Transcript PETE 411 Drilling Engineering Lesson 16 - Lifting Capacity of Drilling Fluids -

PETE 411
Drilling Engineering
Lesson 16
- Lifting Capacity of Drilling Fluids - Slip Velocity -
1
Lifting Capacity of Drilling Fluids
- Slip Velocity  Fluid Velocity in Annulus
 Particle Slip Velocity
 Particle Reynolds Number
 Friction Coefficient
 Example
 Iterative Solution Method
 Alternative Solution Method
 API RP 13D Method
2
Read:
Applied Drilling Engineering, Ch. 4 - all
HW #8:
On the Web - due 10-14-02
3
Messages from Darla-Jean Weatherford
The seniors were supposed to have submitted the
drafts of their papers for the Student Paper Contest
to me last Friday; a few more than half did. Will you
please remind the rest that I need those papers to
complete their grades for 485?
We also are looking for recruiters for the fairs in
Houston, which will be 18 to 22 November this
year. If they can go with us any evening or Friday
morning, they need to let Larry Piper know soon so
we can get t-shirts and transportation (and meals!)
arranged.
4
Lifting Capacity of Drilling Fluids
Historically, when an operator felt that
the hole was not being cleared of cuttings
at a satisfactory rate, he would:
Increase the circulation rate
Thicken the mud
(increase YP/PV)
5
Lifting Capacity of Drilling Fluids
More recent analysis shows that:
 Turbulent flow cleans the hole better.
 Pipe rotation aids cuttings removal.
 With water as drilling fluid, annular
velocities of 100-125 ft/min are
generally adequate (vertical wells)
6
Lifting Capacity of Drilling Fluids
 A relatively “flat” velocity
profile is better than a
highly pointed one.
 Mud properties can be
modified to obtain a flatter
profile in laminar flow
e.g., decrease n
7
Density & Velocity
Drilled cuttings typically
have a density of
about 21 lb/gal.
Vslip
_
Since the fluid density is
less than 21 lb/gal the
cuttings will tend to
settle, or ‘slip’ relative
to the drilling mud.
V f luid
Vparticle
Vparticle  Vf luid  Vslip
8
Velocity Profile
The slip velocity can be reduced by
modifying the mud properties such
that the velocity profile is flattened:
Increase the ratio (YP/PV)
(yield point/plastic viscosity)
or
Decrease the value of n
9
Plug Flow
 Plug Flow is good for hole
cleaning. Plug flow refers
to a “completely” flat
velocity profile.
 The shear rate is zero
where the velocity profile
is flat.
10
Participle Slip Velocity
Newtonian Fluids:
The terminal velocity of a small
spherical particle settling
(slipping) through a Newtonian
fluid under Laminar flow
conditions is given by STOKE’S
LAW:
2
138(s   f )ds
vs 

11
Particle Slip Velocity - small particles
138(s   f )ds
vs 

2
Where v s  slip velocity, ft/s
 s  density of solid particle, lbm/gal
 f  density of fluid, lbm/gal
ds  diameter of particle, in
  fluid viscosity, cp
12
Particle Slip Velocity
Stokes’ Law gives acceptable accuracy for a
particle Reynolds number < 0.1
NRe
928 f v s ds


For Nre > 0.1 an empirical friction factor
may be used.
13
What forces act
on a settling
particle?
Nonspherical
particles
experience
relatively
higher drag
forces
14
Sphericities for Various Particle Shapes
Shape
Sphericity
Sphere
Octahedron
Cube
Prism
 *  * 2
1.00
0.85
0.81
 * 2 * 3
Cylinders
h  r/15
h  r/3
h  r
h  2r
h  20r
0.73
0.77
0.25
0.59
0.83
0.87
0.58
Sphericity =
surface area of
sphere of same
volume as particle
surface area of
particle
15
16
Particle Reynolds Number, fig. 4.46
Based on real cuttings


In field units, v s  1.89 ds  s  1.......... .Eq.( 4.104d)

f 

f

17
Slip Velocity Calculation
using Moore’s graph (Fig. 4.46)
1. Calculate the flow velocity.
2. Determine the fluid n and K values.
3. Calculate the appropriate viscosity
(apparent viscosity).
4. Assume a value for the slip velocity.
5. Calculate the corresponding
Particle Reynolds number.
18
Slip Velocity Calculation
(using Moore’s graph)
6. Obtain the corresponding drag coeff., f,
from the plot of f vs. Nre.
7. Calculate the slip velocity and compare
with the value assumed in step 4 above.
8. If the two values are not close enough,
repeat steps 4 through 7 using the
calculated Vs as the assumed slip velocity
in step 4.
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Example
Use (the modified) Moore’s method to
calculate the slip velocity and the net particle
velocity under the following assumptions:
Well depth:
8,000 ft
Yield point: 4 lbf/100ft2
Drill pipe: 4.5”, 16.6 #/ft
Density of Particle: 21 lbm/gal
Mud Weight:
Particle diameter:
9.1 #/gal
Plastic viscosity: 7 cp
Hole size:
5,000 m
Circulation rate: 340 gal/min
7-7/8”
20
Solution - Slip Velociy Problem
1. Calculate the flow velocity
_
q
340
v

