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Transcript
```Viscosity
Select
one
subject!!
Forfatter
Fornavn
Etternavn
Institusjon
Viscous Fluids
Fluid Flow
Characteristics
Dependency of
Viscosity on
Temperature
Non-Newtonian
Fluids
Examples
Laboratory
exercise
Viscometer
Home Up
Section 1:
Viscous Fluids
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
Horizontal flow
of viscous fluids
Continuity equation
for viscous flow
-Examples
-Laboratory
exercise
Select one
subject
Viscous flow through a
up of a bundle of
identical tubes
Viscous flow in a
cylindrical tube
Exercise: capillary
tube viscosity
measurement
Developers
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Horizontal flow of Viscous Fluids
Two large parallel
plates of area A
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
Y
t<0
-Non-Newtonian
fluids
Lower plate is set
in motion with
constant velocity
-Examples
-Laboratory
exercise
t=0
The fluid start to
move due to
motion of the plate
t is small
After a while the
state velocity
profile
t is large
To maintain this
motion, a constant
force F is required
V
v x  y, t 
Y – a very small
distance
V
vx  y 
y
x
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V
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-Viscous fluids
-Fluid flow
characteristics
The force may be expressed:
-Dependency of
viscosity on
temperature
F
V

A
Y
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
The force F per unit area A is proportional to the
velocity V in the distance Y; the constant of
proportionally  is called the viscosity of the fluid.

Developers
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F Y
A V
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
Also used is the so called
Newton`s equation of viscosity:
The shear
stress
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
dv x
 
dy
The fluid
velocity in
x- direction
The fluid
viscosity
The dimensions:
 N  dv  m   1 
   2 ,  
 

 m  dy  s  m   s 
By definition:
Explicit:
Developers
1Pa  s  1
And

N s

 2 
dv / dy  m 
N s
3

10
p
(
poise
)

10
cp.
2
m
1cp  1mPa  s
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Typical values of viscosity of some fluids:
Temperature
Celsius
Viscosity Castor
Oil, Poise[p]
Viscosity Water,
centiPoise[cp]
Viscosity Air,
Micro Poise[  p]
0
53,00
1,792
171
20
9,86
1,005
181
40
2,31
0,656
190
60
0,80
0,469
200
80
0,30
0,357
209
100
0,17
0,284
218
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Continuity equation for viscous flow
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
The flow will have the
biggest velocity at the
top (the surface )
A liquid is flowing in a
open channel
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
y
y
X
Vx  y 
Fluid flows in an open channel
A graphic presentation
of this phenomena is
shown here
The flow velocity will approximately be zero at
the bottom, due to retardation when liquid
molecules colliding with the non-moving
bottom
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-Viscous fluids
Jp – is often characterised as
the momentum intensity or
as the shear stress,
The unit :
2
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
N / m 
Transferred momentum pr.
time pr. area
-Examples
-Laboratory
exercise
vx
jp  
y
Viscosity is here defined as s
proportionality constant, similar to what
was done in the case of defining
absolute permeability
Developers
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References
The change in fluid flow
velocity pr. distance between
the two layers
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dy is the width of the box and S
= S` is the cross-section
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
j`p
y
-Examples
-Laboratory
exercise
vx
S`
S
z
x
jp
Momentum transfer between layers
S and S` in Newtonian viscous flow
The change of momentum intensity the box:
j p S  jp S    j p  jp S 
Developers
j p
y
Sdy
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The momentum density inside
the box is defined by P .
-Viscous fluids
-Fluid flow
characteristics
Combining with the equation
from last page gives us:
The change of momentum pr.
time inside the box is :
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
P
-Examples
t
-Laboratory
exercise
P
t
Sdy

j p
y
P / t is the sum of all
forces acting on the box.
Let f symbolise a force
of more general origin.
