HW_2_soln

HW_2_soln - .w _. WWW—fl. . -w r7..:fi“-‘—~m~ $5!...

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Unformatted text preview: .w _. WWW—fl. . -w r7..:fi“-‘—~m~ $5! , . \1 2 mm}: m fig [3 ,7, 9.3.“?‘3Lnaoe/k *4 20.1}? 5-0 “Wu/k MEN“ .1 élQ-fLN“-Aj!/3/\ J ' We 7 I. 1: 1W 0W~ r1. ( 1 n , 1.1.4,". a . 936 -“~~n F. D LIFEOKOW5J: 3}: Q a ‘ r v I E a "” :1“ r1 _ “5,8 1~.Q.5‘I;Ir~"~.ol/l,1 . x ‘ Mug 9 wrdaw-saor 4” Cw?” “PM” ' 4 ,— I n “ma «inflows look I“? ="“L" ,i p _f‘ v—- -’ .Jfl' ,L“ '3 ‘. 'r ~. ‘ x_. .1. - : Lr _.r;t, w“ L” ,1. , 3 Emma ‘— 3551? 3mm! Cw" 95010; —,’- +27%}! é: n-mif‘, (grammes. 4x1 leui: QM 3F? if $16? 0 u - raga-«5:26 Walt, {:{D‘HEC'H fflmfi me {WWW} w W , A ODQIQ ca. Q‘flaflCCfE-Mg Pt ‘ - ‘ is Glow-we Cwmgowc. a I - y% a w ‘7 -.; .\ 5 ~ E) K a mo fY 04153 cfigtlm/% mi2 (caimmj, mpg 0.05%6 Car “10% 00050.6 ‘QAtggagg /@ I, 7 a Ofl'v‘zcag may; KN ' Mi \IWI -\ fl ( n - ‘ ‘r‘~‘\ a .33" ‘fldepewoa ” "or :C 3’ fine Tor mm ’ SPECH 3) .W—fl ' ' g. F A- 3 «1.11 9 ® Utal- iT'ic ‘ Um ME A W}? 011:5 (4430‘) = 13.0% Phi + @905 0:11 5 ‘9‘” OrE’B'f HBU‘. « o at; gag +49%: bag ' n3]Q—% {+005 [HE J—nal: A 2m}; 6‘ :mfiloo)’ 0.0M " C '— f““=?£ c ’— 0 {I x a -“ ._ I WM} mt , . rs FIij — ‘ $056305?) — ' ia‘Dgy i , “INA” mm } ’ I mg . l “m‘ H; O m}? mt. r 05m (Ll-r50} T 4W fi/mifl © 0415(400) 4w 9 '4 M Fumw ‘ T of IQ J-h =wa§ wimp—«Facemi’ :5: 56ngth Cu$fi D‘Y‘L _ A [14! -. v 1;; mmr ; ch 33% «W I L» $19 4‘1 £13m EEO CU iumfl .. T ‘m. N} 4': n. I if. um“ /‘:-i In A ~ r13 (Wk) in}? 1, O 0433 Mai?» 337%)! ’"Ogo’ib’b'mof, QaHg i'mo’ Or a l‘z‘f/hnflf ‘ I 0 TY‘D‘, all? /f‘rw E @«ZE WIN OQ/fuoj i 0779 m0! gag/nu); “v r—W , -l c D!" 3 HILL (Q G. {1,} i I} 1'] 3‘/ —-Q I'm! I I gj r J. ’ r f " q‘ 1'39 Fri-’3' {1.353 f3 ' n: {fl'vifiiloaj -1; {If r Dial? ,r‘ : “LC-71;“ " ‘ , m2) 0 II _,!.‘3—1-DJ—J El 3- BOOlbm/h 0.55 10,, 152504 / 05m 0.4510m 5530 x 10m 2513(lbm/h) 0.9010m sto4 / 05m 0.1010m 1420/ 10m 0.7510m H2504 / 05m 0.2510,“ 1-120 / 05m Overafi balance: BOO-Hire = WC (1} Mim: 0.5 5(300) ~i~ 0.902513 = OjSJiIC (2) Solve (l) and {2) simultaneously :9th 2 400mm / l1,n'rC = 7001bm l h 4.21 (cont’d) M wtll ' __ I b. J. '_ 300—130 ‘4 “ . mi“, “150 = —h~~»—~T——(R4 ~25) :5 m_,, a 7.7834 — 44.4 ill-«2'5 . 800400 ' ' #7 3% an? I : J .