2
2
2
2
2
.
448
(
7
.
875

4
.
5
)
2.448(d2  d1 )
 3.325 ft/sec
2. Determine the fluid n and K values
 y   300   p

 300   y   p  4  7  11
 p   600   300

 600   p   300  7  11  18
21
Solution - Slip Velociy Problem - cont’d
2. Determine the fluid n and K values - cont’d
 600
n  3.32 log 
 300



(ADE)
 3.32 log (18/11)
n  0.7101
(510 )300
K
511n
(510 )11

5110.7101
K  66.94 eq.cp
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Solution - Slip Velociy Problem - cont’d
3. Calculate the appropriate viscosity
1n
K  d2  d1 
a 
_

144 
 v 
1 

 (2  ) 
n 

 0.0208 




n
10.7101
66.94  7.875  4.5 
a 


144  3.325 
 a  17.94 cp
.......... Eq. (4.107)
1


(
2

)


0.7101 

 0.0208 




0.7101
  p  7 cp



 K  66.94 eq cp
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Solution - Slip Velociy Problem - cont’d
4. Assume a value for the slip velocity
___
V 3.325
Vs 

 1.663 ft / sec
2
2
5. Calculate the corresponding Particle Reynolds No.
NRe 
928ρ f v s ds
μa


cm
in


928(9.1)(1 .663) 5000 μ m
4


10
μ
m
2.54
cm



17.94
N Re  154
 92.8 vs 
{d s  0.1969 in }
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Solution - Slip Velociy Problem - cont’d
6. Obtain the drag coeff., f, from the plot of f vs. Nre.
From graph,
ds
v s  1.89
f
f = 2.0
ρ s


 1
ρ f

0.1969  21.0

 1.89
 1

2.0  9.1

v s  0.678 ft/s
 1.663
Eq. (4.104d)
 0.959 


f


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Solution - Slip Velocity Problem - cont’d
vs  0.678
4 (ii) Assume
5 (ii) Particle
NRe  92.7 * 0.678  62.9
6 (ii) From graph,
7 (ii)
f  2.7
0.959
vs 
 0.58 ft / s..... etc.
2.7
Subsequent iterations yield 0.56 ft/s and
0.56 ft/s again…...
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Slip Velocity - Alternate Method
1. Fully Laminar:
NRe  3 :
40
f
;
NRe

ds  ρ s

v s  1.89
 1
f ρ f

_
2
ds
ρ s  ρ f 
v s  82.87
μa
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Slip Velocity - Alternate Method
2. Intermediate;
3  NRe  300 :
22
f
;
NRe
_
vs 
2.90 ds (ρ s  ρ f )2/3
(ρ f μ a )
1/3
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Slip Velocity - Alternate Method
3. Fully Turbulent:
N Re  300 :
f  1.5;
NOTE:
Check NRe
d s (ρ s  ρ f )
vs  1.54
ρf
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Slip Velocity - Alternate Method
For the above calculations:
NRe
928ρ f v s ds

μa

ds  ρ s
  1
v s  1.89
f ρ f

NOTE: Check NRe
.........E q.(4.104 d)
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Slip Velocity - Alternate Method_2
If the flow is fully laminar, cuttings transport is
not likely to be a problem.
Method:
1. Calculate slip velocity for Intermediate
mode
2. Calculate slip velocity for Fully Turbulent
Mode.
3. Choose the lower value.
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Example
2.90d s (ρ s  ρ f )2/3
vs 
(ρ fμ a )1/3
_
(i) Intermediate:
2.90 * 0.1969 * (21  9.1) 2/3
vs 
 0.545 ft/sec
1/3
(9.1 * 17.94)
_
d s (ρ s  ρ f )
v s  1.54
ρf
_
(ii) Fully Turbulent:
_
v s  1.54
0.1969 (21  9.1)
9.1
 0.781 ft/sec
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Example - cont’d
Intermediate:
Fully Turbulent:
Vs = 0.545 ft/sec
Vs = 0.781 ft/sec
The correct slip velocity is 0.545 ft/sec
{ agrees reasonably well with iterative method on p.12 }
Check :
N Re
928 * 9.1* 0.545 * 0.1969

