1. 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.

3. {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.

4. 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

5. 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

6. 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.

7. 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.

8. •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.

10. 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.

11. 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

12. 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

14. 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.

15. 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.

16. 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

17. 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

18. 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

19. 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

20. Le/D = 0.06 Re

21. LAMINAR FLOW – VELOCITY PROFILE
TURBULENT FLOW – VELOCITY PROFILE

23. 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?

24. 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).

25. 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

27. 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.

28. 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.

29. 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).

30. 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).

31. 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

32. 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

33. (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

34. 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.

35. (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

36. a
u = 0 at y = 0:
c2 = 0
u = 0 at y = a:

38. a = 1; dp/dx = 2
Why velocity
negative? a

39. u(y) for fully developed laminar flow between two infinite plates
y = a
y = 0
negative

40. (next want to determine shear stress profile,yx)
yx = (du/dy)

41. SHEAR STRESS?
Flow direction

42. dp/dx = negative y = a - + y = 0
Does + and – shear stresses imply that
direction of shear force is different on
top and bottom plates?

43. Sign convention for stresses White
Positive stress is defined in the + x direction
because normal to surface is in the + direction

44. 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

46. 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?

47. very large shear
stresses at start-up

49. 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

50. (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

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

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

53. UPPER PLATE MOVING WITH CONSTANT SPEED U

54. Velocity distribution

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

56. Shear stress distribution

57. 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]

58. Average Velocity l

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

61. EXAMPLE:

62. 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

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

64. 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

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

67. 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

68. 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

69. 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

70. 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

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

72. 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)

73. 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?

74. 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

75. 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]

77. 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

78. 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

79. = 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)

80. VOLUME FLOW RATE – PIPE FLOW

81. 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)

82. 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)

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

84. 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)

85. MAXIMUM FLOW RATE – PIPE FLOW

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

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

89. THE END

90. END

91. 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

92. 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