D 7KHUPRVWDWLFEDWKLPSRVHVLWVWHPSHUDWXUHTRQWKHV\VWHP E &RQWDLQHULPSRVHVFRQVWUDLQWRIFRQVWDQWYROXPH7KHUPDOLVRODWLRQLPSOLHV WKDWKHDWIORZPXVWEH]HURZKLOHPHFKDQLFDOLVRODWLRQ DQGFRQVWDQWYROXPH LPSOLHV WKHUH LV QR ZRUN IORZ &RQVHTXHQWO\ WKHUH LV QR PHFKDQLVP IRU DGGLQJ RU UHPRYLQJ HQHUJ\ IURP WKH V\VWHP 7KXV V\VWHP YROXPH DQG HQHUJ\DUHFRQVWDQW F 7KHUPDOO\LVRODWHGDGLDEDWLF )ULFWLRQOHVVSLVWRQSUHVVXUHRIV\VWHPHTXDOVDPELHQWSUHVVXUH RUDPELHQW SUHVVXUH wg/A LI SLVWRQF\OLQGHU LQ YHUWLFDO SRVLWLRQ +HUH w Z HLJKWRISLVWRQ A LWVDUHDDQGgLVWKHIRUFHRIJUDYLW\ G 7KHUPRVWDWLFEDWKFRQVWDQWWHPSHUDWXUHT )ULFWLRQOHVVSLVWRQFRQVWDQWSUHVVXUH VHHSDUWFDERYH H 6LQFH SUHVVXUH GLIIHUHQFH LQGXFHVDPDVVIORZSUHVVXUHHTXLOLEUDWHVUDSLGO\ 7HPSHUDWXUHHTXLOLEUDWLRQZKLFKLVDUHVXOWRIKHDWFRQGXFWLRQRFFXUVPXFK PRUH VORZO\ 7KHUHIRUH LI YDOYH EHWZHHQ WDQNV LV RSHQHG IRU RQO\ D VKRUW WLPHDQGWKHQVKXWWKHSUHVVXUHLQWKHWZRWDQNVZLOOEHWKHVDPHEXWnotWKH WHPSHUDWXUHV D :DWHULVLQDSSURSULDWHDVDWKHUPRPHWULFIOXLGEHWZHHQq&DQGq&VLQFH WKH YROXPH LV QRW D XQLTXH IXQFWLRQ RI WHPSHUDWXUH LQ WKLV UDQJH LH WZR WHPSHUDWXUHVZLOOFRUUHVSRQGWRWKHVDPHVSHFLILFYROXPH V T q & a V T q & V T q & a V T q & HWF 9+2 7 TLQ R &DQGVLQFF J &RQVHTXHQWO\ ZKLOH T XQLTXHO\ GHWHUPLQHV V V GRHV QRW XQLTXHO\ GHWHUPLQHT E $VVXPLQJ WKDW D PHUFXU\ WKHUPRPHWHU LV FDOLEUDWHG DW q& DQG q& DQG WKDW WKH VSHFLILF YROXPH RI PHUFXU\ YDULHV OLQHDUO\ EHWZHHQ WKHVH WZR WHPSHUDWXUHV\LHOGV Chapter 1 V T &h V cT &h c h V cT cT &h & & V R & R R R V R R TV ZKHUH T LV WKH DFWXDO WHPSHUDWXUH DQG TV LV WKH WHPSHUDWXUH UHDG RQ WKH WKHUPRPHWHU VFDOH $W q& V T q & FF J +RZHYHU H[S WKHVFDOHWHPSHUDWXUHIRUWKLVVSHFLILFYROXPHLVIURPHTQ TV VH[S T u u '7 L F 7 L 7KH WHPSHUDWXUH HUURU SORWWHGKHUH UHVXOWV IURP WKH QRQOLQHDU GHSHQGHQFH RI WKHYROXPHRIPHUFXU\RQWHPSHUDWXUH,QDUHDOWKHUPRPHWHUWKHUHZLOODOVR EHDQHUURUDVVRFLDWHGZLWKWKHLPSHUIHFWERUHRIWKHFDSLOODU\WXEH 'L 'V A c h M wV wT 'T A $ VPDOO DUHD A DQG D ODUJH PDVV RI IOXLG M PDJQLILHV 'L REWDLQHG IRU D JLYHQ 'T 7KXVZHXVHDFDSLOODU\WXEH VPDOOA DQGEXOE ODUJHM WRJHW DQDFFXUDWHWKHUPRPHWHUVLQFH wV wT LVVRVPDOO c :KHQZHXVHDIOXLGILOOHGWKHUPRPHWHUWRPHDVXUH 'T ZHUHDOO\PHDVXUH 'L ZKHUH q & q & 5HSHDWLQJ FDOFXODWLRQ DW RWKHU 7KXV T TV DW q& WHPSHUDWXUHV\LHOGVILJXUHEHORZ DERYH h 1 2 2.1 d C V dM = =M IN -M OUT dt dt a) at steady state g M IN =1 M OUT C volumetric flow rate out min g 10m3 1 g g 1 0.1 3 Cu C min min 10 m3 m dC dC m3 m3 V M C u 10 V M C 10 b) dt min min dt 3 V 6 u 6 u 3 108 m V 6 u 6 u 3 108 m3 3 dC g 10m Cu 108m3 1 min min dt dC g 1 10 1 C m 3 dt 108 m min 108 or for C in g 3 m dC dC 108 10C 1 10.8 dt dt C 0.1 10.8 ln C-0.1 -0.1 t § -0.005 · 10.8 ln ¨ ¸ © -0.1 ¹ t = 32.35 minutes For C = 0.95 * 0.1 = 0.095 108 ln 20 = t 2.2 dC dt kC If Cf = 0.5 Ci ln Cf Ci kt ln 0.5 = -kt or k ln 2 t for a half life of seven years k = 0.099 years-1 so that C ln f k 25 0.099 u 25 2.475 22 Cf 22 exp (-2.475) 1.85 ppm 2.3 Basis: 2-liter vessel = 2000 cm3 = 2×103 ×10-6 Initial moles of phosphine = 5 m3 = 2 × 10-3 m3 cm3 kmol mol ×103 ×2×10-5 m3 = 10 mol kmol m3 t 2 a) Mass balance from stoichiometry Species Initial Final PH3 10 10 – 4X = 7 P4 0 X H2 0 6X Since 10 - 4X = 7 3 = 4X X=¾ P4 = 0.75 moles H2 = 6 moles 10-4X moles moles b) CPH3 at any time = = 5-2X ×10-3 2×10-3 m3 m3 Therefore dCPH3 =3.715×10-6 CPH3 dt dX -2×10-3 =-3.715×10-6 5-2X ×10-3 dt dX =3.715×10-6 2.5-X dt dX d(2.5-X ) = =-3.715×10-6 dt 2.5 - X 2.5-X ln 2.5-X t -ln 2.5-X 0 =-3.715×10-6 t 2.5-X -2.715u10 t =e 2.5 -6 2.5-X =2.5 e 2.715u10 t -6 -6 X =2.5 1-e-2.715×10 t Therefore Moles PH 3 =10-4X =10-10 1-e-2.715u10 t = 10 e-2.715u10 t -6 -6 Moles PH 4 =2.5 1-e-2.715u10 t -6 Moles H 2 = 15 1- e-2.715u10 t -6 Time required for 3 moles to react = time for X = 0.75 -6 X =2.5 1-e-2.715×10 t = 0.75 T = 1.314 × 105 sec = 36.49 hours 2.4 2 NO + O2 ĺ122 a) 0.5 moles NO produces 0.5 moles NO2 b) Initial Species Conc NO O2 NO2 1 3 0 dCNO 2 CO2 =-1.4×10-4 CNO dt d 1- 2 X dX 2 =-1.4×10-4 1-2X 3-X =-2 dt dt Conc. at X 1–2X 3–X 6X 4–X 3 dX 2 =0.7×10-4 1-2X 3-X dt 2X-1 1 1 0.7 × 10-4 t = + ln 5 1-2X 25 X-3 Now for 0.5 mole of NO to react 1 – 2X = 0.5 X = 0.25 1 1 0.7×10-4 t = + ln 0.5/2.75 =0.332 5×0.5 25 0.332 t= ×104 =0.474×104 sec = 1.317 hrs 0.7 3 3.1 (a) By an energy balance, the bicycle stops when final potential energy equals initial kinetic energy. Therefore 1 2 mvi 2 vi2 mgh f or h f F 20 km u 1000 m u 1 hr I H hr km 3600 sec K 2g 2 u 9.807 2 m sec2 or h=1.57 m. (b) The energy balance now is 1 2 1 2 mv f mvi mghi or v 2f vi2 2 ghi 2 2 3600 sec 2 km 2 m km v 2f 20 2 u 9.807 u u u 70 m 1000 m hr hr sec2 v f = 134.88 km/hr. Anyone who has bicycled realizes that this number is much F H I K F H I K too high, which demonstrates the importance of air and wind resistance. 3.2 The velocity change due to the 55 m fall is 3600 sec I km u c'v h 2 u 9.807 secm u 55 m u FH 1000 m hr K 2 2 2 v f = 118.24 km/hr. Now this velocity component is in the vertical direction. The initial velocity of 8 km/hr was obviously in the horizontal direction. So the final velocity is km v vx2 v 2y 11851 . hr 3.3 (a) System: contents of the piston and cylinder (closed isobaric = constant pressure) M.B.: M 2 M1 'M 0 M 2 M1 M 0 0 E.B.: M 2U 2 M1U1 'M H Q Ws PdV z d i c h Q z PdV Q Pz dV Q PaV V f M cU U h Q PM cV V h Q M cU U h M c PV PV h M cU PV h cU PV h M c H H h M U 2 U1 2 2 1 2 2 1 2 1 2 1 1 1 2 2 1 1 Solutions to Chemical and Engineering Thermodynamics, 5h ed P 1013 . bar | 01 . MPa V 1.6958 1.9364 T 100 T 150 Linear interpolation 1.8161 T 125q C Final state P 01 . MPa , V2 3.565 T 500q C 4.028 T 600q C Linear interpolation W U 2506.7 2582.8 Chapter 3 H 2676.2 2776.4 2544.8 2726.3 3 m /kg . 36322 3488.1 3704.7 Initial state T2 500 3565 36322 . . T2 514.5q C 4.028 3565 . 600 500 514.5 500 . H 2 34881 H 2 3519.5 600 500 3704.7 34881 . Q 1 kg 3519.5 2726.3 kJ kg 7932 . kJ z PdV a 1 bar u V2 V1 1 bar u 100,000 18161 m kg . . f 1 bar u 36322 3 Pa 1 kg 1J u u u 18161 m3 kg . bar m s2 Pa m2 s2 kg 1816 . kJ kg (b) System is closed and constant volume M.B.: M 2 M1 M E.B.: M 2U 2 M1U1 Q c M U 2 U1 d i 0 Q W 0 – z PdV 0 'M H h s Here final state is P 2 u 1013 . bar ~ 0.2 MPa ; V2 (since piston-cylinder volume is fixed) . P 0.2 MPa ; V2 18161 T(q C) V U 500 1.7814 3130.8 600 2.013 3301.4 17814 . . 18161 . 2.013 17814 T 500 0.0347 600 500 0.2316 T 515q C V1 18161 m3 kg . . ~ 015 . 01498 U 2 31308 . . U 2 3156.4 kJ kg 01498 . 31308 . 33014 Q 1 kg u 3156.4 2544.8 kJ kg 6116 . kJ (c) Steam as an ideal gas—constant pressure N PV PV 1 1 RT RT1 P2V 2 but V 2 RT2 2V 1 ; P1 P2 Solutions to Chemical and Engineering Thermodynamics, 5h ed P12V 1 T2 T2 PV 1 1 T1 T1 T2 27315 . 125 39815 . K 2 u T1 796.3 K 52315 . qC 760.9 kJ z PdV W 2 u T1 1000 g kg 1 kJ u 34.4 J mol K u 796.3 39815 . Ku 18 g mol 1000 J N' H Q Chapter 3 P'V P FG NRT NRT IJ NRaT T f H P P K 2 1 2 1 1 1000 . u 8.314 u 39815 18 . kJ 1839 (d) Ideal gas - constant volume PV 1 1 RT1 So again 2 P1 V1 ; T2 T2 PV 1 1 T1 Q P2V 2 here V 1 V 2 ; P2 RT2 N'U 2T1 2 P1 796.3 K . 1 1000 g kg . u u 34.4 8.314 u 796.3 39815 1000 18 g mol CP R ; Q CV 577.0 kJ 3.4 M wU w, f M wU w, i Mw M weight a 1 kg u CP Tf Ti 3.5 M weight u g u 1 m 1 kg f 1 kg u 9.807 m s u 1 m u m 1kgJ s 1 kg u 4184 . J g Ku 'T Ws 2 2 1000 g u 'T kg 9.807 K 4184 . u 1000 2 9.807 J 9.807 2.344 u 103 K a f a f From Illustration 3.2-3 we have that H T1, P1 H T2 , P2 for a Joule-Thomson expansion. On the Mollier diagram for steam, Fig. 3.3-1a, the upstream and downstream conditions are connected by a horizontal line. Thus, graphically, we find that T ~ 383 K . (Alternatively, one could also use the Steam Tables of Appendix III.) Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 For the ideal gas, enthalpy is a function of temperature only. Thus, becomes which implies that H T1, P1 H T2 , P2 H T1 H T2 , T1 T2 600q C . a 3.6 f a af a f f System: Contents of Drum (open system) mass balance: M t2 M t1 'M energy balance: MU MU t2 z 'MH in Q Ws PdV t1 steam but Q 0 by problem statement, Ws and PdV P'V 80q C . u 103 m3 kg ). Also from the Steam Tables 1029 V T z H in 0 (Note V T is negligible. H T 300q C, P . bar 30 . u 103 m3 kg , 1003 25q C 300 kPa 3069.3 kJ kg and recognizing that the internal energy of a liquid does not depend on pressure gives U t1 U T 25q C, 1.013 bar U sat., T 25q C 104.88 kJ kg U T 80q C, 1.013 bar U sat., T 80q C 334.86 kJ kg and U t2 Now using mass balance and energy balances with M t1 100 kg yields Solutions to Chemical and Engineering Thermodynamics, 5h ed M t2 u 334.86 kJ 100 u 104.88 kJ Chapter 3 M t2 100 u 3069.3 kJ Thus M t2 3069.3 334.86 M t2 3.7 108.41 kg , and 'M 100 u 3069.3 104.88 M t2 M t1 8.41 kg of steam added. (a) Consider a change from a given state 1 to a given state 2 in a closed system. Since initial and final states are fixed, U1 , U 2 , V1 , V2 , P1 , P2 , etc. are all fixed. The energy balance for the closed system is U 2 U1 where W z Ws PdV adiabatic. Thus, U 2 U1 z Q Ws PdV total work. Also, Q Q W 0 since the change of state is W. Since U1 and U 2 are fixed (that is, the end states are fixed regardless of the path), it follows that W is the same for all adiabatic paths. This is not in contradiction with Illustration 3.5-6, which established that the sum Q W is the same for all paths. If we consider only the subset of paths for which Q 0 , it follows, from that illustration that W must be path independent. (b) Consider two different adiabatic paths between the given initial and final states, and let W * and W ** be the work obtained along each of these paths, i.e., Path 1: U 2 U1 W * ; Path 2: U 2 U1 W ** Now suppose a cycle is constructed in which path 1 is followed from the initial to the final state, and path 2, in reverse, from the final state (state 2) back to state 1. The energy balance for this cycle is U 2 U1 W * U 2 U1 W ** 0 W * W ** a f Thus if the work along the two paths is different, i.e., W * z W ** , we have created energy! 3.8 System contents of tank at any time mass balance: M 2 M1 'M energy balance: MU MU 'MH c h c h 2 1 (a) Tank is initially evacuated M1 in 0 Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 H in H 5 bar, 370q C 3209.6 kJ kg interpolation). Then U P 5 bar, T ? 3209.6 kJ kg . 2 interpolation, using the Steam Tables (Appendix A.III) T 548q C Thus U 2 U 'M , and M2 V P V V Therefore M 5 bar, T 548q C # 0.756 m3 kg c . kg . h 13228 1 m3 0.756 m3 kg (by By (b) Tank is initially filled with steam at 1 bar and 150qC and V1 V P 1 bar, T 150q C 194 . m3 kg U1 2583 kJ kg , M V V 1 V 05155 . kg . Thus, M 05155 . 'M kg . Energy balance 1 2 is M 2U 2 05155 . u 2583 T2 , using T2 and P2 M 05155 . u 3209.6 . Solve by guessing value of 5 bar to find V and U in the Steam Tables 2 2 1 m3 V2 are satisfied. By trial and error: T2 ~ 425q C and M 2 # 1563 . kg of which 1.323 kg was present in tank intially. Thus, 'M M 2 M1 0.24 kg . (Appendix A.III). See if energy balance and M 2 3.9 a) Use kinetic energy = mv2/2 to find velocity. 1 kg u v 2 m2 2 sec2 1000 J = 1000 kg so v= 44.72 m/sec m2 sec2 b) Heat supplied = specific heat capacity u temperature change, so 1 mol J u 2510 . u 'T 1000 J so 'T=2.225 K. 1000g u 55.85g mol K 3.10 System resistor Energy balance: dU dt Ws Q where W E I , and since we are interested only in steady state dU dt s Thus Q Ws 1 amp u 10 volts and 1 watt 1 volt u 1 amp 1 J s . 10 watt u 1 J s watt 25q C T 0.2 J s K 3.11 System 0.2 u T 25q C J s . qC 750 gas contained in piston and cylinder (closed) 0 Energy balance: U t2 U t1 Q Ws PdV (a) V constant, z z PdV 0. 0 , Q U t2 U t1 d N U t2 U t1 i NC aT T f V 2 1 From ideal gas law N Thus PV RT 114,367 Pa u 0120 m3 . 3 8.314 Pa m mol K u 298 K 5539 mol (see note following) . Solutions to Chemical and Engineering Thermodynamics, 5h ed Q NCV 298 K 298 630 . 3610 . K T1 T2 Chapter 3 10,500 J 5.539 mol u 30.1 J mol K Since N and V are fixed, we have, from the ideal gas law, that P2 P1 3610 . u 114.367 kPa 298.0 T2 P1 T1 T2 or P2 T1 1385 u 105 Pa . 114367 . u 105 Pa ; 301 . 8.314 38.414 J mol K (b) P constant CP CV R Energy balance U t2 U t1 a Q P'V , since P constant f Q PaV V f Q N aRT RT f Q NC aT T f NCV T2 T1 P 2 T2 2 1 2 1 1 T1 Q NCP 298 10,500 5539 u 38.414 . 347.35 K and NR'T P 'V 5539 . mol u 8.314 Pa m3 mol K u 49.35 K 114,367 Pa 012 . 0.0199 V Note: The initial pressure P Patm 1013 . bar 0.01987 m3 01399 . m3 Patm Pwt of piston 1013 . u 105 kPa 200 kg 1 Ns2 u u 9.8 m s2 2 kg m 0.15 m kPa 13067 . Thus, intial pressure 114.367 kPa . Pwt piston 13,067 N m2 3.12 System contents of storage tank (open system) Mass balance: M 2 M1 'M Energy balance: 'M H MU MU c h c h 2 1 in since Q 13,067 Pa W 0 and steam entering is of constant properties. Initially system contains 0.02 m3 of liquid water and 40 0.02 39.98 m3 of steam. Since vapor and liquid are in equilibrium at 50qC, from Steam Tables, Also from the Steam Tables V L 0.001012 m3 kg , P 12.349 Pa . V V 12.03 m3 kg , H V 25921 . kJ kg , H L 209.33 kJ kg , U L 209.32 kJ kg , and U V 24435 . kJ kg . Solutions to Chemical and Engineering Thermodynamics, 5h ed 0.02 m3 0.001012 m3 kg M1L Chapter 3 U| |V M || W 19.76 kg; M1L M1V . kg . 2308 39.98 m . kg; 332 12.03 m3 kg . u 24435 . 12,248.6 kJ U1 19.76 u 209.32 332 1 3 M1V Also H in . 010 . u 419.04 0.90 u 26761 2450.4 kJ kg Possibilities for final state: 1) vapor-liquid mixture, 2) all vapor, and 3) all liquid. First possibility is most likely, so we will assume V-L mixture. Since P 1013 . bar , T must be 100qC. Thus we can find properties of saturated vapor and saturated liquid in the Steam Tables: V L 0.001044 m3 kg , and U L 418.94 kJ / kg , V V 16729 . m3 kg , H V 26761 . kJ kg , U V 25065 . kJ / kg. 1 x 0.001044 V x 16729 . 2 0.001044 16719 . x m3 kg , where x quality U 2 x 25065 . 1 x 418.94 418.94 2087.56 x kJ kg Substituting into energy balance M 2 418.94 2087.56 x 12,248.6 . f 2450.4 a M 2308 2 where M2 V V2 40 m3 0.001044 16719 . x (quality), M 2 46.36 kg , and Solving by trial and error yields x 05154 . 'M 2328 . kg . Also the final state is a vapor-liquid mixture, as assumed. 3.13 System = tank and its contents (open system) (a) Steady state mass balance dM M M 0 M 1 2 3 dt M M M 10 kg min 3 1 2 b g . . M1 T1 M2 T2 . M3 , T3 Steady state energy balance dU H M H M H 0 M 1 1 2 2 3 3 dt H 3 H exit stream H at temperature of tank contents Solutions to Chemical and Engineering Thermodynamics, 5h ed Also T3 T Chapter 3 temperature of tank contents H C T T , assuming C is not a function of temperature a f 0 m a fr m a 5T 5T 1 T aT T f 65q C 10 2 Now H 0 P 0 P a fr m 5 H 0 CP T1 T0 5 H 0 CP T2 T0 10 H 0 CP T T0 1 2 1 fr 2 dM M M (no useful information here) 0 M 1 2 3 dt dU H M H M H energy balance: M 1 1 2 2 3 3 dt dU d dU dT dT since CP | CV but MU M MCV ~ MCP dt dt dt dt dt dT liquids. Thus MCP 3 5CPT1 5CPT2 10CPT3 and M 50 kg . dt (b) mass balance: c h 10 dT3 2T3 dt At t o f , T3 80 50 130 T3 Ae t 5 C t 0 , T3 C 65q C A C 25q C A 40q C So finally T3 65q C 40q Ce t 5 t minutes At t E.B: M FU F M iU i 'MH in M FU F M FU F M i U i M i U i c L L V V Also known is that V h c 60 m3 L L V V h minutes . H L, in 0.9 H V, in M LF M VF 01 M LFVLF M VFVVF . 2 equations and 2 unknowns V M VFVVF VLF M LF FG V M V U M U IJ c M U M U h K H V LMV M V M OP 01. H 0.9 H Q N V F F F V V L F L F F V V F L F V F V F V i L L, in i L i V V, in 3.14 Thermodynamic properties of steam from the Steam Tables Initial conditions: Specific volume of liquid and of vapor: m3 m3 ; VVi 08857 . kg kg Specific internal energy of liquid and of vapor kJ kJ U Li 3139 . ; U Vi 24759 . kg kg VLi . 1061 u 103 M.B: M f M i 'M i for i V Solutions to Chemical and Engineering Thermodynamics, 5h ed 200 liters = 194.932 kg; VLi Mi M Li M Vi ; M Li M Vi 60 m3 200 liters =14.476 kg and so Mi=209.408 kg VVi E.B. c M f U f M iU i 'MH in M f U f M f U f M i U i M i U i L V L V Chapter 3 h c L L V V h . H L, in 0.9 H V, in M Lf M Vf 01 Total internal energy of steam + water in the tank 194.932u313.0 + 14.476u2475.9 = 9.686u104 kJ Properties of steam entering, 90% quality Specific volume = Vin = 0.1u1.061u10-3+ 0.9u0.8857 = 0.797 m3/kg Specific enthalpy = H =0.1u504.70 + 0.9 u 2706.7 = 2.486 u103 kJ/kg in 60 m3 Also have that V M LfVLf M VfVVf . This gives two equations, and two unknowns, M Lf and M Vf . The solution (using MATHCAD) is M Lf = 215.306 kg and M Vf = 67.485 kg. Therefore, the fraction of the tank contents that is liquid by weight is 0.761. 3.15 System contents of both chambers (closed, adiabatic system of constant volume. Also Ws 0 ). Energy balance: U t2 U t1 0 or U t2 U t1 (a) For the ideal gas u is a function of temperature only. Thus, U t2 U t1 T t2 T t1 500 K . From ideal gas law af af af af af af af af PV 1 1 PV 2 2 N1RT1 N 2 RT2 but N1 N 2 since system is closed T1 T2 see above and V2 2V1 see problem statement. 1 . MPa T2 500 K, P2 05 . MPa P1 5 bar 05 2 (b) For steam the analysis above leads to U t2 U t1 or, since the system is closed U t U t , V t 2V t . From the Steam Tables, Appendix III, P2 af af af af af af U at f U T 500 K, P 1 MPa U T 22685 . q C, P 1 MPa 2 1 2 1 1 # 2669.4 kJ kg V t1 V T 22685 . q C, P af af af af 1 MPa # 0.2204 m3 kg Therefore U t2 U t1 2669.4 kg kg and V t 2V t 0.4408 m3 kg . By, essentially, trial and error, find that af 2 1 T ~ 216.3q C , P ~ 05 . MPa . Solutions to Chemical and Engineering Thermodynamics, 5h ed af af af af Chapter 3 af af af (c) Here U t2 U t1 , as before, except that U t1 U I t1 U II t1 , where superscript denotes chamber. Also, M t M I t1 M II t1 {mass balance} and 2V M t 2V M I t M II t V t af 2 1 af 2 af 1 1 af 1 For the ideal gas, using mass balance, we have a f P V P V 2P I 1 1 T1I P1I P1II T1I T1II (1) U 0 NCV T T0 , and cancel terms, use N PV RT and get P2 2V1 T2 Energy balance: N 2U 2 Substitute U a II 1 1 T1II Eqns. (1) T2 N1IU 1I N1IIU 1II f 2 P2 Using 2 and (2) P1I P1II get P2 (2) 7.5 u 10 Pa 5 0.75 MPa and 529.4 K 256.25q C . T2 (d) For steam, solution is similar to (b). Use Steam Table to get M1I and M1II in terms of V. Chamber 1: U1I 2669.4 kJ kg ; V1I 0.2204 m3 kg ; M I V V I 4.537V 1 1 1 a Chamber 2: U1II U T 600 K, P V II 05483 . m3 kg ; M II 1824 . V V V II 1 1 2V1 Thus, V2 I M M II U M IU I M IIU II 2 c 1 1 1 2V1 4.537V1 1824 . V1 . MPa 05 f 28459. kJ kg ; 1 0.3144 m3 kg ; h b M M g 2720.0 kJ kg I 1 II 1 By trial and error: T2 ~ 302q C 575 K and P ~ 0.76 MPa . 3.16 System: contents of the turbine (open, steady state) (a) adiabatic dM M M 0 M M mass balance: 1 2 2 1 dt 0 dU H M H Q 0 W P dV 0 M energy balance: 1 1 2 2 s dt dt Ws M1 H1 H2 M1 3450.9 28656 . kJ kg c c h 5853 M u 10 1 . But Ws 5 h J kg 7.5 u 105 watt 7.5 u 105 J s 7.50 u 105 J s . kg s 4.613 u 103 kg h 1281 . 5853 u 105 J kg (b) Energy balance is 0 dU H M H Q W P dV 0 M 1 1 2 2 s dt dt M 1 Solutions to Chemical and Engineering Thermodynamics, 5h ed a Chapter 3 f 60 kJ kg H 2 H 150q C, 0.3 MPa 27610 . kJ kg where Q M 1 Thus Ws 1281 . kg s 3450.9 27610 . 60 kJ kg 807 kJ s 8.07 u 105 watt 807 kW 3.17 System: 1 kg of water (closed system). Work of vaporization PdV P dV z z P'V since P is constant at 1.013 bar. Also, from Steam Tables m3 kg ; 'V 16719 . m3 kg V L 0.001044 m3 kg ; V V 16729 . Energy balance for closed system (1 kg): U U Q PdV Q 1013 . . u 105 Pa u 16719 m3 kg 2 z 1 Q 16945 . u 105 J kg U 2 U 1 . kJ kg 25065 2.5065 u 106 J kg 418.94 kJ kg . u 105 J kg 41894 Thus Q U 2 U1 W u 105 16945 u 105 . . 2.5065 u 106 41894 2.2570 u 106 J kg z PdV W 16945 u 105 J kg . . So heat needed to vaporize liquid 2.2570 u 106 J kg of which 016945 . u 106 is recovered as work against the atmosphere. The remainder, 2.088 u 106 kJ kg , goes to increase internal energy. 3.18 System = Contents of desuperheater (open, steady state) Superheated steam T=500ºC P=3 MPa Desuperheater Water 25ºC M 1 M 2 500 kg hr ; H 1 34565 . kJ kg H sat’d liq., T 25q C ?; H 2 Mass B: 0 Saturated steam 2.25 MPa 104.89 kJ kg M M M 1 2 3 H M H M H Q 0 W 0 P dV M 1 1 2 2 3 3 s dt 500 M 2 kg hr ; H3 H sat’d steam, P 2.25 MPa 0 Energy B: 0 M 3 Thus, b g . kJ kg 28017 Solutions to Chemical and Engineering Thermodynamics, 5h ed b 0 500 u 34565 . M 2 M 2 Chapter 3 g u 28017 104.89 500 M . 2 1214 . kg hr 3.19 The process here is identical to that of Illustration 3.5-3, so that we can use the equation P2 P1 T1 CV CP P2 P1 Tin a T2 2.0 MPa , Tin 20.99 J mol K . developed in the illustration. CP 29.3 J mol K , CV Here, P2 CP R CP Tin 5488 . K CV Cylinder 1: P1 0 , T2 Cylinder 2: P1 01 . MPa , T1 f 20q C Cylinder 3: P1 1 MPa , T1 20q C 39315 . K, 27565 . qC 29315 . K 2.0 . 29315 . 20.99 29.3 2.0 01 . 39315 . 01 T2 120q C 29315 . K ; T2 252.7q C . K 52587 38216 . K 109.01q C 3.20 System: Gas contained in the cylinder (closed system) M piston g 4000 kg 9.8 m s2 (a) P 01013 10133 u 105 u . . MPa A 2.5 m2 1 kgm Ns2 11701 . u 105 Pa moles of N gas initially in system 0117 . MPa PV RT 11701 . u 105 Pa u 25 m3 8.314 Pa m3 mol K u 29315 . K 1200 kmol . . u 103 mol 1200 (b) Energy balance: U 2 U1 Q PdV z 'V 3 m u 2.5 m Final temperature: 2 T2 PV2 NR U 2 U1 7.5 m ; P'V Q P'V since P is constant. 11701 . u 105 Pa u 7.5 m3 3 11701 u 105 Pa u 25 7.5 m3 . 3 . u 10 mol u 8.314 Pam3 mol K 12 a N U 2 U1 8.7758 u 105 J 3812 . K 108.05q C f NC aT T f V 2 1 12 . u 10 mol u 30 8.314 J mol K u 3812 . 29315 . K 3 2.291 u 106 J (c) Q 'U P'V 'T of gas work 2.291 u 106 8.7758 u 105 work 27.7% of energy absorbed 'T 72.3% . 3169 u 106 J . 3169 MJ (d) System: Gas contained within Piston + Cylinder (open system). Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 [Note: Students tend to assume dT dt 0 . This is true, but not obvious!] dN mass balance: N dt dV d 0 NU N H out Q P energy balance: dt dt Here (1) P is constant, (2) Ideal Gas Law V NRT P , (3) T and P of Gas Leaving Cylinder T and P of gas in the system. Thus, d NRT dN dN dU N U P H dt P dt dt dt dT dN d NCV NT H U R dt dt dt dN dT dT dN dT RT 0 NCV NR RT N CV R dt dt dt dt dt dT 0 Q. E.D. dt Thus T3 T2 3812 . K Now going back to F H I K a N dN dU U dt dt H dN dt H U f dT dU dV dN and using P 0 dt dt dt dt dN P dV dV dN P RT or dt dt dt RT dt (**) Since P and T are constants N3 N2 V3 V2 25 m3 25 7.5 m3 0.7692 Thus N 3 0.7692 u 1200 mol 923 mol ; 'N 277 mol 0.277 kmol 3.21 (a) System: Gas contained within piston-cylinder (closed system) [neglecting the potential energy change of gas] energy balance: d NU dt But T Thus PV dT NR dt N dU dt dV dT Q P ; NCV dt dt F I PA dh . H K NR dt P dV NR dt dh Q PA dt Solutions to Chemical and Engineering Thermodynamics, 5h ed F H Chapter 3 I K ACV P dh dh C dh PACP dh PA V 1 AP R dt dt R dt R dt 30 J mol K . u 11701 u 105 Pa u 2.5 m2 u 0.2 m s 8.314 J mol K Q 2111 . u 105 J s (b) System: Gas contained within piston and cylinder (open system). Start from result of Part (d), Problem 3.20 (see eqn. (**) in that illustration) dN dt P dV RT dt PA dh with P and T constant RT dt (See solution to Problem 3.20) u 105 Pa u 2.5 m2 11701 . u 0.2 m s 8.314 J mol K u 3812 . K 0.01846 kmol s a dN dt [check: 18.46 mol sec u 15 sec Problem 3.20] f 18.46 mol s 276.9 mol compare with part d of 3.22 System: gas contained in the cylinder (open system) Important observation . . . gas leaving the system (That is, entering the exit valve of the cylinder) has same properties as gas in the cylinder. dN ½ Note that these are mass balance N °° dt ¾ Eqns. (d) and (e) of d NU ° Illustration 3.5-5 energy balance NH °¿ dt dN Note that these are N mass balance dt Eqns. (d) and (e) of d NU energy balance N H Illustration 2.5- 5 dt Proceeding as in that illustration we get Eqn. (f) |VU |W FG T t IJ HT 0 K CP R FG P t IJ or T t HP0K Pt R CP 320 10a8.314 30f 169.05 K (1) where we have used a slightly different notation. Now using the mass balance we get dN dt F I V d a P T f N H K R dt d PV dt RT or a4.5 28f mol s u 8.314 Pa m mol K a f NR 8.908 Pa K s d PT dt 3 V 3 . m 015 Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 and P Tt P 8.908 u 105 t Tt 0 Using t 5 minutes and (2) yields bar K for P in bar and t in secs. (2) 300 secs in Eqn. (2) and simultaneously solving Eqns. (1) T 5 min 152.57 K , P 5 min 0.6907 bar Computation of rates of change from mass balance d ln P d ln T F I 1 F dP Pd ln T I NR H K T H dt dt K V or dt dt d P dt T NRT PV (3) From energy balance (using 2 eqns. above and eqn. (f) in Illustration (3.5-5)) a f CV d ln T R dt d ln P T C d ln T or P R dt dt d ln P dt (4) Now using Eqn. (4) in Eqn. (3). Thus, CV d ln T CV dT NRT or R dt RT dt PV dT N RT 2 . K sec 1151 dt t 5 min PVCV t 5 min and CP P dT RT dt 5 min dP dt 5 min 0.0188 bar s 3.23 Consider a fixed mass of gas as the (closed) system for this problem. The energy balance is: d NU dt N From the ideal gas law we have P CV N dT dt dU dt NCV dT dt P dV dt NRT V . Thus NRT dV C d ln T V R dt V dt d ln V dt or CV T2 ln R T1 FG IJ H K V T ln 2 2 V1 T1 CV R FG V IJ HV K 1 2 (*) Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 or V2T2 CV R V1T1CV R VT CV R constant Substituting the ideal gas law gives PV CP CV PV J constant. Note that the heat capacity must be independent of temperature to do the integration in Eqn. (*) as indicated. 3.24 System: Contents of the tank (at any time) (a) Final temperature T 330 K and pressure P c h . 1013 u 105 Pa are known. Thus, there is no need to use balance equations. 1013 u 105 Pa u 0.3 m3 . 8.314 J mol K 330 K PV RT Nf 1108 . mol 0.01108 kmol (b) Assume, as usual, that enthalpy of gas leaving the cylinder is the same as gas in the cylinder . . . See Illustration 3.5-5. From Eqn. (f) of that illustration we have Pf Pi Thus Nf 2136 . mol FG T IJ HTK CP R f i Tf or Tf Ti FG P IJ H PK R CP f FG 10133 . u 10 I H 10. u 10 JK 5 8.314 29 . 05187 6 i 05187 . u 330 K 17119 . K, 0.02136 kmol . Pf 1013 . bar , and 3.25 Except for the fact that the two cylinders have different volumes, this problem is just like Illustration 3.5-5. Following that illustration we obtain 2 P1i T1i 2 P1i 2 P1f P2f f T1f T2 2 P1f P2f or P f for Eqn. (a') 2 i P1 3 for Eqn. (c') and again get Eqn. (f) FG T IJ HT K f 1 i 1 Then we obtain P f 1333 . bar , T1f CP R FG P IJ HP K f 1 i 1 2234 . K , and T2f 328.01 K . 3.26 From problem statement P1f P2f P f and T1f T2f T f . Mass balance on the composite system of two cylinders N1f N 2f N1i or 2 P1f P2f f T1f T2 3P f Tf 2 Pi Ti Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 Energy balance on composite system N1iU 1i and T f FI HK 3P f i T1 2 P1i 3 2 i T1 2 3 T1i 2 u 200 bar 3 2 Pi 3 N1f U 1f N 2f U 2f P f 1333 . bar (as before) 250 K . 3.27 Even though the second cylinder is not initially evacuated, this problem still bears many similarities to Illustration 3.5-5). Proceeding as in that illustration, we obtain 2 P1i P2i i T1i T2 2 P1i P2i 2 P1f P2f f instead of Eqn (a') T1f T2 2 P1f P2f 3 P f instead of Eqn. (c) 2 u 200 1 u 20 / 3 140 bar ] and again recover Eqn. (f) for [Thus, P f Cylinder 1 FG T IJ HT K f 1 i 1 Solution is P1f P2f 140 bar , T1f FG P IJ HP K CP R f 1 i 1 226.47 K , T2f Eqn. (f) 28651 . K. 3.28 (a) System = Gas contained in room (open system) dN N mass balance: dt d NU dN N H Q H Q energy balance: dt dt Thus, Q dN d NU H dt dt UH dU dN N dt dt F I V d F PI H K R dt H T K V d P NT d F P I dT dT Q RT F I F I NC RT NC H R K dt H T K P dt H T K dt dt For the ideal gas, H U PV RT ; dN dt d PV dt RT V Q Since P NRT dP dT dT NR NCV dt dt P dt constant, dP dt 0 , Q NCPdT or dt V Solutions to Chemical and Engineering Thermodynamics, 5h ed dT dt Q RT CP PV Chapter 3 15 . u 103 W 8.314 J mol K 28315 . K 5 29 J mol K 10133 u 10 Pa 35 . . u 5 u 3 m3 0.0229 K s 137 . K min (b) System = Gas contained in sealed room (closed system) N d NU dU dT N NCv Q Energy balance: dt dt dt Q NCV dT dt sealed room CP dT CV dt unsealed room 0 29 . K min u 137 29 8.314 1925 . K min 3.29 In each case we must do work to get the weights on the piston, either by pushing the piston down to where it can accept the weights, or by lifting the weights to the location of the piston. We will consider both alternatives here. First, note that choosing the gas contained within piston and cylinder as the system, 'U Q W . Also But 'U 0 , since the gas is ideal and T constant. W PdV NRT ln Vf Vi , for the same reasons. Thus, in each case, we have z a f that the net heat and work flows to the gas are W work done on gas u 10 . FG V IJ 2479 ln 1213 HV K 2.334 u 10 NRT ln 2 f i and Q W 2 1622.5 J 1622.5 J (removed from gas) If more work is delivered to the piston, the piston will oscillate eventually dissipating the addition work as heat. Thus, more heat will be removed from the gas + piston and cylinder than if only the minimum work necessary had been used. Note that in each case the atmosphere will provide Watm P'V 1013 . u 105 kPa u 2.334 1213 . u 102 m3 11356 . J and the change in potential energy of piston mg'h 5 kg u 9.8 m s2 u . 