Introduction
Advanced Transport Phenomena
Introduction
• (Video) lectures (3 hrs) + tutorial sessions (3 hrs)
• Book: ‘Transport Phenomena’, second edition by Bird,
Stewart and Lightfoot  BSL: SLIDES + ADDITIONAL
MATERIAL AVALABLE ON CANVAS
• Assessment
• Exam/assignments (75%) + assignment on generalized Fourier
methods (25%) performed in groups of 2 persons
• follow lectures (development of mathematical toolkit, CANVAS)
• serious preparation of tutorials essential (CANVAS)
J.A.M. Kuipers
• Prior knowledge necessary:
• Introduction to Physical Transport Phenomena
• Mathematics: knowledge about and ability to solve ordinary
differential equations
Introduction
• Further study
• ‘Boundary Layer Theory’ (Schlichting)
• ‘Conduction of Heat in Solids’ (Carslaw/Jager)
• ‘Mathematics of Diffusion’ (Crank)
Introduction
• BSL: structure of content in table 0.2-1 (p. 4)
• Columns: quantity that is transported
• Momentum
• Energy
• Mass
• Rows: type of transport
•
•
•
•
•
•
•
•
Transport by molecular motion
Transport in one dimension
Transport in arbitrary continua
Transport with two independent variables
Transport in turbulent flow
Transport across phase boundaries
Transport in large systems
Transport by other mechanisms
BSL 1.1: Newton’s law of viscosity
• Viscosity characterizes the
resistance of a fluid against
flow
• Fluid contained between
two flat plates.
• At t=0, the lower plate is
suddenly brought into
motion
• Definition for μ:
Advanced Transport Phenomena
BSL 1: Viscosity and the
mechanisms of momentum
transport
J.A.M. Kuipers
dvx
dy
• Only valid for Newtonian
fluids
τ yx = − µ
BSL 1: Non-Newtonian fluids
BSL 1.1: Newton’s law of viscosity
Rheological models: two types of empirical models
• two-parameter models
• three-parameter models
Newton’s law from a different point of view
Shear stress
Momentum flux
Rheological behavior of most Newtonian fluids can be
described as:
τ yx = −η
dvx
dy
where η is the effective viscosity which depends on dvx/dy or
τyx
6
BSL 1: Non-Newtonian fluids
BSL 1: Non-Newtonian fluids
• Pseudoplastic materials:
η decreases with increasing shear rate (-dvx/dy)
• Dilatant materials:
η increases with increasing shear rate (-dvx/dy)
BSL 1: Non-Newtonian fluids
Two-parameter models:
• Bingham model (e.g. suspensions):
dv
0 dy
x
τ yx =
−µ
± τ if τ yx > τ
0
0
dvx
=
0 if τ yx < τ
0
dy
• Ostwald-De Waele model (power-law model):
τ yx = −m
dvx
dy
n −1
dvx
dy
BSL 1: Non-Newtonian fluids
• Eyring-model:
 1 dvx 

 B dy 
τ yx =
A ⋅ arcsinh  −
this model follows from Eyring’s kinetic theory for fluids
Three-parameter models:
• Ellis-model:
α −1
dv
− x = ϕ0 + ϕ1 τ yx
τ yx
dy
(
)
BSL 1: Non-Newtonian fluids
• Reiner-Philippoff model:

dv 
1
− x=

τ
dy  µ∞ + ( µ0 − µ∞ ) / (1 + (τ yx / τ s ) 2 )  yx
• Rheological behavior in unsteady state:
Advanced Transport Phenomena
BSL 2: Shell momentum balances
and velocity distributions in laminar
flow
J.A.M. Kuipers
especially polymers often exhibit time-dependent
rheological behaviour
BSL 2: Velocity distributions in laminar flow
• In general: theory has in fact already been treated in
Introduction to Physical Transport Phenomena
 assumed to be known
BSL 2.1: Shell momentum balance and BC’s
• Momentum balance in words (Newton’s 2nd law):
rate of
accumulation


of momentum
in system

sum of

rate
of
rate
of





 the forces 
 
 
 

 = momentum  − momentum  + 

 in
 out
 acting on 
 
 
 
 the system 
• Boundary conditions (S: solid wall, L: liquid and G: gas)
• SL-interface: fluid velocity equals the velocity of the solid wall (noslip condition)
• LG-interface: stress-tensor components are assumed to be zero
• LL-interface: continuity of velocity (no-slip condition) and the
stress components
BSL 2.2-2.4
BSL 2.5: Flow of two adjacent non-miscible fluids
• BSL 2.2 Flow of a falling film: Introduction to Physical
Transport Phenomena
• BSL 2.3 Flow through a circular tube: Introduction to
Physical Transport Phenomena
• BSL 2.4 Flow through an annulus: Introduction to
Physical Transport Phenomena
Differential z-momentum balance (valid for phase I and II):
dτ xz  dp  p0 − pL
=
−  =
dx  dz 
L
BSL 2.5: Flow of two adjacent non-miscible fluids
Integration of differential momentum balance:
p − pL 
I
τ xzI  0
 x + C1
 L 
p − pL 
II
τ xzII  0
 x + C1
 L 
Continuity of stress at the L-L interface:
=
x 0=
: τ xzI τ xzII
In other words:
I
II
C=
C=
C1
1
1
BSL 2.5: Flow of two adjacent non-miscible fluids
Combination with (2.5-2) and (2.5-3) and applying Newton’s
viscosity law gives:
dvzI  p0 − pL 
−µ = 
 x + C1
dx  L 
dvzII  p0 − pL 
− µ II =

 x + C1
dx  L 
Integration of (2.5-5) and (2.5-6) gives:
I
C
p −p 
vzI =
−  0 I L  x 2 − 1I x + C2I
µ
 2µ L 
C
p −p 
vzII =
−  0 II L  x 2 − II1 x + C2II
µ
 2µ L 
BSL 2.5: Flow of two adjacent non-miscible fluids
Three integration constants  three additional BC’s:
: vzI vzII
=
x 0=
: vzI 0
=
x -b=
0
x=
+b : vzII =
Application of the boundary conditions gives:
C2I = C2II
C
p −p 
−  0 I L  b 2 + 1I b + C2I
0=
µ
 2µ L 
C
p −p 
0=
−  0 II L  b 2 − II1 b + C2II
µ
 2µ L 
BSL 2.5: Flow of two adjacent non-miscible fluids
Velocity profiles:
2
p0 − pL ) b 2  2 µ I   µ I − µ II   x   x  
(
v
=
+
 −  

2 µ I L  µ I + µ II   µ I + µ II   b   b  
2
p0 − pL ) b 2  2 µ II   µ I − µ II   x   x  
(
v
=
+ I
−  
 I
II
II 
II  
2µ L
This results in:
I
II
 p − pL   µ − µ 
C1 = −  0
b
  I
II 
 2L   µ + µ 
 p − p   2µ I 
C2I =
C2II
+  0 I L  b2  I
=
II 
 2µ L   µ + µ 
Momentum flux or shear stress profile:
p − pL   x  1  µ I − µ II  
=
τ xz  0
 b   −  I
II  
 L   b  2  µ + µ  
BSL 2.6: Flow around a sphere
• From (2.6-13) follows for the kinetic force:
I
z
II
z
BSL 2.5: Flow of two adjacent non-miscible fluids
 µ + µ   µ + µ   b   b  
Limiting case of idential phase viscosities:
Velocity profiles in I and II are the same (parabolic profile)
Fk = 6πµ Rv∞
which is Stokes’s law, introduced during Introduction to
Physical Transport Phenomena
• Condition: Re < 0.1
• Importance of Stokes’ law:
• Colloidal systems
• Fluidization
• Sedimentation
BSL 3: Introduction
Advanced Transport Phenomena
BSL 3: The equations of change for
isothermal systems
• Derivation of general microbalances for isothermal
systems consisting of a single component  very
important chapter!
• Non-isothermal systems?  BSL 11
J.A.M. Kuipers
• Reactive (multi-component) systems?  BSL 19
• In BSL 3 derivation of
• Micro balances for mass
• Micro balances for momentum
• Micro balances for mechanical energy
BSL 3.1: Equation of continuity
• Mass balance over a volume element ΔV=ΔxΔyΔz, fixed
in space:
• Mass balance in words:
 rate of  rate of  rate of 

 
 

increase  =  mass  −  mass 
 of mass   in   out 

 
 

BSL 3.1: Equation of continuity
BSL 3.1: Equation of continuity
• Mass balance:
∂ρ
=∆y∆z [( ρ vx ) |x − ( ρ v ) |x+∆x ] +
∂t
∆x∆z ( ρ v y ) | y − ( ρ v y ) | y +∆y  + ∆x∆y [( ρ vz ) |z − ( ρ vz ) |z +∆z ]
∆x∆y∆z
• Dividing by ΔV=ΔxΔyΔz and taking the limit as Δx, Δy
and Δz approach zero gives:
∂ρ
∂
∂
∂

=
−  ( ρ vx ) + ( ρ v y ) + ( ρ vz ) 
∂t
∂y
∂z
 ∂x

BSL 3.1: Equation of continuity
• Alternative way of writing the equation of continuity using
the substantial derivative towards time:
∂v
∂v
∂
∂ρ
∂ρ
( ρ ) + vx + ρ x + v y + ρ y
∂t
∂x
∂x
∂y
∂y
∂vz ∂
∂ρ
+ vz
+ ρ=
⋅v ) 0
( ρ ) + ( v ⋅∇ρ ) + ρ ( ∇=
∂z
∂z ∂t
or:
Dρ
1 Dρ
= − ρ ( ∇ ⋅ v ) or:
= − (∇ ⋅ v )
Dt
ρ Dt
• Continuity equation for incompressible fluid:
(∇ ⋅ v ) = 0
BSL 3.1: Equation of continuity
• Written in compact vector notation:
∂ρ
= − (∇ ⋅ ρv )
∂t
with (∇ ⋅ ρ v ) the divergence of ρ v
• Physical meaning: −(∇ ⋅ ρ v ) is the net rate of mass
addition per unit volume by convection
BSL 3.2: Equation of motion
Momentum balance over a volume element ΔV=ΔxΔyΔz,
fixed in space:
• Momentum balance in words:
accumulation
of momentum


in ∆V per
unit of time
 inflow of
 momentum
 
= 
 per unit
 of time
 outflow of  sum of
 momentum   the forces
 
 
−
+
per
unit
 
 acting
 of time
 on ∆V






BSL 3.2: Equation of motion
BSL 3.2: Equation of motion
• Note: momentum is a vectorial quantity thus momentum
balance is a vector equation with x-,y- and z-components
• We consider the x-component only (analogy for y- and zdirection)
• Net convective transport of x-momentum:
∆y∆z ( ρ vx vx |x − ρ vx vx |x +∆x ) +
∆x∆z ( ρ v y vx | y − ρ v y vx | y +∆y ) +
∆x∆y ( ρ vz vx |z − ρ vz vx |z +∆z )
BSL 3.2: Equation of motion
• Net molecular transport of x-momentum:
∆y∆z (τ xx |x −τ xx |x +∆x ) +
∆x∆z (τ yx | y −τ yx | y +∆y ) +
∆x∆y (τ zx |z −τ zx |z +∆z )
• Net force in the x-direction acting on ΔV:
• Pressure force:
∆y∆z ( p |x − p |x+∆x )
• Force due to gravity
ρ g x ∆x∆y∆z
BSL 3.2: Equation of motion
• Accumulation of momentum per unit of time:
∂
∆x∆y∆z  ρ vx 
 ∂t

• Filling in all the terms, dividing by ΔV and taking the limit
as Δx, Δy and Δz approach zero gives:
∂
∂
∂
∂

−  ( ρ vx vx ) + ( ρ v y vx ) + ( ρ vz vx ) 
( ρ vx ) =
∂t
∂y
∂z
 ∂x

∂
∂
∂
 ∂p
−  (τ xx ) + (τ yx ) + (τ zx )  − + ρ g x
∂y
∂z
 ∂x
 ∂x
BSL 3.2: Equation of motion
• Analogously for the y- and z-direction gives:
∂
∂
∂
∂

ρ vy ) =
−  ( ρ vx v y ) + ( ρ v y v y ) + ( ρ vz v y ) 
(
∂t
∂y
∂z
 ∂x

BSL 3.2: Equation of motion
• Notation in compact vector notation + physical
interpretation of the terms (BSL (3.2-9)):
∂
∂
∂
 ∂p
−  (τ xy ) + (τ yy ) + (τ zy )  − + ρ g y
∂y
∂z
 ∂x
 ∂y
and:
∂
∂
∂
∂

−  ( ρ vx vz ) + ( ρ v y vz ) + ( ρ vz vz ) 
( ρ vz ) =
∂t
∂y
∂z
 ∂x

∂
∂
∂
 ∂p
−  (τ xz ) + (τ yz ) + (τ zz )  − + ρ g z
∂y
∂z
 ∂x
 ∂z
BSL 3.2: Equation of motion
• Alternative way of writing the equation of x-momentum
using the substantial derivative towards time:
ρ
∂τ
Dvx
∂τ 
∂p  ∂τ
=
− −  xx + yx + zx  + ρ g x
Dt
∂x  ∂x
∂y
∂z 
• Analogous equations are valid for the y- and z-direction
• Compact vector notation + physical interpretation:
BSL 3.2: Equation of motion
• From this form it is clear that the Navier-Stokes
equations are a representation of Newton’s 2nd law:
mass per unit of volume x accelaration = sum of the
forces per unit of volume
• For closure of the micro momentum balance a
specification of the components of the tensor τ is
required
BSL 3.2: Equation of motion
BSL 3.2: Equation of motion
Components of τ (for Newtonian fluids):
• Normal components:
• Tangential components:
∂v 2
τ xx = −2 µ x + µ ( ∇ ⋅ v )
∂x 3
∂v 2
τ yy = −2 µ y + µ ( ∇ ⋅ v )
∂y 3
∂v 2
τ zz = −2 µ z + µ ( ∇ ⋅ v )
∂z 3
τ yz =
τ zy =
−µ  y + z 
 ∂z ∂y 
 ∂v
 ∂v
Combination of the equations for τ with the micro balances
for x-, y- and z-momentum gives:
• x-component
∂v 
∂
∂
∂
∂
ρ vx + ( ρ vx vx ) + ( ρ v y vx ) + ( ρ vz vx ) =
∂t
∂x
∂y
∂z
∂v 
τ xy =
τ yx =
−µ  x + y 
 ∂y ∂x 
∂v
 ∂x

τ zx =
τ xz =
−µ  z +
∂vx 

∂z 
−
∂p ∂  ∂vx 2
 ∂   ∂v ∂v  
+  2µ
− µ ( ∇ ⋅ v ) +  µ  x + y 
∂x ∂x 
∂x 3
 ∂y   ∂y ∂x  
+
∂   ∂vx ∂vz  
µ
+
 + ρ gx
∂z   ∂z ∂x  
16
BSL 3.2: Equation of motion
• y-component
∂
∂
∂
∂
ρ v y + ( ρ vx v y ) + ( ρ v y v y ) + ( ρ vz v y ) =
∂t
∂x
∂y
∂z
−

∂p ∂   ∂v y ∂vx   ∂  ∂v y 2
+ µ 
+
+  2µ
− µ ( ∇ ⋅ v )


∂y ∂x   ∂x ∂y   ∂y 
∂y 3

+
∂   ∂vz ∂vx  
µ
+
 + ρgy
∂z   ∂x ∂z  
BSL 3.2: Equation of motion
• z-component
∂
∂
∂
∂
ρ vz + ( ρ vx vz ) + ( ρ v y vz ) + ( ρ vz vz ) =
∂t
∂x
∂y
∂z
∂p ∂   ∂vz ∂vx   ∂   ∂vz ∂v y  
µ
µ
+
+
+
 +
∂z ∂x   ∂x ∂z   ∂y   ∂y ∂z  
∂  ∂v 2

+  2µ z − µ ( ∇ ⋅ v ) + ρ g z
∂z 
∂z 3

−
• In practice, drastic simplification of the general micro
balances is often allowable
BSL 3.2: Equation of motion
Limiting situations
• Constant density (ρ) and viscosity (μ):
∂
( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p + µ∇ 2v + ρ g
∂t
(Navier-Stokes equations) with ∇ 2 the Laplace-operator:
2
2
2
∂
∂
∂
+ 2+ 2
2
∂x ∂y ∂z
Notation of convective transport term:
2
∇=
 ρ vx vx
ρ vv =  ρ v y vx
 ρ vz vx
ρ vx v y
ρ vy vy
ρ vz v y
ρ vx vz 
ρ v y vz 
ρ vz vz 
BSL 3.3: Equation of mechanical energy
Micro balance for momentum in vector form:
D
v = −∇p − ( ∇ ⋅τ ) + ρ g
Dt
Take the dot product with velocity v :
ρ
D 1 2 
 v  = − ( v ⋅∇p ) − ( v ⋅ [∇ ⋅τ ]) + ρ ( v ⋅ g )
Dt  2 
Physical meaning:
ρ
 distance 
force per  


 coverered 
)
 unit of  × 
=( v=
 volume   per unit 

 
 of time 
 work done 
per unit of 




 time and 
 volume 
BSL 3.2: Equation of motion
• Ideal fluid
[ ∇ ⋅τ ] = 0
∂
( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p + ρ g (Euler equation)
∂t
• Description of the compressible flow of a isothermal
flowing fluid:
•
•
•
•
Micro balance for mass
Micro balance for momentum
Equation of state p = p ( ρ )
Equation for pressure dependency of viscosity µ = µ ( ρ )
BSL 3.3: Equation of mechanical energy
Using the definition of the substantial derivative towards time
and manipulating ( v ⋅ [∇ ⋅τ ]) and ( v ⋅∇p ) :
BSL 3.3: Equation of mechanical energy
For Newtonian fluids, the term ( −τ : ∇v ) can be written as a
sum of squares and is therefore always positive:

1 3 3  ∂v ∂v  2
( −τ : ∇v ) = µΦ v = µ=i∑1 =j∑1  i + j  − ( ∇ ⋅ v ) δ ij 
2
 ∂x j ∂xi  3

