At cnstant temperature dμ vd (4.-) If the gas s deal, v R/, Eq. (4.-) becmes d dμ R dμ Rd(ln) (4.-) Fr a real gas, v R/. Hwever we defne a functn f called fugacty s that the change n chemcal ptental wth respect t s gven by an expressn smlar t equatn (4.-) dμ Rd(lnf) (4.-3) In the lmt as 0, f r f/ snce deal gas law apples as 0. Integratng Eq. (4.-3) between a reference state and the state f the system yelds μ μ Rln f f (4.-4) 4-3

87 In ths equatn μ s the chemcal ptental at the reference pressure,, and at the same temperature as the chemcal ptental μ f nterest. Smlarly, the fugacty,, f a speces n slutn s defned as fˆ μ μ Rln f f ˆ ˆ (4.-5) We nw derve the crtera fr phase equlbrum usng fugacty. Fr phases α and β n equlbrum we have μ α μ β Substtutng Eq. (4.-5) nt the abve equatn yelds μ,α + Rln f f, ˆ ˆ α α μ,β + Rln f f, ˆ ˆ β β (4.-6) Rearrangng Eq. (4.-6) gves μ,α μ,β Rln f f,, ˆ ˆ β α + Rln α β f f ˆ ˆ (4.-7) he frst three terms f the equatn are parts f the defntn f fugacty gven by Eq. (4.- 5). μ,α μ,β Rln f f,, ˆ ˆ β α herefre 0 Rln α β f f ˆ ˆ (4.-8) α fˆ β fˆ Equatn (4.-8) frms the crtern fr phase equlbrum n terms f fugacty. 4-4

88 Example Determne the fugacty f lqud water at 30 C and at the saturatn pressure, 0 bar, and 00 bar. Slutn At 30 C, sat bar. In the lmt as 0, f, therefre f sat sat bar. Frm the defntn f fugacty, we have dμ Rd(lnf) vd Integratng the abve equatn frm sat t yelds Rln f f L sat v( sat ) f L f sat v( exp R sat ) We have assumed the vlume f lqud water remans cnstant at the saturatn temperature f 30 C. v m 3 /kg ( )(8.0) m 3 /kml he fugacty f lqud water at 30 C and 0 bar s then f L ( ) exp (834.3)(303.5) bar he fugacty f lqud water at 30 C and 00 bar s f L ( ) exp (834.3)(303.5) bar Hence the fugacty f a pure lqud vares lttle ver a mderate pressure range. Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

89 Example Determne the fugacty f CO at 30 K and a usng the Van der Waals equatn f state (EOS). R v b a v Data: a a m 6 /ml, b m 3 /ml Slutn Frm the defntn f fugacty, we have dμ Rd(lnf) vd (E-) Snce the equatn f state s gven explctly n terms f pressure, we need t rearrange the abve equatn s that the term vd can be easly ntegrated. Fr an deal gas we have Rd(ln) vd R d (E-) Subtractng Eq. (E-) frm Eq. (E-) gves Rd(lnf/) v R d Integratng the equatn frm 0 t gves R f / R d (lnf/) v d 0 ln f R R v d 0 (E-3) We nw want t change the ntegratng varable frm t v usng the prduct rule d(v) dv + vd d v d(v) v dv v Usng the defntn f the cmpressblty factr, Z, we have R d(v) RdZ d R dz dv dz dv v v Z v Substtutng d frm the abve equatn t Eq. (E-3) gves 4-6

90 f v R ln R v v Z dz v dv (E-4) Expandng the rght hand sde f the equatn yelds ln f v R Z v R R dv + v v R dz Z Z Z (E-5) We can ntegrate the secnd ntegral n the RHS f Eq. (E-5). Z v R R Z dz Z Z Z Z v R Z Z dz Z dz Z Z v R R Z Z Z Z dz (Z ln Z ) Z Z ln Z Hence ln f v R R dv ln Z + (Z ) v v We wll nw ntegrate the ntegral wth respect t v usng the Van der Waals equatn f state. R v b a v R R R v v v b a + v v R v R v v dv v dv dv v v v v b v + v adv Rv v R v R v v dv ln v b v v a Rv v v ln v v b ln v v b v R v R v dv ln v v b ln v v b a Rv v R v R v dv ln v b v a Rv b a ln v Rv herefre 4-7

91 f b a ln ln ln Z + (Z ) (E-6) v Rv We need t evaluate the cmpressblty factr Z frm the Van der Waals equatn f state. R v b v va R v v b a Rv Z b / v a Rv Snce v ZR Z b ZR a Z(R ) (E-7) b a Let B and A R (R ), equatn (E-7) becmes Z B Z Z A We nw wll rearrange the equatn nt the plynmal frm B B Z Z Z A Z Z Z BZ Z A + AB Z Z 3 ( + B)Z + AZ AB 0 (E-8) In terms f A and B, equatn (E-6) becmes ln ln f f B A ln ln Z + (Z ) Z Z A (Z ) ln (Z B) (E-9) Z he fugacty can be determned frm Eq. (E-9) wth the values f Z btaned frm the slutn f Eq. (E-8). able 4.- lsts the Matlab prgram and the results fr the fugacty calculatn. Frm the prgram, we have f a 4-8

92 Chapter 4 able % Example 4.-, cmpressblty usng Van der Waals EQS % R 8.34; % a.m3/(ml.k) a0.3658; % a.m6/ml b4.86e-5; % m3/ml %.4e6; % a 30; % K % Aa*/(R*)^;Bb*/(R*); fprntf('a %8.5e, B %8.5e\n',A,B) b-(+b);ba;b0-a*b; fprntf('b %8.5e, b %8.5e, b0 %8.5e\n',b,b,b0) % % Slve fr the cmpressblty factr Z usng Newtn methd % Z; fr :0 fz((z+b)*z+b)*z+b0; dfz(3*z+*b)*z+b; ezfz/dfz;zz-ez; f abs(ez)<.0000; break; end end fexp((z-)-a/z-lg(z-b)); f*f; fprntf('z %8.5f, f/ %8.5f\n',Z,f) fprntf('f(a) %8.5e\n',f) >> e4dd A e-00, B.384e-00 b -.038e+000, b e-00, b e-003 Z , f/ f(a).3499e

93 Example Use data frm the steam table t calculate the fugacty f steam at 300 C and a. Slutn Frm the defntn f fugacty fr pure cmpnent we have dμ dg Rd(lnf) We ntegrate ths equatn frm the pressure lw enugh s that the fugacty s essentally the same as the pressure. g dg IG g R f d (ln f ) g f exp g R IG (E-) Frm the steam table, at 300 C r K and 0 4 a, the pressure s lw enugh s that deal gas law apples, we have h IG 3077 kj/kg and s IG 9.8 kj/kg K herefre g IG h IG s IG 3077 (573.5)(9.8) 43 kj/kg g IG (43)(8) 40,409 J/ml At K and a, we have h 785 kj/kg and s 5.79 kj/kg K g h s 534 kj/kg 9,68 J/ml Substtutng the values f Gbbs free energy nt equatn (E-) yelds g f exp g R IG 0 4 9,68 ( 40,409) exp a 4-0

94 Example Determne the change n Gbbs energy, Δg, f CO at 30 K, a and at 30 K, a usng the Van der Waals equatn f state (EOS). R v b a v Data: a a m 6 /ml, b m 3 /ml (a) Evaluate Δg usng the fugacty, Δg Rln(f /f ). (b) Evaluate Δg usng drect ntegratn, Δg v d. Slutn (a) Evaluate Δg usng the fugacty, Δg Rln(f /f ). he fugacty usng Van der Waals EOS s gven by ln f (Z ) Z A ln (Z B) he cmpressblty Z s the largest rt f the equatn Z 3 ( + B)Z + AZ AB 0 where a A (R ) (at 30 K, a) b B (at 30 K, a) R able 4.- lsts the Matlab prgram t determne Z, f, and Δg wth the fllwng results at 30 K: he change n Gbbs energy s then gven by (a) Z f(a) Δg Rln(f /f ) (8.34)(30)ln(6.680/.35) J 4-

95 (b) Evaluate Δg usng drect ntegratn, Δg v d. We can use 5 pnts Smpsn s rule t evaluate the ntegral where Δg Δ (v + 4v + v 3 + 4v 4 + v 5 ) 3 Δ a he specfc vlume can be determned frm v Z. ZR usng the fllwng values fr and he change n Gbbs energy s then Δg (a) Z v(m 3 /ml) Δ (v + 4v + v 3 + 4v 4 + v 5 ) 447. J 3 he result f the ntegral v d can be mprved by usng mre pnts fr the Smpsn s rule as shwn n the fllwng table n Δg(J) At n 35, the value f v d s almst the same as the value btaned frm the expressn: Δg Rln(f /f ) J. 4-

96 able % Example 4.-4, change n Gbbs energy usng Van der Waals EOS % R 8.34; % a.m3/(ml.k) a0.3658; % a.m6/ml b4.86e-5; % m3/ml % nsnput('odd number f data pnt, ns '); 30; R R*; % K vzers(ns,); % vlnspace(.4e6,.4e7,ns);; % a Z; fvzers(ns,); fr n:ns v(n); Aa*/(R)^;Bb*/(R); b-(+b);ba;b0-a*b; fr :0 fz((z+b)*z+b)*z+b0; dfz(3*z+*b)*z+b; ezfz/dfz;zz-ez; f abs(ez)<.0000; break; end end v(n)z*r/; f*exp((z-)-a/z-lg(z-b)); fprntf('z %8.5f, f(a) %8.3e\n',Z,f) fv(n)f; end dgr**lg(fv(ns)/fv()); fprntf('frm fugacty, dg(j) %8.f\n',dg) cn(v()-v())/3; sumv()+v(ns); fr ::ns sumsum+4*v(); end fr 3::ns- sumsum+*v(); end dgsum*cn; fprntf('frm ntegratn f vd, dg(j) %8.f\n',dg) 4-3

97 >> e4dd4 Odd number f data pnt, ns 35 Z , f(a).35e+006, v(m3/ml).737e-003 Z , f(a).65e+006, v(m3/ml).350e-003 Z 0.97, f(a).966e+006, v(m3/ml).097e-003 Z , f(a).7e+006, v(m3/ml) 9.77e-004 Z , f(a).565e+006, v(m3/ml) 7.840e-004 Z , f(a).849e+006, v(m3/ml) 6.803e-004 Z , f(a) 3.3e+006, v(m3/ml) 5.97e-004 Z , f(a) 3.386e+006, v(m3/ml) 5.9e-004 Z , f(a) 3.638e+006, v(m3/ml) 4.79e-004 Z , f(a) 3.879e+006, v(m3/ml) 4.30e-004 Z , f(a) 4.0e+006, v(m3/ml) 3.806e-004 Z 0.79, f(a) 4.39e+006, v(m3/ml) 3.43e-004 Z 0.708, f(a) 4.537e+006, v(m3/ml) 3.095e-004 Z , f(a) 4.73e+006, v(m3/ml).788e-004 Z , f(a) 4.96e+006, v(m3/ml).50e-004 Z , f(a) 5.086e+006, v(m3/ml).5e-004 Z , f(a) 5.40e+006, v(m3/ml).943e-004 Z 0.483, f(a) 5.377e+006, v(m3/ml).64e-004 Z , f(a) 5.484e+006, v(m3/ml).48e-004 Z , f(a) 5.569e+006, v(m3/ml).06e-004 Z 0.38, f(a) 5.648e+006, v(m3/ml) 9.597e-005 Z 0.394, f(a) 5.75e+006, v(m3/ml) 9.38e-005 Z , f(a) 5.80e+006, v(m3/ml) 8.975e-005 Z , f(a) 5.875e+006, v(m3/ml) 8.767e-005 Z , f(a) 5.949e+006, v(m3/ml) 8.596e-005 Z , f(a) 6.0e+006, v(m3/ml) 8.450e-005 Z , f(a) 6.095e+006, v(m3/ml) 8.34e-005 Z , f(a) 6.68e+006, v(m3/ml) 8.e-005 Z , f(a) 6.4e+006, v(m3/ml) 8.e-005 Z , f(a) 6.34e+006, v(m3/ml) 8.0e-005 Z , f(a) 6.387e+006, v(m3/ml) 7.939e-005 Z , f(a) 6.460e+006, v(m3/ml) 7.863e-005 Z , f(a) 6.533e+006, v(m3/ml) 7.793e-005 Z , f(a) 6.606e+006, v(m3/ml) 7.78e-005 Z , f(a) 6.680e+006, v(m3/ml) 7.667e-005 Frm fugacty, dg(j) Frm ntegratn f vd, dg(j) >>

98 Chapter Fugacty f Speces n a Gas Mxture v he fugacty, fˆ, f speces n a gas mxture depends n temperature, pressure, and cmpstn f the mxture. v fˆ In ths equatn, v (,, n, n,... n m ) y (,, n, n,... n m ) (4.3-) v ϕˆ ϕˆ s the fugacty ceffcent and m s the number f cmpnent. cmpletely specfy the reference state fr a mxture, we need t knw,, and cmpstn. he reference pressure s a lw enugh pressure, lw, s that the mxture behaves as an deal gas. he reference temperature s that f the system f nterest, sys and the reference cmpstn s that f the system n,sys. he fugacty, fˆ, f a speces n slutn s defned as μ μ Rln At cnstant temperature fˆ fˆ (4.3-) dμ V d μ μ lw V d Rln fˆ fˆ (4.3-3) Many equatns f state are explct n but nt V, s t s cnvenent t express the partal mlar vlume n terms f dervatve n. V V n,, n j # (4.3-4) Usng the cyclc rule at cnstant, we have V n, n V, V n, (4.3-4) Snce the reference state and the system have the same cmpstn, the ntegral n equatn (4.3-3) s carred ut wth bth n and cnstant. herefre the secnd partal dervatve n equatn (4.3-4) s replaced wth a ttal dervatve. V n, d dv n, V Rearrangng the equatn yelds 4-5

99 V n, d n, V dv (4.3-5) Equatn (4.3-3) becmes Rln fˆ fˆ fˆ Rln y lw V V IG n,, n j dv (4.3-6) Equatn (4.3-6) can be evaluated wth an equatn f state such as the eng-rbnsn equatn: R v b aα + bv b v (4.3-7) ( Rc ) R c In ths equatn, a , b , and c α [ + ( ω 0.699ω )( )] r ω s the acentrc factr. he eng-rbnsn equatn f state can be expressed n terms f the cmpressblty Z as Z 3 + (B )Z + (A 3B B)Z + (B 3 + B AB) 0 (4.3-8) m m m v In ths equatn, Z, A, B, where R x x j Aj x B r, A r, j c [ + ( ω 0.699ω )( )] r A j A j ( k j )(A A jj ) /, k j bnary nteractn parameter r B , r, he smallest rt f equatn (4.3-8), Z L, s dentfed wth the lqud phase whle the largest rt, Z V, s dentfed wth the vapr phase. Usng the eng-rbnsn equatn, the fugacty ceffcent f speces n a mxture s gven by ln ϕˆ m AB B B j Z + ( + ) B χ j Aj ln ( ) Z + B + B B (Z ) ln(z B) (4.3-9) 4-6

100 Fr the lqud Z Z L L,, and χ x mle fractn f speces n the lqud phase. Fr ϕˆ the vapr Z Z V v,, and χ y mle fractn f speces n the vapr phase. ϕˆ ϕˆ ϕˆ Example Determne the fugacty ceffcent f each speces n a gas mxture f 0% ethane n prpane at a pressure f 50 bar and a temperature f 5 C usng the eng-rbnsn EOS. Slutn he fugacty ceffcent f each speces n a gas mxture mght be btaned frm the free.k. Nguyen s prgram VLE. he results are dsplayed as fllws he prgram hermslver by Kretsky can als determne the fugacty ceffcent as shwn 4-7

101 able 4.3- lsts the Matlab prgram and the results fr the fugacty ceffcent calculatn. Nte: In ths prgram A am, B bm, A j aj, and B b. able % Example 4.3-, Fugacty ceffcent usng eng-rbnsn EOS % Bnary mxture f 0 mle % ethane () n prpane at 50 bar and 5 C % c[ ]; % Degree K c[ ]; % bar w[ ]; m; 50; ;y[..8]; r./c;r./c; b.0778 * r./ r; xk ( * w).* w; alfa + xk.* ( - sqrt(r)); a.4574 * alfa.* alfa.* r./ (r.* r); fr :m aj(, ) a(); end fr : m - fr j + : m aj(, j) sqrt(a() * a(j));aj(j, ) aj(, j); end end am0;bm0; fr : m bm bm + y() * b(); fr j : m am am + y() * y(j) * aj(, j); end end b bm - ; b am - 3 * bm * bm - * bm; b0 bm * (bm * bm + bm - am); Z; fr :0 fz((z+b)*z+b)*z+b0; dfz(3*z+*b)*z+b; ezfz/dfz;zz-ez; f abs(ez)<.0000; break; end end fprntf('z %8.5f\n',Z) ssqrt(); em (Z + (+s )* bm) / (Z + (-s ) * bm); ag lg(em)/(*s*bm); fr : m suma 0; fr j : m suma suma + y(j) * aj(, j); end 4-8

102 fuc() exp(ag * (am*b() / bm - * suma) + b() * (Z - ) / bm - lg(z - bm)); fprntf('speces %g, fugacty cef. %8.5f\n',,fuc()) end >> e4d3d Z Speces, fugacty cef Speces, fugacty cef Fugacty n the Lqud hase In the lqud phase we als need t chse a sutable reference state wth a crrespndng reference chemcal ptental and reference fugacty t cmplete the defntn defned by μ μ Rln Fr an deal slutn, we have fˆ fˆ (4.4-) μ deal g deal Rln x Rln f deal deal fˆ (4.4-) Equatn (4.4-) ndcates that the fugacty f an deal slutn s lnear n mle fractn t the pure speces fugacty deal fˆ x f deal (4.4-3) If the reference state s the pure fugacty f speces, we have the Lews/Randall rule: f deal f f (4.4-4) he actvty ceffcent, γ, s defned as the rat f the fugacty n the actual slutn t the fugacty n the deal slutn at the cmpstn f the mxture. γ l f deal fˆ ˆ l x f fˆ (4.4-5) he actvty ceffcent, γ, s smlar t the fugacty ceffcent, phase v ϕˆ, defned fr the vapr v ϕˆ fˆ p v, sys v fˆ y sys (4.4-6) 4-9

103 he fugacty ceffcent s a dmensnless quantty expressng hw the fugacty n the vapr phase cmpares t hw t wuld hypthetcally behave as an deal gas whle the actvty ceffcent represents a dmensnless quantty f hw the fugacty n the lqud phase cmpares relatve t the deal reference state. Bth f these ceffcents tell us hw far the system s devatng frm deal behavr. Fr gases, the deal state s a unque state where the ntermlecular nteractns are zer. Fr lquds, n the ther hand, the reference state s a state where all the ntermlecular nteractns are the same. he actvty f speces n the lqud, a, s ften used fr system n chemcal equlbrum. It s defned as fllws: a f l fˆ (4.4-7) f a l Actvty f a dea f a x f a a Actvty ceffcent 0 x a Fgure 4.4- Fugacty f a bnary lqud mxture. he actvty cmpares the fugacty f speces n the slutn t the fugacty f the pure speces n ts reference state. On the ther hand, the actvty ceffcent cmpares the fugacty f speces n the slutn t the fugacty f speces n an deal slutn as shwn n Fgure Cmparsn f Equatns (4.4-5) and (4.4-7) gves a relatn between the actvty and the actvty ceffcent. a x γ (4.4-8) We wll nrmally use actvty ceffcents t descrbe nn-deal slutn n phase equlbrum. Hwever actvty wll be mre cnvenent t use when we encunter chemcal equlbrum. In summary, the reference state fr speces n the lqud (r sld) phase s just a defned state, real r hypthetcal, at a gven and x (usually that f the system) and at the temperature f the system. Sme text wll defne the reference state that has the temperature f the system as a standard state t dstngush frm the reference state that s nt at the system temperature. A pure gas can be a nn-deal gas, whle a pure lqud must be an deal slutn snce all ntermlecular frces n a pure lqud are the same. An ncrease n pressure wll lead t devatn frm deal gas law, whereas devatns frm deal slutn can ccur at lw pressure snce nn-deal behavr s manly due t dssmlar ntermlecular frces between speces n a mxture. 4-0

104 Chapter 5 Appled hase Equlbrum 5. Vapr-Lqud Equlbrum fr Ideal Systems he mst cmmn prblems requrng cntact between phases chemcal engneers encunter n the chemcal, petrleum, and related ndustres nvlve vapr-lqud equlbrum. At equlbrum, the fugacty f speces n the vapr phase s equal t that n the lqud phase v fˆ l fˆ (5.-) he fugacty f speces n the vapr phase can be expressed n terms f the mle fractn, y, fugacty ceffcent,, and ttal pressure,, as v fˆ v ϕˆ v y (5.-) ϕˆ he fugacty f speces n the lqud phase can be expressed n terms f the mle fractn, x, actvty ceffcent,, and fugacty f pure cmpnent,, as herefre l fˆ v ϕˆ x γ f (5.-3) v y ϕˆ x γ f (5.-4) If the vapr phase beys deal gas law, (5.-4) becmes v ϕˆ, and the lqud slutn s deal, γ, Eq. y x f (5.-5) At lw pressure, f sat, Eq. (5.-5) becmes Rault s law y x sat (5.-6) he equlbrum rat r K-value s defned as K y x (5.-7) When Rault s law apples we have K sat (5.-8) 5-

105 In general, the K-values depend n temperature, pressure, and the cmpstn n bth phases. Fr lght hydrcarbn system (methane t decane), the K-values have been determned sem-emprcally and can be evaluated frm the equatns gven n able 5.-. In general, K s a functn f temperature, pressure, and cmpstn. able 5.- Equlbrum K values fr lght hydrcarbn systems () ln K A/ + B C ln() + D/ () ln K A/ + B C ln() + D/ (3) ln K A/ + B C ln(), where s n psa, s n R Cmpund A B C D Frm Methane () Ethylene () Ethane () rpylene () rpane () -Butane () n-butane () -entane () n-entane () n-hexane () n-heptane () n-octane (3) n-nnane () n-decane (3) In flash dstllatn, a lqud mxture s partally vaprzed and the vapr s allwed t cme t equlbrum wth the lqud. he prcess flw dagram s shwn n Fgure 5.-. he vapr and lqud phases are then separated. V, y F, x F Q L, x Fgure 5.- Flash dstllatn. Wankat,. C., Equlbrum Staged Separatns, Elsever,

106 Besdes flash calculatn, whch wll be dscussed later, there are fur types f vapr-lqud equlbrum calculatns: () Bubble pnt temperature calculatn, () Bubble pnt pressure calculatn, (3) Dew pnt temperature calculatn, and (4) Dew pnt pressure calculatn. Bubble pnt temperature calculatn In a bubble pnt temperature calculatn, the pressure and lqud phase cmpstn are specfed. We wll slve fr the temperature and the vapr cmpstn. he slutn prvdes the cmpstn f the frst bubble f vapr that frms when heat s suppled t a saturated lqud. Snce the vapr mle fractns are unknwn, we start wth the equatn m y (5.-9) Usng the K-values: K y, Eq. (5.-9) becmes x m K x (5.-0) If the system cntans mre than tw cmpnents, we mght want t slve the lg frm f equatn (5.-0) fr better cnvergence ln m 0 (5.-) K x Example Determne the temperature and cmpstn f the frst bubble created frm a saturated lqud mxture f benzene and tluene cntanng 45 mle percent benzene at 00 ka. Benzene and tluene mxtures may be cnsdered as deal. Data: Vapr pressure, sat, data: ln sat A B/( + C), where sat s n ka and s n K. Cmpund A B C Benzene () luene () Slutn We start wth the equatn y + y (E-) Substtutng y x sat / nt equatn (E-) yelds x sat + x sat (E-) 5-3

107 Wth the numercal values fr mle fractns and pressure, equatn (E-) becmes 0.45exp( /( )) exp( /( )) 00 (E-3) he bubble pnt temperature shuld be between the blng pnts f benzene and tluene gven by bl bl lg(00) lg(00) K K he slutn f the nnlnear algebrac equatn (E-3) can be determned usng Matlab functn fslve wth nlne functn as fllws: funnlne('0.45*exp( /( ))+ 0.55*exp( /( )) - 00'); >> fslve(fun,400,ptmset('dsplay','ff')) he bubble pnt temperature f the benzene-tluene mxture s 39.8 K. At ths temperature, the vapr pressure f benzene s sat exp( /( )) ka he mle fractn f benzene n the vapr phase s then y sat x (0.45)(89.45) he mle fractn f tluene n the vapr phase s y y Bubble pnt pressure calculatn In a bubble pnt pressure calculatn, the temperature and lqud phase cmpstn are specfed. We wll slve fr the pressure and the vapr cmpstn. Snce the vapr mle fractns are unknwn, we start wth the equatn m y (5.-) 5-4

108 Usng the K-values: K m y, Eq. (5.-) becmes x K x 0 (5.-3) Example Determne the pressure and cmpstn f the frst bubble created frm a saturated lqud mxture f benzene and tluene cntanng 45 mle percent benzene at 400 K. Benzene and tluene mxtures may be cnsdered as deal. Data: Vapr pressure, sat, data: ln sat A B/( + C), where sat s n ka and s n K. Cmpund A B C Benzene () luene () Slutn We start wth the equatn y + y (E-) Substtutng y x sat / nt equatn (E-) yelds x sat + x sat (E-) Wth the numercal values fr mle fractns and temperature, the bubble pnt pressure s 0.45exp( /( )) exp( /( )) ka At 400 K, the vapr pressure f benzene s sat exp( /( )) ka he mle fractn f benzene n the vapr phase s then y sat x (0.45)(35.60) he mle fractn f tluene n the vapr phase s y y

109 Dew pnt temperature calculatn In a dew pnt temperature calculatn, the pressure and vapr phase cmpstn are specfed. We wll slve fr the temperature and the lqud cmpstn. he slutn prvdes the cmpstn f the frst drp f dew that frms frm a saturated vapr. Snce the lqud mle fractns are unknwn, we start wth the equatn m x (5.-4) Usng the K-values: K y, Eq. (5.-4) becmes x m y K (5.-5) Example Determne the temperature and cmpstn f the frst dew created frm a saturated vapr mxture f benzene and tluene cntanng 45 mle percent benzene at 00 ka. Benzene and tluene mxtures may be cnsdered as deal. Data: Vapr pressure, sat, data: ln sat A B/( + C), where sat s n ka and s n K. Cmpund A B C Benzene () luene () Slutn We start wth the equatn x + x (E-) Substtutng x y / sat nt equatn (E-) yelds y + sat sat y (E-) Wth the numercal values fr mle fractns and pressure, equatn (E-) becmes 90/exp( /( )) + 0/exp( /( )) (E-3) he dew pnt temperature shuld be between the blng pnts f benzene and tluene gven by 5-6

110 bl lg(00) K bl lg(00) K he slutn f the nnlnear algebrac equatn (E-3) can be determned usng Matlab functn fslve wth nlne functn as fllws: >> funnlne('90/exp( /( ))+ 0/exp( /( )) - '); >> fslve(fun,400,ptmset('dsplay','ff')) he dew pnt temperature f the benzene-tluene mxture s 398. K. At ths temperature, the vapr pressure f benzene s sat exp( /( )) ka he mle fractn f benzene n the lqud phase s then x y sat (0.45)(00) he mle fractn f tluene n the lqud phase s x x Dew pnt pressure calculatn In a dew pnt pressure calculatn, the temperature and vapr phase cmpstn are specfed. We wll slve fr the pressure and the lqud cmpstn. Snce the lqud mle fractns are unknwn, we start wth the equatn m x (5.-6) Usng the K-values: K y, Eq. (5.-6) becmes x m y K (5.-7) 5-7

