# Fluid Flow Concepts and Basic Control Volume Equations

```Basic Governing Differential
Equations
CEE 331
March 16, 2016
Monroe L. Weber-Shirk
School of Civil and
Environmental Engineering
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)
Why Differential Equations?
A droplet of water
Clouds
Wall jet
Hurricane
Conservation of Mass in
Differential Equation Form
 IF v I
F
  y J
v  y J
G
G
H y KH y Kxz
Mass flux out of differential volume

 y xz
t
y
x
vxz
Rate of change of mass in
differential volume
z
Mass flux into differential volume
Continuity Equation
Mass flux out of differential volume
v

v 
F
I
v   y  v y 
y J
xz
G
H y y y y K
2
out
in
Higher order term
Rate of mass decrease
v
 I

F
v   y  v yJ
xz  vxz   yxz
G
H y y K
t
v


 v  
y
y
t
v 

0
y t
1-d continuity equation
Continuity Equation
af af a f
  u  v  w



0
t
x
y
z
divergence

    V  0 Vector notation
t
3-d continuity equation
u, v, w are
velocities in x, y,
and z directions
If density is constant...
u v w
 
0
x y z
or in vector notation
True everywhere! (contrast with CV equations!)
V  0
Continuity Illustrated
u v w
 
0
x y z
y
What must be happening?
v
&lt;0
y
\
u
&gt;0
x
x
Navier-Stokes Equations
 Derived by Claude-Louis-Marie Navier in 1827
 General Equation of Fluid Motion
 Based on conservation of ___________
momentum with forces…
Gravity
 ____________
 ___________________
Pressure
Shear
 ___________________
solution of the Navier-Stokes Equations a top priority
Navier-Stokes Equations
V a   F
a   g  p   V
2
a    p   g   2 V
Navier-Stokes Equation
g is constant
a is a function of t, x, y, z
a  Inertial forces [N/m3], a is Lagrangian acceleration
Is acceleration zero when V/  t = 0?
NO!
  p   g   Pressure gradient (not due to change in elevation)
p   g  0 then _____
V 0
If _________
 V  Shear stress gradient
2
du
 
dx
  V
Notation: Total Derivative
Eulerian Perspective
D
 dt  dx  dy  dz
(t , x, y, z ) 



Dt
t dt x dt y dt z dt
D




(t , x, y, z ) 
u
v
w
Dt
t
x
y
z
DV
V
V
V
V
(t , x, y, z ) 
u
v
w
Dt
t
x
y
z
V
V
V
V
a
u
v
w
t
x
y
z
V
a
 V V
t
Total derivative
(chain rule)
Material or
substantial derivative
Lagrangian acceleration
 ()  ()  ()
() 
i
j
k
x
y
z
 ()
 ()
 ()
 V   ()  u  v  w
x
y
z
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.
Navier-Stokes Equations: A
Simple Case
No acceleration and no velocity gradients
a    p   g   2 V
0    p   g 
p    g
p
  gx
x
g
xyz could have any orientation
p
  g y
y
p
p
0
  g
x
y
p    gy  C
p
   g z Let y be vertical upward
z
Component of g in the x,y,z direction
p
0
z
For constant 
Infinite Horizontal Plates:
Laminar Flow
Derive the equation for the laminar, steady, uniform flow
between infinite horizontal parallel plates.
a    p   g   2 V
y
0    p   g    2 V
  2u 
p
0    2 
x
 y 
v =0
2
2
2
Hydrostatic in y
 v  v  v
p
y 0    gy    2  2  2 
p
y

x

y

z
0    g


y
w
=
0
2
2
2


p

w

w

w
z 0     gz   
 2  2
00
2
z
y
z 
 x
  2u  2u  2u 
p
x 0     gx    2  2  2 
x
z 
 x y
x
Infinite Horizontal Plates:
Laminar Flow
  2u 
p
0    2 
x
 y 
 d 2u  d
dp
   2 
dx
 dy  dy




  2 
dp
d u

dy     2  dy

dx
  dy 

 y


  du 
dp

 A dy     dy
dx

  dy 
du
 
dy
Pressure gradient in x balanced by
No a so forces must balance!
 du 
dp
y  A   
dx
 dy 
y 2 dp
 Ay  B   u
2 dx
Now we must find A and B… Boundary Conditions
Infinite Horizontal Plates:
Boundary Conditions
y
No slip condition
u = 0 at y = 0 and y = a

u
x
y 2 dp
 Ay  B   u
2 dx
dp
let
negative
be___________
dx
What can we learn about ?
