che210exam202 - ChE 210 Fall 2002 Name: Section B Exam 2...

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Unformatted text preview: ChE 210 Fall 2002 Name: Section B Exam 2 SHOW YOUR WORK! Good luck! 1. (15 points) Real Gas. A certain gas has a molecular weight of 30.0, a critical temperature of 3 10 K, and a critical pressure of 4.5 MPa. Calculate the density in kg/m3 of this gas at 465 K and 9.0 MPa if the gas obeys the law of corresponding states. W230 TL: 3\0K T: 469K PC: 4.5”MPA— P: qDMPK m Tr: .11; —. L3” = 1.6‘ 7 we“ '0 " {*‘a 2:0.94 29m Pr; 2. 1 = 3-0 J 3P6 Fe 4.5” 59*5 “I... P: MN P = 30“ 9.0MPA K 'o‘d’“ 2.91- 0.8 km! 465K "3008209m334hhgyol3ufl6‘ f coin/2'31 1;“ u 00; EN u x q» calquaQ. 1?“. ChE 210 Fall 2002 Name: Section B Exam 2 2. (25 points) (Combustion Reactions — Material Balances). The product gas from a solid fuel combustion reaction has the following dry-basis molar composition: ' 72 % C02 2.57% CO 0.0592% 802 25.4% 02 Pure oxygen is fed to the furnace in 20% excess (of that required to burn the fuel completely). There is no oxygen in the fuel. Assume no solid or liquid products. Use a basis of 100 mol dry gas. (a). (3 points) Draw a flow chart. (b). (22 points). Calculate the elemental composition (mole% of the various elements) of the fuel. Note: This problem gave me three equations and three unknowns to solve when I did it. The most important part is to come up with the correct balances and equations. Solve completely if you have the time (solving for remaining 3 unknowns and giving elemental composition worth 5 points). Wading} 3,5“, we maul-s. D/j‘éds W7 e... ‘i‘ 6;“ “1.2 mt ) Wwa LN”? PM «6 5 n; C ",r' 2, m 0‘ «E, COL n5 5 ii 3.57 mob; CD “14 H 20% excess DZ 0'05“); mo“. 502 Who”?a no). 1 nw W\01‘1f2 3") ’3 3055mm? ac : 73‘ + =' WIS 39*5 6 eaahm : 0 beis 393/13 3 Bunkmowng “MW nH = znwi ' * ~7 “H‘ ‘8” . -* n " ‘ "T iiiiiiiiiiiiii m -7 7 01: M44. ‘ i 1 4- . 5 ’ MO)! 39501 0 he. 4- n5 1' : nu, 0 2 , / hw: 446 a; not = lilzhmear 0.25 n”) = no K TOW WW 3‘,“ 0 Gala mm: X 9110; I 3(7)? + 2.57 + (0.059234 sf 5.4) + nw'l nc+ns+nu . . WW. . ,. : 3.53.4 Mg garage“: 28.51900 0 7H 70 H 0.0234,? ChE 210 Fall 2002 Name: Section B Exam 2 3. (30 points) Air conditioner. (Gas-Liquid systems) An air conditioner is designed to keep a room cool at 19 °C and 50% relative humidity. It draws air from the outside, where the temperature is 35 °C and relative humidity is 90%. The capacity of the air conditioner is 15,000 L/h. What is the rate of water condensation? Assume atmospheric pressure at the inlet and outlet. 15,000 L/h T=19° C T235 0C ht=50% h,=90% P=1atm P=1 atm Condensed water F0, wa-[af‘c FYUW 1“b\€ [3-3 p"( 350c3= 42,176 mm”? 19" F1(‘q°cb = mm Mug 1|” tin g inlet: hr: ot‘i = PHzO ' Puzo‘ = 37.01bwmu1, 14°“ «ms 4 Mafia) 3er WW1 = Pd; z 31% = 0.049% ‘1’; 150 apt-5 \éDA‘l = oflSotL 19+ obi-i" . VJ.