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PGE 421K Spring 2010 HWSolution1

# PGE 421K Spring 2010 HWSolution1 - PGE 421K SPRING 2010 HW...

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PGE 421K SPRING 2010 HW 1 Solutions 120 pts 1) Balance equations. Drilling mud flows from a pit, down a length of drill pipe and returns to the pit through the annulus between pipe and earth formation. Large volumes of mud sometimes flow into one troublesome formation located at 6600 ft below ground surface. a. Sketch an appropriate control volume that will be useful for quantifying the possible loss of mud with a cumulative mass balance. b. Write down all relevant transport terms in the balance, with units. If there are none for your control volume, state this. Mass transport rate out into the formation from leaking, m leakout , ft 3 /min. Since it is assumed that the mud is a compressible fluid under this condition, mud density is constant then and canceled out from the mass balance equation and volume unit ft 3 can be used as the unit. c. Write down all relevant generation terms in the balance, with units. If there are none for your control volume, state this. There is no generation term in this case. 30 pts

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Write down all relevant accumulation terms in the balance, with units. If there are none for your control volume, state this. Mud staying in the whole system at t 1 , m Mud | 1 , ft 3 . Mud staying in the whole system at t 2 , m Mud | 2 , ft 3 . e. During the day, your crew drilled a 7 1/2” hole through the troublesome formation without incident, and continued to drill ahead steadily, reaching a depth of 7250 ft at the midnight shift change. The mud pit level stayed constant at 6.5 ft during the day and still read 6.5 ft at the shift change. At 15 minutes after midnight, the new crew observed that the mud pit level had dropped to 5.5 ft. Use the equation resulting from (a)-(d) to determine how long will it be before the mud pit is empty. g G±²³´µ¶ × ·¸g¹ ¸º·¹»¼½¾ = (g ¿µÀ ÁÂ − g ¿µÀ ÁÃ ) g G±²³´µ¶ = (g ¿µÀ ÁÂ − g ¿µÀ ÁÃ ) ·¸g¹ ¸º·¹»¼½¾ g G±²³´µ¶ = (5.5)(200) − (6.5)(200) 15 = −13.33 Ä¶ Å ¿ÆÇ Volume of the mud pit = 200*6.5 = 1300 ft 3 The time to empty the mud pit from the full level = 1300/13.33 = 97.5 min It takes 97.5 – 15 = 82.5 min to empty the mud pit after the mud level drops to 5.5 ft. f. You suspect the troublesome zone is taking fluid and tell the crew to start preparing lost circulation material (LCM). Revise your balance equation to account for this situation. At what rate must you add LCM to the mud system to avoid losing control of the well? The revised mass balance is: m leakout * time interval + LCM rate * time interval = 0 (no accumulation) LCM rate = -m leakout = - (-13.33 ft 3 /min)=13.33 ft 3 /min 2) Balance equations. “Pressure on annulus” is a common and vexing well construction problem, especially when the reservoir is deep, hot and overpressured, or when the field is offshore. An example case is shown below. The pressure in the annulus increases as the mass of gas in the annulus increases.
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PGE 421K Spring 2010 HWSolution1 - PGE 421K SPRING 2010 HW...

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