111 If the system cntans mre than tw cmpnents, we mght want t slve the lg frm f equatn (5.-7) fr better cnvergence m ln y 0 (5.-8) K Example Determne the temperature and cmpstn f the frst dew created frm a saturated vapr mxture f benzene and tluene cntanng 45 mle percent benzene at 400 K. Benzene and tluene mxtures may be cnsdered as deal. Data: Vapr pressure, sat, data: ln sat A B/( + C), where sat s n ka and s n K. Cmpund A B C Benzene () luene () Slutn We start wth the equatn x + x (E-) Substtutng x y / sat nt equatn (E-) yelds y + sat sat y sat y y + sat (E-) Wth the numercal values fr mle fractns and temperature, equatn (E-) becmes exp( /( )) exp( /( )) ka At 400 K, the vapr pressure f benzene s sat exp( /( )) ka he mle fractn f benzene n the lqud phase s then x y sat (0.45)(09.98)

112 Chapter 5 Example Cnstruct a xy dagram fr a mxture f benzene and tluene at 00 ka. Benzene and tluene mxtures may be cnsdered as deal. Data: Vapr pressure, sat, data: ln sat A B/( + C), where sat s n ka and s n K. Cmpund A B C Benzene () luene () Slutn he temperature fr the xy dagram shuld be between the blng pnts f benzene and tluene gven by bl lg(00) K bl lg(00) K he smplest prcedure s t chse a temperature between K and K, evaluate the vapr pressures, and slve fr x and y frm the fllwng equatns: y x sat (E-) At 400 K, the vapr pressure f benzene and tluene are gven by sat exp( /( )) ka sat exp( /( )) ka herefre y 35.6 x 00 y K x.7608x (E-) y x y K ( x ) ( x ) (E-) Substtutng Eq. (E-) nt (E-) yelds K x K ( x ) 5-9

113 herefre x K K K y K x.7608x he Matlab prgram lsted n able E5- plts the xy dagram shwn n Fgure E Saturated lqud Saturated vapr (K) x,y Fgure E5- Calculated xy dagram f benzene and tluene at 00 ka. able E % Example 5.-5, xy dagram fr benzene-tluene mxture at 00 ka % 00; % ka A[ ]; B[ ]; C[ ]; % Blng pnt at 00 ka bb./(a-lg())-c; fprntf('blng pnt f Benzene at %g, b %6.f C\n',,b()) fprntf('blng pnt f luene at %g, b %6.f C\n',,b()) lnspace(b(),b(),50); Kexp(A()-B()./(+C()))/; Kexp(A()-B()./(+C()))/; x(-k)./(k-k); y K.*x; 5-0

114 ymnrund(b()-);ymaxrund(b()+); plt(x,,y,,':') axs([0 ymn ymax]); grd n xlabel('x,y');ylabel('(k)') legend('saturated lqud','saturated vapr') Example We want t cndense the fllwng vapr mxture at 00 F. Ethylene C H 4 : 0% Ethane C H 6 : 0% rpane C 3 H 8 : 40% n-buane C 4 H 0 : 0% Determne the dew pnt pressure and the pressure requred t ttally cndense the vapr usng the K-values. Data: Equlbrum K values fr lght hydrcarbn systems ln K A/ + B C ln() + D/ + E/, where s n psa, s n R Cmpund A B C D E Ethylene Ethane rpane n-butane Slutn Fr dew pnt pressure calculatn, the lqud mle fractns are unknwn. We start wth the equatn m x (E-) Usng the K-values: K y, Eq. (E-) becmes x m y K (E-) We wll slve the lg frm f equatn (E-) fr better cnvergence m ln y 0 (E-3) K Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

115 We wll use Newtn s methd t slve Eq. (E-3). A revew f Newtn s methd s gven n Appendx A. he prcedure fr the calculatn s as fllws: ) Estmate vapr pressure, sat sat, by settng K and slve fr cnstants A, B, and C n the expressn fr the equlbrum rat. usng nly the three sat exp B C ) Assume a pressure 0.5 3) Evaluate K K (, ) m y 4) Evaluate f() ln K m A sat y 5) Evaluate K K (, + d) and f( + d) 6) Snce the slpe f the curve f() versus s apprxmated by Slpe f ( + d) f ( ) d f ( ) cal he calculated cal s gven by f ( ) d cal f ( + d) f ( ) Steps (3-6) are repeated untl cal errr tlerance In a bubble pnt pressure calculatn, the temperature and lqud phase cmpstn are specfed. We wll slve fr the pressure and the vapr cmpstn. Snce the vapr mle fractns are unknwn, we start wth the equatn m y (E-4) Usng the K-values: K m y, Eq. (E-4) becmes x K x 0 (E-5) We wll als use Newtn s methd t slve Eq. (E-5). he prcedure fr the calculatn s as fllws: 5-

116 ) Estmate vapr pressure, sat sat, by settng K and slve fr usng nly the three cnstants A, B, and C n the expressn fr the equlbrum rat. sat exp B C ) Assume a pressure x 3) Evaluate K K (, ) m 4) Evaluate f() K x m A 5) Evaluate K K (, + d) and f( + d) 6) Snce the slpe f the curve f() versus s apprxmated by sat Slpe f ( + d) f ( ) d f ( ) cal he calculated cal s gven by f ( ) d cal f ( + d) f ( ) Steps (3-6) are repeated untl cal errr tlerance. he fllwng Matlab prgram can be used t determne the dew and bubble pnt pressure calculatn: % Example 5.-6, % Dew pnt pressure calculatn % ; A[ ]; B[ ]; C[ ]; D[ ]; E[ ]; % Estmate vapr pressure by settng K satexp((b-a/(*))./c); y[...4.]; 5-3

117 ey*sat'; % Estmate pressure frm vapr pressure and cmpstn.5*e; fk'exp(-a/(*)+b-c*lg()+d/+e/(*))'; d;e; whle abs(e)>. Keval(fK);flg(sum(y./K)); sav;+d; Keval(fK);fdlg(sum(y./K)); ef*d/(fd-f); sav-e; end Keval(fK);xy./(K); fprntf('dew pnt pressure, (psa) %8.f\n',) dsp('speces: Ethylene Ethane rpane n-butane') fprntf('y ');dsp(y) fprntf('x ');dsp(x) dsp(' ') % Bubble pnt pressure calculatn xy;e;e; whle abs(e)>. Keval(fK);fx*K'-; sav;+d; Keval(fK);fdx*K'-; ef*d/(fd-f); sav-e; end Keval(fK);yx.*K; fprntf('bubble pnt pressure, (psa) %8.f\n',) dsp('speces: Ethylene Ethane rpane n-butane') fprntf('x ');dsp(x) fprntf('y ');dsp(y) >> e5dd6 Dew pnt pressure, (psa) Speces: Ethylene Ethane rpane n-butane y x Bubble pnt pressure, (psa) Speces: Ethylene Ethane rpane n-butane x y

118 5. Isthermal Flash Calculatn Usng K-values In flash dstllatn, a lqud mxture s partally vaprzed and the vapr s allwed t cme t equlbrum wth the lqud. he prcess flw dagram s shwn n Fgure 5.-. he vapr and lqud phases are then separated. V, y F, x F Q L, x Fgure 5.- Flash dstllatn. Makng a cmpnent balance gves Fx F Vy + Lx Vy + (F V)x (5.-) Defnng f V/F, Eq. (5.-) becmes x F fy + ( f)x (5.-) he abve equatn can be slved fr y, r fr x, y K x f f x + x F (5.-3) f x xf f ( K ) + (5.-4) We wll dscuss the slutn fr tw cases f sthermal flash calculatn. In the frst case, the fractn f the feed vaprzed, f, s specfed and n the secnd case f s determned. 5-5

119 Case : Fractn f the feed, f, vaprzed s specfed he feed cmpstn x F and the fractn f f the feed vaprzed are gven at a specfed separatr pressure, the temperature and cmpstns x and y can be calculated by slvng the equatn: m x xf f ( K (, ) ) + he prcedure fr the calculatn s as fllws: ) Determne the bubble, b, and dew, d, pnt temperatures at the feed cmpstn. ) Assume a temperature f d + ( f) b 3) Evaluate K K (, ) xf 4) Evaluate f() f ( K (, ) ) + 5) Evaluate K K ( + d, ) and f( + d) 6) Snce the slpe f the curve f() versus s apprxmated by Slpe f ( + d ) f ( ) d f ( ) cal he calculated cal s gven by cal f ( f ( ) d + d ) f ( ) Steps (3-6) are repeated untl cal errr tlerance 5-6

120 Chapter 5 Case : Fractn f the feed, f, vaprzed s determned If the feed cmpstn x F, temperature and pressure f separatr are gven, then the fractn f the feed vaprzed V/F and cmpstns x and y can be calculated. Eqs. (5.-3) and (5.-4) can be arranged s that f V/F s the nly unknwn. y x 0 (5.-5) K xf f K ) + ( F f K ) + ( x 0 (5.-6) ( K ) xf F 0 (5.-7) f ( K ) + Equatn (5.-7), whch s knwn as the Rachfrd-Rce equatn, has excellent cnvergent prpertes and can be slved by Newtn s methd. ake the dervatve f the functn F wth respect t V/F (r f), df ( K ) xf df [ f ( K ) + ] (5.-8) he fllwng prcedure can be used t slve fr V/F: ) Evaluate K K (, ) ) Check t see f s between b and d. If all K-values are greater than, the feed s a superheated vapr abve the dew pnt 3. If all K-values are less than, the feed s a subcled lqud belw the bubble pnt. If ne r mre K-values are greater than and ne r mre K-values are less than we need t evaluate Eq. (5.-7) at f 0 and at f. a) If ( K ) x < 0, the feed s belw ts bubble pnt. F ( K b) )x K F > 0 the feed s abve ts dew pnt. 3) Assume f Seader J. D., and Henley E. J., Separatn rcess rncples, Wley, 998, pg

121 ( K ) xf 4) Evaluate F f ( K ) + 5) Evaluate df df ( K ) xf [ f ( K ) + ] df 6) Let ER F/, and df f f ER 7) If abs(er) >.00 g t step 4, therwse x xf f ( K ) + and y K x Example he fllwng mxture s fed t a flash drum at 00 psa. Ethylene C H 4 : 0% Ethane C H 6 : 0% rpane C 3 H 8 : 40% n-buane C 4 H 0 : 0% Determne the temperature and the cmpstn f the vapr and lqud streams leavng the flash drum f 40% f the feed s vaprzed. Data: Equlbrum K values fr lght hydrcarbn systems ln K A/ + B C ln() + D/ + E/, where s n psa, s n R Cmpund A B C D E Ethylene Ethane rpane n-butane Slutn % Example 5.-, Isthermal Flash wth fv/f specfed % Dew pnt temperature calculatn % 00; % psa f0.4; A[ ]; B[ ]; C[ ]; D[ ]; E[ ]; % Estmate the saturatn temperature by settng K 5-8

122 satsqrt(a./(b-c*lg())); z[...4.];yz; ey*sat'; % Estmate temperature frm saturatn temperature and cmpstn e; fk'exp(-a/(*)+b-c*lg()+d/+e/(*))'; d;e; whle abs(e)>. Keval(fK);fsum(y./K)-; sav;+d; Keval(fK);fdsum(y./K)-; ef*d/(fd-f); sav-e; end d; fprntf('dew pnt temperature, (R) %8.f\n',) dsp(' ') % Bubble pnt temperature calculatn xz;e;e; whle abs(e)>. Keval(fK);flg(x*K'); sav;+d; Keval(fK);fdlg(x*K'); ef*d/(fd-f); sav-e; end b; fprntf('bubble pnt temperature, (R) %8.f\n',) dsp(' ') f*d+(-f)*b; e; fr :0 Keval(fK);fsum(z./(f*(K-)+))-; sav;+d; Keval(fK);fdsum(z./(f*(K-)+))-; ef*d/(fd-f); sav-e; f abs(e)<., break, end end Keval(fK) xz./(f*(k-)+);yx.*k; fprntf('flash temperature, (R) %8.f, V/F %8.4f\n',,f) dsp('speces: Ethylene Ethane rpane n-butane') fprntf('feed z ');dsp(z) fprntf('x ');dsp(x) fprntf('y ');dsp(y) 5-9

123 >> e5dd Dew pnt temperature, (R) Bubble pnt temperature, (R) Flash temperature, (R) 53.98, V/F Speces: Ethylene Ethane rpane n-butane Feed z x y K Example he fllwng mxture s fed t a flash drum at 00 psa. Ethylene C H 4 : 0% Ethane C H 6 : 0% rpane C 3 H 8 : 40% n-buane C 4 H 0 : 0% Determne the fractn f the feed vaprzed and the cmpstn f the vapr and lqud streams leavng the flash drum f the drum s at 400 R, 600 R, and 55 R. Data: Equlbrum K values fr lght hydrcarbn systems ln K A/ + B C ln() + D/ + E/, where s n psa, s n R Cmpund A B C D E Ethylene Ethane rpane n-butane Slutn % Example 5.-, Isthermal Flash wth specfed % Dew pnt temperature calculatn % 00; % psa nput('drum temperature (R) '); A[ ]; B[ ]; C[ ]; D[ ]; E[ ]; z[...4.]; fk'exp(-a/(*)+b-c*lg()+d/+e/(*))'; Keval(fK);KmK-; f0km*z';fsum(km.*z./k); f f0<0 5-0

124 dsp('feed s subcled lqud') elsef f>0 dsp('feed s superheated vapr') else fr :0 KmK-; Fsum((Km.*z)./(f*Km+)); df-sum((km.^).*z./(f*km+).^); erf/df;ff-er; f abs(er)<.000, break, end end xz./(f*km+);yx.*k; fprntf('flash temperature, (R) %8.f, V/F %8.4f\n',,f) dsp('speces: Ethylene Ethane rpane n-butane') fprntf('feed z ');dsp(z) fprntf('x ');dsp(x) fprntf('y ');dsp(y) fprntf('k ');dsp(k) end >> e5dd Drum temperature (R) 400 Feed s subcled lqud >> e5dd Drum temperature (R) 600 Feed s superheated vapr >> e5dd Drum temperature (R) 55 Flash temperature, (R) 55.00, V/F 0.46 Speces: Ethylene Ethane rpane n-butane Feed z x y K K Nguyen s prgram (KEQUI) can perfrm vapr-lqud equlbrum fr lght hydrcarbns usng K-values. 5-

125 5-

126 Chapter Vapr-Lqud Equlbrum wth Nn-deal Lqud In ths sectn we wll cnsder the vapr lqud equlbrum calculatn f sme real bnary system where the pressure s lw enugh fr the vapr phase t be an deal gas. At equlbrum, the fugacty f speces n the vapr phase s equal t that n the lqud phase v fˆ l fˆ (5.3-) he fugacty f speces n the vapr phase can be expressed n terms f the mle fractn, y, fugacty ceffcent,, and ttal pressure,, as v fˆ v ϕˆ v y (5.3-) ϕˆ he fugacty f speces n the lqud phase can be expressed n terms f the mle fractn, x, actvty ceffcent, γ, and fugacty f pure cmpnent, f, as herefre l fˆ x γ f (5.3-3) v y ϕˆ x γ f (5.3-4) v If the vapr phase beys deal gas law, ϕˆ, and at lw pressure, f sat, Eq. (5.3-4) becmes y x γ sat (5.3-5) he actvty ceffcent, γ, can be btaned frm expermental data and ftted t dfferent actvty ceffcent mdels. he better-knwn mdels are the Margules, Van Laar, Wlsn, NRL, and UNIQAC equatns. he actvty ceffcent mdel s nrmally btaned frm the excess Gbbs energy, g E, defned by g E x G E (5.3-6) In ths equatn, the partal mlar excess Gbbs energy s gven by E G G fˆ fˆ deal G μ μ deal R ln deal l R ln γ herefre, the excess Gbbs energy s gven n terms f actvty ceffcent as g E R x ln γ (5.3-7) 5-3

127 he actvty ceffcent mdel must satsfy the fllwng tw cndtns: ) It must prvde an deal slutn as the mle fractn f ne speces apprach ne. hat s g E 0 fr x r x ) It must satsfy the Gbbs-Duhem equatn x lnγ x, lnγ + x x 0, he smplest nn-deal equatn fr the excess Gbbs energy that satsfes these tw cndtns s g E Ax x (5.3-8) Frm the defntn f the partal excess Gbbs energy we have E G G n E,, n E ( ng ) n,, n nn n n A + n,, n E n nn G A n + + n ( n n ) n A ( n + n ) herefre E n G R ln γ A ( n + n ) he equatn fr actvty ceffcent n terms f mle fractn s then Rln γ Ax (5.3-9a) Smlarly Rln γ Ax (5.3-9b) he parameter A can be btaned by fttng Eq. (5.3-9) wth expermental data. hs parameter may change wth temperature and pressure, but t s ndependent f the cmpstn f the system. Mre sphstcated mdel fr the excess Gbbs energy wll yeld actvty ceffcent equatn wth mre than ne parameter that mght better ft the expermental data. 5-4

128 he fur cmmn equatns fr actvty ceffcent are lsted n able he Wlsn mdel s preferred due t ts mprved ablty t crrect fr temperature changes 4. Hwever the Wlsn mdel shuld nt be used fr vapr-lqud equlbrum system that cntans tw lqud phases (LLE). able 5.3- Cmmn Bnary Actvty Ceffcent Mdels Mdel Recmmended fr Margules ln γ x [a + (b a)x ] ln γ x [b + (a b)x ] Armatcs, alchl, ketnes, and ethers Van Laar A ln γ [ ] + ( Ax / Bx ) B ln γ [ ] + ( Bx / Ax ) Wlsn ln γ ln[x + x Λ ] Armatcs, alchl, ketnes, Λ Λ ethers, C 4 -C 8 hydrcarbns + x x + xλ x + xλ ln γ ln[x + x Λ ] Λ Λ + x x + xλ x + xλ NRL ln γ x τ ( ) ( ) Λ τ Λ Aqueus rgancs, LLE + x + xλ x + xλ ln γ x τ Λ + τ Λ ( x + Λ ) ( + Λ ) x x x Example Fnd the bubble pnt pressure and vapr cmpstn fr a lqud mxture f 4. ml% ethanl () and n-hexane () at 33 K. 5 Data: Actvty frm Van Laar equatns: A ln γ [ ] + ( Ax Bx / ) B ln γ [ ] + ( Bx Ax / ) A.409, and B.970 Vapr pressure: ( sat n ka and n K) ln sat ln sat /7/04 5 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

129 Slutn At 33 K sat exp[ /( )] 4.9 ka sat exp[ /( )] ka Fr the lqud slutn wth x 0.4 and x 0.588, we have ln γ [ ] + (.409x.409 /.970x ) ln γ [ ] + (.790x.790 /.409x ) γ γ.5 he partal pressure f ethanl s evaluated frm x γ sat (0.4)(.0)(4.9) ka Smlarly, the partal pressure f n-hexane s gven by x γ sat (0.588)(.5)(70.54) ka he bubble pnt pressure s then ka Mle fractn f ethanl n the vapr phase s calculated frm y / 35.55/ he actual data fr ths system s exp 0.3 ka, and y,exp

130 Example Estmate the bubble pnt temperature and vapr cmpstn fr a acetne () and water () lqud mxture wth x 0.0 at a ttal pressure f 0.3 ka 6. Use the Wlsn mdel wth the parameters Λ 0.73, Λ 0.47 Vapr pressure: ( sat n ka and n K) ln sat ln sat Slutn Snce the vapr mle fractns are unknwn, we start wth the equatn y + y (E-) Substtutng y x γ sat / nt equatn (E-) yelds x γ sat + x γ sat (E-) Fr x 0.0, x 0.99, Λ 0.73, and Λ 0.47 we have ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ.5703 γ ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ.0007 Wth the numercal values fr mle fractns and pressure, equatn (E-) becmes (0.0)(3.069)exp( /( 34.58)) + (0.99)(.0007)exp( /( )) 0.3 (E-3).3069exp( /( 34.58)) exp( /( )) 0.3 (E-4) Snce the blng pnt f pure water at 0.3 ka s 373 K and acetne s mre vlatle than water we wll use 370 K as the frst guess t the rt f the nnlnear algebrac equatn (E- 4). he slutn can be determned usng Matlab functn fslve wth nlne functn as fllws: >> funnlne('.3069*exp( /( )) *exp( /( )) - 0.3') 6 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

131 >> fslve(fun,370,ptmset('dsplay','ff')) he bubble pnt temperature f the acetne-water mxture s 36.7 K. At ths temperature, the vapr pressure f acetne s sat exp( /( )) ka he mle fractn f benzene n the vapr phase s then y sat x (0.45)(89.45) he mle fractn f tluene n the vapr phase s y y Example Estmate the dew pnt temperature and lqud cmpstn fr a acetne () and water () vapr mxture wth y 0.40 at a ttal pressure f 0.3 ka 7. Use the Wlsn mdel wth the parameters Λ 0.73, Λ 0.47 Vapr pressure: ( sat n ka and n K) ln sat ln sat Slutn Snce the lqud mle fractns are unknwn, we start wth the equatn x + x (E-) Substtutng x y /γ sat nt equatn (E-) yelds y /γ sat + y /γ sat (E-) Snce the actvty ceffcents depend n the lqud mle fractns, we assume a value fr x : Let x 0., then x x Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

132 Start Iteratn Wth Λ 0.73, and Λ 0.47 we have ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ.0484 Wth the numercal values fr mle fractns, pressure, and actvty ceffcents, equatn (E- ) becmes (0.)(0.3)/5.493/exp( /( 34.58)) + (0.9)(0.3)/.0484/exp( /( )) (E-3) We slve equatn (E-3) fr temperature and btan K wth an ntal guess f 370 K. We nw need t calculate the lqud mle fractns x y /γ sat , and x y /γ sat We nw need t update values fr the actvty ceffcents: ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ 0.35 ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ he lqud mle fractns are then reevaluated: x y /γ sat , and x y /γ sat.039 he summatn f the lqud mle fractns s gven by x + x Snce the summatn s nt equal t, we need t recalculate the lqud mle fractns usng the nrmalzed mle fractns as the guessed values x /

133 x.039/ We then g back t the Start Iteratn and repeat the prcedure usng K as the ntal temperature fr the slutn f the nn-lnear equatn (E-3). he fnal results after tw teratns are x 0.097, x , and K able 3.5- lsts the Matlab prgram used fr the calculatn. able 5.3- Matlab prgram fr dew pnt calculatn % Example % y.4;y-y; 0.3; % Assume a value fr x x.;x-x; lamda 0.73;lamda 0.47; % d; % Assume 370; % Slvng fr usng Newtn's methd fr k:0 Ax + x*lamda; Ax + x*lamda; gamaexp(-lg(a)+x*(lamda/a-lamda/a)); gamaexp(-lg(a)+x*(lamda/a-lamda/a)); fprntf('x %g, x %g\n',x,x) fprntf('gama %g, gama %g\n',gama,gama) fr :0 fy/gama/exp( /(-34.58)) + y/gama/exp( /( ))-/; +d; fy/gama/exp( /(-34.58)) + y/gama/exp( /( ))-/; ef*d/(f-f); % New value fr -d-e; fprntf('(k) %g\n',) f abs(e)<.00, break, end end xy*/gama/exp( /(-34.58));xy*/gama/exp( /( )); % Update the values f gama and gama at new values f x and x Ax + x*lamda; Ax + x*lamda; gamaexp(-lg(a)+x*(lamda/a-lamda/a)); gamaexp(-lg(a)+x*(lamda/a-lamda/a)); fprntf('x %g, x %g\n',x,x) 5-30

134 fprntf('gama %g, gama %g\n',gama,gama) xy*/gama/exp( /(-34.58));xy*/gama/exp( /( )); % Check f x+x; sumxx+x; fprntf('x %g, x %g, sumx %g\n',x,x, sumx) % New values fr x and x xx/sumx;xx/sumx; f abs(sumx-)<.000, break, end end >> e5d3d3 x 0., x 0.9 gama 5.493, gama.0484 (K) (K) (K) (K) (K) x , x gama 0.35, gama x , x.039, sumx.0777 x , x gama.08, gama.006 (K) (K) (K) x , x gama.5979, gama.006 x 0.097, x , sumx Example Cnstruct a xy dagram fr a mxture f ethanl () wth hexane () at a ttal pressure f 0.3 ka. Use the Wlsn mdel wth the parameters Λ 0.095, Λ 0.73 Vapr pressure: ( sat n ka and n K) ln sat ln sat Slutn cnstruct a xy dagram, we start wth a value f x. he prblem becmes a bubble pnt temperature calculatn. Snce the cmpstn n the vapr phase s nt knwn, we have y + y (E-) 5-3

135 Substtutng y x γ sat / nt equatn (E-) yelds x γ sat + x γ sat (E-) Fr a gven value f x, the actvty ceffcents can be btaned frm ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ.5703 γ ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ γ.0007 Wth the knwn values fr mle fractns, actvty ceffcent, and pressure, equatn (E-) becmes x γ exp( /(-55.75)) + ( x )γ exp( /( )) 0.3 (E-3) Equatn (E-3) can be slved by Newtn methd fr the temperature at a gven value f x. he blng pnt f pure ethanl and hexane can be btaned frm the vapr pressure equatn. We start the calculatn wth x 0.00 and use the blng pnt f hexane as the ntal guess fr the bubble pnt temperature f the mxture. After each bubble pnt calculatn, we ncrease x by a small amunt and repeat the calculatn usng the prevus temperature as the ntal guess. he fllwng Matlab prgram can be used t cnstruct a xy dagram at 0.3 ka. 5-3

136 % Example 5.3-4: Cnstruct a xy dagramfr ethanl () and hexane () mxture % at a ttal pressure (ka) f 0.3; % Use the Wlsn equatn wth parameters G 0.095; G 0.73; % Vapr pressure data: (ka), (K) psat 'exp( /(-55.75))'; psat 'exp( /( ))'; % Estmate blng pnts b /(6.95-lg())+55.75; b 85.4/( lg()) ; x[ ]; xlnspace(.0,.9,46); x3[ ]; xp[x x x3];nplength(xp); ypxp;pxp; p()b;p(np)b; d.0; % Wll nt wrk wth ths guess: y*b+y*b b; fr :np xxp();x-x; % Evaluate actvty ceffcents tem x + x*g; tem x + x*g; gam exp(-lg(tem)+x*(g/tem-g/tem)); gam exp(-lg(tem)+x*(g/tem-g/tem)); fr k:0 fx*gam*eval(psat)+x*gam*eval(psat)-; +d; fx*gam*eval(psat)+x*gam*eval(psat)-; df(f-f)/d; ef/df; -d-e; f abs(e)<.00, break,end end p(); yp()x*gam*eval(psat)/; fprntf('(k) %8.f, x%8.4f;y %8.4f, teratn %g\n',,x,yp(),k) end plt(xp,p,yp,p,':') axs([ ]) xlabel('x,y');ylabel('(k)'); grd n legend('x','y') 5-33

137 355 x y (K) x,y Fgure E-: xy dagram fr ethanl-hexane mxture at 0.3 ka. 5-34

138 Chapter Fttng Actvty Ceffcent Mdels wth VLE Data One vapr-lqud equlbrum data pnt s suffcent t determne the tw parameters f he actvty ceffcent equatns. hs data pnt prvdes,, x, and y s that the actvty ceffcents can be determned frm γ y /x sat (5.4-) he equatns fr the Margules mdel are ln γ x [a + (b a)x ], ln γ x [b + (a b)x ] (5.4-a,b) Slvng fr the tw parameters a and b frm Eqs. (5.4-a) and (5.4-b) yelds a lnγ x x ln +, b xγ lnγ x x lnγ + x (5.4-3a,b) he equatns fr the Van Laar mdel are A ln γ [ ] + ( Ax Bx / ), ln γ [ ] B + ( Bx Ax / ) (5.4-4a,b) Slvng fr the tw parameters A and B frm Eqs. (5.4-4a) and (5.4-4b) yelds A x + lnγ x lnγ lnγ, B + x lnγ lnγ x lnγ (5.4-5a,b) If mre than ne data pnt s avalable, curve fttng shuld be used t determne the parameters fr the actvty ceffcent equatns. As an example, Fgure 5.4- plts the mdel equatn that predcts temperature as a functn f tme usng tw parameters ε and h. hs s smlar t the actvty ceffcent mdel where the actvty ceffcent depends n lqud mle fractn and tw parameters. he relatnshp between the temperature btaned frm the mdel equatn and the expermental value,exp can be expressed generally as,exp (t; ε, h) + e (B.-) In ths equatn e s a randm errr that can be negatve r pstve. s a functn f the ndependent varable t and the parameters h and ε. he randm errr s als called the resdual, whch s the dfference between the calculated and measured values. 5-35

