HW_4_Solution

HW_4_Solution - VIM =‘ 3 ml/Mm «WT—M . ma rl %:...

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Unformatted text preview: VIM =‘ 3 ml/Mm «WT—M . ma rl %: 45W3/Mih O ‘2 thr > who r A “Wang Yea. nu?) .-.~ emu; not in Q “lg/2; \iw Mo; = \* 0-” “‘10:” Va; “lam .91: u 5 Li‘ng Yozflnm‘ftfi‘i 'T- -—n—— E 5 [mm species) @ mom memes (c, M09 N90“) \npfi' ” 9A1me M2 Una-‘0'. \IIM2 fig 7’“ 3W0 no; a: mole ,qM +émac ,q3 = #002061 \md ($50! ImblclfH-m W Jé W353er and O mum a: away 45:2 7r n 3 \qDU‘Wi/W“ C- 69: YCOLQQUO) 2i? YWZ c\'. 3 = WUCIDD) ‘5‘? 0.‘ Qrio5L = 0.19. (HUGH * 0; How Mamsmoo Plum? fiwm N2 balance'. non“ (MOO/3.7L») ‘31 7 €35.32 \IMl \= Hay, “* Na + W20 me Yep; \= V02 + Ym + on 4: (3.00m; + 6.0%?) * = \in; Wm mam \[01 g 0:18.21- VH2. \lNz : Orig!“ m \{02 ’lemm D bm\afi£€ I aLwa.sz)Lo.1g11—x{oz)= 21% + 3300402 +~ Emma 7910»- toama 70; 5‘13?! + @nwo1 we»; 481049 V07, YO} : 04H)? rm : 2mm va'm CRaC‘ +35”. 02 ---n> 002 1— HID-r Hm MM #2 on ---a 4 w2+5 H10 Mimi ‘01-. 39/31)“?— ‘1'5 03/2) 3’ $2017 K"(\‘D‘L/I‘rixn Ola EXQJ‘E-gqé’ m 2 OIL) am Jc’cwfi4 97. Basks m0 “W:- W‘ WV {Lead “1-. CH4. H in 60E. rugde and ‘Wrz'. H01 in (-31 Were-Foray Arcflzgr'Afi—L“ Jnfl "In G; magma Mme. F“ and R. are 02% Gaming *Muenab 9'0 i233. = 04m M 0.44% 0H and 044% H2, sq e, E Q '5 a? 2. H (9 Reader lei rpm: o-uqutfi -* 3‘ ’ 9%2 , r1“; “ QW‘JIUDU) ’g“ 3'1 W: 40H -' U-WUUD) +5“ +332, we: “mm-— 04%, Q : v10 —— o *3; 'wam w. 31.... __,_ _ 1m " 080 ‘17 g" 3% 23:2; = 040% =3? §$=O«4’+L’r4 ‘H “Bub inlay balances I “Hui ’ {-33 3 Rh- yUh/: 55‘“ 1W7 54.19. '13, 1 (33:9) “ID: 0.44 I rmxer 4% 133+ aw = w 7% M F 2 mg moi/h, w. 133+ was: + rig Hm fink—.- 93.301 mm - r‘1 VH1- Hz a. 3641 ———-—-—~— : o. a 3 Wmqu 35-\\+S“s-mm 0‘ i Yew "' Vsz ~ O (2me Mom um: ymfle’osng fhal Mw: 7Ll2)’r‘8= Cm ' Geode. f 34513WM)/022 1. 5.8-3 01.18 ‘39: t 3533 15.23 t a. \M - t ) mm W J RL=3.g3(\‘33)Pg,\lg—§L 4.72 a. b. . i3.86*l2.97 __.,Q‘. Say r23 mols fuel gas constitute the sample in}ected into the QC. If xcm and xclm are the mole fractions of methane and ethane in the fuel, than usimoljxczfilmol CgHz/molfll mol Cf] mol Gill‘s) 20 r1x(moi)xCHd(mol CPL/mol)“ mol Cf! mol CH4) :0? ll XczH,_(mol Cal-imfmol fuel) xCH‘ (moi Cer’mol fuel) =01 I76 mole Calla/mole CH4 in fuel gas mole H30 [L134 g HZOXI mol/08.02 3,) mole producz gas Condensation measurement: —--;--—-—---—~——-— £26 w 0.30 mol product gas Basis: 100 mol product gas, Since we have the most information about the product streatn composition, we choose this basis now, and would subsequently scale to the given fuel and air flOW rates if it were necessary (which it is not). CH4 +202 —+ CO1 +2l~130 7 100 mol dry gas I h n.