Viscous Stress and Force
The Navier-Stokes Equation
Equation of motion
Boundary/initial conditions
Applications of the Navier-Stokes Equation
Couette flow
Poiseuille flow
Combined Couette-Poiseuille flow
Creeping (Stokes) flow
Unsteady laminar flow
A Quick Look
Fluid stress is the sum of
Pressure stress Viscous stress (in normal direction) (proportional to )
) shear (
tangential normal
n
p
Translation Rotation
(rigid body)
Shear Compression
yield viscous stress
Fluid Stress
Viscous stress is proportional to the rate of motions
Fluid element V
xy x
xy
yx
yx
xy
Fluid element V
yxy xy
yx
yx
y V x
xy ?
?
direction surface
gradient V μ
fluid stress
shear
Viscous Stress
D erivation
dy
y
z dx dz x
o
xz
xy xx
xz xx
xy dy
y
xy
?
?
x dx
xx
?
?
z dz
xz
?
?
constant ,
0 V
if
) z , y , x k
( V
i
f k 2 k
3. Obtain the Navier-Stokes equation as V g
1 p t
D V
D 2
p * ν V
ρ V 1
t V
V 2
Navier - Stokes Equation
1. Start from the Euler equation + viscous term
f
g 1 p
t D
V
D
2. Determine the viscous force per unit mass
Fluid-solid boundary condition: no slip
Immiscible fluid-fluid boundary condition
solid
V V
2 1
2 1
2 1
τ τ
) n p ( )
n p ( inviscid If
stress σ
σ
n 2
n 1
1
2
p 2
p 1
Fluid 1
Fluid 2
Boundary Conditions
Navier-Stokes Equation
x x
z y
y
h
V
pw
w
xy xy
V xy
* 1 p
t D
V
D 2
?
?
0 0
Boundary condition
p x
x ( 0 ) 0 , V ( h ) V
V
x p w
p x
h i μ V τ
h V y ) y ( V
2 V V
2 h V W
Q
p p
Plane Couette Flow
x y
z
high
p h low
p
w
w
y=h
y=0
V x
L
w
w
p (x) p (x+L)
C.V.
2 1
2 x
2
C y C x y
d
* p d 2 μ V 1
ν V
* ρ p
1 t
D V D
Plane Poiseuille Flow
Navier-Stokes Equation
Boundary condition
x d
* p d 2
) h (
* p ) (
* L p 2 τ h
x i d
* p d 2
i h x d
* p d μ
2 μ h y i
μ V τ
3 V 2 x
d
* p d 12
h h W V Q
x d
* p d 12
h W
Q
x d
* p d μ
8 h 2
h x
d
* p d μ
2 1 2
y h V V
) y h ( x y
d
* p d μ
2 y 1 x d
* p d μ 2 y h
x d
* p d μ 2 V 1
W
x x
x 0 y x W
max x 2
3
2 2 x
max x
2 x
L x x
0 ) h ( V 0
) 0 (
V x x
Results
Plane Poiseuille Flow
x y
z
high
p h low
p
w
w
y=h
y=0
V x
V p
V
p) y h ( x y d
* p d 2 μ
) 1 y ( V : Poiseuille plane
h V y ) y ( V : Couette plane
x
p x
Superposition of the two solutions
x xp W PP
W PC W
3 p
PP PC
p x
x i d
* p d 2
i h h μ V τ
τ τ
x d
* p d μ
12 h 2
h V W
Q W
Q W
Q
) y h ( x y
d
* p d μ
2 1 h V y ) y ( V
Combined Plane Couette + Poiseuille Flow
Individual solutions
z r
R w
w
) r ( V z
z 0 0 V
θ V
0 V 0
V
z z
θ r
D μ V τ 8
z or d
* p d 4
L π D ) π DL τ (
z d
* p d μ
32 D A
V Q
z d
* p d μ
128 π D dr π r 2 V Q
) r R z (
d
* p d μ
4 V 1
z dr d
* p d μ 2 dV r
W 2
W
2 R
0
4 z
2 2 z
z
z2 z
z
r 2 2 θ θ θ 2
r θ θ
θ 2 2
r r 2 2
θ r
r
ν V r
* p ρ V 1
t V V
θ V r
2 r
V V θ ν
* p ρ
1 r
V V V
t V V
θ V r
2 r
V V r ν
* p ρ
1 2
V V t V
V
Circular Poiseuille Flow
Navier-Stokes Equation
Boundary condition
Results
?