139 ,exp e t Fgure Relatnshp between the mdel equatn and the data Nnlnear regressn s based n determnng the values f the parameters that mnmze the sum f the squares f the resduals called an bjectve functn F bj. N e N ( ) F bj exp exp (B.-3) N, (, ) In ths equatn N s the number f data pnts r measured temperatures n ths case. he tw parameters are adjusted untl the bjectn functn reaches a mnmum. hs nnlnear regressn prcedure s presented n mre detals n Appendx B. In sme cases, the actvty ceffcent equatns can be rearranged s that a lnear curve fttng can be used t btan the parameters. he equatns fr the Margules mdel are ln γ x [a + (b a)x ], ln γ x [b + (a b)x ] (5.4-a,b) hese equatns can be rearranged t lnγ x ax + (b a)x x, lnγ x bx + (a b)x x (5.4-6a,b) Addng Eq.(5.4-6a) t (5.4-6b) yelds lnγ x + lnγ x ax + bx We nw defne a new functn Q x ln γ + x ln γ, s that lnγ Q x x x + lnγ x ax + bx a( x ) + bx 5-36

140 Q a + (b x x a)x (5.4-7) Q he plt f versus x x x wll yeld a straght lne wth slpe (b a) and ntercept a. If lnγ lnγ data s avalable fr nly ne cmpnent, we can plt vs. x r vs. x t btan x x numercal values fr the parameters a and b. If data s avalable fr bth cmpnents, Eq. (5.4-7) shuld be used t average ut the expermental errr. he Van Laar equatns can als be arranged s that the data can be ftted by a lnear equatn. he equatns fr the Van Laar mdel are A ln γ [ ] + ( Ax Bx / ), ln γ [ ] B + ( Bx Ax / ) (5.4-4a,b) akng the nverse f each term we have lnγ [ + ( Ax Bx ] / ) A, lnγ [ + ( Bx Ax ] / ) B (5.4-8a,b) akng the square rt f each term yelds lnγ A + A B x, x lnγ B + B A x x (5.4-9a,b) Agan, f there s nly data fr ne cmpnent, we can plt ether Eq. (5.4-9a) r (5.4-9b) t btan the numercal values fr A and B. If there are tw sets f data, we wll use the fllwng lnear frm x x A + x (5.4-0) Q A ABB Example Use the gven VLE data fr the ethanl ()-benzene () system at 40 C t determne Margules and Van Laar parameters 6. he vapr pressures fr ethanl and benzene at 40 C are 34.0 mmhg and 8.78 mmhg, respectvely. x y ,mmhg Slutn Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

141 Frst, we need t determne the actvty ceffcents frm the expermental data frm the fllwng equatn y x γ sat Applyng the equatn fr the frst data pnt (x.00, y.45, and 08.4) yelds γ y x sat (0.45)(08.4) (0.0)(34.0).74 γ ( y ) sat ( x ) (0.855)(08.4) (0.98)(8.78) Frm equatn (5.4-7) Q a + (b x x a)x (5.4-7) Q he plt f versus x x x wll yeld a straght lne wth slpe (b a) and ntercept a. able 5.4-a lsts the Matlab prgram used t determne the Margules parameters usng lnear curve fttng. Fgure E- shws the ftted lne and the expermental data. Fgure E- shws the calculated and measured mle fractn f ethanl. he tw btaned Margules parameters are a.73, and b

142 able 5.4-a % Example 5.4-, Determnatn f Margules parameters a and b x[ ]; y[ ]; [ ]; vap34.0;vap8.78; % Evaluate gamma,g frm y* x*g*vap x-x; gy.*./(x*vap);g(-y).*./(x*vap); % arameters fr Margules equatn ym(x.*lg(g)+x.*lg(g))./(x.*x); cplyft(x,ym,); % c() slpe, c() ntercept ac();bc()+a; fprntf('margules parameters: a %g b %g \n',a,b) xp0:.0:;xp-xp; gcexp(xp.*xp.*(a+*(b-a)*xp));gcexp(xp.*xp.*(b+*(a-b)*xp)); cxp.*gc*vap;cxp.*gc*vap;ycc./(c+c); xc[0 ]; ycplyval(c,xc); fgure() plt(xc,yc,x,ym,'') grd n legend('ftted','data') xlabel('x');ylabel('q/(x*x)') fgure() plt(xp,yc,x,y,'') grd n xlabel('x');ylabel('y') legend('ftted','data') >> e5d4d Margules parameters: a.778 b

143 .4 Ftted Data. Q/(x*x) x Fgure E- Determnatn f Margues parameters 0.9 Ftted Data y x Fgure E- Calculated and measured mle fractn f ethanl. 5-40

144 Chapter 5 Fr Van Laar mdel, the plt f x x versus x wll yeld a straght lne wth slpe Q A B AB and ntercept A. able 5.4-b lsts the Matlab prgram used t determne the Van Laar parameters usng lnear curve fttng. Fgure E-3 shws the ftted lne and the expermental data. Fgure E-4 shws the calculated and measured mle fractn f ethanl. he tw btaned Van Laar parameters are A.86, and B.555 able 5.4-b % Example 5.4-, Determnatn f Van Laar parameters A and B x[ ]; y[ ]; [ ]; vap34.0;vap8.78; % Evaluate gamma,g frm y* x*g*vap x-x; gy.*./(x*vap);g(-y).*./(x*vap); ym(x.*lg(g)+x.*lg(g))./(x.*x); xp0.0:.0:.98;xp-xp; % arameters fr van Laar equatn yv.0./ym; cplyft(x,yv,); % c() slpe, c() ntercept A/c();BA/(A*c()+); fprntf('van Laar parameters: A %g B %g \n',a,b) gvexp(a./(+a*xp./(b*xp)).^);gvexp(b./(+b*xp./(a*xp)).^); xp.*gv*vap;xp.*gv*vap;yv./(+); xc[0 ]; ycplyval(c,xc); fgure() plt(xc,yc,x,yv,'') grd n legend('ftted','data') xlabel('x');ylabel('(x*x)/q') fgure() plt(xp,yv,x,y,'') grd n xlabel('x');ylabel('y') legend('ftted','data') >> e5d4db Van Laar parameters: A.8558 B

145 0.75 Ftted Data (x*x)/q x Fgure E-3 Determnatn f Van Laar parameters 0.9 Ftted Data y x Fgure E-4 Calculated and measured mle fractn f ethanl. 5-4

146 Example Use the gven VLE data fr the methanl ()-carbn tetrachlrde () system at 0 C t determne Wlsn parameters 7. he vapr pressures fr methanl and carbn tetrachlrde at 0 C are mmhg and 9.08 mmhg, respectvely. x y ,mmhg Slutn he actvty ceffcents can be calculated frm the expermental data usng the fllwng equatn y x γ sat Applyng the equatn fr the frst data pnt (x.00, y.344, and 36.6) yelds γ y x sat (0.344)(36.6) (0.0)(96.87) 4.3 γ ( y ) sat ( x ) (0.656)(36.6) (0.98)(9.08) he expermental Q exp functn s evaluated as Q exp x ln γ + x ln γ (E-) he calculated Q cal functn s btaned by substtutng Eqs. (E-) and (E-3) fr ln γ and ln γ nt equatn (E-) ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ (E-) ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ (E-3) Q cal x ln[x + x Λ ] x ln[x + x Λ ] (E-4) he parameters Λ and Λ can be btaned by mnmzng the fllwng bjectve functn N exp cal F bj ( Q Q ) 7 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

147 able 4.5- lsts the Matlab prgram used t determne the Wlsn parameters Fmnsearch. Fgure E-5 shws the calculated and measured mle fractn f methanl. he tw btaned Wlsn parameters are Λ , and Λ 3.8 able 5.4-b % Example 5.4- % Determne parameters n the Wlsn equatn by mnmzng the bjectve % functn sum[(qexp-qcal)]^, usng fmns (Matlab functn) % Intal guesses G.; G.4; gcalfmnsearch('fq',[g G]); fprntf('g %g, G %g \n',gcal(),gcal()) Ggcal();Ggcal(); x[ ]; y[ ]; vap96.87;vap9.08; xp0.0:.0:.98;xp-xp; temg./(xp+xp*g)-g./(xp+xp*g); gexp(xp.*tem)./(xp+xp*g);gexp(-xp.*tem)./(xp+xp*g); xp.*g*vap;xp.*g*vap;yw./(+); plt(xp,yw,x,y,'') grd n xlabel('x');ylabel('y') legend('ftted','data') functn yyfq(g) % Example 5.4-, functn t evaluate the bjectve functn Fbj % fqsum[(qexp-qcal)]^ x[ ]; y[ ]; [ ]; vap96.87;vap9.08; % Evaluate gamma,g frm y* x*g*vap x-x;gg();gg(); gy.*./(x*vap);g(-y).*./(x*vap); % Qexp(x.*lg(g)+x.*lg(g)); Qcal-x.*lg(x+G*x)-x.*lg(x+G*x); yysum((qexp-qcal).^); >> e5d4d G , G

148 Ftted Data y Azetrpes x Fgure E-5 Calculated and measured mle fractn f methanl. When devatns frm Rault s law are large enugh, the x and y curves can g thrugh a maxmum r a mnmum. he extreme pnt (ether mnmum r maxmum) s called azetrpe where the lqud mle fractn s equal t the vapr mle fractn fr each speces: x y (5.5-) A system that exhbts a maxmum n pressure (pstve devatns frm Rault s law) wll exhbts a mnmum n temperature called mnmum blng azetrpe as shwn the tp part f Fgure 5.5- fr a mxture f chlrfrm and hexane. he xy dagram s pltted at 38 K and the xy dagram s pltted at atm. hs s the case when the lke nteractn s strnger than the unlke nteractn between the mlecules. he mxture wll requre less energy t g t the vapr phase and hence wll bl at a lwer temperature that that f the pure cmpnents. If the unlke nteractn s strnger than the lke nteractn we have negatve devatns frm Rault s law and the system wll exhbt a mnmum n pressure r a maxmum n temperature called maxmum blng azetrpe. A mxture f acetne and chlrfrm shws ths behavr n the bttm part f Fgure he xy dagram s pltted at 38 K and the xy dagram s pltted at atm. he data fr vapr pressure and Wlsn mdel are frm the hermslver prgram by Kretsky. hs prgram can als plt xy and xy dagrams fr dfferent mxtures. able 5.5- lsts the Matlab prgram used t prduce Fgure

149 Chlrfrm-Hexane Chlrfrm-Hexane 0.55 x y x y (atm) 0.5 (K) x,y x,y (atm) x y Acetne-Chlrfrm (K) Acetne-Chlrfrm x y x,y x,y Fgure 5.5- p: mnmum blng azetrpe fr chlrfrm and n-hexane system. Bttm: maxmum blng azetrpe fr acetne and chlrfrm system able % Fgure 5.5-: Cnstruct a xy dagramfr chlrfrm () and hexane () mxture % at a ttal pressure (atm) f ; % Use the Wlsn equatn wth parameters G.04; G ; % Vapr pressure data: (atm), (K) psat 'exp( /(-46.4))'; psat 'exp( /(-48.78))'; % Estmate blng pnts b /( lg())+46.4; b /(9.034-lg())+48.78; x[ ]; xlnspace(.0,.9,46); x3[ ]; xp[x x x3];nplength(xp); ypxp;pxp; p()b;p(np)b; d.0; b; 5-46

150 fr :np xxp();x-x; % Evaluate actvty ceffcents tem x + x*g; tem x + x*g; gam exp(-lg(tem)+x*(g/tem-g/tem)); gam exp(-lg(tem)+x*(g/tem-g/tem)); fr k:0 fx*gam*eval(psat)+x*gam*eval(psat)-; +d; fx*gam*eval(psat)+x*gam*eval(psat)-; df(f-f)/d; ef/df; -d-e; f abs(e)<.00, break,end end p(); yp()x*gam*eval(psat)/; fprntf('(k) %8.f, x%8.4f;y %8.4f, teratn %g\n',,x,yp(),k) end subplt(,,); plt(xp,p,yp,p,':') axs([ ]) xlabel('x,y');ylabel('(k)');ttle('chlrfrm-hexane') grd n legend('x','y') % % Cnstruct a xy dagramfr chlrfrm () and hexane () mxture % at a temperature (K) f 38; pvapeval(psat);pvapeval(psat); xp-xp; tem xp + xp*g; tem xp + xp*g; gam exp(-lg(tem)+xp.*(g./tem-g./tem)); gam exp(-lg(tem)+xp.*(g./tem-g./tem)); pxp.*gam*pvap;pxp.*gam*pvap; pp+p;ypp./p; subplt(,,); plt(xp,p,yp,p,':') axs([ ]) xlabel('x,y');ylabel('(atm)');ttle('chlrfrm-hexane') grd n legend('x','y',) % % Fgure 5.5-: Cnstruct a xy dagramfr acetne () and chlrfrm () mxture % at a ttal pressure (atm) f ; % Use the Wlsn equatn wth parameters G.34; G.734; % Vapr pressure data: (atm), (K) psat 'exp( /(-35.93))'; psat 'exp( /(-46.4))'; % Estmate blng pnts 5-47

151 b /(0.079-lg())+35.93; b /( lg())+46.4; x[ ]; xlnspace(.0,.9,46); x3[ ]; xp[x x x3];nplength(xp); ypxp;pxp; p()b;p(np)b; d.0; b; fr :np xxp();x-x; % Evaluate actvty ceffcents tem x + x*g; tem x + x*g; gam exp(-lg(tem)+x*(g/tem-g/tem)); gam exp(-lg(tem)+x*(g/tem-g/tem)); fr k:0 fx*gam*eval(psat)+x*gam*eval(psat)-; +d; fx*gam*eval(psat)+x*gam*eval(psat)-; df(f-f)/d; ef/df; -d-e; f abs(e)<.00, break,end end p(); yp()x*gam*eval(psat)/; fprntf('(k) %8.f, x%8.4f;y %8.4f, teratn %g\n',,x,yp(),k) end subplt(,,4); plt(xp,p,yp,p,':') axs([ ]) xlabel('x,y');ylabel('(k)');ttle('acetne-chlrfrm') grd n legend('x','y') % % Cnstruct a xy dagramfr acetne () and chlrfrm () mxture % at a temperature (K) f 38; pvapeval(psat);pvapeval(psat); xp-xp; tem xp + xp*g; tem xp + xp*g; gam exp(-lg(tem)+xp.*(g./tem-g./tem)); gam exp(-lg(tem)+xp.*(g./tem-g./tem)); pxp.*gam*pvap;pxp.*gam*pvap; pp+p;ypp./p; subplt(,,3); plt(xp,p,yp,p,':') axs([ ]) xlabel('x,y');ylabel('(atm)');ttle('acetne-chlrfrm') grd n legend('x','y',) 5-48

152 Chapter 5 he ttal pressure, deal, fr an deal mxture s gven by deal x sat + ( x ) sat (5.5-) Fr a nn-deal system, the ttal pressure s gven by x γ sat + ( x )γ sat (5.5-) If γ and γ > we have > deal r a pstve devatn frm Rault s law. Smlarly, f γ and γ < we have < deal r a negatve devatn frm Rault s law. Fgure 5.5- shws bth pstve and negatve devatn f the x curves. Ideal sat stve devatn sat Negatve devatn x 0 Fgure 5.5- stve and negatve devatn frm Rault s law. he azetrpe can be used t determne the tw parameters fr an actvty ceffcent mdel. Fr vapr-lqud equlbrum at lw pressure we have y x γ sat (5.5-3) Snce the vapr and lqud cmpstns are equal at the azetrpe, the cndtn fr equlbrum becmes γ sat (5.5-4) Snce bth the saturatn pressure and the azetrpe pressure are measurable, the actvty ceffcent can smply be btaned frm Eq. (5.5-) at the azetrpc cmpstn. 5-49

153 Example At 64.3 K and 760 mmhg, the system methanl ()-methyl ketne () frms an azetrpe cntanng 84. ml% f methanl 8. he vapr pressures fr methanl and carbn tetrachlrde at 0 C are mmhg and 9.08 mmhg, respectvely. Determne the parameters n Wlsn equatn. Vapr pressure: ( sat n mmhg and n C) lg 0 sat lg 0 sat Slutn he saturatn pressures at 64.3 K can be determned frm the gven Antne s equatns t gve sat mmhg, and sat mmhg. We then use Eq. (5.5-4) t determne the actvty ceffcents γ sat γ sat (5.5-4) Hence γ , γ he Wlsn equatns are gven by ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ (E-) ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ (E-) Eqs (E-) and (E-) can be cmbned as x (E-) + x (E-) t gve x ln γ + x ln γ x ln[x + x Λ ] x ln[x + x Λ ] (E-3) Cmbnng the lg s terms yelds x [ ] x ( x + x Λ ) [ γ ( x + x Λ ] γ (E-4) ) 8 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

154 Slvng fr Λ n terms f Λ and substtutng nt (E-) gves an equatn wth just ne unknwn Λ x Λx + Λ x γ x x + x ln [ γ ( x + Λ x )] x / x [ ( x + Λ )] γ (E-4) x Equatn (E-4) can be slved wth the knwn values f actvty ceffcents and lqud mle fractns: γ.03, γ.667, x 0.84, and x 0.58 Once Λ s evaluated, we can determne Λ frm Eq. (E-) ln γ ln[x + x Λ ] + x x Λ + x Λ x Λ + x Λ (E-) Rearrangng Eq. (E-) s that nly the unknwn Λ s n the left sde f the equatn yelds x Λ + x Λ x Λ + x Λ [ lnγ + ( x + x Λ )] ln x (E-5) he RHS f Eq. (E-5) s a knwn quantty C, therefre x Λ + x Λ C Λ Cx + Cx Λ he value f Λ s then evaluated Λ Cx Cx (E-6) able 5.5- lsts the Matlab prgram that slves equatn (E-4) fr Λ by Newtn methd and evaluates Λ frm equatn (E-6). he fllwng values are btaned Λ.088, C , and Λ

155 able % Example 5.5-, Determnatn f Wlsn parameters frm azetrpc cmpstn g.03;g.667;x.84;x-x; gxg*x;xxx/x; f'(gx/(-g*x/(x+g*x)+lg(g*(x+g*x))/xx))^xx*g*(x+g*x)-'; G;dG.0; fr :0 fgeval(f);gg+dg;fgeval(f); dfg(fg-fg)/dg;egfg/dfg; GG-dG-eG; f abs(eg)<.00, break, end end fprntf('g %8.4f\n',G) cg/(x+x*g)-(lg(g)+lg(x+x*g))/x; Gc*x/(-c*x); fprntf('g %8.4f\n',G) >> e5d5d G.088 G Estmatn f Actvty Ceffcents We smetme need t estmate actvty ceffcents at the desred cndtns frm data avalable at ther cndtns. Example We need equlbrum data fr the desgn f an absrber t remve acetne frm an ar stream usng water as a slvent 9. he equlbrum data fr the acetne ()-water () system are gven at 5 C as fllws: x y (mmhg) Use the gven data t determne the vapr equlbrum mle fractns f acetne n the system where ar s als present s that the ttal pressure s mantaned at 760 mmhg. At 5 C the vapr pressures are: sat mmhg and sat 3.76 mmhg. Slutn (Vapr phase) (Vapr phase) Acetne + water Acetne + water + ar System A System B Acetne + water Acetne + water (Lqud phase) (Lqud phase) (A) Data are avalable (B) Data are needed Fgure E- Ar s present n the system B where data are needed. 9 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

156 In ths prblem we need t estmate actvty ceffcents at the desred cndtn (Fgure E- B) frm data avalable at ther cndtns (Fgure E-A). We frst determne the actvty ceffcents frm the data avalable frm system A usng the equlbrum relatn y x γ sat (E-) Fr the frst data pnt (x 0.094, y 0.534, and 50. mmhg) we have γ x y sat (0.534)(50.) (0.094)(30.05) 5.88 Smlar calculatns lead t the fllwng table x γ (mmhg) At equlbrum n system B where ar s present we als use the equlbrum relatn fr acetne y x γ sat (E-) In equatn (E-), s the ttal pressure f system B, 760 mmhg, and γ s the actvty ceffcent btaned frm the data f system A. We have assumed that ar s essentally nsluble n acetne-water slutn and therefre the actvty ceffcents f system B are the same as thse f system A. he vapr mle fractn f acetne n system B s then calculated frm y xγ sat x γ (30.05) 760 Fr the frst data pnt (x and γ 5.88) we have y (0.094)(5.88)(30.05) he fllwng table lsts the mle fractns f acetne n the vapr phase n system B at the gven mle fractns f acetne n the lqud phase: x y (mmhg)

157 Example We want t estmate the slublty (mle fractn) f ethane () n n-heptanl () at 5 C and 0 atm 0. At 5 C the slublty f ethane n n-heptanl at atm s x he actvty ceffcents n ths system can be represented by ln γ B( x ) he K values (K f L /f G ) fr ethane are gven and n-heptanl may be cnsdered nnvlatle. At 5 C and atm: K 7.0 and at 5 C and 0 atm: K.6 Slutn At 5 C and 0 atm we have G fˆ ˆ L y f G x γ f L y x γ (f L /f G ) y x γ K f Snce n-heptanl may be cnsdered nnvlatle, y, the abve equatn becmes.6x γ (E-) We can use equatn (E-) t fnd the slublty f γ s knwn. We nw evaluate γ at 5 C and atm y x γ K (0.059)γ (7) γ (0.059)(7).394 We then determne the parameter B frm the actvty ceffcent equatn ln γ B( x ) ln(.394) B ( 0.059) If we assume that B s the same fr the system at 0 atm then we can slve equatn (E-). akng the lg f equatn (E-) yelds 0 ln(.6) + ln(x ) + ln(γ ) ln(.6) + ln(x ) ( x ) (E-) he slutn f the nnlnear algebrac equatn (E-) can be determned usng Matlab functn fslve wth nlne functn as fllws: >> ffnlne('lg(.6)+lg(x)+0.873*(-x)^') >> xfslve(ff,0.5,ptmset('dsplay','ff')) x Hence the slublty f ethane n n-heptanl at 5 C and 0 atm s x hase Behavr n artally Mscble Systems 0 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

158 When lke nteractns are sgnfcantly strnger than unlke nteractns, lquds can splt nt tw dfferent partally mscble phases t lwer the ttal Gbbs phases energy f the system. he separatn methd f lqud-lqud extractn, fund n the chemcal, petrleum, and related prcess ndustres, uses the partal lqud mscblty t separate a cmpnent frm a lqud stream. he applcatn f the extractn methd nvlves the use f slvent t preferentally dsslve ne f tw r mre cmpnents frm anther lqud phase. hs prcess requres a system f three r mre cmpnents. We wll frst fcus ur attentn t understand the behavr f bnary systems. Cnstant pressure B A V β L & V 3π a L α Q α L & V C E L & L D α β L β R G H x α s x Fgure 5.7- Cnstant-pressure dagram fr a partally mscble bnary system. x β s he phase behavr f a partally mscble bnary system at cnstant pressure s shwn n Fgure nt E dentes the state where tw lqud phases and a vapr phase are n equlbrum. Accrdng t the phase rule fr m cmpnents, we have F m π here s nly ne degree f freedm fr ths system at E. Fgure 5.7- s depcted at a specfed pressure, therefre the temperature, 3π, and cmpstns are fxed by the equlbrum cnstrants. he lqud phase α cntans mstly speces wth a mle fractn f speces equal t x α s,c where the subscrpt s dentes the saturated r maxmum slublty f speces n the slutn. he secnd lqud phase β cntans mle fractn f speces equal t x β s,d. At temperature less than 3π a three-phase system wuld have a vapr pressure less than the pressure specfed by Fgure 5.7-, and therefre the vapr phase cannt ext at ths specfed pressure. Belw the temperature 3π, Fgure 5.7- s dvded nt three lqud regns label L α, L β, and L α & L β. In the regn L α r L β nly a sngle lqud phase can exsts and n the regn L α & L β tw lqud phases exst. he regn L α represents lqud phase α that s rch n speces and the regn L β represents lqud phase β that s rch n speces. At the temperature a the lqud phase L α can have a cncentratn f speces n the range 0 < x α x α s, and the lqud phase L β can have a cncentratn f speces n the range x β s x β <. If the verall mle fractn f speces s n the range x α s < x < x β s, there wll be tw Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

159 separate lqud phases wth mle fractn f speces n each phase t be x α s and x β s respectvely. Example If ne mle f speces s mxed wth ne mle f speces at a temperature a such that x α s 0.8 and x β s 0.8, determne the amunt f lqud n each phase. Slutn he verall mle fractn f speces n the mxture s 0.5, whch s n the range x α s < x < x β s. herefre we wll have tw lqud phases. Let n α and n β be the mles f lqud n each phase α and β respectvely, we have n α + n β Applyng the mle balance fr speces yelds x α sn α + x β sn β 0.8n α + 0.8n β 0.8( n β ) + 0.8n β n β n α (0.8)() Hence we have mles f lqud α wth 0.8 mle fractn f speces and.059 mles f lqud β wth 0.8 mle fractn f speces Example If ne mle f speces s mxed wth 9 mles f speces at a temperature a such that x α s 0.8 and x β s 0.8, determne the amunt f lqud n each phase. Slutn he verall mle fractn f speces n the mxture s 0. whch s less than x α s 0.8. herefre all ten mles are n lqud phase α wth x

160 Chapter 5 L α α β L & L L β B 3π C E D α L & V V β L & V Cnstant temperature x Fgure 5.7- Cnstant-temperature dagram fr a partally mscble bnary system. he phase behavr f a partally mscble bnary system at cnstant temperature s shwn n Fgure nt E dentes the state where tw lqud phases and a vapr phase are n equlbrum. Snce the temperature specfed by Fgure 5.7- satsfes the requrement fr ne degree f freedm, the pressure, 3π, and cmpstns are fxed by the equlbrum cnstrants. At pressure hgher than 3π the vapr phase cannt exst. Abve the pressure 3π, Fgure 5.7- s dvded nt three lqud regns label L α, L β, and L α & L β. In the regn L α r L β nly a sngle lqud phase can exsts and n the regn L α & L β tw lqud phases exst. he regn L α represents lqud phase α that s rch n speces and the regn L β represents lqud phase β that s rch n speces. he bundares fr the lqud phases are almst vertcal because f neglgble change n mutual slublty wth pressure. Fr a bnary mxture f speces and n LLE (lqud-lqud equlbrum) we have α fˆ β fˆ (5.7-) he fugacty f speces n each lqud phase can be expressed n term f the mle fractn, x, actvty ceffcent, γ, and fugacty f pure cmpnent, f, as x α γ α f x β γ β f (5.7-) We wll assume that the fllwng equatns descrbe the behavr f the actvty ceffcents ln γ A( x ), and ln γ A x (5.7-3a,b) Substtutng Eqs. (5.7-3a,b) nt Eq. (5.7) yelds x α exp[a( x α ) ] x β exp[a( x β ) ] (5.7-4) Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

