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Unformatted text preview:  ., n . i
t
t _, P3.9 A laboratory test tank contains
seawater of salinity S and density ,0. Water
enters the tank at conditions (S 1, ,01, A1,
V1) and is assumed to mix immediately in
the tank. Tank water leaves through an
outlet A2 at velocity V2. If salt is a
“conservative” property (neither created
nor destroyed), use the Reynolds transport
theorem to ﬁnd an expression for the rate of change of salt mass Msalt Within the
tank. Solution: By deﬁnition, salinity S = psalt/p. Since salt is a “conservative” substance
(not consumed or created in this problem), the appropriate control volume relation is dMsal d . .
dt tlsystem: p5 + 81112 — Slml = 0 dMs
dt P3.13 The cylindrical container in Fig. P3. 13 1s 20 cm in diameter and has a conical contraction
at the bottom with an exit hole 3 cm in diameter.
The tank contains fresh water at standard sealevel conditions. If the water surface is falling at the Mt) nearly steady rate dh/dt z —0.072 m/s, estimate the average ve1001ty V from the bottom exit. _______________________ __
Fig. P 3. l 3 IV? Solution: We could simply note that dh/dz‘ is the same as the water velocity at the surface and use Q1 = Q2, or, more instructive, approach it as a control volume problem. Let the
control volume encompass the entire container. Then the mass relation is dm d d 7: 7r
7W. = 0 = 91 deu) + = EwﬂwlDzh) + pzDthV,
CV ' 72' 2 072 72' 2 72' D 2 6172
or p—D — + p—D ~V= 0 Cancel —: V = —— 4 dt 4 ex” ’0 4 (Dem) ( dt)
Introduce the data: V = (206m)2[—(~0.072ﬂ)] = 3.23 Ans. 3cm 5 s a, 9 , , e , P3.14 The open tank in the ﬁgure (3) * 3
contains water at 20°C. For i QrOOlm /S incompressible flow, (a) derive an +0) (2)
analytic expression for dh/dz‘ in terms of h +
(Q1, Q2, Q3). (b) If h is constant, D1=50m D2:7cm determine V2 for the given data if V1 = 3
m/s and Q3 = 0.01 m3/s. Solution: For a control volume enclosing the tank, d _ £01111: _ _
E[C_I[pd0]+P(Q2_Q1_Q3)—P 4 dt+p(Q2 Q1 Q3),
solve E=M Answa)
dt (ml/4) If h is constant, then
Q2 = Q1+ Q3 2 0.01 +%(0.05)2 (3.0) = 0.0159 = €(007)2 V2, solve V2 = 4.13 m/s Ans. (b) P3.26 A thin layer of liquid, draining
from an inclined plane, as in the ﬁgure,
will have a laminar velocity proﬁle u =
Uo(2y/h — yZ/hz), where U0 is the surface
velocity. If the plane has Width [3 into the
paper, (a) deter—mine the volume rate of
ﬂow of the ﬁlm. (b) Suppose that h = 0.5
in and the ﬂow rate per foot of channel
width is 1.25 gal/min. Estimate Uo in ft/s. Solution: (a) The total volume ﬂow is computed by integration over the ﬂow area: h 2
Q= IVndA= ]U,[2—y—y—]bdy=§Uobh Ans. (a) (b) Evaluate the above expression for the given data: 3 2 2 0.5
Q=1.25 2520002785 L=_Uobh=—Uo(i.0ﬁ) —ft ,
mm s 3 3 12 ft
solve for U0 2 0.10 — Ans. (b) \ S (/54, é , é/0%/// P3.36 The jet pump in Fig. P336 injects water at U1 2 40 m/s through a 3in pipe and
entrains a secondary ﬂow of water U2 = 3 m/s 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 U3 in m/s. Solution: First modify the units: D1 = 3 in = 0.0762 m, D2 = 10 in = 0.254 m. For
incompressible ﬂow, the volume ﬂows at inlet and exit must match: Q1+ Q2 = Q3, or: §(0.0762)2(40)+§[(0.254)2 —(0.0762)2](3) = §(0.254)2U3 Solve for Ans. U3 z 6.33 m/s P3.40 The water jet in Fig. P3.40 strikes
normal to a ﬁxed plate. Neglect gravity and
friction, and compute the force F in
newtons required to hold the plate ﬁxed. Solution: For a CV enclosing the plate
and the impinging jet, we obtain: ZFX : "F = mupuup + l’l’ldownudown — = ’mjur mi = ijVj
Thus F = ijvj2 = (998)7r(0.05)2(8)2 z 500 N ( Ans. P3.41 In Fig. P3.41 the vane turns the water jet completely around. Find the maximum jet
velocity V0 for a force Fo. Solution: For a CV enclosing the vane and the inlet and outlet jets, Fig. P3.41
2 Fx : #Fo : Iilloutuout _ 1hinuin : mjet (—VO) — mist (+V0) F0 Ans. Or: ———2‘
mot/4m. F0 = 2pOA V2 0 O, solve for V0: ...
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 Fall '08
 Schwartz,L

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