Chapter 8: 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)


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 pressureM ~ 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

atm. press. = static pressure(what moving fluid particle “sees”)

Hand in steady wind –Felt by hand = stag. press.

for incompressible flowpo = p + ½ V2

Static pressure

Stagnation pressure

Dynamic pressure


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%


•Compressibility requires work, may produce heat andchange 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

The density is not changingas follow fluid particle.

Volume is not changing.

(5.1a)


Flows: Laminar (some have analytic solutions) Turbulent (no analytic solutions)

Depends of Reynolds number,

Re = I.F./V.F

Reynolds Number ~ ratio of inertial to viscous forces -- hand waving argument --

controlvolume

f

Inertial Force ~ Upstream Force on Front Fluid Volume FaceInertial 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 ForceRe = f u l2 u / [ul/k] = kf ul/Re = f ulc/

Where lc is a characteristic length.

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

REYNOLDS NUMBER


Internal = “completely bounded” - FMP

Internal Flows can be Fully Developed Flows: • mean velocity profile not changing in x;• “viscous forces are dominant” - MYO

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.

Fully Developed Laminar Pipe/Duct Flow

Laminar Pipe FlowEntrance Length for

Fully Developed Flow

L/D = 0.06 Re

{L/D = 0.03 Re, Smits}

White

As “inviscid” core accelerates, pressure must drop

Pressure gradientbalances wall

shear stress

Le = 0.6D, Re = 10Le = 140D, Re = 2300

Turbulent Pipe FlowEntrance Length for

Fully Developed Flow

White

As “inviscid” core accelerates, pressure must drop

Pressure gradientbalances wall

shear stress

Le/D = 4.4 Re1/6

MYO

20D < Le < 30D104 < Re < 105

Note – details ofturbulence maytake longer than mean profile

Hydrogen Bubble Flow Visualization

Parallel Plates - Re = UD/ = 140Water Velocity = 0.5 m/s

Circular Pipe – Re = UD/ = 195Water Velocity = 2.4 m/s

FULLY DEVELOPEDLAMINAR PIPE &

DUCT FLOW

Hydrogen Bubble Flow Visualization

Parallel Plates - Re = 140Water Velocity = 0.5 m/s

2-D Duct Flow

a b c

Where takena,b, or c?

Le/D = 0.06 Re

LAMINAR FLOW – VELOCITY PROFILE

TURBULENT FLOW – VELOCITY PROFILE

VELOCITY = 0 AT WALLNO SLIP CONDITION

(DUST ON FAN)

What happens if wall is made of water?Or what happens to fluid particles next

to no-slip layer?

Upper plate moving at 2 mm/sec Re = 0.03 (glycerin, h = 20 mm)

Duct flow, umax = 2 mm/secRe = 0.05(glycerin, h = 40 mm)

No Slip Condition: u = 0 at y = 0

Stokes (1851) “On the effect of the internal friction of fluids on themotion of pendulums” showed that no-slip condition led to remarkableagreement 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.


y=0

y=a

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


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 alongx since fully developed).


Assumptions: (1) steady, incompressible, (3) no body forces, (4) fully developed flow, no changes in z variables, v=w=0,

= 0(4)= 0(3) = 0(1)

+

++ = 0

FSx + FBx = /t (cvudVol )+ csuVdAEq. (4.17)

FSx = 0

* Control volume not accelerating – see pg 131

+

+ = 0

(Want to know what the velocity profile is.)

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

N.S.E. for incompressible flow with and constant viscosity.

Since the pressure does not vary in the span-wise or vertical

direction, streamlines are straight : p/x = dp/dx

(v/t + uv/x + vv/y + wv/z) = gy - p/y + (2v/x2 + 2v/y2 + 2v/z2

Eq 5.27b, pg 215

v = 0 everywhere and always, gy ~ 0 so left with: p/y = 0

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

(Want to know what the velocity profile is.)

For Newtonian fluid*

substitute

integrate

USE 2 BOUNDARY CONDITIONS TO SOLVE FOR c1AND c2

p/x = dp/dx = dxy/dy

u = 0 at y = a:

u = 0 at y = 0: c2 = 0

a

0

u = {(y/1)^2 -(y/1)}; channel height=1m

0

0.2

0.4

0.6

0.8

1

1.2

-0.3 -0.25 -0.2 -0.15 -0.1 -0.05 0

u (m/s)

y (m

)

a

a = 1; dp/dx = 2

Why velocity negative?

u(y) for fully developed laminar flow between two infinite plates

negative

y = 0

y = a

(next want to determine shear stress profile,yx)

yx = (du/dy)

tau = [(y/1)-1/2]; a=1, dp/dx=1

00.10.20.30.40.50.60.70.80.91

-0.6 -0.4 -0.2 0 0.2 0.4 0.6

tau

y

SHEAR STRESS?

