PETE 411 Well Drilling Lesson 13 Pressure Drop Calculations API Recommended Practice 13D Third Edition, June 1, 1995

Homework HW #7. Pressure Drop Calculations Due Oct. 9, 2002 The API Power Law Model

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

Power Law Model K = consistency index n = flow behaviour index

Fluid Flow in Pipes and Annuli

Fluid Flow in Pipes and Annuli Laminar Flow Turbulent LOG (SHEAR STRESS) (psi) n 1

Rotating Sleeve Viscometer

Rotating Sleeve Viscometer (RPM * 1.703) SHEAR RATE sec -1 5.11 170.3 511 1022 VISCOMETER RPM 3 100 300 600 ANNULUS BOB DRILL STRING SLEEVE API RP 13D

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.

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 (at 3 RPM) R100 = 20 (at 100 RPM) R300 = 39 (at 300 RPM) R600 = 65 (at 600 RPM)

Pressure Drop Calculations PPUMP Q = 280 gal/min r = 12.5 lb/gal PPUMP = DPDP + DPDC + DPBIT NOZZLES + DPDC/ANN + DPDP/ANN + DPHYD

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity in Pipe (Vp):

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft Effective Viscosity in Pipe (mep): Reynolds Number in Pipe (NRep):

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft NOTE: NRe > 2,100, so Friction Factor in Pipe (fp): So,

Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft Friction Pressure Gradient (dP/dL)p : Friction Pressure Drop in Drill Pipe : DPdp = 665 psi

Pressure Drop In Drill Collars OD = 6.5 in ID = 2.5 in L = 600 ft Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity inside Drill Collars (Vdc):

Pressure Drop In Drill Collars OD = 6.5 in ID = 2.5 in L = 600 ft Pressure Drop In Drill Collars Effective Viscosity in Collars(mec): Reynolds Number in Collars (NRec):

Pressure Drop In Drill Collars 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): So,

Pressure Drop In Drill Collars OD = 6.5 in ID = 2.5 in L = 600 ft Pressure Drop In Drill Collars Friction Pressure Gradient (dP/dL)dc : Friction Pressure Drop in Drill Collars : DPdc = 227 psi

Pressure Drop across Nozzles DN1 = 11 32nds (in) DN2 = 11 32nds (in) DN3 = 12 32nds (in) DPNozzles = 1,026 psi

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

Pressure Drop in DC/HOLE Annulus DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity in DC/HOLE Annulus (Va):

Pressure Drop in DC/HOLE Annulus DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft Effective Viscosity in Annulus (mea): Reynolds Number in Annulus (NRea):

Pressure Drop in DC/HOLE Annulus DHOLE = 8.5 in ODDC = 6.5 in L = 600 ft NOTE: NRe < 2,100 Friction Factor in Annulus (fa): DPdc/hole = 31.6 psi So,

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

Pressure Drop in DP/HOLE Annulus DHOLE = 8.5 in ODDP = 4.5 in L = 11,400 ft Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity in Annulus (Va):

Pressure Drop in DP/HOLE Annulus Effective Viscosity in Annulus (mea): Reynolds Number in Annulus (NRea):

Pressure Drop in DP/HOLE Annulus NOTE: NRe < 2,100 Friction Factor in Annulus (fa): DPdp/hole = 153.2 psi So, psi

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

PPUMP = 1,918 + 185 = 2,103 psi PPUMP = DPDS + DPANN + DPHYD DPDS = DPDP + DPDC + DPBIT NOZZLES = 665 + 227 + 1,026 = 1,918 psi P = 0 DPANN = DPDC/ANN + DPDP/ANN = 32 + 153 = 185 DPHYD = 0 PPUMP = 1,918 + 185 = 2,103 psi

What is the BHP? BHP = 7,985 psig BHP = 185 + 7,800 BHP = DPFRICTION/ANN + DPHYD/ANN 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

DRILLPIPE DRILL COLLARS BIT NOZZLES ANNULUS 2103 DRILL COLLARS BIT NOZZLES ANNULUS

BHP DRILLSTRING ANNULUS

CIRCULATING 2103 STATIC

DRILLSTRING ANNULUS BIT 2103 DRILLSTRING ANNULUS (Static) BIT

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 and

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: Then and

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?)

Critical Circulation Rate In 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

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.

n = 1.0

Importance of Pipe Size Eq. 4.66e or, *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% !!

Dpf = 11.41 v 1.75 turbulent flow Dpf = 9.11 v laminar flow Use max. Dpf value