prelim_pract1_solns_recitation - EngrD 219 — Prelim 1...

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Unformatted text preview: EngrD 219 — Prelim 1 Initials l. 15 Oilll’S' su ested time: 10 minutes You are asked to find flow rates and compositions of 1L1 streams in the distillation process below. a) How many variables must you find? Give them symbols and label them on the flow chart. h, mo! /h 0.8 mol Cz/mol 0.2 mol Cglmol @ 0.1 mol Cgfnio] 100 mol/h 0.5 mol Cglmol 0.3 mol Cgfmol 0.2 mol Cdmol 0.6 mol Czlmol 0.4 mol C3/mol 100 mol/h 0.3 mol Cy’mol 0.2 mol C3/mol 0.5 mol C4/mol 0.2 mol szmol 0.8 mol Cgfmol Xi Muffgfuol ' moi/h x1 moms/Mag 0.3 monglmol 0.7 molC4/rnol I— x, -x~; mol C'q)M-q b) How many independent material balance equations can you construct? Explain how you obtained that number of equations. Bellman : 3 comm” meM gram/id (33’ 2 n H H (p g 3 p n H G) c) What is the number of degrees of freedom of this system? HDP : s-g :o EngrD 219 — Prelim I Initials 2. 20 ohms su ested time: 15 minutes Silicon is purified for semiconductor fabrication by chemically separating it from its impurities, Metal Si(s) reacts with HCl gas to form several polychlorinated silanes. Trichlorosilane can be separated fiom the other gases by distillation. If 37560 kgmoles of Si (5) are fed to the process, how many kgmoles of trichlorosilane are produced? 0.2142 mOlH23iC12/mfll ,7 0.1429 mo] SiCl4/mol 1 0.6429 mol Hzimol n: 100% HSlC13 (trichlorosilane) HC] (g) no Al'DIMIL bn-{OMLLO WW MAXIM Ween. : Si (3) g‘r' 3.5120 :- Vl; + n; (0.245% 4 011921;] H n“ 5 "1+ r71, E 0. 99%? x2 .+ 0.219”. x ljflz U Vie = 3n,+ n1 [DJ-('19 3(th + 0. NHL-3:131:11 -: nl4.,35‘lly1—L 3-27!) ‘—‘ 0.7N1H1 no '3 n, + Ljru‘l. not 172,: ‘4'"?‘35’ harm-(1,. “‘3 D 3m+ Leno m. m : (swimmers) 3-9;, :- rm «mm-L "a = "75”" law "Tcs 3". f Looo v11. -'-n.1- [FIND/n. K1 \ 1’": ‘ l-7?v x (351er la. VII “4' 0.33“): ml _ .__ ‘3 1”“ It; mewflogma EngrD 219 — Prelim l Initials 3. [30 points; suggested time: 20 minutes: The diagram below shows a two-stage process to produce 503. SulfiJr reacts with 100% excess air in a reactor that achieves 90.0% conversion of S to SO; (no 803 is formed in the reactor). In the converter the conversion of SO; to $03 is 950% complete; Air“: Converter S n: H N2 Take as a basis 100. kg of sulfur in the feed. a) How many kgmoles of S react in the reactor? final-trim! WWlin'x k3 mac-(2 S kiwi-rd _ Md 0 ? : - : . ‘0 0+ SIH’YW (fu.k&$)(ilrfi.MI-'Q) FMth '. 32" a has 0- ?N ‘ 2 tr, mat, b) How manfkgofair are required per 100. kg of sulfur feed? "1“) mm 2 5W. " O 914530.195, mm; at mace [(3, ma 02 - Stuak pmpm‘z-«z 3 14¢ (.55 . . -- 1 Shots pwhfim mums mi“ MW ’- H Sim.prth : fig-25:4 ks Mir-G. S combru UIJ'Ls -: #- gfura‘g 01 Actual L3 “rut oz -' 3-??- 71 :.> an”,qu : 6.291: nuao 3"?- 1 d. a j l \h , .. 