Name: SID: Dscusson Sesson: Chemcal Engneerng Thermodynamcs 4 Fall 2008 Thursday, November 6, 2008 Mdterm II - 70 mnutes 00 onts Total Closed Book and Notes (20 ponts). Evaluate whether the statements are true or false. You may enter your answers on the exam sheet. a) Most compressors and turbnes have mechancal effcences of less than 0.5. FALSE. er SV&A p269: Values of η [for turbnes] usually range from 0.7 to 0.8. er SV&A p274: Compressor effcences are usually n the range of 0.7 to 0.8. b) The regeneratve cycle has a hgher effcency than the standard Rankne cycles because the latent heat that would be lost n the condenser s partally used to preheat the water enterng the boler. TRUE. Ths s a correct descrpton of the regeneratve cycle. c) The thermal effcency of a vapor compresson cycle depends on the choce of the workng flud. TRUE. The vapor compresson cycle contans an rreversble adabatc expanson; what fracton of vapor s lquefed n ths step depends on the thermodynamc propertes of the workng flud. d) In a refrgeraton cycle, the lower the evaporator temperature, the hgher the CO, retanng the rest of the cycle unchanged. FALSE. In fact, ths s exactly backward. e) For pure speces, partal molar propertes equal molar propertes. TRUE. Ths works even for entropy, because the extra -R ln x term n the entropy of a mxture drops out when x. f) At constant temperature and pressure, partal molar propertes can vary ndependently, based on Gbbs-Duhem relatonshp. FALSE. At constant T and, the Gbbs-Duhem relatonshp s x d 0 for any partal molar property M M. What ths equaton means s that partal molar propertes cannot vary ndependently, because f M changes for one speces, all the other necessarly change n a way that keeps the sum stll equal to zero. g) As pressure approaches zero, fugacty coeffcent approaches partal pressure. FALSE. Trck queston. Fugacty approaches partal pressure; fugacty coeffcent approaches. h) Fugacty of pure condensed phase at temperature T and low pressure may be approxmated by uraton pressure at temperature T. M must j
TRUE. Credt was gven for FALSE to those students who commented that at extremely low temperatures, the oyntng correcton can no longer be neglected, even at low pressures. Remember that all gases become deal gases n the low pressure lmt, so t s not necessary to make an added assumpton about deal gas behavor here: that assumpton s already bult n. ) Raoult s law descrbes well an deal gas n coexstence wth an deal soluton at low pressure. TRUE. Ths s exactly what Raoult s law does. j) Actvty coeffcents descrbe devatons from deal soluton behavor. TRUE. Ths s exactly what actvty coeffcents do.
(30 ponts) 2. Freon 34a s to be used n a refrgerator that operates wth an evaporator temperature of - 0 C and a condenser temperature of 30 C. Saturated lqud refrgerant from the condenser flows through an expanson valve nto the evaporator, from whch t emerges as urated vapor. The thermodynamc dagram for Freon 34a s gven on the next page. 4 Condenser 3 Evaporator 2 a) Draw the actual process on the enclosed -H dagram. For a refrgeraton capacty of 5000 J/s, what s the crculaton rate of the refrgerant? Readng from the -H dagram, we fnd: H H 4 240 kj/kg H 2 390 kj/kg Q& c m& H H 2 5kJ / s 0.033 kg/s 390kJ / kg 240kJ / kg b) Suppose the cycle n part (a) s modfed by the ncluson of a countercurrent heat exchanger between the condenser and throttle valve n whch heat s transferred to vapor returnng from the evaporator. The lqud from the condenser enters the exchanger at 30 C and the vapor from the evaporator enters the exchanger at -0 C and leaves at 20 C. Draw the process on the enclosed -H dagram. What s the crculaton rate of the refrgerant?
