1. Overview
 Continuity Equation
 Navier-Stokes Equation
(a bit of vector notation...)
 Examples (all laminar flow)
Flow between stationary parallel horizontal
plates
Flow between inclined parallel plates
Pipe flow (Hagen Poiseuille)

2. Why Differential Equations?
 A droplet of water
 Clouds
 Wall jet
 Hurricane

3. Conservation of Mass in
Differential Equation Form
F+ ∂ρ δyIF+ ∂v δyIδxδz
G ∂y Jv ∂y J
G K
H K
ρ
H Mass flux out of differential volume
∂ρ Rate of change of mass in
δyδxδz
δy ∂t differential volume
δz
δx
ρvδxδz Mass flux into differential volume

4. Continuity Equation
Mass flux out of differential volume
Fv + ρ ∂v δy + v ∂ρ δy + ∂v ∂ρ δy Iδxδz
G ∂y ∂y ∂y ∂y J
H
ρ
K Higher order term
2
out in Rate of mass decrease
Fv + ρ ∂v δy + v ∂ρ δyIδxδz − ρvδxδz = − ∂ρ δyδxδz
G ∂y ∂y J
H
ρ
K ∂t
∂v ∂ρ ∂ρ
ρ +v =−
∂y ∂y ∂t
∂ρv ∂ρ
+ =0 1-d continuity equation
∂y ∂t

5. Continuity Equation
af af a f
∂ ρ ∂ ρu ∂ ρv ∂ ρw
+ + + =0 3-d continuity equation
∂t ∂x ∂y ∂z
divergence
∂ρ u, v, w are
+ ∇ ×ρ V = 0 Vector notation velocities in x, y,
∂t
and z directions
If density is constant...
∂u ∂v ∂w
+ + =0 or in vector notation ∇ ×V = 0
∂x ∂y ∂z
True everywhere! (contrast with CV equations!)

6. Continuity Illustrated
∂u ∂v ∂w y
+ + =0
∂x ∂y ∂z
What must be happening?
∂v ∂u
<0 ∴ >0
∂y ∂x
x

7. Navier-Stokes Equations
 Derived by Claude-Louis-Marie Navier in 1827
 General Equation of Fluid Motion
momentum
 Based on conservation of ___________ with forces…
Gravity
 ____________
Pressure
 ___________________
Shear
 ___________________
 U.S. National Academy of Sciences has made the full
solution of the Navier-Stokes Equations a top priority

8. Navier-Stokes Equations
V ρa = ∑ F
Navier-Stokes Equation
ρ a + ρ g = −∇p + µ∇ V 2
g is constant
ρ a = − ( ∇p + ρ g ) + µ∇ 2 V a is a function of t, x, y, z
ρa = Inertial forces [N/m3], a is Lagrangian acceleration
Is acceleration zero when ∂V/ ∂ t = 0? NO!
− ( ∇p + ρ g ) = Pressure gradient (not due to change in elevation)
∇p + ρ g ≠ 0 V ≠0
If _________ then _____
du
µ∇ V = Shear stress gradient
2 τ =µ τ = µ∇V
dx

9. Notation: Total Derivative
Eulerian Perspective
Dα ∂α dt ∂α dx ∂α dy ∂α dz Total derivative
(t , x, y, z ) = + + +
Dt ∂t dt ∂x dt ∂y dt ∂z dt (chain rule)
Dα ∂α ∂α ∂α ∂α Material or
(t , x, y, z ) = +u +v +w
Dt ∂t ∂x ∂y ∂z substantial derivative
DV ∂V ∂V ∂V ∂V
(t , x, y, z ) = +u +v +w
Dt ∂t ∂x ∂y ∂z Lagrangian acceleration
∂V ∂V ∂V ∂V ∂ () ∂ () ∂ ()
a= +u +v +w ∇() = i+ j+ k
∂t ∂x ∂y ∂z ∂x ∂y ∂z
∂V ∂ () ∂ () ∂ ()
a= +V× V
∇ ( V ×∇ ) () = u + v + w
∂t ∂x ∂y ∂z

10. Application of Navier-Stokes
Equations
 The equations are nonlinear partial
differential equations
 No full analytical solution exists
 The equations can be solved for several
simple flow conditions
 Numerical solutions to Navier-Stokes
equations are increasingly being used to
describe complex flows.

