Shell Momentum Balances
Outline
1.Convective Momentum Transport
2.Shell Momentum Balance
3.Boundary Conditions
4.Flow of a Falling Film
5.Flow Through a Circular Tube
Convective Momentum Transport
Recall: MOLECULAR MOMENTUM TRANSPORT
Convective Momentum
Transport: transport of
momentum by bulk
flow of a fluid.
Outline
1.Convective Momentum Transport
2.Shell Momentum Balance
3.Boundary Conditions
4.Flow of a Falling Film
5.Flow Through a Circular Tube
Shell Momentum Balance
 rate of momentum   rate of momentum

 

in
by
convective

out
by
convective

 

 transport
  transport


 

 rate of momentum   rate of momentum 

 
  force of gravity 
 in by molecular    out by molecular    acting on system   0

 transport
  transport
 

 

1. Steady and fully-developed flow is assumed.
2. Net convective flux in the direction of the flow
is zero.
Outline
1.Convective Momentum Transport
2.Shell Momentum Balance
3.Boundary Conditions
4.Flow of a Falling Film
5.Flow Through a Circular Tube
Boundary Conditions
Recall: No-Slip Condition (for fluid-solid interfaces)
Additional Boundary Conditions:
For liquid-gas interfaces:
“The momentum fluxes at the free liquid
surface is zero.”
For liquid-liquid interfaces:
“The momentum fluxes and velocities at the
interface are continuous.”
Flow of a Falling Film
z
y
x
Liquid is flowing down an inclined
plane of length L and width W.
δ – film thickness
Vz will depend on x-direction only
Why?
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only Vz component is significant
4. At the gas-liquid interface, shear rates are negligible
5. At the solid-liquid interface, no-slip condition
6. Significant gravity effects
Flow of a Falling Film
τij  flux of j-momentum in
the positive i-direction
y
z
x
y
z
L
x
τxz ǀ x + δ
τxz ǀ x
δ
W
Flow of a Falling Film
τij  flux of j-momentum in
the positive i-direction
y
z
x
y
z
L
x
τyz ǀ y=0
δ
τyz ǀ y=W
W
Flow of a Falling Film
τij  flux of j-momentum in
the positive i-direction
y
z
x
τzz ǀ z=0
y
z
L
x
δ
τzz ǀ z=L
W
ρg cos α
Flow of a Falling Film
 rate of momentum   rate of momentum 

 
  force of gravity 
in
by
molecular

out
by
molecular

 
   acting on system   0

 transport
  transport
 

 

P(W∙δ)|z=0 – P(W∙δ)|z=L +
(τxzǀ x )(W*L) – (τxz ǀ x +Δx )(W∙L) +
(τyzǀ y=0 )(L*δ) – (τyz ǀ y=W )(L∙δ) +
(τzz ǀ z=0)(W* δ) – (τzz ǀ z=L)(W∙δ) +
(W∙L∙δ)(ρgcos α) = 0
Dividing all the terms by W∙L∙δ and noting that the direction of flow is along z:
𝜏𝑥𝑧|𝑥+𝛿 − 𝜏𝑥𝑧|𝑥
= 𝜌𝑔 cos 𝛼
𝛿
Flow of a Falling Film
𝜏𝑥𝑧|𝑥+𝛿 − 𝜏𝑥𝑧|𝑥
= 𝜌𝑔 cos 𝛼
∆𝑥
If we let Δx  0,
𝑑(𝜏𝑥𝑧 )
= 𝜌𝑔 cos 𝛼
𝑑𝑥
Integrating and using the boundary conditions to evaluate,
Boundary conditions:
@ x = 0 𝜏𝑥𝑧 = 0
x = x 𝜏𝑥𝑧 = 𝜏𝑥𝑧
𝜏𝑥𝑧 = 𝜌𝑥𝑔 cos 𝛼
Flow of a Falling Film
𝜏𝑥𝑧 = 𝜌𝑥𝑔 cos 𝛼
For a Newtonian fluid, Newton’s law of viscosity is
𝜏𝑥𝑧
𝑑𝑣𝑧
= −𝜇
𝑑𝑥
Substitution and rearranging the equation gives
𝑑𝑣𝑧
𝜌𝑔 cos 𝛼
=−
𝑥
𝑑𝑥
𝜇
Flow of a Falling Film
𝑑𝑣𝑧
𝜌𝑔 cos 𝛼
=−
𝑥
𝑑𝑥
𝜇
Solving for the velocity,
Boundary conditions:
@ x = δ, v z = 0
𝜌𝑔 cos 𝛼 2
𝑣𝑧 = −
𝑥 + 𝐶2
2𝜇
𝜌𝑔𝛿 2 cos 𝛼
𝑥 2
𝑣𝑧 =
(1 − ( ) )
2𝜇
𝛿
Flow of a Falling Film
𝜌𝑔𝛿 2 cos 𝛼
𝑥 2
𝑣𝑧 =
(1 − ( ) )
2𝜇
𝛿
Compute for the following:
Average Velocity:
v z  v z , ave
W