2.334 1213 u 102 m3 2 1 u 10 m 54.9 J The remainder 1622.5 11356 . 54.9 432.0 J must be supplied from other sources, as a minimum. (a) One 100 kg weight. An efficient way of returning the system to its original state is to slowly (i.e., at zero velocity) force the piston down by supplying 432.0 J of energy. When the piston is down to its original location, the 100 kg is slid sideways, onto the piston, with no energy expenditure. An inefficient process would be to lift the 100 kg weight up to the present location of the piston and then put the weight on the piston. In this case we would supply Solutions to Chemical and Engineering Thermodynamics, 5h ed Mg'h Chapter 3 b g 2 u 102 m3 . 'V m 2.334 u 10 1213 100 kg u 9.8 2 u 2 2 A s 1 u 10 m 2 2 1098.6 kg m s 1098.6 J Mg This energy would be transmitted to the gas as the piston moved down. Thus 11356 . J 54.9 J 1098.6 J W on gas atmosphere PE of piston W J Q J Efficient 1622.5 Inefficient 2289.1 22891 . PE of weight Wcycle Qcycle 1622.5 11905 . 432.0 22891 . 11905 . 1098.6 (b) Two 50 kg weights In this case we also recover the potential energy of the topmost weight. mg'h 50 kg u 9.8 1213 u 102 m3 m 1597 . . u 2 0.01 m2 s 188.2 J Thus in an efficient process we need supply only 1622.5 11356 . 54.9 188.2 2438 . J An efficient process would be to move the lowest weight up to the position of the piston, by supplying 50 kg u 9.8 u 102 m3 . m 2.334 1213 u 2 2 1 u 10 m2 s 549.3 J Slide this weight onto the piston and let go. The total work done in this case is 11356 . atmosphere 54.9 'PE of piston 2438 . 'PE of weight 549.3 supplied by us 19836 . J Therefore W J Q Wcycle Qcycle 1622.5 1378.7 2438 . J Efficient 1622.5 19836 . 1378.7 604.9 J Inefficient 1983.6 (c) Four 25 kg weights. In this case the recovered potential energy of weights is 25 kg u 9.8 m s2 u FG 1897 1213 1597 1213 1379 1213 . . . . . . u 10 I JK m H 1 u 10 302.3 J Thus in an efficient process we need supply only 2 2 Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 1622.5 11356 . 54.9 302.3 129.7 J An inefficient process would be to raise the lowest weight up to the piston, expending 25 kg u 9.8 m s2 u . u 102 m3 2.334 1213 1 u 102 m 274.6 J Thus the total work done is 11356 . 54.9 302.3 274.6 1767.4 J and W Q Wcycle Qcycle . 129.5 Efficient 1622.5 1622.5 14930 . 274.4 Inefficient 1767.4 1767.4 14930 (d) Grains of sand Same analysis as above, except that since one grain of sand has essentially zero weight W 1622.5 J , Q 1622.5 J , Wcycle Qcycle 0 . 3.30 System = Gas contained in the cylinder (closed system) d NU dU dT dV P N NCV energy balance: dt dt dt dt ideal gas equation of state} Since CV and CP are constant CV 1 dT R T dt T2 FG IJ FG V IJ H K HV K 1 dV T or 2 V dt T1 . u 25 27315 P2 NRT dV V dt R CV 1 FG L IJ HL K {Using the R CV 1 2 FG 0.03 m IJ H 0.03 0.6 u 0.05 m K 3 2 8.314 30 8.314 3 . K 44.58q C and 22857 . 1 22857 V T2 20 u u 7.666 bar P1 1 . 2 29815 V2 T1 FG IJ FG IJ H KH K From the difference (change of state) form of energy balance 'U Q 0 W a NCV T2 T1 f z PdV PV 20 bar u 0.03 m3 kmol K 0.0242 kmol RT 298.15 K u 8.314 u 102 bar m3 . 22857 . K W 'U 0.0242 kmol u 30 8.314 kJ kmol K 29815 3652 . kJ Where has this work gone? and N Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 (a) To increase potential energy of piston (b) To increase kinetic energy of piston (c) To push back atmosphere so system can expand (d) Work done against friction (and converted to heat). To see this, write Newton’s 2nd Law of Motion for the piston Patm u A f Fr Frictional Force mg f Fr Pressure of gas (P) u A a MA PA Patm A mg f fr f Thus, P m dv mg f fr Patm A dt A A 'U 36,520 J PdV z z PatmdV Now f m dvdt ; v velocity of piston 1 dV A dt dh dt v (h z z piston height) and v 36,520 J z 1 mg dv m dv dV dt dV f fr dt A A dt A dt dt Patm 'V 3000 J Work against atmosphere mv 2 2 since vinitial 0 dv dt (1) c h 1 d 2 v 2 dt mg'h z f fr vdt 1760 J Work used to increase potential energy of piston z mv 2 1760 f fr vdt . 2 (a) If there is no friction f fr 0 then Thus 36,520 J 3000 36520 3000 1760 J u 2 300 kg v2 (b) If we assume only sliding friction, f fr z f fr vdv 2117 . m2 s2 v 14.55 m s kv z k v 2dt 36520 3000 1760 z m 2 v k v 2dt 2 In order to determine the velocity now we need to know the coefficient of sliding friction k, and then would have to solve the integral equation above (or integrate successively over small time steps). It is clear, however, that v with friction v without friction 14.55 m s System for part a 25q C, 3.0 u 106 Pa = 3Mpa 125 kg/s System for part b Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 3.31 25qC, 30 . u 106 Pa 3 MPa 125 kg s (a) mass balance (steady-state) M 0 M 1 2 M M 125 kg s 1 2 Energy balance (neglecting PE terms) 0 IJ K FG H FG H 2 2 H v1 M H v2 M 1 1 2 2 2 2 IJ K UvA mnvA ; U mass density, n molar density, M v velocity, A pipe area, m molecular weight. M P vA m RT 125 kg s 30 . u 106 Pa u v m s u S u 0.09 m2 16 kg kmol 298.15 K u 8.314 u 103 Pa m3 v 22.83 m s a f a f mv 2 16 kg kmol u 22.83 m s 2 4170 u 103 J kmol . 2 2 u 1 kg m Ns2 Back to energy balance, now on a molar basis a mv22 mv12 C p T1 T2 2 2 As a first guess, neglect kinetic energy terms . . . Cp T1 T2 0 T1 T2 25q C H1 H 2 a 417 . kJ kmol f f Now check this assumption . u 106 v1 n1v1 Pv 30 1 1 34.24 m s n2 P2 2.0 u 106 Recalculate including the kinetic energy terms m 2 2 16 Cp T1 T2 v1 v2 34.242 22.832 5209 J kmol 2 2 5209 J kmol T2 T1 T1 014 . qC . J mol u 1000 mol kmol 368 Thus the kinetic energy term makes such a small contribution, we can safely ignore it. v2 a (b) f c h c Mass balance on compressor (steady-state) 0 2.0 u 106 Pa T1=25q C compressor h N 1 N 2 3.0 u 106 Pa T2= ? Energy balance on compressor, which is in steady-state operation Solutions to Chemical and Engineering Thermodynamics, 5h ed 0 Ws Ws N 1 H 1 N 2 H 2 Q 0 Chapter 3 N 1Cp (T2 T1) 25qC adiabatic compressor Can compute Ws if T2 is known or vice versa. However, can not compute both without further information. 2.0 u 106 Pa T2= ? Gas cooler 3.0 u 106 Pa T3=25q C Analysis as above except that Q z 0 but W 0 N 2 N 3 Here we get Q N 1Cp ( T3 T2 ) R| S| T 0. 25q C Can not compute Q until T2 is known. See solution to Problem 3.10. 3.32 a) Define the system to be the nitrogen gas. Since a Joule-Thomson expansion is isenthalpic, H T1, P1 H T2 , P2 . Using the pressure enthalpy diagram for nitrogen, Figure 3.3-3, we have H 135 K,20 MPa 153 kJ / kg and then T2 T P2 0.4 MPa, H 153 kJ / kg a f a f c h From which we find that T = 90 K, with approximately 55% of the nitrogen as vapor, and 45% as liquid. b) Assuming nitrogen to be an ideal gas (poor assumption), then the enthalpy depends only on temperature. Since a Joule-Thomson expansion is isenthalpic, this implies that the temperature is unchanged, so that the final state will be all vapor. 3.33 Plant produces 136 . u 109 kwh of energy per year Plant uses 136 . u 109 u 4 544 . u 109 kwh of heat 6 1 kwh 36 . u 10 J J year . u 106 u 544 . u 109 kwh 19.584 u 1015 J year Plant uses 36 kwh 'H of rock (total) M Cp Tf Ti a 12 10 f kg u 1 J g K u 1000 g kg u 110 600 K = 490 u 1015 J 19.58 u 1015 J year u x years 490 u 1015 J x 2502 . years 3.34 a) Body temperature is 37oC. From Appendix A.III, ice at 0oC has a volume of 0.0010908 m3/kg and an enthalpy (and internal energy) of -333.43 kJ/kg. For water at 35oC Û = 146.67 kJ/kg and at 40oC Û = 167.56 kJ/kg, so that at 37oC Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 Û = 155.03 kJ/kg. Also, 1 L = 10-3 m3, so that 1 L of water 10-3 m3/0.0010908 m3/kg = 0.9168 kg. So that the amount of energy needed to melt 1 L of ice is 0.9168 kg ¯(155.03-(-333.43))kJ/kg = 447.8 kJ. b) 447.8 kJ/(42 kJ/kg fat) = 10.66 g of fat (= 0.023 lbs of fat) c) For water at 0oC, Û = 0 kJ/kg, and V̂ = 0.001 m3/kg. Therefore, 1 L of water = 1 kg, and the energy required to warm up the water is 1 kg¯155.03 kJ/kg = 155.03 kJ, and only 155.03/42 = 3.69 g of fat would be consumed. 3.35 a) M 2 M 3 M M 1 2 Mass balance M M 10 kg/s M 3 1 2 Energy balance Stream 120°C (1 atm) M 3 M 10 kg/s M 1 2 ˆ H ˆ M 1 1 M2H2 10 M 3 10 kq/s kg ˆ ˆ ˆ H H 3 M1H1 M 2 2 s 10 kq ˆ H3 s Ĥ1 = 83.96 J/g Stream 2450°C, 2.5 MPa Ĥ 2 = 3344.0 J/g Ĥ 3 = 376.92 J/g Stream 390°C (1 atm) M 10-M u 83.96 10-M M 1 MB EB M 1 M 2 b) 3344 10 x 376.92 9.101 kg/s M 1 9.101 kg/s M 1 9.101 kq/s 0.899 kq/s M 10 kg/s, T=90°C 10 kg/s, T=20°C O N P 10 kg/s, T=20°C 100 kg/s, quality = 50% Ĥ Steady balance: 1 1 2676.1 419.04 2 2 12547.6 J/g Solutions to Chemical and Engineering Thermodynamics, 5h ed Chapter 3 H ˆ H ˆ ˆ ˆ M M 1 1 M2H2 3 3 M4H4 M M 10 kg/s But 1 3 M 2 M 4 ? kg/s x3344 = 10 × 376.92 + M × 1547.6 10 × 83.96 + M 2 2 M ¯ (3344 – 1547.6) = 10 × (376.92 – 83.96) 2 =1.63 kg/s M 2 3.36 ǻ vap H at 37°C ~ 2412 J/g Amount evaporated = 4184 u 103 J hr 1 2412 J 1.735 u 102 g g hr 1.735 kg/hr If only 75% evaporates M 3.37 1.735 0.75 N 1.631 kg/hr of sweat produced 25°C, 0.8 bar –50°C 0.1 bar Mass balance – steady state N N O N N 1 2 2 1 Energy balance – steady state W W H Q N H H Q O N 1 1 2 1 2 W Q N 1 H 2 H1 H 2 H1 ³ C dT Cp T2 T1 p T1 30 W Q N1 T2 J u 75k mol K T2 ³ C dT p 2250 J mol Cp T2 T1 T1 30 J u 75k mol K PV NRT 100 m3 N N 1 2250 J Mol 0.8bar u 100 m3 u 298.15 M bar m3 8.314 u 105 mol K 3228 mols/min PV RT 0.03228 u 105 mol 3228 mols Solutions to Chemical and Engineering Thermodynamics, 5h ed W Q W Q Chapter 3 7.262 u 106 J/min 1.210 u 105 J/s 1.210 u 105 Watts 121.0 kw $ Cost 0.2 u 121.0 kw 24.2 $/hr kw hr 7.262 u 106 J/min 1.210 u 105 J/s 1.210 u 105 Watts 121.0 kw $ Cost 0.2 u 121.0 kw 24.2 $/hr kw hr 3.38 In the folder Aspen for Textbook>Chapter 3>Problem 3.38 (Prob 3.5) 3.39 In the folder Aspen for Textbook>Chapter 3>Problem 3.39 (Prob 3.16a) 3.40 In the folder Aspen for Textbook>Chapter 3>Problem 3.40 (Prob 3.22) Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 4 4.1 Ball (1) + Water (2) Energy balance: M U f M U f M U i M U i (a) System 1 c h 1 2 c M1CV,1 T1 f T1i M 2CV,2 T2f T2i Tf M1CV,1T1i M 2CV,2T2i M1CV,1 M 2CV,2 2 1 1 2 0 2 h 0 ; also T T . Thus f f 2 1 . u 75 12 u 103 u 4.2 u 5 5 u 103 u 05 . 12 u 103 u 4.2 5 u 103 u 05 8.31q C [Note: Since only 'T s are involved, q C were used instead of K)]. dT (b) For solids and liquids we have (eqn. 4.4-6). That 'S M CP T z which CP is a constant. Thus Ball: 'S . 5 u 103 g u 05 RS T . 8.31 27315 J u ln . 75 27315 gK . s 53161 12 u 103 g u 4.2 Water: 'S RS T T MCP ln 2 for the case in T1 J UV 53161 W . K . 8.31 27315 J u ln . 5 27315 gK UV 596.22 J K W and J J 64.61 K K Note that the system Ball + Water is isolated. Therefore J 'S Sgen 64.61 K Energy balance on the combined system of casting and the oil bath 'S Ball Water 4.2 c 596.22 53161 . h c h 0 since there is a common final temperature. kJ kJ T 450iK 150 kg u 2.6 20 kg u 0.5 d dT 450iK 0 kg K kg K M cCV,c T f Tci M oCV,o T f Toi f f This has the solution Tf = 60oC = 313.15 K Since the final temperature is known, the change in entropy of this system can be calculated kJ . 60 . 60 27315 27315 . . u ln 4135 150 u 2.6 u ln from 'S 20 u 05 K . 50 . 450 27315 27315 F H 4.3 I K Closed system energy and entropy balances dU dV dS Q ; Sgen ; Q Ws P dt dt dt T dS Thus, in general Q T TSgen and dt F H I K Solutions to Chemical and Engineering Thermodynamics, 5th ed dU dS dV T TSgen P dt dt dt dU dV Q P dt dt Ws c Reversible work: WsRev h dUdt T dSdt P dVdt WsRev Sgen c S W s gen (b) System at constant S & P 0 dU dt (a) System at constant U & V dS dt Chapter 4 0 and dV dt h W 0 dS dt 0 Rev S T 0 and dP dt 0 P dV dt d PV dt d U PV dt dH dt so that c h W Ws Sgen rev S 0 dU d PV dt dt 4.4 700 bar, 600oC 10 bar, T = ? Steady-state balance equations dM M 0 M 1 2 dt 0 dU H M H H M H Q 0 W 0 P dV M 0 M 1 1 2 2 1 1 2 2 s dt dt or H 1 H 2 Drawing a line of constant enthalpy on Mollier Diagram we find, at P At 700 bar and 600q C V 0.003973 m3 kg H 3063 kJ kg S 5522 . kJ kg K 10 bar, T # 308q C At 10 bar and 308q C V | 0.2618 m3 kg H | 3063 kJ kg S 7145 . kJ kg K Also dS dt c 0 S S M 1 2 1 Sgen h 0 S M S Q Sgen 0 M 1 1 2 2 T Sgen kJ or S2 S1 7145 1623 . 5522 . . M1 kg K 4.5 1 System 2 Ws Energy balance 'U cU U h cU U h Q f 2 i 2 f 1 i 1 adiabatic z constant WS PdV volume Solutions to Chemical and Engineering Thermodynamics, 5th ed Ws c h c T2f Tf MCp T2f T2i MCp T1 f T1i but T1 f Ws MCP Chapter 4 h MC cT T h cT T h p f 2 i 2 f 1 i 1 2T f T1i T2i Entropy balance adiabatic 'S f 2 i 2 f 1 i 1 cS S h cS S h 0 R T T UV 0 ; T T or lnS TTT W cT h cT T h or T f 2 z cS S h cS S h i 2 f 1 i 1 f 1 i i 2 1 0 Q dt Sgen T 0 for maximum work T Tf MCP ln 2i MCP ln 1 i T2 T1 f f f 2 f 2 1 i i 1 2 f 2 T1i T2i ; but T1 f f T1i T2i and Ws MCP 4.6 2T f T1i T2i 1 bar 10 bar 290 K 575 K T2f Tf 2 T1i T2i T1i T2i (a) Entropy change per mole of gas T P 'S CP ln 2 R ln 2 eqn. (4.4-3) T1 P1 J J J 10 575 Thus 'S 29.3 ln ln 0.9118 8.314 mol K 290 mol K 1 mol K (b) System contents of turbine (steady-state system) dN Mass balance 0 N 1 N 2 N 2 N 1 N dt dV 0 dU 0 Energy balance Ws P 0 N 1 H 1 N 2 H 2 Q dt dt T T Ws N H 2 H 1 NC 1 P 2 J Ws W CP T2 T1 29.3 u 575 290 K mol K N a a f a f f J mol (c) In Illustration 4.5-1, W 7834.8 J mol because of irreversiblitities 'S z 0 , more work is done on the gas here. What happens to this additional energy input? It appears as an increase of the internal energy (temperature) of the gas. 83505 . 4.7 Heat loss from metal block dU dT CP Q dt dt Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 RS T Q heat out of metal T T2 Q 1 T Q heat into heat engine W a T f W z Wdt C z F1 I dT H TK T T O L C MaT T f T ln P W C aT T f C T ln T TQ N LF T I T O W C T MG1 J ln P NH T K T Q F TI Q z C dT C aT T f C T G1 J H TK CP dT T T2 dt T T2 t 2 P T1 0 2 P 2 1 2 2 P 2 P 1 2 1 1 2 2 1 1 P 2 T2 1 P P 2 1 P 2 2 T1 Alternate way to solve the problem T2 System is the metal block + heat engine (closed) dU dT E.B.: CP Q W dt dt dS Q Sgen S.B.: dt T 2 0 for maximum work Q W W W 4.8 dS dU ; T2 dt dt dU dS T2 dt dt z Wdt z FH T2 CP dS dT W ; dU CPdT ; dS T dt C T CPdT T2 P dT CP 1 2 dT T T T2 F H CP 1 T1 a I K T2 dT T f T CP T2 T1 T2CP ln 2 T1 z FH I K I K LF T I T O C T MG1 J ln P NH T K T Q T2 CP T1 1 T2 dT T 1 2 2 1 P 2 This problem is not well posed since we do not know exactly what is happening. There are several possibilities: (1) Water contact is very short so neither stream changes T very much. In this case we have the Carnot efficiency K W Q Thigh Tlow Thigh 22 27 273 22 300 0.0733 7.33% (2) Both warm surface water (27qC) and cold deep water (5qC) enter work producing device, and they leave at a common temperature. TH TL TO Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 b dM M M M M 0 M M H L 0 0 H L dt dU H M H M H W 0 E.B.: 0 M H H L L 0 0 dt H M M H H M W M M.B.: H c H L h L b c H L H H M H H M 0 0 H H L L a f a h g C T T M C T T M H P 0 H L P 0 L S.B.: g 0 f 0 0 dS S M S M S Q Sgen 0 M H H L L 0 0 T dt M H SH M L SL M H M L S0 0 b c h FG T IJ FG T IJ HT K HT K g c S S M S S M H H 0 L L 0 M H H H L H P L P 0 M L T0 M H M L f a 1 or TH M H TL M L L 0 h 0 M C ln TT M C ln TT 0 0 0 M TH H b M M gT M b M M g H L L H T0 L From this can calculate T0 . Then L a W C T T M C T T M H P 0 H L P 0 L f This can be used for any flow rate ratio. (3) Suppose very large amount of surface water is contacted with a small amount of deep water, i.e., !! M . Then T ~ T M H L 0 H W a a f a f C T T M C T T ~ M C T T M H P H H L P H L L P H L f (4) Suppose very large amount of deep water is contacted with a small amount of surface water, i.e., M , T ~T . M H L 0 L C T T M C T T ~ M C T T W M H P 4.9 a L H f L P a L L f H P a L H f System contents of the turbine. This is a steady-state, adiabatic, constant volume system. dM M or M M (a) Mass balance 0 M 2 1 1 2 dt Energy balance constant dU H M H Q adiabatic W P dV volume 0 M s 1 1 2 2 dt dt Entropy balance dS S Q Sgen S M 0 M 1 1 2 2 T dt 0, by problem statement Thus M M 4500 kg h 2 1 W M H H M.B. S2 S.B. S 1 S1 c 1 2 h E.B. Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 State T1 500q C 1 P1 60 bar Steam o Tables H 1 3422.2 kJ kg S1 68803 . kJ kg State 2 Steam o Tables T2 # 240.4q C P2 10 bar S2 S1 kJ kgK . 68803 H 2 | 29205 . kJ kg kg kJ kJ 2257650 u 29205 . 3422.2 h kg h (b) Same exit pressure P2 10 bar , and still adiabatic H H . W M Ws 4500 a c s 1 1 2 6271 . kW f h Here, however, Ws c 08 . Ws Part a H 2 P Thus 08 . 2.258 u 106 2 T2 # 286.7 K S | 7.0677 kJ kg K Steam o Tables . kJ kg 30208 10 bar Sgen h kJh 4500c H 3422.2h kJh c 2 kJ kJ . 7.0677 8433 . h 4500 kgh u 68803 kg K Kh S S M 1 1 2 (c) Flow across valve is a Joule-Thompson (isenthalpic expansion) ... See Illustration 3.4-1. Thus, H into valve H out of valve , and the inlet conditions to the turbine are H 1 P1 H out of valve H into valve 3422.2 kJ kg 30 bar T1 | 484.8q C S1 | 71874 . kJ kg K Steam o Tables Flow across turbine is isentropic, as in part (a) S2 P2 S1 71874 . kJ kg K 10 bar Steam o Tables T2 # 3181 . qC H | 3090.4 kJ kg 2 kg kJ kJ u 3090.4 3422.2 1493 . u 106 414.8 kW h kg h 4.10 Since compression is isentropic, and gas is ideal with constant heat capacity, we have Ws 4500 FG T IJ FG P IJ HT K H PK 2 2 1 1 FG 3 u 10 IJ H 2 u 10 K aT T f Problem 3.31, that W NC So that T2 T1 FG P IJ HPK 6 8.314 36.8 R CP 2 29815 . 6 1 s R CP P 2 1 326.75 K . Now using, from solution to Solutions to Chemical and Engineering Thermodynamics, 5th ed Ws 125 Chapter 4 kg 1 mol J 1000 g u u 368 u 326.75 29815 . . Ku mol K kg s 16 g 8.23 u 106 J s The load on the gas cooler is, from Problem 3.31, Q a T T NC p 3 2 f 125 kg s u 1000 g kg J . . 326.75 K u 368 u 29815 16 g mol mol K 8.23 u 106 J s 4.11 (a) This is a Joule-Thomson expansion H 70 bar, T ? H 10133 . bar, T 400q C | H 1 bar, T 400q C 3278.2 kJ kg and T 447q C , S 6.619 kJ kg K (b) If turbine is adiabatic and reversible S S 6.619 kJ kg K and P 0 , then S c h gen out in 1013 . bar. This suggests that a two-phase mixture is leaving the turbine Let x fraction vapor S V 7.3594 kJ kg K . S L 13026 kJ kg K Then x 7.3594 1 x 13026 . fluid leaving turbine is H . u 08788 . 26755 H V sat’d, 1 bar 6.619 kJ kg K or x . 1 08778 u . 08778 . 417.46 H L sat’d, 1 bar 2399.6 Energy balance 0 H M H Q 0 W P dV M s in in out out dt 0 but M M in out Ws kJ 3278.2 2399.6 878.6 M in kg (c) Saturated vapor at 1 bar S Ws M in Actual Efficiency % Sgen M in 7.3594 kJ kg K ; H . 3278.2 26755 602.7 u 100 878.6 7.3594 6.619 . kJ kg 26755 602.7 kJ kg 68.6% 0.740 kJ Kh Therefore the enthalpy of kJ kg Solutions to Chemical and Engineering Thermodynamics, 5th ed (d) M M M 1 2 2 0 W Steam 70 bar 447q C c c M 1 h H H W Q P dV M 1 1 2 s dt Q S S S M 1 1 2 gen T 0 Water 1 bar 25q C Chapter 4 0 Q h Simplifications to balance equations dV Sgen 0 (for maximum work); P 0 (constant volume) dt Q Q where T0 25q C (all heat transfer at ambient temperature) T T0 kJ kJ ; S sat' d liq, T 25q C 0.3674 H sat' d liq, 25q C 104.89 kg kg K Q Ws T0 S2 S1 ; H 1 H 2 T0 S2 S1 H 1 T0 S1 H 2 T0 S2 M M c h c h c h c h max Ws M . u 6.619 104.89 29815 . u 0.3674 3278.2 29815 max 1304.75 4.65 1309.4 kJ kg 4.12 Take that portion of the methane initially in the tank that is also in the tank finally to be in the system. This system is isentropic S f Si . (a) The ideal gas solution S i Tf Sf FP I TG J HPK f R Cp i N= PV Ni RT 'N N f Ni PV i RTi 0.0195 35.90 kg u 1000 28 g mol 150.2 K 1964.6 mol; N f Si m3 , so that mi kg Ni 8.314 36 Pf V RTf 1768.4 mol (b) Using Figure 2.4-2. 70 bar | 7 MPa, T = 300 K Vi F 35. I H 70 K 300 i g kg 505 . kJ kg K 0.7 m3 m3 0.0195 kg 1282 mol S f . kg. 3590 196.2 mol Solutions to Chemical and Engineering Thermodynamics, 5th ed At 3.5 bar = 0.35 MPa and S f 3 Vf . 0192 . 3646 kg u 1000 Nf 28 'N 4.13 g kg . 3646 kg. 130.2 mol g mol N f Ni 505 . kJ kg K T | 138 K. Also, 0.7 m3 m3 . 0192 kg m , so that m f kg Chapter 4 130.2 1282 11518 . mol dS C dV dT eqn. (4.4-1) R V T 'S z LMN a R bT cT 2 dT 3 e dT dV R 2 T V T f a 2 OP Q z so that a S T2 , V 2 S T1, V 1 f a R ln TT baT T f 2c cT T h 2 2 2 1 2 1 1 c h c h V d 3 e T2 T13 T22 T12 R ln 2 V1 3 2 Now using PV a f a S T2 , P2 S T1, P1 RT V2 V1 T2 P1 T1 P2 f a ln TT baT T f 2c cT T h d e P cT T h cT T h R ln P 3 2 2 2 2 2 1 2 1 1 3 2 2 2 3 1 2 1 2 1 Finally, eliminating T2 using T2 a f a S P2 ,V 2 S P1,V 1 T1 P2 V 2 PV 1 1 yields f a lnFGH PPVV IJK Rb a P V PV f c a P V f a PV f 2R d a P V f a PV f 3R 2 2 2 1 1 2 1 1 2 2 2 2 2 3 2 2 1 1 3 3 1 1 eR 2 P 2 P2V 2 2 PV R ln 2 1 1 P1 2 d i d i Solutions to Chemical and Engineering Thermodynamics, 5th ed 4.14 Chapter 4 System: contents of valve (steady-state, adiabatic, constant volume system) Mass balance 0 Energy balance 0 N 1 N 2 0 dV N 1 H 1 N 2 H 2 Q 0 Ws P dt H1 H 2 0 Q 0 N 1 S 1 N 2 S 2 Sgen T Sgen 'S S 2 S 1 N (a) Using the Mollier Diagram for steam (Fig. 3.3-1a) or the Steam Tables Entropy balance 0 T1 P1 H 1 'S 7 bar T2 | 293q C . J g 30453 S2 7.277 J g K H 2 30453 . J g . Thus S1 S S 0.717 J g K 2 H 2 T1 a 293q C p T2 600 K f a f C ln TT R ln PP S T2 , P2 S T1, P1 P R ln 2 P1 'S . 65598 J g K ; Texit 1 (b) For the ideal gas, H 1 'S 4.15 600 K P2 35 bar H 2 2 2 1 1 . J mol K 1338 0.743 J mol K From the Steam Tables P 15538 . MPa L V 0.001157 m3 / kg VV 012736 . m3 / kg U L 850.65 kJ / kg U V 25953 . kJ / kg At 200oC, L L H H 852.45 kJ / kg 27932 . kJ / kg L V S 2.3309 kJ / kg K S 6.4323 kJ / kg K . kJ / kg K 'H vap 1940.7 kJ / kg 'S vap 41014 (a) Now assuming that there will be a vapor-liquid mixture in the tank at the end, the properties of the steam and water will be P 0.4578 MPa V L 0.001091 m3 / kg VV 0.3928 m3 / kg o At 150 C, U L H L 63168 . kJ / kg 632.20 kJ / kg L S 18418 . kJ / kg K vap 'H 2114.3 kJ / kg U V H V 2559.5 kJ / kg 27465 . kJ / kg V S 68379 . kJ / kg K vap 'S 4.9960 kJ / kg K (b) For simplicity of calculations, assume 1 m3 volume of tank. Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 Then Mass steam initially = 0.8 m3 0.12736 m3 / kg 6.2814 kg 0.2 m3 172.86 kg 0.001157 m3 / kg 6.2814 Weight fraction of steam initially = 0.03506 179.14 6.2814 Weight fraction of water initially = 0.96494 179.14 The mass, energy and entropy balances on the liquid in the tank (open system) at any time yields L L L L dM L L ; dM U L H V ; and dM S L SV M M M dt dt dt L dU L L dM L L H V H V dM ML M U or dt dt dt L L dU dM ML H V U L dt dt Also, in a similar fashion, from the entropy balance be obtain dS L dM L V L dM L vap ML 'S S S dt dt dt There are now several ways to proceed. The most correct is to use the steam tables, and to use either the energy balance or the entropy balance and do the integrals numerically (since the internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature. This is the method we will use first. Then a simpler method will be considered. Using the energy balance, we have dM L dU L , or replacing the derivatives by finite differences L V M H U L MiL1 MiL U iL1 U iL U L U iL or finally MiL1 MiL 1 i V1 L V L Mi H i U i H i U iL So we can start with the known initial mass of water, then using the Steam Tables and the data at every 5oC do a finite difference calculation to obtain the results below. Mass water initially = c c h h FG H i 1 2 3 4 5 6 7 8 9 10 11 T (oC) 200 195 190 185 180 175 170 165 160 155 150 U iL (kJ/kg K) 850.65 828.37 806.19 784.10 762.09 740.17 718.33 696.56 674.87 653.24 631.68 IJ K H iV (kJ/kg K) 2793.2 2790.0 2786.4 2782.4 2778.2 2773.6 2768.7 2763.5 2758.1 2752.4 2746.5 MiL (kg) 172.86 170.88 168.95 167.06 165.22 163.42 161.67 159.95 158.27 156.63 155.02 So the final total mass of water is 155.02 kg; using the specific volume of liquid water at 150oC listed at the beginning of the problem, we have that the water occupies 0.1691 m3 leaving 0.8309 m3 for the steam. Using its specific volume, the final mass of steam is found to be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.09%, the Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 final volume fraction of water is 16.91%, and the fraction of the initial steam + water that has been withdrawn is (172.86+6.28-155.02-2.12)/(172.86+6.28) = 0.1228 or 12.28%. A total of 22.00 kg of steam has withdrawn, and 87.7% of the original mass of steam and water remain in the tank. For comparison, using the entropy balance, we have dM L dS L , or replacing the derivatives by finite differences L V M S S L SiL1 SiL MiL1 MiL SiL1 SiL L L of finally 1 M M 1 i i L M 'S vap 'S vap i FG H i i IJ K So again we can start with the known initial mass of water, then using the Steam Tables and the data at every 5oC do a finite difference calculation to obtain the results below. i T (oC) 1 2 3 4 5 6 7 8 9 10 11 200 195 190 185 180 175 170 165 160 155 150 SiL (kJ/kg K) 2.3309 2.2835 2.2359 2.1879 2.1396 2.0909 2.0419 1.9925 1.9427 1.8925 1.8418 SiL (kJ/kg K) 6.4323 6.4698 6.5079 6.5465 6.5857 6.6256 6.6663 6.7078 6.7502 6.7935 6.8379 MiL (kg) 172.86 170.86 168.92 167.02 165.17 163.36 161.60 159.87 158.18 156.53 154.91 So the final total mass of water is 154.91 kg; using the specific volume of liquid water at 150oC listed at the beginning of the problem, we have that the water occupies 0.1690 m3 leaving 0.8310 m3 for the steam. Using its specific volume, the final mass of steam is found to be 2.12 kg. Using these results, we find that the final volume fraction of steam is 83.10%, the final volume fraction of water is 16.90%, and the fraction of the initial steam + water that has been withdrawn is (172.86+6.28-154.91-2.12)/(172.86+6.28) = 0.1234 or 12.34%. A total of 22.11 kg of steam has withdrawn, and 87.7% of the original mass of steam and water remain in the tank. These results are similar to that from the energy balance. The differences are the result of round off errors in the simple finite difference calculation scheme used here (i.e., more complicated predictor-corrector methods would yield more accurate results.). A simpler method of doing the calculation, avoiding numerical integration, is to assume that the heat capacity and change on vaporization of liquid water are independent of temperature. Since liquid water is a condensed phase and the pressure change is small, we can make the following assumptions U L | H L and H V H L 'H vap dT L dS L CPL dT L dU L dH L | | CPL | ; and dt dt T dt dt dt With these substitutions and approximations, we obtain from the energy balance Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 dH L dM L vap dU L dM L V L 'H o ML H U dt dt dt dt dT dM L vap 'H M LCPL dt dt Now using an average value of CPL and 'H vap over the temperature range we obtain c h 1 dM L M L dt or ML CPL dT 'H vap dt CPL 150 200 'H vap FG M IJ HM K L f L i ln and from the entropy balance dS L dM L vap C L dT dM L vap ML 'S 'S o ML P dt dt T dt dt Now using an average value of CPL and 'S vap over the temperature range we obtain CPL dT T'S vap dt F H 1 dM L M L dt 150 27315 CPL . ln 200 27315 . 