tech drilling pressdropcalc
TRANSCRIPT
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PETE 411Well Drilling
Lesson 13Pressure Drop Calculations
API Recommended Practice 13D
Third Edition, June 1, 1995
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Homework
HW #7. Pressure Drop Calculations
Due Oct. 9, 2002
The API Power Law Model
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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
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Power Law Model
K = consistency index
n = flow behaviour index
SHEAR STRESS
psi
= K n
SHEAR RATE, , sec-1
0
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Fluid Flow in Pipes and Annuli
LOG(PRESSURE)
(psi)
LOG (VELOCITY) (or FLOW RATE)
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Fluid Flow in Pipes and Annuli
LOG
(SHEAR STRESS)
(psi)
Laminar Flow Turbulent
)secor RPM ( ), RATE SHEAR (LOG 1
n1
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RotatingSleeve
Viscometer
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Rotating Sleeve Viscometer
VISCOMETERRPM
3100
300600
(RPM * 1.703)
SHEAR RATE
sec -1
5.11170.3 511
1022
BOB
SLEEVE
ANNULUS
DRILLSTRING
API RP 13D
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API RP 13D, June 1995for 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.
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Example: Pressure Drop Calculations
ExampleCalculate 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 (at 3 RPM)
R100 = 20 (at 100 RPM)
R300 = 39 (at 300 RPM)
R600 = 65 (at 600 RPM)
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PPUMP = PDP + PDC
+ PBIT NOZZLES
+ PDC/ANN + PDP/ANN
+ PHYD
Q = 280 gal/min
= 12.5 lb/gal
Pressure DropCalculations
PPUMP
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Power-Law Constant (n):
Pressure Drop In Drill Pipe
Fluid Consistency Index (K):
Average Bulk Velocity in Pipe (Vp):
OD = 4.5 in ID = 3.78 in L = 11,400 ft
737.039
65log32.3
R
Rlog32.3n
300
600p
2
n
737.0n600
p cm
secdyne017.2
022,1
65*11.5
022,1
R11.5K
p
sec
ft00.8
78.3
280*408.0
D
Q408.0V
22p
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Effective Viscosity in Pipe (ep):
Pressure Drop In Drill Pipe
Reynolds Number in Pipe (NRep):
OD = 4.5 in ID = 3.78 in L = 11,400 ft
ppn
p
p
1n
ppep n4
1n3
D
V96K100
cP53737.0*4
1737.0*3
78.3
8*96017.2*100
737.01737.0
ep
616,653
5.12*00.8*78.3*928VD928N
ep
pRep
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NOTE: NRe > 2,100, so
Friction Factor in Pipe (fp):
Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft
So,
bRe
p
pN
af
0759.050
93.3737.0log
50
93.3nloga p
2690.07
737.0log75.1
7
nlog75.1b p
007126.0616,6
0759.0
N
af
2690.0bRe
p
p
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Friction Pressure Gradient
(dP/dL)p :
Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft
Friction Pressure Drop in Drill
Pipe :400,11*05837.0L
dL
dPP dp
dpdp
Pdp = 665 psi
ft
psi05837.0
78.3*81.25
5.12*8*007126.0
D81.25
Vf
dL
dP 22pp
dp
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Power-Law Constant (n):
Pressure Drop In Drill Collars
Fluid Consistency Index (K):
Average Bulk Velocity inside Drill Collars (Vdc):
OD = 6.5 in ID = 2.5 in L = 600 ft
737.039
65log32.3
R
Rlog32.3n
300
600dc
2
n
737.0n600
dc cm
secdyne017.2
022,1
65*11.5
022,1
R11.5K
p
sec
ft28.18
5.2
280*408.0
D
Q408.0V
22dc
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Effective Viscosity in Collars(ec):
Reynolds Number in Collars (NRec):
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
ppn
p
p
1n
ppedc n4
1n3
D
V96K100
cP21.38737.0*4
1737.0*3
5.2
28.18*96017.2*100
737.01737.0
edc
870,1321.38
5.12*28.18*5.2*928VD928N
edc
dcRedc
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OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
NOTE: NRe > 2,100, so
Friction Factor in DC (fdc):b
Re
dc
dcN
af
So,
0759.050
93.3737.0log
50
93.3nloga dc
2690.07
737.0log75.1
7
nlog75.1b dc
005840.0870,13
0759.0
N
af
2690.0bRe
dc
dc
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Friction Pressure Gradient
(dP/dL)dc :
Friction Pressure Drop in Drill
Collars :
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
ft
psi3780.0
5.2*81.25
5.12*28.18*005840.0
D81.25
Vf
dL
dP 2
dc
2dcdc
dc
600*3780.0LdL
dPP dc
dcdc
Pdc = 227 psi
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Pressure Drop across Nozzles
DN1 = 11 32nds
(in) DN2 = 11
32nds (in) DN3 = 12 32nds (in)
2222
2
Nozzles121111
280*5.12*156P
PNozzles = 1,026 psi
223N
22N
2
1N
2
Nozzles
DDD
Q156P
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Pressure Dropin DC/HOLE
Annulus
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Q = gal/min
= lb/gal 8.5 in
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Power-Law Constant (n):
Fluid Consistency Index (K):
Average Bulk Velocity in DC/HOLE Annulus (Va):
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Pressure Dropin DC/HOLE Annulus
5413.03
20log657.0
R
Rlog657.0n
3
100dca
2
n
5413.0n100
dca cm
secdyne336.6
2.170
20*11.5
2.170
R11.5K
dca