~ Ood’ie‘i’ ; My: 05: PHIB »t Fufc : 8,2355W‘MH3. 1‘04. 3 ‘PB P‘Ni‘fic) 3.9-: @102 : PH°I‘;*_ 3,2385= 0,0108 ’ ‘T; ' 71:0 31"“ 5A.: = oaeqz 19* . nu W1. = = 505,434 «pi; i}; h 0.052% Lv arm 305K hm: seaee‘lz 1W “W1 = 20.0140 11°47 otdlel- 0:;A z Sum—n7. ’0 “7': “DA/3890'?” 1.0+ nwz : igniplflr n1 ~f nW'L: 'évlwn— . nm 5 6"593 an W bpnflv ‘3‘ n?) : “Wl _ nwz- I =- : W's K. ChE 210 Fall 2002 Name: 1 Section B Exam 2 4. (30 points) (Multiple Reactions: Extent Balances) Chlorobenzene is produced by bubbling chlorine gas through liquid benzene in the presence of a catalyst. In undesired side reaction, the product is further chlorinated to dichlorobenzene and in a third reaction the dichlorobenzene is chlorinated to trichlorobenzene: C6H6 + C12 9 C6H5C1+ HCl extent of rxn: £1 C6H5C1 + C12 9 C6H4C12 + HCl extent of rxn: £2 undesired C6H4C12 + C12 9 C6H3Cl3 + HCl extent of rxn: £3 undesired The liquid output from the reactor has the molar composition: 65% C6H6 , 32% C6H5C1 , 0.025% C6H4C12 , and 0.005 % C6H3C13 . The gaseous output contains only HCl. Assume a basis of 100 mol of product gas. The feed to the reactor contains liquid benzene and chlorine gas. The process is specified. (a). (3 points) Draw a flowchart. Keep liquid and gas streams separate. (b). (18 points) Calculate the extents of rxn l, rxn 2 and rxn 3: £1 , £2 and £3 (c). (3 points) Calculate feed flows and liquid product flow. ' an we rs : an (d). (3 points) Calculate the fractional yiafi of benzene. (e). (3 points) Calculate the fractional yield of monochlorobenzene from benzene La) C LL lDOmolS fl (0 ; HQ” > Product 60x3 1L ———__§ 8 c\;[ L\. “up pmdud' Y“; 7 n5 3 n3 (3‘31 n5 C6 HS~Cl too more (av W5 \s “0* W 0.025m3 chqul ekSlQS‘i' basis -FJ\/ Solunj Frvbwm, 0.005’ n C u m $0.,‘ f3‘0CCC’xQ basis +0 5 b 3 3 Stream:%. Wk Scale {lam/5 .thlawtcl. Dicky: dues ~01 all-v an; 1g» (d): 531] National ®and 42'” l’ '5' Foe, \oo mu LC . 97.03 33.07 3) .07 (‘5‘ S\ § \ £2 3 1 ,, Ht : 32.05511 n1 = 33.068” 1 W°¢ 5mm 3 302,519 at "‘3 Loo ZCLFAM NW 100 g 4,,5 8 *5 201.19 P w W 50“]? 0‘07?» 0.9le ‘Mfi g OJ??? as (Marsh 3 E E 5: W =5 £ 1‘ FVM‘M buzene. 2 Q7.03“6$‘: 0L33’ 3.65 Q7433 m" = 27502.55 ‘- am. \l-n—u—I—u 3’ O\ 303$). LC) fiddz MMMM be.on ; 5a; 332g “QM! 5‘2 3 fl‘ - 49'qu (AS ((MLfiAS Veranda-«4T 33‘07 MW *9 FMZQ3 0L» ~0+ :' 1°C « Decw. ; 1 10° : An C5 $00,342, chug-C! n; 11:04 1 \33\o‘7 Pro-46: W :9» L’Ast :L :00 18 mmmw m 1:45.12, mg ma“; Lemma“? 3+3ch u; prawm sWM: W. Sidd mane £60m : 3—); : 0‘33 ken?!“ (ASSM'Kj 5.1.7.03 2 lea—«7.4.4. VJW‘O 1.9. [mm-.3 data—we) = [OD . 0333) 303.57: '2: ...
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che210exam202 - ChE 210 Fall 2002 Name: Section B Exam 2...

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