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

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

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3 Contents u The Power Law Model u The Rotational Viscometer u A detailed Example - Pump Pressure l Pressure Drop in the Drillpipe l Pressure Drop in the Bit Nozzles l Pressure Drop in the Annulus u Wellbore Pressure Profiles

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4 Power Law Model K = consistency index n = flow behaviour index 0

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5 Fluid Flow in Pipes and Annuli

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6 LOG (SHEAR STRESS) (psi) Laminar FlowTurbulent n 1

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7 Rotating Sleeve Viscometer

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8 VISCOMETER RPM (RPM * 1.703) SHEAR RATE sec BOB SLEEVE ANNULUS DRILL STRING API RP 13D

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9 API RP 13D, June 1995 for Oil-Well Drilling Fluids u API RP 13D recommends using only FOUR of the six usual viscometer readings: u Use 3, 100, 300, 600 RPM Readings. u The 3 and 100 RPM reading are used for pressure drop calculations in the annulus, where shear rates are, generally, not very high. u 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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10 Example: Pressure Drop Calculations u Example Calculate the pump pressure in the wellbore shown on the next page, using the API method. u The relevant rotational viscometer readings are as follows: u R 3 = 3 (at 3 RPM) u R 100 = 20 (at 100 RPM) u R 300 = 39 (at 300 RPM) u R 600 = 65 (at 600 RPM)

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11 P PUMP = P DP + P DC + P BIT NOZZLES + P DC/ANN + P DP/ANN + P HYD Q = 280 gal/min = 12.5 lb/gal Pressure Drop Calculations P PUMP

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

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13 Effective Viscosity in Pipe ( ep ): Pressure Drop In Drill Pipe Reynolds Number in Pipe (N Rep ): OD = 4.5 in ID = 3.78 in L = 11,400 ft

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14 NOTE: N Re > 2,100, so Friction Factor in Pipe (f p ): Pressure Drop In Drill Pipe OD = 4.5 in ID = 3.78 in L = 11,400 ft So,

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15 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 : P dp = 665 psi

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

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17 Effective Viscosity in Collars ( ec ): Reynolds Number in Collars (N Rec ): OD = 6.5 in ID = 2.5 in L = 600 ft Pressure Drop In Drill Collars

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18 OD = 6.5 in ID = 2.5 in L = 600 ft Pressure Drop In Drill Collars NOTE: N Re > 2,100, so Friction Factor in DC (f dc ): So,

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19 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 P dc = 227 psi

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20 Pressure Drop across Nozzles D N1 = 11 32nds (in) D N2 = 11 32nds (in) D N3 = 12 32nds (in) P Nozzles = 1,026 psi

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21 Pressure Drop in DC/HOLE Annulus D HOLE = 8.5 in OD DC = 6.5 in L = 600 ft Q = gal/min = lb/gal 8.5 in

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22 Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity in DC/HOLE Annulus (V a ): D HOLE = 8.5 in OD DC = 6.5 in L = 600 ft Pressure Drop in DC/HOLE Annulus

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23 Effective Viscosity in Annulus ( ea ): Reynolds Number in Annulus (N Rea ): D HOLE = 8.5 in OD DC = 6.5 in L = 600 ft Pressure Drop in DC/HOLE Annulus

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24 So, D HOLE = 8.5 in OD DC = 6.5 in L = 600 ft Pressure Drop in DC/HOLE Annulus NOTE: N Re < 2,100 Friction Factor in Annulus (f a ): P dc/hole = 31.6 psi

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25 q = gal/min = lb/gal Pressure Drop in DP/HOLE Annulus D HOLE = 8.5 in OD DP = 4.5 in L = 11,400 ft

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26 Power-Law Constant (n): Fluid Consistency Index (K): Average Bulk Velocity in Annulus (V a ): Pressure Drop in DP/HOLE Annulus D HOLE = 8.5 in OD DP = 4.5 in L = 11,400 ft

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27 Effective Viscosity in Annulus ( ea ): Reynolds Number in Annulus (N Rea ): Pressure Drop in DP/HOLE Annulus

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28 So,psi Pressure Drop in DP/HOLE Annulus NOTE: N Re < 2,100 Friction Factor in Annulus (f a ): P dp/hole = psi

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29 Pressure Drop Calculations - SUMMARY - P PUMP = P DP + P DC + P BIT NOZZLES + P DC/ANN + P DP/ANN + P HYD P PUMP = + + + + + P PUMP = psi

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30 P PUMP = 1, = 2,103 psi P HYD = 0 P PUMP = P DS + P ANN + P HYD P DS = P DP + P DC + P BIT NOZZLES = ,026 = 1,918 psi P ANN = P DC/ANN + P DP/ANN = = 185 2,103 psi P=0P=0

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31 BHP = ,800 What is the BHP? BHP = P FRICTION/ANN + P HYD/ANN BHP = P DC/ANN + P DP/ANN * 12.5 * 12,000 = ,800 = 7,985 psig 2,103 psi P=0P=0 BHP= 7,985 psig

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32 DRILLPIPE DRILL COLLARS BIT NOZZLES ANNULUS 2103

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33 BHP DRILLSTRINGANNULUS

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34 STATIC CIRCULATING 2103

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35 DRILLSTRING ANNULUS (Static) BIT 2103

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36 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 N Re < 2,100, then Friction Factor in Pipe (f p ): Thenand

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37 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 N Re > 2,100, then Friction Factor in the Annulus: Thenand

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38 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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39 Critical Circulation Rate In the Drillpipe/Hole Annulus: Q, gal/minV, ft/sec N re , , , , , , ,100

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40 Optimum Bit Hydraulics u Under what conditions do we get the best hydraulic cleaning at the bit? l maximum hydraulic horsepower? l 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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42 n = 1.0

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43 Importance of Pipe Size or, *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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44 p f = v 1.75 turbulent flow p f = 9.11 v laminar flow Use max. p f value

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