Fluid mechanics (wb1225)

Transcription

1 Fluid mechanics (wb225) Lecture 8: flows through pipes and ducts

2 Examples of pipe flows [] [2] [3] [4] 2

3 Flow states in pipe flow [5] Re crit

4 Laminar vs turbulent [5] [6] 4

5 Pipe flow transition [5] 5

6 Pipe entrance length laminar flow: L e d 0.06 Re turbulent flow: L e d 4.4 Re 6 [5] 6

7 Flow with energy losses p + ρv ρgz p 2 ρv ρgz 2 = 0 [5] p + ρv ρgz p 2 ρv ρgz 2 = p loss 7

8 Flow in a circular pipe [5] 8

9 Control volume analysis continuity: Q = Q 2 V = V 2 steady-flow energy equation: p + ρv ρgz = p 2 + ρv ρgz 2 + ρgh f [5] h f = z + p ρg z + p 2 2 ρg = z + p ρg h f = z + p ρg momentum equation: p π R 2 + ρg( π R 2 ) L sinφ τ w ( 2π R) L = &m ( V 2 V )= 0 z + p ρg = h f = 2τ w ρg L R with: z = L sinφ 9

10 Dimensional analysis h f = 2τ w ρg L R, τ w 8τ w ρv = f = F Re, ε 2 d = F ( ρ,v, µ, d,ε), ε = wall roughness ρgh f = f L d ρv 2 2, f = Darcy friction factor 0

11 Equations of motion continuity equation: ( r r ru r )+ ( r θ u θ )+ u x x = 0 ( r r ru r )= 0 ru r = const BC: u r = 0, r = R u r = 0 momentum equation: u = u x (r) ρ u t r + ρu u x = dp r rτ ( ) = d dx τ (r) = 2 r d p dx dx + ρg x + r r rτ ( ) ( p ρgx sinφ) = d dx p 4 2 ρgz 43 p ( )

12 Laminar flow solution τ = µ du dr = rk K = d p 2 dx integrate: u(r) = 4 r 2 K µ + C BC: u = 0, r = R C = R 2 K 4µ u(r) = K 4µ R 2 r 2 ( ) u max = KR 2 4 µ [5] Q = u(r)2π r dr Q = 0 R Kπ R 4 8µ V = Q A = Q π R 2 = 2 u max 2

13 Laminar flow solution (cont d) Q = Kπ R 4 8µ Q = π R 4 8µ p L p = 8µ LQ π R 4 τ w = µ du dr r= R = 2µu max R = 2 R p L u(r) = u ( max r 2 R 2 ) u max = KR 2 4 µ = pr 2 4 µl p = f L d ρv 2 2 p = 8µLQ = 8µL π R 4 π R π R 2 V = 8 2 2µ L 4 2 R 2R = 64 L ρvd µ d 2 ρv 2 f = 64 Re π R 2 π R 2 ρv 2 2 ρv 2 3

14 Example 6.4 An oil with ρ = 900 kg/m 3 and ν = 2x0-4 m 2 /s flows upward through an inclined pipe as shown in the diagram. The pressure and elevation ar given at sections and 2, 0 m apart. Assuming steady laminar flow: a) verify that the flow is up; b) compute h f between and 2; c) compute Q, V, and Re d. d) Is the flow laminar? [5] 4

15 Bore diameter u = K 4µ R 2 r 2 ( ) Q = u(r) 2π r dr a 0 = Kπ 8µ R 4 = P / L 28µ / π D 4 P = 28 π = 64 Re µ L D L D Q D = 28 3 π µ L D ρu 2 2 π / 4 D 2 U = 28 2 D 3 4 ν UD L D ρu 2 2 Q = P / L 28µ / π D 4 δq ~ 0.% δ P ~ 0.% δ D D 4 δq Q + δ P P 0.05% D = 50 µm δ D 75 nm 5

16 Stenosis [5] 6

17 Vascular networks [7] [9] [8] [0] 7

18 Tube networks n= n=2 n=3 D L D 2 L 2 D 3 L3 D 2 = α D D 3 = α D 2 = α 2 D L 2 = β L L 3 = β L 2 = β 2 L Q = Q = 2 Q 2 = 4 Q 3 = P = P + P 2 + P 3 + 8

19 P = 28ηQ L π D ηQ 2L 2 π D ηQ 3L 3 π D = 28η π = 28ηQ π Q L + Q 2 L 2 + Q 3L D D 2 D 3 L D = 28ηQ L π D L 2 D L 2 L D D 2 4 L 3 D L 3 L D D 3 4 = 28ηQ L β π D α + β α = P + 2 β + ( β ) 2 α α 4 = P 2 β 2 α 4 β α 4 < 9

20 Optimal Value volume β surface pressure α 20

21 Vascular network Distance (mm) Distance (mm) [5] Distance (mm) Distance (mm) 2

22 Summary Chapter 6: , 6.4 Examples: Problems: see BlackBoard 22

23 Source. Trans Alaska Oil Pipeline, photo courtesy of Daniel Acker/Bloomberg 2. Suncor refinery, Montreal, photo courtesy of manuel crank 3. Heart Disease, 4. Waterleiding verdeler (waterworks distributor), 05/slides/Waterleiding%20verdeler.html, photo courtesy of Het Kapotte Huis 5. Frank M. White, Fluid Mechanics, McGraw-Hill Series in Mechanical Engineering 6. Multimedia Fluid Mechanics DVD-ROM, G. M. Homsy, University of California, Santa Barbara 7. The vascular system of a bat wing, photo courtesy of Internet Archive 8. Lungs, photo courtesy of Mikael Häggström 9. Poplar leaf - Feuille de tremble, photo courtesy of monteregina 0. Structure in leaf, photo courtesy of Rob Chesney 23