1
PETE 661Drilling Engineering
Lesson 4
Wellbore Hydraulics,
Pressure Drop Calculations
2
Wellbore Hydraulics
• Hydrostatics• Buoyancy• Pipe Tension vs. Depth• Effect of Mud Pressure• Laminar and Turbulent Flow• Pressure Drop Calculations
– Bingham Plastic Model– API Power-Law Model
3
Assignments:
READ: ADE Ch. 4
HW #3: On the Web. - Axial Tension
Due 09-20-04
4
)DD(052.0pp 1ii
n
1ii0
Fig. 4-3. A Complex
Liquid Column
D052.0p
pD052.0p 0
5Fig. 4-4. Viewing the Well as a Manometer (U-Tube)
PPUMP = ?
6
Figure 4.4
})000,10(0.9)000,1(7.16
)700,1(7.12)300(5.8)000,7(5.10{052.00
ppa
psig 00 p
psig 266,1p a
D052.0p
7
Buoyancy Force = weight of fluid displaced (Archimedes, 250 BC)
Figure 4-9. Hydraulic forces acting on a submerged body
8
Effective (buoyed) Weight
s
fe 1WW
Buoyancy Factor
Valid for a solid body or an open-ended pipe!
sf
f
be
W-W
V-W
FWW
We = buoyed weightW = weight in airFb = buoyancy forceV = volume of bodyf = fluid densitys = body density
9
Example
For steel,
immersed in mud,
the buoyancy factor is:
gal/lbm .s 565
)gal/lbm .( f 015
771.05.65
0.1511
s
f
A drillstring weighs 100,000 lbs in air.
Buoyed weight = 100,000 * 0.771 = 77,100 lbs
(= 490 lbm/ft3 )
10
Axial Forces in Drillstring
Fb = bit weightF1 & F1 are pressure forces
11
Simple Example - Empty Wellbore
Drillpipe weight = 19.5 lbf/ft 10,000 ft
OD = 5.000 inID = 4.276 in
22 IDOD4
A
A = 5.265 in2
W = 19.5 lbf/ft * 10,000 ft = 195,000 lbf
AXIAL TENSION, lbf
DE
PT
H,
ft
0 lbf 195,000 lbf
12
Example - 15 lb/gal Mud in Wellbore
Drillpipe weight = 19.5 lbf/ft 10,000 ft
OD = 5.000 inID = 4.276 in
22 IDOD4
A
A = 5.265 in2
W = 195,000 - 41,100 = 153,900 lbf
AXIAL TENSION, lbf
DE
PT
H,
ft
0 195,000 lbf
Pressure at bottom = 0.052 * 15 * 10,000 = 7,800 psiF = P * A= 7,800 * 5.265= 41,100 lbf
153,900- 41,100
13
Axial Tension in Drill String
Example A drill string consists of
10,000 ft of 19.5 #/ft drillpipe and
600 ft of 147 #/ft drill collars
suspended off bottom in 15#/gal mud
(Fb = bit weight = 0).
• What is the axial tension in the
drillstring as a function of depth?
14
Example
Pressure at top of collars
= 0.052 (15) 10,000 = 7,800 psi
Pressure at bottom of collars
= 0.052 (15) 10,600 = 8,268 psi
Cross-sectional area of pipe,
22
2
31 in73.5ft
in144*
ft/lb490
ft/lb5.19A
A1
10,000’
10,600’
15
Cross-sectional area of collars,
22 in2.43144*
490
147A
2
1
537735243 in...
AAarea alDifferenti 2
A2
A1Example – cont’d
16
1. At 10,600 ft. (bottom of drill collars)
Compressive force = p A
= 357,200 lbf
[ axial tension = - 357,200 lbf ]
22
in2.43*in
lbf268,8
4
32
1
Example - cont’d
17
Example - cont’d
2. At 10,000 ft+ (top of collars)
FT = W2 - F2 - Fb
= 147 lbm/ft * 600 ft - 357,200
= 88,200 - 357,200
= -269,000 lbf
4
32
1
Fb = FBIT = 0
18
3. At 10,000 ft - (bottom of drillpipe)
FT = W1+W2+F1-F2-Fb
= 88,200 + 7800 lbf/in2 * 37.5in2 - 357,200
= 88,200 + 292,500 - 357,200
= + 23,500 lbf
4
32
1
Example - cont’d
19
4. At Surface
FT = W1 + W2 + F1 - F2 - Fb
= 19.5 * 10,000 + 88,200
+ 292,500 - 357,200 - 0
= 218,500 lbf
Alternatively: FT = WAIR * BF
= 283,200 * 0.7710 = 218,345 lbf
4
32
1
Example - cont’d
20Fig. 4-11. Axial tensions as a function of depth for Example 4.9
21
Example - Summary
1. At 10,600 ft FT = -357,200 lbf [compression]
2. At 10,000 + ft FT = -269,000 lbf [compression]
3. At 10,000 - ft FT = +23,500 lbf [tension]
4. At Surface FT = +218,500 lbf [tension]
22
Axial Load with FBIT = 68,000 lbf
23
24
For multiple nozzles in parallel
Vn is the same for each nozzle even if the dn varies!