— mg -00 60‘20 (RB “0)::m3 15.0223 500 l/_ —1 :2 . " line—«11520 mi‘liggym, —4) =5 lnx 2 0.263212, +1923: x = 6.841e”'25m= ' "0 44.4 r - M nus-.3002RA:LQL—=443,ma=400:5R8=w:~3:r/ 7.78 [5.0 w' .-—- \ m55%:5R,== 1 10(33—lz733 0268 6.841 W BJSQE balance: 0.01in“ + 0.901% = 0.750% = 0.7507504 +013) z . 3 c. Overall balance: n31],a +1205 : mc [0.75 — 0.0 l(6.84 580353310 )](7.78 RA ~ 44.4) 0.1 5 :5 RB = (259 — 0236901532135 )RA + 13590-35510 —— 8.13 W 25550103400: (120023.73) _ 8 13 _ 333 Check: RA = 443,10, 2 7,78 2 R3 2 (2.59 - 0.2369°'16“l7~“})443 +1353 f 423 W'Vic‘i ‘O‘D‘fl “Um-WW"? ‘okcool i mofih > i H5 mL7’m\'r} mm £0?qu I WE“ magi/ml.- mm (Digit; “3 "3 D'a‘x 1mm j / H,L,c:\ H3 ‘9 A "—a" k i 1am) mvm x: PTO f‘r w/rfwbav I‘fg‘fl‘r‘lfit/ _ 7L— m av‘zQCr-i-c‘r V'thw’sfj "hm 3,1 {‘03 :2: A [A m C g . “MM W) Uua i 100/151 (9,00% 3.75% (19.50%, 3353mm . mm :— 3 P W. iJ-‘Cfl/ ‘ may-3 . W . L " I‘LL—z L— ‘ "‘= E GED Imxm \qrj [mm 3 it? ~D I; . k x M 1‘ 3T. 1 — in: H 1‘91:— 0 dxcd 51 de 5“) 1h ‘1’" ‘f-EWW'J‘ 5 r‘mw ’ ‘Cs—Ofimm EFDHEDDUXMTQQ/Dy "" " Q 0 - 1603 ML. =] 00-26% “’13 (Mg? ‘ rpm ,. .T-“L'w' ‘m CF") 011+EV) ‘ New (é) urea &r?¥l"fl— ng: “MOM k rm. 1“ 51% rgmovadv ‘ was, 3‘5-w“03 +mm V ~ 7:3? L. min :3“ .1 j. 5‘) ’13 /L" 3gb 4.34 a. “ M 175 kg H,O I 3(45‘3’9 ofwaler fed to evaporator) mt: (kgKfiOslS-l F ilrercakc , ’57: (kg H10] 5} Crysiallizer i Filler lfllfiflkgi‘1150; 15} n}: (kgsolnls) I {0.400 kg K2304 i kg} @600 kg nzor kg Jihl'kui Sl m4(kngsov 1‘5} 0.l96 kg atso‘. f’xg walks H;Ols) l Evaporator 9.304 kg HIGIkg : L..____.s 2' 513(kgfs) 0.400 kg K2304 .’ kg 0.600 kg H20 1' kg Let K a K3304, W = H: Basis: 175 kg W evaporated/5 Overall grocess: 2 unknowns (2521,1731) Mixing ooint: 4 unlcnowm (anagrams) - 2 balances ~ 2 balances 0 DP 2 DP Evaporator: 4 unknowns (Inhrirsnnfigir?) Crvstallizer: 4 unknowns (m2,1i13,1525,l577) — 2 balances w 2 balances w l oercent evagoration 2 DF 1 BF Strategy: Overall balances 23151:, rirz verify that each % evaporation ::> 2515 chosen subsystem involves Balances around mixing point :9 mg, on no more than two Balances around evaporator ::> r2525, an unknown variables Overall mass balance: =175 +1032 1H513 Overall K balance: 0396161I = 1017':2 + 0.4001232 Production rate of crystals 2 102512 45% evagoration: 175 kg evaporated / min = 0.450.615 “ W balance around mixlna goint: 0.804173, + 0.6001523 z #15 Mass