)
V F
R
a
V R
V j
j
t V D
V D
V
* 1 p
t D
V D
2
2
Results
j
j
j
3 3 3 3 R
R 4
a 5 R
4 a 1 3
sin V
V
R 2
a R
2 a 1 3
cos V
V
a V 6
F
R 4
a 5 R
4 a 1 3
R sin V
R 2
cos aV
p 3
* p
3 3 R
2
*
j
j
j
Creeping Flow (having a very small Re)
Navier-Stokes Equation
const
x with slowly
changes a
pipe converging
x d
p d
) x ( a a
x x 2
r x
x
x V
x p 1 r
V V x
V V t
V ?
?
?
??
?
?? ?
?
?
?
?
?
?
?
?
) steady (
0 x
p
?
?
x p
?
?
) a r
x ( p 1 4 V 1
. C . B apply
r r V d x dr
p r
2 2
x
x
r
a o a Q x
①
①
x=0 p=p
1steady
x=L p=p
2Converging Pipe
Navier-Stokes Equation
Solution
Volume flow rate
Pressure drop
4
4
a 0
2 a 2
0 x
a Q 8 x
p
x p 1 8
a
dr ) a r
( x r
p 1 4 dr 2
r 2 V Q
) 1 1
( 1 a
Q p 8
Δ
L x 1
a ) x ( a
3 4
o o
Converging Pipe
Rayleigh problem (1-D)
V V 1 p * V
t
V 2
0
x y z
V
U
? ) 0 t ( V
0 ) 0 t ( V
Infinite large plate ( no x,y dependence) No asymptotic profile!
- momentum diffuses into fluid via viscosity
- if the fluid were inviscid there would be no fluid motion
t > 0 ( time ↑)
V z
Unsteady Flow - Example
Navier-Stokes Equation
x z
y fluid z ≥0
V x (z, t)
U 0 at t=0
x-comp. Navier-Stokes eq.
x 2
x z x
y x
x x
x V p 1
z V V
y V V
x V V
t V
?
?
?
?
?
?
?
?
?
?
?
0 0 0
0
very large plate
(continuity)
(no external pressure applied)
mass
everywhere 0
V
z 0 V y
V x
V V
z
y z x
?
?
?
?
?
?
?
?
0 0
thus diffusion eq .
z V t
V
2 x 2
x
Rayleigh Problem
Solution
initial condition boundary condition variables change 0
) 0 t , z (
V x
0 ) 0 t , z
( V
U )
0 t , 0 z ( V
x x
t z
V 0 2 V
V t
1 z
and V V
t 2 t
now V
x 2
x 2
2 x 2
t 2
x 2 x
z x
similarity
solution
U V x
t z
All solutions are similar so that they are combined into 1 solution.
Rayleigh Problem
f
V x
1 2 V x
exp 4 C
f
t
erf z t U U
) t , z ( V x
Let
then Apply IC + BC to finally obtain
the desired velocity profile as
Head Loss in Pipes & Ducts
Straight pipes: h given Q; Q given h
Straight noncircular ducts
Loss from area change, bend, fittings, etc.