sec
ft808.3
5.65.8
280*408.0
DD
Q408.0V
2221
22
dca
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Effective Viscosity in Annulus (ea):
Reynolds Number in Annulus (NRea):
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Pressure Dropin DC/HOLE Annulus
cP20.555413.0*3
15413.0*2
5.65.8
808.3*144336.6*100
5413.015413.0
ea
600,1
20.55
5.12*808.3*5.65.8928VDD928N
ea
a12Rea
aa n
a
a
1n
12
aaea n3
1n2
DD
V144K100
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So,
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Pressure Dropin DC/HOLE Annulus
NOTE: NRe < 2,100 Friction Factor in
Annulus (fa): 01500.0600,1
24
N
24f
aRea
ft
psi05266.0
5.65.881.25
5.12*808.3*01500.0
DD81.25
Vf
dL
dP 2
12
2aa
a
600*05266.0LdL
dPP hole/dc
hole/dchole/dc
Pdc/hole = 31.6 psi
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q = gal/min
= lb/gal
Pressure Dropin DP/HOLE Annulus
DHOLE = 8.5 inODDP = 4.5 in L = 11,400 ft
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Power-Law Constant (n):
Fluid Consistency Index (K):
Average Bulk Velocity in Annulus (Va):
Pressure Dropin DP/HOLE Annulus
DHOLE = 8.5 inODDP = 4.5 in L = 11,400 ft
5413.03
20log657.0
R
Rlog657.0n
3
100dpa
2
n
5413.0n100
dpa cm
secdyne336.6
2.170
20*11.5
2.170
R11.5K
dpa
sec
ft197.2
5.45.8
280*408.0
DD
Q408.0V
2221
22
dpa
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Effective Viscosity in Annulus (ea):
Reynolds Number in Annulus (NRea):
Pressure Dropin DP/HOLE Annulus
aa n
a
a
1n
12
aaea n3
1n2
DD
V144K100
cP64.975413.0*3
15413.0*2
5.45.8
197.2*144336.6*100
5413.015413.0
ea
044,1
64.97
5.12*197.2*5.45.8928VDD928N
ea
a12Rea
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So, psi
Pressure Dropin DP/HOLE Annulus
NOTE: NRe < 2,100 Friction Factor in
Annulus (fa): 02299.0044,1
24
N
24f
aRea
ft
psi01343.0
5.45.881.25
5.12*197.2*02299.0
DD81.25
Vf
dL
dP 2
12
2aa
a
400,11*01343.0LdL
dPP hole/dp
hole/dphole/dp
Pdp/hole = 153.2 psi
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Pressure DropCalculations
- SUMMARY -
PPUMP = PDP + PDC + PBIT NOZZLES
+ PDC/ANN + PDP/ANN + PHYD
PPUMP = + +
+ + +
PPUMP = psi
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PPUMP = 1,918 + 185 = 2,103 psi
PHYD = 0
PPUMP = PDS + PANN + PHYD
PDS = PDP + PDC + PBIT NOZZLES
= 665 + 227 + 1,026 = 1,918 psiPANN = PDC/ANN + PDP/ANN
= 32 + 153 = 185
2,103 psi
P = 0
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BHP = 185 + 7,800
What is the BHP?
BHP = PFRICTION/ANN + PHYD/ANN
BHP = PDC/ANN + PDP/ANN
+ 0.052 * 12.5 * 12,000
= 32 + 153 + 7,800 = 7,985 psig
2,103 psi
P = 0
BHP= 7,985 psig
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"Friction" Pressures
0
500
1,000
1,500
2,000
2,500
0 5,000 10,000 15,000 20,000 25,000
Distance from Standpipe, ft
"Fri
ctio
n" P
ress
ure,
psi
DRILLPIPE
DRILL COLLARS
BIT NOZZLES
ANNULUS
2103
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Hydrostatic Pressures in the Wellbore
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
9,000
0 5,000 10,000 15,000 20,000 25,000
Distance from Standpipe, ft
Hyd
rost
atic
Pre
ssur
e, p
si
BHP
DRILLSTRING ANNULUS
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Pressures in the Wellbore
0
1,000
2,000
3,000
4,000
5,000
6,000
7,000
8,000
9,000
10,000
0 5,000 10,000 15,000 20,000 25,000
Distance from Standpipe, ft
Pre
ssur
es,
psi
STATIC
CIRCULATING
2103
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Wellbore Pressure Profile
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
0 2,000 4,000 6,000 8,000 10,000
Pressure, psi
De
pth
, f
t
DRILLSTRING
ANNULUS
(Static)
BIT
2103
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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):
pRep N
16f
D81.25
Vf
dL
dP2
pp
dp
Then and
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Annular Flow - TurbulentIn 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:
bRe
a
aN
af Then and
50
93.3nloga a
7
nlog75.1b a
12
2aa
a DD81.25
Vf
dL
dP
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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?)
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Critical Circulation RateIn the Drillpipe/Hole Annulus:
Q, gal/min V, ft/sec Nre
280 2.197 1,044 300 2.354 1,154 350 2.746 1,446 400 3.138 1,756 450 3.531 2,086 452 3.546 2,099
452.1 3.547 2,100
ea
a12Re
VDD928N
a
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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.
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42d 8.25
vf
dL
dp_2
f
n = 1.0
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Importance of Pipe Size
or,
25.1
25.075.1_
75.0f
d1800
v
dL
dp
75.4
25.075.175.0f
d624,8
q
dL
dp
*Note that a small change in the pipe diameter results in large change in the pressure drop! (q = const.)
Eq. 4.66e
Decreasing the pipe ID 10% from 5.0” to 4.5” would result in an increase of frictional pressure drop by about 65% !!
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pf = 11.41 v 1.75
turbulent flow
pf = 9.11 vlaminar flow
Use max. pf value