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

This preview shows pages 1–5. 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 is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Page1 / 5

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online