2
Describing the degradation of mechanical energy into
thermal energy (viscous dissipation heating)
• Importance at flow with very high velocities (“space
shuttle”) and other situations in which very large
velocity gradients occur
• If the term ( −τ : ∇v ) would not be there, full reversible
interconversion of the various forms of energy would be
possible
BSL 3: Micro balances for momentum and mass in
curvilinear coordinates
Important coordinates for practical applications:
• Cartesian coordinates
• Cylindrical coordinates
• Spherical coordinates
Mathematical details of the transformation of coordinates
 BSL Appendix A
BSL 3.3: Equation of mechanical energy
The term( p ( −∇ ⋅ v ) ) describes the reversible conversion of
mechanical energy into thermal energy per unit of volume
and time
• Important for compressible fluids that undergo sudden
and/or large pressure changes (f.i. compressors and
turbines)
BSL 3: Micro balances for momentum and mass in
curvilinear coordinates
What is needed in order to transform the equations?
• Equations for the gradient-operator
• Equations for the derivatives of the unit vector towards the new
coordinates
Micro balances in different coordinate systems
• Micro balance for mass  BSL §B.4
• Micro balance for momentum in terms of τ  BSL §B.5
• Micro balance for momentum in terms of velocity gradients  BSL
§B.6
• Components of the viscous tension tensor τ  BSL §B.1
• Viscous dissipation term ( −τ : ∇v ) =µΦ v for Newtonian fluids  BSL
§B.7
BSL 3: Analysis of flow phenomena using general
micro balances
BSL 3: Applications
BSL 2: analysis of flow phenomena based on differential
micro balances
• Flow of a falling film
Disadvantages
• Flow of a Newtonian fluid through a circular tube
• Momentum balance has to be set up for every single problem
• Limited application (rectilinear flow patterns!)
Alternative: analysis of flow phenomena based in (reduced)
micro balances
Advantages:
• Generally applicable
• Reduction of general micro balances ‘automatically’ generates a list
of assumptions made
• Flow of a Newtonian between two concentric cylinders
(Couette viscosity meter)
• Flow of an Newtonian fluid between two concentric
spheres
BSL 3: Flow of a falling film
Use z-momentum balance based on geometry:
∂v
∂v
∂v 
 ∂v
∂p
z +v
z +v
z=
ρ z +v
− +
 ∂t
x ∂x
y ∂y
z ∂z 
∂z


 ∂ 2v
∂ 2v
∂ 2v 
z+
z+
z  + ρg
µ
z
2
2
 ∂x 2
∂y
∂z 


BSL 3: Flow of a falling film
Reduced momentum balance:
0= µ
∂ 2 vz
∂ 2 vz
g
+
=
+ ρ g cos ( β )
ρ
µ
z
∂x 2
∂x 2
Since vz is merely a function of x, the reduced momentum
balance can be rewritten into:
0 µ
=
d 2 vz
+ ρ g cos ( β )
dx 2
Boundary conditions:
dvz
=
at x 0:=
0
dx
=
at x δ=
: vz 0
Solution:
ρ gδ 2 cos ( β )   x 2 
=
vz
1 −   
2µ
 δ  
31
BSL 3: Flow of a Newtonian fluid through a circular
tube
Use z -momentum balance based on geometry:
∂v
∂v
∂v 
 ∂v z
∂p
z +v
z +v
z=
+v
− +
 ∂t
θ ∂θ
r ∂r
z ∂z 
∂z



∂ 2v
∂ 2v 
∂
v
∂
1
1


z
z  + ρg
z
+ 2
+
µ
r


z
 r ∂r  ∂r  r ∂θ 2 ∂z 2 


BSL 3: Flow of a Newtonian fluid through a circular
tube
Reduced momentum balance:
1 ∂  ∂vz 
∂p
0=
− +µ
r
 + ρ gz
∂z
r ∂r  ∂r 
ρ
Since vz is merely a function of r, the reduced momentum
balance can be rewritten into:
1 d  dvz 
dP
−
+µ
0=
r

dz
r dr  dr 
Boundary conditions:
dvz
=
at r 0:=
0
dr
=
at r R=
: vz 0
Solution:
2
P0 − PL ) R 2   r  
(
=
vz
1 −   
4µ L

R 
33
BSL 3: Flow between two concentric cylinders
• Applicable for e.g. bearings, Couette viscosity meter
BSL 3: Flow between two concentric cylinders
r -momentum balance:
 ∂vr
ρ
+ vr
2
∂vr vθ ∂vr vθ
∂v 
∂p
1 ∂ 2 v 2 ∂v ∂ 2 v 
 ∂ 1 ∂
+
− + vz z  =
− +µ 
( rvr )  + 2 2r − 2 θ + 2r  + ρ g r
∂r r ∂θ r
∂z 
∂r
r ∂θ ∂z 
 r ∂θ
 ∂r  r ∂r
 ∂t
reduced r -momentum balance:
vθ2
∂p
=
−
r
∂r
θ -momentum balance:
−ρ
∂v 
∂vθ
∂v v ∂v v v
1 ∂p
1 ∂ 2v
2 δ v ∂ 2v 
∂ 1∂
−
+ µ  
+ vr θ + θ θ + r θ + vz θ  =
( rvθ )  + 2 θ2 + 2 r + 2θ  + ρ gθ
∂z 
r
r ∂θ
r δθ ∂z 
∂r r ∂θ
 r ∂θ
 ∂t
 ∂r  r ∂r
reduced θ -momentum balance:
ρ 
d 1 d
( rvθ ) 

dr  r dr

z -momentum balance:
0=
∂vz
∂v v ∂v
∂v 
∂p
 1 ∂  ∂vz  1 ∂ 2 vz ∂ 2 vz 
+ vr z + θ z + vz z  =
− +µ
+ 2  + ρ gz
r
+ 2
2
∂z 
∂r r ∂θ
∂z 
∂z
 ∂t
 r ∂r  ∂r  r ∂θ
ρ 
• Flow in the θ-direction where vθ is a function or radius r
reduced z -momentum balance:
∂p
0=
− + ρ gz
∂z
BSL 3: Flow between two concentric cylinders
Boundary conditions for the θ -momentum balance:
=
R: vθ 0
at r κ=
at r = R: vθ = Ω0 R
With Ω0 the rotation speed of the outer cylinder
Integration of the reduced θ -momentum balance gives:
κR − r 


r κR
vθ = Ω0 R 
κ − 1 


κ

BSL 3: Flow between two concentric cylinders
Application: Couette viscosity meter
• From measurements of M and at known geometry (κ, R
and L) and known rotation speed Ω0, the viscosity of the
fluid can be determined
Pressure as a function of r and z?
• Integration of the equation for the total differential of p in
which first the reduced momentum balances in the rand z-direction are substituted
BSL 3: Flow between two concentric cylinders
BSL gives the momentum density in the radial direction:
 ∂ v 
τ rθ = − µ  r  θ  
 ∂r  r  
Substitution in the equation for vθ gives:
1  κ 2 

τ rθ =
−2 µΩ0 R  2  
2 
 r 1− κ 
2
Moment M that must be applied on the outer cylinder:
=
M 2π RL ( −τ rθ |r = R ) R
BSL 3: Flow through two concentric spheres
assumption: creeping
flow conditions
BSL 3: Flow between two concentric spheres
θ -momentum balance:
2
vφ
 ∂vθ
∂v v ∂v
∂vθ vr vθ vφ cot (θ ) 
+ vr θ + θ θ +
+
−
=
sin
θ
θ
φ
∂
∂
∂
∂
t
r
r
r
r
r
(
)


ρ
BSL 3: Flow between two concentric spheres
Equation of continuity (with vr = 0):
=
0
∂
1 ∂  2 ∂vr 
1
( vθ sin (θ ) )
r
+
2
r ∂r  ∂r  r sin (θ ) ∂θ
∂
1
 1 ∂  2 ∂vθ  1 ∂  1

=
∂
=
( vθ sin (θ ) ) 0
sin
θ
v
(
)
(
)
+ 2
 2 r

θ


∂
r
θ
θ
sin
(
)
∂
∂
∂
∂
sin
θ
θ
θ
r
r
r
r
(
)


1 ∂p

 
−
+µ
2
Note: the quantity vθ sin (θ ) is independent of θ and is therefore


r ∂θ
∂ vθ 2 ∂vr 2 cot (θ ) ∂vφ
1
+
+
−
+
g
ρ
 2 2
θ
2
merely a function of r!
r 2 ∂θ r 2 sin 2 (θ ) ∂φ
 r sin (θ ) ∂φ

=
u ( r ) vθ ( r ,θ ) ⋅ sin (θ ) and substitute this result in the
Define
reduced θ -momentum balance:
 1 ∂  ∂v
1 ∂p
−
+ µ  2  r2 θ
0=
r ∂θ
 r ∂r  ∂r

 1 ∂  1 ∂
( vθ sin (θ ) )  
+ 2

 r ∂θ  sin(θ ) ∂θ

Note: vθ is function or r as well as θ!
θ -momentum balance:

1 ∂p
1
d  2 du  
−
+µ 2
0=
r

r ∂θ
 r sin (θ ) dr  dr  
BSL 3: Flow between two concentric spheres
In this situation, p is merely a function of θ and u is merely
a function of r which means that the reduced momentum
balance can be separated!
Separated equations:
dp
sin (θ )
=B
dθ
and:
µ d  2 du 
r
=B
r dr  dr 
where B is a separation constant that has yet to be determined
BSL 3: Flow between two concentric spheres
Integration of the latter two equations using cos (θ ) = x gives:
1 − cos ( ε ) 
 1- x 
∆p = B ⋅ ln 
=
⋅
ln
B
1 + cos ε  = − B ⋅ E ( ε )
( )
1 + x 

Boundary conditions:
at θ = ε : p = p0 and at θ = π − ε : p = p1 ( ∆p = p0 − p1 )
and:
=
u
R ⋅ ∆p  r 
 R 
1 −  + κ 1 −  

2 µ E ( ε )  R 
 r 
Boundary conditions:
at
r κ R=
: u 0 and at
r R=
:u 0
=
=
BSL 3: Flow between two concentric spheres
Equation for the volume flow rate
( ∑i ( 2π ri ∆r ) ui ):
π
π R 3 ∆p
3

2π rdr
1− κ )
Q ∫=
vθ  r ,θ
=
=
(

κR
2
6µ E (ε )

R
Advanced Transport Phenomena
BSL 4: Velocity distributions with
more than one independent
variable
J.A.M. Kuipers
BSL 4: Velocity distributions with more than one
independent variable
In practice, velocity profiles depend on more than one
independent variable!
•Unsteady flow
•Two-dimensional flow
Contents of this chapter:
• Unsteady flow: unsteady flow near a wall suddenly set in motion
and unsteady laminar flow between two parallel plates
• Stationary viscous flow in two dimensions (stream function)
• Stationary ideal flow in two dimensions (potential flow)
• Boundary-layer theory
BSL 4: Velocity distributions with more than one
independent variable
By looking at certain examples, some of the widely used
techniques for solving (complex non-linear) partial
differential equations (PDE) are presented:
•Method of combination of variables (similarity solutions)
•Method of separation of variables
•Stream function and velocity potential
•Boundary layer approaches (approximate solutions)
BSL 4.1: Time-dependent flow of Newtonian fluids
Flow near a wall suddenly set in motion
Initially the fluid and the wall
are at rest.
At t=0, the wall is suddenly
set in motion.
BSL 4.1: Time-dependent flow of Newtonian fluids
Reduced x-momentum balance:
∂vx
∂ 2 vx
=ν 2
∂t
∂y
with ν = kinematic viscosity
Initial and boundary conditions:
at t ≤ 0, vx =
0 for all y
=
vx V for all t > 0
at y 0,=
∞, v x =
at y =
0 for all t > 0
BSL 4.1: Time-dependent flow of Newtonian fluids
Try a solution of the form:
vx ( y , t )
y
φ=
=
(η ) with η
V
4ν t
Transformation of the derivative towards time
and derivative towards spatial co-ordinate:
∂  vx  ∂
1η
  = (φ ) = − φ '
∂t  V  ∂t
2t
and:
∂ 2  vx  ∂ 2
1
=
=
φ
φ"
(
)


∂y 2  V  ∂y 2
4ν t
BSL 4.1: Time-dependent flow of Newtonian fluids
Substitution of this result in in the PDE:
0
φ "+ 2ηφ ' =
Integration using φ ' = F :
F= φ=' C1e −η
2
and again integration:
η
2
φ C1 ∫ e −η dη + C2
=
o
Application of the boundary conditions:
η
∫e
−η 2
1 − ∞0 2
φ=
∫e
0
−η
dη
dη
2 η −η 2
1−
erfc (η )
=
∫ e dη =
π
0
BSL 4.1: Time-dependent flow of Newtonian fluids
BSL 4.1: Time-dependent flow of Newtonian fluids
or:
vx
 y 
 y 
=
1 − erf 
erfc 
=

V
 4ν t 
 4ν t 
In correspondence with solution presented in Introduction
to Physical Transport Phenomena
Graphical solution in which the power-law model is also
displayed with
m =µ
ρ
BSL 4.1: Time-dependent flow of Newtonian fluids
Unsteady laminar flow between two parallel plates
Initially the fluid is at rest.
At t=0, a constant pressure
 dp p0 − pL 
gradient  − =

 dx
L


is imposed over the plate.
The fluid will accelerate and
eventually, a stationary velocity
profile will be established
BSL 4.1: Time-dependent flow of Newtonian fluids
Use dimensionless quantities:
Dimensionless velocity φ:
∂v
∂v
∂p
ρ x=
− + µ 2x
∂t
∂x
∂y
vx
vx
=
φ =
2
∞
( p0 − pL ) d ( vx )max
2µ L
Initial condition:
Dimensionless time τ :
at=
t 0, - d ≤ y ≤ d : =
vx 0
v ⋅t
τ= 2
d
Boundary conditions
Dimensionless y -coordinate ξ :
at t > 0, y = -d : vx = 0
y
at t > 0, y =
+ d : vx =
0
ξ=
d
Reduced x-momentum balance:
2
BSL 4.1: Time-dependent flow of Newtonian fluids
Combine with the dimensionless quantities gives:
steady state
∂φ
∂φ
= 2+ 2
∂τ
∂ξ
2
φ → φ∞
d 2φ∞
0
+2=
dξ 2
with dimensionless initial and boundary conditions:
at τ = 0, − 1 ≤ ξ ≤ 1: φ = 0
BSL 4.1: Time-dependent flow of Newtonian fluids
Try a solution of the following form:
φ (=
ξ ,τ ) φ∞ (ξ ) − φt (ξ ,τ )
solution = "steady solution" - "unsteady solution"
Substitute this “trial” solution in the dimensionless PDE
• ODE for steady solution
d 2φ∞
0
+2=
dξ 2
• PDE for unsteady solution
∂φt ∂ 2φt
=
∂τ ∂ξ 2
−1: φ =
at τ > 0, ξ =
0
+1: φ =
at τ > 0, ξ =
0
BSL 4.1: Time-dependent flow of Newtonian fluids
BSL 4.1: Time-dependent flow of Newtonian fluids
The boundary conditions for the stationary solution φ∞ and
the unsteady solution φt remain unchanged, however…….
Solution of PDE for φt: try a solution of the form:
Initial condition for φt:
Combination with dimensionless PDE for φt and division by
gives:
at τ = 0, − 1 ≤ ξ ≤ 1:
φt = φ∞
F ' (τ ) G " (ξ )
=
F (τ ) G (ξ )
Solution of ODE for φ∞:
φ∞ = 1 − ξ
2
check:
φ (ξ=
,τ ) F (τ ) ⋅ G (ξ )
d 2φ∞
+ 2 =−2 + 2 =0
dξ 2
Note: left side of the equation is independent of ξ and
identically equal to the right side of the equation which
is independent of τ
BSL 4.1: Time-dependent flow of Newtonian fluids
BSL 4.1: Time-dependent flow of Newtonian fluids
Left and right side of the equation equal a constant (the
so-called separation constant):
Because of symmetry: C1=0
F ' (τ ) G " (ξ )
=
= −µ 2
F (τ ) G (ξ )
Separated equations:
Based on the boundary conditions for ξ=±1:
1
C2 cos ( µ ) =0 → µn = n +  π
2

with: n = 0,1, 2 and µn the so-called eigenvalues and
F '+ µ 2 F =
0
G "+ µ 2G =
0
the corresponding eigenfunctions cos(µnξ )
General solution for φt:
e − µ τ [C1 sin ( µξ ) + C2 cos ( µξ )]
where C1 and C2 are integration constants
φt
2
BSL 4.1: Time-dependent flow of Newtonian fluids
Resulting solution for φt :
(φt )n K n e ((
) )
2
− n+ 1 π τ
2
1


cos   n +  πξ 
2 

1


with cos   n +  πξ  the eigenfunctions and K n constants
2 

that follow "in theory" from the initial condition:
1


1 ξ 2 =K n cos   n +  πξ 
φ∞ =−
2
 

Problem: the initial condition φ∞ = 1 − ξ 2 can not be described
1


by a single eigenfunction cos   n +  πξ !
2 

BSL 4.1: Time-dependent flow of Newtonian fluids
Solution: try to realize a description of the initial condition based
on the sum of all available eigenfunctions:
n =∞ 
1


φ∞ =−
1 ξ 2 =∑  K n cos   n +  πξ  
n =0
2




1 

Multiply the right and left side by cos   m +  πξ  d ξ ( m = 0,1, 2,..)
2 

and integrate from ξ =
−1 to ξ =
+1:
1
∫
-1
=
(1 − ξ ) cos   m + 12  πξ dξ
2



1
1
 1



 
K n ∫ cos   n +  πξ  ⋅ cos   m +  πξ d ξ 

n =0
2 
2  

 −1  
n =∞
∑
BSL 4.1: Time-dependent flow of Newtonian fluids
BSL 4.1: Time-dependent flow of Newtonian fluids
Evaluation: use orthogonality of eigenfunctions:
Solution to the original problem:
1 
1 


∫ cos   n +  πξ  ⋅ cos   m +  πξ d ξ =0 if m ≠ n
−1
2 
2 


and:


