prelim1solns - Name Problem 1(15 minutes Formaldehyde(HCHO...

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Unformatted text preview: Name: Problem 1: (15 minutes) Formaldehyde (HCHO) is formed when methane and oxygen react in the presence of a catalyst. Two reactions occur simultaneously: g CH4 + 02 1+ HCHO + H20 CH4 + 202% C0; + 2 H20 The feed to the reactor contains guimolar amounts of methane and oxygen. (3.) Draw and label a flowchart for this process. Perform a degree of freedom analysis based on extents of reaction. Are any additional pieces of information required to fully specify the problem? If so, how many process variable values must be specified for the remaining variable values to be calculated. (7 pts) [00 Mal/hr Em.) V] we H Mam” ) ’1 mo 05‘ mfg; VI ['91. ””0 ’1 cm; aria/M31: now... er //____/ Dofi: mtmvvni ”Haw, fir/~99, ”501-, ”all” 140:. 3” g :7 f i' I L— Spficiaoi r _____________,__.__._-——_ 7~5‘—: 2 00:5 2 MM. Spec; Mei/WM. Name: (Problem 1 continued) (b) Derive expressions for the product strwm component flow rates in terms of the extents of reactions. (5 pts) ”my: natty " g; ” gL : YOMM,‘ g, —§L ”01- "— ’19pr " E -—'L'§7__ : 579mg“? “22.. (c) If the selectivity of formaldehyde production relative to C02 production is 19.00 mol HCHO! mol C02, and the fractional yield of HCHO is 0.855, determine the fractional conversion of methane and the molar compositions of the reactor output stream. (12 pts) 0.35")“ = ”we ——a mm, : q1.75’M/1.,— 5'0 14.00 1 ”mm; ..—--——_' .....-, m= tam/i. rt. 1/1”” = LIL73‘ ,u t.(2.u'} : WU” WNW Vin, =~ 57’— cfrnr -247..sz -. 2.7? Mil/lir- Vlcup $7?» v1.7?» 2. 25" :rw/tr firmflml CmMm'w-z '- 57? “”6th m -- ".100 03 /{UC'L’ 0 Ollr/ eff-Hwy 5.00.0173; 93M: “00/? mbhfiwy‘ ’\ it “- Name: Problem 2: (35 minutes) Acetone is a common solvent used in many industrial applications. It release into the environment must be carefully monitored. In order to minimize the release of acetone, you are asked to design an acetone recovery system, illustrated below. The concentrations on the flowsheet are in weight%. Water (100%) Exiting gas g m. Wkglh : of?” distillate 0. on“ 0.62M 001w Dlstlllatlon column Enlen'ng gas Mr Goff: G kglh : Air 0.95 g Bottoms product . Acetone 0.03 5 53‘76 gem! Water 0.02 """""""""""""""""""""""""""""""""""""""""" F kgm - - Acetone 0-19 Separation unlt Water 0.81 (a) Perform a degree of freedom analysis of the absorber column and calculate the flowrates W, G, L, and F. (Hint: chose G = lOleggwhen defining the basis) (8 pts) r" anthem: [456 1,}; «a Mom 19m fool‘s/4r :{2 3,0 lit/hit 33 tfibwfl 7'3 .- 41‘F,Mk’fi M79”, 00¢ :34 >0 Name:—_—___ (Problem 2 continued) LUZ/711V 1951mm: Wi‘ (“NJ/0.01) 3 L (coo?) Fri/aw) W mm: mm/m = L Maw-J ——v A 2 wit/r219. h fla [Ma/mac‘- {0.04){(oo) 7— PKaF/J ~—) F- :- WWW (We m/t/ 91$me (W; H. 17 lg/Ar Name: (Problem 2 continued) F is the flowrate of acetone/water that has been scrubbed from the air and is now being sent to the distillation column to maximize recovery of acetone. The weight fraction of acetone in the overhead vapor stream is 0.98. This stream is fed into a condenser where it is fully condensed. Part of the condensate is sent back to the column and part is withdrawn from the process as the overhead product stream (distillate). The reflux ratio (the ratio between the amount of reflux that goes back down the distillation column and the amount of reflux that is collected as distillate) is 0.6. The recovery of acetone in the distillate is 95% of the acetone in the feed. The bottoms stream leaves the column and flows to a reboiler where a fraction is recycled back to the column and a fraction is withdrawn as the bottoms product stream. The bottoms product is 55% of the feed to the reboiler. The bottoms stream leaving the column contains 96wt% of water. The bottoms product stream contains 99 wt% water. (b) Perform a degrees of freedom analysis for each system in the separation unit, including the overall separation unit, specifying the unlmowns, the number of independent balances, and any specifications or other information available. (12 pts) OHM unit'- UnlL1mlI/Vly :1 belt [493)»qu 21’ 0M : 2:?