Transcript Old Lesson 17B - University of Stavanger

PETE 411
Well Drilling
Lesson 13
Pressure Drop Calculations
API Recommended Practice 13D
Third Edition, June 1, 1995
1
Homework

HW #7. Pressure Drop Calculations

Due Oct. 9, 2002
 The API
Power Law Model
2
Contents
 The
Power Law Model
 The Rotational Viscometer
 A detailed Example - Pump Pressure
Pressure Drop in the Drillpipe
 Pressure Drop in the Bit Nozzles
 Pressure Drop in the Annulus

 Wellbore
Pressure Profiles
3
Power Law Model
K = consistency index
n = flow behaviour index
=K n
SHEAR
STRESS

psi
0
SHEAR RATE,  , sec -1
4
Fluid Flow in Pipes and Annuli
LOG
(PRESSURE)
(psi)
LOG (VELOCITY) (or FLOW RATE)
5
Fluid Flow in Pipes and Annuli
Laminar Flow
Turbulent
LOG
(SHEAR
STRESS)
(psi)
n
1
LOG ( SHEAR RAT E),
( RPM or sec1 )
6
Rotating
Sleeve
Viscometer
7
Rotating Sleeve Viscometer
(RPM * 1.703)
VISCOMETER
RPM
SHEAR RATE
3
100
ANNULUS
5.11
170.3
300
600
DRILL
STRING
511
1022
sec -1
BOB
SLEEVE
API RP 13D
8
API RP 13D, June 1995
for Oil-Well Drilling Fluids
API RP 13D recommends using only FOUR of
the six usual viscometer readings:
 Use 3, 100, 300, 600 RPM Readings.
 The 3 and 100 RPM reading are used for
pressure drop calculations in the annulus,
where shear rates are, generally, not very high.
 The 300 and 600 RPM reading are used for
pressure drop calculations inside drillpipe,
where shear rates are, generally, quite high.

9
Example: Pressure Drop Calculations

Example
Calculate the pump pressure in
the wellbore shown on the next page, using the
API method.
 The
relevant rotational viscometer readings
are as follows:
 R3
= 3
 R100 = 20
 R300
= 39
 R600 = 65
(at 3 RPM)
(at 100 RPM)
(at 300 RPM)
(at 600 RPM)
10
Pressure Drop
Calculations
PPUMP
Q = 280 gal/min
r = 12.5 lb/gal
PPUMP = DPDP + DPDC
+ DPBIT NOZZLES
+ DPDC/ANN + DPDP/ANN
+ DPHYD
11
Pressure Drop In Drill Pipe
OD = 4.5 in
ID = 3.78 in
L = 11,400 ft
Power-Law Constant (n):
 R 600
np  3.32 log 
 R 300

 65 
  3.32 log 
  0.737
 39 

Fluid Consistency Index (K):
Kp 
5.11R600
np
1,022
5.11* 65
dynesecn

 2.017
0.737
1,022
cm2
Average Bulk Velocity in Pipe (Vp):
0.408Q
Vp 
D2
0.408 * 280
ft

 8.00
2
3.78
sec
12
Pressure Drop In Drill Pipe
Effective Viscosity in Pipe (mep):
mep
 96Vp 

 100 K p 
 D 
np 1
 96 * 8 
mep  100 * 2.017 

 3.78 
 3np  1


 4n 
p 

0.737 1
OD = 4.5 in
ID = 3.78 in
L = 11,400 ft
np
 3 * 0.737  1


 4 * 0.737 
0.737
 53 cP
Reynolds Number in Pipe (NRep):
NRe p 
928 D Vpr
mep

928 * 3.78 * 8.00 * 12.5
 6,616
53
13
Pressure Drop In Drill Pipe
NOTE: NRe > 2,100, so
Friction Factor in Pipe (fp):
a
b
log np  3.93

