Homework 11 Solutions F11

Homework 11 Solutions F11 - PROBLEM €3.13 EMOQZQ! An...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PROBLEM €3.13 EMOQZQ! An Air-shmral Ofio add: Ls \mth'm‘i bu ro‘flacin Hm. lseM'ro-m'c Mastic-v!- MAJ. cthsferu. Process” w't Pour hvp'u. fwasW—S lam?“ n=g.3. 11m. wwvessfm raft-lo, sheet. a. +1”. btflium'yfi e+ WrectLo-M.’ and K2. maximum «Add: ‘l‘cwp- enflwre art 0.“ kmum. Elmo: magma“; (a) 'fluz: had" Musk/r owul wovk Per uuld- Mas, {brand/L Prone-$5 m We. (Add.ch (lo) flu MM( c‘F-FCJCncj.’ and Cc.) +4“. mcan EFFGQ—‘h—V‘C— \VEAJ T ‘. ,prcsrwre. ENGINEER! g. . W: (“The adu- Ls +1.4 clasul s shun. (21 The. co rcssion. “,4 {m (egocegns an. ‘DOLmopic J tack nix: virus, (yAu 9t9ccssas arc [Manual reVWsMe. (Ln-rt». owr behaves as am '1 col tacit. 1:- (5) Inhe’dc Mpofim-Hd enqutJ eHccfs are neqngilale. ANALYSIS: Begin ‘oa #Im‘ns ecu-k 17b M {JIM-r WuuPa-L Shfis. wHkT. = 3003:) Tat-chw?- tanu-us/ u. '= 2M0? Polka, Hod, u.st 65- 3-“: {w arts-L poiavopu‘c. Cow-Imps‘h'm 3 h-l O. Tz:n[%] = C30n§)[°f] =saol< . tom TAM—M, ulz 449.32%} Also 3mm “131-10an) L4? = lLH?kJ/t,. Fur 4H» pal-athut “paw! 5m. I n_ 0.3 ‘- T+= T3 1-; toooE'cIi'] : Ion-K , and Q4: Elsie) pair} (a) Coufr-Ln-r 140L- {u-ousg L..,.- an“) hymn“: Lua'hn (Mom: l-Z, . (vi-fisEIIJJ? w . ‘ = an —'r. = (3.3:): 2.8-?1- K“! 1: sac-Emma __ .33 4. w ._ - lea-miss «— Am emu/,3 bruit-H... 63‘} 2. {Mt-(404'- Ufl' :. C4H-ff- I'm-07') —‘ LL13: _ “2'4; no"? ‘— y... Roux-31's: Wuao and uh wbgla—wu 4_ .. , = 1 JV l a 0% -. (urn) 1 (Ina-7 m n) ' 3‘1 ‘ he) 4_ Pwu83‘3-‘i-l USP-«5 Efi-B-S? Y - I new) = 121mg] , min—Ta) a (9 sums-fix on: p3 *_ E35 " I: Pd“- m... W'— fim emr‘ iguana arm Clay/M =(H‘-H1) awn/M: (mayor- “333] 4 cu“: 333.5%? PmmCSNL-(l W4n‘0 and an WW a G) Q3»! 2 ul-uq = 211.07— 'ie‘r‘m'= -S?“-‘” FUN-'3 *— 'PBOGLEM 943 (Cum‘d- )- Pagez Lb} T1“ +1-w~t C<Fa,£1‘c4'eMO? Lg. n =LWryclo/hJ/(Qiu/M) 3 «gnu. WHO"- : \3:_L+ my 2 -1573 4423.1: @5322. walk; M UM. I». COM ._ <D_23 + 9n 1 nnlr+ S‘s-~45: k212.brfilffi Wh- Uln. Illn- Tin-LI. ‘. (95.2 0 __ 0&0? (Sb-7 / vl’ new 0‘ ‘JA (a) W W-u-n W L. Mae's: rm mam/M 4.!"er 1r.[_t- 3- VI “Au-A... .. T _ ?3l ”'_“.“ U- - Lfi' - (.1352 K5.nc)(”“""‘) :0.er “Fir: {orN/ML Tun _ (95m. n was "5"“ '“‘,”J= 35c- tur «4"? M1) - q" Tyke (o.