6 - jarpoint 20 cm from the tube the WW pressure drop in...

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Unformatted text preview: jarpoint 20' cm from the tube . the WW pressure drop in The first 20 "ofthg mbe'lensth. . Determine the total 'appmnt pressme drop (friction plus acceleration) in the first 20 cm of the tube length. NRE 4214 HW Set # 6 — Single Phase Fluid Mechanics Solutions 1. Pmblcm 9~5 of'l'odneas and Kazjmi, Voi. 1, pp. 409. a a . Av" “w are“ “'7 P "’ 11"“ D aim-1" mam" . -: - .. 2 3:- ‘ bta‘u'di‘ Jamel-tr _. figs: g:II-CF‘H- bl Care MSNM a film" “NB "12M 3E¢¢9iJlmLafl3 . -3 Rt: E'UmD,//A 1r 72m*:;&:ils.3zma & 4‘q3mu‘5 m :. Fla» is. What lead- (aybmuflcxHM Prafime Dray: M '5 cc}- 1.1-, 25.11 km CAM.) Use Chums-«Teawm Model _; E1, 9*” 9A “39 gTaMfi-a P/D :: Magma -.-. 1-3125“ . a Frm Table 01—3 PM aw Carer?” aha-«M a 1:..- a-H'fi‘? ’ bl :: *5. Mesa J log, a -- a-o‘i‘fifi I .. .. -_.- udsz‘i E FPS?) CfI-T C 06 ,- ’_ . ’ s ‘FiT ‘5 QF‘T m =- 5. M44- [Reir *5 2* fifi'HSX-m] 4" a. ‘ I ($2427) [when RqT refus ‘ro Reunaids» numb-er waflbare" ) ” bwml'e on; Cafwn'aheai AEoUQ] 1 I. L (3..” - @5144- A 3'7 "3*72“ 3 13.37%”: 3* : 31.15 14"; Guns.) W “P 2/7 Check: 5mm? "H4" {IN-f is hrbulcud-J owe m1.) 95-h.“ 0"” “PProx;mA‘-& “an” 4W‘HM 15%ch 'qu-or _ . _ 3.7 (4H,) - A .. — D .013 i- __ a: "725 { PP“ Chm) MeoAU ‘3 37 X [03 fi- 2. (c) Enll'rqutce we} Exii' Pressure Lassa; [Scc' Foo+w111 .1 Apcnb’MQ + A'PeyLL : (KC+ KC) 2... F53 9—3:, (@333 @341; Kcto-S an“! 146:“; 1 . . 444-) _ . .. -.APe/wmu+ Ape-M -__-: L.g 1* L 2 _ _ kPCi A Pressuu‘t’ Loss a}? Spacer; A5: 2 (Pf—t2) _ 23.4 WU“)- 2. LA : é, #720x<4‘°“’) 23"? Elam-e“ 715 * f 2. (117.4 $2.5m”- Cu 2 8.70 kPa (AM) (E) Prcsawr: Lac-55 ;n Fi'fFJst- '. L59 Rahal/HE 5 c: r m At! chit:- md Fij civil?) _ F‘ - ‘- 3‘12? Rc_4‘i§xrp ———} CUT—(=- a Aazcht—tf> = 4-195 MAI-Ml 1. AP ‘ 5 2 *4 i 72:” Ghana) 1‘ 41:2: 1 Fui‘hH-JL 2' HTl L) z: 111:“: laPc: CAM») 4’4/7 .3 h. Problem EH: othl'ndreas and Kztzimi, Vul. 1, pp. 4139. 2— 2.. .5 —'_TL DE 7* LBJ- Lm D:I'-DCH “nil-D :I: r“ 631- W15 1 U5! wfiflfirdePtcu MEAELI (Esta—'32 ED m “3—33; um Taxable (LL-3) H P-‘I-gfiw Fur IQM’IHEV 1’“ Q91 iflf'flr‘rfl'l’ C—Lanmzl DUI-I11 : {~50 {1:35.55 J 51:2L3J J max b1:—-I‘io.z :' = 39-557 4- ELZTI fine-5.3- __ [smug #leg)L 'L = 1151.??? v= mm «may '2 am CAM.) b. For Ra: IDS 33¢] b : G-Gfibg‘q J (Lu-J. bz:_fl.nfifi'z¢ a: DJ J I Cf. .; 045-433 ET I Gib—1+3? DJQQ‘K- @n; JET : flag 0 ) [D C. Far LAMIfl-flr‘ / LLJ'C CM MDT W m C-IKMQI 1,wa Team-1+; flfU {'iu. E-fiufuazlcfl'} or. “I mfian-J +5 acacia-:11 W {rich-4‘... Cl bull-Ly- +urb9 ID I {5 15 MM! mt. half HM Um (Dian) HM fling] m +11; {41:31.11} filtwi’ I Far R: ; [[35 J .HM M; A0].qu h: fa+fm LECL ctfl—Iq) JEW Ctkrwiar fl'PLS. LanLHL‘. (5mm HM M95914 aha/(‘1' llfiEHf-Ld) 1C: 0.134 (so-r5“ = Mrm Eifi—‘I'U 'nus um 15. W13 N 5/. W {rm MH atLMfi/ti um Ca MCI-4%) WQILJ W HA Tc, amid-3U: ‘FMllijIWF-tvi Ian/drum? 'F‘f’H-fg “HQ. rfid-bu‘IHA mhlfllimum 161.6% J l—M;n J 15 aux—.4 '51:) Lum‘m a: a.o§ RE 1:, 3 3. "Pi/'2 Water at ETC {[1 -— IflflU Jig-"n13 and |.|. — {mm kg’m-s} flows through a 1.0 cm IE") circular tuht: HI 3 rate; of 15.? gs. u. [km-mini: th' frictional pruHHun: gratiiunl, {tip-“Lidfrm'mn, al :1 puirn EU L'rn From the lube L3 [III‘HI I CC . h. Determine the frictional pressure drop in Line first 2H cm ofthe tube length. rs. Utili'l'l'l'lil'tc the: total uppurcril [JTLIJSSLJTL‘ tlrop (Friction pins aucciurutiori] in tin.~ first EU crit oi" the: who Iurigih. . —3 V“ i5.'?£i'.0 baa Via-(,5 : flrd = ————-—--——""—: = UM 8’" IL Off-toil woo -* fi- 4 —1 RE ..__ # Inna. fi-D.2D+ [Kin = Bash 1-! i453“ U515 ‘Ffj q—i'} J 4.21: 31'4- anq fixi— Zcrtm ,. Fair E: 1" CM .-——? 1m] '[Dfi I x "3 igiic r“- I9 ———=} {3% —- firsxmia IEIRCtZL __-,. .[3’ fileto F —-E " -—-:—:r I 9x“: Re i” 3"" 'F ’1“: Farr “*7 L ’ ._L W» T?- ‘thrCH-M : 4 f} D 3- 1 ark” ~3 Lm (Ci-1) = '76s- AHE- .___—-—-—-—______________._-—'—'_—-—-—-_'___‘_- 2' U 1 ‘é‘P‘FrififihM : % 1 355—3399“ -z 5.2:: a; [m (ELL) .. D '* -————-" 4:: ED '47 Pa (AM) ...
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6 - jarpoint 20 cm from the tube the WW pressure drop in...

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