Homework 4 solution - PETE 661 Homework 4 solution HSP =...

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PETE 661 Homework 4 solution Given TD 11000 ft Hole size 9 in DP OD 5.5 in For Laminar Flow Wt 21.9 lb/ft DP ID 4.778 in DC OD 7 in DC ID 2.5 in DC L 500 ft Q 600 gpm MW 13 ppg For Turbulent Flow Casing 6000 ft Casing ID 10 in Nozzles 14 3 Viscometer Data R600 95 R300 70 R100 28 R3 8 Solution Hydrostatic Pressure Depth HSP 6000 4056 10500 7098 11000 7436 Annulus Pressures Depth n K v Mu ea Nre a b f dp/dl dP Pressure 6000 0.357453 22.80953 3.509677 129.8707 1467.101 24 1 0.016359 0.022554 135.3255 4191.326 10500 0.357453 22.80953 4.823645 90.08233 2260.972 0.069664 0.313826 0.006171 0.020663 27.76889 7261.094 11000 0.357453 22.80953 7.65 46.75014 3948.207 0.069664 0.313826 0.00518 0.076351 38.17556 7637.27 201.27 ECD 13.35187 ppg bhp/.052/TD Pressure drop across the bit Pnozzles 2111.62 psi Pressure just inside the bit is the BHP plus the bit pressure drop 9748.89 psi Pump Pressure is equal to all the pressure drops in the system +/- and difference in HSP In this case the mud density is constant through out. Now we calculate frictional pressure inside the drillstring and add this to the annular friction and the bit pressure drop
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This note was uploaded on 10/30/2011 for the course PETROLEUM 661 taught by Professor Dr.jeromej.schubert during the Fall '11 term at Texas A&M University-Galveston.

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