50
1.75  log np
So,
a
NRe p
a
NRe p
b
log 0.737  3.93
 0.0759
50
1.75  log 0.737

 0.2690
7
7
fp 
fp 
OD = 4.5 in
ID = 3.78 in
L = 11,400 ft
b
0.0759

 0.007126
0.2690
6,616
14
Pressure Drop In Drill Pipe
Friction Pressure Gradient (dP/dL)p :
fp Vp r
 dP 

 
 dL  dp 25.81D
2
OD = 4.5 in
ID = 3.78 in
L = 11,400 ft
0.007126 * 8 2 * 12.5
psi

 0.05837
25.81 * 3.78
ft
Friction Pressure Drop in Drill Pipe :
 dP 
DPdp  
 DLdp
 dL dp
 0.05837 * 11,400
DPdp = 665 psi
15
Pressure Drop In Drill Collars
Power-Law Constant (n):
ndc
 R 600
 3.32 log 
 R 300
OD = 6.5 in
ID = 2.5 in
L = 600 ft

 65 
  3.32 log 
  0.737
 39 

Fluid Consistency Index (K):
K dc 
5.11R600
np
1,022
5.11* 65
dynesecn

 2.017
0.737
1,022
cm2
Average Bulk Velocity inside Drill Collars (Vdc):
Vdc
0.408Q

D2
0.408 * 280
ft

 18.28
2
2 .5
sec
16
Pressure Drop In Drill Collars
Effective Viscosity in Collars(mec):
m edc
medc
 96Vp 

 100 K p 
 D 
np 1
 3np  1


 4n 
p 

 96 * 18.28 
 100 * 2.017 

 2.5 
0.737 1
OD = 6.5 in
ID = 2.5 in
L = 600 ft
np
 3 * 0.737  1


 4 * 0.737 
0.737
 38.21 cP
Reynolds Number in Collars (NRec):
NRe dc
928 D Vdc r

m edc
928 * 2.5 * 18.28 * 12.5

 13,870
38.21
17
Pressure Drop In Drill Collars
NOTE: NRe > 2,100, so
Friction Factor in DC (fdc):
fdc 
a
NRe dc
b
log ndc  3.93
a
50
log 0.737  3.93

 0.0759
50
1.75  log ndc
b
7
1.75  log 0.737

 0.2690
7
So,
fdc 
a
NRe dc
b
OD = 6.5 in
ID = 2.5 in
L = 600 ft
0.0759

 0.005840
0.2690
13,870
18
Pressure Drop In Drill Collars
Friction Pressure Gradient (dP/dL)dc :
fdc Vdc r
 dP 



dL

dc 25.81Ddc
2
OD = 6.5 in
ID = 2.5 in
L = 600 ft
0.005840 * 18.282 * 12.5
psi

 0.3780
25.81* 2.5
ft
Friction Pressure Drop in Drill Collars :
 dP 
DPdc  
 DL dc
 dL dc
 0.3780 * 600
DPdc = 227 psi
19
Pressure Drop across Nozzles
DPNozzles 
D
156 r Q
2
N1
DPNozzles 
2
 DN2  DN3
2
2

2
DN1 = 11 32nds (in)
DN2 = 11 32nds (in)
DN3 = 12 32nds (in)
156 * 12.5 * 2802
11
2
 11  12
2

2 2
DPNozzles = 1,026 psi
20
Pressure Drop
in DC/HOLE
Annulus
Q = 280 gal/min
r = 12.5 lb/gal
8.5 in
DHOLE = 8.5 in
ODDC = 6.5 in
L
= 600 ft
21
Pressure Drop
in DC/HOLE Annulus
Power-Law Constant (n):
ndca
 R100
 0.657 log 
 R3
DHOLE = 8.5 in
ODDC = 6.5 in
L
= 600 ft

 20 
  0.657 log 
  0.5413
 3 

Fluid Consistency Index (K):
K dca
5.11R100
5.11* 20
dynesecn


 6.336
ndca
0.5413
170.2
cm2
170.2
Average Bulk Velocity in DC/HOLE Annulus (Va):
Vdca
0.408Q