B0
a 2 dp
 Aa  0
2 dx
y  y  a  dp
u
2  dx
a
 a dp
A
2 dx
 du 
dp
   y  A
dx
 dy 


a  dp
 y 
2 dx

Laminar Flow Between Parallel
Plates
a    g  p  2 V
0    g  p  2 V
No fluid particles
are accelerating
  2u  2u  2u 
p
0   gx     2  2  2 
x
z 
 x y
  2u 
p
0   gx     2 
x
 y 
q
Write the x-component
Flow between Parallel Plates
  2u 
p
0   gx     2 
x
 y 
 d 2u 
dp
0   gx     2 
dx
 dy 
 d 2u 
dp
  2    gx 
dx
 dy 
u is only a function of y
g x  g  ˆi
General equation describing laminar
flow between parallel plates with the
only velocity in the x direction
Flow Between Parallel Plates:
Integration
d 2u
dp
 2   gx 
dy
dx
dp 
 d 2u

  2 dy     g x   dy
dx 

 dy
du
dp 


 y   g x    A t
dy
dx 


 
dp 
 du

dy

y

g


A
dy

 x




dx 
 dy
 

y2 
dp 
u    g x    Ay  B
2 
dx 
q
Boundary Conditions
y2 
dp 
u     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 
U     g x    Aa
2
dx 
Uy y 2  ay 
dp 
u

  gx  
a
2 
dx 
U a 
dp 
A
   gx  
a
2
dx 
Discharge
y
y 2  ay 
dp 
u U
  gx  
a
2 
dx 
a
 y
y 2  ay 
dp  
q   udy    U 
  g x    dy
2 
dx  
 a
0
a
0
Ua a 3 
dp 
q


g

 x

2 12 
dx 
Discharge per unit width!
Example: Oil Skimmer
An oil skimmer uses a 5 m wide x 6 m long
moving belt above a fixed platform (q=60&ordm;) to
skim oil off of rivers (T=10 &ordm;C). 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
3
 = 860 kg/m
60&ordm;
 = 1x10-2 Ns/m2
l
h
g
Example: Oil Skimmer
Ua a 3 
dp 
q

  gx  
2 12 
dx 
dp
0
dx
60&ordm;
x
g x  g  ˆi  g cos(60)  0.5 g
a  0.002 m
U  3 m/s
(3 m/s)(0.002 m)
(0.002 m)3
2
3
q

0.5
9.806
m/s
860
kg/m
)




-2
2
2
12 1x10 N  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
Example: Oil Skimmer Power
Requirements
How do we get the power requirement?
Power = Force x Velocity [N&middot;m/s]
___________________________
What is the force acting on the belt?
 Shear
___________________________
force ( &middot; L &middot; W)
Remember the equation for shear?
 = (du/dy)
_____________
Evaluate at y = a.
U a 
dp 
du
dp 

A
   gx  

 y   gx    A
a
2
dx 
dy
dx 

a 
dp  U 

   y    g x   
2 
dx  a

Example: Oil Skimmer Power
Requirements
a 
dp  U 

   y    g x   
2 
dx  a

dp 

  g x     g cos 60
dx 

a
U
   g cos 60 
2
a
 3 m 
2 N  s 
1x10


2 
0.002
m


860
kg
N
s
m




2 

9.8 m/s  
 19.2 2

  0.5  
3
2
m
 0.002 m 
 m 
Power   LWU
FV
N
3 m 

Power  19.2 2 6 m 5 m 
s
m 

(shear by belt on fluid)
= 3.46 kW
How could you reduce the power requirement? Decrease
__________

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   Qh Potential energy
N 
m3 

P   8430 3  0.0136 3 m 
s 
m 

h=3m
P  344 W
Velocity Profiles
y
y 2  ay 
dp 
u  U  

g

 x

a
2 
dx 
and gravity have
the same effect.
