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Lecture07 - ChE 210 Meeting 6 Handout A liquid mixture of...

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Unformatted text preview: ChE 210: Meeting 6 January 24, 2011 Handout A liquid mixture of benzene and toluene contains 58.0 mole % benzene. This mixture is to be partially evaporated to produce a vapor containing 85.0% benzene by mass and a residual liquid stream containing 10.6% benzene by mass. This process will be carried out continuously, at steady- state, with a feed rate of 1200 moles/h. What are the mass flow rates of the liquid (th ) and vapor (rhv) streams exiting the evaporator in units of kg/h? (Note: You will need to use information from Table 8.1 in Felder and Rousseau) a) Draw and label the process flow diagram. Use consistent units for all of the streams. Xn \l’ ‘— C3; (CBC x; ' i" ‘ m :0“ “3D . s00- ; / . -" \l‘ "V..._'--_' V L:\J’T3,.\.\l“ 5k.) D'kaj’ ’ , im__| __ ‘7 film 033‘); \0 mm: Mu { \J‘h XFJJI : 0.540 K \ M g o 'Luoo . mm____.__.____., L (\\\ l '0. k8|h V‘, J le‘Vu‘ ‘ntw «fl 1 SH'L'kS'h fl\.- 5/10 ChE 210: Meeting 6 January 24, 2011 **Need to convert feed stream to mass flow rate and mass fraction** 6mm wacfiwo 78 5.1" ILA 51w 6H0 whom h men 10013:} TQM“ O “4&0 \abbmox C151 \' \kC‘ ° ' ’ts— : ‘ ~ ,9 ‘ \ h \ 0138\m ‘\ Hm.uk3‘rlt\ ann 8Com 21mg We, >8 M351 “HEEL—J . 0 I, 511x56h+1b1 LSTIh fig \/ c 1.000 . _ 1&0; 'T\ X‘mn - ”JEEP/L. 001190 5+1 156% wept! 19% min ‘ mm + Ninth : 51.115111 1:- 11.1137“): 161 {cs/h 6/10 ChE 210: Meeting 6 January 24, 2011 b) Write the General Balance Equation, and then write the simplified , total balance equation and the balance equation for each species. in q Gen — Ow - (“on ”3' O (V \/ ® ’TQ-iC/LL (I m€)_if‘\ ’1‘ (hwi‘ufi .4 {hv __m\d : (:1 \l , a \/ ,/ @ (benfine‘ : x .. . l i (new ”X631 (‘0‘, * XELmL‘:O \/ 6’) Telqu {Y3 , ~/ J b “in ‘- XTJVmVI w XTILm_’O c) Decide on a solution strategy. and solve for mL and rhv. Don’t forget to check your solution. 7‘ OW 6681(111 (20:5 contain .5? cmhwowfli'u, rixv and 0fo X , my “ .‘ V‘ o - . T C' - mUb‘ «Ewe 1“; rm an ‘it‘m‘is o? m L l"T (Or we Val-flu) (all) m : 101:ng n @ 5”” ifiih “ 0.850(ioikwlh-rixg « {O‘lOL§\l‘f‘-L_;CJ [$1. " ”=5? {<3th 7/10 ChE 210: Meeting 6 A January 24, 2011 ‘ d) Due to processing conditions upstream, the flow rate into the evaporator increases to 150 kg/h. Calculate the new mass flow rates of the vapor and liquid streams exiting the evaporator. 7kdegoe 6F lgY (7..» ” liov M _ Hg Em . 3r , q 7- 1 i . . r ‘ “ (if ’v lei lam am was) : L, um, 1:75-13, k; wily ;: E7 ; mix gum, , asexu; will) .— us will 4) \ .l M 8/10 ChE 210: Meeting 6 January 24, 2011 Flow Chart Scaling Once a process flow diagram has been balanced, it can easily be scaled to new flow rates. The composition of the streams (x’s and y’s) DO NOT ' CHANGE when a process is scaled. w “l \\ .2 C rt ( 2. Calculate new flow rates for ALL other streams using the scale factor. list Paces m afivxbws example Scales go “AN (new) : loo. “Nb SF : mvmew‘) : loo mm 0an (oncjrull 58ft kglh ring I. SO 1 (‘0‘ <\ UNQLQ) ll :5: mm (ongmj l) =(l.;'7Q)(l©l Pill?) 4-713") let-ill) (human : :grm’imefim113) : (1579' ”M3 WMWMQ ‘5'“ X 5Q\\l€ port cl 09 example pmblem 9/10 ...
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