HW07sols - Problem*4.127 1[Difficulty 3[p127 Ajiet ofwater...

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Unformatted text preview: Problem *4.127 1 [Difficulty: 3] [p127 Ajiet ofwater directed agahrst a vane, whidhoouilidbe a Q) Madam a turbine or many other of hydrantic manifinem e _ _ _ CS (moves The water Teams.- stafionary 4D=mm=diameter nozzie a W, I at speed U) speed. of 25 mis and, enters fire vane tangent to the surface at ' I A.. The inside surface of the vane at B? makes angie 5‘ a 150’” with. the x direction. Compute the force that be appiied to I vane speed, constant at U = 5? mr'si -’~’ I W Given: Water jet striking moving vane Find: ‘ Force needed to hold vane to speed U = 5 m/s Solution: Basic equations: Momentum flux in x and y directions RX = p(V — U)2.A.(cos(e) — 1) Using given data a . 2 2 k _ N. Rx = 1000-—g x [(25 — 3-2] x 1.26 x 10 3.m2x (cos(150~deg) — 1) x S v 3 s kg-m m Then' Ry = vl-(—p-V1-A1)+ VZ-(p-VZ-AZ) = —o + (v — U)-sin(9).[p-(V — U)-A] \- I 2 k m 2 ' 3 2 ‘ N52 R = p(V— U) -A-sin(9) Ry = 1000-—g x .(25 — 5)-—. i x 1.26x 10— cm x sin(150-deg) x R = 252N y 3 s kg'm y m Hence the force required is 940 N to the left and 252 N upwards to maintain motion at 5 m/s v Problem 4.129 [Difficulty: 2] H- @329 The fistula! dish “thus-e: muss: senfibn f5; 3131mm has an outside diameter of 0,211 m. A watezr jet wiflm. speed of 3.5 m 9 the dish mnmflmfimflgfi'l‘he dish to: the at ){e' L U = 40" '15 rinst The; jifli dimmer 10 The dish a Twila at ifs. d = l :1 “m Denied: that a straw of "Natal" ‘lfl mm diametm‘ m ( V = 3 {M I #2» pass. wiflhom The mnmiinder :01? the jet ' «- «.— daflecbed and flows. a‘flmg dfis’PL Cakflfiat’e the foama Y 1 '6) :mquired m mammal, the dish muticm. X T U =‘1-‘Tnvs I Ix CV D': '20 ’ / — — = "H" l v e W! flpp fa wnimwfi/ {in-d 9L ’5. _,., mome‘nfnom. egagfion ‘l‘b CV mow/79 w (#9 C/Iél’) mag/gown. Assam/2,0750 [75? (1)5teadg-me unfit. CV gzgg (Z) No pmsar‘c 76725.5 0/? CV é HOF'igonfa/Jo =0 . ‘ 0+) Unl'flmzm flow at COL—Ch Sfpflofl (.55 No change, In space! 01‘ch rt/aé/{xc 73-0 Vac/76 (G) Incompmsété/e flaw ." #771? . [132-6(2) = {fl—(bxbzo)a ~-(0,0/o)"]m/2"- a 2.34. x Io‘qth =.— «pH/*Uyg (0z “CHM/5‘ cos 940°) = _. 9'96? gigssvsfimjx 2.3m/o-wa(/+ dos¥o°)x N6“ m3 3‘1 kj :M Ifx = “IE7 N ($763. waif ééappl’édfl Pig/M) {Note : E1 = A4? ) Show fhm ('5 no {267" momen7‘m 75/44“ M 7%: W; 57 "dime My), ' Problem 4.135 ' [Difficultyz3] $13.5 "libs: whose Ems is. shown». has, an unmifle dimmfier Daf.’ [£15 A. WEEK jet strikes the; dish 1—,- mmantfimfly than flows. flumm aliasing surface 0f 1 Elia dish. jet :35, and. the dish. mm'es- to afi .10 the (El-f the jet Shea-t? at? a radius. 3'5 mm the jet What flame on the: disyl'm. is — tor 1E1?!in mnfi‘omf?’ V angks- an :2 30’ and, 433: = 45", showni 331mm Emile Problem 4.136‘ 4411355: fleasfidfl a Sexies GE “335% a milk fimom jflt of “9am: leaves a 5”“th dfimfltfl nmfle at canstmrnt speed, V 5-.“ 3&6 ms. The Win35- mm'e “flh comm speed. U‘ a 50 ms; Narnia fihat an the fi-W fem-“g WM crosses. fihfi vanes“ The curvature of, 7311*? vanes-is descfibe’d angfie‘, m to; mum flmfi Elia jet enters tangent to the Eeadiimg edge 0f Each «may IC-a‘ficuffiate the; fflrw that must to: mafimtain the: vane: Spam mmflan‘fi" [Difficultyr 3] Problem 4.139 4.139 A plane jet of. water sttikes a splitter vane and divides into two flat swamps, as shown. Find the massfiow rate ratio; :fegfritg, required to produce zero net vertical fame on: the splitter vane. If. there is, a resistive force of ‘16 N applied. to the splitter vane; find, the steady speed U of vane Given: ‘ Jet impacting a splitter vane Find: Mass flow rate ratio; new speed U _ solution: Apply momentum equationlto inertial CV : No pressure force; neglect water mass on vane; steady flow wrt vane; uniform flow; no change of speed wrt the vane Assumption Basic equation . — - k Gwen data V = 25-2 A = 7.8510 5-m2 U = 10-2 e = 30-deg _ p = 999—g s , 5 m3 For constant speed wrt the vane, the jet velocity at each location is V _ U f velocity and mass flow rates, respectively, at the inlet and exits, wrt the vane » ' coordinates i Hence 0 = 0 + (V — U)-m2 — (V — U)-sin(9)-m3 01' m2 = m3-sin(9) ,FortnOXelitiQalfolTCG,,XmngILfllmbEGQmQS, , ,0,,,,=,,,v,1;,(:m1): 312,111,217 X31113"; , , 23:6)!in miarethe‘YeIii9élcom139nentS —=sin9 =— Note that m1 '= p.A.(V _ and and using x momentum . . . / ' 005(9) l Wr1t1ng Interms ofm R = V — .m - —-— — 1 = —7.46N 1 x ( U) 1 (1 + sin(9) ) RX , 005(9) \ ~ ' Instead, the force 15 now R = —16-N but R = V— -m . —-- — 1 and m = -A- V— X X ( U) 1(1+sin(9) j 1 p ( U) Hence Solving for U ...
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HW07sols - Problem*4.127 1[Difficulty 3[p127 Ajiet ofwater...

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