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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} (twillJ 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; ringh
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 freefalling 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. [Iimm 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
BlotsMailman 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 sedimentwater 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 Mir3) =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 * “‘éEM tag0.1) ' = 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 52:012th {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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 Summer '07
 Gallivan
 Gravel, maximum growth rate, following initial conditions, following measured population

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