P
t

j p
y
vx   2vx f


2
t
 y

Substituting:
P  v x
v
j   x
y
f
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Viscous flow in a cylindrical tube
-Viscous fluids
To develop the
continuity equation
for this example, dr
(a thin layer), of
liquid is considered
Cross-section of
viscous flow through a
cylindrical tube
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
s`
r
r + dr
R
Viscous flow in a tube is
decreasing flow velocity,
because of the boundary
effect of the tube wall
Developers
s
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
The momentum flux through the cylindrical volume:
 j p jp 
j p S  jp S   j p 2rl  jp 2 r  dr l  
 2rdrl
r 
 r
-Examples
-Laboratory
exercise
Here the flux area S is
varying and S  S .
Defining the change in the momentum density inside the cylindrical volume,
as P
, the general equation for laminar flow is:
t
2rdrl
jp is redefined
according to the
geometric conditions
in this example:
Developers
 j p j p 
P
 
  f
t
r 
 r
The minus sign show that
the flow velocity vx is
r is increasing
v x
j p  
r
FAQ
The momentum
intensity
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
The continuity equation for viscous flow in a cylindrical tube is:
v    2v 1 v  f
 
  2 
t   r
r r  
This is under stationary
conditions.
-Examples
-Laboratory
exercise
At a constant pressure drop  p along the tube, no velocity variation is
observed v / t  0.
 2v 1 v f

 0
2
r
r r 
  v 
f
  r   r
r  r 

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-Viscous fluids
The general solution of the previous equation, found by integrating twice, is:
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
v
-Non-Newtonian
fluids
1 f 2
r  C1 ln r  C2
4
The general constants are
found by considering the
boundary conditions. For
this example the flow is
directed along the x-axis
-Examples
-Laboratory
exercise
1.
2.
Since the maximum flow velocity, vx(r=0) in
the centre of the tube is less then infinity;
C1=0
Since the flow velocity is zero along the tube
C2  1 / 4( f /  ) R 2
wall; vx(r=R)=0 and
The solution is then:


1 2 2 f
v  R r
4

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-Viscous fluids
-Fluid flow
characteristics
If the outer force, driving the liquid through the tube is the pressure drop p
along a tube length l then f  p / l and:
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
v
-Examples
Defines a the laminar viscous
flow pattern
1
p
R2  r 2
4
l


-Laboratory
exercise
R
VX
-R
This flow velocity profile in the
tube is given by the formula
above.
Developers
Velocity profile in cylindrical tube flow
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Viscous flow through a porous medium made up of a bundle
of identical tubes
The incremental flow rate through a fraction of the cross-section of a
capillary tube can be expressed:
2r 2 2 p
dq(r )  v(r )  2r  dr 
R r
dr
4
l


The total flow rate can be found by integration:
4
dq
p

R
p
2
2
q   dr 
R  r dr 

dr
2l 0
8 l
0
R
R


For the sake of convenience we may present the last equation in the
following form:
R 2 p
qA
8 l
Developers
FAQ
This is also known as the Poiseuille`s
equation, where A is the cross-section of
the capillary tube
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Considering a porous medium as a bundle of identical capillary tubes, the
total flow qp through the medium is:
Ri2 p
k p
q p   qi   Ai
A
8 l
 l
i
i
Where Ai  Ri2 is the capillary tube cross-section and A is the crosssection of the porous medium. From the equation above the permeability of
the medium where n m capillary tubes are packed together is found:

2
R
k  R2  
32
8
Where the porosity of the bundle
of capillary tubes is given by:
  /4
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Exercise: Capillary tube viscosity measurement
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
Fluid viscosity may also be estimated by the measuring the volume of fluid
flowing through a capillary tube pr. time (as the figure below).
Rewriting
the Poiseuille`s equation an expression for the dynamic fluid viscosity is
4
written:

-Non-Newtonian
fluids
R p t
8
-Examples
l V
-Laboratory
exercise
h ~ p
L
V
2R
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
For horizontal flow through a capillary tube of length L and radius R, the
time t it takes to fill a certain volume V is measured. The flow pressure  p
, is fixed during the process, by maintaining a constant fluid level at a certain
elevation above the capillary tube.