2
n
1




 4 ( −1)
−   n + π  τ
n=
∞
1   
2

 2 
φ (ξ ,τ ) =(1 − ξ ) − ∑ 
e
cos
n
+
 πξ  

3
n =0
2

 

1




 n+ π




  

2 
1
1 
1 


cos   n +  πξ  ⋅ cos   m +  πξ d ξ =1 if m =n
−1
2 
2 


Combination of results gives the following expression for K n :
1
∫
Note: for t→∞, the stationary solution (parabolic velocity
profile) is obtained!
4 ( −1)
1 

Kn =
∫ (1 − ξ ) cos   n +  πξ d ξ =
3
−1
2 

1 

  n + π 
2 

1
n
2
BSL 4.1: Time-dependent flow of Newtonian fluids
BSL 4.4: Boundary-layer theory
Absence of molecular momentum transport (μ=0) in the
bulk of a real fluid is often a reasonable approximation but
is not applicable for the analysis of transport phenomena
in the immediate neighborhood of a wall
boundary-layer theory is used
Boundary-layer theory: molecular transport dominates in
the immediate neighborhood of the wall and the complete
velocity gradient is situated in a very thin boundary layer
near the wall (thickness=δ)
BSL 4.4: Boundary-layer theory
BSL 4.4: Boundary-layer theory
Unsteady boundary-layer flow
Viscous effects only play a role inside the boundary layer
A fluid is set into motion by a suddenly moving solid wall
Assumption:the dimensionless velocity profiles are
“self similar” during development in time t:
Qualitative development of the velocity profile and
approximation with the boundary-layer approach:
vx
y
=
φ=
(η ) with: η
V
δ (t )
Reduced x-momentum balance:
∂vx
∂ 2v
= ν 2x
∂t
∂y
23
BSL 4.4: Boundary-layer theory
Evaluation of the derivatives in terms of the yet to be
determined functions φ and η:
∂vx
y dδ
η dδ
=
V φ '  − 2 
=
V φ '  − 
∂t
 δ  dt
 δ  dt
and:
∂ vx
1
= V φ " 2 
2
∂y
δ 
Combination with the reduced x-momentum balance gives:
2
{δ δ }{ ηφ }
d
dt
−
' =
vφ "
BSL 4.4: Boundary-layer theory
Integrate this equation from η=0 to η=1:
{ }{
δ
dδ
dt
1
∫ −ηφ 'dη
0
}
1
=
v ∫ φ "dη
0
Define the constants M and N as follows:
1
1
1
1
0
0
0
M=
−η φ + ∫ φ dη =
∫ −ηφ ' dη =
∫ φ dη
0
N
=
1
η
∫ φ " d=
0
1
' φ ' (1) − φ ' ( 0 )
φ=
0
Combine this with the found ODE for δ gives after
integration with initial condition δ=0 at t=0:
δ = 2 ( N M )ν t
BSL 4.4: Boundary-layer theory
BSL 4.4: Boundary-layer theory
Before proceeding, an assumption must be made regarding
the dimensionless velocity distribution ! vx V = φ (η )
Try a polynomial expression of the following form:
vx
=φ (η ) =a0 + a1η + a2η 2 + a3η 3
V
Combination of the equation for φ(η) with the equations for
M and N gives M=3/8 and N=3/2 and finally:
δ = 2 2ν t
Compare this result to the results found at Introduction to
Physical Transport Phenomena:
Boundary conditions:
δ = 2 πν t
=
φ ( 0 ) 1,=
φ (1) 0, =
φ ' (1) 0 and =
φ "( 0) 0
tutorials: extension of boundary layer approach to a
power-law fluid
This gives:
φ (η ) =1 − 32 η + 12 η 3 ( 0 ≤ η ≤ 1)
dv
τ yx = −m x
dy
BSL 4.4: Boundary-layer theory
n −1
dvx
dy
BSL 4.4: Boundary-layer theory
Steady two-dimensional boundary layer flow along a flat plate
Mathematical description
• Equation of continuity
∂vx ∂v y
+
=
0
∂x ∂y
• x-momentum balance
vx
∂vx
∂v
 ∂ 2v ∂ 2v 
+ v y x = ν  2x + 2x 
∂x
∂y
∂y 
 ∂x
This leads to 2 equations for vx and vy which can in principle
be solved after specification of suitable boundary conditions:
In this steady system, the thickness of the boundary layer
increases with increasing x
29
=
at x 0:=
vx v∞
=
y 0:=
vx 0
at
=
at y δ=
( x ): vx v∞
BSL 4.4: Boundary-layer theory
BSL 4.4: Boundary-layer theory
Using the equation of continuity, vy can be expressed in
terms of vx using vy=0 at y=0:
Note: PDE is non-linear and 1st order in x and 2nd order in
y  find an approximate solution!
y ∂v


v y = − ∫  x  dy
0  ∂x 
Substitution of this in in the x-momentum balance gives:
Assumption: the dimensionless velocity profiles are “self
similar” with increasing distance x:
vx
∂vx  y  ∂vx   ∂vx
∂ 2 vx
− ∫ 
=
dy
ν
 
∂x  0  ∂x   ∂y
∂y 2
vx
y
φ=
=
(η ) with: η
v∞
δ ( x)
in which convective transport of momentum in the xdirection is assumed to dominate the molecular transport
of x-momentum in the x-direction
BSL 4.4: Boundary-layer theory
Evaluation of the terms from the x-momentum balance in
terms of φ(η):
∂vx
y dδ
η dδ
v∞φ '  − 2 
v∞φ '  − 
=
=
∂x
 δ  dx
 δ  dx
∂vx
∂ 2 vx
1
1
'
v=
v∞φ " 2 
φ
 
∞
2
∂y
∂y
δ 
δ 
Substitution of these equations in the x-momentum balance:
η dδ  η 
η dδ 
1


v∞φ v∞φ '  −   −  ∫  v∞φ '  − 
δ dη  v∞φ '  

 δ  dx   0 
 δ  dx 
δ 



 1 
= ν v∞φ "  2  
 δ 

BSL 4.4: Boundary-layer theory
or:
ν
φ ' η∫ ηφ ' dη − ηφφ ' δ d δ =
φ"
 0
 dx v
∞
Integrate this equation with respect to η:
ν 1
 1∫ φ ' η∫ uφ 'du  dη − 1∫ ηφφ ' dη  δ d δ =
φ " dη

 0  0
 dx v∞ 0∫
0
or in compact notation:
dδ ν
C
( B − A) δ =
dx v∞
1
1
with: =
A ∫ ηφφ ' dη
C ∫ φ=
"dη φ ' (1) − φ ' ( 0 )
=
0
0
η
η
η =1
1
1
1
B =∫ φ '  ∫ uφ ' du dη =φ  ∫ uφ ' du  − ∫ ηφφ ' dη =∫ ηφ ' dη − A




0
0
0

0
 η =0 0
BSL 4.4: Boundary-layer theory
Integration of the equation for the boundary layer thickness
with boundary condition δ=0 at x=0:
C ν x

 B − A  v∞
δ ( x ) = 2 
Before proceeding, an assumption regarding the
dimensionless velocity vx v∞ = φ (η ) must be made!
Try a polynomial expression of the following form:
vx
=φ (η ) =a0 + a1η + a2η 2 + a3η 3
v∞
Boundary conditions:
=
φ ( 0 ) 0,=
φ (1) 1,=
φ ' (1) 0 and =
φ "( 0) 0
BSL 4.4: Boundary-layer theory
Using the approximate solution for vx, the drag force Fx
can be obtained from:
Fx =−
2 ∫ ∫ ( τ yx )
W L
0 0
y =0
dxdz =
1.292 ρµ LW 2 v∞3
Exact numerical solution (Blasius): replace the constant
1.292 by 1.328!!!
BSL 4.4: Boundary-layer theory
This gives:
φ (η=
)
3
2
η − 12 η 3
Substitution of this equation for φ(η) in the equations for A,
B and C gives A=9/35, B=33/280 and C=-3/2
Boundary layer thickness:
=
δ
280 ν x
νx
= 4.64
13 v∞
v∞
Velocity profile within the boundary layer (0≤y≤δ(x)):
vx
=
v∞

 1

y
y
3
−




2
2
 4.64 ν x v∞   4.64 ν x v∞ 
3
BSL 4.4: Boundary-layer theory
Generalization of the boundary-layer theory to flows with a
more complex geometry (e.g. airplane wing)
BSL Fig. 4.4-1
Orthogonal co-ordinate system !
x in tangential direction along
the surface
y in normal direction to
the surface
38
BSL 4.4: Boundary-layer theory
Qualitative: in contrast (due to curvature) to the flat plate
(no curvature), pressure changes occur outside the
boundary layer that are imposed on the boundary layer
 dp∞ 
 − ∂p 
=


−

 ∂x  boundary layer  dx  y =δ ( x )
Mathematical description on basis of micro-balances
•equation of continuity
∂vx ∂v y
+
=
0
∂x ∂y
•x-momentum balance
∂vx
∂vx
1 ∂p
 ∂ 2 vx ∂ 2 vx 
vx
+ vy
=
−
+ν  2 + 2 
∂x
∂y
ρ ∂x
∂y 
 ∂x
BSL 4.4: Boundary-layer theory
Eliminate vy from the x-momentum balance by integration of
the equation of continuity and neglect molecular transport in
comparison with convective transport in the x-direction:
∂vx  y  ∂vx   ∂vx
1 ∂p
 ∂ 2 vx 
vx
− ∫ 
=
−
+ν  2 
 dy 
ρ ∂x
∂x  0  ∂x   ∂y
 ∂y 
Outside the boundary layer: ideal flow for which the Bernoulli:
equation holds:
1
2
ρ v∞2 ( x ) + p∞ ( x ) =
constant
Differentiation with respect to x gives:
dv dp
ρ v∞ ∞ + ∞ =
0
dx
dx
BSL 4.4: Boundary-layer theory
Combination of these equations with the x-momentum
balance in the boundary layer gives the Prandtl boundary
layer equation:
∂vx  y  ∂vx   ∂vx
dv∞
1 ∂p
 ∂ 2 vx 
 ∂ 2 vx 
vx
v∞
− ∫ 
=
−
+ν  2  =
+ν  2 
 dy 
ρ ∂x
dx
∂x  0  ∂x   ∂y
 ∂y 
 ∂y 
Limitation: curvature of surface should not be too large
because in this case separation occurs and therefore the
boundary layer concept will break down
This is due to the fact that the velocity profile in the
boundary layer ceases to satisfy the “similarity” condition
(flow reversal occurs)
BSL 4.4: Boundary-layer theory
Integration of Prandtl boundary layer equation with respect
to y gives the famous von Kármán momentum balance:
1
− τ yx
ρ
y =0
=
dv
d
δ 2 v∞2 ) + δ1v∞ ∞
(
dx
dx
with:
δ1
=
∞

0

∫ 1 −
vx 
dy ("displacement thickness")
v∞ 
and:
=
δ2
vx  vx 
1 − v  dy ("momentum thickness")
0 v
∞ 
∞ 
∞
∫
BSL 4.4: Boundary-layer theory
• Solution of Prandtl boundary layer equation  “exact”
solution
• Solution of von Kármán momentum balance based on
uniformity of velocity profiles  approximate solution
• Further study: “Boundary Layer Theory” (Schlichting)
BSL 5: velocity distributions in turbulent flow
• Theory treated so far only valid for laminar flow
• Laminar flow: streamlines do not intersect (“orderly
movement”)
• Description of laminar flow:
Pressure and velocity profiles can be derived by solving
the micro balances for momentum and mass by
analytical or numerical techniques (Computational Fluid
Dynamics)
• In practice in process equipment, turbulent flow
(“eddies”) often occurs!
• Turbulent flow: the streamlines intersect continuously,
movement of fluid particles takes place in all directions
(“chaotic movement”)
Advanced Transport Phenomena
BSL 5: Velocity distributions in
turbulent flow
J.A.M. Kuipers
BSL 5: velocity distributions in turbulent flow
coherent structures in a turbulent channel flow
BSL 5: velocity distributions in turbulent flow
presence of vortical
flow structures does
not necessarily imply a
turbulent flow condition:
vortices can also
prevail in laminar
flows
BSL 5: velocity distributions in turbulent flow
Consequence of this way of describing is that the micro
balances of (time-averaged or time-smoothed) momentum
contain terms for the “turbulent momentum flux”:
(τ ) (τ )   ρ v v
 
(τ ) (τ ) =  ρ v v
(τ ) (τ )   ρ v v
• Local pressure and velocity components contain a
strongly fluctuating component
• Description of turbulent flow:
equation of continuity and the Navier-Stokes equations
are valid!
• Solving conservation laws gives the pressure and
velocity components as a function of (x,y,z) and t
• Practical problem: the scale (in space and time) on
which changes occur in turbulent flow are very small!
Directly solution of the conservation laws is not possible
Use approximation methods
von Karman vortex street behind cylinder
in cross-flow operation (Das et al., 2018)
(τ xxt )
 t
(τ yx )
τt
 ( zx )
BSL 5: velocity distributions in turbulent flow
t
xy
t
xz
' '
x x
ρ vx' v 'y
ρ vx' vz' 
t
yy
t
yz
' '
y x
ρ v 'y v 'y
ρ v 'y vz' 
t
zy
t
zz
' '
z x
ρ vz' v 'y

ρ vz' vz' 
which have to be determined empirically in terms of the
time-smoothed velocity components (“closure problem”) !
BSL 5.1: Comparisons of laminar and turbulent
flows
Time-smoothed velocity profiles
• Radial velocity profile for laminar flow in a circular tube:
  r 2 
v
1
dp
and: Φ v ≈  − 
= 1 −    and: z =
vz ,max   R  
vz ,max 2
 dz 
Condition: Re<2100
• Time-smoothed radial velocity profile for turbulent flow
in a circular tube:
vz
vz
vz ,max
r 

= 1 −   
  R 
17
and:
vz
vz ,max
Condition: 104 < Re < 105
4
 dp 
=
and: Φ v ≈  − 
5
 dz 
47
BSL 5.1: Comparisons of laminar and turbulent
flows
Qualitative comparison of radial velocity profiles for flow in
a circular tube
Turbulent flow
compared with laminar
flow: velocity profile is
flatter

Eddy transport: turbulent
momentum transport
•With increasing distance to the tube wall,
we can distinguish:
Laminar sublayer
Buffer zone
Fully developed turbulent flow zone
Velocity fluctuations in turbulent flows
where t0 is large in
comparison with time scale
for turbulent fluctuations but
small in comparison with
time scale at which the
macroscopic flow
patterns change
Velocity distribution in a circular tube near the solid wall:
•Center of the tube: random velocity fluctuations
•Near the tube wall: velocity fluctuations in axial direction
exceed velocity fluctuations in radial direction
•Turbulent flow has a structure that changes gradually
BSL 5.1: Comparisons of laminar and turbulent
flows
Definition of time-smoothed
velocity:
1 t + t0
vz = ∫ vz dt
t0 t
BSL 5.1: Comparisons of laminar and turbulent
flows
=
vz
1 t + t0
vz dt O=
(v z ) v z
∫=
t0 t
BSL 5.1: Comparisons of laminar and turbulent
flows
The velocity in terms of the mean value and the fluctuation:
v=
vz + vz'
z
where
=
vz' 0 and ( vz' ) ≠ 0
2
Intensity of turbulence I:
I=
(v )
'
z
2
vz
Flow in a circular tube: I=1% ↔ I=10%
BSL 5.1: Comparisons of laminar and turbulent
flows
Turbulent flow in ducts
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
Every quantity is composed of the sum of its mean value and
its fluctuating component (Reynolds decomposition)
•Equation of continuity:
∂
∂
∂
vx + vx' ) + ( v y + v 'y ) + ( vz + vz' ) =
0
(
∂x
∂y
∂z
•Equation of motion for x-component (y- and z-component in
similar way):
∂
∂
∂
ρ ( vx + vx' ) ) =
− ( p + p ' ) − ( ρ ( vx + vx' )( vx + vx' ) )
(
∂t
∂x
∂x
∂
∂
− ( ρ ( v y + v 'y ) ( vx + vx' ) ) − ( ρ ( vz + vz' )( vx + vx' ) )
∂y
∂z
+ µ∇ 2 ( vx + vx' ) + ρ g x
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
O(
∂
∂
∂
vx + vx' ) + ( v y + v 'y ) + ( vz + vz' )) =
O(0)
(
∂x
∂y
∂z
∂
∂
∂
O(vx + vx' ) ) + ( O(v y + v 'y ) ) + ( O (vz + vz' ) ) =
0
(
∂x
∂y
∂z
∂
∂
∂
O(vx ) + O(vx' ) ) + ( O(v y ) + O(v 'y ) ) + ( O (vz ) + O(vz' ) ) =
0
(
∂x
∂y
∂z
∂
∂
∂
0
( vx ) + ( v y ) + ( vz ) =
∂x
∂y
∂z
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
∂
∂
∂
ρ ( vx + vx' ) )) =O(− ( p + p ' ) − ( ρ ( vx + vx' )( vx + vx' ) )
(
∂t
∂x
∂x
∂
∂
− ( ρ ( v y + v 'y ) ( vx + vx' ) ) − ( ρ ( vz + vz' )( vx + vx' ) )
∂y
∂z
O(
+ µ∇ 2 ( vx + vx' ) + ρ g x )
focus on this term:
∂
∂
'
'
ρ ( vx + vx' )( vx + v=
ρ O(( vx + vx' )( vx + v=
(
(
x ) ))
x )) )
∂x
∂x
∂
∂
vx' vx' ) )
ρ O(vx vx + 2vx vx' +=
ρ (O(vx vx ) + O(2vx vx' ) + O(vx' vx' )) )
(
(
∂x
∂x
∂
∂
similar for two other
ρ vx vx + ρ vx' vx' =
ρ vx vx + τ xxt
=
convection terms
∂x
∂x
O(
(
)
(
)
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
O(
∂
∂
'
'
ρ ( v y + v 'y ) ( vx + v=
ρ O(( v y + v 'y ) ( vx + v=
(
(
x ) ))
x )) )
∂y
∂y
∂
ρ O(v y vx + v y vx' + v 'y vx + v 'y vx' ) ) =
(
∂y
∂
∂
ρ v y vx + ρ v 'y vx' =
ρ v y vx + τ yxt )
(
∂y
∂y
(
∂
∂
∂
0
( vx ) + ( v y ) + ( vz ) =
∂x
∂y
∂z
∂
∂
'
'
O( ( ρ ( vz + vz' )( vx + v=
ρ O(( vz + vz' )( vx + v=
(
x ) ))
x )) )
∂z
∂z
∂
ρ O(vz vx + vz vx' + vz' vx + vz' vx' ) ) =
(
∂z
∂
∂
ρ vz vx + ρ vz' vx' =
ρ vz vx + τ zxt )
(
∂z
∂z
)
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
Vector notation
• Equation of continuity
∂
∂p ∂
∂
∂
− − ( ρ vx vx ) − ( ρ v y vx ) − ( ρ vz vx )
( ρ vx ) =
∂t
∂x ∂x
∂y
∂z
∂
∂
∂

