Shell Momentum Balances
Outline
1.Flow Through a Vertical Tube
2.Flow Through an Annulus
3.Exercises
Flow Through a Vertical Tube
The tube is oriented
vertically.
What will be the
velocity profile of a
fluid whose direction
of flow is in the +zdirection
(downwards)?
Flow Through a Vertical Tube
Same system, but
this time gravity will
also
cause
momentum flux.
Flow Through a Vertical Tube
 ra te of m om en tu m   ra te of m om en tu m 

 
  force of gra vity 
in
by
m
olecu
la
r

ou
t
by
m
olecu
la
r

 
   a ctin g on system   0

 tra n sport
  tra n sport
 

 

pressu re : P A
z0
 PA
zL
net m om entu m flu x :  rz A1
r
  rz A 2
r  r
gravity : +  gV (w h y positive?)
Adding all term s together:
P  2 r  r 
z 0
 P  2 r  r 
z L
  rz  2  rL    rz  2  rL 
r
r  r
  g (2  r  rL )  0
Flow Through a Vertical Tube
P  2 r  r 
z 0
 P  2 r  r 
z L
  rz  2 rL    rz  2 rL 
r
D ividin g b y 2  L  r :
P


z0
P
zL
L
 rz r

r 

r
  rz r
r  r
r
Let  r  0 :
d
 P0  PL 
 rz r    gr  0

r 
L
dr


  gr  0
r  r
  g (2 r  rL )  0
Flow Through a Vertical Tube
 P0  PL

L

d

 rz r    gr  0
r 
dr

R ew ritin g:
 g (0)   gL
 P0  PL

 P0  PL
 g r  

 rz r   
dr
L
L
L



d

r

 P0   g (0) PL   gL

 rz r   
dr
L
L


r

W e let:  z  Pz   g z
 0  L
 rz r   
dr
L

d
d

r

Flow Through a Vertical Tube
 P0  PL
 rz r   
dr
L


r

Flow through a
circular tube
 0  L
 rz r   
dr
L


r

Flow through a
vertical tube
d
d
Flow Through a Vertical Tube
 0  L  2
2
vz  
 R  r 
 4L 
v ave
 0  L

 32L
 2
D

Hagen-Poiseuille
Equation
Outline
1.Flow Through a Vertical Tube
2.Flow Through an Annulus
3.Exercises
Flow Through an Annulus
Liquid is flowing upward
through an annulus (space
between two concentric
cylinders)
Important quantities:
R : radius of outer cylinder
κR : radius of inner
cylinder
Flow Through an Annulus
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only Vz component is
significant
4. At the solid-liquid interface,
no-slip condition
5. Significant gravity effects
6. Vmax is attained at a
distance λR from the
center of the inner cylinder
(not necessarily the center)
Flow Through an Annulus
 ra te of m om en tu m   ra te of m om en tu m 

 
  force of gra vity 
in
by
m
olecu
la
r

ou
t
by
m
olecu
la
r

 
   a ctin g on system   0

 tra n sport
  tra n sport
 

 

pressu re : P A
z0
 PA
zL
net m om entu m flu x :  rz A1
gravity :
r
  rz A 2
r  r
  gV (w h y n ega tive?)
Adding all term s together:
P  2 r  r 
z 0
 P  2 r  r 
z L
  rz  2  rL    rz  2  rL 
r
r  r
  g (2  r  rL )  0
Flow Through an Annulus
 P0  PL

L

d

 rz r    gr  0
r 
dr

R ew ritin g:
 g (0)   gL
 P0  PL

 P0  PL
 g r  

 rz r   
dr
L
L
L



d

r

 P0   g (0) PL   gL

 rz r   
dr
L
L


r

W e let:  z  Pz   g z
 0  L
 rz r   
dr
L

d
d

r

Flow Through an Annulus
 0  L
 rz r   
dr
L

d
S olvin g:
 0  L
 rz r   
dr
L

d

r

 0  L  2
 r  C1
2L


 rz r  
 rz
 0  L

2L

C1

r 
r


r

BOUNDARY CONDITION!
At a distance λR from the center of
the inner cylinder, Vmax is attained in
the annulus, or zero momentum flux.
 0  L
0
2L