— lemn I Lnlb. ma [m0] We; ”ll/"3 _5. but: ga,ku‘kf 3L Dot; 1' fel :3 Emmy/[Jflt'rkr “qty fh, m3) m1, = 3 b4! 1 (Wolf) 6(2ch 1 f N’th'Lari (mic! Mojgwwimqjmr :L/ ha}: [Vinita/9w 3L: 5%: S‘Y’u 7'! 00F :- Lf-L-‘l 3'- I Nunez (Problem 2 continued) (c) Solve for the unknowns. (16 pts) Mange: F: mp +1445” F:- 15:77 / Mkri FUN?!) ’3 ”I... {0.013 1‘ /01 W) Mr (1.7? Z 0.0?JML 1—0.W( rim») m], : 2.61; raj/1.,» (Mr/«aoé dam/dc, w“ Mbofwr: aha/wen my -_~, We ,tmr 5m! x»; : ‘95qu -—7 my: Lawn/4r- vaofi Cal. 5.1%“ 80:} 1.40 1’43““ f M”: ”WW" 3 “”0 M M4; m9) —w/M 6W1”: Ml 72M; {rm—7: V” 0"" ' ”’ a{ m; ’ 17¢ {cg/4r? Nflax m - ‘r— ONVbtt I ‘1‘7161/ flay—‘89} Name: Problem 3: (35 minutes) To make raisins fi'om grapes, the grapes must be dried in a stream of hot air containing at least 10 wt% moisture so that the skins do not become too tough. The original fresh air feed contains 3 wt% water. Grapes are dried fi‘om 90 wt% water to 20 wt% water. The wet air leaving the drier contains 16 “1% water. How much wet air must be returned to the drier if 2000 kg/hr of grapes are to be dried? (3) Label the flowsheet below with the known and unknown information, with the correct units “9th Recycle K ifigihr‘ Wit/HM attached. (7 pts) Air ' n 3 KJ/hr- 3 Wt% water 10 wt% 16 wt% water 47mg [1 “a“??? slum 4 h.— Grapes "3".” 2000 kglh Ralsms 90 wt% water 20 wt% water (0 Midi. my. 80 (#7.: saw. L g i. {a {fin-Va i MEUL I’M-M WI”,- {flaw VIM _‘ 7:91: "fl Ml“ m}- [9/41” _/'-'9 rum , m} icy/hr“ EC _. _l_ ; "Min lo [Cg (ream )(m _ ¥rwv1 ! 1" if» :0. M10 (Qt/Arr a” ‘ 1‘5 2"— i“0.[- x”. *4.“ ""I. Name:__—______— (Problem 3 continued) (b) Perform a degree of freedom analysis of the process, write out the equations, and solve for the appropriate unknowns to obtain the recycle flowrate. (10 pts) WWW: MIC" 14, W5 ”2, M : kakr, fix/ I ilflflff [fa/.1) fig gDDW S of mm: {0.I)(L00o) : (w) m. fl (m, : 2w [Ci/Aw? Mar: /o.o”s)(m\ He‘d/L00“) :{0'w)(n:) imam [qr/‘3 [0,677)m : [oft/)0} n5 : 1305725 Icy/1v- (it: “303’ k'J/irh/ {Witty}: n: 1" fl. ‘— P15- Name: (Problem 3 continued) (c) Suppose the rais' are then sent to a vessel to make rum-raisin ice cream. To make the ice cream, input f sugar, milk, and mm are also sent to _I.he vessel, which mixes the ingredients and chills them so that an output stream of iceml containing 10 wt% raisins. The mass ratio of the rum to milk input is 1/10 and the volume flow rate of milk is 20 Umin. Milk is a solution that is primarily water and can be assumed to have the same properties. If you did not find a solution for the raisin flow rate above, assume a value of 300 kg/h and state that in the problem. Draw the additional process unit on the flowsheet above and label it. In this section take raisins to be a whole component — do not separate into solids and water. (7 pts) ((1) Perform a degree of freedom analysis on the ice cream-making vessel. (8 pts) Dw‘fioevw—z :- (e) Determine the weight fi'actious of sugar, rum, and milk in the ice cream and the final ice cream production rate. (8 pts) W‘Y‘ P"? l‘ 56*” m»; 3 2% WEE-M (12:22) : ”20973152: {pic by Md“: Wham L (“in 3 11> Mr“... 2: ”0.1.51— hr" wzfilf-g/hr— ”'1" ” 0" ”’Ic I’M-4:5 : arm/3,44... _ findwfflt'mraje- ” 0/30 Mulr— hi? Name: MEI/(— " $414 I '. mm; 7—. \Kyz‘ {21“UQ) m}- z: w {2903 XI“ : 0.0% M rum/k? I“ V‘M “"lr -O.} ll 5 2 0.37216; MM ...
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  • Spring '08
  • HINES, M

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