1PETE 411Well Drilling
Lesson 11 Laminar Flow
2Lesson 11 - Laminar Flow
Rheological Models Newtonian Bingham Plastic Power-Law
Rotational Viscometer Laminar Flow in Wellbore
Fluid Flow in Pipes Fluid Flow in Annuli
3ReadADE Ch. 4 to p. 138
HW #5ADE Problems 4.3, 4.4, 4.5, 4.6
Due Friday, Sept. 27, 2002
4Newtonian Fluid Model
Shear stress = viscosity * shear rate
AF ,
LVallyExperiment =
=
5Laminar Flow of Newtonian Fluids
AF
LV=
6Newtonian Fluid Model
In a Newtonian fluid the shear stress is directly proportional to the shear rate (in laminar flow):
i.e.,The constant of proportionality, is the viscosity of the fluid and is independent of shear rate.
=
sec1
2 cmdyne
= .
7Newtonian Fluid Model
Viscosity may be expressed in poise or centipoise.
poise 0.01 centipoise 1
scmg1
cms-dyne1 poise 1 2
=
==
2cmsecdyne
= .
8Shear Stress vs. Shear Rate for a Newtonian Fluid
Slope of line ====
. =
9Example - Newtonian Fluid
10
Example 4.16
Area of upper plate = 20 cm2
Distance between plates = 1 cm
Force reqd to move upper plate at 10 cm/s= 100 dynes.
What is fluid viscosity?
11
Example 4.16
poise 5.0cm
sdyne5.0105
2 =
==
1-
2
sec 10/1dynes/cm 20/100
//
rate shearstressshear
===
LVAF
cp 50=
=
12
Bingham Plastic Model
13
Bingham Plastic Model
- if
- if 0
if
yyp
yy
yyp
+=
and y are often expressed in lbf/100 sq.ft
14
Bingham Plastic Model
2
2
22
48.30
sec980454
*100100
1
=
ftcm
cmlbfg
ftlbf
ftlbf
22 dyne/cm 79.4100
1 =ft
lbf(p.134)
1 dyne is the force that, if applied to a standard 1 gram body, would give that body an acceleration of 1 cm/sec2
15
Example 4.17{parallel plates again!}
Bingham Plastic FluidArea of upper plate = 20 cm2
Distance between plates = 1 cm
1. Min. force to cause plate to move = 200 dynes
2. Force reqd to move plate at 10 cm/s = 400 dynes
Calculate yield point and plastic viscosity
16
Example 4.17
Yield point,
22y
y cmdynes10
cm20dynes200
AF
===
22 cmdynes79.4
ft100lbf1 but =
79.4
10y == 2ftlbf/10009.2
+= py
17
Example 4.17
cp 100 .e.i p =
poise 1110
10202 =
=
=cm
sdynep
+=
cm 1cm/s 10
cm 20dynes 200
cm 20dynes 400
22 p
Plastic viscosity, p
+= py
bygiven is
18
Power-Law Model
19
Power-Law Model
n = flow behavior indexK = consistency index
0 if K
0 if K1n
n
20
Power-Law Model
2
2
2
n
2
n
ftcm48.30
seccm980
lbfg454
*ft
slbfft
slbf1
=
.poise .eq 479cm/sdyne 479ft
slbf1 2n2n
==
.cp .eq 900,47ft
slbf1 2n
=
21
Example 4.18Power-Law Fluid
Area of upper plate = 20 cm2
Distance between plates = 1 cmForce on upper plate = 50 dyne if v = 4 cm/sForce on upper plate = 100 dyne if v = 10 cm/s
Calculate consistency index (K) and flow behavior index (n)
22
Example 4.18
v = 4 cm/s
( )n
n
n
44
4K5.2
14K
2050
K
=
=
=
Area of upper plate = 20 cm2
Distance between plates = 1 cm
Force on upper plate= 50 dyne if v = 4 cm/s
((((i))))
23
Example 4.18
v = 10 cm/s
( )n
n
n1010
10K5
110K
20100
K
=
=
=
Area of upper plate = 20 cm2
Distance between plates = 1 cm
Force on upper plate= 100 dyne
if v = 10 cm/s
((((ii))))
24
Example 4.18
Combining Eqs. (i) & (ii):
5.2 log n2 log
5.24K
10 K5.2
5 nn
n
=
==
7565.0n =
( )nK 45.2 = ((((i))))((((ii))))( )nK 105 =
25
Example 4.18
From Eq. (ii):
poise eq. 8760.010
510
5K 7565.0n ===
cp. eq. 6.87K =
( )nK 105 = ((((ii))))
26
Apparent Viscosity
Apparent viscosity = ( / )is the slope at each shear rate,
.,, 321
27
Apparent Viscosity
Is not constant for a pseudoplastic fluid
The apparent viscosity decreases with increasing shear rate
(for a power-law fluid)
(and also for aBingham Plastic fluid)
28
Typical Drilling Fluid Vs. Newtonian, Bingham and Power
Law Fluids
0000
(Plotted on linear paper)
29
Rheological Models
1. Newtonian Fluid:
2. Bingham Plastic Fluid:
= rate shear
viscosityabsolutestressshear
=
=
=
+= *)( py viscosityplastic
point yield
=
=
p
y
What if y ==== 0?