balance around mixing goint: n'r, + m3 = mquies K balance around evagorator: #16 z #14 w W balance around evagorator: airs == l75 +1737 the Mole fraction of K in strearn entering evaporator = ‘——7— m4 +1715 1). Fresh feed rate: mg = 221 kg! 5 Production rate of crystals 2 lOriz2 = 41.6 kg 1((3) /s n13(lcg recycle/s) 3523 60 kg recycle Recycle ratio: _ . ——-———————— m}(§<g fresh feed/s) 220.8 kg fresh feed c. Scale to 75% of capacity. GWSl-ml . Flow rate of stream enteringmeaperzator E 0.756% kg/ 5) = 299 kg 1’ s 463% K, 533% W d. Drying. Principal costs are likely to be the heating cost for the.__evaporaror and the dryer and the cooling cost for the crystallizer. 4.37 ‘ 11. 171] (mm din) 98 lbmdry shins 3 lbm Whizo E00 lb 2 lb,n din v v 98 lbm dry shins Tub ‘ m2 (mmW'i'liZD) 11:3(lbm) I n14{lbm) 091 m (fin I lb “ I H I t' ' ‘ m m 0.6}; lbmdm flbm 0.}.2 ibmdm- lbm‘ 0‘08 lb Whiz“ ; Ibfl I 0.97 lthllizzo Ill:m E187 ibm M11220 I lbm m . I "1&8an l—x {ibmaimlbmi .r (lbwI Whizzo’lbm) Strateav I . 93% dirt removal 2: m, {a 5% of the dirt entering) Overall balances: 2 allowed (we have implicitly used a clean shirt balance in labeling the chart) :> m1, ms (solves Part (a)) Balances around the mixing point involve 3 unknowns (m3, ma, x), as do balances around the filter (n14, m5, 2:), but the rub only involves 2(m3, m4) and 2 balances are allowed for each subsystem. Balances around tub :> m3, m4 Balances around mixing point 2.51216, .1: (solves Part 03)) 95% dirt removal: mi =(0.05)(2. ) Overall din balance: 2.0: 0.10 + (o 92):;25 :> ms = 2.065 lbm am Overall Whizzo balance: m2 m[3+((}.08)(2.065)](1bm Whizzo) = 3.17 lbm Whizzo Tub dint balance: 2 + 0.03m3 : 0.1 O + 013m4 ' (1) Tab Whizzo balance: 0.971713 2 3 + 0.87m4 {2) Solve (i) 8:: (2) simultaneously :> 1213 = 20.4 lbwmq = 19.3 lbm Mixina Qt. mass balance: 317+m‘S = 20.4 lbm 31216 = 173 lbm Mixing gt. Whizzo balance: 3.i7+x(17.3)= (O.97)(20.4) :3» x 2 9.96% lbw \PVhizzo/ibIn :> 96% Whizzo, 4% din: M ...
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This homework help was uploaded on 04/05/2008 for the course CHE 100 taught by Professor Monbouquette during the Fall '07 term at UCLA.

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HW_2_soln - .w _. WWW—fl. . -w r7..:fi“-‘—~m~ $5!...

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