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Homework 5 Solutions F11

Homework 5 Solutions F11 - PROBLEM 4.” 4.11 As shown in...

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Unformatted text preview: PROBLEM 4.” 4.11 As shown in Fig. P.4.11, air with a volumetric flow rate of 15,000 ft3/rnin enters an air-handling unit at 35°F, 1 arm. The air-handling unit delivers air at 80°F, 1 arm to a duct system with three branches consisting of two 26-in.-diameter ducts and one 50-in. duct. The velocity in each 26-in. duct is 10 ft/s. Assuming ideal gas behavior for the air: determine at steady state (a) the mass flow rate of air entering the air-handlin unit, in 11313. (b) the volumetric flow rate in each 26-in. duct, in ft /min. (c) the velocity in the 50-in.duct, in ft/s. T2 : T3 = T4 = 80°F D2=D3=26 in. V2=V3=10 ft/S Air handling unit .............. P1=P2 2P3 =P4 =1atm T, = 35°F (AV); = 15,000 ft3/min Fig. P.4.11 KNOWN: At steady state, air enters an air-handling unit and exits from three ducts in an attached duct system FIND: Determine the mass flow rate of air entering the air handling unit; in lb/s; the volumetric flow rate in each 26-in. duct, in mein; and velocity in the 50-in.duct, in ft/s. SCHEMATIC AND GIVEN DATA: See Figure P.4.11 ENGINEERING MODEL: (1) The control volume shown on the schematic is at steady state. (2) The air behaves as an ideal gas. PROBLEM 4.1: (Can'lfnued) ANALYSIS: _ (a) Determine the mass flow rate of air entering the air handling unit, in lb/s, as follows: f“ n] lbf 15000—_— (latm) 14 72_ _ 2 _ "'1‘ Z (“)1 = (AER = [ ftbe 1 tin. 1:? 12:: ' 20 05— F— v a m s ' ‘ ——1545 (495°R) 28.97 lb-° R (b) Determine the volumetric flow rate in each 26411. duct, in ft3/min, as follows: D 2 ' ‘ 2 2 ' 3 (AV)2 =(AV)3 = L 2 V2 = L6”) [10—125] ”T 2 60? =2212. 25t— ‘— 4 4 5 144m. 1mm min (c) Determine the velocity in the 50-in.duct, in ft/s, as follows: dmn . . . . dt =0=m1—m2 —m3 —m4 3 [2212 23+ (latm) 14 711% 2 m _ m (AV)2 _ (AV)2 p2 mm in} 144111 1111111 2 71113 2— 3— _—‘—:"—-—4_—__——2——'—_ - — RT 1 tm 1ft 60 V2 2 l__545 ft- lbf (54010 a S 3 28.97 1b-° R m,— — m1— m2— m3 =20. 05—3— 2(2. "Ill—b] =14. 631t-J s s ”'14 = (AV)! v4 [14 631:][_ 1545 ft 1be(5 40 R) M4124 mil-T4 - 28. 97 lb R latm ft ‘___ V4 _ =- '— — _.146 —' ‘A 7:132 _ . 7:50'in.2 E 5 4 p4[ 44] (latm{ ( 4 ) 14-7in.2 PROBLEM 4.2 ISM—I: Dad?“ are Prawded Fer a; Cylcndn‘aJ “funk. L114- “J drfll‘HEJ of Loafer EIND! 132+"an +51: fi‘me Jr" m'n.’ when flu. “Eur. hold: 900 any of carrier. ScHzmA'rlc :1 GIVEN DATA: -—————____L___________._ EUG 9- MO DEL.