2
2
D2  D1
0.408 * 280
ft

 3.808
2
2
8.5  6.5
sec
22
Pressure Drop
in DC/HOLE Annulus
Effective Viscosity in Annulus (mea):
m ea
mea
 144 Va 

 100 K a 
 D 2  D1 
na 1
 144 * 3.808 
 100 * 6.336

 8.5  6.5 
 2na  1


 3na 
0.5413 1
DHOLE = 8.5 in
ODDC = 6.5 in
L
= 600 ft
na
 2 * 0.5413  1


 3 * 0.5413 
0.5413
 55.20 cP
Reynolds Number in Annulus (NRea):
NRe a 
928 D2  D1  Var
mea

928 8.5  6.5* 3.808 * 12.5
 1,600
55.20
23
Pressure Drop
in DC/HOLE Annulus
NOTE: NRe < 2,100
Friction Factor in Annulus (fa):
fa 
24
NRe a

24
 0.01500
1,600
fa Va r
 dP 

 
 dL a 25.81D2  D1 
2
DPdc / hole
0.01500 * 3.8082 * 12.5
psi

 0.05266
25.818.5  6.5
ft
 dP 

DL dc / hole

 dL dc / hole
So,
DHOLE = 8.5 in
ODDC = 6.5 in
L
= 600 ft
 0.05266 * 600
DPdc/hole = 31.6 psi
24
Pressure Drop
in DP/HOLE Annulus
q = 280 gal/min
r = 12.5 lb/gal
DHOLE = 8.5 in
ODDP = 4.5 in
L
= 11,400 ft
25
Pressure Drop
in DP/HOLE Annulus
DHOLE = 8.5 in
ODDP = 4.5 in
L
= 11,400 ft
Power-Law Constant (n):
ndpa
 R100
 0.657 log 
 R3

 20 
  0.657 log 
  0.5413
 3 

Fluid Consistency Index (K):
K dpa 
5.11R100
ndpa
170.2
5.11* 20
dynesecn

 6.336
0.5413
170.2
cm2
Average Bulk Velocity in Annulus (Va):
Vdpa
0.408Q
 2
2
D2  D1
0.408 * 280
ft

 2.197
2
2
8.5  4.5
sec
26
Pressure Drop
in DP/HOLE Annulus
Effective Viscosity in Annulus (mea):
m ea
mea
 144Va 

 100 K a 
 D2  D1 
 144 * 2.197 
 100 * 6.336

8
.
5

4
.
5


na 1
0.5413 1
 2na  1


 3na 
na
 2 * 0.5413  1


3
*
0
.
5413


0.5413
 97.64 cP
Reynolds Number in Annulus (NRea):
NRe a
928 D2  D1  Var

mea
928 8.5  4.5* 2.197 * 12.5

 1,044
97.64
27
Pressure Drop
in DP/HOLE Annulus
NOTE: NRe < 2,100
Friction Factor in Annulus (fa):
fa 
24
NRe a