'S vap or I lnFG M IJ K HM K L f L i From the Steam Table data listed above, we obtain the following estimates: U (T 200o C) U (T 150o C) 852.45 632.20 kJ CPL 4.405 kg K 200o C - 150o C 50 or using the ln mean value (more appropriate for the entropy calculation) based on FG T IJ SaT f SaT f HTK CPL ln 2 2 1 1 S(T kJ . 200o C) S(T 150o C) 2.3309 18418 4.3793 . . 200 27315 47315 kg K ln ln . . 150 27315 42315 Also, obtaining average values of the property changes on vaporization, yields 1 kJ 1 u 'H vap T 150o C 'H vap T 200o C u 2114.3 1940.7 2027.5 'H vap 2 2 kg 1 1 kJ 4.5487 u 'S vap T 150o C 'S vap T 200o C u 4.9960 41014 'S vap . 2 2 kg K With this information, we can now use either the energy of the entropy balance to solve the problem. To compare the results, we will use both (with the linear average Cp in the energy balance and the log mean in the entropy balance. First using the energy balance M fL CPL 4.405 u 50 ln 150 200 . 010863 L vap M 2027.5 'H CPL F H F H I K b g b g I K b b g g FG IJ H K i M L f L i exp 010863 . . 089706 M Now using the entropy balance M fL . CPL 150 27315 ln ln L vap . M 200 27315 'S FG IJ . I . I 42315 F I 4.3793 lnF 42315 0.9628 lnF H K H K H . . K 4.5487 47315 47315 H K M . I F 42315 . 089805 H 47315 . K M i L f L i 0.9628 Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 Given the approximations, the two results are in quite good agreement. For what follows, the energy balance result will be used. Therefore, the mass of water finally present (per m3) is . . kg u M L initial 15506 M L final 0897 L L o . u 0.001091 01692 . occupying V M final u V 150 C 15506 m3 b g Therefore, the steam occupies 0.8308 m3 , corresponding to 0.8308 m3 2115 . kg m3 0.3928 kg So the fraction of liquid in the tank by mass at the end is 155.06/(155.06+2.12) = 0.9865, though the fraction by volume is 0.1692. Similarly the fraction of the tank volume that is steam is 0.8308, though steam is only 2.12/(155.06+2.12) = 0.0135 of the mass in the tank. 0.8308 m3 V V 150o C M V final b g (c) Initially there was 6.28 + 172.86 = 179.14 kg of combined steam and water, and finally from the simpler calculation above there is 155.06 + 2.12 = 157.18 kg. Therefore, 87.7% of the total amount of steam + water initially in the tank are there finally, or 12.3% has been withdrawn. This corresponds to 21.96 kg being withdrawn. This is in excellent agreement with the more rigorous finite difference calculations done above. 4.16 (a) dN dt 0 N 1 N 2 ; dU dt 0 dV N 1 H 1 N 2 H 2 WS Q P dt dS dt 0 or N 2 N 1 Q N 1 S 1 N 1 S 2 Sgen T0 z Tf WS = H 2 - H1 = CPdT N 1 298.15K Sgen N W WS N 1 H 1 N 1 H 2 or S = H 2 - H 1 N 1 S 2 S1 CP 37151 . 1 c h CP Tf 29815 . K J mol K if the heat capacity is independent of temperature. First consider the reversible case, z Tf S 2 S1 0 gives CP dT T T i WSrev N 1 Wact CP 49914 . 29815 . K 125 . WSrev (b) dP P 1 7467 c The J . mol The solution is 499.14K. actual work is 25% Then greater h J CP Tf 29815 . K mol 549.39 K 9334 The solultion is Tf z 10 R Repeat the calculation with a temperature-dependent heat capacity CP T 22.243 5977 . 102 T 3499 . 105 T 2 7.464 109 T 3 Assuming reversibility Tf = 479.44K. Repeating the calculations above with the temperaturedependent heat capacity we find Wact = 9191 J, and Tf =520.92K. So there is a significant difference between the results for the constant heat capacity and variable heat capacity cases. Solutions to Chemical and Engineering Thermodynamics, 5th ed 4.17 Chapter 4 Ti = 300 K, Tf = 800 K, and Pi = 1.0 bar CP (T ) = 29.088 - 0.192 u 10-2 T + 0.4 u 10-5 T 2 - 0.870 u 10-9 T 3 z T f 800K J mol K z Pf CP (T ) dT T T 300K P i dP P P 1 i Calculated final pressure Pf = 3.092 u 106 Pa. z T f 800K Wrev 1458 u 104 . CP (T )dT Ti 300K 4.18 J mol Stage 1 is as in the previous problem. Stage 2 Following the same calculation as above. Stage 2 allowed pressure Pf ,2 = 9.563 u 107 Pa Wrev = 1.458 u 10-4 J = Stage 2 work mol Stage 3 Following the same calculation method Pf ,3 = 2.957 u 10-9 Pa = Stage 3 allowed pressure. J = Stage 3 work mol Question for the student: Why is the calculated work the same for each stage? Wrev = 1.458 u 104 4.19 The mass, energy and entropy balances are dM M 0, M M M 1 2 2 1 dt dU H M H Q W ; M H H W 0 M s s 1 1 2 2 1 1 2 dt H H Ws M 1 2 1 c c dS dt 0 300q C, 5 bar 05 . MPa 100q C, 1 bar 01 . MPa c gen M 1 2676.2 3064.2 S2 S1 h 0 S S S M gen 1 1 2 h H 1 3064.2 kJ kg S1 7.4599 kJ kg K H 2676.2 kJ kg 2 S2 Ws M 1 S 0; h S M S Q S M gen 1 1 2 2 T S S Sgen M 1 2 1 c h 7.3614 kJ kg K 388 kJ kg satisfied the energy balance. 7.3614 7.4599 0.0985 kJ kg K can not be. Therefore the process is impossible. Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 4.19 N M 0.5 kg/s 100°C, steam 1 kg/s e 300°C 1 MPa Streams 1 2 3 0.5 kg/s 100°C, sat’d liq T 300°C 100°C 100°C P 1 MPa Condition sup heated Ĥ kJ/kg 3051.2 Ŝ kJ/kg 7.1229 0.10135 MPa sat’d liq 419.04 1.3069 Energy balance kg ˆ 0.5 u 1344.0 W 0 1 u 3051.2 0.5 u H 2 s 1.15 MW 1500 kJ but W s ˆ 0.5 u 419.04 1500 3051.2 0.5 u H 2 Ĥ 2 2683.4 kJ kg ˆV Now at at 0.1 MPa and T 100o C H 2676.1 kJ kJ and Sˆ 7.3549 kg kg K So steam is slightly superheated and Quality is 1 Entropy balance (using Ŝ at saturation) 1 u 7.1229 – 0.5 u 1.3069 – 0.5 u 7.3549 + S gen = 0 S 0.5 1.3069 7.3549 7.1229 gen 2.792 kJ / K S So process is not possible (using better estimate, by interpolation, for exit entropy of steam would have no effect on conclusion). 4.20 Steam 20 bar 2 MPa and 300q C H 30235 . kJ kg S 6.7664 kJ kg U 2772.6 kJ kg Final pressure = 1 bar. For reference saturation conditions are P 01 . MPa, T 99.63 L H L 417.46 U 417.36 S L 13026 . V V 25061 26755 S V 7.3594 U . H . (a) Adiabatic expansion valve W 0 and Q 0 (from Steam Tables) Solutions to Chemical and Engineering Thermodynamics, 5th ed dM ; M M 0; M M 2 1 1 2 dt dU H M H E.B.: 0 ; H 2 H 1 M 1 1 2 2 dt From Steam Tables T 250q C . H2 30235 H 2974.3 kJ kg Chapter 4 M.B.: P By interpolation T kJ kg K S 81245 . dS dt S gen M gen 300q C H 3074.3 kJ/kg 275q C gives H = 3023.5 kJ / kg all vapor T 01 . MPa S 8.0333 kJ kg K S 8.2158 kJ/kg K M S S MS gen 1 2 2 0 S2 S1 = 8.1254 6.7664 kJ kg K 1359 . (b) Well designed, adiabatic turbine H M H W E.B.: M 1 1 2 2 cH H h 0 ; W 2 1 S M S 0 ; S S ; S 6.7664 kJ kg K S.B.: M 1 1 2 2 2 2 1 Two-phase mixture. Solve for fraction of liquid using entropy balance. . 6.7664 x 7.3594 1 x 13026 x 0.902 not good for turbine! . 0.098 u 417.46 0.902 u 26755 H 2 W M . 2454.2 30235 569.3 kJ kg W M 2454.2 kJ kg 569.3 kJ kg (c) Isorthermal turbine superheated vapor T 300q C H 3074.3 kJ kg final state P 01 . MPa S 8.2158 kJ kg K H M H Q W 0 E.B.: M OP Q 1 1 2 2 s 0 S M S Q Sgen S.B.: M 0 1 1 2 2 T Q S M S S S M M 1 1 2 2 1 2 1 T Q . 8.2158 6.7664 kJ kg K T S2 S1 300 27315 M c c W s M h h 830.7 kJ kg Q H 1 H 2 M c h 830.7 30235. 3074.3 779.9 kJ kg get more work out than in adiabatic case, but have to put in heat. Solutions to Chemical and Engineering Thermodynamics, 5th ed 4.21 Chapter 4 contents of the compressor (steady-state, constant volume). Also, gas is ideal. (a) Mass balance 0 N N o N N System 1 Energy balance 2 2 1 N 1 H 1 N 2 H 2 0 dV 0 Ws P Q dt adiabatic 0 Entropy balance reversible 0 Q 0 compressor Sgen S 1 N 1 S 1 N 2 S 2 T 0 From the energy balance Ws T T or Ws NC P 2 1 N a f From the entropy balance S 1 S 2 T2 T1 FG P IJ HPK S2 a CP T2 T1 f R CP 2 1 Thus Ws N CPT1 LMF P I MNGH P JK R CP 2 OP PQ 1 1 (b) Two stage compression, with intercooling, so that gas is returned to initial temperature, before entering 2nd compressor LF P I work in stage 1 NC T MG J MNH P K LP NC T MF I work in stage 2 W NH P K L T MFG P IJ Total work W W NC MNH P K R CP WsI 1 II s R CP 2 P 1 * I s R CP II 1 To find P* for minimum work, set d Ws dP R R F P I aR C f1 1 T |S G J b g 0 NC |T C H P K cP h aP P f d Ws dP P P 1 P P1 1 2 R CP 1 2 R CP 2 P 1 s OP PQ O 1P Q P F I HP K 1 ; where P P 1 2 0. F Ia H K R P2 CP P pressure after 1st compressor. OP W PQ s f F P I U| R CP 1 H P K V|W 2 2 R CP or P P1 P2 Students should check that this results in minimum, and not maximum work. 4.22 System: nitrogen contained in both tanks (closed, adiabatic, constant volume) (1) Mass balance: M1i M1f M 2f f f f f (2) Energy balance: M U M U M U 1 1 1 P1 f 1 P2f 2 2 (3) Final pressure condition: For the entropy balance, the nitrogen in the first tank that remains in the tank will be taken as the system. Then (4) S1i S1f Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 Equation (1) –(4), together with eqn. of state information of the form S S T , P , U U T , P and V V T , P which we can get from Fig. 3.3-3 provides 4 eqns. for the 4 unknowns T1 f , P1 f , T2f and P2f . Procedure to be followed in solution (iv) Guess a final pressure P f Use eqn. (4) + Fig. 3.3-3 to compute T1 f , caluculate U1f Use Fig. 2.4-3 to get V1 f , compute M1f V f V1 f f f f V M M M M , and V (v) Use P and V2f to get T2f and U 2f (i) (ii) (iii) 2 2 1 1 2 f (vi) See if energy balance, Eqn. (2), is satisfied. If it is, guessed P f is correct. If it is not, guess new p f , go back to (ii), and repeat calculation. Some preliminaries Figure H 1i 368 kJ kg T1i 250 K o 2.4 - 3 V1i | 0.0037 m3 kg P1i 200 bar Thus M1i V / V1i 0.01 m3 / 0.0037 m3 kg 2.703 kg As a first guess, use ideal gas solution for pressure. (Also try some neighboring pressures.) My solution is P1 f P2f . bar same as ideal gas solution 1333 c 285 K cideal gas solution: T T1 f 226 K ideal gas solution: T1 f T2 f f 2 h 330.4 Kh 222.8 K 4.23 (a) Set up just as in Problem 3.22 above. Solution after a number of iterations is P1 f T1 f 2756 . q C and T2f P2f | 5 bar . 497.7q C . (b) Since now there is heat exchange between the two chambers we have T1 f T2f . This equation is used instead of entropy balance. Solution procedure is to guess a final pressure, and then compute final temperature using first the mass balance M1f M 2f M1i V1 V1i V1 V2 Vf V f V1 V2 V1 f V2 f V1 V2 i V1 V1 (1) That is, choose T f until eqn (1) is satisfied. Then compute T f from energy balance i.e. M1iU1i M1f U1f M 2f U 2f f c M M hU U f 1 f 2 f 2 i 1 U f f (2) When guessed P is correct, T computed from eqns. (1) and (2) using the Steam Tables will be identical. My solution is P f 5 bar and T f # 366q C . 4.24 System contents of turbine (open, constrant volume, steady-state) dN Mass balance: 0 N 1 N 2 N 2 N 1 dt constant volume Energy balance: dN dt 0 N 1 H 1 N 2 H 2 dV 0 0 Ws P Q dt adiabatic Solutions to Chemical and Engineering Thermodynamics, 5th ed a Ws N 1 H 1 H 2 Chapter 4 f N C aT T f for the ideal gas 1 P 1 2 Q 0 dN Entropy balance: Sgen 0 N 1 S 1 N 2 S 2 T dt P T N 1 CP ln 2 R ln 2 Sgen N 1 S 1 S 2 P1 T1 or R CP Sgen P exp T2 T1 2 P N C a LM N FG IJ H K 1 RS FG IJ T H K f 1 P FG IJ UV H KW OP Q (a) For T2 to be a minimum, since Sgen t 0 and N 1 ! 0 , Sgen must be zero. Thus the minimum outlet temperature occurs in reversible operation. (b) Ws N 1CP T2 T1 . Since T1 ! T2 , the maximum work occurs when T2 is a minimum. Thus, W is a maximum (in magnitude) for a reversible process. a f s 4.25 (a) For any system: LM M S Q S OP T Q N dS dt i i gen S t 0 or ¦ M S d 0 . Also, S depending 0n the process Q t 0 or Q d 0 and ¦ M gen i i i i 0 or Sgen ! 0 , depending on whether or not the process is reversible. Thus, dS dt for a system can be greater than, less than, or equal to zero. Since, by definition, the universe contains everything, it must be a closed system and adiabatic, since there is nothing for the universe to exchange mass or heat with. Therefore dS dt dS 0 0 Sgen dt Sgen t 0 Thus the entropy of the universe can not decrease, and the statement is true. (b) Consider the change from any state 1 to any state 2 in a closed system. The energy and entropy balances for this transformation are: 1 2 U2 U2 S2 S1 Q W Sgen RS T W Since the process is adiabatic If the transformation is possible, then Sgen t 0 now consider the transformation from state 2 to state 1. Here 3 4 U1 U1 S1 S2 W Sgen Comparing eqns. (1) and (3) we have W W (This is ok). Comparing eqns. (2) and (4) we have Sgen Sgen (5) Separately we have, if the processes are possible, that Sgen t 0 and Sgen t 0 . The only way that all these three equations for Sgen and Sgen can be satisfied is if Sgen Sgen 0 , that is, both processes are reversible. Generally, processes are not reversible. However, eqn. (5) requires that only one of Sgen and Sgen can be greater than zero. Thus, Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 If Sgen ! 0 1 o 2 is possible, but 2 o 1 is not possible. If Sgen ! 0 2 o 1 is possible, but 1 o 2 is not possible. 4.26 W 1 § T Tamb · ¨ c ¸Q 2 © TC ¹ Q q k Tc Tamb Heat flow into collector Convective heat loss from collector 1 § Tamb · ¨1 ¸ q k Tc Tamb 2 © Tc ¹ W 1 1 1 Tamb 1 1 T2 q k Tc Tamb q kTamb k amb 2 2 2 Tc 2 2 Tc dW dTc 1 k 2 kTc2 1 Tamb 1 T2 q k amb 2 2 Tc 2 Tc2 2 qTamb kTamb Tc 4.27 0 (for a maximum) amb qT 2 Tamb k q 1 kTamb Tamb System: contents of the tank at any time (open, adiabatic, constant volume system). dN (a) Mass balance: N dt PV PV ideal gas law N ; N where V volumetric flow rate. RT RT d P P d PV PV Thus, V V since V and V are both constant. dt T T dt RT RT d P V P P Vt bar or or exp 1082 . u 103 ln dt T V T 5 min T 0 V K F I H K F I H K F I H K P 5 min Energy balance: or N dU dt FG IJ H K 1082 . u 103 T 5 min d NU dt dN dU N dt dt dT NCV dt N H U dN dN H U RT dt dt ^P K` bar, T H dN dt FG IJ H K RC P P dP P dT or T2 T1 2 CP P1 dt T dt [Note: could have gotten this result from the entropy balance also!] using N PV RT yields R (1) Solutions to Chemical and Engineering Thermodynamics, 5th ed T 5 min Chapter 4 F P 5 min I H 1 bar K 8.314 39 (2) 340 simultaneously solving equations (1) and (2) yields 0.281 bar and T 5 min P 5 min 259.3 K S out or T3 (b) Since pump is adiabatic and reversible, S in T2 FG P IJ HPK R CP 3 since P3 P1 . This 2 equation implies that T3 4.28 T1 340 K . Number of moles of gas in tank initially N 0 N0 N 0 15 bar u 0.2 m3 u 105 J m3 bar 8.314 J mol K u 22 27315 . K P 0V RT 0 122.26 mol dN N 4.5 mol min N t 122.26 4.5t mol t min dt (a) Entropy balance on an element of gas that remains in the tank (see Illustration 3.5-2) yields S t S 0 0 Tt T0 FG P t IJ HP0K R CP T (t ) FPt I H 15 K 0.3779 29515 . (1) From the ideal gas equation of state V R Tt N t Pt PtV N t RT t Tt Pt 29515 . K 122.26 u 15 bar 122.26 4.5t T0N 0 P0 Thus 19.68 1 0.03681t (2) Now using eqn. (1) in eqn. (2) to solve for P t and T t yields Pt 15 u 1 0.03681t 1.6075 Tt . u 1 0.03681t 0.6075 29515 But T t is temperature in the tank. What about temperature of gas leaving the throttling valve? Gas going thru valve undergoes a Joule-Thomson expansion H in H out . Since gas is ideal, this implies Tin Tout . Thus, T t out of valve 29515 . u 1 0.03681t 0.6075 (b) If tank is isothermal, then, instead of eqn. (2), we have Pt N t RT V P0 Pt N 0 15 1 0.03681t bar Solutions to Chemical and Engineering Thermodynamics, 5th ed and T t constant Chapter 4 29515 . K Summary T K P bar 0.6075 . 1 0.03681t 15 1 0.03681t 1.6075 Adiabatic 29515 . 15 1 0.03681t Isothermal 29515 4.29 This is a tough problem! Subscript 1 denotes properties in initally filled tank Subscript 2 denotes properties in initially "evacuated" tank We will use i and f (superscripts) to denote initial and final properties, and we will assume negligible mass hold-up in engine. 1) Mass balance on closed system consisting of both tanks N1i N 2i N1f N 2f P1iV1 P2iV2 i T1i T2 P1 f V1 P2f V2 T1 f T2f but P1 f Pf Pf P2 f FG 0.3 0.75IJ 14.0 u 0.3 0.35 u 0.75 H T T K 97315 29815 . . f f 1 2 bar m3 u 10 5196 . K (1) 3 2) Entropy balance on gas contained in tank 1 initially and finally. This is a closed, adiabatic, reversible system dS dt 0 S1i FG T IJ R lnFG P IJ HT K HP K (2) F P IJ 97315 . G H 14.0K (3) S1f CP ln f f 1 1 i 1 i 1 Thus FG P IJ HPK f T1 f T1i R CP f 1 27 1 i 1 Equation (2) implies that T1 and P1 are related as follows d ln T1 dt1 R d ln P1 CP dt (4) 3) Mass and energy balances on tank 1 or N1CV dN1 dt d N 1 { N ; dt dT1 dt dN1 H1 U 1 dt a a N U f N H 1 1 f RT dNdt 1 1 1 RT1 N U1 dU 1 dN1 N1 dt dt Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 using eqn. (4) gives N1 CV d ln P1 CP dt dN1 dt N (5) 4) balances on the engine: adiabatic, reversible (for maximum work) and since no hold-up of mass, dN1 dt 0 Subscript eng refers to gas leaving engine and going into tank 2. 0 H 1 H eng N Ws d i 0 dS S i N 1 eng S1 S eng ; Also P2 Peng c Teng T1 Peng P1 h a R CP f T1 P2 P1 R CP (6) (Note that Teng z T2 ) and c Ws T T NC P 1 eng h (7) 5) balances on tank 2 [Note, irreversible mixing occurs unless, fortuitously, Teng T2 at all times (can this occur?). Thus, Sgen ! 0 , and entropy balance gives no useful information] N Mass balance: dN 2 dt a f U dN N dU N H dN H But, dt dt dt H C aT T f UV where T reference temperature U C aT T f RT C T C T W Energy balance: P 0 V 0 d N 2U 2 dt 2 2 2 2 2 eng eng 0 0 dN 2 H eng U 2 dt or d V P 0 i dNdt mC T C T C T C T r N C dTdt 2 2 P eng P 0 m N CPTeng CVT2 V 2 P 0 2 V r N C dTdt 2 (8) 2 V and N IJ PV dT V dP a f dtd FGH PV RT dt RT K RT dt d N2 dt 2 2 2 2 2 2 2 dT2 dt 2 2 2 2 RT22 N T2 dP2 PV P2 dt2 2 2 (9) using eqns. (6) and (9) in eqn. (8) R| F I S| GH JK T P N CPT1 2 P1 R CP CVT2 U| N C RT N N C T dP V| PV P dt W 2 2 2 V 2 2 2 2 2 V 2 2 PV T dP CVT2 N 2 2 CV 2 2 RT2 P2 dt2 Solutions to Chemical and Engineering Thermodynamics, 5th ed FG IJ H K T P2 NC P 1 P1 R CP Chapter 4 CVV2 dP2 R dt and using eqn. (5) N FG IJ H K CV P CPT1 2 CP P1 R CP d ln P1 dt CV V2 dP2 R dt or, finally FG P IJ HPK R CP dP1 dt 1 dP V2 P2 R CP 2 dt V1 2 V2 dP2 dt V1 P1 R CP dP1 dt or z dP2 V2 R CP P Pi 2 2 z P1 f P2f V1 dP1 R CP P Pi 1 1 {c h 2.5 P f Pf CV CP 0.35 CV CP } {c P h 3053 . bar: using Eqn. (3), T1 f f CV CP 14.0 CV CP } 629.8 K . f Now using eqn. (1), T2 6119 . K Finally, to get the total work, we do an overall energy balance (system closed, constant volume). N f U 1f N 2f U 2f N1i U 1i N 2i U i2 Ws Ws 4.30 m a f two tanks; adiabatic, r CV P f V1 V2 P1iV1 P2iV2 R 5 ^3053 . 0.3 0.75 14 u 0.3 0.35 u 0.75` 2 3142 u 105 J 314.2 kJ bar m3 3142 . . Note: be careful about coordinate system. A mass flow in the negative u direct is negative! L Heat exchanger is in steady-state operation +x 'L Case I: Concurrent flow Mass balance on shaded volume Solutions to Chemical and Engineering Thermodynamics, 5th ed mass in element at time t 't mass in element at time t Chapter 4 mass in at face at L in 't b mass out at face at L 'L in 't g M M 0 L L 'L 't steady - state M M M L 'L L Energy balance on the shaded volume energy in energy in element element at t 't at t energy flow in by heat flow in time 't energy flow in energy flow out by mass flow by mass flow at at L in 't L 'L in 't (steady-state) H 't M 0 M L L L 'L H L 'L 't Q'L't H H T M MC T Q 'L a L 'L L f P a L 'L L f dividing by 'L , taking limit as 'L o 0 , and using subscript 1 to denote fluid 1 a C dT1 M 1 P ,1 dC Q N T2 T1 f Q heat flow rate per unit length of exchanger. Similarly, for fluid 2 (other part of exchanger) C dT2 M 2 P ,2 dL a Q N T2 T1 f and M are both + for concurrent flow) (M 1 2 Adding the 2 equations C dT1 M C dT2 0 M 1 P ,1 2 P ,2 dL dL From problem statement, M 1 a and C M 2 P ,1 f dT1 dT2 d T1 T2 0 or T1 T2 dL dL dL and T1 C T2 ; T2 C T1 now going back to dT1 dL and integrating N MC P CP,2 constant C 2T f aT T f N aCMC 1 2 1 P Solutions to Chemical and Engineering Thermodynamics, 5th ed FG C 2T IJ 2NL H C 2T K MC f T1 f 15q C , T1i 1 ln i 1 35q C , T2f P 5q C , T2i Chapter 4 MC P 2N L ; L0 L0 15q C . Also, C T1i T2i T1 f T2f 20q C [ i initial conditions, conditions at L L* length of exchanger] Using this in equation above gives 0; f final conditions, conditions at L * where F 20 70I ln 5 1609 . H 20 30K L* L0 ln And, more generally, at any point in the exchanger FG L IJ T L C FG1 expFG L IJ IJ T expFG L IJ H L KK H LK H LK 2H F LI T L 10 25 expG J q C H LK F LI T L C T L 10 25 expG J q C H LK C 2T1 L C 2T1i exp i 1 1 0 0 0 1 0 2 1 0 Now writing an entropy balance S 't M M L L L 'L S L 'L 't 0 M dS dS dL Q T1 a N T2 T1 f Q 'L't T N 50 exp L L0 . 10 25 exp L L0 27315 a T1 a a f f need absolute T here FG IJ a fH K 25 expa L L f F LI dG J C 28315 . 25 expa L L f H L K 28315 . 25 expa L L f U S L S L 0 C lnRS VW 30815 . T f L 50NL0 exp L L0 d M 28315 L0 . 25 exp L L0 0 p 0 0 0 p Case II Countercurrent flow M M 2 1 dT 1 C (1) M N T2 T1 1 P dL 35 15q C dT2 (2) M 2CP N T2 T1 -15 5q C dL C dT2 N T T (3) M 1 P 2 1 dL . Subtracting eqn. (1) from eqn. (3) gives Eqn. (3) comes from eqn. (2) using M M 2 1 a f a a f f Solutions to Chemical and Engineering Thermodynamics, 5th ed a f d T1 T2 0 T1 T2 dL T2 T1 30q C Chapter 4 constant 30q C C Thus a f C dT1 N T T 30N M 1 P 2 1 dL 30NL T1i T1 MC P 30NL L L 35 35 15 T2 5 15 MC L L P 0 0 MC P and the entropy balance 2N where L0 dS1 M dL dS Q T1 N 30 T1 N 30 27315 M (35 30NL MC . ) P 1 dL absolute temperature needed here 15C F LI FG IJ dG J HL K K H L L . 30815 15 g L 30NL0 d 30815 M L . 30NL MC 0 P dS1 b CP dx . x 2054 FG H L S 0 CP ln 1 0.048 L0 S L P 0 IJ K Summary Concurrent flow FG L IJ H LK F LI T 10 25 expG J H LK T1 10 25 exp 0 2 0 S L S L 0 CP ln . 25 expa L L f U RS 28315 VW . 30815 T 0 Countercurrent flow T1 35 15 T2 5 15 S L 4.31 (a) L L0 L L0 FG H L S 0 CP ln 1 0.048 L0 dU dt dV WS Q P dt JIK a f dV dV WS Q P0 P P0 dt dt 0 Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 and dS dt Q Sgen T0 Now let a f dV dV P P0 WS P0 dt dt W and Wu a dV W P0 dt f dV dU WS P P0 dt dt dV Wu Q P0 dt or a f U 2 U1 Wu Q P0 V2 V1 S2 S1 Q Sgen Q T0 U 2 U1 Wu T0 S2 T0 S1 T0 Sgen P0 V2 V1 T0 S2 T0S1 T0Sgen and Wu a f 0 1 0 gen aU PV T S f aU PV T S f T S 2 0 2 0 2 1 0 1 since T0 Sgen t 0 A 2 A1, where A Wumax (b) 0 0 Here WS U PV 0 T0 S H M H Q W M S 1 2 Q S M S S M gen 1 2 T0 Wu Wu H M H Q M 2 1 Wu H T S M H T S T S M 0 2 0 1 0 gen 2 1 a Since T0 Sgen t 0 Wumax b f a g H M H MT S MT S T S M 0 1 0 2 0 gen 2 1 B B where B M 2 1 f H T0 S (c) Using the Steam Tables we find i) at 30 bar 3 MPa and 600qC U 32850 . kJ kg , S 7.5085 kJ kg K , V 013243 . m3 kg Solutions to Chemical and Engineering Thermodynamics, 5th ed 1 A Chapter 4 U PV 0 T0 S 32850 bar 013243 m3 kg u 102 kJ bar m3 29815 . 1013 . . . u 7.5085 1059.76 kJ kg ii) . MPa and 300qC at 5 bar 05 U 2802.9 kJ kg , S 7.4599 kJ kg K , V 2 A 05226 . m3 kg 2802.9 1013 . u 05226 . u 102 29815 . u 7.4599 Wu 2A 1 A (63167 . 1059.76) kJ kg 63167 . kJ kg 428.09 kJ kg This is the maximum useful work that can be obtained in the transformation with the environment at 25qC and 1.013 bar. It is now a problem of clever engineering design to develop a device which will extract this work from the steam in a nonflow process. bar, any component which (d) Since the inlet and exit streams are at 25qC and P 1013 . passes through the power plant unchanged (i.e., the organic matter, nitrogen and excess oxygen in the air, etc.) does not contribute to the change in availability, or produce any useful work. Therefore, for each kilogram of coal the net change is: 0.7 kg of carbon 58.33 mol of C 58.33 mol of O2 to produce 58.33 mol CO2 also 015 . kg of water 8.33 mol of H 2O undergoes a phase change from liquid to vapor Therefore B in M ¦ b N i Bi gin i B out M 58.33 u 0 58.33 u 0 8.33 u 68.317 29815 . u 0.039 carbon oxygen liquid water 1976 kJ kg coal 58.33 u 94.052 8.33 u 57.8 29815 . u 0.0106 ¦ N B b i i g i out carbon dioxide water vapor 24858 kJ kg coal Wumax Wuactual 24858 1976 kJ kg coal 22882 kJ kg coal 2.2 kW - hr kg coal 7920 kJ kg coal 7920 u 100 Efficiency in % 34.6% 22882 Thus a coal-fired electrical power generation plants converts slightly more than 1/3 of the useful work obtainable from the coal it consumes. This suggests that it would be useful to look for another method of generating electrical power from coal . . . for example, using an electro-chemical fuel cell. Considering the amount of coal consumed each year in power generation, and the consequences (strip mining, acid rain, greenhouse effect, etc.) the potential economic savings and environmental impact of using only 1/3 as much coal is enormous. Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 4.32 Three subsystems: unknowns T1 f , P1 f , T2f , P2f , T3f , P3f (6 unknowns) After process P1 f P2f P3f (2 equations) Subsystem 1 has undergone a reversible adiabatic expansion S 1f FG P IJ HPK f S 1i , or T1 f T1i R CP 1 (1 equation) i 1 (#1) Subsystem 3 has undergone a reversible adiabatic compression S 3f S i3 , or T3f FG P IJ HPK f 3 i 3 T3i R CP FG P IJ HPK f T3i R CP (1 equation) i 3 (#2) Mass balance subsystems 1 + 2 N1f N 2f or Pf N1i N 2i P1 f V1 P2f V2 T1 f T2f P1iV1 P2iV2i i T1i T2 FG 05. V IJ 10 u 05. 1 u 0.25 0.017909 (1 equation) 29315 . . H T T K 29315 f 1 f 2 f 2 (#3) Energy balance on subsystems 1 + 2 + 3 N1f U 1f N 2f U 2f N 3f U 3f N1i U 1i N 2i U i2 N 3i U i3 P fV P fV P1 f V1 CVT1 f 2 f2 CVT2f 3 f3 CVT3f f RT3 RT2 RT1 c P f V1 V2f V3f but V1 V2f V3f V1 V2i V3i h PV PV PV i 1 1 05 . 0.25 0.25 i 2 2 in eqn. (1) o T1 f 252.45 K in eqn. (2) o T3 448.93 K f P3i T3f P3f T3i 0.25 u i 3 3 1 m3 10 u 05 . 1 u 0.25 1 u 0.25 1 Pf using this result PiV PiV P1iV1 C T i 2 2i CVT2i 3 i3 CVT3i i V 1 RT3 RT2 RT1 1 448.93 u . 29315 . 55 V3f V3i V2f 0.25 u 2 0.06961 0.4304 m3 55 . bar 0.06961 m3 Now using Eqn. (#3) Pf FG 05. V IJ 55. FG 05. 0.4304 IJ 0.017909 T H T T K H 252.45 T K f 1 f 2 f 2 f f 2 2 Thus the state of the system is as follows T1 P1 Initial 293.15 K 10 bar Final 252.45 K 5.5 bar 337.41 K Solutions to Chemical and Engineering Thermodynamics, 5th ed T2 P2 V2 T3 P3 V3 Chapter 4 293.15 K 1 bar 337.41 K 5.5 bar 0.25 m3 293.15 K 0.4304 m3 448.93 K 1 5.5 bar 3 0.25 m 0.0696 m3 Work done on subsystem 3 Energy balance N 3f U 3f N 3i U i3 W z PdV P3f V3f P3iV3i f CVT3i C T V 3 RT3i RT3f W P3f V3f CV C P3iV3i V R R W CP R f f . u 0.0696 1 u 0.25 P3 V3 P3iV3i 3 55 R 0.3984 bar m3 39.84 kJ c h From simple statics the change in atmospheric pressure dP accompanying a change in height dh is dP Ugdh where U is the local mass density and g is the gravitational constant. Assuming a packet of air undergoes an altitude change relatively rapidly (compared to heat transfer), the entropy change for this process is CP R dS dT dP 0 since both Q and Sgen equal zero. T P Combining the two equations above we have CP R R R N M dT dP Ugdh Mgdh gdh T P P PV T Mg dT or dh CP K dT dT For dry air # 9.7 is referred to as the adiabatic lapse rate. . Note that km dh dh Also, its value will be less than that above as the humidity increases. In fact, if the humidity is 100%, so water will condense as the pressure decreases, the adiabatic lapse rate will be almost zero. 4.33 4.34 W -10 kW M i 300°C 1 MPa 1 kg/s Mass balance 100°C Steady State N 300°C Sat’d L Solutions to Chemical and Engineering Thermodynamics, 5th ed 0 Chapter 4 0.5 1 M 1 M M M i 1 2 0 M 1 0.5 kg/s Energy balance 0 assume no heat loss Q 1344.0 kJ/kg Ĥ 2 10 kw 10 kJ s O 3.2534 kJ/kg K Sˆ i 1344.0 kJ/kg Sˆ 2 Ĥ i 0 1 W H ˆ ˆ ˆ M i i 0.5 H1 0.5 H 2 W 3.2534 kJ/kg K kg kJ ˆ 0.5 u 1344.0 10 kJ u 1344.0 0.5 u H 1 s kg s ˆ 0.5 H 1 1344.0 0.5 u 1344.0 10 Ĥ1 =1324.0 kJ/kg 662 at 100°C HV = 2676.1 At 100°C HL = 419.04 1324 = x (2676.1) + (1-x) 419.04 1324 – 419.04 = x (2676.1 – 419.04) x = 0.401 quality where x = quality Ŝ1 = 0.401 × 7.3549 + 0.599 × 1.3069 = 3.7322 kJ/kg.K Entropy balance 0 Sˆ M Sˆ M Sˆ S M i i 1 1 2 2 gen where S gen t 0 1u 3.2534 0.5 u 3.7322 0.5 u 3.2534 S gen 0.2394 S gen O S gen kJ kg K s 0.2394 0 so device is possible 4.35 a) Basis: 1 mole air Energy balance Uf Ui WQ 0 C v Tf Ti Entropy balance Sf Si 0 Sf Si S Tf , Vf S Ti , Vf ln C*V ln Tf V R ln f Tf Vi 0 8.314 § 1 · ln ¨ ¸ 1.2743 21 © 25 ¹ Tf Ti Tf Ti e1.2743 W 8.314 303.15e1.2743 1084.1 K J u 1084.1 303.15 K mol K 6493 J mol 6.493 kJ mol 6.493 kJ / mol Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 b) ln Tf Ti 8.314 § 1 · ln ¨ ¸ 21 © 10 ¹ 0.9116 Tf 303.15 u e0.9116 754.33 W 8.314 u 754.33 303.15 3751.1 J mol 3.751 kJ mol 4.36 N M 2 MPa, 200°C 200 kPa, 100°C Find initial number of moles in each compartment PV=NRT N PV RT 200 kPa u 100 u 103 cm3 u N1 kPa m3 8.314 u 10 u 373.15 K mol K a) 8.314 u 103 6.447 moles 3 2 u 103 kPa u 2 u 105 cm3 u N2 1 m3 106 cm3 1 m3 106 cm3 kPa m3 u 473.15 K mol K Mass balance: Nf – N1 – N2 = 0 101.68 moles Nf = N1 + N2 = 6.45 + 101.68 Nf = 108.13 Energy balance: Nf Uf – N1 U1 – N2 U2 = 0 6.448 × CV (Tf – 373.15) + 101.68 × CV (Tf – 473.15) = 0 (6.447 + 101.68) Tf = 6.447 × 373.15 + 101.68 × 473.15 6.447 u 373.15 101.68 u 473.15 = 467.19 K Tf 6.447 101.68 Final Pressure kPa m3 u 467.19 K NRT mol K P V m3 3 u 105 cm3 u 106 cm3 3 2 2 1.0813 u 10 u 8.314 u 10 u 4.6719 u 10 14.0 u 102 kPa 1.4 MPa 3 u 101 108.16 mol u 8.314 u 103 b) Entropy Change N f Sf N1S1 N 2 S2 Sgen t 0 ­ ­ T P ½ T P ½ N1 ®CP ln f R ln f ¾ N 2 ®CP ln f R ln f ¾ Sgen T1 P1 ¿ T2 P2 ¿ ¯ ¯ Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 467.9 1.4 ½ J 467.9 1.4 ½ ­ ­ 8.314 ln 101.68 ®30 ln 8.314 ln 6.45 mol u ®30 ln ¾ ¾ Sgen 373.15 0.2 ¿ mol K 473.15 2 ¿ ¯ ¯ 0 6.45 u ^30 u 0.2263 8.314 u 1.9459` 101.68 ^30 u 0.01116 8.314 0.35667 ` = 60.56 J J 267.47 K K Sgen Q 4.37 206.9 J t 0! (as it should be) K 0 Assume steady state operation N M 2 bar = 0.2 MPa 20 bar 400°C W M M 0 1 2 2 bar = 0.2 MPa M 2 M 1 Entropy balance 0 0 for maximum work Q ˆ ˆ S M S S 0 M 1 1 2 2 gen T Sˆ 2 Sˆ 1 Sˆ 1 Sˆ 2 MPa, 400qC 7.1271 Ĥ1 Ŝ2 0.2MPa, T at T ? kJ kg K 3247.6 kJ / kg 7.1271 kJ / kg K 200qC Sˆ 7.5066 150qC Sˆ 7.1272 kJ / kg K ˆ H 2870.5 kJ / kg kJ / kg Ĥ 2768.8 kJ / kg T2 = 150°C Energy Balance 0 ˆ H ˆ 0 M 1 1 M1 H 2 W Q W 5000 kg kJ 9247.6 2768.8 hr kg 2.394 u 106 kJ hr 665 kJ S 665 kW 4.38 20°C T=? 