Internal Incompressible Viscous Flow 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) Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

{p =  RT} M2 = V2/(kp/) = [2/k][1/2 V2/p] 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] Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

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

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

Internal Incompressible Viscous Flow 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% Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

Internal Incompressible Viscous Flow 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 Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

•V = -(1/)D/Dt = 0 Volume is not changing. CONSERVATION OF MASS & INCOMPRESSIBLE •V = -(1/)D/Dt = 0 (5.1a) Volume is not changing. Note that stratified flow can be incompressible, but not constant density. Note – viscosity can cause temperature changes. This introduces T and could couple the momentum and energy equations. This happens for incompressible flows. But for viscous flows the change in T can usually be neglected. The density is not changing as follow fluid particle.

Internal Incompressible Viscous Flow 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 Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

REYNOLDS NUMBER Reynolds Number ~ ratio of inertial to viscous forces -- hand waving argument -- REYNOLDS NUMBER Inertial Force ~ Upstream Force on Front Fluid Volume Face Inertial Force = momentum flux = f u l2 x u (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] = kf 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

Internal Incompressible Viscous Flow 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 Mean free path of air molecules at standard atm and pressure is 0.05mm. Gas is typically 104 time more compressible than water 1% change in pressure ~ change of altitude of 85 feet.

Fully Developed Laminar Pipe/Duct Flow LAMINAR Pipe Flow Re< 2300 (2100 for MYO) LAMINAR Duct Flow Re<1500 (2000 for SMITS) Uo = uavg UD/nu = 140; 50D/.01 = 140; D =140/5000 ~ 0.024cm (need small gaps so approximates infinite plates (D<<Length of plate) With great care laminar to turbulent transition has been pushed to Re = 100,000! Mu = absolute viscosity, viscosity, dynamic viscosity Nu = kinematic viscosity (mu/rho), introduced by Maxwell since this parameter alone (given U and D) influences flow pattern. OUTSIDE BLUNDARY LAYER TREAT AS INVISCID, CAN USE B.E.

L/D = 0.06 Re 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 UD/nu = 140; 50D/.01 = 140; D =140/5000 ~ 0.024cm (need small gaps so approximates infinite plates (D<<Length of plate) With great care laminar to turbulent transition has been pushed to Re = 100,000! Mu = absolute viscosity, viscosity, dynamic viscosity Nu = kinematic viscosity (mu/rho), introduced by Maxwell since this parameter alone (given U and D) influences flow pattern. Hydrogen Bubble Flow Visualization

Where taken a,b, or c? 2-D Duct Flow a b c Parallel Plates - Re = 140 Hydrogen Bubble Flow Visualization Where taken a,b, or c? UD/nu = 140; 50D/.01 = 140; D =140/5000 ~ 0.024cm (need small gaps so approximates infinite plates (D<<Length of plate) Parallel Plates - Re = 140 Water Velocity = 0.5 m/s

Le/D = 0.06 Re

LAMINAR FLOW – VELOCITY PROFILE TURBULENT FLOW – VELOCITY PROFILE

What happens if wall is made of water? 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 du/dx + dv/dy + dw/dz = 0 du/dx and dw/dz = o on plates because of no slip so dv/dy = 0 on plates. But v = 0 on plate so 0 everywhere (dv/dy=0). 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 Why isn’t this dp/dx and not p/x? Since v=w=0 then dp/dy=dp/dz=0 so p not function of y and z. Since u(y) can write du/dy and not u/y.

Assumptions: steady, incompressible, no changes in z variables, v=w=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 du/dx + dv/dy + dw/dz = 0 du/dx and dw/dz = o on plates because of no slip so dv/dy = 0 on plates. But v = 0 on plate so 0 everywhere (dv/dy=0).

no changes in z variables, w = 0 ~ 2-Dimensional, symmetry arguments 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). du/dx + dv/dy + dw/dz = 0 du/dx and dw/dz = o on plates because of no slip so dv/dy = 0 on plates. But v = 0 on plate so 0 everywhere (dv/dy=0).