161 ( x α )exp[a(x α ) ] ( x β )exp[a(x β ) ] (5.7-5) Equatns (5.7-4) and (5.7-5) can be slved fr the tw unknwns x α and x β (r x α s and x β s as dented n Fgure 5.7-). he reasn fr partal lqud mscblty can be understd f we cnsder the Gbbs free energy change n frmng a slutn frm n ml f pure lqud and n ml f pure lqud. We have ΔG G n g n g (5.7-6) In ths equatn, G s the Gbbs free energy f the slutn and s gven by G n μ + n μ (5.7-7) Cmbnng Eqs. (5.7-6) and (5.7-7) gves ΔG n (μ g ) + n (μ g ) (5.7-8) Dvdng Eq. (5.7-8) by the ttal number f mles (n + n ) yelds nly ntensve prpertes ΔG n + n x (μ g ) + x (μ g ) (5.7-9) Frm the defntn f the fugacty, we have μ μ R ln f l fˆ R ln x γ (5.7-0) If the standard state s pure lqud μ g and f f, Eq. (5.7-9) becmes Δg x R ln x γ + x R ln x γ Rearrangng the equatn gves Δ g x ln x + x ln γ + x ln x + x ln γ (5.7-) R If Eqs. (5.7-3a,b: ln γ A( x ), and ln γ A x ) are used fr the actvty ceffcent, we have Δ g x ln x + x ln x + x A( x ) + x A x R (5.7-) Equatn (5.7-) s pltted n Fgure fr three values f A: A,, and 3 fr the bttm, mddle, and tp curve respectvely. he sld prtn f the tp curve between C and D represents a hypthetcal sngle lqud phase. Snce the pnts n ths curve have a hgher free energy than pnts n the dash lne CD, the lqud wll exst n tw lqud phases wth cmpstns crrespndng t the pnts C and D t lwer the free energy f the system. 5-58

162 C D dg/r x Fgure Cnstant-pressure dagram fr a bnary system wth ln γ A( x ) and ln γ A x. able 5.7- lsts the Matlab prgram used t plt Fgure able % Fgure clf Av:3; nvlength(av); x0.0:0.0:0.98;x[ x ]; x-x; hld n; fr :nv AAv(); dgx.*lg(x)+x.*lg(x)+a*x.*x.^+a*x.*x.^; plt(x,dg) end x[ ];x-x; dgx.*lg(x)+x.*lg(x)+a*x.*x.^+a*x.*x.^; plt(x,dg,':');xlabel('x');ylabel('dg/r'); grd n

163 he cmpstns crrespndng t the pnts C and D can be determned by slvng Eqs. (5.7-4) and (5.7-5) wth A 3: x α exp[3( x α ) ] x β exp[3( x β ) ] (5.7-3) ( x α )exp[3(x α ) ] ( x β )exp[3(x β ) ] (5.7-4) We can fnd the slutn t Eqs. (5.7-3) and (5.7-4) by mnmzng the fllwng bjectve functns In ths equatn, we have y f + f (5.7-5) f x α exp[3( x α ) ] x β exp[3( x β ) ] 0 (5.7-6) f ( x α )exp[3(x α ) ] ( x β )exp[3(x β ) ] 0 (5.7-7) he bjectve functn, y, wll be mnmzed (equal t zer) when f 0 and f 0. able 5.7- lsts the functn LLE representng the bjectve functn y. he Matlab cmmand fmnsearch s then used t mnmze the bjectve functn. able functn ylle(x) fx()*exp(3*(-x())^)-x()*exp(3*(-x())^); f(-x())*exp(3*x()^)-(-x())*exp(3*x()^); yf*f+f*f; >> fmnsearch('lle',[..9]) ans he cmpstns crrespndng t the pnts C and D are x α and x β 0.99 Any mxture wth an verall cmpstn n the range < x < 0.99 wll exst as a LLE system wth lqud phase α that s rch n speces and lqud phase β that s rch n speces. 5-60

164 Example Calculate the equlbrum cmpstn f the tw lqud phases n a bnary mxture f methyl dethylamne () and water () at bar and 0 C 3. he fllwng Margules mdel can be used fr ths bnary system: ln γ ( A + 3B) R ( x ) 4B R ( x ) 3 ( A 3B) ln γ R x 4B + R x 3 In these equatns, A 6349 J/ml and B 384 J/ml Slutn We have fr LLE system x α γ α f x β γ β f (E-) herefre x α ( + 3 exp[ R A B) ( x α ) 4B R ( x α ) 3 ] x β ( + 3 exp[ R A B) ( x β ) 4B R ( x β ) 3 ] ( x α )exp[ ( A 3B) R (x α ) 4B + R (x α ) 3 ] ( x β )exp[ ( A 3B) R (x β ) 4B + R (x β ) 3 ] We can fnd the slutn t the abve equatns by mnmzng the fllwng bjectve functns y f + f (E-) In ths equatn, we have f x α ( + 3 exp[ R A B) ( x α ) 4B R ( x α ) 3 ] x β ( + 3 exp[ R A B) ( x β ) 4B R ( x β ) 3 ] f ( x α )exp[ ( A 3B) R (x α ) 4B + R (x α ) 3 ] ( x β ( 3 )exp[ R A B) (x β ) 4B + R (x β ) 3 ] he bjectve functn, y, wll be mnmzed (equal t zer) when f 0 and f 0. able E- lsts the functn E5d7d3 representng the bjectve functn y. he Matlab cmmand fmnsearch s then used t mnmze the bjectve functn. 3 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

165 able E functn ye5d7d3(x) R8.34*93.5; A6349;B-384; c(a+3*b)/r;c(a-3*b)/r;c34*b/r; fx()*exp(c*(-x())^-c3*(-x())^3)-x()*exp(c*(-x())^-c3*(-x())^3); f(-x())*exp(c*x()^+c3*x()^3)-(-x())*exp(c*x()^+c3*x()^3); yf*f+f*f; >> fmnsearch('e5d7d3',[..9]) ans he equlbrum cmpstns are x α 0.04 and x β he slutns can als be btaned frm the ntersectns f the curves f x γ and ( x )γ versus x as shwn n Fgure E-... x*gamma (-x)*gamma x Fgure E- lt f x γ and ( x )γ versus x

166 Chapter 5 Appled hase Equlbrum 5.8 Vapr-Lqud-Lqud Equlbrum: VLLE Cnstant pressure B A V, β L & V Vapr phase Lqud phase α 3π a L α Q α L & V C E L & L D α β L β R Lqud phase β G x α s x x β s H Fgure 5.8-a VLLE prblem. Fgure 5.8-b A partal mscble system. We wll cnsder a system as shwn n Fgure 5.8-a when three phases are n equlbrum: a vapr phase and tw lqud phases, α and β. he phase behavr f a partally mscble bnary system at cnstant pressure s shwn n Fgure 5.8-b. nt E dentes the state where tw lqud phases and a vapr phase are n equlbrum. Accrdng t the phase rule fr m cmpnents, we have F m π here s nly ne degree f freedm fr ths system at E. Fgure 5.8-b s depcted at a specfed pressure, therefre the temperature, 3π, and cmpstns are fxed by the equlbrum cnstrants. Belw the temperature 3π, Fgure 5.8-b s dvded nt three lqud regns label L α, L β, and L α & L β. In the regn L α r L β nly a sngle lqud phase can exsts and n the regn L α & L β tw lqud phases exst. he regn L α represents lqud phase α that s rch n speces and the regn L β represents lqud phase β that s rch n speces. At the temperature a the lqud phase L α can have a cncentratn f speces n the range 0 < x α x α s, and the lqud phase L β can have a cncentratn f speces n the range x β s x β <. If the verall mle fractn f speces s n the range x α s < x < x β s, there wll be tw separate lqud phases wth mle fractn f speces n each phase t be x α s and x β s respectvely. At the temperature 3π nly lqud phase α s present at lw x and nly lqud phase β s present at hgh x. Hwever, at mle fractn x n between these tw sngle-phase regns, bth α and β lqud phases can cexst alng wth the vapr. he rdnates f pnt C, D, and 5-63

167 E prvde the mle fractns f speces n lqud phase α, n lqud phase β, and n vapr phase, respectvely. Fr a bnary mxture f speces and n VLLE (Vapr-lqud-lqud equlbrum) we have v fˆ α β fˆ fˆ (5.8-) he fugacty f speces n each lqud phase can be expressed n term f the mle fractn, x, actvty ceffcent, γ, and fugacty f pure cmpnent, f, as v fˆ x α γ α f x β γ β f (5.8-) Fr system at lw pressure v fˆ y and f sat. herefre y x α γ α sat x β γ β sat ( y ) ( x α )γ α sat ( x β )γ β sat (5.8-3) (5.8-4) We wll assume that the fllwng equatns descrbe the behavr f the actvty ceffcents ln γ A( x ), and ln γ A x (5.8-5a,b) Substtutng the expressn fr actvty ceffcents nt Eqs. (5.8-3) and (5.8-4) yelds y x α exp[a( x α ) ] sat x β exp[a( x β ) ] sat ( y ) ( x α )exp[a(x α ) ] sat ( x β )exp[a(x β ) ] sat (5.8-6) (5.8-7) At a gven temperature, the fur unknwns y, x α, x β, and can be determned frm the slutns f the abve fur nnlnear equatns. Example A bnary mxture exhbts vapr-lqud-lqud equlbrum at 300 K. he fllwng Margules mdel can be used fr ths bnary system: ln γ R b ( x ) and ln γ R b x, where b 635 J/ml. sat sat he saturatn pressures are gven by 00 ka and cmpstn f the three phases and the ttal pressure. 50 ka. Determne the Slutn Fr the tw lqud phases n equlbrum we have 4 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

168 α fˆ β fˆ (E-) he fugacty f speces n each lqud phase can be expressed n term f the mle fractn, x, actvty ceffcent, γ, and fugacty f pure cmpnent, f, as x α γ α f x β γ β f (E-) We are gven fllwng equatns fr the behavr f the actvty ceffcents ln γ A( x ), and ln γ A x (E-3) Substtutng Eqs. (E-3) nt Eq. (E-) yelds x α exp[a( x α ) ] x β exp[a( x β ) ] (E-4) ( x α )exp[a(x α ) ] ( x β )exp[a(x β ) ] (E-5) We can fnd the slutn t Eqs. (E-4) and (E-5) by mnmzng the fllwng bjectve functns y f + f (E-6) In ths equatn, we have f x α exp[a( x α ) ] x β exp[a( x β ) ] 0 f ( x α )exp[a(x α ) ] ( x β )exp[a(x β ) ] 0 (E-7) (E-8) he bjectve functn, y, wll be mnmzed (equal t zer) when f 0 and f 0. able E- lsts the functn e5d8d representng the bjectve functn y. he Matlab cmmand fmnsearch s then used t mnmze the bjectve functn. able E functn ye5d8d(x) R8.34*300; A635/R; fx()*exp(a*(-x())^)-x()*exp(a*(-x())^); f(-x())*exp(a*x()^)-(-x())*exp(a*x()^); yf*f+f*f; >> fmnsearch('e5d8d',[..9]) ans he equlbrum cmpstns are x α and x β

169 We have A R (8.34)(300) he partal pressures are calculated as x α exp[a( x α ) ] sat (0.448)exp[(.4998)(0.855) ](00) 90. ka ( x α )exp[a(x α ) ] sat (0.855)exp[(.4998)(0.448) ](50) ka ka he mle fractn f speces n the vapr phase s then y / 90./ Example At 70 C the system - exhbts partal lqud mscblty wth x α 0.3 and x β 0.7. he vapr pressures are gven by sat 600 mmhg and sat 500 mmhg. Calculate: ) he vapr mle fractn f speces and the ttal pressure when tw lqud phases and a vapr phase are n equlbrum. ) he vapr mle fractn f speces and the ttal pressure when a vapr phase s n equlbrum wth a lqud phase cntanng a mle fractn f speces f 0.. he fllwng Margules mdel can be used fr ths bnary system: ln γ A( x ) and ln γ Ax Slutn L α α β L & L L β 3π α L & V V β L & V Cnstant temperature x α s Fgure E- Cnstant-temperature dagram fr a partally mscble bnary system x x β s ) he vapr mle fractn f speces and the ttal pressure when tw lqud phases and a vapr phase are n equlbrum. Fr the tw lqud phases n equlbrum we have 5 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

170 α fˆ β fˆ (E-) he fugacty f speces n each lqud phase can be expressed n term f the mle fractn, x, actvty ceffcent, γ, and fugacty f pure cmpnent, f, as x α γ α f x β γ β f (E-) Applyng Eq. (E-) t speces yelds x α γ α x β γ β ln x α + ln γ α ln x β + ln γ β (E-3) Snce x α 0.3 and x β 0.7 we have ln A( 0.3) ln A( 0.7) A( ) ln(7/3) A.8 he values f the actvty ceffcents are then γ α exp[a( x ) ] exp[(.8)(0.7) ].834 γ α exp[ax ] exp[(.8)(0.3) ].00 he partal pressure f each speces s calculated as he ttal pressure s x α γ α sat (0.3)(.834)(600) 508. mmhg x α γ α sat (0.7)(.00)(500) 43.5 mmhg mmhg he vapr mle factn s then y / ) he vapr mle fractn f speces and the ttal pressure when a vapr phase s n equlbrum wth a lqud phase cntanng a mle fractn f speces f 0.. Snce x 0. x α 0.3, we have lqud phase α. he values f the actvty ceffcents are then γ α exp[a( x ) ] exp[(.8)(0.9) ] γ α exp[ax ] exp[(.8)(0.) ].04 he partal pressure f each speces s calculated as 5-67

171 he ttal pressure s x α γ α sat (0.)(5.560)(600) mmhg x α γ α sat (0.9)(.04)(500) mmhg mmhg he vapr mle factn s then y / Example At 5 C a bnary system cntanng cmpnents and s n a state f vapr-lqud-lqud equlbrum. he cmpstns f the saturated lqud phases are: x α s 0.0 and x β s he vapr pressures are gven by sat 0. atm and sat.0 atm. Calculate: ) he vapr mle fractn f speces and the ttal pressure when tw lqud phases and a vapr phase are n equlbrum. ) he vapr mle fractn f speces and the ttal pressure when a vapr phase s n equlbrum wth a lqud phase cntanng a mle fractn x f ) At 5 C a vapr cntanng 0% speces and 80% speces ntally at 0. atm s cmpressed sthermally. Fnd the dew pnt pressure and lqud cmpstn. Slutn ) he vapr mle fractn f speces and the ttal pressure when tw lqud phases and a vapr phase are n equlbrum. Snce the system s at lw pressure, we have y x α γ α sat x β γ β sat ( y ) ( x α )γ α sat ( x β )γ β sat (E-) (E-) Snce we d nt have the data fr actvty ceffcent and x β x β s 0.98, we assume γ β (Rault s law). herefre Smlarly he ttal pressure s y x β γ β sat (0.98)()(0.) atm ( y ) ( x α )γ α sat (0.98)()() 0.98 atm atm 6 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

172 he vapr mle factn s then y / ) he vapr mle fractn f speces and the ttal pressure when a vapr phase s n equlbrum wth a lqud phase cntanng a mle fractn x f 0.0. L α α β L & L L β B A 3π α L&V β L & V V x α s Fgure E- Cnstant-temperature dagram fr a partally mscble bnary system. x x β s Frm part () we have atm. Snce x α 0.0 we assume Henry s law wll apply fr speces n lqud phase α atm x α H α H α Fr x 0.0 x H α (0.0)(4.9) atm 4.9 atm Snce x α 0.99 we assume Rault s law wll apply fr speces n lqud phase α he ttal pressure s x α sat (0.99)() 0.99 atm atm he vapr mle factn s then y / 0.049/ ) At 5 C a vapr cntanng 0% speces and 80% speces ntally at 0. atm s cmpressed sthermally. Fnd the dew pnt pressure and lqud cmpstn. Frm the phase dagram (Fgure E-), we have lqud phase β at the dew pnt fr the vapr mxture wth 0% speces. 5-69

173 he mle fractn f speces n lqud phase β wll be clse t and the mle fractn f speces n lqud phase β wll be clse t zer. herefre we can apply Rault s law t speces and Henry s law t speces. y x β sat y x β H β (E-3) (E-4) he Henry s law cnstant can be determned frm the data gven n part () 0.98 atm x β sh β 0.0H β H β atm Snce the mle fractn n the lqud phase must add up t, we have x β + x β Substtutng the mle fractns frm Eqs. (E-3) and (E-4) nt the abve expressn yelds y + sat sat H y sat y y + H sat he dew pnt pressure s then sat y y + sat atm he mle fractn f speces n lqud phase β s gven by x β y sat (0.)(0.496)

174 Chapter 5 Example A hydrcarbn has a vapr pressure f atm at 0 C. he slublty f water n ths lqud hydrcarbn s x ws Estmate the necessary equlbrum data t desgn a dstllatn clumn t remve water frm the hydrcarbn. Slutn Vapr phase Hydrcarbn phase α Water phase β Fgure E- Vapr-lqud-lqud equlbrum system. Let α dentes the hydrcarbn phase and β dentes the water phase. We have the fllwng equlbrum relatns h x h α γ h α h sat x h β γ h β h sat w ( x h α )γ w α w sat ( x h β )γ w β w sat (E-) (E-) he slublty f lqud hydrcarbn n lqud water s als neglgble s we have γ α h, γ β w, and x β w. herefre h x α h γ α h sat h ()() sat sat h h w x β w γ β w sat w ()() sat sat w w he mle fractn f water n the vapr phase s y w w + h w Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

175 Fr the estmatn f equlbrum data fr the hydrcarbn lqud cntanng less than the saturatn water cntent, we assume that Henry s law apples t the water n ths α phase: w x w α γ w α w sat x w α H w α he Henry s law cnstant, H w α, can be evaluated frm the saturatn cndtn H w α 7.5 x w α w mmhg he partal pressure f water vapr s then w α x w he partal pressure f hydrcarbn s essentally the vapr pressure h x α h γ α h sat h ()() sat h sat h ()(760) 50 mmhg he vapr mle fractn f water s y w w + h w x x α w α w + 50 he relatve vlatlty f water t hydrcarbn, α wh, s a measure f the ease f separatn by dstllatn. he relatve vlatlty s defned as α wh y y w h / x / x w h y w /x w herefre α wh y w /x w α x w Fr x w α < 0-4, α wh α x w he hgher the value f α wh cmpares t unty, the better the separatn. he value f 55 fr relatvty vlatlty fr ths system ndcates that separatn by dstllatn wuld be very easy t accmplsh. 5-7

176 5.9 he hermdynamcs f Osmss We wll cnsder the equlbrum state f lqud mxtures n tw regns separated by a membrane that s permeable t sme f the speces present and mpermeable t thers. hs stuatn s llustrated n Fgure 5.9- where a sem-permeable membrane separates regns A that cntans a nndffusng slute and regn B that cntans nly water. Water and nn-dffusable slute A A h water B B Fgure 5.9- Osmtc pressure π A B ρ A gh Water wll dffuse frm regn B nt regn A untl the chemcal ptental r fugacty f water n each sde f the membrane s the same. hs phenmenn s called smss and the pressure dfference between regns A and B at equlbrum s the smtc pressure f regn A. he chemcal ptental f speces, μ, s defned by the fllwng relatns μ G n,, n j H n, S, n j U n S, V, n j A n V,, n j In these expressns, the subscrpt j dentes the mles f every speces except s a cnstant. he defntn can be btaned frm the fllwng dagram G H S Great hyscsts Have Study Under Very Able eachers A V U 5-73

177 We nw want an expressn that gves us the slutn smtc pressure as a functn f the slute cncentratn. At equlbrum f (, A B, x s ) (, B ) (5.9-) A W f W B where (, B A f W ) s the fugacty f water as a pure cmpnent and f W (, A, x s ) s the fugacty f water as t exsts n slutn wth slute at mle fractn x s. A smlar equatn s nt wrtten fr the slute snce t cannt dffuse thrugh the membrane. Equatn (5.9-) can A be expressed n terms f the pure water fugacty usng the actvty ceffcent A γ W A x W A f W (, A B ) (, B ) (5.9-) he fugacty s a thermdynamc functn defned by f W γ W g(, ) g f(, ) exp R IG (, ) R exp v d R 0 In ths expressn g(, ) s the mlar Gbbs free energy and g IG (, ) s the mlar Gbbs free energy as the flud apprached deal gas state. he water fugactes at states (, A ) and (, B ) are then ln A f (, ) A R A 0 v R d ln B f (, ) B R B 0 v R d Snce A > B ln A f (, ) A R B 0 v R d + R A B v R d ln A f (, ) A B f (, ) ln B + R A vd B A d B v s the mlar vlume f water, an ncmpressble lqud ln A f (, ) A B f (, ) ln B + A v( R B ) ln A B ln A f (, ) A B f (, ) ln + A A v( R B ) 5-74

178 A f (, ) A B f (, ) A A v( exp R B ) A A A Frm the equalty f fugacty, equatn (5.9-) (, A B ) (, B ), we have γ W x W f W f W A γ W A x W A fw (, B v ) exp A ( R L B W ) B f (, B W ) A Snce (, B B ) (, B ) pure water fugacty f W A γ W f W A xw v exp he smtc pressure s then A ( R L B W ) π A B R ln A A ( ) L v W γ (5.9-3) W xw Fr an deal aqueus slutn at 98 K wth x W 0.98, γ W, the smtc pressure s π A B R ln(xw ) L v W π 3 5 bar m (98 K) ml K ln(0.98) 7.8 bar m / ml A A A Fr deal slutn and small slute cncentratn,, and ln( ) ( ) x W x W x W R Hence π L ln( v W x A W ) R ( A x ) L v W W R L v W A x S (5.9-4) A x S Mles Mles slute slute + Mles slvent Mles Mles slute slvent x v A S L W ( Mles Mles slute Vlume slvent) Mles slvent slvent Mles Vlume slute slvent C S he deal dlute slutn smtc pressure, descrbed by equatn (5.9-4), s knwn as van t Hff s law. hs equatn can als be wrtten n terms f the mass cncentratn ρ S 5-75

179 π R x A L S R C S R v W ρ S Mw S (5.9-5) Mass slute where ρ S and Mw S mlecular weght f slute. Equatn (5.9-5) can be Vlume slvent used t determne slvent actvty ceffcent n a slvent-slute system prvded a sempermeable membrane can be fund. π A B R ln ( γ A A ) L v W x W W Osmtc pressure measurements are mre cmmnly used t determne the mlecular weghts f prtens and ther macrmlecules usng an smmeter shwn n Fgure At equlbrum the smtc pressure π s equal t ρgh, where ρ s the slutn densty and h s the dfference n lqud heghts. Equatn (5.9-5) s then slved fr the mlecular weght f the slute. Mw S R π ρ S h Slvent slute mxture Slvent Sempermeable membrane Fgure 5.9- A graphcal depctn f a smple smmeter. Example he plymer plyvnyl chlrde (VC) s sluble n slvent cyclhexanne. At 5 C t s fund that f a g f a specfc batch f VC per lter f slvent s placed n an smmeter, the heght h t whch the pure cyclhexanne rses s 0.85 cm. Use ths nfrmatn t estmate the mlecular weght f the VC plymer. Densty f cyclhexanne s 0.98 g/cm 3. Slutn kg m π ρgh m s m 8.7 a Mw S R π ρ S 8 Sandler, Chemcal and Engneerng hermdynamcs, Wley, 999, p

180 3 a m g Mw S K,000 3 /8.7 a 60,670 g/ml ml K m If the dlute slutn cntans N deal slutes then π R N C S, he term smle s defned as ne mle f a nndffusng and nndsscatng substance. One mle f a dsscatng substance such as NaCl s equvalent t tw smles. he number f smles per lter f slutn s called smlarty. Fr physlgcal slutns, t s cnvenent t wrk n terms f mllsmles (mosm) r mllsmlar (mosm). he number f partcles frmed by a gven slute determnes smtc pressure. Each nndffusng partcle n the slutn cntrbutes the same amunt t the smtc pressure regardless f the sze f the partcle. he smtc pressure dfference between the ntersttal and plasma fluds s due t the plasma prtens snce the prtens d nt readly pass thrugh the capllary wall. he smtc pressure created by the prtens s gven the specal name f clld smtc pressure r nctc pressure. Fr human plasma, the clld smtc pressure s abut 8 mmhg; 9 mmhg caused by the plasma prtens and 9 mmhg caused by the catns wthn the plasma that are retaned thrugh electrstatc nteractn wth the negatve surface charges f the prtens. Fgure Osmss f water thrugh red bld cell 9. If a cell such as red bld cell s placed n a hyptnc slutn that has a lwer cncentratn f slutes r smlarty, then the establshment f smtc equlbrum requres the smss f water nt the cell resultng n swellng f the cell. If the cell s placed n a hypertnc slutn wth a hgher cncentratn f slutes r smlarty, then smtc equlbrum requres smss f water ut f the cell resultng n shrnkage f the cell. An stnc slutn has the same smlarty f the cell and wll nt cause any smss f water as shwn n Fgure A 0.9 weght percent slutn f sdum chlrde r a 5 weght percent slutn f glucse s just abut stnc wth respect t a cell. 9 Seeley R.R, Stephens.D., ate., Anatmy & hyslgy, McGraw Hll,

181 Example Experments shw that at 0 C a 0. mlarty sucrse slutn has an smtc pressure (relatve t pure water) f 4.76 atm whereas a 0. mlarty NaCl slutn has an smtc pressure f 8.75 atm. Estmate the fractn f the NaCl mlecules that are dsscated at ths temperature and cncentratn. Slutn h Sucrse slutn h NaCl slutn Water Sempermeable membrane Assume that all the sucrse dsslves and let α be the dsscated fractn f NaCl. When a salt dsscates each n cntrbutes t the smtc pressure NaCl Na + + Cl If α s the fractn f NaCl that s dsscated then α s the fractn that s nt dsscated. Snce each dsscated NaCl mlecule cntrbutes ns and each undsscated mlecules cntrbutes mlecules, the smlarty f the NaCl slutn s N C NaCl, [α + ( α )]C NaCl he smlarty f the sucrse slutn s herefre, N C Sucrse, C Sucrse π π NaCl Sucrse R ( α + )C RC Sucrse NaCl α Hence α Wess.F., Cellular Bphyscs ransprt, MI ress, 996, pg

182 Chapter Dstrbutn f a slute between tw lqud phases 5.0a Slublty f a Sld n a Lqud hase he slublty f a sld n a lqud slvent and the dstrbutn f a slute between tw lqud phases wll be cnsdered n ths sectn. When a slute s transprted frm ne phase t anther, the slute must crss the nterface between phases as shwn n Fgue We assume that the slute at the nterface s n phase equlbrum. If the mle fractn y s knwn at a gven temperature, x can be determned frm the equlbrum relatn and vce-versa. Vapr y A Lqud x Vapr y A Lqud y x y x x Mass transfer frm the lqud Mass transfer frm the gas t the gas phase t the lqud phase Fgure 5.0- Slute transprt acrss the nterface. 5.0a Slublty f a sld n a lqud slvent Cnsder a bnary system wth slute () n equlbrum wth slvent () as shwn n Fgure We assume that the slvent s nt sluble n the sld s that the sld slute wll exst as a pure phase. At equlbrum, the s fugacty f the slute n the sld phase, f, s equal t the fugacty f the slute n slutn, sl sl L f. he fugacty f s related t the fugacty f pure lqud slute, f, at the equlbrum temperature and pressure f the slutn by sl L f γ x s f f (5.0-) slute () Fgure 5.0- Slute () n slvent (). In equatn (5.0-) x s the slublty r the equlbrum mle fractn f slute n slutn and γ s the actvty ceffcent f the slute. Slvng fr the slublty gves x γ f f s L (5.0-) 5-79

183 s L he rat f / f may be estmated by f f s L m ΔH exp (5.0-3) R m In ths expressn, ΔH m s the enthalpy f fusn at the nrmal meltng temperature m. Fr an deal slutn the actvty γ s equal t. Fr nn-deal slutns, an apprprate actvty ceffcent mdel must be used t calculate the slublty. Fr nn-plar slutes and slvents, the actvty ceffcent mght be btaned frm the Scatchard-Hldebrand equatn: L V ( δ δ ) Φ γ exp R (5.0-4) L In ths equatn V s the mlar vlume f the slute as a subcled lqud at the temperature L f the slutn. Hwever, V s usually assumed t be the same as the mlar vlume f the slute as a lqud at the meltng pnt. he δ s are the slublty parameters fr the slute and the slvent, and Φ s the vlume fractn f the slvent defned by the fllwng equatn. L xv Φ L x V + xv L (5.0-5) he slublty parameter s gven by δ / vap ΔH R L V (5.0-6) he heat f evapratn can be btaned frm the heat f sublmatn and the heat f fusn. vap Δ H sub Δ H m Δ H (5.0-7) If the vapr pressure f the sld s knwn as a functn f temperature, the heat f sublmatn can be estmated. ln sat A B (5.0-8) If the sld s n equlbrum wth the vapr we have dg V S V d + V V d sat S S d + V S d sat dg S (5.0-9) Rearrangng ths equatn gves Furner, R. L., Basc ransprt henmena n Bmedcal Engneerng, aylr & Francs, 007, p