This follows since p is the same across each nozzle.
tn A117.3
qv
2
2
t2d
-5
bit AC
q10*8.311Δp
10*074.8
pcv
4dn
&
Cd = 0.95
25
Hydraulic Horsepower
… of pump putting out 400 gpm at 3,000 psi = ?
Power, in field units:
1714
000,3*400 HHP
1714
pq HHP
Hydraulic Horsepower of Pump = 700 hp
26
What is Hydraulic Impact Force
… developed by bit?
If:
psi 169,1Δp
lb/gal 12
gal/min 400q
95.0C
n
D
pqc01823.0F dj
27
Impact = rate of change of momentum
lbf 820169,1*12400*95.0*01823.0F
pqc01823.0F
60*17.32
vqv
t
m
t
mvF
j
dj
nj
28
Laminar Flow
Rheological Models Newtonian Bingham Plastic Power-Law (ADE & API)
Rotational Viscometer
Laminar Flow in Wellbore Fluid Flow in Pipes Fluid Flow in Annuli
29
Laminar Flow of Newtonian Fluids
A
F
L
V
Experimentally:
30
Newtonian 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.
sec
12
cm
dyne
31
Newtonian Fluid Model
Viscosity may be expressed in poise or centipoise.
poise 0.01 centipoise 1
scm
g1
cm
s-dyne1 poise 1
2
2cm
secdyne
32
Shear Stress vs. Shear Rate for a Newtonian Fluid
Slope of line
.
33
Apparent Viscosity
Apparent viscosity =
is the slope at each shear rate, .,, 321
/
34
Typical Drilling Fluid Vs. Newtonian, Bingham and Power Law Fluids
(Plotted on linear paper)
35
Rheological Models
1. Newtonian Fluid:
2. Bingham Plastic Fluid:
viscosityplastic
point yield
p
y
What if
y
py
rate shear
viscosity absolute
stress shear
36
RotatingSleeve
Viscometer
37
Figure 3.6Rotating Viscometer
Rheometer
We determine rheological properties of drilling fluids in this device
Infinite parallel plates
38
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(
39
Rheometer - base case
N (RPM) sec-1) 3 5.11 6 10.22 100 170 200 340 300 511 600 1022
RPM * 1.703 = sec-1
40
Example
A 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
= 20 = 12
See Appendix A
41
Example
8-12
p300y
2y ft lbf/100 4
= 20 = 12
(See Appendix A)
42
Gel Strength
43
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)
44
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
45
Velocity Profiles(laminar flow)
Fig. 4-26. Velocity profiles for laminar flow: (a) pipe flow and (b) annular flow
46
“It looks like concentric rings of fluid telescoping down the pipe at different velocities”
3D View of Laminar Flow in a pipe - Newtonian Fluid
47
Table 4.3 - Summary of Equations for Rotational Viscometer
Newtonian Model
Na N
300
Nr
066.52
300a
or
48
Table 4.3 - Summary of Equations for Rotational Viscometer
300
N
or
1pNy 1
rpm 3 atmaxg
Bingham Plastic Model
300600p )(NN
300
or
12 NN12
p
p300y
or
or
49
Example 4.22
Compute the frictional pressure loss for a 7” x 5” annulus, 10,000 ft long, using the slot flow representation in the annulus. The flow rate is 80 gal/min. The viscosity is 15 cp. Assume the flow pattern is laminar.