FSx + FBx = /t (cvudVol )+ csuVdA = 0(3) = 0(1) = 0(4) FSx + FBx = /t (cvudVol )+ csuVdA 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 + Why isn’t this dp/dx and not p/x? Since v=w=0 then dp/dy=dp/dz=0 so p not function of y and z. Since u(y) can write du/dy and not u/y. + + = 0 * Control volume not accelerating – see pg 131

p/x = dxy/dy = constant (Want to know what the velocity profile is.) + = 0 + p/x = dxy/dy Explain sign convention p/x = dxy/dy = constant Since the pressure does not vary in the span-wise or vertical direction, streamlines are straight : p/x = dp/dx

(v/t + uv/x + vv/y + wv/z) = 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 + uv/x + vv/y + wv/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.

(Want to know what the velocity profile is.) integrate p/x = dp/dx = dxy/dy (Want to know what the velocity profile is.) For Newtonian fluid* substitute * Note that there are additional terms for turbulent flow integrate USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1AND c2

a u = 0 at y = 0: c2 = 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

dp/dx = negative y = a - + 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

yx = (du/dy) (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? What’s up? 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

(next want to determine average velocity) A = la = uavg

(next want to determine maximum velocity) (a2/4)/a2 – (a/2)/a = -1/4

UPPER PLATE MOVING WITH CONSTANT SPEED U

Velocity distribution

+ UPPER PLATE MOVING WITH CONSTANT SPEED U Boundary driven Pressure driven

Shear stress distribution

Volume Flow Rate = Uy2/(2a) + (1/(2))(dp/dx)[(y3/3) – ay2/2]; y = a = Ua/2 + (1/(2))(dp/dx)[(2a3 – 3a3)/6] = Ua/2 + (1/(12))(dp/dx)[– a3]

Average Velocity l

Maximum Velocity y = a umax = a/2 y = 0

EXAMPLE:

FSx + FBx = /t (cvudVol )+ csuVdA FSx + FBx = /t (cvudVol )+ csuVdA Eq. (4.17) Assume: (1) surface forces due to shear alone, no pressure forces (patm on either side along boundary) (2) steady flow and (3) fully developed Fsx + FBx = 0 Fs1 – Fs2 - gdxdydz = 0 F1 = [yx + (dyx/dy)(dy/2)]dxdz F2 = [yx - (dyx/dy)(dy/2)]dxdz dyx/dy = g

d yx/dy = g yx = du/dy = gy + c1 du/dy = gy/ + c1/ u = gy2/(2) + yc1/ + c2

At y=h, u = gh2/(2) - gh2/ +U0 u = gy2/(2) + yc1/ + c2 u = gy2/(2) + yc1/ + c2 = gy2/(2) - ghy/ +U0 At y=h, u = gh2/(2) - gh2/ +U0 At y=h, u = -gh2/(2) +U0

FULLY DEVELOPED LAMINAR PIPE FLOW APPROACH JUST LIKE FOR DUCT FLOW Note however, that direction of positive shear stress is opposite.

dFR = -(p + [dp/dx]dx) 2rdr dFI = -rx2rdx dFL dFR dFL = p2rdr dFR = -(p + [dp/dx]dx) 2rdr dFI = -rx2rdx dFO = (rx + [d rx/dr]dr) 2(r + dr) dx

dFR = -(p + [dp/dx]dx)2rdr dFL + dFR = -[dp/dx]dx2rdr dFL = p2rdr dFR = -(p + [dp/dx]dx)2rdr dFL + dFR = -[dp/dx]dx2rdr

dFO = (rx + [d rx/dr]dr) 2(r + dr) dx dFL dFR dFI = -rx2rdx dFO = (rx + [d rx/dr]dr) 2(r + dr) dx dFO+ dFI = -rx 2rdx + rx 2rdx + rx 2drdx + [drx/dr)]dr2rdx + [drx/dr]dr 2dr dx dFO + dFI = rx 2drdx + [drx/dr]dr2rdx ~ 0

r r r dFL dFR dFL + dFR + dFI + dFO = 0 -[dp/dx]dx2rdr+rx 2drdx(r/r)+(drx/dr)dr2rdx = 0 [dp/dx] = rx/r + drx/dr = (1/r)d(rxr)/dr

dp/dx = (1/r)(d[rrx]/dr) Because of spherical coordinates, more complicated than for duct. dp/dx = dxy/dy

dp/dx = (1/r)(d[rrx]/dr) rx is at most a function of r, because fully developed, rx  f(x), symmetry, rx  f(). p is uniform at each section, since F.D., so not function of r or . dp/dx = constant = (1/r)(d[rrx]/dr)

dp/dx = constant = (1/r)(d[rrx]/dr) d[rrx]/dr = rdp/dx integrating….. rrx = r2(dp/dx)/2 + c1 rx = du/dr rx = du/dr = r(dp/dx)/2 + c1/r What we you say about c1?