1 bar 5 bar Heater a) Assume adiabatic and reversible operation of compressor Isentropic S (20°C, 1 bar) – S (T = ?, 5 bar) = 0 C*p ln T 5 R ln 293.15 1 0 400°C 5 bar Solutions to Chemical and Engineering Thermodynamics, 5th ed R Chapter 4 § 5 · Cp T 293.15 ¨ ¸ 462.7 K since C*p C*v R ©1¹ E.B. H (20°C, 1 bar) – H (462.7 K, 5 bar) + W = 0 W = H (462.7 K, 5 bar) – H (293.15 K, 1 bar) Cpx 462.7 293.15 4971.2 J / mol * 21 8.314 29.314 J mol K b) Heater H (462.7 K, 5 bar) – H (673.15 K, 5 bar) + Q = 0 Cpx 673.15 462.7 6169.1 J / mol 4.39 From Fig. 3.3-2 ˆ At 160 K H 260 kJ / kg L ŜL 2.05 kJ / kg K ˆ # 0.00295 m3 / kg V L If adiabatic and reversible, Sgen 0 Sˆ initial Sˆ initial ?, P ? 0.25 Sˆ V T 2.05 kJ / kg 0.75 Sˆ L T Sˆ final ?, P ? but since it is an equilibrium V-L mixture T & P are connected by the saturation line ˆ Sˆ V 5.03 Sˆ Guess T = 120 K SL 1.03 Ŝ 0.75 u 1.03 0.25 5.03 2.03 ˆ H L ˆ V L ˆ H V 3 ˆ 0.0025 m / kg VV 105 kJ / kg ˆ #H ˆ U L L ˆ U V ˆ U f 0.3 m3 / kg P 0.2 MPa 105 kJ / kg ˆ PV ˆ H L L 595 kJ Pa 1J kJ 0.2 MPa u 106 u 3 u 103 kg MPa m Pa J 595 kJ kJ 60 kg kg ˆ 0.25 U ˆ 0.75 U L V ˆ H ˆ U i i 595 kJ / kg Close Enough 535 kJ kg 212.5 kJ / kg 260 kJ / kg Energy balance on piston-cylinder (closed system) o ˆ U ˆ U Q W 212.5 260 47.5 kJ / kg f i (Negative sign system does work on surroundings) W 4.40 a) 20 bar 550°C 20 bar = 2 MPa = 2,000 kPa From the superheated steam tables 1 bar 0.1 MPa Adiabatic and Isentropic Sˆ IN Sˆ OUT Vertical line of Fig. 3.3-1b Solutions to Chemical and Engineering Thermodynamics, 5th ed Ĥ (20 bar, 600°C) = 3690.1 Chapter 4 kJ/kg kJ/kg K Ŝ (20 bar, 600°C) = 8.6451 Estimate From Fig. 3.3-1b (vertical line down to 1 bar) ˆ 2820 kJ / kg Also all vapor T # 172qC H Steady state energy balance W H ˆ ˆ ˆ ˆ 0 M H IN IN M OUT H OUT W OUT H IN M IN W kJ 870.1 kJ u 5000 kg 1327.2 kJ 3690.16 2820 870.1 W kg kg hr s M 1327.2 kW IN If only 90% of the work generated, W 0.9 u 870.1 783.1 kW kJ All vapor so Ĥ out 3690.1 783.1 2907.0 at 1 bar (0.1 MPa) kg By interpolation using the Superheated Steam Tables kJ T 216o C and S=7.8983 kg K b) 0 Entropy balance dS Sˆ M Sˆ Q S 0 M in in out out gen dt T S gen kJ 7.8983 7.7024 0.1959 or S gen kg K M in 979.5 kJ hr K 4.41 The gas in the tank undergoes a uniform adiabatic expansion, which is isentropic, so that its temperature and pressure are related by R / C*p §P · TT,f TT,i ¨ T,f ¸ ¨P ¸ © T,i ¹ The gas in the cylinder to the right of the piston undergoes a uniform adiabatic compression, which is also isentropic, so its temperature and pressure are related by TCR ,f §P · TCR ,i ¨ CR ,f ¸ ¨P ¸ © CD,i ¹ R / C*p Also PT,f = Pc,f. An energy balance on the whole system gives (UT,f + UCL,f + UCR,f) – (UT,I + UCR,i) = 0 NT,f UT,f + NCL,f UCL,f + NCR,f UCR,f = NT,i UT,i + NCR,i UCR,i Also, the mass balance NT,f + NCL,f + NCR,f = NT,i + NCR,i Now using the ideal gas law V V V M.B. Pf TT Pf CL,f Pf CR ,f RTCL,f RTCR ,f R Tf E.B. PT,i V VT PCR ,i CR ,i RTT,i RTCR ,i PV PV Pf VT CV TT,f t CL C v TCL,f f CR ,f C V TCR ,f RTT,f kTCL,f RTCR ,i These equations reduce to PT,i VT RTT,i C V TT,i PCR ,i VCR ,i RTCR ,i CV TCR ,i Solutions to Chemical and Engineering Thermodynamics, 5th ed ªV V V º Pf « T CL,f CR ,f » ¬« TTf TCL,f TCR ,f ¼» and Pf ª¬ VT VCL,f VCR ,f º¼ but VCL,f VCR ,f Pf TT,i PCR ,i VCR ,i TCR ,i PT,i VT PCR ,i VCR ,i 0.5 m3 VCR ,i Pf ª¬ VT 0.5 m3 º¼ PT,i VT Chapter 4 0.4 VT 0.1u 0.5 m3 0.4VT 0.1u 0.5 VT 0.5 0.4 u 0.25 0.1u 0.5 0.25 0.5 0.2 MPa 8.314 TT,f § 0.2 · 29.3 473.15 ¨ ¸ © 0.4 ¹ TCR ,f § 0.2 · 288.15 ¨ ¸ © 0.1 ¹ 8.314 388.7 K 29.3 350.8 K Mass balance on region to right of piston P V PCR ,f VCR ,f N CR ,i N CR ,f CR ,i CR ,i RTCR ,i RTCR ,f 0.1u 0.5 288.15 0.2 u VCR ,f 350.8 VCR ,f 0.1u 0.5 350.8 u 288.15 0.2 0.3043 m3 VCL,f = 0.5 – 0.3043 = 0.1557 m3 Moles in tank initially PT,i VT,i 0.4 u 0.25 N T,i 25.42 moles RTi 8.314 u 106 u 473.15 Moles in tank finally 0.2 u 0.25 N T,f 15.47 moles 8.314 u 106 u 388.7 N CL,f 25.42 15.47 9.95 moles TCL,f PCL,f VCL,f R N CL,f 0.2 u 0.1557 8.314 u 106 u 9.95 376.43 0.1u 0.5 20.87 moles 8.314 u 106 u 288.15 (QWURS\FKDQJH ǻ6of N2 finally in tank ǻ6 of N2 finally left of piston ǻ6of N2 finally right of piston N CR 388.7 0.2 · 376.43 0.2 · 350.8 0.2 · § § § 15.47 ¨ C*P ln R ln 9.95 ¨ C*P ln R ln N CR ¨ C*P ln R ln ¸ ¸ 473.15 0.4 ¹ 473.15 0.4 ¹ 288.15 0.1 ¸¹ © © © 9.26 [Note that the first and third terms should be zero, since the gas in the tank under went a uniform, adiabatic expansion, and the gas to the right of the piston under went a uniform, adiabatic compression. So only the middle term is important.] 4.42 Initial To the left of the piston TCL,f = 376.43 K VCL,f = 0.1557 P = 0.2 MPa J K Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 NCL,f = 9.95 Mols TCR,f = 350.8 K VCR,f = 0.3043 m3 NCR,f = 20.87 K Mass balance after conduction of heat Pf VCL Pf VCR N CL,f 9.95 N CR 20.87 RTf RTf but Pf is the same on both sides of piston, and Tf is the same on both sides of piston 9.95 20.87 VCL VCR VCL 0.5 VCL 0.5 – VCL = 2.097 VCL 0.5 VCL 20.87 VCL 9.95 0.5 3.097 0.1614 m3 VCL VCR = 0.5 –0.1614 = 0.3386 m3 Energy balance 9.95 UL,2 + 20.87 UR,2 = 9.95 UL,1 + 20.87 UR,1 9.95 (T–376.43) + 20.87 (T–350.8) = 0 30.82 T = 9.95 u376.43 + 20.87 + 350.8 T = 359.1 K Pf 9.95 moles u 8.314 u 10 5 9.95 u R u T VCL bar m3 u 359.1 K mol K 0.1614 m3 1.840 bar 184.0 kPa 4.43 110 bar 80oC V=0.1 m3, initial conditions 25oC and 1 bar a) Mass balance: N f N i 'N with N i PVi RT 1 bar 0.1m3 bar m3 8.314 10-5 298.15 K mol K 4.034 mol Energy balance: NfU f Ni U i 'N H in For simplicity in the calculation of thermal properties, will use 25oC as Tref so that Uref = 0 and Href=-298.15R, so that 100 bar 0.1 m3 1 bar 0.1 m3 CV T Tref CV Tin Tref R Tf R Ti § 100 bar 0.1 m3 1 bar 0.1 m3 · ¨ ¸ CP Tin CV Tref ¨ ¸ R Tf R Ti © ¹ which reduces to Solutions to Chemical and Engineering Thermodynamics, 5th ed 100 CV T f CP Tin Tf Chapter 4 1 CV Ti CP Tin Ti So that Tf 100 CP Tin 489.86 K = 216.71o C T 100 CV CV CP in Ti 100 bar 0.1 m3 245.54 mol 3 -5 bar m 489.86 K 8.314 10 mol K 241.51 mol so that 'N 241.51 mol and the fill time is t 12.075 sec 20 mol/s c) Final pressure, since volume and number of moles is constant, can be computer from Pf ' Pf Pf 100 Tf ' 298.15 60.86 bar so that Pf ' Tf ' Tf Tf 489.86 b) Number of moles finally N f 4.44 c 1 bar 20°C 10 bar e d Liquid, 85°C 10 bar = 1000 kPa =1 MPa 200°C ­1 bar, 20qC L® ¯1 bar,85qC ˆ H 1 ˆ H 3 ˆ V ^10 bar, 200qC H 2 M M M 1 2 3 M.B. E.B. 0 355.90 Sˆ 1 Sˆ 3 1.1343 2827.9 Sˆ 2 6.6940 83.96 0 M 3 0.2966 kg M1 M 2 s M 1 M 1 2 1 ˆ H ˆ ˆ M 1 1 M 2 H 2 M3 H3 83.96 M 2827.9 1 355.90 1 M 2 2 2827.9 83.96 355.90 83.96 M 2 M 2 M 1 355.90 83.96 0.0991 2827.9 83.96 1 0.0991 0.9009 S.B. 0 Sˆ M Sˆ M Sˆ S M 1 1 2 2 3 3 gen 0 0.9091u 0.2966 0.0991u 6.6940 1u 1.1343 S gen 0 Solutions to Chemical and Engineering Thermodynamics, 5th ed S gen 0.2013 Chapter 4 J K s Reactor is I, initially evacuated tank is II. Mass balance is 4.45 N I N II , which when using the ideal gas law becomes Ni 200 bar V I 673.15 K P I V I P II V II TI T II 200 673.15 4 · § 1 P I ¨ I II ¸ T T © ¹ or § V I 4V I · P I ¨¨ I II ¸¸ T ¹ ©T since P I P II The energy balance of the reactor and tank is N I U I N II U II , which when using the ideal gas law and constant heat capacity becomes NiU i 200 bar CV 673.15 K 673.15 K or 200=P I 4 P I P I CVT I 4 P I CVT II TI T II 200 and therefore P I 40 bar 5 The entropy balance on just the contents of the reactor that undergo a uniform expansion leads to S constant which leads to R T I § P I · CP Ti ¨¨ ¸¸ © Pi ¹ 8.314 § 40 · 73.2 673.15 ¨ ¸ © 200 ¹ 560.7 K Using this result in the mass balance gives 200 673.15 4 º ª 1 II » 40 « 560.7 T ¼ ¬ which gives T II 708.7 K 4.46 N steam 1 bar, 250°C M steam 1 bar, 100°C 1 bar, liquid water e ˆ H 2676.1 ˆ O liquid 1 bar, 100qC O H ˆ N steam 1 bar, 250qC N H 419.04 kJ Sˆ 7.3549 kg K ˆS 1.3069 2974.3 Sˆ 8.0333 M steam 1 bar, 100qC M M.B. E.B. M M M 0 1 2 3 H ˆ M H ˆ M H ˆ M 1 1 2 2 3 kJ kg M M 2 3 3 0 M 1 1 Solutions to Chemical and Engineering Thermodynamics, 5th ed M 2676.1 M 2974.3 M 419.04 M 2 3 2 3 Chapter 4 0 2974.3 2676.1 M 419.04 2676.1 M 2 3 M 2 2676.1 419.04 7.5689 2974.3 2676.1 M 3 M 2 M M 2 3 but M 1 7.5689 M 3 3 1 0.1167 8.5689 Sˆ M Sˆ M Sˆ Sˆ M 0 M 3 S.B. 1 1 2 2 3 3 1 M 2 1 0.1167 0.8833 gen 1u 7.3549 0.8833 u 8.0333 0.1167 u 1.3069 Sˆ gen 0.1066 Ŝgen 7.5689 M 3 0.1066 kJ ˆ Sgen K 0 0 kJ process is not possible K 4.47 4.48 The maximum temperature difference will occur if the process is carried out reversibly. In this case the energy balance is 1 1 1 H1 H 2 H 3 0 or H1 H 2 H 3 and since CP is constant T1 T2 T3 2 2 2 The entropy balance is T P T P 0 CP ln 1 R ln 1 CP ln 1 R ln 1 T2 P2 T3 P3 0 29.3ln 273.15 25 4 273.15 25 4 8.314 ln 29.3ln 8.314 ln T2 T3 1.013 1.013 The solution to these equations using MATHCAD is T2 298.15 T3 298.15 Given 298.15 § 1 · ( T2 T3) ¨ ¸ © 2¹ 29.3 ln §¨ 298.15 298.15 · © T2 T3 ¸ ¹ 8.314 2 ln §¨ 4 · ¸ © 1.013 ¹ z Find ( T2 T3) z § 78.788 · ¨ ¸ © 517.512 ¹ Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 T2=78.8 K and T3=517.5 K, a difference of more than 438 K. However, in fact a HilschRanque vortex tube is very inefficient and very irreversible 4.49 In the folder Aspen for Textbook>Chapter 4>Problems 4.50 In the folder Aspen for Textbook>Chapter 4>Problems 4.51 In the folder Aspen for Textbook>Chapter 4>Problems 4.52 In the folder Aspen for Textbook>Chapter 4>Problems 4.53 In the folder Aspen for Textbook>Chapter 4>Problems 4.54 In the folder Aspen for Textbook>Chapter 4>Problems 4.55 In the folder Aspen for Textbook>Chapter 4>Problems 4.56 In the folder Aspen for Textbook>Chapter 4>Problems 4.57 In the folder Aspen for Textbook>Chapter 4>Problems 4.58 In the folder Aspen for Textbook>Chapter 4>Problems 4.59 Equation 4.6-14 is to be used to solve this problem Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 M ª Uˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1 Uˆ T2 , P2 PambVˆ T2 , P2 Tamb Sˆ T2 , P2 º ¬ ¼ a) So for part a, assuming the exiting stream is a vapor WS ,max ª Uˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa WS ,max M « « Uˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar «¬ ª§ kJ m3 kJ · º - 298.15 K 8.1765 «¨ 3657.0 +(1 bar)0.2467 ¸» kg kg kg K ¹ » © « 100 kg « 3 § ·» «- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ » kg kg kg K ¹ »¼ «¬ © ª§ kJ m3 kJ kJ · º u100 3657.0 +(1 bar)0.2467 - 298.15 K 8.1765 «¨ ¸» 3 kg kg bar m kg K ¹ » © =100 kg «« 3 § ·» «- ¨104.88 kJ +(1 bar)43.36 m u100 kJ 3 - 298.15 K 8.5580 kJ ¸ » kg kg bar m kg K ¹ »¼ «¬ © º » » ¼» =100 kJ ª¬ 3657.0+24.7-2437.8 - 104.88+4336-2551.6 º¼ =100 kJ >1243.9-1889.3@ 64540 kJ b) WS ,max ª Uˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa M« « Uˆ 673.15 K, 0.6 MPa (1 bar)Vˆ 673.15 K, 0.6 MPa 298.15 K Sˆ 673.15 K, 0.6 MPa «¬ ª§ kJ m3 kJ · º - 298.15 K 8.1765 «¨ 3657.0 +(1 bar)0.2467 ¸ » kg kg kg K ¹ » © « 100 kg « 3 § ·» «- ¨ 2962.1 kJ +(1 bar)0.5137 m - 298.15 K 7.7079 kJ ¸ » kg kg kg K ¹ »¼ «¬ © 100 kJ ª¬ 3657.0+24.67-2437.8 - 2962.1+51.37-2298.1 º¼ 100 kJ >1243.9-715.4@ 52850 kJ c) The simplest way to calculate this is as WS ,max WS ,max part a WS ,max part b 64540 52850 kJ 11690 kJ 4.60 In the first printing, Problem 4.60 inadvertently duplicates Problem 4.59 4.61 In first printing, this problem duplicates parts a and b of Problem 4.59 4.62 º » » ¼» Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 The starting point is Wˆ S ,max Hˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1 Hˆ Tamb , Pamb PambVˆ Tamb , Pamb Tamb Sˆ Tamb , Pamb o For the 800 C and 2 MPa stream WˆS ,max Hˆ 800o C, 2 MPa (1 bar)Vˆ 800o C, 2 MPa 298.15 K Sˆ 800o C, 2 MPa Hˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar ª§ kJ m3 kJ · º 4150.3 +(1 bar)0.2467 - 298.15 K 8.1765 «¨ ¸» kg kg kg K ¹ » © « = « § 3 ·» «- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ » kg kg kg K ¹ »¼ «¬ © For the 900o C and 1.4 MPa stream WˆS ,max Hˆ 900o C,1.4 MPa (1 bar)Vˆ 900o C,1.4 MPa 298.15 K Sˆ 900o C,1.4 MPa Hˆ 298.15 K),1 bar (1 bar)Vˆ 298.15 K),1 bar 298.15 K Sˆ 298.15 K),1 bar ª§ kJ m3 kJ · º 4391.5 +(1 bar)0.3861 - 298.15 K 8.5556 «¨ ¸» kg kg kg K ¹ » © « = « § 3 ·» «- ¨104.88 kJ +(1 bar)43.36 m - 298.15 K 8.5580 kJ ¸ » kg kg kg K ¹ ¼» ¬« © 4.63 Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 The starting point is WS ,max Hˆ T1 , P1 PambVˆ T1 , P1 Tamb Sˆ T1 , P1 Hˆ Tamb , Pamb PambVˆ Tamb , Pamb Tamb Sˆ Tamb , Pamb o For the 1000 C and 25 bar stream WS ,max H 1000o C, 25 bar (1 bar)V 1000o C, 25 bar 298.15 K S 800o C, 25 bar H 298.15 K),1 bar (1 bar)V 298.15 K),1 bar 298.15 K S 298.15 K),1 bar For the 900o C and 40 bar stream WS ,max H 900o C, 40 bar (1 bar)V 900o C, 40 bar 298.15 K S 800o C, 25 bar Toevaluate which stream has the potential to do more work, calculate the difference in available work 'WS ,max WS ,max 1000o C, 25 bar WS ,max 900o C, 40 bar = H 1000o C, 25 bar H 900o C, 40 bar 1 bar u V 1000o C, 25 bar V 900o C, 40 bar 298.15 K S 1000o C, 25 bar S 900o C, 40 bar Now since the fluid is an ideal gas § 1273.15 K 1173.15 K · CP 1000o C 900o C 1 bar u R ¨ ¸ 40 bar ¹ © 25 bar 1273.15 25 · § 298.15 K ¨ CP ln R ln ¸ 1173.15 40 ¹ © 3 bar m 1273.15 25 · § CP 100 1 bar u 8.314 u 105 R ln ¸ 29.329 298.15 K ¨ CP ln mol K 1173.15 40 ¹ © 'WS ,max 4.64 A U PambV Tamb S § wA · ¨ ¸ © wT ¹V § wA · ¨ ¸ © wV ¹T §C · § T · § wU · § wS · CV Tamb ¨ V ¸ CV ¨1 amb ¸ ¨ ¸ Tamb ¨ ¸ T ¹ © wT ¹V © wT ¹V © T ¹ © § wU · § wS · § wS · § wP · § wU · ¨ ¸ Pamb Tamb ¨ ¸ ; but ¨ ¸ ¨ ¸ and ¨ ¸ © wV ¹T © wV ¹T © wV ¹T © wT ¹V © wV ¹T § wA · so ¨ ¸ © wV ¹T § wP · § wP · T¨ ¸ P Pamb Tamb ¨ ¸ © wT ¹V © wT ¹V § wP · T¨ ¸ P © wT ¹V § wP · T Tamb ¨ ¸ Pamb P © wT ¹V Solutions to Chemical and Engineering Thermodynamics, 5th ed Chapter 4 A U PambV Tamb S § wA · ¨ ¸ © wT ¹V § wA · ¨ ¸ © wV ¹T §C · § T · § wU · § wS · CV Tamb ¨ V ¸ CV ¨1 amb ¸ ¨ ¸ Tamb ¨ ¸ T ¹ © wT ¹V © wT ¹V © T ¹ © § wU · § wS · § wS · § wP · § wU · ¨ ¸ Pamb Tamb ¨ ¸ ; but ¨ ¸ ¨ ¸ and ¨ ¸ © wV ¹T © wV ¹T © wV ¹T © wT ¹V © wV ¹T § wA · so ¨ ¸ © wV ¹T § wP · § wP · T¨ ¸ P Pamb Tamb ¨ ¸ © wT ¹V © wT ¹V § wP · T¨ ¸ P © wT ¹V § wP · T Tamb ¨ ¸ Pamb P © wT ¹V 7KLV LV OLNH ,OOXVWUDWLRQ H[FHSW WKDW WKH 5DQNLQH UDWKHU WKDQ YDSRU FRPSUHVVLRQ UHIULJHUDWLRQ F\FOH LV XVHG 2QO\ SURSHUWLHV RI SRLQW DQG SDWK IURPoFKDQJHV 3RLQWLVHQWURSLF S N- NJ. 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C 2 H 4 C 2 C 2 u 11 . u 1023 C2 H 4C 2 J C 2 H 4 C 2 Reac Va N a C 2 H 4C 2 J C2 H 4 C 2 Ka N 50q C b ca e b : NB NBq N N A NBq u N q N A N B N B2 N A NBq N A N B N B2 ef d S C ca a d E T I a be N d e a c ,5 F a be e Chap e 13 d M e f ac NA N A N Bq N B2 A NA NA B N Bq N Bq 2 X N Bq 2 N B2 N A N Bq N B2 N B2 X B2 N A N Bq N B2 N A N Bq X N A N Bq N B2 S g h e a aB2 q B q B A a 2 A q 2 B A 2 N A N Bq a 2 A 4Ka 1 , a d 2 1 B2 c N N h cN N h 4K c N h 2 N B2 a dG 2 q B g e N B2 he e c h cN 2N h N B2 N A N Bq N B2 aB2 Ka B e ba c B c B G 2 12 a B a a 2 2 B 2 h 2 12 a B a 2 B 2 2 G h cN N h 2 G ; N N cN N h 2 G 2 N N N a N N qfFH 2G IK a d N N N N Nq 2G A q B A N B2 B B2 B A q B A q B A B A B B B2 Th N A N Bq N A N B N B2 JA JB c N B N A N Bq A B 13.14 (a a a ab e a a Ma hcad Reac B G a d h cN N N h MNL1 2 G OPQ 2G 2G c 2 N Bq 2 B2 B 2 A A B h 2 12 B A hee ) 1 C2 H5OH O2 CH 3CHO H 2O 2 C2H5OH CH3CHO H2 1 : 2 S C ca a d E T d a c ,5 Chap e 13 d S ec e ba a ce ab e: S ec e I 1 O2 0.75 C2H5OH O 1 X1 X2 075 . 05 . 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Q N 1 f RS K K T R . 29815 K S K K T T X1 . 29815 Ka1,12 T 12 a ,1 12 2 a ,2 X2 12 a ,2 12 a ,1 12 2 a ,2 ab e be i 2 Ka1,12 Ka1,22 1 1 D X1 T . 29815 . X 12 1 1 D X 1 T 29815 Ka ,1 1 1 D X1 1 1 D X1 a eg e 2 1 2 1 1 a Re f . Th Ka ,1 1 Ka ,1 PV Ka ,2 2 1 2 Ka1,12 Ka1,22 a fa a P 1 ba f X a P 1 ba f X 1 1 1 1 X X D D a fa f 1 1 D X e a eg e a X2 X1 def e D Ka,1 (a) X 22 1 X1 X 2 1 X1 X 2 P 1 ba 2 X 12T . 1 1 D X1 29815 UV W U 4T K K V . 29815 W 4T Ka1,12 Ka1,22 . 29815 12 a ,1 12 a ,2 . 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Le 05 1 G he e G a be e he + 1 2 X X ea 2 2 1 X 2 1 2X 2 X 4 1 2 X 1 X 9 1 3X X F a e eeded! a he de a e F H 2 2 he e X ec X 8X 2 1 2 X X 8X 1 2 X 3 3 1 2 X 1 3X X 1 3X 0 c a cP NH 3 1 ba h 2 c P 1 ba hc P 1 ba h H2 H2 3 ge a added a affec ed, a d he S C a a d c ea da c 13.1 ca a d E (a T d a c ea ha he eac . Th a a a e e h ca e. a a ab e a a Ma hcad Reac C4H8 S ec e ba a ce ab e I da a g he d ec c be ee eac d a 1 F a C4H10 1 1 X1 C4H8 0 X1 X2 C4H6 0 X2 H2 0 X1 X2 e f c ea ed a c a hee ) 1 X1 1 X1 X 2 X1 X 2 1 X1 X 2 X2 1 X1 X 2 X1 X 2 1 X1 X 2 1 X1 X2 6 S ce Chap e 13 d C4H10 C4H8 H2 C4H6 H2 2 : S ec e c ,5 e P 1 ba 1 ba , a . Th a X X fa X X f a1 X fa1 X X f K c1 X X X X h X X C4 H8 H 2 Ka,1 1 2 C 4 H 10 a, 1 1 1 2 1 2 1 2 1 2 2 1 2 2 2 (1) a d C4 H 6 H 2 Ka,2 c a C4 H 8 a Ka,2 X1 X12 X 2 X 22 13.1 U g CHEMEQ (f T Ka,1 e b c f X 2 X1 X 2 X1 X 2 1 X1 X 2 a fa h X X X 1 2 f 2 2 (2) ) a d Ma hcad (f )I ba Ka,2 X1 X2 C4 H10 C4 H8 C4 H6 H2 900 1000 0.9731 0.1191 5.814 0.5575 0.724 0.951 0.147 0.464 0.147 0.020 0.308 0.202 0.079 0.192 0.466 0.586 We c de each f he eac e aae . S C 2Ag + ' 1 O 2 o Ag 2 O; K1 2 G ,1 e K1 ca a d E T aAg2O 2 0.5 Ag O 2 a a d a c ,5 1 aO0.52 Chap e 13 d 1 ce he ac 0.5 O2 1 J 'Gf,Ag 2O 2'Gf,Ag 'Gf,O2 'Gf,Ag 2O 9.33 2 § ' ,1G · 9330 § · f h ¨ ¸ e ¨ ¸ 43.15; RT ¹ © 8.314 * 298.15 ¹ © 1 0.000538 he e b 43.152 f ge he a 0.21. S ge ca ce a The e eac e c de aAg2O aH2 aH 2 2Ag + H 2 O o Ag 2O + H 2 ; K 2 2 aAg aH 2O aH 2O he ' K2 O2 ,2 G e Th e 'Gf,Ag 2O 'Gf,Ag 2'Gf,Ag 'Gf,H2O § ' ,2G · ¨ ¸ e RT ¹ © b c a e f ac a H2 h d . eac eea e f O2. H e e , he e f ac f each d e b , a e. ce he ac . 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F he da a ,4 f he eac , f h g he e. a ee f he eac f each d cc he aAg2S aO2 1 1 a a aSO2 SO 2 a 0.03 2 Ag SO2 b e , SO 2 c ce 'Gf,Ag2S 'Gf,O2 2'Gf,Ag 'Gf,SO2 G ce he ac , ha 31.80 299.91 3 u10 8 SO 2 268.11 § ' ,4G · 268110 § · ¨ ¸ e ¨ ¸ e RT ¹ © 8.314* 298.15 ¹ © The ef e, ba ed he e a a e f he e 108.16 1.063 u10 47 b a , h c S b e he a h g f he f eac c e. b e h g f 13.20 U a e e , Tab e 3.4 a d A K4 g eac 2Ag + SO 2 o Ag 2S + O 2 ; K 4 ' Chap e 13 d 'Gf,Ag2S 'Gf,H 2 2'Gf,Ag 'Gf,H2S G H2 c ,5 5 e K3 a de 2Ag + H 2S o Ag 2S + H 2 ; K 2 S d e g he da a he be Ka,1 750 K aCaOS O 2 aCO 2 aCaCO 3 aS O 2 he a 1481 . aCO 2 c e d e. A.IV, I f d CO 2 P 1 ba eac , (1) J . S C ca a d E ce he ac fa T d a c ,5 he d ae . aS3 O 2 aFe 3O 4 aCO Ka,2 750 K Chap e 13 d 0.0277 3 a FeO S O 2 aCO 2 aCO aCO 2 CO , (2) CO 2 a d aFe 3O 4 aS3 O 2 Ka,3 750 K 12 3 aFeOS O 2 aO 2 F e CO 2 ,f O2 P F 1 ba I GH P JK 1 aO1 22 12 O2 (3) . 1242 u 1028 1 ba . (2) e ha e CO A 08973 u 1014 . e 0.0277 , h e f ec c CO 2 148 ba , h e f . (1), PCO2 CO c be a he be, he # 104 a a he c e e be ee 75 a d 105 ba . F a , f E . (3), e c c de he e O2 he a he e, c a ed a ace f ec c c be a . C c ? S e ha a b g ! Ca c a a d da a a e a a e ag ee e , b a e ce a a a e ag ee e . C de he ce a a he ea e e , he a he c de d b ed a ea ab e e, a d ca be e ec ed. 13.21 (a) The c he e ' . G G The ha e e F ha d be e he eac H TS , a d f f che ca e K aV e § ' G V · ¨¨ ¸¸ RT © ¹ K aL e § ' G L · ¨¨ ¸¸ © RT ¹ b V aEB aSV aHV L aEB aSL aHL ; a ; ha e (1) d ha e (2) he a da d a e G bb f ee e e g cha ge b ae e e e e eac ha e ae L EB V EB , L S ' GV 83.0 J g , e e a , d e he ' a H V S a d L H V H (3) . u 1014 . Th , a d K V 3482 c e he ga ha e. N T' a S e ha e e a J/ . S C A , ce, f ca a d E T d a c ,5 Chap e 13 d he ab e ha e, G LH 2 ! G V H 2 . The ef e, h d ge , he a L ee V be f ab he a e e a d g a ' G . C e e , he ha ' G d ha e che ca e b c a a be a ge, a d he h d ge a eac e e a g c e he d ha e. e f ac f e e be e a b h ha e . The b e he ed ce de e g he b f he e ce h d ge he d e h be e e, a d de e g he a f e h be e e he a . Th , he e a be ed ae L V L EB EB EBJ EB EB EB P a d L V L H H HJ H H HP He e e ha e a ed ha he a ha e dea . A af g e , e a e ha e e h d ge d ed he d g eg a he ha e. Th , J EB 1 , I EB 1 , a d, e a V LH G EB G H RT JH f 2 2 31 . cc u 8.8 3.25 ca cc u 4.184 J / ca K u 29815 8.314 J . K 01612 . J H 1175 . Ne e ha e e a e he f gac f h d ge he d ha e. A b a ceed e Sha c ea , Sec. 11.1. H e e , h d ge a ed de e g h c ea ,a dPa a aga ef gh ga e ch a h d ge a d he ; ce e e e a da a a e a a ab e, e ha e e ch ce b . K a d PC 12.97 ba e h c ea . N e, h e e , ha f h d ge , TC 332 , c ha T T ea 898 . , h ch T TC 898 . (a e a d H U g he a 1 T , e f d ha PEBa J a ), he a L H H e e da a f e h 1273 . Pa a 25qC. EB J EB PEBa EB P . a d H 0996 [S ce he ga ha e eed eae a ]. 0.049 H H EB #0 A a e a e ca c a e e f h d ge a e g e S c ca 0.951 a d e e 0.049 , a d be e e, EB c PC 5188 . ba 3 ba e ha he ga - ha e 3 ba 1.175 u 51.88 ba P e ba 1013 . ba , 25q C 4 u 12.97 ba g e e c ec f h e e H ee L e L [N e he he P eg g b e.] A af g e , The ef e, ff he ca e f F g. 11.1-1. If EB ed 0.004 a T ae h 1.013 ba a e 4, e a e h d ge . 1 H 0951 . . he f f 0.951 u 1 u 1273 . # 0.004 3 u 100 Pa a e h d ge , a a 0.996 h a Pa ed, he e 25q C a d P 3 ba # 0.0 e he Pe g-R b e a f a e. The Tab e 6.6-1. The a e f e h be e e a e S S C ca a d E T d a c ,5 Chap e 13 d TC = 617.2 K, PC = 36 ba , Z=0.302, a d TB=409.3 K. The e a e b a e ac a a e e f h d ge h he c e Tab e 9.4-1, e a e ha a e e . U g he he a f a h ca c a he g a VLMU e b a he f g e 0.9952 0.0018 H H a T 25q C 0.9982 a d EB 0.0048 EB a d P 3 ba S # 0.0 S # 0 Th a be a e acc a e ca c a ha g eg a he h ch e ed a e a a f he P a -Sha c e a .. H e e , he e ba ed he a ha = 0. I d be be e ha e e e e e a da a ge a be e e ae f h aa ee. (b) A 150qC a d 3 ba . U g he da a he P b e a e e , A e d ce A.II a d A.IV e f d Ka T 150q C 31 . u 108 . S aga e ca e e ha a he e e he a a d d ha e c e ed e h be e e. A af a a ( ea ), e a e ha he d ha e e e a e e h be e e. Th e ba . 3.25 88 . 2 31 JH e . 112 . . 1987 u 42315 He e aga , e f d, e a a g F g. 11.1-1, ha 3 L 5188 . ba H 0.052 H 5188 . u 112 . . 0948 . . a d EB 1 0052 RS T N ,h e e , PEBa 1303 . ba H 1303 . u 0.948 0.412 3 a he e a , e ba T 150q C . 0.029 0586 H EB 0.971 N a g he e a e f Aga ha UV W EB EB 0.414 3 ba ST 0.0 H 0.0016 H . 05034 EB 0.9984 EB 0.4966 3 ba 0 ST g he Pe g-R b = 0, e b a he f e a g f a e, he ga T VLMU a d he 150q C 0 0.0 ST ST I b h a a a d b e ee ha he e f he e a f a e ca c a ae a a e, b a a e ag ee e h he P a -Sha c e a . H e e , he a e ed c ch h ghe b e f h d ge he d ha e. The e a f a e ca c a ch ea e d (g e he a a ab f he ga VLMU). If e e e e a da a e e a a ab e f h d ge b e h be e e ( he a a c ), he a e f c d be ad ed e d ce ha da a. The e d ha e e c f de ce he e a f a e ed c f he b e he e. If ch e e e a da a e e a a ab e, c ea h e d ad he P a Sha c e a a ch ha ch da a. S C ca a d E 13.22 (a) T d E e g ba a ce U c ,5 a f ed 0 dV Q P d a . 0 f ga FG w U IJ FG w U IJ FG w T IJ H w K HwTK Hw K Q W Chap e 13 d V V V FwU I FG IJ GH w T JK . Hw K Q wT CV b a V V e ea U T ¦NU ¦N U , U N e a d a f ce e e c e ed he e. A he e TR ec e e efe e ce e e e a e he ga dea a N ,0 Q X N he a d T U TR CV T ea e. TR FG w U IJ FG w IJ ¦ N U H w T K Hw TK Fw X I ¦ Q U GH w T JK ¦ N C CV ,eff V ¦Q U FwU I V V 'U V, V he e 'U Fw N I ¦ GH w T JK U ¦ N GH w T JK T V FG w X IJ ¦ N C T H wT K V, V e a e e g cha ge eac . S ec e ba a ce ab e S ec e I O N2O4 1 1 X NO2 0 2X 1 X 6 Ka he 1 X 1 X 2X 1 X c c 2 aNO 2 aN 2O 4 B he dea ga a a c d PV NO2 P 1 ba h NRT P NT 2 N 2 O 4 P 1 ba 2 NO2 P h N 2 O 4 1 ba P0 N 0T0 a f bc 0 de 4 X 2 P 1 ba 1 X 1 X he e he . ba 1 X T FG N IJ FG T IJ 1013 H N KH T K 300 P P0 Ka 4 X 1 X T 1013 . 1 X 1 X T0 X2 1 X T0 Ka 4 1013 . T 0 2 0 D 4 1013 . X 2T 1 X T0 X 2 D 1 X 0 e S C ca a d E T d D b V Ka 1 T T T ¦Q 1 ' F he d d a c che ca eac e e . Th ' U RT 2 0 X D U ' g e he c ; he a e e ha e' ¦Q (2) 1 «X « 1 4 D ¬ º » (3) » ¼ ' H RT U 2 ª H RT ' ( e f ac he e ha ce e U de e de ce RT 2 ) e gh ed hea ca ac f f he hea ca ac d e he ce ha a fe eg ab bed a he eac e b h f , a d a ec d ' U de e de ce, de e e he e e fa hf he e b h e e a e. (b) U g CHEMEQ a d he da a A e d A.II (f CP* ), de e E ed a each e ea ea . (1) a d (3) X a d CV,eff a T (K) 300 350 400 450 500 550 600 700 D g h CV f 0.044275 0.1891 1.0016 0.6177 10.0613 0.9163 58.883 0.9835 236.25 0.9958 721.36 0.9986 1796.25 0.9994 7220.4 0.99986 ce h Ka b h NO2 a d N2O4 . The , f e . The e a e ab a ed a d X (1) . A 1 X CV,N 2O4 XCV,NO2 CV,eff Chap e 13 d ¦Q U ¦Q H RT U ' c ,5 RS 1 4 1UV 2T D W LMD RS 1 4 1UVOP LM X 1 OP D T N2 T D W Q N 1 a4 D f Q T X DX D F XI H TK a ed be . P (ba ) CV,NO2 CV,N2O4 CV,eff 1.205 1.912 2.588 3.014 3.370 3.712 4.051 4.727 37.11 38.94 40.65 42.22 43.67 45.00 46.23 48.36 78.83 84.47 89.55 94.06 98.01 101.39 104.20 108.12 J/ K 410.5 546.9 195.9 69.11 49.23 46.49 46.71 48.45 a S C ca a d E T d a c ,5 Chap e 13 d 13.23 a) U g he De b gh e h d, e ha e 5C + 12H = C5 (1) 5C + 12H = C5 (2) 5C + 12H = e C5 (3) S ce C H e e , b ac E . (1) f E . (2 a d 3) ba C5 = C5 a d C5 = e C5 The e a e he de e de eac .I h e , he e a e 3 c e ( he h ee e a e ), C=3, e ha e, P=1 a d de e de eac M=2. The ef e he be f deg ee f f eed a e F=C-P-M+2=3-1-2+2=2 deg ee f f eed . b) F h b e , ce T a d P a e ec f ed, he be e -def ed a d h be a e . The e b ea ha ha e be ed a e: a C5 a e C5 C5 P /1 ba C5 e C5 P /1 ba e C5 . K a,1 ; a d K a,2 a C5 a C5 C5 P /1 ba C5 C5 P /1 ba C5 T ca c a e he e b c a , e e he V f g a e Ka,1=0.18240 a d Ka,2=0.06737. a BASIC ga , e ba he S C ca a d E K1 0.18240 T d a c ,5 Chap e 13 d K2 0.06737 c5 ( X1 X2) 1 X1 X2 c5( X1 X2) X1 X1 0.1 e c5 ( X1 X2) X2 X2 0.1 G e X1 K1 K2 1 X1 X2 f d( X1 X2) X1 0 X2 § 0.146· ¨ ¸ © 0.054¹ 1 c5 ( X1 X2) 0.8 c) P e N e ha 0. 13.24 The eac ¦ e b c Ka )F X2 1 X1 X2 c5( X1 X2) effec he e PCH 4 a CH 4 a 5.75 H 2O ac e c5 ( X1 X2) b c ce RT A 283 K, K a 'G f Ka a f he H 2O J H 2O f ce he h ch he 5.75 1 ba f he be , ae he 1 ba . The ef e K a . A 278 K, K a PCH 4 a 8.314 278 (0.0238) 8640 1 ba 6.8 MPa e 1.0 ha 'G f 0.054 CH4 g 5.75H2O( ) R CH 4 5.75H2O( ) a ah d a e h 0.146 1 ba 0.1 MPa ee ,a d 1 ba 1 ba 4.2 MPa 0.1 MPa e 0.0238 . 