Flow direction

tau = [(y/1)-1/2]; a=1, dp/dx=1

00.10.20.30.40.50.60.70.80.91

-0.6 -0.4 -0.2 0 0.2 0.4 0.6

tau

y

y = 0

y = a

Does + and – shear stresses imply that direction of shear force is different on top and bottom plates?

dp/dx = negative+

-

White

Positive stress is defined in the + x direction because normal to surface is in the + direction

Sign conventionfor stresses

(next want to determine shear stress profile,yx)

yx = (du/dy)y = 0

y = a

For dp/dx = negativeyx on top is negative & in the – x directionyx on bottom is positive & in the – x direction

Shear force+

+ + sheardirection

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’ = y – a/2; y = y’ + a/2

(y’2 + ay’ + a2/4 –y’a – a2/2)/ a2 = (y’/a)2 – 1/4

y’=0

y = 0

y’ = a/2

y’ = -a/2

y = a

(next want to determine volume flow rate, Q)

[y3/3 – ay2/2]oa = a3/3 – a3/2 = -a3/6

If dp/dx = const

y=0

y=a

(next want to determine average velocity)

= uavg

A = la

(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

+

Pressure drivenBoundary driven

Shear stress distribution

= 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]

Volume Flow Rate

Average Velocity

l

Maximum Velocity

umax = a/2y = 0

y = a

EXAMPLE:

FSx + FBx = /t (cvudVol )+ csuVdAEq. (4.17)

0

Assume: (1) surface forces due to shear alone, no pressure forces (patm on either side along boundary)(2) steady flow and (3) fully developed

0

Fsx + FBx = 0Fs1 – Fs2 - gdxdydz = 0 F1 = [yx + (dyx/dy)(dy/2)]dxdzF2 = [yx - (dyx/dy)(dy/2)]dxdzdyx/dy = g

d yx/dy = gyx = du/dy = gy + c1

du/dy = gy/ + c1/u = gy2/(2) + yc1/ + c2

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.

r

dFL = p2rdr dFR = -(p + [dp/dx]dx) 2rdrdFI = -rx2rdxdFO = (rx + [d rx/dr]dr) 2(r + dr) dx

r

rdFL dFR

r

rr

dFL = p2rdr

dFR = -(p + [dp/dx]dx)2rdr

dFL + dFR = -[dp/dx]dx2rdr

dFL dFR

r

rr

dFI = -rx2rdxdFO = (rx + [d rx/dr]dr) 2(r + dr) dx

dFO+ dFI = -rx 2rdx + rx 2rdx + rx 2drdx + [drx/dr)]dr2rdx + [drx/dr]dr 2dr dx

dFO + dFI = rx 2drdx + [drx/dr]dr2rdx

dFL dFR

r

~ 0

rr

dFL + dFR + dFI + dFO = 0-[dp/dx]dx2rdr+rx 2drdx(r/r)+(drx/dr)dr2rdx = 0

[dp/dx] = rx/r + drx/dr = (1/r)d(rxr)/dr

dFL dFR

r

dp/dx = dxy/dy

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

dp/dx = (1/r)(d[rrx]/dr)

p is uniform at each section, since F.D., so not function of r or .

rx is at most a function of r, because fully developed, rx f(x),symmetry, rx f().

dp/dx = constant = (1/r)(d[rrx]/dr)

dp/dx = constant = (1/r)(d[rrx]/dr)d[rrx]/dr = rdp/dx

integrating…..rrx = r2(dp/dx)/2 + c1

rx = du/drrx = du/dr = r(dp/dx)/2 + c1/r

What we you say about c1?

rx = du/dr = r(dp/dx)/2 + c1/r c1 = 0 or else rx =

rx = du/dr = r(dp/dx)/2

For dp/dx negative, get negative shear stress on CVbut positive shear stress on fluid/wall outside control volume

Shear forces on CV

du/dr = r(dp/dx)/2u = 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]

SHEAR STRESS PROFILE

rx = r(dp/dx)/2 TRUE FOR LAMINAR AND TURBULENT FLOW

du/dr = r(dp/dx)/2TRUE ONLY FOR LAMINAR FLOW

rx = -du/dr

SHEAR STRESS PROFILE

FULLY DEVELOPED DUCT FLOW

FULLY DEVELOPED PIPE FLOW

= direction of shear stress on CV

- for flow to right

VOLUME FLOW RATE – PIPE FLOW

Q = A V • dA = 0

R u2rdr = 0

R [ r2 - R2] (dp/dx)/(4) 2rdr

Q = [(dp/dx)/(4)][ r4/4 - R2r2/2 ]0R (2)

= (-R4dp/dx)/(8)

VOLUME FLOW RATE – PIPE FLOW

VOLUME FLOW RATE – PIPE FLOW as a function of p/L

p/x = constant = (p2-p1)/L = -p/L

p2 = p + p

Lp1

Q = (-R4dp/dx)/(8) = R4p/(8L) = D4(p/(128L)

AVERAGE FLOW RATE – PIPE FLOW

uAVG = Q/A = Q/(R2) = R4p/(R28L)= R2p/(8L) = -(R2/(8)) (dp/dx)

Q = R4p/(8L)

AVERAGE FLOW RATE – PIPE FLOW

uAVG = V = Q/A = Q/(R2) = R4p/(R28L)uAVG = R2p/(8L) = -(R2/(8)) (dp/dx)

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

u/umax = 1 – (r/R)2


r/R

u/umax


r/R

u/umax

(r)/w

Shear stress CV exerts

THE END

END

L/D = 0.06 ReRe = 2300L = 140 D

u = Uavg

at A at BFlux of K.E. per unit volume = u{½ u2(r2)dr}

u = Uavg2[1-(y/r)2]

A

B

pA-po = ½ Uavg2 what is p between A and B?

poPrandtl & Tietjens

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Chapter 8: Internal Incompressible Viscous Flow