ChBE 2110 Final Exam Solution Fall 2007

ChBE 2110 Final Exam Solution Fall 2007 - NAME...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NAME ‘4‘”‘0 a; 27" 0 N ChBE 2110 Professor Gallivan, Fall 2007 Final Exam ..mw«~wmmmmmwm.——_mm (l) (25 pts) A process has been proposed whereby 1.5 kmol of an ideal gas (Cp = 30 kJ/kmol—K) is taken from P =10 bar and T = 300 Kto P = 1 bar and T = 500 K in a closed system. During the process the system does work, and receives 50,000 k] of heat reversibly from the surroimdings at 300 K. a. How much work would need to be done by the system on the surroundings? b. Is this process possible? ' 0. Would it be possible for the gas to undergo the same change in state adiabatically? Explain. d. Would it be possible for the gas to return to its original state adiabatically? Explain. ‘ t1 .3 T9 [/9243 x1073 wdk (imam WWW ‘ “BX/0733 A S S ~Nle N'PlRQi-fl ald‘] a T. 3-?de p? 79 fl (EM: 1500 m! (30 In f—% 43314 in E _, «r J E s » “SW73 .1 A- fiW " “I 11/05 fl _ S Imfbfilflé ‘ Ang 3 Ag” 45”} *USx/O <O a" ‘ 6- Ass-so :2 Asmajmsyfiamtg pOSS’EZf gf at. A5»...;v: «6.3/0ng fmflaifififiéj 1 L g \ €551.31 am- (2) (25 pts) Consider entropy as a function of temperature and pressure, S(T,P). a. Derive an expression for the change of entropy in terms of measurable variables. b. Evaluate the expression from part a. for a Censtant temperature process, using the ' virial equation of state Z21+£ RT .. ~ PW? gimfiit {Vii/{61 :7 7119 are: ens/£17 9x ‘1 D ‘ fwi 4’ i v as -_; 3V 1 09 W t air (at); (El 6 E3 CMSMM‘ It. ML 6130313): %) at p - s La- Nua’ flu"; {m £03 2: i + B? m 75 . 0 U" fiiS‘Srma 5J3 Mafia/1% [ME/tit] + B ‘ ENE/i 140- fgflWir-VZMVVKLJ r WW0? I"; 1. K . 37‘ r3 . P 013:- algae? a P d (' S‘s-A31“ lifih‘ot DBL“ (3) (25 pts) Calculate the enthalpy departure function for the equation of state P2 P(V—B)=RT+a—E~ Use whichever version of the departure formulas is most cenvenient: R V ,0 _ P H—= TEE] —P dV+Z~1=I—T[a—Z] f393-4»z—1=—J"r[a—Z] if: RT” 61",, 0 anp 0 (3pr Hardfa €oh"? 6‘" V V‘Bn‘g-Iai 7 p T : pV_ P U a J fl ’02- 5‘23(B+"F”T “ MM‘L‘X"? PT {- P m“ run/Q 92 ,BP‘ ~7- H: _ 10*“ ~89 15001 p ' 4- 0J0; '3 1U" '£!(Q?—"fi1% ENQT - *’ B 2%) +9 . . «f “9+,” 6113 J a RT 1 Wm ' max-1. __ ._ . . (4) (25 pts) A vapor compression refrigeration process using ammonia as the Working fluid is to operate between cold and hot reservoirs at 20 and 80 C. Determine the coefficient of performance for (a) reversible operation (except for the isenthalpic throttle valve); and for (b) irreversible operation. In the irreversible operation, entropy is generated because the working fluid is 10 degrees lower for evaporation and 10 degrees ‘ higher for the condenser, i.e. the irreversible cycle working fluid operates between 10 and 90 C. Assume for the irreversible operation that the compressor is only 75% efficient. Compare your COP’s from (a) and (b). How significant is the irreversible expansion in ' the throttle valve? Should one consider an isentropic valve, which is more expensive? 85’s. fQ 3 .. .. .. ._ .. . . mmxmvaur‘. -_ COP: GQS—g‘tO 33g ' 770-928 ‘ H5 “@163 ........ ,IQQ_.M__MEL Wig? + I 9 0 5'0 52 7 0 O. St}! _ ,. I W 3,. 750 35,0 770 {.14 3 7S0 I 820 M flu. W»— ; CW rt’SSWI I _ ‘ 0 79, cos AH; H3 441 770 4220 H3 = 8 1.1,, a J K 3 “fl " 3 I g m I (/05 “3"”; 2/05" COP: H;— H. c 20'170 1m) us 250 5-3-15? MN {1% {SEW’C WM/ 820 -920 A (210 {ISO .. Cop- a /.8 mm— .. . .W, ., m ‘mmm .w EEEE .flfi ~‘Ihinha. §§ ‘ . . A65 .mcom am. 3:3 52. mo "35$:an 5:5 .mma £2“? .fl tum .mmfiufii 2.82% 33530 Sammy“ .< .m E; £833 .2 .M .auwaom .<. .0 58nd .3558“ BM SENS“. Robsoégufiomfioh mé manna V ...m:£:3m. .m .‘ .3 ma .3 m6 . #6 ad \ I’d r“z P“;' 08 n a ....>4?fimwflm :, . . 7 @flwflofinv 4.1. . an. 7 .yliusal. _ \1 .1. «fig», a“. F £5.21?» n r. (man .2335 fl n . .Euflmfiflfluaw IIIIIIIIIIIIIIII. "l-llll $rllflfll‘lllli'.llll.‘ flaw"?Iflumflflmuhummflflfimmmmaw .w huh JAWS»... 9., . ,, fl ._ flflmqmmrflflwfiwflflflmmmmfl. Amwgy Iii ‘iflafifikmfiflflflfiflfifiq flfluuwyfi 2., W mflwMM ,2? MMWMflmmmnuy m}... .. '1 AN «wu~df“‘l.“lrll§Elli—II.Ilsa. g I_ w. ...
View Full Document

Page1 / 7

ChBE 2110 Final Exam Solution Fall 2007 - NAME...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online