0.0147 a d 8.314 838 (0.0147) 9929 b) Need c e 10 % CH3OH e gh f a e be 18, a d e ha J d J ae e f ac . Ta g he ec a be 32, a d f 100 g f ha 10 % 90 18 0.9412 . A e ha , he a e e f ac e ha he e ha H 2O 90 10 18 32 + ae e ca be de c bed b he Va Laa de h aa ee g e Tab e 9.5-1 ( h ch a e a d a 25 C, b e a e a e e e a e de e de ), D a d E=0.46 he e ha c e 1. Th ead a ac c eff c e f a e f 1.002. (We c d c de he e f he UNIFAC de ead, h e e , S C ca a d E T he a e ac c eff c e he e b ea ah d a e Ka PCH 4 a CH 4 a 5.75 H 2O 1 ba d c a e , 1.0 H 2O J H 2O c ,5 e d ffe e ce PCH4 5.75 Chap e 13 d 1 ba d be ee .) S 1.0 0.9412 1.002 ha 1.0 ba PCH 4 0.714 5.75 0.1 MPa 0.1 MPa ha a 278 K PCH4 5.885 MPa K a 0.714 0.0238 0.714 0.1 MPa 9.528 MPa a d a 283 K PCH4 0.0147 0.714 1 1 H 2 g I 2 g a d I2 g I2 . 13.2 The eac a e HI g 2 2 U g he da a he C a E Ha e ha e PCH4 ,1 G ' 1.95 ca =8.159 J °­ ' ,1G °½ ® ¾ ¯° RT ¿° e K a ,1 e K a ,2 S ec e I a HI 1 H2 0 I2 0 6 N e: a da d a e e e 2 Ka , 1 1 Th a a ed) e ha e e e ( 2478.3 2 e 2 aHI aI 2 aI 2 g e a de he a c d e b e , ha , a cha ged. eac 1 . a 1 X 1 X 1 X 2 1 X 2 1 he C 1 X 2 1 X 2 a1 2f X a P 1 a f 1 X aP 1 a f 2 Ka , 1 g a1H 2 a1I 2 2 F a 12 2 . u 102 372 3.72 u 10 ­° ' ,2 G ½° ® ¾ ¯° RT ¿° S d ec a f e c a he e e a h ch he f b f d a ea The ef e, he f e c e he a c 4.63 ca =-19.37 J ,2 G ' 1 X P 1a 1 2 XP 1a 1 2 XP 1a a f a f a E 1a Ha X 2 1 X Ka,1 0.069248 ( de e de f e e!) gh ha ha e c ec dea . a ca be S C ca a d E T d H2 h ch cc . N I2 he c c d 0.03462 ; HI he f 1 aI 2 I2 2478.3 1 I2 P 1 a Th , f P ! 0.01181 ba 2NaHCO3 ec a ha e a d N H 2 O P CO2 f 0.01181 ba . cc . H2O UV W b aNaHCO 3 1 aNaCO 3 1 Ka aH 2 OaCO 2 PH 2 O PCO 2 1 ba 1 ba he e N a aNaCO 3 aCO 2 aH 2 O aNaHCO 2 N P N N CO2 PH 2 O PCO2 The ef e d I2 f 1a Ka , 2 I 2 P d I2 f 3 P a F PI H1 a K 0.01166 a a NaCO3 Ka b ec : 1a 0.03462 u 2478.3 P 093075 . he de he ec d eac Ka , 2 Chap e 13 c ,5 f he a aI 2 13.2 (a) a e d ha e, N ga e f ga 1 P. 2 Ka LM a1 2f P OP N 1 ba Q 2 a d LM a1 2f u 0.826 Pa OP 1706 N 100 Pa Q . u 10 a1 2f u 166.97 Pa O 0.697 K 110q C LM N 100 Pa PQ 2 Ka 30q C 5 2 a Ka 30q C 10.979; 0.3610 Ka 110q C N K a T2 K a T1 K a T2 K a T1 ' R H§ 1 1· ¨ ¸ © T2 T1 ¹ (1) S C ca a d E T d Th a e ha ' H ed da a he be h he a c ,5 de e de ae e Chap e 13 d f T, he a e ca a e ' H 128.2 J (b) Ka T Ka G g bac E . (1) ab e e ha e § ' H§1 1 ·· 5 ¨ ¨ ¸ ¸ 1.706 u10 e R © T 303.15 ¹ ¹ © 15420 15420 10.9788 508659 39.8871 . T T Ka T 30 C e Ka (T ) T 13.2 Reac U 10 . ba PH 2O PCO2 (c) LM a1 2f u 2.0 OP N 1 Q 2 g he ga P PCO2 1 ba . CH4 CHEMEQ e ha e Ka T S ec e I a (ga ) 1000 K H2 1 1 2 X CH4 0 X 6 1 X aCH 4 aCaH2 2 aCH 4 aH2 2 009838 . F a (ga ) C Ka 2.0 ba 1 3866 . K 11345 . qC f C 2H2 10 . ba § 1 ·· §1 ¨ 15420 ¨ ¸¸ © T 303.15 ¹ ¹ © 1 2 X 1 X X 1 X F H XP 1 X 2 1 ba P 1 X 1 ba 1 2 X 2 I K 2 X 1 X X 1 X 1 ba 1 2 X 2 P 1 2 X 2 h e a a e X 0.0769 a d X 0.9231 . a a e f he e b c a , he X 0.0769 0.917 a d CH 4 0.083 e H2 01004 . c The W h ch a ec e. Th Th ( bab h e e e a a d ca c a a e )e e a e a he e b c . The ef e, he eac ce he d a ca ed, a a fe ed. C e e c ea e ba eb g he h d ge f ha e effec he 13.2 Reac (a a a ab e a a Ma hcad hee ) he a he g he ce . S C ca a d E CO H 2 C 2H 2O CO2 C CO2 2H 2 2CO CO H 2O CO2 H 2 c , e e eac CO H 2 C H 2O CO2 2H 2 CO2 C C 2H 2O 2CO CO2 H 2 CO H 2O N eed Sa b de g f he CO CO E 2H O C 2O H 2O CO2 eac ce H2 H 2H e O d C H 2O F N T 4 che C 2O CO2 ce e e e ca eac CO H 2O aeO Chap e 13 d CO H2 O ee e c ,5 1, 2 a d 4 de e de ae 1 1 CO2 C 2 2 a a c 1 CO2 2 1 H 2 CO2 2 1 C CO 2 1 C H 2O 2 C CO 2 UV Th e e f C 2H O W de e de eac C 2H 2 CO 2 . ge ee . 2CO 2 . Add he e 2H 2 C 2CO 2 We 2CO C 2H 2O H 2 CO2 CO H 2O e H 2 CO 2 2H 2 CO 2 UV a he C 2H O W de e de eac CO H 2 O 2 . S C S ec e ca a d E I T a d 1 X1 X 2 H2 1 1 X1 2 X 2 CO 0 X1 H 2O 0 X1 2 X 2 C 0 aa Ka,2 aCO aH 2 O f 2 2 fa 1 2 1 2 CHEMEQ, I f d he f 2 1 2 g Ka,2 758.6 48.43 5.950 1.137 0.2974 1 03665 . u 10 0.1110 0.2493 0.4596 0.7387 X2 0 de Ka,1 a f 2 2 Ka,1 600 700 800 900 1000 N ca b f a X 2 X f a2 X f 1 ba a1 X 2 X f a1 X X f P 1 a H2 2 O C a a aCO 2 aH 2 T K de 1 1 X 1 X1 2 X 2 1 X1 X 2 1 X1 2 X 2 a H2 2 aCO 2 ga g he e e 2 1 X1 2 X 2 ¦ 0 2 1 ga ha e) ¦ 2 X2 Ka,1 S LM1 X X OP P N ¦ Q 1 ba LM1 X 2 X OP P N ¦ Q 1 ba LM X OP P N ¦ Q 1 ba LM X 2 X OP P N ¦ Q 1 ba 1 X1 X 2 ¦ 1 X1 2 X 2 ¦ X1 ¦ aX f ( (a) Chap e 13 d a 1 g he c ,5 F a CO 2 U a X 12 a1 X f 1 X 12 2 1 ba 1 X1 3 P a d Ka,2 2 a f ,If d TK 600 700 800 900 1000 P ba 1151 . u 104 6111 . u 103 0.126 1.357 9.237 If he . e ef ag e e ea e ab e he e e ca c a ed, ca b S C (b) ca a d E E ac T 30% f ca b Ka,1 a TK X1 P ba a de d a c ,5 ed X 2 03 . f Chap e 13 d f a 0.6f 17. 1 ba a0.7 X fa0.4 X f P 600 700 800 900 1000 0.0157 0.0410 0.0750 0.1104 0.1427 0.0084 0.170 1.972 15.266 85.419 X 1 X 1 0.6 a d Ka,2 0.7 X 1 0.4 X 1 fa 2 1 1 1 2 13.2 The eac he e g ee c ce ed ab T ( ) + S O 2( ) = T O 2( ) + S ( ) Th e ae he f eac he be ae e eac . The ef e J J J ' G 674 (644) 30 30, 000 a d he e Ka C e b c a f h he ec d eac § ' G · § 30000 · ¨¨ ¸¸ e ¨ ¸ 17.02 R T ¹ © 8.314 1273 ¹ © e e , a he e g ee fea , he a e a e c ac h c d de. e be affec ed b h gh 13.30 a) R 8.314 T 80 273.15 1203.828 · § ¨ 4.01816 ¸ T 53.229 ¹ P a B( T) 10© 1209.299 · § ¨ 3.98022 ¸ T 49.623 ¹ P a C( T) 10© B 0.2 P( T) P a C ( T) 0.991 2 ª (1 ) º « 100 » R T ¼ ¬ ga C( T) e 2 ª ( ) º « 100 » R T ¼ ¬ ga B( T) P a B ( T) ( 1 ) ga C( T) P a C( T) P( 0.2 353.15) 1 ba ( ga B( T) P a B( T) ) P( T) So b=0.206 and nC=0.794 ) The 1.01 C 1 B ga B( T) e ( T) P a B ( T) a ba a ce ab e : ( 0.2 353.15) 0.206 S C ca a d E T a 0.794 0.206 0 f a 0.794-X 0.206+X 3X S ec e C6 be e e h d ge a S d a c ,5 Chap e 13 d e f ac (0.794-X)/(1+3X) (0.206-X)/(1+3X) 3X/(1+3X) 1+3X ha he e b ea a 3H a B aC Ka 3 3 § H P · § BP · ¨ ¸ ¨ ¸ © 1 ba ¹ © 1 ba ¹ § CP · ¨ ¸ © 1 ba ¹ § 3X · § 0.206 X · ¨ ¸ ¨ ¸ © 1 3X ¹ © 1 3X ¹ § 0.794 X · ¨ ¸ © 1 3X ¹ From the Vis al Basic chemical eq ilibri m program Ka 0.064608 X 0.1 3 1 1 º · ª § ¨ 3 X ¸ «( 0.206 X) » 1 3 X ¹ ¬ 1 3 X¼ © G e Ka ( 0.794 X) X f d( X) cC H X ( 0.799 X) 1 1 3 X 0.248 cC 0.316 1 3 X ( 3 X) cB ( 0.201 X) 1 3 X cB 0.257 H 0.427 1 3 X c) Since h drogen is so far abo e its critical point, onl a negligible amo nt of h drogen ill appear in the liq id phase,. PP 1 g e B ga B( B T) P a B ( T) 0.257 PP cC ga C( B T) P a C ( T) 0.32 PP B cC 1 f d( B cC PP) § 0.44 · ¨ 0.56 ¸ ¨ ¸ © 1.747 ¹ So the de point press re is 1.77 bar, and the liq id that forms has a mole fraction of ben ene of 0.448, a mole fraction of c clohe ane of 0.552, and (b ass mption), a negligible amo nt of h drogen. S C ca a d E T d 13.31 The che ca eac CH3-CHOH-CH3 = CH3-CO-CH3 + H2 A g e a h e ace e, he (g e he h gh e e a e a d e S ec e ace e h d ge 6 B X = .564, Ka aace aH a P 0.8043 c ,5 Chap e 13 d a ba a ce ab e e) ha ec e a a a 1-X (1-X)/(1+X) (1-X)u95.9/(1+X) )u100 X X/(1+X) Xu95.9/(1+X) )u100 X X/(1+X) Xu95.9/(1+X) )u100 1+X aI-P=0.2673, a d aace = aH =0.3458. The ef e 1 0 0 0.3458 u 0.3458 0.2673 ' a G ; RT ' G 0.4474 § ' G · ¨¨ ¸ RT ¸¹ © e 8.314 u 452.2 u 0.8043 3023.8 J 13.32 The eac ae C6H6 + H2 = 1,3-c c he ad e e C6H6 + 2H2 = c c he e e C6H6 + 3H2 = c c he a e The G bb f ee e e g ff J ' f G (be e e) 124.5 ' f G (c c he a e) 26.9 a da a eeded e h ' f G (c c he e e) 106.9 J be J ' f G (1, 3 c c he d e e) 178.97 J The G bb f ee e e g f f a f 1,3-c c he ad e e a a ab e IV, Pe The Che ca E g ee Ha db he Ha db f Che Ph c . The a e a f d g da a he WWW e h :// ebb . .g /che . Th Web e c a he Na a I S a da d a d Tech g (NIST) che da a b . The a e f d ea e J ' f H (1,3 c c he d e e, 298.15 K) 71.41 S (1,3 c c he d e e, 298.15 K) 197.3 S (C,g a h e, 298.15 K) S (H 2 , 298.15 K) 130.68 5.88 J J A e d a d e f h Web J K K K N e ha he e e e ae h e ec he e c e a d 0 K. A , he e cha ge f eac The ef e e a e f he a 0K e f a e eac . S C ca a d E T d a c ,5 Chap e 13 d ' f S (1,3 c c he d e e, 298.15 K) S (1, 3 c c he d e e, 298.15 K) 6 S (C, 298.15 K) 4 S (H 2 , 298.15 K) 197.3 6 5.88 4 130.68 360.75 ' f G (1,3 c c he d e e, 298.15 K) J K ' f H T 'f S 178967 J 178.97 71410 298.15 360.75 J The a ba a ce ab e a g a he ga c a e e e he d ha e, a d ha he h d ge ee g ea e ce ee a a e e f ed a 1 ba . A , ce a he ga c a e a, e a e he f a dea d e, a d ha he e h d ge he d ha e. Ma ba a ce ab e f d ha e: S ec e I O Be e e 1,3-c c he c c he e e c c he a e T a 1 0 0 0 1-X1-X2-X3 X1 X2 X3 1 The e ea b a1,3 c c Ka,1 abe aH 2 e Ka,2 ( 21973 . ) Ka,3 X2 1 X1 X 2 X 3 (7.1002) 1212.2 ac cha e X3 1 X1 X 2 X 3 abe aH3 2 e (39.374) 1-X1-X2-X3 X1 X2 X3 1-X1-X2-X3 X1 X2 X3 e F 178970 124500 I H 8.314 u 29815 K . e F 106900 124500 I H 8.314 u 29815 K . e F 26900 124500 I H 8.314 u 29815 K . 2.866 u 1010 ac che e abe aH2 2 e ae X1 1 X1 X 2 X 3 ac 12587 u 1017 . B e a g he a e f he e b c a he e e a , e ee ha X3 ab e a f he e e a he h d, e ha e 10 X 1 2.866 u 10 2.27 u 1027 | X 1 X 3 12587 . u 1017 a d b a g he a f he ec d f he e e a X2 1212.2 9.63 u 1015 | X 2 17 X 3 12587 u 10 . , e d ec b a . The b a g he a he h d Th gge ha X3 1, X2 f he de 10-17, a d X1 f de -27 10 . Th he be e e eac f e e a a c c he a e. 13.33 (a a a ab e a a Ma hcad hee ). g a f f he f S C ca a d E T G 2400 e R 8.31451 aA A J A AJ A aC A J C 1 Ka e K e T Chap e 13 d 298.15 K V 4 e 1 Ka 2.633 a D 0.5 10 Pa a C A 1 R T c ,5 e a D( P ) G a e aB A J C d P 5 10 Pa Part (a): A (initial g ess) 0.5 5 G e Ka C 1 A 0.16 a B a A F d A A 1 A A C 0.84 Part (b): Recogni ing that the partial molar Gibb's e cess is in the form of the one constant Marg les e pression ields: 2 0.3 C J A e G e J C e 1 A 0.132 2 5 a D 0.5 10 Pa a C Ka C 0.3 A a B a A A C 0.868 A e A e 0.3 A 0.3 1 2 A 2 A F d A S C ca a d E T d a c ,5 Chap e 13 d Part (c): Mass Balance Table: Species A B C D In 1 2 0 0 O t 1-X 2-X X X Ass ming that D is an ideal gas: P=(n*R*T)/V P( X ) e X R T a hee 13.3 (a a a ab e a Da Ma hcad C( X). 1 ) V F e he S ea TabKae P a = 12.349 Pa G F CHEMEQ Ka = 171.2 a B a A ( 1 X 1 ) HNH3 (f be a e e ) = 384.5 Pa/ e f ac a NA 1 X A 1 X NA 0.575 A 0.575 V (initial g ess) X 0.5 13.34 Need X The 0.425 X R T a e e P ( Pa) NB 0.10 0.25 0.50 1.00 2.00 4.00N B 6.00 8.00 10.00 12.00 14.00 16.00 18.00 18.40 18.50 19.00 20.00 25.00 30.00 35.00 40.00 50.00 2 X 1.575 X F d( X) X NC X 0.0899 0 0.2017 C X 0 0.3467 0 0.5478 0 0.7868 0 1.0322 N C 0.425 0 1.1670 0 1.2559 C 0.425 0 1.3204 0 1.3701 0 1.4099 0 1.4429 0 1.4707 0 1.4757 2.02u10-05 1.4797 0.0114 1.4996 0.0680 1.5339 0.1589 1.6437 0.3989 1.7063 0.5068 1.7488 0.5713 1.7806 0.6162 1.8267 0.6792 ND PD X X R T e V ND 0.425 PD 5 2.633 10 Pa S C ca a d E T d 60.00 70.00 80.00 90.00 100.00 104.50 a c ,5 Chap e 13 d 1.8602 1.8869 1.9093 1.9291 1.9472 1.9449 0.7261 0.7661 0.8030 0.8388 0.8745 0.8908 2.5 X and f 2.0 1.5 X, molar e tent of reaction f, fraction liq id 1.0 0.5 0.0 0 20 40 60 80 100 HCONH 3 R NH 3 CO , f h ch he Press re, kPa 13.34 ab e The che S ec e HCONH2 NH3 CO ca eac a 0 0 f a 1 X X T a A , P NRT V a NH3 a CO Ka a HCONH3 = 1-X X2 ba 3 1 8.314 u105 u K 25 1 X 1 X 8.314 u 105 T 25 u 10 3 ba 2 RT § X · ¨ ¸ 1 X V © 1+X ¹ 1 X § · ¨ ¸ © 1+X ¹ u 103 3 uT ba a ce e f ac (1-X)/(1+X) X/(1+X) X/(1+X) 1+X 1 X RT 1 X P V § NH3 P · § CO P · ¨ ¸¨ ¸ © 1 ba ¹ © 1 ba ¹ § HCONH3 P · ¨ ¸ © 1 ba ¹ a X 2 RT 1 X V S X2 1 X C ca a d E ', he e ' = T d ' X'; X 2 'X ' 0, a d X U g he che N ca e b Ka 7.8184u105 4.5987u105 2.8598u105 1.8660u105 1.2695u105 8.9571u104 c) I a PV RT be c e HCONH2 fa ba u 10-3 3 A d he T(K) P(ba ) e CO 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1.022 e a u 298.15 K ge a d 0.807 ce he e b e c f a ge . 3.022 u 8.315 u 105 u T 25 u 10 3 ha 480 4.82 500 5.02 NH3=0.331 CO=0.331 O2=0.071 N2=0.267 e P NRT V 400 4.02 P(ba ) 2.66 2.794 2.927 3.06 3.193 3.326 3 K The ef e, he e be 0.215 e f The c a a e ea e ( eac g c e ) 1.0 e NH3 1.0 e CO 0.215 e O2 0.807 e N2 NH3 0 0 0 0 0 0 1.013 ba u 25 8.314 u 10 5 Chap e 13 d ' ' 2 4' 2 ga c e K a, e b a a X 1 1 1 1 1 1 f c ,5 K a u 25 8.314 u 102 X2 T(K) 400 420 440 460 480 500 a 420 4.22 440 4.42 460 4.62 a ge ha he S C ca a d E T d a c ,5 Chap e 13 d 13.3 (a a a ab e a a Ma hcad hee ). F he S ea Tab e P a = 12.349 Pa F CHEMEQ Ka = 171.2 HNH3 (f be a e e ) = 384.5 Pa/ e f ac The a a e e P ( Pa) X 0.10 0.0899 0 0.25 0.2017 0 0.50 0.3467 0 1.00 0.5478 0 2.00 0.7868 0 4.00 1.0322 0 6.00 1.1670 0 8.00 1.2559 0 10.00 1.3204 0 12.00 1.3701 0 14.00 1.4099 0 16.00 1.4429 0 18.00 1.4707 0 18.40 1.4757 2.02u10-05 18.50 1.4797 0.0114 19.00 1.4996 0.0680 20.00 1.5339 0.1589 25.00 1.6437 0.3989 30.00 1.7063 0.5068 35.00 1.7488 0.5713 40.00 1.7806 0.6162 50.00 1.8267 0.6792 60.00 1.8602 0.7261 70.00 1.8869 0.7661 80.00 1.9093 0.8030 90.00 1.9291 0.8388 100.00 1.9472 0.8745 104.50 1.9449 0.8908 2.5 X and f 2.0 1.5 X, molar e tent of reaction f, fraction liq id 1.0 0.5 0.0 0 20 40 60 Press re, kPa 80 100 S C ca a d E T d a c ,5 Chap e 13 d 13.3 a) The eac 2H 2S CH 4 R CS2 4H 2 F he he che ca e b c a g a , e ha e he a e be e b c a (A , a e ace, I h e f he a e ca c a ed T(K) Ka X 773.15 863.15 923.15 1013.15 1073.15 4.7915u10-7 2.9923u10-5 3.0602u10-4 6.0722u10-3 3.3960u10-2 0.023 0.052 0.080 0.137 0.183 b) S ec e H2S CH4 CS2 H2 1 1 0 0 a CH4 CS2 H2 0.466 0.426 0.388 0.319 0.268 0.477 0.451 0.426 0.379 0.345 0.011 0.025 0.037 0.060 0.077 0.045 0.099 0.149 0.242 0.310 f a 1-2X 1-X X 4X T a ( e f ac ) (1-2X)/(2+2X) (1-X)/(2+2X) X/(2+2X) 4X/(2+2X) 2+2X A 1 ba Ka H2S f he a b) 4 4X X 2 2X 2 2X 4 44 X 5 2 64X5 2 1 2X 1 X 2 2X § 1 2X · 1 X ¨ ¸ 2 2X 2 2X © ¹ The f X a d he e f ac a eg e c) A 10 ba (a g he a ha e dea ) 2 1 2X 2 he ab e ab 1 X 2 2X 2 e. 4 Ka 5 4X X § 10 ba · ¨ ¸ 2 2X 2 2X 4 © 1 ba ¹ 2 3 § 1 2X · 1 X § 10 ba · ¨ ¸ ¨ ¸ © 2 2X ¹ 2 2X © 1 ba ¹ The e ae he ab e be T(K) X 773.15 863.15 923.15 1013.15 1073.15 9.611u10-3 0.021 0.033 0.059 0.082 100 64 X5 1 2X 2 1 X 2 2X H2S CH4 0.486 0.469 0.452 0.416 0.386 0.490 0.479 0.468 0.444 0.424 S he eac e ed b he h ghe LeCha e e c e a d he eac ch he a ha e). 2 CS2 e 4.76u10-3 0.010 0.016 0.028 0.038 e H2 0.019 0.042 0.065 0.112 0.151 e, a d be e ec ed f ( ha c ea e he be f e S C ca a d E T 13.3 (a a a ab e a a Ma hcad d a Chap e 13 c ,5 d P2 10 Pa hee ). Gi en: 298.15 K T1 5 10 Pa P1 e ' G C3H8 24300 ' H C3H8 104700 e e e 650 K T2 ' G CH4 50500 ' H CH4 74500 e e e e 8.31451 R 6 ' G C2H4 68500 ' H C2H4 52500 e K e e (at 298.15 K) e e Mass Balance Table: Species In O t C3H8 CH4 C2H4 Total 1 0 0 1-X X X 1+X (1-X)/(1+X) X/(1+X) X/(1+X) Calc lation of mole fractions and acti ities: C3H8 ( X ) 1 X 1 X CH4( X ) P aC3H8( X P ) C3H8( X ) aC2H4( X P ) C2H4( X ) X 1 CH4( X ) aCH4( X P ) 5 10 Pa C2H4( X ) X 10 Pa P 5 10 Pa ' G CH4 ' G C3H8 'G 4 4.23 10 e 'H ' H C2H4 ' H CH4 ' H C3H8 'H 4 8.27 10 e 'G e Ka 298.15 R T1 Part (a): X G e Ka 298.15 C3H8( Xa ) 1 10 4 3.885 10 1 e 1 e 8 (initial g ess) aC2H4( X P1 ) aCH4( X P1 ) aC3H8( X P1 ) C2H4( Xa ) 1.971 10 4 Xa F d( X) CH4( Xa ) 1.971 10 Xa 1.971 10 Xb 0.854 4 Part (b): From eq ation 9.1-22b: Ka 650 X Ka 298.15 e .5 'H R 1 T2 1 Ka 650 T1 2.704 (initial g ess) G e Ka 650 C3H8( Xb ) 0.079 X P ' G C2H4 Ka 298.15 X 1 5 'G aC2H4( X P1 ) aCH4( X P1 ) aC3H8( X P1 ) C2H4( Xb ) 0.461 Xb CH4( Xb ) F d( X) 0.461 e 4 S C ca a d E T d a c ,5 Chap e 13 d Part (c): X (initial g ess) .5 G e aC2H4( X P2 ) aCH4( X P2 ) Ka 650 C3H8( Xc ) Xc aC3H8( X P2 ) 0.369 C2H4( Xc ) 13.3 (a R 8.31451 aA A J A AJ A aC A J C 1 Ka e 0.316 CH4( Xc ) a a ab e a a Ma hcad e K e T V 4 e P 5 10 Pa Ka 2.633 a D 0.5 10 Pa a C A 1 RT 298.15 K 1 a D( P ) G 0.461 0.316 e aB A J C Xc hee ). 2400 G e F d( X) Part (a): A (initial g ess) 0.5 5 G e Ka C 1 A 0.16 a B a A A A 1 F d A A 0.84 C Part (b): Recogni ing that the partial molar Gibb's e cess is in the form of the one constant Marg les e pression ields: J A e 2 0.3 C J C e 0.3 A 5 G e a D 0.5 10 Pa a C Ka C 1 A 0.132 a B a A A C 0.868 A e 2 A e 0.3 A 0.3 1 2 A 2 A F d A S C ca a d E T d a c ,5 Chap e 13 d Part (c): Mass Balance Table: Species A B C D In 1 2 0 0 O t 1-X 2-X X X Ass ming that D is an ideal gas: P=(n*R*T)/V X P( X ) X R T V (initial g ess) 0.5 G e Ka X 0.425 NA 1 X A 1 X NA 0.575 A 0.575 aD X R T V a Ba A ( 1 NB NB 2 X 1.575 e a C( X 1 ) X X 1) F d( X) NC X ND X C X PD NC 0.425 ND 0.425 C 0.425 PD 5 2.633 10 Pa X R T e V S C 13.3 (a ca a d E T d a a a ab e a a Ma hcad c ,5 Chap e 13 d hee ). Witho t dissociation: a h drogen gas is eq al to the press re of h drogen gas: Ass ming it ofh molec A _Adthe acti bed_W K1lar H2 A _Ad K1 P H2 bed_W h With dissociation: A _Ad bed_W h K1 a H2 1 2 K3 a H Using the eq ilibri m constant for the reaction H2 = 2H, the acti it of atomic h drogen can be sol ed for in terms of the acti it of molec lar h drogen: 2 G e K2 aH F d aH a H2 Using the positi e root for the acti it of h drogen ields: A _Ad bed_W h K1 a H2 1 2 K3 K2 a H2 Ass ming the acti it of molec lar h drogen gas is eq al to the press re of h drogen gas: A _Ad bed_W h K1 P H2 1 2 K3 K2 P H2 If the amo nt adsorbed aries linearl ith the partial press re of molec lar h drogen then no dissociation is occ rring. If the amo nt adsorbed aries as the sq are root of the partial press re, then dissociation is occ rring. 13.40 F he da a he be G ' H T' S ' ' The e b c a § ' G · K e ¨ ¨ ¸¸ © R 298.15 ¹ The e b c a e ae h be ae e H ' e ha e ha G T' S 148 a 25 C e 19600 § · ¨ ¸ © 8.314 298.15 ¹ a af c f e 2716 ea e b a ed f E .13.1-22b; he S C ca a d E T d a c ,5 Chap e 13 d Eq ilibri m constant 5 1 10 4 1 10 3 1 10 K 100 10 260 280 f he eac The a ba a ce ab e f S ec e M e D e I a M 0 Ka aD 2 aM b h h ch ca be 2 eac F a M(1-2D) DM ea DM 1 2D 320 2A U A2 Th The e 300 T D M 2 1 2D 2 M e a 1 · § 4D 2 ¨ 4 ¸ 1 0 M K ¹ © a d ha he 2 D The 1 · 1 · § § ¨4 ¸ ¨4 ¸ 16 K ¹ M K M © ¹ © 8 f he f ac d e ed a a f c f e ea e g e be : S C 13.41 (a ca a d E H N2 N2 e a aN2 ga P N2 H N2 A P N2 ge Process #2: aN e a Chap e 13 d hee ). here A 2 Ka 1 H N2 N2(gas) =2*N(metal) 2 H N N e a aN2 ga 2 P N2 Ka 2 P N2 N ea % c ,5 Ka 1 P N2 N2 e a Ka 2 a N2(gas) = N2(metal) aN2 e a % d a a ab e a a Ma hcad Process #1: Ka 1 T HN ge B P N2 here B Ka 2 HN The empirical e pression gi en in the problem is s pported b process #2. 13.42 a) F he che ca e b a CO 2 P 1 ba CaCO3 R CaO CO2 Ka 1.1395 a CaO a CO 2 a CaCO3 The ef e, he e b a d e f ha a d ba e e a a ed N e e c a ed. The c a ga a 1150 K f he eac ce a CaO =a CaCO3 =1 a he a e e d . he c de 1.1395 ba f a CaCO3 ee , e f CO2 ha d be he ga ha e f 1.1395 1.1395 ba u 10 u10-3 3 0.119 e 3 5 ba 8.314 u10 1150 K K ha 0.801 e f d CaCO3 e a a 1150 K. PV RT b) If a 100 e e e ed a 1150 K, 1.19 e f CO2 c d be acc da ed a a e e f 1.1395 ba , h ch e ha a he ca c ca b a e d ha e d c a ed d ce 1 e f CO2 a d he e e d be S C NRT V P ca a d E 1 T u 8.314 u 10 5 d 3 ba 10 u10-3 3 K a c ,5 1150 K Chap e 13 d 0.965 ba 2H2Se g R 2H2 g Se2 The a ba a ce ab e S ec e a f a H2Se 1 1-2X 1-2X H2 0 2X 2X Se2 0 (X) (X) 13.43 The eac T a S ha K a 1 4X 4X 2 F each X 1000 1050 1100 1150 1200 1250 0.239 0.271 0.299 0.324 0.346 0.364 13.44 K a,1 S ec e N2 O2 N 2O NO NO2 a 0.79 0.21 0 0 0 S e f Se2 f ed e a 2H2Se f Se2 f a N 2O ; K a,2 0.5 2 0 e H2Se ha eac = X/2 a N 2 a O2 a 2NO ; a d K a,3 a N 2 a O2 f a 0.79-X-Y-0.5Z 0.21-0.5X-Y-Z X 2Y Z 4Y 2 0.79-X-Y-0.5Z 0.21-0.5X-Y-Z c 1 2X ed. a f he che a NO2 0.5 a N2 a O2 e f ac (0.79-X-Y-0.5Z)/(1-0.5X-0.5Z) (0.21-0.5X-Y-Z)/ (1-0.5X-0.5Z) X/ (1-0.5X-0.5Z) 2Y/ (1-0.5X-0.5Z) Z/ (1-0.5X-0.5Z) 1 1 1 1 a d K a,3 ca e b 1 X § ·§ 1-0.5X-0.5Z · 2 § 1 · 2 ¨ ¸¨ ¸ ¨ ¸ © 0.79-X-Y-0.5Z ¹© 0.21-0.5X-Y-Z ¹ © 1.013 ¹ 0.79-X-Y-0.5Z § 0.21-0.5X-Y-Z · 2 § 1.013 · 2 ¨ ¸ ¨ ¸ 1-0.5X-0.5Z © 1-0.5X-0.5Z ¹ © 1 ¹ a , e ba b 2 2X 1 0.119 0.135 0.150 0.162 0.173 0.182 X (1-0.5X-0.5Z) K a,2 E e a 2H2 a Se2 4X 2 4 K a 1 X 2 4K a X K a e f H2Se ha eac , T(K) K a,1 Ka 1 1 Z § 1-0.5X-0.5Z · 2 § 1 · 2 ¨ ¸ ¨ ¸ 0.21-0.5X-Y-Z © 0.79-X-Y-0.5Z ¹ © 1.013 ¹ c a ga S C ca a d E T d a c ,5 Chap e 13 d T(K) 1000 1200 1400 1600 1800 2000 Ka,1 6.7142u10-9 3.6675u10-8 1.2688u10-7 3.3069u10-7 7.1595u10-7 1.3675u10-6 Ka,2 7.5363u10-9 2.8345u10-7 3.7886u10-6 2.6496u10-5 1.2000u10-4 4.0052u10-4 Ka,3 1.0131u10-5 1.9428u10-5 3.1098u10-5 4.4434u10-5 5.8845u10-5 7.3931u10-5 T(K) 1000 1200 1400 1600 1800 2000 X 1.719u10-9 1.336u10-8 4.605u10-8 1.200u10-7 2.588u10-7 4.909u10-7 Y 1.986u10-5 1.084u10-4 1.002u10-4 1.044u10-3 2.216u10-3 4.026u10-3 Z 1.902u10-6 3.648u10-6 5.839u10-6 8.300u10-6 1.092u10-5 1.359u10-5 T(K) 1000 0.79 0.21 1.719u10-9 3.972u10-5 0.79 0.21 1.336u10-8 2.168u10-4 0.79 0.21 4.605u10-8 2.004u10-4 0.789 0.209 1.200u10-7 2.089u10-3 0.788 0.208 2.588u10-7 4.432u10-3 0.786 0.206 4.909u10-7 8.052u10-3 1200 1400 1600 1800 2000 N2 1.902u10-6 3.648u10 -6 5.839u10 -6 8.300u10 -6 1.092u10 -5 1.359u10 -5 N e ha a he e he a e ( de O2 N2O f ge (f e e ) a ge. efe ed NO a NOX) a e a NO2 a S C 13.4 (a ca a d E T d a a ab e a a Ma hcad a c ,5 Chap e 13 d hee ). Gi en: 'GAgC 108700 'GAg 77110 e 'GT C 186020 R 8.31451 e e 'GT 32450 e e K e 'GC 131170 e e e e e T 298.15K e Part (a): 'G AgC 'GAg 'GC 'GAgC 'G § 'G AgC · ¨ ¸ R T © ¹ KAgC KAgC e AgC 4 -1 5.464 u 10 e e 10 2.676 u 10 The sol bilit prod ct gi en in ill stration 9.3-2 is 1.607E-10. This e perimental al e is of the same order of magnit de as the theoretical al e calc lated abo e. Part (b): 'G T C 'GT 'GC 'GT C 'G § 'G T C · ¨ ¸ R T © ¹ KT C KT C e TC 4 2.24 u 10 -1 e e 4 1.19 u 10 The sol bilit prod ct gi en in ill stration 9.3-2 is 1.116E-2. This e perimental al e is t o orders of magnit de greater than the theoretical al e calc lated abo e. C2H4 H2O R C2H5OH 13.4 The eac N e ha e h e e ha a b g f 169.3 K a d a c ca e e a e f 282.4 K. The ef e, e h e e a ga a 298.15 K, h e a e a d e ha ae d.I a ,I a e ha a e a d e ha f a dea , ha he e ae e ha he a , a d eh e e he d. The c ec f he e a b ea . a C2H5OH C2 H5OH J C2 H5OH Ka 1 ba a C 2 H 4 a H 2O H 2 O J H 2O 1.013 ba a , a 298.15 K, f he P e ga a a PH O 0.03171 ba a d PC H OH 0.07889 ba 2 The S ,a 2 a ba a ce ab e S ec e Wa e C2H5OH T a d g dea 1 0 5 a beha , f a 1-X X 1 , 1-X X d e f ac S C ca a d E Ka C 2H5OH 1 ba H 2O 1.013 ba S he a e aef 0.9578 a d C2H5OH T 22.89 he H 2O d a c ,5 Chap e 13 d X 22.89 , a d X= 1 ba 22.89+1.013 1 X 1.013 ba d e f ac ae 0.0422. 0.9578 H e e, e ha e ha + ae e ae dea . U g he V a Ba c UNIFAC ga e a e he ac c eff c e g e JC2H5OH ( 0.9578) 1.0011 a d J H2O ( 0.0422) 2.4802 . The ef e, Ka X 1.0011 22.89 , 1 ba 1 X 2.4802 1.013 ba 22.89 a d X= 0.9824 0.0176 C2 H5OH a d H 2O 22.89+0.4089 Reca c a g he ac c eff c e h he UNIFAC ga g e JC2H5OH ( 0.9824) 1.0002 a d J H2O ( 0.0176) 2.5557 a d eca c a g he e b X=0.9830 C2H5OH a d H 2O h ch dea ea c c g e 0.0170 . ee gh he e e ha I : C2H5OH 0.9578 a d H2O 0.0422. : C2 H5OH =0.9830 a d c a e e a e: 0.0170 H 2O N eed c ec f e ha a d ae he a PC 2 H5OH 0.9830 u 1.0002 u 0.07889 0.0776 ba PH 2O e a e aga . S f ha e. The a a 0.017 u 2.5557 u 0.03171 0.00138 ba S he a a e e f e h e e 1 - 0.0776 - 0.00138 = 0.9210 ba a d ac 0.9210 a C2 H 4 0.9092 ( 1/1.013 a ha fa bee a ed). 1.013 S eca c a g he e b ( g he a e f he e b c a f a ea ) e ha e X 1.0002 X K a 22.89 0.4360 , 1 ba 1 X 1 X 2.5557 u 0.9092 1.013 ba 22.89 a d X= 0.9811 C2H5OH a d H 2O 0.0189. 22.89+0.44 A h e e d ffe e f he e ea ,I e a e aga . 13.4 F he che ca e K. The a ba a ce ab e S ec e a f a SO2 1 1-X O2 0.5 0.5-0.5X SO3 0 X T a (1.5-0.5X) b c a ca c a ( e f ac ) (1-X)/(1.5-0.5X) (0.5-0.5X)/(1.5-0.5X) X/(1.5-0.5X) he g a , Ka = 4.5905 a 1000 S C ca a d E T d a c ,5 Chap e 13 d X P ISO3 1.5 0.5X 1.013 Ka 0.5 0.5 a SO2 a O 1 X P § 0.5 0.5X P · 2 ISO2 ¨ IO 2 ¸ 1.5 0.5X 1.013 © 1.5 0.5X 1.013 ¹ he e he I e a e f gac c eff c e . The e a be ed f XISO3 1 4.5905 0.5 1 X ISO2 § 0.5 0.5X P · I ¨ O ¸ © 1.5 0.5X 1.013 2 ¹ a) A P = 1.013 ba , e ca a e ha he a e f a f gac , ha a SO3 4.5905 X § 1.5 0.5X · ¨ ¸ 1 X © 0.5 0.5X ¹ 0.304, SO2 0.5 X 0.152 a d O2 a aa dba e c eff c e ae 0.642 SO3 =0.544. b) A P=101.3 ba , he dea ga ha e beha ( ha , he f gac c eff c e ca be eg ec ed. The ced e I e f e he f gac c eff c e ca c a e he e b c , e he e c he Pe g-R b e a f ae ca c a e he f gac c eff c e , eca c a e he e b c h e f gac c eff c e , a d c e ea g c e ge ce ach e ed. · X § 1.5 0.5X ¨¨ ¸¸ 1 X © 100 0.5 0.5X ¹ 4.5905 0.089, SO2 0.045 a d O2 0.5 X 0.907 SO3 =0.866. A g ha he b a e ac a a e e f he e f gac c eff c e a e ISO2 1.0056, IO2 1.0405 a d ISO3 =1.0090. The ef e KI ISO3 1.0090 0.5 ISO2 IO 2 The X 0.908, 1.0056 u 1.0405 he SO2 S he cha ge f e e a e be g 0.088, O2 he dea ga h gh, ha 0.5 ga c ac d a e , he 0.984 0.048 a d ha e ca e SO3 =0.867. h ea e g. a , a ge a a e f he ) , h S C 13.4 ca a d E (a T d a a ab e a a Ma hcad a c ,5 Chap e 13 d hee ). Pa a 'G R T ( Ka ) 57.33 d d T R T 0.17677 'H R R T 2 57.33 ' H R T 2 R T 2 57.33 'H J e Part (b): 'G ( T) Ka( T ) e e 57330 'G 176.77 e e T eK R In 1 0 1 O t 1-X 2*X 1+X 8.31451 K e e ( T) R T Mass Balance Table: Species N2O4 NO2 Total Calc lation of mole fractions and acti ities: N2O4( X ) a N2O4( X P ) KaX( X P ) 1 X 1 X NO2( X ) N2O4( X ) a NO2( X P ) 2 a N2O4( X P ) P 5 10 Pa a NO2( X P ) 2X 1 X NO2( X ) P 5 10 Pa S C ca a d E T Pa b X 0.7 (initial g ess) G e Ka( T ) KaX( X P ) Xb 0.1 4 X 323.15 K 10 Pa d a c ,5 Chap e 13 d X( T P ) Xb 1 F d( X) 5 X 323.15 K 10 Pa Xb 10 6 X 323.15 K 10 Pa NO2 Xb 0.1 0.91 NO2 Xb 1 0.605 NO2 Xb 10 0.261 N2O4 Xb 0.1 0.09 N2O4 Xb 1 0.395 N2O4 Xb 10 0.739 Part (c): Xc 0.1 4 X 473.15 K 10 Pa NO2 Xc 0.1 13.4 N2O4 Xc 0.1 1 1.246 10 4 Xc 1 5 X 473.15 K 10 Pa NO2 Xc 1 0.999 N2O4 Xc 1 1.243 10 3 Xc 10 6 X 473.15 K 10 Pa NO2 Xc 10 0.988 N2O4 Xc 10 0.012 S C ca a d E T d a c ,5 Chap e 13 d 13.49 Gi en: Ta e 58620 'H 138.2 'S e 298.15 K Tb 373.15 K K e R e 4 10 Pa P1 P2 8.31451 5 10 Pa K P3 e e 6 10 Pa Mass Balance Table: Species In O t M D Total 2 0 2-2*X X 2-X (2-2*X)/(2-X) X/(2-X) Acti ities, Eq ilibri m Constant, and Eq ilibri m E pression: a M( X P ) 'G ( T) KaX( X P ) X 0.999 2 2 2X X P a D( X P ) 5 10 Pa T ' S 'H Ka( T ) X P 5 2 X 10 Pa e 'G ( T) R T a D( X P ) a M( X P ) 2 (initial g ess for sol er) G e KaX( X P ) Ka( T ) Part (a): Deg eeOfD e a ( Ta P1 ) 0.953 Deg eeOfD e a ( Ta P2 ) 0.985 Deg eeOfD e a ( Ta P3 ) 0.995 X 1 Deg eeOfD e a ( T P) F d( X) S C 13. 0 (a ca a d E T d a a a ab e a a Ma hcad c ,5 Chap e 13 d hee ). Gi en (All Units are SI): T1 267 T2 255 6 2 10 P1 CH4 Acti ities: ah d ae 1 a H2O 1 6 1.5 10 P2 CH4 R 8.31451 P CH4 a CH4 P CH4 5 10 Part (a): ah d a e Ka P CH4 5.75 a H2O 'G a CH4 P CH4 'G T1 P1 CH4 3 6.65 10 'G T2 P2 CH4 3 5.742 10 R T T P CH4 Ka P CH4 Part (b): 3 'H 10 G e a 'G F d 'H 'H (initial g esses) 'S 10 T1 P1 CH4 'H 'S T1 ' S 'H 4 1.357 10 'S a 'G 0 T2 P2 CH4 'S a 1 75.737 Part (c): Ka 273 P273 CH4 G e P273 CH4 'H Ka P1 CH4 e 6 10 R 1 1 273 T1 Ka 273 0.044 (initial g ess) Ka P273 CH4 6 2.288 10 Ka 273 P273 CH4 F d P273 CH4 'H T2 ' S S C ca a d E 13. 1 The eac The T d a c ,5 Chap e 13 d ae ea a B1 B ba a ce ec e A a NA be e O ea a f a e a d A1 B1 A B ¦ N ¦¦ N A f A1 ba a ce NB A e A1 A F I ¦ GH N ¦ N JK AB A AB ec e B be f B1 f ¦ N ¦¦ N B F I ¦ GH N ¦ N JK AB B AB b 0; GA GA1 GB GB1 c 0 ; GA B G A1 GB1 h 0; a d G A e b G a ¦ N A G A ¦ N B GB ¦ ¦ N A B G A B a h e ec 0 each e e f eac . ¦ GA dNA ¦ GB dNB ¦ ¦ GA B dNA B dGtotal ¦ N A dGA ¦ N B dGB ¦ ¦ N A B dGA B 0 0 b the GibbsD hem eq ation A , G g he e a 0 0 b e a ¦ G A1 N A ¦ GB1 N B ¦ ¦ b G A1 GB1g N A B G A1 ¦ N A GB1 ¦ N B ¦ G A1¦ N A B ¦ GB1 ¦ N A B LM F N ¦ N I OP G LM¦ e N ¦ N jOP JK Q N N GH Q G A1 ¦ A AB B1 GA . A FG w G IJ H wN K AB G A1 N A GB1 N B FG w G IJ H wN K a A N BT , P { GA a B N A, T, P { GB GB S C ca a d E T G A1 The ef e A d eg a c ,5 Chap e 13 d GA , e ha e ha , b def Th , a GB1 a d FG w G IJ H w PK g be ee a RT T FG w IJ H wP K a e , e ha e GB (1) (2) T a f a f RT aaTT,, PP ff 2 G T , P2 G T , P1 N g E a f ed a a T a d P . (3) h E e ha A T, P A e a e e ha E . (1) . (1), a d ec g A1 T , P A1 P b A P a 0 a d Po0 A1 P 0 A1 P ha 13. 