−
ρ vx' vx' +
ρ v 'y vx' +
ρ vz' vx'  + µ∇ 2 vx + ρ g x
∂y
∂z
 ∂x

(
)
(
)
(
)
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
1, 2 and 3
vi ↔ vi i =
p ↔ p
1, 2 and 3
τ ij ↔ τ ij(l ) + τ ij(t ) i, j =
• Equation of motion
∂
( ρ v ) + ( ∇ ⋅ ρ vv ) = −∇p − ∇ ⋅τ (l )  − ∇ ⋅τ (t )  + ρ g
∂t
with τ(t) the turbulent momentum flux tensor:
τ (t )
• Time-smoothed equation of motion for x-component (yand z- component in a similar way):
Correspondence rules:
((∇ ⋅ v )) = 0
(τ ) (τ ) (τ ) 
 t

(τ yx ) (τ yyt ) (τ yzt )
=
τt
t
t 
 ( zx ) (τ zy ) (τ zz ) 
t
xx
Time-smoothed equations
• Time-smoothed equation of continuity:
)
(
BSL 5.2: Time-smoothed equations of change for
incompressible fluids
t
xy
t
xz
ρv v
 ' '
 ρ v y vx
 ' '
 ρ vz vx
' '
x x
Equations in BSL Table B.5 can be adapted for timesmoothed turbulent flow systems by changing all
' '
x y
ρv v
ρv v 
ρ v 'y v 'y
ρ v 'y vz' 
ρv v
'
z
'
y
' '
x z

ρ v v 
' '
z z
• these quantities are usually referred to as the Reynolds
stresses
vi to vi and p to p as well as τ ij to τ=
τ ij(l ) + τ ij(t )
ij
in any of the coordinate systems given
Components of τ(t) need to be specified in terms of timesmoothed velocity components  empirical approach 
behavior of turbulent flow cannot be predicted a priori
BSL 5.4: Empirical expressions for the turbulent
momentum flux
BSL 5.4: Empirical expressions for the turbulent
momentum flux
Solving turbulent flow problems  specification of
turbulent stress model is needed  empirical information
necessary
Boussinesq-model:
4 empirical models
in which μ(t) represent the so-called turbulent viscosity (or
eddy viscosity) (in contrast with μ, this is not at property of
the fluid)
τ yx(t ) = − µ (t )
•Boussinesq-model
•Prandtl-model
•Von Karman-model
•Deissler-model
Qualitative: μ(t) is large in the center of the turbulent flow
and small near the solid walls
BSL 5.4: Empirical expressions for the turbulent
momentum flux
τ yx
dvx dvx
= − ρl
dy dy
2
BSL 5.4: Empirical expressions for the turbulent
momentum flux
Von Karman-model:
s
Prandtl-model:
(t )
dvx
dy
wall
(t )
τ sz
dvz dvz
=
− ρl
=
−τ rz(t )
ds ds
(t )
τ yx = − ρκ
r
2
R
where l is the so-called mixing length for which Prandtl
proposed the following expressions:
Wall turbulence: l = κ1 y (y =distance from wall)
Free turbulence: l = κ 2b (b =width of mixing zone)
The model assumes that the behavior of eddies is similar
to that of molecules in a low-density gas (l ↔ λ)
r=0
2
2
3
( dvx dy ) dvx
(d
2
2
vx dy 2 ) dy
where κ2 is a “universal” constant (κ2≈0.36-0.40)
Deissler-model:
 n 2v z s 
 n 2v x y 


 −
 
 −
 
dv
ν  dvz
ν 

 x τ sz(t ) =
− ρ n 2 vz s  1 − e
− ρ n 2 vx y 1 − e
τ yx(t ) =

 ds

 dy




where n is an empirical constant (Deissler: n=0.124)
In contrast to Prandt-model and Von Karman-model, the
Deissler model is applicable in the area near the wall
BSL 5.4: Velocity distributions in turbulent pipe
flow
Derivation of the radial velocity profile for turbulent in a
circular tube
BSL 5.4: Velocity distributions in turbulent pipe
flow
∂vz
∂v v ∂v
∂v 
+ vr z + θ z + vz z  =
∂r r ∂θ
∂z 
 ∂t
ρ 
•far from wall region (turbulent transport dominates)
z-momentum balance
1 ∂
∂p 1 ∂
∂
(rτ rz ) −
(τ θ z ) − (τ zz ) + ρ g z
−
r ∂θ
∂z r ∂r
∂z
−
Distinguish two different areas:
BSL §B.5
apply correspondence rules
stress components are now
•near wall region (laminar transport dominates)
sum of laminar and turbulent
∂v
∂v v ∂v
∂v
ρ  z + vr z + θ z + vz z  =
terms due to time-smoothing
∂r r ∂θ
∂z 
 ∂t
=
τ ij τ ij( l ) + τ ij( t )
−
BSL 5.4: Velocity distributions in turbulent pipe
flow
•
•
•
•
•
1 ∂
∂p 1 ∂
∂
(τ rz ) −
(τ θ z ) − (τ zz ) + ρ g z
−
r ∂θ
∂z r ∂r
∂z
“steady state” time-smoothed flow
rotation symmetry in time-smoothed flow
fully developed flow in z-direction
pressure gradient driven pipe flow
no gravity acting in the z-direction
simplification
p − pL 1 d
∂p 1 ∂
0=
(τ rz ) ⇒ 0 =0
(rτ rz )
− −
−
L
r dr
∂z r ∂r
reduced z-momentum equation
BSL 5.4: Velocity distributions in turbulent pipe
flow
velocity profile “far from wall”
Time-smoothed equation of motion for the steady flow in a
circular tube in separated form:
∂v
∂v v ∂v
∂v
ρ  z + vr z + θ z + vz z  =
∂r r ∂θ
∂z 
 ∂t
−
1 ∂
∂p 1 ∂
∂
(τ rz ) −
(τ θ z ) − (τ zz ) + ρ g z
−
∂z r ∂r
r ∂θ
∂z
d (rτ rz ) =
( p0 − pL ) rdr
L
=
rτ rz
( p0 − pL ) 1 r 2 + K
L
2
1
τ rz(l ) + τ rz(t )
with: τ=
rz
τ rz ≈ τ rz(t )
Far from wall: turbulent transport dominates for which we
take the Prandtl mixing length model:
(t )
τ rz
 dv 
= ρκ s  z 
 ds 
2 2
1
2
s= R − r (distance from the wall)
BSL 5.4: Velocity distributions in turbulent pipe
flow
Integration of the equation of motion with boundary
condition τrz=0 at r=0 gives:
r
s
( p0 − pL ) R=
=
τ rz
τ 0 1 − 
2L
R
 R
where τ0 is the shear wall stress at the tube wall (s=0):
( p0 − pL )π R 2 =
τ 0 2π RL integral force balance z-direction
Center of the turbulent flow  turbulent momentum flux
dominates so:
2
 s
2 2  dv z 
ρκ1 s  =
 τ 0 1 − 
 R
 ds 
Prandtl’s assumption:
s
τ 0 1 −  ≈ τ 0
 R
BSL 5.4: Velocity distributions in turbulent pipe
flow
Calibration of turbulence-model:
Experimentally determined radial velocity profile fits the
model best if:
κ1
0.36
=
and: s1+ 26 for=
which: v1+ 12.85
This gives the so-called logarithmic velocity distribution
which for Re>2·104 describes the measured velocity
profiles for turbulent flow reasonably:
+
v=
1
ln ( s + ) + 3.8 s + ≥ 26
0.36
BSL 5.4: Velocity distributions in turbulent pipe
flow
Using this approximation gives:
τ 1
dvz
dvz
1
v*
=
± 0
=
>0
because :
ds
ds
ρ κ1 s
κ1 s
Integration from s=s1 (end of the buffer layer seen from
the wall) to an arbitrary position s:
s
=
vz − vz ,1
ln   s ≥ s1
κ1  s1 
v*
turbulent
core
s
or in dimensionless variables:
=
v + − v1+
buffer
zone
wall
viscous s=s1
sub-layer
 s+ 
ρ v* s
vz
=
ln  +  s + ≥ s1+ =
with: s +
and: v +
κ1  s1 
µ
v*
1
BSL 5.4: Velocity distributions in turbulent pipe
flow
velocity profile “near wall”
Superposition of the laminar and turbulent momentum flux:
dv
dv
τ rz = τ rz( l ) + τ rz( t ) = τ rz( l ) − τ sz( t ) = − µ z − τ sz( t ) = µ z − τ sz( t )
dr
ds
dv
dv
= µ z + ρ n 2 vz s (1 − exp {− n 2 vz s ν } ) z
ds
ds
combined with:
( p0 − pL ) R r = τ 1 − s  ≈ τ
τ rz =
 0
0
2L
R
R

wall

gives with s=R-r:
(1)
τ0 = µ
buffer
zone
dvz
dv
+ ρ n 2 vz s (1 − exp {− n 2 vz s ν } ) z
ds
ds
turbulent
core
s
viscous s=s1
sub-layer
BSL 5.4: Velocity distributions in turbulent pipe
flow
Integration from the tube wall (s=0) to an arbitrary value of
s gives in terms of the already mentioned dimensionless
distance s+ and dimensionless velocity v+:
s+
ds +
+
0 ≤ s + ≤ 26
v
(2)
∫
2 + +
2 + +
0 1 + n v s (1 − exp {− n v s } )
with n=0.124 for long smooth tubes. Note that we have
found an implicit equation for v+
For small values of s+, the implicit equation reduces into:
+
v=
s + 0 ≤ s+ ≤ 5
BSL 5.4: Empirical expressions for the turbulent
momentum flux
Graphical representation of the velocity profiles of the two
examples
(1)
(2)
(3)
(3)
which follows from integration of Newton’s law of viscosity
over the viscous sublayer
BSL 7: Macroscopic balances for isothermal flow
systems
Advanced Transport Phenomena
BSL 7: Macroscopic balances for
isothermal flow systems
J.A.M. Kuipers
BSL 3: micro balances for conservation of mass, momentum
and mechanical energy
BSL 7: macro balances for conservation of mass,
momentum and mechanical energy
Derivation of macro balances
from micro balances:
• integration over an
arbitrary macroscopic
system volume
• macroscopic system with
one inflow and one outflow
integral theorems
required as tools
BSL 7: Macroscopic balances for isothermal flow
systems
BSL 7: Macroscopic balances for isothermal flow
systems
Integral Theorems
Integral Theorems
f
Volume V enclosed
by surface S
n
S
unit outward
normal on S
f ⋅n
V
vs
Volume V(t) enclosed
by surface S(t)
ρ = ρ ( x, y , z , t )
n
S (t )
V (t )
f = f ( x, y , z , t )
dV ∫∫ ( f ⋅ n )dS
∫∫∫ (∇⋅ f )=
V
d
ρdV=
dt V∫∫∫
(t )
S
Gauss divergence theorem
Gottfried Wilhelm
Leibnitz
Carl Friedrich Gauss
BSL 7: Macroscopic balances for isothermal flow
systems
Averaging
A2
r
θ
a
x
b
A=
∫ f ( x)dx =< f > (b − a)
a
extension to 2D:
A: area below the
curve defined by f(x)
< f
∫∫ f ( x, y)dxdy =
∫∫ f (r ,θ )rdrdθ =
∫∫ fdS =
S1
S1
Leibnitz Theorem
(1)
∂ρ
+ (∇ ⋅ ρv ) = 0
∂t
(2)
∂
( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g = 0
∂t
b
x
∂ρ
dV + ∫∫ ρ (vs ⋅ n )dS
∫∫∫
∂
t
V (t )
S (t )
BSL 7: Macroscopic balances for isothermal flow
systems
average value of f ( x)
on interval [a,b] ≡< f >
A1
y
surface expands with
local velocity v s
micro balances for mass (1), momentum (2) and mechanical energy (3)
A1 = A2
f ( x)
vs ⋅ n
unit outward
normal on S(t)
S1
> S1
(3)
kg / (m3 s )
N / m3
∂ 1 2
2
1
(
2 ρ v ) = − ( ∇ ⋅ 2 ρ v v ) − ( ∇ ⋅ pv ) − p ( − ( ∇ ⋅ v ) )
∂t
− ( ∇ ⋅ (τ ⋅ v ) ) − ( −τ : ∇v ) + ρ ( v ⋅ g )
W / m3
BSL 7.1: Macroscopic mass balance
Conservation law for mass on micro scale:
∂ρ
+ (∇ ⋅ ρv ) = 0
∂t
Integration of the micro balance for mass:
∫∫∫
V
{
• Convection term: (using Gauss’s divergence theorem):
∫∫∫
V
}
∂ρ
+ ( ∇ ⋅ ρ v ) dV = 0
∂t
d
∂ρ
dV
=
V ( t ) ∂t
dt
{
∫∫∫
V (t )
}
ρ dV −
assume
constant
volume
∫∫
S (t )
n )dS
ρ (vs ⋅ =
d
( mtot )
dt
BSL 7: Macroscopic balances for isothermal flow
systems
surface integral evaluation for mass convection
S
∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS =
Sw
S1
S2
S2
S1
∫∫ (ρ v ⋅ n )dS =ρ ∫∫ −vdS =−ρ
1
S1
1
+ρ
ρ ∫∫ +vdS =
∫∫ (ρ v ⋅ n )dS =
S2
< v1 > S1
S1
2
S2
∆
= "2"− "1"
BSL 7.2: Macroscopic momentum balance
Conservation law of momentum on micro scale:
∂
( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g = 0
∂t
Integration of the microscopic momentum balance:
∂
∫∫∫
( ρ v ) + ( ∇ ⋅ ρ vv ) + ∇p + ( ∇ ⋅τ ) − ρ g dV = 0
V ∂t
only valid
Working out the terms:
for constant
volume
• Accumulation term:
}
{
∫∫ (ρ v ⋅ n )dS + ∫∫ (ρ v ⋅ n )dS
S1
d
=
( mtot ) ρ1 v1 S1 − ρ 2 v2 S2
dt
d
( mtot ) = w1 − w2 = −( w2 − w1 ) = −∆w
dt
with mtot the total mass in the system at time t
∫∫ (ρ v ⋅ n )dS =
( ∇ ⋅ ρ v )dV = ∫∫S ( ρ v ⋅ n )dS = − ρ1 v1 S1 + ρ 2 v2 S2
Combination of the results gives the final expression for
the macroscopic mass balance:
Working out the terms:
• Accumulation term (Leibnitz theorem):
∫∫∫
BSL 7.1: Macroscopic mass balance
2
< v2 > S 2
n
S2
v
∫∫∫
V
v
n
{
}
∂
d
( ρ v=
) dV
∂t
dt
{
∫∫∫ ρ vdV
V
}−
∫∫
S (t )
ρ v (vs ⋅=
n )dS
d
( Ptot )
dt
where Ptot is the total amount of momentum in the system
at time t
BSL 7.2: Macroscopic momentum balance
• Convection term (using Gauss’s divergence theorem):
∫∫∫
V
( ∇ ⋅ ρ vv=
)dV
∫∫
S
( ρ vv ⋅ n )dS
BSL 7: Macroscopic balances for isothermal flow
systems
surface integral evaluation for momentum convection
∫∫ (ρ vv ⋅ n )dS= ∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS=
S
=
− ρ1 v12 S1 + ρ 2 v22 S 2
∫∫ ( ρ v ) vn dS =
S
where S1 and S 2 are vectorial quantities that are defined
as follows:
Sw
S1
S2
∫∫ (ρ vv ⋅ n )dS + ∫∫ (ρ vv ⋅ n )dS
S1
S2
S1
∫∫ (ρ vv ⋅ n )dS =ρ e ∫∫ −v dS =−ρ
2
1 1
• Magnitude:
=
S1 S=
S2
1 and S 2
S1
• Direction: corresponds with the direction of the flow at S1 and S2
respectively
1
ρ 2 e2 ∫∫ +v 2 dS =
+ ρ 2 < v22 > S 2
∫∫ (ρ vv ⋅ n )dS =
n
S2
v
S2
e1 and e2 : unit vectors in direction of the flow
BSL 7.2: Macroscopic momentum balance
• Pressure force (using Gauss’s divergence theorem):
∫∫∫
V
− p1S1 + p2 S 2 + Fp
∫∫ p ndS =
{∇p}dV =⋅
S
• Viscous friction force:
∫∫∫
V
( ∇ ⋅τ )dV = ∫∫S (τ ⋅ n )dS = + Fv
• Gravitational force:
∫∫∫
V
ρ g )dV g=
mtot g
∫∫∫ ρ dV
(=
V
pressure force
exerted by fluid
on walls between
S1 and S2
viscous force
exerted by fluid
on walls between
S1 and S2
e1
< v > S1
S1
S2
v
2
1
e2
n
BSL 7.2: Macroscopic momentum balance
• Substitution of the results gives the macroscopic
momentum balance in vector notation:
d
=
( Ptot ) ρ1 v12 S1 − ρ 2 v22 S2 + p1S1 − p2 S2 − F + mtot g
dt
F
= Fp + Fv
or:
d
( P=
tot )
dt
v12
v1
w1 −
v22
v2
w2 + p1S1 − p2 S 2 − F + mtot g
where:
w1 = ρ1 v1 S1
w2 = ρ 2 v2 S 2
vectorial mass flow rates (kg/s) !
BSL 7.2: Macroscopic momentum balance
Notation of the macroscopic momentum balance in terms of Δ:
 v2

d
∆
= "2"− "1"
( Ptot ) = −∆  w + pS  − F + mtot g
dt
v


For steady systems:
 v2

F = −∆ 
w + pS  + mtot g
 v

This equation is useful for the calculation of forces exerted on
walls (structures) by fluids
Approximation for turbulent flowing fluids in comparison with
“flat” velocity profile:
 v2 
 v 2
∆
 ≈ ∆
=∆ v
v
v