Total head loss
Head Changes in Systems with Pumps and Turbines
Matching pumps & hydroturbines to flow systems
Complex Networks
Serial components
Parallel components
A Quick Look
Head (incompressible, 1-phase flow)
Straight Pipes
viscous shear on a pipe causes a Δp ;
) gz p
R g p
* p g (
2 V g
* p g
R g g
2 V g
p
2
2
D V L
2 p 1
p
*in *out 2
f
If there is no change in velocity Dimensionless loss coefficient
g 2 V D
L g
p
p
*in *out 2out
in
f
f
g 2 K V D ,
K L thus
V g 2 g 2 / K V
2 2
2
f f
f
f
Head Loss in Pipes and Ducts
bulent tur
a 1 r 49 60 V
V
laminar
a 1 r
V 2 V
1/7 z
2 z
a g
z V V
2 . 49 1
60 V
V V 2 V : 0 r
tur max, z
lam max, z
Velocity Profiles
Basic Formula:
laminar:
turbulent:
) Re (
64
D
2300 Re
f
D
) D / , Re ( D
f f
) 2300 Re
f (
f
D
Re
D51 . 2 7
. 3
D log /
0 . 1 2
0.008
10
3 230010
410
510
610
710
80.10
0.04
0.01
0 D /
10
6D /
10
-510
-410
-310
-2
V
D Re
Df
(Darcy) / D
① Critical Zone, Discontinuous
smooth pipes
rough pipes
① Transition Zone
① Laminar Flow Zone
① Complete Turbulence Zone
The Pipe Darcy-Weisbach Friction Factor
Turbulent Skin Friction and Drag
Wall shear stress is much larger than in a laminar flow
Much less dependent upon the flow Re number
Even independent of Re in some circumstances
Turbulent Pipe Flow
Friction factor
Including roughness 51 . 2
f log Re
0 . 2 f
1 D
f Re
51 . 2 3.7
log /D 2 f
1
D
10
-210
-410
-6ε/D=0 log f
4 6 8
-1.5 -2.0 -2.5
Pipe Flow
x
y z
2 2 2 2
x 2 2
z V y
μ V dx
* 0 dp
: direction x
ν V
* ρ p
1 t
D V
D
) 1 ( ρ V
2 1
D ) dx /
* dp f (
Perimeter Area D 4
: D
2 h h
h
; factor friction
Darcy
diameter hydraulic
l
; l dimension stic
characteri
x d
* p d μ V
D Re 2
f
gives )
1 (
) 2 ( ρ V
2 1 D f L x L
d
* p p d
Δ
h D
2 h
h
; difference pressure
Poiseuille Flow in Noncircular Ducts
a
b
a
b b
2a
h D
2 2D D
b 2
L V b
/ D
Re
* f p ) 2 ( from Then
) tube circular
a for ( 64 Re
f from 25%
than more
by differ not
does Re
f
h
h h
b/a
Ellipse Rectangle Isosceles Triangle
D
hRe
f D h / b f Re D
hD h / b f Re D
hD h / b
1.0 64.00 1.0 56.91 1.0 52.61 0.828
0.5 67.29 1.298 62.19 1.333 50.49 0.944 0.2 74.41 1.497 76.28 1.667 48.63 0.990 0.1 77.26 1.548 84.68 1.818 48.31 0.998
Friction Factor for Noncircular Ducts
Friction factor & hydraulic diameter for noncircular tubes
Note the discontinuity at
2 T
2 L
D
10 728 . 4
10 783 . 2 2300
Re
f f
Actually the increase in f is spread over a range of Re near 2300. This numerical
discontinuity seldom causes trouble in pipe flow calculation. In most of the computer codes, they have continuous Darcy friction factor curves interpolated in the critical zone.
Head loss for given Q:
Calculate
Calculate or depending on
Calculate
2 2
D Q V 4
4 V D
Q
D Q 4 D
Re
DV
f
Lf
TRe
D 2300 or Re
D 2300
f f
f
f
, p * g
g 2 K V D ,
K L
2
The Pipe Darcy-Weisbach Friction Factor
8 inch Φ Commercial Steel Pipe, ε = 5×10 -5 m
gpm 1000 Q km 2
m 22
m 100
Tank Storage
f
f
55
5 D 6
2
3 2
10 95 . 3
51 . 2 )
2032 . 0 ( 7 . 3
10 log 5
0 . 1 2
turbulent :
10 95 . 10 3
1
) 2032 . 0 ( ) 944 . 1 Re (
s / m 944 . D 1
Q V 4
s / m 10 303 . 6 gpm 1000 Q
m 2032 . 0 D . 1
2 -2
10 1.616 iterations
more 3
after obtain
finally to
10 1 assuming
by Begin .