‘ -_——‘_—'———- l-AS shown in 1"”.- “sci-ch} a. Can he! Vowmc {Kalashna flu fink. i: bang Considered. 2'. a: $8: 5-.[31 3 TM densn'i—l afwnfer to (OaFj/hp‘? ANALYSIS: Mas: rad'e b‘lanu; ddlfgcv : ““29, w Pcne 6+9 3.0440“;- r‘ak :4 1'“ 241-12“ 1115+ LI.) “112: fiACV; (EI' +40). Heard-Ina!) ‘tcpnsflaat JJ’ X. 3:151: - 31*:fo ,muu Q: (7.32) =- 32- A: (23%).“ (a) Alw2MH M Wadi): \fAZ Junk-Dre Ali-HM Area 01014-4 Lua'kr 7439 Tamara. wnut-4:, 2 : MwIt)/J3A ant! E‘b-C') be M 421‘: -= ~fiAe [7. wv(+)]y‘___ _[ ngflchi (matey/z «Lt fA M" \ av-W_ [La 31%;] V'- ‘Ia. '_—' "‘ _" 2- Qmw‘) elk A 16...: )‘I‘T-Tf- -[ZSPAEJVZ'C‘F . 0r- Ink1rnh'n5 from 1.2013 12+ 3 “a" jet V o— A ”z 1/ '1; +£=~ 72A '[cwcqn :chvcon J (2) 9f he; I? 3‘134-“13 Maw“): ZSDOKL 0“; '3 A : 2. 1 .=10 F3)»- 1 c : :‘bawi: ’f Enserfinj A ”“3 ‘3 N,(—q‘)_~_900t51 we Set 1 "z. .11 1 1m“ 1(1‘5) fiUIOOr-fl -- [251136] ] 60.5. t; — — ’ am ?L)L‘°"-E5)(“'5*"")z :- ‘ 1539mm :PROBLEM ‘HZ IigjhethgM: 5‘th dad-o. an, provded {m‘ a +kro++lm5 Valve. u; SQHES ‘I-ilt‘f‘fi.. ' a s44”... hvbgnt, each. ¢+ Pied? Sflffi- Mpg-acr- eiwclureil 1'. Iva-Hunkrhang. per lb ‘9. :i-exw‘. {Haw-‘5, 1's Sp-mfi'd. FIND; ' fig 14”, Siam-'5': Id: 'Hvu ‘l'vv-bu'uc Wat}. g$¢mmhMm-1i GIVE“ “Tm enausegwa HoIDEQ' Q‘W‘hflli. . 1- A Coni'n-l Valuing c,fl¢(o.tsn,_¥f§g—;._I . 4Q. \ . . 7 1" V‘WC And {-Hrbt‘ne It's ‘ can “Jinn-d. \ z. “The con'h-ul V‘ntvmc 1‘; .1- . 'L ,l 2 sway... \ 5R4; sir-dc. ‘ I. «_‘ I 3. Fori-k-r Can‘hml Volt/Me, <2ch {.3 .. ' I ,. n: has me am: Hm an-fi 9F \ 1 3‘5 . f. an: qr WW—BSDELW Innate. andpoi-M H ”.7 C ‘\ 1-7;- lb “as-sand. where man ¢r°55'5+“1- \ I Hana“: aé-Hau min—J valumc. s __ 3 _. P3 AM ALTSJS 41:3 P3 v; Known. To £1: Hm vim-h: re, mam; an $4164.41 Duo-Q. Draperf-‘g Value at 5.15: W: “-9-“: 9*“ ha. ' TD befi'”) Mass raft balance: 3““: MI- - M1= M3 (= M). An enmrgy (bl-c ban-lav“; .F. 1%.: cm‘hn-l V’cilku {adage-5. as follaws: 0:;6—qu*QLCh‘nnh3)*(/V11f+1(z—zj)] =K> “32h h‘HQNcr/Q) “1"?” 1'“:- fi'm'fa-blf H'+EJ L13 : (1.97.0; BM... '59 Eli» :- "05?“: 3451/6 ‘b 5 “1‘4" {duh {TM Thbk A_3E .1- 5 lbf/fp'z J “.9: 130.11} Bh/I‘J ha: NW 55/ I. f My ‘II‘FCI “53% 3 +15”: in “HM +wu- Lek-43¢; A'1U.I-J-va_porreamh.Accord-“51» w; - ' ' x3. haw-f :91”; “a“? = 9.“; (1.34%) ..________e g; “1- h; HBII- non- . F. . w ML: Skadr-E'f-n'f: Operafi'flj dad‘a. are PrU'Vr'dJZU' Jan.