4 4 Condenser 3 Evaporator 2 2 Usng an energy balance on the heat exchanger gves: H 4 + H 2 H 4 + H 2 Solvng for H 4 gves H 4 H 4 + H 2 H 2 The heat exchanger operates at constant pressure. Thus, H 2 420 kj/kg. H 4 H 240 kj/kg + 390 kj/kg 420 kj/kg 20 kj/kg Q& c 5kJ / s m& 0.028 kg/s ' H 2 H 390kJ / kg 20kJ / kg
Blue part a urple part b
(30 ponts) 3. a) Show that the fugacty of a vapor that obeys the volumetrc equaton of state: s gven by: V b b f V exp. The general equaton for fugacty n terms of volumetrc propertes s f exp V d 0 Our equaton of state can be rewrtten V + b Insertng ths equaton nto the general fugacty equaton gves f V exp 0 b f V exp d 0 b f V exp + b d b) Derve an equaton for the fugacty coeffcent of the lqud at uraton, functon of uraton pressure. ln φ L, as a At uraton, we have f L f V. We already know the fugacty of the vapor phase: b f V exp Evaluatng ths at the uraton pressure, we have b f f L V exp Rewrtng to solve for the fugacty coeffcent, we fnd ln φ SAT b c) When the substance used n part (a) s sothermally compressed, an ncompressble lqud s produced. Develop an equaton for the fugacty of the lqud. The equaton for the fugacty of a lqud can be wrtten
f L φ exp Vd And snce the lqud s ncompressble, V s not a functon of. Thus, we have f L φ exp V0 ( ) Fnally, we already know f from part (b). luggng ths n, we have b f L exp exp V0 ( ) Or V 0 V0 + b f L exp d) Now let s pass supercrtcal CO 2 over our compressed lqud. Develop an expresson for the solublty of our lqud n SCF CO 2? Assume a neglgble amount of CO 2 dssolves nto compressed lqud. The gaseous mxture obeys the followng equaton of state: V b mx wth mxng rule b y b. mx Solublty s a measure of how much of our lqud, call t speces (), wll dssolve nto the supercrtcal CO 2 phase at equlbrum. We assume the amount of CO 2 that dssolves nto the lqud phase s neglgble. Thus, the fugacty of our pure compressed lqud must be equal to the fugacty of the same substance n the CO 2 supercrtcal phase: f lqud f,supercrt Assumng the Lews-Randall rule apples n the supercrtcal phase, we have v v f φ y We have assumed that the vapor phase obeys the equaton V v To get φ, we start by applyng the defnton of partal molar fugacty: v nln φ d ln φ where ln φ ( ) n Z T, 0 We know (V-b mx ), from whch we can show ( V bmx) V bmx b mx
V bmx Z d bmx d bmx ln φ ( Z ) d 0 0 0 (Note: havng done parts (a)-(c) already, you could jump rght to ths step as a startng pont.) b mx ln φ nbmx ( nb + n2b2 ) nln φ v nln φ b ln φ n T, v b ( ln ) y v f exp φ exp Now that we have the fugacty for substance n the supercrtcal phase, we equate t wth the fugacty of substance n the lqud phase, whch we already know from part (c): L v f y f V0 V0 + b exp b y exp ( V b )( ) 0 y exp Ths s essentally the equaton from lecture nvolvng the enhancement factor, only wth expressons for the fugacty coeffcents substtuted n. (20 ponts) 4. It has been dscovered that a 20 wt% soluton of acetc acd n water wll effectvely kll the majorty of common speces of weeds, whle actually encouragng the growth of desrable plants. Vnegar has the added beneft of also dscouragng ants from eatng your flowers. The molar volume of a water-acetc acd mxture at 25 C s well descrbed by the equaton V 4.4857 x² + 34.96 x + 8.02, where x s the mole fracton of acetc acd n the soluton. (a) Fnd the partal molar volumes of water and acetc acd n a 20 wt% acetc acd soluton. The molecular weghts of water and acetc acd are 8.05 g/mol and 60.053 g/mol, respectvely. We begn by convertng weght percent nto mole fracton: w / MW x w / MW + w / MW ( ) H2O
0.20 / 60.053 x 0.20 / 60.053 + 0.80 /8.05 x 0.070 To fnd partal molar volumes, we use the formulas V V V + ( x ) x T, V VH 2O V ( x ) x T, Observng that V 4.4857 x² + 34.96 x + 9.02, we fnd V 8.974x + 34.96 x T, luggng n that x 0.070 gves V (8.974)(0.070) + 34.96 35.589 mol/ml x T, At x 0.070, V (4.4857)(0.070)² + (34.96)(0.070) + 8.02 20.489 mol/ml V 20.489 + ( 0.070)(35.589) 53.587 ml/mol V 2 20.489 (0.070)(35.589) 7.998 ml/mol H O (b) After weedng your garden, you decde to make a salad, topped (of course!) wth a vnegar-and-ol dressng. What volume of 20 wt% acetc acd should be combned wth pure water to make 00 ml of a 5 wt% acetc acd soluton (eg, vnegar)? To complete ths problem, we must recognze that for non-deal solutons, volume s not addtve. If you smply add 25 ml of 20 wt% acetc acd to 75 ml of water, you wll not get a 00 ml soluton of 5 wt% acetc acd. Luckly, mass (or equvalently, number of moles) s addtve (after all, conservaton of mass s a fundamental precept of (non-relatvstc non-nuclear) physcs). We wsh to make 00 ml of 5 wt% acetc acd soluton. Agan, we begn by convertng to mole fracton: 0.05 / 60.053 x 0.05 / 60.053 + 0.95 /8.05 x 0.055 To determne how many moles we actually have, we need the molar volume.
V 4.4857*(0.055)² + 34.96*0.055 + 8.02 8.563 ml/mol The total number of moles s thus 00 ml / 8.563 ml/mol 5.387 mol N 0.055*5.387 0.0835 moles N H2O ( 0.055)*5.387 5.3036 moles All 0.0835 moles of acetc acd must come from the 20 wt% (7 mol%) soluton. Thus, we need 0.0835mol.929 mol of the 20 wt% (7 mol%) soluton. The molar 0.070mol molsoluton volume of that soluton was 20.489 ml/mol, so we need.929 mol * 20.489 ml/mol 24.44 ml of 20 wt% acetc acd.