2
f f
g z
z z
z g p
p p
p
64 . ) 30 807 . 9 ( 2
) 944 . 1 10 ( 59 . g 1
2 K V
10 59 . 2032 1 . 0 10 2000 616
. D 1 f L K . 3
out in
out in
out in
* out
* in out
in
2 2
2 f f
2 2
f
Pipe Friction Factor - Example
A
lA
SV
SSudden Contraction
A
lA
SSudden Expansion A 0
when A 4
. A 0
1 A 4 . 0
K c s s
l l
A 0 when A A 1
1 A
K s
2
e s
l l
Total Head Loss
i
ii i
f i
j
i j i
f i
g p
p
D f L where
g 2 K V
* out
* in
2
Head Changes in System with Pumps / Turbines
t p
in
out
l
Loss from Area Changes
A Heat Collecting Circuit for a Solar Panel
K e = 1.0
K c = 0.4 K b = 1.0
header diameter
>> pipe diameter
D = 1 cm L = 22 m Q = 4 l /min
70 . 53 ( 10 K K K ) V 3 . 01 m
g 2 K V
D h L
2 2
j j
l
f
489 , 8 /
D V Re
s / m 8489 .
0 A / Q V
m 10 854 . 7 4 / D A
s / m 10
667 . 6 Q
D
2 5 2
3 5
Choose 10 6 m
2
D
10 205 . 3
Re 51 . 2 7
. 3
D log /
0 . 1 2
f
f f
Loss from Area Changes - Example
(Turbulent)
max power
Q max
/
Q
) pump
(
system piping
0 0.2 0.4 0.6 0.8 1.0
0.2 0.4 0.6 0.8 1.0
h max
h
x
max
max h
Q
h Q
x
? ?
?
?
??
?
?
?
0 h
Q Q at
h h
) .
e . i ( 0 Q at
pump
max max
off colsed
discarge
Matching Pumps to Flow Systems (1)
Q will be determined for given pump and the piping system .
Q as h
Q as h p
l
h g
Q h
g
m
Procedure
Ideal power delivered to fluid stream passing through the pump [W]
Power required to through the pump:
/ p
h g
Q
where p : pump efficiency
Pump are designed such that is reached at i.e. the maximum power.
max ,
p ( Q h ) max
Matching Pumps to Flow Systems (2)
Pump:
Assume
Calculate Q 20
reservoir water
80 m pump Q ?
? ?
?
?
? ?
?
?
? ?
?
? ?
?
?
2 p
5
) gpm ( 1000
) gpm ( 1 Q
) m 150 ( h
m 10 5
. in 6 D
km 5 . 1 L
) loss head
viscous (
h m
80
h
p
l
m 20 h
l
4 2
3
10 043 . L 3
D h g
Re 2
lf
10
5292 . 2
Re 51 . 2 7
. 3
D log /
Re 0
. 2 Re
f f
1 435 / 1000 121 . 6 m ( 80 20 ) m
150 h
, gpm 4 435
Re
Q
20 D
p
2
Now assume h l 40 m , Q 40 = 626 gpm
1 626 / 1000 91 . 22 m ( 80 20 ) m
150
h
p
2
Iterate more in this way, Q 527 . 7 gpm and h p 108 m
Matching Pumps to Flow Systems - Example
Hydroturbine
s / m 659 . D 5
Q V 4
2
turbulent :
10 488 . D 8
Re
DV
6
366 . Re 9
51 . 2 7
. 3
D log /
0 .
1 2
f
f
281 . D 2
K L
f
f
0 . 1 K
K , 4 . 0 K
K
inlet
C
outlet
e
m 010 . 2 6
) V K
K K
2 ( K V h
2 out
in
2
l
fm 99 . 93 h
100 h
Now
t
l
kW 501 h
g Q )
Power Turbine
(
) 85 . 0 (
turbine
t
h = 100 m
D = 1.5 m
m 10 1 ?
4
L = 300 m
Q = 10 m
3/s
Matching Pumps to Flow Systems - Example
pipe A
pipe B
Q Q A Q B Q
Q A = Q B = Q
Series Parallel
Q
AQ Q
pipe B pipe A
Q
BQ
A= Q
B= Q
B , l A
,
l