final_practice_solns

# final_practice_solns - Page 2 Problem I[30 points 25...

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Unformatted text preview: Page 2 Problem I [30 points, 25 minutes) A ﬁre! stream containing 85.0 mol% methane and 15.0 mol% ethane at 25°C is combined with a stream of pure oxygen at 25°C and completely burned. The combustion products are cooled so that they exit the reactor at 25°C. The process is continuous with a fuel feed rate of 1 mall's. Oxygen is fed at 10% excess. Calculate the rate at which heat must he removed {tom the combustion reactor. Be sure to include an enthalpy table in your calculations. How is your result affected by the percent excess oxygen? ##ﬂ—t—ﬁﬁ [tr/Mal CW1“- 202 (31—, C02.(91+ “Lo (v) on: -— ‘Moféta C;Hs(97+ ionﬁ) ~42 ZLo-Lta) + sit-Low) ma"L v- -ls§ﬁ.q Wmol L_________—J -—9 These Jam‘meT-abl; B-l Pti’t‘alia +1: liiuid Woof-w aspndud'. w“ WHMHHMM vim-I, as biplmwhbili‘ka 1'2 m: 035 molth ,5 0'”; not czar/5 Vi; mol 501,15 Ii} mo] legal; “Lme Shah. be. HLDl-Q hi, Mar 0,, I; and hula-L? Io‘L Mum 01. 15°C. mt» momma} mHv. Mira: l:de : mum D 1m“ 01 1 x 0.35 W‘CHH + 3—.__..'S'“°'°" a: BIS “"51”!- Mﬂl (All! S MIC-Ll,” 5 :- (2)(o.is) + (smug): 2.11; 5‘19 7 5 10'1. Menus = (1.11§)H.Io) : ZJ-PS “4°”; 5 n+DW/HL in h; ,Yig J 1:1»;- . . Mlcu C: n; = (0.893(1?+ (049(1) :7) h1=L|S 37-; H1 2|"); 3 (0.1:)LQ) '1' (OJ?) (6) :7) I53 : Z_f§ Illa—[SE- o: riu= (0.1)(2.1-'L'5} :- 0.11%;01 EYWan! £1: 0.89".“%J :hc“ ELL- o.IS"‘”T" :{ncwb l A l A 'n m H [N 1"] Du: Hour §ubshw u— tth (kaml) twin {warmly C “W 0 g; -Q—— -— —‘ abut OJ? '9"— 01 2.43" «9' 0-7-1 ‘9 _. ._ t. IS“ COL __ __ 2.r§‘ H] HLO{V) L? we Hum) nsnmw W53 Iva-Emu. shut-c. . , n . A0 C3: mCHH‘(AH0°)CHu + VIC-UH, (19qu r A "' gnLHL + :rijﬁj J IN obn- = (’o 3: “4°?” )( £990.39 [‘3 mica-w - 1;:- mr) + (ms 5 ‘)(-Is§7,qﬁ ‘9‘”: kw + mun'SH‘lDlv3)( #410] : I13 8614MB, ﬂu Md 012 Fax/M'm WWJ, ‘wa ‘HuL MEL LS" /\ 9410 SM u (awn (\$15.1) [£123 (ﬁg-Jar) CHL; 0.g5‘ ~75+V U CL,“ 0.x? —€L+.L7 UL 2.45— —8~ all 9‘ COL H a LI? -3‘?Z.§ “low; _ u 2.1? Ll‘ﬂ‘ﬂ A . . . A“ Q : nolﬁm (A H; )OL + VILOLDW (AH; 3CDL 1 A . A0 + mﬁwlﬂmf (AH‘fOJH'LDw') * noun (DHf )01, A . A“ - o _ mm,“ (MEDCW » mm“, (AH; )CIHL : or: (e) + i.;g‘{#1ﬂ.§)+ Lifhwmﬂ - 2.x (9) * oxﬁ-Hm) H emu—MU) ‘1 I 1375 aw You are checking the performance of a reactor in which acetylene is produced from methane in the reaction: 2CH4 (g) -) C'sz (g) + 3Hg(g) An undesired side reaction is the decomposition of acetylene: CqHz (g) -) 2C(s) + H2(g} Methane is fed to the reactor at [500°C at a rate of 100 mol CH4fsr Heat is transferred to the reactor at a. rate of 975 kW‘ The product temperature is 1500°C and the fractional conversion of methane is 0.600, A ﬂowchart of the process and an enthalpy table are shown below. 975 kW 10.0 mol Cl-l.(g).‘s 1500°C Product at ISOODC N1 (mol CHJS) N2 [m0] CzHy’S) N3 (mol H215} N4 (mol C(s)fs) References: C s , H2 - at 25"C, 1 atm Substance - (A) [30 points] Using heat capacities (which can be taken as constants) and reference conditions given above, write and solve material balances and an energy balance to determine the product component flow rates and the yield of acetylene (mole (:sz produced! mol CH4 consumed), cm (g): Cp : 0,079 kJumoPC) (:sz (g): Cp : 0.052 kmmoPC) H2(g): Cp = 0 031 kJ/(moPC) C(s): Cp = 0,022 kJ.’