BSL 7.4: Macroscopic mechanical energy balance
(Bernoulli’s law)
W is amount of work exerted by the fluid on the surroundings
per unit of time
EV is the amount of mechanical energy that is converted
(dissipated) into thermal energy per unit of time:
EV =
∫∫∫
V
( −τ : ∇v )dV
Reduction of the general macroscopic energy balance:
Steady system
d
( Ktot + Φ tot + Atot ) = 0
dt
 from the mass balance): w=
w=
w
1
2
BSL 7.4: Macroscopic mechanical energy balance
(Bernoulli’s law)
Conservation law for mechanical energy on micro scale:
∂ 1 2
2
1
(
2 ρ v ) = − ( ∇ ⋅ 2 ρ v v ) − ( ∇ ⋅ pv ) − p ( − ( ∇ ⋅ v ) )
∂t
− ( ∇ ⋅ (τ ⋅ v ) ) − ( −τ : ∇v ) + ρ ( v ⋅ g )
Integration over macroscopic volume of flow system gives:
 1 v 3
 
d
ˆ + Gˆ  w − W − EV
+Φ
( Ktot + Φ tot + Atot ) = −∆ 
dt
 2 v
 
With Ktot, Φ tot and Atot the total kinetic, potential and (Helmholtz) free
energy respectively of the system:
ˆ
ˆ dV Atot = ∫∫∫ ρ AdV
K tot = ∫∫∫ 12 ρ v 2 dV Φ tot = ∫∫∫ ρΦ
V
V
V
ˆ , Aˆ and Gˆ represent respectively potential energy, free energy and
Φ
free enthalpy per unit of mass
BSL 7.4: Macroscopic mechanical energy balance
(Bernoulli’s law)
• Isothermal system
p2
dp
p1
ρ
∆G =
∫
• Define:
EV
W
and: EˆV
=
Wˆ
=
w
w
Using these assumptions/definitions, the final form of
the steady macroscopic mechanical energy balance is:
p2
 1 v3 
ˆ + ∫ dp + Wˆ + EˆV = 0 Bernoulli’s equation
∆
 + ∆Φ
p1 ρ
2 v 
BSL 7: Macroscopic balances for isothermal flow
systems
Summary of macroscopic balances in ∆ formulation
• Macroscopic mass balance:
d
( mtot ) = −∆w
dt
• Macroscopic momentum balance:
BSL 7: Applications
Applications of macroscopic balances
• Pressure rise and friction loss in a sudden enlargement
• Performance of a liquid-liquid ejector
 v2

d
( Ptot ) = −∆  w + pS  − F + mtot g
dt
 v

• Thrust on a pipe-bend
• Macroscopic mechanical energy balance:
 1 v 3
 
d
ˆ + Gˆ  w − W − EV
+Φ
( Ktot + Φ tot + Atot ) = −∆ 
dt
 2 v
 
BSL 7: Applications
BSL 7: Applications
Pressure rise and friction loss in a sudden enlargement
Macroscopic mass balance:
=
w1 w=
ρ 2 v2 S2
2 or: ρ1 v1 S1
for an incompressible fluid (β=S1/S2):
v1 1
=
v2 β
Macroscopic momentum balance for the main flow direction:
Medium: fluid in turbulent flow
Question: find an expression for the pressure rise
between “1” and “2” and the friction loss
0 = v1w1 − v2 w2 + p1S1 − p2 S 2 − F
Components of the force F:
• Viscous force on the cylindrical surface  negligible
• Pressure force acting on (S2-S1): (S2-S1)p1
21
BSL 7: Applications
Substitution of result in the momentum balance gives:
p2 − =
p1 ρ v2 ( v1 − v2 )
Combining the momentum balance with the mass balance:
1 
p2 −=
p1 ρ v22  − 1
β 
since β<1  p2>p1
BSL 7: Applications
Eliminating the pressure difference using the momentum
balance:
2

1  1  2 1   v2 
ˆ
EV =
−  − 1 ρ v2 + 2    − v22 
 β 

ρβ 


2

  1 2  1  2
1  1  1 
2 
= v2 1 −  + 2    − 1 = 2 v2  − 1


β 
 
 β    β 
Mechanical energy balance:
1
2
(v
2
2
− v12 ) +
1
ρ
0
( p2 − p1 ) + EˆV =
23
BSL 7: Applications
Performance of a liquid-liquid ejector
Medium: fluid in turbulent flow
Total area in plane “1” and “2” equals S (constant)
v=
0 and: S1
1
3
ρ 2 v2 S2
Macroscopic mass=
balance: w1 w=
2 or: ρ1 v1 S1
1
2 1
2
for an incompressible fluid: v1 =
v2 =
3 v0 + 3 ( 2 v0 ) =
3 v0
Macroscopic momentum balance:
 v2

General form: F = −∆ 
w + pS  + mtot g
 v

Working out the general form:
 v22
  v12

0= 
w2 + p2 S 2  − 
w1 + p1S1 
 v2
  v1

Assumption: force F acting on the system wall is negligible
Central stream at “1”:
v1
BSL 7: Applications
S
Annular stream at “1”:
v0
and: S1 23 S
v1
=
2
24
25
BSL 7: Applications
Evaluation of average quantities:
v22
=
v2
and:
v12
=
v1
v=
2
2
3
v0
v + 23 ( )
=
1
2 v0
v
+
3 0
3( 2 )
1
3
+ 16
v0
=
2
1
3
3
4
v0
3
Using this result in the momentum balance gives:
2
1
p2 − p1 =
18 ρ v0
Pressure rise occurs due to the mixing of the two fluids:
kinetic energy decreases from “1” to “2”
26
Substitution in macroscopic mechanical energy balance:
1
0
( 92 − 165 ) v02 + ( 181 ρ v02 ) + EˆV =
ρ
EˆV =
( 165 − 92 − 181 ) v22 =
5
144
v22
Results are valid for liquid-liquid ejectors
Gas-gas ejectors: ρ varies considerably and the overallenergy balance and an equation of state must be used
28
 v3
v3  1
Application of general form:  12 2 − 12 1  + [ p2 − p1 ] + EˆV =
0
v1  ρ
 v2
v23
Evaluation of average quantities:=
v2
1
2
v13
1
and:=
2
v1
v + 32 ( v20 )
=
v + 32 ( v20 )
3
1 3
1 3 0
2 1
3 0
+ 2
=
v0
1
1
12
1 3
2
2
3
5
16
1
2
2
v2 ) 12=
( 23 v0 )
(=
2
2
9
v02
v02
27
BSL 7: Applications
or:
Macroscopic mechanical energy balance:
p2 dp
 1 v3 
ˆ
+
∆Φ
+
+ Wˆ + EˆV = 0
General form: ∆  2
∫

p1 ρ
v


v0 2
2
2
0
BSL 7: Applications
BSL 7: Applications
Thrust on a pipe bend
Water at 95 °C is flowing at a rate of 2.0 ft3/s through a 60°
bend, in which there is contraction from 4 inch to 3 inch
internal diameter
Force exerted on bend if downstream pressure is 1.1 atm ?
ρwater=0.962 g/cm3
μwater=0.299 cp
Conversion factors:
1ft3 = 28.3·10-3 m3
1 inch = 0.0254 m
ft = 0.305 m
1 cp = 10-3 kg/(m·s)
29
BSL 7: Applications
Calculation of Re:
ρ v D 4Q ρ 4 ⋅ ( 2.0 ⋅ 28.3 ⋅10 ) ⋅ 962
=
=
=
=
Re
3.04 ⋅106
−3
µ
π Dµ π ⋅ 0.0762 ⋅ 0.299 ⋅10
−3
assumption of flat velocity profiles OK (turbulent flow)
Macroscopic mass balance:
v S
w1 =w2 → 1 = 2 =β ( β < 1)
v2 S1
Macroscopic momentum balance:
 v2

F = −∆ 
w + pS  + mtot g
 v

BSL 7: Applications
Working out the momentum balance for the x- and ycomponent respectively:
 v12
  v22

Fx= 
w1x + p1S1x  − 
w2 x + p2 S 2 x 
 v1
  v2

= v1 ( ρ v1S1 ) + p1S1 − v2 ( ρ v2 S 2 cos (θ ) ) − p2 S 2 cos (θ )
= ρ v22 S 2 ( β − cos (θ ) ) + ( p1 − p2 ) S1 + p2 ( S1 − S 2 cos (θ ) )
and:
 v12
  v22

Fy= 
w1 y + p1S1 y  − 
w2 y + p2 S 2 y  − mtot g
 v1
  v2

=
−v2 ( ρ v2 S 2 sin (θ ) ) − p2 S 2 sin (θ ) − mtot g
=
− ρ v22 S 2 ( sin (θ ) ) − p2 S 2 sin (θ ) − mtot g
30
BSL 7: Applications
Mechanical energy balance:
 1 v23 1 v13  1
0
−2
2
 + [ p2 − p1 ] + EˆV =
v
v
ρ
1 
 2
1
2
v − 12 v +
2
1
1
ρ
Inserting this into the mechanical energy balance:
1
p1 − =
p2 ρ v22 ( 12 − 12 β 2 + =
ρ v22 ( 107 − 12 β 2 )
5)
Inserting this into the x-momentum balance gives:
Fx = ρ v22 S 2 ( β − cos (θ ) ) + ρ v22 ( S 2 β ) ( 107 − 12 β 2 )
+ p2 S 2 (1 β − cos (θ ) )
Simplification:
2
2
BSL 7: Applications
[ p2 − p1 ] + EˆV
=
0
Estimation of ÊV using friction charts or tables and the
definition for ÊV :
2
1 2 2
1 2
Eˆ v 12 ( v ) ev gives: Eˆ=
5 v2
v ≈ 2 v2 ( 5 )
Summary:
x-momentum balance:
=
Fx ρ ( Q 2 S 2 ) {107 β −1 − cos (θ ) + 12 β } + p2 S 2 ( β −1 − cos (θ ) )
y-momentum balance:
Fy =
− ρ ( Q 2 S 2 ) sin (θ ) − p2 S 2 sin (θ ) − π R 2 L ρ g
where R and L are the radius and length of a cylinder with
the same volume as the chosen control volume
33
BSL 7: Applications
Now, the magnitude and direction of the force F can be
derived from the given equations:
•Magnitude:
=
F
Fx2 + Fy2
BSL 9 & 10: Thermal conductivity and the
mechanism of energy transport & shell energy
balances and temperature distributions in solids
and laminar flow
J.A.M. Kuipers
•Direction:
α
Advanced Transport Phenomena
arctan ( Fx − Fy )
where alpha is the angle that the force makes with the
vertical
BSL 10: Shell heat balance + BC’s
BSL 9.1: Fourier’s law of heat conduction
Development of steady-state temperature
profile for a solid slab exposed to
temperature difference
d ( ρ C pT )
dT
k d ( ρ C pT )
qy =
−k
=
−
=
−α
dy
ρC p
dy
dy
Fourier’s law of
heat conduction
Heat balance in words:
rate of heat

accumulation
in system


net heat 
 heat flow  heat flow  

= 
−
 + production 
 rate in  rate out  in system 



boundary
conditions
required
+ Prescribed temperature (Dirichlet boundary condition)
T = Tw
Jean-Baptiste
Joseph Fourier
+ Prescribed heat flux (Neumann boundary condition)
isotropic material
∂T
∂T 
 ∂T
 ∂T ∂T ∂T 
, ky
, kz
q = ( qx , q y , qz ) = −  k x
= − K ⋅∇T = −k  , ,  = −k ∇T

∂y
∂z 
 ∂x
 ∂x ∂y ∂z 
Anisotropic materials: thermal conductivity is different in x-, y- and z-direction.
Note: anisotropic thermal conductivity also occurs in disperse systems
−k
∂T
∂n
w
=
qw
+ Mixed boundary condition (Robin boundary condition)
−k
∂T
∂n
w
=h(T∞ − Tw )
BSL 10.2: Heat conduction with an electrical heat
source
System: copper wire with radius R (cross sectional area S),
length L and electrical conductivity ke=(1/ρe) [Ω-1cm-1] through
which an electrical current with current density J [A/cm2] is fed
Qualitative: electrical energy is converted into heat  a radial
temperature profile is present inside the wire: T=T(r)
Amount of heat produced per unit of volume and time:
I 2 Re ( JS ) 2
S L
J2
2
ρ
Se =
Re J 2 ρ=
J
=
=
=
(W / m3 )
e
e
SL
SL
L S
ke
Re: electrical resistance of wire in Ω, ρe is the
specific electrical resistance (material property)
Assumptions:
• thermal conductivity k is independent of T and the surface
temperature (at r=R) of the wire is constant and equals T0
• there is no heat transport (conduction) in axial direction
BSL 10.2: Heat conduction with an electrical heat
source
Inserting this into the heat balance and dividing by dV gives:
0
( rqr ) r − ( rqr ) r +∆r
+ Se
r ∆r
Take the limit as Δr → 0:
1d
0=
−
( rqr ) + Se
r dr
Boundary conditions:
rqr
Integration:=
1
2
=
at r 0=
: qr 0
=
at r R=
: T T0
Se r + C1
2
BSL 10.2: Heat conduction with an electrical heat
source
Description of the temperature
profile ?  set up a differential
heat balance over a differential
volume element dV=2πrLΔr
Net conductive radial heat
transport:
( 2π rL ⋅ qr ) r − ( 2π rL ⋅ qr ) r +∆r
Heat production in dV:
( 2π rL∆r ) Se
BSL 10.2: Heat conduction with an electrical heat
source
Based on the boundary conditions at r=0, it follows that
C1=0. Inserting Fourier’s law gives:
−k
dT Se r
=
2
dr
Integration using the boundary condition at r=R gives:
2
Se R 2   r  
T −=
T0
1 −   
4k   R  
Note that the temperature profile is parabolic (analogous to
laminar tube flow): correspondence rules are given in BSL
BSL 10.4: Heat conduction with a viscous heat
source
System: steady incompressible flow between two concentric
cylinders
Qualitative: due to friction between the rotating fluid layers,
mechanical energy will be converted into internal energy
(ΔT) (viscous dissipation)
BSL 10.4: Heat conduction with a viscous heat
source
Sv: amount of heat produced per unit of volume and time
due to viscous dissipation: Sv=Sv(μ,(dvz/dx)):
 dv 
Sv =−
( τ : ∇v ) =µ  z 
 dx 
2
check with BSL §B.7 !
Velocity profile in idealized system:
 x
vz =   V
b
Idealization:
neglect the
curvature (flat
plate approach)
V 
Sv = µ  
b
Differential heat balance over a volume element dV=A·dx
gives:
2
V

0=
qx x A − qx x + dx A + µ   Adx
b
Dividing by A·dx en taking the limit as dx → 0 gives after
combination with Fourier’s law:
2
d
d
dT 
V
−  k
µ  
( qx ) =
=
dx
dx  dx 
b
=
− kT
2
1
2
V
µ   x 2 + C1 x + C2
b
0=µ
d 2 vz
dx 2
Combining these two relations:
BSL 10.4: Heat conduction with a viscous heat
source
Integration gives:
follows from reduced
z-momentum equation
2
=
vz K1 x + K 2
=
x 0=
: vz 0
=
x b=
: vz V
BSL 10.4: Heat conduction with a viscous heat
source
Integration constants follow from the boundary conditions:
at x 0=
: T T0 and=
at x b=
: T Tb
=
Solution:
T − T0  x  1  x    x  
=
  + 2 Br   1 −   
Tb − T0  b 
 b    b 
where Br is the dimensionless Brinkman number:
=
Br
µV 2
internal rate of heat production
=
k (Tb − T0 )
heat flow by conduction
In general: viscous heating is not important, exceptions are encountered
in system with extremely large velocity gradients:
•Bearings
•Extruders (rheology)
•High velocity flow around objects
BSL 10.7: Heat conduction in a cooling fin
BSL 10.7: Heat conduction in a cooling fin
Aim of applying of cooling fins:
enlarging the heat transfer area
Differential heat balance over control volume (2BW)dz:
Problem: efficiency is not 100%
due to heat transfer limitation in
the cooling fin
Analogy: effectiveness of a
porous catalyst
Dividing by 2BWdz en taking the limit as dz→0 gives:
dq
h
− z =
(T − Ta )
dz B
Combination with Fourier’s law gives:
d 2T h
=
(T − Ta )
dz 2 kB
Boundary conditions:
=
at z 0=
: T Tw
qz z 2 BW − qz
Actual situation
Model representation
T is a function of x,y and z but the dependence
on z is dominant
T is function of z alone
Some heat is lost from the fin at the end (area
2BW) and at the edges (2BL+2BL)
No heat is lost from the end or from the edges
The heat transfer coefficient is a function of
position
External heat flux is given by q=h(T-Ta), where h
is constant and T depends on z
BSL 10.7: Heat conduction in a cooling fin
Make the equations dimensionless using:
T − Ta
• Dimensionless temperature: Θ =
Tw − Ta
z
• Dimensionless distance: ξ =
L
hL2
• Dimensionless heat transfer coefficient: N =
kB
The differential equation then takes the form:
d 2Θ
N 2Θ
=
2
dξ
see: matfys.doc (canvas)
Θ =e mξ
try
m = ±N
=
Θ K1e Nξ + K 2 e − Nξ
z + dz
2 BW − h (T − Ta ) 2Wdz =
0
dT
=
at z L=
:
0
dz
BSL 10.7: Heat conduction in a cooling fin
Dimensionless boundary conditions:
e x + e− x
at ξ= 0 : Θ
= 1
cosh( x) =
2
dΘ
x
at ξ 1:= 0
=
e − e− x
sinh( x) =
dξ
2
Solution:
sinh( x)
cosh ( N (1 − ξ ) )
tanh( x) =
Θ=
cosh( x)
cosh ( N )
The relation for the “effectiveness” of the cooling fin follows
from the (dimensionless) temperature
Definition of η:
η=
actual rate of heat loss from the fin
rate of heat loss from an isothermal fin at Tw
BSL 10.7: Heat conduction in a cooling fin
BSL 10.8: Forced convection
or:
W L
η
∫ ∫h
(T − Ta ) dzdy
=
(Tw − Ta ) dzdy
0 o
W L
∫ ∫h
0 o
1
∫ Θd ξ
0
1
∫ dξ
0
Inserting the expression for Θ gives after integration:
=
η
{
}
tanh ( N )
1
1
1
− sinh ( N (1 − ξ ) )=
0
cosh ( N ) N
N
Qualitative comparison
between forced and free
convection
Application: error in thermocouple measurement due to
parasitic heat conduction
BSL 10.8: Forced convection
Example: steady laminar flow of a viscous fluid in a circular
tube with radius R
Assumption: constant physical properties (ρ, μ, k and Cp)
• z<0: uniform fluid temperature T0
• z>0: constant radial heat flux q1 imposed at the wall
Radial velocity profile:
  r 2 
=
vz vz ,max 1 −   
 R 
with:
p0 − pL ) R 2
(
vz ,max =
4µ L
BSL 10.8: Forced convection
Analysis of heat transport in developed laminar flow
Qualitative: T=T(r,z)  set up a differential energy balance
over an annular ring with volume dV=2πrΔrΔz
Thermal energy balance in words:
net convective   net conductive 
accumulation  
 

