Michigan State University DEPARTMENT OF CHEMICAL ENGINEERING AND MATERIALS SCIENCE CHE 201: Material and Energy Balances, Section 001 Fall, 2021 Written Homework Problem Set 4; 115 points Due Date: Friday, October 1st by 5:00 PM EST on Crowdmark 1. (40 pts.) Ethanol is produced by the hydration (addition of water) of ethylene, and some of the product is converted to diethyl ether in a side reaction. A feed to a reactor contains 2.5 times as much steam as ethylene. a. (10 pts.) Write out the correctly balanced chemical equations for the two reactions occurring in the reactor. b. (30 pts.) If the fractional conversion of ethylene is 0.65 and the yield is 30%, find the composition of the product gas. (*Hint: you will need to choose an initial feed basis) Use an atomic species approach when solving this problem to receive full credit. A) Cztty 2CzHs 5- pts for , nie not moi cette tho OH → GHSOH ( cztts)z0 each reaction b) hi → Hzo t Hzo t ( properly ☒ balanced) - rig ri , Mol Cztty not the cztls-OHrismollGHs-koweknowiriz-2.tn# n' 5 pts 1 diagram 5hr01 / setup laltomic species) DOF #Unkn_ ato-zmspe.ie # - 6 Cñ , , the we ② 's ,ñ•) Choose 100 + Ñz= hi = +2.5 = , yield , molls } ) spts =0 . can = also 10.0 use : . a basis Choosingusing and ' 65 molls yield N proyl e hhfprl h t stat appropr:# Mfo 71.4m¥ fc2.eu things ) not feed rii-m.tn#--28r6zg-.rishis we , Ñi Ñr= use can , 2. 5Ñ 28.6 therefore 0.65 , Ñz few a Goo basis Ñ, = for info our a vi. = we , ñz= 2.5N , use can since ③ C§-qµ n 100m¥ ② 3 (c) HD ) rising , > #otheo - } fracti o nal cow of ✓ g- pts 30% → ← w§,¥¥¥y yield -0.3 2 all ✗ Ctu limiting 0.3 can C? §y¥y§µ getup of ( 28.6) 2 = his ( 2 up balances : 17.74) +4 rig 2 Kitts)z0 zany , + 1015.43) SP " 25.8m¥ molls m¥ = is %, + 5.43 = . maximum so atomic 10.0) rxtntcons product 7.74 yyg.gg , yay, y , , , .o, t 5 , (7.7-4) " Ñy goneYanks atomy set now ↳ , n' = A. g- = g- pts limiting reactant , desired of amount Therefore: we nmgt.saesdfsedrpdrod.PT?'nfYFardx-n@ = is vis Now that calculate Yczthe = his = - 621m04s tho rig are known , composition of ñztÑ÷-ñstn• = we can the exit stream : t.tt#z7.-uTs-.y3 =° Ipt 62.1 Ipt Ytho-ioF.tt#l+s-.Es ¥3 Yarkon Davis)zo = ?I?¥ , = 1- , - Ipt nuts .u5 0.12-0.73-0.09 = 0.0-6 t.pt 2. (40 pts) In the Wacker process, ethylene (C2H4) is oxidized by oxygen gas (O2) to produce acetaldehyde (C2H4O). However, acetic acid is also produced (C2H4O2) as a side product. The selectivity for acetaldehyde over acetic acid with the reactor you are using is 12, both for overall and single-pass. The product stream is then fed to one condenser that removes all of the acetic acid, followed by another condenser that removes all of the acetaldehyde. The remaining components comprise a vapor stream, 10.0% of which is purged to the outside and the remainder of which is recycled back to a mixing point prior to the reactor entrance. Your boss wants a 85.0% overall yield of acetaldehyde from ethylene and is willing to allow a molar flow rate of oxygen in the feed stream that is 15 times larger than the molar flow rate of ethylene in the feed stream. a. (5 pts) Write balanced chemical reactions for this process. b. (10 pts) Draw and label a diagram for this process. c. (10 pts) Determine the composition of the purge stream. d. (10 pts) Determine the single-pass conversion of ethylene and single-pass yield of acetaldehyde. e. (5 pts) How would a change in the fraction of vapor stream that is purged impact the overall conversion of ethylene? a) Zczttytoz Cztly + → Oz Czttyoz → b) Ñ -¥÷ , (02,41-14) ÑZ (044+4) 2. 5- pts Zcztty 0 ( Cztluo a. city, , Gaya) f n' u Ñg ( 4%0) 02,41kg 2. 5- pts (02,41-4) f ¥1m) ( ¥a) his Ccattyo) 10ptsforcorreq(G%0 diagram 2 r c) need Will Ñr Ñ' = Ogg , , we of composition Determine to assume 150 mods catty can do now Lboundaries stream ? → identifying overall want thats → overall balance %%F i an pt , tÉ green) in purge basis a mots 1500 = the purge composition : Ña Oz , eww - 2 {, - correct getup {2 ¥+0T balance " air Czttyoz Czttyo '" = - { - ' of NN balance : Au = 1- 0 eqns {z balance : rig we 10m¥ = of reaction ) (extent entry balance = can yield yield O t 2 zpts {, information use some 0.85 =moYYµ%YmedtorFÉay , : = mainly yield ,y÷ } one - - correct / use setup 9 basis, the maximum amount Gwen our chosen of litho rib Therefore : could form that ; agg ¥¥f getup qÉs " I÷ riu= can we purge {, 4.25 is equal 14 to : {z = 0.59m¥ { 2=0.59 , solve now •n ñy= for flow rates in stream : Mia ,czµq = via of , 2 = I¥s%%¥e% = = since = selectivity know selectivity , 10m04s is 8. 