7” 5” 1”
6
50
Example 4.22
The average velocity in the annulus,
)52.448(7
80
)d2.448(d
qv
2221
22
_
ft/s 1.362v_
212
_
f
dd1000
vμ
dL
dp
51
Example 4.22
51.0750 psi 51
)57(1000
)000,10()362.1()15(D
dL
dpΔp
2f
fp
212
_
f
dd1000
vμ
dL
dp
52
Total Pump Pressure
• Pressure loss in surf. equipment
• Pressure loss in drill pipe
• Pressure loss in drill collars
• Pressure drop across the bit nozzles
• Pressure loss in the annulus between the drill collars and the hole wall
• Pressure loss in the annulus between the drill pipe and the hole wall
• Hydrostatic pressure difference ( varies)
53
Types of flow
Laminar
Fig. 4-30. Laminar and turbulent flow patterns in a circular pipe: (a) laminar flow, (b) transition between laminar and turbulent flow and (c) turbulent flow
Turbulent
54
Turbulent Flow - Newtonian Fluid
We often assume that fluid flow is
turbulent if Nre > 2100
cp. fluid, ofviscosity μ
in I.D., piped
ft/s velocity,fluid avg. v
lbm/gal density, fluid ρ where_
μ
dvρ928N
_
Re
55
Turbulent Flow - Newtonian Fluid
25.1
25.075.1_
75.0f
d1800
v
dL
dp
Turbulent Flow - Bingham Plastic Fluid
25.1
25.0p
75.1_75.0
f
d1800
v
dL
dp
25.112
25.0p
75.1_75.0
f
dd396,1
v
dL
dp
25.112
25.075.1_
75.0f
dd396,1
v
dL
dp
In Annulus
In Pipe
56
API Power Law Model
K = consistency index
n = flow behaviour index
SHEAR STRESS
psi
= K n
SHEAR RATE, , sec-1
0
API RP 13D
57
Rotating Sleeve Viscometer
VISCOMETERRPM
3100
300600
(RPM * 1.703)
SHEAR RATE
sec -1
5.11170.3 511
1022
BOB
SLEEVE
ANNULUS
DRILLSTRING
58
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)
59
PPUMP = PDP + PDC
+ PBIT NOZZLES
+ PDC/ANN + PDP/ANN
+ PHYD
Q = 280 gal/min
= 12.5 lb/gal
Pressure DropCalculations
PPUMP
60
Power-Law Constant (n):
Pressure Drop In Drill Pipe
Fluid Consistency Index (K):
Average Bulk Velocity in Pipe (V):
OD = 4.5 in ID = 3.78 in L = 11,400 ft
737.039
65log32.3
R
Rlog32.3n
300
600
2737.0600 sec
017.2022,1
65*11.5
022,1
11.5
cm
dyneRK
n
n
sec
ft00.8
78.3
280*408.0
D
Q408.0V
22
61
Effective Viscosity in Pipe (e):
Pressure Drop In Drill Pipe
Reynolds Number in Pipe (NRe):
OD = 4.5 in ID = 3.78 in L = 11,400 ft
n1n
e n4
1n3
D
V96K100
cP53737.0*4
1737.0*3
78.3
8*96017.2*100
737.01737.0
e
616,653
5.12*00.8*78.3*928VD928N
eRe
62
NOTE: NRe > 2,100, soFriction Factor in Pipe (f):
Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft
So,
bReN
af
0759.050
93.3737.0log
50
93.3nloga
2690.07
737.0log75.1
7
nlog75.1b
007126.0616,6
0759.0
N
af
2690.0bRe
63
Friction Pressure Gradient (dP/dL) :
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*05837.0LdL
dPP
Pdp = 665 psi
ft
psi05837.0
78.3*81.25
5.12*8*007126.0
D81.25
Vf
dL
dP 22
64
Power-Law Constant (n):
Pressure Drop In Drill Collars
Fluid Consistency Index (K):
Average Bulk Velocity inside Drill Collars (V):
OD = 6.5 in ID = 2.5 in L = 600 ft
737.039
65log32.3
R
Rlog32.3n
300
600
2
n
737.0n600
cm
secdyne017.2
022,1
65*11.5
022,1
R11.5K
sec
ft28.18
5.2
280*408.0
D
Q408.0V
22
65
Effective Viscosity in Collars(e):
Reynolds Number in Collars (NRe):
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
n1n
e n4
1n3
D
V96K100
cP21.38737.0*4
1737.0*3
5.2
28.18*96017.2*100
737.01737.0
e
870,1321.38
5.12*28.18*5.2*928VD928N
eRe
66
OD = 6.5 in ID = 2.5 in L = 600 ft
