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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:15bf/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 xmomentum 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 = EpAlVi 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 xmomentum 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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 Fall '08
 Schwartz,L

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