184 sat d V S d S S V S V V ΔS ΔV SV SV (5.0-0) Snce ΔS SV ΔH SV /, and we can neglect the vlar vlume f the sld n cmparsn t the vlume f the vapr, equatn (5.0-0) becmes sat d SV d ΔH R sat (5.0-) We assume deal gas law n equatn (5.0-), whch may be wrtten as fllws: dln d sat ΔH R SV (5.0-) akng dervatve f equatn (5.0-8), ln sat A B, we btan dln d sat B ΔH R SV ΔH SV sub Δ H RB (5.0-3) Example A drug has a mlecular weght f 30 and a meltng temperature f 55 C. Estmate the slublty f ths drug n benzene and n n-hexane at 5 C assumng a) Ideal slutn b) Nndeal slutn usng the Scatchard-Hldebrand equatn Data: Heat f fusn f the drug 4300 cal/ml Densty f the drug.04 g/cm 3 at 5 C Vapr pressure f the sld drug ln sat (mm Hg) (K) Mlar vlume f benzene 89.4 cm 3 /ml Slublty parameter fr benzene 9. (cal/cm 3 ) / Mlar vlume f n-hexane 3.6 cm 3 /ml Slublty parameter fr n-hexane 7.3 (cal/cm 3 ) / Slutn a) he deal slublty f the drug s gven by Furner, R. L., Basc ransprt henmena n Bmedcal Engneerng, aylr & Francs, 007, p

185 s m f ΔH x exp L γ f R m where γ 4300 cal/ml - x exp K.987 cal/(ml K) he slublty s the same whether the slvent s benzene r n-hexane. b) Nndeal slutn usng the Scatchard-Hldebrand equatn ln sat (mm Hg) (K) dln d sat 896 ΔH R SV ΔH SV sub Δ H 896R Δ H (896 K)(.987 cal/ml K) cal/ml sub he heat f evapratn f the drug s then estmated: vap Δ H sub Δ H he slublty parameter f the drug s gven by m Δ H 7, ,300 3,436.9 δ δ ΔH vap L V R / 3,436.9 cal/ml (.987 cal/ml K)(98 K) cm 3 / g ( 30 g/ml ).04 / 7.6 (cal/cm 3 ) / he drug slublty s calculated frm x γ f f s L γ m ΔH exp R m V Substtutng γ exp have ( δ δ ) Φ R L L xv and Φ L x V + xv 5-8 L nt the abve equatn we

186 x V exp m ΔH exp R m ( δ δ ) ( x ) V L L L L R ( x) V + xv he abve equatn s mplct n the slublty x and can be wrtten as f(x ) x V exp m ΔH exp R m ( δ δ ) ( x ) V L L L L R ( x) V + xv 0 he nnlnear equatn f(x ) 0 can be slved usng Matlab functn fzer wth the deal slutn as the ntal guess. he slublty f drug n benzene s evaluated usng the fzer('slublty',0.) wth the functn slublty representng f(x ) 0. >> xfzer('slublty',0.) x e-00 functn yslublty(x) dhm4300; % cal/ml R.987; % cal/(ml*k) K73.5; % K m55+k; dhsubr*896; % cal/ml dhvapdhsub-dhm; VL30/.04; % mlar vlume f drug (cm3/ml) 5+K; del((dhvap-r*)/vl)^0.5; % slublty parameter fr the drug (cal/cm3)^0.5 VL89.4; % mlar vlume f slvent, benzene (cm3/ml) del9.; % slublty parameter fr benzene (cal/cm3)^0.5 temvl*(del-del)^/(r*); tem(-x)*vl/((-x)*vl+x*vl); gammaexp(tem*tem^); yx-exp(dhm*(/m-/)/r)/gamma; he slublty f drug n n-hexane s evaluated usng the fzer('slhexane',0.) wth the functn slhexane representng f(x ) 0. >> xfzer('slhexane',0.) 5-83

187 x.0748e-00 functn yslhexane(x) dhm4300; % cal/ml R.987; % cal/(ml*k) K73.5; % K m55+k; dhsubr*896; % cal/ml dhvapdhsub-dhm; VL30/.04; % mlar vlume f drug (cm3/ml) 5+K; del((dhvap-r*)/vl)^0.5; % slublty parameter fr the drug (cal/cm3)^0.5 VL3.6; % mlar vlume f slvent, n-hexane (cm3/ml) del7.3; % slublty parameter fr n-hexane (cal/cm3)^0.5 temvl*(del-del)^/(r*); tem(-x)*vl/((-x)*vl+x*vl); gammaexp(tem*tem^); yx-exp(dhm*(/m-/)/r)/gamma; he drug and n-hexane frm an deal slutn snce the slublty calculated usng nn-deal mdel (x 0.075) s very clse t the deal mdel (x,deal 0.). 5.0b Dstrbutn f a slute between lqud phases One r mre f the cmpnents n a lqud mxture mght be separated by cntactng the mxture wth anther lqud n the prcess f lqud extractn. he separatn s due t the unequal dstrbutn f a slute between tw partally mscble lqud phases. hrugh the prcess f lqud extractn, a prduct such as penclln prduced n fermentatn mxtures can be extracted nt a sutable slvent and purfed frm the fermentatn brth. Chce f slvent extractn wuld depend n txcty, cst, degree f mscblty wth the fermentatn brth, and selectvty fr the slute. We nrmally have three cmpnents, A, B, and C and tw phases n equlbrum n a lqudlqud system. Frm the phase rule, the degree f freedm F s gven by F C (5.0-4) he varables are temperature, pressure, and fur cncentratns. Fur cncentratn ccur snce the mle fractn f the cmpnents n a phase must be equal t ne: x A + x B + x C If pressure and temperature are fxed, whch s the usual case, then, at equlbrum, settng ne cncentratn n ether phase defnes the system. Cnsder an equlbrum system frm the mxng f N mles f slutes, N mles f slvent, and N 3 mles f slvent 3. At equlbrum, the fugacty f cmpnent n phase I s equal t ts fugacty n phase II. I γ I x I f (, ) II γ II x II f (, ) (5.0-5) 5-84

188 I II In ths equatn f (, ) f (, ) f (, ) fugacty f pure cmpnent at the same temperature and pressure f the system. Equatn (5.0-5) becmes I γ I x II γ II x (5.0-6) he dstrbutn ceffcent r the equlbrum cnstant K s defned as the rat f the mle fractn f cmpnent n the tw phases. K x x I II γ γ II I (5.0-7) In lqud-lqud equlbrum system, the tw partally mscble lqud phases usually frm a nn-deal slutn. he actvty ceffcents shuld be determned frm mult-cmpnent actvty mdels that can descrbe lqud-lqud equlbrum system. Equlateral trangular crdnates can be used t represent the equlbrum data fr a threecmpnent system as shwn n Fgure Each f the three crners represents a pure cmpnent, A, B, r C. he pnt M represents a mxture wth x A 0.4, x B 0., and x C 0.4.he perpendcular dstance frm the pnt M t the base AB represents the mle fractn x C f C n the mxture at M, the dstance t the base CB represents the mle fractn x A f A, and the dstance t the base AC represents the mle fractn x B f B. C x B x C M A x A B Fgure Equlateral trangular crdnates. 5-85

189 A cmmn phase dagram where cmpnents A and B are partally mscble s shwn n Fgure In ths system, lqud C dsslves cmpletely n A r B. Lqud A s nly slghtly sluble n B and B slghtly sluble n A. he phase dagram s separated nt tw regns by a curved r phase envelpe. he regn utsde the curved envelpe s the ne phase regn and the regn nsde the curved envelpe s the tw-phase regn. Any rgnal mxture wth cmpstn n the tw-phase regn wll separate ut nt tw phases wth the equlbrum cmpstns cnnected by the te lne. Fr example, a mxture wth 5 mles f A, 3 mles f B, and mles f C wll nt exst at equlbrum as a slutn at pnt M. Instead ths mxture wll separate nt lqud phase I and lqud phase II wth cmpstns gven by pnt a and b respectvely. Lqud phase I has the cmpstns x A,I 0.79, x B,I 0.03, and x C,I 0.8. Lqud phase II has the cmpstns x A,II 0.08, x B,II 0.68, and x C,II 0.4. he mles f lqud n each phase can be determned frm the materals balance. x A,I L I + x A,II L II 5 mles C ne-phase regn a M b equlbrum te lne tw-phase regn 0 A B Fgure Lqud-lqud phase dagram where cmpnents A and B are partally mscble. Snce L I + L II 0 mles, we have 0.79L I + (0.08)(0 L I ) 5 L I L II mles mles 5-86

190 he lqud-lqud equlbrum system wll be smplfed f the tw slvents are mmscble. Usually slvents are selected t mnmze ther mutual slublty s mmscble system s a reasnable assumptn. In many blgcal applcatns the slute cncentratn s s lw that the actvty ceffcents apprach ther nfnte dlutn values and the dstrbutn ceffcent s a cnstant. Cnsder an equlbrum system frm the mxng f N mles f slutes, N mles f slvent, and N 3 mles f slvent 3. If the slute cncentratn s lw then N << N and N << N 3. Fr a system wth tw mmscble lqud phases at equlbrum, a mle balance f the slutes s gven by N x,i L I + x,ii L II (5.0-8) In ths equatn L I s the mles f slvent and sme slute n lqud phase I and L II s the mles f slvent 3 and sme slute n lqud phase II. Snce the slute s present at such small quantty the value f L I and L II are assumed t be cnstant and equal t N and N 3 respectvely. Equatn (5.0-8) becmes N x,i N + x,ii N 3 (5.0-8) If the dstrbutn ceffcent s knwn, x,i K x,ii, then the mle fractn f the slute n phase II can be slved: N N K x,ii N + x,ii N 3 x,ii K N + N 3 (5.0-9) Example We have 0.0 mles (N ) f drug dsslved n 00 mles f water (N 3 ). We then add t ths phase 00 mles (N ) f ctanl. he ctanl-water parttn ceffcent fr the drug s 89. I x K 89 II x Fr ths system, ctanl (phase I) and water (phase II) are mmscble. Estmate the mle fractns f the drug n the tw phases nce equlbrum has been attaned and the % extractn f the drug frm the aqueus phase. Slutn Makng a mle balance fr the drug gves N x,i N + x,ii N 3 Substtutng x,i K x,ii nt the abve equatn we btan N K x,ii N + x,ii N 3 he mle fractn f the drug n the water phase s 3 Furner, R. L., Basc ransprt henmena n Bmedcal Engneerng, aylr & Francs, 007, p

191 N x,ii KN + N he mle fractn f the drug n the ctanl phase s x,i K x,ii (89)(. 0-6 ) he % extractn f the drug frm the aqueus phase s gven by % extractn 5 x, I N N % 0.0 Example urfcatn f an Antbtc Benzylpenclln s an lder antbtc effectve aganst pneumcccal and menngccal nfectns, anthrax, and Lyme dsease. As part f a purfcatn prcess, 00 mg f benzylpencln s mxed wth 5 ml f n-ctanl and 5 ml f water. After equlbrum s establshed, there s a water-rch phase that cntans essentally n n-ctanl and an ctanlrch phase that cntans 74 ml % n-ctanl and 6 ml % water. Determne the cncentratns f benzylpenclln n each f these phases. Data: he mlecular weght f benzylpenclln s 334.5, that f n-ctanl s 30.3, the lqud densty f n-ctanl s 0.86 g/cm 3, and K OW,B C C O B W B mg B/(ml n-ctanl) 65.5 mg B/(ml water) Slutn Snce n-ctanl s nsluble n water, the number f mles f n-ctanl n the ctanl-rch phase s 5 ml 0.86 g/ml 30.3 g/ml ml he amunt f water n the ctanl-rch phase s ml ctanl 0.6 ml water 0.74 ml ctanl ml water Assumng n change n vlume upn mxng, the vlume f the ctanl-rch phase s 4 Sandler, S. I., Chemcal, Bchemcal, and Engneerng hermdyanmcs, Wley, 006, p

192 V O ml 30.3 g/ml 0.86 g/ml ml 8 g/ml g/ml ml he vlume f the water-rch phase s V W 50 ml ml ml he ttal number f mles f benzylpenclln s 0. g g/ml ml Makng a mle balance n f benzylpenclln gves ml V W O + V O W V W W + K OW,B V O C W B C B W W ml ( ) 77.5 ml C B C B C B C B W C B O C B O C B ml/ml g/ml 0.58 mg/ml W K OW,B ml/ml C B g/ml mg/ml c Sngle-Stage Equlbrum Extractn Aqueus Stream II L, x II,n L I Extractr I L, xi,ut II L, xii,ut Fgure Sngle stage equlbrum lqud-lqud extractn. Fgure shws a sngle stage lqud extractr where a pure flwng slvent stream at mlar flw rate L I s cntacted wth an aqueus stream flwng at L II wth a slute f mle fractn x II,n. We assume that the slvent and water are mmscble and there s n change n 5-89

193 flwrates f L I r L II. he streams leavng the lqud extractr are at equlbrum s that x I,ut Kx II,ut. Makng a mle balance n the slute gves L II x II,n L I x I,ut + L II x II,ut L I Kx II,ut + L II x II,ut L II x II,n (L I K + L II ) x II,ut (5.0-0) Slvng fr x II,ut / x I,ut we btan x x II, ut II, n L KL II I I + L I I KL + II L + E (5.0-) I KL In equatn (5.0-), E s defned as the extractn factr,. he amunt f slute II L enterng the extractr s L II x II,n. he amunt f slute extracted frm the aqueus phase II s L II x II,n L II x II,ut. he % extractn f the slute frm phase II s then gven by % extractn LxII, n LxII, ut LxII, n 00 x x II, ut II, n 00 Example A drug s n an aqueus (phase II) stream flwng at 00 mles/mn at a drug mle fractn f 0.0. he aqueus stream s then cntacted wth an extractr wth a pure slvent (phase I) flwng at 00 mles/mn. he dstrbutn ceffcent fr ths partcular drug s gven by I x K 6 II x Determne the equlbrum mle fractn f the drug n the streams extng the extractr and the % extractn f the drug frm the aqueus stream. Slutn Aqueus Stream II L 00 mles/mn, x 0.0 II,n I L 00 mles/mn Extractr I L, xi,ut II L, xii,ut 5 Furner, R. L., Basc ransprt henmena n Bmedcal Engneerng, aylr & Francs, 007, p

194 Makng a mle balance n the slute gves L II x II,n L I x I,ut + L II x II,ut L I Kx II,ut + L II x II,ut L II x II,n (L I K + L II ) x II,ut Slvng fr x II,ut / x I,ut we btan x II, ut x II, n L KL II I I + L I I KL + II L + E (6)(00) x II,ut (0.0)( ) , x I,ut (6)( ) % extractn LxII, n LxII, ut LxII, n 00 x II, ut x 00 ( ) % II, n 5-9

195 Chapter 6 Chemcal Equlbrum 6. Intrductn Chemcal reactn engneerng and separatn prcesses are the tw felds unque t chemcal engneerng. he desgn f chemcal reactr ften requres extensve repettve calculatns wth the help f sphstcated cmputer mdels nw avalable t desgn engneer. Analyss f chemcal reactns s central t the desgn f chemcal reactr. We wll frst revew the stchmetry f the reactns. Cnsder the fllwng reactn n whch α mles f speces A react wth β mles f speces B t prduce ρ mles f R and ν mles f S. αa + βb ρr + νs r ρr + γs αa βb 0 4 ν r I 0 where ν stchmetrc ceffcent f speces I ν s pstve f I s a prduct ν s negatve f I s a reactant ν s zer f I s an nert Fr example, cnsder the reactn A + B C r C A B 0, then ν C, ν A, and ν C. At any tme the mles N f speces I can be related t the ntal mles N f speces I by the fllwng relatn N N + ν ζ where the quantty ζ, whch has the same unt as N, s called the mlar extent f reactn. he defntn f ζ ζ N N ν s nt the same as the defntn fr the fractnal cnversn f 6-

196 f N N N mles reacted mles fed he relatn between the mlar extent f reactn and the fractnal cnversn s N ζ N fn ν ν ζ s nt restrcted t le between 0 and. In fact, the mlar extent f reactn may be negatve f the reactn prceeds n the reverse drectn t that ndcated (e. g., f C s dsscated t frm A and B). Example Cnsder the fllwng gas phase reactn at 400 K and atm. A + B C he mle fractns f the reactve speces at equlbrum satsfy the relatn y y C A yb.4 Startng wth equmlar quanttes f A and B, and n C, calculate y A, y B, and y C, the mlar extent f reactn, and the fractnal cnversn f A and B f the reactn prceeds t equlbrum. Slutn A B C tal Intal 0 Fnal ζ ζ ζ ( ζ) Snce N A N A ζ and N B N B ζ. he mle fractn f each speces s then calculated y A ζ 0.5, y B ( ζ ) ζ, and y C ( ζ) ζ ( ζ) Frm the equlbrum relatn y C.4 y A y B r ζ ( ζ ) ζ ( ζ ) ζ ( ζ) 0.556( ζ) 6-

197 4.48 ζ 4.48ζ here are tw rts: ζ 0.56 and ζ If ζ then N B N B ζ ζ < 0 : nt pssble herefre the mlar extent f reactn ζ must be equal t 0.56, and y B ζ ( ζ ) 0.408, and y C ζ ( ζ ) 0.09 f A N A N N A A ζ N A 0.56, and f B N B N N B B ζ N B 0.3 ********************************************************************* Frm the relatn N N + ν ζ, the rate f change f the number f mles f speces can be wrtten as fllws: dn ν dt dζ dt Fr chemsts and chemcal engneers, the rate f change f the number f mles s usually wrtten n terms f the reactn rate per unt vlume r: dn dζ ν r V dt dt where V s the reactr vlume. Order f Reactns When reactn rates are determned expermentally, s ften fund that the expressn fr the reactn rate can be expressed n the fllwng way: ν d r ζ V dt α β kc A C B In ths equatn C A and C B are the reactant cncentratn, and the cnstant f prprtnalty k s the reactn rate cnstant. he expnents f the cncentratns, α and β, are called the rders f the reactn. In ths case the reactn wuld be called f rder α wth respect t A, f rder β wth respect t B, r f rder α plus β verall. Fr a reactn wth stchmetry A prducts, a frst rder reactn wll gve the rate f cnsumptn f A per unt reactr vlume as r A kc A, and a secnd rder reactn wll gve r A k(c A ) 6-3

198 Smetmes the rate f cnsumptn f A s ndependent f the cncentratn f A, r r A k Nte that the unt f k s dependent n the rder f the reactn. he zer-rder reactn s cntrlled by smethng ther than the cncentratn f A. In fact, zer-rder knetcs are very cmmn n the case f catalyzed reactns. ype f Reactrs Batch reactr: here s nether nflw f reactants nr utflw f prducts whle the reactn s beng carred ut. Sembatch reactr: Whle the reactn s beng carred ut, there s ether nflw f reactants r utflw f prducts but nt bth. Cntnuus-strred tank reactr CSR: Whle the reactn s beng carred ut, there are nflw f reactants and utflw f prducts. Except fr startup and shutdwn, the CSR s nrmally run at steady state, and s well mxed. As a result f mxng, the temperature and cncentratn are assumed t be dentcal everywhere wthn the reactr ncludng the ext pnt. ubular reactr: It cnssts f a cylndrcal ppe and s nrmally perated at steady state. Fr the systems n whch the flw s hghly turbulent, the flw feld may be mdeled by that f plug flw. he reactr s then called cntnuus plug-flw reactr FR. he reactants are cntnuusly cnsumed as they flw dwn the length f the reactr. he cncentratns vary cntnuusly nly n the axal drectn thrugh the reactr but nt n the radal drectn. BACH SEMIBACH CSR Fllng FR Dranng Fgure 6.-. Cmmn reactr types 6-4

199 Example he fllwng secnd rder, rreversble gas phase reactn AB A + B, where k cm 3 /ml mn s allwed t decmpse sthermally n a cnstant pressure batch reactr. he reactr ntally cntans pure AB wth a vlume f.0 m 3 at.5 atm and 500 C. Assumng deal behavr, determne the tme fr the reactn t reach 90% cnversn. Slutn Let X fractnal cnversn, n ntal mles f AB n the reactr. he mles f A and B n the reactr at any tme are gven by n A n B n X We have AB A B tal Intal n 0 0 n Fnal n ( X) n X n X n ( + X) Applyng the mass balance fr AB n the reactr gves d dt [ n ( X )] k(c AB ) V kn ( X ) V (E-) Snce the pressure s a cnstant, the vlume must change snce the number f mles change. Frm the deal gas law V n ( + X ) R V 0 nr V V 0 ( + X) Equatn (E-) becmes dx kn n ( X ) dt V ( + X ) 0 t V X 0 + X kn ( X ) 0 dx We can use partal fractn t btan X ( + X ) a X b ( + X ) + + X ( + X ) herefre, t V 0 kn dx 0.9 dx + X ( X ) V 0 kn ln( X ) + X 0 6-5

200 t V 0 kn ln( 0.) + ( ).9 V 0 [ ln( 0.) + 8 ] kn Frm the deal gas law V 0 R t n k R [ ln( 0.) + 8 ] Substtutng the numercal values gves t [ 0.) 8] ln( +.685[ ln( 0.) + 8] 9.9 mn We use the deal gas cnstant R 8.05 atm cm 3 /ml K 6. Chemcal Reactn and Gbbs Energy We apply the same prncple t slve phase equlbrum prblems t chemcal equlbrum. Cnsder the fllwng deal gas reactn at a ttal pressure f bar. A + B AB (6.-) Fr a system n whch we ntally have mle f A and mle f B, the mles f A and B n the reactr at any tme are gven by n A n B ζ (6.-) In ths expressn ζ s the mlar extent f reactn. We have A B AB tal Intal 0 Fnal ζ ζ ζ he ttal Gbbs energy s gven by G n G n nμ Aμ A + n B μ B + n AB μ AB (6.-3) he chemcal ptental f each cmpnent n an deal gas mxture s gven by μ g + Rln bar (6.-4) Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

201 In ths expressn, the reference state s the deal gas state at the temperature f the reactn and a partal pressure f bar. Snce we specfed the pressure f the reference state, g s a functn f nly temperature. Substtutn f equatn (6.-4) nt equatn (6.-3) yelds G n A g A + n B g B + n AB g AB + R[n A ln A + n B ln B + n AB ln AB ] (6.-5) In ths expressn, the partal pressures must have unts f bar. Snce y, we have G n A g A + n B g B + n AB g AB + R[n A lny A + n B lny B + n AB lny AB ] + R(n A + n B + n AB )ln In terms f the mlar extent f reactn, we have G ( ζ)(g A + g B) + ζg AB + R[( ζ)lny A + ( ζ)lny B + ζ lny AB ] + R ln We wll cnsder the system wth the fllwng numercal values: 000 K, bar, g A g B 0, and g AB 9.5 kj/ml. Wth a gas cnstant value R 8.34 J/ml K, we have R 8.34 kj/ml. he Gbbs energy s pltted as a functn f ζ n Fgure 6.-. he pure speces Gbbs energy s labeled erm. erm ( ζ)(g A + g B) + ζg AB 0 G erm -5-0 G(kJ) Zeta Fgure 6.- Gbbs energy f the system A + B AB. 6-7

202 able 6.- lsts the Matlab prgram used t plt Fgure 6.-. able % Fgure 6.-: A, B, and 3 AB % g3-9.5; R8.34; zeta.0:.0:.98; y(-zeta)/;yy; y3zeta; G*zeta*g3+R*((-zeta).*lg(y)+(-zeta).*lg(y)+*zeta.*lg(y3)); G*R*lg(.5);Gn*g3; zetap[0 zeta ];Gp[G G Gn]; plt(zetap,gp,[0 ],[0 Gn],':') grd n xlabel('zeta');ylabel('g(kj)'); legend('g','erm ') he prduct, speces AB, has lwer Gbbs energy than the reactants, speces A and B. Hwever, the equlbrum cnversn s nt pure speces AB, but rather the cmpstn at whch the Gbbs energy s a mnmum. hs s due t the fact that a mxture has much hgher entrpy than a pure cmpnent. he system can lwer the Gbbs energy f the reactn s nt cmpleted. Startng wth pure reactants r prduct, the system wll have a mnmum Gbbs energy when all three speces are present. he cntrbutn t the Gbbs energy f the system due t mxng s gven by the expressn ΔG mx R[( ζ)lny A + ( ζ)lny B + ζ lny AB ] We nly need t specfy the amunt f each f the elements that are present and the system temperature and pressure t predct the equlbrum state f the system. he change n Gbbs energy wth respect t temperature, pressure, and the number f mles s gven by C dg Vd Sd + μ dn (6.-6) When the cmpstn change s the result f a sngle chemcal reactn we have n n + ν ζ dn ν dζ At cnstant temperature and pressure, the change n Gbbs energy s gven by dg, μν dζ here s n change n Gbbs energy at equlbrum: dg, 0 G ζ, ν μ 0 6-8

203 Chapter he Cndtn f Equlbrum fr a Chemcal Reactn At cnstant temperature and pressure, the system cmes t chemcal equlbrum when the Gbbs energy s a mnmum. dg, μ ν dζ (6.3-) At equlbrum: dg, 0 G ζ, ν μ 0 (6.3-) Equatn (6.3-) s the cndtn f equlbrum fr a chemcal reactn that can be slved f we relate chemcal ptental t fugacty. Frm the defntn f the fugacty n a mxture we have dμ R d ln (6.3-3) fˆ Integratng ths equatn between a standard state, dented by the superscrpt, and the equlbrum state yelds μ g + R ln f fˆ (6.3-4) In ths equatn, the standard state s defned as the pure speces at the temperature f the reactn and a pressure f bar (r atm, when apprprate). Snce the Gbbs energy s nt a strng functn f pressure we wll freely nterchange atm and bar as the reference state pressure. We shuld nte the dfference between a standard state and a reference state. In a standard state, nly pressure and physcal frm (gas, lqud, r crystal structure fr sld) are specfed whle n reference state temperature, pressure and physcal frm are specfed. Fr the reference state the temperature can be specfed t be dfferent frm the reactn temperature. Fr the standard state the temperature must be the temperature f the reactn. Snce the pressure s specfed n the standard state, the pure speces mlar Gbbs energy, g, s a functn f temperature nly. Substtutng equatn (6.3-4) nt equatn (6.3-) gves ν g fˆ + R ν ln 0 (6.3-5) f he rat f fugactes s called actvty, a new thermdynamc functn: â â f fˆ (6.3-6) 6-9

204 Whle the fugacty f depends nly n the state f the system, the numercal value f the actvty als depend n the chce f standard state. Rearrangng equatn (6.3-5) and usng the mathematcal dentty that the sum f lgarthms s equal t the lg f the prducts gves ln Π f ˆ f ν ν R g rxn Δg R (6.3-7) g rxn We have defned a new term, Δ ν g, whch s called the standards Gbbs free energy change f reactn r smply the Gbbs energy f reactn. he prduct n the left hand sde f equatn (6.3-7) s called the equlbrum cnstant K K Π f ˆ f ν (6.3-8) Equatn (6.3-7) can nw be wrtten as ln K rxn Δg R (6.3-9) he reactn equatn fr a sngle reactn may be wrtten as ν I 0 (6.3-0) When mre than ne chemcal reactn s requred t descrbe chemcal equlbrum n a system we need ne equatn fr each reactn j r ν j I j 0 (6.3-) he change n Gbbs energy f the system s then μ j ν j dg dζ j (6.3-) j At equlbrum G s a mnmum, and all partal dervatves wth respect t extent f reactn ζ must equal zer G ζ j, μ j ν j 0 (6.3-3) Equatn (6.3-3) apples t each reactn, and at equlbrum the equlbrum cnstant fr each reactn must be satsfed. 6-0