0
0
v dA   v dxdy





  dA   dxdy
z
z
W
0
0
How does
this profile
look like?
Flow of a Falling Film
𝜌𝑔𝛿 2 cos 𝛼
𝑥 2
𝑣𝑧 = −
(1 − ( ) )
2𝜇
𝛿
Compute for the following:
Mass Flowrate:
m   v z dA  W v z
Flow Between Inclined Plates
z
x
θ
L
δ
Derive the velocity profile of the fluid inside the two stationary plates.
The plate is initially horizontal and the fluid is stationary. It is suddenly
raised to the position shown above. The plate has width W.
Outline
1.Convective Momentum Transport
2.Shell Momentum Balance
3.Boundary Conditions
4.Flow of a Falling Film
5.Flow Through a Circular Tube
Flow Through a Circular Tube
Liquid is flowing across a pipe of
length L and radius R.
Assumptions:
1. Steady-state flow
2. Incompressible fluid
3. Only Vx component is significant
4. At the solid-liquid interface, no-slip condition
Recall: Cylindrical Coordinates
Flow Through a Circular Tube
 rate of momentum   rate of momentum 

 
  force of gravity 
in
by
molecular

out
by
molecular

 
   acting on system   0

 transport
  transport
 

 

pressure : PA z  0  PA z  L
net momentum flux :  rz A1 r   rz A2 r  r
Adding all terms together:
P  2 r r  z  0  P  2 r r  z  L   rz  2 rL  r   rz  2 rL  r  r  0
Flow Through a Circular Tube
P 2 r r  z 0  P 2 r r  z L   rz 2 rL  r   rz 2 rL  r r  0
Dividing by 2 Lr :
 P z  0  P z  L   rz r r   rz r r  r
0

 r 
L
r


Let x  0 :
d
 P0  PL 

 r   rz r   0
dr
 L 
Flow Through a Circular Tube
d
 P0  PL 

 r   rz r   0
 L  dr
Solving:
d
 P0  PL 
 rz r   
r
dr
 L 
P P 
 rz r   0 L  r 2  C1
 2L 
C1
 P0  PL 
 rz  
r 
r
 2L 
BOUNDARY CONDITION!
At the center of the pipe, the flux is
zero (the velocity profile attains a
maximum value at the center).
C1
00
r
C1 must be
zero! 
Flow Through a Circular Tube
 P0  PL 
 rz  
r
 2L 
From the definition of flux:
dv z
 rz   
dr
Plugging in:
 P0  PL 
dv z
 
r
dr
 2L 
 P0  PL  2
vz   
 r  C2
 4 L 
BOUNDARY CONDITION!
At r = R, vz = 0.
P P 
0    0 L  R 2  C2
 4 L 
P P 
C2   0 L  R 2
 4 L 
 P0  PL  2  P0  PL  2
vz   
r  
R
 4L 
 4L 
Flow Through a Circular Tube
 P0  PL  2 2
vz  
 R  r 
 4L 
Compute for the following:
Average Velocity:
v z  v z , ave
2
R
0
0
v dA   v rdrd





  dA   rdrd
z
z
2
R
0
0
Hagen-Poiseuille Equation
vave
 P0  PL  2

D
 32L 
Describes the pressure drop and flow of
fluid (in the laminar regime) across a
conduit with length L and diameter D
What if…?
The tube is oriented
vertically.
What will be the
velocity profile of a
fluid whose direction
of flow is in the +zdirection
(downwards)?