{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture_25_notes

# Lecture_25_notes - ChE 374—Lecture 25—Integral Momentum...

This preview shows pages 1–6. Sign up to view the full content.

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

View Full Document

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ChE 374—Lecture 25—Integral Momentum Balance 0 Integral Momentum Balance — Find Forces or Accelerations . 213 : %fovp17dV-l- fOSpJU-ﬁ’d/l. — For uniform properties: Xi“ = aw?) + (2W? 0... — (2W7 m — For uniform properties and Steady State: 2E = (; mam M (2 me in >l< TAKE THIS EQUATION LITERALLY. - Vector components are positive in the positive coordinate direction. - m = pAl'UI is a scalar quantity, always positive. — Note: m6 is momentum ﬂow rate. — Note: p171? is momentum ﬂux. — Note: poser—f is x—momentum ﬂux. 0 Choose a control volume — Not limited to the ﬂuid — Choose it perpendicular to inlets and outlets — The momentum balance is WITH RESPECT TO THIS CV. * So velocities are RELATIVE to the control volume. * Also, 7%; is also relative to the control volume. 0 Forces are (1) BODY; (2) SURFACE; (3) OTHER — Body forces are gravity (mg). — Surface forces are Pressure (always normal to CV, and towards the CV); and viscous forces (usually neglect at inlets and outlets where the ﬂows often cross the boundary) — Other forces are external, like bolts, the ground, and other anchoring forces, etc. o Momentum ﬂux correction factor corrects for nonuniform ﬂow at inlets and outlets. Similar to kinetic energy correction factor. ,8? % fv2dA. ,6 : 4/3 for laminar a? % f'usdA. a : 2 for laminar 0 Example: Flow through a nozzle. — Restraining force is F = A1(P1 + pu§(1 — Al/Ag), where (1) is inlet, (2) is outlet, and F is in the direction opposite the ﬂow. 0 Example: Flow deﬂected 90 degrees: — Restraining force is F = 7m}, with force directed opposite the ﬂow. 0 Example: Flow deﬂected 180 degrees: e Restraining force is F 2 2m, with force directed opposite the ﬂow. 0 Example: Flow deﬂected 6 < 90 degrees: F1 2 inv(1 * cos 0), x is ﬂow inlet direction, and FI is opposite the ﬂow direction. F,, = mv(sin 6), y is perpendicular to inlet direction, and Fy is with the direction of deﬂection. 7% XXL.ch M1 >56 7 ‘Plu\e’ §{-a416§ ' Wag ROLa/M( true/777 gala—wt Xg mantwﬂ‘w 'Ealﬂmcf CLP "l (bch r1amon¢v~> \léloolxl/ T’TO'QI'IUA aw" Foffcs ,——-_». ,Vtow,’ F ': M (L '; MJU ’ ;: ' : I7.o}f 0Q IﬂDMt‘A7me' D‘C/N‘Jt’fj Lav/Jam PIP? UleC“!'/ “'3'!” , 4 F076! (MOMCwJ'VV‘) 734mg “ Use) a Diﬁozm‘p‘ol APV(ML_ * — 0wa; DO low-{'6 {4’ APP/oku 71‘0" ﬁv‘gcm ((\$55) Dch/uwﬂo' Kola;ch {imman/ 600‘ (:04; F) 0- ’ (a él‘OO‘A #mcw I tr! .3 I 6‘}C ' f/f‘ RCYnoU5 5T}a"9fw)% 2E5” : lgfbév 4 g (“9'0 13A J E:M‘\J}, 13:?) 7? “L s W A F; DIF F a ’M’ as Uw-‘wax FWWH’C’” ’1‘ , a ~ + (757 -( Mg). , f MV _,, ow} 2: n’\ N M ( 3 Kid/V._____._—J .4 _, «4 A O a! §.§ (U kmls we? (Ulfvowf’vtﬁs ~ vmar F4 / 7 can .4» x, 7 QMPMML; N941, “)1 Q )3 Momw;v* )75‘46 .‘ Z I; Vol,qu '- \Qc’xk [MPHCH 71-.- kg( or a. , Vclacfhw at? gala-hut {a a“, ﬁ \) f v ,. v_ Va", ’ Bf ‘. f ’@ r V « ’—>> @9H‘t/7 CPD» Viva 7L: 'Ddzw‘l; )Mv‘Jr \$3...) my» aaon 76.” a” 9mm 0? 7“ «Wk, my, M LIMCJV) ‘éhooér m1, _L to amid/MM. V C. .V, ’l‘ml‘¥<’3 ({D F’L—u‘CJ ' In 6) I &»’1}—/ 11M “0"{2070 Xv chC/g {’1 71w §L\“Jc<) 41762 I‘M‘ ‘— }(_ luv-x « Tqu can ‘€/<+Cﬂno)l Fact/J ExamPlt I Flak: {Lrwak Na'ulf "ﬂ. 8. 7; . ping Fm“ of. TsoHs, * 6516.“ (a) \J‘IAUAII éouCD/lfnj I i ) v. I ’ _.\ ' PA ’ r — — , 'i ( "L [M‘Jld * («wk I) t VJ) z/oAlUl F0366(WVOW\ bfaﬁlowx limrcs ¥ra- Fl 30H; 25’ “Z’VL A(ﬁ,ﬁm ‘ 45—... ﬁ' \ ( l"9 “LAmj‘P/HZKQ)‘: ’VDAN} - m, :lel as : 'Ely'Al + 79A”)? “‘ PA‘L (gi/A') A if F5- AI?) +fv'1(‘~ﬁj] ' Flo“) have , ‘ an! (now/“3 Foec X' D‘ftcya“ ‘" A+~\ (ﬂank rJh *4 \ v—a ’5,“ : "’l’vl X " mom‘cu'hdm ', *F .: y:/1 e .__. M V t "Nter 7\ \'M0M(¢\+W\ ; -1 , ~ - MM F / — W \J 949 Thus'kg x F”: MU([—'(¢Jﬁer) Tfy jg F7 :- M \JS‘I'ﬂC—D '2‘ ___, w—b s ——L I0 : Irx‘t 4 r7] LN . ND+¢ éOu'tI 14M vx)(I'H'5.,\ x_ M .. %_7;\)¢0\$Q ——-—> Fx -—)>. Fx : M (059 _ Mv J— » é \ i —> C brad rx I 4" 6’71 1*“?!ch (\- mk a 3‘ I; E M €7vx / . . 1-) : 3—? Fx #9 I a”) {2’7" IMP’DC’] F», 71%“ 6 )K A f n I; #9.. 7 99 7'3: ‘J m(4v) ‘6‘“) ﬂ“? X-“p’1/7 '1, #23 mi 6; ' 65‘ T3001: ' WI“ L! ur‘gjrz'ac’y? ’ BC 0‘44»? “90—3 iH+M,°rr,Lr‘7 Wr~;# M 330a};- éf—i + *1» Wk) 9 i , ‘7 a? a (W) Mm W .1“ H v I l 1" {S 71% Mtg/074412“ TQM, Dr“)? “’7 m 124% g WM” “Nam/c1), 0 KW “L94 {0 b0 If no? 0,.)Lgpury £3. ans) JnJ’7IQH f Z ‘SfﬁilUMcJA Q [.9 C ' . ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

Lecture_25_notes - ChE 374—Lecture 25—Integral Momentum...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online