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## Chapter 8: Internal Incompressible Viscous Flow

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**Chapter 8:**Internal Incompressible Viscous Flow Flows: Laminar (some have analytic solutions) Turbulent (no analytic solutions) Incompressible: For water usually considered constant For gas usually considered constant for M (~100m/s < 0.3)**Chapter 8: Internal Incompressible Viscous Flow**M2 = V2/c2 {c2 = kRT} M2 = V2/kRT {p = RT} M2 = V2/(kp/) = [2/k][1/2 V2/p] M2 = 1.43 dynamic pressure / static pressure M ~ 1.20 [dynamic pressure / static pressure]**What are static, dynamic and stagnation pressures?**The thermodynamic pressure, p, used throughout this book chapters refers to the static pressure (a bit of a misnomer). This is the pressure experienced by a fluid particle as it moves. The dynamic pressure is defined as ½ V2. The stagnation pressure is obtained when the fluid is decelerated to zero speed through an isentropic process (no heat transfer, no friction). For incompressible flow: po = p + ½ V2**Static pressure**atm. press. = static pressure (what moving fluid particle “sees”) Hand in steady wind – Felt by hand = stag. press. Stagnation pressure for incompressible flow po = p + ½ V2 Dynamic pressure**Chapter 8:**Internal Incompressible Viscous Flow At 200C the speed of sound is 343m/s; If M=V/c=0.3, V=103m/s p = 1/2 (V22 - 0) = 6400Pa = 6% of 1 atm. p = RT; assume isothermal (wrong) p/p = / = 6% p/k = const; assume isentropic (right) / ~ 5%**Chapter 8:**Internal Incompressible Viscous Flow • Compressibility requires work, may produce heat and • change temperature (note temperature changes due • to viscous dissipation usually not important) • Need “relatively” high speeds (230 mph) for • compressibility to be important • Pressure drop in pipes “usually” not large enough to • make compressibility an issue**CONSERVATION OF MASS**& INCOMPRESSIBLE •V = -(1/)D/Dt= 0 (5.1a) Volume is not changing. The density is not changing as follow fluid particle.**Chapter 8:**Internal Incompressible Viscous Flow Flows: Laminar (some have analytic solutions) Turbulent (no analytic solutions) Depends of Reynolds number, Re = I.F./V.F**REYNOLDS NUMBER**Reynolds Number ~ ratio of inertial to viscous forces -- hand waving argument -- Inertial Force ~ Upstream Force on Front Fluid Volume Face Inertial Force = momentum flux = f u l2xu (mass flux x velocity) Viscous Force ~ Shear Stress Force on Top Fluid Volume Face = (du/dy) ~ (u/[kl]) Viscous Force = (u/[kl])l2 = ul/k Re = Inertial Force / Viscous Force Re = f u l2 u / [ul/k] = kf ul/ Re = f ulc/ Where lc is a characteristic length. f control volume**REYNOLDS NUMBER**Reynolds conducted many experiments using glass tubes of 7,9, 15 and 27 mm diameter and water temperatures from 4o to 44oC. He discovered that transition from laminar to turbulent flow occurred for a critical value of uD/ (or uD/), regardless of individual values of or u or D or . Later this dimensionless number, uD/, was called the Reynolds number in his honor. ~ Nakayama & Boucher**Chapter 8:**Internal Incompressible Viscous Flow Internal = “completely bounded” - FMP • Internal Flows can be Fully Developed Flows: • mean velocity profile not changing in x; • “viscous forces are dominant” - MYO**Fully Developed Laminar Pipe/Duct Flow**LAMINAR Pipe Flow Re< 2300 (2100 for MYO) LAMINAR Duct Flow Re<1500 (2000 for SMITS) Uo= uavg OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE B.E.**As “inviscid” core accelerates, pressure must drop**Laminar Pipe Flow Entrance Length for Fully Developed Flow L/D = 0.06 Re {L/D = 0.03 Re, Smits} White Pressure gradient balances wall shear stress Le = 0.6D, Re = 10 Le = 140D, Re = 2300**As “inviscid” core accelerates, pressure must drop**Turbulent Pipe Flow Entrance Length for Fully Developed Flow Le/D = 4.4 Re1/6 MYO Note – details of turbulence may take longer than mean profile White Pressure gradient balances wall shear stress 20D < Le < 30D 104 < Re < 105**FULLY DEVELOPED**LAMINAR PIPE & DUCT FLOW Parallel Plates - Re = UD/ = 140 Water Velocity = 0.5 m/s Circular Pipe – Re = UD/ = 195 Water Velocity = 2.4 m/s Hydrogen Bubble Flow Visualization**2-D Duct Flow**a b c Hydrogen Bubble Flow Visualization Where taken a,b, or c? Parallel Plates - Re = 140 Water Velocity = 0.5 m/s**LAMINAR FLOW – VELOCITY PROFILE**TURBULENT FLOW – VELOCITY PROFILE**Upper plate moving at 2 mm/sec**Re = 0.03 (glycerin, h = 20 mm) Duct flow, umax = 2 mm/sec Re = 0.05(glycerin, h = 40 mm) VELOCITY = 0 AT WALL NO SLIP CONDITION (DUST ON FAN) What happens if wall is made of water? Or what happens to fluid particles next to no-slip layer?