%l-a;§)£l~+,] Hr: to I" F11? flan-l Calm QCYQL‘ :. LUCQCL. In 444...: final—4..., Bicycle/M: GlT-lKD/IEJ can-é an. m __ _ t.qu 4, [13-9.]; + 33.41" —-§"W.'l3’ = ‘51-’1- *3/55 HA- www Mm +1 (Jo-I90; M «Mth Obi-wand w: pad-(a), 9.22 Consider a cold air-standard Diesel cycle. At the beginning of compression, p1 : 14.0 lbf/inf1 and T1 = 520°R. The mass of air is 0.145 lb and the compression ratio is 17. The maximum temperature in the cycle is 4000°R. Determine (a) the heat addition, in Btu. (b) the thermal efficiency. (0) the cutoff ratio. KNOWN: An air-standard Diesel cycle has a known compression ratio and a specified state at the beginning of compression. The mass and the maximum cycle temperature are given. FIND: Determine (a) the heat addition, (b) the thermal efficiency, and (c) the cutoff ratio. SCHEMATIC AND GIVEN DATA: T3 = 4000°R m = 0.145 lb 19] : 14.01bfi’in.2 Tl = 520°R r= V11V2= 17 ENGINEERING MODEL: See Example 9.2. ANALYSIS: Begin by fixing each principal state of the cycle (Table A-22E). State 1: T1= 520°R —» u] = 88.62 Btu/lb, vrz = 158.58 M: For the isentropic compression vrz : (V2/V1)-v,1= (1/17)(158.58) = 9.3282 Thus, interpolating in the table: T2 = 1534.5“R, h; = 378.32 Btu/lb State 3: T3 = 4000°R —> h3 = 1088.3 Btu/1b, 171-3 = 0.4518 State 4: For the isentropic expansion Problem 9.22 (Continued) — Page 2 (Vi/V3) : (VI/V2)-(V2/V3) = (VI/V2)-(T2/T3) fl (17)-(1534.5/4000) = 6.522 and 12,4 = (V4/V3)-v,4 = 2.9466 Thus, interpolating in the table: T4 = 2253.7°R, m = 421.25 Btu/1b (a) The heat addition is determined from an energy balance on Process 2—3, as follows. Q23 = m(u3 — 112) + W23 = m(u3 v 21;) + mp2(v3 — v2) = m(h3 — I12) Inserting values Q23 = (0.145 lb)(1088.3 — 378.32) Btu/lb I 102.9 Btu (b) To determine the thermal efficiency, first evaluate the net work of the cycle. chcle = Qcycle 3 Q23 — Q4: = Q23 — m(u4 * 111) = 102.9 — (0.145)(421.25 + 88.62) = 54.67 Btu Thus, the thermal efficiency is 17 = chcle/Q23 = 5467/1029 = 0.531 (53.1%) (c) Since [92 = p3, the cutoff ratio is rc = V3/V2 = Tng2 = 4000/1534.5 = 2.61 9.23 Solve Problem 9.22 on a cold air-standard basis with specific heats evaluated at 300 K. KNOWN: An air-standard Diesel cycle has a known compression ratio and a specified state at the beginning of compression. The mass and the maximum cycle temperature are given. FIND: Determine (a) the heat addition, (b) the thermal efficiency, and (c) the cutoff ratio. SCHEMATIC AND GIVEN DATA: air T3 = 4000°R m t 0.145 lb p1 = 14.010f/‘m.2 r. = 520°R r= WV2 : 17 V ENGINEERING MODEL: See Example 9.2. Also, the specific heats are constant evaluated at 300K. ANALYSIS: First, determine the temperature at each principal state of the cycle (Table