2 A P ee be B1 T , P e e ha A P be A1 . (1) P 0 a d he A P 0 g ha E B T, P e c d eg a e E . (2) be ee be a f ed a a T a d P. Th A1 P (3) 1 ePa d 0 (LeCha e e c . A1 A1 A2 A1 A2 A3 # A1 A A 1 e c. T a f A1 N0 NT NT T ¦ he e N T a ¦N be f e a1 d¦ i ¦ ¦ N a h ch N1 2 N 2 3 N 3 " A " N ¦ N N0 NT e a ¦¦ e ha ¦ 1 aa 1 ¦ a d ¦¦ a1 2 e e) S (a) C a ca a d E a1 T FG N IJ a d HN K 2 0 1 T A K 1 d a c ,5 Chap e 13 d FG N IJ HN K K 0 T a f a P 1 ba f a 1 I 1 1 P 1 ba a a1 I I1 1 I 1 1 1 ba I I1 1 P 2 The I I1P I 1 F he e e 1K a S f he 1K 1 e ha I I1P I 1 de e de I 1 I I1P 1 1 1K de D a d he 2 1D K 2 1 2D K 3 4 3 4 1 DK a D Kf 1 h 2 ha h deed he ca e. The a D Kf D K 2 1 1 a f 2 3 1 D K a a f # aD Kf 1 1 # e c. NT The N0 ¦ A N f he N0 ¦ KD 1 1 . ¦ 1 e e f ge f 1 1 T ¦T 0 ¦ KD 1 1 1 e c e ha e f ¦ KD 1 1 1 f 1 ¦ a KD 1f 1 1 f 1 ¦ a KD 1f 0 1 1 KD 1 1 (*) S C ca a d E T d a c ,5 Chap e 13 d ha 1 1 KD 1 1 1 ; KD 1 1 KD 1 1 A FG ¦ T IJ K TH f F 1I T H1T K f ¦ T 1 0 0 f 1 1 T 2 ¦ T 1 1 ha f f ¦ KD 1 1 1 1 ¦ a KD 1f 1 1 1 1 F 1 NT N0 1 he a de Waa e 0 1 2 1 FG IJ H 1 KD K 2 FG N IJ HN K 2 1 1 E . (*) ab e a d 1 a FG N IJ HN K 0 T 1 1; a a1 0 T 1 1 12 2 1 a1 f a e e ha e RT a 2 V V P ¦ a KD 1f 1 NT N0 a1 KD f f f 1 NRT N 2a VN V a d I B ZB B1 ZB ZB h ch he e bec I 2¦ a RTV e I ha ZB 2¦ a1 B1 2a ZB ¦ RTV ZB RTV B 2 a1 N 0 ZB ZB RTV N T S C ca a d E T I I1 I 1 d a I 1 de e de RT V 2a1 N 0 RTV N T ZB 1 ZB RT V 1 N0 NT a 1 ZB 1 RT P V N T RT N TV N 0 1 f h ch de e de f he de f he de . We ca e a D a a N T RT NTV N0 1 1 a dD 1 1 DK B1 1 1 1 B 1 I I1 I 1 PI I1 Chap e 13 d I I1 I 1 1 (b) c ,5 ed. N ha RT V N0 NT a 1 1 1 2 1 1 2 1 1 1 1 1 1 1 aV RTK f r aV RTK f 4 V 1 1 RT a 2 V V 2 RT a 1 21 V 1 V 1 2 2 1 1 21 ha P a RT a 12 V 1 V a de Waa e a . 1 f a e f he h ch de e d aV f r dV 4 V i aV f aV f 1 1 2 (***) E a (**) a d (***) a e he e h ch f he e a a c a g a de Waa f d. N ce ha e f V e eed 1 ge c b c. V ; he ef e, he e a N e ha f he f d -a c a g, he K 0 h he 1 21 (**) A h ch 1 1 1 (c) P RT V 1 f FG1 RTK IJ 1 aV RTK f V H V K aV RTK f V 0 aV RTK f V 0 1 ha D e 1 21 1 S C ca a d E T d a c ,5 Chap e 13 d 13. 3 The de c f HF c a g e de c bed he a c e C ec f Pha e E b Da a f Se a a Tech g b W a Sch e I . E C .P D . D . (1980), 19, 432 439. B a ca ef e a a f he de a d he da a, he ed ha HF a c a e he a ha e acc d g he eac . 2HF HF 2 6HF HF 6 a d, e he e 8HF HF 8 e a ge f 195 240 K, he e ea K2 2 2 1 K6 6 6 1 K8 8 8 1 b c a ae LM 6429.4 24.1456OP N T Q 21100.9 O e LM N T 69.7292PQ 25224.5 e LM 83.4689OP N T Q e 1 f a ,a d he f gac f ec e . Ne , Sch e ed a ( e b Ta a d W a [C . E . S (1978), 33, 651] P f he I e , d e , he a e a d ha he f gac c eff c e ca e aea a a e e a , a d ca be ca c a ed f he f gac f e HF. Th a e a ed he eed ec f he ec a a a e e f he a c a c e e . U g h de , Sch e b a ed e g d ag ee e f he a c a fac (de ) f e HF a d a HF-F e e g a e e a f a e ch a Pe gR b ( h ch be ed e a e ce che ca e b e ed he ha e e b ca c a ). A a e a e, ead f g he Ta -W a a f e a f f gac c eff c e e he de P b e 9.52 he e K ha e ab e) a g a d he ea HF a a che HF 8 . S a , HF + e : -a ca gc -a ca a 2 ca eac gc a1 a d 1 e e h HF, HF 2 , HF 6 a d d be ea ed a a f e c e e + HF, HF 2 , HF 6 a d HF 8 . I each f he e ca e he c f he HF c e cha ge a he e b cha ged. The be h h a ha he a a a e e f he a c a c e e bec e ea ca a ge. F e a e, a8 64 u a1 . C e e , c ee he e ca -c ec de f he HF a c a g e e , h gh he de ea e bab a fac f e g ee g ca c a . S C 13. 4 Sa af c 13. ca a d E KW KW gf § K T2 ¨¨ © K T1 a a ge f e fa C H 1) c ,5 Chap e 13 d 7.085 25 7.0 30 6.915 40 6.77 50 6.63 60 6.51 g ha ' 1 KW 2 ' H ( J/ e, a d ha 20 º H e ha e ha H ) 59.22 58.79 56.89 58.83 52.70 54.24 49.46 K W o H OH BOH o B OH a d H 2 O m ªM - º ¬ OH ¼ W ª¬ M H+ º¼ W ª¬ MOH- º¼ W ` ª¬M º J 2r OH- ¼ W M H+ ^M h ha º BOH ª ¬ MOH- ¼ W ` ª¬M º OH- ¼ W J r2 ha M H E ea ed ea a H§ 1 1 · ¸ a da ¨ R © T2 T1 ¹ 7.27 OH- ¼ BOH M OH a ' 10 ªM º ¬ OH ¼ W 13. H 7.47 de he eac S · ¸¸ ¹ d T( C) 0 M H M OH J r2 b ^ª¬M T a e M BOH M 2BOH 4K W / J r2 M 2BOH 4K W / J r2 M BOH 2 2 ª º ¬ M OH ¼ W M BOH KW M 2BOH 4K W / J r2 M BOH 2K W M OH J 2r M 2BOH 4K W / J r2 M BOH J r2 (15.1-10) e aced b K a,W , e aced b MH J r . B g M H J r , e cha ged. 1 2 ce aga ha e H | KW M BOH J 2r M 2H J 2r M HA, M 1 he e e MH ha h he def 1 1 K a,HA g M HA, 2 2 ha , e . (15.1- S C ca a d E 13. The eac T d a c ,5 Chap e 13 d ae K a ,HA K a ,W o H A a d H 2O m o H OH HA m M H J r M A J r h K a,HA M HA b The ef e M H J r M OH J r a d K a,W M HA, M H a d M A K a,HA ha M HA M 1 M 1 2 M H M 2H J r2 M HA, M H J r M 1 he e e MH a ea ed e ½° Ka,HA ­° 4M HA, 1¾ a d ® 1 2 °¯ Ka,HA °¿ e . 15.1-10 cha ged. M H J r 13. Wea ba e + f ae g ac d, eg ec e aced b MH J r . , g H g M H J r g H+ a d OH- d ced f he a K BOH ZZZZ X B OH HA o H A a d BOH YZZZZ h K a,BOH MBOH M B MOH M BOH M 1 U g D be he MBOH, D; MB+ ( a f BOH d c a ed, D; a d MOH- D MHA ce each e f OH- f ed b + fH f ed f he d ca D MHA D K a,BOH MBOH, D M 1 he d ca f he ac d f ( h ch e e e cha ge f BOH (ab e I a e ca ed a e e a h HA.) S K a,BOH M HA D K a,BOH M HA 2 15.1-7) e ce e f e he 4K a,BOH M BOH, 2 M OH D M HA K a,BOH M HA K a,BOH M HA M H KW M OH The ef e 4K a,BOH M BOH, 2K W K a,BOH M HA 2 4K a,BOH M BOH, K a,BOH M HA Add g e ec e dea M B J r M OH J r K a,BOH K a,BOH M BOH M 1 J r2 ha e e 2 2 a d 13. f BOH eac ha a e .) S ha he e e ace K a,BOH M B M OH M BOH M 1 h K a,BOH / J r2 S D C ca a d E §K · ¨¨ a,BOH M HA ¸¸ 2 © Jr ¹ M OH T d 13. 0 2 § K a,BOH · K M HA ¸¸ 4 a,BOH M BOH, ¨¨ 2 J 2r © Jr ¹ 2 KW M OH J 2r 2K W 2 §K · §K · K M HA ¸¸ J 2r ¨¨ a,BOH M HA ¸¸ 4 a,BOH M BOH, ¨¨ a,BOH 2 2 2 Jr © Jr ¹ © Jr ¹ Add g e ec e K a,1M HA M A ; J r M A 2 M H J 2r ST S ST S S 13. 1 The e Chap e 13 d 2 a d M H c ,5 § K a,BOH · K M HA ¸¸ 4 a,BOH M BOH, ¨¨ 2 J 2r © Jr ¹ 2 §K · ¨¨ a,BOH M HA ¸¸ 2 © Jr ¹ D M HA a K a,1M HA M H J 2r dea K a,2 M A J r M H J r K a,1K a,2 M HA M H 2 3 Jr ha M A2 K a,1Ka,2 M HA MH J 3r ª º K K K ... « a,1 a,1 a,22 ...» M HA «M »N M H ¬« H ¼» S ª K º K1 K 2 10 1 10 « .... « H ....»» 2 3 H 2 M H J r2 10 J r M H J r 10 Jr «¬ »¼ K a,1 b K a,1K a,2 ea ae H NH3 CH 2 COO NH3 CH 2COO H NH 2CH 2COO K1 = ; K2 = ; KD = NH3 CH2COOH NH3 CH 2 COO NH 2CH 2COOH The ef e S C ca a d E T d a c ,5 Chap e 13 d NH 3 CH 2COO K D NH 2CH 2COOH=NH 2CH 2COOH 10 K D NH 3 CH 2COOH H NH 3 CH 2COO K1 H NH 2 CH 2 COOH 10 KD K1 NH 2 CH 2 COO K 2 NH3 CH 2 COO H NH 2 CH 2COOH 10 H K 2 K D NH 2 CH 2 COOH NH 2 CH 2 COOH NH 2 CH 2 COOH NH CH 2 COO NH 3 CH 2COOH NH 2CH 2COO 3 1 1 10 KD 10 K1 H K D 10 H K2 KD 10 K D NH 3 CH 2 COO NH 3 CH 2 COOH NH 2 CH 2 COO A H=3 NH 2 CH 2 COOH A 1 10 K D 10 K1 H K D 10 H K 2 K D 10 K1 H KD 1 10 KD 10 K1 H K D 10 H K2 KD 10 H K2 KD 1 10 KD 10 K1 H KD 10 H K2 KD | 0; NH 3 CH 2COO 0.818; NH3 CH 2 COOH ; 0.182; NH 2 CH 2 COO |0 H=10 NH 2 CH 2 COOH NH 2CH 2COOH 10 K1 H K D | 0; NH 3 CH 2COO 0424; NH 3 CH 2COOH | 0; NH 2CH 2 COO | 0.576 S C ca a d E 13. 2 T d a c ,5 Chap e 13 d a) The eac NH 3 + 2O 2 + (7.5238N 2 ) o HNO 3 + H 2O+ (7.5238N 2 ) The e gh a e HNO 3 =63.01; H 2O= 18.01; NH 3 =17.03; H 2O=18.01 ec a S 100 g f a 60 ae, HNO3 = The e % f HNO 3 c he f g g a e f ac 0.9522 e he d c 0.9522 =0.0979, H 2 O =0.1285 a d 0.9522 1.25 7.5238 he feed a e e f ac N2 f HNO3 a d 1.25 c d g S f ge : 0.773576 1 =0.0950 O2 =0.1900 a d N 2 0.71493 1 2 7.5238 ; 'Gf (H 2O, ) 237.1 J/ ; 'Gf (HNO 3 , ) 'Gf (NH3 ,g) 16.5 J/ NH3 e = 'G = 111.3 237.1 ( 16.5) Th he a a S effec he f d c 331.9 J/ ha c e 111.3 J/ a eac ed, fa d be b a ed h c de g effec . e NH 3 eac ed 'GP = -RT (1 0.0979+1 0.1285+7.7238 0.7736)=1.710 J/ S effec he feed 'Gf = -RT (1 0.0950+2 0.1900+7.7238 0.71493)=2.714 J/ The S e a 'G f he a ce ha 50 g=64.7 he a ce ='G f 'GP 'Gf ha ca be effec a e eg g b e c =0.446 MPa A 50 g=64.7 a =0.446 MPa, I 331.9 1.7 2.7 d ced a ed a 25 C H =104.89 J/ g a d S =0.3674 J/ g-K 332.9 J/ he f ee e e g T 147.9 C, H =2743.9 J/ g a d S =6.8565 J/ g-K S ce 1 = 2000 b =4410 g, J/ g 332.9 J/ NH 3 NH 3 eac ed. A f eac . G=104.89-0.3674 u 298.15=-4.65 e 0.45 MPa 'G f ea = -143.03-(-4.65)=-138.4 J/ g NH 3 ce a e ca be ead f ea ab e G =2743.9-6.8586 u (147.9+273.15)=-143 S C ca a d E 13. 3 See f 13. 4 g T d a c ,5 c ce a Chap e 13 d be S gf he h d ge a he ec e b gf a de a g he c ce a f e ge 1 § ·2 K K a1 ª¬ H º¼ ¨10 14 PCO2 ¸ KH © ¹ h ch ead PCO2 0 H=7 PCO2 2.7 u 104 H=4.233 PCO2 3.85 u 104 H=4.156 PCO2 5.5 u 104 H=4.079 13. Pa a Ma bala ce able S b a ce I CH3OH F a M e f ac 1 1-X (1-X)/(1+X) CH2O 0 X X/(1+X) H2 0 X X/(1+X) T a 1 1+X Ka aCH 2O aH 2 CH 2 O H 2 aCH3OH Ka 1 X X a 2 P / 1 ba X2 1 X 1 X CH 3OH 2 X2 ; 1 X 2 (1 K a ) X 2 ; X ; Ka Ka 1 Ka ' G ( J/ ) ' H ( J/ CH 2O -102.5 -108.6 H2 0 0 CH 3OH -162.0 -200.7 Reac 59.5 92.1 K 298.15 K e § ' G· ¨ ¸ e © RT ¹ ) 59000 § · ¨ ¸ e ( 24.0034); K © 8.314 u 298.15 ¹ 3.7625 u1011 S C ca a d E Pa c Ma T bala ce able S b a ce CH3OH CH2O H2 N2 T a Ka aCH2 O aH2 I a 1 0 0 0.5 1.5 CH 2 O H 2 aCH3OH P / 1 ba CH3OH K a 1.5 0.5 X X 2 2 S B e a c ,5 Chap e 13 d F a 1-X X X 0.5 1.5+X M e f ac (1-X)/(1.5+X) X/(1.5+X) X/(1.5+X) 0.5/(1.5+X) X2 1.5 X 1 X X2 ; 1.5 0.5 X X 2 X 2 ; 1.5K a 0.5 K a X K a X 2 (1 K a ) X 2 0.5 K a 1.5 K a 13. 13. d X 2; 0 he a e a P b e 13.39 , h he a e a be h ge e ac g h d ge . 13.38, ee he ha be . S C ca a d E a) Reac ac fBa dDa e § 'G · ¨ ¸ © RT ¹ Th ha he a c ,5 Chap e 13 d A he e a 0.5 CJ C J A c) U e N A S ec e § P · aC ¨ D ¸ © 1 ba ¹ aA aC aD aA aB ha e a e 0.5 C 0.5 1 A A A ed e ee he =0.166 a d C =0.834. he be d ha e a a f e e be c each e f A eac ed. D M e a NA NA N B NB NB N C NA N D N S he e b K aC aD aA NA N f A added a a . N e ha he a h g he c e f a A be , e ha e he f g e f ac Pa a be uN / NA ed u R uT / V NA N , h e a be ed f >Cha >Cha >Cha >Cha >Cha >Cha >Cha >Cha e 13>P e 13>P e 13>P e 13>P e 13>P e 13>P e 13>P e 13>P N . 13. 13. h h h h h h h h be be be be be be be be f f f f f f f f de A de A de A de A de A de A de A de A e e e e e e e e f f f f f f f f Te Te Te Te Te Te Te Te b b b b b b b b e / NA N ea f e f he eac N e be dC NA N ce a e f d / NA A N A , R, T a d V a e a 13. S 13. 0 S 13. 1 S 13. 2 S 13. 3 S 13. 4 S 13. S 13. S e, he =0.195 a d C =0.805 be Th ha he d ced f d a d ga e 0.5 C e (0.3(1 C ) 2 ) (1 C ) e (0.3 C2 ) A A e g he . The ef e 2.058 K K d A( )+B( ) o C( )+D(g). S a e b) N T be be be be be be be be 13.69 13.70 13.71 13.72 13.73 13.74 13.75 13.76 u R uT / V e S 13. C ca a d E T d a c ,5 d Chap e 13 S C 13. 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F K M AH 2 = M AH M H 2 M AH 2 M H M AH 2 , 2 H= K a1 K a 2 b) He e he e a e M A M H K a2 M A M H 2 ha a he ha c H= K a1 K a 2 2 de K K a1 a2 a3 AH3 o AH 2 H , AH 2 o AH H , a d AH o A H f h ch he e K a1 b M AH M H 2 M AH ea ae M AH M H , K a2 M AH 3 he ab e 3 e ec c H M AH M H M AH M 2 H M AH M AH e e K a1 K a2 M A ec e = 2M AH M AH M A . The ef e, a he 3 M AH 2 M AH K a3 a d 2 M H M AH 2 The a e age e cha ge a M A MH a d K a3 M H K a2 3 K a1 K a2 + . S M H Ka 2 2M AH M AH , 3 e ec c N e ha d ffe e , b h ch cha ge ec e The ec d b e e ha 3 M AH + M H M AH K a1 K a2 3 M AH M H , 3 +K a1 M H 2 = M H 3§ K · ¨ 2 a1 ¸ ¨ M H ¸¹ © 3 g M H K a1 K a 2 K a3 g 2 10 H K a1 ae ,e a d be b a ed de e d g a ed g he cha ge e a c d . c de 3 H 2 M AH 2 M AH 3 K a1 K a2 K a3 =2 M H § K · K a1 K a 2 K a3 g ¨ 2 a1 ¸ ¨ M H ¸¹ © 2M AH M H K a3 2 M AH 2 § K · g K a1 g K a2 g K a3 g ¨ 2 a1 ¸ ¨ M H ¸¹ © ha 2 M AH M H 2 ha a he 2 S C ca a d E T d K a c ,5 Chap e 13 d K K a1 a2 a3 AH3 o AH 2 H , AH 2 o AH H , a d AH o A H f h ch he e b M AH M H K a1 2 M AH ea , K a2 M AH M H 3 The a e age e cha ge The ef e, a he ha H K a3 ae M AH he MA M AH M H M M 3 - AH H 2M AH 2M AH K a3 a d ha a he M H M AH 3 1 M AH M AH , 3 2 M AH M H M 3 - H 2M AH 2 2 M AH M H 2 M AH ha 3 M H M AH K a1 K a2 3 , M H M H 3 1 1 K a1 K a2 K a3 = M H - K a1 K a 2 M H = 2 2 2 - 2 § · 1 K K g K a1 g K a2 g K a3 g ¨ a1 a22 ¸ ¨2 ¸ ¨ 2 M H ¸ © ¹ § · 1 K K K a1 K a 2 K a3 g ¨ a1 a22 ¸ ¨2 ¸ ¨ 2 M H ¸ © ¹ 2 M AH e ec c 3 2K a1 K a2 M A M H ec e = M AH M AH 2M A . 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(a) At the bubble point we have xi Pi vap P where P yi 5 bar xET 0.05 ; xP 010 . , xNB 0.40 and xMP 0.45 . Procedure used was i) Guess T, ii) Compute each yi , and the sum ¦ yi . If ¦ yi 1 , guessed T is correct; if ¦ yi ! 1 , guessed T is too high; if ¦ yi 1 , guessed T is too low. If ¦ yi z 1 , we correct T and K (bubble point) yET 0.4167 , yP 01730 , recalculate. Solution: T 29366 . . and yMP 0.2502 . yNB 01601 . (b) The dew point calculation is similar. Here, yET 0.05 , yP 010 . , yNB 0.40 and yMP 0.45 . P 5 bar, and T and the xi 's are the unknowns. Thus, here we guess the dew point temperature, compute each of the xi 's from xi Pyi Pi vap and evaluate ¦ xi . If ¦ xi guessed temperature is the dew point temperature; if 1 , the ¦ xi ! 1 , guessed T is too low; if ¦ xi 1 , guessed T is too high. Solution (obtained using the computer) T 314.23 K (dew and xMP 0.4409 . point) xET 0.0039 , xP 0.0337 , xNB 05215 . (c) The advantage of the Mathcad worksheet for the isothermal flash calculation is that one can use the initial flash equations directly, rather than having to make the substitutions below. For the isothermal flash vaporization calculation, we proceed as in Illustration 8.1-3. First, we calculate the K factors, i.e. . 4.402229 10 817.08 30315 , KET 10185 . 5 and, similarly KP 2.238 , KNB 0546 and KMP 0.743 . Thus, the equations to be solved . are: (1) xET xP xNB xMP 1 (2) yET yP yNB yMP 1 10185 . xET 2.238 xP 0546 . xNB 0.743xMP 1 Also, xET L 1 KET KET 0.05 xET 10185 . 9185 . L 0.05 (3) and, similarly (4) xP 2.238 1238 . L 010 . (5) xNB 0546 . 0.454 L 0.40 (6) xMP 0.743 0.257 L 0.45 Solution procedure I used was to guess L, compute the xi 's from eqns. (3 to 6), and then ascertain whether eqns. (1) and (2) were satisfied. After a number of iterations, I obtained the following solution: a a f f Solutions to Chemical and Engineering Thermodynamics, th ed L xET xP xNB xMP ¦ xi 086667 . 0.0225 0.0858 0.4258 0.4659 . 1000 Section 10.1 V 013333 . 0.2289 01921 . 0.2326 0.3464 1000 . yET yP yNB yMP ¦ yi (d) For an adiabatic flush vaporization, shown below, the energy balance must also be satisfied liquid liquid X vapor pressure reducing valve or device This is a (two-phase) Joule-Thomson expansion, so that the energy balance yields H in H out , or ¦ xi H i T , P, x inlet conditions L ¦ xi H i T , P, x L L c V ¦ yi H i T , P, y V h outlet liquid conditions outlet vapor conditions This equation must be satisfied, together with the mass balances and phase equilibrium equations of part c. Thus, we have one new unknown here, the outlet temperature, and an additional equation from which to find that unknown. 10.1-2 (a) We start with eqn. (8.2-12b): temperature SdT ¦ xi dGi SdT VdP 0. And note that at constant 0 and dGi RTd ln f i a a ff RTd ln xiJ i f i T1 P , so that RT ¦ xi d ln xi RT ¦ xi d ln J i RT ¦ xi d ln fi VdP 0 However, for the pure fluid fugacity, we have, from eqn. (7.2-8a) RTd ln f i dGi V i dP Thus b g RT ¦ xi d ln xi RT ¦ xi d ln J i ¦ xiV i V dP Also 0 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 ¦ xiV i V ¦ xiV i ¦ xiV i ¦ xiV i V ex RT ¦ xi d lna xiJ i f V exdP ex Now assuming i) Ideal gas-phase behavior: xiJ i Pi vap yi P or xiJ i 0 yi P Pi vap and ii) That PV ex / RT 1 we obtain ex a f ¦ x d lnc y P P h PVRT d ln P ¦ xi d ln xiJ i vap i i i or ¦ xi d ln yi ¦ xi d ln P ¦ xi d ln Pi vap d PV ex RT id ln P Now noting that ¦ xid ln Pi vap 0 , since Pi vap is a function of temperature only and b g d ln P , yields ¦ x d ln y FGH PVRT 1IJK d ln P or ¦ xid ln P a y1 y2 x1 1 x1 y1 1 y1 i ex 1 2 1 1 2 dy2 dy1 . 1 1 1 1 1 1, f x a1 y f y a1 x f x y y a1 y f y a1 y f a y x f dy d ln P y a1 y f dx dx 1 i FG x x IJ dy FG PV 1IJ d ln P | d ln P H y y K H RT K x x1 dy1 2 dy2 y2 y1 Since ex d ln P ¦ xi 1 1 1 1 1 1 y2 1 1 1 Also 1 y , and x2 1 x1 , so 1 1 To obtain the x y diagram, I used the equation above in a finite difference form. Using the argument i to denote the ith data point, the equation above becomes y1 i x1 i y1 i y1 i 1 y1 i 1 y1 i a f ln P i ln P i 1 y1 i is unknown, however, P1 i , P1 i 1 , x1 i are known. Also y1 1 is either 0 or 1 depending on which end of the data one starts with. In fact, I started at both ends, in two separate calculations, to check the results. I solved this problem using the equation above rewritten as y1 i where B B r B 2 4C 2 x1 i y1 i 1 ' ln P and C 1 ' ln P x1 i y1 i 1 1 ' ln P and averaged the results from starting at the x1 0 and x1 1 ends. Once x1 and y1 , were known, the activity coefficients were calculated from J1 y1 P x1 P1vap and J 2 y2 P x2 P2vap . Results are given below. Solutions to Chemical and Engineering Thermodynamics, th ed b. CCl4 + n-Heptane System = exprmntl x-y data c. Ethylene bromide + 1-nitropentane System Azeotrope = exprmntl x-y data Section 10.1 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 (also available as a Mathcad worksheet) 10.1-3 T 69 p5 10.422 p6 26799 8.314 ( 273.15 p5 exp ( p5 ) p5 2.721 8.314 ( 273.15 T) p6 exp ( p6 ) p6 1.024 35200 8.314 ( 273.15 T) p7 exp ( p7 ) p7 0.389 x7 29676 10.456 p7 T) 11.431 x5 0.25 x6 0.45 P x5 p5 x6 p6 x7 p7 ( x5 p5 ) y5 0.3 P 1.258 ( x6 p6 ) y6 P y7 ( x7 p7 ) P Bubble point pressure P Bubble point compositions P 1.258 y5 0.541 y6 0.366 x6 0.6 y7 0.093 Now on to dew point calculation Initial guesses z5 0.25 GIVEN x5 x6 soln x5 P z6 1 0.45 x5 z7 x7 0.3 0.3 x5 p5 z5 P x7 1 0.1 x6 p6 z6 P x7 p7 z7 P x7 soln 2 soln 3 P 0.768 FIND( x5 x6 x7 P ) soln 0 x6 soln 1 Dew point pressure Dew point compositions x5 0.071 P x6 0.338 x7 0.592 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 (also available as a Mathcad worksheet) 10.1-4 Solving for the bubble point pressure T P 69 p5( T ) 26799 8.314 ( 273.15 exp 10.422 p6( T ) exp 10.456 p7( T ) y5 1.013 exp 11.431 0.33 y6 p5( T ) T) 35200 8.314 ( 273.15 T) y7 p5( T ) P T) 29676 8.314 ( 273.15 0.33 K5( T P ) 2.721 p6( T ) p6( T ) P p7( T ) K7( T P ) 0.389 p7( T ) P z5 0.33 K6( T P ) 1.024 0.25 z6 0.45 z7 0.3 GIVEN K6( T P ) z6 K5( T P ) z5 soln y6 K6( T P ) z6 y5 K5( T P ) z5 K7( T P ) z7 1 y7 K7( T P ) z7 FIND( y5 y6 y7 P ) y5 soln 0 y5 0.541 y6 y7 soln 1 y6 soln 2 0.366 y7 P soln 3 0.093 P 1.258 This is the bubble-point pressure solution. Now on to the dew-point pressure problem. x5 0.33 x6 0.33 x7 Note that xi=yi/Ki 0.33 GIVEN z5 z6 z7 K5( T P ) K6( T P ) K7( T P ) soln 1 x5 soln 2 P z5 K5( T P ) x6 z6 K6( T P ) x7 z7 K7( T P ) FIND( x5 x6 x7 P ) x5 soln 0 x5 0.071 x6 soln 1 x6 x7 0.338 x7 0.592 soln 3 P 0.768 This is the dew-point pressure solution. So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, at a temperature of 69 C, the mixture will be all liquid at pressures above 1.258 bar, and all vapor at pressures below 0.768 bar. Vapor-liquid equilibrium will exist at this temperature only between 0.768 and 1.258 bar, so this is the pressure range we will examine. Solutions to Chemical and Engineering Thermodynamics, th ed T 69 L 0.99 P 1.2 K5( T P ) x5 GIVEN K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 soln 0 Section 10.1 soln 1 K7( T P ) x7 soln soln 2 L x7 0.32 0.223 x6 0.456 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.507 y6 0.389 0.104 P 1.1 L K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 soln 0 soln 1 0.362 0.458 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 y5 0.445 y6 0.427 y7 1 L L 0.906 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) soln 3 V 1 L V 0.264 L 0.736 K7( T P ) x7 0.128 0.60 K5( T P ) x5 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 soln x7 x6 GIVEN K7( T P ) x7 L 0.18 L V V 0.094 soln 2 x5 1.0 x7 ) 0 0.80 K5( T P ) x5 GIVEN P y7 x6 FIND( x5 x6 x7 L) soln 3 x5 ( x5 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.414 soln 3 x5 0.141 x6 0.445 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.383 y6 0.456 0.161 y7 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 V 0.451 L L 0.549 Solutions to Chemical and Engineering Thermodynamics, th ed P 0.9 L K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 P soln 0 soln soln 2 L x7 0.481 x6 0.412 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.323 y6 0.469 0.208 L 0.8 y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.563 x6 0.359 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.267 y6 0.459 0.274 y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.59 x6 0.339 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.251 y6 0.451 0.298 y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 L L 0.337 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 L L 0.09 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 L V 0.995 L 5.309 10 0.95 K5( T P ) x5 GIVEN ( x5 soln 3 0.071 L 1 V 0.91 x5 1.25 V 0.10 K5( T P ) x5 x5 FIND( x5 x6 x7 L) soln 3 0.078 GIVEN x7 ) 0 V 0.663 x5 L x6 0.20 K5( T P ) x5 0.77 ( x5 soln 3 0.107 x5 P soln 1 K7( T P ) x7 x5 GIVEN P 0.40 K5( T P ) x5 GIVEN Section 10.1 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.303 soln 3 x5 0.246 x6 0.451 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.536 y6 0.369 0.094 y7 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 V 0.013 L L 0.987 3 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 0 1 9 i 0.768 0.071 0.338 0.592 0 .77 0.071 0.339 0.590 0.0053 .8 0.078 0.359 0.563 0.090 .9 0.107 0.412 0.481 0.337 1.0 PP xx5 0.141 xx6 0.445 xx7 0.414 0.549 LL 1.1 0.180 0.458 0.362 0.736 1.2 0.223 0.456 0.320 0.906 1.25 0.246 0.451 0.303 0.987 1.258 0.250 0.450 0.300 1.0 (also available as a Mathcad worksheet) 10.1-5 Solving for the bubble point temperature T 69 p5( T ) p6( T ) p7( T ) P 1.013 exp 10.422 exp 10.456 exp 11.431 26799 8.314 ( 273.15 p5( T ) K5( T P ) T) 29676 8.314 ( 273.15 2.721 P p6( T ) 1.024 K6( T P ) T) T) p6( T ) P 35200 8.314 ( 273.15 p5( T ) p7( T ) 0.389 K7( T P ) p7( T ) P Solutions to Chemical and Engineering Thermodynamics, th ed y5 0.33 y6 0.33 y7 0.33 z5 Section 10.1 0.25 z6 0.45 z7 0.3 GIVEN K5( T P ) z5 K6( T P ) z6 K7( T P ) z7 1 y5 K5( T P ) z5 y6 K6( T P ) z6 y7 K7( T P ) z7 FIND( y5 y6 y7 T ) soln y5 soln 0 y5 0.548 y6 soln 1 y6 y7 soln 2 0.363 T y7 soln 3 0.088 T 61.788 This is the bubble-point temperature solution. Now on to the dew-point temperature problem. x5 0.33 x6 0.33 x7 Note that xi=yi/Ki 0.33 GIVEN z5 z6 z7 K5( T P ) K6( T P ) K7( T P ) 1 x5 soln 2 T z5 x6 K5( T P ) z6 K6( T P ) x7 z7 K7( T P ) FIND( x5 x6 x7 T ) soln x5 soln 0 x5 0.074 x6 soln 1 x6 x7 0.346 x7 soln 3 0.579 T 77.436 This is the dew-point pressure solution. So for a mixture of the composition z5=0.25, z6=0.45 and z7=0.30, the mixture will be all liquid at temperatures below 61.79 C, and all vapor at temperatures above 77.44 C. Vapor-liquid equilibrium will exist only between 61.79 and 77.44 C, so this is the temperature range we will examine. T 62 L 0.99 P 1.013 K5( T P ) x5 GIVEN K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.303 soln 3 x5 0.246 x6 0.451 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.543 y6 0.367 0.09 T 65 L 0.80 y7 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 V 0.012 L L 0.988 Solutions to Chemical and Engineering Thermodynamics, th ed x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 T soln 0 x7 0.343 0.459 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 y5 0.476 y6 0.41 L y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 soln 0 soln 1 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 y5 0.411 y6 0.444 y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 soln 0 ( x5 L L 0.813 x6 x7 ) 0 FIND( x5 x6 x7 L) soln 3 V 1 L V 0.365 L 0.635 K7( T P ) x7 0.144 soln 1 K7( T P ) x7 soln soln 2 L x7 0.443 0.124 x6 0.432 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.352 y6 0.464 0.183 y7 ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) soln 3 x5 V 1 L V 0.551 L 0.20 K5( T P ) x5 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 GIVEN 1 0.40 K5( T P ) x5 L V V 0.187 soln 0.389 0.453 74 FIND( x5 x6 x7 L) soln 3 K7( T P ) x7 x7 x6 x5 x7 ) 0 0.114 L 0.157 GIVEN x6 K7( T P ) x7 soln 2 x5 L ( x5 0.60 K5( T P ) x5 71 soln L x6 68 K7 ( T P ) x7 soln 2 0.198 x5 T soln 1 x5 GIVEN T K6 ( T P ) x6 K5 ( T P ) x5 GIVEN Section 10.1 K7( T P ) x7 soln ( x5 x6 x7 ) 0 FIND( x5 x6 x7 L) 0.449 Solutions to Chemical and Engineering Thermodynamics, th ed x5 T soln x7 soln 2 L x7 0.504 soln 3 0.098 x6 0.398 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.301 y6 0.467 0.232 76 L y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.548 x6 0.369 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.27 y6 0.46 0.27 L y7 L L 0.251 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 L V 0.891 L 0.109 0.10 K5( T P ) x5 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 GIVEN soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.57 0.077 x6 0.353 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.256 y6 0.453 0.291 L y7 x6 x7 ) 0 FIND( x5 x6 x7 L) V 1 L V 0.966 L 0.05 K5( T P ) x5 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 GIVEN ( x5 soln 3 x5 78 ( x5 soln 3 0.083 x5 1 V 0.749 x5 77 V 0.10 K5( T P ) x5 x5 T soln 1 x5 GIVEN T x6 0 Section 10.1 K7( T P ) x7 ( x5 x6 x7 ) 0 0.034 Solutions to Chemical and Engineering Thermodynamics, th ed x7 ( L ( 1 x5 T K7( T P ) ) z7 x6 x7 soln 0 soln 1 FIND( x5 x6 x7 L) soln soln 2 L x7 0.592 soln 3 x5 0.071 x6 0.337 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.242 y6 0.445 0.313 61 L y7 K6( T P ) x6 x5 ( L ( 1 K5( T P ) ) K5( T P ) ) z5 x6 ( L ( 1 K6( T P ) ) K6( T P ) ) z6 x7 ( L ( 1 K7( T P ) ) K7( T P ) ) z7 x6 x7 x5 V 1 L V 1.045 L 0.045 0.9 K5( T P ) x5 GIVEN i K7( T P ) ) Section 10.1 soln 0 soln 1 K7( T P ) x7 soln soln 2 L x7 0.29 0.264 x6 0.446 y5 K5( T P ) x5 y6 K6( T P ) x6 y7 K7( T P ) x7 y5 0.566 y6 0.351 0.083 x6 x7 ) 0 FIND( x5 x6 x7 L) soln 3 x5 y7 ( x5 V V 1 L 0.047 L 1.047 0 1 9 TT 61.788 0.25 0.45 0.3 1.0 62 0.246 0.451 0.303 0.988 65 0.198 0.459 0.343 0.813 68 0.157 0.453 0.389 0.635 71 xx5 0.124 xx6 0.432 xx7 0.443 LL 0.449 74 0.098 0.398 0.504 0.251 76 0.083 0.369 0.548 0.109 77 0.077 0.353 0.570 0.034 77.436 0.074 0.346 0.570 0.0 Solutions to Chemical and Engineering Thermodynamics, th ed 10.1-6 Section 10.1 Clearly, it is only water that condenses out at the dew point, since O2 and N 2 are far above their critical temperatures. Thus, at the dew point PHvap 2O yH 2 O P PH 2 O partial pressure of H 2O in air. [In writing this expression, all fugacity coefficients have been assumed equal to unity.] From the data in Problem 5.12 we have, at the dew point, 5432.8 7.8086 26.3026 ln PHvap 2O 27315 . 256 . and PHvap dew point PH 2 O 24618 . Pa 2O at the air conditions PHvap T 2O 256 . qC 3354.9 PH 2 O PHvap T 2O 20.6q C relative humidity PHvap T 2O 256 . qC u 100% 7338% . Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 10.1-7 Start by determining the pressure if only vapor was present using ideal gas law PV=NRT 5 1 u 8.315 10 P 298.15 P 3 8.264 bar 3 10 which is much higher than the vapor pressure of either pure component. Therefore, a vapor-liquid mixture must be present. The molecular weight of butane is 58.124 and its density is 0.575 g/cc, so that 0.1 moles of liquid butane would occupy 10.1 cc; the molecular weight of hexane is 86.178, and its density is 0.655 g/cc so that 0.9 moles of hexane would occupy 118.4 cc. Therefore to proceed, I will assume that finally there will be about 2875 cc of vapor and 125 cc of vapor. LL 0.9 xb 0.075 P 0.5 yb xb V 1 LL 2.428 yh xh P xh 1 xb 0.2 P Given xb 2.428 yb P xh 0.2 P equilibrium relation for butane equilibrium relation for hexane yh P xb LL yb V 0.1 xh LL yh V 0.9 xb xh 1 yb yh 1 5 V 8.314 10 mass balance on butane mass balance on hexane sum of liquid mole fractions = 1 sum of vapor mole fractions = 1 Pressure from ideal gas equation of state 298.15 3 2.875 10 z find ( xb xh yb yh LL V P) § 0.081 · ¨ ¸ ¨ 0.919 ¸ ¨ 0.516 ¸ z ¨ 0.484 ¸ ¨ ¸ ¨ 0.956 ¸ ¨ 0.044 ¸ ¨ 0.38 ¸ © ¹ xb z xh z 0 V z 2 yh z LL z 3 4 P z 5 6 yb 0.516 LL 0.956 Volume of liquid yb z 1 yh 0.484 VL LL §¨ xb © 58.124 0.575 xb xh 0.081 86.178 · ¸ 0.655 ¹ xh 0.919 VL 123.415 Close enough to the guess value of 125 cc that it is not worth iterating. Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 10.1-8 p5 2.755 p6 1.021 p7 0.39 a) Bubble point pressure P 0.55 p5 0.25 p6 0.2 p7 y5 0.55 y7 0.2 p5 P 1.8485 p6 y5 0.81972 y6 0.25 y7 0.0422 y5 y6 y7 P p7 P y6 P 0.13808 1 b) Dew point pressure P 2 Initial guess x5 0.55 P x6 0.25 p5 P x7 0.2 p6 P p7 Given 0.55 x5 P x6 p5 x5 x6 x7 0.25 x5 z x5 0.20854 x6 z x6 0.25578 x7 0.53569 1 x7 z 2 x5 x6 x7 p6 x7 P 0.2 p7 1 z find ( x5 x6 x7 P) 0 P 1 0.20854 · ¨§ ¸ 0.25578 ¸ ¨ z ¨ 0.53569 ¸ ¨ 1.04459 ¸ © ¹ P z 3 P 1.04459 bar Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 10.1-9 p5 2.755 a) p6 1.021 p7 0.39 0.55 x5( L p ) x6( L p ) § 1 p5 · L p5 ¨ ¸ p p ¹ © 0.25 p6 §1 L ¨ ¸ p p ¹ © p6 · 0.20 x7( L p ) p7 §1 ¨ ¸ L p p ¹ © p7 · y6( L p ) x6( L p ) y5( L p ) x5( L p ) p6 y7( L p ) x7( L p ) p p 0.9775 Given L 0.1 x5( L p ) §¨ 1 0 © p5 p p7 p initial guess for P, fix L p6 · p7 · x6( L p ) §¨ 1 x7( L p ) §¨ 1 ¸ ¸ ¸ p ¹ p ¹ p ¹ © © p5 · p Find ( p ) p 1.129 x5( L p ) 0.24 x6( L p ) y5( L p ) 0.584 y6( L p ) 0.274 0.247 x5( L p ) x6( L p ) x7( L p ) 1 y5( L p ) y6( L p ) y7( L p ) 1 x7( L p ) 0.487 y7( L p ) 0.168 b) p 0.9775 Given 0 L 0.9 x5( L p ) §¨ 1 © initial guess for P, fix L p6 · p7 · x7( L p ) §¨ 1 x6( L p ) §¨ 1 ¸ ¸ ¸ p ¹ p ¹ p ¹ © © p5 · p Find ( p ) L 0.9 p 1.789 x5( L p ) 0.522 x6( L p ) y5( L p ) 0.804 y6( L p ) 0.261 0.149 x7( L p ) y7( L p ) 0.217 0.047 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 10.1-10 p5 2.755 a) x5( L p ) p6 1.021 p7 0.39 0.55 x6( L p ) § 1 p5 · L p5 ¨ ¸ p p ¹ © 0.25 p6 §1 L ¨ ¸ p p ¹ © p6 · 0.20 x7( L p ) § 1 p7 · L p7 ¨ ¸ p p ¹ © y6( L p ) x6( L p ) y5( L p ) x5( L p ) p6 y7( L p ) x7( L p ) p p 0.9775 Given 0 L 0.5 x5( L p ) §¨ 1 © p5 p p7 p initial guess for P, fix L p6 · p7 · x6( L p ) §¨ 1 x7( L p ) §¨ 1 ¸ ¸ ¸ p ¹ p ¹ p ¹ © © p5 · p Find ( p ) p b) x5( L p ) 0.386 x6( L p ) y5( L p ) 0.714 y6( L p ) y5( L p ) y6( L p ) y7( L p ) 1 0 x5( L p ) §¨ 1 © p 0.317 y7( L p ) 0.203 1 L 0.27152 x7( L p ) 0.297 x5( L p ) x6( L p ) x7( L p ) p 0.9775 Given 10.1-11 a) 1.491 0.083 Guess p, iterate on L p6 · p7 · x6( L p ) §¨ 1 x7( L p ) §¨ 1 ¸ ¸ ¸ p ¹ p ¹ p ¹ © © p5 · 1.285 x5( L p ) 0.3 x6( L p ) y5( L p ) 0.643 y6( L p ) 0.294 0.234 x5( L p ) x6( L p ) x7( L p ) 1 y5( L p ) y6( L p ) y7( L p ) 1 x7( L p ) y7( L p ) 0.406 0.123 Number of unknowns x5, x6, x7, y5, y6, y7, TI, PI, TII, PII, V and L = 12 (4 constraints) Number of constraints TI = TII, PI = PII[i \i = 1 VAP x i Pi yi P i 5, 6 and 7 (3 constraints) Mass balances xi L + yi V = zi (3 constraints) where z is the feed composition p Find ( p ) Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 (Note summing the three mass balances gives L + V = 1, so that is not an independent equation) So the number of degrees of freedom is 12 – 4 – 3 – 3 = 2 Setting T = 69°C fixes one degree of freedom. Therefore, can not also set one liquid phase composition and one vapor phase composition! The system would be overspecified. b) Joe has to release one degree of freedom. Let T vary, then can find a T and P to meet the composition specifications. Pivap 10.1-12 yi P 4.35 kPa and ln P (MPa) = 8.347 – 2644/T T = 191.8 K 10.1-13 xBZ = 0.99 98% of benzene in the feed xBZ = 0.4 xTMP = 0.6 105 mols/hr xBZ = ? Benzene mass balance TMP mass balance Product restriction Also xC6 = 1 - xBZ 0.4 × 105 = 0.99 D + xBZB 0.6 × 105 = 0.01 D + xTMPB 0.98 × 0.4 × 105 = 0.99 D D = 3.9596 × 104 moles/hr B = 105–D = 6.0404 × 104 moles/hr 0.6 × 105 = 0.01 × 3.9596 × 104 + xTMP × 6.0404 × 104 6 0.039596 ×104 x TMP 0.98676 6.0404 u 104 x BZ 0.01324 10.1-14 Since the pressure in the smaller vessel is 0.6 bar, that is the partial pressure of the butane in the vapor phase of the larger vessel. That is 0.6 yB P 0.6 xB PBvap xB u 2.583 so that xB 0.232 and xH 1 xB 0.768 2.583 Now P xB PBvap xH PHvap 0.232 u 2.583 0.768 u 0.218 0.767 bar so that finally 0.6 0.6 yB P 0.767 0.782 and yB 1 0.782 0.218 10.1-15 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.15 Solutions to Chemical and Engineering Thermodynamics, th ed Section 10.1 10.1-16 The Gibbs energy of a mixture consisting of N1 moles of species 1 and N 2 moles of species 2 is G N1 G1 RT ln x1 N 2 G 2 RT ln x2 o o § o § o N1 · N2 · N1 ¨ G1 RT ln ¸ ¸ N 2 ¨ G 2 RT ln N1 N 2 ¹ N1 N 2 ¹ © © So the change in Gibbs energy between a state I and a state II (without chemical reaction) is 'G N1II G1 RT ln x1II N 2II G 2 RT ln x2II N1I G1 RT ln x1I N 2I G 2 RT ln x2I o o o o Since there is no chemical reaction, N1II = N1I and N 2II = N 2I so this equation reduces to 'G N1II RT ln x1II N 2II RT ln x2II N1I RT ln x1I N 2I RT ln x2I Based on 1 mole of feed that is 0.5 moles of species 1 and 0.5 moles of species going to one stream of 0.5 moles of a mixture 0.495 moles of species 1 and 0.005 of species 2, we have 'G RT 0.495ln 0.99 0.005ln 0.01 0.5ln 0.5 0.5ln 0.5 = 0.495 u 0.01005 0.005 u 4.605 0.5 u 0.693 u 2 = 0.0050 0.0230 0.693 0.665 J kJ 'G 0.665 RT 0.665 u 8.314 u 298.15 K=1.648 mol K mol This increase in free energy must occur as the result of work be supplied to the system. Therefore, the work required for the separation is 1.648 kJ per mole of feed. 10.1-17 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.17 10.1-18 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.18 10.1-19 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.1 10.1-20 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.2 10.1-21 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.3 10.1-22 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.4 10.1-23 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.5 10.1-24 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.8 10.1-25 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.9 Solutions to Chemical and Engineering Thermodynamics, th ed 10.1-26 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.10 10.1-27 In the folder Aspen for Textbook>Chapter 10.1>Problems>Prob 10.13 Section 10.1 Section 10.2 Solutions to Chemical and 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[RIZDWHU ZDWHU GLR[DQH -------------COMPONENT DATA -------------Molecular Subgroups ---------------------------------------------------------------------Comp Name Number Description Frequency ---- -------------------------------- -----------------------1 water 19 H2O 1 2 1,4-dioxane 81 c-CH2OCH2 ---------------------------------------------------x(1) GAM(1) GAM(2) GEX(J/MOL) ---------------------------------------------------0.0000 3.5435 1.0000 0.00 0.0250 3.4073 1.0005 83.65 0.0500 3.2761 1.0020 164.59 0.0750 3.1496 1.0047 242.75 0.1000 3.0277 1.0085 318.05 0.1250 2.9105 1.0135 390.41 0.1500 2.7976 1.0200 459.73 0.1750 2.6891 1.0278 525.92 0.2000 2.5849 1.0372 588.89 0.2250 2.4847 1.0484 648.55 0.2500 2.3885 1.0613 704.77 0.2750 2.2963 1.0763 757.47 0.3000 2.2078 1.0935 806.52 0.3250 2.1231 1.1132 851.81 0.3500 2.0419 1.1355 893.21 0.3750 1.9643 1.1608 930.59 0.4000 1.8901 1.1894 963.83 0.4250 1.8192 1.2218 992.76 0.4500 1.7517 1.2583 1017.25 0.4750 1.6873 1.2995 1037.15 0.5000 1.6260 1.3461 1052.27 0.5250 1.5677 1.3988 1062.46 0.5500 1.5124 1.4584 1067.53 0.5750 1.4601 1.5260 1067.28 0.6000 1.4105 1.6029 1061.52 0.6250 1.3638 1.6905 1050.04 0.6500 1.3199 1.7908 1032.60 0.6750 1.2787 1.9058 1008.97 0.7000 1.2402 2.0384 978.91 0.7250 1.2044 2.1920 942.13 0.7500 1.1713 2.3706 898.37 0.7750 1.1409 2.5797 847.32 0.8000 1.1132 2.8257 788.67 2 Solutions to Chemical and Engineering Thermodynamics,5th ed 0.8250 0.8500 0.8750 0.9000 0.9250 0.9500 0.9750 1.0000 1.0883 1.0662 1.0469 1.0308 1.0178 1.0081 1.0021 1.0000 3.1174 3.4656 3.8845 4.3930 5.0160 5.7872 6.7527 7.9759 722.07 647.17 563.58 470.90 368.69 256.47 133.76 0.00 ZDWHUDQGGLR[DQHDW. [RIZDWHU ---------------------------------------------------x(1) GAM(1) GAM(2) GEX(J/MOL) ---------------------------------------------------0.0000 3.5435 1.0000 0.00 0.0250 3.4073 1.0005 83.65 0.0500 3.2761 1.0020 164.59 0.0750 3.1496 1.0047 242.75 0.1000 3.0277 1.0085 318.05 0.1250 2.9105 1.0135 390.41 0.1500 2.7976 1.0200 459.73 0.1750 2.6891 1.0278 525.92 0.2000 2.5849 1.0372 588.89 0.2250 2.4847 1.0484 648.55 0.2500 2.3885 1.0613 704.77 0.2750 2.2963 1.0763 757.47 0.3000 2.2078 1.0935 806.52 0.3250 2.1231 1.1132 851.81 0.3500 2.0419 1.1355 893.21 0.3750 1.9643 1.1608 930.59 0.4000 1.8901 1.1894 963.83 0.4250 1.8192 1.2218 992.76 0.4500 1.7517 1.2583 1017.25 0.4750 1.6873 1.2995 1037.15 0.5000 1.6260 1.3461 1052.27 0.5250 1.5677 1.3988 1062.46 0.5500 1.5124 1.4584 1067.53 0.5750 1.4601 1.5260 1067.28 0.6000 1.4105 1.6029 1061.52 0.6250 1.3638 1.6905 1050.04 0.6500 1.3199 1.7908 1032.60 0.6750 1.2787 1.9058 1008.97 0.7000 1.2402 2.0384 978.91 0.7250 1.2044 2.1920 942.13 0.7500 1.1713 2.3706 898.37 0.7750 1.1409 2.5797 847.32 0.8000 1.1132 2.8257 788.67 0.8250 1.0883 3.1174 722.07 0.8500 1.0662 3.4656 647.17 0.8750 1.0469 3.8845 563.58 0.9000 1.0308 4.3930 470.90 0.9250 1.0178 5.0160 368.69 0.9500 1.0081 5.7872 256.47 0.9750 1.0021 6.7527 133.76 1.0000 1.0000 7.9759 0.00 Section 10.2 Section 10.2 Solutions to Chemical and Engineering Thermodynamics,5th ed x(1) 0.0000 0.0250 0.0500 0.0750 0.1000 0.1250 0.1500 0.1750 0.2000 0.2250 0.2500 0.2750 0.3000 0.3250 0.3500 0.3750 0.4000 0.4250 0.4500 0.4750 0.5000 0.5250 0.5500 0.5750 0.6000 0.6250 0.6500 0.6750 0.7000 0.7250 0.7500 0.7750 0.8000 0.8250 0.8500 0.8750 0.9000 0.9250 0.9500 0.9750 1.0000 y(1) 0.0000 0.0645 0.1196 0.1671 0.2084 0.2446 0.2764 0.3046 0.3296 0.3520 0.3719 0.3898 0.4058 0.4202 0.4332 0.4449 0.4554 0.4649 0.4734 0.4811 0.4881 0.4944 0.5001 0.5054 0.5103 0.5149 0.5193 0.5238 0.5284 0.5335 0.5392 0.5460 0.5543 0.5650 0.5791 0.5983 0.6250 0.6639 0.7232 0.8204 1.0000 y(2) 1.0000 0.9355 0.8804 0.8329 0.7916 0.7554 0.7236 0.6954 0.6704 0.6480 0.6281 0.6102 0.5942 0.5798 0.5668 0.5551 0.5446 0.5351 0.5266 0.5189 0.5119 0.5056 0.4999 0.4946 0.4897 0.4851 0.4807 0.4762 0.4716 0.4665 0.4608 0.4540 0.4457 0.4350 0.4209 0.4017 0.3750 0.3361 0.2768 0.1796 0.0000 P(bar) 0.156 0.163 0.169 0.175 0.179 0.184 0.187 0.191 0.194 0.196 0.198 0.200 0.202 0.203 0.204 0.205 0.205 0.205 0.206 0.206 0.206 0.206 0.205 0.205 0.205 0.204 0.204 0.204 0.203 0.202 0.201 0.200 0.198 0.196 0.193 0.189 0.183 0.175 0.164 0.147 0.124 ZDWHUDQGGLR[DQHDW. 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/ŶƚŚĞĨŽůĚĞƌƐƉĞŶĨŽƌdĞdžƚŬхŚĂƉƚĞƌϭϬ͘ϮхWƌŽďůĞŵƐхWƌŽďϭϬ͘ϮͲϰϲ /ŶƚŚĞĨŽůĚĞƌƐƉĞŶĨŽƌdĞdžƚŬхŚĂƉƚĞƌϭϬ͘ϮхWƌŽďůĞŵƐхWƌŽďϭϬ͘ϮͲϱϮ Section 10.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed 10.3-1 Section 10.3 Using the critical properties in the text, the program VLMU and the following interaction parameters kC2 C3 0.001 ; kC2 C4 0.010 ; kC3 C4 0.003 I obtain the following Bubble point, P bar 1 5 10 15 20 25 30 35 40 42 43 10.3-2 T K . 23131 280.26 . 30817 . 32712 . 34199 354.47 . 36583 37527 . 384.56 388.31 390.31 yC2 0.3444 0.2222 . 01738 . 01460 . 01261 . 01102 0.0932 0.0835 0.0699 0.0632 0.0588 yC3 . 05907 0.6576 0.6693 0.6691 0.6642 0.6565 0.6434 0.6329 0.6133 0.6011 05919 . yC4 0.0649 . 01202 . 01569 . 01849 0.2097 0.2333 0.2634 0.2836 0.3168 0.3357 0.3493 Using the same program and information as above, I obtain the following Dew point, P bar T K xC2 xC3 xC4 1 5 10 15 20 25 30 32 35 38 40 42 43 25393 . 299.65 32527 . 342.32 35543 . 36615 . 37515 . 378.44 38300 . 387.09 389.57 39176 . 392.64 0.0038 0.0081 0.0117 0.0148 0.0180 0.0213 0.0257 0.0271 0.0296 0.0331 0.0361 0.0402 0.0434 0.2309 0.3128 0.3595 0.3924 0.4198 0.4446 0.4724 0.4806 0.4936 05103 . 05231 . 05389 . 05499 . 0.7653 0.6791 0.6289 05928 . 05623 . 05341 . 05019 . 0.4923 0.4768 0.4566 0.4408 0.4209 0.4068 See figure on following page. 10.3-3 Again we use the program VLMU and the data from Problem 10.3-1. Also, since, at 20 bar, the bubble point of the mixture is 341.99 K and the dew point is 355.43 K, we only need to consider temperatures between these two extremes. The results follow: Solutions to Chemical and Engineering Thermodynamics, 5th ed P 20 bar T K 342.5 344.0 346.0 348.0 350.0 352.0 354.0 3550 . x1 0.0478 0.0418 0.0352 0.0299 0.0258 0.0224 0.0197 0.0185 x2 . 05670 . 05564 . 05382 . 05170 0.4928 0.4668 0.4397 0.4259 x3 0.3852 0.4019 0.4263 0.4531 0.4814 0.5107 05407 . 05557 . y1 0.0120 . 01069 0.0914 0.0789 0.0690 0.0609 0.0542 0.0512 Section 10.3 y2 0.6648 0.6641 0.6580 0.6464 0.6304 0.6106 05878 . 05756 . y3 0.2143 0.2290 0.2506 0.2746 0.3007 0.3285 0.3580 0.3732 V L split 0.0303 0.9697 . . 01265 08735 0.2635 0.7365 . 0.4095 05905 . 05612 0.4388 0.7175 0.2825 08796 . 01204 . 0.9628 0.0372 Note to instructor: Re: Problems 10.3-1, 2 and 3 You should take time to discuss how the compositions are changing with pressure in each of the cases above. For example, in Problem 10.3-1, at low bubble point pressures the vapor composition is very different than the liquid, with the vapor greatly enriched in the light C2 component. However, as the pressure increases, and the critical point is approached, the vapor composition becomes similar to that of the liquid. Analogous comments apply to the dew point case of Problem 10.3-2. a f Vapor Liquid Equlibrium 40 30 P_bp P in bar i P_dp i Bubble point curve 20 Dew point curve 10 0 220 240 260 280 300 320 340 360 380 400 T_bp T_dp i i T in K 10.3-4 See solution to Problem 6.2. The derivation of Eqs. 10.3-8 is identical to the derivation of eqns. 6.4-29 & 30. 10.3-5 (a and b) These algorithms are incorporated into the program VLMU Examine that program to see the algorithms used. 10.3-6 (a) The equations to be used to solve this problem are the mass balances, the equilibrium condition (equality of fugacities) and the energy balance. Writing these equations for an open, steadyflow system, we have (for 1 mole of feed of compositions zi ). Solutions to Chemical and Engineering Thermodynamics, 5th ed Section 10.3 Mass balances zi xi L yiV i 1, 2, ! , N 1 L V (summing equations above over all species) phase equilibrium condition f i L f iV xi P fi L xi P yi P fi V yi P or L xi I i T , P, x V i 1, 2, ! , N c h yi I i T , P, y (1) (2) or Ki energy balance L Ii yi xi Ki T , P, x V Ii i 1, 2, ! , N 0 ¦ b N H g ¦ b N H g ¦ b N H g 0 ¦ b z H g V ¦ b x H g L¦ b y H g or V i i in i i i out i i in i i Also, we have the summation conditions ¦ xi 1 and ¦ yi i L out V i (3) L (4) i (5) 1 (b) With the exception of Eq. (4), the other equations are the ones used in the isothermal flash calculation (see Eqs. 10.1-14 & 15, and 10.3-4). Therefore the easiest algorithm to implement is to use the one for the isothermal flash with an extra, outer loop which iterates on final temperature. This is done by adding an enthalpy calculation (see eq. 10.3-8a) to the program, and calculating the enthalpy of the feed stream, and the liquid and vapor streams. If eq. (4) is not satisfied, the exit temperature is adjusted, and the calculation repeated. (c) A Mathcad worksheet (10.3-6.MCD) is available for this calculation. This worksheet is also available as the Adobe PDF file 10.3-6.pdf. 10.3-7 This problem is probably most easily solved using an equation of state, such as the Peng-Robinson equation. Using the program VLMU, the critical properties in Table 6.6-1 and k12 0.01 (from Table 9.4-1) we obtain the following mole % ethane Calculated Measurement in liquid 7.8 vapor mol % 6.62 P bar 39.68 vapor mol % 6.2 P bar 39.73 22.8 19.35 37.08 19.7 37.07 30.3 2581 . 3579 . 255 . 3560 . 59.0 5213 . 30.94 531 . 3213 . 89.0 8528 . 2589 . 854 . 2545 . Thus we see, using the program VLMU, we obtain very accurate predictions in a simple manner (though, of course, much work went into preparing the program). With the exception of the 59 mol % ethane liquid, the compositions are predicted to about 0.004 mole fraction and 0.4 bar accuracy. 10.3-8 Again the program VLMU will be used with the critical properties in Table 6.6-1 and the value k12 0.055 given in Table 9.4-1. The only question is how to use the program to get K values. I have used the isothermal flash, since that is the only option that allows me to specify T and P. One then has to choose a feed composition that assures that one is in the two-phase region (this can be checked by a collection of bubble point or similar calculations, if desired). This corresponds to methane compositions in the range Solutions to Chemical and Engineering Thermodynamics, 5th ed Section 10.3 0.0711 d xC1 d 0.9916 In this range we find the following x y K methane 0.0711 0.9916 1395 . benzene 0.9289 0.0084 0.009083 A asked for in problem statement 10.3-9 We use the program VLMU, critical properties in Table 6.6-1, and a value of k12 0.003 given in Table 9.4-1. The problem must be solved by trial-and-error. The easiest way is to start with the bubble-point P program. This leads to the result that the bubble-point pressure is 8.47 bar. Therefore, a lower pressure than 8.47 bar must be used to produce appreciable vapor. Likewise, using the dew point pressure program gives a dew point pressure of 6.12 bar. Therefore, the pressure must be between 6.12 bar and 8.47 bar to produce appreciable vapor. By trial-and-error . 05 . 1 and . K , P 7.234 bar V L 05 we find at T 31315 C3 nC4 Phase compressibility, Z Feed Liq 05 . 0.3699 05 . 0.6301 0.0268 Vapor K 0.6301 1703 . 0.3699 05871 . 08594 . 10.3-10 Using the program VLMU, critical properties in Table 9.4-1 and the following interaction parameters kC2 C3 0.001 ; kC2 nC4 0.010 ; kC2 iC 4 0.007 ; kC3 nC4 0.003 ; kC3 iC4 0.007 and k nC4 iC4 0.0 we obtain the following Component ethane propane n - butane i - butane moles compressibility Feed 0.3100 0.3400 0.2100 01400 . 10 . Liquid . 01638 0.3372 0.3084 01907 . 0.4676 0.0548 Vapor K 0.4384 2.677 0.3425 1016 . 01236 . 0.4008 0.0955 05008 . 05324 . 0.7952 Therefore, 53.24 mol % of tank contents will be vapor and the remaining 46.76 mol % will be liquid. [If we believe the compressibilities, even though the liquid compressibility has some error, then we have that 5.71% of the volume of the tank is filled with liquid, and the remaining 94.29% is filled with vapor.] 10.3-11 The analysis is similar to that of Problem 10.3-6 except that the energy balance 0 ¦ b z H g L¦ b x H g V ¦ b y H g L i i in i V i i i used there is now replaced with the entropy balance 0 ¦ c z S h L¦ e x S j V ¦ e y S j i i in i L i i V i A Mathcad worksheet (10-3-11.MCD) is available for this calculation. This worksheet can be viewed as the Adobe PDF file 10-3-11.pdf. Solutions to Chemical and Engineering Thermodynamics, 5th ed Section 10.3 10.3-12 A reasonably good fit is obtained with kij = 0.11 Bubble point pressure at 273.13 K 35 30 25 20 15 10 5 0 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 mole fraction of carbon dioxide Vapor Liquid Bubble point pressure calculation using the Peng Robinson equation of state Temperature = 273.13 K T (K) 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 P (bar) 1.5559 2.0182 2.4785 2.9347 3.3918 3.8465 4.2990 4.7491 5.1970 5.6425 6.0857 6.5263 6.9645 7.4002 7.8332 8.2635 8.6911 9.1159 9.5379 9.9569 10.3729 10.7860 11.1959 11.6026 12.0062 12.4065 12.8034 13.1970 13.5872 13.9739 14.3570 14.7366 15.1126 15.4849 15.8534 16.2183 16.5794 16.9366 17.2900 17.6396 17.9852 18.3269 x1 0.0000 0.0100 0.0200 0.0300 0.0400 0.0500 0.0600 0.0700 0.0800 0.0900 0.1000 0.1100 0.1200 0.1300 0.1400 0.1500 0.1600 0.1700 0.1800 0.1900 0.2000 0.2100 0.2200 0.2300 0.2400 0.2500 0.2600 0.2700 0.2800 0.2900 0.3000 0.3100 0.3200 0.3300 0.3400 0.3500 0.3600 0.3700 0.3800 0.3900 0.4000 0.4100 y1 PV/RT(liq) PV/RT(vap) 0.0000 0.0064 0.9485 0.2237 0.0083 0.9497 0.3650 0.0102 0.9491 0.4623 0.0120 0.9478 0.5334 0.0138 0.9458 0.5876 0.0156 0.9435 0.6303 0.0173 0.9409 0.6648 0.0190 0.9382 0.6933 0.0207 0.9354 0.7172 0.0224 0.9325 0.7375 0.0240 0.9295 0.7551 0.0257 0.9264 0.7703 0.0273 0.9233 0.7837 0.0288 0.9202 0.7956 0.0304 0.9170 0.8062 0.0319 0.9139 0.8157 0.0334 0.9107 0.8242 0.0349 0.9075 0.8320 0.0363 0.9043 0.8391 0.0377 0.9011 0.8456 0.0391 0.8978 0.8515 0.0404 0.8946 0.8570 0.0418 0.8914 0.8621 0.0431 0.8882 0.8668 0.0444 0.8850 0.8712 0.0456 0.8818 0.8753 0.0469 0.8786 0.8791 0.0481 0.8754 0.8827 0.0492 0.8722 0.8861 0.0504 0.8690 0.8893 0.0515 0.8659 0.8923 0.0526 0.8627 0.8951 0.0537 0.8596 0.8978 0.0547 0.8565 0.9003 0.0557 0.8533 0.9028 0.0567 0.8503 0.9051 0.0577 0.8472 0.9073 0.0586 0.8441 0.9094 0.0596 0.8411 0.9114 0.0604 0.8381 0.9133 0.0613 0.8351 0.9151 0.0621 0.8321 Solutions to Chemical and Engineering Thermodynamics, 5th ed 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 273.13 18.6648 18.9986 19.3286 19.6545 19.9766 20.2947 20.6088 20.9191 21.2255 21.5280 21.8267 22.1216 22.4129 22.7004 22.9843 23.2648 23.5417 23.8153 24.0856 24.3528 24.6169 24.8780 25.1364 25.3921 25.6452 25.8960 26.1446 26.3912 26.6360 26.8791 27.1208 27.3612 27.6007 27.8394 28.0751 28.3156 28.5511 28.7918 29.0282 29.2704 29.5088 29.7537 29.9979 30.2444 30.4934 30.7453 31.0006 31.2596 31.5228 31.7907 32.0637 32.3424 32.6272 32.9189 33.2178 33.5246 33.8398 34.1638 34.4965 0.4200 0.4300 0.4400 0.4500 0.4600 0.4700 0.4800 0.4900 0.5000 0.5100 0.5200 0.5300 0.5400 0.5500 0.5600 0.5700 0.5800 0.5900 0.6000 0.6100 0.6200 0.6300 0.6400 0.6500 0.6600 0.6700 0.6800 0.6900 0.7000 0.7100 0.7200 0.7300 0.7400 0.7500 0.7600 0.7700 0.7800 0.7900 0.8000 0.8100 0.8200 0.8300 0.8400 0.8500 0.8600 0.8700 0.8800 0.8900 0.9000 0.9100 0.9200 0.9300 0.9400 0.9500 0.9600 0.9700 0.9800 0.9900 1.0000 0.9169 0.9186 0.9202 0.9218 0.9234 0.9248 0.9263 0.9276 0.9290 0.9303 0.9316 0.9328 0.9340 0.9352 0.9363 0.9375 0.9386 0.9396 0.9407 0.9418 0.9428 0.9438 0.9448 0.9458 0.9468 0.9478 0.9488 0.9498 0.9508 0.9518 0.9528 0.9538 0.9548 0.9558 0.9568 0.9578 0.9589 0.9600 0.9611 0.9622 0.9634 0.9646 0.9659 0.9672 0.9685 0.9699 0.9714 0.9730 0.9746 0.9764 0.9782 0.9802 0.9823 0.9846 0.9872 0.9899 0.9929 0.9963 1.0000 0.0630 0.0637 0.0645 0.0653 0.0660 0.0667 0.0673 0.0680 0.0686 0.0692 0.0698 0.0703 0.0708 0.0714 0.0718 0.0723 0.0727 0.0732 0.0736 0.0740 0.0743 0.0747 0.0750 0.0753 0.0756 0.0758 0.0761 0.0763 0.0765 0.0767 0.0769 0.0771 0.0772 0.0773 0.0774 0.0775 0.0776 0.0776 0.0777 0.0777 0.0777 0.0777 0.0776 0.0776 0.0775 0.0774 0.0772 0.0771 0.0769 0.0766 0.0764 0.0761 0.0758 0.0754 0.0750 0.0746 0.0742 0.0736 0.0731 Section 10.3 0.8291 0.8262 0.8233 0.8204 0.8175 0.8147 0.8119 0.8091 0.8063 0.8035 0.8008 0.7981 0.7955 0.7928 0.7902 0.7876 0.7850 0.7825 0.7800 0.7775 0.7750 0.7725 0.7701 0.7677 0.7653 0.7629 0.7606 0.7583 0.7559 0.7536 0.7514 0.7491 0.7468 0.7446 0.7424 0.7401 0.7379 0.7357 0.7335 0.7312 0.7291 0.7268 0.7246 0.7224 0.7202 0.7180 0.7158 0.7136 0.7114 0.7092 0.7070 0.7049 0.7027 0.7006 0.6986 0.6966 0.6947 0.6930 0.6915 Solutions to Chemical and Engineering Thermodynamics, 5th ed Section 10.3 Points are experimental data, and lines are correlation using the Peng-Robinson equation of state with k12 = 0.11 40 25 0.9 35 0.8 30 20 0.7 0.6 15 20 15 k2 k1 P. bar 25 0.5 0.4 10 0.3 10 0.2 5 5 0.1 0 0 0 0.2 0.4 0.6 0.8 1 0 0 0.2 x 1, y 1 0.4 0.6 0.8 1 0 0.2 0.4 x1 0.6 0.8 1 x1 10.3-13 Clearly, because high pressures and a simple carbon dioxide + hydrocarbon system is involved, and equation of state, such as the Peng-Robinson equation, should be used. Using the program VLMU with kCO 2 nC6 011 . (Table 9.4-1) results in no solution at 140 bar and 75qC. However, trying the bubble point and dew point pressure programs we obtain the following results (at T 34815 . K ). bubble point P, bar 12.96 37.93 64.15 89.14 91.36 93.49 95.52 97.43 99.17 xCO 2 0.1 0.3 0.5 0.7 0.72 0.74 0.76 0.78 0.80 yCO 2 0.001 0.05 0.1 0.3 0.4 0.5 0.6 0.7 0.73 0.75 0.78 0.80 0.82 Program doesn’t converge at higher CO2 concentrations dew point P, bar 1.21 1.28 1.35 1.76 2.07 2.51 3.17 4.31 4.83 5.25 6.03 6.70 7.56 yCO 2 0.9 0.92 0.93 0.94 0.945 0.947 0.949 0.9495 dew point P, bar 15.06 20.25 24.64 32.02 38.56 42.62 49.39 52.75 Program doesn’t converge at higher CO2 concentrations 120 100 80 bubble curve 60 dew curve 40 20 0 0 0.2 0.4 0.6 0.8 1 Since the program doesn’t converge at higher concentrations of CO2 along either the bubble point or dew point curves, we have to make an estimate of the CO2 concentration based on the data Solutions to Chemical and Engineering Thermodynamics, 5th ed Section 10.3 above. There are two possibilities: (1) The CO2 saturation of the liquid at 140 bar is in the retrograde region at somewhere between xCO 2 of 0.8 and 0.95 [Note, simple equation of state programs, such as VLMU typically do not converge in the retrograde region, and more sophisticated algorithms and numerical methods must be used]; (2) at 140 bar only the vapor exists, that is, all the hexane vaporizes. Note the enormous solubility of carbon dioxide in hexane and, indeed, in reservoir crude! That is why carbon dioxide has been used in enhanced oil recovery (crude oil swells so more is recovered, and viscosity drops so the trapped oil in the earth matrix flows more easily.) 10.3-14 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-14 10.3-15 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-15 10.3-16 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-16 10.3-17 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-17 10.3-18 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-18 10.3-19 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-19 10.3-20 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-20 10.3-21 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-21 10.3-22 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-22 10.3-23 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-23 10.3-24 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-24 10.3-25 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-25 10.3-26 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-26 10.3-27 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-27 10.3-28 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-28 10.3-29 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-29 10.3-30 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-30 10.3-31 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-31 10.3-32 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-32 10.3-33 In the folder Aspen for Textbook>Chapter 10.3>Problems>Prob 10.3-33 Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 11.1 11.1-1 (a) H V data are obtained from the Superheated Steam Tables. Also H L T , P 01 . MPa | H L T , any pressure since, at low and moderate pressures, the liquid enthalpy is independent of pressure to an excellent approximation. Therefore, H L T , P is taken along the saturation line, and then extrapolated to high temperature, as shown below. 11.1-2 General: The starting point for solving this problem is f NL2 f NV2 xN 2 J N 2 f NL2 yN 2 P PN 2 where PN 2 is the partial pressure of nitrogen in the gas phase. Also, f NL2 "liquid" nitrogen, is 1000 bar according to the problem statement. (a) Ideal solution J N 2 1 ; also PN 2 1 bar (problem statement) thus xN 2 (b) Non-ideal solution PN 2 f NL2 1 1000 0.001 fugacity of pure Solutions to Chemical and Engineering Thermodynamics, 5th ed. c Here ln J N 2 0526 . 