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HW Set No.4 Solutions - Homework Set No 4 ChE 317 Spring...

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Unformatted text preview: Homework Set No. 4 ChE 317 Spring 2010 RB. Eldridge Problem No. 1 — F&R Problem 4.40 Problem No. 2 - A well known reaction to generate hydrogen (currently in much demand for your fuel cell operated Porsche) from steam is the so called water gas shift reaction: C0 + H20 <—> C02 + H2. If the gaseous feed to the reactor consists of 30 moles of CO per hour, 12 moles of C02 per hour, and 35 moles of steam per hour at 800 °C, and 18 moles of H2 are produced per hour, calculate: The limiting reactant The excess reactant The fraction conversion of steam to H2 The degree of completion of the reaction The kg of H2 yielded per kg of steam fed The moles of C02 produced by the reaction per mole of CO fed The extent of reaction (zoning-99‘s» Problem No. 3 — F&R Problem 4.51 HINT: Use element balances Problem No. 4 — F&R Problem 4.71 pMeOH = 0.792 g/ml, HINT: Use element balances Problem No. 5 —— F &R Problem 4.73 HINT: Use element balances Problem No. 6 — F&R Problem 4.75 HINT: Use element balances Problem No. 7 — Exam No. 2 Fall 2009 Fuel to a furnace is 200 moles/hr of 50 mole % CH4 and 50 mole % H2. Combustion is incomplete with 60 % of the C oxidized to C02 and 40 % to CO. Excess air of 30 % is used. Determine: Moles / hr of 02 passing up the stack. HINT: Carefully consider the definition of excess air. mutatign Pad. 5 "“17 811‘: fl :1 ngineer’s C013 A O 1317? '"NT “9 'l "7 :3 1 . ./L‘-fl-L1. ._ Q—A'Euu fiw fé7” Na. 1; FTP/WP»? ”:1; V I: fry/ale” 7!7E 9,90] 4) flq'fw .0 f ”(0/ 02. : LR! ‘1 M7} N0 - b) /00/(we//\,UTI ‘f/(wcl 0L > : /o?/ km?) &A /(C w +17%, W, W, 2 / ‘//\IH3»+ {0;\ —> 9N0 +6J/A0 /r (/0 (/3; 6W5, ) :- /7/ [Shaw] 0L . WLWWJWW o) fakj/Mf'( //<zm/ Mp ) = 2,77 kjwa)/\IHJ /7/<j)/HI’ mag/c1 { 24343;) = 7.22; xmwpég.‘ 3515'. 137:0] NH} Heap ( ; égwalol. { ”—W l, Sm‘l M11 2/22” < 33¢?!) 0% ,; Awe”; fle7/Vfl NH} if C‘cvv fuy‘ra'lm a5; \7 i/Zf. 5% 2.; :ywy’ Mg = 267.5 M 01 s) 50” ‘ ”‘7 Oz. 2.! £ij NH} 2 5005‘; ”H3 = {EL—~31; /00 Ema—47‘ 05 flxv 2/2; ~ ('5); g ’ f/ffnwi 0)- : /7,¢ Z M02 — ‘Q/ZLG 0' 6:3; ltd/WV, Mqrf W {/fojna‘og{qk5mc1A/0)/ 70 k; W ijc’ W ,- 7750 ng AL» Jus a? 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