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Final Exam Solution CHE 2120 Fall 2004

# Final Exam Solution CHE 2120 Fall 2004 - Name ChE 2 | 20...

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Unformatted text preview: Name ChE 2 | 20 Final Exam December 8, 2004 In all problems, you should use numerical methods you learned in class to solve the problems, and must show all your work. You do not need to wait for methods to converge, but instead may perform three iterations only. There are five problems, and each problem is worth 20 pts. l. The logistic model is used to simulate population as in where p = population, kg,” = the maximum growth rate under unlimited conditions, and pum- = the carrying capacity. Simulate the world’s population from 1960 to 2000. Employ the following initial conditions and parameter values for your simulation: pa (in 1960) = 3,040 million people. kg,n = 0.026fyr. and pnm : 12,000 million people. Have the function generate output corresponding to the dates for the following measured population data. Use the results to compute the sum of the squares of the residuals between the data and the simulation output. I965 m '- 1990. 2000 _ 5276 5686 6079 _ Omit zﬁwﬁ'm} (twill-J t/a/(ZMJL Wait/1% ewe/H100! (/git afoul/t. guys? {(47%) (“140036 ‘ Ptltizpi+km({’ Eiar>FiIh htgi’ a AFB #0 pi pm £5,110; ring-h 130% mozéﬂa Egg 309/0. 5 = 3335 Pr 3335+0.026(/-. 3:353:35. 5= 36%) I00 P3‘3978 (91,-.— was? 31,: (ff: (pl- : fol-,m)?‘ p6: 5055 : (ﬁlm—304024 (3335;33t/GJL +(3938—87OSJZ‘L ' " + (5822~GO79)L :238x/05 Ex) A free-falling object such as a bungee jumper is subject to the upward force of air resistance. Assume that this force is proportional to the square of velocity as in 2 ‘FH = adv where F” = the upward force of air resistance [N = kg mfsz]. Cd = a drag coefficient (kgr’m), and v = velocity. Estimate the drag coefficient given the data in the table. [Ii-mm m- - Do you think this is a good or bad fit? Describe how you are assessing the fit. L90 3% 3%W PQjV‘QSSI M waits) tell 0501 WW?— K‘ V 4’ 25 /00 Y“ 70 Z= r00 380 300 550 M00 6:0 .2500 M10 3600 830 4900 H50 62/00 2‘2: 88 W >_) ZT \f: 52.0 x /O7 .n 8 ‘3: ﬂ Sr: 3 I x/05 3'0 2.0 30 (/0 5'0 607080 V 380 ’ 2).] [5! 580 a 374 TM; is 0 i HUI/WW) 100081;? 600 a 585 M +0 m 0% V; G “220- 891 W V: 70. m. WWW 7% “50,500 M ire/Ni, wit/7’1 SM 3. A civil engineer involved in construction requires 6000, 5000, and 8000 m3 of sand, fine gravel, and coarse gravel, respectively, for a building project. There are three pits from which these materials can be obtained. The composition of these pits is - Pit 2 P113 How many cubic meters must be hauled from each pit in order to meet the engineer’s needs? some: (5000: 03219, + 025:0, + 0.35093 ﬁmenli 5000 = 0.3079’ 4 MO ’02 * 0J5 P, (\oaV‘g—éérnwp78000 : 0.38 P, + 0. 35 f; + 0.50 )0, P, is HM vofum of grad W W pill 6' SVS‘FW 0; NW 169917001435 Him mtji‘mw m0? 