11. Navier-Stokes Equations: A
Simple Case
 No acceleration and no velocity gradients
ρ a = − ( ∇p + ρ g ) + µ∇ 2 V
0 = − ( ∇p + ρ g )
∇p = − ρ g xyz could have any orientation
−ρg
∂p ∂p ∂p
= −ρ gx = −ρ g y = − ρ g z Let y be vertical upward
∂x ∂y ∂z
∂p ∂p ∂p Component of g in the x,y,z direction
=0 = −ρ g =0
∂x ∂y ∂z
p = − ρ gy + C For constant ρ

12. Infinite Horizontal Plates:
Laminar Flow
Derive the equation for the laminar, steady, uniform flow
between infinite horizontal parallel plates. y
ρ a = − ( ∇p + ρ g ) + µ∇ 2 V
0 = − ( ∇p + ρ g ) + µ∇ 2 V x
∂p  ∂ 2u ∂ 2u ∂ 2u  ∂p  ∂ 2u 
x 0 = − − ρ gx + µ  2 + 2 + 2 ÷ 0 = − +µ 2 ÷
∂x  ∂x ∂y ∂z  ∂x  ∂y 
v=0 Hydrostatic in y
∂p ∂ v ∂ v ∂ v
2 2 2
y 0 = − − ρ gy + µ  2 + 2 + 2 ÷ ∂p
∂y  ∂x ∂y ∂z  0 = − − ρg
w=0 ∂y
∂p  ∂2w ∂2w ∂2w 
z 0 = − − ρ gz + µ  2 + 2 + 2 ÷ 0=0
∂z  ∂x ∂y ∂z 

13. Infinite Horizontal Plates:
Laminar Flow
∂p  ∂ 2u  du
0 = − +µ 2 ÷ τ =µ
∂x  ∂y  dy
dp  d 2u  dτ Pressure gradient in x balanced by
= µ  2 ÷=
dx  dy  dy shear gradient in y
No a so forces must balance!
⌠  2   du 
⌠ dp  µ d u dy dp


dy =   2 ÷ y + A = µ ÷=τ
⌡ dx ⌡  dy 
 dx  dy 
⌠ ⌠  du 
dp  y 2 dp
 y

+ A dy =  µ  ÷

dy + Ay + B = µ u
⌡ dx  ⌡  dy  2 dx
Now we must find A and B… Boundary Conditions

14. Infinite Horizontal Plates:
Boundary Conditions
y
No slip condition
u = 0 at y = 0 and y = a a τ u
y 2 dp x
+ Ay + B = µ u dp
2 dx let negative
be___________
dx
B=0 What can we learn about τ?
a 2 dp − a dp  du  dp
+ Aa = 0 A= µ  ÷= y + A
2 dx 2 dx  dy  dx
y ( y − a ) dp  a  dp
u= τ =y− ÷
2µ dx  2  dx

15. Laminar Flow Between Parallel
Plates
y
a U
ρ a = − ρ g − ∇p + µ∇ 2 V
x
u
0 = − ρ g − ∇p + µ∇ V
2 No fluid particles
are accelerating θ
∂p  ∂ 2u ∂ 2u ∂ 2u 
0 = −ρ gx − + µ  2 + 2 + 2 ÷ Write the x-component
∂x  ∂x ∂y ∂z 
∂p  ∂ 2u 
0 = −ρ gx − + µ  2 ÷
∂x  ∂y 

16. Flow between Parallel Plates
∂p  ∂ 2u 
0 = −ρ gx − + µ  2 ÷ u is only a function of y
∂x  ∂y 
dp  d 2u 
0 = −ρ gx − +µ 2 ÷ g x = g ׈
i
dx  dy 
 d 2u  dp General equation describing laminar
µ  2 ÷= ρ gx +
flow between parallel plates with the
 dy  dx
only velocity in the x direction

17. Flow Between Parallel Plates:
Integration
2
y
d u dp a U
µ = ρ gx +
dy 2 dx
x
⌠ d u 2
⌠ dp 
 µ 2 dy =   ρ g x + ÷dy u
⌡ dy ⌡ dx 
du  dp  θ
µ = y  ρ g x + ÷+ A = τ
dy  dx 
⌠ du ⌠  dp  
 µ dy =   y  ρ g x + ÷+ A ÷dy
⌡ dy ⌡  dx  
y2  dp 
µ u =  ρ g x + ÷+ Ay + B
2  dx 

