cee218A-fa08-mt1-Harley-soln

cee218A-fa08-mt1-Harley-soln - Page 1 Midterm Exam CEE 218A...

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Unformatted text preview: Page 1 Midterm Exam CEE 218A — Air Quality Engineering University of California at Berkeley Fall Semester 2008 NAME: g0 0 VLS an Instructions: answer the questions that follow directly on these pages in the spaces provided. Use the back of the page if you need more room for your answer. If you believe there is insufficient information provided to answer a question completely, state reasonable additional assumptions and proceed from there. The exam is closed-book, but see last page for aid sheet. You may use a calculator. Please write your name in the space provided above! Time: 90 minutes (2: lO—3:40 PM) Question SCORE: OUT OF: 1 10 2 8 3 7 Total 25 Page 2 1. ETHANOL CONIBUSTION (21) Suppose the molar ratio of CO to C02, Q = all) : 0.3 is measured in the exhaust plume of a passing car that is burning pure ethanol as fuel. Calculate the air—fuel mass ratio for these conditions. For fuel—rich conditions, the combustion stoichiometry can be approximated as follows: CH3CHEOH + m E 027+ 3.76 N2] —> a C0 + 1) C02 + 3 H20 + 3.76111 N2 Mew (Ta-H522. (Rafi/13:0,? - S: 2., £75:- a:2~—£=O,L{-é ,m t .ECD,%-+2H.W+3»I) _'= 2.7? ma 2 .13 x 26: «ii/mi slim-7(a) a (2M; 3W1». . WE I K Lita 3/W\ law (b) Estimate the percentage increase in fuel consumption for conditions of part (a), relative to the ideal case of complete combustion. See next page for needed thermodynamic data. EM:( “’4’ ) ,4 Ewe/Mi % flic- .._. mmw .i+ a/b age/w [few 3M E} .W“(’ mo“ ‘32 CO 1 “132.1 “‘1 3‘?‘t06v Hweo 2 23351 HOE: aw! (3. Mi. saw was taut-1M 2 395:“ 028300 < he? 3v 3M 5’32 :: 271k? ‘ % IWW. in M vawwkvm {It ,3; :: "Flour/’- Page 3 (c) The lower heating value for ethanol is 30 kJ g_1. Calculate the adiabatic flame temperature for ethanol, assuming the following revised stoichiometry: CH3CH20H+3.3[02+3.76N2]*2C02+3H20+0.3 02+12.4N2 WW. cuwraww W G" Taiilqu, — Standard Enthaiy Heat Caaci (J 11101— K‘ ): C}: = a + bT M00 moi" ) — (:02 (g) —394 000 44.3191 0 00730 co () H110 700 29.6127 0.00301 —242 000 32.4766 0.00862 __ 30-5041 0 00349 29.2313 0.00307 a ARC -:.—. :11; 121% (rs-1;)? + [EfCTz—Tffl + 3E6WCT-Ta) + 336203)] “F 9‘3 [C361, CTJFQ) + CTZP'Eifl + W [an CT «719) + 19:: [TR-T30]. ‘ 0:086 (Linda), 4" 313111104" [301* h’i‘ / $307.7 CTFHT") Bicep“? 30111.04r 030161;" [2* q‘aue] “i” Ago T" 0 i 2. acacia-T2" + 597:7"? “1649722 :2- 0 9.0Ler ana$b W} T: 023762K _ EM Page 4 2. PARTICLE COLLECTION IN A DUCT (a) Derive a general expression for the critical particle size DPC at which 50% of particles flowing through a horizontal section of an air duct will be collected by gravitational settling, Neglect all other particle collection mechanisms, assume air flows through the duct at a uniform horizontal velocity U0, that particles are spherical, Stokes law applies, and neglect slip correction factors. Dimensions of the rectangular duct arc length L, width W, and height H. The flow is in the L direction, and particle settling occurs in the H direction. “a 7:13 3.: l: 3 BW/“Drvs ’2‘ :59 DP 3} Page 5 (b) Plot a particle collection efficiency curve 7] vs. DP/DPC for the duct in part (21).” Calculate and show values of n for Dp/DPC = 0.5, 1, and 2 on your plot. q as M CD?) .3 DP “’- (my) V93 FDR: 3 D? /DPc 71’ imi— 715;! W . Page 6 3. REFRIGERANTS R—134a is tetrafluoroethane (CH2FCF3), and is replacing other refi‘igerants used in automotive air conditioning systems. (a) Global background concentrations of R—l34a were 35 parts per trillion (ppt; 1 trillion = 1012) on a mol fraction basis as of mid—2005. Estimate the total mass of R—l34a that had been emitted to the atmosphere prior to 2005, assuming constant emissions for 10 years starting in 1995, and an atmospheric half-life of 14 years. A significant portion of the R— ; 134a emitted between 1995 and 2005 had already reacted away by 2005. Recall the first order reaction rate constant can be written as k : 0.7/1: where ’E is half-life. Poi: may ma: WEIR; ‘ SEX/0,57% ‘frml 9“ 2 “$32 : ligwoowf emf/mt _ nrm" E x ha: éix/o WI [iv {o m 2: layanxiozi : @254“) pwol (b) Why is R—134a preferred to the previously-used R—12 (CClgFg) - rwo ma, 5’0 SJ'th-fifbhflfi/I‘c 032w. (LLB/6A, "- WM W, vs five/2 Z) W GWP ...
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cee218A-fa08-mt1-Harley-soln - Page 1 Midterm Exam CEE 218A...

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