194W MMka/OVLS’ " (ll/130W EM [AN/K EL CW“ aw‘ém “(mm 0,32 0.25 0,35 0000 0.32 0.05 0.35 P: 6000 l 0 01% —0.I78 P z #625 0.30 0th 0.6 pa = 5000 I O 003310.08? 375 0:38 of?" _ 0‘ 90 p3 5000 as 0. 053i _ __ ___ _ ._..: .._._—_-— 30.320 an OJéé Gangs JINWM" 0.531 0,08H’Ll 875 224-; 05.9 , 0,9375: 0'30 “O (“:8 55:32 {0.531 @0570 #200 H 0.32, —-(O‘30 0. Z 0H“! 0 04003 0/78 ~625 032 0‘2; 035 0000 333i; 0i ,{18751 0‘33 0'35 Q50 9000 0 0.100» 20,173 lo" “629 an 0.32. «(0,30 0,197 ow 7,25) 0 0 mini 2075 O 0.053 0.085“;Z 875 Blots-Mailman j P,:7000,i%:L/VOO,QZ7000 4. Fick’s first diffusion law states that Mass ﬂux = —D£ dx where mass ﬂux = the quantity of mass that passes across a unit area per unit time (gfcmzfs), D = a diffusion coefficient (cmgi’s), c 2 concentration, and x :3 distance (cm). An environmental engineer measures the following concentration of a pollutant in the sediments underlying a lake (x = 0 at the sediment-water interface and increases downward): x, cm 0 1 3 c, 10 gi’cm‘ | 0.1 0.4 0.9 Derive the best finite difference approximation possible to estimate the derivative at x = 0. Employ this estimate in conjunction with the equation above to compute the mass flux of pollutant out of the sediments and into the Overlyin g waters (D = 2 x 10'6 cmzfs). For a lake with 3 x 106 m2 of sediments, how much pollutant would be transported into the lake over a year’s time? NM to redarim 001’“) ﬁrm/ta dx'ﬂz/umci MM), Sinu we NW2 amiswdba 210W 5mm W m I) an we) (to) ammo); +0; 9 Mir-3) =f(x=0)4f’(x:0)(3'0 * if’(x=o)(5—o)z+og ) Solve 0m aﬁwﬁm far {yérso}: ["{r:0)=f(x:;)-f wolf/day Pita we Mom/t «ﬂotsam f0 W m: 0;“ WM Wmit'm few/WIS: I ttxt3)= (x=0)+f’(x=0)-3 + a :27 (t/XaJ‘r/iaydmza) Solvf 1%? ('04)): m) : ( rm) * any» «git—{m2} I “\$0.9 -* “‘é-EM tag-0.1) ' = 0.31 63AM? x/O 5 CM? 0441.? WWWMM 062i 31/09m1(’%99’1“1 39005 290w WW “"23 M Mr ‘ Tm? , saga/gs: I7 Mei Anew CHOPWO‘OLK 1L0 M MW» 115 aﬁmpoiaﬁm. Name/1, 14w; {Wu/(A aim/fwd {MLLUL {0001; is «(W W {by 52:012-th {W MVaﬁ'm of unué WWW. .27 tam/d he an appnp/iaﬁ nae/ﬁxed W m pm, M? if {S 2101‘ a W01 Wad, 6Z4 NXM in {pm pm ﬂab/Jaw. 5. You are designing a spherical tank to hold water for a small village in a developing country. The volume of liquid it can hold can be computed as ) R _ h v = 7m- [3 1 _ 3 v It where V = volume [ft3], h = depth of water I I in tank [ft], and R = the tank radius [ft]. If R = 10 ft, what depth must the tank be . filled to so that it holds 1000 ft]? Nanazyrtam ‘ (OOZL'ﬂi/d/V‘g - 2 . .. TL" l ' _. mo- M1 [igl] a [(14): /000— 3 l» [3044]- 8isediM5 ( j 0 1.1455090 bYﬂM: [4:0 2"‘2—0 0 .._____.__ ___ mm? /0 y/095'_ 5 331/3__'_____>'OE 0 l5 Est/5410 i-/095i75 V326; <0 ‘ i l 5<x< 7.5 ...
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Final Exam Solution CHE 2120 Fall 2004 - Name ChE 2 | 20...

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