30
3. Power Law Fluid:
When n = 1, fluid is Newtonian and K = We shall use power-law model(s) to
calculate pressure losses (mostly).
n
)(K
= K = consistency index
n = flow behavior index
Rheological Models
31
Figure 3.6Rotating Viscometer
Rheometer
We determine rheologicalproperties of drilling fluids in this device
Infinite parallel plates
32
Rheometer (Rotational Viscometer)
Shear Stress = f (Dial Reading)Shear Rate = f (Sleeve RPM)Shear Stress = f (Shear Rate)
)(f =BOB
sleeve
fluid
Rate Shear the (GAMMA), of value the on depends Stress Shear the ),TAU(
33
Rheometer - base case
RPM sec-13 5.116 10.22
100 170200 340300 511600 1022
RPM * 1.703 = sec-1
34
ExampleA rotational viscometer containing a Bingham plastic fluid gives a dial reading of 12 at a rotor speed of 300 RPM and a dial reading of 20 at a rotor speed of 600 RPM
Compute plastic viscosity and yield point
12-20
300600p
=
=
cp 8p =
600600600600 = 20300300300300 = 12
See Appendix A
35
Example
8-12
p300y
=
=
2y ft lbf/100 4=
600600600600 = 20300300300300 = 12
(See Appendix A)
36
Rotational Viscometer, Power-Law Model
Example: A rotational viscometer containing a non-Newtonian fluid gives a dial reading of 12 at 300 RPM and 20 at 600 RPM.
Assuming power-law fluid, calculate the flow behavior index and the consistency index.
37
Example
cp eq. 67.61511
12*510511
510
7370.0
1220 log 322.3 log 322.3
7372.0300
300
600
===
=
=
=
nK
n
n
600600600600 = 20300300300300 = 12
38
Gel Strength
39
Gel Strength
= shear stress at which fluid movement begins
The yield strength, extrapolated from the 300 and 600 RPM readings is not a good representation of the gel strength of the fluid
Gel strength may be measured by turning the rotor at a low speed and noting the dial reading at which the gel structure is broken(usually at 3 RPM)
40
Gel Strength
In field units,
In practice, this is often approximated to
06.1g =2ft 100/lbf
2ft 100/lbf
The gel strength is the maximum dial reading when the viscometer is started at 3 rpm.
g = max,3
41
Velocity Profiles(laminar flow)
Fig. 4-26. Velocity profiles for laminar flow: (a) pipe flow and (b) annular flow
42
It looks like concentric rings of fluid telescoping down the pipe at different velocities
3D View of Laminar Flow in a pipe - Newtonian Fluid
43
Table 4.4 - Summary of Laminar Flow Equations for Pipes and Annuli
Fictional Pressure Loss Shear Rate at Pipe Well
Newtonian
Pipe Pipe
Annulus Annulus
2
_
f
d500,1v
dLdp
=
212
_
f
)dd(000,1v
dLdp
=
dv96_
w =
)dd(v144
12
_
w
=
44
Table 4.4 - Summary of Laminar Flow Equations for Pipes and Annuli
Fictional Pressure Loss Shear Rate at Pipe Wall
Bingham PlasticPipe Pipe
Annulus Annulus
d225d500,1v
dLdp y
2
_
pf +=
)dd(200)dd(000,1v
dLdp
12
y
12
_
pf
+
=
p
y
_
w 7.159dv96
+=
p
y
12
_
w 5.239)dd(v144
+
=
45
Table 4.4 - Summary of Laminar Flow Equations for Pipes and Annuli
Fictional Pressure Loss Shear Rate at Pipe Well
Power-LawPipe Pipe
Annulus Annulus
n
n
nf n
dvK
dLdp
+=
+ 0416.0/13
000,144 1
_
n
n
nf n
ddvK
dLdp
+
=+ 0208.0
/12)(000,144 112
_
)n/13(d
v24_
w +=
)n/12(ddv48
12
_
w +
=
46
Table 4.3 - Summary of Equations for Rotational Viscometer
Newtonian Model
Na N300 =
Nr066.5
2=
300a =
or
47
Table 4.3 - Summary of Equations for Rotational Viscometer
300N
or
1pNy 1 =
rpm 3 atmaxg =
Bingham Plastic Model
300600p = )(NN
300or
12 NN12
p
=
p300y =
or
or
48
Table 4.3 - Summary of Equations for Rotational Viscometer
Power-Law Model
=
1
2
N
N
NNlog
logn 1
2
n300
)511( 510K =
nN
)N703.1( 510K
or
=
or or )log( 322.3n
300
600
=
or