{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

basic_law_finite_lecture_1

# basic_law_finite_lecture_1 - Basie Laws for Finite Systems...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}