A-22E). T] = 520°R and T3 2 4000°R are given. State 2: For the isentropic compression with k = 1.401 from Table A—20E T2 = (WI/21"“ T1 = (17)“0‘ (520) = 1619.6 °R State 3: Similarly, for the isentropic expansion Var/3 : (14/sz V2/V3) = (r)(T2/T3) = (l7)(1619.6/4000) = 6.8833 and T4 = (Vs/V4)k'l 73 = (1/6.8833)°‘4°1 (4000) = 1845.5°R (a) The heat addition is determined from an energy balance on Process 2-3, as follows. Q23 = m(u3 - 112) t W23 = "1013 — 112) + mp2(V3 - V2) = "1013 H h2) = me(T3 e T2) Problem 9.23 (Continued) — Page 2 With cp = 0.240 from Table A-ZOE Q23 = (0.145 1b)(0.240 Btu/lb-°R)(4000 — 1619.6) °R = 82.84 Btu (b) To determine the thermal efficiency, first evaluate the net work of the cycle. chcle = Qcycle = Q23 - Q41 = Q23 - "1(114 - 141) = Q23 - mcv(T4 — T1) With CV : 0.171 from Table A-20E chcle = 82.84 — (0.145)(0.171)(1845.5 — 520) = 49.97 Btu Thus, the thermal efficiency is 27 = chm/ng = 4997/8284 = 0.603 (60.3%) (0) Since [72 =p3, the cutoff ratio is 1"c = V3/V2 I 73/2"; = 4000/1619.6 = 2.47 Women “1.34 m: Opamh‘nfl Mn. am. pmle -Fcr an afir—SW M (Adda. mg: Debt-mm: (a)+he+¢m.pwaturs 4+ “Ha. and 3+ each heaj-afigm-iom process) Clo) Hm. he} movie parawizf‘ wuss, CC)+L|£ {Mer dFC-t'ciwvig’ WM) #4. mm awed-M- prefim- 5W5 d, amen Dim: ELJCLLJEQNG MDEL: 596 Exam-1312 $3. M52156: USI-nQ‘dai'n/fi'nm Table. 5-21) ONO-in dmod‘ cam fn‘ndpad SW . 321L945“. 1 300K, a =100Id’a =b ul: 2.14.01 [cu—{kalmfibuz 5H2. 11 For-W1. Isad‘mpic mwmin ’U-rzfi-Trl(%%) :_ = =>TL= 102.1 KJ M1: 5:496] LEI/[:3 V 3%}: Fbr process 2:33 Qtizmcuzvuz) a) 14,3: $33.3 +147. The mass {3 5 z _ fiV‘ :QOOkPA)C0rOI4-M') 10mm 11:: ‘ W‘ ‘ r211 3.3m IJ z . emu“? M M _ was; ' H 73=Jso7m< ’3 3 — ——-—Co‘olbu’)+ 514.40‘ = 0.1.1.52 k3”; ${L‘f [6453' Lair M: For-+143 mm msswmheqzt WW6“, 6234: mC hq-IE).lew P .. l. in - 9%+ in; = 0—153— +ua+sal = 2343.34 6/6 (0.0 67.6 m t ) % T4=2°7231<4——-—Ii v V (‘15) v V uh: 2.463? 21932.18 =>T5=usa.sl<, usaefie.bk-g =(2.2.'r) —( 0.0902er 6%.9—1Wm1) = 22.1 --H.o‘£8 =11.tao IJ Wagolelm=7l3fl lJ/Lj 4___—__'§Eflj_m (c) K: wage/a1.“ a who/“"—' =o.sn CSI.I°/o)‘.____.____._._v.‘« Kat] The mm awed-We Pressuma is W cLe. W“? V.(l—V=-Iv.) H (“.60 L3) " (o-onJMI — é.) tog»: «M L {:3 ...
View Full Document

This note was uploaded on 04/03/2012 for the course ENES 232 taught by Professor Hines during the Spring '12 term at Maryland.

Page1 / 7

Homework 11 Solutions F11 - PROBLEM €3.13 EMOQZQ! An...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online