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ChBE 2120 Exam 2 Summer 07 Solution

ChBE 2120 Exam 2 Summer 07 Solution - NAME SOgWUTiOKI‘ r...

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Unformatted text preview: NAME SOgWUTiOKI‘ ' r ChBE 2120 Professor Gallivan, Summer 2007 Exam 2 (I) You need to purchase a new pump for your plant, and have the following heuristic relating initial purchase cost C toannual maintenance cost M: 1 . ‘ 109 M:— ' C $10,000 < C < $100,000 . Additionally, the final salvage cost is 0. l C. Consider a lifetime of 4 years, and do not perform annual maintenance just prior to salvaging. A Use an (effective) interest rate of 1 m 0. 2 and derive an expression for the net 1 - present value of the various design options. ' B. Set up an optimization problem to find the best design (lowest cost). Clearly state your design variables, your constraints, and your objective function C Solve the optimization problem using Newton’ 8 method, beginning with the purchase cost of $30,000. Perform 2 iterations. 1 1119 low ~1- . 1 1 l 1 ii iii ‘ 9 11111. _' 19:11 31111 if L' e , 1 1 1" l 1 i i J (2': a!” (H's ("'2‘ M 11111131111931.1151 ’ ‘ i i7 .H 11:11: #111 , 1631\1‘111‘0 * 0 11111911115 10‘) i i M 153111 1111111111,“ .. H151 ‘5 ”f” 222222222, 2‘ 13 23222 232532 $222222 22222222} . \.22;G‘21;§L2 4Q 4 321002000 02 329222 222 2 22222222 2‘35 22222322 22‘2"; 2 32202242222 2212442 2662000 22:” _& { Q2 2 ¥ (C2 22 2 22 .' ‘3 2232,2222. 322-5222 2222222222 2 {u S a 22.222 : 222222222222: '1 ,‘2 F '2 .(x 12 21,2 ‘2 2‘2 N {223222223 2222 2 222-2 {0, 22 29222 2‘22 22 262 7L.M\— x\“"\:\ LL\ 222222 222222222 2222222022. 2(2): 2222 . 22"} 2/0 m— : I I; ‘1 '2 i- : —-*" 222226 {232 ‘2 2.“ (2‘3 2" “220222222 (2) Minimize the following function: f(x,y) = sine) + xsino) ' 05x53 I OSySZH point (myg): i 1,3). You do not have to actually solve the 1 D optimization, but compute the constraints on 11, and estimate the value of ham using your constraints and the plot for (2)13. Then based on this estimate of kept, compute (X! y!) ,5 fit A. Perform one iteration of the gradient steepest descent method beginning at the f : (LO-$131.? Skxmtér' "Si-E“ :MoefiQ- H..,...-m...,.W———mm__..._w.,—u..mmw wxmwlep Mug limesfiofi 11.- Dgnafi - ‘ ilx®5%€lW % Oi; numbing“ “in :7 05: @111th s: a? \L)\£95\“r\\5\f\3\ s. T i ’27 h htv'agni UT"3 \ was it . i . 310363“th 29—3:st mus ““5 W W €05 -C",“[\+ WHWQ E heifj‘wme (3*)‘i‘twv “mu—W.- HM WWW} B. Sketch three iterations of the gradient steepest descent method beginning at the same point. “so . . . ”K ' (3) Two fluids a® temgeratures enter‘a mixer and WE}; temperamreQmTihe heat capaeity __QffluidAis given by; cp=3.381+1.‘804x1(r"21."-—zi.300x1041"-2 and the heat capacity of fluid B is given by: cp=sflu+12dpd04T—AUMX105T2 Where CI, is in units of cal/mol-K and T is in units of K. Note that where His in cal/r1101. Fluid A enters the mixer “at .400 Cr B enters the mixer at 700 C. There __are equal amounts of A_ and B entering the mixer (based 0'11 mj).Atwl—1at temperature d0 the“ two fluids exit the m1xer‘? - Lacrr/ « .. - w-w ---—~ m“? B. Perforr.‘_"_s of [email protected] to find this outlet temperature. Use initial guesses of 400 C and 700 C. BBQ. ,_ w. ., Q W led-a o “__ _‘ ORR-Vain Q1 2 CW9 1“ My *0 “aJle :7 AH :0. “L5 Hf: Z Ahi-Edi‘oi — O : AHA ”’f éHB A " T'F EHA‘ 367314 (3) 3“ *thESoLl-flb‘l—V ~Huo”6t")m T O x ..2. >9 —¢ '9 r [:3 3e\x*V ”t l‘fiiié: (Tl ‘ 1+ 3 :3 T§:X6¢sk r; % 3% ‘L ‘ yo“ ' 2— y - 1 T1. if Leos; ((1%) _ 1+ s L%[email protected]§ —. 80123 012» QM ‘2 m. ~tl sqao \ KT "4.07%xl0“§ FILM“ : {315sz + {‘qul‘f‘fi‘u‘i “*ivmwwilgyib . 3 3 . W39 _= 8 s23— "ff 7“ WWW? —- Lt ofi‘firxiflfifi ——~ 960101‘5 "‘2’ r 3 l \ ' I III—IImmIWWIIMWIIIIIIIM—wWWII ..... aw“— \ , D—k . a. _ I __2 ('5 3.35;. +8,ng )TF Jr (balm 9—1—\1C1L Tf‘ .,. (WW my, IW')T‘ I ‘ - , - \ ——' 6:50., :16 , 2;; :_ O . :— f (1%) . ‘ fl»~“‘““ / TA : Gtoo +2173) k: 6’73 {4 Rafi/200, 4v.“— 3) (4—:I0M3K- “n z n — M in? a; £334 —-€vL) N) - L.u—_....__.., _ V,..._. 7.7 H..7V,,,7,,__. " """"" ‘h - vilI.I—--h N Nifll—I—n-Iir 7 I I I I." - -~ ,_ .m_ a" ' i ' g PW) “ I “MT? 3“ o III_II.3IIII Hf); —_ leoszSSWJI" 61%15Jl 30- 3 30(1)) 10(01’15La) 43‘6qu}. {(13): flank“): — ”MOVLS i ‘In 2-“43 _ Qggkql3( 673~4qn3) //fI IHE“ I II IWWV :r'[email protected]) =- 6550C. D-MQ [{flcfltqm . Ho, . ”J T~1=0ms\<. - Tor—011m. {C T4) .3 f(01r}3 K) I uléOHLS {\(To) —_—_ f{q;%\¢) :3. _ L\_[g f ______u...,.._,_..u.m r?”— T\ :. 0‘1?).... {I —- LHQ (Q Owl—5“ q1gj- } if}; :m\\\ _ , I: Olin 3 K 3-1-684BDL \ \ \ ‘\ . / V “n 11 éowq ‘g i. LH O / m..._~.._._,__._._r.fi , ...
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