205 6.4 Calculatn f Equlbrum Cnstant frm Data he Gbbs energy f reactn may be btaned frm the Gbbs energy f frmatn whch s the Gbbs energy f reactn when the speces f nterested s frmed frm ts pure elements, as fund n nature, that s, Δg Elements f Speces he Gbbs energy f frmatn f a pure element, as t s fund n nature, s equal t zer. he Gbbs energy f frmatn s avalable fr many speces at 5 C and bar. he Gbbs energy f reactn s then gven by Δg rxn,98 ν Δ g f, 98 (6.4-) In ths equatn, the stchmetrc ceffcents fr prducts are pstve and thse fr reactants are negatve. he equlbrum cnstant s then evaluated: ln K Δ g rxn,98 R (6.4-) Example Calculate the equlbrum cnstant fr the fllwng reactn usng the data at 98 K H O(g) + CH 3 OH(g) CO (g) + 3H (g) H O(g) CH 3 OH(g) CO (g) H (g) Δg f (kj/ml) Slutn Δg rxn,98 ν Δ g f, 98 Δg rxn,98 Δg rxn,98 Δg rxn,98 (Δg f) CO + 3(Δg f) H (Δg f) HO (Δg f) CH3OH ( ) CO + 3(0) H ( 8.57) HO ( 6.96) CH3OH 3.83 kj/ml he equlbrum cnstant s then Δg rxn K exp R,98 3,830 exp (8.34)(98.5) 4.69 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

206 Example Fnd the equlbrum cncentratn f N O 4 (g) [Ntrgen ntrxde] due t the chemcal reactn at 5 C and atm. N O 4 (g) NO (g) he fllwng data are avalable: () N O 4 (g) N (g) + O (g) Δg rxn, Δg f,no4 3.4 kcal/ml () 0.5N (g) + O (g) NO (g) Δg rxn, Δg f,no.4 kcal/ml Slutn he reactn N O 4 (g) NO (g) may be btaned by cmbnng reactns () and () herefre () + () N O 4 (g) NO (g) Δg rxn Δg rxn, + Δg rxn, Δg f,no4 + Δg f,no 3.4 +(.4) Δg rxn. kcal/ml 4644 J/ml R ln K Δg rxn K exp R,98 4,644 exp (8.34)(98.5) he equlbrum cnstant K can als be expressed n terms f the actvtes: K a a NO NO4 fˆ f fˆ f NO 0 NO NO4 0 NO4 (E-) 0 Assume deal gas fˆ, NO NO4 atm, equatn (E-) becmes f K fˆ f fˆ f NO 0 NO NO4 0 NO4 NO 0 NO NO 0 N O 4 4 yno atm yn O 4 atm (E-) Snce atm, equatn (E-) becmes 6-

207 yno K atm yn O 4 atm y y NO NO (E-3) We have N O 4 NO tal Intal 0 Fnal ζ ζ + ζ he mle fractns f N O 4 and NO are gven by y NO4 ζ + ζ y NO ζ + ζ Substtutng y NO4 and y NO nt equatn (E-3) gves ζ + ζ ζ + ζ ζ ( ζ4 )( + ζ ) ζ 0.563( ζ ) ζ ζ 0.93 he equlbrum cncentratn f N O 4 (g) s then ζ y NO ζ If the Gbbs energes f frmatn are avalable at the reactn temperature, the equlbrum cnstant can be calculated drectly frm Δ K exp R g rxn Hwever mst reactns we want t study are at temperatures dfferent than 5 C where data are nt avalable. We need an expressn t determne the equlbrum cnstant at any temperature frm ne set f Gbbs energy data. 6-3

208 6.5 Varatn f Equlbrum Cnstant wth emperature h Snce c p and expressns fr c p as a functn f temperature are avalable fr many speces, f we culd relate Gbbs energy t enthalpy we mght fund an equatn t determne the temperature dependence f the equlbrum cnstant. G H G H S S A V U A V U Frm the thermdynamc dagram we have S G r s g (6.5-) akng the dervatve f g / wth respect t yelds g g g (6.5-) Substtutng g s nt equatn (6.5-) gves g s g (6.5-3) Snce g h s, we have g h s s g h (6.5-4) Cmparng equatns (6.5-3) and (6.5-4) yelds g h (6.5-5) 6-4

209 Snce ln K ν R g Δg rxn R (6.5-6) Δg rxn R R ν g (ln K) (6.5-7) (ln K) R ν h Δh rxn R (6.5-8) In equatn (6.5-8) we have defned the enthalpy f reactn Δh rxn ν h. Fr exthermc reactn (Δh rxn < 0), the equlbrum cnstant decreases as temperature ncreases, snce R s always pstve. Fr endthermc reactn (Δh rxn < 0), the equlbrum cnstant ncreases as temperature ncreases. he enthalpy f reactn can be determned at any temperature based n the enthalpy f reactn at 98 K and heat capacty data frm the equatn Δh rxn Δh rxn,98 + ( c p ) 98 ν d (6.5-9) c Let p R A + B + C, equatn (6.5-9) becmes Δh rxn Δh rxn,98 + R ( A + ν B + ν C ) Integratng equatn (6.5-0) yelds 98 Δh rxn Δh rxn,98 + R[ΔA( 98) + ν d (6.5-0) Δ B ( 98 ) + Δ C ( ) ] (6.5-) 3 In ths equatn we have defned ΔA ν A, ΔB ν B, and ΔC ν C. Equatn (6.5-) can als be wrtten as where Δh rxn Δh cn + R[ΔA + ΔB + ΔC 3 ] (6.5-) 3 Δh cn Δh rxn,98 R[98ΔA + 98 Δ B Δ C ] (6.5-3) 3 Substtutng Δh rxn Δh cn + R[ΔA + ΔB + ΔC 3 ] nt equatn (6.5-8) 3 6-5

210 (ln K) Δh R rxn (6.5-8) We btan (ln K) Δh R cn Δ A Δ Δ (6.5-4) B 3C Integratng equatn (6.5-4) frm 98 K (r any temperature ) t yelds ln K K Δh cn R + ΔAln + Δ B ( ) + Δ C ( ) (6.5-5) 6 c If the heat capacty has the frm p R A + B + C + D -, equatn (6.5-9) becmes Δh rxn Δh rxn,98 + R ( ΔA + ΔB + ΔC + ΔD ) Integratng equatn (6.5-6) yelds Δh rxn Δh rxn,98 + R[ΔA( 98) + 98 Δ B ( 98 ) + Let Δh cn Δh rxn,98 R[98ΔA + 98 Δ B ΔC Δ D 3 98 reactn s then Δh rxn Δh cn + R[ΔA + ΔB + We need t ntegrate the fllwng equatn ΔC 3 3 Δ D ] d (6.5-6) Δ C ( ) ΔD 3 98 ], the expressn fr heat f ] (ln K ) Δh R cn Δ A Δ Δ ΔD B 3C he equatn fr equlbrum cnstant as a functn f temperature s fnally ln K K Δh cn R + ΔAln + Δ B ( ) + Δ C ( ) + 6 ΔD 6-6

211 Chapter 6 Example Calculate the equlbrum cnstant fr the fllwng reactn at 60 C usng the data at 98 K H O(g) + CH 3 OH(g) CO (g) + 3H (g) H O(g) CH 3 OH(g) CO (g) H (g) Δg f,98 (kj/ml) Δh f,98 (kj/ml) Slutn he equlbrum cnstant as a functn f temperature s gven by (ln K) Δh R rxn (E-) Snce the data are avalable at 5 C and K s needed at 60 C, we can ntegrate equatn (E-) assumng cnstant Δh rxn Δh rxn Δ hrxn, 98 ν Δ f h, 98 Δ Δ hrxn, 98 hrxn, 98 (Δh f) CO + 3(Δh f) H (Δh f) HO (Δh f) CH3OH ( 393.5) CO + 3(0) H ( 4.8) HO ( 00.66) CH3OH Δh rxn Δ kj/ml hrxn, 98 he equlbrum cnstant K at 98 K s evaluated Δg rxn,98 Δg rxn,98 Δg rxn,98 (Δg f) CO + 3(Δg f) H (Δg f) HO (Δg f) CH3OH ( ) CO + 3(0) H ( 8.57) HO ( 6.96) CH3OH 3.83 kj/ml he equlbrum cnstant at 98 K s then Δg rxn K exp R,98 3,830 exp (8.34)(98.5) Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

212 Integratng equatn (E-) frm 98 K t 333 K yelds ln K K Δh rxn R 98 ln K , K Example Frmaldehyde, CH O, s prduced by the gas-phase pyrlyss f methanl, CH 3 OH, accrdng t the reactn CH 3 OH(g) CH O(g) + H (g) ) Determne the equlbrum cnstant at rm temperature. ) Determne the equlbrum cnstant at 600 C and bar assumng (a) Δh rxn cnstant, and (b) Δh rxn Δh rxn (). c he heat capacty has the frm: p A + B + C + D - R CH O(g) CH 3 OH(g) H (g) Δg f,98 (kj/ml) Δh f,98 (kj/ml) ν A B C D Slutn ) Determne the equlbrum cnstant at rm temperature. Δg rxn,98 ν Δ g f, 98 Δg rxn,98 Δg rxn,98 Δg rxn,98 (Δg f) CHO + (Δg f) H (Δg f) CH3OH ( 0.0) CHO + 3(0) H ( 6.0) CH3OH 5 kj/ml 4 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

213 he equlbrum cnstant s then Δg rxn K exp R,98 5,000 exp (8.34)(98.5) Snce K a CH O a a H, very lttle frmaldehyde wll frm at 98 K. CH3OH a) Determne the equlbrum cnstant at 600 C and bar assumng he equlbrum cnstant as a functn f temperature s gven by Δ h rxn cnstant (ln K) Δh R rxn (E-) We need the enthalpy f reactn Δh t calculate K at 600 rxn C. Δh rxn Δ hrxn, 98 ν Δ f h, 98 Δ Δ hrxn, 98 hrxn, 98 (Δh f) CHO + (Δh f) H (Δh f) CH3OH ( 6.0) CHO + (0) H ( 00.7) CH3OH Δh rxn Δ 84.7 kj/ml hrxn, 98 Integratng equatn (E-) frm 98 K t 873 K yelds ln K K Δh rxn R 98 K ln 0 84, K b) Determne the equlbrum cnstant at 600 C and bar assumng Δ () h rxn Δh rxn he equlbrum cnstant at 600 C can be evaluated frm 6-9

214 ln K K Δh cn R + ΔAln + Δ B ( ) + Δ C ( ) + 6 ΔD In ths expressn Δh cn R Δh cn,98 R [98ΔA + 98 Δ B ΔC Δ D ] (E-) 3 98 ΔA ν A 3.30, ΔB ν B , ΔC ν C , ΔD ν D Substtutng the numercal values nt equatn (E-) Δh cn R [(98)(3.30) + ( ) + ( ) ] Δh cn R K erfrmng the calculatn fr the equlbrum cnstant gves K ln K exp(3.5) 8.8 K Hmgeneus Gas hase Reactn he equlbrum cnstant s gven by the expressn K Π f ˆ f ν (6.6-) Fr gas phase we chse the standard state pressure t be deal gas at bar, therefre f bar (6.6-) he fugacty n equatn (6.6-) must have unt f bar and becmes K Π ν ( [ bar] ) ˆ ( ) ν f Π y ˆ ϕ [ bar] (6.6-3) 6-0

215 Fr the calculatn f the equlbrum cncentratn we usually assume equatn (6.6-3) can be wrtten as ϕˆ ϕ s that ν K Π ( y ϕ ) (6.6-4) Fr an deal gas ϕˆ, equatn (6.6-3) can be smplfed t ν y ν K Π (6.6-4) Example Ethylene s prduced frm the decmpstn f ethane accrdng t the reactn C H 6 (g) C H 4 (g) + H (g) Determne the equlbrum cmpstn at 000 C and bar assumng Δh rxn C H 6 (g) C H 4 (g) H (g) Δg f,98 (kj/ml) Δh f,98 (kj/ml) ν cnstant. Slutn Determne the equlbrum cnstant at rm temperature Δg rxn,98 ν Δ g f, 98 Δg rxn,98 Δg rxn,98 Δg rxn,98 (Δg f) CH4 + (Δg f) H (Δg f) CH6 (68.5) CH4 + 3(0) H ( 3.84) CH kj/ml he equlbrum cnstant s then Δg rxn K exp R,98 00,990 exp (8.34)(98.5) he equlbrum cnstant as a functn f temperature s gven by (ln K ) Δh R rxn (E-) 5 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

216 We need the enthalpy f reactn t calculate K at 000 C. Δh rxn Δh rxn rxn h 98, Δ Δ f h 98, ν (Δh f) CH4 + (Δh f) H (Δh f) CH6 rxn h 98, Δ (5.6) CH4 + (0) H ( 84.68) CH6 rxn h 98, Δ kj/ml Δh rxn rxn h 98, Δ Integratng equatn (E-) frm 98 K t 873 K yelds ln K K R Δh rxn 98 ln K , K he equlbrum cnstant K can als be expressed n terms f the actvtes: K 6 4 H C H H C a a a ˆ ˆ ˆ H C H C H H H C H C f f f f f f (E-) Assume deal gas, f CH6 CH4 H bar, equatn (E-) becmes fˆ K ˆ ˆ ˆ H C H C H H H C H C f f f f f f H C H C H H H C H C bar bar bar 6 4 y y y H C H H C (E-3) Snce bar, equatn (E-3) becmes 6-

217 yc H y 4 H K bar bar yc H6 bar y C H y 4 C H y 6 H (E-4) We have C H 6 (g) C H 4 (g) + H (g) C H 6 C H 4 H tal Intal 0 0 Fnal ζ ζ ζ + ζ he mle fractns f C H 6, C H 4, and H are gven by y CH6 ζ, y CH4 + ζ ζ, and y H + ζ ζ + ζ Substtutng the mle fractns n terms f extent f reactn nt equatn (E-4) gves ζ + ζ ζ + ζ ζ ( ζ )( + ζ ) ζ 4.933( ζ ) ζ ζ 0.95 he equlbrum cncentratn f C H 4 (g) s then y CH4 ζ ζ Example Calculate the partal pressure f mnatmc hydrgen n hydrgen gas at 000 K and atm fr 0.5H (g) H(g) Data: Δh 7,990 J/ml, J/ml rxn, 98 Δsrxn, 98 K, heat capacty f H(g).5R, and heat capacty f H (g) 3 J/ml K. Slutn

218 000 K 0.5H H(g) Δh Δh 0.5H 98 K H(g) Δg rxn,000 Δh rxn,000 Δs rxn,000 he heat f reactn Δh rxn,000 and the entrpy f reactn Δs rxn,000 can be calculated usng the heat capacty data as fllws: Δh rxn, c p,h (98 000) + Δh rxn,98 + c p,h (000 98) Δh rxn,000 Δh rxn, (c p,h 0.5c p,h ) 7, ( ),830 J Δs rxn, c p,h d + Δs 000 rxn,98 + c p,h d Δs 000 rxn,98 + (c p,h 0.5c p,h )ln Δs 000 rxn, ( ) ln J/ml K 98 Δg rxn,000 Δh rxn,000 Δs rxn,000, (43.58) 5,670 J/ml he equlbrum cnstant s gven by Δ K exp g rxn,000 R 5,670 exp (8.34)(000) Fr deal gas, the fugacty can be replaced by pressure wth standard state pressure equal t atm. H atm K 0. 5 H atm H ( ) 0. 5 H H ( ) 0. 5 H he partal pressure f mnatmc hydrgen s then H atm 6-4

219 Chapter 6 Example Ammna s prduced frm a stchmetrc feed f ntrgen and hydrgen. N (g) + 3H (g) NH 3 (g) Determne the maxmum pssble cnversn at 500 C and 300 bar usng (a) deal gas law and (b) the Van der Waals equatn f state. cp he heat capacty has the frm: A + B + C + D - R NH 3 (g) N (g) H (g) Δg f,98 (kj/ml) Δh f,98 (kj/ml) ν 3 A B C D Slutn Determne the equlbrum cnstant at rm temperature Δg rxn,98 ν Δ g f, 98 Δg rxn,98 Δg rxn,98 (Δg f) NH3 (Δg f) N 3(Δg f) H ( 6.45) NH3 (0) N 3(0) H 3.9 kj/ml he equlbrum cnstant s then Δg rxn K exp R,98 3,900 exp (8.34)(98.5) he standard heat f reactn s needed t evaluate equlbrum cnstant at 500 C Δ hrxn, 98 ν Δ f h, 98 Δ hrxn, 98 (Δh f) NH3 (Δh f) N 3(Δh f) H 6 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

220 Δ hrxn, 98 ( 46.) NH3 (0) N 3(0) H 9. kj/ml he equlbrum cnstant at 500 C can be evaluated frm ln K K Δh cn R + ΔAln + Δ B ( ) + ΔD (E-) In ths expressn Δh cn R Δh cn,98 R [98ΔA + 98 ΔB Δ D ] (E-) 98 ΔA ν A 5.87, ΔB ν B , ΔC ν C 0, ΔD ν D Substtutng the numercal values nt equatn (E-) Δh cn R 9, [(98)( 5.87) + ( ) ] Δh cn R K erfrmng the calculatn fr the equlbrum cnstant gves K ln K K exp( 4.39) he equlbrum cnstant K can als be expressed n terms f the actvtes: anh3 K 3 a a N H NH3 3 f NH 3 fˆ f fˆ N N fˆ f H H (E-3) We chse the standard state pressure t be deal gas at bar, therefre f bar 6-6

221 he fugacty f speces n the vapr phase can be expressed n terms f the mle fractn, y, fugacty ceffcent,, and ttal pressure,, as ϕˆ fˆ y ϕˆ Hence (a) Ideal gas law y ˆ NH ϕ 3 NH 3 K 3 3 y ϕ y ˆ ϕ N N H H - (E-4) Fr deal gas, ϕˆ, the equlbrum cnstant s smplfed t ynh3 K deal 3 y y N H - (E-5) We have N (g) + 3H (g) NH 3 (g) N H NH 3 tal Intal Fnal ζ 3( ζ) ζ 4 ζ he mle fractns f N, H, and NH 3 are gven by y N ζ, y H 4 ζ 3 3ζ, and y NH3 4 ζ ζ 4 ζ Substtutng the mle fractns n terms f extent f reactn nt equatn (E-5) yelds ( ) ( ) ζ 4 ζ ( ζ )( 3 3ζ ) 3 - Fr 300 bar, we have (ζ) (4 ζ).3347( ζ)(3 3ζ) 3 We can use the fllwng Matlab statements t slve the nnlnear equatn >> ffnlne('(*z)^*(4-*z)^-.3347*(-z)*(3-3*z)^3'); >> zetafslve(ff,.5,ptmset('dsplay','ff')) zeta

222 he extent f reactn at 500 K and 300 bar s ζ (b) he Van der Waals equatn f state. We assume that ϕˆ ϕ fugacty f pure speces that can be evaluated frm the Van de Waals equatn f state lnϕ (Z ) Z A ln (Z B) (E-6) Frm Example 4.-, the Van der Waals equatn f state R v b a v can be wrtten n v terms f the cmpressblty factr Z as R Z 3 ( + B)Z + AZ AB 0 (E-7) b a In ths equatn B and A R (R ) the crtcal prpertes: where the cnstant a and b can be evaluated frm 7 a 64 ( R c ) c and b R 8 c c (E-8) able E- lsts the crtcal prpertes, the parameters a and b evaluated frm equatn (E-8), the cmpressblty frm equatn (E-7), and the fugacty ceffcent frm equatn (E-6). able E- Summary f Fugacty Ceffcent Calculatn NH 3 (g) N (g) H (g) c [ K] c [atm] a [a m 6 /ml ] b [m 3 /ml] Z ϕ he fugacty ceffcents can be substtuted nt equatn (E-4) y ˆ NH ϕ 3 NH3 K 3 3 y ϕ y ˆ ϕ N N H H - (E-4) ynh3 K y y N H - 6-8

223 In terms f the extent f reactn the abve equatn becmes ( ) ( ) ζ 4 ζ ( ζ )( 3 3ζ ) 3 - Fr 300 bar, we have (ζ) (4 ζ).5868( ζ)(3 3ζ) 3 We can use the fllwng Matlab statements t slve the nnlnear equatn >> ffnlne('(*z)^*(4-*z)^-.5868*(-z)*(3-3*z)^3'); >> zetafslve(ff,.5,ptmset('dsplay','ff')) zeta 0.43 he extent f reactn at 500 K and 300 bar s ζ able E- lsts the Matlab prgram t evaluate the fugacty ceffcent able E % Example 6.6-c, cmpressblty usng Van der Waals EQS % R 8.34; % a.m3/(ml.k) c[ ] ; % Degree K c[ ]*.035e5 ; %a av7*(r*c).^./(64*c); % a.m6/ml bvr*c./(8*c); % m3/ml name'nh3*n*h'; % 3e7; % a 773; % K fr c:3 b3*(c-)+;eb+; nname(b:e); aav(c);bbv(c); % dsp(n) fprntf('a(a.m6/ml) %g, b(m3/ml) %g\n',a,b) Aa*/(R*)^;Bb*/(R*); fprntf('a %8.5e, B %8.5e\n',A,B) b-(+b);ba;b0-a*b; fprntf('b %8.5e, b %8.5e, b0 %8.5e\n',b,b,b0) Z; fr :0 fz((z+b)*z+b)*z+b0; 6-9

224 dfz(3*z+*b)*z+b; ezfz/dfz;zz-ez; f abs(ez)<.0000; break; end end fexp((z-)-a/z-lg(z-b)); fprntf('z %8.5f, f/ %8.5f\n',Z,f) end >> e6d6dc NH3 a(a.m6/ml) 0.458, b(m3/ml) e-005 A e-00, B.74434e-00 b e+000, b e-00, b e-00 Z , f/ *N a(a.m6/ml) , b(m3/ml) e-005 A e-00, B.80364e-00 b e+000, b e-00, b e-00 Z.05, f/.0979 *H a(a.m6/ml) , b(m3/ml).6683e-005 A.8096e-00, B.4557e-00 b -.456e+000, b.8096e-00, b e-003 Z.007, f/ Hetergeneus Reactn We wll cnsder an equlbrum system wth a gas phase and ne r mre cndensed phases. he mle fractns n the equlbrum relatns refer t the mle fractn n a gven phase, nt the ttal mle fractn. When the hetergeneus system s n equlbrum, there wll be chemcal equlbrum n the gas phase and als phase equlbrum between speces n the gas phase and the pure cndensed phases. When a pure cndensed phase s present, the partal pressure f that cmpnent n the gas phase wll equal the vapr r saturatn pressure f the pure cndensed phase. Hence a pure cndensed phase cannt exst f the partal pressure f that cmpnent s less than the vapr pressure. he actvty f a pure cndensed phase s gven by a f f In ths defntn, f s the fugacty f the pure sld n the equlbrum system and f s the fugacty f the pure cndensed phase at bar. Fr a pure substance at gven and, we have At cnstant temperature dμ vd sd (6.7-) 6-30

225 dμ vd Rd(lnf) (6.7-) Integratng equatn (6.7-) frm the standard pressure bar t the pressure f the system yelds R f d (lnf) f v d Rln f f Rln a v( ) (6.7-3) We have assumed ncmpressble fr the vlume f the cndensed phase. he actvty s gven by v ( ) a exp R (6.7-3) Except fr very hgh pressures the actvty s clse t unty. he state f the pure cndensed phase at equlbrum s nt sgnfcantly dfferent frm the standard state. Example Determne the actvty f lqud water at 30 C and at the saturatn pressure, 0 bar, and 00 bar. Slutn At 30 C, sat bar. We wll assume the vlume f lqud water remans cnstant at the saturatn temperature f 30 C. v m 3 /kg ( )(8.0) m 3 /kml he actvty f lqud water at 30 C and bar s v ( ) (0.044 ) 0 a exp R exp (834.3)(303.5) he actvty f lqud water at 30 C and 0 bar s then ( ) 0 a exp (834.3)(303.5) he actvty f lqud water at 30 C and 00 bar s 7 Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

226 0.0809( ) 0 a exp (834.3)(303.5) Hence the fugacty f a pure lqud vares lttle ver a mderate pressure range. Example Determne the equlbrum cnversn fr the smerzatn reactn f methylcyclpentane (CH 3 C 5 H 9 ) t cyclhexane (C 6 H ) at 98 K. Gbbs energes f frmatn are gven at 98 K as: Δg f,ch3c5h9 3.7 [kj/ml] and Δg f,c6h 6.89 [kj/ml] Slutn Determne the equlbrum cnstant at rm temperature Δg rxn,98 ν Δ g f, 98 Δg rxn,98 (Δg f) C6H (Δg f) CH3C5H kj/ml he equlbrum cnstant s then Δg rxn K exp R,98 4,830 exp (8.34)(98.5) 7.08 he equlbrum cnstant s als gven by the expressn K Π f ˆ f ν ν xγ f Π f (E-) Except fr very hgh pressure the rat becmes f f v ( ) exp R s clse t unty. Equatn (E-) ν ( ) K Π γ (E-) x ( x ) ν Fr deal slutn: K Π x x C H 6 CH C H ζ 7.08 ζ ζ % f the lqud exsts as cyclhexane. 8 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

227 Chapter 6 Example Calcum carbnate can dsscate accrdng t the fllwng reactn: CaCO 3 (s) CaO(s) + CO (g) Determne the equlbrum pressure fr a clsed system wth pure CaCO 3 n vacuum at 000 K. Assume that the tw sld phases are cmpletely mmscble. CaCO 3 (s) CaO(s) CO (g) Δg f,000 (kj/ml) Slutn he equlbrum cnstant K can be expressed n terms f the actvtes: K a CaO a a CO CaCO3 fˆ f CaO CaO fˆ f fˆ CaCO3 f CaCO3 CO CO (E-) Snce we have pure sld phases: acao fˆ f CaO CaO f CaO fcao v ( ) exp R at lw pressure acaco 3 fˆ f CaCO 3 CaCO 3 f CaCO fcaco 3 3 v ( ) exp R at lw pressure Fr the gas phase aco fˆ f CO CO y CO bar CO herefre K CO Δ exp g rxn,000 R he Gbbs energy f reactn s gven by 9 Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

228 Δg rxn,000 ν Δ g f, 000 Δg rxn,000 Δg rxn,000 Δg rxn,000 (Δg f) CaO + (Δg f) CO (Δg f) CaCO3 ( 53.09) CaO + ( 395.8) CO ( 95.5) CaCO kj/ml Δ K exp g rxn,000 R 4,350 exp (8.34)(98.5) herefre, calcum carbnate wll dsscate at 000 K untl the pressure reaches bar r untl all the avalable CaCO 3 (s) reacted. 6.8 hermdynamcs f ack Cementatn ack cementatn s a prcess where a pure element r master ally s depsted n the surface f a superally t extend ts lfe n crrsve and xdzng envrnments at hgh temperature. here are fur cnsttuents t ths prcess: a fller, a pure element r master ally, an actvatr, and a substrate. he nert r fller prvdes a medum fr vapr transprt, e.g., alumnum xde Al O 3. he pure element r master ally wll be depsted n the substrate. he actvatr s used t transprt the master ally thrugh the fller t the substrate, whch s the surface f the superally. We wll cnsder the case where alumnum wth AlF 3 actvatr wll be mxed wth alumnum xde pwder n a pack cementatn prcess at 400 K. A schematc f the prcess s shwn n Fgure 6.8- where the system s mantaned at atm n an envrnment f Argn gas. he bulk pack s the regn where alumnum and actvatr exst wthn the fller. In the depleted zne, there s n alumnum r actvatr. Fr ths prcess alumnum s transferred frm the bulk pack t the substrate n the frm f alumnum flurde vapr, under the actn f the thermdynamc actvty gradent that exsts between the pack and substrate. Substrate Bulk pack Depleted zne Catng Fgure 6.8- Schematc f pack alumnzng prcess. At the bulk pack the fllwng reactns wll ccur AlF 3 (s) AlF 3 (g) (6.8-) 6-34

