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HWK 3 solution

# HWK 3 solution - torque at a given rotating speed The top...

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PGE 203 HWK 3 Solution Due at start of lecture: Feb. 22 The hoist that raises pipe out of a well or lowers pipe into a well is called the draw works. The power that a draw works must have is related to the weight being picked up and the speed at which the pipe is raised or lowered. (1) Compute the weight suspended from the derrick if the pipe weighs 26 lbf/ft and the length of pipe in the well is 20,000 ft and the drilling mud in the well has a density of 16 lbm/gal. ( 29 ( 29 ( 29 f 26 20,000 1 0.0153 16 392,704 lb W = - = (2) Assuming the draw works can raise pipe 90 ft in one minute and that the total pipe weight is 1,000,000 lbf, what is the horsepower requirement for the draw works? Recall that Hp x 33,000 = lbf-ft/min. ( 29 f f 1,000,000 90 lb ft hp-min 2,727 hp 33,000 min lb ft P = = Torque is required to overcome friction in the well and cause the bit to turn at the bottom of the drill string. A machine called a top drive provides the power necessary to create the

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Unformatted text preview: torque at a given rotating speed. The top drive is suspended in the derrick and connected directly to the top of the drill pipe. (3) What is the maximum torque a 375 KW top drive can produce at 125 revolutions/minute (recall that there are 2π radians/revolution)? Please show the torque in units of ft-lbf. ( 29 f f f f hp 375 KW 502 hp 0.7457 KW lb ft lb ft 502 hp 33,000 16,566,000 min min lb ft rev min 16,566,000 21,093 ft-lb min 2 radians 125 rev P P T π = = = = = = The mud pumps are required to circulate drilling mud from the surface through the inside of the drill string and back to the surface in the annulus between the hole and the outside of the drill pipe. (4) A pump is required to circulate mud at 150 gal/min with a pressure of 5,000 psi. What is the hydraulic horsepower of the pump? 3 2 f 2 2 f 5,000 lb 150 gal ft 144 in min hp min 7.48 gal 33,000 lb ft in ft 438 hp P P = =...
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