=

  inflow of heat  +  inflow of heat 
 of heat in dV  
 

 in z -direction  in r- and z-direction 
BSL 10.8: Forced convection
BSL 10.8: Forced convection
Net heat conduction in axial direction:
Net convective inflow of energy:
{ρ h
z
− ρ h z +∆z } vz 2π r ∆r
where h is the enthalpy per unit of mass:
T
h= h0 + ∫ C p dT
T0
Net heat conduction in radial direction:
( 2π r ∆zqr ) r − ( 2π r ∆zqr ) r +∆r
BSL 10.8: Forced convection
Inserting Fourier’s law and the expression for the radial
velocity profile gives:
  r  2  ∂T
 1 ∂  ∂T  ∂ 2T 
ρ C p vz ,max 1 −    = k 
r
+ 2 
 r ∂r  ∂r  ∂z 
  R   ∂z
where k is the (constant) thermal conductivity
Reduction of micro balance: axial heat transport is
dominated by convection:
  r  2  ∂T
 1 ∂  ∂T  
ρ C p vz ,max 1 −   
=
k
r

R
∂
z
r
∂
r


 ∂r  



( 2π r ∆rqz ) z − ( 2π r ∆rqz ) z +∆z
Combination with the energy balance in words gives after
dividing by dV=2πrΔrΔz :
ρ vz
0=
h z − h z +∆z ( rqr ) r − ( rqr ) r +∆r qz z − qz
+
+
∆z
∆z
r ∆r
z +∆z
Taking the limit as Δr→0 and Δz→ 0 gives:
ρ vz
∂q
∂h
∂T
1∂
=
ρ C p vz
=
−
( rqr ) − z
∂z
∂z
r ∂r
∂z
BSL 10.8: Forced convection
Boundary conditions:
r = 0: for all z: T =finite
∂T
=
−q1
r=
R: for all z: − k
∂r
=
z 0:=
for all r: T T0
Introducing dimensionless quantities:
• Dimensionless temperature:
T − T0
Θ=
q1 R k
r
• Dimensionless radial coordinate: ξ =
R
• Dimensionless axial coordinate: λ =
zk
ρ C p vz ,max R 2
BSL 10.8: Forced convection
• Dimensionless energy balance:
1 ∂  ∂Θ 
=
ξ
(1 − ξ 2 ) ∂Θ
∂λ ξ ∂ξ  ∂ξ 
• Dimensionless boundary conditions:
ξ
0 : for all λ: Θ=finite
∂Θ
ξ 1:=
for all λ:
1
∂ξ
λ 0 : for all ξ : Θ=0
Solution: R. Siegel et al., Applied Scientific Research, A7,
386-392 (1958) using separation of independent variables:
ˆ (ξ , λ ) G ' + µ 2G =
Θ(ξ , λ ) = Θ∞ (ξ , λ ) − Θ
0
ˆ (ξ , λ ) =
(ξ F ' )' + µ 2ξ (1 − ξ 2 ) F =
0
Θ
F (ξ )G (λ )
BSL 10.8: Forced convection
Asymptotic solution for large λ (thermally developed flow):
Θ = Θ (ξ , λ ) = C0 λ + Ψ (ξ )
where C0 is a yet to be determined constant
Physical motivation: for large λ, Θ increases linearly with λ
and the shape of the dimensionless (radial) temperature
profiles do not change with increasing λ (self-similarity)
Problem: boundary condition at tube inlet can not be met 
replace it therefore with an integral energy balance:
=
2π Rzq1
R
∫
0
ρ C p (T − T0 ) vz ( r ) 2π rdr
Sturm-Liouville problem for F: see matfys.doc (canvas)
BSL 10.8: Forced convection
or in dimensionless form:
λ=
∫ Θ (ξ , λ ) (1 − ξ ) ξ d ξ
1
2
0
Inserting the asymptotic solution into the dimensionless
energy balance:
1 d  dΨ 
C0 (1 − ξ 2 )
ξ =


ξ dξ  dξ 
Integration gives:
dΨ
ξ = C0 ( 12 ξ 2 − 14 ξ 4 ) + C1
dξ
Based on the “natural boundary condition” (Θ and
therefore Ψ must be finite)  C1=0
BSL 10.8: Forced convection
Further integration:
=
Ψ C0 ( 14 ξ 2 − 161 ξ 4 ) + C2
Total asymptotic solution:
=
Θ C0 λ + C0 ( 14 ξ 2 − 161 ξ 4 ) + C2
From the boundary condition at the tube wall, it follows that
C0=4 and from the integral energy balance, it follows that
C2=-7/24
Expression for the dimensionless temperature profile:
Θ
= 4λ + ξ 2 − 14 ξ 4 − 247
BSL 10.8: Forced convection
BSL 10.9: Free convection
Using the expression for the temperature profile, all
important secondary quantities can be derived:
• Radial arithmetic average temperature:
2π R
T =
∫ ∫T
0 0
( r , z ) rdrdθ
2π R
∫ ∫ rdrdθ
collect fluid
in stirred vessel
and measure
temperature
0 0
• Flow-averaged or “cup-mixing” temperature:
vzT
=
vz
2π R
∫ ∫ vz
0 0
( r ) T ( r , z ) rdrdθ
2π R
∫ ∫ vz
0 0
( r ) rdrdθ
vz ( r )
T (r , z )
∞
Fluid with (variable) density ρ
and dynamic viscosity μ is
located between two parallel
plates a distance 2B apart
Left plate:
T=T2
Right plate: T=T1
Qualitative: fluid rises near left
(hot) wall en descends near the
(cool) right wall: however no net
convection in de z-direction
BSL 10.9: Free convection
Thermal energy balance and reduced form:
∂T
 ∂ 2T ∂ 2T 
ρ C p vz ( =
y)
k 2 + 2 
∂z
∂z  temperature profile
 ∂y
is fully developed
Boundary conditions:
=
y -b=
: T T2
T2 =
− K1b + K 2
+ K1b + K 2
y=
+b : T =
T1
T1 =
Solution of the thermal energy balance:
1
y
T = Tm − ∆T  
2
b
with:
∆T =
(T2 − T1 ) and Tm =
T2 + T1
2
BSL 10.9: Free convection
d 2T
k 2 =0
dy
T K1 y + K 2
=
density variations are rather small
∂vx ∂v y ∂vz
∂ρ ∂
∂
∂
+ ( ρ vx ) + ( ρ v y ) + ( ρ vz ) = 0 ⇒
+
+
=0
∂t ∂x
∂y
∂z
∂x ∂y ∂z
z-momentum balance with “constant” physical properties
∂v
∂v
∂v 
∂p
 ∂vz
 ∂ 2v ∂ 2v ∂ 2v 
+ vx z + v y z + vz z  =
− + µ  2z + 2z + 2z  + ρ g z
∂x
∂y
∂z 
∂z
∂y
∂z 
 ∂t
 ∂x
ρ
•
•
•
•
•
steady state flow
only flow in z-direction
no x-dependence (infinite size in x-direction)
flow in z-direction is fully developed
gravity acting in negative z-direction
reduced z-momentum equation
BSL 10.9: Free convection
BSL 10.9: Free convection
Reduced momentum balance:
d 2 vz
∂p
0=
− + µ 2 − ρg
∂z
dy
where ρ is the temperature dependent ρ(T) for which the
following Taylor-expansion is used:
∂ρ
ρ = ρ m +   (T − Tm ) + = ρ m − ρ m β m (T − Tm ) +
 ∂T T =Tm
where ρm= ρ(Tm) and β the volumetric expansion coefficient
1 ∂ρ
1
defined as: β =
−   ⇒ β = for ideal gas
ρ  ∂T  p
T
Regarding the pressure in the z-momentum balance, we
assume that the pressure is only a function of z
Inserting these results into the z-momentum balance gives:
µ
d 2 vz  dp
= + ρ m g  − ρ m β m (T − Tm ) g
2
dy
 dz

Substitute the expression for the temperature profile:
d 2 vz  dp
y 
 1
µ 2=  + ρ m g  − ρ m β m g  − ∆T   
dy
 dz

 b 
 2
Boundary conditions:
y=
−b : vz =
0
y=
+b : vz =
0
∂p
dp
→
∂z
dz
BSL 10.9: Free convection
Solution:
vz
A
B
ρ m β m gb 2 ∆T 3 b 2  dp
η +  + ρ m g η 2 + C1η + C2 η = y b
12 µ
2 µ  dz

vz = Aη 3 + Bη 2 + C1η + C2
) vz (1=) 0
Boundary conditions: vz ( −1=
0 =− A + B − C1 + C2
0 =+ A + B + C1 + C2
vz
BSL 10.9: Free convection
Solution:
ρ m β m gb 2 ∆T 3
b 2  dp
vz
η − η ) +  + ρ m g  (η 2 − 1)
=
(
12 µ
2 µ  dz

Demand: net volume flow rate of the fluid equals zero:
1
∫
C2 =
− B C1 =
−A
ρ m β m gb 2 ∆T 3
b 2  dp
 2
η
−
η
+
(
)
 + ρ m g  (η − 1)
12 µ
2 µ  dz

−1
vz (η ) dη = 0
Inserting the expression for the velocity profile gives:
b 2  dp
4
dp

= − ρm g
 + ρm g  × − = 0 →
2 µ  dz
3
dz

BSL 10.9: Free convection
Final result:
vz
3
ρ m β m gb 2 ∆T   y   y  
  −  
12 µ
 b   b 
or in terms of the dimensionless velocity φ = bvz ρ m µ
and the dimensionless length η = y b
Gr 3
φ
=
(η − η )
12
where Gr is the dimensionless Grashof number
gb3
gb3 ∆ρ
Gr
βm∆
T
=
=
vm2
vm2 ρ m
and:
∆ρ = ρ1 − ρ 2
BSL 11: The equations of change for
nonisothermal systems
BSL 10: analysis of heat transport based on simple balances
 temperature profiles + related quantities
BSL 11: derivation of energy equations (PDE that describes
transport and interconversion of different forms of energy)
When are equations of energy needed?  for transport in
nonisothermal systems (reactive systems with heat effects)
Equations of energy
• Mechanical energy equation (BSL 3)
• Total energy equation (BSL 11)
• Thermal energy equation (BSL 11)
Advanced Transport Phenomena
BSL 11: The equations of change
for nonisothermal systems
J.A.M. Kuipers
BSL 11.1: The energy equation
Law of conservation of energy in words:
accumulation of



internal and kinetic 
=
energy per unit of time 


conductive inflow

+ of internal energy
per unit of time

convective inflow of  convective outflow of 

 

internal and kinetic  − internal and kinetic 
energy per unit of time  energy per unit of time 

 

 conductive outflow
 
 − of internal energy
 per unit of time
 
 net work done by the

 

 − system on the surroundings 
 per unit of time

 

Dimension of all terms: (J/s=W)
Note: the potential energy term is part of the work term
BSL 11.1: The energy equation
BSL 11.1: The energy equation
ρ evx x+∆x
ρ evx x
Balance for a differential volume element with size ∆x, ∆y and
∆z in the x-, y- and z-direction respectively and fixed in space
Accumulation of internal and kinetic energy:
e
∂
∂
of energy
∆x∆y∆z ( ρUˆ + 12 ρ v 2 ) =∆x∆y∆z ( ρ e ) accumulation
per
unit
of
volume
∂t
∂t
Net convective inflow of internal and kinetic energy per unit of
time:
∆y∆z vx ( ρUˆ + 12 ρ v 2 ) − vx ( ρUˆ + 12 ρ v 2 )
+
{
{
∆x∆z v y ( ρUˆ + 12 ρ v 2 ) − v y ( ρUˆ + 12 ρ v 2 )
y
{
∆x∆y vz ( ρUˆ + 12 ρ v 2 ) − vz ( ρUˆ + 12 ρ v 2 )
z
{−∇ ⋅ ( ρUˆ +
BSL 11.1: The energy equation
x + dx
{−(∇ ⋅ q )} ∆x∆y∆z
} + ∆x∆z {q
y y
− qy
y + dy
} + ∆x∆y {q
z z
− qz
z + dz
}=
Net work done by the system on surroundings consists of:
}=
net convective transport of
e per unit of volume and time
{
net work done against gravity
per unit of volume and time
}
∆y∆z {vx p x+dx − vx p x } + ∆x∆z v y p y +dy − v y p y +
∆x∆y {vz p z +dz − vz p z } =
{(∇ ⋅ pv )} ∆x∆y∆z
net work done against
pressure forces per unit
of volume and time
Work done by viscous forces per unit of time:
{
∆y∆z (τ xx vx + τ xy v y + τ xz vz ) x+dx − (τ xx vx + τ xy v y + τ xz vz ) x
{
Work done against forces acting on the control volume (body
forces: gravity)
Work done against forces acting on the walls of the control
volume (surface forces: pressure forces + viscous forces)
− ρ∆x∆y∆z {vx g x + v y g y + vz g z } =− ρ (v ⋅ g )∆x∆y∆z
z + dz
}+
Work done by pressure forces per unit of time:
net conductive heat transport per unit of volume and time
Gravitational work per unit of time:
ρ v 2 )v } ∆x∆y∆z
y + dy
BSL 11.1: The energy equation
Net contribution due to conduction per unit of time:
∆y∆z {qx x − qx
1
2
}
x + dx
x
}
∆x∆z (τ yx vx + τ yy v y + τ yz vz ) y +dy − (τ yx vx + τ yy v y + τ yz vz ) y
{
∆x∆y (τ zx vx + τ zy v y + τ zz vz ) z +dz − (τ zx vx + τ zy v y + τ zz vz ) z
=
{( ∇ ⋅ (τ ⋅ v ) )} ∆x∆y∆z
net work done against
viscous forces per unit
of volume and time
}
}
BSL 11.1: The energy equation
Inserting this into the balance and taking the limit as Δx→0,
Δy→0 and Δz→0 gives after dividing by ΔxΔyΔz:
∂
∂
∂
∂
  ∂q ∂q ∂q 
−  ( ρ evx ) + ( ρ ev y ) + ( ρ evz )  −  x + y + z 
( ρe) =
∂t
∂y
∂z
 ∂x
  ∂x ∂y ∂z 
∂
∂
∂