**No Slip Condition: u = 0 at y = 0**Stokes (1851) “On the effect of the internal friction of fluids on the motion of pendulums” showed that no-slip condition led to remarkable agreement with a wide range of experiments including the capillary tube experiments of Poiseuille (1940) and Hagen (1939).**VELOCITY = 0 AT WALL**NO SLIP CONDITION Each air molecule at the table top makes about 1010 collisions per second. Equilibrium achieved after about 10 collisions or 10-9 second, during which molecule has traveled less than 1 micron (10-4 cm). ~ Laminar Boundary Layers - Rosenhead**FULLY DEVELOPED LAMINAR FLOW**BETWEEN INFINITE PARALLEL PLATES Perform force balance on differential control volume to determine velocity profile, from which will determine volume flow, shear stress, pressure drop and maximum velocity.**FULLY DEVELOPED LAMINAR FLOW**BETWEEN INFINITE PARALLEL PLATES y=a y=0**FULLY DEVELOPED LAMINAR FLOW**BETWEEN INFINITE PARALLEL PLATES Assumptions: steady, incompressible, no changes in z variables, v=w=0, fully developed flow, no body forces No Slip Condition: u = 0 at y = 0 and y = a**FULLY DEVELOPED LAMINAR FLOW**BETWEEN INFINITE PARALLEL PLATES no changes in z variables, w = 0 ~ 2-Dimensional, symmetry arguments v = 0 du/dx + dv/dy = 0 via Continuity, 2-Dim., Fully Dev. du/dx = 0 everywhere since fully developed, therefore dv/dy = 0 everywhere, but since v = 0 at surface, then v = 0 everywhere along y (and along x since fully developed).**= 0(3)**= 0(1) = 0(4) FSx + FBx = /t (cvudVol )+ csuVdA Eq. (4.17) Assumptions: (1) steady, incompressible, (3) no body forces, (4) fully developed flow, no changes in z variables, v=w=0, FSx = 0 + + = 0 + * Control volume not accelerating – see pg 131**(Want to know what the velocity profile is.)**+ = 0 + p/x = dxy/dy p/x = dxy/dy = constant Since the pressure does not vary in the span-wise or vertical direction, streamlines are straight : p/x = dp/dx**Since the pressure does not vary in the span-wise or**vertical direction, streamlines are straight : p/x = dp/dx N.S.E. for incompressible flow with and constant viscosity. (v/t + uv/x + vv/y + wv/z) = gy - p/y + (2v/x2 + 2v/y2 + 2v/z2 v = 0 everywhere and always, gy ~ 0 so left with: p/y = 0 Eq 5.27b, pg 215**Important distinction because**book integrates p/x with respect to y and pulls p/x out of integral (pg 314), can only do that if dp/dx, which is not a function of y.**integrate**p/x = dp/dx = dxy/dy (Want to know what the velocity profile is.) For Newtonian fluid* substitute integrate USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1AND c2**a**0 c2 = 0 u = 0 at y = 0: u = 0 at y = a:**a = 1; dp/dx = 2**Why velocity negative? a**u(y) for fully developed laminar flow**between two infinite plates y = a y = 0 negative**(next want to determine shear stress profile,yx)**yx = (du/dy)**SHEAR STRESS?**Flow direction**y = a**- dp/dx = negative + y = 0 Does + and – shear stresses imply that direction of shear force is different on top and bottom plates?**Sign convention**for stresses White Positive stress is defined in the + x direction because normal to surface is in the + direction**(next want to determine shear stress profile,yx)**Shear force + yx = (du/dy) y = a y = 0 + + shear direction For dp/dx = negative yx on top is negative & in the – x direction yx on bottom is positive & in the – x direction**Question?**Given previous flow and wall = 1 (N/m2) Set this experiment up and add cells that are insensitive to shears less than wall Yet find some cells are dead. What’s up?**very large shear**stresses at start-up**u(y) for fully developed laminar flow**between two infinite plates y = a y’ = a/2 y’=0 y = 0 y’ = -a/2 y’ = y – a/2; y = y’ + a/2 (y’2 + ay’ + a2/4 –y’a – a2/2)/ a2 = (y’/a)2 – 1/4**(next want to determine volume flow rate, Q)**y=a y=0 [y3/3 – ay2/2]oa = a3/3 – a3/2 = -a3/6 If dp/dx = const