— 0?. ‘fkr-a'fl'l'n V‘Mve H3 Ser-‘es ms+h 4. Mai“ etch-anger. ‘7 F'NDI De‘k‘VH-um +4--‘ runurc at ‘H'u waive “it and-Hum” W In“: mF +Lu1 ltfiu§é i‘lveaw‘. SCHEMATIC '5 GIVEN D7974: Ewen MODEL —-———_£____—_ _ Hem exchanger l . A g :ln-o‘uJ n in +59 shtick; Saturated l s u ‘- cwivol volume ha . . Iv aturate . e 8 . 25:”35233 l “W H“ W J: "’ '" :41 £626sz I ‘ 2‘ TM wnle-UW: i . 3 4| ‘1" gut-candy 9:1. I- . I . fir-fu CWM W‘U / Liquld Water ' ‘ Liquid water , 3 and a£T5=IO°C T I mnzwc Fig. 104.95 it:{ch:‘:+;:1:fzu (new-37 *E1C1C“"‘=""‘r can H 1 [sh-0 (Ed- Y 4, Theoxflans.m Muir" AMALTSI 5-; HM ware v3 c. +qurrlch3 PfaLOSS‘;h-1_Zhl . ggpz Tknfllffi G-nd 115x134“ (a) ‘Stm-Pbumdgnns.m fkaA-W’ \jn'lVi L'S‘ 5‘- pmss, htxhl. fi‘mTr‘L—R—IA'IUJ hi: 100.25 Mle-j- 31 - : —' . rt: 1.: TumamwthVQTwhfidv«1'35WJk h=w°z 3 SM ”L w} {mi/k. i—wu—PMH ”3“” Taurus-sure w Yin-u w; "Sac: P: 2.1?o+b4r=2'7'°4kf’a. «——-—-a,_ 2. ’“KJ Sabrahm {WQJHLIN “j- L‘.) M4139 5:43 EWWM btlahm {WV—“‘4“ (—5 “171-4 'VDIVMN—sox 05:42.64” mama-m + Mam—w] Ithi'h‘] EA'“r] 3-5]: 7.42.54 K?/JCJ.(_To—{r—LA-to)_ 5r Liquid wk“: )-mww-+‘"Jv~’rv~ {3mm "Raul-and) h+:8‘3.w53/P5 :3 m4 -_- In; Sm“ [3:94; “3"?“ “4.3: #41274) , hr: Ins-LT; an; \n)’ 1’ «2'0. hilt?- . -- .. .— . SJ . _ . .591 1.42.54 Ian 2 M4 ‘ to M AL ng,e‘,...4z.m] - 08% F515 .- PaoE'LEM ‘1. 103 etchgn;9r- tn serves Two scare; an. invoiced: “2. end helium. FIND: Defer-mug '5'“. qur {-(ow qu'c MC «Hue 11¢!»ij a... heft...“ __.!. SCHth-nc r' was» DATA: swan-tummy; MODEL: I. A Card-r5! Vo‘U'IMP eh;!b$"n§I Helium '——I_ - 4 T4=250C both Cnmponm'h!‘ k'a <29"="J""‘l Pcwfifin=50kw ‘ I ‘ ' 2- The con-fro! volume is” at I Sic-A, I+A+€. L 3. Fig.- +l~p comm! Valvmea ch'é 350K Heya'jable and lr-c'neh'c and =100kP H h t - ' TI2280Ka eatexc anLer PORN-{‘54 .c "9 r5: gfif¢¢+8 arc rill=025kgfs ignored- Fig- P4403 4" The «Rica-9 34$ Modal BARF/fr: “I? each 54:. Far heh-‘agm; K=b6i ANALYSIS} Apptyn'n, Mass: and {nu-53 ran bglnncr: 4a a. can‘hro-t‘ VON-"Hf EHqQfinS bfi-h camronrn'h' Wt 391:) ' D I n c 0: Id, -ch -1- “flgtha-h31-i-‘M1-[hfi-hr] .g Y§I+= M3] 3 Lulu-CH: (-Ww) :S‘OK'W. " ' [HS—3'13 Ean-‘h‘v‘g propqrfif‘h dad-s firm Thick 4- -_-2.3 99 (.1ch For ”3' f“. _ k3 (3:41..qu80) “1"”?- q 5;;‘9 ”Inf = ‘-_"—'—"'—-a—q - E . 33): [T ] 2.3m who: ' F9? bclqu’w” 'i I ' Chs— he) = Q? [Tr- T4'] =. (' a? "—"XH 8'3”- LI X'WK) =fié'5' ENE} 0‘5?“ 4- 903. K: —K Cr“) gauge-5.15. remH-S, MI: . ":4 2 50ml?” I +(0.zs_§g)(I-?2.B.KJIH) my" mu, _ _ . g ...
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