+ ρ ( g x vx + g y v y + g z vz ) −  ( pvx ) + ( pv y ) + ( pvz ) 
∂y
∂z
 ∂x

∂
∂
∂

−  (τ xx vx + τ xy v y + τ xz vz ) + (τ yx vx + τ yy v y + τ yz vz ) + (τ zx vx + τ zy v y + τ zz vz ) 
∂y
∂z
 ∂x

where e is the sum of internal and kinetic energy per unit of
mass: e= Uˆ + 12 v 2
Alternative compact vector notation by removing ΔxΔyΔz
from all terms in vector representation:
∂
( ρ e ) = − ( ∇ ⋅ ρ ev ) − ( ∇ ⋅ q ) + ρ ( v ⋅ g ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) )
∂t
BSL 11.1: The energy equation
This equation describes the change in total energy per unit
of volume and time that a stationary observer would notice
Using definition of substantial derivative towards time gives:
ρ
This equation describes the change in total energy per unit
of volume and time from the perspective of an observer
moving with the local fluid velocity
Corresponding equation of mechanical energy (BSL 3):
D
ρ ( 12 v 2=
) p ( ∇ ⋅ v ) − ( ∇ ⋅ pv ) + ρ ( v ⋅ g ) − ( ∇ ⋅ (τ ⋅ v ) ) + (τ : ∇v )
Dt
BSL 11.1: The energy equation
Subtracting this balance of mechanical energy from the
balance of total energy gives the thermal energy balance:
D
ρ (Uˆ ) = − ( ∇ ⋅ q ) − p ( ∇ ⋅ v ) − (τ : ∇v )
Dt
The terms p ( ∇ ⋅ v ) and (τ : ∇v ) are present in both equations
but with opposite signs  these terms describes therefore the
interconversion of mechanical and thermal energy!
Term p ( ∇ ⋅ v ) can be either >0 or <0 for an incompressible
fluid and therefore describes reversible interconversion of
energy
Term ( −τ : ∇v ) is always >0 and therefore describes an
irreversible conversion of mechanical into thermal energy
D ˆ 1 2
(U + 2 v ) = − ( ∇ ⋅ q ) + ρ ( v ⋅ g ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) )
Dt
BSL 11.1: The energy equation
Potential energy in the equation of change for total energy
ˆ where Φ̂ is the
External force per unit of mass g = −∇Φ
potential energy per unit of mass. Using the definition of the
substantial derivative towards time gives:
D ˆ
∂ ˆ
ˆ )= ∂ ( Φ
ˆ ) − (v ⋅ g )
Φ )=
Φ ) + ( v ⋅∇Φ
(
(
Dt
∂t
∂t
If Φ is independent of time, the equation of total energy can be
rewritten as:
ρ
D ˆ ˆ 1 2
(U + Φ + 2 v ) = − ( ∇ ⋅ q ) − ( ∇ ⋅ pv ) − ( ∇ ⋅ (τ ⋅ v ) )
Dt
BSL 11.1: The energy equation
Modification of equation of change for thermal energy
Formulation in terms of temperature and heat capacity
instead of internal energy:
=
Uˆ Uˆ (Vˆ , T ) →
 ∂Uˆ  ˆ  ∂Uˆ 
∂p 

dUˆ = 
dV + 
dT =  − p + T    dVˆ + CˆV dT


 ∂T Vˆ 

 ∂Vˆ T
 ∂T Vˆ
This gives:
D
DT
∂p  DVˆ

ρ (Uˆ ) =  − p + T    ρ
+ ρ CˆV
Dt
Dt
 ∂T Vˆ  Dt

Using the continuity equation gives (BSL 3):
DVˆ
D 1
1 Dρ
ρ
= ρ  = −
= (∇ ⋅ v )
Dt
Dt  ρ 
ρ Dt
BSL 11.1: The energy equation
Inserting the results gives the equation of change for thermal
energy in terms of T:
∂p
DT
ρ CˆV
= − ( ∇ ⋅ q ) − T   ( ∇ ⋅ v ) − (τ : ∇v )
Dt
 ∂T Vˆ
Simplified forms of the equation of thermal energy
Newtonian fluid with constant heat conductivity:
∂p
DT
ρ CˆV
= k ∇ 2T − T   ( ∇ ⋅ v ) + µΦV
Dt
 ∂T  ρ
where ΦV is the viscous dissipation function (usually neglected):
 ∂vx 2  ∂v y  2  ∂v y  2   ∂v y ∂vx  2
=
ΦV 2  
+
 +
 +
 +

 ∂x   ∂y   ∂y    ∂x ∂y 
2
 ∂vz ∂v y   ∂vx ∂vz  2  ∂vx ∂v y ∂vz 
+
+
+
+
+
 − 
 +

 ∂y ∂z   ∂z ∂x  3  ∂x ∂y ∂z 
2
BSL 11.1: The energy equation
Other source terms for the equation of change of thermal
energy:
• Chemical reactions
• Nuclear reactions
• Electrical phenomena
2
BSL 11.2: Special forms of the energy equation
i.
Ideal gas
DT
ρ CˆV
= k ∇ 2T − p ( ∇ ⋅ v )
Dt
ii. Fluid flowing in constant pressure system
DT
ρ Cˆ p
= k ∇ 2T
Dt
iii. Fluid with constant density
DT
ρ Cˆ p
= k ∇ 2T
Dt
iv. Stationary solid
∂T
ρ Cˆ p = k ∇ 2T
∂t
Energy equation
Energy equation for curvilinear coordinate systems
Important coordinate systems:
see matfys.doc and appendix A
of BSL for details of transformation
to general orthogonal co-ordinates
BSL 11.3: The equation of motion for forced and
free convection
Forced convection
Flow occurs due to action of some external force such as a
pressure force or a gravitational force
ρ
• Cartesian coordinates
Dv
= −∇p − ( ∇ ⋅τ ) + ρ g
Dt
with
• Cylindrical coordinates
• Spherical coordinates
See BSL §B.8 and §B.9 for detailed tables
BSL 11.3: The equation of motion for forced and
free convection
Free convection
Flow occurs due to internally generated forces due to
density differences (“buoyant forces”)
Free convection requires a temperature dependent density
Consider a medium of which the temperature varies around
a certain average Tm:
Pressure gradient in stagnant medium: ∇p =ρ m g
{(
)
2
3
τ = − µ ( ∇v ) + ( ∇v ) − µ ( ∇ ⋅ v ) I
T
}
Non-isothermal flow: ρ and μ must be given as function of
pressure p and temperature T  linking equation of motion
and equation of change for thermal energy
BSL 11.3: The equation of motion for forced and
free convection
Assumption: velocity gradients due to ΔT are small and
pressure gradient is virtually constant in comparison with
stagnant medium:
Dv
ρ
= − ρ m g − ( ∇ ⋅τ ) + ρ g
Dt
Using ρ=ρm in the left side and ρ- ρm=-ρmβm(T-Tm) in the
right side gives:
Dv
ρm
= − ( ∇ ⋅τ ) − ρ m β m (T − Tm ) g
Dt
Where βm is the cubic expansion coefficient at Tm:
 1 ∂ρ 
β m = −    
 ρ  ∂T  p  mean
BSL 11.3: The equation of motion for forced and
free convection
Forced convection ↔ Free convection
Forced convection: direct effect of pressure and
gravitational forces
BSL 11.4: Use of the equations of change to solve
steady-state problems
Formulation in terms of different reference frames 
extensive table on BSL p. 340-341
BSL Example 11.4-2: Tangential flow in an annulus with
viscous heat generation
Free convection: indirect effect of pressure and
gravitational forces via the density
Situations are limiting cases: in reality there is a smooth
transition between the two transport mechanisms
Assumption: physical properties ρ, μ and κ are constant
BSL 11.4: Use of the equations of change to solve
steady-state problems
ρ CˆV
DT
∂p
= −(∇ ⋅ q ) − T   ( ∇ ⋅ v ) − (−τ : ∇v )
Dt
 ∂T  ρ
• incompressible flow
• Newtonian fluid with constant viscosity
• isotropic heat conduction with constant k
DT
ρ CˆV
= k ∇ 2T + µΦ v
Dt
∂T
∂T vθ ∂T
∂T 
 1 ∂  ∂T  1 ∂ 2T ∂ 2T 
k
ρ CˆV  + vr
+ =
+ vz
 r ∂r  r ∂r  + r 2 ∂θ 2 + ∂z 2  + µΦ v
∂r r ∂θ
∂z 
 ∂t


• steady state conditions, only flow in θ-direction
• rotational symmetry, no θ-direction of state variables
• infinite size in z-direction: no z-dependence
BSL 11.4: Use of the equations of change to solve
steady-state problems
=
0 k
1 ∂  ∂T 
r
 + µΦ v
r ∂r  ∂r 
reduced thermal
energy equation
• steady state conditions, only flow in θ-direction
• rotational symmetry, no θ-dependence of state variables
• infinite size in z-direction: no z-dependence of state variables
 ∂vr  2  1 ∂vθ vr  2  ∂vz  2   ∂  vθ  1 ∂vr 
=
Φ v 2 
+  +
 +
  + r   +

 ∂r   r ∂θ r   ∂z    ∂r  r  r ∂θ 
2
 1 ∂vz ∂vθ   ∂vr ∂vz  2
+
+
+
+
− [ (∇ ⋅ v ) ]


 r ∂θ ∂z   ∂z ∂r  3
2
2
 ∂ v
Φ v = r  θ
 ∂r  r



2
reduced viscous
dissipation function
2
BSL 11.4: Use of the equations of change to solve
steady-state problems
Velocity profile (BSL 3):
 r −κR


κ
R r 

vθ = Ω0 R
 1 −κ 


κ

Equation of change for thermal energy:
2
1 d  dT 
 d  vθ  
0 k
r
 + µ r  
r dr  dr 
 dr  r  
Substitution of the equation for vθ(r) in the thermal energy
equation gives:
0
1 d  dT  4 µΩ02 R 4κ 4 1
k
r
+
r dr  dr  (1 − κ 2 )2 r 4
BSL 11.4: Use of the equations of change to solve
steady-state problems
with:
N
Dimensionless quantities:
• Dimensionless radius ξ:
r
ξ=
R
• Dimensionless temperature Θ:
T − Tκ
Θ=
T1 − Tκ
• Dimension thermal energy equation:
=
0
1 d  dΘ 
1
ξ
+ 4N 4


ξ dξ  dξ 
ξ
BSL 11.4: Use of the equations of change to solve
steady-state problems
Boundary conditions:
µΩ02 R 2
κ4
κ4
Br
=
2
k (T1 − Tκ ) (1 − κ 2 )2
(1 − κ 2 )
and Br the dimensionless Brinkman number:
Br =
BSL 11.4: Use of the equations of change to solve
steady-state problems
µΩ02 R 2
k (T1 − Tκ )
Integration (twice) of the dimensionless thermal energy
equation:
Θ=−
N
ξ2
+ C1 ln (ξ ) + C2
ξ κ Θ
=
= 0
ξ= 1 Θ
= 1
Solution (radial temperature profile):
N 
N  ln (ξ )

Θ= ( N + 1) − 2  − ( N + 1) − 2 
ξ  
κ  ln (κ )

For sufficiently high values of N (for sufficiently strong
viscous dissipation) a maximum in the temperature
distribution occurs:
2 ln(1/ κ )
ξm =
(1/ κ 2 ) − (1 + 1/ N )
BSL 12.1: Unsteady heat conduction in solids
Energy equation for solids:
∂T
= − ( ∇ ⋅ q ) = ( ∇ ⋅ k ∇T )
ρC p
∂t
For constant thermal conductivity k the heat diffusion
equation is obtained with α=k/(ρCp):
Advanced Transport Phenomena
BSL 12: Temperature distributions
with more than one independent
variable
k
∂T
=
∇ 2T =α∇ 2T
∂t ρ C p
J.A.M. Kuipers
Reference for solving heat diffusion equation for different
geometries and boundary conditions: Conduction of Heat
in Solids by H.S. Carslaw and J.C. Jaeger (CJ)
Laplace transformation is used in CJ to obtain analytical solutions,
we will use separation of independent variables requiring concept of
orthogonality of functions
Orthogonality of functions
Orthogonality
• ORTHOGONALITY OF VECTORS
⋅ g)
(f =
n
g
∑ f=
i =1
j
j
b
0
• FUNCTION VALUES OF f AND g AT n EQUALLY SPACED POINTS
[ f ( h)
f (nh) ]
[ g ( h)
g (nh) ] h =
• ORTHOGONALITY RELATION FOR f AND g
n
∑ f ( jh) g ( jh) = 0
multiply by h
j =1
n
h∑ f ( jh) g ( jh) = 0
j =1
h→0
n→∞
a
f and g are orthogonal
on the interval [a,b]
∫ f ( x) g ( x)dx = 0
a
b
dx
∫ φ ( x)φ ( x)=
i
j
0 i≠ j
a
b−a
n +1
b
∫ f ( x) g ( x)dx = 0
orthogonality of functions
f(x) and g(x) and functions
of the same set
(basis functions)
b
orthogonality of functions
with respect to a
weight function w(x)
∫ f ( x) g ( x)w( x)dx = 0
a
b
)dx
∫ φ ( x)φ ( x)w( x=
i
j
0
i≠ j
a
concept of orthogonality is very important for the solution of PDE’s
where a prescribed function needs to be represented in basis functions
satisfying the spatial part of the PDE leading to Sturm-Liouville problems
Sturm-Liouville Theory
d 
dy 
0
 p ( x)  + (q ( x) + λ w( x)) y =
dx 
dx 
Orthogonality + Sturm-Liouville Theory
i =∞
differential equation for
y(x) on interval [a,b]
β1 y (b) + β 2 y (b) =
0
b
∫
boundary conditions
a
solution exists only for distinct values of parameter λi=
( i 1,.., ∞ )
termed the eigenvalues, the corresponding solutions are the
=
eigenfunctions yi ( x) which comprise an orthogonalset with
respect to the weight function w( x)
b
)dx 0
∫ yi ( x) y j ( x)w( x=
a
BSL 12.1: Unsteady heat conduction in solids
BSL Example 12.1-2: Heating of a finite solid slab
Heat diffusion equation
∂T
∂T
=α 2
∂t
∂y
2
Initial condition
=
t 0 -b ≤ y ≤ b: =
T T0
Boundary conditions
t > 0 y= -b : T= T1
t >0 y=
+b : T =
T1
b
 i =∞

f ( x)φ j ( x) w( x)dx = ∫  ∑ K iφi ( x) φ j ( x)( w( x)dx

a  i =1
b
b

2
(
)
(
)
(
)
K
φ
x
φ
x
w
x
dx
K
=

∑
i ∫ i
j
i ∫ φi ( x ) w( x ) dx
i =1
a
a

i =∞
b
i≠ j
note: parameter λ=µ2 is the separation constant introduced
through the method of separation of independent variables
in the original PDE
the Sturm-Liouville problem description
i =1
α1 y ( a ) + α 2 y ' ( a ) =
0
homogeneous two-point
'
expand f ( x) in eigenfunctions φi ( x) satisfying
f ( x) = ∑ K iφi ( x)
∫ f ( x)φ ( x)w( x)dx
i
a
⇒ Ki =
b
∫φ
2
i
( x) w( x)dx
multiply with φ j ( x) w( x)dx
and integrate over[a,b]
use orthogonality of
eigenfunctions with
respect to w(x)
integrals can be evaluated using
standard expressions for the
specific eigenfunctions: see book
on Fourier Analysis by
M.R. Spiegel (canvas)
a
BSL 12.1: Unsteady heat conduction in solids
Dimensionless temperature:
T −T
Θ= 1
T1 − T0
Dimensionless distance:
η=
y
b
Dimensionless time:
τ=
αt
b2
BSL 12.1: Unsteady heat conduction in solids
BSL 12.1: Unsteady heat conduction in solids
Solving the problem using separation of variables
Dimensionless problem description
∂Θ ∂ 2 Θ
=
∂τ ∂η 2
=
Θ f (η ) ⋅ g (τ )
Try a solution of the following form:
Inserting this in the dimensionless PDE and dividing by
=
Θ f (η ) ⋅ g (τ ) gives after separation:
Dimensionless initial & boundary conditions:
=
Θ 1
τ 0 -1 ≤ η ≤ 1:=
dg
= −µ 2 g
dτ
and:
and:
τ >0=
η -1: =
Θ 0
τ > 0 η = +1: Θ = 0
d2 f
= −µ 2 f
2
dη
where µ2 is the separation constant
BSL 12.1: Unsteady heat conduction in solids
Solving the ordinary DE’s gives the general solution:
=
Θ exp ( − µ 2τ ) [ B ⋅ sin ( µη ) + C ⋅ cos ( µη )]
Based on the symmetry of the problem, B must equal 0!
Based on the boundary condition at η=1:
C ⋅ cos ( µ ) =
0
From these equations, the eigenvalues µn emerge:
µn= ( n + ) π n= 0,1, 2,…, ∞
1
2
and the corresponding eigenfunctions:
cos =
( µnη ) cos ( ( n + 12 ) πη )
BSL 12.1: Unsteady heat conduction in solids
Applying the superposition principle:
=
Θ
∞
∑
n =0
(
)
K n exp − ( n + 12 ) π 2τ ⋅ cos ( ( n + 12 ) πη )
2
Applying the initial condition:
=
1
∞
∑
n =0
K n cos ( ( n + 12 ) πη )
Using the Sturm-Liouville’s theorem gives:
1
∫ cos
( ( n + 12 ) πη )dη
2 ( −1)
=
Kn =
2
1
( n + 12 ) π
∫ cos ( ( n + 2 ) πη )dη
0
1
0
n
BSL 12.1: Unsteady heat conduction in solids
BSL 12.1: Unsteady heat conduction in solids
Inserting this gives the following solution in terms of the
original variables:
T1 − T
T1 − T0
( −1) exp  − n + 1 2 π 2 α t  ⋅ cos  n + 1 π y 
2∑
 (

(

2)
2)
n =0 ( n + 1 ) π
b2 
b 


2
∞
n
Derived quantities like average temperature and the heat
flux at the wall can be obtained respectively by integration
and differentiation term by term of this series
BSL 12.1: Unsteady heat conduction in solids
BSL 12.1: Unsteady heat conduction in solids
Unsteady heat conduction in an infinitely long cylinder
Dimensionless temperature:
Heat diffusion equation:
T −T
Θ= 1
T1 − T0
∂T
1 ∂  ∂T 
=α
r

∂t
r ∂r  ∂r 
Dimensionless distance:
Initial condition:
η=
t=
T0
0 r ≤ R: T =
Dimensionless time:
Boundary conditions:
t > 0=
r 0:
τ=
T (r ) ≤ M
t > 0 r= R : T= T1
r
R
this boundary condition is equivalent to:
∂T
∂r
=0
r =0
αt
R2
BSL 12.1: Unsteady heat conduction in solids
BSL 12.1: Unsteady heat conduction in solids
Solving the problem using separation of variables
Dimensionless problem description:
∂Θ 1 ∂  ∂Θ 
η
=
∂τ η ∂η  ∂η 
=
Θ f (η ) ⋅ g (τ )
Try a solution of the following form:
Inserting this in the dimensionless PDE and dividing by
=
Θ f (η ) ⋅ g (τ ) gives after separation:
Dimensionless boundary conditions:
=
Θ 1
τ 0 η ≤ 1: =
dg
= −µ 2 g
dτ
and:
and:
1 d  df 
η  = −µ 2 f

