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HW_7_solutions - P3.46 When a jet strikes an inclined plate...

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Unformatted text preview: P3.46 When a jet strikes an inclined plate, it breaks into two jets of equal velocity V but unequal fluxes aQ at (2) a) and (1 — a)Q at (3), as shown. Find a, assuming that the tangential force on the plate is zero. Why doesn’t the result depend upon the properties of the jet flow? Solution: Let the CV enclose all three jets and the surface of the plate. Analyze the force and momentum balance tangential to the plate: 2Ft = F, = 0 = m2V+m3(—V)—m1Vcos6’ =arhV—(l—a)mV—chos6=0, solvefor a=%(1+c0s0) Ans. The jet mass flow cancels out. Jet (3) has a fractional flow (1 — a) = (1/2)(1 — cosé’). P3.49 The horizontal nozzle in Fig. P3.49 has D1 = 12 in, D2 = 6 in, with p1 = 38 psia and V2 = 56 ft/s. For water at 20°C, find the force provided by the flange bolts to hold the nozzle fixed. Solution: For an open jet, p2 = pa = 15 psia. Subtract pa everywhere so the only nonzero pressure is p1 = 38 — 15 = 23 psig. A1 ‘ Pa:15|bf/in2abs 386, [AI]— (19’ Open 1:: L f I P3.61 A 20°C water jet strikes a vane on a tank with frictionless wheels, as shown. The jet turns and falls into the tank without spilling. If 6?: 30°, estimate the horizontal force F needed to hold the tank stationary. Solution: The CV surrounds the tank and wheels and cuts through the jet, as shown. We should assume that the Splashing into the tank does not increase the x-momentum of the water in the tank. Then we can write the CV horizontal force relation: (a __ _ g . . . 23,2 ZFX — F — dt ( I up do)tank ‘mmum = 0—mVjet independent of (9 Th F _ A 2 _ slug 72' 2 2 ft 2 us —,0 jvj _ 1.94 —3— — — ft 50 — z106lbf Ans. ft 4 12 s JUUUU 7 \:fi,fi, l... IAI' P3.51 A liquid jet of velocity Vj and area Aj strikes a single 180O bucket on a turbine Wheel rotating at angular velocity 9. Find an expression for the power P delivered. At what 9 is the power a maximum? How does the analysis differ if there are many buckets, so the jet continually strikes at least one? *— uckcl Wheel, fadluS R Solution: Let the CV enclose the bucket and jet and let it move to the right at bucket velocity V = QR, so that the jet enters the CV at relative speed (Vj — QR). Then, 2 Fx = _Fbucket : rhuout _ rhuin = m[—(vj — QR)] — m[vj — QR] _-—a-- Relative l velocixy : V l :_ or: PMct = 2mmj — QR) = 2 pAJ- (vj — QR)2, and the power is P = QRFbucket = 2 pA jQR(vj — QR)2 Ans. Maximum power is found by differentiating this expression: V. (1—1? = if QR = —i Ans. whence Pmax = i pA -V3. d9 3 27 J J If there were many buckets, then the full jet mass flow would be available for work: . ' 1 V. Inavailable = ijVj9 P : ZijVjQRCVj _QR)9 Pmax = EpAl-Vi at QR = Y] Ans. P3.73 A pump in a tank of water directs a a jet at 45 fi/s and 200 gal/min against a vane, as shown in the figure. Compute the force F to hold the cart stationary if the jet follows (a) path A; or (b) path B. The tank holds 550 gallons of water at this instant. Solution: The CV encloses the tank and passes through jet B. (a) For jet path A, no momentum flux crosses the CV, therefore F = 0 Ans. (a) (b) For jet path B, there is momentum flux, so the x-momentum relation yields: ZE=F=m out _ jetuB Now we don’t really know uB exactly, but we make the reasonable assumption that the jet trajectory is frictionless and maintains its horizontal velocity component, that is, uB z Vjet cos 60°. Thus we can estimate 3 F=mu3= 1_94 ”lg fl 1: ff 448.8 s l A ’kafi ngjane/l/ t The mass balance yields the inlet velocity: 5 2— 5 2 V=14E v14(12)—(56)4(6), 1 S The density of water is 1.94 slugs per cubic foot. Then the horizontal force balance is 72' Compute 11b,lts = 2601 —(1.94)—4—(1 £02 {14 (4SCOS6OO) z 19.5 lbf Ans. (b) 2Fx = 43,301,, +(23 psig)%(12 in)2 = m2u2 —m sll 56—14 —j z 17001bf Ans. ...
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HW_7_solutions - P3.46 When a jet strikes an inclined plate...

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