C1

 R 
R

2
   L 
 C1    0

R


2
L


Flow Through an Annulus
 0  L    R 
 0  L 
r  
2L



 rz  
2L


2
r
R ew ritin g:
 rz 
 0
 L  R   r 
2  R 
      
2L
 r 
 R 
F rom th e defin ition of flu x:
 rz   
dv z
dr
dv z
dr
 
 0
 L  R  r 
2  R 
      
2L
 r 
 R 
Flow Through an Annulus
dv z
dr
 
 0
 L  R  r 
2  R 
      
2L
 r 
 R 
Solvin g:
2





R
 0
1r 
L 
2
vz  
 
   R ln  r   C 2 
2L
R  2 

Flow Through an Annulus
vz  
 0

 L  R  1  r 2 
2
 
   R ln  r   C 2 
2L
R  2 

Rew riting:
2
2





R
 0
 r 
 R
L 
2
vz  
    2  ln  r    C 2 
4L
 R
  R 

2

 0   L  R  r 
2
vz  
   2
4L
  R 
2
Take out R/2
Multiply r in log term
by R/R (or 1)

  r 

Expand log term
 ln    ln ( R )   C 2 
 R


2
2





R
 0
 r 
 r 
L 
2
vz  
    2  ln    C 2 
4L
R
  R 

Lump all constants
into C2
Flow Through an Annulus
2


 0   L  R  r 
 r 
2
vz  
    2  ln    C 2 
4L
R
  R 

2
We have two unknown constants: C2 and λ
We can use two boundary conditions:
No-slip Conditions
At r = κR, vz = 0
At r = R, vz = 0
Flow Through an Annulus
vz  
 0
2
2

 L  R  r 
 r 
2
    2  ln    C 2 
4L
R
  R 

U sin g B .C . # 1 :
0
0
 0
 L  R
2
  2  2  2 ln     C 2 


4L
2
 2  ln     C 2
2
 C2  1
U sin g B .C . # 2 :
0  
 0
0  1  C2
 L  R
4L
2
2
1  C 2 
2
 2 
 1
ln  

Flow Through an Annulus
vz  
 0
2
2

 L  R  r 
 r 
2
    2  ln    C 2 
4L
R
  R 

2
2




R
 1
 0
 r 
L 
vz  
   
4L
  R 
 ln 
2
 C2  1
2
2
 2 
 1
ln  


  r 
 ln    1 

 R
Shell Balances
1. Identify all the forces that influence the flow
(pressure, gravity, momentum flux) and their
directions. Set the positive directions of your axes.
2. Create a shell with a differential thickness across the
direction of the flux that will represent the flow
system.
3. Identify the areas (cross-sectional and surface areas)
and volumes for which the flow occurs.
4. Formulate the shell balance equation and the
corresponding differential equation for the
momentum flux.
Shell Balances
5. Identify all boundary conditions (solid-liquid, liquidliquid, liquid-free surface, momentum flux values at
boundaries, symmetry for zero flux).
6. Integrate the DE for your momentum flux and
determine the values of the constants using the BCs.
7. Insert Newton’s law (momentum flux definition) to
get the differential equation for velocity.
8. Integrate the DE for velocity and determine values of
constants using the BCs.
9. Characterize the flow using this velocity profile.
Shell Balances
Important Assumptions*
1. The flow is always assumed to be at steadystate.
2. Neglect entrance and exit effects. The flow is
always assumed to be fully-developed.
3. The fluid is always assumed to be
incompressible.
4. Consider the flow to be unidirectional.
*unless otherwise stated
Design Equations for Laminar
and Turbulent Flow in Pipes
Outline
1.Velocity Profiles in Pipes
2.Pressure Drop and Friction Loss (Laminar
Flow)
3.Friction Loss (Turbulent Flow)
4.Frictional Losses in Piping Systems
Velocity Profiles in Pipes
Recall velocity profile in a circular tube:
 P0  PL  2
2
vz  
 R  r 
 4L 
1. What is the shape of this profile?
2. The maximum occurs at which region?
3. What is the average velocity of the fluid
flowing through this pipe?
Velocity Profiles in Pipes
Velocity Profiles in Pipes
Velocity Profile in a Pipe:
 P0  PL  2
2
vz  
 R  r 
 4L 
Average Velocity of a Fluid in a Pipe:
v ave
 P0  PL

