che210exam302 - IOWA STATE UNIVERSITY of Science and...

Info iconThis preview shows pages 1–6. 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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: IOWA STATE UNIVERSITY of Science and Technology DEPARTMENT OF CHEMICAL ENGINEERING ChE 210B Exam 3 1. (35 points) VLE heaven. One mole of a liquid mixture containing 60.0 mole % benzene and 40.0 mole o/o toluene is fed to a continuous flash evaporator. Vapor and liquid streams leave the unit in equilibrium at 1.00 atm. The molar flow rate of the vapor product stream is 30% of the molar flow rate of the feed stream. 115 110‘ 105 8100 i; 95 g .90 a 85 E) 80 75 7o O 0.2 0.4 0.6 0.8 1.0 Mole fraction benzene P = 1 atm - (a) Txy diagram (a) Using the Txy diagram, estimate the operating temperature of the evaporator and the compositions of the liquid and vapor product streams. (b) If instead of using the Txy diagram, you had to assume ideality and use Raoult’s Law to solve the problem. Write down the equations that you must use to solve the problem and the solution strategy — but do not solve! You may assume that you have Antoine’s data for vapor pressure (do not write Antoine’s equations out). ' #1. Vapor Fugzb Lt. wd) was; X37- 0 b F X16: 0 4 > BYAWM‘I‘D‘ 4,5 A=V+L = 03+}, =vo.”7 mo‘s 69*5 B baJamuz'. 0 (0(1) = 9; (0.3) 4— £510.?!) 10,-”. M 7* 43°C. 6g: o‘er] (rwmmu .cr) X; 2 045 PLUG, uM-o B “mum: Docs ‘hvt balance. Web? T=q3°Q (0.3) + K;(O~7> : 0.52 no T: «2°C 53" :ms XQL= 0.46: cvfi 0-53 mo 5 ‘P‘B 9‘” 2:. 0.15— by. = 0.51 c4039. ‘— ms’s‘ x9 = WV: 0.2:; 9:190 WIMH% w\+5m+0m% C330 3 3' kg: 045’ v V b ~5-V' ' ’ or :1 : HT“? LJ'B Ramg'sméij Egg - x; PB”. CT3/7eomHé ( V: P» P1, 4 p3 9cm «WWW fw V ’1‘.» v: a L r K L 'r “r 6302. 130.9 : 0.3 33 + 0.7%: KW) 1‘ "quEof (I; " *B'gii‘igk Same as Qw mm 1= <1 «3% 211:) + was 'I j- ’( ) VGOQHZI 7(aOrva? P ' PT 1 * “a P275") 'Pr'ir) 1"” ®6u€$5 T 7km“? 750mm} zm® cochOasz YgL {ram «543) 1,5@ ciachng ljgv £769“ a)“ (1) 2rs@ G‘oés ebn (2) cum? 11; we» loop wwwgm “galvv. 2. (30 points) Energy Balances. Write and simplify the appropriate energy balance for each of the following processes and state whether nonzero heat and work terms are positive or negative. Begin by defining the system. (a) Air is heated from 25°C to 150°C in a heater. The flowrate of air at the heater outlet is 1.25 m3/min and the air pressure at this point is 122 kPa absolute. (b) A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa. A total of 84 J of heat is transferred to the system, causing the steam temperature to rise and the cylinder volume to increase. The pressure exerted by the piston on the steam remains constant at 125 kPa. (c) You are watching your pennies, so before ChE 210 you mix cold leftover coffee at 25°C and hot coffee at 39°C in a ratio (1 kg cold coffee/2 kg hot coffee) in a thermos mug. You use a spoon to get a uniform temperature before you drink it. The coffee suffices to keep you awake during Dr. Shanks’ lecture. [.15m3 LCQ Ark. “lode? at my; 25°C [SD P: I73. 