Formulas and Calculations
6.
Surge and Swab Pressures
Method 1 1. Determine n:
n = 3.32 log φ600 φ300
2. Determine K:
K= φ600 511n
3. Determine velocity, ft/mm: For plugged flow:
v = [ 0.45 + Dp2 ] Vp 2 2 Dh — Dp
For open pipe:
v = [ 0.45 + Dp2 — Di2 ] Vp Dh2 — Dp2 + Di2
4. Maximum pipe velocity:
Vm = 1.5 x v
5. Determine pressure losses:
Ps = (2.4 Vm x 2n + 1)n x KL . Dh — Dp 3n 300 (Dh — Dp)
Nomenclature: n = dimensionless K = dimensionless φ600 = 600 viscometer dial reading φ300 = 300 viscometer dial reading v = fluid velocity, ft/min Vm = maximum pipe velocity, ft/mm
Di = drill pipe or drill collar ID, in. Dh = hole diameter, in. Dp = drill pipe or drill collar OD, in Ps = pressure loss, psi Vp = pipe velocity, ft/min L = pipe length, ft
Example 1: Determine surge pressure for plugged pipe: Data:
Well depth = 15,000 ft Drill pipe OD Hole size = 7-7/8 in. Drill pipe ID Drill collar length = 700 ft Mud weight Average pipe running speed = 270 ft/mm Drill collar = 6-1/4” OD x 2-3/4” ID Viscometer readings: φ600 = 140 φ300 = 80
1. Determine n:
n = 3.32 log 140 80 n = 0.8069
2. Determine K:
K= 80 5110.8069 K= 0.522
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= 4-1/2 in. = 3.82 in. = 15.0 ppg
Formulas and Calculations
3. Determine velocity, ft/mm:
v = [ 0.45 +
4.52 ] 270 2 2 7.875 — 4.5
v = (0.45 + 0.484)270 v = 252 ft/min 4. Determine maximum pipe velocity, ft/min:
Vm = 1.5 x 252 Vm = 378 ft/min
5. Determine pressure losses, psi: Ps =[ 2.4 x 378 x 2(0.8069) + 1]0.8069 x (0.522)(14300) 7.875 — 4.5 3(0.8069) 300 (7.875 — 4.5) Ps = (268.8 x 1.1798) 0.8069 x 7464..6 1012.5 Ps = 97.098 x 7.37 Ps = 716 psi surge pressure Therefore, this pressure is added to the hydrostatic pressure of the mud in the wellbore. If, however, the swab pressure is desired, this pressure would be subtracted from the hydrostatic pressure. Example 2: Determine surge pressure for open pipe: 1. Determine velocity, ft/mm: :
v = [ 0.45 + 4.52 — 3.822 ] 270 2 2 2 7.875 — 4.5 + 3.82 v = (0.45 + 5.66 ) 270 56.4 v = (0.45 + 0.100)270 v = 149 ft/mm
2 . Maximum pipe velocity, ft/mm:
3 . Pressure loss, psi:
Vm = 149 x 1.5 Vm = 224 ft/mm
Ps = [ 2.4 x 224 x 2(0.8069) + 1 ]0.8069 x (0.522)(14300) 7.875 — 4.5 3(0.8069) 300(7.875 — 4.5) Ps = (159.29 x 1.0798)0.8069 x 7464.5 1012.5 Ps = 63.66 x 7.37 Ps = 469 psi surge pressure
Therefore, this pressure would be added to the hydrostatic pressure of the mud in the wellbore.
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Formulas and Calculations
If, however, the swab pressure is desired, this pressure would be subtracted from the hydrostatic pressure of the mud in the wellbore.
Method 2 Surge and swab pressures Assume:
1) Plugged pipe 2) Laminar flow around drill pipe 3) Turbulent flow around drill collars
These calculations outline the procedure and calculations necessary to determine the increase or decrease in equivalent mud weight (bottomhole pressure) due to pressure surges caused by pulling or running pipe. These calculations assume that the end of the pipe is plugged (as in running casing with a float shoe or drill pipe with bit and jet nozzles in place), not open ended. A. Surge pressure around drill pipe: 1. Estimated annular fluid velocity (v) around drill pipe:
v = [ 0.45 + Dp2 ] Vp Dh2 — Dp2
2. Maximum pipe velocity (Vm):
Vm = v x 1.5
3. Determine n:
n = 3.32 log φ600 φ300
4. Determine K:
K= φ600 511n
5. Calculate the shear rate (Ym) of the mud moving around the pipe: Ym = 2.4 x Vm Dh — DP 6. Calculate the shear stress (T) of the mud moving around the pipe: 7. Calculate the pressure (Ps) decrease for the interval:
T = K (Ym)n
Ps = 3.33 T x L Dh — Dp 1000
B. Surge pressure around drill collars: 1. Calculate the estimated annular fluid velocity (v) around the drill collars: v = [ 0.45 + Dp2 ] Vp 2 2 Dh — Dp 2. Calculate maximum pipe velocity (Vm):
Vm = v x 1.5
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Formulas and Calculations