1 xN 2 Section 11.1 h . Since, from the ideal solution case above, x 2 N2 is quite small (and will be even smaller here), it is reasonable to assume that 1 xN 2 | 1 , ln J N 2 J N2 , and 0526 . . Thus 16922 . PN 2 xN 2 J N 2 f NL2 1 . u 1000 16922 . u 103 0591 The Henry's Law Constant is the constant in the expression Hi xi thus 1 bar xN 2 HN 2 f i V , here, HN 2 xN 2 1 bar 1692 bar. Also, since HN 2 { J N 2 f NL2 and V ex 0 , we have that the activity coefficient J N 2 is independent of pressure. Thus, the only pressure variation of the Henry's law constant will be through the Poynting pressure correction to the "liquid" phase of pure nitrogen. 11.1-3 (a) To compute the ideal solubility, we will use Shair's correlation, Fig. 11.1-1 first we need to compute the reduced temperature 29815 . Tr 1564 . 190.6 thus, fL # 365 . and f L PC Fig. 8.3-1 365 . u 46.0 bar 167.9 bar Therefore at 1 atm partial pressure of methane PCH 4 ideal xCH 4 (b) To compute J CH 4 L f CH 4 the L 1013 . xCH 4 f CH 1013 . bar 167.9 bar activity 4 CH 4 - benzene: J CH 4 CH 4 - C6 H12: J CH 4 6.03 u 103 (all solvents) coefficients, we use L xCH 4 J CH 4 f CH 4 1013 . bar or 6.03 u 103 xCH 4 The following results are obtained 2.91 CH 4 - CCl4: J CH 4 213 . CH 4 - C6 H14: J CH 4 211 . 191 . McDaniel 142 . Guerry (c) The regular solution model gives, for the CH 4 - C6 H14 mixture J CH4 R|V cG G h I expS RT |T L CH 4 2 CH 4 C6H14 U| expR52 u 568 . 7.3 U . , V| ST 1987 V 126 u 298.2 W . W 2 2 C6H14 Since I C6 H14 | 1 . Note: CH 4 parameters from Table 11.1-1 C6 H14 parameters from Table 9.61. This result is not in agreement with either set of data, but is distinctly closer to Guerry's result! Solutions to Chemical and Engineering Thermodynamics, 5th ed. f i G xiJ i f i L 11.1-4 (a) Start from f i L yi P FfI H PK Section 11.1 i Determination of nitrogen properties: Tr N From Fig. 37315 . K 126.2 K 2 2.96 ; Pr N F f I #1. H PK 7.4-1 2 75 bar 33.94 bar 2.21 using Shair's Now correlation (Fig. 11.1-1) N2 f NL2 P 1013 . bar 61 . ; however, at 75 bar we have PC f NL2 P f NL2 P 75 bar PC 1013 . PC exp R|V S| T L N2 . P 1013 RT 6.6 f NL2 P 61 . exp 0.0784 U| V| W 75 bar 224 bar Determination of benzene properties: Tr C H 6 6 From Fig. 9.8-1 . K 37315 . K 5621 0.664 and Pr C H 6 6 75 bar 48.96 bar 1532 . f # 0.23 and, for the liquid, we need the vapor pressure. From the P vapor bar at T 801 Chemical Engineers Handbook we have P vap 1013 . . q C , and P vap 2.026 bar A at T 1038 . . q C using ln P vap bar. B as the interpolating formula, we find P vap 1823 T f From Fig. 7.4-1, we have (with a little extrapolation) that | 0.96 . [Along the saturation P sat line]. Thus F I H K f CL6 H 6 PCvap 6H 6 . UV 2159 R 89 u 73173 F f I expR|SV c P P h U|V 1823 . bar H P K |T RT |W . u 0.96 u expST8314 . u 37315 . W L C6 H 6 vap sat Thus, the equations to be solved are xN 2 J N 2 224 bar yN 2 75 bar xC6 H 6 J C6 H 6 2.159 bar xN 2 xC6 H 6 1 yN 2 yC6 H 6 1 together with J N2 and yC6 H 6 75 bar R|V cG G h I expS RT |T L N2 N2 C6 H 6 2 2 C6 H 6 U| V| W Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 R|V cG G h I U| expS V| RT |T W L C6 H 6 J C6 H 6 N2 C6 H 6 2 2 N2 where the N 2 parameters are gotten from Table 11.1-1 and the benzene parameters from Table 9.6-1. Because of the nonlinear nature of the equations (due to the composition dependence of the activity coefficient), this problem is best solved by trial and error. I chose xN 2 0.0 as the initial guess. Solution obtained was: xN 2 0.047 and yN 2 0883 . xN 2 0.045 and yN 2 0.944 Measured values: (b) At 100 bar the calculation is similar, however the numbers are a little different. I find f NL2 f V | 0175 . ; | 6.76 and f CL6 H 6 | 2.290 P C6 H 6 PC Solution obtained was: xN 2 0.061 and yN 2 0874 . Measurement yields xN2 0.0595 y N 2 0.968 In both cases the liquid compositions are in better agreement with experiment than the vapor compositions! Note: I have found that some students try to make a large extrapolation of the vapor pressure, rather than using Shair's correlation ... it is a large extrapolation here, since the nitrogen critical temperature is 126.2 K . If we extrapolate the low temperature vapor pressure, and make the (small) fugacity coefficient correction, we obtain f NL2 1195 bar compared to 224 bar here at 75 bar. This leads to xN 2 0.0082 , compared to 0.047 calculated here and 0.045 observed experimentally. Moral: Use Shair's correlation instead of making large extrapolations of the vapor pressure. 11.1-5 (a) Suppose a small amount of liquid, 'N is, vaporized, then there are y'N moles of dissolved gas in the vapor, and x 'x N 'N moles of dissolved gas left in the liquid we are interested in a different distillation, i.e., the case where 'x and 'N will be very small. Thus x 'x N 'N xN x'N N'x 'x 'N | xN x'N N'x Now writing a mass (mole) balance on the dissolved gas yields xN Thus 'x 'N y'N x 'x N 'N | y'N xN x'N N'x yx dx and taking the limit as 'N o 0 yields N dN yx N Solutions to Chemical and Engineering Thermodynamics, 5th ed. (b) Using y ax , N f 0 0 z Hx P x dx Hx yields dN P and x, N yields x x0 d ln x N F H x HP N P Section 11.1 I or d ln x K d ln N z HP N d ln N or ln x ln x0 N0 P HP . Now integrating between P HP ln N ln N 0 P which can be rewritten as x x0 FG N IJ HN K ( H P) P or 0 N N0 FG x IJ Hx K P HP 0 x N 0.01 1 266 0.983 . Thus, only 1.7% of initial number of moles of 0.01 we have x0 N0 liquid need be vaporized in a differential distillation (i.e., no violent boiling, otherwise equilibrium will not be obtained) to remove 99% of the CO2 . x N For 0.0001 0.0001 1 266 0.966 x0 N0 (c) For Thus only 3.4% of initial number of moles of liquid need be vaporized to remove 99.99% of the CO2 . 11.1-6 I used the bubble point option of program VLMU treating the liquid mole fraction of CO2 as an adjustable parameter, until a CO2 partial pressure of 1.013 bar in the vapor phase was obtained. The results appear below: xCO 2 0.006 0.020 0.022 0.0221 0.0222 Ptot 0.31 0.96 105 . 105 . 106 . yCO 2 PCO 2 yCO 2 Ptot 08680 . 0.269 bar 0.9562 0.918 0.960 1008 . 0.9601 10081 . 0.9603 1018 . Therefore, the predicted solubility is xCO 2 | 0.02215 . 11.1-7 (a) This problem is treated in the same manner as the previous problem. The results are given below: k12 0 xCO 2 Ptot yCO 2 0.015 141 . 0.6729 0.016 148 . 0.6869 0.0159 147 . 0.6856 PCO 2 yCO 2 Ptot 0.9488 101662 . 10078 . Therefore, the predicted solubility is | 0.01595 experimental value of xCO 2 0.00328 xCO 2 . This is considerably higher than the Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 (b) k12 0.2 xCO 2 Ptot yCO 2 PCO 2 yCO 2 Ptot 0.015 4.94 0.9011 4.4514 0.003 136 0.6566 089298 . . 0.0033 145 0.6775 0.9824 . 0.0034 148 0.6839 10121 . . In this case xCO 2 | 0.0034 which is quite close to the experimental value of 0.00328. This illustrates the importance of the binary interaction parameter k12 . 11.1-8 (a) N m . K u 27315 N G 8.314 u 105 bar mol K 3 PV RT V . 1013 bar D u 1013 . 1 m3U W g m3 N W 8.314 u 105 u 27315 . 18.015 g mol NG LM1 N OP N NQ 1 1 NNWG NG NG NW x D 1 W G . KO U O L U u 8.314 u 10 u 27315 LM1 1244 . u 10 x M1 P . D u 1013 D PQ N 18.015 Q N 1 5 3 W 1 W volume gas ; for simplicity, assume 1 m3 liquid volume liquid (b) L N G 8.314 u 105 T ; NG 1013 . V 1 m3 L . V 1013 8.314 u 105 T LM1 N OP LM1 U u 8.314 u 10 T OP . Q N N Q N 18.015 L u 1013 LM1 4.555 u 10 T U OP L Q N 1 5 W x 1013 . L 8.314 u 105 T 1 W G 6 (c) S V 11.1-9 . Ku N G 8.314 u 105 bar m3 mol K u 27315 106 cm3 m3 1013 . bar 1013 . 1 ; NW 8.314 u 27315 . 18.015 1.013 S u 8314 NG . u 27315 . NG Su x NG NW (d) xKH 1013 x . (e) N G S0 ; NW MS NG NG NW x 1 W S u1.013 181.05 . u 27315 . 8314 S u 1013 . S u 1013 1260599 . . 1013 . KH 100 18.015 S0 MS 18.015S0 18.015S0 100 MS S0 18100 .015 MS Condition for equilibrium as the solubility limit is f i L 'G eq 0 RT ln fi L fi V S S 1244 RT ln a f RT ln x RT ln xisol f i L T , P, xi 1 f i V T , P, yi 1 a f f i V which implies sol i a f f i L T , P, xi 1 f i T , P 1013 . bar V Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 since the fugacity of the species in the vapor phase as this low pressure is given by the Lewis-Randall rule, and just equal to the fugacity of the pure species at 1.013 bar. Now the last term on the right of this equation is equal to the Gibbs free energy change of transferring the gas from the ideal gas state to a solution of unit mole fraction. Therefore f L T , P, xi 1 RT ln xisol 'G 'G eq 0 RT ln xisol RT ln V i . bar f i T , P 1013 a f where 'G is the free energy changed asked for in the problem statement . Therefore 'G A ln xisol w GT wT B C ln T DT ET 2 , then 'G T G H 2 wT ' H T wT T2 w B 'H R A C ln T DT ET 2 wT T 'H 'G H G 'S G H TS S T T RS LM T N 'S CP LM N RT A OP Q RT ln xisol , but wG B C ln T DT ET 2 , also T1 wT T OPUVcT h R B CT DT 2 ET QW 2 2 H or 3 R B CT DT 2 2 ET 3 AT B CT ln T DT 2 ET 3 T R AT CT ln T CT 2 DT 2 3ET 3 T R A C ln T C 2 DT 3ET 2 F wH I 'C F w' H I R C 2 DT 6ET H wT K H wT K 2 P P 11.1-10 a) We expect the solubility of bromine in water to be quite low, so that the molar concentration of water will be essentially unchanged. Then the change in free energy for the bromine dissolution process is aq 'G aq f H O( M) f Br2 M 554 . RT ln RT ln aq 2 f Br2 M 554 M 554 . . f ( M 0) H 2O M 147 M 5555 5555 . . M RT ln RT ln M 5555 M 5555 28.4 5555 . . . where 55.55 is the molar concentration of water. Assuming M will be very small, the second term can be neglected and we obtain 'G M 147 M RT ln M 5555 . 28.4 The results are plotted below. Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 1 0.307734 0 1 G i 2 3 3.167164 4 0 0.01 b) 0.05 0.1 M i 0.15 0.2 0.2 At saturation, the fugacity of liquid bromine equals that of bromine in solution. Therefore aq f Br2 f BrL2 ; or 147 M 28.4; so M 01932 . Either by using this value in the equation above, or using the graph above, we see that the Free energy change is zero if the liquid is saturated with bromine. Z 0.225 TB 194.7 K 11.1-11 CO2 TC 304.2 K PC 7376 . bar Toluene 591.7 41.13 0.257 383.8 From Prausnitz-Shair Correlation fL Tr 0.98 ; # 0.6 . Assume very little toluene in vapor yCO 2 ~ 1 PC xCO 2 a L T, P J CO 2 f CO 2 At 10 bar Tr 0.98 , Pr a f 1013 . bar f expn V 10 1013 . bar RTs yCO 2 P f P CO 2 10 7376 . 0136 . f Figure 7.4-1 | 0.96 P L f CO 0.6 u 73.76 44.26 bar 2 Now calculate the Poynting correction 55 cc mol u 8.987 bar 1 m3 u exp 298.15 K u 8.314 u 105 bar m3 mol cc 106 cc LM N LM 55. u 8.987 u 10 OP exp 0.02 102 . N 2.9816 u 10 u 8.314 u 10 Q 1 exp 2 1 So ideal “Prausnitz-Shair” solubility 10 u 0.96 xCO2 0.213 44.26 u 102 . OP Q Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 Now consider solution nonideality Table 9.6-1 Table 11.1-1 G CO 2 6.0 G T 8.9 VT 107 VCO 2 55 ln J CO2 IT a55 cc molfI 6.0 8.9 cal cc 2 2 T 0.781I 2T 8.314 J mol K u 29815 . K u 0.239 cal J VT xT VT xT VCO2 xCO2 c 107 1 xCO2 c 107 1 xCO2 c h h 107 1 xCO2 55xCO2 h c 107 1 xCO2 107 107 55 xCO2 h 107 52 xCO2 So to find the nonideal solubility must solve 0.213 0.213 xCO2 exp 0.781I 2T exp 0.781 107 1 xCO 107 52 xCO c { which has the solution xCO2 2 hc h} 2 2 (using MATHCAD) 0102 . From Program VLMU with kCO 2 T 0.0 (using flash with equimolar feed) Pressure 10.05 bar x y 0.2062 0.9952 0.7938 0.0048 yCO 2 u P 0.9952 u 10.05 ~ 10 bar CO2 T Note PCO 2 kCO2 T 010 . CO2 T PCO 2 ~ 10 bar x 0.1174 0.8826 y 0.9949 0.0051 Which is reasonably close to the Prausnitz-Shair correlation result, especially given the difference in the methods. 11.1-12 (a) From the Steam Tables T 25q C , P 3169 . kPa (b) 1 atm = 1.-10 kPa 3169 . yW 0.0313 bar ; remainder is oxygen and nitrogen. 1013 . Initial partial pressure of N 2 0.79 u 1013 . 32 . 77.5 kPa Initial partial pressure of O2 0.21 u 1013 . 32 . 20.6 kPa Mole fraction N 2 in water xN 2 HN 2 PN 2 xN 2 77.5 kPa 100 kPa bar 8.48 u 104 bar mole fraction 0.775 8.48 u 104 0.0914 u 104 914 . u 106 Mole fraction O2 in water 20.6 100 bar xO 2 4.35 u 104 bar mole fraction 20.6 100 u 4.35 u 104 4.74 u 106 11.1-13 a) Since the vapor pressures are low, we will assume the fugacity coefficients at saturation are unity. Then the fugacity of methyl acetate (MA) and methanol (M) is Solutions to Chemical and Engineering Thermodynamics, 5th ed. 2 · vap 2 § A x MA exp ¨ 1 x MA ¸ PMA x MA exp 1.06 1 x MA 1.1260 bar © RT ¹ 2· 2 § A x M exp ¨ 1 x M ¸ PMvap x M exp 1.06 1 x M 0.8465 bar © RT ¹ vap x MA J MA x MA PMA fMA fM Section 11.1 x M J M x M PMvap 1.5 1 fugma fugm 0.5 0 0 0.2 0.4 0.6 0.8 1 xx methyl acetate mole fraction b) The partial pressure of species i is gotten from P P fi x i J i x i Pivap yi P Pi so that i J i x i Pivap and lim i x o0 x i xi lim J i x i Pivap x o0 § A · Pivap exp ¨ ¸ © RT ¹ § A · Pivap exp ¨ ¸ , so that the hypothetical standard state of a component in a mixture © RT ¹ § A · that obeys the one0constant Margules equation is Pivap exp ¨ ¸ , while the actual pure component © RT ¹ Therefore, Hi fugacity is simply Pivap . xair 11.1-14 xair Pair H Henry’s law constant xH 2 O mole fraction of air in water (liquid) 01333 . kPa u 103 Pa kPa # 0.3 u 107 4.3 u 104 bar u 105 Pa bar 1 (Do not have to consider air trapped in water) At equilibrium xH 2 O PHvap 2O f HL2 O f HS2 O f HV2 O ; PHsub 2O yH 2 O P since xH 2 O PHvap 2O 1 [Note: Because of the low pressures involved, we have neglected f P terms and Poynting corrections.] For comparison, in an air-free measurement, we have PHvap PHsub PTP 2O 2O where PTP is the true triple point temperature. Solutions to Chemical and Engineering Thermodynamics, 5th ed. Since PHvap 2O Section 11.1 has to be satisfied in both cases, we would obtain the same triple point PHsub 2O temperature in both the air-free experiment, and the measurement with air. In the air-free experiment we measure P PH 2 O and get the triple point pressure. In the experiment with air we measure P PH 2 O Pair and, mistakenly, assume this is the triple point pressure; actually PH 2 O is the triple point pressure. The error, 'P , is equal to the partial pressure of air; here 0.1333 kPa. Thus, we have % error 'P u 100 PHTP2 O 01333 . u 100 0.6113 218% . [Note: From the Steam Tables, triple point pressure is 0.6113 kPa.] 11.1-15 The fugacity of a species 1 at infinite dilution in species 2 is, from eq. 9.4-10 b Pº ª L Z 1 2 2 » ª f1L b1 L a2 b1 º « 2 § L b2P · RT lim x1o0 ln Z2 1 ln ¨ Z2 » « 2 » ln « ¸ RT ¹ 2 2b 2 RT ¬ b 2 ¼ « ZL 1 2 b 2 P » x1P1vap b 2 © 2 ¬ RT ¼ so that fL Lim x1o0 1 H1 x1 we have that ª « b P1vap exp « 1 « b2 H1 b P ºº ª L Z 1 2 2 »» ª a2 b1 º « 2 RT »» « 2 » ln « RT ¹ 2 2b 2 RT ¬ b 2 ¼ « ZL 1 2 b 2 P » » 2 ¬ RT ¼ »¼ b P· § Z2L 1 ln ¨ Z2L 2 ¸ © ¬« 11.1-16 We start with eqns. (9.9-11, 12 and 13) and have the following at infinite dilution of species 1 § wND · a1 ln J f *1 ¨ ¸ C © wN1 ¹T,N ,x o0 b1RT 2 1 1 § wN 2Q · ¨ ¸ N ©¨ wN1 ¹¸T,N ,x o0 2 1 D x o0 1 a · § 2 ¨ b2 2 ¸ RT © ¹ a2 b 2 RT § wNb · ¨ ¸ © N1 ¹T,N 2 ,x1o0 § b 2 RT a · b 22 a1 ln J f · § 2 ¨ b 2 2 ¸ RT *1 ¸¸ ¨¨1 b 2 RT a 2 © RT ¹ b 2 a 2 RT © b1RT C ¹ 2b 2 § b 22 a RT ln J1f · ¨¨ RT 1 ¸ b 2 RT a 2 © b1 C* ¸¹ Solutions to Chemical and Engineering Thermodynamics, 5th ed. 1 § wN 2a · ¨ ¸ N ¨© wN1 ¸¹T,N ,x o0 2 1 Section 11.1 § ª a RT ln J1f º a 2 ª b 2 RT b 22 a ln J f · º 2 b 2 RT a 2 RT « ¨¨1 1 *1 ¸¸ » b 2 « 1 » b 2 «¬ b 2 RT a 2 b 2 RT a 2 © b1RT C ¹ »¼ C* ¼ ¬ b1 ª a1 RT ln J1f º § a2 ª b 22 a ln J f · º « 2b 2 RT RT ¨¨1 1 *1 ¸¸ » b 2 « » b 2 «¬ b 2 RT a 2 © b1RT C* ¼ C ¹ »¼ ¬ b1 These expressions are substituted into eq. (9.9-11) to obtain ª 1 § wNb · º b P· § ZL2 1 ln ¨ ZL2 2 ¸ « ¨ » ¸ RT ¹ © « b1 © wN1 ¹T,N 2 ,x1o0 » « » f1 b2P ·» H1 Lim x1o0 P« § L ­ 1 1 § wN 2a · ½ ¨ Z2 1 2 x1 a2 1 § wNb · ° ° « RT ¸ » ln ¨ ¨ ¸ ¸» ® ¾ ¨ ¸ « 2 2b RT a N ¨ wN ¸ b w b N 2P ¸ 1 ¹T,N ,x o0 2© 1 ¹T,N 2 ,x1 o0 ° ¨ ZL 1 2 2 °¯ 2 © « » ¿ 2 1 2 © RT ¹ ¼ ¬ Solutions to Chemical and Engineering Thermodynamics, 5th ed. Section 11.1 11.1-17 I will use the Prausnitz-Shair correlation From Table 9.6-1 From Table 11.3-1 From Table 6.6-1 Tr ( 25 273.15) Tr TcO2 From Fig. 11.1-1 ) ( x) GBr 11.5 VBr 51 GO2 4.0 VO2 33.0 TcO2 154.6 PcO2 50.46 1.929 fO2 5.6 PcO2 f/Pc=5.6 so that x VO2 )B( x) 1 ) ( x) x VO2 ( 1 x) VBr ª ª¬ VO2 ) ( x) 2 4.184 GO2 GBr 2º¼ º » 8.314 298.15 ¬ ¼ J ( x) exp « Initial guess given x 0.002 x J ( x) fO2 1 x find ( x) x 3 3.539 u 10 fO2 282.576 Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed D )URPHTQ ZHKDYHDWHTXLOLEULXPWKDW xiI J iI x I xiII J iII x II IRUDOOVSHFLHVi d i DQGIURPUHJXODUVROXWLRQWKHRU\ZHKDYH DQG 7KHFULWLFDOVROXWLRQWHPSHUDWXUHLVIRXQGIURP FG w G IJ H wx K FG w G IJ H wx K ,0 f IJ K TP FG w G IJ H wx K a ,0 a xV xV H[ a f xxV V G G ax V x V f TP I I a f f V V G G xV xV a E 7RILQGWKHXSSHUFRQVROXWHWHPSHUDWXUHWKDWLV TCPD[ ZHXVH FG wT IJ H wx K RU C P FG wT IJ V V aG G f RSx x x x aV V f UV H wx K a x V x V f T ax V x V f W FOHDUO\WKHWHUP k p PXVWEH]HURDWWKHXSSHUFRQVROXWHWHPSHUDWXUH7KXVZHREWDLQ R C f ZHREWDLQ RTC f V V G G 7KXVVHWWLQJ TP FG H RT w G H[ DQG xx wx T P FG w G IJ H wx K T P a T P %\WDNLQJGHULYDWLYHVZHILQGWKDW ,0 xG T P x G T P RT x OQ x x OQ x FG w G IJ H wx K ZKHUH FG w G IJ H wx K H[ T P G ,0 d i P Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed x 1RWH WKDW WKLV FRPSRVLWLRQ GHSHQGV RQO\ RQ WKH PRODU YROXPHV RI VSHFLHV DQG QRW WKHLU DV VKRXOG EH H[SHFWHG VROXELOLW\ SDUDPHWHUV 1RWH DOVR WKDW DV V o x o 6XEVWLWXWLQJWKLVUHVXOWLQWRHTQ f cV V V V h V aV V f x x a 2QO\WKHQHJDWLYHVROXWLRQLVUHDOLVWLF7KXVWKHFRPSRVLWLRQDWWKHXSSHUFRQVROXWHSRLQWLV V V V V V r V V V V RTC DERYH\LHOGV a f {a fd i V V } aV V f {dV V V V i aV V f} V V G G V V V V V V F wd N G iI GH wN JK H[ Axx ZHKDYHIURP GiH[ RT OQ J i DQG , i ,, i ,, i ,, i DQGZHKDYHWZRHTXDWLRQVIRUWKHWZRXQNQRZQV xiI xiII E /LPLWRIVWDELOLW\FULWHULRQLV G FG w G IJ H wx K ZKHUH TP a f xG x G RT x OQ x x OQ x Axx G ,0 FG w G IJ H wx K VHWWLQJ T P N j zi ,, ,, i , i , i Ax DQG R| Ac x h U| R x h U| V| S| RT V| x H[S|S| AcRT W W T T R U R U c x h H[S|S| AxRT |V| c x h H[S|S| AxRT |V| T W T W xi, H[S 7KDW RT OQ J Ax , D 7KH HTXLOLEULXP FXUYH LV RQH IRU ZKLFK Gi Gi RU xi,J i, xi,,J i,, IRU ERWK VSHFLHV LH i 7KXVWKHHTXLOLEULXPSKDVHHQYHORSHLVWKHVROXWLRQWRWKHHTXDWLRQV i RT OQ J )RU G H[ TP VROYLQJIRU x \LHOGV x \LHOGV xx G ex RT A a x x RT r RU xi, A f x x RT DQG xi,, A RT A Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed D 5HJXODU VROXWLRQ WKHRU\ VXSSRVH WKDW S H[ RU VLQFH G H[ H H[ T S H[ WKDW G H[ H H[ 7KLV LV WKH FDVH IRU WKH & + &&O V\VWHP EXW QRW IRU WKH & + &6 V\VWHP 7KHUHIRUH UHJXODU VROXWLRQ WKHRU\ LV QRW DSSOLFDEOH WR WKH & + &6 V\VWHP 7R WHVW WKH +LOGHEUDQG 6FDWFKDUGPRGHOZHXVH xV xV G H[ G G xV xV ZHREWDLQ DW x u u u G H[ FDO PRO - PRO u FRPSDUHGZLWK | - PRO H[SHULPHQWDOO\7KXVZHFRQFOXGHGWKDWZKLOHWKH &&O & + a f V\VWHPKDV S H[ DQGWKXVPD\VDWLVI\WKHUHJXODUVROXWLRQmodelLWLVQRWZHOOUHSUHVHQWE\ 6FDWFKDUG+LOGHEUDQGUHJXODUVROXWLRQtheory E 6LQFH G ex LVDV\PPHWULFIXQFWLRQRIFRPSRVLWLRQIRUWKH &6 & + V\VWHPZHZLOOUHSUHVHQW WKHFRPSRVLWLRQGHSHQGHQFHRI G H[ E\WKHRQHFRQVWDQW0DUJXOHVH[SUHVVLRQ G H[ a H[ f Axx ZLWK A - PRO VRWKDW G x - PRO DVLVREVHUYHGH[SHULPHQWDOO\)RUWKH RQHFRQVWDQW0DUJXOHVHTQE\HTQ A T8& | . 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7 L L L L L 1XPEHURIGDWDSRLQWV Q URZV GDWD Q ,QWHUFHSW E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORWV I ; E E ; I ; D ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORWV I ; E E ; I ; E ; < DL L = EL L Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed ª ª W * º ª º º» »º » ¬ [ [ *¼ «¬ > [ [ *@ »¼ »¼ »¼ * JDP [ * * W W H[S« [ « W ª« « ¬ « ¬ ª ª W * º ª º º» º» » ¬ [ [ *¼ «¬ > [ [ *@ »¼ »¼ »¼ * JDP [ * * W W H[S« [ « W ª« « ¬ * « ¬ * [ W OQ * W [ OQ * [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * OQ * W ] )LQG * * W W * ] * ] W [ OQ * [ W §¨ ·¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ OQ * [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W ] )LQG * * W W OQ * · ¨§ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W § · ¨ ¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W [ JDP [ * * W W OQ * W ] )LQG * * W W [ § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ ] [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W [ JDP [ * * W W OQ * ] ] )LQG * * W W [ § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W ] )LQG * * W W [ OQ * ] § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W ] )LQG * * W W [ OQ * [ [ JDP [ * * W W W OQ * [ [ § · ¨ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ *LYHQ [ JDP [ * * W W W [ [ [ [ [ JDP [ * * W W [ JDP [ * * W W OQ * ] )LQG * * W W ] [ [ § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ Section 11.2 Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W ] )LQG * * W W § · ¨ ¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W ] )LQG * * W W § · ¨ ¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¸ GDWD ¨ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ 7 GDWD L * GDWD L L ; 7 L 1XPEHURIGDWDSRLQWV Q E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORWV I ; E E ; * GDWD L L < * L L = * L Q URZV GDWD ,QWHUFHSW L L L u I ; * ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORW I ; E E ; u I ; * ; Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed JDP [ $ % H[Sª¬ $ % [ % [ º¼ JDP [ $ % H[Sª¬ $ % [ % [ º¼ $ % [ [ [ [ [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] [ § · ¨ ¸ © ¹ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] [ *LYHQ § · ¨ ¸ © ¹ [ [ [ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % $ ] ] [ [ § · ¨ ¸ © ¹ % ] [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ ] § · ¨ ¸ © u ¹ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] § · ¨ ¸ © ¹ Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ § · ¨ ¸ © ¹ ] [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ § · ¨ ¸ © ¹ ] [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ § · ¨ ¸ © ¹ ] [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ § · ¨ ¸ © ¹ ] [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] § · ¨ ¸ © ¹ Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¸ GDWD ¨ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ 7 GDWD L < $ L $ GDWD L L ; 7 L L % GDWD L L L = % L L L 1XPEHURIGDWDSRLQWV Q URZV GDWD Q ,QWHUFHSW E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORWV I ; E E ; I ; $ ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORW II ; E E ; II ; % ; Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed U / T U U T U T [ I [ / [ T [ U I [ [ U [ U U / T / [ T T [ [ T [ T T U T U [ T [ T [ U [ U [ U W W ª § I [ · u T OQ§ T [ · I [ § / U / · º H[Sª ª T OQ T [ T [ W T T [ § ·ºº ¸ ¨ ¸ ¨ ¸» «« ¨ ¸»» U ¹ ¼ © ¬ © [ ¹ © I [ ¹ ¬¬ © T [ T [ W T [ W T [ ¹ ¼ ¼ JDP [ W W H[S« OQ¨ W W ª § I [ · T OQ§ T [ · I [ § / U / · º H[Sª ª T OQ T [ W T [ T u T [ § ·ºº ¸ ¨ ¸ ¨ ¸» «« ¨ ¸»» U ¹ ¼ © ¬ © [¹ © I [ ¹ ¬¬ © T [ T [ W T [ W T [ ¹ ¼ ¼ JDP [ W W H[S« OQ¨ W W [ [ [ [ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W § · ¨ ¸ © ¹ ] [ [ [ [ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W ] [ [ § · ¨ ¸ © ¹ [ [ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W [ ] [ § · ¨ ¸ © ¹ [ [ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W ] [ [ § · ¨ ¸ © ¹ [ [ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W ] [ [ [ [ § · ¨ ¸ © ¹ [ [ *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] ] )LQG W W [ [ [ [ § · ¨ ¸ © ¹ [ [ Solutions to Chemical and Engineering Thermodynamics, 5th ed *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W [ [ W ] § · ¨ ¸ © ¹ ] [ [ [ [ W ] *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W [ [ W ] § · ¨ ¸ © ¹ ] [ [ [ [ W ] *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W § · ¨ ¸ © ¹ ] [ [ [ [ [ [ W ] W ] *LYHQ [ JDP [ W W [ JDP [ W W [ JDP [ W W [ JDP [ W W ] )LQG W W ] § · ¨ ¸ © ¹ L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¸ GDWD ¨ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ 1XPEHURIGDWDSRLQWV 7 GDWD L W GDWD L ; 7 L Q URZV GDWD E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORW I ; E E ; L L < W L L L = W L Q ,QWHUFHSW W GDWD L L L u I ; W ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORW I ; E E ; u I ; W ; Section 11.2 Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed DOI JDP [ DOI EHW H[Sª« »º JDP [ DOI EHW H[Sª« » «ª DOI [ º «¬ «¬ EHW [ »¼ »¼ DOI EHW »º » «ª > EHW [ @ º «¬ «¬ DOI [ »¼ »¼ EHW [ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW ] [ § · ¨ ¸ © ¹ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW [ ] § · ¨ ¸ © ¹ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW [ ] § · ¨ ¸ © ¹ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW ] § · ¨ ¸ © ¹ Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW § · ¨ ¸ © ¹ ] [ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW § · ¨ ¸ © ¹ ] [ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW § · ¨ ¸ © ¹ ] [ [ [ [ [ [ *LYHQ [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW [ JDP [ DOI EHW ] )LQG DOI EHW § · ¨ ¸ © ¹ ] L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ GDWD ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ < DL L ; 7 L 7 GDWD L L E L GDWD L = EL L L D L D L GDWD E 7 7 1XPEHURIGDWDSRLQWV Q URZV GDWD Q ,QWHUFHSW E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORWV I ; E E ; I ; D WUDFH WUDFH ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORWV I ; E E ; I ; E WUDFH WUDFH ; Section 11.2 Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed ª ª W * º ª º º» »º » ¬ [ [ *¼ «¬ > [ [ *@ »¼ »¼ »¼ * JDP [ * * W W H[S« [ « W ª« « ¬ « ¬ ª ª W * º ª º º» º» » ¬ [ [ *¼ «¬ > [ [ *@ »¼ »¼ »¼ * JDP [ * * W W H[S« [ « W ª« « ¬ * « ¬ * [ W OQ * W [ OQ * [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * OQ * W ] )LQG * * W W * ] * ] W [ OQ * [ W § · ¨ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ OQ * [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W ] )LQG * * W W OQ * · ¨§ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W §¨ ·¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W * ] * ] [ OQ * W [ OQ * W [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W §¨ ¸· ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W OQ * § · ¨ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W [ [ * ] W OQ * [ [ OQ * W *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W W OQ * § · ¨ ¸ ¸ ¨ ] ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W [ [ * ] * ] W [ [ OQ * [ [ W *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W W OQ * W OQ * OQ * ] )LQG * * W W ] [ [ * ] § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W [ JDP [ * * W W OQ * W OQ * W § · ¨ ¸ ¸ ] ¨ ¨ ¸ ¨ ¸ © ¹ ] )LQG * * W W L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ GDWD ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ 1XPEHURIGDWDSRLQWV 7 GDWD L * GDWD L ; 7 L Q Q URZV GDWD ,QWHUFHSW E LQWHUFHSW ; < E 6ORSH E VORSH ; < E 3ORWV I ; E E ; * GDWD L L L < * L L L = * L L L u I ; * ; ,QWHUFHSW E LQWHUFHSW ; = E 6ORSH E VORSH ; = E 3ORW I ; E E ; I ; * u 1RWDYHU\DFFXUDWHILW VKRXOGXVHDTXDGUDWLFIXQFWLRQRI7 ; Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed JDP [ $ % H[Sª¬ $ % [ % [ º¼ JDP [ $ % H[Sª¬ $ % [ % [ º¼ $ % [ [ [ [ [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] [ § · ¨ ¸ © ¹ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] [ *LYHQ § · ¨ ¸ © ¹ [ [ [ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % [ ] [ [ § · ¨ ¸ © ¹ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % ] § · ¨ ¸ © ¹ Section 11.2 Solutions to Chemical and Engineering Thermodynamics, 5th ed [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % § · ¨ ¸ © ¹ ] [ [ [ [ [ [ [ [ [ [ [ [ [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % § · ¨ ¸ © ¹ ] % ] $ ] [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % § · ¨ ¸ © ¹ ] [ [ *LYHQ [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % [ JDP [ $ % ] )LQG $ % § · ¨ ¸ © ¹ ] L § · ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ GDWD ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ ¨ ¸ © ¹ 7 GDWD L L D L GDWD L E L GDWD L < DL L = EL L ; 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