‘ (moi cm.) 0-1175 '1- (m' CzHal 0125 moi no J mol “2 (mm C01} (1.874 mol dry gas J mol F 0.119 mol colmoi D.G. | [h x(mo|Nz!mo£) “3 (m0 0" l (0881-14) {mol glmul no.) 3756 FL. {mol N.,.’h} Strateav: H balance : n}; C balance 22> n7; m"— “ 0 balance H balance: 4m +(6)(O.l17621,)z(100)(0.126)(2)2> nl =5.356 mol (3e:4 in fer 2:» 0,1E76(5.356)= 0.630 mol C385 in fuel C balance: 5.356+(2)(0.630)+112 = (100)(O.874)(0.l 19) :> MI =3.784 mol CO2 in fuel N2 belanco :> n3, :c Comgosition offuel: 1.156 mol CH4, 0.630 mol Cal-la, 3.784 mols CO2 m 0.548 CH4, 0.054 clufi, 0,333 co2 N2 balance: 3.76t73=(100)(0.874)x 0 balance: (2)0734) +2n3 :(100)(0.126)+(100)(0.874)(2)[0.119 + (0.881—x)] Solve simultaneously: 173:1836 mols 02 fed, x = 0.8%3 _ 5.356 mol CH4 2 mol 07 0.630 mol Cali6 Theoretical 07: ' + ' W E mol CH4 = 12.92 mo] 02 tequii'ecl 3.5 mo] 02 I mol CH4 0.21m0103 (5.356 + 0.630 +3.784) mol fueil 7 mol air molair [1 mol fuel Desired Ozflz = l4.3§ mol 0: 4." — 2. 2 Mxtomzlm Desired "/9 excess air: Actuai % excess alr: ' x100% = 46% i797 (13.36/02?) mol air 9.77 mol feed =2 Actual molar feed ratio ofair to fuel: GAD—3'. Dig-+0.00% C \WHL {ml (Dz :71244 mi wa = (H 12-013C [mic H20: D-IDUWMH \W‘ H WW” =- ArHBmm hot nu, Luigi} lmbH-X S0; 033000368 “WAS I‘m“ 802 r 0.001% M\Q60L “(15 52.1)[933 \ml 8 4.1% new? . Y“ = QQDIO QLQQ/SG 02- W J % WW1“ 8%02 Wmd + 95% mmi n}: \-ar5(‘149'-14+ 3241‘s +0.0q3u23 = lava? WW 02. WW ELng 518‘ 01 bahnca'. n3; §Q~9‘:l—mu\ 07, - 013! ml 02 =” QMW ML Oz (led WM N7. bdmw" (\Z : an“ Lil-11‘) ‘“ %-\4- muk M1. Mk .3 SO; mums '5: &\—§, 80:, in SW Kn‘s Jr O-XGUfifina "3 302' {h m 0.0%me— O.DO°13L9 ~ 0.0mm: x r: CG-UOfii‘zw 0.D%%1Qr $07, In mm. Wk “1.9% + a«%m% “rm-N + womamcfl’Maax) : max M0596 W2 02 H2. ‘30; 5533 +0. 99%ax :1 WW‘ d/flj 055v; mlmm <30; Lam was) in war, “fig 55? W .~ tam-9 -. “" 5‘3-‘35‘ T134: a w Mlouj N 1 WHO Kz'élfig v> We ...
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HW_4_Solution - VIM =‘ 3 ml/Mm «WT—M . ma rl %:...

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