(mol°C) GDVOMMIM'. 10.: loll-04:) = 4mm] CHols 6541mm: io{1) = LHi)+ 3N1+ N4 :5 gml+ml+2e (1:) l-lbnimu. 10!”) '= Ll-ll-t)+ 3N1+QN3 11> 1M1+1N521L~1Rﬂ l—l'L : (AHEJL + CPL (.0500 age-3 ‘— ﬁHq C;HL C H9. 1 acmTwle 8-1 A GHQ AME) + CPL.” (N773) : -7485+ (00.)?)(lujg) : L“. 6g wq — 0.051 11475) = loll-u" CZH’I— 5 — 215.75 + i ) r” ‘3 + (0.031)(m7~.’)= 98-77.. I47, .' = e + (0011) (NW) 2 324\$ C: A A Q=AH=97S\$§: 2mm ~E HALHd; DIM: in 97S: (u)(t+1.e~:)+(303»w31\11_+MSFDJMC: + (Ba-LA“)qu — (LU-(SX'O) illqﬁ : 30%.Lh’ M1+ 4:,71m3 + 3.1.wa 93%) Solve KLQUUJ (NH) Swmwmglﬁ ZNzl-qub ZNL‘J (PMLP 1~¢+2m=94 :3 G-M4+QN3=Q|+ :3 Dug—IQ =M+ “'0” = “Z. r Own; + 0.107 m, C 302!“— 3N7, + 0.35»; 1 0.1mm E (04“, + 0.313 N3 + 01¢qu 2 o? 2 0.30 m — 01786 M; 0% N1 —— 0-735 (1N3 er?) = (034541.),‘11 T ILA/5:7..Dg N3” I'.2?'2. N3 : ,207 % N3 ‘ gifMoszé M” 2(‘fSJ—ﬂ’c u” :5 Musimrca N“ 2N1.” érl 3—) ML: 1.:er1HL/S (B) [10 points] The reactor efﬁciency is deﬁned as the ratio (actual acetylene yield I acetylene yield with no side reaction). What is the reactor efficiency for this process? NIH; no guilt ryn» M [l IOLIrOJQ 3 mei cm): Nu = D ML 3 Ema! (iHL f; M : 9mm Hm %N"{’l 5 MS 3 0 g— MWGHI 6m} CHle Jim/qu WHM‘t _ (WU/I7 Kwaﬁmﬁ * 7? : 0.9354 ‘ 0| f ﬁw’lww = 2-90 m Cw“ 6m»: CH9 MMMJJ‘S U 0‘ Ly}? Mi CLH-L Mlﬁuv consuqu A liquid stream contains 12.5 mol% n~butane. The rest ot‘ the stream is a heavy nonvolatile hydrocarbon. The stream is fed to the top of a shipping column, where it contacts an upward- fiowing nitrogen stream. Residual liquid leaving the bottom of the column contains all of the heavy hydrocarbon, 5% of the butane that entered the columrL and a negligible amount of dissolved nitrogen. 100 mol/s @30°C, latm m (mol!s)@30°C. latm xB=12.5 moles 87.5 mol other HCls yt (molemol) l-ya (molNzi'mol) 88.125 mol/s n1 (molNyS) (J 625 me] His (5% ofB fed) 87.5 mol HCls Basis: 100 mol/s liquid feed stream, Let B: n—butane. HC=other hydrocarbons. (A) [15 points]. The highest possible butane mole fraction in the exiting gas would be that in equilibrium with the butane in the entering liquid {a condition that would require an inﬁnitely tall column to achieve). Using Cox charts and Raoult’s law to relate the mole fractions ofbutane in the entering liquid and exiting gas, calculate the molar feed stream ratio (mol gas fed 1' mol liquid fed) corresponding to this limiting condition. Lb; ?; (30°C) 1:. 4| 1 = ZIIOMMHé (Fig. 6.1-?) I'm Rama‘s 1w y“ P = «3 33mm y“: mgPﬂtso-c) (aizsxzilo) = Z O. P ZHo 3H 667. Math , leawsmwhud ‘ n‘l' (o‘mﬂ = ULSN'QS‘) ml 3 31rd, mt)th 5 7:111 mote. Ioo+ A3 : 31L] +‘3%.I Lpttwu. '- 15; = 22.1 L15“?- Male 3:! 11.1 0.111 nghéd Malt. +Cd Loo {rd H (B) [15 points} Suppose the actual mole fraction of butane in the exit gas is 80% ofits theoretical maximum value, and the percentage stripped (95%} is the same as in part (a) Calculate the ratio (mol gas fed I moi liquid fed) for this case. IF m= Oshawa = 0.174, "glint-i 5am slaps a; in LA) C’s—lbw 1m (0.279) : CIL§)(D.7S) ﬂnppm 'm, : Lem, “LN—WM S TQM Mob ion-(.qu “30+ V113 '= LVL L 4» QqJ Vin: 30.1mm: 5 mufﬂed C 30.7 : 020.7 Melt. kw; M L00 ...
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• Spring '08
• HINES, M
• Mole, Tom, OJ, Oshawa, MEL LS

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