18. Boundary Conditions
y2  dp 
u µ =  ρ g x + ÷+ Ay + B
2  dx 
Boundary condition u = 0 at y = 0
0 = 0+0+ B
Boundary condition u = U at y = a
a2  dp  Uµ a  dp 
U µ =  ρ g x + ÷+ Aa A= −  ρ gx + ÷
2 dx  a 2 dx 
Uy y 2 − ay  dp 
u= +  ρ gx + ÷
a 2µ  dx 

19. Discharge
y y 2 − ay  dp 
u= U+  ρ gx + ÷
a 2µ  dx 
a
a
⌠ y y 2 − ay  dp  
q = ∫ udy =   U +  ρ g x + ÷÷dy
0 ⌡ a 2µ  dx  
0
Ua a 3  dp 
q= −  ρ gx + ÷ Discharge per unit width!
2 12 µ  dx 

20. Example: Oil Skimmer
An oil skimmer uses a 5 m wide x 6 m long
moving belt above a fixed platform (θ=60º) to
skim oil off of rivers (T=10 ºC). The belt travels
at 3 m/s. The distance between the belt and the
fixed platform is 2 mm. The belt discharges into
an open container on the ship. The fluid is
actually a mixture of oil and water. To simplify
the analysis, assume crude oil dominates. Find
the discharge and the power required to move
the belt. g x h
ρ = 860 kg/m 3
60º l
µ = 1x10-2 Ns/m2

21. g x
Example: Oil Skimmer 60º
Ua a 3  dp  dp
q= −  ρ gx + ÷ =0 g x = g ׈ = g cos(60) = 0.5 g
i
2 12 µ  dx  dx
a = 0.002 m U = 3 m/s
(3 m/s)(0.002 m) (0.002 m)3
q= − ( 0.5) ( 9.806 m/s 2 ) ( 860 kg/m3 ) )
2 12 ( 1x10-2 N × 2 )
s/m
dominates
q = 0.0027 m2/s (per unit width) In direction of belt
Q = 0.0027 m2/s (5 m) = 0.0136 m3/s

22. Example: Oil Skimmer Power
Requirements
 How do we get the power requirement?
___________________________
Power = Force x Velocity [N·m/s]
 What is the force acting on the belt?
 Shear force (τ·L · W)
___________________________
 Remember the equation for shear?
τ=µ(du/dy)
_____________ Evaluate at y = a.
du  dp  Uµ a  dp 
µ = y  ρ g x + ÷+ A A= −  ρ gx + ÷
dy  dx  a 2 dx 
 a  dp  U µ
τ =  y − ÷ ρ g x + ÷+
 2  dx  a

23. Example: Oil Skimmer Power
Requirements
 a  dp  U µ  dp 
τ =  y − ÷ ρ g x + ÷+  ρ g x + ÷ = ρ g cos 60
 2  dx  a  dx 
a Uµ
τ = ρ g cos 60 +
2 a
 3 m  −2 N × 
s
( 0.002 m )  ÷ 1x10 ÷
 860 kg  m2  N
τ=
2
( 9.8 m/s )  m3 ÷( 0.5) +
2
 
 s 
( 0.002 m )
= 19.2 2
m
Power = τ LWU FV (shear by belt on fluid)
19.2 N ( 6 m )( 5 m ) ( 3 m )
Power =  2 = 3.46 kW
 m  s
How could you reduce the power requirement? Decrease τ
__________

24. Example: Oil Skimmer
Where did the Power Go?
 Where did the energy input from the belt
go?
Potential and kinetic energy
Heating the oil (thermal energy)
P = γ Qh Potential energy
 8430 N  0.0136 m ( 3 m )
3
P= 
3 

 m  s 
P = 344 W
h=3m

25. Velocity Profiles
Pressure gradients y y 2 − ay  dp 
uµ = U µ +  ρ gx + ÷
and gravity have a 2  dx 
3
the same effect.
2
In the absence of 1
u (m/s)
pressure gradients 0
oil
and gravity the -1
water
velocity profile is -2
0 0.0005 0.001 0.0015 0.002
________
linear
y (m)

26. Example: No flow
 Find the velocity of a vertical belt that is 5
mm from a stationary surface that will
result in no flow of glycerin at 20°C (m =
0.62 Ns/m2 and ρ =1250 kg/m3)
 Draw the glycerin velocity profile.
 What is your solution scheme?
Ua a 3  dp 
q= −  ρ g y + dy ÷
2 12 µ  

27. Laminar Flow through Circular
Tubes
 Different geometry, same equation
development (see Munson, et al. p 327)
 Apply equation of motion to cylindrical
sleeve (use cylindrical coordinates)