229 Al(l) + AlF 3 (s) 3AlF(g) (6.8-) Al(l) + AlF 3 (s) 3AlF (g) (6.8-3) AlF 3 (g) Al F 6 (g) (6.8-4) Snce the meltng pnt f pure alumnum s K, alumnum wll exst n the bulk pack as a lqud. he fve partal pressures ( AlF, AlF, AlF3, AlF6, and Ar ) n the bulk pack can be btaned frm the fur equlbrum cndtns abve and the assumptn that AlF + AlF + AlF3 + AlF6 + Ar atm (6.8-5) able 6.8- prvdes data fr the Gbbs energes f frmatn fr the speces present n the prcess. able 6.8- Δg n kcal/ml rxn,400 K hermdynamc Data fr Mneral echnlgy, L. B. ankrats, J. M. Stuve, and N. A. Gkcen Speces AlF 3 (c) AlF(g) AlF (g) AlF 3 (g) Al(c,l) Al F 6 (g) Δ g rxn, he equlbrum cnstant K s then calculated frm K exp( Δg rxn,400 /R) (6.8-6) In ths equatn, Δg rxn,400 ν Δ g f, 400 he equlbrum cnstant K fr reactn (6.8-): [Al(l) + AlF 3 (s) 3AlF(g)] can be expressed n terms f the actvtes: 3 a AlF K a a AlF 3 Al (6.8-7) Snce we have pure sld and lqud phase: a AlF fˆ f AlF AlF f AlF f AlF v ( ) exp R a Al fˆ f Al Al f Al f Al v ( ) exp R Fr the gas phase a AlF 3 fˆ f AlF 3 AlF 3 y AlF atm 3 AlF

230 he equlbrum cnstants fr the fur reactns (6.8-) - (6.8-4) are then related t the partal pressures by the fllwng relatns K AlF3 (6.8-8) 3 K AlF 3 K 3 AlF Al K 4 F 6 AlF 3 (6.8-9) (6.8-0) (6.8-) he fve partal pressures can be easly determned frm equatns (6.8-8)-(6.8-) and equatn (6.8-5). he results are AlF , AlF 0.34, AlF 0.037, AlF , and Ar he fllwng Matlab prgram s used t evaluate the partal pressures % Alumnum packng wth AlF3 actvatr % 400 K R.987; 400;R R*;; % Data at 400 K dgalf3c -74.5;dGAlF ; dgalf -88.4; dgalf ; dghf ; dgalf ; % dgr dgalf3 - dgalf3c; K exp(-000*dgr/r); dgr 3*dGAlF- dgalf3c; K exp(-000*dgr/r); dgr 3*dGAlF - *dgalf3c; K3 exp(-000*dgr/r); dgr dgalf6 - *dgalf3; K4 exp(-000*dgr/r); % Evaluate the partal pressure at the bulk pack % AlF3 K; AlF K^(/3); AlF K3^(/3); AlF6 K4*AlF3^; Ar - (AlF3 + AlF + AlF + AlF6); dsp('artal pressure n atm, dffusn n argn') dsp('alf3 AlF AlF AlF6 Ar') fprntf('%0.3e %0.3e %0.3e %0.3e %0.3e %0.3e\n',AlF3,AlF,AlF,AlF6,Ar) >> c6d8 artal pressure n atm, dffusn n argn AlF3 AlF AlF AlF6 Ar.043e e e e e

231 6.9 Equlbrum n Electrchemcal Systems In electrchemcal systems we can btan wrk by applyng an electrc ptental t tw electrdes. A galvanc cell s an electrchemcal cell that uses a spntaneus reactn t prduce useful wrk. When cpper and rn ppes are cnnected, a galvanc cell s created as shwn n Fgure Electrn wll flw frm rn wth lwer electrn affnty t cpper wth hgher affnty untl a few vlts cunterbalancng electrcal ptental s establshed between them. An electrchemcal cell als needs an electrlytc slutn n cntact wth the tw metals that allws ns t be transprted frm ne metal t the ther. Water s an electrlyte snce t cntans H +, OH -, and ther ns. Electrn flw - Metallc cnnectn + Anns (-) mgrate tward ande Fe Fe(OH) 3 H + H(g) OH - Cu Water: electrlyte; cntans ns Catns (+) mgrate tward cathde Irn ppe: ande; metal xdzng reactn ccurs here Cpper ppe: cathde; reducng reactn ccurs here Fgure 6.9- Schematc f a galvanc cell. An electrlytc cell s an electrchemcal cell that requres electrcal wrk t nduce a reactn that wuld nt ccur spntaneusly. An example f an electrlytc cell s shwn n Fgure 6.9- where bth electrdes are mmersed n a cmmn electrlyte. Electrn flw E - + Anns (-) mgrate tward ande t O(g) Cu + H + SO - 4 Cu Catns (+) mgrate tward cathde Fgure 6.9- Schematc f an electrlytc cell. At the cathde we have reductn f cuprc ns t grw sld cpper: 6-37

232 Cu + (l) + e - Cu(s) (6.9-) At the ande we have xdatn f the water: H O(l) 0.5O (g) + H + (l) + e - (6.9-) he reactn (6.9-) ccurrng n the cathde and reactn (6.9-) ccurrng n the ande are called half-cell reactns. he verall reactn s btaned by addng the xdatn and reductn half reactns: H O(l) + Cu + (l) 0.5O (g) + H + (l) + Cu(s) (6.9-3) Electrchemcal cells can cntan dfferent electrlyte cmpstns at the ande and the cathde. An example shwn n Fgure s an alternatve electrplatng prcess fr cpper n whch znc s xdzed at the ande and ges nt slutn as Zn +. hs cell s galvanc snce the cpper grwth ccurs spntaneusly wthut the nput f electrcal wrk. he tw electrlytes are separated by a salt brdge that allws a net charge t be transferred frm ne electrlyte slutn t the ther but des nt allw undesred mxng f the electrlyte. A salt brdge n mpermeable t the ns. It can be a smple prus dsk r a gel saturated wth a strng electrlyte such as KCl. Electrn flw - + Anns (-) mgrate tward ande ZnSO 4 CuSO 4 catns anns Zn Cu Salt brdge Electrlyte Electrlyte Catns (+) mgrate tward cathde Fgure A galvanc cell wth a salt brdge. A shrthand ntatn has been develped t descrbe electrchemcal cells. Startng at the ande, we pass thrugh the electrlyte t the cathde and ndcate the actve speces n chemcal ntatn. A vertcal bar separates tw phases, that s, phase phase. A duble bar ndcates a salt brdge r an mpermeable separatn f the phases, that s, phase phase. he shrt hand ntatn fr the electrlytc cell shwn n Fgure 6.9- s: t O (g) H SO 4 (l), CuSO 4 (l) Cu(s) (6.9-4) he shrt hand ntatn fr the galvanc cell shwn n Fgure s: Zn(s) ZnSO 4 (l) CuSO 4 (l) Cu(s) (6.9-5) 6-38

233 he dfferental electrcal wrk, δw *, can be related t the electrc ptental dfference between the cathde and the ande, E, and the dfferental amunt f charge transferred, dq, by the equatn: δw * EdQ (6.9-6) he sgn cnventn fr equatn (6.9-6) s chsen s that when the cathde has a pstve ptental wth respect t the ande, the prcess s spntaneus, whle a negatve ptental ndcates that wrk s requred fr the prcess t prceed. he xdatn half-reactn at the ande causes a flw f electrns n the external crcut; therefre the dfferental charge transferred can be related t the extent f reactn as fllws: mle e - lberated charge {charge transferred} - mle speces reactng mle e {extent f reactn} dq zfdζ (6.9-7) In ths equatn, z s the number f mles f electrn lberated per mle f speces that reacts and F s Faraday s cnstant, 96,485 C/(mle e - ), whch represents the charge f mle f electrns. Substtutng dq frm equatn (6.9-7) nt equatn (6.9-6) yelds δw * zefdζ (6.9-8) Fr reversble reactn wth nn-v wrk we have δw * (dg), μ ν dζ (6.9-9) Cmparng equatns (6.9-8) and (6.9-9) we btan zef μ ν fˆ g + R ln ν (6.9-0) f Assumng the actvty f the slds n equatn (6.9-0) s unty, we btan ν ν zef Δ g rxn + Rln ( yϕˆ ) ( xγ ) (6.9-) vaprs lquds Electrchemcal cells typcally perate at lw pressure s we can assume ϕˆ. Fr the lqud n the electrlyte, the standard state s -m slutn where m dentes mlalty (mles per kg f slvent). Equatn (6.9-) becmes ν ν zef Δ g rxn + Rln ( ) ( cγ ) (6.9-) vaprs lquds 6-39

234 In ths equatn, the cncentratn c s n mlalty. If a speces n the lqud cncentratn f mle per kg f slvent and ts nteractns crrespndng t an deal slutn, t has c γ. Dvdng equatn (6.9-) by zf yelds E E rxn R ν ln ( ) ( ) (6.9-) zf ν cγ vaprs lquds In ths equatn, the standard ptental f reactn s defned as E rxn Δg rxn R (6.9-3) he standard ptental f reactn s btaned frm thermchemcal data and s measured wth reference t hydrgen-hydrgen n xdatn reactn, whse ptental s defned as zer: H (g) H + (l) + e - E 0.0 V (6.9-4) Example he reverse cpper dsprprtn reactn has been prpsed t etch sld cpper: Cu + Cu + (l) Cu + (l) Determne the equlbrum cnstant f the dsprprtnatn reactn usng the fllwng half-cell reactns: Cu + (l) + e - Cu + (l) Cu Cu + (l) + e - E 0.53 V E 0.5 V Slutn he sum f the half-cell reactns gves the cpper dsprprtn reactn, therefre E rxn V Δg rxn zf ( )(96,485)( 0.368) 35.5 kj/ml E rxn Snce Δg rxn >0, ths reactn wll nt ccur spntaneusly. he equlbrum K cnstant s then Δg rxn 35,500 K K exp exp R (8.34)(98.5) Kretsky M.D., Engneerng and Chemcal hermdynamcs, Wley, 004, pg

235 Chapter Cmplex Chemcal Equlbrum We want t determne the number f ntensve varables that must be specfed t defne a system cnsstng f m dentfable chemcal speces dstrbuted amng π phases. descrbe each phase we need m ndependent cmpstns,, and t a ttal f m+ ntensve varables. he ttal number f varables requred t specfy π phases s π(m+). Fr phase equlbrum we have the fllwng set f equatns: α β π α β π α μ α μ β μ β μ π μ π μ M α β π μ m μ m μ m Each rw n the set f equatns abve cntans (π ) ndependent equatns. hus there are a ttal f (π )(m+) ndependent equatns between the varables. In addtn t these equatns there are R equatns frm the number f ndependent reactns μ j ν j 0 (j R) here mght be addtnal equatns, expressble n terms f ntensve varables, whch apply t the system. hese equatns wll be called addtnal cnstrants and ther number wll be desgnated s. he ttal number f ndependent equatns nvlvng ntensve varables s therefre (π )(m+) + R + s he number f varables we can ndependently pck (the s-called degree f freedm, F) s btaned by subtractng the ttal π(m+) varables we need t specfy by the [(π )(m+) + R + s] ndependent equatns. F π(m+) [(π )(m+) + R + s] m + π R s We can ndependently specfy (m + π R s) t cmplete defne a system wth m cmpnents and π phases. he cmbnatn f (m R s) s called the phase rule cmpnents, C. We nw need t determne the number f ndependent reactn, R, amng 6-4

236 the chemcal speces n a system. he prcedure wll be llustrated by the fllwng examples. Example In a reactng system at equlbrum, the fllwng speces are cnsdered t be present: CO, CO, C, CH 4, H, H O, and N. Determne the number f phase rule cmpnents and the number f ndependent reactns. Slutn We wll frst create a ceffcent matrx called the β matrx as fllws: C O H N CO 0 0 CO 0 0 C CH H N H O 0 0 he number f phase rule cmpnents, C, s equal t the rank f the β matrx. he rank f a matrx s the rder f the largest determnant havng a nnzer value that can be created frm the matrx. he rder f a determnant, r a square matrx, s the number f rws r clumns. Fr ths example, the rank f the matrx s equal t r less than 4. A furth rder nnzer determnant can be frmed frm the β matrx. herefre C 4. he rank f the β matrx can als be funded frm the fllwng Matlab statement: >> rank([ 0 0; 0 0; 0 0 0; 0 4 0;0 0 0;0 0 0 ;0 0]) ans 4 Snce C m R s, and s 0. he number f ndependent reactn R s R m C We nw need t fnd the three ndependent reactns. Frst we wrte the frmatn reactn fr each f the 4 mlecular speces (CO, CO, CH 4, and H O). C + O CO () C + 0.5O CO () C + H CH 4 (3) H O H O (4) Snce free xygen s nt amng the m speces, we can elmnate O frm the abve set by usng the secnd equatn Kyle, B.G., Chemcal and rcess hermdynamcs, rentce Hall, 999, pg

237 Equatn () becmes C + 0.5O CO 0.5O CO C C + CO C CO CO C +CO Equatn (4) becmes H + CO C H O H + CO C + H O We nw have a set f three ndependent reactns CO C +CO C + H CH 4 H + CO C + H O (E-) (E-) (E-3) It shuld be nted that there are mre than ne set f 3 ndependent reactns. Example In a reactng system at equlbrum, the fllwng speces are cnsdered t be present: H, ethane (C H 6 ), CO, H O, and methanl (CH 3 OH). Determne the number f phase rule cmpnents and the number f ndependent reactns. Slutn he β matrx s created as fllws H C H 6 CO H O CH 3 OH C 0 0 H O he rank f the β matrx s 3 C (number f phase rule cmpnents) Snce C m R s, and s 0. he number f ndependent reactn R s R m C

238 Appendx A Slvng Algebrac Equatns A. he Newtn-Raphsn Methd he Newtn-Raphsn methd and ts mdfcatn s prbably the mst wdely used f all rt-fndng methds. Startng wth an ntal guess x at the rt, the next guess x s the ntersectn f the tangent frm the pnt [x, f(x )] t the x-axs. he next guess x 3 s the ntersectn f the tangent frm the pnt [x, f(x )] t the x-axs as shwn n Fgure A.. he prcess can be repeated untl the desred tlerance s attaned. f(x) f(x ) B x 3 x x Fgure A. Graphcal depctn f the Newtn-Raphsn methd. he Newtn-Raphsn methd can be derved frm the defntn f a slpe f (x ) f ( x x x ) 0 x x f ( x) f ' ( x ) In general, frm the pnt [x n, f(x n )], the next guess s calculated as x n+ x n f ( xn ) f ' ( x ) n he dervatve r slpe f(x n ) can be apprxmated numercally as f (x n ) f ( x n + Δx) f ( xn ) Δx A-

239 Example A Slve f(x) x 3 + 4x 0 usng the Newtn-Raphsn methd fr a rt n [, ]. Slutn f ( xn ) Frm the frmula x n+ x n f ' ( x ) 3 f(x n ) f (x n ) 3 + 8x n x n x n n x n x n+ x n 3 xn + 4xn 0 3x + 8x n n Usng the ntal guess, x n.5, x n+ s estmated as x n A Matlab prgram fr the Newtn-Raphsn methd s lsted n able A. where the functn f(x) s an nput t the prgram. he statement eval(f) s used t evaluate the functn at a f ( xn + Δx) f ( xn ) gven value f x. he dervatve s evaluated numercally usng f (x n ) Δx wth Δx 0.0. A sample result s gven at the end f the prgram. able A % Example A.: Newtn methd wth numercal dervatve % fnput('f(x)','s'); tlnput('errr tlerance e-5, new tlerance'); f length(tl)0,tle-5;end xnput(' Frst guess'); xx; fxeval(f); fr :00 f abs(fx)<tl, break,end xx+.0; ffeval(f); fdx(ff-fx)/.0; xx-fx/fdx; xx; fxeval(f); fprntf(' %g, x %g, fx %g\n',,x,fx) end A-

240 >> ed f(x)x^3+4*x^-0 errr tlerance e-5, new tlerance Frst guess.5, x.3739, fx , x.3653, fx , x.3653, fx 6.399e A. Newtn s Methd fr Systems f Nnlnear Algebrac Equatns 0 0 Cnsder tw equatns f (x, x ) and f (x, x ) fr whch the rts are desred. Let p, p be 0 0 the guessed values fr the rts. f (x, x ) and f (x, x ) can be expanded abut pnt ( p, p ) t btan p p 0 0 f (x, x ) f (, ) f (x, x ) f (, ) + p p f x f x 0 (x p ) + 0 (x p ) + f x f x 0 (x p ) 0 0 (x p ) Let (x ) and (x ), the abve set can be wrtten n the matrx frm y p y p f x f x f x f x y y f( p, p ) 0 0 f( p, p ) r J(p (0) )y (0) F(p (0) ) In general, the superscrpt (0) can be replaced by (k ) J(p (k-) )y (k-) F(p (k-) ) J(p (k-) ) s the Jacban matrx f the system. he new guessed values x at teratn k are gven by x p (k) p (k-) + y (k-) A-3

241 Example A.- Use Newtn s methd wth the ntal guess x [ ] t btan the slutns t the fllwng equatns f (x, x, x 3 ) 3x cs(x x 3 ) 0 f (x, x, x 3 ) 8(x + 0.) + sn x x f (x, x, x 3 ) + 0x 3 + x x e π 0 Slutn he fllwng tw frmulas can be appled t btan the rts J(p (k-) )y (k-) F(p (k-) ) J(p (k-) ) s the Jacban matrx f the system. J(p (k-) ) x f x f x f x f x f x f x f x f x f F(p (k-) ) s the clumn vectr f the gven functns F(p (k-) ) ),, ( ),, ( ),, ( x x x f x x x f x x x f he new guessed values x at teratn k are gven by x p (k) p (k-) + y (k-) able A.- lsts the Matlab prgram t evaluate the rts frm the gven ntal guesses. able A.- Matlab prgram fr Example A % Newtn Methd fr set f nnlnear equatns % f'3*x()-cs(x()*x(3))-.5'; f'x()*x()-8*(x()+.)^+sn(x(3))+.06'; f3 'exp(-x()*x())+0*x(3)+0*p/3-' ; Numercal Analyss by Burden and Fares A-4

242 % Intal guess % x[ ]; fr :5 f[eval(f) eval(f) eval(f3)]; Jt[3 *x() -x()*exp(-x()*x()) x(3)*sn(x()*x(3)) -6*(x()+.) -x()*exp(-x()*x()) x()*sn(x()*x(3)) cs(x(3)) 0]'; % dxjt\f'; xx-dx'; fprntf('x ');dsp(x) end >> ed5d x x x x x Matlab can als evaluate the Jacban matrx f the system analytcally as shwn n able A.- able A.- Matlab prgram fr Example A % Newtn Methd wth Jacban matrx evaluated analytcally by Matlab % syms x x x3 F[3*x-cs(x*x3)-.5 x^-8*(x+.)^+sn(x3)+.06 exp(-x*x)+0*x3+(0*p-3)/3]; Jac[dff(F,x) dff(f,x) dff(f,x3)]; x.;x.;x3-.; k0; dsp(' k x x x3') fprntf('%3.0f %0.7f %0.7f %0.7f\n',k,x,x,x3) fr k:0 Ameval(Jac);Bceval(F); ykam\bc; xx-yk(); xx-yk(); x3x3-yk(3); fprntf('%3.0f %0.7f %0.7f %0.7f\n',k,x,x,x3) f max(abs(yk))<.0000, break, end end >> ed5db A-5

243 Jac [ 3, sn(x*x3)*x3, sn(x*x3)*x] [ *x, -6*x-8/5, cs(x3)] [ -x*exp(-x*x), -x*exp(-x*x), 0] k x x x Slvng set f nnlnear equatns wth Excel Use the ntal guess x [ ] t btan the slutns t the fllwng equatns f (x, x, x 3 ) 3x cs(x x 3 ) 0 f (x, x, x 3 ) x 8(x + 0.) + sn x π 3 f (x, x, x 3 ) x x e + 0x A-6

244 Appendx B Curve Fttng B. Nnlnear Curve Fttng A sample cnssts f a layer f alumnum and a layer f a cmpste catng s tested n a vacuum chamber by measurng ts temperature as a functn f tme. he behavr f the sample temperature has a nnlnear dependence n the emssvty ε f the sample and the heat transfer ceffcent h between the catng and the vacuum ar. Insulatn Cmpste catng Alumnum Vacuum chamber Fgure B.-. A sample enclsed wthn a testng chamber. he unknwn parameters h and ε may be btaned by fttng the mdel equatn t expermental data as shwn n Fgure B.. where the curve represents the mdel equatn and the crcles represent the data emperature(k) me(mn.) Fgure B.-. ransent temperature f a typcal sample. B-

245 llustrate hw ths s dne, frst cnsder a prtn f the graph n Fgure B.- that s repltted n Fgure B.-3. he relatnshp between the temperature btaned frm the mdel equatn and the expermental value,exp can be expressed generally as,exp (t; ε, h) + e (B.-) where e s a randm errr that can be negatve r pstve. s a functn f the ndependent varable t and the parameters h and ε. he randm errr s als called the resdual, whch s the dfference between the calculated and measured values.,exp e t Fgure B.-3. Relatnshp between the mdel equatn and the data Nnlnear regressn s based n determnng the values f the parameters that mnmze the sum f the squares f the resduals called an bjectve functn F bj. N e N ( ) F bj exp exp (B.-3) N, (, ) Where N s the number f data pnts r measured temperatures n ths case. he temperature frm equatn B.- can be expanded n a aylr seres arund h and ε and curtaled after the frst dervatve.,j+,j +, j Δε +, j Δh (B.-4) ε h Where j s the guess and j+ s the predctn, Δε ε j+ ε j, and Δh h j+ h j. We have lnearzed the rgnal mdel wth respect t the parameters h and ε. Equatn (B.-4) can be substtuted nt Eq. (B.-) t yeld,exp,j, j Δε +, j Δh + e (B.-5a) ε h r n matrx frm B-

246 {D} [Z j ]{ΔA} + {E} (B.-5b) where [Z j ] s the matrx f partal dervatves f the functn (called the Jacban matrx) evaluated at the guess j, the vectr {D}cntans the dfferences between the measure temperature and the calculated temperature at the guess j, the vectr {ΔA}cntans the changes n the parameter values, and the vectr {E} cntans the resduals. It shuld be nted that as the fnal values f the parameters are btaned after the teratns vectr {D} s the same as vectr {E}. ε [Z j ] ε M N ε h,exp, j Δε h, {D},exp, j, {ΔA}, {E} M M Δh N N,exp N, j h e e M e N We mnmze the bjectve functn N N e (, exp ) F bj (B.-3) by takng ts dervatve wth respect t each f the parameters and settng the resultng equatn t zer. bj ε F N (,exp ) 0 (B.-6a) ε F bj h N (,exp ) h 0 (B.-6b) hs algrthm s Gauss-Newtn methd fr mnmzng the sum f the squares f the resduals between data and nnlnear functns. Equatns (B.-6a) and (B.-6b) can be cmbned n a matrx frm [Z j ] {E} 0 (B.-7) where [Z j ] s the transpse f [Z j ]. Let cnsder N 3 s we can see the cmbnatn frm (B.-6a) and (B.-6b) t (B.-7). ε h ε h 3 ε 3 h,exp,exp 3,exp 3 0 Substtute {E} {D} [Z j ]{ΔA} frm Eq. (B.-5b) nt (B.-7) B-3

247 [Z j ] {{D} [Z j ]{ΔA}} 0 r [Z j ] [Z j ]{ΔA} {[Z j ] {D}} (B.-8) he Jacban matrx [Z j ] may be evaluated numercally fr the mdel equatn (B.-). ε ( ε + δε, h) ( ε, h) δε (B.-9a) h ( ε, h + δh) ( ε, h) δh (B.-9b) ypcally, δε can be chsen t be 0.0 and δh can be chsen t be 0.0 W/m K. hus, the Gauss-Newtn methd cnssts f slvng Eq. (B.-8) fr {ΔA}, whch can be emplyed t cmpute mprved values fr the parameters h and ε. ε j+ ε j + Δε (frm {ΔA}) h j+ h j + Δh (frm {ΔA}) hs prcedure s repeated untl the slutn cnverges that s untl Δε and Δh fall belw an acceptable crtern. he Gauss-Newtn methd s a cmmn algrthm that can be fund n many numercal methd texts. Hwever, ths descrptn fllws the ntatns and develpment by Chapra and Canale. Example B.- Ft the functn (t; ε, h) ε( e ht ) t the data. t Use ntal guesses f h and ε fr the parameters. Slutn he partal dervatves f the functn wth respect t the parameters h and ε are e ht and ε h εte ht B-4

248 ε [Z j ] ε M N ε h h M N h he matrx multpled by ts transpse results n [Z j ] [Z j ] [Z j ] [Z j ] he vectr {D} cnssts f the dfferences between the measurements and the mdel predctns (t; ε, h) ε( e ht ) {D} he vectr {D} s pre-multpled by [Z j ] t gve [Z j ] {D} he vectr {ΔA} can be calculated by usng MALAB statement dazjzj\zjd ({ΔA} {[Z j ] [Z j ] \[Z j ] {D}}) 0.74 {ΔA} B-5

249 he next guesses fr the parameters ε and h are ε h able B.- lsts the MALAB prgram wth the results f tw teratns. able B.- % Gauss-Newtn methd % t[ ]'; [ ]'; e;h; mdel'e*(-exp(-h*t))'; dde'-exp(-h*t)';ddh'e*t.*exp(-h*t)'; fr : Zj[eval(dde) eval(ddh)]; ZjZjZj'*Zj D-eval(mdel) ZjDZj'*D dazjzj\zjd ee+da(); hh+da(); fprntf('iteratn #%g: e %8.4f, h %8.4f\n',,e,h) end >> Gauss ZjZj D ZjD da Iteratn #: e 0.785, h.509 ZjZj B-6

250 D ZjD da Iteratn #: e 0.790, h.6777 able B.- lsts the Matlab prgram that slves fr the tw parameters ε and h untl the maxmum change f the parameters between tw successve teratns s less than 0.0. he standard errr and the crrelatn ceffcent f the ftted values are als evaluated. he standard errr stde s determne frm stde N N ( exp ), S N N, where S (,exp ) he standard errr quantfes the spread arund the regressn curve. A crude measure f the gdness-f-ft s gven by the crrelatn ceffcent r r N S, where S t ( ) S t,exp,exp In ths expressn,exp s the mean f the expermental dependent varable. able B.- Gauss Newtn methd fr nnlnear curve fttng % Gauss-Newtn methd % t[ ]'; [ ]'; e;h; mdel'e*(-exp(-h*t))'; dde'-exp(-h*t)';ddh'e*t.*exp(-h*t)'; fr :0 Zj[eval(dde) eval(ddh)]; ZjZjZj'*Zj; D-eval(mdel); ZjDZj'*D; dazjzj\zjd; ee+da(); hh+da(); B-7

251 f max(abs(da))<.0,break, end end fprntf('# f Iteratn %g: e %8.4f, h %8.4f\n',,e,h) Ssum((-eval(mdel)).^); stdesqrt(s/(length(t)-)); avemean();sdevsum((-ave).^); crsqrt(-s/sdev); fprntf('standard errr %8.4f, Crrelatn ceffcent %8.4f\n',stde,cr) >> gauss # f Iteratn 3: e 0.799, h.6753 Standard errr 0.049, Crrelatn ceffcent he Matlab functn fmnsearch can als be used t ft the data t an expressn wth mre than ne parameter, (t; ε, h) ε( e ht ). able B.-3 lsts the functn requred by fmnsearch. able B.-4 lsts the prgram that calls fmnsearch t fnd the tw parameters ε and h. he prgram als plts the ftted results wth the expermental data shwn n Fgure B.-4. able B.-3 Matlab prgram t defne the bjectve functn functn ynln(p) t[ ]; [ ]; ep();hp(); ce*(-exp(-h*t)); ysum((-c).^); clf t[ ]; [ ]; pfmnsearch('nln',[ ]) tp.5:.:.5; ep();hp(); ce*(-exp(-h*tp)); plt(tp,c,t,,'') grd n xlabel('t');ylabel('') legend('ftted','data') >> nlnear p able B.-4 Matlab prgram t fnd ε and h B-8