η dη  dη 
τ ≥ 0 for all η ≤ 1: Θ ≤ N
τ ≥0 =
η 1: =
Θ 0
where µ2 is the separation constant
BSL 12.1: Unsteady heat conduction in solids
Solving the ordinary DE’s gives the general solution:
=
Θ exp ( − µ 2τ ) [ B ⋅ J 0 ( µη ) + C ⋅ Y0 ( µη )]
where J0 is the 0th order Bessel function of the 1st kind and
Y0 the 0th order Bessel function of the 2nd kind (Neumann
function)
Bessel’s differential
equation
0
x 2 y '' + xy ' + ( µ 2 x 2 − n 2 ) y =
non-integer n: y(x)=K1 J n ( µ x) + K 2 J − n ( µ x)
for integer n: y(x)=K1 J n ( µ x) + K 2Yn ( µ x)
see book of
M.R. Spiegel
“Fourier Analysis”
(canvas)
BSL 12.1: Unsteady heat conduction in solids
Based on the natural boundary condition, C equals 0
Based on the boundary condition at η=1 we obtain:
B ⋅ J0 ( µ ) =
0
These equations give the eigenvalues µn:
=
µn β n =
n 1, 2,3,…, ∞
and the corresponding eigenfunctions:
J 0 ( β nη ) = 0
where βn are the roots of J0
J0 ( x) = 1−
x2
x4
x6
+
−
+ ⋅⋅⋅
22 22 42 22 42 62
first six roots of J0
∆β n ≈ π for large n
BSL 12.1: Unsteady heat conduction in solids
This result can easily be obtained using standard relations
for Bessel functions (handout + book of Spiegel)
Applying the superposition principle:
Θ
∞
∑
n =1
K n exp ( − β n2τ ) J 0 ( β nη )
Inserting gives the following solution in terms of the
original variables:
Applying the initial condition:
1=
∞
∑
n =1
K n J 0 ( β nη )
Use orthogonality of Bessel functions:
1
Kn
∫η J0
( β nη )dη
2
=
2
β n J1 ( β n )
∫ η J 0 ( β nη )dη
0
1
0
BSL 12.1: Unsteady heat conduction in solids
1
( µ x)dx
∫ xJ =
0
2
n
d n
( x J n ( x) ) = x n J n−1 ( x)
dx
d −n
( x J n ( x) ) = − x − n J n+1 ( x)
dx
1 '
n2
2
2
 ( J n ( µ )) + (1 − 2 )( J n ( µ )) 
2
µ

∞
T1 − T
αt   β r 
1

=
2∑
exp  − β n2 2  ⋅ J 0  n 
n =1 β J ( β )
T1 − T0
R 

 R 
n 1
n
Derived quantities like average temperature and the heat
flux at the wall can be obtained respectively by integration
and differentiation term by term of this series
see “Fourier Analysis” by M.R. Spiegel
BSL 12.1: Unsteady heat conduction in solids
BSL 12.1: Unsteady heat conduction in solids
Unsteady heat conduction in a sphere
Heat diffusion equation:
∂T
1 ∂  ∂T 
= α 2  r2

∂t
r ∂r  ∂r 
Initial condition:
t=
T0
0 r ≤ R: T =
Boundary conditions:
t > 0 for all r ≤ R :
t > 0 r= R : T= T1
T (r ) ≤ M
BSL 12.1: Unsteady heat conduction in solids
Dimensionless temperature:
T −T
Θ= 1
T1 − T0
Dimensionless distance:
η=
r
R
Dimensionless time:
τ=
αt
R2
BSL 12.1: Unsteady heat conduction in solids
Solving the problem using separation of variables
Θ f (η ) ⋅ g (τ )
Try a solution of the following form:=
Inserting this in the dimensionless PDE and dividing by
=
Θ f (η ) ⋅ g (τ ) gives after separation:
dg
= −µ 2 g
dτ
and:
1 d  2 df 
η
= −µ 2 f


2
η dη  dη 
where µ2 is the separation constant
BSL 12.1: Unsteady heat conduction in solids
Dimensionless problem description:
∂Θ 1 ∂  2 ∂Θ 
=
η
∂τ η 2 ∂η  ∂η 
Dimensionless boundary conditions:
=
τ 0 η ≤ 1: =
Θ 1
and:
τ ≥ 0 for all η ≤ 1: Θ ≤ N
τ ≥0 =
η 1: =
Θ 0
BSL 12.1: Unsteady heat conduction in solids
Solving the ordinary DE’s gives the general solution:
[ B ⋅ sin ( µη ) + C ⋅ cos ( µη )]
=
Θ exp ( − µ 2τ )
η
Based on the natural boundary condition, C equals 0
Based on the boundary condition at η=1 we obtain:
B ⋅ sin ( µ ) =
0
These equations give the eigenvalues µn:
µn = n ⋅ π n = 1, 2,3,…, ∞
and the corresponding eigenfunctions:
sin ( nπη ) η
BSL 12.1: Unsteady heat conduction in solids
Applying the superposition principle:
=
Θ
∞
∑
n =1
(
K n exp − ( nπ ) τ
2
) sin (ηnπη )
Applying the initial condition:
1=
∞
∑
n =1
Kn
BSL 12.1: Unsteady heat conduction in solids
Inserting gives the following solution in terms of the
original variables:
∞ ( −1)
T1 − T
2∑
=
n =1
T1 − T0
nπ
n +1
R
nπ r 
2 αt
exp  − ( nπ ) 2  sin 

R r

 R 
Derived quantities like average temperature and the heat
flux at the wall can be obtained respectively by integration
and differentiation term by term of this series
sin ( nπη )
η
Using Sturm-Liouiville’s theorem gives:
1
Kn
∫ η ⋅ sin
0
( nπη )dη
( −1)
2
=
1
2
nπ
∫ sin ( nπη )dη
0
n +1
BSL 12.1: Unsteady heat conduction in solids
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
General procedure for analysis of heat conduction in
flowing fluids
•Solve microscopic momentum balance  velocity profile
•Combine velocity profile with thermal energy equation and
solve the resulting equation  temperature profile
Condition: temperature dependency of viscosity can be
neglected.
Due to the T-dependence of viscosity, this is a critical
condition for liquids
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Heat conduction in laminar tube flow
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Solution of this PDE: three solutions can be distinguished:
1. Heat conduction in laminar tube flow due to a stepwise
change in the wall temperature (Graetz-Nusselt problem)
a. Complete solution of the PDE by separation of variables
(BSL Example 12.2-1)
2. Heat conduction in laminar tube flow due to a constant
heat flux imposed at the wall
b. Asymptotic solution of the PDE for short distances down
the tube by combination of variables (BSL Example 12.22)
We will focus on problem type 2.
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Asymptotic solution of the PDE for short distances down
the tube by combination of variables (BSL Example 12.2-2)
Microscopic balance for thermal energy in original form:
  r  2  ∂T
1 ∂  ∂T 
vz ,max 1 −   
α
=
r

R
z
r
r
∂
∂


 ∂r 


where vz,max is the velocity in the centre of the tube:
vz ,max
p0 − pL ) R 2
(
=
4µ L
c. Asymptotic solution for large distances down the tube
(BSL §10.8)
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Simplification for working out the problem:
a. Curvature is neglected; instead of radial coordinate r, here
s=R - r (distance to the tube wall) is used as a new variable
b. Fluid is unbounded in the s-direction (s=0 to s=∞)
c. The velocity profile near the tube wall is linearized, due to
the small heat penetration depth in the s-direction:
2s
s
 dv 
vz ( s ) =
vz ( 0 ) +  z  s =
vz ,max   =
v0
R
R
 ds  s =0
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Inserting the approximations in the microscopic balance for
heat:
∂T
∂ 2T
s
=α 2
v0
R ∂z
∂s
with boundary conditions:
( )
s=
T0
∞: T =
∂T
0=
: -k
q1
∂s
: T T0
z 0=
=
s
Note that two out of three boundary conditions are identical
which is essential for the method of combination of
independent variables to work!
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Dimensionless problem description
Dimensionless thermal energy equation:
∂Θ ∂ 2 Θ
η
=
∂λ ∂η 2
Dimensionless boundary conditions:
η =∞: Θ =0
∂Θ
η 0=
: 1
=
∂η
λ 0: =
=
Θ 0
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Dimensionless variables
Dimensionless temperature:
Θ=
T − T0
q1 R k
Dimensionless radial coordinate:
η=
s
R
Dimensionless axial coordinate:
zα
λ=
v0 R 2
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
For this problem it is easier to generate the solution in
terms of the dimensionless heat flux Ψ:
∞
Θ(∞, λ ) − Θ(η , λ ) =
−Θ(η , λ ) =
− ∫ Ψ (η , λ ) dη
η
Inserting this into the dimensionless PDE:
η
∂ ∞
 − ∂Ψ
∫ Ψ (η , λ ) dη =

∂λ η
∂η
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Dividing by η and differentiation to η gives:
∂Ψ ∂  1 ∂Ψ 
=
∂λ ∂η  η ∂η 
with the boundary conditions:
assuming =
ω η a λ b ⇒ a + 3=
b 0 to obtain ODE for F(ω )
Since F=1 at ω=0 it follows that K2=1
Using the boundary condition F=0 at ω=∞ gives:
−1
K1 = ∞
3
∫ u ⋅ exp ( −u ) du
0
∫ u ⋅ exp
( −u ) du − ∫ u ⋅ exp ( −u ) du
∫ u ⋅ exp ( −u ) du
ω
3
( −u ) du
∫ u ⋅ exp ( −u ) du
∫ u ⋅ exp
F = Ψ = ω∞
0
3
3
0
∞
F = K1 ∫ u ⋅ exp ( −u 3 ) du + K 2
ω
0
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Rewriting in standard form using gamma-functions (x=u3):
∞
Ψ=
∫
ω
3
x 2 3−1e − x dx
∞
2 3−1 − x
e dx
∫x
= 1−
3
3
Γin ( 23 , ω 3 )
Γ ( 23 )
where Γ(z) is the complete gamma-function and Γin(z,y) is
the incomplete gamma-function:
0
∞
or:
Separation and integration yields:
0
Inserting the constants gives:
0
dF
=K1ω ⋅ exp ( −ω 3 )
dω
where K1 is an integration constant
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
F=
ω F " + ( 3ω 3 − 1) F ' =
0
G=
Combination of independent variables: Try a solution of
the form: Ψ =F (ω ) where ω is given by:
∞
Transformation of PDE gives the following ODE:
Lowering the order F ' = G and integration of the ODE gives:
=
λ 0: =
Ψ 0
=
η 0: =
Ψ 1
η =∞: Ψ =0
η
ω=
13
( 9λ )
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
∞
z −1 − x
Γ( z) =
∫ x e dx
0
y
z −1 − x
Γin ( z , y ) =
∫ x e dx
0
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Resulting expression for the dimensionless heat flux:
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Working out the integral gives:
 2 η3 
Γ in  3 , 
qs
9λ
Ψ=
= 1−  2 
q1
Γ( 3)


 −η 3 
 2 η 3  
 exp  9λ  η  Γ in  3 , 9λ  

−

 
3
9λ 
=
Θ
1 −

3
2
2
Γ ( 3 ) 
9λ 
 Γ( 3)

Derivation of the expression for the temperature profile can



easily be obtained from the expression for Ψ by integration
(Fourier’s law):
This is the final expression for the dimensionless
temperature profile in the entrance zone of the tube
2 u3 

use integration by
Γin ( , ) 
∞
∞
parts and Leibnitz
3
due to “thermal saturation” the wall
3 9λ du formulae
9λ
Θ(η , λ ) =
∫ Ψ ( u , λ ) du =−
∫ 1
to differentiate

Θ
(0,
λ
)
=
temperature
(at η=0) must increase in order
2
η
η
the incomplete
Γ ( 23 )

Γ( ) 
to
maintain
the
imposed constant heat flux
gamma function
3


BSL 12.2: Steady heat conduction in laminar,
incompressible flow
The solution for the case that the tube wall is at a constant
temperature T1 (at z>0) can be derived analogously:
 1 η3 
Γin  3 , 
T1 − T
 9λ 
=Θ=
T1 − T0
Γ ( 13 )
The heat flux at the tube wall:
−k
∂T
∂r
r=R
k
k
3
∂Θ
−1 3
=
− (T1 − T0 )
=
− (T1 − T0 ) 1 ( 9λ )
R
∂η η =0
R
Γ(3)
BSL 12.2: Steady heat conduction in laminar,
incompressible flow
Using the definition of the Nusselt number gives (with
Γ(1/3)=2.679):
∂T
−1 3
2R
−k
6
241 3  α z 
∂r r = R
−1 3
Nu= = =
( 9λ )


k (T1 − T0 )
Γ ( 13 )
Γ ( 13 )  v d 2 
z

= 1.08  Gz ⋅ 
L

−1 3
which is the well-known solution for the Nusselt number in
the thermally non-developed regime
BSL 12.4: Boundary layer theory for nonisothermal
flow
Extension of the boundary layer theory to simultaneous
momentum and heat transport
Model system: laminar flow along a flat heated plate
BSL 12.4: Boundary layer theory for nonisothermal
flow
Quantitative description
Continuity equation:
∂vx ∂v y
+
=
0
∂x ∂y
Momentum equation x-direction:
∂vx
∂vx
∂ 2 vx
vx
+ vy
=
ν 2
∂x
∂y
∂y
Thermal energy equation:
∂T
∂T
∂ 2T
vx
+ vy
=
α 2
∂x
∂y
∂y
BSL 12.4: Boundary layer theory for nonisothermal
flow
Solving the micro balances using a boundary layer approach
Dimensionless velocity φ:
v0 − vx vx
y
η
= = φ (η ) with: =
v0 − v∞ v∞
δ ( x)
Dimensionless temperature Θ:
T0 − T
y
=
Θ (ηT ) with: ηT =
T0 − T∞
δT ( x )
where δ is the thickness of the momentum boundary layer
and δT is the thickness of the thermal boundary layer
BSL 12.4: Boundary layer theory for nonisothermal
flow
Assume δ T = δ∆ and distinguish Δ≤1 and Δ ≥1
Solve the heat transfer problem for Δ≤ 1:
combination of the continuity equation and the thermal
energy equation (elimination of vy):
vx
φ
{
}
{
}
∂T y ∂vx
∂T
∂ 2T
− ∫
=
α 2
( x, yˆ )dyˆ
∂x 0 ∂x
∂y
∂y
∂Θ y ∂φ
∂Θ α ∂ 2 Θ
− ∫ ( x, yˆ )dyˆ
=
∂x 0 ∂x
∂y v∞ ∂y 2
BSL 12.4: Boundary layer theory for nonisothermal
flow
Evaluation of terms (chain rule)
η dδ
1 dδT
∂Θ
= −Θ' (ηT ) T T = −ηT Θ' (ηT )
∂x
δ T dx
δ T dx
η dδ
1 dδ
∂φ
=
−φ ' (η )
=
−ηφ ' (η )
δ dx
δ dx
∂x
∂Θ 1 '
=
Θ (ηT )
∂y δ T
BSL 12.4: Boundary layer theory for nonisothermal
flow
Insertion into the thermal energy equation:
−φ (η ) {ηT Θ' (ηT )}
{
}
1 dδT η '
dδ 1 '
α 1 "
+ ∫ uφ du
Θ (ηT ) =
Θ (ηT )
0
dx δ T
v∞ δ T2
δ T dx
Multiply by δTδT and use:
δT =
δ∆ and: η =
ηT ∆
Insertion gives:
−φ (ηT ∆ )ηT Θ' (ηT )∆ 2δ
∂Θ 1 "
=
Θ (ηT )
∂y 2 δ T2
2
dδ
+
dx
{
ηT ∆
∫
0
}
uφ ' du Θ' (ηT )∆δ
dδ α "
= Θ (ηT )
dx v∞
BSL 12.4: Boundary layer theory for nonisothermal
flow
BSL 12.4: Boundary layer theory for nonisothermal
flow
Multiplication of equation with dηT and integration of the
equation over the thermal boundary layer gives:
Earlier obtained result for hydrodynamics (BSL 3)
( E − D)δ
dδ α
F
=
dx v∞
in which the numbers D, E and F follow from:
1
E =∆ ∫ Θ
0
'
{
ηT ∆
∫
0
}
uφ du dηT
'
and:
dδ ν
C
=
dx v∞
in which the numbers A, B and C follow from:
1
B = ∫φ'
and:
0
{ uφ du}dη
η
∫
1
0
1
A = ∫ φφ 'η dη
0
and:
1
"
∫Θ
0
dηT
'
0
1
D =∆ 2 ∫ φ (ηT ∆ ) Θ'ηT dηT
and
F=
( B − A) δ
C = ∫ φ "dη
0
BSL 12.4: Boundary layer theory for nonisothermal
flow
BSL 12.4: Boundary layer theory for nonisothermal
flow
Evaluation of the numbers A to F
Division of the left hand and right hand side of the two ODE’s:
dδ α F
dx = v∞
dδ ν
C
( B − A) δ
dx v∞
( E − D )δ
( E − D)
F
C
α 1
= f (∆) =
=
( B − A)
ν Pr
rate of momentum diffusion
Pr =
rate of heat diffusion
The ratio between the thicknesses of the boundary layers is a
function of the Prandtl-number only !!!
BSL 12.4: Boundary layer theory for nonisothermal
flow
Inserting into the equation for A to F gives:
Hydrodynamic boundary layer thickness δ:
1260 ν x
νx
δ = 5.84
37 v∞
v∞
Ratio parameter Δ:
1 3
3 5
1 6 37 1
∆ −
∆ +
∆ =
15
280
360
630 Pr
Approximating equation for ratio parameter Δ:
∆ =Pr −1 3
Temperature profile in thermal boundary layer:
3
T0 − T
y
y
y
=2   − 2   +  
T0 − T∞
 δ∆   δ∆   δ∆ 
4
polynomial representation of the dimensionless velocity and
temperature profiles within their respective boundary layers:
coefficients follow from the boundary conditions and PDE’s!
Dimensionless velocity distribution:
φ =2η − 2η 3 + η 4
Dimensionless temperature distribution:
=
Θ 2ηT − 2ηT3 + ηT4
BSL 12.4: Boundary layer theory for nonisothermal
flow
Local heat transfer coefficient αx:
−k
α=
x
∂T
∂y
1
1
2k
2k  v∞ x  2 ν  3
k 12 13
= =
= 0.342 Re x Pr
T0 − T∞
x
δ∆ 5.84 x  ν  α 
y =0
Local Nusselt-number Nux:
Nu
=
x
αxx
1
1
= 0.342 Re 2x Pr 3
k
Extension to more complicated situations:
• “Boundary layer theory”, Schlichting
• “Convective Heat and Mass Transfer”, Kays & Crawford