Lecture 15: Capillary motion
Capillary motion is any flow governed by forces
associated with surface tension.
Examples: paper towels, sponges, wicking fabrics. Their
pores act as small capillaries, absorbing a comparatively
large amount of liquid.
Water absorption by
paper towel
Capillary flow in a brick
Height of a meniscus
The meniscus will be approximately hemispherical with
a constant radius of curvature,
R
h0
a
R1  R 2 
cos
θ
Applying the Young-Laplace equation we obtain
a
Hence,
patm  pliq  gh 0 
2c
 cos
h0  2
 2 cos
g a
a
2 cos
a

 
g
2
c
is the capillary
length.
h0 may be positive and negative, e.g. for mercury θ~1400 and the
meniscus will fall, not rise.
For water, α=73*10-3N/m, and in 0.1mm radius clean glass capillary,
h0=15cm.
Let us calculate the rate at which the meniscus rises to
the height h0.
Assume that the velocity profile is given by the Poiseuille
a
profile,
A 2 2
Aa 4
v  a  r ; Q  2  vrdr 
4
8
0
The average velocity is
Q
Aa 2 p1  p 2 a 2 dh
v  2


a
8
h 8 dt
Here h  h 0 is the
instantaneous distance of
the meniscus above the
pool level.
The pressure difference at the pool level, p1, and at the top of the
capillary (just under the meniscus) , p2, is
2 cos
p1  p 2 
Thus,
2
dh  2 cos
a


 gh 
dt  a
 8h
a
 gh
Or, separating the variables,
8
h dh
8 h d h

 dt
2
2
ga  2 cos
 ga h 0  h
 g a  h 


For integration, it is also continent to rearrange the terms in the rhs

h0  h  h0
h0 
h dh

d h   1 
dh

h0  h
h0  h
 h0  h 
Integration gives
8
 h  h 0 lnh 0  h   t  c
2
ga
The constant of integration c can be determined from initial condition,
h  0 at t  0 . Hence,
8
0  h 0 ln h 0   c
2
ga
Finally,
h 0  8h 0 
h0
8 
h
t
 h  h 0 ln

ln
 
2 
2 

ga 
h 0  h  ga  h 0  h h 0 
Or, introducing  
8h 0
, we obtain
2
ga

h0
h
t   ln
 
 h0  h h0 
As h  h 0 ,
h/h0
t

 ln
h0
h0  h

 t 
h  h 0 1  exp  
  

For water in a glass capillary
of 0.1mm radius,
t/τ
  12s
For this solution, we assumed the steady Poiseuille flow profile.
This assumption is not true until a fully developed profile is attained,
which implies that our solution is valid only for times
a2
t 
For water in a capillary tube of 0.1mm radius, a 2

~ 10 2 s

Lecture 16: Non-isothermal flow
• Conservation of energy in ideal fluid
• The general equation of heat transfer
• General governing equations for a singlephase fluid
• Governing equations for non-isothermal
incompressible flow
Conservation of energy in ideal
fluid

 v 2


 e  -- total energy of unit volume of fluid
2


kinetic internal energy,
energy
e is the internal energy per unit mass
2

Let us analyse how the energy varies with time:   v  e  .

t  2

For derivations, we will use the continuity and Euler’s equation (NavierStokes equation for an inviscid fluid):


 div v  0
t


v 
p
 v   v  
t





  v 2
 v 2 v 2 

e



e





e



1: 
(differentiation of a product)

t  2

t
2
2

t

t

t




 v

e  v 2
(use of continuity equation)
 v 

   e  div v 
t
t  2



  
 p 
 (use of Euler’s
e  v 2
 v   v   v 





e

div

v
 2


equation)


t




Next, we will use the following vector identity (to re-write the first term):


 


v2
v2
v  v   v  v iv k kv i  v k k
 v   
2
2
and the 1st law of thermodynamics (applied for a fluid particle of unit mass,
V=1/ρ):
p
 1
d e  T d S  p d   T d S  2 d 


Equation (1) takes the following form:





 v2 

  v 2
S p   v 2

 e    v    v  p  T

   e  div v 
2:
t  2
2
t  t  2


(use of
2
2
 v


S  v
p
continuity
  v    v  p  T
   e   div v
2
t  2

equation)
We will also use the enthalpy per unit mass (V=1/ρ) defined as
h e 
p

dh  de 
dp


p
2
d   T dS 
dp


dp

 d h  T dS
Equation (2) will now read





 v2



  v 2
S  v 2
 e    v    v  h  Tv  S  T
   h  div v 
3: 
t  2
2
t  2



  v 2

 S 

  div  v   h   T 
 v   S 
 t


  2
If a fluid particle moves reversibly (without loss or dissipation of energy),
then
d S S 

 v   S  0
dt
t
Finally,


  v 2


  v 2
conservation of energy

 e    div  v   h 
for an ideal fluid
t  2


  2


 v 2
v   h  -- energy flux
 2

In integral form,
using Gauss’s theorem



  v 2


v 2

  v 2

 e  dV    div  v   h  dV    v n   h  d S 
t V  2


 2

  2
V
S

v 2

   v n   e  d S   pvn dS
 2

S
S
energy
transported
by the mass
of fluid
work done by
the pressure
forces
The general equation of heat
transfer


  v 2


  v 2

 e    div  v   h 
t  2


  2
conservation of energy
for an ideal fluid
The conservation of energy still holds for a real fluid, but the energy
flux must include