252 Appendx C rcess Smulatr rcess smulatr s ne f the mst cmmn types f sftware used n prcess desgn. It can be used t perfrm steady r unsteady materals and energy balances fr a wde range f prcess equpment. A user usually must fllw the fllwng steps t set up a prblem n a smulatr :. Select all f the chemcal cmpnents that are requred n the prcess frm the cmpnent database.. Select the thermdynamc mdels requred fr the smulatn. hese may be dfferent fr dfferent peces f equpment. Fr example, the eng Rbnsn equatn f state can be used t smulate a pump. Hwever, fr a lqud-lqud extractr n the same prcess, t s necessary t use a thermdynamc mdel (e.g. Wlsn mdel) that can predct lqudphase actvty ceffcents and the exstence f tw lqud phases. 3. Select the tplgy f the flw sheet t be smulated by specfyng the nput and utput streams fr each pece f equpment. 4. Select the prpertes (temperature, pressure, flw rate, and cmpstn) f the feed streams t the prcess. 5. Select the equpment specfcatns (parameters) fr each pece f equpment n the prcess. 6. Select the frmat f the utput results. 7. Select the cnvergence methd and run the smulatn. here are bascally three types f slutn algrthm fr prcess smulatrs: sequental mdular, equatn slvng (smultaneus nn-mlar), and smultaneus mdular. In the sequental mdular apprach the prcess s slved equpment pece by equpment pece. In the equatn slvng technque, all the equatns fr the prcess are wrtten ut tgether and then the resultng matrx f nnlnear smultaneus equatns s slved t yeld the slutn. he fnal smultaneus mdular technque cmbnes the mdularzng f the equatns relatng t specfc equpment wth the effcent slutn algrthms fr the smultaneus equatn slvng technque. We wll g ver a sequence f events needed t smulate a tw-stage cmpressr system usng rvsn. hs prcess smulatr uses the sequental mdular algrthm n whch each pece f equpment s slved n sequence, startng wth the frst then fllwed by the secnd, and s n. hs methd s nt as effcent as the equatn slvng technque. Hwever the user can easly detect the lcatn where cnvergence s nt btaned. Example: A 000 lb-mle per hur mxture f 5% methane, 0% ethane, 5% prpane, 30% n- butane, and 0% n-pentane at 4.7 psa and 90 F s t be cmpressed frst t 50 psa, then t 00 psa usng a tw-stage cmpressr system. An nner-stage cler s used t cl the cmpressed gas frm the frst stage t ts rgnal temperature by passng the gas stream thrugh a heat exchanger n whch water s used as a clant. hs s dne t mantan a lwer energy requrement fr the secnd stage cmpressr. A knckut drum s nstalled after the frst cler t remve any lqud present n the cmpressed gas befre feedng t t the secnd cmpressr. If yu have lqud present n a gas stream, t wll damage the mpeller f the cmpressr. urtn et al, Analyss, Synthess, and Desgn f Chemcal rcesses, rentce Hall, 998, pg. 480 Dr. ang s rvsn Student Manual C-

253 pen rvsn, g t the Start menu, clck n rgrams, Smsc, and then rii 8.0. he fllwng screen shuld appear: Nte the dfferent clred bxes and ther meanngs. hey wll be very mprtant t remember later n n the smulatn. Clck n OK at the bttm f the bx t cntnue nt the smulatn envrnment. Next, clck n Fle and then New. hs shuld brng yu t the fllwng screen: C-

254 hs s the basc smulatn envrnment frm whch yu wll begn each tme yu use rvsn. It s called the FD screen. Befre yu begn wth the prcedure gven t yu n the assgnment, yu need t enter the cmpnents that yu wll be usng and yur equatn f state. Frst clck n the Cmpnent Selectn buttn n the tp tlbar. hs buttn lks lke ths:. he fllwng screen shuld appear: Frm here yu can ether type n the names f yur desred cmpnents r yu can select them frm a lst already nsde rvsn. We wll select ur cmpnents frm a lst by clckng n the Select frm Lsts buttn. he fllwng screen shuld appear: In ths assgnment, all f the needed cmpnents can be fund n the All Cmpnents - ROCESS Bank fle. Clck n the desred speces and clck Add Cmpnents, the selected speces wll appear n the bttm text bx Addtns t Cmpnent Lst. After yu selected all speces (Methane, Ethane, rpane, N-butane, entane, and Water) clck OK t return t the fllwng screen. C-3

255 Clck OK t return t the FD screen. Next clck n the hermdynamc Data buttn,. Once nsde, clck n Mst Cmmnly Used and then clck n eng-rbnsn n rmary Methd: bx. hen clck Add and then OK t return back t the FD screen. We are nw ready t begn the prcedure t smulate a w-stage Cmpressr System. Usng yur drawn FD as a gude, select needed unts and place them n the FD screen ne by ne. All f the chces fr equpment are lcated n the tlbar n the rght sde f the screen. It s the thn vertcal tlbar ttled FD. Clck n the Cmpressr buttn and then clck n the space n the FD where yu wsh t place the cmpressr. Fllw the same steps fr Smple HX (), Flash, Cmpressr (), and Smple HX (). Yu wll ntce that each unt s named by ROVISION, autmatcally. Yu wll als ntce that each name has a red brder meanng that sme cndtns have t be suppled. Clck n the Stream buttn n the buld palette t draw the stream lnes. All the pssble EXI prts appear n each unt as sn as the Stream buttn s selected. he requred utlet prt are clred red, whle the ptnal exts are gven n green. Add streams by clckng n the rgn f the stream (ether a unt ext prt r an external feed stream, n a unccuped part f the FD), then clck agan at the destnatn f the stream. Fr example: t apply a feed stream t C, clck n the FD that s nt ccuped by any unt, and then clck agan at cmpressr. Ntce the name s gven as S. C-4

256 Cnnect all the streams n rder shwn n the Fgure. ay attentn t the number f stream. Each stream s numbered n the rder they are created. Fgure A tw-stage cmpressr system It s always safe t save frequently. save, clck n Fle n the menu bar and clck n Save n the Fle menu. ype n the fle name, Case r any name yu lke, and clck n the OK buttn t g back t the FD. Yu nw need t supply the requred Stream and Unt Data. Feed Stream S: Yu can select a stream r unt by duble clckng n the name n FD. Let s frst start by cmpletng the begnnng feed stream, S. Duble clck n S t select that stream. Yu can gve t a name, Feed Stream, by typng t n the bx next t Descrptn: Clck n the Flwrate and Cmpstn... buttn t supply the flw rate and cmpstns. Snce we knw the ttal flw rate, clck n tal Flud Flwrate and type n 000 lb-ml/hr. he brder becmes blue as sn as yu nput the data. Fll n the cmpstn snce they have red brders (0.05, 0., 0.5, 0.3, 0.). Clck n OK buttn when dne. C-5

257 satsfy the phase rule, yu need tw mre varables. Fr ths case yu knw temperature and pressure. enter temperature, clck n the red bx named Frst Specfcatn: chse emperature and type n the temperature, 90 F. enter the pressure, clck n the red bx named Secnd Specfcatn: chse ressure and type n 4.7 psa. Clck n the OK buttn snce all the requrements are met. S nw has a black brder. Cmpressr C: Duble clck n the cmpressr, C, n the FD. Supply the Outlet ressure as 50 SIA n the red bx. Change the Adabatc Effcency t 80%. Default s 00%. Ntce the bx s green, meanng that the user can nput data. Clck n the OK buttn t g back t the FD. Heat Exchanger E: Duble clck n the heat exchanger, E, n the FD. Yu wll see the RO/II-Heat Exchanger wndw. Check fr ht and cld streams by clckng n the rcess Stream... buttn. S and S3 shuld be ht, and S8 and S9 shuld be cld streams. If the sdes are nt set crrectly, yu can crrect t by clckng n the apprprate crcle. C-6

258 Clck n the OK buttn t g back t RO/II-Heat Exchanger wndw. Clck n the Specfcatn... buttn and clck n the red bx. Chse Ht rduct emperature whch s the prcess stream gng ut f the heat exchanger and type n 90 F fr the value. Clck n the OK buttn t g back t RO/II-Heat Exchanger wndw. Input 3 ps fr Ht Sde ressure Drp and ps fr Cld Sde ressure Drp. Clck n the OK buttn t g back t the FD. Stream gng t E, S8: Duble clck n S8 n the FD. Clck n the Flwrate and Cmpstn... buttn t supply the flw rate and cmpstn. Chse tal Flud Flwrate. Snce the set unt s lb-ml/hr, and yu have mass flw rate, lb/hr, yu must d the cnversn. C-7

259 Instead f dng t by yurself, yu can clck n the UOM buttn n the menu bar. Clck n Mass and chse lb. Clck n the Change Unts buttn t change the unt. Nw enter lb/hr fr the flw rate. Enter.00 fr the water cmpstn. Clck n OK buttn when dne. Clck n Frst Specfcatn: chse emperature and type n the temperature, 70 F. enter the pressure, clck n Secnd Specfcatn: chse ressure and type n 00 psa. Clck n the OK buttn t g back t the FD. Flash F: Select the flash drum, F, by duble clckng n t. Clck n rduct hase... t make sure the phases f streams, S4 and S5, are crrectly set. S4 shuld be Vapr, and S5 shuld be Lqud. Make necessary changes by clckng n ncrrect bxes and selectng the rght phase. Clck n the OK buttn t return t the RO/II-Flash Drum wndw. Input 3 ps fr the ressure Drp. Clck n the Unt Specfcatn: bx. Select Duty and type n 0.00 fr the value snce t s an adabatc flash drum. Clck n the OK buttn when dne. C-8

260 Secnd Cmpressr C: Duble clck n the cmpressr, C, n the FD. Supply the Outlet ressure as 00 psa n the red bx. Change the Adabatc Effcency t 80%. Clck n the OK buttn t g back t the FD. Secnd Heat Exchanger E: Duble clck n the heat exchanger, E, n the FD. Check fr ht and cld streams by clckng n the rcess Stream... buttn. S6 and S7 shuld be ht, and S0 and S shuld be cld streams. Clck n the OK buttn when dne. C-9

261 Clck at the Specfcatn and clck n the red bx. Chse Ht rduct emperature and type n 90 F fr the value. Clck n the OK buttn. Input 3 ps fr the Ht Sde ressure Drp and ps fr the Cld Sde ressure Drp. Clck n the OK buttn t g back t the FD. Stream gng t E, S0: Duble clck n S0 n the FD. Clck n the Flwrate and Cmpstn... buttn t supply the flw rate and cmpstn. Chse tal Flud Flwrate. Yu need t change the unt frm lb/hr t lb-ml/hr. Clck n the UOM buttn n the menu bar. Clck n Mass and chse lb. Clck n the Change Unts buttn t change the unt. Nw enter lb/hr fr the flw rate. Enter.00 fr the water cmpstn. Clck n OK buttn when dne. Clck n Frst Specfcatn: chse emperature and type n the temperature, C-0

262 70 F. enter the pressure, clck n Secnd Specfcatn: chse ressure and type n 00 psa. Clck n the OK buttn t g back t the FD. It s nw tme t run yur smulatn. he buttn fr runnng yur smulatn lks lke ths:. ROVISION starts the calculatn frm the begnnng f the smulatn and ges thrugh every unt peratn ne by ne. Durng the calculatn, each unt ges thrugh clr changes. Yellw Red Green Blue Dark Blue urple - n calculatn perfrmed - unt peratn has nt been slved - calculatn s beng perfrmed - unt peratn has been slved - unt peratn has been calculated - ndcate the break pnt When the calculatn s fnshed, all the unt peratns shuld be blue, ndcatng that all the calculatns are fnshed fr the entre case. Once the calculatns are perfrmed, yu d nt need t run the calculatns agan as lng as yu d nt change anythng. Revewng and rntng the Results Once the calculatns are perfrmed, yu are ready t check the results by generatng a reprt. generate a reprt, fllw these steps: ) Clck n Output frm the menu bar n the man FD. ) Clck n Generate Reprt n the Output menu and supply a name (cmpressr r any name yu lke) t save the utput fle. After yu clck n the OK buttn, the prgram wll C-

263 brng up the rgrammer s Fle Edtr wndw. Yu can scrll up and dwn the reprt by usng the scrll bars. he reprt lsts nfrmatn lke RO/II keywrd nput cmmands, cmpnent prpertes, unt peratn results, and stream results. Yu can als edt the reprt by usng the Edt menu. Yu can prnt the entre reprt r part f the reprt by selectng rnt n Fle menu. 3) clse the rgrammer s Fle Edtr wndw, duble clck n the bar lcated n the tp rght crner f the screen. Yu can vew the results f any stream r unt peratn by clckng n the rght buttn f yur muse when the cursr s pntng n t. rntng Materal and Energy Balance n FD In rder t nclude materal and energy balances n yur FD, yu must defne a prperty lst frst. he prperty lst tells ROVISION what t nclude n the materal and energy balance table. Let s defne a prperty lst that ncludes Stream Name, Stream hase, tal Mlar Rate, emperature, ressure, tal Enthalpy, and tal Mlar Cmp. Fractns. C-

264 Clck n Optn n the menu bar n the FD wndw and chse Stream rperty Lsts. Clck n New buttn n Defne Stream rperty Lst wndw t name the new lst and type case fr the Lst Name. Clck n the OK buttn t g back t Defne Stream rperty Lst wndw and chse the prpertes t nclude n the lst frm Select rpertes bx. Select Stream Name and clck n Add buttn t nclude the selected prperty nt the lst. Repeat the prcedure fr Stream hase, tal Mlar Rate, emperature, ressure, tal Enthalpy, and tal Mlar Cmp. Fractns. Save the lst and clck OK. Once the prperty lst has been defned, yu can nw place the materal and energy balance n the FD. Fllw these steps t place prperty table n the FD:. Clck n Output n the menu bar n the FD wndw. Clck n Stream rperty able n the Output menu 3. Clck n the FD where yu want t place the prperty table 4. Duble clck n the prperty table t select the table 5. Chse case n rperty Lst t be used 6. Yu can select the streams yu want by clckng n each stream r yu culd select all by clckng n Add All. Let s select all the streams by clckng n Add All. C-3

265 7. When yu clck n the OK buttn, the FD shuld lk lke ths: rntng FD: Clck n Fle n the menu bar n the man FD wndw and chse rnt. Clck n the OK buttn t start prntng. Yu can draw r prnt text n the FD by chsng Draw/text n the menu bar. After typng text n the bx, use muse t place n the FD. C-4

266 Appendx D revus Exam CHE 303 (Wnter 005) LAS NAME, FIRS Quz # Nte: Yur answers must be crrect t 3 sgnfcant fgures and have the apprprate unts. I. Crcle the crrect relatn (A) h v v + (B) h v v (C) h v v + (D) h v v II. Crcle the crrect relatn (A) s v c (B) s v c (C) v s v c (D) s v c III. A gas mxture cntans 30 ml % H O, 30 mle % N, and 40 ml % H. Mass fractn f H O s IV. Ntrgen gas flwng at the rate f 400 kg/hr enters a cmpressr at bar and 300 K and leaves at 0 bar and 450 K. Clng water enters the cmpressr at 90 K and leaves at 340 K. Clng water flw rate s 500 kg/hr. Ntrgen (MW 8) s an deal gas wth Cp.05 kj/kg. K. Heat capacty f water s 4.84 kj/kg. K. Gas cnstant R 8.34 J/ml. K. ) (4 pts) he entrpy change f ntrgen (kj/hr K) s (Shw all the wrk) D-

267 ) he entrpy change f water (kj/hr K) s 3) he wrk prvded by the cmpressr s 4) If the entrpy change f ntrgen s 5 kj/hr K and the entre prcess s reversble, the temperature f the ext water s V. Cnsder the pstn-cylnder assembly shwn belw; 00 mles f gas expand sthermally after the remval f a 0,000 kg blck. R a Assume that the v behavr can be descrbed by the Van der Waals equatn v b v wth a 0.5 [J m 3 /ml ] and b [m 3 /ml] and that the deal gas heat capacty has a cnstant value f c 35 J/(ml K), R 8.34 J/ml K. Weghtless, frctnless pstn atm Isthermal A 0. m atm expansn A 0. m 0.4 m ) (4 pts) Determne the temperature f the system (Shw all the wrk) D-

268 CHE 303 (Wnter 005) LAS NAME, FIRS Quz # Nte: Yur answers must be crrect t 3 sgnfcant fgures and have the apprprate unts. I. Crcle the crrect relatn (A) ds c p d + dv v c (B) ds p d dv v (C) ds c p d + c dv (D) ds v d + v dv v R a II. Fr the van der Waals equatn:, we have v b v 3 3 v Rv ( v b) v Rv ( v b) (A) (B) Rv 3 + a v b Rv 3 a v b ( ) ( ) (C) v Rv Rv 3 + ( v b) a( v b) (D) v Rv Rv 3 3 ( v b) a( v b) III. he fllwng data are avalable at 5 C fr graphte and damnd: Δg(5 C, atm) g damnd g graphte 866 [J/ml], ρ damnd 3.5 [g/cm 3 ], and ρ graphte.6 [g/cm 3 ]. (3) he pressure at whch these tw frms f carbn are n equlbrum at 5 C can be determned frm the fllwng equatn J ml J ml (A) 866 ( vgraph vdam )( ) (B) 866 ( vdam vgraph )( ) J (C) ( v J 866 graph vdam ) (D) 866 ( vdam vgraph ) ml ml (4) In the crrect equatn t determne the pressure n [a] where Δg(5 C, atm) 866 [J/ml], the numercal value f v graph (wth the crrect unt) s D-3

269 IV. At 5 C, the vapr pressure f water s 3.69 ka and at 35 C the vapr pressure f water s 5.68 ka. Estmate the enthalpy f vaprzatn f water n kj/ml. v l d h h Nte:, gas cnstant R 8.34 J/ml d v l ( K v v ) 5) V. (6) Calculate the pwer (kj/hr) fr cmpressng 5,000 kg/hr f methane frm 0.8 Mpa, 0 K t 4 Mpa. he adabatc effcency f the cmpressr s 75%. (-h dagram s gven fr methane) VI. (7) Saturated lqud methane at 3 Ma s reduced t Ma thrugh an expansn (let-dwn) valve. he vapr fractn f methane leavng the valve s (-h dagram s gven fr methane) VII. he equlbrum cndtns fr tw phases I and II f a sngle cmpnent are A) I II, I II, h I h II B) I II, I II, s I s II C) I II, I II, u I u II D) Nne f the abve D-4

270 D-5

271 VIII. Cnsder the pstn-cylnder assembly shwn belw; 00 mles f gas expand sthermally after the remval f a 0,000 kg blck. R a Assume that the v behavr can be descrbed by the Van der Waals equatn v b v 3 wth a 0.5 [J m /ml ] and b [m 3 /ml] and that the deal gas heat capacty has a cnstant value f c 35 J/(ml K), R 8.34 J/ml K. Weghtless, frctnless pstn atm Isthermal A 0. m atm expansn A 0. m 0.4 m 9) (4 pts) Determne the wrk [J] dne by the gas n the surrundng (Shw all the wrk) D-6

272 CHE 303 (Wnter 005) LAS NAME, FIRS Quz #3 Nte: Yur answers must be crrect t 3 sgnfcant fgures and have the apprprate unts. I. Crcle the crrect relatn (A) u v + v (B) u v v (C) u v v (D) u v + v II. A vapr-cmpressn refrgeratn prcess usng ammna as the wrkng flud s t perate between 3 F and 86 F. In step (4) () heat s suppled t the flud at 3 F under the pressure. he saturated vapr at s then cmpressed sentrpcally t, where t becmes superheated vapr, state (). Remval f heat frm ths vapr leads t clng at cnstant pressure fllwed by cndensatn at 86 F, step () (3). he cycle s clsed by passng the saturated lqud at (3) thrugh a let-dwn valve t the lwer pressure at (4). he fur states are gven (nt n any partcular rder) as fllw ( F) (psa) h(btu/lb) s(btu/lb R) ) he heat transferred (Btu/lb) n step (4) () s Q L ) he heat transferred (Btu/lb) n step () (3) s Q H 3) he wrk suppled (Btu/lb) by the cmpressr s W 4) he entrpy (Btu/lb R) at state () s D-7

273 III. he free energy change fr the fllwng change n state s 685 cal/ml at 5 C carbn (graphte, atm) carbn (damnd, atm) he pressure at whch these tw frms f carbn are n equlbrum at 5 C s 4,959 atm. he denstes f graphte and damnd are.6 and 3.5 g/cm 3, respectvely, at 5 C ndependent f pressure. Determne Δg (cal/g) fr the fllwng prcesses at 5 C (cal 4.3 cm 3 atm) ) carbn (graphte, atm) carbn (graphte, 4,959 atm) ) carbn (graphte, 4,959 atm) carbn (damnd, 4,959 atm) IV. he vapr pressure f lqud water at 0 C s bar. Specfc vlume f water s m 3 /kml. Gas cnstant R a m 3 /kml K. bar 0 5 a. he fugacty f water at 0 C and 00 bar s V. (4 pts, shw all yur wrk) What fractn f f the vlume f a rgd tube must be ccuped by lqud water at 00 C (the remander beng water vapr) s that when the tube s heated the cntents wll pass thrugh the crtcal state? Data: crtcal state v c m 3 /kg, at 00 C: v G m 3 /kg, v L.673 m 3 /kg. f D-8

274 CHE 303 (Wnter 005) LAS NAME, FIRS Quz #4 Nte: Yur answers must be crrect t 3 sgnfcant fgures and have the apprprate unts. I. A. he fugacty f a cmpnent n a perfect gas mxture s equal t ts partal pressure. B. At lw pressure the fugacty f a lqud s equal t ts vapr pressure.. A and B are true. Only A s true 3. Only B s true 4. A and B are false II. Calculate the fugacty f steam at 450 C and 0 Ma. f Data: At 450 C and 0.0 Ma: g kj/kg At 450 C and 0 Ma: g kj/kg Gas cnstant R 8.34 J/(ml. K), mlecular weght f water 8 dg R d(ln f) III. w cmpunds A and B are knwn t frm deal lqud slutns. A vapr mxture cntanng 40 ml% f A and 60 ml% f B s ntally at 00 F and atm. he vapr pressures sat sat f A and B at ths temperature are A 0.90 atm and B.50 atm. hs mxture s cmpressed sthermally.. he dew pnt pressure s. he bubble pnt pressure s D-9

275 IV. A vessel ntally cntanng prpane at 50 R s cnnected t a ntrgen cylnder, and the pressure s ncreased t 300 psa. he vapr pressure f prpane at 50 R s 55 psa. Ntrgen s nsluble n lqud prpane.. Assume deal slutn, ml fractn f prpane n the vapr phase s. Usng K value frm table, ml fractn f prpane n the vapr phase s able. Equlbrum K values fr lght hydrcarbn systems (Ref: Equlbrum Staged Separatns by Wankat) () ln K -A/ + B - C ln() + D/ () ln K -A/ + B - C ln() + D/, where s n psa, s n R cmpund A B C D Frm rpane () n-butane () D-0

276 V. A lqud cntanng 60 ml% prpane (C3) and 40 ml% n-butane (C4) s subjected t a flash vaprzatn prcess at a pressure f 0 psa where the temperature nsde the flash drum s 475 R. At these cndtn K C3.484, K C (frm able ).. he ext ml fractn f prpane n the vapr phase s. If the ext ml fractn f prpane n the lqud phase s 0.3, V/F 3. he blng pnt f pure n-butane at 0 psa s VI. A. In a clsed system at cnstant and, a spntaneus prcess wll ncrease Gbbs free energy, G, untl a maxmum value fr G s reached at equlbrum. B. Fr an slated system, a prcess that wuld mve the system t a less cnstraned equlbrum state wll decrease the entrpy f the system.. A and B are true. Only A s true 3. Only B s true 4. A and B are false D-

277 CHE 303 (Wnter 005) LAS NAME, FIRS Quz #5 Nte: Yur answers must be crrect t 3 sgnfcant fgures and have the apprprate unts. I. At 5 C a bnary system cntanng cmpnents A and B s n a state f lqud-lqud vapr equlbrum. he mxture s nt deal hwever Rault s and Henry s laws can be appled. he cmpstns f the saturated lqud phases are I I II II x As 0.0, x Bs 0.98 and x As 0.97, x Bs 0.03 he vapr pressure at 5 C are: 0. atm, 0.8 atm A ) he vapr mle fractn f A at three-phase equlbrum s B ) he vapr mle fractn f A n equlbrum wth x A 0.0 s 3) he vapr mle fractn f A n equlbrum wth x A 0.99 s 4) At 5 C a vapr cntanng 0 ml % A and 90 ml % B ntally at 0. atm s cmpressed sthermally untl lqud phase I s frmed. he dew pnt pressure s D-

278 II. In the system A-B, actvty ceffcents can be expressed by ln γ A 0.5 and ln γ B 0.5. A he vapr pressure f A and B at 80 C are: 650 mm Hg, 900 mmhg. here s an azetrpe n ths system. 5) If the azetrpe mle fractn x A s the azetrpe pressure s B x B x A 6) he azetrpe mle fractn x A can be determne frm A) exp{[ ( x A ) ]} / B) exp{[ ( x A ) x ]} / x A x A x A B A A B C) exp{[ ( )]} / D) Nne f the abve A A B III. A system cntans 40 ml % tluene, 30 ml % ethylbenzene, and 30 ml % water. Mxtures f ethylbenzene and tluene bey Rault's law and the hydrcarbns are cmpletely mmscble n water. he system pressure s atm. (Nte tluene, EB ethylbenzene, W water) 7) he bubble pnt temperature can be determne by slvng the fllwng equatn: EB A).3 w atm B) w atm 4 3 C) w + + EB atm D) Nne f the abve 7 7 EB 8) If the hydrcarbn phase cndenses frst, the dew pnt temperature can be determne frm.4atm A) 4atm C) 7.3atm + EB 3atm + 7 EB.3atm B) W.4atm + C) Nne f the abve.3atm + EB D-3

279 IV.(9) A slute dffuses thrugh a membrane that separates tw cmpartments A and B that have dfferent ntal cncentratns. he slute cncentratns n the tw cmpartments as a functn f tme, C A and C B are shwn n Fgure. he vlumes f the tw cmpartments are V A and V B. (A) V A < V B (B) Slute dffuses frm cmpartment B t A. a. A and B are true b. Only A s true c. Only B s true d. A and B are false 0 C A 5 C B 0 t Fgure. Cncentratn f slute as a functn f tme n cmpartments A and B. V.(0) Measurements shw that the smlarty f human tears s 30 msml/l. Determne the value f the smtc pressure f human tears n pascals at 300 K. N 3 a m π R C S,, gas cnstant R 8.34 ml K π D-4

280 Answers t CHE303 Quzzes 005 Quz # I. (D) II. (B) III IV. ) 03.9 kj/hr K ) kj/hr K 3) 67,600 kj/hr 4) K Quz # I. (D) II. (D) III. 3) (A) 4) m 3 /ml IV kj/ml V kj/hr VI VII. (D) VIII J Quz #3 I. (C) II. ) 48.5 Btu/lb ) 54.6 Btu/lb 3) 60. Btu/lb 4).734 Btu/lb R III. ) 06.3 cal/g ) 0 IV bar V Quz #4 I. () II Ma III. ).84 atm ).60 atm IV. ) ) V. ) 0.59 ) ) 503R VI. (4) Quz #5 I. ) ) ) ) atm II. 5) mmhg 6) (D) III. 7) (C) 8) (A) IV. (d) V a D-5