24
 0.02299
1,044
fa Va r
 dP 
  
 dL a 25.81D2  D1 
2
0.02299 * 2.1972 * 12.5
psi

 0.01343
25.818.5  4.5
ft
 dP 
DPdp / hole  
DLdp / hole

 dL dp / hole
So,
 0.01343 * 11,400
psi psi
DPdp/hole = 153.2
28
Pressure Drop
Calculations
- SUMMARY PPUMP = DPDP + DPDC + DPBIT NOZZLES
+ DPDC/ANN + DPDP/ANN + DPHYD
PPUMP = 665 + 227 + 1,026
+ 32 + 153 + 0
PPUMP = 1,918  185  2,103 psi
29
PPUMP = DPDS + DPANN + DPHYD
DPDS = DPDP + DPDC + DPBIT NOZZLES
= 665 + 227 + 1,026 = 1,918 psi
2,103 psi
P
=
0
DPANN = DPDC/ANN + DPDP/ANN
= 32 + 153 = 185
DPHYD = 0
PPUMP = 1,918 + 185
= 2,103 psi
30
What is the BHP?
BHP = DPFRICTION/ANN + DPHYD/ANN
2,103 psi
P
=
0
BHP = DPDC/ANN + DPDP/ANN
+ 0.052 * 12.5 * 12,000
= 32 + 153 + 7,800 = 7,985 psig
BHP = 185 + 7,800
BHP = 7,985 psig
31
"Friction" Pressures
"Friction" Pressure, psi
2,500
DRILLPIPE
2103
2,000
1,500
DRILL COLLARS
1,000
BIT NOZZLES
500
ANNULUS
0
0
5,000
10,000
15,000
20,000
25,000
Distance from Standpipe, ft
32
Hydrostatic Pressures in the Wellbore
Hydrostatic Pressure, psi
9,000
BHP
8,000
7,000
6,000
5,000
DRILLSTRING
ANNULUS
4,000
3,000
2,000
1,000
0
0
5,000
10,000
15,000
20,000
25,000
Distance from Standpipe, ft
33
Pressures, psi
Pressures in the Wellbore
10,000
9,000
8,000
7,000
6,000
5,000
4,000
3,000
2,000
1,000
0
CIRCULATING
2103
STATIC
0
5,000
10,000
15,000
20,000
25,000
Distance from Standpipe, ft
34
Wellbore Pressure Profile
2103
0
2,000
DRILLSTRING
Depth, ft
4,000
6,000
ANNULUS
8,000
10,000
(Static)
12,000
BIT
14,000
0
2,000
4,000
6,000
Pressure, psi
8,000
10,000
35
Pipe Flow - Laminar
In the above example the flow down the
drillpipe was turbulent.
Under conditions of very high viscosity,
the flow may very well be laminar.
NOTE: if NRe < 2,100, then
Friction Factor in Pipe (fp):
Then
16
fp 
NRe p
fp Vp r
2
and
 dP 

 
 dL  dp 25.81D
36
Annular Flow - Turbulent
In the above example the flow up the annulus
was laminar.
Under conditions of low viscosity and/or high
flow rate, the flow may very well be turbulent.
NOTE: if NRe > 2,100, then Friction Factor in the Annulus:
log na  3.93
a
50
Then
fa 
a
N bRe a
b
and
1.75  log na
7
fa Va r
 dP 



 dL a 25.81D2  D1 
2
37
Critical Circulation Rate
Example
The above fluid is flowing in the annulus
between a 4.5” OD string of drill pipe
and an 8.5 in hole.
The fluid density is 12.5 lb/gal.
What is the minimum circulation rate that
will ensure turbulent flow?
(why is this of interest?)
38
Critical Circulation Rate
In the Drillpipe/Hole Annulus:
NRe a
Q, gal/min
280
300
350
400
450
452
452.1
928 D2  D1  Var

mea
V, ft/sec
Nre
2.197
2.354
2.746
3.138
3.531
3.546
1,044
1,154
1,446
1,756
2,086
2,099
3.547
2,100
39
Optimum Bit Hydraulics

Under what conditions do we get the best
hydraulic cleaning at the bit?


maximum hydraulic horsepower?
maximum impact force?
Both these items increase when the circulation
rate increases.
However, when the circulation rate increases, so
does the frictional pressure drop.
40
41
n = 1.0
_
2
dpf fr v

dL 25 .8 d
42
Importance of Pipe Size
_ 1.75
Eq. 4.66e
or,
dp f
r 0.75 v m 0.25


1.25
dL
1800d
dp f
r 0.75q1.75 m 0.25

4.75
dL
8,624d
*Note that a small change in the pipe diameter results in
large change in the pressure drop! (q = const.)
Decreasing the pipe ID 10% from 5.0” to 4.5” would result in an
increase of frictional pressure drop by about 65% !!
43
Dpf = 11.41 v 1.75
turbulent flow
Dpf = 9.11 v
laminar flow
Use max. Dpf value44