 32L
 2
D

Maximum vs. Average Velocity
Outline
1.Velocity Profiles in Pipes
2.Pressure Drop and Friction Loss (Laminar
Flow)
3.Friction Loss (Turbulent Flow)
4.Frictional Losses in Piping Systems
Recall: Hagen-Poiseuille
Equation
v ave
 P0  PL

 32L
 2
D

Describes the pressure drop and flow of
fluid (in the laminar regime) across a
conduit with length L and diameter D
Hagen-Poiseuille Equation
P0  PL 
3 2  Lv ave
D
2
Pressure drop / Pressure loss (P0 – PL):
Pressure lost due to skin friction
Friction Loss
P0  PL 
3 2  Lv ave
D
In terms of energy
lost per unit mass:
2
Ff 
PO  PL


3 2  L v ave
D
2
Mechanical energy lost due to friction in
pipe (because of what?)
Friction Factor
Definition: Drag force per wetted surface
unit area (or shear stress at the surface)
divided by the product of density times
velocity head
f 
S
 v
2
2


  P0  PL  AC  A S


 v
2
2

Friction Factor
Ff
gc
 4 fF
L v
2
D 2gc
Frictional force/loss head is proportional
to the velocity head of the flow and to
the ratio of the length to the diameter of
the flow stream
Friction Factor for Laminar Flow
Consider the Hagen-Poiseuille equation
(describes laminar flow) and the
definition of the friction factor:
v ave
 P0  PL

 32L
Prove: fF 
 2
D

16
N Re
Ff
gc

PO  PL
g
 4 fF
L v
2
D 2gc
Valid only for laminar flow
Outline
1.Velocity Profiles in Pipes
2.Pressure Drop and Friction Loss (Laminar
Flow)
3.Friction Loss (Turbulent Flow)
4.Frictional Losses in Piping Systems
Friction Factor for Turbulent
Flow
Ff
gc
 4 fF
L v
2
D 2gc
1. Friction factor is dependent on NRe and
the relative roughness of the pipe.
2. The value of fF is determined
empirically.
Friction Factor for Turbulent
Flow
How to compute/find the value of the friction factor for
turbulent flow:
1. Use Moody diagrams.
- Friction factor vs. Reynolds number with a series of
parametric curves related to the relative roughness
2. Use correlations that involve the friction factor f.
- Blasius equation, Colebrook formula, Churchill
equation (Perry 8th Edition)
Moody Diagrams
Important notes:
1. Both fF and NRe are plotted in logarithmic scales.
Some Moody diagrams show fD (Darcy friction
factor). Make the necessary conversions.
2. No curves are shown for the transition region.
3. Lowest possible friction factor for a given NRe in
turbulent flow is shown by the smooth pipe line.
Friction Factor Correlations
1. Blasius equation for turbulent flow in smooth
tubes:
fF 
0.079
N Re
0.25
4000  N Re  10
2. Colebrook formula
1
fD
 
2 .5 1
  2 log 1 0 

 3 .7 D N
f
Re
D





5
Friction Factor Correlations
3. Churchill equation (Colebrook formula explicit in fD)
1
fD
  2 log 1 0

0 .2 7   7


N
 D
 Re

4. Swamee-Jain correlation
0 .2 5
fD 
2 log 1 0
 
5 .7 4 


0 .9 
N Re 
 3 .7 D



0 .9




Equivalent Roughness, ε
Materials of Construction
Copper, brass, lead (tubing)
Commercial or welded steel
Wrought iron
Ductile iron – coated
Ductile iron – uncoated
Concrete
Riveted Steel
Equivalent Roughness (m)
1.5 E-06
4.6 E-05
4.6 E-05
1.2 E-04
2.4 E-04
1.2 E-04
1.8 E-03
Frictional Losses for Non-Circular
Conduits
Instead of deriving new correlations for f, an approximation
is developed for an equivalent diameter, Deq, which may be
used to calculate NRe and f.
D eq  4 R H  4
where
S
Pw
RH = hydraulic radius
S = cross-sectional area
Pw = wetted perimeter: sum of the length
of the boundaries of the cross-section
actually in contact with the fluid
Equivalent Diameter (Deq)
D eq  4 R H  4
S
Pw
Determine the equivalent diameter of the
following conduit types:
1. Annular space with outside diameter Do and
inside diameter Di
2. Rectangular duct with sides a and b
3. Open channels with liquid depth y and liquid
width b