The accuracy in these measurements is strongly related to the fabrication
accuracy of the capillary tubes, as can be seen from the formula above. If the
relative uncertainty is considered,


4
R
R
Which means that if the relative accuracy in the tube-radius is +/- 2-3 %, then
the relative accuracy in the viscosity is about +/- 10 %. A small variation in
capillary tube fabrication induces large uncertainty in the viscosity
measurement.
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Section 2:
Fluid Flow Characteristics
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
For laminar flow in a cylindrical tube we have derived Poiseuille`s equation:
q R 2 p
v 
A 8 l
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
For turbulent flow in a cylindrical tube, an empirical law, called Fanning`s
equation has been found:
R p
v 
F l
2
F – the Fanning
friction factor
Developers
FAQ
References
This factor is dependent of
the tube surface
roughness, but also on the
flow regime established in
the tube
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
The flow regime is correlated to the Reynolds number, which characterises
the fluid flow in the tube. The Reynolds number, Re, is defined as a
dimensionless number balancing the turbulent and the viscous (laminar)
flow [ p / l T / p / l L ].
2R is the spatial dimension where the flow
2 Rv
occur – the diameter of a capillary tube or
Re 
width of an open channel.

v – the average flow velocity
 – density
– the fluid viscosity
From experimental studies an upper limit for laminar flow has been defined
at a Reynolds number; Re =2000. Above this number, turbulent flow will
dominate. (This limit is not absolute and may therefore change somewhat
depending on the experimental conditions.)
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
In the case of porous flow, the velocity v in the previous equation should be
the pore flow velocity, where v pore  q /(A(1  S or )) and R is the pore
radius. The Reynolds number for flow in a porous medium is:
-Non-Newtonian
fluids
Re 
-Examples
-Laboratory
exercise
2v B 
 1  S or 
vB- the bulk velocity
Typical parameters for laboratory
liquid flow experiments:
R

vB


S or
pore dimension
bulk velocity
10m  10 106 m
1cm / s  0,01m / s
1,0 g / cm3  1000kg / m3
porosity
0,25
1cP  1103 Kg / m  s
viscosity
residual oil saturation 10%  0,1
fluid density
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Using the previous table we find Reynolds number Re = 1 (ca.), for
laboratory core flow, which is far below the limit of turbulent flow.
In case of reservoir flow, the ”normal” reservoir flow velocity is ca. 1
foot/day or 3,5m / s , which indicate that turbulent liquid flow under
reservoir conditions is not very likely to occur.
For gas, turbulent flow may occur if the potentials are steep enough. If the
formula for Reynolds number and Poiseuille`s law is compared:
Re 
2 Rk p 
l  2
The only fluid dependent parameters are
the density and the viscosity
Comparing the Reynolds number for typical values of gas and oil :
Re gas
Re oil

(  /  2 ) gas
(  /  2 ) oil
0,2 g / cm3 /( 0,02cP) 2

 2500
3
2
0,8 g / cm /( 2,0cP)
Demonstrates the possibilities for turbulent when gas is
flowing in a porous medium
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Section 3:
Dependency of Viscosity on Temperature
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
The Viscosity of liquids decreases with increasing temperature. For gas it`s
the opposite; viscosity increase with increasing temperature.
-Non-Newtonian
fluids
-Examples
Viscosity, cp
-Laboratory
exercise
Heavy oil
Light oil
Temperature, K
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
T 3/ 2
K
T C
Temperature depending viscosity of gases
expressed by the Satterland`s equation:
Where K and C are constants
depending on the type of gas.
Another commonly equation:
Where n depends on the type
of gas (1 < n < 0,75)
Developers
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T 
   0  
 T0 
Summary
n
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Section 4:
Non-Newtonian Fluids
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Viscous-Plastic fluids
Bingham (1916) and Shvedov (1889) investigated the rheology of viscousplastic fluids. These fluids also feature elasticity in addition to viscosity.