Pressure Drop In Drill Collars
NOTE: NRe > 2,100, soFriction Factor in DC (f): b
ReN
af
So,
0759.050
93.3737.0log
50
93.3nloga
2690.07
737.0log75.1
7
nlog75.1b
005840.0870,13
0759.0
N
af
2690.0bRe
67
Friction Pressure Gradient (dP/dL) :
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 22
600*3780.0LdL
dPP
Pdc = 227 psi
68
Pressure Drop across Nozzles
DN1 = 11 32nds
(in) DN2 = 11
32nds (in) DN3 = 12 32nds (in)
2222
2
121111
280*5.12*156P
PNozzles = 1,026 psi
22
3N2
2N
2
1N
2
DDD
Q156P
69
Pressure Dropin DC/HOLE
Annulus
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Q = gal/min
= lb/gal 8.5 in
70
Power-Law Constant (n):
Fluid Consistency Index (K):
Average Bulk Velocity in DC/HOLE Annulus (V):
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
Pressure Dropin DC/HOLE Annulus
5413.03
20log657.0
R
Rlog657.0n
3
100
2
n
5413.0n100
cm
secdyne336.6
2.170
20*11.5
2.170
R11.5K
sec
ft808.3
5.65.8
280*408.0
DD
Q408.0V
2221
22
71
Effective Viscosity in Annulus (e):
Reynolds Number in Annulus (NRe):
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
cP20.555413.0*3
15413.0*2
5.65.8
808.3*144336.6*100
5413.015413.0
e
600,1
20.55
5.12*808.3*5.65.8928VDD928N
e
12Re
n1n
12e n3
1n2
DD
V144K100
Pressure Dropin DC/HOLE Annulus
72
So,
DHOLE = 8.5 inODDC = 6.5 in L = 600 ft
NOTE: NRe < 2,100 Friction Factor in Annulus (f):
01500.0600,1
24
N
24f
Re
ft
psi05266.0
5.65.881.25
5.12*808.3*01500.0
DD81.25
Vf
dL
dP 2
12
2
600*05266.0LdL
dPP
Pdc/hole = 31.6 psi
Pressure Dropin DC/HOLE Annulus
73
q = gal/min
= lb/gal
Pressure Dropin DP/HOLE Annulus
DHOLE = 8.5 inODDP = 4.5 in L = 11,400 ft
74
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
100
2
n
5413.0n100
cm
secdyne336.6
2.170
20*11.5
2.170
R11.5K
sec
ft197.2
5.45.8
280*408.0
DD
Q408.0V
2221
22
75
Effective Viscosity in Annulus (e):
Reynolds Number in Annulus (NRe):
Pressure Dropin DP/HOLE Annulus
n1n
12e n3
1n2
DD
V144K100
cP64.975413.0*3
15413.0*2
5.45.8
197.2*144336.6*100
5413.015413.0
e
044,1
64.97
5.12*197.2*5.45.8928VDD928N
e
12Re
76
So, psi
Pressure Dropin DP/HOLE Annulus
NOTE: NRe < 2,100 Friction Factor in Annulus (f):
02299.0044,1
24
N
24f
Re
ft
psi01343.0
5.45.881.25
5.12*197.2*02299.0
DD81.25
Vf
dL
dP 2
12
2
400,11*01343.0LdL
dPP
Pdp/hole = 153.2 psi
77
Pressure Drop Calcs.- SUMMARY -
PPUMP = PDP + PDC + PBIT NOZZLES
+ PDC/ANN + PDP/ANN + PHYD
PPUMP = 665 + 227 + 1,026
+ 32 + 153 + 0
PPUMP = 1,918 + 185 = 2,103 psi
78
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
79
"Friction" Pressures
0
500
1,000
1,500
2,000
2,500
0 5,000 10,000 15,000 20,000 25,000
Cumulative Distance from Standpipe, ft
"Fri
ctio
n" P
ress
ure,
psi
DRILLPIPE
DRILL COLLARS
BIT NOZZLES
ANNULUS
80
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
Cumulative Distance from Standpipe, ft
Hyd
rost
atic
Pre
ssur
e, p
si
BHP
DRILLSTRING ANNULUS
81
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
Cumulative Distance from Standpipe, ft
Pre
ssur
es,
psi
STATIC
CIRCULATING
82
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
83
Pipe Flow - LaminarIn 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, thenFriction Factor in Pipe (f):
ReN
16f
D81.25
Vf
dL
dP2
Then and
84
85d 8.25
vf
dL
dp_2
n = 1.0
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