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Solution Manual For An Introduction To Combustion: Concepts And Applications, 3rd Edition

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    SOLUTIONS MANUAL to accompany AN INTRODUCTION TO COMBUSTION: Concepts and Applications Third Edition Stephen R. Turns Propulsion Engineering Research Center and Department of Mechanical and Nuclear Engineering The Pennsylvania State University

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    PREFACE This solutions manual has been developed to supplement the third edition of AN INTRODUCTION TO COMBUSTION: Concepts and Applications by S. R. Turns. The solution to each problem is presented in the same format to encourage students to develop an organized approach to problem solving. Although a more detailed format could be adopted, the approach used here is, first, to present a list of given or readily obtained information; second, to list what is to be determined; third, to list the major assumptions; fourth to present the solution, being careful to indicate equation numbers and other references; and, lastly, to comment on the solution. This last step, frequently neglected by students, is quite important in that it can result in students reflecting on what they have learned from the problem, questioning their assumptions, and, in general, providing a wider context for the problem. The solution to each problem starts on a new page to facilitate copying and posting. Solutions to the following problems have been corrected from previous editions: 4.7,13.3, 13.4, 13.5, 13.9, and 15.4.

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    PeoBLEM 2-1 oven: Aw with a molar composition of 79% 0, and UT 0, FIND: The mass Gacion of 0, and Ny om the our APPROACH This « Sivaple Conversion problem Using Mw, x EG MW; and Y= 7K fo Gon tla mila composibion Kp: 079 ad X,7 029 mixture Molecular weight mw, = =X; Mwy = Xo, MW, + Xo, MWe, = 0.71 (28.013) + (0.2)(32) = 23.35 %. Mass Feackion of o, and. N+ Mido 32 [A x (55) = 0% (7%) = 0.233 Mp, 23.03 \ _ To Xan (5) : 01 2385 ) = 0.77 Commendss Mole thot Yor >, Shee MW, > Mu, and Fhat zYi=1 as would [3 expect.

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    PROBLEM 2-2 Gen: The Folloramg rare Species. * Mowes xX; Y MW, co 0.095 0.002 0.Coz 28.010 C02 6 0.127 0.495 gy. 0n Hyo 7 0.47 0.0% Loi No 34 07722 0.797 23.013 mo @oos | wwe wxigt 30.00 Tw: Hl Ie Io Frode a) The mole fochon | mde 7, and ppm of No m Fhe mine b Defermme tle Mw of ha mixchye 9 Dekrmme the mass Cacho of each cons bhuent Ne No 6.00S AKT TENG TR + Wy, thy + Ny 00S + C+ 7 34 40.008 Ws = 106 1G ENV mmox #10 NocA per = orm (100) - Treg (i) = Wo { 1 516°) = [106 ppm | Whee A 2 Avogadro's Muamsber B) Mp. = ZG Mw; = Xp MW, + Xoo, Mo, F Kyo UG 4X, MU), +X pM, where Xe, Ke, | Koy and Ky, a fod m the same Mame as XK, was found Myc = (0.062)(28.01) + (0,17) ddan) + (0.149)().00¢) + om(18.0%)4 (1 oY 33000) Mo = 23:6 50 MR; a Yi X% mo, Tc: Y= (o-0oR)aw01 52.) = 0.002 o paix Co, Y = (0.1727) (44.01 /2.6) = 0.198 ( V/A Hoot Y= (od (rag/angy =0.0M Nyi Y= (722) (zon/zve) 20797 Not Y = (Josie) (30.006¢/2%.€) = 11x16 COMMENTS = Mole Fhat PpMmE x); (1x16) anal that 2%: and ZY: Cam olen be used + check your “

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    FRoBLem 2-3 cwEL 2 Mixture with ST mele Hy and ZB Emel Op FIND Yop 5 Yo, 5 MW Ni, 7 SOLUTION Ne Bs A, = — A = > Zo. 62 a) ¢ Noe 4 tH, +3 [o. ¢e5] Uy = /= Ty = Imo eT =|237° 5) Myr, = Z Fe MG = Toy My, + Ao, Mil = 0.615 (2.006) + ,375 (31.999) : CMe 2.006 0.095] < —— Ee . = zi =— = ) Y; 5 Max b fe 0,625 13,260 0.09 Y,, =/- Yu = J —0.095" = 0,705 ommerT Een Thoagh Te pote “rachon pA Ho lig re ls mitbeeedon we ght" results in its tory “ small pases rach,