3. Convert the equivalent velocity of the mud due to pipe movement to equivalent flow rate (Q): Q = Vm [(Dh)2 — (Dp)2] 24.5 4. Calculate the pressure loss for each interval (Ps): Ps = 0.000077 x MW0.8 x Q1~8 x PV0.2 x L (Dh — Dp)3 x (Dh + Dp)1.8 C. Total surge pressures converted to mud weight: Total surge (or swab) pressures:
psi = Ps (drill pipe) + Ps (drill collars)
D. If surge pressure is desired:
SP, ppg = Ps ÷ 0.052 ÷ TVD, ft “+“ MW, ppg
E. If swab pressure is desired:
SP, ppg = Ps ÷ 0.052 ÷ TVD, ft “—“ MW, ppg
Example:
Determine both the surge and swab pressure for the data listed below:
Data: Mud weight = 15.0 ppg Yield point = 20 lb/l00 sq ft Drill pipe OD = 4-1/2 in. Drill collar OD = 6-1/4 in. Pipe running speed = 270 ft/min
Plastic viscosity = 60 cps Hole diameter = 7-7/8 in. Drill pipe length = 14,300 ft Drill collar length = 700 ft
A. Around drill pipe: 1.Calculate annular fluid velocity (v) around drill pipe:
v = [ 0.45 +
(45)2 ] 270 2 2 7.875 — 4.5
v = [0.45 + 0.4848] 270 v = 253 ft/mm 2. Calculate maximum pipe velocity (Vm):
Vm = 253 x 1.5 Vm = 379 ft/min
NOTE: Determine n and K from the plastic viscosity and yield point as follows: PV + YP = φ300 reading Example: PV = 60
φ300 reading + PV = φ600 reading
YP = 20
60 + 20 = 80 (φ300 reading)
80 + 60 = 140 (φ600 reading)
3. Calculate n:
n = 3.32 log 80 140 80 n = 0.8069
4. Calculate K:
K = 80 5110.8069 K = 0.522
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Formulas and Calculations
5. Calculate the shear rate (Ym) of the mud moving around the pipe:
Ym = 2.4 x 379 (7.875 — 4.5) Ym = 269.5
6. Calculate the shear stress (T) of the mud moving around the pipe:
7. Calculate the pressure decrease (Ps) for the interval:
T = 0.522 (269.5)0.8069 T = 0.522 x 91.457 T = 47.74
Ps = 3.33 (47.7) x 14,300 (7.875 — 4.5) 1000 Ps = 47.064 x 14.3 Ps = 673 psi
B. Around drill collars: 1. Calculate the estimated annular fluid velocity (v) around the drill collars: v = [ 0.45 + (6.252 ÷ (7.8752 — 6.252))] 270 v = (0.45 + 1.70)270 v = 581 ft/mm 2. Calculate maximum pipe velocity (Vm):
Vm = 581 x 1.5 Vm = 871.54 ft/mm
3. Convert the equivalent velocity of the mud due to pipe movement to equivalent flow-rate (Q): Q = 871.54 (7.8752 — 6.252) 24.5 Q = 20004.567 24.5 Q = 816.5 4. Calculate the pressure loss (Ps) for the interval: Ps = 0.000077 x 150.8 x 8161.8 x 600.2 x 700 (7.875 — 6.25)3 x (7.875 + 6.25)1.8 Ps = 185837.9 504.126 Ps = 368.6 psi C. Total pressures:
psi = 672.9 psi + 368.6 psi psi = 1041.5 psi
D. Pressure converted to mud weight, ppg:
ppg = 1041.5 psi ÷ 0.052 ÷ 15,000 ft ppg = 1.34
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Formulas and Calculations
E. If surge pressure is desired:
Surge pressure, ppg = 15.0 ppg + 1.34 ppg Surge pressure = 16.34 ppg
F. If swab pressure is desired:
Swab pressure, ppg = 15.0 ppg — 1.34 ppg Swab pressure = 13.66 ppg
7.
Equivalent Circulation Density (ECD)
1. Determine n:
n = 3.32 log φ600 φ300
2. Determine K:
K= φ600 511n
3. Determine annular velocity (v), ft/mm: v = 24.5 x Q Dh2 — D2 4. Determine critical velocity (Vc), ft/mm: Vc = (3.878 x 104 x K)(1÷ (2— n)) x ( 2.4 x 2n +1) (n÷ (2— n)) MW Dh—Dp 3n 5. Pressure loss for laminar flow (Ps), psi: Ps = ( 2.4v x 2n +1 )n x KL . Dh — Dp 3n 300 (Dh — Dp) Ps = 7.7 x 10—5 x MW0.8 x Q1.8 x PV0.2 x L (Dh — Dp)3 x (Dh + Dp)1.8
6. Pressure loss for turbulent flow (Ps), psi:
7. Determine equivalent circulating density (ECD), ppg: ECD, ppg = Ps — 0.052 TVD, ft + 0MW, ppg Example: Data:
Equivalent circulating density (ECD), ppg:
Mud weight = 12.5 ppg Yield point = 12 lb/100 sq ft Drill collar OD = 6.5 in. Drill collar length = 700 ft True vertical depth = 12,000 ft
Plastic viscosity Circulation rate Drill pipe OD Drill pipe length Hole diameter
= 24 cps = 400 gpm = 5.0 in = 11,300 ft = 8.5 in.
NOTE: If φ600 and φ300 viscometer dial readings are unknown, they may be obtained from the plastic viscosity and yield point as follows: 24 + 12 = 36 Thus, 36 is the φ300 reading. 36 + 24 = 60 Thus, 60 is the φ600 reading.
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