4’ ""1" at} WMNTMW‘ “ é'wfiw‘ 0” " Mr X ELL :i 96; 13ml marginsth 12:!“th qr EngrD 219 - Prelim l Initials W o) Calculate the Romposition of the gas leaving the reactor. _ Reader 5: 1m [:3th n50: '. Hsnzfo-f“ g ’1' 0-K- : me, = mung = I'M—2.8: : 3.93kj1m/02 "m"; file! i? = n = , 77 Mar-6N1 =27; “2' MLIL-Jill 6 01 x [g i 13'8‘ “1((1 N1 thjmdh 5} Tm “ :— 2. 5 35°}, 0' 5m: 23‘: : 0-3“?! 3:17,: 3-5“; : O-HS- (ll-'7— 27.7 I Eff—“W, WK d) Calculate the composition of the gas leaving the convener‘ “Sharr- 7. . Lawn-0.3 $01. Hid-Cd H Lam; SOLWJ O a: adv-M walnutm " k7 Suit“ 2-“ r . lab/“ms 5'31de ‘-= 2.6?- _ = I.1... : Quitmso: hg - nSolio $5 1 a q 50 _ Lb? lug 3 n“: 1050310 +£ -: O + 1-“? _l -_ 1.1% Mfg DZ '01: no“. '14.; 7 su‘s “IL-(Lin) la. it h% = H1 ___________.__ 12: q : Tomemru. TF/f—fi 0.0+ ‘ 2.10 _ S -.- ——e— =o.oo~m3 gol= ' a 0.073? “of [3“ 73% 18 ‘4 W19“th lg“) —— 24-? - o o'ilfo 5m: 1—1"- 0.21?— no 15.4 EngrD 219 — Prelim 1 Initials 4. 35 ointssu estedtime: 25 minutes H2 and CD can be combined to form methanol according by the reaction C0 + 2H2 —> CH3OH . The diagram illustrates a process to carry this out. mm n1 moles CH30H — -------------- —- Reactor ------------- -- Separator g- ---------------- ~—-—I- [film ____ Purge 100 moles W? 67.3 mol% H; 32.5 mol% CO 0.2 mol% CH4 Recycle :13 moles x1 mo] Hglmol x2 mol CO/mol . 032 mol CI-Ldmol The mole fraction of CI-L in the purge stream is set to 0.032. The fractional conversion in the reactor (per pass) is known to be [8%. Compute: moles of recycle, moles of CH30H produced. moles of purge, and the purge gas composition. 73: 0.032. “42 2 ’4- " = Not-2;: Cf-Jq is an lhfl- Hm; m? mer'u’pdc m -Hu_ rm. Do Wédmlam CHI-f1 (/00)(.001) = n1(.032) :0 Jill: 6.2.3" males! (5—- 7?qu moles Now do MAMA Wet lute-«cu (Figuring, CW) : 6.2.5 Q C: 0009(0325‘? = n. + rmL x1 n, + (flat-1,)g25’ [l ll'l- @ H1 (1003(1) 0:673) r; the, {- Zflnqx’ : 1H2, + 115‘“, (‘7. o: 0130)“)~ 1728‘) *- HI+Ht><12T {54m as C qua-1761») ‘x- ("I 5‘83 no help. g-lb‘t n] Moi Ix" G) 32.3 = n, + (9.05— (9.15%, (2) Wm: um + mun MuH‘lflfl 0b? 2 and add in ® ______—____—_._.—«————————_ 197.5 —._ 6n. => In, 2 3?.ZS‘MukSCHgoH/ m 9410me “4+0 (3) +9 1cm; ,- ‘X: = 0.7% “My”, / 3 7L. a1: 0359*1, = [0.200 "NICO Wu] _..________________ Nona flack n3. Camrth submst 1% MW + Sagan/{m Frawm‘m? mmszfl H Qm'f head“! S" 09 CO GM‘f‘ F41 : 0"? and rech = 0-19 [(00 (. 325’) + "3 “1.3 :2 0.1% [10° (31:) 4*”3 (01:09)] 3 9W5 +-03é I43 )H '* OM ‘5 MM Winn/ted m YEW ODO)(c_7>'2—§) + 10306-2, —" [nz+wj)fl'1 = S‘srr—‘cSé n3 31" 4‘ 0—1142. ' (¢'7«f+ n3) (rs-too) = 515‘ +.o3é=n3 SD/ch-a-r n3 ‘4' 3 “GEL: V33 “'3 r“ 793 [Mufcs manta, ...
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prelim_pract1_solns_recitation - EngrD 219 — Prelim 1...

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