 51
17.94
Range OK
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Slip Velocity - API RP 13D
Iterative Procedure
Calculate Fluid Properties, n & K
Calculate Shear Rate
Calculate Apparent Viscosity
Calculate Slip Velocity
Example
34
Settling
Velocity
of Drilled
Cuttings
in Water
From
API RP 13D
p.24
35
Calculation Procedure
1. Calculate ns for the settling particle
2. Calculate Ks for the particle
3. Assume a value for the slip velocity, Vs
4. Calculate the shear rate,
gs
5. Calculate the corresponding apparent viscosity, es
6. Calculate the slip velocity, Vs
7. Use this value of Vs and repeat steps 4-6 until the
assumed and calculated slip velocities ~“agree”
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Slip Velocity - Example
ASSUMPTIONS:
3 RPM Reading
100 RPM Reading
R3
R100
3
lbf/100 ft 2
20
lbf/100 ft 2
Particle Density
Mud Density
p

22.5
12.5
lb/gal
lb/gal
Particle Dia. =
Dp
0.5
in
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Slip Velocity - Example
1. Calculate ns for the settling particle
 R100 

nS  0.657 log 
 R3 
 20 
nS  0.657 log    0.5413
 3 
2. Calculate Ks for the particle
5.11 R100
KS 
170.2ns
5.11 * 20
dyne sec n
KS 
 6.336
0.5413
170.2
cm2
38
Slip Velocity - Example
3. Assume a value for the slip velocity, Vs
Assume
Vs = 1 ft/sec
4. Calculate the shear rate,
12 VS
g S 
Dp
gs
12 * 1
g S 
 24.0 sec 1
0.5
39
Slip Velocity - Example
5. Calculate the corresp. apparent viscosity:
es  100 K s g s
 es  100 * 6.336 * 24
ns 1
0.5413 1
 147.5 cp
6. Calculate the slip velocity, Vs
2



D



 p 



Vs  0.0002403 e 5.03   es   1  (920,790 e 5.03  Dp  p  1
D  
 
  es 
 p 


 1


40
Slip Velocity - Example
6. Calculate the slip velocity, Vs
If
  0.80, then:
2
  es  
 p
 Dp  



Vs  0.01344
1  16,465 Dp   1
 D   
 
  es 
 p 


 1


2
 147.48  
 22.5
 0.5 * 12.5 
Vs  0.01344 
 1
  1  16,465 * 0.5

0
.
5
*
12
.
5
12
.
5
147
.
48





Vs = 0.8078 ft/sec

 1


Repeat steps 4-6
41
Slip Velocity - Example
Second Iteration - using
4. Shear rate:
5. Apparent viscosity:
6. Slip velocity:
gs = 19.386
sec-1
es = 162.65 cp
Vs = 0.7854 ft/sec
Third Iteration - using
4. Shear rate:
5. Apparent viscosity:
6. Slip velocity:
Vs = 0.8078 ft/sec
Vs = 0.7854 ft/sec
gs = 18.849
sec-1
es = 164.75 cp
Vs = 0.7823 ft/sec
42
Slip Velocity - Example
Fourth Iteration - using
4. Shear rate:
5. Apparent viscosity:
6. Slip velocity:
Vs = 0.7823 ft/sec
gs = 18.776
sec-1
es = 165.04 cp
Vs = 0.7819 ft/sec
Slip Velocity, Vs = 0.7819 ft/sec
{ Vs = 1.0, 0.808, 0.782, 0.782 ft/sec }
43
Transport Ratio
particle velocity
Transport Ratio 
fluid velocity
particle velocity
Transport Efficiency 
* 100%
fluid velocity
Example : Particle velocity
 90 ft/min
Fluid velocity
 120 ft/min
Transport Efficiency  ?
44
Transport Ratio
Transport efficiency  (90 / 120) *100%
 75%
A transport efficiency of 50% or higher is desirable!
Note: Net particle velocity = fluid velocity - slip velocity.
In example, particle slip velocity = 120 - 90 = 30 ft/min
With a fluid velocity of 120 ft/min a minimum particle
velocity of 60 ft/min is required to attain a transport
efficiency of 50%
45
Potential Hole-Cleaning Problems
1. Hole is enlarged. This may result in
reduced fluid velocity which is lower
than the slip velocity.
2. High downhole temperatures may
adversely affect mud properties
downhole.
[ We measured these at the surface.]
46
Potential Hole-Cleaning Problems
3. Lost circulation problems may preclude
using thick mud or high circulating
velocity. Thick slugs may be the
answer.
4. Slow rate of mud thickening - after it has
been sheared (and thinned) through the
bit nozzles, where the shear rate is very
high.
47
The End
Lesson 16
- Lifting Capacity of Drilling Fluids - Slip Velocity -
48