11%. Sflskmt 0.x}- W35 {51 “(a-ten” OPe mt 5S. E— \oaUbeLn. 49‘s AH+AEP+ AER: Q—wé 1 AEP = 0 no okay»? .i. \vt . :1 AER 20 no stamffcgm ck“ r... VQOC'MB, axumhfl Pu; act rwmmkb \Ou) P 1; W5 : o M ska-{'4' work (m: ckamfuu FM‘H) \AH': (i a A = eni‘l' -' l;\\(.'¥ 51V " A -- A . (2rd = o . mmklum} L‘soec) Hat-v J Sham 5H0...» 559cm: Ska-n Ilgtol‘ (Khan, T74w‘ b \aKLL \ 18$“sz V>78$Cm3 CLO‘Qd E “ “cm” ' 7: 1252.? = Q—w 4’ Au + Na +— AEK ‘ ‘ OP Lhwizon‘hi) )1 55.‘ = O (m+ manna) zw= 5m firm-v.) :— v| CNSfP A0 = 62 - PLVw'V\) " MM. a. M we 2- mu L SouoTlON ‘ EXAM $3 1(9). . Shw cm» m “mob 3 a}, fl, 0. ‘5 ° . gm, 2C .m mm.” Wed (lh5u|M> Hzfimsfm ' ‘Md SW ssskm: ufiae CLoSLA 5354mm: 4 AU + AEP + AEK 2 Q - V\/ 1 AEP= 0 no ohms; u‘. m. 1 AEK: O “04" mowvxa, 2 Q = o (mum.de Z, W 4 0 Well done 9;: sash.“ Sfitrs‘ ‘B E $3 E H k j EXAM #3 j £3. 3 6"COM S+¢a W —-? . 151.9% 3°° C— F: Slaau/ 0.5st B 51“; V’ 5 ‘om & ~— H (lS‘\'8°C.) -' 27475 twin} 'fiLzooec, snow) 2 30(45— as”? Ag: Ell-LS- : 5"1l5" K‘s- “ks-w ‘4 w \ A g AH = (3°14— z-nl.) -_ zqg 51:3: 5‘. 3.4 gf’m\ rGM =1— l: E “‘3 “0-5, “0‘8 L 41 3 '1? LC) CPHL‘N" = 334‘" “ °"‘89° ‘1' + 6-7604 1 ~ 35413 T 5‘ éb N 300% AH =‘ g (0.03341: 4- 0.6880551” + aneuxlu’g—{L _. 35%“; 1:3)dT 1516 3 ° 3 = “334le o «r 0.6881m's-1") n + onLoA-wo'gT.’ r" __ .5593 “II—film ln-$ 7—- (Si-‘8 3 \fi-% 5pm Lab Taku 15.8 Tit3w'c3 = (1.57. almo\ ’E’Llsm‘o ’3 34254 + 6.01) —. 4.275 53‘ l mo‘ ME = ‘157 — 4.17? = S‘z‘is hr _ 7:14 :51 74:1 ‘3‘ (93'). Cd) axe, 0.x ‘1— bw 9 ASQKN N 'PreS‘Su-IK a 3. (35 points) (Enthalpy Calculations, Short Answer) 3945 ?-= 5.0 3M 5 lwat‘ bk} M1- i;\-_.,\j (3.“ 44¢; opm‘hr-L IF T: l§L8°C M P>S‘,o low , :Ewtd WOuJA undo-5‘ (Joel «4' Ta&‘€' 3.7) r? wo-AA ‘ozw a.” LZsu ‘ AT [Ow P J H (5 “0" 4’? fiat/hit» Viz P . MS Ks< F: 3' bay Jafc‘, I\ A " : 3‘1 5 1 1°" Houd‘ — 27‘7- 5' ‘95“‘3 Vou- ° 5' m lka‘ 296 - 3 m- SZSMIk I Make“, 2?“ Fag THE 2’47) A wet steam at a pressure of 5.0 bar with a quality of 0.85 is isothermally “dried” by evaporating the entrained liquid. The flow rate of the dried steam is 52.5 m3/h. (a) Use the steam tables to determine the temperature at which this operation occurs, the specific enthalpies of the wet and dry streams, and the total mass flow rate of the process stream. What can you say about the exit pressure or range of exit pressures? The dried steam is now heated fiirther to 300°C. (If you did not find the answer to (a), use a temperature of 200°C for the initial state). Calculate the specific enthalpy change of this dried steam using: (b) Steam tables (c) Heat capacity formulas (d) Tabulated enthalpies (‘0 WET STEAM ‘30 I IL 52.5-m3/h “Drum—:D"SrrEAm (rt-ms \TLS STEAM VA?0< P: 5.0 Baa DC = 0.85 T= .7 T: 7, TABLE 3. b v49”. A-l' mos? W ...
View Full Document

This note was uploaded on 03/26/2008 for the course CHEM E 210 taught by Professor Shanks during the Fall '02 term at Iowa State.

Page1 / 6

che210exam302 - IOWA STATE UNIVERSITY of Science and...

This preview shows document pages 1 - 6. Sign up to view the full document.

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