28. Laminar Flow through Circular
Tubes: Equations
r 2 − R2  dp 
vl =  ρ gx + ÷ R is radius of the tube
4µ  dx 
R2  dp 
vmax =−  ρ gx + ÷ Max velocity when r = 0
4µ  dx 
Velocity distribution is paraboloid of
R2  dp 
V =−  ρ gx + ÷ average velocity
revolution therefore _____________
8µ  dx  (V) is 1/2 vmax
_____________
π R4  dp 
Q=−  ρ gx + ÷ Q = VA = VπR2
8µ  dx 

29. Laminar Flow through Circular
Tubes: Diagram
r 2 − R2  dp 
vl =  ρ gx + ÷
4µ  dx 
dvl r  dp  Velocity
=  ρ gx + ÷
dr 2 µ  dx 
Shear (wall on fluid)
Laminar flow
dvl r  dp 
τ =µ =  ρ gx + ÷
dr 2  dx 
Shear at the wall
Next slide!
r  ρ ghl  True for Laminar or ρ ghl d
τ =−  ÷ τ0 = −
2  l  Turbulent flow 4l
Remember the approximations of no shear, no head loss?

30. Relationship between head loss
and pressure gradient for pipes
p1 V12 p2 V22
+ z1 + α1 + hp = + z2 + α 2 + ht + hl cv energy equation
ρ1 g 2g ρ2 g 2g
p1 p2 Constant cross section
+ z1 = + z 2 +hl
ρg
1 ρ2 g
In the energy equation
ρghl =−( p2 − p1 ) −( ρgz2 − ρgz1 )
the z axis is tangent to g
ρg
hl
=−
∆p
− ρg
∆z x is tangent to V
∆x ∆x ∆x z
∆z x
g = gx
hl ∆p  ∆x
ρg =− + ρg x ÷
∆x ∆x 
l is distance between control
hl  dp 
ρg =− + ρg x ÷ surfaces (length of the pipe)
l  dx 

31. The Hagen-Poiseuille Equation
hl  dp  Relationship between head loss
ρg =− + ρg x ÷
l  dx  and pressure gradient
Hagen-Poiseuille Laminar pipe flow equations
π R4  dp  From Navier-Stokes
Q=−  ρ gx + ÷
8µ  dx 
What happens if you double the
π R4  hl  pressure gradient in a horizontal
Q=−  −ρ g l ÷
8µ   tube? ____________
flow doubles
π D 4 ρ ghl D 2 ρ ghl
Q= V= V is average velocity
128µ l 32 µ l

32. Example: Laminar Flow (Team
work)
Calculate the discharge of 20ºC
water through a long vertical section of 0.5
mm ID hypodermic tube. The inlet and outlet
pressures are both atmospheric. You may
neglect minor losses.
What is the total shear force?
What assumption did you make? (Check your
assumption!)

33. Example: Hypodermic Tubing
Flow
p1 V12 p2 V22
+ z1 + α1 + Hp = + z2 + α 2 + H t + hl
γ1 2g γ2 2g
γπD 4 hl
Q=
128µ L
γhl d
afa f
τ0 = −
c h
9806 N / m3 π 0.0005m
4
Q = 158 x10−8 m3 / s
.
4l
Q=
c
128 1x10−3 Ns / m2 h Q = 158µL / s
.
Fshear = −
2πrlγhl d
4l
4Q
V= 2 V = 0.0764m / s
πd Fshear = −π r 2 hl γ
Vd ρ ( 0.0764m / s ) ( 0.0005m ) ( 1000kg / m3 ) = weight!
Re = Re =
µ ( 1x10 −3
Ns / m 2 )
Re = 38

34. Summary
 Navier-Stokes Equations and the Continuity
Equation describe complex flow including
turbulence
 The Navier-Stokes Equations can be solved
analytically for several simple flows
 Numerical solutions are required to describe
turbulent flows

35. Glycerin
Ua a 3  dp 
Q= −  ρ g y + dy ÷
2 12 µ  ρgy +
dp
= ρg
 dy
Ua a 3 ρ g y
0= −
2 12 µ
a2 ρ g
U=
6µ
( 0.005m ) ( 12300 N / m3 )
2
U= = 0.083m / s
6 ( 0.62 Ns / m )
2
ρVl c kg / m h.083m / sfa005mf
a 3
1254 0 0.
R= = = 0.8
µ 0.62 Ns / m2