Equation describing viscous-plastic fluids:
dv
   0   ` , when   0
dy
when   0

•Here 0is the breaking shear
stress
• `is the so-called structural
viscosity

there is no fluid flow
Some oils, drilling mud and cements slurries represent viscous-plastic
fluids.
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
Pseudo-Plastic fluids
Some fluids do not have breaking shear stress but rather, their apparent
viscosity depends on a shear rate:
-Examples
n
 dv 
  k   , n  1
 dy 
-Laboratory
exercise
Which means that their apparent viscosity decreases when dv/dy grows:

 dv 

 k  
dv / dy
 dy 
Developers
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n 1
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Section 5:
Examples
-Viscous fluids
-Fluid flow
characteristics
6.2.1
Water viscosity at reservoir
conditions
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
6.2.2
Falling sphere
viscosity measurement
6.2.4
Rotating cylinder
viscosity measurement
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Water viscosity at reservoir conditions
-Viscous fluids
-Dependency of
viscosity on
temperature
Water viscosity is primarily a function of temperature. Salinity has also a slight
influence on  w . The pure water viscosity is listed in viscous fluids section.
Due to correction foe salinity and reservoir temperature, the normal range of
viscosity at reservoir conditions is from 0,2 to 1,0 cP.
-Non-Newtonian
fluids
Correlation for estimation of water viscosity at res. temp. :
-Fluid flow
characteristics
-Examples
-Laboratory
exercise
 w  4,33  0,07T  4,73 104 T 2  1,415 106 T 3  1,56 10 9 T 4
Water viscosity is measured in cP
and temperature in Fahrenheit
At a res. temp. of 110 C (230 F ) , the water viscosity will be:
 w  (4,33  0,07  230  4,73 104  2302  1,415 106  2303  1,56 109  2304 )cP
 0,40cP

NB! Even if the temperature is above
, water
100
C at
reservoir condition will still be in a liquid phase since
the res. Pressure is quite high as
compared to
surface condition.
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Falling sphere viscosity measurement
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
A metal sphere falling in viscous fluid
reaches a constant velocity vs
Then the viscous retarding force plus
the buoyancy force equals the weight of
the sphere
–s density of the metal sphere
 f– density of the fluid
The force Fs is given by
Stoke`s law:
Fs  6rvs
t0
h
Vs – thermal speed
 – viscosity
t1
The weight of the sphere will
balance the viscous force plus
the buoyancy force at the
terminal sphere velocity when
the sum of forces acting on
the sphere is zero
4 3
4 3
r g s  6rvs  r g f
3
3
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g – the constant
of gravitation
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-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
Viscosity can be estimated by measuring the falling speed of the metal sphere
in a cylindrical tube: ( vs  h / t )
2 2
1
  r g (s   f )
9
vs
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
Alternatively; measuring the time
t  t1  t0 :
  C (  s   f )t
C – characteristic
constant, determined
through calibration with a
fluid of known viscosity
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Rotating cylinder viscosity measurement
-Viscous fluids
Here we will also Look at the
motion of a fluid between two
coaxial cylinders
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples

One cylinder is
rotating with an
angular velocity
-Laboratory
exercise
The inner cylinder:
– Constant
Angular velocity
Due to the elastic force (in
the fluid) a viscosity shear
will exist there
The other one is
kept constant
The outer cylinder:
Held stationary by a
spring balance
which measures the
torque M
h

M
 1
1 
M  4h  2  2 
 rinner router 
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1
h – the fluid height
level on the two
cylinders
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-Viscous fluids
-Fluid flow
characteristics
For a certain viscometer, the viscosity as function of angular momentum and
angular velocity is:
-Dependency of
viscosity on
temperature
 C
-Non-Newtonian
fluids
-Examples
M

-Laboratory
exercise
C – the characteristic constant
for the viscometer
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Section 6:
Laboratory exercise
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
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Summary
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Section 6:
-Viscous fluids
-Fluid flow
characteristics
-Dependency of
viscosity on
temperature
-Non-Newtonian
fluids
-Examples
-Laboratory
exercise
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Summary
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```
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