3
In the absence of
and gravity the
velocity profile is
________
linear
u (m/s)
2
1
0
oil
water
-1
-2
0
0.0005
0.001
y (m)
0.0015
0.002
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&deg;C (m = 0.62
Ns/m2 and  =1250 kg/m3)
Draw the glycerin velocity profile.
Ua a 3 
dp 
q

gy  

2 12 
dy 
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)
Laminar Flow through Circular
Tubes: Equations
r 2  R2 
dp 
vl 
  gx  
4 
dx 
vmax
R2 
dp 

  gx  
4 
dx 
R is radius of the tube
Max velocity when r = 0
R2 
dp 
V 
  gx  
8 
dx 
Velocity distribution is paraboloid of
average velocity
revolution therefore _____________
(V) is 1/2 vmax
_____________
 R4 
dp 
Q

g

 x

8 
dx 
Q = VA = VR2
Laminar Flow through Circular
Tubes: Diagram
r 2  R2 
dp 
vl 
  gx  
4 
dx 
dvl
r 
dp 

  gx  
dr 2  
dx 
Velocity
Shear (wall on fluid)
dvl r 
dp 
 
   gx  
dr 2 
dx 
Next slide!
r   ghl  True for Laminar or
  

2  l  Turbulent flow
Laminar flow
Shear at the wall
0  
 ghl d
4l
Remember the approximations of no shear, no head loss?
p1
V12
p2
V22
 z1  1
 hp 
 z2   2
 ht  hl cv energy equation
1 g
2g
2 g
2g
p1
p2
 z1 
 z2  hl
1 g
2 g
 ghl
In the energy equation
   p2  p1     gz2   gz1 
the z axis is tangent to g
hl
p
z
g

 g
x
x
x
g
g
Constant cross section
hl
 p

 
  gx 
x
 x

hl
 dp

 
  gx 
l
 dx

x is tangent to V
z
z
x
g
 gx
x
l is distance between control
surfaces (length of the pipe)
The Hagen-Poiseuille Equation
hl
 dp

g
 
  gx 
l
 dx

Hagen-Poiseuille Laminar pipe flow equations
 R4 
dp 
Q
  gx  
8 
dx 
hl 
 R4 
Q
  g 
8 
l 
 D 4  ghl
Q
128 l
From Navier-Stokes
What happens if you double the
tube? ____________
flow doubles
D 2  ghl
V
32 l
V is average velocity
Example: Laminar Flow (Team
work)
Calculate the discharge of 20&ordm;C
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!)
Example: Hypodermic Tubing
Flow
V12
p2
V22
 z1  1
 Hp 
 z2   2
 H t  hl
1
2g
2
2g
D 4 hl
Q
128 L
p1
9806 N / m h
 0.0005mf
afa
c
Q
128c
1x10 Ns / m h
4
3
3
4Q
V 2
d
Re 
Vd 

Re  38
Q  158
. x10 m / s
3
2
V  0.0764m / s
Re 
8
Q  158
. L / s
Fshear
4l
2rlhl d

4l
Fshear   r 2 hl 
3
0.0764
m
/
s
0.0005
m
1000
kg
/
m




1x10
0  
hl d
3
Ns / m 2 
= weight!
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
Glycerin
Ua a 3 
dp 
Q

gy  

2 12 
dy 
Ua a 3  g
0

2
12
dp
gy   g
dy
y
a2  g
U
6
2
0.005
m

 12300 N / m3 
U
 0.083m / s
2
6  0.62 Ns / m 
1254 kg / m h
0.083m / sfa
0.005mf
a
Vl c
R

 0.8
3

0.62 Ns / m2
```