chapter 8: internal incompressible viscous flow flows: laminar (some have analytic solutions)...
TRANSCRIPT
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 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
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 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)
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 ~ 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
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
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.
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
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
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 alongx since fully developed).
FULLY DEVELOPED LAMINAR FLOW BETWEEN INFINITE PARALLEL PLATES
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
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-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
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-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
FULLY DEVELOPED LAMINAR PIPE FLOW
r/R
u/umax
FULLY DEVELOPED LAMINAR PIPE FLOW
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