basic_law_finite_lecture_1

basic_law_finite_lecture_1 - Basie Laws for Finite Systems...

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Unformatted text preview: Basie Laws for Finite Systems Moving-System 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) war-Ln 51's rEM m" 47 DB lim Q}; (I + or) + 32 (r + an) — (B; (r) + 32 (.0) Dr air—H} A: +_Ii1_u__ mama-slams) arms-U 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 [adv-v 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+ IpV-r’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 jV-fidA—l— [V*fidA=U A1 A2 — [ndA-i- 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—Vet-q 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+ jV-FtctAa- W-r’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“: Eff-V: 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 nchGn-rszz) 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 :r-le :VA 4R2)fl 2 Cr 2 (.1 ll 14:214. ...
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basic_law_finite_lecture_1 - Basie Laws for Finite Systems...

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