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Unformatted text preview: P3.12 The pipe ﬂow in Fig. P3.12 ﬁlls a cylindrical tank as shown. At time t = 0= the water
depth in the tank is 30 cm. Estimate the time required to ﬁll the remainder of the tank. D=rs T
_ _L — —
V1=2.5WS Ed=126ﬂl V2=L9WS Fig. P3.12 Solution: For a control volume enclosing the tank and the portion of the pipe below the tank= g] pdv] +rirow. —n'1m = 0
d}
.erz immom 4pm... = 0 dr ﬂ _ +
d: 9980:)(0752)
Ar = 0.300153 2 46 .5“ Am. [993[%](0.122)(2.5—1.9)] = 0.0153 mfs. P3.14 The open tank in the ﬁgure contains water at 20°C. For (3)1; I ngonlmi‘is
incompressible ﬂow, (a) derive an (I) {2)
analytic expression for 0’}:de in terms of + h (Q1. Q2, Q3). (13) If h is constant= D1=5cm D25; determine V2 for the given data if V1 = 3
me’s and Q3 2 0.01 mgr’s. Solution: For a control volume enclosing the tank= d d2 d}
EiideJer‘LQZ _Ql ‘93): p%?:+p(Q2 _Q1 ‘Qsla all 2—{214—{23 _QZ Am. (a) solve — dt (#dzfrt) If h is constant, then Q2 = Q1+Q3 = 0.01+§(0.05}2(3.0) = 0.0159 = $00321»; solve V2 2 4.13 rats Am. (b) P3.19 Water ﬁ'om a storm drain ﬂows over an outfall onto a porous bed which absorbs the water at a uniform vertical velocity of 8 IDIIU'S, as shown in Fig. P3.19. The system is S
m deep into the paper. Find the length L of bed which will completely absorb the storm
water. Initial depﬂi = 40 cm Fig. P3.19 Solution: For the bed to completely absorb the water, the ﬂow rate over the outfall
must equal that into the porous bed, Q = QPB; or (2 mfs)(0.2 m)(5 m) = (0.008 mfs)(5 m)L L = 50 in Am. P320 Oil (SG0.91) enters the thrust r imam—1
bearing at 250 thr and exits radially Wﬁﬂ’ﬁ
through the narrow clearance between  _..
thrust plates. Compute (a) the outlet volume ﬂow in mLfs, and (b) the average ”t outlet velocity in cmfs. (9%10 3
Solution: The speciﬁc weight of the ml 15 F'g' P32" (0.91)(9790)= 8909 me3. Then 2508600 Nrs_ as m_3 mL
2 1 8909 Nrm3 s 5 But also Q2 =v2 rr(0.l m)(0.002 m) = F.8x10_6= solve for v2 = 1.24 E Ans. (b)
S P3.33 In some Wind tunnels the test section is perforated to suck out ﬂuid and provide a thin viscous boundary layer. The test section wall in Fig. P333 contains 1200 holes of 5
mm diameter each per square meter of wall area. The suction velocity through each hole is Fr 2 8 ms, and the testsection entrance velocity is V1 2 35 mfs. Assuming incompressible
steady ﬂow of air at 20°C, compute (a) V0, (b) V2. and (c) Vf, in mx’s. Test section
.I' = 2.5 m =.U 3 In
/Unil'orm suction !
VjI‘I— 2 I H—H—Ivl — Lam—J Fig. P3.33 0.52.21“ Solution: The test section wall area is (x)(0.8 m)(4 m) = 10.053 m2, hence the total
number of holes is (1200)(10.05 3) = 12064 holes. The total suction ﬂow leaving is om =NQhole =(l2064)(9r/4)(0.005 Inf (8 mfs) a 1.895 m3/s (a) Find v0: Qonl or véesqusﬁmsﬁ solve for V0 2 3.58 E Am. (a) S
(b) Q2 = Q1 pm. 43930.3? 4.895 =V2 EMF. or: vzmsm E Ansfb)
S (c) Finde: Qf=Q2 or magma)?281.2)?03?= solve for V} m 4.13 E Am. (C)
s P136 Thejet pump in Fig. P336 injects water at U1 2 40 mfs through a 3in pipe and
entrains a secondary ﬂow of water U2 2 3 ms in the annular region around the small pipe.
The two ﬂows become fully mixed downstream, where U3 is approximately constant. For
steady incompressible ﬂow, compute Us in ms. Mining FulIy
0 = 3 1n Inkl region mind Solution: First modify the units: D1 = 3 in = 0.0?62 111, D2 = 10 in = 0.254 m. For
incompressible ﬂow, the volume ﬂows at inlet and exit must match: Q1+Q2 =Q3, or: g(U.0?62}2(40)+§[(0.254)2 —(0.0?62}2](3) = §(0.254)2U3 Solve for U3 = 6.33 Im‘s Ans. ...
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 Spring '08
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