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    PROBLEM 2-4 GEN Lt Oy —Cly mixture @ F00 k FF Joo BT; Np, | = 0.2 FIND ¢ Tei, 5 Mo, | ASSOMPTIONS © dead. as mixture. SoLoTI0M) Mls a = Mere ) Tet, Hat, MWyary, = det, - MW csby Yuk, MW, + (1—- ety y MR, =o, 1e2%3 6,2 (16.043) 0.2 (16.043) +0,8(5/.999) 28.508 » - , 5) Ry 7 = My, TT Zu, = Ken, P Jw fp = Te 2 Ro T° . 2 =(62)Pot” | ia 5 bet 83/5 (30a) m3 Comme: Gretel Treatment of vats is reguired, mn. part +,

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    PROBLEM 2-5 GWEN: Ny = Ar pyxclure. witn Moo = 3 Mgrs T= Stok , P=zSl RA Foe 1 No, MU, B ve B) Mow 17 AssomeTiin i cdend gas mixture ’ 20¢uT0ad 2 Vy, = J SMe 3 2p 75] Apx BN, + Np 4 _ Bp = 1-077 = |0.25 5) Min, = 27 Mu = 5,75 (28.004) + 0.25(39.94%) Mpg = 30.998 Mop, 2g. 0d Dh Gn = 077 Fm To. Yoo =~ Yo = 0.322 d) RY =n, RT 5 Fo = Fu Boe Ny, [Y= Ton Fe = (0.75) 2s0-10° = 0,045] ®.T 23s” (500) | kro Jm® LommaisT ¢ Cored Freatment of units « pret in pert od.

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    PROGLEN. 2-6 Give) + C95 0, mixctuwe wide Ay, =o) Thy =0.9 T= ok FIND Standardzed enthalpy o£ puxcture ASSOM PATIO 4 Tdeal-gar be havior socuT7on t+ This is a 5Trarght fo read appli otro of Za 2.454 combined with, TA afin frre 4 ctoadoad ped enthalpy (Za. 2.34) . heo. = bo, + bh, = ~393 sar + 2003 (Table 2) =-389, 5¢3 £T/omet hy, = k + sl, = 6 +533) = 3030 KTlbpets > ves (Tube Alt) Prone = 2% k = oo, A, + “Fo A, (5p 2.150) = 0.1(-3%9,5¥3) +0.9 (3631) hoiy= -36226 £T only COMMENT: The use of Fre Appeadss A Sables male Trie problem Simple. Nyte_ Toat—"The stm information. b aoadable as carvefit op wotrons in Table A-13

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    PRO®RLEM 2-7 Glen: A stoichiometric midue of methane ond air Fivd = The mixhae malecular weight AssupTions: Arie consists of Ny and Oy and has the Followwng Composition © U7 Op and 79% Ky by volume APPROACH: Delermine the Stoichiometric oho of ar and Fel ond Ahern find the conshifuent mole Fachans and Mpa Shichiomehre relation Coby +a 02 + 3Uakdy => XC0, + Vo HO +3%aN, ethane © xl Bak E— a= xt VA =Z sa the air - fuel stoichiome dic Mixture 1c | CHy + 20; + 72.820, n Nog + i Le chy = <Hy Nor NagtNo thy 052 ~ BE No, = 2 Ko, ® Mor = 0.52 oO. Oy, 7.52 MNy,® 7:52 Koy* TNT Tres = O71 Mw, = TX MW; = Key, MW, + Xo, MIs, +X, MW, = (0.098) (16.043) 1 (0.198) (32) + (0.715) (29.013) Mp = 22.6 Viemal COMMENTS * I§ ths as a fuel-rich or {uel - leon mctare xX+ Yi a= ® Where P= equivalences roe

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    PROBLEM ~~ 2-®& GIVER: A stoichiometric air -prapace (CaM) rite FD: The shichiomebric AF mtio (mass) ASSUMPNONS® Air js comprised of 79%N; and 2176, by volume APPROACH * Delermme the molar A rio and covert b mass AF radio Shoichiomebric relation: Glg + a Oy *37%a Ny —> 3c0,t YH0 +3, xt YH i . - a= "5 x=3 y= $= a= 5 A + 3 «Ka 4. Toa Malar Af ratio AJ = [ : / A Mair far) Mass AF rtio = (Ene (fees ) = 4% (Fs 27.85 4) - 5 ] (? MASS 4t(9) 4.0% A = (OF) = 15.6

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    PROBLEM 2-9 GIVER: Propane (Hs) Buening at an aw fod pho (mass) of 19%] ZIShE The equivalence oko ) ¢ ASSumpPnons: Aw Js comprised, of 797% Ny and 27% 0, by volume APPROACH + Dedermme the sStoichiomedrrc HE radio and then the equivalence mio Sthichiomebric relation: 4G Hg + a0, + 3%aly, => 3¢0,7 HHa +3260, Lak xe . . as ra x3, v8, ¢ «| a=§ A , Mac \ 2) = s YE) = Ya [ 5 5) 4% )( Gio) 6 Equivalence robo: (4) 56. ¢- BHC go © 0.87 ( GY. ’ ¢-0% | Comments: Smee P| this Combustion process is fuel -lean. Also nole that @ oloes not depend” on whether the AF ratios are expressed in berms of mofes or mass See 1] is also a ho