5m04s = { we also yield of Zpt -24 10 150 - {, - - {2=10-214.251-0.59 = 0.91m¥ { 2--1500-4.25-0.51=145 Therefore: Yeah ,9=µ¥÷ 0.006 = , You = d) single and 0.995 - - pass yield - Single pass - ñ"""n÷?÷;_} setup Ñ3,czl Yield acetaldehyde pass . correct ( fatty single acetaldehyde of t.pt → catty ethylene of conversion single pass conversion }2pts . amount Cztluo prod no side rxh ideal . ) Coessumrng Letisstartwithfq.tl# First , city we fed is at Balance Ñz need , qµy to to determine how ( Ñz ,czµu ) reactor mixing point : = Ñ never + much Ñ 8. catty rig 2¥} gown µoÑ mind that know we rig a = : ( e. g. 10% of vapor stream is purged) he therefore of tie > Ñooczuy port Ñ8czH Therefore 9 = 19m01s : = , need we catty , ( 0.91m¥ ) 8. = , Ñzgµ Now 9 Ña = cute +8.19m¥ 10m¥ = his determine to -18.19m¥ > city all we " ne yggfOÑ find go " Amir leaves { the catty m Ñ 3. Catty purge / recycle Ñt = , cette a n°3 Ña Czthe = Ñz,czH4 = , n' 3) that leaves goes straight through reactor the he that know " Carey , cute, 8.19 = 9.1 + Ñ 8. t 0.91 mobs catty , e. condenser g. Therefore single , fczteu Now not Yield correct s¥Yidd ' ↳ §rgµp%µced µ¥¥¥f For Ñ3o_ the the fully that cztheo reacted to ¥21440 2m01 of c~HÑ Therefore = the we is ✗ Catty single pass %÷ - = in that stream 8.5m¥ = Czthe m°¥H4 1219 ÑG = of - know we , leaves denominator, 18.19m¥ 0.5 produced Cztheo . produced n' g. so , amt numerator Cztluo = yield pass ideal all 6 single pass , Ipt : 1%Yg.,-g = = ↳ For ethylene - single the , conv . pass know fed , that so if cztheo form __ rxt 18.19m¥cztkeo yield : 0.47 - Ipt e) if is you increased the vapor that purged of amount system , , you'd Cztly this have a leaving higher overall decreasing few overall , 5-pts identifying between purge relationship / corn . 3. (35 pts.) Methane (CH4) reacts with Cl2 to produce methyl chloride (CH3Cl) and HCl. However some of the CH3Cl undergoes further chlorination to form CH2Cl2 and HCl. The balanced reactions are shown below: CH4+ Cl2 → CH3Cl + HCl CH3Cl + Cl2 → CH2Cl2 + HCl A fresh feed consisting of 150 mol/h of methane with 125% excess Cl2 is joined by a recycle stream and the mixture is fed to a reactor where it reacts to produce CH3Cl, CH2Cl2 and HCl. The products from the reactor are sent to a separator where 75% of the unreacted CH4 and Cl2 are recycled, while the remaining CH4, Cl2 and the products are sent to another part of the plant to be processed. If the overall conversion is 85% and the selectivity is 3, find: a) (25 points) the molar flow rates of HCl, CH3Cl and CH2Cl2 b) (10 pts) the single-pass conversion of CH4 sptsdiagram To receive full credit, use the extent of reaction approach in this problem. ÷É☒Éss 4¥) 4¥¥÷¥D ¥7,9 ñ :* .=¥ ri.ae ✓ 337.5m¥ n' ( " ' 4 k¥:) a) the molar are ethan he 5 flow rates the same . 3 of in HCl , ctlzd.be streams 3 we can use flow some Ell our to info given some rates Ctke conversion 85% → tianya.at?s-ci-ia:rred- { pts use of overall eÑ We 0.85 = 0.85 n' -22.5 thot /h = 5cm , Clz know also l502goÑ5c = fed is 125% in excess , zpf.wfrom s?xngniciz--2.25l150.mot-n)(Ynm%fek-*.)- m# 337.5 excess We now can write balances on OLÉ lb.0.undaro.es CHU : Clz Ñ5aµµ= Ñicµ n' : CHM " sick n' 5 = , Ñiciz = age , { - 0 - { ) in blue g-pts i { correct {i + the - {z - , {z extent of Rlhenquatrons ris.eu HCl : Cttzclz Ñ5oµµz : the to { We :{ ¥ 3 also = 3 we { ' of sweetie "Ñow we 2 = that can values known - { , selectivity = Cttzclz t2§ = MIN the Ñ5sctts# , our 127.5 = , know ÑS {z in 150 = {z balance: cthe 22.5m¥ + 0 + = plug can we {, 1- 0 = - =3 , so : %_g{÷ p 31.9 we find know the 9, he {z flow rates 3 ÑS Cuza , = 127.5-31.9 = 95.6 Ipt moly 159.TW#1pth5,CtlzClz-- Ñsttcl = 127.51-31.9 = - m¥ 31.9 Ipt - b) find single to need we { setup zpts rect we £+14 Csingle Ñ3cµu sepor §and galant We Ñ2cµy-ñs = I know also of Ctty conversion : pass) from separator zpts pass a that = balance " over the : Ñscthe know that treaty + Ñ5cµu= 22.5m¥ that -_ " " he Ñhlcthl "* C problem " " the statement) Therefore : n' zcµ = 225 +0.75mi , Ñ3cµy = 90m¥ sexy hire We zp§ get the balWHY ance of need now can = my 67.5 rizotey from rizane-n.am Ñ2cµ pi 150m¥ = , Ñrctiu Therefore £0144 : ñaoµ , + 217.5 = 67.5m¥ m£ , )= ↳ ingle pass faey + which balance a point mixing , = 21752,29¥ 0.59-a Ipt we at