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Unformatted text preview: Basie Laws for Finite Systems MovingSystem Control—Volume Relation
System = deﬁned quantity of matter
Control volume = fixed volume in the flow domain [3 = quantity (mass, momentum, angular momentum, or
energy) per unit volume. Total amount of property in a finite system
30‘) z [)3 (x, y, a, want?“ do“ = 1roiume element Time rate of change of the total property of a moving system
E _ lim B(r+&t)— BU)
D! at —> U A:  Control Volume and System our. System at time t: ‘ ($121“; so) = 31(3) + 32 (r)
3r
3*” “Tm” System at time t+ At:
BU + or) 2 310+ at) + 33 (r + or) warLn 51's rEM m" 47 DB lim Q}; (I + or) + 32 (r + an) — (B; (r) + 32 (.0) Dr air—H} A: +_Ii1_u__ mamaslams)
armsU A: DB;r D: = Rate of change in CV + Inﬂowf Om'ﬁow through CS Time rate of change of the property (Bl + Bﬂin the control
volume . BCV(T+ﬂf)—BCK_I_)*38CV __6__ aprva or # a: _arcy Inflow = 810‘ + or)" and Outﬂow =33 (r + or) Flovv Through the Control Surface 17% =component of velocity normal to the control surface
i7  as: = length of ﬂuid crossing through (11% in time A! l7  sum = volume through dA in time or. Bl?  cam = the property through CIA in time a: Total amount of property flow rate crossing the control surface um was a: —a~0 a: CS l7  ii = VcosEJ = positive for out flow (0 E 9 a: 90) and
negative for inflow (90 «:1 I9 3180) Reynolds Transport Theorem 23:3 [advv lﬁV—ﬁdA Dr 6‘ CV cs Conservation of Mass
Total property = B : jﬁd‘v’ = mass = j'pd‘c’ and ﬂ = ,0 Conservation of mass P—(mnss) =D—B~ :2 Iﬁd‘v’Jr fﬁ?r’id4=0
DI DE ('3ch CS
:3 _. IpdV+ IpVr’idA=U
530V (:3 For steady ﬂow [p.17  era: 2 0
CS
Incompressible (constant density and ﬁlled CV)
If} ' EMA = 0 steady or unsteady
CS Example: Steady incompressible flow in a 1rariaole area channel I7  F: is zero at a solid surface
jVﬁdA—l— [V*ﬁdA=U A1 A2 — [ndAi jndA=G ﬁ=(lndA)fA
A1 A2 A
E1141 2 ﬁzAE 01' VIA} = V2142 Compressible Flow: pit/1X11 = pal/2142 Flow through a cross section l
W
h—Vetq th = distance front advances in time dt
Ath = volume through A time dt
dvfdt = AV=tj(m3fs) = volume ﬂow rate through the cross section
pA V = aims“ at: = or (kg/s) = mass ﬂow rate down the duct Example 2.1 Find the velocity at the liquid surface in the
tank given the volume flow rate cjﬂ from the nozzle Control volume 1:
[I7 am = D for steady ﬂow
CS
—V1A1+qn =0 arid V} Equals/41
Control volume 2:
3 lpdV+ jVFtctAa Wr’tdxtﬂ)
A2 afar A1
a .
E[pa(H h)«41+ phAtl— poAlV1+ pa” = 0 dh nth
 —A+#A— AV+'=D
padrl Jodi} pd ll logo Surface velocity Vi = ﬂdhf d: and then V1 2 on f A} Example: Determine the maximum velocity V... at section 2
for incompressible ﬂuid ﬂow in the length of circular pipe. umFoam Penn an“:
EffV: v=vc(l_H1)/R1) Flow enters the pipe with a nearly uniform velocity proﬁle. Friction slows the ﬂow nearer the wall and, consequently,
speeds up the flow nearer the center. Aﬁer some distance, the entry length, the velocity profile
no longer changes and the flow is fully developed. There is a balance between pressure and friction forces.
The velocity proﬁle is parabolic nchGnrszz) W—r’idA=—V;At+ [feet:0
A: es
s r2
KAI = Indxi'=l’;j(l——2)2irrdr
A; U R
'5' 4
rs r R l 2 1
229:1”. —*—— — =Vn:R :rle :VA 4R2)ﬂ 2 Cr 2 (.1 ll 14:214. ...
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 Spring '09

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