rx = du/dr = r(dp/dx)/2 + c1/r rx = du/dr = r(dp/dx)/2 c1 = 0 or else rx =  rx = du/dr = r(dp/dx)/2 Shear forces on CV For dp/dx negative, get negative shear stress on CV but positive shear stress on fluid/wall outside control volume

du/dr = r(dp/dx)/2 u = r2(dp/dx)/(4) + c2 u=0 at r=R, so c2=-R2(dp/dx)/(4) u = r2(dp/dx)/(4) - R2(dp/dx)/(4) u = [ r2 - R2] (dp/dx)/(4) u = -R2(dp/dx)/(4)[ 1 – (r/R)2]

rx = r(dp/dx)/2 du/dr = r(dp/dx)/2 SHEAR STRESS PROFILE rx = -du/dr rx = r(dp/dx)/2 TRUE FOR LAMINAR AND TURBULENT FLOW du/dr = r(dp/dx)/2 TRUE ONLY FOR LAMINAR FLOW

FULLY DEVELOPED PIPE FLOW SHEAR STRESS PROFILE FULLY DEVELOPED PIPE FLOW = direction of shear stress on CV FULLY DEVELOPED DUCT FLOW - for flow to right

= 0R [ r2 - R2] (dp/dx)/(4) 2rdr VOLUME FLOW RATE – PIPE FLOW Q = A V • dA = 0R u2rdr = 0R [ r2 - R2] (dp/dx)/(4) 2rdr Q = [(dp/dx)/(4)][ r4/4 - R2r2/2 ]0R (2) = (-R4dp/dx)/(8)

VOLUME FLOW RATE – PIPE FLOW

p/x = constant = (p2-p1)/L = -p/L VOLUME FLOW RATE – PIPE FLOW as a function of p/L p/x = constant = (p2-p1)/L = -p/L p2 = p + p p1 L Q = (-R4dp/dx)/(8) = R4p/(8L) = D4(p/(128L)

Q = R4p/(8L) uAVG = Q/A = Q/(R2) = R4p/(R28L) = R2p/(8L) AVERAGE FLOW RATE – PIPE FLOW Q = R4p/(8L) uAVG = Q/A = Q/(R2) = R4p/(R28L) = R2p/(8L) = -(R2/(8)) (dp/dx)

AVERAGE FLOW RATE – PIPE FLOW uAVG = V = Q/A = Q/(R2) = R4p/(R28L) uAVG = R2p/(8L) = -(R2/(8)) (dp/dx)

du/dr = (r/[2])p/x At umax, du/dr = 0; which occurs at r = 0 MAXIMUM FLOW RATE – PIPE FLOW du/dr = (r/[2])p/x At umax, du/dr = 0; which occurs at r = 0 umax = R2(p/x)/(4)

MAXIMUM FLOW RATE – PIPE FLOW

FULLY DEVELOPED LAMINAR PIPE FLOW u/umax = 1 – (r/R)2 u/umax r/R

FULLY DEVELOPED LAMINAR PIPE FLOW (r)/w u/umax Shear stress CV exerts r/R

THE END

END

Is 140 D the biggest L? Np laminar flows with care can exist up to Re~100,000. L/D = 0.06 Re Re = 2300 L = 140 D

pA-po = ½ Uavg2 what is p between A and B? Prandtl & Tietjens A B Is 140 D the biggest L? Np laminar flows with care can exist up to Re~100,000. pA-po = ½ Uavg2 what is p between A and B? Flux of K.E. per unit volume = u{½ u2(r2)dr} at B at A u = Uavg2[1-(y/r)2] u = Uavg