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    PROBLEM 2-10 GIVEN: An equivalence rds of 0.4 FIND: The correspond me Alc rakes (mass) for methane ( CHy) , propane (CaHg) ond ~ decane (¢, Haz) ASsumPnons: fic is comprised of 797% Ky ord 2% 0, by volume APPROACH + Use the relationships : Yu _ MW pir ax = and Ars ® 4.7% MW gel methane (cH) x=, y= , Mw = 6.043 “rele 283¢ ac Lo gm HB) (Gn) Oo. A mass 28.50 5 Yg-fel propane (CHe) x= 3, y=®, Mw = 44.0% 5 mole 3+ % 28.95 a= Sg cam Hh) can (im) 22) Af) = 2674 5 Ve Fel decone (Coll) 23.95 z 2335 as ot Hh © 2ST > Af), HH (257) (m= 299 ) ~ Ks-aiv i Mh). = HT hg -oel COMMENTS © Pele hao the AF robo (mass) Changes arly shaktly [SS Ayetro carbon fel hb ansther. While Fhe AF radi (molas) Vames from /S9 ( methane) + 123 (lecane), This difference behavior is Jue b the Mp, Mcreasivg as the malas AF ratia MCveages.

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    ProRLEM 2-1] owen: 37 (by volume) Or measured in the exhaust of a propane (Hd - Fueled ruck Fob: The air-fuel roo (mass) supplied fo the engine ASSUMPTIONS © Complete combustion with no dissociation AppRoAcd + Use conservahon of 0 atoms fo determme the AF be fon the exhaust Oxygen mole Fraction assumed, Combus Hon reaction © GH +a 0, +37%al, => 3C0, +9H,0 + hOy +3 %aly conservation of O atoms: Ra = 3(z) + 4+ 5(2) a= $rh b=a-$ exhaust 0, mole Gractvon * SS Xo, TN + Rye +R, +R, TOR +Y +b + 37% using b=a-6: a-£ $+ 2K, Koy * Ya + 2 AT Tux, Tz ohact 0; 3% (by vile) —> X= 0.03 2 . $+ 2005 AT xm = 1 =4.7%6 (0.03) S90 — MU, 23.%5 ) Ase = 47a (Fs) = 4%( 5:70) (Fim AF oss = 137 CommenTs + This engine [5 running at a fuel-lean condition Lh IA EN SE EE) — $= oq TOE

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    PROBLEM 2-12 GIVEN: | mole of aleahol (cy Hy Cz) Undergoing complele combustion Fon: Shichiameteie balance equation and number of moles of air to buen | moe of alcahd ASSUMPTIONS © no dissociation and air that is 77M. and 212.0, (vol) APPROACH + Use consecvation of elements or The combughon of one male of aleohs! Stoichiomehric helamce : Ci HyOp + a0, +3%a ly, = h CO, + CHO + 37a lly conservation of carbon x = b — b=x conservahion of MH: y= 2c — | C= % Conservation of ©: 2+ 2a= 2b +C= 2x + Ya a= X+W-% number of moles of air PB buen mole of aleshol ¢ Nae - a+ 3 Xa - 4.2% Neel / Mae oogae[ xs Y= Zh] Neel COMMENTS © oe that storchsometric combustion of an alcohol (cetyay) requires less Oxygen Hhan the combushon of a comparble hyd cathon Luel Cex ty) due TB the presence of eyqom nthe fuel |

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    PROBLEM 2/3 Given: Metharol (cH oH) and the results of problem 2-3 Fmd: The shichiomedric AF ratio (Mass) awd compare with That of methane . APPROACH: (Use +he relabonship developed n problem 2% For Ath me than! and methane methanol (CH, OH) © Cy Hy Og X= yy 2 Mw = 32 Ymole 2 MWg, Ys * 48 [x he A ) (i) sa [14%- 4] (£F) = CH A = 7) 6 methane ( cy) : Cx Hy Oy x= , y=, 2:0 MW = 16 Hkmole Mai = _Z “ Dons = 4 [ x + 2] (Fos) _ uw 223¢ w3c [14% -0) (5) A - %),.- 17.2 Comments: The lame Afforence mn the AF rabies for methand and methane is primarily due to the differences in fuel mw, On a molar basis the AE cabo for methane js T4 and 9.52 for methane.

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