(a) the flux due to processes of internal friction (viscous heating), v 
(b) the flux due to thermal conduction (molecular transfer of energy
from hot to cold regions; does not involve macroscopic motion).
For (b), assume that

q
(i) is related to the spatial variations of temperature field;
(ii) temperature gradients are not large.
Heat flux due to thermal q  T
conduction:
thermal conductivity
12
The conservation of energy law for a real fluid


  v
  v


 
 e    div  v   h   v   T 

t  2


 2

2
2
viscous
heating
heat
conduction
We will re-write this equation by using
(1)
(2)
(3)


 div v  0
t


v 
p 1
 v   v  
  
t
 
-- continuity equation
-- Navier-Stokes equation
p
 1
d e  T d S  p d   T d S  d 


(4) d h  d e 
2
dp


p

2
d   T dS 
-- 1st law of thermodynamics
d p -- 1st law of thermodynamics
 in terms of enthalpy h  e  p

e, h and
S are the
internal
energy,
enthalpy
and
entropy
per unit
mass
13
 v2
 vivi



 vi vi  v  v
(5)
t 2
t 2
t
t
2

v
v2
 


(6) v  v   v  v ivk kv i  vk k  v   
2
2

 
(7) a div A  A  a  ai Ai  Ai ia  iaAi  div aA
1st term in the lhs:
Differentiation
of product
(1+5)
(2)
 v
v 
 v
v
v



div v  v 

t 2
2 t
t 2
2
t
(6)


v2
p 1

div v  v    v    v 
    
2



 (4)
2
2
2
2
v2
v2

div v  v  
 v  p  v       
2
2
v2
v2

div v  v  
 v  h  Tv  S  v      
2
2
14
2nd term in the lhs:
Differentiation
of product
(3)
(1)


e

S p 

 e   e    e  T 
t
t
t
t
t  t
S p
S
 e div v  T
 div v  h div v  T
t 
t
15
LHS (1+2):
 v 2

 e  

 2

v2

v2

 S

 
 h  div v  v   
 h   T 
 v    S   v       
 t

 2

 2


t
 v2

 S

  div  v 
 h    T 
 v    S   v      
 t

 2


(7)
RHS:
 v2


 div  v 
 h   v    T 

  2

LHS=RHS (canceling like terms):
 S

 v    S   v        div v    T 
 t

T 
16
In the lhs, v        vk i ik
In the rhs, div v     ivk ik  vk i ik   ik ivk
Finally,
vi general equation of
 S

 v    S   div T    ik
xk heat transfer
 t

T 
heat gained
by unit
volume
heat
conducted into
considered
volume
energy
dissipated
into heat by
viscosity
17
Governing equations for a general
single-phase flow


 div v  0
t
-- continuity equation


 v 
   v   v   p   
 t

-- Navier-Stokes equation
vi
 S

T 
 v    S   div T    ik
xk
 t

-- general equation of heat
transfer
+ expression for the viscous stress tensor
+ equations of state: p(ρ, T) and S(ρ, T)
18
Incompressible flow
To define a thermodynamic state of a single-phase system, we need
only two independent thermodynamic variables, let us choose pressure
and temperature.
Next, we wish to analyse how fluid density can be changed.
    p ,T 
2
  
  
d    d p 
 dT
 T 
 p 
T
d 

c
2
d p   d T
 p 
c  
  
p
-- sound speed
S
c
1
 p 
a    c ,  
c
   
2
2
p
T
T

1  V 
1   






V  T 
  T 
p
V
p
-- thermal
expansion
coefficient
19
1. Typical variations of pressure in a fluid flow, p  v
2
2. Variations of density,     v    T
c 
2
3. Incompressible flow ≡ slow fluid motion,  v 
   1
c 
2
4. Hence, we can neglect variations in density field caused by pressure
variations     T
5. Similarly, for variation of entropy.
In general,
 S 
 S 
dS    d p  
 dT
 T  p
 p T
but for incompressible flow,
c
 S 
dS  
d
T

dT

T
 T 
p
p
 S  -- specific heat (capacity)
c T


T

 under constant pressure
p
p
20
For incompressible flow, the general equation of heat transfer takes the
following form:
v

T

c 
 v   T   div T   
p
 t

ik
x
i
k
Frequently,
(i) the thermal conductivity coefficient κ can be approximated as being
constant;
(ii) the effect of viscous heating is negligible.
Then, the general equation of heat transfer simplifies to

T
 v   T  T
t


c
p
-- temperature
conductivity
21
Boundary conditions for the temperature field:
1. wall:
a) given temperature, T  T
q
b) given heat flux, T

n

c) thermally insulated wall, T
0
n
wall
n
2. interface between two liquids:
T
T
T1  T2
and 1 1   2 2
n
n
22
Governing equations for
incompressible non-isothermal fluid
flow

divv  0
-- continuity equation



v 
p
 v   v  
v
t

-- Navier-Stokes equation

T
 v   T  T
t
-- general equation of heat transfer
Thermal conductivity and viscosity coefficients are assumed to be constant.
+ initial and boundary conditions
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